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Fossil fuel emissions release billions of tons of carbon into the atmosphere each year, which is changing the climate and threatening the sustainability of life on planet Earth. In Brazil, the demand for alternative energy sources has led to an increase in biofuel crops. A new "News and Views" paper in Nature Climate Change, co-authored by Woods Hole Research Center scientists Marcia Macedo and Eric Davidson, reviews new research conducted by Brazilian colleagues demonstrating the high carbon costs of converting intact Brazilian savanna compared to the carbon gains obtained from converting underutilized pastureland for biofuel crops. Compared to corn, soy and palm oil, the rapid growth rate of sugar cane has put it at the forefront of biofuel crops. Brazil's national commitment to reduce greenhouse gas emissions, along with rising gasoline prices, has led to the world's largest fleet of flex-fuel vehicles, fueled by the over 36 million tonnes of sugar cane currently grown in the country. This number is expected to climb with new technologies and greater global demand. The challenge for Brazil lies in identifying optimal lands for expanding sugarcane while still meeting demands for food crops and conserving native forests and savannas. The Cerrado, Brazil's 2 million km2 savanna region, is the most biologically rich savanna on Earth. Unlike the Amazon, which remains over 80% forested, over half of the Cerrado has been cleared for agriculture, including sugar cane biofuel crops. As Drs. Macedo and Davidson note, the new research shows that it would take 17 years of sugar cane production to make up for the carbon losses caused by clearing the Cerrado. In contrast, converting already cleared pastures to sugar cane production provides a nearly immediate carbon payback when ethanol is burned in lieu of gas and oil. With over 2.5 million square kilometers of existing cleared lands in Brazil, much of which is degraded pasture lands, there is already a large potential area for biofuel crop expansion. For Dr. Macedo, "Because Brazil has a large supply of under-used, low productivity pastures that are suitable for sugar cane, there is no reason to clear additional native Cerrado for sugar cane production." Dr. Davidson adds "A study commissioned by the World Bank shows that there is likely room for an all-of-the-above future land-use strategy, which includes using degraded pastures for a combination of reforestation, expansion of biofuel and food crops, and intensification of cattle production." In agreement with their Brazilian colleagues, Macedo and Davidson conclude that Brazil can meet today's demands for food, fiber, feed and fuel with no further biodiversity loss, minimal carbon costs, and even a carbon gain, which would help slow climate change. |Contact: Eunice Youmans| Woods Hole Research Center

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# 69. Sqrt(x) Leetcode Solution ## Sqrt(x) Leetcode Problem : Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well. You must not use any built-in exponent function or operator. For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python. ## Sqrt(x) Leetcode Solution : ### Constraints : • 0 <= x <= 2^31 – 1 ### Example 1: • Input: x = 8 • Output: 2 Intuition : We want to find the square root of a given non-negative integer x. Instead of using a traditional approach like repeatedly subtracting numbers until we reach 0 or using a library function, we’ll use a smarter method called “Binary Search.” Binary Search helps us quickly find the square root by repeatedly narrowing down the search range. Approach : 1. We first check if x is 0 or 1. If it is, we know that the square root of 0 and 1 is 0 and 1 respectively, so we directly return x. 2. For any other value of x, we set up a search range between 1 and x. We initialize two variables start and end to represent the range. 3. Now comes the clever part: We use a while loop to repeatedly divide the search range in half (Binary Search) to find the square root. 4. In each iteration of the loop, we calculate the middle value mid using the formula start + (end – start) / 2. This formula ensures that we don’t encounter any integer overflow when dealing with large values of x. 5. Next, we calculate the square of mid and compare it with x. 6. If the square of mid is greater than x, we know the square root lies in the lower half of the search range. So, we move the end pointer to the left to narrow down the search range. 7. If the square of mid is equal to x, we have found the square root! So, we return mid as the answer. 8. If the square of mid is less than x, we know the square root lies in the upper half of the search range. So, we move the start pointer to the right to continue the search. 9. We repeat steps 4 to 8 until the start pointer becomes greater than the end pointer. At this point, we have found the floor value of the square root, and end holds that value. 10. To ensure that we return the correct floor value of the square root, we round down the value of end to the nearest integer using the Math.round() method. ### Related Banners Get PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription ## Get over 200+ course One Subscription Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others

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Flip-flops, also called bistable gates, are digital logic circuits that can be in one of two states. Flip-flops maintain their state indefinitely until an input pulse called a trigger is received. When a trigger is received, the flip-flop outputs change state according to defined rules and remain in those states until another trigger is received. Flip-flop circuits are interconnected to form the logic gates for the digital integrated circuits (IC s) used in memory chips and microprocessors. Flip-flops can be used to store one bit, or binary digit, of data. The data may represent the state of a sequencer, the value of a counter, an ASCII character in a computer's memory or any other piece of information. There are several different kinds of flip-flop circuits, with designators such as T (toggle), S-R (set/reset) J-K (possibly named for Jack Kilby) and D (delay). A flip-flop typically includes zero, one, or two input signals as well as a clock signal and an output signal. Some flip-flops also include a clear input signal to reset the current output. The first electronic flip-flop was invented in 1919 by W. H. Eccles and F. W. Jordan. It used vacuum tubes and was initially called the Eccles-Jordan trigger circuit.

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# Vectors on a plane • Dec 4th 2009, 03:48 AM Drdumbom Vectors on a plane Hi I really need some assistance with the following problem: "Find all unit vectors in the plane determined by u = (3,0,1) and v = (1,-1,1) that are perpendicular to the vector w = (1,2,0)". The problem for me is that I donīt really know how to approach planes in general. For example, how do I calculate what plane is determined by the two vectors u and v? What would u and v look like if they were serparate planes? If someone know a website that explains the basics of vector planes I would appreciate the adress. • Dec 4th 2009, 07:25 AM Drdumbom Bump • Dec 5th 2009, 05:11 AM HallsofIvy Quote: Originally Posted by Drdumbom Bump You waited less than 4 hours before a bump? People are not sitting waiting for you to post questions, you know! Some people only look in here every few days. • Dec 5th 2009, 05:15 AM HallsofIvy Quote: Originally Posted by Drdumbom Hi I really need some assistance with the following problem: "Find all unit vectors in the plane determined by u = (3,0,1) and v = (1,-1,1) that are perpendicular to the vector w = (1,2,0)". The problem for me is that I donīt really know how to approach planes in general. For example, how do I calculate what plane is determined by the two vectors u and v? Use the cross product: the vector $\displaystyle \vec{u}\times \vec{v}$ is perpendicular to any linear combination of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ and, so, is perpendicular to the entire plane. You need to find all unit vectors such that their dot product with that cross product and (1, 2, 0) is 0. That is, take (a, b, c) to be the vector. On condition is that $\displaystyle (a, b, c)\cdot(1, 2, 0)= a+ 2b= 0$. So you know that a= -2b. Now do the same with (a, b, c) and $\displaystyle (3,0,1)\times (1,-1,1)$. What would u and v look like if they were serparate planes? If someone know a website that explains the basics of vector planes I would appreciate the adress.

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# Unit Length Vector and Dot Product Suppose $\vec a$ = [4, 6] and $\vec b$ = [1, 2]. Determine: a) A vector with unit length in the opposite direction to $\vec b$ For this question I understand I would have to use the $\vec a$ = k ($\vec b$) equation since we are a talking about opposite direction which I would consider collinear and from there using the magnitude equation to equal $1$ since that is the unit length and I would substitute the result of $\vec a$ = k ($\vec b$) like so.. $$1=\sqrt (k^2+2k^2)$$ $$1=5k^2$$ $${ 1 \over\sqrt 5} = k$$ But now I have no idea what to do next because the final answer comes to [$-\sqrt 5 \over 5$,$-2\sqrt 5 \over 5$]. Have I done everything correct so far? What do I need to do next? b) The components of a vector with the same magnitude as $\vec a$ making an angle of $60^\circ$ with the positive x-axis. I have no idea how to do this question but I feel like I would have to use the dot product for it This is really simple, Dunja. First, normalize the vector $b$ by dividing each of its coordinates by the length of $b$ (which is $\sqrt 5$, here). By this, you shrink the vector to length one. Then put a minus in front of this normalized vector to make it opposite direction. Done. • @FriedrichPhillip How do you know to do that? I understand what is to be done but I don't understand why. Commented Mar 3, 2016 at 1:43 • Well, let $|x|$ denote the length of a vector $x$. It is easy to prove that for a positive number $a$ we have $|ax| = a\cdot|x|$. Now, let $a = |x|$. Then the length of the vector $(1/a)x$ is $|(1/a)x| = (1/a)|x| = 1$ since $a = |x|$. Commented Mar 3, 2016 at 1:49 Recall that the formula for the angle between two vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is $$\cos(\theta) = \frac{\overrightarrow{a} \bullet \overrightarrow{b}}{||\overrightarrow{a}||\cdot||\overrightarrow{b}||}$$ Now, let's define our two vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$. We know that $$\overrightarrow{a}$$ is $$\begin{pmatrix} 4 \\ 6 \end{pmatrix}$$so therefore its magnitude is $$\sqrt{4^2+6^2} = 2\sqrt{13}$$. We know that the x-axis is a horizontal line, and we can represent any vector along it as $$\begin{pmatrix} k \\ 0 \end{pmatrix}$$ where $$k \in \mathbb{R}$$. For simplicity's sake, lets let $$k = 1$$. Now, we define an arbitrary vector $$\overrightarrow{c}$$ such that $$\overrightarrow{c} = \begin{pmatrix} x \\ y \end{pmatrix}$$ Then, $$||\overrightarrow{c}|| = \sqrt{x^2+y^2}$$ We now have all the information we need to solve this problem. We want the angle between our two vectors to be $$60^\circ$$, so the LHS of our first equation becomes $$\cos(60^\circ) = \frac{1}{2}$$ $$\frac{1}{2} = \frac{\overrightarrow{c} \bullet \overrightarrow{x}}{2\sqrt{13}}$$ The dot product of $$\overrightarrow{c} \bullet \overrightarrow{x} = x$$, so our after cross-multiplying, our equation becomes $$2x = 2\sqrt{13} \Longrightarrow x = \sqrt{13}$$ Remember, we're solving for the vector $$\overrightarrow{c}$$, and so far, we only know the value of the $$x$$ component of that vector. We know that $$||\overrightarrow{c}|| = 2\sqrt{13}$$, so $$\sqrt{x^2+y^2} = 2\sqrt{13}$$ Squaring both sides:$$x^2+y^2=52 \Longrightarrow 13 + y^2 = 52$$ From there we can solve for $$y$$, leaving us with $$y = \sqrt{39}$$. Therefore, our vector $$\overrightarrow{c} = \begin{pmatrix} \sqrt{13} \\ \sqrt{39} \end{pmatrix}$$

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# Casino Math: Understanding the Odds and Probabilities Introduction: Behind the glittering lights and vibrant atmosphere of a casino lies a world governed by mathematics. Understanding the odds and probabilities associated with various casino games is crucial for making informed decisions and maximizing your chances of success. In this exploration of casino math, we nhà cái Hi88 will unravel the intricacies of odds and probabilities, shedding light on the mathematical principles that shape the gaming experience. 1. The House Edge: The Casino’s Built-In Advantage: • Definition: The house edge represents the statistical advantage the casino holds over players in any given game. It ensures that, over time, the casino will profit. • Calculation: Expressed as a percentage, the house edge is derived by subtracting the player’s expected return from 100%. For example, a game with a 5% house edge implies the casino expects to retain 5% of all wagers made. 2. Odds vs. Probabilities: Clarifying Terminology: • Odds: Typically presented as a ratio (e.g., 3:1), odds represent the likelihood of a particular outcome occurring. For instance, odds of 3:1 mean there are three chances of success to one chance of failure. • Probabilities: Expressed as a percentage (e.g., 25%), probabilities convey the likelihood of an event happening. The relationship between odds and probabilities is mathematical, and understanding one helps grasp the other. 3. Understanding Probability in Dice Games: • Craps Example: In the game of craps, understanding the probability of rolling specific numbers informs strategic decisions. The likelihood of rolling a seven, for instance, is approximately 16.67%, as there are six combinations that result in a sum of seven (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) out of a possible 36 combinations (6 sides on two dice). • Expected Value: Calculating the expected value involves multiplying the probability of each outcome by its associated payoff and summing these values. A positive expected value suggests a potentially favorable outcome for the player. 4. Slot Machine Math: Symbols, Reels, and Payouts: • Symbol Distribution: Slot machines rely on specific symbol distributions across reels to determine winning combinations. Understanding the likelihood of landing certain symbols provides insights into the odds of hitting a winning spin. • Payout Percentages: The Return to Player (RTP) percentage represents the portion of wagered money a slot machine is expected to return over time. For example, a slot with a 95% RTP should theoretically return \$95 for every \$100 wagered. 5. Blackjack Probability: Card Counting and Strategy: • Card Counting: In blackjack, card counting involves tracking the ratio of high-value to low-value cards remaining in the deck. A higher ratio favors the player, as it increases the likelihood of receiving favorable hands. • Basic Strategy: Understanding basic blackjack strategy, which involves making optimal decisions based on the player’s hand and the dealer’s upcard, minimizes the house edge. Charts and tables are available to guide players in making statistically sound choices. 6. Roulette Odds: The Wheel of Chance: • European vs. American Roulette: Understanding the differences between European and American roulette wheels is crucial. The addition of a double zero in the American version increases the house edge, impacting the odds of winning on specific bets. • Betting Strategies: Employing various betting strategies, such as the Martingale or Fibonacci systems, can influence the probability of short-term successes. However, it’s essential to recognize the limitations and potential risks associated with these strategies. 7. Poker Mathematics: Calculating Pot Odds and Expected Value: • Pot Odds: Calculating pot odds involves comparing the current size of the pot to the cost of a contemplated call. If the odds of completing a drawing hand are higher than the pot odds, it may be a statistically favorable decision. • Expected Value in Poker: Evaluating the expected value of different plays in poker incorporates probabilities and potential payouts. Decisions with positive expected value contribute to long-term success. Conclusion: Casino math serves as the foundation for strategic decision-making in various games. Whether it’s calculating the house edge in traditional table games, understanding symbol distribution in slot machines, or applying probability concepts in poker, a grasp of the mathematical principles enhances the player’s ability to navigate the gaming landscape. By combining knowledge of odds and probabilities with sound strategic choices, players can approach the casino experience with a greater understanding of the underlying mathematics that govern the games of chance.

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Here is the original puzzler about how to measure 1/4th full mark in a sideways cylindrical tank. Sometime ago you entertained a caller who wanted to know how to measure the fuel level in the cylindrical tank of his diesel truck. These tanks are cylinders that lie on their side and the filler is on the top. Specifically, he wanted to know if he inserted a broomstick through the tank’s filler opening, where on the stick should he put the ¼-full mark? The ‘answer’ (although wrong) to this puzzler is essentially to take a circular piece of cardboard, cut it in half. Then use a pencil to find where this cardboard balances. This (they claim) will be the 1/4th mark. They even have a video of this technique. So, that is their answer. It is wrong. Wait. Let me remind you how much I love Car Talk. Actually, I suggested the names “Car” and “Talk” for two of our children. These names were rejected from the Allain-family naming committee. Ok, let me get on with this. Why is this wrong. First, let me skip to the problem of finding the point on a flat circle such that one fourth of the area below that point is one fourth of the area. Can we all agree that this is the real problem and that it is equivalent to finding the height where the volume of the cylindrical tank is one fourth full? Great. Here is my main problem Ray (from Car Talk) found the center of mass (center of area) of a half circle. I suspect his reasoning was something like this: “Ok, so the half circle is balanced on this pencil. This means that half of the cardboard (and thus half the area) are on each side of that point. Expanding this to a full circle would mean that the location is the fourth-full mark.” The mistake is thinking that the center of mass means that equal masses (or areas) are on each side of this point. BOOOGUS. (Ray likes to say that). Ray is confusing torque and weight. Let me give an example of where Ray’s method works. Here a line going through the center of mass would also be a line splitting the object into two equal areas. Suppose the above shape is cardboard. Suppose also I have an extra piece of cardboard that I attach to each side with a coat hanger wire like this: For this case, the dotted line still splits the object into two equal areas. However, it would not balance here. If something balances, what does that mean? That means that the net torque on the object about that balance point is zero (technically a vector). You could say that the torque from the stuff on the left of the balance point is equal and opposite to the torque on the right side. Here is the key: torque depends on the weight AND its distance from the balance point. Let me write torque like this. The torque about some point is: The vector r is from the balance point to the mass (the center of the mass) and F is obviously the force. θ is the angle between these two, for simple cases (like here) θ is π/2. But how does this relate to the half-circle cardboard thing. Suppose I find the balance point and then fold it in half along the radius. This would be a side view. I drew those rectangles so that you could imagine them as individual masses. On the left, you need more of these rectangles because they get shorter (however, they are also farther away). The point is that just because it is balanced does not also mean equal areas. One more point. This is probably close to the correct answer. However, taking 1/4th the diameter is pretty close to the correct answer also. Warning: complicated math For completeness, let me calculate the center of mass (even though this is in just about every single calculus textbook) and compare that to point to indicate a fourth of the tank. Center of mass (area) for a half circle Here is my object and my coordinate system: Clearly, I just need to look at the x-direction for the center of mass (the y-coordinate of the center of mass would be zero). The x-coordinate of the center of mass is: This just says that the center of mass is the weighted average of the masses of these rectangles that I drew. They are weighted by the distance from the origin. The dmi is the mass of these rectangles and x Here I have the variable x, but an integration variable dm. That needs to be fixed. So, what is the mass of the little tall rectangle in terms of x? Well, suppose the surface area density is: This means the area and mass of the rectangle is: (The 2 comes from the height of the rectangle) Great, I removed the dm but now I have a y. Well, there is a relationship between x and y since it is the equation of a circle. I can write: Putting this together, I get the following integral: This isn’t too difficult of an integral. It can be evaluated by doing a substitution. Anyway, if you do that, you get (or you can try this on Wolfram Alpha). Actually, Wolfram Alpha will even show the steps in this integration and even let you save it as an image. Good job. Here is that image. (but don’t cheat and use this for your homework) Now, I just need to evaluate the limits of integration. I get: Check in your calc book or google it. This is the same answer. Also, it has the right units (distance) and it is negative (for this case). There are three answers for this problem. First, the real answer (determined using calculus). This gives the area as a function of distance from the bottom as: Note, this is the area for a half-circle partially filled. Put in h = R and you get the area of half of a circle. But, what I want is the h that gives half of a half of a circle. This means I need to solve for h in the following: Solving that for h does not look like fun. Good thing I already did this (see previous post). For 1/4th full mark, it is 0.298 times the diameter from the bottom. Let me call this 0.596R The next method is the Car Talk Balance Method. From above this gives a distance from the bottom of the tank for 1/4th as: (remember the x-center of mass from above was from the center of the circle) Putting in values for π this gives a height of 0.5756 R. There is a third method. What if I just measure 1/4th of the height of the tank? This would give a height of 0.5R. To summarize: here are the percent differences from the actual answer - Correct method = 0.596R. This is a 0% difference from the correct answer. - Balancing pencil method = 0.5756R. This is a 3.4% difference from the correct answer. - The fourth is a fourth method = 0.5R. This is a 16.1% difference from the correct answer. I still love Car Talk and it is still a very clever method that gives a fairly close approximation for a fourth full tank. This doesn’t work for any other measurements though (well, I guess you would have to think of some other clever method).Go Back to Top. Skip To: Start of Article.

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Photorespiration (also known as the oxidative photosynthetic carbon cycle, or C2 photosynthesis) is a process in plant metabolism which attempts to ameliorate the consequences of a wasteful oxygenation reaction by the enzyme RuBisCO. The desired reaction is the addition of carbon dioxide to RuBP (carboxylation), a key step in the Calvin-Benson cycle, however approximately 25% of reactions by RuBisCO instead add oxygen to RuBP (oxygenation), producing a product that cannot be used within the Calvin-Benson cycle. This process reduces efficiency of photosynthesis, potentially reducing photosynthetic output by 25% in C3 plants. Photorespiration involves a complex network of enzyme reactions that exchange metabolites between chloroplasts, leaf peroxisomes and mitochondria. The oxygenation reaction of RuBisCO is a wasteful process because 3-Phosphoglycerate is created at a reduced rate and higher metabolic cost compared with RuBP carboxylase activity. While photorespiratory carbon cycling results in the formation of G3P eventually, there is still a net loss of carbon (around 25% of carbon fixed by photosynthesis is re-released as CO2) and nitrogen, as ammonia. The ammonia must be detoxified at a substantial cost to the cell. Photorespiration also incurs a direct cost of one ATP and one NAD(P)H. While it is common to refer to the entire process as photorespiration, technically the term refers only to the metabolic network which acts to rescue the products of the oxygenation reaction (phosphoglycolate). - 1 Photorespiratory reactions - 2 Substrate specificity of RuBisCO - 3 Conditions which increase photorespiration - 4 Biological adaptation to minimize photorespiration - 5 Photorespiration may serve some purpose - 6 See also - 7 References Addition of molecular oxygen to ribulose-1,5-bisphosphate produces 3-phosphoglycerate (PGA) and 2-phosphoglycolate (2PG, or PG). PGA is the normal product of carboxylation, and productively enters the Calvin cycle. Phosphoglycolate, however, inhibits certain enzymes involved in photosynthetic carbon fixation (hence is often said to be an 'inhibitor of photosynthesis'). It is also relatively difficult to recycle: in higher plants it is salvaged by a series of reactions in the peroxisome, mitochondria, and again in the peroxisome where it is converted into glycerate. Glycerate reenters the chloroplast and by the same transporter that exports glycolate. A cost of 1 ATP is associated with conversion to 3-phosphoglycerate (PGA) (Phosphorylation), within the chloroplast, which is then free to re-enter the Calvin cycle. There are several costs associated with this metabolic pathway; one being the production of hydrogen peroxide in the peroxisome (associated with the conversion of glycolate to glyoxylate). Hydrogen peroxide is a dangerously strong oxidant which must be immediately broken down into water and oxygen by the enzyme catalase. The conversion of 2x 2Carbon glycine to 1 C3 serine in the mitochondria by the enzyme glycine-decarboxylase is a key step, which releases CO2, NH3, and reduces NAD to NADH. Thus, 1 CO 2 molecule is produced for every 3 molecules of O 2 (two deriving from the activity of RuBisCO, the third from peroxisomal oxidations). The assimilation of NH3 occurs via the GS-GOGAT cycle, at a cost of one ATP and one NADPH. Cyanobacteria have three possible pathways through which they can metabolise 2-phosphoglycolate. They are unable to grow if all three pathways are knocked out, despite having a carbon concentrating mechanism that should dramatically reduce the rate of photorespiration (see below). Substrate specificity of RuBisCO - RuBP + O 2 → Phosphoglycolate + 3-phosphoglycerate + 2H+ During the catalysis by RuBisCO, an 'activated' intermediate is formed (an enediol intermediate) in the RuBisCO active site. This intermediate is able to react with either CO 2 or O 2. It has been demonstrated that the specific shape of the RuBisCO active site acts to encourage reactions with CO 2. Although there is a significant "failure" rate (~25% of reactions are oxygenation rather than carboxylation), this represents significant favouring of CO 2, when the relative abundance of the two gasses is taken into account: in the current atmosphere, O 2 is approximately 500 times more abundant, and in solution O 2 is 25X more abundant than CO 2. The ability of RuBisCO to specify between the two gasses is known as its selectivity factor (or Srel), and it varies between species, with angiosperms more efficient than other plants, but with little variation among the vascular plants. A suggested explanation into Rubisco's inability to discriminate completely between CO 2 and O 2 is that it is an evolutionary relic: The early atmosphere in which primitive plants originated contained very little oxygen, the early evolution of RuBisCO was not influenced by its ability to discriminate between O 2 and CO Conditions which increase photorespiration Photorespiration rates are increased by: Altered substrate availability: lowered CO2 or increased O2 Factors which influence this include the atmospheric abundance of the two gases, the supply of the gases to the site of fixation (i.e. in land plants: whether the stomata are open or closed), the length of the liquid phase (how far these gases have to diffuse through water in order to reach the reaction site). For example, when the stomata are closed to prevent water loss during drought: this limits the CO 2 supply, while O 2 production within the leaf will continue. In algae (and plants which photosynthesise underwater) gases have to diffuse significant distances through water, which results in a decrease in the availability of CO 2 relative to O 2. It has been predicted that the increase in ambient CO 2 concentrations predicted over the next 100 years may reduce the rate of photorespiration in most plants by around 50%. At higher temperatures RuBisCO is less able to discriminate between CO 2 and O 2. This is because the enediol intermediate is less stable. Increasing temperatures also reduce the solubility of CO 2, thus reducing the concentration of CO 2 relative to O 2 in the chloroplast. Photorespiration may be used as a mechanism to dissipate excess energy at high irradiance levels. Biological adaptation to minimize photorespiration Certain species of plants or algae have mechanisms to reduce uptake of molecular oxygen by RuBisCO. These are commonly referred to as Carbon Concentrating Mechanisms (CCMs), as they increase the concentration of CO 2 so that Rubisco is less likely to produce glycolate through reaction with O Biochemical carbon concentrating mechanisms Biochemical CCMs concentrate carbon dioxide in one temporal or spacial region, through metabolite exchange. C4 and CAM photosynthesis both use the enzyme Phosphoenolpyruvate carboxylase (PEPC) to add CO 2 to a 3-Carbon sugar. PEPC is faster than Rubisco, and more selective for CO C4 plants capture carbon dioxide in their mesophyll cells (using an enzyme called Phosphoenolpyruvate carboxylase which catalyzes the combination of carbon dioxide with a compound called Phosphoenolpyruvate (PEP)), forming oxaloacetate. This oxaloacetate is then converted to malate and is transported into the bundle sheath cells (site of carbon dioxide fixation by RuBisCO) where oxygen concentration is low to avoid photorespiration. Here, carbon dioxide is removed from the malate and combined with RuBP by Rubisco in the usual way, and the Calvin cycle proceeds as normal. The CO 2 concentrations in the Bundle Sheath are approximately 10-20 fold higher than the concentration in the mesophyll cells. This ability to avoid photorespiration makes these plants more hardy than other plants in dry and hot environments, wherein stomata are closed and internal carbon dioxide levels are low. Under these conditions, photorespiration does occur in C4 plants, but at a much reduced level compared with C3 plants in the same conditions. C4 plants include sugar cane, corn (maize), and sorghum. CAM (Crassulacean acid metabolism) CAM plants, such as cacti and succulent plants, also use the enzyme PEP carboxylase to capture carbon dioxide, but only at night. Crassulacean acid metabolism allows plants to conduct most of their gas exchange in the cooler night-time air, sequestering carbon in 4 carbon sugars which can be released to the photosynthesizing cells during the day. This allows CAM plants to reduce water loss (transpiration) by maintaining closed stomata during the day. CAM plants usually display other water-saving characteristics, such as thick cuticles, stomata with small apertures, and typically lose around 1/3 of the amount of water per CO There have been some reports of algae operating a biochemical CCM: shuttling metabolites within single cells to concentrate CO2 in one area. This process is not fully understood and may not be biologically widespread. Biophysical carbon concentrating mechanisms This type of carbon concentrating mechanism (CCM) relies on a contained compartment within the cell into which CO2 is shuttled, and where RuBisCO is highly expressed. In many species, biophysical CCMs are only induced under low carbon dioxide concentrations. Biophysical CCMs are more evolutionarily ancient than biochemical CCMs. There is some debate as to when biophysical CCMs first evolved, but it is likely to have been during a period of low carbon dioxide, after the great oxidation event (2.4 billion years ago). Low CO 2 periods occurred around 750, 650, and 320-270 million years ago. In nearly all species of eukaryotic algae (Chloromonas being one notable exception), upon induction of the CCM, ~95% of RuBisCO is densely packed into a single subcellular compartment: the pyrenoid. Carbon dioxide is concentrated in this compartment using a combination of CO 2 pumps, bicarbonate pumps, and carbonic anhydrases. The pyrenoid is not a membrane bound compartment, but is found within the chloroplast, often surrounded by a starch sheath (which is not thought to serve a function in the CCM). Certain species of Hornwort are the only land plants which are known to have a biophysical CCM involving concentration of carbon dioxide within pyrenoids in their chloroplasts. Cyanobacterial CCMs are similar in principle to those found in Eukaryotic algae and Hornworts, but the compartment into which carbon dioxide is concentrated has several structural differences. Instead of the pyrenoid, cyanobacteria contain Carboxysomes, which have a protein shell, and linker proteins packing RuBisCO inside with a very regular structure. Cyanobacterial CCMs are much better understood than those found in Eukaryotes, partly due to the ease of genetic manipulation of prokaryotes. Photorespiration may serve some purpose Reducing photorespiration may not result in increased growth rates for plants. Some research has suggested, for example, that photorespiration may be necessary for the assimilation of nitrate from soil. Thus, a reduction in photorespiration by genetic engineering or because of increasing atmospheric carbon dioxide due to fossil fuel burning may not benefit plants as has been proposed. Several physiological processes may be responsible for linking photorespiration and nitrogen assimilation: one is that photorespiration increases availability of NADH, which is required for the conversion of nitrate to nitrite; another link is that certain nitrite transporters also transport bicarbonate, and elevated CO2 has been shown to suppress nitrite transport into chloroplasts. Although photorespiration is greatly reduced in C4 species, it is still an essential pathway - mutants without functioning 2-phosphoglycolate metabolism cannot grow in normal conditions. One mutant was shown to rapidly accumulate glycolate. Although the functions of photorespiration remain controversial, it is widely accepted that this pathway influences a wide range of processes from bioenergetics, photosystem II function, and carbon metabolism to nitrogen assimilation and respiration. The photorespiratory pathway is a major source of H 2 in photosynthetic cells. Through H 2 production and pyridine nucleotide interactions, photorespiration makes a key contribution to cellular redox homeostasis. In so doing, it influences multiple signalling pathways, in particular, those that govern plant hormonal responses controlling growth, environmental and defense responses, and programmed cell death. Another theory postulates that it may function as a "safety valve", preventing the excess of reductive potential coming from an overreduced NADPH-pool from reacting with oxygen and producing free radicals, as these can damage the metabolic functions of the cell by subsequent oxidation of membrane lipids, proteins or nucleotides. - Sharkey, Thomas (1988). "Estimating the rate of photorespiration in leaves". Physiologia Plantarum 73 (1): 147–152. doi:10.1111/j.1399-3054.1988.tb09205.x. - Leegood, R. C. (2007). A welcome diversion from photorespiration. Nature Biotechnology, 25(5), 539–540. - Peterhansel, C.; Krause, K.; Braun, H. -P.; Espie, G. S.; Fernie, A. R.; Hanson, D. T.; Keech, O.; Maurino, V. G.; Mielewczik, M.; Sage, R. F. (2012). "Engineering photorespiration: Current state and future possibilities". Plant Biology 15 (4): n/a. doi:10.1111/j.1438-8677.2012.00681.x. - Eisenhut, M.; Ruth, W.; Haimovich, M.; Bauwe, H.; Kaplan, A.; Hagemann, M. (2008). "The photorespiratory glycolate metabolism is essential for cyanobacteria and might have been conveyed endosymbiontically to plants". Proceedings of the National Academy of Sciences 105 (44): 17199. doi:10.1073/pnas.0807043105. - Griffiths, H. (2006). Designs on Rubisco" Nature 441(7096), 940–941. Retrieved from http://www.rsbs.anu.edu.au/profiles/graham_farquhar/documents/233TcherkezPNASNaturecommentary.pdf - Ehleringer, J.R.; Sage, R.F.; Flanagan, L.B.; Pearcy, R.W. (1991). "Climate change and the evolution of C4 photosynthesis". Trends in Ecology & Evolution 6 (3): 95–99. doi:10.1016/0169-5347(91)90183-x. - Plant Physiology, Fifth Edition - Sinauer Associates, Inc. Taiz and Zeiger. 2010. Chapter 8, Page 222 - Giordano, Mario; Beardall, John; Raven, John A. (June 2005). "CO2 CONCENTRATING MECHANISMS IN ALGAE: Mechanisms, Environmental Modulation, and Evolution". Annual Review of Plant Biology (Annual Reviews) 56: 99–131. doi:10.1146/annurev.arplant.56.032604.144052. PMID 15862091. Retrieved 26 August 2013. - Raven, J. A.; Giordano, M.; Beardall, J.; Maberly, S. C. (2012). "Algal evolution in relation to atmospheric CO2: Carboxylases, carbon-concentrating mechanisms and carbon oxidation cycles". Philosophical Transactions of the Royal Society B: Biological Sciences 367 (1588): 493. doi:10.1098/rstb.2011.0212. - Villarejo, A.; Martinez, F.; Pino Plumed, M.; Ramazanov, Z. (1996). "The induction of the CO2 concentrating mechanism in a starch-less mutant of Chlamydomonas reinhardtii". Physiologia Plantarum 98 (4): 798. doi:10.1111/j.1399-3054.1996.tb06687.x. - Rachmilevitch S, Cousins AB, Bloom AJ (August 2004). "Nitrate assimilation in plant shoots depends on photorespiration". Proc. Natl. Acad. Sci. U.S.A. 101 (31): 11506–10. doi:10.1073/pnas.0404388101. PMC 509230. PMID 15272076. - Bloom, A. J.; Burger, M.; Asensio, J. S. R.; Cousins, A. B. (2010). "Carbon Dioxide Enrichment Inhibits Nitrate Assimilation in Wheat and Arabidopsis". Science 328 (5980): 899–903. doi:10.1126/science.1186440. PMID 20466933. - Zabaleta, E.; Martin, M. V.; Braun, H. P. (2012). "A basal carbon concentrating mechanism in plants?". Plant Science 187: 97–104. doi:10.1016/j.plantsci.2012.02.001. PMID 22404837. - Foyer, C.H.; Bloom, A.J.; Queval, G.; Noctor, G. (2009). "Photorespiratory Metabolism: Genes, Mutants, Energetics, and Redox Signaling". Annual Review of Plant Biology 60 (1): 455–484. doi:10.1146/annurev.arplant.043008.091948. PMID 19575589. - Stuhlfauth, t.; Scheuermann, R; Fock, H. P. (1990). "Light Energy Dissipation under Water Stress Conditions". Plant Physiology 92: 1053–1061. doi:10.1104/pp.92.4.1053. PMC 1062415.

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What are muscle cramps? There are 3 different types of muscles in your body. There is cardiac muscle, which is what your heart is made out of, there is smooth muscle, which is what lines your blood vessels, gastrointestinal tract, and some of your organs, and there is skeletal muscle. Skeletal muscles attach to your bones and contract in order to allow voluntary movements of your body to happen. Sometimes these skeletal muscles can contract involuntarily and forcefully, this is called a muscle spasm. If this muscle contraction sustains and does not relax, this becomes a muscle cramp. These muscle contractions are often painful. The most common area of the body to have muscle cramps is the muscles of the legs. The most likely muscles to crap are your quadriceps (thigh), hamstrings (back of the thigh), and gastrocnemius (calf). While these are the most common, cramps can happen in any skeletal muscle. Other common places for cramps to occur are in the arms, hands, feet, and abdominal wall. If a cramp happens in your leg it can make it difficult to walk, and the pain of cramps can often wake people in the middle of the night. In some cases there can also be a bulging lump of muscle tissue beneath the skin, that accompanies the pain of the cramp. There may be some twitching of the muscle. Muscle cramps can last anywhere from a few seconds to over 15 minutes long. In most cases muscle cramps are harmless, however they can cause a lot of pain and discomfort for the person experiencing them. What causes muscle cramps? There are many things that can cause muscle cramps to happen. These include: - ♦ Overuse of the muscle, usually due to exercising. - ♦ An injury. This could be directly to the muscle, or could be to a nerve or a disc, causing the surrounding muscles to spasm. - ♦ Excessive loss of fluids in the body / dehydration. - ♦ Low levels of minerals that contribute to healthy muscle function. These include magnesium, calcium, potassium, and sodium. - ♦ Low blood supply to the muscles, this usually happens when walking or exercising. - ♦ Physical deconditioning of the muscle. - ♦ An issue with the spine or spinal nerves, such as spinal nerve compression. - ♦ A medical condition. Some examples are kidney failure, liver cirrhosis, atherosclerosis, amyotrophic lateral sclerosis (ALS), thyroid disorders such as hypothyroidism. - ♦ Certain medications or supplements. - ♦ Poor posture. How can chiropractic help with muscle cramps? If you are suffering from regular muscle cramps, it is a good idea to talk to a chiropractor and see how they can help. There are many ways your chiropractor may be able to help you with your muscle cramps, these include: - ♦ Helping you figure out what is causing your cramps. The first step to stopping your cramps from happening is to figure out what is actually causing them in the first place. Chiropractors are very knowledgeable when it comes to health and how the body works, so they can work with you to figure out what be causing the cramps, so you can find a solution for them from there. - ♦ Improving muscle health. If your muscles aren’t functioning well and have a lot of extra stress placed on them, this is going to increase the chance of you having muscle cramps. Having misaligned joints throughout your body places excess stress on the muscles in that area, as they are having to do extra work to compensate for the misaligned joint. Too much stress on that muscle for a long period of time may cause it to start cramping. Chiropractors are able to locate these joint misalignments, and perform a chiropractic adjustment on them to get them moving in the proper way again. When the joint is moving properly it allows all the surrounding muscles to function properly, reducing the likelihood of muscle cramps happening. - ♦ Improving posture. Poor posture can place stress on the muscles in your body, and in some cases this can cause those muscles to start cramping. Poor posture happens due to weakness of the small stabilising muscles that surround your spine. They become weak due to lack of spinal movement. Chiropractors are able to locate the areas of the spine that aren’t moving properly and adjust them. This helps to get the spine moving again, and this movement helps to strengthen up those small spinal muscles. As those muscles get stronger, your posture will start to improve, which takes stress off of your bigger muscles. Less stress on these muscles will reduce the likelihood of muscle cramps happening. If you are wanting to improve your posture, it is a good idea to see a structural chiropractor, as they specialise in improving the structure of the spine and improving posture. - ♦ Improving nerve health. All of your muscles are controlled by nerves, that are in charge sending information to the muscles to make them contract or relax. If the nerves aren’t working at their best then they may be sending the wrong messages to your muscles causing them to contract when they shouldn’t be. Nerve health is largely influenced by spinal health, as the nervous system runs directly through the spine. Chiropractors are able to adjust the spine in order to improve the function of your nervous system, which will ensure that the nerves that connect to your muscles are functioning the way they are supposed to. - ♦ Injury recovery. It may be an injury that is causing your muscles cramps. Chiropractors are able to assist your body in healing your injury, whether it is an injury directly to the muscles, or to something else like a disc or a joint. - ♦ Exercise and stretching advice. Chiropractors can provide you with stretches and exercises to help with muscle health, which will reduce muscle cramps. They can also give you stretches to do when a muscle cramp is happening, to help the muscle relax and stop the cramp. - ♦ Other advice. Chiropractors can also provide you with other advice to help with your muscle cramps. This may be advice on supplements, what you are eating, and what to do during a cramp, such as applying heat to the muscle.

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# Advice 1: How to find the diagonal right prism Finding a diagonal right prism is often used as an intermediate step for more complex tasks. The General formula is easily derived by consideration of two right triangles. Instruction 1 To find the diagonal right prism you need to understand just a few definitions. A prism is called a polyhedron, having as bases two equal polygon (triangle, quadrangle, etc.) lying in parallel planes and the lateral faces parallelograms. Direct the prism is called prism, whose lateral faces rectangles. A right prism is called a straight prism, the base of which are regular polygon (equilateral triangle, square, etc.) АА1В1В - the side faces of the regular quadrangular prism. All four side faces of the prism are equal. ABCD and А1В1С1D1 -base of the prism (squares lying in parallel planes). A diagonal of a polyhedron is the line segment joining two not adjacent vertices, i.e. vertices that do not belong to the same face. The figure shows that the point A and the point 1 does not belong to one face and therefore the cut of AC1 - diagonal of the prism. 2 To find the diagonal of prism necessary to consider the triangle АСС1. This triangle is rectangular. Diagonal prism AC1 in the triangle will be the hypotenuse, and the segments as SS1 and the other two sides. From the Pythagorean theorem (in a right triangle the hypotenuse squared is equal to the sum of the squares of the legs) it follows that: АС12 = AC2 + СС12 (1); 3 Next, you should consider the triangle АСD. Triangle АСD is also rectangular (because the base of the prism is square). For convenience, you can designate the base of the letter. Therefore, by the Pythagorean theorem: AC2 = A2 +A2 AC = √2A (2); 4 If we denote the height of the prism with the letter h, and substitute the expression (2) into the expression (1), we get: АС12 = 2A2+h2, AC1 = √(2a^2+h^2 ), where a - side of the base, h is the height. This formula is valid for any right prism. # Advice 2: How to find the diagonal Each polyhedron, a rectangle, a parallelogram and a diagonal. It usually connects the corners of any of these geometric shapes. The value of the diagonal has to find the solution of problems in elementary and higher mathematics. Instruction 1 A diagonal is any straight line connecting the corners of the polyhedra. Its location depends on the type of shapes (rhombus, square, parallelogram) and the data given in the problem. The easiest way to find the diagonal of a rectangle is the following.The two sides of the rectangle a and b. Knowing that all of its angles equal to 90°, and its diagonal is the hypotenuse of the two triangles, we can conclude that the diagonal of this fighti can be found by using the Pythagorean theorem. In this case, the sides of the rectangle are the legs of triangles. It follows that the diagonal of a rectangle is:d=√(a^2+b^2)a special case of applying this method to finding diagonals is a square. Its diagonal can also be found by the Pythagorean theorem, but given that all its sides are equal, the diagonal of a square is equal to a√2. The value a is the square side. 2 If given a parallelogram, its diagonal find, as a rule, cosine theorem. However, in exceptional cases, for a given value of the second diagonal can be found from the first equation:d1=√2(a^2+b^2)-d2^2Теорема of cosines applies when not given a second diagonaland are given only the sides and corners. It is a generalized Pythagorean theorem. For example, given a parallelogram whose sides are b and c. Through two opposite corners of a parallelogram is the diagonal of a. Since a, b and c form a triangle, we can apply the theorem of cosines, which can be computed diagonal:a^2=√b^2+c^2-2bc*cosα When given the area of a parallelogram and one of the diagonals, and the angle between the two diagonals, the diagonal can be calculated in the following way:d2=S/d1*cos arambam is called a parallelogram whose all sides are equal. Let him have the two sides equal to a, and, the unknown diagonal. Then, knowing the theorem of cosines, the diagonal can be calculated by the formula:d=a^2+a^2-2a*a*cosα=2a^2(1-cosα) 3 The diagonal of a trapezoid is several ways. To calculate you need to know, as a rule, three values - top and bottom base, and at least one lateral side. This can be seen on the example of a rectangular trapezoid.For example, given a rectangular trapezoid. First you need to find a small segment that is a leg of a right triangle. It is equal to the difference between the upper and lower bases. As a rectangular trapezoid, the drawing shows that the height is equal to the side of the trapezoid. As a result, you can find another lateral side of the trapezoid. If you know the upper base and lateral side, the cosine theorem can be found first diagonal:c^2=a^2+b^2-2ab*soavtory diagonal is based on values of the first lateral side and the upper base on the Pythagorean theorem. In this case, this diagonal is the hypotenuse of a right triangle. Search

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A certain junior class has 1,000 students and a certain : GMAT Problem Solving (PS) - Page 2 Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 17 Jan 2017, 05:15 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # A certain junior class has 1,000 students and a certain Author Message TAGS: ### Hide Tags Intern Joined: 24 Feb 2012 Posts: 33 Followers: 0 Kudos [?]: 15 [0], given: 18 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 24 Feb 2012, 13:10 $$60/1000 * 1/800$$ $$=3/40000$$ Alternate Technique: $$\frac{Desired}{Total} = \frac{60}{1000 * 800}$$ $$= 3/40000$$ Solution to Bunuel's extension: $$\frac{120}{10000} * \frac{2}{800}$$ $$=\frac{3}{10000}$$ VP Status: Final Lap Up!!! Affiliations: NYK Line Joined: 21 Sep 2012 Posts: 1096 Location: India GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31 GPA: 3.84 WE: Engineering (Transportation) Followers: 37 Kudos [?]: 526 [0], given: 70 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 05 Oct 2012, 06:58 Can anyone clear my doubt i a m struggling TOtal number of outcome = 1000*800 Sibbling with Juniors = 60 so 60 out of 1000 Sibling with Senior = 60 so 60 out of 800 we have to find probability of 2 sibbling 1 from each 60 C 1= Selection from Senior 60 C 1 = From Junior total fav = 1000*800 Prob ={ 60 C 1 *60 C 1}/1000*800 =9/2000 pls tell me where i am wrong in my approach and what correction needed Intern Joined: 16 Jan 2013 Posts: 33 Concentration: Finance, Entrepreneurship GMAT Date: 08-25-2013 Followers: 0 Kudos [?]: 23 [0], given: 8 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 12 Jul 2013, 22:12 eschn3am wrote: $formdata=\frac{60*1}{1000*800}+=+\frac{60}{800000}=\frac{3}{40000}$ Let's say you're picking out of the Junior class first and the senior class second (although the order doesn't make any difference). There are 1000 juniors and 60 of them have a sibling in the senior class, so you have a $formdata=\frac{60}{1000}$ shot of choosing one of the siblings. Then you move onto the senior class. There are 800 seniors and only one sibling of the person you chose from the junior class. Thus, you have a $formdata=\frac{1}{800}$ chance of choosing the sibling. Hi , Don't we need to consider the case where we pick the senior class first and then the junior class. Plz clarify. Math Expert Joined: 02 Sep 2009 Posts: 36531 Followers: 7071 Kudos [?]: 92969 [0], given: 10541 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 12 Jul 2013, 23:16 Countdown wrote: eschn3am wrote: $formdata=\frac{60*1}{1000*800}+=+\frac{60}{800000}=\frac{3}{40000}$ Let's say you're picking out of the Junior class first and the senior class second (although the order doesn't make any difference). There are 1000 juniors and 60 of them have a sibling in the senior class, so you have a $formdata=\frac{60}{1000}$ shot of choosing one of the siblings. Then you move onto the senior class. There are 800 seniors and only one sibling of the person you chose from the junior class. Thus, you have a $formdata=\frac{1}{800}$ chance of choosing the sibling. Hi , Don't we need to consider the case where we pick the senior class first and then the junior class. Plz clarify. These posts might help: a-certain-junior-class-has-1-000-students-and-a-certain-58914.html#p778756 a-certain-junior-class-has-1-000-students-and-a-certain-58914.html#p812392 _________________ Intern Joined: 07 Jan 2013 Posts: 43 Location: India Concentration: Finance, Strategy GMAT 1: 570 Q46 V23 GMAT 2: 710 Q49 V38 GPA: 2.9 WE: Information Technology (Computer Software) Followers: 1 Kudos [?]: 23 [0], given: 23 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 23 Jul 2013, 18:22 i just wanted to validate this answer the other way around i,e to find the no. of ways at least one sibling or no sibling is present in the chosen 2 members and then subtracting from 1 i.e Req prob =1- (prob of one sibling chosen from either class) + no sibling chosen from either class = 1-(prob of one sibling from junior) + (prob of one siblng from senior ) + (no sibling) = 1-$$(\frac{60}{1000}*\frac{799}{800})+(\frac{60}{800}*\frac{999}{1000})+(\frac{740}{800}*\frac{940}{1000})$$ which is coming out to be -ve which obviously is wrong ,, i want to know as to what i am adding extra as a result the answer is -ve _________________ Manager Joined: 11 Jan 2011 Posts: 69 GMAT 1: 680 Q44 V39 GMAT 2: 710 Q48 V40 Followers: 0 Kudos [?]: 18 [0], given: 3 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 11 Nov 2013, 18:44 Bunuel wrote: blog wrote: A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs , each consisting of 1 junior and 1 senior. If i student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair? A. 3/40,000 B. 1/3,600 C. 9/2,000 D. 1/60 E. 1/15 There are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior. What is the probability of choosing ANY sibling from junior class? $$\frac{60}{1000}$$ (as there are 60 of them). What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be $$\frac{1}{800}$$ (as there is only one sibling pair of chosen one). So the probability of that the 2 students selected will be a sibling pair is: $$\frac{60}{1000}*\frac{1}{800}=\frac{3}{40000}$$ This problem can be solved in another way: In how many ways we can choose 1 person from 1000: $$C^1_{1000}=1000$$; In how many ways we can choose 1 person from 800: $$C^1_{800}=800$$; So total # of ways of choosing 1 from 1000 and 1 from 800 is $$C^1_{1000}*C^1_{800}=1000*800$$ --> this is total # of outcomes. Let’s count favorable outcomes: 1 from 60 - $$C^1_{60}=60$$; The pair of the one chosen: $$C^1_1=1$$ So total # of favorable outcomes is $$C^1_{60}*C^1_1=60$$ $$Probability=\frac{# \ of \ favorable \ outcomes}{Total \ # \ of \ outcomes}=\frac{60}{1000*800}=\frac{3}{40000}$$. Let’s consider another example: A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair? The same way here: What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them). What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one). So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000 Another way: In how many ways we can choose 1 person from 1000=1C1000=1000 In how many ways we can choose 1 person from 800=1C800=800 So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes. Favorable outcomes: 1 from 120=120C1=120 The pair of the one chosen=1C2=2 So total favorable outcomes=120C1*1C2=240 Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000 Also discussed at: probability-85523.html?hilit=certain%20junior%20class#p641153 Hi Bunuel, In your second example, how is it possible to have the equation bolded above: 1c2? Thanks, Rich Intern Joined: 05 Dec 2013 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 12 Dec 2013, 04:58 jeeteshsingh wrote: blog wrote: A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs , each consisting of 1 junior and 1 senior. If i student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair? A. 3/40,000 B. 1/3,600 C. 9/2,000 D. 1/60 E. 1/15 No of ways of choosing 1 sibling pair out of 60 pairs = 60c1 No of ways of choosing 1 student from each class = 1000c1 x 800c1 Therefore probability of having 2 students choosen as a sibling pair = 60c1 / (1000c1 x 800c1) = 60 / (1000 x 800) = 3 / 40000 = A This was What I thought when I solved this question Intern Status: Researching for Schools Joined: 21 Apr 2013 Posts: 21 Location: United States GMAT 1: 640 Q45 V34 GMAT 2: 730 Q49 V40 WE: Project Management (Computer Software) Followers: 0 Kudos [?]: 7 [0], given: 76 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 31 Jan 2014, 13:18 Archit143 wrote: Can anyone clear my doubt i a m struggling TOtal number of outcome = 1000*800 Sibbling with Juniors = 60 so 60 out of 1000 Sibling with Senior = 60 so 60 out of 800 we have to find probability of 2 sibbling 1 from each 60 C 1= Selection from Senior 60 C 1 = From Junior total fav = 1000*800 Prob ={ 60 C 1 *60 C 1}/1000*800 =9/2000 pls tell me where i am wrong in my approach and what correction needed For the selection from Junior we would have only 1 and not 60 C 1. This is because since we have chosen someone from the senior class and now the only choice we have is the sibling of the senior student we chose. GMAT Club Legend Joined: 09 Sep 2013 Posts: 13423 Followers: 575 Kudos [?]: 163 [0], given: 0 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 07 Feb 2015, 02:51 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 22 Sep 2014 Posts: 38 Followers: 1 Kudos [?]: 23 [0], given: 12 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 29 Jul 2015, 23:04 Archit143 wrote: Can anyone clear my doubt i a m struggling TOtal number of outcome = 1000*800 Sibbling with Juniors = 60 so 60 out of 1000 Sibling with Senior = 60 so 60 out of 800 we have to find probability of 2 sibbling 1 from each 60 C 1= Selection from Senior 60 C 1 = From Junior total fav = 1000*800 Prob ={ 60 C 1 *60 C 1}/1000*800 =9/2000 pls tell me where i am wrong in my approach and what correction needed Mistake that you are doing is that you considering 60 sibling pairs are present in each junior and senior class. But the question says 60 sibling pairs are present and out of these pair one is present in junior and its corresponding sibling in senior. That means 30 in junior and its opposite pair 30 in senior. _________________ Thanks & Regards, Vikash Alex (Do like the below link on FB and join us in contributing towards education to under-privilege children.) Manager Joined: 03 Apr 2015 Posts: 74 Followers: 0 Kudos [?]: 0 [0], given: 9 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 15 Apr 2016, 23:45 another easy understanding total ways to select the= 1000 X 800 (as each of 1000 juniors can be selected with 800 options) now selecting a sibling = 60 ways X 1 way (as there are 60 siblings so we have 60 ways of selecting on student from junior group and as soon as we select one we are left with only one way to select from senior group as that person has only one sibling in the senior group ) req prob=(desired)/total =60x1/1000x800=3/40000 ...(A) hope this helps Intern Joined: 23 Sep 2016 Posts: 8 Followers: 0 Kudos [?]: 0 [0], given: 369 Re: A certain junior class has 1,000 students and a certain [#permalink] ### Show Tags 06 Nov 2016, 22:21 Bunuel wrote: blog wrote: A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs , each consisting of 1 junior and 1 senior. If i student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair? A. 3/40,000 B. 1/3,600 C. 9/2,000 D. 1/60 E. 1/15 There are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior. What is the probability of choosing ANY sibling from junior class? $$\frac{60}{1000}$$ (as there are 60 of them). What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be $$\frac{1}{800}$$ (as there is only one sibling pair of chosen one). So the probability of that the 2 students selected will be a sibling pair is: $$\frac{60}{1000}*\frac{1}{800}=\frac{3}{40000}$$ This problem can be solved in another way: In how many ways we can choose 1 person from 1000: $$C^1_{1000}=1000$$; In how many ways we can choose 1 person from 800: $$C^1_{800}=800$$; So total # of ways of choosing 1 from 1000 and 1 from 800 is $$C^1_{1000}*C^1_{800}=1000*800$$ --> this is total # of outcomes. Let’s count favorable outcomes: 1 from 60 - $$C^1_{60}=60$$; The pair of the one chosen: $$C^1_1=1$$ So total # of favorable outcomes is $$C^1_{60}*C^1_1=60$$ $$Probability=\frac{# \ of \ favorable \ outcomes}{Total \ # \ of \ outcomes}=\frac{60}{1000*800}=\frac{3}{40000}$$. Let’s consider another example: A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair? The same way here: What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them). What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one). So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000 Another way: In how many ways we can choose 1 person from 1000=1C1000=1000 In how many ways we can choose 1 person from 800=1C800=800 So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes. Favorable outcomes: 1 from 120=120C1=120 The pair of the one chosen=1C2=2 So total favorable outcomes=120C1*1C2=240 Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000 Bunuel Is my approach correct? There are 60 sibling pairs. Probability of Selecting One Junior AND their corresponding Senior sibling OR another Junior and their corresponding sibling and so on: Probability of Selecting John is: 1/1000 Probability of Selecting John's brother: 1/800 Number of Pairs: 60 So, 1/1000 x 1/800 x 60 = 30/40000 ways to select any one particular Junior and their corresponding Senior Sibling [ With the breakdown being: P(John and his sibling) + P (Mary and her sibling) + P(Tom and his sibling) + ..... = (1/1000 x 1/800) + (1/1000 x 1/800) + (1/1000 x 1/800) + ..... and so on, until you add the 60 different possibilities ] Re: A certain junior class has 1,000 students and a certain   [#permalink] 06 Nov 2016, 22:21 Go to page   Previous    1   2   [ 32 posts ] Similar topics Replies Last post Similar Topics: 2 At a certain high school, the junior class is twice the size 6 21 May 2012, 04:29 7 A certain freshman class has 600 students and a certain sophomore 18 17 May 2011, 05:36 29 In a certain class consisting of 36 students, some boys and 6 08 Feb 2011, 03:33 20 A certain junior class has 1000 students and a certain 18 13 Aug 2010, 02:15 8 A certain junior class has 1000 students and a certain 8 12 Dec 2009, 08:37 Display posts from previous: Sort by

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tripeptidyl peptidase 1 The TPP1 gene provides instructions for making an enzyme called tripeptidyl peptidase 1. This enzyme is produced as an inactive enzyme, called a proenzyme, which has an extra segment attached. This segment must be removed, followed by additional processing steps, for the enzyme to become active. The active tripeptidyl peptidase 1 enzyme is found in cell structures called lysosomes, which digest and recycle different types of molecules. Tripeptidyl peptidase 1 acts as a peptidase, which means that it breaks down protein fragments, known as peptides, into their individual building blocks (amino acids). Specifically, tripeptidyl peptidase 1 cuts (cleaves) peptides into groups of three amino acids. At least 115 mutations in the TPP1 gene have been found to cause CLN2 disease. This condition impairs motor and mental development, typically starting in early childhood, causing gradually worsening movement disorders and a decline in intellectual function. In addition, affected children often develop recurrent seizures (epilepsy) and vision impairment. In some cases, signs and symptoms of CLN2 disease do not appear until later in childhood, usually after age 4. Most of the TPP1 gene mutations that cause CLN2 disease change single amino acids in tripeptidyl peptidase 1, resulting in a severe decrease in enzyme activity. A reduction in functional enzyme results in the incomplete breakdown of certain peptides. CLN2 disease is characterized by the accumulation of proteins or peptides and other substances in lysosomes. These accumulations occur in cells throughout the body; however, nerve cells seem to be particularly vulnerable to their effects. The accumulations can cause cell damage leading to cell death. The progressive death of nerve cells in the brain and other tissues leads to the signs and symptoms of CLN2 disease. Individuals who are diagnosed with CLN2 disease later in childhood likely have TPP1 gene mutations that result in the production of an enzyme with a small amount of normal function. Protein function in these individuals is higher than in those who have the condition beginning earlier in childhood. As a result, it takes longer for peptides and other substances to accumulate in the lysosomes and damage nerve cells. Mutations in the TPP1 gene have also been found to cause spinocerebellar ataxia, autosomal recessive 7 (SCAR7), which is a condition characterized by progressive problems with movement. During childhood, individuals with SCAR7 develop walking difficulties; impaired speech (dysarthria); and eye movement problems, such as involuntary movement of the eyes (nystagmus), rapid eye movements (saccades), and trouble moving the eyes side-to-side (oculomotor apraxia). People with SCAR7 have progressive loss of cells (atrophy) of various parts of the brain, particularly within the cerebellum, which is the area of the brain involved in coordinating movements. Compared to individuals with CLN2 disease (described above), individuals with SCAR7 likely have a higher level of normally functioning tripeptidyl peptidase 1. As a result, SCAR7 is associated with milder signs and symptoms than CLN2 disease and tends to develop in late childhood or adolescence. When examined, cells from some individuals with SCAR7 showed lysosomal accumulations while cells from other affected individuals did not. - cell growth-inhibiting gene 1 protein - growth-inhibiting protein 1 - lysosomal pepstatin insensitive protease - tripeptidyl aminopeptidase - tripeptidyl-peptidase 1 - tripeptidyl-peptidase 1 preproprotein - tripeptidyl peptidase I

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+0 # Find Point 0 344 1 Find the point with y-coordinate -28 on the line through (-62, 4213) with slope -69! Thanks Jul 18, 2017 #1 +94544 +2 Find the point with y-coordinate -28 on the line through (-62, 4213) with slope -69! So we can solve this for the x coordinate of the point in the slope "formula" [ 4213 - -28 ] / [ -62 - x] = -69    simplify 4241 / - [ 62 + x ]  =  - 69       multiply both sides by  - [ 62 + x ] 4241 = - 69  *  - [ 62 + x ] 4241 =  69  *   [ 62 + x ] 4241 =  4278 + 69x -37 = 69x -37/69   = x So the point is (-37/69, -28) Jul 18, 2017

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with extensive loss of life due to trauma —earthquakes, storms, human conflict, etc.— many resources are often expended on burying the dead quickly, and applying disinfectant to bodies, to prevent disease . This is an inappropriate use of scarce resources and manpower . The health risks from dead bodies in such cases are minimal. According to health professionals, the fear of spread of disease by bodies killed by trauma rather than disease is not justified. Among others, Steven Rottman, director of the UCLA Center for Public Health and Disasters , said that no scientific evidence exists that bodies of disaster victims increase the risk of epidemics , adding that cadavers posed less risk of contagion than living people. In disasters involving trauma where there is competition for resources, they should be going into establishment of water supply, sanitation , shelter, warmth and hygienic food for the survivors, not digging mass graves. Spraying is a waste of disinfectant and manpower. Indiscriminate burial of corpses demoralises survivors and the lack of death certificates can cause practical problems to survivors . Countervailing considerations include religious and cultural practices, the stench, and the effect on morale. This advice does not apply in the case of a health disaster such as an epidemic where the victims are affected by diseases which can be communicated by dead bodies. Roots of incorrect notion The incorrect notion that all dead bodies inherently cause diseases is probably a combination of: - The incorrect miasma theory of disease, which holds that diseases are spread by foul air—in this case fouled by the stench of decomposing corpses. - Confusion between normal decay processes and signs of disease and the idea that microorganisms responsible for decomposition are dangerous to living people. - After a major disaster disease among survivors living in harsh conditions with poor sanitation is common. - Noting that corpses of those who died from certain contagious diseases (for example, in epidemics) do, indeed, spread disease, such as the case with smallpox and the 1918 flu. Contamination of water supplies by unburied bodies, burial sites, or temporary storage sites may result in the spread of gastroenteritis from normal intestinal contents. According to an article on the Infectious Disease Risks From Dead Bodies Following Natural Disasters by the Pan American Health Organization There is little evidence of microbiological contamination of groundwater from burial [...] Where dead bodies have contaminated water supplies, gastroenteritis has been the most notable problem, although communities will rarely use a water supply where they know it to be contaminated by dead bodies. To those in close contact with the dead, such as rescue workers, there is a health risk from chronic infectious diseases which spread by direct contact, for example: hepatitis B, hepatitis C, HIV, enteric (intestinal) pathogens and tuberculosis. The substances cadaverine and putrescine are produced during the decomposition of animal bodies, and give off a foul odor. They are toxic if massive doses are ingested (2g per kg of body weight in rats for putrescine, a larger dose for cadaverine) , causing adverse effects. If these figures are assumed to apply to humans, a 60kg (132lb) person would be significantly affected by 120g (4oz) of pure putrescine, and would show no effects at all for a tenth of that dose. By way of comparison the similar substance spermine, found in semen, is over 3 times as toxic.

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Increasing Earth temperatures and rising sea levels. Both of these are effects of climate change. The current concern is that human activity is changing our climate at a rate well above the natural climate cycling. Understanding how the Earth’s climate system works and responds to human impact is therefore of uttermost importance. “To predict future climate change we must first go back in the archives to understand a bit more about the natural cycles. This will enable us to decipher between natural changes and what’s out of the ordinary,” explained Gilbert Camoin, Centre National de la Recherche Scientifique (CNRS), at the European Geosciences Union (EGU) General Assembly in Vienna in April 2007. The last deglaciation (23,000 – 6,000 years ago) is generally seen as a potential recent analogue for today’s environmental changes. This period was characterised by abrupt climatic change and rapid sea-level rise due to polar ice sheets melting, similar conditions to what we see the Earth facing now and which are predicted for the future. Reconstructing past environments may lead to a better estimate of the climate sensitivity and hence of future climate change. Fossil coral reefs can be used to accurately reconstruct past sea level variations, climate change and environmental perturbations. According to Camoin, they provide the most precise records of sea-level changes. This is because corals always live within very strict ecological requirements. They need clear and oxygenated water, only live in the first 50m of the sea column, at temperatures ranging from 18-35 C and within the very narrow salinity range of 35-36 per ml. Any ecological changes affecting the narrow requirements of the coral reef environment will lead to changes in coral reef growth and composition. Thus, a fossil coral reef drill sample gives accurate information about the sea level, salinity and temperature at different times in history. Interpreting this information provides a way to reconstruct past climate change. Using coral reefs as climatic archives started about 20 years ago. Initially, due to technical constraints, it was only possible to capture the last 10,000 years through coral reef drilling. To get datasets covering the entire period of the last deglaciation, it was crucial to find another technique. This became possible during a project where the Integrated Ocean Drilling Program (IODP) provided Camoin with the technology to drill down to 1100m recovering more than 600m of reef cores from 37 holes at depths ranging from 40 – 117m from the area around the island of Tahiti. Based on this initial success, the CHECREEF project was developed. CHECREEF is part of the European Science Foundation EUROCORES Programme EuroMARC. CHECREEF will look at data from both the Tahiti drilling site and an additional site in the Great Barrier Reef. “We think that we will be able to reconstruct the sea level change going back 16-17,000 years in Tahiti and even further back in time in the Great Barrier Reef. With this we will be able to reconstruct sea surface temperature and salinity which are good indicators of the climatic changes at that time to see what happened to the reefs. We are pretty sure that we will achieve this goal within three years based on the datasets we have already collected,” said Camoin. The two sites provide a good basis for creating a picture of past climate change. Firstly, they are away from the regions of the world covered by ice during the last ice age. Secondly, the core sites are located in the tropical pacific, a crucial point of the globe where many climatic anomalies are born e.g. El Nino. In addition, the two sites are located in zones which are tectonically quiet unlike volcanic islands and continental margins. This is the rational behind the CHECREEF proposal. However, Camoin goes on to say that to get a clear picture and a very good dataset, it is necessary to investigate other sites in the Indian Ocean and the Caribbean. This is Camoin’s plan for the future. Camoin also highlighted that it is crucial to collaborate and compare the coral reef data with other techniques, from ice cores to ocean and lake sediments, to verify datasets. “The project has just started and we are trying to get a good chronological frame to make sure where we are. The ice core records are well advanced and we will go back to compare our data to theirs in a year of two. We also have geophysical modellers waiting for our results to enter into their models,” explained Camoin. Many of the European Science Foundation projects work towards understanding the Earth processes, both past and future. With its emphasis on a multidisciplinary and pan-European approach, the Foundation provides the leadership necessary to open new frontiers in European science. Cite This Page:

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On this day in 1941, the German invasion of the Soviet Union begins a new stage, with Hitler’s forces capturing Mariupol. The Axis power reached the Sea of Azov. Despite the fact that Germany and Russia had signed a “pact” in 1939, each guaranteeing the other a specific region of influence without interference from the other, suspicion remained high. Despite warnings from his advisers that Germany could not fight the war on two fronts (as Germany’s experience in World War I proved), Hitler became convinced that England was holding out against repeated German air assaults, refusing to surrender, because it had struck a secret deal with Russia. Fearing he would be “strangled” from the East and the West, he created, in December 1940, “Directive No. 21: Case Barbarossa”–the plan to invade and occupy the very nation he had actually asked to join the Axis only a month before. On June 22, 1941, after having postponed the invasion of Russia when Italy’s attack on Greece forced Hitler to bail out his struggling ally in order to keep the Allies from gaining a foothold in the Balkans, three German army groups struck Russia hard by surprise. The Russian army was larger than German intelligence had anticipated, but they were demobilized. Stalin had shrugged off warnings from his own advisers, even Winston Churchill himself, that a German attack was imminent. By the end of the first day of the invasion, the German air force had destroyed more than 1,000 Soviet aircraft. And despite the toughness of the Russian troops, and the number of tanks and other armaments at their disposal, the Red Army was disorganized, enabling the Germans to penetrate up to 300 miles into Russian territory within the next few days. Hitler’s battle for Stalingrad and Moscow still lay ahead, but the capture of Mariupol, at the sea’s edge, signaled the beginning of the end of Russia-as least as far as Hitler’s propaganda machine was concerned. “Soviet Russia has been vanquished!” Otto Dietrich, Hitler’s press chief, announced to foreign journalists the very next day.

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In a previous post, we discussed how researchers at the Illinois Sustainable Technology Center (ISTC), on the campus of the University of Illinois at Urbana-Champaign have developed an energy-efficient, non-toxic, nondestructive chemical process to recover polymers from the complex plastic blends found in items like cellphone cases. But that’s not the only exciting news this Earth Month related to innovations in reclaiming materials from electronic scrap (commonly referred to as “e-waste”). In a GreenBiz article dated 4/18/18, Heather Clancy highlights an electrochemical process developed by Canadian venture EnviroLeach Technologies, which is similar to the conventional method of leaching gold and other metals out of ores, concentrates and tailings. The difference is that “instead of using cyanide, the patent-pending formula uses five non-toxic, FDA-approved ingredients that are combined with water at ambient temperatures.’The process does not require pressure, elevated temperatures, complex process circuits, intensive gas monitoring or costly detoxification systems,’ explained EnviroLeach on its website.” Read the full story on the GreenBiz web site. You can also check out the EnviroLeach web site for further information. This development is particularly encouraging considering a recent article from Environmental Leader reporting that n a study by researchers from Tsinghua University in Beijing and Macquarie University in Australia, which suggests extracting metals from e-waste costs 13 times less than mining ore. Perhaps the new process will make the economic benefit even more striking, while minimizing environmental impacts. Elsewhere in Canada, researchers at the University of British Columbia “have perfected a process to efficiently separate fibreglass and resin – two of the most commonly discarded parts of a cellphone – bringing them closer to their goal of a zero-waste cellphone.” As UBC News reports, “Most e-waste recycling firms focus on recovering useful metals like gold, silver, copper and palladium, which can be used to manufacture other products. But nonmetal parts like fibreglass and resins, which make up the bulk of cellphones’ printed circuit boards, are generally discarded because they’re less valuable and more difficult to process. They’re either fed to incinerators or become landfill, where they can leach hazardous chemicals into groundwater, soil and air.” But UBC mining engineering professor Maria Holuszko, along with PhD student Amit Kumar, has developed a process using gravity separation “and other simple phycial techniques to process cellphone fibreglass and resins in an environmentally neutral fashion.” The next step in pursuing this innovation is developing a large-scale commercial model of the process with their industrial partner and recycling company Ronin8. Read the full UBC article on the UBC News web site. Read more at https://ifixit.org/recycling on why electronics recycling, though of course important, should not be considered the answer to the problem of ever-growing amounts of e-waste, due to the difficulty in reclaiming materials (eased slowly by new innovations like the ones described above) and energy use. While these developments in electronic scrap recycling are heartening, it’s important to remember to keep your electronics in service as long as possible through repair and upgrades, and when you no longer want or need a functioning device, sell or donate it so someone else can use it. Recycling should only come at the ultimate end of a device’s useful life.

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In this tutorial, we will learn the Support Vector Machine algorithm and implement it in Python. Support Vector Machine: Support Vector Machine is a discriminative classifier which finds the optimal hyperplane that distinctly classifies the data points in an N-dimensional space(N - the number of features). In a two dimensional space, a hyperplane is a line that optimally divides the data points into two different classes. How the Algorithm Works: Let's say you need to classify two different classes of data points in a two-dimensional space. Look at the following illustration. Here we see two classes of data points, one in the red and other in green. Now, what can we do to separate these classes? We can simply draw a line that separates them. This line could be drawn anywhere in the plane. Here, any of the lines can separate the classes. But our task is to find the best fit or optimal line that classifies the data points most accurately. Here the Support vector machine can help us to do so. This algorithm finds us the optimal line/hyperplane. It does so by finding the line with the maximum margin(i.e. the highest distance between data points of both classes). Here support vectors are those two data points that are supporting the decision boundary(the data points which have the maximum margin from the hyperplane). ThatÃÆââââ¬Å¡Ã¬Ã¢ââ¬Å¾Ã¢s why this algorithm is called support vector machine. Note: In higher dimensional space(more than two dimensions), the classes cannot be represented as single data points, so they are represented as vectors. This is one of the simplest but yet a powerful algorithm to solve classification problems. SVM in python: Now we will implement this algorithm in Python. For this task, we will use the dataset Social_Network_Ads.csv. Let's have a glimpse of that dataset. This dataset contains the buying decision of a customer based on gender, age and salary. Now, using SVM, we need to classify this dataset to predict the decision for unknown data points. You can download the whole dataset from here. First of all, we need to import the essential libraries to our program. import numpy as np import matplotlib.pyplot as plt import pandas as pd Now, lets import the datset. dataset = pd.read_csv('Social_Network_Ads.csv') In the dataset, the Age and EstimatedSalary columns are independent and the Purchased column is dependent. So we will take both the Age and EstimatedSalary in our feature matrix and the Purchased column in the dependent variable vector. X = dataset.iloc[:, [2, 3]].values y = dataset.iloc[:, 4].values Now, we will split our dataset in training and test sets. from sklearn.model_selection import train_test_split X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.25, random_state = 0) We need to scale our dataset for getting a more accurate prediction. from sklearn.preprocessing import StandardScaler sc = StandardScaler() X_train = sc.fit_transform(X_train) X_test = sc.transform(X_test) Well, its time to fit the SVM algorithm to our training set. For this, we use the SVC class from the ScikiLearn library. from sklearn.svm import SVC classifier = SVC(kernel = 'linear', random_state = 0) classifier.fit(X_train, y_train) Note: Here kernel specifies the type of algorithm we are using. You will know about it in detail in our Kernel SVM tutorial. For simplicity, here we choose the linear kernel. Our model is ready. Now, let's see how it predicts for our test set. y_pred = classifier.predict(X_test) To see how good is our SVM model is, let's calculate the predictions made by it using the confusion matrix. from sklearn.metrics import confusion_matrix cm = confusion_matrix(y_test, y_pred) The output of the confusion matrix will be Now, let's visualize our test set result. # Visualising the Test set results from matplotlib.colors import ListedColormap X_set, y_set = X_test, y_test X1, X2 = np.meshgrid(np.arange(start = X_set[:, 0].min() - 1, stop = X_set[:, 0].max() + 1, step = 0.01), np.arange(start = X_set[:, 1].min() - 1, stop = X_set[:, 1].max() + 1, step = 0.01)) plt.contourf(X1, X2, classifier.predict(np.array([X1.ravel(), X2.ravel()]).T).reshape(X1.shape), alpha = 0.75, cmap = ListedColormap(('red', 'green'))) for i, j in enumerate(np.unique(y_set)): plt.scatter(X_set[y_set == j, 0], X_set[y_set == j, 1], c = ListedColormap(('red', 'green'))(i), label = j) plt.title('SVM (Test set)') The graph will like the following From the above graph, we can see that our model tries to find the optimal line that separates the data points accurately. This tutorial only explains SVM in two-dimensional space, in the next tutorial we will see SVM in higher dimensional spaces.

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# Engineering Technology Mathematics -Limits and Continuity- Topics: Continuous function, Limit of a function, Mathematical analysis Pages: 14 (1279 words) Published: January 31, 2012 FKB20203 Engineering Technology Mathematics 2 Lecture 2: Limits and Continuity Lecturer: Norhayati binti Bakri ( ([email protected]) WEEK 2 Objective: To evaluate limits of a function graphically and algebraically To determine the continuity of a function at a point Limits (a) (b) A 1. in everyday life in mathematics Limits – Graphical Approach Examples f(x) = x + 2  x+2 , x ≠ 2 h(x) =  , x=2  3 7 6 5 4 3 2 1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 4 5 g(x) = x2 − 4 x −2 7 6 5 4 3 2 1 0 -3 -2 -1 0 7 6 5 4 3 2 1 0 -3 -2 -1 0 1 2 3 4 5 Finding limits: at x= -4 at x= -3 at x= -2 at x= -1 at x= 0 at x= 1 at x= 2 at x= 3 at x= 4 [email protected] | LIMITS and CONTINUITY CONTINUIT 1 2. One Sided Limits (a) (b) x approaches c from the right side x approaches c from the left side oaches x → c+ lim f(x) = L x → c− lim f(x) = L 3. Two Sided Limits Two sided limits exists if and only if the one sided limits exist and are equal. x → c+ lim f(x) = lim− f(x) = L x →c 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 then lim f(x) = L x→c 2.5 3 3.5 Finding one-sided limits: at x= 1 at x= 2 [email protected] | LIMITS and CONTINUITY CONTINUIT 2 4. Infinite Limit As x approaches a number, the limit is infinity oaches x → c− lim f(x) = ∞ , lim f(x) = ∞ x→c , x → c+ lim f(x) = ∞ x → c− lim f(x) = ∞ , lim f(x) = ∞ x→c , x → c+ lim f(x) = ∞ Finding limits: at x= 1 at x= 2 at x= 3 at x= 4 [email protected] | LIMITS and CONTINUITY CONTINUIT 3 5. Limit at Infinity As x approaches infinity (positive or negative), the limit is a numerical value x → +∞ lim f(x) = L , x → −∞ lim f(x) = L Finding limits: at x= 4 at x= -4 6. Limits and Asymptote Asymptotes are defined with respect to limits at infinity and infinite limits. [email protected] | LIMITS and CONTINUITY CONTINUIT 4 (a) Horizontal Asymptote x → −∞ lim f(x) = L or x→ ∞ lim f(x) = L As x decreases or increases without bound, the function approaches a numerical value. The horizontal line, y=L is an asymptote. (b) Vertical Asymptote x → c− lim f(x) = ∞ or x → c+ lim f(x) = ∞ As x approaches the value c (from left or right), the function increases without bound x → c− lim f(x) = −∞ or x → c+ lim f(x) = −∞ As x approaches the value c (from left or right), the function decreases without bound 7. When do limits fail to exist? B 1. Limits – Algebraic Approach Principal Limit Theorem (Main Limit Theorem) Let c and k be real numbers, n≥0 and let f and g be functions with limits at c such that x→ c lim f(x) = L lim k = k or x→ c lim g(x) = M (a) (b) (c) Constant Rule: Identity Rule: Coefficient Rule: x→ c x→ c lim x = c lim k f(x) = k lim f(x) = k L x→ c x→ c [email protected] | LIMITS and CONTINUITY 5 (d) (e) (f) Sum Rule: Difference Rule: Product Rule: x→ c lim (f(x) + g(x) ) = lim f(x) + lim g(x) = L + M x→ c x→ c x→ c x→ c x→ c lim (f(x) − g(x) ) = lim f(x) − lim g(x) = L − M x→ c x→ c x→ c lim (f(x) . g(x) ) = lim f(x) . lim g(x) = LM (g) Quotient Rule: lim f(x)  f(x)  x → c L lim   g(x)  = lim g(x) = M  x→ c   x→ c lim provided that x → c g(x) ≠ 0 n   =  lim f(x)  = Ln x→ c    (h) Power Rule: x→ c lim (f(x) ) n 2. Examples Example 1 Assume that lim f(x) = 7 and lim g(x) = −3 , find x →b x →b x →b lim (f(x) + g(x) ) lim (3 f(x)g(x) ) lim 4 x g(x) Ans : [ 4 ] Ans : [ − 63 ] Ans : [ − 12b ] x →b x →b x →b lim g(x) f(x) Ans : [ − 3 ] 7 Example 2 Example 3 lim (2x 3 − 3x 2 + x + 1) x →2 Ans : [ 7 ] Ans : [ 216 ] x →3 lim 2x 3 x 2 + 7 Example 4  2x − 1  lim  3  x →1 x − 2   Ans : [ − 1 ] Example... Please join StudyMode to read the full document

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# proof of pigeonhole principle ###### Proof. It will first be proven that, if a bijection exists between two finite sets, then the two sets have the same number of elements. Let $S$ and $T$ be finite sets and $f\colon S\to T$ be a bijection. The claim will be proven by induction on $|S|$. If $|S|=0$, then $S=\emptyset$, and $f\colon\emptyset\to T$ can only be surjective if $T=\emptyset$. Assume the statement holds for any set $S$ with $|S|=n$. Let $|S|=n+1$. Let $x_{1},\dots,x_{n+1}\in S$ with $S=\{x_{1},\dots,x_{n+1}\}$. Let $R=S\setminus\{x_{n+1}\}$. Then $|R|=n$. Define $g\colon R\to T\setminus\{f(x_{n+1})\}$ by $g(x)=f(x)$. Since $R\subset S$, $f(x)\in T$ for all $x\in R$. Thus, to show that $g$ is well-defined, it only needs to be verified that $f(x)\neq f(x_{n+1})$ for all $x\in R$. This follows immediately from the facts that $x_{n+1}\notin R$ and $f$ is injective. Therefore, $g$ is well-defined. Now it need to be proven that $g$ is a bijection. The fact that $g$ is injective follows immediately from the fact that $f$ is injective. To verify that $g$ is surjective, let $y\in T\setminus\{f(x_{n+1})\}$. Since $f$ is surjective, there exists $x\in S$ with $f(x)=y$. Since $f(x)=y\neq f(x_{n+1})$ and $f$ is injective, $x\neq x_{n+1}$. Thus, $x\in R$. Hence, $g(x)=f(x)=y$. It follows that $g$ is a bijection. By the induction hypothesis, $|R|=|T\setminus\{f(x_{n+1})\}|$. Thus, $n=|R|=|T\setminus\{f(x_{n+1})\}|=|T|-1$. Therefore, $|T|=n+1=|S|$. ∎ Title proof of pigeonhole principle ProofOfPigeonholePrinciple 2013-03-22 13:31:09 2013-03-22 13:31:09 Wkbj79 (1863) Wkbj79 (1863) 8 Wkbj79 (1863) Proof msc 03E05

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The European earwig was known only from a few localities east of the Mississippi River in 1940. These sites were in the coastal areas of Massachusetts and Rhode Island and in upstate New York near the Great Lakes -- a total of 12 observations. By 1970, only a few scattered counties in New York had not been reported as having serious infestations. The insect is also reported in neighboring counties of Pennsylvania, New Jersey, Connecticut, Massachusetts and Vermont. Reports of annoyance and damage increase each year. The European earwig has been known widely on the West Coast since the early 1900's and has moved eastward to the Plains states. European earwigs generally feed as scavengers on dead insects and rotting plant material, but they are also reported as feeding on flower blossoms, lettuce, and other succulent garden plants, especially when populations are numerous. A few cases of earwigs feeding on aphids or other insects have been reported. In addition to their feeding activities, earwigs often occur in close proximity to people, even getting into houses and garages, especially during periods of warm wet weather. Once indoors they seek out moist areas and thus may be found in basements, kitchens, and occasionally in bathrooms. Inside they are nuisance pests, and they may feed on stored paper or fiber products especially if they are stored in moist situations. The earwigs are nocturnal and during the day they rest in dark, moist places. The most distinctive feature of earwigs (Fig. 1) is the pair of forceps on the tip of the abdomen. On the male the forceps are strongly curved, in the female they are nearly straight. The adult is about 18 mm (5/8") long, a somewhat flattened elongated insect, dark red-brown in color, with short wing covers. It seldom flies. The young (nymphs) are similar to the adults, gray-brown in color and lacking wings. The female earwig deposits 20 to 60 white, nearly spherical eggs in a cell in the soil at a depth of 15 mm. Depending on temperature, incubation lasts from 12 to 85 days, eggs produced early in the spring requiring the longest to hatch. The female guards the eggs and newly hatched young, sometimes for a year or longer. A year or more is required for development, and there is one generation per year. Both eggs and young require moisture although heavy rains are not tolerated. The adults can survive extended periods of dryness. If a sudden invasion occurs, earwigs can be vacuumed up or swept up and disposed of. Infestation of the home can be limited by removing damp articles and debris and by taking measures to dry out moist areas. Physically block openings through which earwigs may enter with screening or caulk. Moist leaf mold and mulches should be kept at least 3 feet away from the foundation, window wells and doorways. For long-term management look for sources of moisture and correct them. Trim back vegetation that shades and contributes to moisture retention around the foundation or other parts of the structure. Ground covers may need to be removed from along foundation walls. Clean gutters and repair broken downspouts as needed. Move log piles away from the structure. Grade the property so that water drains away from the foundation, not toward it. Be sure crawl spaces are well ventilated to remove moisture. People have successfully made traps of rolled newspaper or other tubular containers. Traps are placed outside prior to darkness, and checked; earwigs emptied out and disposed of, the following morning. Traps may also be useful as monitoring devices to help you determine where control efforts should be focused. When earwigs are numerous outdoors, invasions of the home or of buildings can be expected. If vacuuming is not enough, household formulations of the insecticides silica gel plus pyrethrum, Baygon, bifenthrin, cyfluthrin, resmethrin, or tetramethrin are registered for use indoors. Spot crack and crevice treatments can be made in difficult to reach harborages. Pesticides, however, only provide temporary control, and if the conditions conducive to infestation are not corrected, reinfestation may occur. Outdoor harborages may need to be addressed. Correcting situations that provide hiding places and harborage should be stressed. Insecticides registered for earwig control outdoors include Baygon, bifenthrin, permethrin or resmethrin. They are used around foundation walls as a spot or perimeter treatment. Remember that the insecticide treatment is only a temporary measure. 2/03, Revised by: Carolyn Klass. This publication contains pesticide recommendations. Changes in pesticide regulations occur constantly and human errors are still possible. Some materials mentioned may no longer be available and some uses may no longer be legal. Download a PDF of this tip.

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A common running injury is a calf strain or a tear. The calf muscles, Gastrocnemius and soleus, are loaded repetitively and heavily during running. With every stride we take when running, the calf gets loaded, firstly to absorb the shock of our body weight landing, then to help propel us forward into the next stride. When running, we take roughly 1500 strides per mile. Which makes it easy to see that if there is a weakness in the calf complex, or a fault elsewhere in the kinetic chain or running technique leading to increased load on the calf, injury is almost inevitable. The calf is classified as a global mobilizer muscle, meaning that its main anatomical function is to absorb and then create large motions and forces. It is accompanied above and below by stabiliser muscles which are responsible for keeping the joints stable – so that it can carry out its main function. However, if stability is compromised, particularly at the foot and ankle complex, leading to excessive pronation, the calf will begin to try and take on a stabilising role also – leading to loading it is not positioned well to cope with. An example of which would be an overpronating foot or weak glutes causing excessive inward rotation of the knee. Often a poor warm-up is cited as a reason why athletes sustain calf injuries. Most of us appreciate the necessity for a thorough warm-up. I often use ‘blue tack’ as an example when describing how muscles and tendons respond to a warm up. When you try and stretch cold blue tack it is tough and usually breaks, whereas when is has been warmed up it stretches nicely. It is also important to note that as we age, these elastic properties of tendons and muscles diminish – thus accounting for the increased occurrence of calf strains in the more senior of our athletic population. A final contributor to soft tissue injuries in runners, especially long-distance runners is dehydration. Dehydration negatively impacts muscle function by reducing blood flow to muscles and decreasing muscle elasticity or flexibility and endurance. Grades of strain or tear: Muscular strains are classified according to their severity in terms of how many fibres have been disrupted or ruptured |Grade 1||Grade 2||Grade 3| |This is the least severe of calf injuries. A small number of muscle fibres have been damaged within the muscle. Signs and symptoms of this type of less serious strain may not be noticed until cessation of the activity. Tightness, cramping feelings and slight soreness are common when the muscle is stretched.||This is sometimes referred to as a partial calf tear. A greater number of muscle fibres have been torn, but the muscle remains largely intact. More immediate localised calf pain is present during activity, especially walking and running. Often the area is sore to touch.||Total rupture. All the muscle fibres have been torn, losing continuity throughout the muscle. This is a serious injury and highly disabling. The athlete will be unable to walk pain free. Often bruising will appear below the tear site and there may well be a palpable bulge where the calf muscle has recoiled upon itself.| Treatment and Rehabilitation: Initially, the Rest, Ice, Compression, Elevation (R.I.C.E) principal should be followed. Therapeutic Ultrasound, Acupuncture, Sports massage and Taping are all methods used to facilitate soft tissue healing. Sports massage, however should not be performed until the acute phase has passed (3 days +). Approximate timescales for rest are; 3 weeks for a grade 1 strain and 4-6 weeks for a grade 2 strain. Grade 3 tears will most likely require surgery followed by a 12 week rehabilitation programme. As with any injury, progressive and comprehensive exercise based rehabilitation is key to avoiding recurrence or secondary injuries. Secondary injuries often occur through compensatory mechanisms which may have become habit during the injured period. The rehabilitation period is also an ideal opportunity to target those areas that get ignored in weekly training routines. Core stability and gluteal muscles are a great place to focus on when activity is restricted. Research has found that the stronger these muscles can become, the more likely a successful outcome is to be reached in terms of injury recovery, injury prevention and most importantly performance. Specific single-leg exercises are important to build the strength in the injured limb and to regain balance which will have been lost on the injured side during the injury period. Here are some suggested exercises. However, I recommend you see a Physiotherapist for a tailored program and appropriate guidance. Single Leg Standing Stand on one leg keeping your bottom squeezed and core engaged. Ensure that your pelvis is level, your knee is facing forward and your trunk is upright. Try not to lock your knee. If you can successfully hold this position on each leg for 15-20 seconds you are ready to progress to a single leg squat. This exercise targets your core and glute muscles. The same principles are applied in this exercise as in the single leg stand. The picture shows the athlete squatting to a chair. The chair provides a nice prompt to ensure that you are squatting correctly (sticking your backside out and not just bending at the knee). I suggest that you start by only squatting down as far as you can control your knee (keep you knee cap over your 2nd toe) and keep your pelvis level. This may only be a tiny dip to start with, but it will improve. Supported Heel Raise This exercise should be pain-free and should therefore not be considered until you are symptom free walking up a flight of stairs. Begin with 50:50 of your body weight in both feet and raise up on to your toes, if you feel the need you may hold onto a rail/kitchen unit for support. Complete 3 sets of 10. If this is easy then you may progress on to 60:40, increasing the load in the injured side. The increases my continue 70:30, 80:20 up to a single leg heel raise. There are TWO main stretches to perform for the calf complex. One is for Gastrocnemius (straight knee) and the other is for Soleus and the Achilles (bent knee). Hold the stretches for 30 seconds as this will promote true lengthening of the soft tissue. Once improvements have been made in single leg strength and balance, low level plyometric exercises may be re-introduced as a precursor to running. Jumping, hopping and skipping are all useful to re-introduce the muscle to the dynamic loading needed for running. You should seek professional advice before starting a plyometric program. Following successfully progressing through the multi-directional plyometric exercises, running may be gradually re-introduced. A sure way to cause re-injury is to do too much too soon at this stage. Running should not be increased by more than 5% per week in intensity, duration or frequency. Running technique should also be monitored at this time to ensure efficency and avoid overloading of any part of the kinetic chain. - Go and see a Physiotherapist! If they are any good they will look at you holistically and identify why you have sustained the injury in the first place. You should then receive bespoke treatment and a rehabilitation program. - Stay hydrated before, during and after sport. - Incorporate Strength and Conditioning into your weekly training regime – focus on Core Stability and Glutes. - Make sure you warm up gradually and cool down properly – cool downs will prevent you getting too tight in your muscles. Excessively tight muscles will lead to injury. - Have the correct footwear. If you are running off road you will need a shoe that is lightweight and provides grip on those unpredictable surfaces. If you are road running your trainer must be up to date, no older that 2 years old or clocked over 500 miles. Cushioning and the correct level of support is vital when you are running on such unforgiving surfaces. Many running specialist stores now provide gait analysis to ensure the correct shoe fitting. - Consider compression gear. Calf guards or compression socks have been proven to reduce muscle vibration and assist with circulation during and after sport.

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When stink bugs are frightened or disturbed, they emit an odor from their glands in their thorax. They are called stink bugs simply because they stink. These bugs can be found in fields, yards and gardens, and at times they can also be found indoors such as in homes, offices, building and hotels. While there are around 221 species of stink bugs, their life cycle is almost the same. Knowing the Stink Bug Life Cycle Waking up from hibernation. It is during winter that these bugs wake up from their hibernation beneath the earth or leaf litter. During spring, these bugs look for food. In most cases, when the female bug comes out, she is carrying eggs with her and is just trying to find a place to where to have them. Depending on the climate and their location, these eggs may be laid as early as mid-April or as late as mid-December. Stink bugs eggs. The eggs are commonly white when laid in rows of clusters, but they change to pink as they get closer to hatching. Their mothers watch over them until it is time for the eggs to hatch and the tiny nymphs emerge. The nymphs. As the nymphs begin to break free through the protective membrane, the other nymphs will also follow. The only purpose of these nymphs once they emerged from their eggs is to just eat. When they come out, they look the same as an adult stink bug, but looks rounder rather than shield-shaped. Through the course of four to five weeks, these nymphs will go through five instars. Instars are the stages between molts in an insect’s life. For nymphs, the first and second instars look like tick-like, but are yellowish or reddish in color. The final three instars will make the nymphs darker and closer in appearance to the adults. Adults. Three weeks after her final molt, the female stink bugs can lay her first pod of eggs, which would then start the whole process again. The adult stink bugs mostly feed on plants or insects. These bugs may mate up to four times per year, which could just take for only a few minutes of just a few days. Their Life Cycle from Fall through Winter Every fall, these bugs thrive even more. They will get inside homes and you may be able to see the bugs near windows and attics. During fall, aggregation of insects on sunny sides of the home are common. As the summer progresses, these insects will seek warmer places to spend their days where you might be able to see them even more. As winter approaches, these bugs will seek permanent shelter from the cold. They may even spread the word about a preferred overwintering site through the use of aggregation pheromones, which most would join in. Although many of these bugs would die off when the cold winter comes, but most of the female stink bugs will leave eggs behind to start a new population for next year. Others would simply migrate to warmer climates. Some would stay in burrows in the leaf litter or hinder under loose bark to protect them from the frost. However, your home is still prone to bed bug invasion as they seek shelter from the cold and a place where they could lay their eggs. Dealing with Stink Bugs at Home The sudden appearance of dozens of even hundreds of insects in your home may cause for an alarm, but there are still known simple remedies to get rid of stink bugs in your home. Keep in mind that stink bugs do not bite. They also do not infest your pantry like other pests do, and they do not do any structural damage to your home. The main reason that they are staying in your home is so they can survive the winter. To prevent pest entry make sure to: - Inspect your home for opening or cracks and repair them once seen. - Install insect screening over chimney caps and gable vents and windows. Once stink bugs enter the structure Do not use insecticide once these insects have invaded the wall voids and the attic areas. Even though insecticides may kill most of these bugs, there is still a chance that other insects may consume the dead bugs, which could then attack other foods and items inside the home. Do not squish these insects because they emit a foul defensive odor when injured or threatened. Vacuum clean the area where the bugs are residing instead of using chemical pesticides, then remove the vacuum bag and dispose the trash properly when done. Don’t let stink bugs stink up your home. If you have an existing stink bug infestation, let us help you! Click New Jersey Stink Bugs Removal.

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October 20, 2011 — If we see a bird or some other creature, and it is close enough, there is something that usually catches our eye early in our observation: the eyes. In nature, there are many adaptations of vision. Earthworms have simple eyes (ocelli) that are able to detect light and dark, while eagles and hawks have single lens eyes with visual acuity five times greater than a human. (An eagle has five times the density of vision cells at the central focus area of the retina than humans do.) Each animal species in its habitat requires different vision adaptations in order to survive and reproduce. Look at the images to get an “eyeful” of these varied vision adaptations.

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Wikijunior:Big Cats/In danger of extinction Extinction is Forever Some big cats throughout history have become extinct because they were replaced with newer species better suited to the environment. The Sabretooth (Smilodon fatalis) is one example of a large Ice-Age predator that died out because the large prey it needed retreated with the glaciers. Pumas and jaguars now roam where the mighty Sabertooth once ruled. Natural extinction is part of the grand drama of life on Earth. However, many more cat species are in danger of dying out due to unnatural extinction, the killing of an entire species by man for reasons having nothing to do with fitness for survival. These species are not replaced with newer ones, their death merely leaves a hole in the fabric of life on Earth. Many big cats have been killed because they either compete with humans for the same prey animals or because they occasionally attack human-raised livestock. Some big cats that become too weak to hunt their own natural prey find domestic livestock much simpler to acquire. Other big cats develop a taste for livestock out of sheer opportunity. There are times when control of individual predators, through moving or killing, appear to be justified. However there is a much more dangerous approach to predator control where an entire population or even an entire species is classified as a "pest" and open to extermination. Extermination is an attempt to kill every last individual of a population or species. There were times when pumas were targeted for extermination in large areas of the American west. Bobcats and jaguars have also been targets of extermination. These days most governments in the world agree that extermination is not a good way to control cats, but sometimes local peoples ignore laws designed to protect species from extermination. The majority of people in western countries no longer give big game hunters the same respect they once held in the writings of Ernest Hemmingway. The cheetah, which was once abundant in India, was hunted to complete extinction there. The Mughal emperor Akbar killed nearly 1000 cheetahs during his lifetime when the number of cheetahs was already declining. The Asian lion met with the same fate. Most outdoorsmen no longer seek trophies for their mantles and entrance halls. However, a number of people still consider locating, outwitting, and defeating large predators to be the ultimate test of courage and a satisfying form of enjoying the out of doors. This practice is losing popularity, though. In all fairness, it should be said that sport hunters support laws and practices that benefit wildlife. In the United States, wildlife populations have increased within the past century. This is largely due to funds generated via an excise tax on hunting equipment known as the Pittman-Robertson Act. In addition, sportsmen contribute hundreds of millions of dollars each year to wildlife conservation through sporting organizations that benefit all wildlife. People who defy existing laws to kill predators for money, animal parts, or personal reasons are called poachers. As outlaws, many poachers are dangerous people who are willing to protect their livelihood through violent means. Famous conservation leaders George Adamson and Diane Fossey were killed by poachers who saw them as a threat. Stopping poaching is very difficult because most big cat habitat is remote land that is difficult to patrol and exists in some of the world's poorest countries without many law enforcement resources. The most effective way to curb poaching is to reduce the demand for the products they provide. A number of people believe, without any scientific evidence, that folk medicines made from parts of big cats can treat or even cure certain illnesses and conditions. Belief in sympathetic magic, that like-causes-like, leads people to seek the attributes they most admire about big cats by using parts of their bodies. People seeking courage, strength, or a greater capacity for physical intimacy attempt to acquire those things through eating, drinking, applying or wearing parts of the animals that are supposed to possess those traits. For nearly everything supposedly treatable with feline folk medicines, there are effective, safe and proven remedies available in modern medicine. The Fur Trade The soft, warm, boldly patterned pelts (skins with fur) of big cats were long considered the ultimate expression of fashion and extravagance. Even today, most fashion items made with real fur come from carnivores such as bobcats and mink. Those legal for sale are produced from animals raised on fur farms rather than taken from the wild. The vast majority of natural leopard, ocelot, lynx and jaguar furs are banned on the international market by laws protecting endangered species. Habitat loss is the silent killer. An animal's habitat is an area where it can live, and for most large predators that means cover, adequate prey, freedom of movement, and water. Due to their predatory lifestyle, most big cats require large areas of land without many manmade barriers where they can hunt and raise young unmolested. Uncontrolled development of wild areas, including such wasteful practices as slash-and-burn agriculture, reduce the number of places where big cats can survive and thrive. To some degree protected areas known as Parks and Wildlife Sanctuaries help preserve endangered species habitat. However in many poor countries there is not adequate law enforcement to prevent poaching or illegal development of land inside park boundaries. In addition, animals need more land than the human race can afford to protect in parks. More enlightened use of habitat by man can increase the number of big cats and preserve their genetic diversity. For instance, a timber plantation can provide both high quality wood and habitat for predators and their prey. Using sustainable management techniques, land can provide a never-ending source of quality wood products while continuing to preserve wildlife. It Is Up To You As someone interested in big cats, you can make your love of big cats known through the way you vote, your lifestyle, and your charitable giving. Governments can only do so much to help stop extinction. For big cats to be saved, they must be saved by all of us working together. Learn what you can do about the challenges facing your favorite animals, and get involved. Always remember: "We appreciate what we understand and save what we appreciate."

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Question Jan's hypothesis H is that her car needs a new transmission belt, and the potential evidence... Jan's hypothesis H is that her car needs a new transmission belt, and the potential evidence E is that her mechanic has said that it probably does. Which of the following cases would be a case where observing E would count as evidence in favour of H? P(E) = 57% and P(H) = 47% P(H given E) = 57% and P(E given H) = 47% P(H) = 57% and P(H given E) = 47% P(H given E) = 57% and P(H) = 47% Say we have two events A and B and we know the probability that they occur is P(A) = 0.4 and P(B) = 0.2 respectively. What condition must be satisfied in order to say that the probability that either A or B occurs is 0.6? The probabilities must be related. The probabilities must be less than one. The events should not both be able to happen at the same time. The probabilities must be independent Match each of the scenarios with the definition of probability it relies on: 123 You decide that you have a 95% chance of passing this unit. 123 Collect 283 Pokemon of which 97 are Zubats. Assume that the probability a random Pokemon is a Zubat is 97/283. 123 Toss a fair coin. Assume the probability of it coming up Tails is 50%. 1 Long run frequency interpretation. 2 Equally likely outcomes. 3 Subjective probability. Problem 1: P(H given E) = 57% and P(E given H) = 47% Problem 2: P(A or B)=P(A)+P(B)-P(A and B)=0.6 or, P(A and B)=0.6-0.6=0 (since P(A)=0.4, P(B)=0.2) Hence A and B are mutually exclusive. Option: The events should not both be able to happen at the same time. Problem 3: You decide that you have a 95% chance of passing this unit 3. Subjective probability Collect 283 Pokemon of which 97 are Zubats. Assume that the probability a random Pokemon is a Zubat is 97/283 2. Equally likely outcomes Toss a fair coin. Assume the probability of it coming up Tails is 50% 1. Long run frequency interpretation. Earn Coins Coins can be redeemed for fabulous gifts.

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Thursday, 31 January, 2013 The importance of the envelope In a project that began with the retinal cells of nocturnal animals and has led to fundamental insights into the organization of genomic DNA, LMU researchers show how the nuclear envelope affects nuclear architecture - and gene regulation. The double-stranded DNA molecules that make up the genetic material are wrapped around protein complexes to form compacted “chromatin”. The active portion of the genome is less densely packed, and thus more easily accessible, than the inactive fraction, and is referred to as euchromatin. Euchromatin is typically located in the inner regions of the cell nucleus, while much of the inactive DNA in “heterochromatin” is associated with the inner face of the nuclear envelope. This type of chromatin organization is found in almost all higher organisms and may have been invented 500 million years ago.

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# Behind Opinion Polls: Fundamentals and Reliability | This article is quite technical and requires a mathematical background for a comprehensive understanding. All technical points have a reference to the corresponding Wikipedia’s article for further explanations. We are in an election time in France. Everyday news is rythmed by polls. Nicolas Sarkozy is falling from 2% in today’s poll from IPSOS, Francois Hollande slipped below 33% after the crisis etc… Everybody seems to give a lot of credit to these polls forgetting most of the time that they are just statistical estimations of the reality. But how reliable are they? The results for the first round of the french presidential election seemed to surprise so many people and medias that it pushed me to dig into my maths lessons from college to look for some scientific explanation if any. ## Organizing the Opinion Poll Opinion polls tend to estimate the percentage of people who will vote for a given candidate. Polling Institutes want to produce accurate polls but they can’t poll a too big sample as it would be time consuming and expensive. They choose a representative sample of the population and ask them for whom they plan to vote. The responses are assumed independant. Once collected, how do we use them build an estimate of the votes and what is the precision of this estimate? The natural and logical estimate is to take the sample mean as an estimator. How do we justify such an estimate and what is its precision? We’ll see that what sounds like common sense has a strong mathematical justification. ## Building the Estimate We are in a situation where we have n people $X_1, X_2,…X_n$ who can choose between k candidates. We define the variable $\delta_{i,j}$ which is equal to 1 if person i votes for candidate j and 0 if not. The real vote intentions we want to estimate are noted : $\pi_1, \pi_2,…\pi_k$ with the constraint: $\sum_j^k \pi_j = 1$ The probability to observe the result of the polling is: $p_{\pi}(X_1, X_2,…,X_n) = \prod_i^n p_{\pi} (X_i)$ as the n people are chosen independantly. $\prod_i^n p_{\pi} (X_i) = \prod_j^k \prod_i^n \pi_j^{\delta_{i,j}} = \prod_j^k \pi_j^{n_j}$ where $n_j = \sum_i^n \delta_{i,j}$ The strategy we adopt is, given the n people, to maximize this propability of observing the output. Maximizing this propability is the same as minimizing the -log of the probability as -log is a convex function. With this transformation, the computation of the extremum is much easier as the product becomes a sum. $max[p_{\pi}(X_1, X_2,…,X_n)] = min[(-log(p_{\pi}(X_1, X_2,…,X_n))] = min[ \sum_j^k n_j log(\pi_j) ]$ To find the minimum with the above constraint we can use the Lagrangian multiplier and just minimize: $\sum_j^k n_j log(\pi_j) + \lambda(\sum_j^k \pi_j -1)$ which gives us the estimates which minimize the above function $\hat{\pi_1}, \hat{\pi_2},…\hat{\pi_k}$ $\hat{\pi_j} = \frac{n_j}{n}$ We just justified the use of the empirical mean as an estimate of the true mean given a set of observations using maximum a posteriori estimation. ## Reliability Given the estimator, which level of trust can we give it? How close is are the estimated and true values. The law of large numbers tells us that when the size of the sample goes to infinity the estimated value tends to the true value. But in reality there is not an infinity of people who get polled. Usually just a few thousands are. From one sample to the other the estimator won’t be the same. So how confident are the polling institutes in their estimations? The reliability of the estimator depends on the statistical fluctuations of the estimate around the true value. The central limit theorem tells us that the random variables (m is the true mean and sigma the variance): $\frac{\hat{\pi_j} -nm}{\sigma \sqrt{n}}$ converge in distribution to a normal random variable with mean 0 and variance 1. From here we have the approximation: $P(|\hat{\pi_j} - \pi_j| \leq \epsilon) = P(\frac{\sqrt{n}|\hat{\pi_j} - \pi_j|}{\sqrt{p(1-p)}} \leq \frac{\sqrt{n}\epsilon}{\sqrt{p(1-p)}}) \approx \int_{-a}^{a} g(x)dx$ where g is the density of the normal distribution and $a = \frac{\sqrt{n}\epsilon}{\sqrt{p(1-p)}}$ from which we can write $\epsilon = \frac{a\sqrt{p(1-p)}}{\sqrt{n}}$ So given a threshold of confidence t we can build a interval of confidence where we can state with t% of confidence that the true value is within this interval: $I_t = [\hat{\pi}_{j,n} - \frac{a_t}{2\sqrt{n}}, \hat{\pi}_{j,n} + \frac{a_t}{2\sqrt{n}}]$ where $\int_{-a_t}^{a_t} g(x)dx = t$ The value for a can be found in a computation table. For t equal to 95% the value of a is 1.96 which gives us for the interval where there is 95% of chance that the true value is: $I_t = [\hat{\pi}_{j,n} - \frac{1}{\sqrt{n}}, \hat{\pi}_{j,n} + \frac{1}{\sqrt{n}}]$ ## In Practice With the simple slider and graphic you can see by yourself the influence of the number of people polled for an opinion poll. The box represents the 95% interval of confidence that means there is 95% of chance that the true vote intention is within this interval. 1000 ## Conclusion This quick refresher on opinion polls should remind us that opinion polls can be quite far from reality. And there is one variable that I didn’t take into account: there are people who don’t dare to name the true candidate they plan to vote for. This is especially true for extremist votes and explains why we can experience such a discrepancy between the estimations and the true results of one election.

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# Minimum Path Sum in Python Suppose we have a m x n matrix filled with non-negative integers, find a path from top left corner to bottom right corner which minimizes the sum of all numbers along its path. Movements can only be either down or right at any point in time. So for example, if the matrix is like below 1 3 1 1 5 1 4 2 1 The output will be 7, the path will be 1,3,1,1,1, this will minimize the sum Let us see the steps − • a := number of rows, b := number of columns • i := a – 1, j := b – 1 • while j >= 0 • matrix[a, j] := matrix[a, j] + matrix[a, j + 1] • decrease j by 1 • while i >= 0 • matrix[i, b] := matrix[i, b] + matrix[i + 1, b] • decrease i by 1 • j := b – 1 and i := row – 1 • while i >= 0 • while j >= 0 • matrix[i, j] := matrix[i, j] + minimum of matrix[i, j + 1] and matrix[i + 1, j] • decrease j by 1 • j := b – 1 • i := i – 1 • return matrix[0, 0] Let us see the following implementation to get better understanding − ## Example Live Demo class Solution(object): def minPathSum(self, grid): row = len(grid)-1 column = len(grid[0])-1 i=row-1 j=column-1 while j>=0: grid[row][j]+=grid[row][j+1] j-=1 while i>=0: grid[i][column]+=grid[i+1][column] i-=1 j=column-1 i = row-1 while i>=0: while j>=0: grid[i][j] += min(grid[i][j+1],grid[i+1][j]) j-=1 j=column-1 i-=1 return(grid[0][0]) ob1 = Solution() print(ob1.minPathSum([[1,3,1],[1,5,1],[4,2,1]])) ## Input [[1,3,1],[1,5,1],[4,2,1]] ## Output 7 Updated on: 04-May-2020 461 Views

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The alluvial forests contain an exceptional diversity of habitats, with a wide range of environmental conditions that offer an "ecological niche" for a large number of species of fauna and flora. The natural habitats and forestland are considered to be of high environmental value, although the considerable changes that have been made in and around the Rhine over the past decades have brought about a modification of this very special environment, which is threatening certain habitats and species. Unique to Strasbourg Strasbourg is the only European city with alluvial forests round its outskirts. Various tree-planting operations (beech, spruce, sycamore maple and walnut) have been carried out over the years, but there are still old forest stands and sectors which have seen no woodcutting since 1984. An ancient and luxurious forest The three great forest areas (Neuhof, Robertsau and Rohrschollen) contain about 70 to 80 woody species (trees, shrubs and lianas), some of which are rare and even threatened, such as the White Elm and Wild Pear and Wild Apple trees. Lianas such as ivy, clematis, hops and the rare, protected wild vine, and the richness of the flora help give the forest the luxurious vegetation reminiscent of a tropical forest. Ivy is a creeper which does no harm to its host tree, which it uses only to climb up to the light. As there has been no woodcutting in certain parts of the forest since the 1980s, old trees can still be found there with trunk diameters that sometimes exceed 200 cm (mainly oak trees), along with a significant amount of dead wood left lying on the ground, which encourages the natural recycling process. Some sectors contain the remains of natural alluvial forests, with ash, oak and lime trees. The Rhine alluvial forest is home to a number of nesting forest birds (54 species in the natural reserve of the island of Rohrschollen), including 6 species of woodpecker – the Black Woodpecker, the Green Woodpecker, the Grey Headed Woodpecker, the Great Spotted Woodpecker, the Middle Spotted Woodpecker and the Lesser Spotted Woodpecker – and the Short Toed Treecreeper, a species which is normally only seen in coniferous mountain forests, but which can be spotted in the forests of the Robertsau and Rohrschollen. Fauna and flora of the waterways and wetlands A vast number of amphibians and remarkable plant species are to be found in and around the internal waterways, wetlands and ponds of the alluvial forests. Among the rare or protected species to be found there is the weird and wonderful carnivorous Utricularia australis, which feeds on microorganisms in the water. The three forests are home to 10 species of amphibians, including the very rare Crested Newt in the Robertsau and Rohrschollen and large populations of the Common Frog and Common Toad. A number of Libellulidae can be seen along the fast flowing Altenheimerkopf waterway in the forest of Neuhof, including the protected Southern Damselfly, the Small Pincertail, the Keeled Skimmer and the Southern Skimmer, and, since 2011, the Golden Ringed dragonfly. The ponds, wetlands and slower-flowing waterways of the Robertsau and Rohrschollen forests are home to many species of nesting waterbirds, including a number of rare ones such as the Tufted Duck and the Little Grebe, whose nest is a floating raft made of plant debris. Other species include the Mallard, Mute Swan and the Eurasian Coot, while the Black Kite can sometimes be spotted flying overhead searching for the dead fish floating on the surface of the water which make up much of its diet. The waterways also contain a rich fish population, which includes eels, pike and the Spined Loach, which is only to be found in the Natural Reserve of the island of Rohrschollen. Flora, butterflies, grasshoppers and crickets The meadowland at the edges of the forests is wet in some sections and dry in others and contains a wide variety of fauna and flora. Some of the meadows have up to 70 or 80 plant species spread over just a few hundred square metres. Five protected species have been found in the meadows so far, along with other remarkable species including orchids. The wet meadows can be home to the Early Marsh Orchid and the Large Pink, while the Bee Orchid, Pyramidal Orchid and Burnt-tip Orchid can be found in the dry sections. The Entomofauna includes 29 species of orthoptera (grasshoppers and crickets) as well as 48 species of diurnal Lepidoptera (butterflies) including a number of remarkable and protected species such as the Dryad, Old World Swallowtail, the Grey Bush Cricket, the Tessellated Shieldback, the Blue Winged Grasshopper and the Sphingonotus caerulans locust, which are to be found in the dry meadows, and on the rare gravel banks of the Natural Reserve of the island of Rohrschollen. The protected Large Copper and Dusky Large Blue butterflies, the Large Marsh Grasshopper and Leek Grasshopper are more likely to be found in wet meadowland, especially in the Robertsau forest or occasionally in the Rohrschollen and Neuhof forests. Facts and figures - 13 locally or nationally protected plant species - Robertsau forest: - 33 species of Lepidoptera and 20 species of Orthoptera recorded since 2008 - 48 species of nesting birds, including 39 forest species and 9 waterbirds - 3 species of reptiles and 6 species of amphibians in 2011 - Neuhof forest: - 32 species of Libellulidae, 17 species of Orthoptera and 31 species of Lepidoptera recorded since 2008 - 4 species of reptiles and 5 species of amphibians in 2011 - 18 species of fish recorded from 1995 to 2007 - Natural Reserve of the island of Rohrschollen : - 54 species of forest-nesting birds and 30 species of meadow-nesting birds - 10 species of amphibians and 4 species of reptiles recorded - 45 species of Lepidoptera and 27 species of Orthoptera recorded in the meadowland - 22 species of fish in the internal waterways on the island and the Vieux Rhin - 10 species of Chiroptera recorded, including 2 species which still need confirmation.

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Middle and high school science teachers looking to stretch their students' gray matter will find plenty of material in these neuroscience lesson plans from the Center for Neurotechnology (CNT). The lessons were designed by science teachers for units lasting between one and five weeks, to be easily integrated into any biology, physics, chemistry, or computer science classroom. Students will be inspired by curricula on topics such as "Neural Engineering and Ethical Implications," "Traumatic Brain Injury," and "Building Artificial Neural Networks." Each unit plan includes complete lesson outlines (downloadable as PDFs), with detailed teacher instructions, outlines of necessary materials and preparation, classroom activities, and post-lesson assessments. Best of all, each unit has been aligned to national curriculum standards from Next Generation Science Standards (NGSS). Lesson plans are easily found by scrolling through the page linked above. For teachers who want more, each lesson page also links to additional resources from the teachers who designed the plans, including academic publications and news articles of interest. The CNT is an NSF-funded Engineering Research Center, located at the University of Washington in Seattle, and partners with the Massachusetts Institute of Technology and San Diego State University.

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  TOPICS Exercise - 1.5 Question-1 :-  Classify the following numbers as rational or irrational: (i) 2 - √5, (ii) (3+√23) - √23, (iii) 2√7/7√7, (iv) 1/√2, (v) 2π Solution :- ```(i) 2 - √5 = 2 - 2.236067... = 0.236067 As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number. (ii) (3+√23) - √23 = 3+√23-√23 = 3 =3/1 therefore, it is a rational no. and it can be represented in p/q. (iii) 2√7/7√7 = 2/7 therefore, it is a rational no. and it can be represented in p/q. (iv) 1/√2 = 1/√2 x √2/√2 = √2/2 = 0.7071067811... As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number. (v) 2π = 2 x 3.1415... = 6.2830... As the decimal expansion of this expression is non-terminating non-recurring, therefore, it is an irrational number. ``` Question-2 :- Simplify each of the following expressions: (i) (3+√3)(2+√2), (ii) (3+√3)(3-√3), (iii) (√5+√2)², (iv) (√5-√2)(√5+√2) Solution :- ```(i) (3+√3)(2+√2) =3(2+√2)+√3(2+√2) =6+3√2+2√3+√6 (ii) (3+√3)(3-√3) =3²-(√3)² =9-3 = 6 (iii) (√5+√2)² =(√5)²+(√2)²+2√5√2 =5+2+2√10 =7+2√10 (iv) (√5-√2)(√5+√2) =(√5)²-(√2)² =5-2 = 3 ``` Question-3 :-  Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d This seems to contradict the fact that π is irrational. How will you resolve this contradiction? Solution :- ``` There is no contradiction. When we measure a length with scale or any other instrument, we only obtain an approximate rational value.We never obtain an exact value. For this reason, we may not realise that either c or d is an irrational. Therefore, the c/d fraction is irrational. Hence, π is an irrational. ``` Question-4 :- Represent √9.3 . on the number line. Solution :- ``` Mark a line segment OB = 9.3 on number line. Further, take BC of 1 unit. Find the midpoint D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular to line OC passing through point B. Let it intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is √9.3 ``` Question-5 :- Rationalise the denominators of the following: (i) 1/√7, (ii) 1/(√7-√6), (iii) 1/(√5+√2), (iv) 1/(√7-2) Solution :- ```(i) 1/√7 Here 1 is the Numenator and √7 is Denominator, so multiply and divide by √7 =1/√7 x √7/√7 =√7/7 (ii) 1/(√7-√6) Here 1 is the Numenator and (√7-√6) is Denominator, so multiply and divide by (√7+√6) =1/(√7-√6) x (√7+√6)/(√7+√6) =(√7+√6)/(7-6) =(√7+√6)/1 = √7+√6 (iii) 1/(√5+√2) Here 1 is the Numenator and (√5-√2) is Denominator, so multiply and divide by (√5-√2) =1/(√5+√6) x (√5-√2)/(√5-√2) =(√5-√2)/(5-2) =(√5-√2)/3 (iv) 1/(√7-2) Here 1 is the Numenator and (√7-2) is Denominator, so multiply and divide by (√7+2) =1/(√7-2) x (√7+2)/(√7+2) =(√7+2)/(7-4) =(√7+2)/3 ``` CLASSES ## Connect with us: Copyright © 2015-16 by a1classes.

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0 # Find the domain and codomain and determine whether T is linear w1=5x1-x2+x3 w2= -x1+x2+7x3 w3= 2x1-4x2-x3 My solution so far: From the system I can tell that it is R3, so domain is R3, since there w1, w2, w3 I can tell that codomain is R3. Is there a way to show work? Thank you! ### 1 Answer by Expert Tutors 5.0 5.0 (1072 lesson ratings) (1072) 0 ---------null--------- Thank you very much, Matt! I have already submitted my assignment where I just explained in my own words. Your explanation made me feel better, since there is no need to actually show work. If you don't mind, I would love to learn how to show that T is linear. Thanks! Being totally rigorous requires having precise definitions, and I don't know if your course gave precise definitions for what it means to be linear or what a vector space is. Most introductions to multivariable calculus and linear algebra (I'm guessing you're in multivariable calculus) aren't totally rigorous; the "informal" argument I gave in my first answer would be fine. But the rigorous definition of a linear function between two vector spaces V and W goes as follows. A function T:V -> W is linear if and only if T(v1+v2)=T(v1)+T(v2) and T(av)=aT(v) where v, v1, and v2 are vectors in the vector space V and a is a scalar. In your example, we're considering a function T:R3->R3 defined by T(x1, x2, x3) = <5x1 - x2 + x3, -x1 + x2 + 7x3, 2x1 -4x-x3 > So the vector space we're working over (the domain vector space) is R3. A vector in this space has the form <x1, x2, x3> So to show that T really is linear, we need to show the two conditions above hold for the equations that define T. Let's start with the first condition: given any two vectors v1=<x1, x2, x3> and v2=<y1, y2, y3>, we need to show that T(v1+v2)=T(v1)+T(v2) What does the left-hand side (I'm going to abbreviate this as LHS) of this equation equal? Well, by the definition of vector addition, we have v1 + v2 = <x1, x2, x3> + <y1, y2, y3> = <x1 + y1, x2 + y2, x3 + y3 > So the LHS is equal to T(x1+y1, x2+y2, x3+y3)=<5(x1+y1) -(x2+y2) +(x3+y3), -(x1+y1) + (x2+y2) + 7(x3+y3), 2(x1+y1) -4(x2+y2) -(x3+y3) > Note that this just amounts to replacing x1 by (x1+y1), x2 by (x2+y2), and x3 by (x3+y3). Now all we've done is compute the LHS of the above equation. We need to show it's equal to the right-hand side (RHS) T(v1)+T(v2)=T(x1, x2, x3) + T(y1, y2, y3) But what does this equal? Well, we know T(x1, x2, x3) is <5x1 - x2 + x3, -x1 + x2 + 7x3, 2x1 -4x2 -x3 > and T(y1, y2, y3) is <5y1 - y2 + y3, -y1 + y2 + 7y3, 2y1 -4y2 - y3 > So to get the RHS we just add these two vectors, which by the definition of vector addition is just <5x1 - x2 + x3+5y1 - y2 + y3, -x1 + x2 + 7x3 -y1 + y2 + 7y3, 2x1 -4x2 -x3 2y1 -4y2 - y3> The entire question is: is the LHS we computed above equal to this RHS? The answer is yes for a very simply reason: just distribute the coefficients from our equation for the LHS above (in bold) and you'll see you get exactly what we got in the line directly above for the RHS. This shows that T(v1 + v2) = T(v1) + T(v2) To finish the rigorous proof that T is linear, you'd need to show the second condition as well, namely that T(av)=aT(v) But the strategy for doing this is exactly the same. What's the point of all this work? The point is that because the w variables were defined just as sums or differences of multiples of the x variables, the distributive property will wind up guaranteeing that T(v1+v2)=T(v1)+T(v2) and T(av)=aT(v). But if the w variables had included something like the square of one of the x variables, these conditions would no longer hold. That's why the informal criterion for being linear is just making sure that the w variables are just simple linear combinations of the x variables and don't have any non-linearities like squares or products or powers.

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NOAA’s Modeling and Mapping Data Enhance Nation’s Ability to Provide Tsunami Warnings Along U.S. Coastlines As we kick off Tsunami Preparedness Week, we pause to remember the 124 Americans who prematurely lost their lives without warning 50 years ago, when a powerful earthquake sent several tsunami waves crashing into coastal towns in Alaska, Oregon and California. On March 27, 1964, a 9.2 magnitude earthquake – the largest recorded earthquake in U.S. history and the second largest in world history – occurred in Alaska’s Prince William Sound. In addition to the lives lost, the tsunamis caused an estimated $1 billion in damage. Since 1964, we’ve been reminded about the power and danger of tsunamis. The devastation and heartbreak of the 2004 Indian Ocean tsunami remains with us a decade later, and images from the Japan tsunami are still fresh in our minds three years later. These events should serve as a reminder that a powerful tsunami can strike anywhere in the world, any time of year, and the U.S. is no exception. Coastal populations and infrastructure have increased significantly over the past 50 years, making the U.S. even more vulnerable to the impacts of a tsunami. However, the nation also has made substantial advancements in earthquake science and the ability to prepare for, detect, forecast, and warn about tsunamis. While we cannot stop a tsunami from happening, we can minimize loss of life and property through preparation. Today, NOAA leads the U.S. Tsunami Warning System, which includes operating two, 24/7 tsunami warning centers; managing a network of tide gauges and tsunami buoys, and monitoring seismic stations throughout the world’s oceans; administering the TsunamiReady program; and leading the National Tsunami Hazard Mitigation Program, a state-federal partnership that works together to prepare America for a tsunami. Effective preparedness depends on accurate hazard assessments. In recent years, NOAA and its state partners have made significant progress in modeling and mapping the tsunami hazard along U.S. coastlines, which has enhanced the nation’s ability to forecast and provide warnings for tsunamis. One key aspect of a hazard assessment is the accurate prediction of where coasts will flood during a tsunami. NOAA builds and updates high-resolution coastal digital elevation models, which depict Earth’s solid surface to further the understanding of ocean processes, like tsunamis, and inform decision-making. The models are incorporated into tsunami models, which simulate tsunami movement across the ocean and the magnitude and location of coastal flooding caused when the tsunami reaches the shore. The results of these simulations enable tsunami warning centers to issue more accurate forecasts, as well as support state-level evacuation mapping, preparedness and mitigation planning.

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# The centripetal force direction on hill and valley A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley? On the top of the hill, the centripetal force is downward. So we have $$F_N-mg-m(\frac{v^2}{R})=0$$ since we don't have motion in the vertical direction. So when $$F_N=0$$ we have $$-g=\frac{v^2}{R}$$ but this is impossible since the right-hand side is positive but the left is not. Why is that? You got the signs wrong - the centripetal acceleration is towards the center of motion and enters the equation from the other side. This is a common mistake in analyzing free-body diagrams - you must include only the forces that act on the object. For every force you include in the free-body diagram you must be able to answer the following question: What object is exerting this force? In case of weight that object is Earth, and in case of normal force that object is the surface. What would be the object that exerts the centripetal force? In the second Newton's law $$\sum_{i} \vec{F}_i = ma$$ the left-hand side is the vector sum of the forces from the free-body diagram, and the right-hand side is the consequence (motion). In your example, the centripetal force is the consequence. If you define the positive direction for the vertical ($$\hat{\jmath}$$) axis to be upwards (away from the Earth's center), then the free-body diagram would show $$\vec{n} + \vec{w} = \vec{F}_\text{net} \quad \rightarrow \quad N \hat{\jmath} - W \hat{\jmath} = \vec{F}_\text{net}$$ where $$N$$ is magnitude of the normal force, $$W$$ is magnitude of the weight, and $$\vec{F}_\text{net}$$ is the net force vector. We know that for the circular motion the net force points towards the center of motion (also known as the centripetal force), hence at the top of the hill the net force is $$\vec{F}_\text{net} = -m\frac{v^2}{R} \hat{\jmath}$$ from which we could easily calculate the magnitude of the normal force $$N = W - F_\text{rad} = mg - m\frac{v^2}{R} = m \Bigl( g - \frac{v^2}{R} \Bigr)$$ The car loses contact with the road (surface) when the normal force is zero, or when the velocity is $$\boxed{v = \sqrt{g \cdot R}}$$ At the bottom of the hill the net force is $$\vec{F}_\text{net} = m \frac{v^2}{R} \hat{\jmath}$$ Assuming the velocity did not change from the top to the bottom of the hill, it is trivial to calculate the normal force at the bottom. • Isn't the weight and the centripetal force in the same direction? – Mina Commented Jan 18, 2022 at 15:47 • It is, but centripetal force is a consequence and not the reason. This is a common mistake - when you draw the free-body diagram, you should not draw the resultant force which is in this case the centripetal force. Commented Jan 18, 2022 at 15:48 • I get it now. Thanks a lot. – Mina Commented Jan 18, 2022 at 15:51 • @Mina When you draw the free-body diagram, for every force you draw you must answer one important question: what other object is exerting this force? In case of weight it is the Earth, and in case of normal force it is the surface. What would be the object exerting the centripetal force? Commented Jan 18, 2022 at 16:05 • If I could give five upvotes to the comments by @MarkoGulin I would. Centripetal force is very commonly misunderstood by novices. Be sure to understand what he is saying. Commented Jan 18, 2022 at 18:26

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Question # What is the difference between the simple interest (S.I) on a sum of ₹ 2,000 being calculated at 6% for 3 years and 7% for 3 years?₹ 50₹ 55₹ 60₹ 40 Solution ## The correct option is C ₹ 60Principal (P) = ₹ 2,000 Time period (T) = 3 years Simple interest (S.I) = P×R×T100 For rate of interest (R) = 6%, S.I = 2,000×6×3100 ⇒S.I = ₹ 360 For rate of interest (R) = 7%, S.I = 2,000×7×3100 ⇒S.I = ₹ 420 Difference in simple interest =₹ 420−₹ 360=₹ 60 Suggest corrections

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# Mythbusters: Debunking the Seven Fold Limit Myth Paper, an unassuming everyday object, holds a mystique of its own. Its tangible nature has ingrained it deeply in our lives, even in this digital age. While technology has made it possible to connect without paper, we still cherish receiving physical letters. Moreover, certain industries, like insurance, insist on hard copies for processing. However, one myth surrounding paper continues to intrigue us all: the seven fold limit. Numerous attempts, including the renowned TV show “Mythbusters,” have tried to debunk it, with varying results. So, what is the truth? ## Unraveling the Seven Fold Limit It is widely believed that a single sheet of paper cannot be folded in half more than seven times, regardless of its finish, size, or weight. The reasons behind this limitation are twofold: For those inclined towards numbers, let’s dive deeper into the seven fold limit: • Each fold doubles the number of paper layers. • The equation 2x = total layers of paper, where x = number of folds. • Here’s a breakdown: Number of Folds Layers of Paper 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 256 9 512 10 1024 11 2048 12 4096 As you can see, reaching 128 layers of paper is no small feat! ## Fact or Fiction? So, is the seven fold limit fact or fiction? Well, it’s a bit of both. For the average person, the limit stands true. However, with a larger sheet of paper, you can indeed surpass the seven fold barrier. But be warned, as you fold more times, the physical resistance builds up, making it increasingly difficult to continue. Additionally, the size of the sheet must be significantly increased before attempting an extra fold. While the seven fold limit has been “busted” by both the Mythbusters and a Californian high school student, there’s more to the story. ## The Mythbusters’ Journey The Mythbusters team successfully managed to fold a sheet of paper eight times without any machinery. However, they faced considerable challenges after the seventh fold. Imagine the difficulty of moving 128 layers of a giant football-field-sized sheet of paper! To achieve 11 folds, they had to resort to a steamroller and a forklift. Here’s a glimpse of their feat: Watch the Mythbusters in action It’s essential to note that their extraordinary achievement required resources far beyond the reach of the average person. ## Britney Gallivan’s Triumph Another individual who challenged the seven fold limit was Britney Gallivan, a determined high school student from California. As part of her math class, she attempted to go beyond the myth for extra credit. Using a 1.2-kilometer-long (approximately ¾ of a mile) sheet of toilet paper, she achieved an astonishing 12 folds. While her sheet lacked the weight and width of a typical paper, Britney undeniably shattered the myth.

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Eruption of Deciduous TeethEruptionEruption is essentially the process whereby a tooth moves from its developmentalposition in the jaw into its functional position in the mouth. There is no evidence tosuggest that eruption entirely ceases once a tooth meets its antagonist in the mouth, andoutward axial movements occurring during the functional phase may also be eruptivemovements. (viz. overeruption following removal of the antagonist tooth in the oppositejaw).While the main direction of the eruptive force is axial (i.e. related to the long axis ofthe tooth), movement also occurs in other planes, accounting for tilting and drifting.Eruption rates of teeth are greatest at the time of crown emergence. Rates also differaccording to tooth type. Permanent maxillary central incisors are reported to erupt atabout 1mm/month; the rates for mandibular second premolars have been determined to beas great as 4.5mm in 14 weeks. For permanent third molars, where space is available,eruption rates of 1mm in 3 months have been recorded. In crowded dentitions, however,the rates are less than 1mm in 6 months.As a tooth approaches the oral cavity, the overlying bone is resorbed and there aremarked changes in the overlying soft tissues. The enamel surface is covered by thereduced enamel epithelium, which is a vestige of the enamel organ.Pre-eruptive tooth movement: Pre-eruptive phase, which starts with the initiationof tooth developmentn.2) Eruptive: Tooth eruption, which begins once the roots begin to form.3) Post-eruptive tooth movement: After the teeth have emerged into the oral cavity,there is a protracted phase concerned with the development and maintenance of occlusion(the functional phase).Eruption of TEETH(I)Patten of tooth movement1. Pre-eruptive tooth movementWhen deciduous tooth germs first differentiate, there is a good deal of spacebetween them. Because of their rapid growth, this available space is utilized and thedeveloping teeth become crowded together, especially in the incisor and canineregion.This crowding is relieved by growth in the length of the infant jaws,which provides room for the second deciduous molars to drift backward and anteriorteeth to drift forward. At the same time the tooth germs also move outward as the jaws increasein width, and upward( downward in the upper jaw) as the jaws increase in height.2. Eruptive tooth movementDuring the phase of eruptive tooth movement the tooth moves from itsposition within the bone of the jaw to its functional position in occlusion and theprincipal direction of movement is occlual or axial .It is important to recognize that jaw growth is normally occurring while mostteeth are erupting, so that movement in planes other than axial is superimposed oneruptive movement.3. Post-eruptive tooth movement(1) maintain the position of theee erupted tooth while the jaw continues to grow(2) compensate for occlusal and proximal wear.The former movement, like eruptive movement, occurs principally in an axialdirection so as to keep pace with the increase in height of the jaws. It involves both thetooth its socket and ceases when jaw growth is completed. The movements compensatingfor occlusal and proximal wear continue throughout life and consist of axialand mesial migration, respectively.Histology of Tooth MovementPre eruptive phase:1. Total bodily movement of the germ2. There is its excentric growth:excentric growth: one part of the developing tooth germ remains stationary while theremainder continues to grow, leading to a shift in its center.Eruptive phase:During the eruptive phase of physiologic tooth movement, significant developmentalchanges occur:1. Formation of root2. Formation of periodontal ligament,3. Formation of dentogingival junction4. Another specialized feature associated with the erupting permanent tooth is thepresence of a gubernacular canal.or cord.Post-eruptive phase:(III) Mechanism of tooth movement1. Root growth2. Vascular pressure Root growth3. Bone growth4. Ligament traction Shedding of deciduous teeth(I) Patten of SheddingFor a deciduous incisor or canine, root resorption initially occurs on the lingualsurface adjacent to the developing permanent tooth. With subsequent movement andrelocation of the teeth in the growing jaws, the developing permanent tooth comes to liedirectly beneath the deciduous tooth and further resorption occurs from the apex.For a deciduous molar, root resorption often commences on the inner surfaceswhere the permanent premolars initially develop.The premolars later come to lie beneath the roots of the deciduous molar andfurther resorption occurs from the root apices. The shift in position of the deciduous toothrelative to the permanent successor may account for the intermittent nature ofroot resorption.(II) Mechanism of resorption and histology of sheddingThe initiation of root resorption may be an inherent developmental process or it maybe related to pressure from the permanent successor against the overlying bone or tooth.Mechanism of TOOTH EruptionTooth eruption is traditionally considered to be a developmental process whereby thetooth moves in an axial direction from its position within the alveolar crypt of the jawinto a functional position within the oral cavity. However, eruption can be regarded as alifelong process since a tooth will often move axially in response to changing functionalsituations (e.g. overeruption resulting from the removal of an antagonist, andcompensatory eruption related to attrition).The rate of eruption represents a balance between forces tending to move the tooth intothe mouth (eruptive force) and forces tending to prevent this movement (resistive force).Resistance may be produced by overlying soft tissues and alveolar bone, the viscosityof the surrounding periodontal ligament and occlusal forces. Thus, changes in the rate oftooth movement may be brought about by changes in either the eruptive forces and/or theresistive forces.At present, little is known about the nature, source and magnitude of either the eruptiveor resistive forces. Furthermore, it is not known whether the forces are of the same natureand value at various stages of the eruptive cycle. By and large, this situation results fromdifficulties encountered in producing experimental systems which isolate for study singlepossible agents associated with the eruptive process.All tissues within the vicinity of the tooth thought capable of generating a force have,at one time or another, been implicated in the eruptive process.The tooth is pushed out as a result of forces generated beneath and around it, eitherby alveolar bone growth, root growth, blood-pressure/tissue fluid pressure, or cellproliferation. Alternatively, the tooth may be pulled out as a result of tension within theconnective tissue of the periodontal ligament. Although no one theory is yet supported bysufficient experimental evidence.

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|Blog||Archive||QR Code||RSS||Archive||Tag Cloud||Videos||Widget||XML| Owen Chamberlain, the Antiproton, and Polarized Targets Owen Chamberlain is "most remembered for his role in the discovery of the antiproton in 1955, ... for which he shared the 1959 Nobel Prize in physics … . The discovery of the antiproton, the mirror image counterpart to the proton in ordinary matter, was made possible through the combination of the Bevatron accelerator … and a unique detector, designed by Chamberlain and his colleague, Clyde Wiegand, that was set off only by particles moving at the speed predicted for antiprotons. In early 1942, at the prompting of [Ernest O.] Lawrence, Chamberlain joined the Manhattan Project, the U.S. government’s secret effort to build an atomic bomb. Working as an assistant to [Emilio] Segrè, first in Berkeley, and then at the laboratory in Los Alamos, New Mexico, he investigated nuclear cross sections for intermediate-energy neutrons and the spontaneous fission of heavy elements."1 "In 1946, he joined Argonne National Laboratory in Chicago, where he conducted research on slow-neutron diffraction in liquids while working toward his Ph.D. in physics, which he obtained in 1948 from the University of Chicago. "2 "His mentor was the great Italian physicist and Nobel Laureate Enrico Fermi, the world’s leading authority on neutrons."1 Chamberlain ' "pioneered the use of polarized targets in high energy scattering experiments, which have helped us to understand the forces acting between particles and allowed us to test the symmetry principles underlying the physics," [colleague Herbert Steiner] said. Symmetry principles state, for example, that some interactions look the same when reflected in a mirror or run backwards in time." One of Chamberlain's gifts was teaching, which he did best one-on-one and with an informality that included his insistence that students call him "Owen." His unique explanations for physical phenomena came to be called "Chamberlainisms" among the students.'2 1Edited excerpts from Berkeley Scientist Great Own Chamberlain … 2Edited excerpts from Owen Chamberlain, Physics Nobelist, UC Berkeley Professor … Additional information about Owen Chamberlain, the antiproton, and polarized targets is available in electronic documents and on the Web. Observation of Antiprotons, DOE Technical Report, October 1955 Antiprotons, DOE Technical Report, November 1955 Experiments on Antiprotons: Antiproton-Nucleon Cross Sections, DOE Technical Report, July 1957 The Early Antiproton Work [Nobel Lecture], DOE Technical Report, December 1959 Personal History of Nucleon Polarization Experiments, DOE Technical Report, September 1984 Additional Web Pages: Observation of Antiprotons The Golden Anniversary of the Antiproton, science@berkeley lab, October 27, 2005 In Memoriam: Owen Chamberlain Selections from the Nobel Laureates' Papers …: Owen Chamberlain Dr. Chamberlain Biography submitted to the Nobel Committee A Man of Civility and Conscience, Berkeleyan, March 8, 2006 Owen Chamberlain …; Discovered Antiproton Nobel Laureate, Berkeley Physicist Owen Chamberlain Some links on this page may take you to non-federal websites. Their policies may differ from this site.

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Raglan castle is one of the finest late medieval buildings in the British Isles and, although now ruined, it remains a striking presence in the landscape of south-east Wales. Much of what remains at Raglan dates from the 15th century, the period of the Wars of the Roses and the rise of the Tudor dynasty. The Great Tower is the most impressive of the buildings from this period, dominating the two courtyards of the castle. It was built either by William Herbert or his father, William ap Thomas, who had purchased Raglan in 1432. William Herbert was a key figure in the politics of the late 15th century. During the War of the Roses he supported Edward IV. The reward for his loyalty was considerable, providing him with the title Earl of Pembroke, and sufficient resources to convert Raglan into a palace-fortress. Earl William's success was, however, to be short-lived. In 1469 he was captured by Lancastrian supporters at the Battle of Edgecote and put to death. The Herberts retained control of Raglan until 1492 when it passed to the Somerset family. William Somerset, the third Earl of Worcester (1526-1589), was the first of his family to significantly alter the castle's buildings. Earl William focused his efforts on upgrading the quality of the hall and service ranges to meet the social expectations of his time. He also established the gardens, including a series of walled terraces, an artificial lake, a fountain, flower beds and herb gardens. By the middle of the 17th century, Raglan's fortunes were at their peak. It had achieved a level of sophistication and opulence that only the greatest country houses could match. However, the English Civil War was to change all this. In 1642, the fifth Earl of Worcester declared his support for the Royalist cause, offering considerable financial support to King Charles I. This was to make Raglan a target for Parliamentarian forces, which subsequently besieged the castle in June 1646. Its defenders held out during the summer, but by mid-August the Parliamentarians had moved their siege works to within sixty yards of the castle. Its defenders surrendered on the 19th of August. In the years that followed Raglan was abandoned and left to decay, becoming a convenient source of building material and a picturesque tourist attraction. Today this decay has been halted and the building conserved through the work of Cadw and its predecessors, guardians of the castle since 1938. Raglan Castle by J. R. Kenyon. Published by Cadw (2003). Article Date: 9 June 2007

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The esophagus is the muscular tube that carries food from out throat to the stomach. Based on current statistics, cancers of this tube kill more than half a million people worldwide every year. There are many factors that have been thought to trigger esophageal cancer; nutritional deficiencies, infections and dietary toxins. Despite the wide speculative belief that it's alcohol and tobacco that cause throat and esophagus cancer, other populations have also had a high rate of this disease; populations that do not drink or smoke. It is with a new study that a new risk has finally come to light. There has been a rise in esophageal cancer among countries where it is traditional to drink black tea above 70ºC (or 158ºF). Testing & Studies in Temperature of Teas Iranian researchers have found that drinking this hot tea can be linked to the increased odds of developing cancer of the esophagus and they have provided what most believe is the most conclusive research; the hypothesis that thermal injury can cause cancer. Northern Iran has one of the highest rates of the most common types of esophageal cancer. It is here that a team from the University of Tehran set up a case controlled study of a single area in northern Iran, the Golestan Province, to better understand the spread of esophageal cancer. The team found that smoking tobacco products and drinking alcohol is an uncommon trait in this population but tea drinking is common. Past studies did not provide suitable data. It was difficult to distinguish whether the negative factors of tea drinking were because of drink temperature or type of beverage consumed. There was also insufficient evidence where the temperature of the tea was rated by the drinker without independent confirmation. A Difficult Test Procedure Finally Pays Off To address the issues of the previous studies, the researchers studied tens of thousands of people in the Golestan Province. The subjects of the study drank, on average, a liter of black tea per day each, and temperatures, along with the speed at which it is drunk, were measured. The study has now been printed in the BMJ (formerly The British Medical Journal). Professor Reza Malekzadeh, the leader of the researchers, explains that "Our results showed a noticeable increase in risk of esophageal squamous cell carcinoma associated with drinking hot tea." The team discovered that the tea itself did not constitute a risk of cancer, but as the temperature increases so does the risk. Common Sense Might Just Save You from Cancer Having tea around 65ºC compared to consuming tea above 65ºC but below 70ºC increases odds of esophagus cancer twice, while consuming tea at over 70ºC increases the risk eight times over. The time in which the hot tea was drunk also plays a part. Drinking tea about four minutes after it's poured decreases the chances of developing esophagus cancer. This can be seen as using common sense and it should be taken to heart. It is best to allow five to ten minutes between making and pouring a hot beverage and this not only applies to tea, but any high temperature food or drink. Any hot beverage or food should be allowed to cool before drinking or eating to better protect your esophagus, and lessen the risk of cancer. Adding milk to hot drinks such as tea and coffee, as is common practice in the West, adequately cools the beverage enough to eliminate the risk.

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When we look out into the universe, we find objects tend to orient themselves in nice, flat planes. For those who don’t know the physical processes behind this phenomena, this seems extremely strange considering the three-dimensional nature of space (special dimensions that is). Before I start, you should be aware that solar systems and galaxies aren’t completely flat. Yes, the bulk of objects in galaxies and solar systems orbit on the same general plane as one another, but then you’ll have objects like Pluto. Pluto has an orbital inclination of about 17 degrees relative to the plane of the solar system. On the galactic level, most galactic nuclei have a cloud of globular clusters in orbit around them. Most of the time, this cloud orbits more as a sphere rather than a disk. OK, on to the actual question. Why do solar systems and galaxies favor relatively flat orbital planes? This mainly has to deal with angular momentum. “In physics, angular momentum (or rotational momentum) is a vector quantity that represents the product of a body’s rotational inertia and rotational velocity about a particular axis.” From here, I’ll pull some small-scale examples of angular momentum. Take a gyroscope as an example. You remember those from your childhood. Wrap a string around it’s center access, pull the string as hard as you can to get the center spinning as fast as possible, then enjoy as it defies gravity. Force, torque, momentum, and angular momentum combine to provide this elegant balance. Swinging a bucket full of water around (in a circle) is another example of angular momentum at work. Another such example is watching a figure skater. In this case, when a skater turns, they can more noticeably alter the values for the forces at work by pulling in to a tight ball (where they will spin very fast) or stretching out long (causing them to spin more slowly). In fact, the same process can be seen in just about any rotating system of particles. Take pizza dough as an example. You know when chefs take the round ball of dough, and throw it into the air adding a flick of the wrist to add some angular momentum, then it magically lengthens into a disk (or, the dough orients around the plane on the axis of rotation). Planets do exactly the same thing. The Earth bulges out in the center due to the same processes. Solar systems and galaxies get their angular momentum from the condensing matter that makes up the system. Use a solar system as an example. As the cloud of interstellar gas condensed to make the solar system, off-center collisions of particles added a little torque. This twisting compounded as more and more off center collisions happened and the cloud elongated in the same way pizza dough does. In exactly the same way, galaxies too were formed from collapsing interstellar gas and other galaxy collisions. Now, if you excuse me, I think I’m going to go order some pizza.

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As far as our information goes, the time of day was noted by the sun and the night by the position of Ursa major, the Seven Stars. The year was designated by the winter, each winter constituting a new year. Two divisions or seasons were recognized; spring and autumn were regarded as originating with the whites. Each season was considered as composed of moons; the period during which the moon was invisible taken as the beginning of another moon. We found little consistency in the nomenclature of moons, our information implying that they were considered more by numerals than by names. The tendency was to count the moons from about October, the beginning of winter or the New Year. Variation seems to have been due to the fact that calendar counts were kept by a few individuals, usually medicine men, who modified the system according to their own theories. One man who kept a calendar gave the following list: – |Winter Moons||Summer Moons| |1. Beginning winter moon||Beginning summer’s moon| |2. Wind moon||Frog moon| |3. Cold moon||Thunder moon| |4. Two-big-Sunday moon||Big-Sunday moon| |5. Changeable moon||Berry moon| |6. Uncertain moon||Chokecherry moon| |7. Geese moon| The references to Sunday are to the Christmas and July holidays of our own calendar. The year is generally regarded as comprising fourteen moons equally divided among the two seasons. As calendars were usually in the keeping of men owning beaver bundles and the number seven was employed in enumerating parts of their rituals, this division of the year into moons may be a matter of convention rather than observation. They claim to have reckoned twenty-six days to a moon. Some, however, assert that thirty days were counted; but in this case the year could not have comprised fourteen moons. From one man we secured a set of 179 sticks used for keeping track of time. Red sticks were used for years. Another, used a bag with two parts; one faced with red, the other with blue. Fourteen pebbles were used to mark the moons; each time the moon became invisible he moved a pebble to the other side. Calendars, or winter counts, were kept by memory rather than by sticks, or paintings. We get the impression, however, that there was less interest in such records than among the Dakota and Kiowa. The following is Elk-horn’s winter count, beginning about 1845: - Camped down at Mouth River; Gambles killed; sun dance at Crow Garden (a place). - Camped near Fort Benton; moved to Yellowstone country; some Crow escaped by letting themselves down from a rock with a rope; Yellow River, the place of the sun dance; camped at a place where Bad-tail killed a Sioux. - Crossed Missouri River to camp; traded at Ft. Benton and spent most of the winter on the Marias; a fight with the Snake; the ice broke up in the winter (unusual); sun dance near this place; some Piegan killed by enemies. - On the Marias; man named Goose killed; in autumn hunted south of Ft. Benton; traded at Ft. Benton. - Wintered on the Teton; spring, moved down the Missouri; killed a man named High-ridge; made two sun dances; went to Bear Paw Mountains; went toward Crow country; John Monroe came up to tell Piegan that soldiers were near to issue ammunition and some Piegan did not go because they were skeptical; six Flathead came there for ammunition, some Nez Perce, two North Blackfoot, a few Blood, four North Piegan and some Gros Ventre, but no Sarcee. - Camped on Two Medicine River. - Missouri River; deep snow winter; sun dance at Yellow River. - Slippery winter; some Piegan killed by the Snake. - Camped on Cut Bank; went toward Missouri; Some-bull killed by fall from a horse (chief of the tribe); traded at Sun River. - Sweet Grass Hills; spent spring on the Marias; in summer went south; Big-snakes (chief) killed; ammunition issued. - South of the Missouri; Blood fought among themselves; first time steamboats came to Ft. Teton. - Camped at Bad Waters; Sioux after Piegan; this camp north of the Missouri; killed 7 Cree; a fight with the Crow and lost two chiefs. Good-raven and Mad-plume. - On the Marias; first fight with Gros Ventre; summer camp on the northeast side of Sweet Grass Hills (Canada). - A few cases of smallpox; fight with the Kootenai in which many were killed; during the summer Mountain-chief was attacked by Sioux; a Piegan was killed by a number of Gros Ventre. - Captured a double barrel shot gun; sun dance at High Ridge. - Flies-low was killed. - Many Piegan visited the Southern Gros Ventre (?); ammunition issued; summer camp above Sweet Grass Hills; a fight with the Flathead; also with the Gros Ventre; returned to Two Medicine River. - Eagle-chief killed; in summer killed Eagle-horse. - Fought with the Crow, Gros Ventre, and Flathead. - Straggling-wolf killed near camp; Piegan killed Crow in revenge. - Assiniboine (name of a chief) killed. - Big-prairies father killed by his own people. - Body-sticking-out killed by his own people. - Three-eagles killed by his own people. - Many-horses (the chief) died. - Many buffalo and many trading posts on the Marias. - Man tried to kill his wife, she (Sarcee woman) stabbed him, he killed her; in summer, Home-chief died. - Chief Old-woman-child dies; an open winter. - Killed seven Assiniboine. - Crossed the Missouri; Sitting-bull killed many Piegan. - Camped south of the Missouri. - Camped on Two Medicine River; White-dry, chief of Assiniboine, killed by Piegan; after this the Piegan were confined to the reservation. - Wolf-eagle shot in the arm by Cree. - Many Indians died of sore throat; Chief Birch-bark died. - Crow-big-foot visited Piegan; Crow came to steal horses. - Eagle-child died. - Many cattle died. - Stallions issued. - Mares issued. - Two Indians arrested and died in prison; in summer cattle were issued. - Wolf-coming-over-hill dies. - Chief Walking-through-the-beach dies. - Crow-big-foot dies. - Yellow-medicine dies. - Three-bulls dies. - Big-nose dies. - Four-bear dies. - Gets-paint dies. - Black-living-over-tail dies. - Old-kicking-woman dies. - Lance-chief dies. - Fat-buffalo-horse dies. - Bites killed in a runaway. - Running-rabbit dies. - White-calf dies. This calendar is given as a type and not for the value of its contents, though it doubtless has its merits from that point of view. The narrator was somewhat uncertain as to the order of many counts and made frequent use of a set of improvised counting sticks. We asked him why in later years the winter counts were designated chiefly by the deaths of the most prominent men, to which he replied that since his people were confined to the limits of the reservation nothing else happened worth remembering, and further, that the count ended with the death of White-calf because there were now no men living of sufficient worth to be honored with such mention. From the human point of view we agreed with him in that the book should be closed, for the old ways have all but gone. If we were interested in the historical aspect of this account the dates could doubtless be checked by certain specific references as Nos. 11, 22, 43, and 56. For completeness, we add the winter count of Big-brave, covering a span of sixty-one years, but not giving full representation to the later years. Since reservation days, there is a general tendency among the older men to fix their counts in units of residence at a given spot; i. e., “for five winters, I lived on Two Medicine, then for eight winters on Cut Bank, etc.: - The fall of the year. Gambler went on the warpath and was killed; Piegan spent the winter on the Marias River. - In the fall of the year. Big-lake, chief of The-don’t-laugh band died; Piegan wintered on the Marias River which was high and flooded their camps. In the summer, they had a sun dance at Sweet Grass Hills; Bob-tail-horse was shot and killed; a woman was also killed. - Leaves-big-lodge-camp-marks clubbed a Flathead but did not kill him; in the summer, Piegan killed some Sioux on the Marias. - Black-tattoo became crazy; in the spring a man named Goose was killed by Sioux; in the summer. Goose’s father went to war and killed some Crow; some of the Crow escaped by letting themselves down a high cliff with a rope. - Still-smoking was killed; the Piegan stole a sorrel race horse from the Flathead. In the summer some Piegan were on the warpath south of the Missouri River. They came to some white settlers and there saw a Sioux Indian whom Last-bull killed with a club. The Sioux had been visiting with the white men. - In the fall, the first treaty was made by the Government at the mouth of Yellow River; there were seven different tribes there. That winter, Mountain-chief spent on Belly River. One of his daughter’s clothes caught fire and she was burnt to death. During the summer Mountain-chief became ill with the hiccoughs which lasted some time. - This winter was called the slippery winter because there was so much ice. In the summer Mountain-chief and his people went to Canada and killed thirty Sioux. - The Piegan camped on Marias, and one by the name of Blood killed a Flathead Indian. Lame-bull, a chief, was killed by falling from his horse in the summer. - Mountain-chief spent the winter on Milk River and found an extra large buffalo dung which was about three feet across when measured. Chief Big-snake was killed in the summer. - Lazy-boy was killed. In the summer, the Blood camped at Yellow Mountains and fought among themselves; Calf-shirt killed some of his own people. - A man named Peace-maker was killed. Eagle-child was killed in the summer; a Blood was shot through the face with an arrow by a Sioux but did not die. - Piegan fought with the Gros Ventre and one, Many-butterfly, was killed. The Piegan killed five Sioux who had a horn spoon. - Chief Coward was killed by Crow Indians. In the summer, the Piegan attacked the camps of the Gros Ventre and killed many of them; also, some Piegan were killed while out hunting. - The Assiniboine attacked Mountain-chief’s camps on Big River in Canada, at night, but did not kill anyone. The Piegan fought with the Gros Ventre in the summer and a Piegan, whose name was Half-breed, was killed. - Piegan had what was called red smallpox; in the summer they attacked the Assiniboine seventy lodges and running them out captured the lodges. - At Fort Benton, the Government gave the Piegan clothes, etc.; the white man who issued the things to them went by the name of Black-horse-owner. At this place they also made peace with the Gros Ventre. In the summer Little-dog was killed and the Piegan fought with a great number of enemies, with the Crow, Assiniboine, and Gros Ventre who helped one another in fighting the Piegan; but the Piegan overpowered or whipped them all. - Bear-chief was killed south of the Missouri and the following summer the Piegan killed Weasel-horse, a chief of the Blood. - Mountain-chief camped south of the Missouri and the Piegan killed two Flathead near the Piegan camps; in the summer the Piegan killed thirty Assiniboine who were picking gum off the pine trees. - Strangle-wolf was killed by the Gros Ventre while out hunting; Chief Crow was killed by Gros Ventre while he was out hunting. He had six women with him. - The Piegan had smallpox and the soldiers attacked seventy camps, killing many old men, women, and children. Running-raven was wounded by a Gros Ventre. - The Piegan fought with the Cree on Belly River in Canada .and killed one hundred of them. In the summer they had a big battle with the Assiniboine and Big-brave and his horse were wounded. - A Piegan, Red-old-man, was killed by the Gros Ventre near Bear Paw Mountain while he was trying to steal some horses from them; Black-eagle, a Piegan, killed an Assiniboine and his wife, in the summer. - In the summer Big-brave moved to Blacktail Creek and wintered there. - Mares were issued to the people and Little-dog received two buck-skin mares. - Big-brave moved to White Tail Creek and lived there two winters and summers. - Big-brave moved to Blacktail and has been living there ever since, nineteen winters and summers he has lived there. Though we failed to find among the Blackfoot such elaborate chronicles as among the Dakota and Kiowa, what did come to hand were obviously of the same type and suggest common origins. Further, we get the impression that in details our material is more like the counts of the Kiowa than the Dakota.

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# Worksheet: Calculating the Energy Stored in a Capacitor In this worksheet, we will practice relating the charge stored in and the voltage applied across a capacitor to the energy stored in it. Q1: A capacitor has a charge of 2.5 µC when connected to a battery of 6.0 V. How much energy is stored in this capacitor? Q2: A 165 µF capacitor is used in conjunction with a dc motor. How much energy is stored in the capacitor when 119 V is applied? Q3: Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for s, with an average light power output of 270 kW. If the conversion of electrical energy to light is efficient (because the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? What is the capacitance of the capacitor if the potential difference across its plates is 125 V when it stores enough energy for one flash? Q4: Suppose that the capacitance of a variable capacitor can be manually changed from 100 pF to 800 pF by turning a dial, connected to one set of plates by a shaft, from to . With the dial set at (corresponding to ), the capacitor is connected to a 500 V source. After charging, the capacitor is disconnected from the source, and the dial is turned to . If friction is negligible, how much work is required to turn the dial from to ? Q5: A prankster applies 300 V to a 65.0 µF capacitor and then tosses it to an unsuspecting victim. The victim’s finger is burned by the discharge of the capacitor through 0.130 g of flesh. What is the temperature increase of the flesh? Take the specific heat of tissues to be 3,500 J/kg⋅K. Q6: On a particular day, it takes J of electrical energy to start a truck’s engine. What is the capacitance of a capacitor that could store that amount of energy at 8.0 V? Q7: How much energy is stored in a 12 µF capacitor whose plates are at a potential difference of 3.0 V? Q8: In open-heart surgeries, a very small amount of energy is used to defibrillate the heart. A 5.00 µF capacitor is used in a heart defibrillator, which stores 30.0 J of energy. What is the potential difference across the capacitor in the defibrillator? How much charge is stored by the capacitor in the defibrillator?

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DACPD86 - January 2022 - Developing Spelling Skills in Learners with Dyslexia Why developing spelling skills? This Level 4 Unit provides the knowledge, skills and understanding for teachers and teaching assistants to support learners with dyslexia to develop spelling skills. Although the focus is on how to nurture the development of spelling skills in learners with dyslexia, the approach will benefit a wide range of learners struggling to acquire effective spelling skills. What can I expect from the course? On completion, participants will be able to: - Identify the sub-skills necessary for learners to acquire spelling skills - Recognise the role of phonological awareness in supporting learners with dyslexia with spelling - Describe a variety of techniques to support learners with spelling - Discuss the components of an effective spelling program for learners |Course Fee (Incs. VAT)||£ 354.00| Event terms and conditions

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Share Explore BrainMass # Seven Problems on Determinants, Cramer's Rule and Matrices Seven practice problems are in the attached file. 1. Use Cramer's rule to solve the system x + y -z = -1 2x - 2y + 3z = -12 X + y - 4z = 13 Solution set is 2. Evaluate the determinate 5 0 0 1 -3 1 1 2 4 3. Use Cramer's rule to solve the system or to determine if it is inconsistent(I) or dependant(D) -2x - 5y = -7 -8x - 20y = -28 The solution set is The system is either I or D 4. Use Cramer's rule to solve the system or to determine if it is inconsistent(I) or dependant(D) x + y = 5 x - y = 1 The solution set is The system is either I or D 5. a) Write the linear system as a matrix equation in the form AX=B b) Solve the system using the inverse that is given for the coefficient matrix X - 4y - 4z = -4 X + 2y - 4z = -10 A-1 1/3 0 2/3 X + 2y + 2z = 8 -1/6 1/6 0 0 -1/6 1/6 AX=B __ __ __ __ __ __ __ __ __ = __ __ __ __ __ __ The solution set is __ __ __ 6. a) Write the linear system as a matrix equation in the form AX=B b) Solve the system using the inverse that is given for the coefficient matrix -2/3 0 -1/3 x + y +z = -2 1/3 -1/3 0 -x - 2y + z = 10 A-1 0 1/3 -1/3 -x - 2y - 2z = -2 AX=B __ __ __ __ __ __ __ __ __ = __ __ __ __ __ __ The solution set is __ __ __ 7. a) Write the linear system as a matrix equation in the form AX=B b) Solve the system using the inverse that is given for the coefficient matrix X - y - z =0 3 3 1 -2y - z = 1 A-1 4 -4 -1 4x = 3y = 2 -8 7 2 AX=B __ __ __ __ __ __ __ __ __ = __ __ __ __ __ __ The solution set is __ __ __ #### Solution Summary Seven practice problems on Determinants, Cramers's Rule and Matrices fully solved. \$2.19

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# Number Sense: Fact Families (Multiplication and Division) Packet includes: 54 practice sets (some of which include multiple problems). Fact families are sets of problems and their inverses (so, 1x2=2, 2x1=2, 2÷1=2, and 2÷2=1). Because multiplication and division problems are related, students really only need to memorize one set of facts to know three sets of facts. However, some students needs practice to see and understand how multiplication and division problems relate to each other. This set of practice problems helps students build that understanding slowly, by first asking students to fill in blanks and then asking students to create entire fact families. Fact families focus on multiplication and division facts up to 100. Sample Problem(s) After reading the multiplication problems below, complete the related division problems. 4 x 5 = 20 So, 20 ÷ 5 = _____ And, 20 ÷ 4 = ______ Based on the one problem shown, write the other three math problems that complete the fact family. 6 x 2 = 12 ___ x ___ = ___ ___ ÷ ___ = ___ ___ ÷ ___ = ___

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Games Problems Go Pro! Seventh grader Lucie from Australia asks, "How do you do scientific notation?" That's a great question, Lucie. There are a couple reasons why we have scientific notation. The first reason is that many numbers are either so big or so tiny that they are hard to write out without scientific notation. For example, would you like to know the distance from the Milky Way to Andromeda galaxy? The distance is: d = 23,600,000,000,000,000,000,000 meters Would you like to know the size of a water molecule? The size is: w = 0.0000000025 meters Both of those numbers are a nuisance to write. The first one because it's so big, and needs a lot of zeroes, and the second one because it's so small and needs a lot of zeroes! If only there was a way to avoid writing all those zeroes! Well, that's the good news about scientific notation; it lets us avoid all that writing. If you're writing a big number with scientific notation, you're going to ask yourself, "How many places would I need to move the decimal to the left in order to have only one digit in front of the decimal? In the case of the Andromeda distance, if you moved the decimal 22 places to the left, you would have the decimal right between the 2 and the 3. Once you've figured that out, you can write the number as: d = 2.36 x 1022 meters Notice that we didn't have to write all those zeroes! We put the decimal after the first digit, and then we wrote the number of places we had to move the decimal as an exponent of 10. We do the opposite for very small numbers: we ask, "How many places would I need to move the decimal to the right in order to have one non-zero digit in front of the decimal?" In the case of the size of a water molecule, if we wanted the decimal to to be after the two (the first non-zero digit), we would need to move it 10 places to the right. Once you've figured that out, you can write the following: w = 2.5 x 10-10 meters Note that we used a negative exponent this time. That's because we were moving the decimal ten places in the other direction. And again, we get to skip all those zeroes, making it easier to write. I said there were a couple reasons for writing numbers in scientific notation. One reason, as shown above, is to make them easier to write. The other reason is to make them easier to read and understand. Let me show you an example. Which number is bigger: 23,500,000,000,000,000,000 or 4,200,000,000,000,000,000? In order to answer my question, you had to count the number of groups in each number to find out which one was bigger. Suppose it was written this way instead: Which number is bigger: 2.35 x 1019 or 4.2 x 1018? This is so much easier! If both numbers are properly written in scientific notation, the number with the larger exponent is bigger! And what if they have the same exponent? Then you compare the numbers in front of the multiplication. For example: Which number is bigger: 3.4 x 1032 or 4.5 x 1032? In this case, both numbers are in scientific notation, and they both have an exponent of 32. So we just compare 3.4 and 4.5. Since 4.5 is bigger, we know that 4.5 x 1032 is the larger number. I hope that helps, Lucie! # Blogs on This Site Reviews and book lists - books we love! The site administrator fields questions from visitors.

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A new controversial energy source, the cold fusion reaction, has been making the scientific community fall off their horses for a while now. While some sceptics think of it as pseudo-science, others actually think of it as the revolutionary solution to the world’s problems. Fusion is a very old process by which the universe has been illuminating its dark recesses since as long as time began. The way we humans have explored it so far, nuclear fusion is a process happening at a very high temperature, in which two nuclei merge into a heavier one releasing a lot of energy and ionising radiation. Here, high temperatures is the key, since these reactions mostly occur in the core of stars. For such a reaction to initiate, a lot of heat is required. Cold fusion, it is claimed, is a Low Energy Nuclear Reaction (LENR) that surprisingly requires no initial radioactive material (unlike nuclear fission, the technology behind nuclear reactors and atom bombs), no high temperature or pressure conditions and it does not release any ionising radiation (unlike nuclear fission or fusion). The basic underlying principle being transmutation of nickel to copper through hydrogen atoms. Transmutation is changing the atom of one element to another either by nuclear reaction or radioactive decay. A person familiar with basic high school physics would tell you that two hydrogen nuclei would under normal circumstances repel each other. The reason why hydrogen atoms fuse to form helium in the sun’s core is due to the extreme temperature and pressure prevailing there. Read more about cold fusion intro and cold fusion and the future. Andrea Rossi, an Italian inventor who has a Doctor’s Degree in Philosophy from Milan University 1975 claims, that he has devised an energy catalyzer that can create energy from the fusion of nickel and hydrogen using the energy provided by the mains supply and under a pressure of 25 atm (read more here). However, Andrea Rossi has not been able to prove his claims. On a blog on scienceblogs.com, Ethan Seigal explores the physics behind why Rossi might be bluffing. In his words, If you want to create copper from any of these elements by adding a proton (hydrogen nucleus) to them, here are the reactions you’re looking for: - 58Ni + 1H → 59Cu*, - 60Ni + 1H → 61Cu*, - 61Ni + 1H → 62Cu*, - 62Ni + 1H → 63Cu*, - 64Ni + 1H → 65Cu*. That doesn’t look so prohibitive, does it? Of course, there is the fact that you’ve got to overcome the tremendous Coulomb barrier (the electrical repulsion between nickel and hydrogen nuclei), which — according to our knowledge of nuclear physics — requires temperatures and pressures not found naturally anywhere in the Universe. Not in the Sun, not in the cores of the most massive stars, and (to the best of our knowledge) not even in supernova explosions! You can read the entire blog here. However, despite the lack of a formal scientific explanation supporting cold fusion, it is happening nonetheless. Andrea Rossi has pushed E.Cat into production. So unless Rossi has found the Philosopher’s Stone, or a tech stolen from the future its difficult to explain how he is doing it. Further read, cold fusion berkeley, cold fusion a revolutionary technology and how is Rossi doing it.

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School supply lists have items like pencils, paper, glue, markers and notebooks – all the tools a child needs to be successful in school. But something important is missing. Students need food to perform their best in class. One in eight children in the U.S. is food insecure, which can threaten a child’s health and well-being. Food insecurity can also impact a child’s education. Not having adequate nutrition makes school a challenge in several ways. And it starts early. Good nutrition is critical for brain development and learning from infancy onwards. When kids go to school, not having enough food or not eating enough nutritious foods puts children at a disadvantage. Experts say poor nutrition can undermine children’s school readiness, which can be a predictor of academic success in the future. Studies have shown that children who experience food insecurity in their first years lack the same opportunities to be successful in school as children who were not food insecure. It’s not just younger kids who are affected when they don’t have enough to eat. The effects follow a child and grow worse if the issue isn’t addressed. Experts say that food-insecure children have a harder time getting along with others, and they are more likely to repeat a grade. We can all relate to being hungry and not being able to concentrate, but what if it’s time to take a test? A study in North Carolina found that when students’ families ran out of SNAP benefits for the month, their test scores dropped. Not only does food insecurity translate to lower math and reading scores, it can also lead to more absences and tardiness. In fact, food-insecure students are less likely to graduate high school. There are also school-related emotional and social setbacks when a child doesn’t have enough to eat. School is more than just reading and writing, it’s a time for social growth and development. Students who are food insecure have a harder time making friends and forming bonds. While the National School Lunch and Breakfast Program is very helpful in combatting students’ hunger, it hasn’t completely solved the problem. Why? Some students’ families may not qualify for the program. Or they didn’t fill out the necessary paperwork. And while students may be eating breakfast and lunch at school, they don’t always know if they’ll have dinner at home. It’s why we need people like you – committed to helping children get the food they need to grow up healthy and successful. Together we are ensuring children and families receive much-needed food and essentials like hygiene products. Your donation reaches the most vulnerable students through our School Resource Rooms and backpacks program, making it possible for kids to start the school year off nourished and ready to learn. More About Back to School Families on the Brink after SNAP Reductions Amidst scorching 2023 summer, US families face tough choices - bills or food... Hunger’s Effect on Education School supply lists have items like pencils, paper, glue, markers and notebooks... The Cost of School Supplies Could Hit Some Families Hard As back-to-school season nears, families like Katrina's and Kim's face daily...

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Lightning is destructive force of nature and quite challenging to predict. University of Washington has done a comprehensive study and says that AI can improve lightening forecasts. Machine learning algorithms can be used to improve predictions in lightning. The University of Washington researchers say the forecasts will be helpful for preparing for potential wildfires, gives advanced safety warnings, and develop reliable long-range climate models. The University of Washington is the first to develop machine learning algorithms and combine it with past data of lightning. The new model brought forward by the University of Washington has combined weather forecasts with machine learning equations studying data of past events of lightning. The university presented its hybrid model at American Geophysical Union’s fall meeting. The new technique has several advanced integrations than the existing technique. Several phenomena were taken into consideration such as amount of precipitation, ascent speed of storm clouds, and several other factors. Further, the researchers hope to improve the technique using more data sources, weather variables, and other advanced techniques making its more reliable and accurate for predictions.

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An exothermic process occurs uniformly through out a 10-cm-diameter sphere (k = 300 W/m · K, cp = 400 J/kg · K, ρ = 7500 kg/m3), and it generates heat at a constant rate of 1.2 MW/m3. The sphere initially is at a uniform temperature of 20°C, and the exothermic process is commenced at time t = 0. To keep the sphere temperature under control, it is submerged in a liquid bath maintained at 20°C. The heat transfer coefficient at the sphere surface is 250 W/m2 · K. Due to the high thermal conductivity of sphere, the conductive resistance within the sphere can be neglected in comparison to the convective resistance at its surface. Accordingly, this unsteady state heat transfer situation could be analyzed as a lumped system. (a) Show that the variation of sphere temperature T with time t can be expressed as dT/dt = 0.5 - 0.005T (b) Predict the steady-state temperature of the sphere. (C) Calculate the time needed for the sphere temperature to reach the average of its initial and final (steady) temperatures.

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Published May 17, 2012 The story begins innocently enough. Fertile soil attracts labor from far and wide. Factories provide employment, farmland is plentiful, and for a time the economy of Ivory Coast booms as a much-desired commodity – cocoa – is exported across the globe. But the story of cocoa has never been an innocent one. So valuable in the Aztec court that it was used as currency, blood has been shed over cocoa profits since Europeans first developed a taste for chocolate. Over the past two centuries, farming and production have moved from country to country, from the Caribbean to West Africa, always dependent on rich farmland and cheap labor. Ivory Coast’s ethnic strife is the most recent chapter in cocoa’s troubled history. Initially migrant workers from across West Africa were invited to the country to share in its farmland, helping Ivory Coast become the world's top producer. (Today it provides some 40 percent of the world's crop.) But once the economy went sour in the 1980s, cocoa profits were more jealously guarded. Land disputes erupted, sparking xenophobic violence that became a 10-year civil war. With the cessation of post-election violence last year and the ascendance of a new government, the war is supposedly over. But new attacks are still carried out between rival factions; thousands of people still live in refugee camps; and those who return to their destroyed homes swear vengeance. As always, cocoa production continues through the strife — but reconciliation and a true end to conflict may still be a long way off.

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7 Details To Include In Setting & 5 Ways To Use It To Advance A Plot Setting is defined as the physical location and time of a story. It acts as a story’s backbone. Nobody exists in a vacuum. We all do things somewhere. Setting includes the basic surroundings, the era, or the moment the story occupies, and it often has its own personality. We should introduce our main settings in the beginning of the book. Readers like to feel comfortable with the places you are going to use. Once we identify our setting, we must flesh it out with sensory detail. The senses allow us to set moods, evoke feelings and trigger memories. 7 Details To Include In Setting - Place. Where does the story take place? This could be a planet, a country, a city, a building, a forest, a ship, a spaceship, an island, or one room in a house. - Culture. How do laws, taboos, social mores, politics, sport, religious practices, education, wars, and technology set the scene? - Era. When does your story happen? During the violence of the anti-apartheid riots of the ‘70s and ‘80s? When the first white settlers arrived in the Cape of Good Hope in the 1650s? - Geography. Mountains, desserts, volcanoes, farmland, vegetation, animals, oceans, lakes, and seas all colour the background. - Things. These include pets, possessions, items in shops, landmarks, road signs – anything your character is able to touch, see, hear, smell, or taste. - Time. This could be an hour, a day, a season, a year, or a lifetime. - Weather. Rain, drought, fog, snow, sunshine, high or low temperatures, storms – all of these affect your story. While it is obvious that setting adds layers to our story, provides the framework for the story, and affects our characters, we often overlook the role setting can have in moving a plot forward. Writers do this by using the ‘change factor’. Human beings don’t like change. Change takes us out of our comfort zones and our primary comfort zone is our environment. 5 Ways To Use Setting To Advance A Plot - Reveal something that was previously hidden. The beautiful mountains behind a town might not be as stable as everyone thinks. An unspoilt park becomes a nightmare when the family walking their dog find a dislodged sign that says ‘Keep Out. Venomous Snakes.’ - Create an outside threat to the environment. You can do this with weather. A bad storm could prevent a character from achieving a goal. You can do this with religion. A cult may have set up its headquarters in the area, forever changing the town. You can do this with technology. A technical error could shut down the electricity supply to hotel. - Remove possessions or pets. If a character loses something or has something stolen, it will affect the environment. The loss of an animal changes the setting completely. - Changes in society. A change in attitudes, laws, or politics could alter a setting. The location of a business that depends on a law being passed may be destroyed when it isn’t. - Move your character into an alien environment. You could move the character out of her family home into a studio apartment or she could simply take the wrong train home. Changes in setting do not have to be pivotal to the plot, but they can help an author who wants to advance a story without using direct confrontations with other characters to do so. © Amanda Patterson If you enjoyed this post, read: - How To Write A Beginning And An Ending That Readers Will Never Forget - 10 Elementary Tips For Writers From Sherlock Holmes - The Daily Word Counts of 39 Famous Authors Become a Writers Write patron: If you’re inspired, educated, or entertained by our posts, please support us with a donation.

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I'm not at all a cook. So don't fear, this is not going to be a cooking demonstration. But I do want to talk to you about something that I think is dear to all of us. And that is bread — something which is as simple as our basic, most fundamental human staple. And I think few of us spend the day without eating bread in some form. Unless you're on one of these Californian low-carb diets, bread is standard. Bread is not only standard in the Western diet. As I will show to you, it is actually the mainstay of modern life. So I'm going to bake bread for you. In the meantime I'm also talking to you, so my life is going to be complicated. Bear with me. First of all, a little bit of audience participation. I have two loaves of bread here. One is a supermarket standard: white bread, pre-packaged, which I'm told is called a Wonderbread. (Laughter) I didn't know this word until I arrived. And this is more or less, a whole-meal, handmade, small-bakery loaf of bread. Here we go. I want to see a show of hands. Who prefers the whole-meal bread? Okay let me do this differently. Is anybody preferring the Wonderbread at all? (Laughter) I have two tentative male hands. (Laughter) Okay, now the question is really, why is this so? And I think it is because we feel that this kind of bread really is about authenticity. It's about a traditional way of living. A way that is perhaps more real, more honest. This is an image from Tuscany, where we feel agriculture is still about beauty. And life is really, too. And this is about good taste, good traditions. Why do we have this image? Why do we feel that this is more true than this? Well I think it has a lot to do with our history. In the 10,000 years since agriculture evolved, most of our ancestors have actually been agriculturalists or they were closely related to food production. And we have this mythical image of how life was in rural areas in the past. Art has helped us to maintain that kind of image. It was a mythical past. Of course, the reality is quite different. These poor farmers working the land by hand or with their animals, had yield levels that are comparable to the poorest farmers today in West Africa. But we have, somehow, in the course of the last few centuries, or even decades, started to cultivate an image of a mythical, rural agricultural past. It was only 200 years ago that we had the advent of the Industrial Revolution. And while I'm starting to make some bread for you here, it's very important to understand what that revolution did to us. It brought us power. It brought us mechanization, fertilizers. And it actually drove up our yields. And even sort of horrible things, like picking beans by hand, can now be done automatically. All that is a real, great improvement, as we shall see. Of course we also, particularly in the last decade, managed to envelop the world in a dense chain of supermarkets, in a chain of global trade. And it means that you now eat products, which can come from all around the world. That is the reality of our modern life. Now you may prefer this loaf of bread. Excuse my hands but this is how it is. But actually the real relevant bread, historically, is this white Wonder loaf. And don't despise the white bread because it really, I think, symbolizes the fact that bread and food have become plentiful and affordable to all. And that is a feat that we are not really conscious of that much. But it has changed the world. This tiny bread that is tasteless in some ways and has a lot of problems has changed the world. So what is happening? Well the best way to look at that is to do a tiny bit of simplistic statistics. With the advent of the Industrial Revolution with modernization of agriculture in the last few decades, since the 1960s, food availability, per head, in this world, has increased by 25 percent. And the world population in the meantime has doubled. That means that we have now more food available than ever before in human history. And that is the result, directly, of being so successful at increasing the scale and volume of our production. And this is true, as you can see, for all countries, including the so-called developing countries. What happened to our bread in the meantime? As food became plentiful here, it also meant that we were able to decrease the number of people working in agriculture to something like, on average, in the high income countries, five percent or less of the population. In the U.S. only one percent of the people are actually farmers. And it frees us all up to do other things — to sit at TED meetings and not to worry about our food. That is, historically, a really unique situation. Never before has the responsibility to feed the world been in the hands of so few people. And never before have so many people been oblivious of that fact. So as food became more plentiful, bread became cheaper. And as it became cheaper, bread manufacturers decided to add in all kinds of things. We added in more sugar. We add in raisins and oil and milk and all kinds of things to make bread, from a simple food into kind of a support for calories. And today, bread now is associated with obesity, which is very strange. It is the basic, most fundamental food that we've had in the last ten thousand years. Wheat is the most important crop — the first crop we domesticated and the most important crop we still grow today. But this is now this strange concoction of high calories. And that's not only true in this country, it is true all over the world. Bread has migrated to tropical countries, where the middle classes now eat French rolls and hamburgers and where the commuters find bread much more handy to use than rice or cassava. So bread has become from a main staple, a source of calories associated with obesity and also a source of modernity, of modern life. And the whiter the bread, in many countries, the better it is. So this is the story of bread as we know it now. But of course the price of mass production has been that we moved large-scale. And large-scale has meant destruction of many of our landscapes, destruction of biodiversity — still a lonely emu here in the Brazilian cerrado soybean fields. The costs have been tremendous — water pollution, all the things you know about, destruction of our habitats. What we need to do is to go back to understanding what our food is about. And this is where I have to query all of you. How many of you can actually tell wheat apart from other cereals? How many of you actually can make a bread in this way, without starting with a bread machine or just some kind of packaged flavor? Can you actually bake bread? Do you know how much a loaf of bread actually costs? We have become very removed from what our bread really is, which, again, evolutionarily speaking, is very strange. In fact not many of you know that our bread, of course, was not a European invention. It was invented by farmers in Iraq and Syria in particular. The tiny spike on the left to the center is actually the forefather of wheat. This is where it all comes from, and where these farmers who actually, ten thousand years ago, put us on the road of bread. Now it is not surprising that with this massification and large-scale production, there is a counter-movement that emerged — very much also here in California. The counter-movement says, "Let's go back to this. Let's go back to traditional farming. Let's go back to small-scale, to farmers' markets, small bakeries and all that." Wonderful. Don't we all agree? I certainly agree. I would love to go back to Tuscany to this kind of traditional setting, gastronomy, good food. But this is a fallacy. And the fallacy comes from idealizing a past that we have forgotten about. If we do this, if we want to stay with traditional small-scale farming we are going, actually, to relegate these poor farmers and their husbands — among whom I have lived for many years, working without electricity and water, to try to improve their food production — we relegate them to poverty. What they want are implements to increase their production: something to fertilize the soil, something to protect their crop and to bring it to a market. We cannot just think that small-scale is the solution to the world food problem. It's a luxury solution for us who can afford it, if you want to afford it. In fact we do not want this poor woman to work the land like this. If we say just small-scale production, as is the tendency here, to go back to local food means that a poor man like Hans Rosling cannot even eat oranges anymore because in Scandinavia we don't have oranges. So local food production is out. But also we do not want to relegate to poverty in the rural areas. And we do not want to relegate the urban poor to starvation. So we must find other solutions. One of our problems is that world food production needs to increase very rapidly — doubling by about 2030. The main driver of that is actually meat. And meat consumption in Southeast Asia and China in particular is what drives the prices of cereals. That need for animal protein is going to continue. We can discuss alternatives in another talk, perhaps one day, but this is our driving force. So what can we do? Can we find a solution to produce more? Yes. But we need mechanization. And I'm making a real plea here. I feel so strongly that you cannot ask a small farmer to work the land and bend over to grow a hectare of rice, 150,000 times, just to plant a crop and weed it. You cannot ask people to work under these conditions. We need clever low-key mechanization that avoids the problems of the large-scale mechanization that we've had. So what can we do? We must feed three billion people in cities. We will not do that through small farmers' markets because these people have no small farmers' markets at their disposal. They have low incomes. And they benefit from cheap, affordable, safe and diverse food. That's what we must aim for in the next 20 to 30 years. But yes there are some solutions. And let me just do one simple conceptual thing: if I plot science as a proxy for control of the production process and scale. What you see is that we've started in the left-hand corner with traditional agriculture, which was sort of small-scale and low-control. We've moved towards large-scale and very high control. What I want us to do is to keep up the science and even get more science in there but go to a kind of regional scale — not just in terms of the scale of the fields, but in terms of the entire food network. That's where we should move. And the ultimate may be, but it doesn't apply to cereals, that we have entirely closed ecosystems — the horticultural systems right at the top left-hand corner. So we need to think differently about agriculture science. Agriculture science for most people — and there are not many farmers among you here — has this name of being bad, of being about pollution, about large-scale, about the destruction of the environment. That is not necessary. We need more science and not less. And we need good science. So what kind of science can we have? Well first of all I think we can do much better on the existing technologies. Use biotechnology where useful, particularly in pest and disease resistance. There are also robots, for example, who can recognize weeds with a resolution of half an inch. We have much cleverer irrigation. We do not need to spill the water if we don't want to. And we need to think very dispassionately about the comparative advantages of small-scale and large-scale. We need to think that land is multi-functional. It has different functions. There are different ways in which we must use it — for residential, for nature, for agriculture purposes. And we also need to re-examine livestock. Go regional and go to urban food systems. I want to see fish ponds in parking lots and basements. I want to have horticulture and greenhouses on top of residential areas. And I want to use the energy that comes from those greenhouses and from the fermentation of crops to heat our residential areas. There are all kinds of ways we can do it. We cannot solve the world food problem by using biological agriculture. But we can do a lot more. And the main thing that I would really ask all of you as you go back to your countries, or as you stay here: ask your government for an integrated food policy. Food is as important as energy, as security, as the environment. Everything is linked together. So we can do that. In fact in a densely populated country like the River Delta, where I live in the Netherlands, we have combined these functions. So this is not science fiction. We can combine things even in a social sense of making the rural areas more accessible to people — to house, for example, the chronically sick. There is all kinds of things we can do. But there is something you must do. It's not enough for me to say, "Let's get more bold science into agriculture." You must go back and think about your own food chain. Talk to farmers. When was the last time you went to a farm and talked to a farmer? Talk to people in restaurants. Understand where you are in the food chain, where your food comes from. Understand that you are part of this enormous chain of events. And that frees you up to do other things. And above all, to me, food is about respect. It's about understanding, when you eat, that there are also many people who are still in this situation, who are still struggling for their daily food. And the kind of simplistic solutions that we sometimes have, to think that doing everything by hand is going to be the solution, is really not morally justified. We need to help to lift them out of poverty. We need to make them proud of being a farmer because they allow us to survive. Never before, as I said, has the responsibility for food been in the hands of so few. And never before have we had the luxury of taking it for granted because it is now so cheap. And I think there is nobody else who has expressed better, to me, the idea that food, in the end, in our own tradition, is something holy. It's not about nutrients and calories. It's about sharing. It's about honesty. It's about identity. Who said this so beautifully was Mahatma Gandhi, 75 years ago, when he spoke about bread. He did not speak about rice, in India. He said, "To those who have to go without two meals a day, God can only appear as bread." And so as I'm finishing my bread here — and I've been baking it, and I'll try not to burn my hands. Let me share with those of you here in the first row. Let me share some of the food with you. Take some of my bread. And as you eat it, and as you try it — please come and stand up. Have some of it. I want you to think that every bite connects you to the past and the future: to these anonymous farmers, that first bred the first wheat varieties; and to the farmers of today, who've been making this. And you don't even know who they are. Every meal you eat contains ingredients from all across the world. Everything makes us so privileged, that we can eat this food, that we don't struggle every day. And that, I think, evolutionarily-speaking is unique. We've never had that before. So enjoy your bread. Eat it, and feel privileged. Thank you very much. (Applause)

We need to feed the whole world

What Is It, Causes, Diagnosis, Treatments, and More Author: Alyssa Haag Illustrator: Jillian Dunbar What is bulbar palsy? Bulbar palsy refers to a set of signs and symptoms linked to the impaired function of the lower cranial nerves, typically caused by damage to their lower motor neurons or to the lower cranial nerve itself. The impacted cranial nerves are a set of nerves that arise straight from the brainstem and include cranial nerves IX (9), X (10), XI (11), and XII (12). Lower motor neurons are the neurons that connect the central nervous system, such as the brain and spinal cord, to the muscles they innervate. Damage to lower motor neurons can result in a wide variety of symptoms based on the cranial nerve that is damaged. For example, cranial nerve IX (the glossopharyngeal nerve) is involved in salivation, swallowing, and the gag reflex. If cranial nerve IX is injured, it can lead to difficulty swallowing (dysphagia) and a reduced gag reflex. Common signs and symptoms of damage to the other cranial nerves include difficulty chewing, nasal regurgitation, slurred speech, difficulty in handling secretions, aspiration of secretions, altered vocal ability (dysphonia) and difficulty articulating words (dysarthria).Other clinical signs associated with bulbar palsy include a nasal speech that lacks in modulation (e.g. controlling or adjusting of one’s speech), difficulty with all consonants, an atrophic (wasting) tongue, drooling, weakness of the jaw and facial muscles, a normal or absent jaw jerk (upward jerk of the jaw upon striking the chin) and an absent gag reflex. Bulbar palsy is sometimes classified as progressive or non-progressive. Progressive bulbar palsy is more common and refers to the escalation of symptoms over time. It can occur in both children and adults. Non-progressive bulbar palsy, on the other hand, refers to bulbar palsy that does not worsen; it is considered very uncommon.Bulbar palsy is sometimes confused with pseudobulbar palsy, which is the result of damage to the upper motor neurons. While the two conditions share many of the same symptoms, pseudobulbar palsy is often characterized by the atypical expression of emotion displayed by unusual outbursts of laughing or crying, called emotional lability. Meanwhile, with bulbar palsy, an individual’s emotions usually remain unaffected. Other differentiating signs and symptoms of psudeobulbar palsy include the absence of facial emotions, a spastic and pointed tongue, and an exaggerated jaw jerk. What causes bulbar palsy? The most common causes of bulbar palsy include brainstem strokes and tumors. The brainstem is the part of the brain where the cranial nerves arise from and where all motor control signals are transmitted. Thus, damage to the brainstem—from strokes or tumors—can damage various cranial nerves and disrupt motor control.Other causes of bulbar palsy include certain degenerative diseases (e.g. amyotrophic lateral sclerosis), autoimmune diseases (e.g. Guillain–Barré syndrome), and genetic diseases (e.g. Kennedy disease). Amyotrophic lateral sclerosis, commonly referred to as ALS, is a rare neurological disorder characterized by a gradual deterioration and death of motor neurons. Guillain–Barré syndrome causes the immune system to attack its own nerves, potentially targeting cranial nerves. Kennedy disease is a lower motor neuron disease that can affect transmission signals between the brain and the spinal cord, potentially resulting in bulbar palsy. Is bulbar palsy hereditary? Although it is not always caused by genetic considerations, bulbar palsy has been linked to several hereditary conditions. Recently, childhood forms of progressive bulbar palsy have been genetically caused by Brown–Vialetto–Van Laere (BVVL) and Fazio–Londe syndromes. BVVL is an autosomal recessive disorder, which means that both parents may be carriers for the disease without expressing the condition themselves, and can pass on the expressive form of the disease to their child. Fazio–Londo syndrome presents similarly and is inherited in a similar manner to BVVL. Additionally, bulbar palsy can be the result of a condition that is considered hereditary. For example, Kennedy disease is an X-linked recessive disorder that can cause bulbar palsy. How do you diagnose bulbar palsy? Reviewing an individual's history of symptoms and clinical features can help to diagnose bulbar palsy. However, additional tests may be required in order to diagnose the underlying cause. An analysis of an individual’s cerebral spinal fluid (fluid that surrounds the brain and spinal cord) may be conducted to rule out multiple sclerosis, which can sometimes present in a similar manner to bulbar palsy. Magnetic resonance imaging (MRI) of the brain may also be performed in order to diagnose a stroke or tumor. How do you treat bulbar palsy? There is currently no known treatment for bulbar palsy. However, supportive treatments are used for the management of symptoms and underlying conditions. Certain medications may be prescribed to control drooling. A feeding tube can also be given to individuals who have severe difficulty swallowing. Finally, speech and language therapy may assist with difficulties in speech, chewing, and swallowing. Additional condition-specific treatments will likely be required for individuals facing degenerative, autoimmune, or genetic diseases that resulted in bulbar palsy. Is bulbar palsy fatal? Bulbar palsy can prove to be fatal in progressive cases. Death from progressive bulbar palsy often occurs 1 to 3 years from the start of the disorder, however, it is often attributed to the development of associated aspiration pneumonia (infection of the lungs). Aspiration pneumonia often occurs in individuals with bulbar palsy as they may have difficulty swallowing and as a result, a large amount of material, or food, may enter the lungs and lead to infection.Extensive bulbar damage may also damage the respiratory center in the brainstem, which is involved in signaling and controlling breathing. This can lead to a life threatening inability to breathe properly. Additionally, progressive bulbar palsy may advance to ALS, or amyotrophic lateral sclerosis, and prognosis is usually poor. With ALS, death of motor neurons interferes with an individual’s ability to breathe and can ultimately result in fatality. What are the most important facts to know about bulbar palsy? Bulbar palsy is a set of conditions that can occur due to damage to the lower cranial nerves. Clinical features of bulbar palsy range from difficulty swallowing and a lack of a gag reflex to inability to articulate words and excessive drooling. Bulbar palsy is most commonly caused by a brainstem stroke or tumor. Certain autoimmune diseases, genetic diseases, and degenerative disorders can also potentially result in the development of bulbar palsy. Treatment primarily focuses on the management of related symptoms. Related linksClostridium botulinum (Botulism) Resources for research and reference Bosch, A. M., Stroek, K., Abeling, N.G. et al. (2012). The Brown-Vialetto-Van Laere and Fazio Londe syndrome revisited: natural history, genetics, treatment and future perspectives. Orphanet Journal of Rare Diseases, 7(83). DOI: 10.1186/1750-1172-7-83 Garg M., Kulkarni, S. D., Hegde, A. U., & Shah, K. N. (2018). Riboflavin Treatment in Genetically Proven Brown-Vialetto-Van Laere Syndrome. Journal of Pediatric Neuroscience, 13(4): 471-473. DOI: 10.4103/JPN.JPN_131_17 Green, P., Wiseman, M., et al. (2010). Brown-Vialetto-Van Laere syndrome, a Ponto-Bulbar Palsy with Deafness, is Caused by Mutations in c20orf54. American Journal of Human Genetics, 86(3): 485-9. DOI: 10.1016/j.ajhg.2010.02.006 Motor Neuron Diseases Fact Sheet. (2015). In National Institute of Neurological Disorders and Stroke (NINDS). Retrieved November 23, 2020, from https://www.ninds.nih.gov/Disorders/All-Disorders/Motor-Neuron-Diseases-Information-Page Saleem, F. & Munakomi, S. Pseudobulbar Palsy. (2020). In StatPearls [Internet]. Retrieved November 23, 2020, from https://www.ncbi.nlm.nih.gov/books/NBK553160/

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No Widgets found in the Sidebar # How does charles law relate to scuba diving #### Bydreamtravel Sep 20, 2023 ## Charles’ Law and Scuba Diving Charles’ Law, also known as the law of volumes, is a gas law that describes the relationship between the volume and temperature of a gas when the pressure remains constant. It states that the volume of a gas is directly proportional to its absolute temperature. ### Formula and Mathematical Explanation The formula for Charles’ Law is: “` V₁/T₁ = V₂/T₂ “` where: V₁ is the initial volume of the gas T₁ is the initial temperature of the gas V₂ is the final volume of the gas T₂ is the final temperature of the gas Explanation: This formula indicates that the ratio of the initial volume to the initial temperature is equal to the ratio of the final volume to the final temperature. This means that if the temperature of a gas increases while the pressure remains constant, the volume of the gas will also increase. Conversely, if the temperature of a gas decreases, the volume of the gas will also decrease. ### Application to Scuba Diving Charles’ Law has important implications for scuba diving because it explains how the volume of air in a diver’s lungs changes as they ascend and descend in the water. Ascent: As a diver ascends, the pressure surrounding them decreases. According to Charles’ Law, this decrease in pressure causes the air in their lungs to expand. This expansion can lead to a condition known as pulmonary barotrauma, which occurs when the expanding air causes damage to the lungs or airways. Descent: Conversely, as a diver descends, the pressure surrounding them increases. This increase in pressure causes the air in their lungs to compress, according to Charles’ Law. If the diver descends too quickly, this compression can cause a condition known as air embolism, which occurs when compressed air enters the bloodstream and causes blockages. ### Implications for Scuba Divers Understanding Charles’ Law is crucial for scuba divers to ensure their safety while diving. To minimize the risks associated with volume changes, divers must: Ascent slowly: Ascending too quickly can cause the air in the lungs to expand rapidly, leading to pulmonary barotrauma. Divers should ascend at a rate of no more than 30 feet per minute. Equalize pressure: Equalizing pressure involves adding air to the middle ear and sinuses to prevent them from collapsing due to increased pressure during descent. Divers can equalize pressure by swallowing, pinching their nose and blowing, or using the Valsalva maneuver. Avoid diving with a cold or congestion: Mucus in the ears or sinuses can make it difficult to equalize pressure, increasing the risk of barotrauma. Divers should avoid diving if they are experiencing any respiratory issues. ### Conclusion Charles’ Law is a fundamental gas law that has significant implications for scuba diving. By understanding how changes in temperature and pressure affect the volume of air in their lungs, divers can avoid the risks associated with volume changes and ensure their safety while diving. Read Post  Is there good scuba diving in puerto rico

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Categories # Factoring You grade this on Sunday. What do you do on Monday? Go over the procedure that this student almost flawlessly executed?  Again?  What do you emphasize?  Checking your answer? Thanks to John Weisenfeld for the submission. ## 4 replies on “Factoring” Begin at the beginning. Check to see if the student knows how to reduce a fraction. They seemed to ignore the factoring of the coefficients entirely. I might consider talking about what on earth ‘factoring’ means again. I would emphasize that factoring is a process of rewriting an expression as an equivalent expression in a particular way. So, you should be checking that the answer is really equivalent to what you started with. So yeah, checking the answer is definitely a part of the Monday conversation. This looks like a student who has some idea that there’s something involving common factors to be done, but not with what that means. To them, maybe (4n^3)/(3n^2) and 4n^3 + 3n^2 and 4n^3 * 3n^2 all look pretty similar, and when you tell them in the first two cases that there’s something to be done with the common factor of n^2, it seems arbitrary to them that the thing to be done is different. So maybe it’s time for a whole bunch of “compare and contrast” style exercises on Monday where they tell you about what’s similar and what’s different in each case. (Also, in response to ccssimath: SIMPLIFY a fraction, please. Again, because I want to emphasize the equivalence.) Sorry, we never use the hackneyed verb “simplify” in any instruction due to its ambiguity. Which is simpler, 1 1/3 or 4/3? We reduce fractions to lowest terms. It’s mathematically unambiguous. We suggested going back to the beginning because asking a student to reduce a fraction to lowest terms by factoring reveals if the students knows how to find the GCF. Understanding reducing fractions involves the same skill set and will lead to a better understanding of the rationale underlying factoring algebraic expressions. Otherwise, students proceed by rote, not by understanding. ccssimath.blogspot.com

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Linear or Non-linear Systems (Linearity Property): A linear system is a system which follows the superposition principle. Let us consider a system having its response as ‘T’, input as x(n) and it produces output y(n). This is shown in figure below: Let us consider two inputs. Input x1(n) produces output y1(n) and input x2(n) produces output y2(n). Now consider two arbitrary constants a1 and a2. Then simply multiply these constants with input x1(n) and x2(n) respectively. Thus a1x1(n) produces output a1y1(n) and a2x2(n) produces output a2y2(n). Theorem for linearity of the system: A system is said to be linear if the combined response of a1x1(n) and a2x2(n) is equal to the addition of the individual responses. T[a1 x1(n) + a2 x2(n)] = a1 T[x1(n)] + a2 T[x2(n)]…………….1) The above theorem is also known as superposition theorem. Linear system has one important characteristic: If the input to the system is zero then it produces zero output. If the given system produces some output (non-zero) at zero input then the system is said to be Non-linear system. If this condition is satisfied then apply the superposition theorem to determine whether the given system is linear or not? For continuous time system: Similar to the discrete time system a continuous time system is said to be linear if it follows the superposition theorem. Let us consider two systems as follows: y1(t) = f[x1(t)] And y2(t) = f[x2(t)] Here y1(t) and y2(t) are the responses of the system and x1(t) and x2(t) are the excitations. Then the system is said to be linear if it satisfies the following expression: f[a1 x1(t) + a2 x2(t)] = a1 y1(t) + a2 y2(t)…………….1) Where a1 and a2 are constants. A system is said to be non-linear system if does not satisfies the above expression. Communication channels and filters are examples of linear systems. How to determine whether the given system is Linear or not? To determine whether the given system is Linear or not, we have to follow the following steps: Step 1: Apply zero input and check the output. If the output is zero then the system is linear. If this step is satisfied then follow the remaining steps. Step 2: Apply individual inputs to the system and determine corresponding outputs. Then add all outputs. Denote this addition by y’(n). This is the R.H.S. of the 1st equation. Step 3: Combine all inputs. Apply it to the system and find out y”(n). This is L.H.S. of equation (1). Step 4: if y’(n) = y”(n) then the system is linear otherwise it is non-linear system. Determine whether the following system is linear or not? y(n) = n x(n) Step 1: When input x(n) is zero then output is also zero. Here first step is satisfied so we will check remaining steps for linearity. Step 2: Let us consider two inputs x1(n) and x2(n) be the two inputs which produces outputs y1(t) and y2(t) respectively. It is given as follows: Now add these two output to get y’(n) Therefore y’(n) = y1(n) + y2(n) = n x1(n) + n x2(n) Therefore y’(n) = n [x1(n) + x2(n)] Step 3: Now add x1(n) and x2(n) and apply this input to the system. We know that the function of system is to multiply input by ‘n’. Here [x1(n) + x2(n)] acts as one input to the system. So the corresponding output is, y”(n) = n [x1(n) + x2(n)] Step 4: Compare y’(n) and y”(n). Here y’(n) = y”(n). hence the given system is linear. You may also like: - Properties of a system - Stable or Unstable System (Stability Property) - Causal and Noncausal System (Causality Property) - Time Variant or Time Invariant Systems - Static or dynamic systems [contact-form][contact-field label=’Name’ type=’name’ required=’1’/][contact-field label=’Email’ type=’email’ required=’1’/][contact-field label=’Website’ type=’url’/][contact-field label=’Comment’ type=’textarea’ required=’1’/][/contact-form]

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By Phil Gardocki The main purpose of the German Jager Divisions was to fight in adverse terrain where using smaller, coordinated units was deemed a better way to fight than employing the brute force offered by the standard infantry divisions. The Jäger divisions were more heavily equipped than a mountain division, but not as well-armed as a larger infantry division. In the early stages of the war, they were the interface divisions between the mountains and the plains. The Jägers (means “hunters” in German) relied on high degrees of training, and superior communications, as well as their not inconsiderable artillery support. In the middle stages of the war, as the standard infantry divisions were downsized, the Jäger model with two infantry regiments came to dominate the standard tables of organization, and, in 1943, in an attempt to boost morale, Hitler declared that all infantry divisions were now Jägers. The 100th Jäger Division was formed in 1940 in the 17th Wherkreis, where it trained intensively until June of 1941. The obvious differences between a Jäger and a standard infantry division were one less infantry regiment and a slightly lighter artillery park. Less obvious was the increase in signals troops. A Jäger division had a normal infantry division’s signals battalion, plus an additional full signal company at the regimental level. This increased the coordination between the regiments and the divisional artillery. On the regimental level, there were other organizational differences. In addition to the standard infantry regiment antitank or artillery section assets, a Jäger regiment had a motorized antitank company and relied on its communications to coordinate all of these assets. Paradoxically, a Jäger, or light, battalion was heavier than its line infantry counterpart. In addition to three companies of infantry, a Jäger battalion had two supporting companies of heavy weapons compared to only one for an infantry battalion. This provided ten 7.5cm guns and six 8.1cm mortars of support for every Jäger battalion, compared to only two 7.5cm guns and six 8.1 mortars for the standard infantry battalion. When the 100th went to war, it was assigned the reinforced 369th Croat Jäger Regiment, which came with its own artillery battalion. Historically, it is uncertain if this battalion was reassigned to the division’s artillery regiment or not. Given the philosophy of devolving firepower to the lower levels, and language problems, it probably remained as a permanent asset of the 369th, split up among the various Croat Jäger battalions. The 100th’s first action was with the 17th Army around Odessa, and thereafter it followed Army Group South’s advance across the Ukraine. In 1942, it was involved in the Second Battle of Kharkov, and eventually was used to subjugate Stalingrad. After participating in the bitter street fighting there, it was encircled along with the rest of the 6th Army and destroyed early in 1943. A new 100th Jäger Division was created in Yugoslavia later in 1943, where it was used for counter insurgency operations there and in Albania. By the end of the winter of 1943–1944, it was transferred to the 2nd SS Panzer Corps in Army Group Center. The division took part in the withdrawal across Poland and ended the war in Czechoslovakia. Equipment and Doctrine The weapons used by the German infantry always involved close team work between the machineguns and the artillery. The machineguns would pin the enemy down, while the artillery would take them out. To facilitate this teamwork, artillery was placed at many levels in the organization so that there was always some artillery support at every level. At the company level, there were the trench mortars. In 1939, these were mostly the feeble 5cm, which could throw a 1kg shell about 800 meters. By 1944, the heavier 8.1cm mortar was being issued to all infantry companies and it could throw a 3kg shell 2.4km. Each Jäger battalion also had a heavy machinegun company. This was a deceptive name, as the heavy machinegun was the same as the light machineguns issued to the squads, but it also was equipped with six 8.1cm mortars, and two light infantry guns. In German artillery parlance, any caliber smaller than 15cm was classified as light, mediums were 15-17 cm, while heavy guns started at 20cm. These light infantry guns were 7.5cm L/12’s, L12 meaning the barrel length was 12 times the diameter. They were rapid fire, short ranged, direct fire weapons. They were very useful for removing enemy strong points and machinegun nests as well as breaking up massed attacks. All told, one of these companies could throw more ordnance, by weight, than the rest of the battalion combined. At the regimental level there was also several support companies. Usually, this would be an antitank company with mostly 3.7cm and sometimes a few 5cm guns. As the war progressed and the enemy tanks got heavier, this weapon’s mix started leaning towards mostly 5cm with the odd 7.5cm antitank gun. There was also a heavy weapons company with more 7.5cm L/12’s and two 15cm L/12’s infantry cannons. Of course, the real backbone of a division is its artillery regiment. For the normal German infantry division, this would consist of 36 × 10cm L/32 and 12 × 15cm guns. The German practice was to dedicate a specific artillery battalion to an Infantry or Jäger Regiment or Battalion that was expected to be involved in combat. As both caliber’s ranged about 14 km, and outranged their Russian counterparts by 5 or more km, German counter-battery fire was quite effective. Once enemy artillery was silenced, the artillery regiment was capable of throwing hundreds of tons of shells per hour onto a target. The division had several supporting battalions. A reconnaissance battalion that was mostly motorized, with many heavy weapons, and engineering support. It was designed to outfight what it could not avoid. The bicycle battalion was used as a fast reaction force to reinforce beleaguered troops. A replacement battalion was provided for front line training experience. The antitank battalion was often the only 100% motorized element in an Infantry or Jäger division. Later in the war, this merged in function with the bicycle battalion to become the “Schnell”, or “Hurry Up” Battalion.

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Ice-walled melt ponds on the surfaces of active valley-floor rock glaciers and Matthes (Little Ice Age) moraines in the southern Sierra Nevada indicate that most of these landforms consist of glacier ice under thin (ca. 1 - 10 m) but continuous covers of rock-fall-generated debris. These debris blankets effectively insulate the underlying ice and greatly reduce rates of ablation relative to that of uncovered ice. Such insulation explains the observations that ice-cored rock glaciers in the Sierra, actually debris-covered glaciers, are apparently less sensitive to climatic warming and commonly advance to lower altitudes than do adjacent bare-ice glaciers. Accumulation-area ratios and toe-to-headwall-altitude ratios used to estimate equilibrium-line altitudes (ELAs) of former glaciers may therefore yield incorrect results for cirque glaciers subject to abundant rockfall. Inadvertent lumping of deposits from former debris-covered and bare-ice glaciers partially explains an apparently anomalous regional ELA gradient reported for the pre-Matthes Recess Peak Neoglacial advance. Distinguishing such deposits may be important to studies that rely on paleo-ELA estimates. Moreover, Matthes and Recess Peak ELA gradients along the crest evidently depend strongly on local orographic effects rather than latitudinal climatic trends, indicating that simple linear projections and regional climatic interpretations of ELA gradients of small glaciers may be unreliable. Additional publication details Debris-Covered Glaciers in the Sierra Nevada, California, and Their Implications for Snowline Reconstructions

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In economics, elasticity is the measurement of how much one thing (such as quantity) changes when another thing (such as price) changes. Elasticity = % Change in Quantity / % Change in Price % Change in Quantity = (Quantity End – Quantity Start) / Quantity Start % Change in Price = (Price End – Price Start) / Price Start 500 units are produced at the start and 600 at the end. In the same period, cost to produce goes from $20 to $25. % Change in Quantity = (600 – 500) / 500 = 100 / 500 = 0.20 % Change in Price = ($25 – $20) / $20 = $5 / $20 = 0.25 Elasticity = 0.20 / 0.25 = 0.80 Therefore, elasticity is 0.80. - Khan Academy – Elasticity Tutorial – Part of a large course on economics, this page is an introduction to different types of elasticity. - Wikipedia – Elasticity (economics) – An overview of the concept of elasticity. It includes examples of different types of elasticity. - Dr. Emma Hutchinson – Principles of Microeconomics – 4.1 Calculating Elasticity – Part of a textbook on microeconomics. How to calculate elasticity. - Lumen Learning – Elasticity – Part of a larger course on microeconomics. This page summarizes elasticity with a number of examples.

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Global resources play a vital role in shaping our world and ensuring its sustainability. From energy and minerals to food and water, these resources are essential for human development, economic growth, and environmental stability. In this article, we will explore the significance of global resources, their distribution, challenges faced, and the importance of sustainable management. Let’s delve into the vast realm of global resources and understand their impact on our lives. - Introduction: Understanding Global Resources Global resources refer to the wide range of materials and substances available on Earth, which are utilized to meet the needs of humanity. These global resources can be classified into various categories, including natural resources (such as forests, water, and minerals), energy resources (like fossil fuels and renewable energy), and agricultural resources (such as crops and livestock). The management and sustainability of these resources are crucial for the well-being of present and future generations. - The Diversity of Global Resources The world is abundant with diverse resources that vary in their types and locations. Natural resources encompass forests, oceans, fertile soils, minerals, and freshwater reserves. Energy resources comprise fossil fuels like coal, oil, and natural gas, as well as renewable sources such as solar, wind, and hydroelectric power. Additionally, agricultural resources encompass the cultivation of crops, livestock, and fisheries, which provide sustenance and contribute to the economy. - Distribution and Accessibility of Global Resources Global resources are not evenly distributed around the world. Some regions are naturally endowed with an abundance of certain resources, while others may face scarcity. Factors such as geological formations, climate conditions, and historical processes influence the distribution of resources. Accessibility to these resources is also influenced by socioeconomic factors, technological advancements, and political dynamics, which can create disparities in resource availability and utilization. - Environmental Impacts and Conservation Efforts The extraction, production, and consumption of global resources can have significant environmental impacts. Activities like mining, deforestation, and fossil fuel combustion contribute to air and water pollution, habitat destruction, and climate change. To mitigate these adverse effects, conservation efforts are crucial. These include sustainable forestry practices, renewable energy adoption, water resource management, and wildlife conservation initiatives. The conservation of global resources ensures their long-term availability and maintains the balance of ecosystems. - Challenges in Resource Management Resource management faces various challenges on a global scale. Overexploitation and unsustainable practices can deplete resources, leading to ecological imbalances and social conflicts. Additionally, population growth, urbanization, and industrialization place increasing demands on resources, putting strain on their availability. Climate change and natural disasters also pose threats to resource availability and require adaptive strategies. Effective resource management requires collaboration, innovative solutions, and a holistic approach. - The Role of Technology in Resource Optimization Technological advancements have revolutionized resource optimization and management. Innovations in renewable energy technologies, precision agriculture, water treatment, and waste management have increased efficiency and reduced environmental impacts. Advanced data analytics, artificial intelligence, and automation play crucial roles in resource monitoring, forecasting, and decision-making processes. Technology enables us to better understand resource patterns, optimize resource utilization, and develop sustainable practices for resource management. - Sustainable Practices for Resource Preservation To ensure the long-term availability of global resources, it is essential to adopt sustainable practices. This includes promoting energy efficiency, transitioning to renewable energy sources, practicing responsible mining and logging, implementing water conservation measures, and supporting sustainable agriculture. Sustainable practices aim to meet present needs without compromising the ability of future generations to meet their own needs. By prioritizing sustainability, we can preserve the integrity of ecosystems and safeguard the well-being of both humans and the planet. - Social and Economic Implications of Global Resources Global resources have profound social and economic implications. The availability and distribution of resources influence geopolitical dynamics, trade relations, and economic development. Resource-rich countries often experience economic growth and attract investments, while resource-dependent communities may face challenges related to environmental degradation, displacement, and social inequalities. Effective resource governance and equitable resource-sharing mechanisms are crucial to ensure the benefits of globl resources reach all segments of society, fostering sustainable and inclusive development. - Collaboration and International Relations Given the global nature of resource utilization, collaboration and international cooperation are essential. Nations must engage in dialogue, negotiate agreements, and establish frameworks for responsible resource management. This includes sharing best practices, promoting technology transfer, and supporting capacity-building initiatives in resource-rich countries. Collaborative efforts can enhance resource efficiency, reduce conflicts over resource access, and foster sustainable development on a global scale. - The Future of Global Resource Management As we look ahead, the future of global resource management holds both challenges and opportunities. It is crucial to embrace a holistic approach that considers environmental, social, and economic dimensions. This includes investing in research and innovation, promoting sustainable consumption and production patterns, and integrating resource management into policy frameworks. By adopting a forward-thinking mindset and taking proactive measures, we can navigate the complexities of global resource management and ensure a sustainable and prosperous future. Global resources are the foundation of human civilization, supporting our daily lives, economic activities, and environmental well-being. Their sustainable management is of utmost importance to address the challenges of resource scarcity, environmental degradation, and social inequalities. By adopting sustainable practices, embracing technology, fostering collaboration, and promoting equitable resource governance, we can pave the way for a future where global resources are harnessed responsibly, benefiting current and future generations alike. - How can individuals contribute to global resource sustainability? Individuals can contribute by practicing energy efficiency, reducing waste, supporting sustainable businesses, and advocating for responsible resource management in their communities. - Are renewable energy sources the key to global resource sustainability? Renewable energy sources play a significant role in reducing our reliance on fossil fuels and mitigating environmental impacts. However, a holistic approach that encompasses all aspects of resource management is crucial for long-term sustainability. - What role does education play in global resource management? Education plays a vital role in raising awareness about resource conservation, sustainable practices, and the importance of responsible consumption. It empowers individuals to make informed choices and contribute to a more sustainable future. - How can international cooperation address resource-related conflicts? International cooperation can foster dialogue, negotiation, and the establishment of frameworks that promote equitable resource-sharing, resolve conflicts, and ensure the sustainable utilization of resources. - Why is sustainable resource management important for future generations? Sustainable resource management ensures that future generations have access to the necessary resources for their well-being and development. It safeguards the environment, reduces social inequalities, and fosters long-term prosperity.

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100−−−√100100100 Use Newton’s method to approximate the given number correct to eight # 100−−−√100100100 Use Newton’s method to approximate the given number correct to eight • MATH sec2(π2−x)−1=cot2(x)sec2(π2-x)-1=cot2(x) Verfiy the identity. First, sec(x)=1cos(x)sec(x)=1cos(x) and cot(x)=cos(x)sin(x).cot(x)=cos(x)sin(x). Second, cos(π2−x)=sin(x)cos(π2-x)=sin(x) and sin(π2−x)=cos(x).sin(π2-x)=cos(x). Therefore the left part is • MATH sin(t)csc(π2−t)=tan(t)sin(t)csc(π2-t)=tan(t) Verfiy the identity. Verify the identity: sin(t)csc(π2−t)=tan(t)sin(t)csc(π2-t)=tan(t) Simplify the left side of the equation by using the cofunction identity csc((pi/2)-t)=sec(t). sin(t)sec(t)=tan(t)sin(t)sec(t)=tan(t) Simplify the left side of the… • MATH sec2(y)−cot2(π2−y)=1sec2(y)-cot2(π2-y)=1 Verfiy the identity. First, sec(y)=1cos(y)sec(y)=1cos(y) and cot(y)=cos(y)sin(y).cot(y)=cos(y)sin(y). Second, cos(π2−y)=sin(y)cos(π2-y)=sin(y) and sin(π2−y)=cos(y).sin(π2-y)=cos(y). Therefore sec2(y)−cot2(π2−y)=1cos2(y)−sin2(y)cos2(y)=sec2(y)-cot2(π2-y)=1cos2(y)-sin2(y)cos2(y)=… • MATH cos2(β)+cos2(π2−β)=1cos2(β)+cos2(π2-β)=1 Verfiy the identity. cos(π2−b)=sin(b)cos(π2-b)=sin(b) for all b. So the initial identity is equivalent to cos2(b)+sin2(b)=1,cos2(b)+sin2(b)=1, which is true. • MATH 1−cos(θ)1+cos(θ)−−−−−−−−−√=1−cos(θ)|sin(θ)|1-cos(θ)1+cos(θ)=1-cos(θ)|sin(θ)| Verfiy the identity. By squaring both sides you are assuming that the equality is true, but this is what was to be established. Better is to multiply left side by sqrt(1-cos)/sqrt(1-cos) resulting in… • MATH 1+sin(θ)1−sin(θ)−−−−−−−−−√=1+sin(θ)|cos(θ)|1+sin(θ)1-sin(θ)=1+sin(θ)|cos(θ)| Verfiy the identity. Verify 1+sin(θ)1−sin(θ)−−−−−−−−−√=1+sin(θ)|cos(θ)|1+sin(θ)1-sin(θ)=1+sin(θ)|cos(θ)| Working from the left side, we show that it is equivalent to the right side: 1+sin(θ)1−sin(θ)−−−−−−−−−√1+sin(θ)1-sin(θ)… • MATH cos(x)−cos(y)sin(x)+sin(y)+sin(x)−sin(y)cos(x)+cos(y)=0cos(x)-cos(y)sin(x)+sin(y)+sin(x)-sin(y)cos(x)+cos(y)=0 Verfiy the… We’ll use the formula (a−b)⋅(a+b)=a2−b2.(a-b)⋅(a+b)=a2-b2. Multiply equation by (sin(x)+sin(y))⋅(cos(x)+cos(y))(sin(x)+sin(y))⋅(cos(x)+cos(y)) (the product of the denominators): • MATH tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x)tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x) Verfiy the identity. Verify the identity: tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x)tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x) Rewrite the left side of the equation as two fractions. tan(x)tan(x)cot(y)+cot(y)tan(x)cot(y)=tan(y)+cot(x)tan(x)tan(x)cot(y)+cot(y)tan(x)cot(y)=tan(y)+cot(x)… • MATH tan(x)+tan(y)1−tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)−1tan(x)+tan(y)1-tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)-1 Verfiy the… Verify the identity: tan(x)+tan(y)1−tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)−1tan(x)+tan(y)1-tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)-1 Divide every term on the left side of the equation by tan(x)tan(y)… • MATH (1+sin(y))[1+sin(−y)]=cos2(y)(1+sin(y))[1+sin(-y)]=cos2(y) Verfiy the identity. Verify (1+sin(y))(1-sin(y))=cos^2(y): Working from the left side we show that it is equivalent to the right side: (1+sin(y))(1-sin(y)) =1-sin^2(y) =cos^2(y) using the pythagorean identity • MATH csc(−x)sec(−x)=−cot(x)csc(-x)sec(-x)=-cot(x) Verfiy the identity. Verify the identity: csc(−x)sec(−x)=−cot(x)csc(-x)sec(-x)=-cot(x) Simplify the left side of the equation using the following even/odd identities: csc(-x)=-csc(x) and sec(-x)=sec(x) −csc(x)sec(x)=−cot(x)-csc(x)sec(x)=-cot(x)… • MATH tan(x)cot(x)cos(x)=sec(x)tan(x)cot(x)cos(x)=sec(x) Verfiy the identity. Verify the identity: tan(x)cot(x)cos(x)=sec(x)tan(x)cot(x)cos(x)=sec(x) Simplify the numerator on the left side of the equation. Since tan(x) and cot(x) are reciprocals their product is 1. 1cos(x)=sec(x)1cos(x)=sec(x) Simplify… • MATH cos[(π2)−x]sin[(π2)−x]=tan(x)cos[(π2)-x]sin[(π2)-x]=tan(x) Verfiy the identity. Verify the identity: cos[(π2)−x]sin[(π2)−x]=tanxcos[(π2)-x]sin[(π2)-x]=tanx Simplify the left side of the equation using the cofunction identity. sin(x)cos(x)=tanxsin(x)cos(x)=tanx Simplify the left side of the equation using… • MATH tan(π2−θ)tan(θ)=1tan(π2-θ)tan(θ)=1 Verfiy the identity. By definition, tan(t)=sin(t)/cos(t). Also, sin(π2−t)=cos(t)sin(π2-t)=cos(t) and cos(π2−t)=sin(t).cos(π2-t)=sin(t). Therefore tan(π2−t)=cos(t)sin(t)tan(π2-t)=cos(t)sin(t) and tan(π2−t)⋅tan(t)=1,tan(π2-t)⋅tan(t)=1, QED. • MATH cos(x)−cos(x)1−tan(x)=sin(x)cos(x)sin(x)−cos(x)cos(x)-cos(x)1-tan(x)=sin(x)cos(x)sin(x)-cos(x) Verify the identity. Verify the identity. cos(x)−[cos(x)1−tan(x)]=sinxcosxsinx−cosxcos(x)-[cos(x)1-tan(x)]=sinxcosxsinx-cosx cos(x)(1−tan(x))−cos(x)1−tan(x)=sin(x)cos(x)sin(x)−cos(x)cos(x)(1-tan(x))-cos(x)1-tan(x)=sin(x)cos(x)sin(x)-cos(x)… • MATH 1cos(x)+1+1cos(x)−1=−2csc(x)cot(x)1cos(x)+1+1cos(x)-1=-2csc(x)cot(x) Verify the identity. Verify the identity: 1cos(x)+1+1cos(x)−1=−2csc(x)cot(x)1cos(x)+1+1cos(x)-1=-2csc(x)cot(x)cos(x)−1+cos(x)+1cos2(x)−1=−2csc(x)cot(x)cos(x)-1+cos(x)+1cos2(x)-1=-2csc(x)cot(x) 2cos(x)cos2(x)−1=−2csc(x)cot(x)2cos(x)cos2(x)-1=-2csc(x)cot(x) Use the pythagorean identity… • MATH cos(θ)cot(θ)1−sin(θ)−1=csc(θ)cos(θ)cot(θ)1-sin(θ)-1=csc(θ) Verfiy the identity. Verify the identity cos(θ)cot(θ)1−sin(θ)−1=csc(θ)cos(θ)cot(θ)1-sin(θ)-1=csc(θ) cos(θ)cot(θ)1−sin(θ)−1=csc(θ)cos(θ)cot(θ)1-sin(θ)-1=csc(θ) Rewrite cot(θ)cot(θ) as a quotient…. • MATH 1+sin(θ)cos(θ)+cos(θ)1+sin(θ)=2sec(θ)1+sin(θ)cos(θ)+cos(θ)1+sin(θ)=2sec(θ) Verfiy the identity. Transform the left part: 1+sin(θ)cos(θ)+cos(θ)1+sin(θ)=1+sin(θ)cos(θ)+cos(θ)1+sin(θ)= (1+sin(θ))2+cos2(θ)cos(θ)⋅(1+sin(θ))=(1+sin(θ))2+cos2(θ)cos(θ)⋅(1+sin(θ))= • MATH 1sin(x)−1csc(x)=csc(x)−sin(x)1sin(x)-1csc(x)=csc(x)-sin(x) Verfiy the identity. Verify the identity: 1sin(x)−1csc(x)=csc(x)−sin(x)1sin(x)-1csc(x)=csc(x)-sin(x) Simplify the left side of the equation using the following reciprocal identities: 1/sin(x)=csc(x) and 1/csc(x)=sin(x)…. • MATH 1tan(x)+1cot(x)=tan(x)+cot(x)1tan(x)+1cot(x)=tan(x)+cot(x) Verfiy the identity. cot(x)=1/tan(x) and tan(x)=1/cot(x). So the left side is cot(x)+tan(x) and the right is tan(x)+cot(x), which are the same. • MATH sec(x)(csc(x)−2sin(x))=cot(x)−tan(x)sec(x)(csc(x)-2sin(x))=cot(x)-tan(x) Verify the identity. Transform the left side: sec(x)⋅(csc(x)−2sin(x))=(1cos(x))⋅(1−2sin2(x)sin(x)=cos2(x)−sin2(x)cos(x)⋅sin(x).sec(x)⋅(csc(x)-2sin(x))=(1cos(x))⋅(1-2sin2(x)sin(x)=cos2(x)-sin2(x)cos(x)⋅sin(x).The right side is… • MATH sec(x)−cos(x)=sin(x)tan(x)sec(x)-cos(x)=sin(x)tan(x) Verfiy the identity. Verify the identity: sec(x)−cos(x)=sin(x)tan(x)sec(x)-cos(x)=sin(x)tan(x) The reciprocal of sec(x)=1/cos(x). sec(x)−cos(x)=sin(x)tan(x)sec(x)-cos(x)=sin(x)tan(x) 1cos(x)−cos(x)=sin(x)tan(x)1cos(x)-cos(x)=sin(x)tan(x) 1−cos2(x)cos(x)=sin(x)tan(x)1-cos2(x)cos(x)=sin(x)tan(x)Use the… • MATH sec(θ)−11−cos(θ)=sec(θ)sec(θ)-11-cos(θ)=sec(θ) Verfiy the identity. Verify the identity: sec(θ)−11−cos(θ)=sec(θ)sec(θ)-11-cos(θ)=sec(θ) Use the reciprocal identity cos(θ)=1sec(θ)cos(θ)=1sec(θ) sec(θ)−11−cos(θ)=sec(θ)sec(θ)-11-cos(θ)=sec(θ)… • MATH cot(x)sec(x)=csc(x)−sin(x)cot(x)sec(x)=csc(x)-sin(x) Verfiy the identity. Verify cot(x)/sec(x)=csc(x)-sin(x): Working from the left side, we show that it is equivalent to the right side: cot(x)/sec(x) =cos(x)sin(x)1cos(x)=cos(x)sin(x)1cos(x) =cos2(x)sin(x)=cos2(x)sin(x)… • MATH sec6(x)(sec(x)tan(x))−sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)sec6(x)(sec(x)tan(x))-sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)Verify the identity. Verify the identity. sec6(x)(sec(x)tan(x))−sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)sec6(x)(sec(x)tan(x))-sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)Factor out the GCF sec4(x)(sec(x)tan(x))sec4(x)(sec(x)tan(x)) sec4(x)(sec(x)tan(x))[sec2(x)−1]=sec5(x)tan3(x)sec4(x)(sec(x)tan(x))[sec2(x)-1]=sec5(x)tan3(x) Use the… • MATH sin12xcosx−sin52xcosx=cos3(x)sin(x)−−−−−√sin12xcosx-sin52xcosx=cos3(x)sin(x) Verify the identity. Verify the identity. sin12(x)cos(x)−sin52(x)cos(x)=cos3(x)sin(x)−−−−−√sin12(x)cos(x)-sin52(x)cos(x)=cos3(x)sin(x) Factor out the GCF sin12(x)cos(x).sin12(x)cos(x). sin12(x)cos(x)[1−sin2(x)]=cos3(x)sin(x)−−−−−√sin12(x)cos(x)[1-sin2(x)]=cos3(x)sin(x) Use the pythagorean… • MATH 1tan(β)+tan(β)=sec2(β)tan(β)1tan(β)+tan(β)=sec2(β)tan(β) Verfiy the identity. Verify the identity: 1tan(β)+tan(β)=sec2(β)tan(β)1tan(β)+tan(β)=sec2(β)tan(β) 1+tan2(β)tan(β)=sec2(β)tan(β)1+tan2(β)tan(β)=sec2(β)tan(β) Use the pythagorean identity 1+tan2(β)=sec2(β).1+tan2(β)=sec2(β)…. • MATH cot2(t)csc(t)=1−sin2(t)sin(t)cot2(t)csc(t)=1-sin2(t)sin(t) Verfiy the identity. cot(t)=cos(t)sin(t),cot(t)=cos(t)sin(t), csc(t)=1sin(t).csc(t)=1sin(t). So cot2(t)csc(t)=cos2(t)sin2(t)⋅⎛⎝11sin(t)⎞⎠=cos2(t)sin2(t)⋅sin(t)=cos2(t)sin(t),cot2(t)csc(t)=cos2(t)sin2(t)⋅(11sin(t))=cos2(t)sin2(t)⋅sin(t)=cos2(t)sin(t),which is the same as the right part… • MATH cot3(t)csc(t)=cos(t)(csc2(t)−1)cot3(t)csc(t)=cos(t)(csc2(t)-1) Verfiy the identity. You need to remember that 1csct=sint1csct=sint , hence, replacing sintsint for 1csct1csct to the left side, yields: sint⋅cot3t=(cost)⋅(csc2t−1)sint⋅cot3t=(cost)⋅(csc2t-1)You need to replace cos3tsin3tcos3tsin3t… • MATH tan2(θ)sec(θ)=sin(θ)tan(θ)tan2(θ)sec(θ)=sin(θ)tan(θ) Verfiy the identity. L.H.S tan2(θ)sec(θ)=tan2(θ)⋅cos(θ)=tan(θ)⋅cos(θ)⋅{tan(θ)}=tan2(θ)sec(θ)=tan2(θ)⋅cos(θ)=tan(θ)⋅cos(θ)⋅{tan(θ)}= tan(θ)⋅cos(θ)⋅{sin(θ)cos(θ)}=tan(θ)⋅sin(θ)=tan(θ)⋅cos(θ)⋅{sin(θ)cos(θ)}=tan(θ)⋅sin(θ)= R.H.S • MATH sin2(α)−sin4(α)=cos2(α)−cos4(α)sin2(α)-sin4(α)=cos2(α)-cos4(α) Verfiy the identity. Verify: sin2(α)−sin4(α)=cos2(α)−cos4(α)sin2(α)-sin4(α)=cos2(α)-cos4(α) Use the pythagorean identity sin2(α)+cos2(α)=1sin2(α)+cos2(α)=1 if sin2(α)sin2(α) is isolated the pythagorean identity is… • MATH cos2(β)−sin2(β)=1−2sin2(β)cos2(β)-sin2(β)=1-2sin2(β) Verfiy the identity. Verify cos2(β)−sin2(β)=1−2sin2(β)cos2(β)-sin2(β)=1-2sin2(β) Working from the left side, we show that it is equivalent to the right side: cos2(β)−sin2(β)cos2(β)-sin2(β) =(1−sin2(β))−sin2(β)=(1-sin2(β))-sin2(β)… • MATH cos2(β)−sin2(β)=2cos2(β)−1cos2(β)-sin2(β)=2cos2(β)-1 Verfiy the identity. Verify the identity: cos2(β)−sin2(β)=2cos2(β)−1cos2(β)-sin2(β)=2cos2(β)-1 Use the pythagorean identity sin2(β)+cos2(β)=1.sin2(β)+cos2(β)=1. If sin2(β)sin2(β) is isolated the pythagorean identity would be… • MATH (1+sin(α))(1−sin(α))=cos2(α)(1+sin(α))(1-sin(α))=cos2(α) Verfiy the identity. First, (1−x)⋅(1+x)=1−x2.(1-x)⋅(1+x)=1-x2. So the left part is 1−sin2(α),1-sin2(α),which is obviously equal to cos2(α).cos2(α). • MATH cos(x)+sin(x)tan(x)=sec(x)cos(x)+sin(x)tan(x)=sec(x) Verfiy the identity. Verify cos(x)+sin(x)tan(x)=sec(x)cos(x)+sin(x)tan(x)=sec(x) : cos(x)+sin(x)tan(x)cos(x)+sin(x)tan(x) =cos(x)+sin(x)⋅sin(x)cos(x)=cos(x)+sin(x)⋅sin(x)cos(x) =cos(x)+sin2(x)cos(x)=cos(x)+sin2(x)cos(x) =cos2(x)+sin2(x)cos(x)=cos2(x)+sin2(x)cos(x) =1cos(x)=1cos(x) =sec(x)=sec(x) as required. • MATH cot2(y)(sec2(y)−1)=1cot2(y)(sec2(y)-1)=1 Verfiy the identity. Verify the identity: cot2(y)[sec2(y)−1]=1cot2(y)[sec2(y)-1]=1 Use the pythagorean identity 1+tan2(y)=sec2(y).1+tan2(y)=sec2(y). If tan2(y)tan2(y) is isolated the identity would be tan2(y)=sec2(y)−1.tan2(y)=sec2(y)-1. cot2(y)[sec2(y)−1]=1cot2(y)[sec2(y)-1]=1… • MATH sec(y)cos(y)=1sec(y)cos(y)=1 Verfiy the identity. By definition, sec(y)=1cos(y).sec(y)=1cos(y). Therefore sec(y)⋅cos(y)=[1cos(y)]⋅cos(y)=1sec(y)⋅cos(y)=[1cos(y)]⋅cos(y)=1 , QED. • MATH earctan(x)=x3+1−−−−−√earctan(x)=x3+1 Use Newton’s method to find all roots of the equation correct to… You need to remember how Newton’s method is used, to evaluate the roots of a transcendent equation. xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) , where xnxn is the approximate solution to the equation… • MATH 4(e−x2)sin(x)=x2−x+14(e-x2)sin(x)=x2-x+1 Use Newton’s method to find all roots of the equation… 4(e−x2)sin(x)=x2−x+14(e-x2)sin(x)=x2-x+1 f(x)=4(e−x2)sin(x)−x2+x−1=0f(x)=4(e-x2)sin(x)-x2+x-1=0 To solve equation using Newton’s method apply the formula, xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn)… • MATH cos(x2−x)=x4cos(x2-x)=x4 Use Newton’s method to find all roots of the equation correct to eight… cos(x2−x)=x4cos(x2-x)=x4 Set the left side equal to zero. 0=x4−cos(x2−x)0=x4-cos(x2-x) To solve using Newton’s method, apply the formula: xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Let the function of the given equation… • MATH xx2+1=1−x−−−−−√xx2+1=1-x Use Newton’s method to find all roots of the equation correct to… xx2+1=1−x−−−−−√xx2+1=1-x f(x)=xx2+1−1−x−−−−−√=0f(x)=xx2+1-1-x=0 f'(x)=(x2+1)−x(2x)(x2+1)2−(12)(1−x)−12(−1)f′(x)=(x2+1)-x(2x)(x2+1)2-(12)(1-x)-12(-1) f'(x)=1−x2(x2+1)2+121−x−−−−−√f′(x)=1-x2(x2+1)2+121-x See the attached graph. From the graph the curve… • MATH x5−3×4+x3−x2−x+6=0x5-3×4+x3-x2-x+6=0 Use Newton’s method to find all roots of the equation… x5−3×4+x3−x2−x+6=0x5-3×4+x3-x2-x+6=0 To solve this, using Newton’s method, apply the formula:  xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Let the function be:  f(x)=x5−3×4+x3−x2−x+6f(x)=x5-3×4+x3-x2-x+6 Then, take its… • MATH x6−x5−6×4−x2+x+10=0x6-x5-6×4-x2+x+10=0 Use Newton’s method to find all roots of the equation… f(x)=x6−x5−6×4−x2+x+10f(x)=x6-x5-6×4-x2+x+10 f'(x)=6×5−5×4−24×3−2x+1f′(x)=6×5-5×4-24×3-2x+1 xn+1=xn−(xn)6−(xn)5−6(xn)4−(xn)2+xn+106(xn)5−5(xn)4−24(xn)3−2(xn)+1xn+1=xn-(xn)6-(xn)5-6(xn)4-(xn)2+xn+106(xn)5-5(xn)4-24(xn)3-2(xn)+1 See the attached graph. From the graph the… • MATH sin(x)=x2−2sin(x)=x2-2 Use Newton’s method to find all roots of the equation correct to six… sin(x)=x2−2sin(x)=x2-2 f(x)=x2−2−sin(x)=0f(x)=x2-2-sin(x)=0 f'(x)=2x−cos(x)f′(x)=2x-cos(x) See the attached graph. From the graph, the curve of the function intersects the x-axis at ≈≈ -1.05 and 1.70. These can be used as initial… • MATH x3=tan−1(x)x3=tan-1(x) Use Newton’s method to find all roots of the equation correct to six… x3=tan−1(x)x3=tan-1(x) f(x)=x3−tan−1(x)=0f(x)=x3-tan-1(x)=0 f'(x)=3×2−(1×2+1)f′(x)=3×2-(1×2+1) To solve using Newton’s method apply the formula, xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Plug in f(x) and f'(x) in the formula… • MATH 1x=1+x31x=1+x3 Use Newton’s method to find all roots of the equation correct to six decimal… 1x=1+x31x=1+x3 Set the left side equal to zero. 0=1+x3−1×0=1+x3-1x To solve this using Newton’s method, apply the formula: xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn)Let the function be: f(x)=1+x3−1xf(x)=1+x3-1x Take… • MATH (x−2)2=ln(x)(x-2)2=ln(x) Use Newton’s method to find all roots of the equation correct to six… (x−2)2=ln(x)(x-2)2=ln(x) f(x)=(x−2)2−ln(x)=0f(x)=(x-2)2-ln(x)=0 f'(x)=2(x−2)−1xf′(x)=2(x-2)-1x See the attached graph. The curve of the function intersects the x-axis at x ≈≈ 1.4 and 3. These can be used as initial approximates…. • MATH x+1−−−−−√=x2−xx+1=x2-x Use Newton’s method to find all roots of the equation correct to six… x+1−−−−−√=x2−xx+1=x2-x Set the left side equal to zero. 0=x2−x−x−1−−−−−√0=x2-x-x-1 To solve using Newton’s method, apply the formula: xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Let the function f(x) be:… • MATH 3cos(x)=x+13cos(x)=x+1 Use Newton’s method to find all roots of the equation correct to six… 3cos(x)=x+13cos(x)=x+1 f(x)=x+1−3cos(x)=0f(x)=x+1-3cos(x)=0 To solve using Newton’s method apply the formula, xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) f'(x)=1+3sin(x)f′(x)=1+3sin(x) Plug in f(x) and f'(x) in the formula,…

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24. Dead zones • Dead zones are large areas in the ocean that have low oxygen concentration. • The marine life in these areas mostly suffocates and dies or if they are mobile like the fish then, they leave the area. • Though at many times, dead zones occur naturally, scientists are also of the opinion that they are created due to increased human activity. • The main cause of the zones created by humans is nutrient pollution. Excess nutrients (nitrogen and phosphorus) can result in the overgrowth of algae, which later decomposes in the water consuming excess oxygen, depleting the supply available for the marine life. Dead zones can be found in virtually every oceanic body, the largest encompassing almost the whole bottom of the Baltic Sea. Another large dead zone is located in the Gulf of Mexico. Other dead zones occur off the western coasts of North and South America and off the coast of Namibia and western coast of India. Global warming triggered by climate change is predicted to lead to an expansion of these dead zones. However, it is not certain whether the climate change would lead to the removal of the last traces of oxygen from the bay as well.

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# math help urgent what di di do wrong in this??? the question is: x-8y=18 -16x+16y=-8 i did : first multiplied the first equation with 2 then 2x-16y=36 -16+16y=-8 ----------- -14x=28 x=28/-14 x=-2 now x-8y=18 (-2)-8y=18 -8y=18+2 -8y=20 y=20/-8 y= 2.5 plz help me what did i do wrong?? 1. I can't see anything besides that y=-2.5 because you divide by a negitive posted by Sally 2. oo....ok thanks sally posted by Anonymous ## Similar Questions 1. ### Math - PreCalc (12th Grade) Which rectangular equation corresponds to these parametric equations? x = 0.2sec t and y = -0.25tan t A) 25x^2 − 16y^2 = 1 B) 25x^2 + 16y^2 = 1 C) x^2 − 16y^2 = 25 D) 25y^2 − 16x^2 = 1 E) 4x^2 − 16x^2 = 1 2. ### math-algebra Simplify: [(4x^3y)(4^2xy)^-1]/(4xy^2) A. x /16y^2 B. 16x/y^2 C. x^2/16y D. y/16x^2 3. ### Algebra 2 16x^2+16y^2-16x+24y-3=0 How do I solve this? 4. ### alg. solve the system by addition: -5x-9y = -5 x+5y=-15 I've tried this one over and over, but I can't seem to get the right answer. I don't even really know where to start. each time i try to do it, i go about it a little differently 5. ### Math 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13 Answer Multiply factors. -15x - 10 - x + 3 = -16x - 20 +13 Group like terms. -16x - 7 = -16x - 7 Add 16x + 7 to both sides and write the equation as follows 0 = 0 6. ### algebra What are the factors of the following trinomial? 16y^2 - 12y + 2 a) (4y - 1) (4y - 2) b) (2y - 1) (8y + 2) c) (16y - 1) (y - 2) d) (8y - 1) (2y - 2) i think a but i can be wrong 7. ### math This is a multiple choice question. What is the solution of the equation 2(x+2)^2-4=28 1)2 only 2)6 only 3)-2 and 6 4)2 and -6 I did (2x+4)(2x+4)-4=28 4x^2+8x+8x+16-4=28 4x^2+16x+12=28 4x^2+16x=16 This is where I get stuck, can 8. ### Pre-Cal Classify the conic and write the equation in standard form: (1) 9x^2 + 9y^2 + 18x - 18y = 0 (2) 16x^2 + 16y^2 - 64x + 32y + 55 = 0 (3) 6x^2 + 2y^2 + 18x - 10y + 2 = 0 9. ### Algebra There are two solutions to | 16x - 5 | = 3. The greatest solution is ___. Since the expression, 16x - 5, can be either positive or negative, solve for both. 16x - 5 = 3 16x = 8 x = .5 -(16x - 5) = 3 -16x + 5 = 3 -16x = -2 x = 1/8 10. ### math x-8y=18 -16x+16y=-8 More Similar Questions

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What is a porcupine? The word porcupine means “quill pig” in Latin; however, porcupines are large rodents and have no relation to pigs. Porcupines are the largest and heaviest of all African rodents. The head is roundish and rather domed, with a blunt muzzle and small eyes and ears. The legs are short and sturdy, and each foot has five toes, all equipped with powerful claws. Their most recognizable feature is, of course, its quills. Quill length varies on different parts of the body, ranging from 2.5 to 30.5 centimeters (1 to 12 inches). Usually, the quills lie flat against the body, but if danger threatens, they raise and spread them. Scales on quill tips lodge in the skin like fishhooks and are difficult to pull out. New quills grow in to replace lost ones. Hystrix cristata (Crested Porcupine); Atherurus africanus (African Brush-tailed Porcupine); Hystrix africaeaustralis (Cape Porcupine) 10 to 30 kilograms (22 to 66 pounds) 60 to 93 centimeters in length (23 to 37 inches) 15 years in the wild; up to 20 years in captivity Hilly, rocky country, woodland savanna 90 to 110 days Pythons, leopards, large owls Human-wildlife conflict threatens porcupines’ existence. As human populations expand, humans and porcupines find themselves in increasingly close quarters. When porcupine populations close to cultivated areas surge, they can become serious agricultural pests. They are smoked out of their burrows and hunted with spears, nets, or dogs. These practices have eliminated them from densely settled areas. They are targeted for their quills. Porcupine quills have long been a favorite ornament and good-luck charm in Africa. The hollow rattle quills serve as musical instruments and were once used as containers for gold dust. In addition to being targeted for their quills, they are illegally hunted for their meat. Our solutions to protecting the porcupine: African Wildlife Foundation educates communities about the importance of sustainable practices for agricultural and settlement growth by providing training on best practices and incentivizing conservation agriculture when appropriate. AWF engages local communities to set aside land for wildlife to live undisturbed. In the Laikipia region of Kenya — which has no formal protected areas — we partnered with the Koija community and a private operator to construct the Koija Starbeds Lodge. Koija Starbeds sets aside land for wildlife while, at the same time, creating jobs and income for the local community. Porcupines like to set up homes in burrows. These large rodents modify natural shelters among roots and rocks, inhabit holes made by other animals, or dig their own hideaways. These burrows are most commonly occupied in family units. Females like to give birth in the comfort of their own homes. The gestation period of the African crested porcupine is about 112 days. Between one and four young are born in the grass-lined burrow. They are well-developed and are born with open eyes. The young leave home for the first time at about two weeks of age, as their quills — soft at birth — begin to harden. They are quite playful and, outside the burrow, they run and chase one another. The young are suckled from six to eight weeks of age and then begin to eat vegetable matter. The porcupine attacks in reverse. They warn potential enemies of their defense system when alarmed. Porcupines will stamp their feet, click their teeth, and growl or hiss while vibrating specialized quills that produce a characteristic rattle. If an enemy persists, then they run backward until they ram their attacker with their quills. The reverse charge is most effective because the hindquarters are the most heavily armed, and the quills are directed to the rear. They will cheat on their vegetarian diets. Porcupines primarily eat roots, tubers, bark, and fallen fruit but also have a fondness for cultivated root crops such as cassava, potatoes, and carrots. Sometimes, they will even take carrion back to the burrow to nibble on. Where do porcupines live? Porcupines are most common in hilly, rocky country, but they can adapt to most habitats — excessively moist forests and the most barren of deserts seem to be the only exceptions. They have even been found on Mt. Kilimanjaro, as high up as 3,500 meters (11,480 feet).

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Finally, let’s look at several rather simple functions, point-max. These give information about the size of a buffer and the location of point within it. buffer-size tells you the size of the current buffer; that is, the function returns a count of the number of characters in the buffer. You can evaluate this in the usual way, by positioning the cursor after the expression and typing C-x C-e. In Emacs, the current position of the cursor is called point. (point) returns a number that tells you where the cursor is located as a count of the number of characters from the beginning of the buffer up to point. You can see the character count for point in this buffer by evaluating the following expression in the usual way: As I write this, the value of point is 65724. The function is frequently used in some of the examples later in this The value of point depends, of course, on its location within the buffer. If you evaluate point in this spot, the number will be larger: For me, the value of point in this location is 66043, which means that there are 319 characters (including spaces) between the two expressions. (Doubtless, you will see different numbers, since I will have edited this since I first evaluated point.) point-min is somewhat similar to it returns the value of the minimum permissible value of point in the current buffer. This is the number 1 unless narrowing is in effect. (Narrowing is a mechanism whereby you can restrict yourself, or a program, to operations on just a part of a buffer. See Narrowing and Widening.) Likewise, the point-max returns the value of the maximum permissible value of point in the current buffer.

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With a little bit of waste energy, a technology called forward osmosis could make salty water drinkable Meny Elimelech: In principle, if energy cost nothing, then you can have unlimited supply of water. You can have seawater desalination endlessly. [sounds of the harbor, water, boats] Lisa Raffensperger: As the winds pick up over Boston Harbor, engineer Rob McGinnis stands on the dock. He sweeps his arm toward the bustling waterfront. Rob McGinnis: …container ships loading up not far from here, cruise ships coming in. We have power plants which use the water for cooling purposes. Lisa Raffensperger: But for all the things this water is good for— Rob McGinnis: It’s not good for drinking. Lisa Raffensperger: But, as chief technology officer of a water company, McGinnis will tell you, that’s not strictly true. Desalination makes seawater drinkable for millions of people every day. But existing technologies consume energy—mostly fossil fuels—that could be used for other things. Not so for the desalination technology McGinnis developed as a Ph.D. student at Yale, in the lab of Meny Elimelech. The revolutionary part of it is the energy source. Rob McGinnis: We can take this very, very low temperature energy, the kind of temperature that you would find in, say, a hot bath, say, 40 degrees Celsius. This is the temperature at which power plants often will reject heat to the atmosphere. We can use that energy source to drive a desalination process. Lisa Raffensperger: The process uses “forward” osmosis—so called to distinguish it from reverse osmosis, the primary desalination technology today. Existing reverse osmosis requires energy to push water through a membrane, against its natural flow. Forward osmosis, on the other hand, doesn’t apply pressure. Seawater goes on one side of a membrane, and even saltier water goes on the other. The saltier side is called the “draw” solution because it pulls water to that side. But the salts in the draw solution perform a chemistry trick: When heated to low temperatures, they bubble out as gases… [brief bubbling sound] …leaving behind pure water. Meny Elimelech: So, the invention here was to come up with a draw solution that you can really separate relatively easy and inexpensively by means of waste heat. Lisa Raffensperger: Waste heat, like the billows of steam you see coming from power plants. If waste heat is available, the only energy input is for pumping. Which means— Rob McGinnis: —that all the water we produce by this method would not require additional fuel, and this is a huge difference in terms of sustainability. Lisa Raffensperger: The systems could be installed alongside power plants to use their discharged heat. Ultimately, the water produced in this way could cost half as much as water produced by reverse osmosis and use just a tenth of the electricity. [sounds of the laboratory] Lisa Raffensperger: It’s at a much smaller scale that testing continues at Elimelech’s lab at Yale. Ph.D. student Laura Hoover is studying the membrane. Traditional membranes are too thick for forward osmosis. She’s trying to make thinner membranes to allow more water through. [sound of machine turning on] Lisa Raffensperger: The test unit is just two big jars of water, connected by tubes and pumps—on one side, seawater; on the other, a “draw” solution of the special ammonium carbonate salts. The water streams pass on either side of a tiny membrane. Laura Hoover: So we have the draw solution here on mass balance so we can measure the weight that’s in this container over time, and so we can see how much water has moved into the draw side of the system from the feed side. Lisa Raffensperger: We watch the numbers climb. Today’s best forward-osmosis membranes can produce the same flow as the older reverse-osmosis ones, and no pumping is required. Lisa Raffensperger: Meanwhile, back in Boston, Rob McGinnis at Oasys Water is looking at the big scale. The company is close to commercializing the technology. And there are applications besides seawater. Forward osmosis could be used for municipal water treatment or water recycling. But desalination is the company’s first goal. And, says Rob McGinnis, there’s really only one question that matters. Rob McGinnis: And the question is, what do we use to do that? Do we use fossil fuels or electricity that can be used for so many other things, or do we find some way to use less resources to do it, and we think that’s what we can do. Lisa Raffensperger: And that’s something all of us can raise a glass to. I’m Lisa Raffensperger, in Boston. [harbor sounds, music] To Probe Further Check out the rest of the special report: Water vs Energy.

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During the Civil War, Louisa May Alcott served as a volunteer nurse, caring for Union soldiers in Washington, D.C., between December 12, 1862, and January 21, 1863. This well-researched biographical vignette explores the brief but pivotal episode in Alcott’s life. An abolitionist, Alcott longed to fight in the Union Army, but she did her part by serving as a nurse. Alcott met the female nursing requirements: She was 30, plain, strong and unmarried. Krull describes her challenging solo journey from Massachusetts by train and ship and her lonely arrival in Washington at the “overcrowded, damp, dark, airless” hospital. For three weeks she nursed and provided “motherly” support for her “boys” before succumbing to typhoid fever, forcing her to return to Massachusetts. Krull shows how Alcott’s short tenure as a nurse affected her life, inspiring her to publish letters she sent home as Hospital Sketches. This honest account of the war earned rave reviews and taught Alcott to use her own experiences in her writing, leading to Little Women. Peppered with Alcott’s own words, the straightforward text is enhanced by bold, realistic illustrations rendered in digital oils on gessoed canvas. A somber palette reinforces the grim wartime atmosphere, dramatically highlighting Alcott in her red cape and white nurse’s apron. An insightful glimpse into a key period in Alcott’s life and women in nursing. (notes on women in medicine and the Battle of Fredericksburg, sources, map) (Picture book/biography. 9-11)

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Environmental policy is the commitment of an organization or government to the laws, regulations, and other policy mechanisms concerning environmental issues. These issues generally include air and water pollution, waste management, ecosystem management, maintenance of biodiversity, the management of natural resources, wildlife and endangered species. For example, concerning environmental policy, the implementation of an eco-energy-oriented policy at a global level to address the issues of global warming and climate changes could be addressed. Policies concerning energy or regulation of toxic substances including pesticides and many types of industrial waste are part of the topic of environmental policy. This policy can be deliberately taken to influence human activities and thereby prevent undesirable effects on the biophysical environment and natural resources, as well as to make sure that changes in the environment do not have unacceptable effects on humans. One way is to describe environmental policy is that it comprises two major terms: environment and policy. Environment refers to the physical ecosystems, but can also take into consideration the social dimension (quality of life, health) and an economic dimension (resource management, biodiversity). Policy can be defined as a "course of action or principle adopted or proposed by a government, party, business or individual". Thus, environmental policy tends to focus on problems arising from human impact on the environment, which is important to human society by having a (negative) impact on human values. Such human values are often labeled as good health or the 'clean and green' environment. In practice, policy analysts provide a wide variety of types of information to the public decision making process. Environmental issues typically addressed by environmental policy include (but are not limited to) air and water pollution, waste management, ecosystem management, biodiversity protection, the protection of natural resources, wildlife and endangered species, and the management of these natural resources for future generations. Relatively recently, environmental policy has also attended to the communication of environmental issues. In contrast to environmental policy, ecological policy addresses issues that focus on achieving benefits (both monetary and non monetary) from the non human ecological world. Broadly included in ecological policy is natural resource management (fisheries, forestry, wildlife, range, biodiversity, and at-risk species). This specialized area of policy possesses its own distinctive features. The rationale for governmental involvement in the environment is often attributed to market failure in the form of forces beyond the control of one person, including the free rider problem and the tragedy of the commons. An example of an externality is when a factory produces waste pollution which may be discharged into a river, ultimately contaminating water. The cost of such action is paid by society-at-large when they must clean the water before drinking it and is external to the costs of the polluter. The free rider problem occurs when the private marginal cost of taking action to protect the environment is greater than the private marginal benefit, but the social marginal cost is less than the social marginal benefit. The tragedy of the commons is the condition that, because no one person owns the commons, each individual has an incentive to utilize common resources as much as possible. Without governmental involvement, the commons is overused. Examples of tragedies of the commons are overfishing and overgrazing. The role of Non-Governmental Organizations Non-Governmental organizations have the greatest influence on environmental policies. These days, many countries are facing huge environmental, social, and economic impacts of rapid population growth, development, and natural resource constraints. As NGOs try to help countries to tackle these issues more successfully, a lack of understanding about their role in civil society and the public perception that the government alone is responsible for the well-being of its citizens and residents makes NGOs tasks more difficult to achieve. NGOs such as Greenpeace and World Wildlife Fund can help tackling issues by conducting research to facilitate policy development, building institutional capacity, and facilitating independent dialogue with civil society to help people live more sustainable lifestyles. The need for a legal framework to recognize NGOs and enable them to access more diverse funding sources, high-level support/endorsement from local figureheads, and engaging NGOs in policy development and implementation is more important as environmental issues continue to increase. International organizations have also made great impacts on environmental policies by creating programmes such as the United Nations Environment Programme and hosting conferences such as the United Nations Earth Summit to address environmental issues. Instruments, problems, and issues Environmental policy instruments are tools used by governments and other organizations to implement their environmental policies. Governments, for example, may use a number of different types of instruments. For example, economic incentives and market-based instruments such as taxes and tax exemptions, tradable permits, and fees can be very effective to encourage compliance with environmental policy. The assumption is that corporations and other organizations who engage in efficient environmental management and are transparent about their environmental data and reporting presumably benefit from improved business and organizational performance. Bilateral agreements between the government and private firms and commitments made by firms independent of government requirement are examples of voluntary environmental measures. Another instrument is the implementation of greener public purchasing programs. Several instruments are sometimes combined in a policy mix to address a particular environmental problem. Since environmental issues have many aspects, several policy instruments may be required to adequately address each one. Furthermore, a combination of different policies may give firms greater flexibility in policy compliance and reduce uncertainty as to the cost of such compliance. Ideally, government policies are to be carefully formulated so that the individual measures do not undermine one another, or create a rigid and cost-ineffective framework. Overlapping policies result in unnecessary administrative costs, increasing the cost of implementation. To help governments realize their policy goals, the OECD Environment Directorate, for example, collects data on the efficiency and consequences of environmental policies implemented by the national governments. The website, www.economicinstruments.com, provides database detailing countries' experiences with their environmental policies. The United Nations Economic Commission for Europe, through UNECE Environmental Performance Reviews, evaluates progress made by its member countries in improving their environmental policies. The current reliance on a market-based framework has supporters and detractors. Among the detractors for example, some environmentalists contend that a more radical, overarching approach is needed than a set of specific initiatives, to deal with climate change. For example, energy efficiency measures may actually increase energy consumption in the absence of a cap on fossil fuel use, as people might drive more fuel-efficient cars. To combat this result, Aubrey Meyer calls for a 'framework-based market' of contraction and convergence. The Cap and Share and the Sky Trust are proposals based on the idea. Environmental impact assessments (EIA) are conducted to compare impacts of various policy alternatives. Moreover, although it is often assumed that policymakers make rational decisions based on the merits of the project, Eccleston and March argue that although policymakers normally have access to reasonably accurate environmental information, political and economic factors are important and often lead to policy decisions that rank environmental priorities of secondary importance. The decision-making theory casts doubt on this premise. Irrational decisions are reached based on unconscious biases, illogical assumptions, and the desire to avoid ambiguity and uncertainty. Eccleston identifies and describes four of the most critical environmental policy issues facing humanity: water scarcity, food scarcity, climate change, and the population paradox. Research and innovation policy Synergic to the environmental policy is the environmental research and innovation policy. An example is the European environmental research and innovation policy, which aims at defining and implementing a transformative agenda to greening the economy and the society as a whole so to achieve a truly sustainable development. Europe is particularly active in this field, via a set of strategies, actions and programmes to promote more and better research and innovation for building a resource-efficient, climate resilient society and thriving economy in sync with its natural environment. Research and innovation in Europe are financially supported by the programme Horizon 2020, which is also open to participation worldwide. UNFCCC research shows that climate-related projects and policies that involve women are more effective. Policies, projects and investments without meaningful participation by women are less effective and often increase existing gender inequalities. Women's found climate solutions that cross political or ethnic boundaries have been particularly important in regions where entire ecosystems are under threat, e.g. small island states, the Arctic and the Amazon and in areas where people's livelihoods depend on natural resources e.g. fishing, farming and forestry. Though the Clean Air Act 1956 in response to London's Great Smog of 1952 was a historical step forward, and the 1955 Air Pollution Control Act was the first U.S. federal legislation that pertained to air pollution, the 1960s marked the beginning of modern environmental policy making. The stage had been set for change by the publication of Rachel Carson's New York Times bestseller Silent Spring in 1962 and strengthened the Environmental movement. Earth Day founder Gaylord Nelson, then a U.S. Senator from Wisconsin, after witnessing the ravages of the 1969 massive oil spill in Santa Barbara, California, became famous for his environmental work. Administrator Ruckelshaus was confirmed by the Senate on December 2, 1970, which is the traditional date used as the birth of the United States Environmental Protection Agency (EPA). Five months earlier, in July 1970, President Nixon had signed Reorganization Plan No. 3 calling for the establishment of EPA. At the time, Environmental Policy was a bipartisan issue and the efforts of the United States of America helped spark countries around the world to create environmental policies. During this period, legislation was passed to regulate pollutants that go into the air, water tables, and solid waste disposal. President Nixon signed the Clean Air Act in 1970 which set the US as one of the world leaders in environmental conservation. The world's first minister of the environment was the British Politician Peter Walker from the Conservative Party in 1970. The German "Benzinbleigesetz" reduced Tetraethyllead since 1972. In the European Union, the very first Environmental Action Programme was adopted by national government representatives in July 1973 during the first meeting of the Council of Environmental Ministers. Since then an increasingly dense network of legislation has developed, which now extends to all areas of environmental protection including air pollution control, water protection and waste policy but also nature conservation and the control of chemicals, biotechnology and other industrial risks. EU environmental policy has thus become a core area of European politics. The German Umweltbundesamt was founded in Berlin 1974. Overall organizations are becoming more aware of their environmental risks and performance requirements. In line with the ISO 14001 standard they are developing environmental policies suitable for their organization. This statement outlines environmental performance of the organization as well as its environmental objectives. Written by top management of the organization they document a commitment to continuous improvement and complying with legal and other requirements, such as the environmental policy objectives set by their governments. Environmental policy integration The concept of environmental policy integration (EPI) refers to the process of integrating environmental objectives into non-environmental policy areas, such as energy, agriculture and transport, rather than leaving them to be pursued solely through purely environmental policy practices. This is oftentimes particularly challenging because of the need to reconcile global objectives and international rules with domestic needs and laws. EPI is widely recognised as one of the key elements of sustainable development. More recently, the notion of ‘climate policy integration’, also denoted as ‘mainstreaming’, has been applied to indicate the integration of climate considerations (both mitigation and adaptation) into the normal (often economically focused) activity of government. Environmental policy studies Given the growing need for trained environmental practitioners, graduate schools throughout the world offer specialized professional degrees in environmental policy studies. While there is not a standard curriculum, students typically take classes in policy analysis, environmental science, environmental law and politics, ecology, energy, and natural resource management. Graduates of these programs are employed by governments, international organizations, private sector, think tanks, advocacy organizations, universities, and so on. Academic institutions use varying designations to refer to their environmental policy degrees. The degrees typically fall in one of four broad categories: master of arts, master of science, master of public administration, and PhD. Sometimes, more specific names are used to reflect the focus of the academic program. Notable institutions include the Balsillie School of International Affairs, SIPA at Columbia, Sciences Po Paris, Graduate Institute Geneva, University of Oxford, University of Warwick, and University of British Columbia, among others. - ^ Eccleston, Charles H. (2010). Global Environmental Policy: Concepts, Principles, and Practice. ISBN 978-1439847664. - ^ Banovac, Eraldo; Stojkov, Marinko; Kozak, Dražan (February 2017). "Designing a global energy policy model". Proceedings of the Institution of Civil Engineers - Energy. 170 (1): 2–11. doi:10.1680/jener.16.00005. - ^ McCormick, John (2001). Environmental Policy in the European Union. The European Series. Palgrave. p. 21. - ^ Bührs, Ton; Bartlett, Robert V (1991). Environmental Policy in New Zealand. The Politics of Clean and Green. Oxford University Press. p. 9. - ^ Concise Oxford Dictionary, 1995. - ^ Loomis, John; Helfand, Gloria (2001). Environmental Policy Analysis for Decision Making. Springer. p. 330. ISBN 978-0-306-48023-2. - ^ A major article outlining and analyzing the history of environmental communication policy within the European Union has recently come out in The Information Society, a journal based in the United States. See Mathur, Piyush. "Environmental Communication in the Information Society: The Blueprint from Europe," The Information Society: An International Journal, 25: 2, March 2009 , pp. 119–38. - ^ Lackey, Robert (2006). "Axioms of ecological policy" (PDF). Fisheries. 31 (6): 286–290. - ^ Rushefsky, Mark E. (2002). Public Policy in the United States at the Dawn of the Twenty-first Century (3rd ed.). New York: M.E. Sharpe, Inc. pp. 253–254. ISBN 978-0-7656-1663-0. - ^ "The Role of NGOs in Global Governance". www.worldpoliticsreview.com. Retrieved 2021-01-27. - ^ "The Role of NGOs in Tackling Environmental Issues". Middle East Institute. Retrieved 2021-01-27. - ^ http://www.oecd.org/about/0,3347,en_2649_34281_1_1_1_1_1,00.html http://www.oecd.org/about/0,3347,en_2649_34295_1_1_1_1_1,00.html - ^ "Environmental Compliance & Corporate Performance - Can You Have It All?". www.emisoft.com. 2016-10-26. - ^ en_2649_34281_1_1_1_1_ 1,00.html Archived June 10, 2015, at the Wayback Machine - ^ "Instrument Mixes for Environmental Policy" (Paris: OECD Publications, 2007) 15–16. - ^ “Environmental Policies and Instruments,” http://www.oecd.org/department/0,3355,en_2649_34281_1_1_1_1_1,00.html - ^ "Economic Instruments". Economic Instruments. 2011-01-26. Archived from the original on 2011-02-07. Retrieved 2012-11-02. - ^ Eccleston C. and Doub P., Preparing NEPA Environmental Assessments: A Users Guide to Best Professional Practices, CRC Press Inc., 300 pages (publication date: March 2012). - ^ Eccleston C. and March F., Global Environmental Policy: Principles, Concepts And Practice, CRC Press Inc. 412 pages (2010). - ^ "The Population Paradox - Our World". - ^ "Population paradox: Europe's time bomb". 2008-08-08. - ^ See Horizon 2020 – the EU's new research and innovation programme http://europa.eu/rapid/press-release_MEMO-13-1085_en.htm - ^ "Development Solutions: How to fight climate change with gender equality". European Investment Bank. Retrieved 2020-09-17. - ^ unfccc.int https://unfccc.int/news/women-still-underrepresented-in-decision-making-on-climate-issues-under-the-un. Retrieved 2020-09-17. - ^ unfccc.int https://unfccc.int/news/5-reasons-why-climate-action-needs-women. Retrieved 2020-09-17. - ^ “Managing the Environment, Managing Ourselves: A History of American Environmental Policy” - ^ Knill, C. and Liefferink, D. (2012) The establishment of EU environmental policy. In: Jordan, A.J. and C. Adelle (ed.) Environmental Policy in the European Union: Contexts, Actors and Policy Dynamics (3e). Earthscan: London and Sterling, VA. - ^ Eccleston, Charles H. (2010). Global Environmental Policy: Concepts, Principles, and Practice. Chapter 7. ISBN 978-1439847664. - ^ Farah, Paolo Davide; Rossi, Piercarlo (December 2, 2011). "National Energy Policies and Energy Security in the Context of Climate Change and Global Environmental Risks: A Theoretical Framework for Reconciling Domestic and International Law Through a Multiscalar and Multilevel Approach". European Energy and Environmental Law Review. 2 (6): 232–244. SSRN 1970698. - ^ Taskforce on Conceptual Foundations of Earth System Governance http://www.earthsystemgovernance.net/conceptual-foundations/?page_id=144

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Using ESA?s Herschel Room Observatory, a group of astronomers has seen primary proof of a noble-gas primarily based molecule in place. A compound of argon, the molecule was detected inside the gaseous filaments belonging to the Crab Nebula, one of the most well-known supernova remnants in our Galaxy. Though argon is definitely a merchandise of supernova explosions, the development and survival of argon-based molecules while in the harsh natural environment of the supernova remnant is really an unexpected shock. Just just like a team of men and women, the periodic table of chemical parts has its share of group players and loners. Though some factors are likely to react a great deal more quickly with other species, forming molecules together with other compounds, other people infrequently take part in chemical reactions and are largely found in isolation. ?Inert? parts par excellence are classified as the noble gases: helium, neon, argon, krypton, xenon and radon. The identify of 1 of these ? argon ? derives in the Greek word for idle, to emphasize its really inert nature. But noble gases usually are not totally inactive. Whereas initially scientists doubted that chemical compounds could even have noble gases, a variety of these types of species are actually regarded and also have been thoroughly examined inside of the laboratory.Factors are more complicated in place. About the many years, astronomers have detected atoms and ions of noble gases in a number of cosmic environments, ranging from the Solar Model on phd in technology management online the atmospheres of stars, from dense nebulae with the diffuse interstellar medium. Even so the search for noble-gas centered compounds experienced until eventually now proved unsuccessful, suggesting that these essentially inert components may need a hard time reacting with other species in area. The group of astronomers has detected emission from argon hydride (ArH+), a molecular ion containing the noble gas argon, within the Crab Nebula. A wispy and filamentary cloud of gas and dust, the Crab Nebula is a remnant of the supernova explosion which was noticed by Chinese astronomers during the year 1054.?With scorching gasoline nevertheless expanding at superior speeds right after the explosion, a supernova remnant is usually a severe, hostile surroundings, and a person for the places in which we minimum envisioned to find a noble-gas centered molecule,? he provides.Argon hydride is produced when ions of argon (Ar+) respond with hydrogen molecules (H2), https://www.phdresearch.net/ but these two species tend to be noticed in different regions of a nebula. Even when ions type inside most energetic locations, in which radiation from a star or stellar remnant ionizes the gasoline, molecules consider shape from the denser, colder pockets of fuel which might be shielded from this amazing radiation. This new picture was supported by the comparison within the Herschel knowledge with observations belonging to the Crab Nebula executed at other wavelengths, which unveiled that the locations wherever that they had discovered ArH+ also show bigger concentrations of both equally Ar+ and H2. There, argon ions can respond with hydrogen molecules forming argon hydride and atomic hydrogen.The identification of such traces was a complicated process. To this close, the astronomers exploited two thorough databases of molecular spectra and, right after lengthy investigation, they matched the observed benefits with two attribute traces emitted by ArH+.?And there?s icing to the cake: from a http://pocketknowledge.tc.columbia.edu/home.php/browse/213551 molecule?s emission, we can easily determine the isotope on the elements that kind it ? an item that we can?t do after we see only ions,? provides Swinyard.

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# Dependent Events Objective: Find the probability of dependent events. Standard Addressed: 2.7.11: D Use theoretical probability distributions to make judgments. ## Presentation on theme: "Dependent Events Objective: Find the probability of dependent events. Standard Addressed: 2.7.11: D Use theoretical probability distributions to make judgments."— Presentation transcript: Dependent Events Objective: Find the probability of dependent events. Standard Addressed: 2.7.11: D Use theoretical probability distributions to make judgments about the likelihood of various outcomes in uncertain situations. KNOWING THE OUTCOME OF ONE EVENT CAN AFFECT THE PROBABILITY OF ANOTHER EVENT If one event does affect the occurrence of the other event, the events are dependent. Probability of Dependent Events Probability of Dependent Events Events A and B are dependent events if and only if P(A and B) = P (A) x P(B). SOMETIMES THE PROBABILITY OF THE SECOND EVENT CHANGES DEPENDING ON THE OUTCOME OF THE FIRST EVENT. Ex. 2 A bag contains 12 blue disks and 5 green disks. For each case below, find the probability of selecting a green disk on the first draw and a green disk on the second draw. a. The first disk is replaced. a. The first disk is replaced. 5/17 * 5/17 = 25/289 = 8.7% 5/17 * 5/17 = 25/289 = 8.7% b. The first disk is NOT replaced. b. The first disk is NOT replaced. 5/17 * 4/16 = 20/272 = 7.35% Ex. 3 A bag contains 8 red disks, 9 yellow disks, and 5 blue disks. Two consecutive draws are made from the bag WITHOUT replacement of the first draw. Find the probability of each event. Ex. 3 A bag contains 8 red disks, 9 yellow disks, and 5 blue disks. Two consecutive draws are made from the bag WITHOUT replacement of the first draw. Find the probability of each event. a. red first, red second a. red first, red second 8/22 * 7/21 = 56/462 = 12.12% b. yellow first, blue second b. yellow first, blue second 9/22 * 5/21 = 45/462 = 9.7% c. blue first, blue second c. blue first, blue second 5/22 * 4/21 = 20/462 = 4.3% d. red first, blue second d. red first, blue second 8/22 * 5/21 = 40/462 = 8.7% Writing Activities Download ppt "Dependent Events Objective: Find the probability of dependent events. Standard Addressed: 2.7.11: D Use theoretical probability distributions to make judgments." Similar presentations

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Titus Lucretius Carus (/ˈtaɪtəs luːˈkriːʃəs/ TY-təs loo-KREE-shəs, Classical Latin: [ˈtɪtʊs lʊˈkreːtɪ.ʊs]; c. 99 BC – c. 55 BC) was a Roman poet and philosopher. His only known work is the philosophical poem De rerum natura, a didactic work about the tenets and philosophy of Epicureanism, and which is usually translated into English as On the Nature of Things. Lucretius has been credited with originating the concept of the three-age system which was formalised in 1836 by C. J. Thomsen. Very little is known about Lucretius’s life; the only certain fact is that he was either a friend or client of Gaius Memmius, to whom the poem was addressed and dedicated. De rerum natura was a considerable influence on the Augustan poets, particularly Virgil (in his Aeneid and Georgics, and to a lesser extent on the Eclogues) and Horace. The work virtually disappeared during the Middle Ages, but was rediscovered in 1417 in a monastery in Germany by Poggio Bracciolini and it played an important role both in the development of atomism (Lucretius was an important influence on Pierre Gassendi) and the efforts of various figures of the Enlightenment era to construct a new Christian humanism. Lucretius’s scientific poem “On the Nature of Things” (c. 60 BC) has a remarkable description of Brownian motion of dust particles in verses 113–140 from Book II. He uses this as a proof of the existence of atoms.

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Mammograms are X-rays of the mammary glands or the breasts, primarily used to detect problems in the breasts, such as lumps, cysts or any kind of solid mass. The basic purpose of mammograms is to detect or screen out breast cancer amongst women. In fact, mammogram tests are able to detect tiny tumors before they can even be felt by women or their physicians. Cancer can be treated in the best manner if it is detected at an early stage. Remember that mammograms cannot treat or prevent breast cancer, but can reduce the risk of a woman developing this horrific disease. Experts are often seen to be at a difference of option regarding how often and when a woman should undergo a mammogram test. According to the U.S Preventive Service Task Forces, women aged between 40 and 49 years should not have regular mammogram tests, but should discuss with their doctors to find out when and how often they need to have mammogram screening, depending on their unique situation, their risk quotient, and their preference. However, there are a few organizations that recommend all women over the age of 40 years to undergo mammograms regularly within a span of 1-2 years. Those women older than 50 years of age, must get mammogram tests carried out regularly. Women who are above the age of 75 years can consult with their doctor to find out whether or not they need to get mammogram screening. Sometimes a doctor may recommend mammograms even at a much younger age to women who are more at risk of developing breast cancer. One thing that is extremely important to remember is that mammograms are not always 100 percent accurate. Sometimes, mammograms detect cancer or lumps even when they are not present, especially in cases of younger women. Thus, about 5 or 10 percent of mammogram screenings often require further testing. These further tests can be carried out through another mammogram of the specific breast tissues or other tests including biopsy or ultrasound. Why Are Mammogram Tests Done? Mammogram tests are carried out to: • Screen women for breast cancer, even if they are not suffering from any symptoms. • Detect the presence of breast cancer amongst women displaying symptoms of the disease. These symptoms may include thickening inside the breast, presence of lumps, dimpling of skin somewhere on the breast, or discharge from the nipples. In case any kind of abnormality is found in the mammograms, then a breast tissue from the suspicious region in the breast is removed and examined under the microscope, through a process known as biopsy. Preparing yourself for the Mammogram Test If you have ever had mammograms done before in your life, ask the previous clinic to send over your results to the current clinic or get them at the time of the examination, if you have the results with you. Besides this, you need to let your health care professional know in case you: • Are or could be pregnant: Mammograms are x-rays with low-dose radiation exposure and should not be carried out in case you are pregnant. • Are breast-feeding: Mammograms do not always provide the right result in women whose breasts contain milk. • Have had breast implants: Women with breast implants need to undergo a different and modified methods of mammogram testing. • Have had breast biopsy in the past: Knowledge about the presence and location of any scar tissues will help your radiologist to read the mammogram more accurately. On the day of getting testing, avoid using any perfume, deodorant, ointment, or powder on the breasts. This is important as the residue that these products leave on the skin can interfere with the x-ray. In case you are having your menstrual periods, it would be ideal to get your mammograms done within 2 weeks of the end of your menstrual periods. This is because the process will be a lot more comfortable, particularly if the breasts become more tender before the beginning of the next period. How Mammogram Testing is done? Mammogram testing is done by mammogram technologists or radiology technologist. Once the testing is done, the x-rays are interpreted by radiologists who specialized in x-ray evaluations. To get the test done, you will have to remove all jewelries that could interfere with the x-rays. You will also have to remove all your clothes from above the waist and wear a paper or cloth gown given to you while the test is performed. If you are anxious about any particular area on your breast, point it out to your technologist so that special attention can be given to that region. Most of the time, mammograms are performed while the patient is standing up, but sometimes you may be asked to lie down or sit down, according to the kind of x-ray equipment being used. Each of your breasts is placed on a plate containing the x-ray films one by one. Another plate is pressed against the breast firmly in order to flatten the breast tissue. Extreme compression is required to get high-quality images. You can often be asked to hold the other breasts away using your hand, or be asked to lift your arms to get them out of the way. You will also have to hold your breath for few seconds when the x-ray will be taken. Most of the time, two pictures of each breast is taken, one from the sides and the other from the top. The entire time taken in the mammogram clinic can go up to an hour, even though the test does not take more than 10 or 15 minutes. You can also be asked to wait for some time till the x-ray is developed, in case they need to taken repeat pictures. Some hospitals and clinics use digital screens where the x-ray images can be viewed instantly. How It Feels? Mammograms can often be uncomfortable but are rarely too painful. Those who have fragile or sensitive skin or are suffering from any skin disorder can inform the technician regarding the same before undergoing the exam. The procedure becomes more comfortable when performed 2 weeks after the end of your periods. The x-ray plate usually feels quite cold when the breast is placed on it. Having the breasts squeezed and flattened is usually quite uncomfortable; but it is very important to flatten out the tissues in order to get the perfect pictures.

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The Fourth of July is on Monday and the United States of America will turn 240 years old! We’ve compiled some fun facts about Independence Day that you can use to impress your friends while watching fireworks and celebrating America’s birthday. - Thomas Jefferson drafted the Declaration of Independence on a “laptop” – that is, a writing desk that could fit in his lap. - Jefferson changed the wording of the Declaration of Independence from “the pursuit of property” to “the pursuit of happiness.” - John Adams and Jefferson, both signers of the Declaration of Independence, died on July 4, 1826. James Monroe also died on July 4th in 1831. - Only two people actually signed the Declaration of Independence on July 4, 1776 – John Hancock and Charles Thompson. - Congress declared July 4th as an official holiday in 1870 as part of a bill to officially recognize other holidays, including Christmas. - At 27, Thomas Lynch, Jr., was the youngest signer of the Declaration of Independence; Ben Franklin, age 70, was the oldest signer. - The oldest, continuous Independence Day celebration in the United States is the 4th of July Parade in Bristol, Rhode Island which began in 1785. - The Pennsylvania Evening Post was the first newspaper to print the Declaration of Independence. - Americans began observing the Fourth of July as early as 1777 with a celebration in Philadelphia that included a parade, a thirteen-shot cannon salute, and fireworks. - Eight of the 56 signers of the Declaration of Independence were born in Britain. Got a fact we missed? Leave it for us in the comments! Learn more about our Reliving U.S. History Programs.

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After traveling upriver in the Peruvian Amazon, researchers from the University of Oklahoma and Peru’s National Health Institute returned with evidence that—when it comes to the bacteria living in your gut, at least—you are what you eat, not where you live. The goal of the three-day mission was to better understand how diet and lifestyle shape the billions of bacteria living in our guts—a complex world known as the gut microbiome. This vast community of microbes plays an active role in digestion and metabolism, with important implications for health, including susceptibility to obesity, diabetes, colon cancer, and other afflictions. (Learn more about the microbiome in National Geographic magazine's "Small, Small World." In a paper published this week in the journal Nature Communications, University of Oklahoma geneticist Cecil Lewis and his colleagues analyzed the bacteria they collected from the digestive tracts of Amazonian hunter-gatherers called the Matsés. Among the last of the world’s hunter-gatherers, the Matsés subsist on a diet of tubers and plantains, supplemented with protein from fish, monkey, sloth, alligator, and other jungle species. The researchers found the bacteria in the guts of these native communities included species that people in the industrialized world lack. The team had also gathered samples from traditional potato farmers in Tunapuco, a village in the highlands of Peru’s Andes mountains. The two groups’ microbiomes were then compared to the microbiomes of residents of Norman, Oklahoma. Rather than sharing gut bacteria species with their neighbors in Tunapuco, the Matsés hunter-gatherers had gut microbiota that matched those of other hunter-gatherers in Africa, an ocean away. But the microbial communities of the two traditional Peruvian groups resembled each other more than that of the Oklahomans, who ate a typical “industrialized diet.” The researchers also found more diverse types of bacteria living in the guts of people in traditional societies. “It’s not about geographic distance; it’s about dietary practices,” says Lewis. On their journey up the Galvez River in the Amazon, Lewis’s team brought along microscopes to first teach the Matsés about microbes before asking them for fecal samples, which were then stored in a massive block of ice for the journey back to the Peruvian capital of Lima. There they were frozen and shipped to the U.S. When they analyzed the samples back in their lab in Oklahoma, the researchers found the bacteria in the guts of these native communities included species that people in the industrialized world lack. Treponema, a bacterial genus that includes species that cause diseases like syphilis and yaws, were present in large numbers in both traditional communities. The treponeme species in the Matsés and the farmers from Tunapuco were more closely related to variants that help break down fiber and digest carbohydrates in swine, cattle, and termites. “They’re good fermenters of fibrous material,” says Lewis. In earlier research on the gut microbiome, conducted mostly in affluent Western populations, these treponemes didn’t show up at all. When they were first found to be common in the microbiomes of traditional peoples in Africa and South America, researchers assumed these indigenous groups were the exception to the rule. As traditional peoples come into contact with industrialized society and change their lifestyle and diet, their gut microbiomes may be endangered. But the new study supports other recently published research suggesting that it’s the industrialized lifestyle typified by the Oklahomans—a diet of processed food, combined with modern sanitation and frequent use of antibiotics—that is the real outlier. In 2012, Lewis’s lab sequenced 1,000-year-old fossilized feces, called coprolites, and found communities of microbes that resembled those of traditional peoples like the Matsés and the Tunapuco villagers more than those of people who eat a modern, Western diet. “There’s nothing that stands out more than these treponemes,” Lewis says. “They’re clearly common in all traditional diets.” In a paper published last year, Stephanie Schnorr, a researcher at the Max Planck Institute for Evolutionary Anthropology in Leipzig, Germany, analyzed the gut microbiome of Hadza hunter-gatherers in Tanzania. What Schnorr saw closely resembles what Lewis reports finding in the Matsés of Peru—including abundant treponemes and other ancient species rarely found in the industrialized world. The new study “really vindicates our findings,” says Schnorr, who was not involved with the research. “Once you start putting it all together, treponemes are popping up all over. We’re realizing that not having treponemes is probably the aberrant thing.” Studies like these may capture a dietary portrait of preindustrial humankind before it vanishes. As traditional peoples come into contact with industrialized society and change their lifestyle and diet, their gut microbiomes may be endangered. Lewis says that since fecal samples were first collected from the Matsés hunter-gatherers in 2012, the remote tribe “has had much more interaction with industrialized foods,” Lewis says. “We might be seeing a moment where this lifestyle is disappearing. If these traditional practices end, it may be that these treponemes are lost to us.” Whether the presence of the treponemes—or their absence—can shed light on so-called diseases of civilization, like obesity and diabetes, remains to be seen. “These bacteria co-evolved with primates for millions of years, and now they’re gone in industrialized people,” Lewis says. “Why are they absent, and does that matter?” Follow Andrew Curry on Twitter.

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|Volcán de Colima| Volcán de Colima is on the left with Nevado de Colima on the right. |Elevation||3820+ m (12,533+ ft) | |Prominence||600 m (2,000 ft) | |Location||Jalisco / Colima, Mexico| |Parent range||Trans-Mexican Volcanic Belt| |Age of rock||5 million years| |Volcanic arc/belt||Trans-Mexican Volcanic Belt| |Last eruption||2013 to 2018 (ongoing)| The Volcán de Colima, 3,820 m (12,533 ft), also known as Volcán de Fuego, is part of the Colima Volcanic Complex (CVC) consisting of Volcán de Colima, Nevado de Colima (Spanish pronunciation: [ne'?aðo ðe ko'lima] ) and the eroded El Cántaro (listed as extinct). It is the youngest of the three and as of 2015 is one of the most active volcanos in Mexico and in North America. It has erupted more than 40 times since 1576. One of the largest eruptions was on January 20-24, 1913. Nevado de Colima, also known as Tzapotépetl, lies 5 kilometres (3.1 mi) north of its more active neighbor and is the taller of the two at 4,271 meters (14,015 ft). It is the 26th-most prominent peak in North America. Despite its name, only a fraction of the volcano's surface area is in the state of Colima; the majority of its surface area lies over the border in the neighboring state of Jalisco, toward the western end of the Trans-Mexican Volcanic Belt. It is about 485 km (301 mi) west of Mexico City and 125 km (78 mi) south of Guadalajara, Jalisco. Since 1869-1878, a parasitic set of domes, collectively known as El Volcancito, has formed on the northeast flank of the main cone of the volcano. In the late Pleistocene era, a huge landslide occurred at the mountain, with approximately 25 km³ (6 cubic miles) of debris travelling some 120 km, reaching the Pacific Ocean. An area of some 2,200 km² was covered in landslide deposits. The currently active cone is within a large caldera that was probably formed by a combination of landslides and large eruptions. The lava is andesite containing 56-61% SiO2. About 300,000 people live within 40 km (25 miles) of the volcano, which makes it the most dangerous volcano in Mexico. In light of its history of large eruptions and situation in a densely populated area, it was designated a Decade Volcano, singling it out for study. In recent years there have been frequent temporary evacuations of nearby villagers due to threatening volcanic activity. Eruptions have occurred in 1991, 1998-1999 and from 2001 to the present day, with activity being characterised by extrusion of viscous lava forming a lava dome, and occasional larger explosions, forming pyroclastic flows and dusting the areas surrounding the volcano with ash and tephra. The largest eruption for several years occurred on May 24, 2005. An ash cloud rose to more than 3 km over the volcano, and satellite monitoring indicated that the cloud spread over an area extending 110 nautical miles (200 km) west of the volcano in the hours after the eruption. Pyroclastic flows travelled 4-5 km from the vent, and lava bombs landed 3–4 km away. Authorities set up an exclusion zone within 6.5 km of the summit. On November 21, 2014, the volcano erupted again. An ash column was sent 5 km into the air, covering towns as far as 25 km away in ash. No fatalities were reported, and no evacuations took place. There were eruptions on January 10, 21 and 25, with the ash from the January 21 eruption falling in towns more than 15 miles (24 km) away. On 10 July 2015, there was another eruption. Another eruption occurred on Sunday, September 25, 2016, sending a plume of ash and smoke 10,000 ft (3048 m) into the sky. During December 2016, ashes plumes occurred once or twice a day. On Sunday, December 18, 2016, there were three eruptions. The biggest columns of ash reached 2 kilometers in height. Colima volcano experienced another strong explosion at 06:27 UTC (00:27 CST) on January 18, 2017. The eruption spewed volcanic ash up to 4 km (13 123 feet) above the crater. The volcano is monitored by the Colima Volcano Observatory at the University of Colima, Mexico. A team analyzes, interprets and communicates every event that occurs at this volcano. Last year a webcam was installed close to the volcano, and volcanic activity can be seen in real time.

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Operation G # Operations with Fractions: Multiplication and Division ## Recognize division is the inverse of multiplication and vise versa (÷4 is the same as × 1 1⁄4) Understanding this foundational concept gives students flexibility when solving multiplication and division questions involving fractions. Students may intuitively select the operation that makes the most sense to them; for example, students may draw upon their understanding of this inverse relationship to solve a division question using multiplication. As educators we want to recognize and value the reasoning involved in the student strategies in these instances. (It's also interesting to note that we do value this strategy in procedural terms; the standard algorithm asks students to use multiplication to solve division questions by inverting and multiplying.) ### Background Multiplication and division involving fractions is widely recognized to be more complex than multiplication and division with whole numbers. Often algorithms are introduced with little emphasis on understanding the mathematics behind the algorithm. There is some research evidence that suggests that early emphasis on procedures over concepts can actually impede students' fractions understanding in the long term (Brown and Quinn, 2007). However, with precise instruction, students can develop both conceptual and procedural knowledge. Multiplication with fractions may increase or decrease a quantity, or leave it unchanged. Multiplication with whole numbers is often connected to repeated addition; however, with fractions, other interpretations, such as the Cartesian product or area mode, support increased understanding. Consider 113×12. Since we are multiplying a quantity by half, we should expect the result to be less than 113. We can carefully labelled array or Cartesian model to solve this. First, we model 113 Then we partition it into half. Not that we could do this a number of ways (including partitioning each third vertically in half) but the example below connects the use of the array (length and width) to model the multiplication Determine the answer from the intersecting area. We can see that each whole has been partitioned into sixths. There are four sixths in the overlapping area. 113 × 12 = 46 Ontario fractions research revealed several effective strategies for teaching multiplication of fractions with an emphasis on meaning making. See Fractions Operations; Multiplication and Division Literature Review for more details. Recommend best practices based on the research include: 1. Take time to focus on conceptual understanding. Instruction should start with contextual problems which allow students to focus on what they are solving rather than how they are solving, and helps them to develop the meaning of the operations. They also should have hands-on experiences with materials such as relational rods, paper folding and visual models like number lines. 2. 3. Recognize and draw on students' informal and formal knowledge as well as prior experiences with fractions. Students have prior experience with fractions, including partitioning different models (which is helpful when solving multiplication problems). If they have had opportunities to engage in learning experiences earlier in the Fractions Learning Pathways, they will also have an understanding of the unit fraction, which helps establish an early understanding of multiplication connected to repeated addition (e.g., ). As well, students can draw upon their understanding of the various constructs of fractions (i.e., number, part-whole, part-part, quotient, operator and linear measure) and their work with equivalent fractions. 4. 5. Draw on student familiarity with whole number operations. Specific connections to multiplication with whole numbers should be made, as the properties hold true for multiplication with fractions as well. 1. The Commutative Property: a × b = b × a 2. The Associative Property: (a × b) × c = a × (b × c) 3. The Identity Property: a × 1 = a 4. The Distributive Property a × (b + c) = a × b + a × c 5. As well, students should understand that multiplication is the inverse of division and vice versa. That is to say that if a × b = c, then a = c ÷ b and b = c ÷ a. This is true for a, b, c ≠ 0. For example, although 3 × 0 = 0, 0 ÷ 0 ≠ 3. Of couse, some generalizations that hold true for whole numbers don't hold true for fractions, and this can be a source of confusion for students. For example, many students hold onto the idea that multiplication "makes a number larger", which is true with whole numbers. However, when we are multiplying by a quantity that is less than 1, as in the case of fractions, we "shrink" the quantity instead. 6. Include carefully selected and multiple representations to convey meaning. There are a few representations which have longevity and are particularly helpful for building meaning across number systems and contexts - the number line and the area model (or Cartesian or array). Using these representations across fractions learning allows students to understand the structure of the representations and use them effectively to solve questions. Models help students understand the mathematics better - both conceptually and procedurally - and enhance their retention of the learning. ### References Brown, G., & Quinn, R.J. (2007). Investigating the relationship between fraction proficiency and success in algebra. Australian Mathematics Teacher, 63(4): 8- 15.

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Copyright © University of Cambridge. All rights reserved. 'Auditorium Steps' printed from http://nrich.maths.org/ This activity has been particularly created for the highest attainging children. (The pupils that you come across in many classrooms just once every few years.) Here's a model of a small auditorium. There are four squares at the bottom inside and there are three sets of steps to the top. To help to visualise it I've coloured each face of the cube blocks according to which way they are facing. Your challenge is to imagine that this is a small model that is a present for someone and so it has to be carefully wrapped up with paper. We do not want to waste paper so the wrapping needs to be from a cut shape so that each visable face (no matter where you are viewing it from) needs to be covered CLOSELY with ONE LAYER OF PAPER ONLY and there is to be just ONE piece of Draw the shape of that wrapping paper which would need to be cut out. Describe how you went about it and say what you can about the paper shape. Then ask yourself "I wonder what would happen if I ...?"

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© Monte Costa For more information about this photo, click here to contact the photograher The Polynesian Voyaging canoe Hōkūle`a Photo © Monte Costa Everyone from Hawai`i knows who Nainoa Thompson is: the Hawaiian navigator who helped bring the ancient Polynesian tradition of wayfinding back from the brink of extinction. Historians believe that over fifteen hundred years ago, people arrived on the isolated archipelago that is now Hawai`i from southern Pacific islands, first the Marquesas, and then Tahiti. They didn’t make only one single trip in double-hulled canoes over twenty-five hundred miles of ocean, but went back and forth between the North and South Pacific. In the old days, long before explorers like Magellan or Captain Cook, islanders were criss-crossing the Pacific using the stars, the ocean, the birds, the winds and other signs that elude the modern eye. Over time and generations, these wayfaring skills were lost, but in the 1970s, Native Hawaiians began the recovery of language and culture in what is now referred to as the Hawaiian Renaissance. Nainoa Thompson was one of those interested in learning how to navigate the ocean the way his ancestors did and became a crew member of the Hōkūle`a, the first voyaging canoe to sail the twenty-five hundred miles of ocean back to Tahiti in 1976. Since then, voyaging canoes have sailed from Hawai`i to all points in the Polynesian triangle, west to Aotearoa (New Zealand) and east to Rapa Nui (Easter Island.) In the 1970s, those who wanted to learn to navigate the old way set out to find teachers, but throughout the Pacific, there were only a few. These founders of the Polynesian Voyaging Society in their search to find a navigator, finally contacted a Micronesian, Mau Piailug from the island of Satawal who agreed to help them. At the time, there were six navigators living in Micronesia, still practicing their traditional skills between islands that had been left alone, mainly because they had no large lagoons that other nations coveted in order to park their battleships. With Mau’s help, the Hawai`i crew was able to learn to sail the voyaging canoes using the old ways. It was Mau who navigated the first voyage to Tahiti thirty years ago and then went on to teach the Hawaiian crewmembers. The irony is that today, Mau is the last remaining navigator in Micronesia where young people, now sent to schools on other islands, are not learning the ancient skills. According to Nainoa, Mau thinks that someday, it will be the Hawaiians who will be re-teaching the Micronesians the non-instrument voyaging techniques. There are others in Hawai`i who have participated in this recovery of the tradition of voyaging canoes, but Nainoa Thompson is seen as the leader, the one who kept at it even when all seemed hopeless. He credits others, especially his father the late Myron “Pinky” Thompson, but many agree that it was Nainoa’s single-minded determination that has made the ancient art of Polynesian wayfaring a modern reality. There is now a Pacific-wide collaboration to provide the training and teaching of these non-instrument navigational skills, restoring connections and pride among the various island groups. Nainoa Thompson refers to the islands linked by the canoes as the Polynesian Nation. Epeli Hau`ofa, a Tongan writer, calls the Pacific and all of its islands Oceania. The voyaging canoes remind us © Monte Costa For more information about this photo, click here to contact the photograher that the islands in the Pacific Ocean and its people are not solitary entities, separated by the colonizing nations of the past two hundred years, but relatives who are re-establishing their ancient connections. To read more about the Polynesian Voyaging Society, go to: www.pvs-hawaii.com. On a Northwest note, a second canoe, Hawai`i Loa was built in the early 1990s in the traditional way, with two Sitka Spruce trees given by the Tlingit and Haida tribes of Alaska because Hawai`i’s forests have been depleted of the massive koa trees needed for the large canoes. In 1996, Hōkūle`a and Hawai`i Loa arrived in Seattle to begin their journey: Hōkūle`a sailed south along the Northwest coast to California to pay tribute to the many people from Hawai`i now living on the West Coast, and Hawai`i Loa went north to Alaska, to thank the Native Alaskans who donated the trees. 9th Circuit grants Kamehameha Schools’ petition for en banc review HONOLULU (www.ksbe.edu) - The U.S. Ninth Circuit Court of Appeals agreed earlier today to rehear the legal challenge to Kamehameha Schools’ 118-year-old policy of offering admissions preference to applicants of Hawaiian ancestry. The rehearing, known as an en banc review, was granted by court order earlier today. Under the court’s order the case will be reargued on a date to be scheduled before a panel of 15 Ninth Circuit judges. “We are pleased to be able to present our arguments to a larger court panel,” said Kamehameha Trustee-chair Robert Kihune. “It signals that the appeals court agrees that this lawsuit raises unique issues of exceptional importance to Native Hawaiians. We are a private school founded by a Native Hawaiian princess for the education of Native Hawaiians and funded entirely by the income from our land holdings and investments. We are hopeful that when the case is reheard the court will affirm the U.S. District Court decision and allow Kamehameha to continue todirect our resources to those children who are in need of our programs and are the intended beneficiaries of this trust.” A three-judge appeals court panel overturned Kamehameha Schools’ admissions policy last August in a 2-1 decision. Schools officials immediately vowed to seek a rehearing. The schools’ petition for en banc review was supported by 12 amicus briefs filed by 44 individuals or organizations, including the State of Hawaii, the City and County of Honolulu, the National Association of Independent Schools and various Hawaiian service organizations and Royal Trusts. The schools preference policy remains in effect pending the outcome of the appeal. “The preference policy is critical to our ability to fulfill our educational mission and we are fully committed to the legal fight ahead,” said Kamehameha Schools CEO Dee Jay Mailer. “At the same time, we are moving forward with plans to extend our educational reach further into Hawaiian communities. We have already embarked on a plan to more than double the number of children we currently serve. “We have committed $55 million dollars this year for community outreach programs that target Hawaiian keiki under the age of eight and their parents and caregivers. We are providing more support to preschools and other early childhood education programs. We have tailored our post-high scholarship programs to serve young mothers and others who are among the neediest in our population. We have provided more funding for charter schools with large Hawaiian students populations. “Pauahi felt a kuleana to provide educational opportunities for the Hawaiian people. She entrusted that kuleana to the leadership of Kamehameha Schools. We will not let her down.” Chairman Kihune added, “Kamehameha is an institution of exceptional importance to both Native Hawaiians and all of Hawai’i. The programs Kamehameha provides to Native Hawaiian children are intended to help remedy the continuing effects of past wrongs suffered by the Native Hawaiian people. Kamehameha has had great success helping Native Hawaiian children but much remains to be done, and the needs of Native Hawaiian children far outstrip Kamehameha’s ability to provide needed services. Last year’s panel decision would have required Kamehameha to offer its programs to children who do not need them, while turning away Native Hawaiian children who need access to Kamehameha’s programs.” Established in 1884 through the last will and testament of Princess Bernice Pauahi Bishop, Kamehameha Schools operates K-12 campuses on O’ahu, Hawai’i and Maui and 32 preschool sites statewide. Thousands of other learners are also being served through a range of educational outreach programs and community collaborations throughout the state. By Darrell Williams With the sort of facilities that compare to that of the original Gold’s Gym in Venice California, Moloka`i's Peter Pale is leading by example to promote an emerging strongman scene on the island. The 33-year-old Pale took the first ever Moloka`i strongman event in 2005, finishing in first place in every discipline. A born and bred native of Moloka`i, Pale works at the Na Pu`uwai fitness center, the only gym currently on the island. Without all the sophisticated machines that mainland competitors have the privilege of using, it's not easy for Moloka`i athletes to compete at the highest level. Pale said he is seeing increased interest within the youth of the Island since his strongman win. "We need to get the young ones coming through," he said. Pale also competed in the Hawaiian Makahiki games of 2006, where he finished first in the Decathlon. Unlike winning Mr. Olympia, where the prize money can mean financial security for life, Pale's prize was a Samoan Crab, and two mullet – the first to be produced by a newly restored ancient fishpond on the East End of Moloka'i. A very humble family man, Pale has no thoughts of taking his strength and power to the next level. "I too old," he said, preferring to encourage the young ones to get involved with the discipline of weight training. As he said at the strongman contest "I thank you everybody, now start working out alright!" Lahainaluna, the “Oldest School West of the Rockies ” was founded by missionaries in 1831 as a seminary boarding school for men on the island of Maui. According to its website, the goal of Lahainaluna from the beginning was “to educate and prepare its students to a way of life in the community, the state and the nation.” In 1849, the Hawaiian government took over the school from the American Board of Mission. In 1923, it went coed and became a public technical school and finally in 1961, Lahainaluna became a comprehensive public high school for boarders and day students. Lahainaluna is celebrating its 175 years as one of the oldest schools in the United States, beginning with an Open House and class parade on April 7th and lū`au on the 8th. For more information, write to Lahainaluna 175th Anniversary, P.O. Box 1684, Lahaina, Hawaii 96767, call the school office at (808) 662-4000 or email: [email protected]. Copyright © 2004-2007 by Northwest Hawai`i Times All Rights Reserved

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blog # Understanding 210mm to Inches: A Comprehensive Guide When it comes to measurements, it’s important to have a clear understanding of different units and their conversions. One such conversion that often arises is the conversion of millimeters (mm) to inches. In this article, we will delve into the topic of converting 210mm to inches, providing valuable insights and practical examples to help you grasp this conversion effortlessly. ## What is a Millimeter (mm)? Before we dive into the conversion, let’s first understand what a millimeter is. The millimeter is a unit of length in the metric system, denoted by the symbol “mm.” It is equal to one-thousandth of a meter, making it a smaller unit of measurement compared to centimeters and inches. ## What is an Inch? An inch, on the other hand, is a unit of length commonly used in the United States and other countries that have not adopted the metric system. It is denoted by the symbol “in” or double quotes (“). One inch is equivalent to 25.4 millimeters, making it a larger unit of measurement compared to millimeters. ## Converting 210mm to Inches Now that we have a basic understanding of millimeters and inches, let’s focus on the conversion of 210mm to inches. To convert millimeters to inches, we need to use the conversion factor of 25.4, which represents the number of millimeters in one inch. To convert 210mm to inches, we can use the following formula: Inches = Millimeters / 25.4 Using this formula, we can calculate the conversion of 210mm to inches: Inches = 210 / 25.4 = 8.2677 inches Therefore, 210mm is approximately equal to 8.2677 inches. ## Practical Examples Understanding the conversion of 210mm to inches becomes more practical when we consider real-life examples. Let’s explore a few scenarios where this conversion can be useful: ### Example 1: Paper Sizes When it comes to paper sizes, the A4 size is widely used around the world. A4 paper measures 210mm by 297mm. If you need to determine the equivalent size in inches, you can convert both dimensions to inches using the conversion factor. Converting the width of A4 paper (210mm) to inches: Width (inches) = 210 / 25.4 = 8.2677 inches Converting the height of A4 paper (297mm) to inches: Height (inches) = 297 / 25.4 = 11.6929 inches Therefore, A4 paper measures approximately 8.2677 inches by 11.6929 inches. ### Example 2: Woodworking In woodworking, precise measurements are crucial for creating accurate and well-fitting pieces. Let’s say you are working on a project that requires a piece of wood with a length of 210mm. If you prefer working with inches, you can convert the measurement to inches to ensure accuracy. Converting the length of the wood (210mm) to inches: Length (inches) = 210 / 25.4 = 8.2677 inches Therefore, the wood piece measures approximately 8.2677 inches in length. ## Q&A Here are some commonly asked questions related to the conversion of 210mm to inches: 1. ### Q: Is 210mm equal to 8.27 inches? A: No, 210mm is approximately equal to 8.2677 inches. Rounding it to two decimal places, it can be represented as 8.27 inches. 2. ### Q: Can I use an online converter to convert 210mm to inches? A: Yes, there are numerous online converters available that can quickly and accurately convert 210mm to inches. However, it’s always beneficial to understand the conversion process to verify the results. 3. ### Q: How can I convert inches to millimeters? A: To convert inches to millimeters, you can use the formula: Millimeters = Inches * 25.4. Multiply the number of inches by 25.4 to obtain the equivalent measurement in millimeters. 4. ### Q: Are millimeters or inches more commonly used? A: The usage of millimeters or inches depends on the country and the field of measurement. Millimeters are commonly used in countries that have adopted the metric system, while inches are more prevalent in countries that still follow imperial measurements, such as the United States. 5. ### Q: Can I convert 210mm to feet? A: Yes, you can convert 210mm to feet by first converting it to inches and then dividing the result by 12 (since there are 12 inches in a foot). However, it’s important to note that 210mm is a relatively small measurement, and converting it to feet may not provide a practical value. ## Summary In conclusion, understanding the conversion of 210mm to inches is essential for various applications, ranging from paper sizes to woodworking projects. By using the conversion factor of 25.4, we can convert millimeters to inches accurately. In the case of 210mm, the conversion results in approximately 8.2677 inches. Remember, having a clear understanding of different units of measurement and their conversions is crucial for accurate and precise work in various fields.

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## Bisecting (and trisecting) lines ### Bisecting (and trisecting) lines In order to make an accurate ruler, map, or blueprint, it will be useful to find the point on a line that is exactly in the middle of two other points. This is called bisection, and you can also trisect a line as well. First, you'll have to make a drawing compass, which should not be difficult compared to some of the other mechanical tasks in the book. Basically you need two metal pieces, one with a very pointy end, the other with something that can hold a pencil or other marking device. They're connected by a hinge or some other method to allow you to change the distance between the two, and a fastener to allow you to keep it in place: Then you'll need a straightedge, which you can construct with the help of pulling a piece of thread taut. Then, let's assume you have a line and two points on that line. You can find the midpoint of the two points by setting your compass to any distance that's longer than the midpoint will be. Then you take it, place the pointy end on one of your endpoints, and draw a circle around it. Then you keep your compass the exact same length, put the pointy side on the other endpoint, and do the same thing. Then you take your straightedge and draw a line between the points where the circles intersect. (The circles above let you find the midpoint E between points A and B.) To trisect a line, draw your two circles. We'll call them circles 1 and 2. Then, draw a line between where the circles intersect at the top to the first endpoint (points A and F in that diagram.) Then draw a line where the last line you drew intersects circle 1 below the line. Then draw a line from that point all the way to where circle 2 intersects your original line. Then draw a line from where the circles intersect at the top to the place the last line you drew intersected circle 2. That point on your line is exactly 1/3 of the way from point A to point B. You can use similar techniques to bisect angles as well. Trisection of angles is impossible except with another tool. To do that, you'll need a neusis -- basically a ruler that allows you to rotate it around a fixed point, e.g. sticking a pin in a paper ruler -- or a string and a cylinder, or a tomahawk. http://en.wikipedia.org/wiki/Bisection http://www41.homepage.villanova.edu/rob ... 20segment/ http://en.wikipedia.org/wiki/Angle_trisection http://en.wikipedia.org/wiki/Neusis_construction (PS the reason for dividing the hour into 12 parts is partly due to the fact that 12 is divisible by 3 and 4. A lot of ancient number choices are derived from the fact that they are divisible by a lot of things. 60 is great because it's divisible by 2, 3, and 5.) illuminatedwax Posts: 3 Joined: Fri May 09, 2014 10:18 pm

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Courses Courses for Kids Free study material Offline Centres More Store How to Solve Linear Differential Equation? Reviewed by: Last updated date: 02nd Aug 2024 Total views: 401.7k Views today: 7.01k Know About How to Solve Linear Differential Equation? Linear differentiation equation is a form of the equation that is linear and polynomial. It is a part of the new derivatives and functions of an equation of a structure. A linear equation usually consists of derivatives of multiple variables. When a result and variable is partial, then it is also stated as a linear partial differential equation. In an equation when its new function and the derivative is not in the linear form, it is then considered to be a nonlinear differential equation. It is used to find a solution for multiple chaoses. In an equation where x dy / dx + Y = 0, it involves variables and a derivative which is dependent on y variable linked to independent variable x. This form of an equation is called a differential equation. Usually, an equation with a result of dependent variable connected to independent variables is known as a differential equation. It involves dependent variable derivatives concerning a single independent variable, also known as an ordinary differential equation. Apart from these, students get to learn about the first-order nonlinear differential equation and first-order linear differential equation. Here the variables like P and Q works as an independent variable or remain constant. The equation dy/dx + Py = Q can be a form of differential equation in a first-order linear differential. Here P and Q can be either function of y, which is an independent variable or constants only. A student has to derive the standard form or a depiction of this solution. How to Solve First Order Nonlinear Differential Equation A first-order differential equation is considered to be separable if it can be written in the form ϕ(y)dy = ψ(x)dx. This suggests the official process of anti differentiating ϕ(y) and ψ(x) concerning y and x. Moreover, results to arrive at an implicit general solution is equated as Z ϕ(y)dy = Z ψ(x)dx + C. It is important to note that here R ϕ(y)dy and R ψ(x)dx stand for definite antiderivatives of ϕ(y) and ψ(x). The nonlinear differential equations definition can be understood from this explanation. In a nonlinear system of equations cannot be written or solved as a linear combination of new function or variable. A system can be termed as nonlinear despite a linear function shows in an equation. In other words, a differential equation can be linear when its the unknown function and derivatives are linear in terms. This will be same even if the other variables in terms of nonlinear appears. Apart from the nonlinear differential equation, students need to practice equation based on linear equations. They should follow quality test papers and exercise materials, offering multiple questions on how to solve a nonlinear differential equation. One can check Vedantu, which is one of the most prominent education portal offering multiple benefits. They provide solutions and an example of the nonlinear differential equation for revision purposes. Moreover, students can check their live classes and training sessions available for a budget-friendly price. To make the best of these features, download the official app today! FAQs on How to Solve Linear Differential Equation? 1. What is the Primary Difference Between Linear and Nonlinear Equation? Ans. A linear equation can be defined as a form of the equation that is linear polynomial in nature. It is a part of the new derivatives and functions of an equation of a structure. A linear equation usually consists of derivatives of multiple variables. When a result and variable is partial, then it is also stated as a linear partial differential equation. It forms a straight line on the graph, and its general form is x = ny + c. Here x and y are variables while n is a slope and c is the constant value. While when a new function and the derivative is not in the linear form or not polynomial, it is then considered to be a nonlinear equation. It forms a curve on the graph as its general form is ax² + by² = c. Here a, b and ca are constant value and x, y is variable. 2. What is the Formula to Solve the First-Order Differential Equation? Ans. A first-order differential equation is a sum in which ƒ(x, y) is a function of two variables. It is generally defined in an XY-plane region. The equation is of the first order because it occupies only the first derivative dy and dx. The first-order differential equation is a form of a sum which can be shown as F(p, y', y ) = 0. The solution for this equation will be a function f(p) which makes F(p, f(p), f′(p)) = 0 for all the value of p. Moreover, F can be considered as a function of three variables which we can show as p, y' and y. It can be seen that ˙y  appears plainly in the equation though p and y are not required. First-order, this term here means the first derivative of y appears, but no higher order of results are seen. 3. What is the Answer for this Differential Equation: dy/dx + (sec x)y = 6? Ans. Evaluating the sum where dy/dx + Py = Q. We consider,  P = sec  x, Q = 6 Here the integrating factor can be found by using the formula e∫Pdx= I.F and e∫secdx  = I.F.I.F. =  eln|secx + tanx|=secx + tanx We can write LHS as d(y × I.F)/dx}, that is d(y × (sec x + tan x ). Now d(y × (sec x + tan x ))/dx =6(sec x + tan x). Incorporating both the sides w, r, t,x, We get ∫d(y × (secx + tanx)) = ∫6(secx + tanx)dx y × (secx + tanx) = 6(ln|secx + tanx|+ log|secx|), y = 6(ln|secx + tanx|+ log|secx|(secx + tanx) + c.

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In a division sum, we have dividend = 199, quotient = 16 and remainder = 7. The divisor is - Mathematics MCQ In a division sum, we have dividend = 199, quotient = 16 and remainder = 7. The divisor is Options • 11 • 23 • 12 • none of these Solution 12 Dividend = 199 Quotient = 16 Remainder = 7 According to the division algorithm, we have: Dividend = divisor × quotient + remainder ⇒ 199 = divisor × 16 + 7 ⇒ 199 − 7 = divisor × 16 ⇒ Divisor = 192 ÷ 16 Concept: Concept for Whole Numbers Is there an error in this question or solution? APPEARS IN RS Aggarwal Class 6 Mathematics Chapter 3 Whole Numbers Exercise 3F | Q 9 | Page 57

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Research-Based Informationto keep in mind while teaching English Language Learners Bibliography: American Educator, Summer 2008, Teaching English Language Learners. Susana Dutro/2008, A Focused Approach to Systematic ELD. Achievement Gap for ELLs 2007 National Assessment of Educational Progress, Fourth Grade ELLs: * scored 36 points below non-ELLs in reading. * scored 25 points below non-ELLs in math. Eighth grade scores: 42 points below in reading 37 points below in math Statistics 1990 – 1 in 20 students 2008 – 1 in 9 students 2028 – demographers project 1 in 4 students 80% of ELLs are Spanish speakers *24% are from families below poverty level before immigrating. *Less than 40% of the adult immigrants have a high school education. *87.5% of ELL population were born in the US. Language and socioeconomic factors put ELLs at risk of poor educational outcomes. CREDE ReviewA report on researched findings about educating English language learners: *Teaching students to read in their L1 promotes higher levels of reading in English. *What we know about good instruction and curriculum in general holds true for ELLs as well; but *When instructing ELLs in English, teachers must modify instruction, taking into account students ‘ language limitations. Overview of English Proficiency Levels Beginning: Student uses gestures, learns high-frequency words and basic phrases. Effective Instructional Practices: *Exposure to abundant basic vocabulary supported with visuals and realia. *Making explicit connections to known vocabulary/concepts in L1 if possible. *Model simple sentence patterns. *Provide many repetitions. Early intermediate: Student learns to use routine expressions independently, responds orally/writing using simple sentences when provided scaffold. Effective Instructional Practices: *Give extensive practice in variety of ways to communicate thoughts. *Provide instructional feedback. *Repetitive and patterned text extends grammatical forms practice. *Begin building standard content vocabulary around standard content themes. Intermediate: Student learns to use a variety of verb tenses and grammatical structures, express ideas, describe events, give information orally/writing, comprehend basic content learning. (All using scaffolds) Suggested Teaching Strategies: *Focus on array of academic purposes. *Explore word relationships: word sorts/graphic organizers. *Increasingly precise vocabulary using frames, stems … Exposure to varied and extended texts with scaffolds. *Build English content vocabulary within thematic content. Early Advanced: Student initiates and sustains spontaneous language interactions Is able to comprehend increasingly complex oral and written material. Uses academic vocabulary to represent thoughts. Effective Instructional Practices: *Consistent exposure to complex narrative and expository text, focused on comprehension. *Develop academic vocabulary and complex tenses. *Address persistent problem areas in grammar. *Discuss and use metaphoric and figurative language. , Advanced: The student comprehends general and implied meanings. Writes for social and academic purposes, although expression is sometimes stilted. Have mastered language conventions for formal and informal use. Effective Instructional Practices: *Repeated opportunities to express thinking about abstract concepts. *Authentic practice opportunities to develop fluency and automaticity in communication. *Direct instruction of finer, more subtle points of usage. Addressing the needs of ELLs through Research- Based Strategies will: Help to close the wide learning gap between Ells and English speaking students. Benefit L1 and L2 students through best Instructional practices. Reduce the costs of large scale under- achievement in the future of US society. If any of these arguments appeals to you, then Advocate for research-based, ESOL instruction.

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# Current Electricity Class 12 Notes Pdf ## Current Electricity Class 12 Notes Current electricity is the most important chapter for the board exam of class 12, so about all students make self-written notes of this chapter or also want to be download notes on the internet. so I create notes for those students who don’t make a note by yourself. let’s start to discuss some parts of current electricity class 12 notes. In Chapter 1 Electric charge and fields, all charges, whether free or bound, were considered to be in rest. Moving charges create an electric current. A similar current is found in many situations in nature. Lightning is a phenomenon in which charges reach clouds from the Earth through the atmosphere, sometimes resulting in terrible consequences. The flow of charge in lightning is not constant. But in our daily life, we see charges flowing in many ways in the same way that water flows in rivers. Torch and cell-powered clock are some examples of such devices. In this study, we will study some basic rules related to an irreversible or constant current. ### Electric field applied on charge In current electricity class 12 notes, Imagine a small area perpendicular to the charge flow. Both positive and negative charges can flow through this region in the forward or backward direction. Suppose the net forward positive current flowing through this region in a time interval is q, (ie the difference between the forward and backward). Similarly, suppose the net forward negative current flowing through this region is q_. Then in this time interval t is the net charge flowing through this region. For a constant current, it is directly proportional to t and the quotient defines the electric current flowing in the forward direction through the field It = Q. (If this number is negative, it gives an indication that the current is in a backward direction.) The electric current is not always irreversible, so more broadly, we define electric current as follows. Suppose the net charge flowing through the cross-section of a conductor in the time interval At is Ag. Then the electric current flowing through this transverse cut of the conductor at time t is defined as the value of the ratio of Ag or At with the limits of Attending towards zero. #### Electric field applied on charge In current electricity class 12 notes, If an electric field is applied to an electric charge, it will experience a force. If it is free to move, it will also move and generate electric current. Like the upper level of the atmosphere, called the ionosphere, free charged particles are found in nature. However, the negatively charged electrons and positively charged electrons in molecules and atoms are not free to move due to their bonding with each other. The gross matter is made up of many molecules, for example, one gram of water contains about 1022 molecules. These molecules are so clustered that the electron is no longer attached to an individual nucleus. Electrons are still bound in some materials, that is, they are not accelerated even when applied electric fields. In some other substances, especially in metals, some electrons are free to move realistically within the material. In these materials, commonly called conductors, an electric current is generated by applying an electric field. If we consider the solid conductor, in fact, these atoms are closely bound in you, due to which the charged electrons carry the electric current. However, there are other types of conductors, such as electrical decomposition solutions, in which both positive and negative charges can move. We will focus our discussion on solid conductors in which negatively charged electrons carry electric current in the background of stable cations. Let us first consider a situation where no electric field is present. Electrons interact with bonded ions while doing thermal motions. After electron impact error: Content is protected !!

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# MCQ Questions Chapter 13 Probability Class 12 Mathematics Please refer to MCQ Questions Chapter 13 Probability Class 12 Mathematics with answers provided below. These multiple-choice questions have been developed based on the latest NCERT book for class 12 Mathematics issued for the current academic year. We have provided MCQ Questions for Class 12 Mathematics for all chapters on our website. Students should learn the objective based questions for Chapter 13 Probability in Class 12 Mathematics provided below to get more marks in exams. ## Chapter 13 Probability MCQ Questions Question. If the chance that a ship arrives safely at a port is 9/10 ; the chance that out of 5 expected ships, at least 4 will arrive safely at the port 2, is (a) 91854/100000 (b) 32805/100000 (c) 59049/100000 (d) 26244/100000 A Question. The probability that a man hits a target is p = 0.1. He fires n = 100 times. The expected number n of times he will hit the target is : (a) 33 (b) 30 (c) 20 (d) 10 D Question. Let three fair coins be tossed. Let A = {all heads or all tails}, B = {atleast two heads}, and C = {atmost two tails}. Which of the following events are independent? (a) A and C (b) B and C (c) A and B (d) None of these C Question. For a loaded die, the probabilities of outcomes are given as under. P(1) = P (2) = 0.2 P(3) = P (5) = P(6) = 0.1 and P(4) = 0.3 The dice is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’’ respectively. Then, match the terms of column I with their respective values in column II. Codes A B C D (a) 4 3 2 1 (b) 4 3 1 5 (c) 4 1 3 5 (d) 2 4 1 3 A Question. If P(A∩B) = 0.15 , P(B’) = 0.10 , then P(A/B) = (a) 1/3 (b) 1/4 (c) 1/6 (d) 1/5 C Question. P(E∩F) is equal to (a) P(E). P(F|E) (b) P(F). P(E|F) (c) Both (a) and (b) (d) None of these C Question. The probability of a man hitting a target is 1/4. The number of times he must shoot so that the probability he hits the target, at least once is more than 0.9, is [use log 4 = 0.602 and log 3 = 0.477] (a) 7 (b) 8 (c) 6 (d) 5 B Question. If P(A) = 2/5, P(B) = 3/10 and P(A ∩ B) = 1/5, then P(A’ | B’) . P(B’ | A’) is equal to (a) 5/6 (b) 5/7 (c) 25/42 (d) 1 C Question. A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Codes A B C (a) 2 3 1 (b) 1 2 3 (c) 2 1 3 (d) 3 1 2 A Question. If X follows Binomial distribution with mean 3 and variance 2, then P(X ≥ 8) is equal to : (a) 17/39 (b) 18/39 (c) 19/39 (d) 20/39 C Question. Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is (a) 0.2 (b) 0.7 (c) 0.06 (d) 0.14 D Question. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0, if he opposed and X = 1, if he is in favour. Then, E(X) and Var (X) respectively are (a) 3/7, 5/17 (b) 13/15, 2/15 (c) 7/10, 21/100 (d) 7/10, 23/100 C Question. The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P (X = 1) is (a) 1/4 (b) 1/32 (c) 1/16 (d) 1/8 B Question. Given, two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Then, match the terms of column I with their respectively values in column II . Codes A B C D (a) 4 3 1 2 (b) 4 2 1 3 (c) 1 2 3 4 (d) 4 3 2 1 D Question. Suppose X follows a binomial distribution with parameters n and p, where 0 < p < 1, if P(X = r)/P(X = n – r) is independent of n and r, then (a) p = 1/2 (b) p = 1/3 (c) p = 1/4 (d) None of these A Question. There is 30% chance that it rains on any particular day. Given that there is at least one rainy day, then the probability that there are at least two rainy days is D Question. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is (a) 1 (b) 2 (c) 5 (d) 8/3 B Question. Two dice are thrown. If it is known that the sum of the numbers on the dice is less than 6, the probability of getting a sum 3 is (a) 1/8 (b) 2/5 (c) 1/5 (d) 5/18 C Question. Match the terms given in column-I with the terms given in column-II and choose the correct option from the codes given below. B Question. A signal which can be green or red with probability 4/5 and 1/5 respectively, is received by station A and then trasmitted to station B. The probability of each station receiving the signal correctly is 3/4. If the signal received at station B is given, then the probability that the original signal is green, is (a) 3/5 (b) 6/7 (c) 20/23 (d) 9/20 A Question. If P(B) = 3/5 , P(A | B) = 1/2 and P(A ∪ B) = 4/5 , then P(A∪B)’ + P(A’∪B) = (a) 1/5 (b) 4/5 (c) 1/2 (d) 1 D Question. The probability of the simultaneous occurrence of two events A and B is p. If the probability that exactly one of the events occurs is q, then which of the following is not correct? (a) P(A’) + P(B’) = 2 + 2q − p (b) P(A’) + P(B’) = 2 − 2p − q (c) P(A ∩ B | A ∪ B) = p/p+q (d) P(A’∩ B’) = 1− p − q A Question. Which one is not a requirement of a binomial distribution? (a) There are 2 outcomes for each trial (b) There is a fixed number of trials (c) The outcomes must be dependent on each other (d) The probability of success must be the same for all the trial C Question. Bag P contains 6 red and 4 blue balls and bag Q contains 5 red and 6 blue balls. A ball is transferred from bag P to bag Q and then a ball is drawn from bag Q. What is the probability that the ball drawn is blue? (a) 7/15 (b) 8/15 (c) 4/19 (d) 8/19 B Question. In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. Then, the probability that he will knock down fewer than 2 hurdles is (a) 59/2×69 (b) 510/2×610 (c) 59/2×610 (d) 510/2×69 D Question. Two dice are thrown, simultaneously. If X denotes the number of sixes, then the expected value of X is (a) E(X) = 1/3 (b) E(X) = 2/3 (c) E(X) = 1/6 (d) E(X) = 5/6 A Question. If A and B are two events such that P(A) ≠ 0 and C Question. Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is (a) 0.024 (b) 0.188 (c) 0.336 (d) 0.452 B Question. A and B are events such that P (A/B) = P (B/A) then (a) A ⊂ B (b) B = A (c) A ∩ B = Φ (d) P(A) = P (B) D Question. Coloured balls are distributed in four boxes as shown in the following table A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black. Probability that the ball drawn from Box III, is (a) 0.161 (b) 0.162 (c) 0.165 (d) 0.104 C Question. Two integers are selected at random from the set {1, 2, …., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is : (a) 7/10 (b) 1/2 (c) 2/5 (d) 3/5 C Question. An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, …, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is: (a) 13/36 (b) 15/72 (c) 19/72 (d) 19/36 C Question. Let A, B, C, be pairwise independent events with P (C) > 0 and P( A ∩ B ∩ C) =0. Then P(Ac ∩ Bc / C). (a) P(Bc) – P(B) (b) P(Ac) + P(Bc) (c) P(Ac) – P(B) (d) P(Ac) – P(B) D Question. If C and D are two events such that C ⊂ D and P(D) ≠ 0, then the correct statement among the following is (a) P(C | D) ≥ P(C) (b) P(C | D) < P(C) (c) P(C | D) = P(D)/P(C) (d) P(C | D) = P(C) A Question. Let EC denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1 ∩ E2 ∩ E3 ) = 0. Then P(E2C ∩ E3C / E1 ) is equal to : (a) P(E2C) + P(E3) (b) P(E3C) – P(E2C) (c) P(E3) – P(E2C) (d) P(E3C) – P(E2) D Question. An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value –1. Then the expected value of X, is: (a) 3/16 (b) 1/8 (c) − 3/16 (d) − 1/8 B Question. Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is: (a) 1/11 (b) 1/10 (c) 1/12 (d) 1/17 A Question. Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is : (a) 5 (b) 6 (c) 8 (d) 7 D Question. Let A and E be any two events with positive probabilities: Statement – 1: P(E/A) ≥ P(A/E) P(E) Statement – 2: P(A/E) ≥ P(A ∩ E) (a) Both the statements are true (b) Both the statements are false (c) Statement-1 is true, Statement-2 is false (d) Statement-1 is false, Statement-2 is true A Question. A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are 3/4, 1/2, 5/8. The probability that the target is hit by A or B but not by C is : (a) 21/64 (b) 7/8 (c) 7/32 (d) 9/64 A Question. The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is : (a) 5 (b) 3 (c) 4 (d) 2 C Question. The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is (a) 2/256 (b) 219/256 (c) 128/256 (d) 37/256 A Question. A random variable X has the probability distribution: (image 66) For the events E = {X is a prime number } and F = {X < 4}, the P(E ∪ F) is (a) 0.50 (b) 0.77 (c) 0.35 (d) 0.87 B Question. In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to : (a) 200/65 (b) 150/65 (c) 225/65 (d) 175/65 D Question. In a game, a man wins ₹ 100 if he gets 5 or 6 on a throw of a fair die and loses ₹ 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : (a) (400/9)loss (b) 0 (c) (400/3)gain (d) (400/3)loss C Question. In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is : (a) 5/31 (b) 31/61 (c) 5/6 (d) 30/61 D Question. In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is _________. 1100 Question. The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4, is ___________. 3.00 Question. A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is : (a) 1/4 (b) 1/3 (c) 1/8 (d) 1/9 D Question. If the probability of hitting a target by a shooter, in any shot, is 1/3, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 5/6, is: (a) 3 (b) 6 (c) 5 (d) 4 C Question. Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X d” 2) = k/216, then k is equal to: (a) 17 (b) 121 (c) 1 (d) 137 D Question. A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is : (a) (1/2)gain (b) (1/4)loss (c) (1/2)loss (d) 2 gain C Question. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P(X = 2) equals: (a) 49/169 (b) 52/169 (c) 24/169 (d) 25/169 D Question. An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red is: (a) 21/49 (b) 27/49 (c) 26/49 (d) 32/49 D Question. A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is (a) 8/729 (b) 8/243 (c) 1/729 (d) 8/9 B Question. A coin is tossed three times is succession. If E is the event that there are at least two heads and F is the event in which first throw is a head, then P(E/F) equal to: (a) 3/4 (b) 3/8 (c) 1/2 (d) 1/8 A Question. Five bad eggs are mixed with 10 good ones. If three eggs are drawn one by one with replacement, then the probability distribution of the number of good eggs drawn, is D Question. If two events A and B are such that P(A’) = 0.3, P(B) = 0.4 and P(A∩B’) = 0.5, then P (B/A ∪ B) = (a) 1/4 (b) 1/5 (c) 3/5 (d) 2/5 A ASSERTION – REASON TYPE QUESTIONS (a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion. (b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion (c) Assertion is correct, Reason is incorrect (d) Assertion is incorrect, Reason is correct. Question. Let A and B be two events associated with an experiment such that P(A∩B) = P(A)P(B) Assertion : P(A|B) = P(A) and P(B|A) = P(B) Reason : P(A∪B) = P(A) + P(B) C Question. Consider the two events E and F which are associated with the sample space of a random experiment. Assertion : P(E / F) = n (E ∩ F)/n (F). Reason : P(E / F) = n (E ∩ F)/P(F). A Question. Consider the following statements Assertion : Let A and B be two independent events. Then P(A ∩ B) = P(A) + P(B) Reason : Three events A, B and C are said to be independent, if P (A ∩ B ∩ C) = P(A) P(B) P(C). D Question. Assertion : For a binomial distribution B(n, p), Mean > Variance Reason : Probability is less than or equal to 1 A Question. Assertion : The mean of a random variable X is also called the expectation of X, denoted by E(X). Reason : The mean or expectation of a random variable X is not sum of the products of all possible values of X by their respective probabilities.

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1. ## Range Please help me to find the range of $sec^{4}(x)+cosec^{4}(x)$. 2. ## Re: Range Originally Posted by kjchauhan Please help me to find the range of $sec^{4}(x)+cosec^{4}(x)$. What have you done? Have you at least graphed this expression? 3. ## Re: Range Originally Posted by kjchauhan Please help me to find the range of $sec^{4}(x)+cosec^{4}(x)$. $\sec^4(x) = (\sec^2(x))^2 = (1+\tan^2(x))^2 = \tan^4(x) + 2\tan^2 + 1$ $\csc^4(x) = (\csc^2(x))^2 = (1+\cot^2(x))^2 = \cot^4(x) + 2\cot^2 + 1$ $\sec^4(x) + \csc^4(x) = \tan^4(x) + 2\tan^2 + 1 + \cot^4(x) + 2\cot^2 + 1$ $= \tan^4(x) + \cot^4(x) + 2(\tan^2(x)+\cot^2(x)) + 2$ The range of $\tan(x)$ is $(-\infty, \infty)$. Likewise, the range of $\cot(x)$ is $(-\infty, \infty)$. Since $\tan^4(x)$ and $\cot^4(x)$ are positive even powers, both have range $[0,\infty)$. BELOW is where others may disagree with me: I contend that the range of $\tan^4(x)+\cot^4(x)$ is $(0,\infty)$ as opposed to $[0,\infty)$, which the sum of the parts may intuitively suggest. The two ranges differ insofar $(0,\infty)$ does not contain $0$, whereas $[0,\infty)$ does contain $0$. I believe that the range of $\tan^4(x)+\cot^4(x)$ should NOT include 0, i.e., it should be $(0,\infty)$. Proof is achieved if we show $\tan^4(x)+\cot^4(x)>0$ on the entire domain (reals that are not multiples of $\pi$). Both terms in $\tan^4(x)+\cot^4(x)$ are non-negative in the reals, so clearly the sum itself is non-negative. $\tan^4(x)+\cot^4(x)$ cannot equal 0 because then either $\tan^4(x)=-\cot^4(x)$ (which per the line above is not possible) OR $\tan^4(x)=\cot^4(x)=0$, which also can't happen because $\cot^4(x)=\tfrac{1}{\tan^4(x)}$ and 0 cannot be a denominator. As such, $\tan^4(x)+\cot^4(x)>0$. And with this new information, $\sec^4(x) + \csc^4(x) = \underbrace{\left[\tan^4(x) + \cot^4(x)\right]}_{\text{always positive!}} + 2\underbrace{(\tan^2(x)+\cot^2(x))}_{\text{can show positive similarly}} + 2 > 2$. It at long last follows that the range of $\sec^4(x) + \csc^4(x)$ is $\left(2,\infty\right)$. If someone has an argument for a left bracket instead of my left open parenthesis, I'd love to hear it! -Andy 4. ## Re: Range Originally Posted by abender [TEX]\sec^4(x) = (\sec^2(x))^2 = I contend that the range of $\tan^4(x)+\cot^4(x)$ is $(0,\infty)$ as opposed to $[0,\infty)$, which the sum of the parts may intuitively suggest. It at long last follows that the range of $\sec^4(x) + \csc^4(x)$ is $\left(2,\infty\right)$. If someone has an argument for a left bracket instead of my left open parenthesis, I'd love to hear it! Take a look at the graph of the function. 5. ## Re: Range $[8,\infty)$, evidently. My way was a fun adventure; I wanted to do it using identities for some bizarre reason. These problems nearly always boil down to the rudimentary sines and cosines. I see the airtight solution now, the symmetry, no plus 2 times "something positive". In retrospect, why the hell did I accept 2 as my best lower bound? Oh well, my Ravens are in the Super Bowl! Usually never post if I'm uncertain.

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# The initial speed of a bullet fired from a rifle is $630 \mathrm{~m} / \mathrm{s}$. The rifle is fired at the centre of a target $700 \mathrm{~m}$ away at the same level as the target. How far above the centre of the target ? Question: The initial speed of a bullet fired from a rifle is $630 \mathrm{~m} / \mathrm{s}$. The rifle is fired at the centre of a target $700 \mathrm{~m}$ away at the same level as the target. How far above the centre of the target ? 1. $1.0 \mathrm{~m}$ 2. $4.2 \mathrm{~m}$ 3. $6.1 \mathrm{~m}$ 4. $9.8 \mathrm{~m}$ JEE Main Previous Year Single Correct Question of JEE Main from Physics Motion in a Plane chapter. JEE Main Previous Year April 11, 2014 Correct Option: 3 Solution: ### Related Questions • Let $\left|\overrightarrow{A_{1}}\right|=3,\left|\overrightarrow{A_{2}}\right|=5$ and $\left|\overrightarrow{A_{1}}+\overrightarrow{A_{2}}\right|=5$. The value of $\left(2 \overrightarrow{\mathrm{A}_{1}}+3 \overrightarrow{\mathrm{A}_{2}}\right) \cdot\left(3 \overrightarrow{\mathrm{A}_{1}}-2 \overrightarrow{\mathrm{A}_{2}}\right)$ is : View Solution • In the cube of side ‘a’ shown in the figure, the vector from the central point of the face $\mathrm{ABOD}$ to the central point of the face BEFO will be: View Solution • Two forces $P$ and $Q$, of magnitude $2 F$ and $3 F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is: View Solution • Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. The magnitude of $(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})$ is ‘ $n$ ‘ times the magnitude of $(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})$. The angle between $\vec{A}$ and $\vec{B}$ is: View Solution • Let $\overrightarrow{\mathrm{A}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})$ and $\overrightarrow{\mathrm{B}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})$. The magnitude of a coplanar vector $\overrightarrow{\mathrm{C}}$ such that $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}$ is given by View Solution • A vector $\vec{A}$ is rotated by a small angle $\Delta \theta \operatorname{radian}(\Delta \theta<<1)$ to get a new vector $\vec{B}$. In that case $|\vec{B}-\vec{A}|$ is : View Solution • If $\vec{A} \times \vec{B}=\vec{B} \times \vec{A}$, then the angle between $\mathrm{A}$ and $\mathrm{B}$ is View Solution • A balloon is moving up in air vertically above a point A on the ground. When it is at a height $h_{1}$, a girl standing at a distance $d$ (point B) from A (see figure) sees it at an angle $45^{\circ}$ with respect to the vertical. When the balloon climbs up a further height $h_{2}$, it is seen at an angle $60^{\circ}$ with respect to the vertical if the girl moves further by a distance $2.464 d$ (point $\mathrm{C}$ ). Then the height $h_{2}$ is (given $\tan 30^{\circ}=0.5774$ ): View Solution • Starting from the origin at time $t=0$, with initial velocity $5 \hat{j} \mathrm{~ms}^{-1}$, a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{i}+4 \hat{j}) \mathrm{ms}^{-2}$. At time $t$, its coordiantes are $\left(20 \mathrm{~m}, y_{0} \mathrm{~m}\right)$. The values of $t$ and $y_{0}$ are, respectively: View Solution • The position vector of a particle changes with time according to the relation $\vec{r}(\mathrm{t})=15 \mathrm{t}^{2} \hat{i}+\left(4-20 \mathrm{t}^{2}\right) \hat{j}$. What is the magnitude of the acceleration at $t=1 ?$ View Solution error: Content is protected !!

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This map shows the South Pole of Enceladus. Enceladus is a medium-size, icy moon of Saturn. Can you see the huge "tiger stripe" cracks in this view? They go from upper left to lower right across the map. The pictures used to make this map came from the Cassini spacecraft. This map is about 220 km (137 miles) across. Click on image for full size Image courtesy of DLR and NASA/JPL/Cassini Imaging Team. The South Pole of Enceladus Enceladus is a medium-sized, icy moon of Saturn. The South Pole of Enceladus is one of the strangest places in the Solar System. Temperatures near the pole are far warmer than anywhere else on the moon. Huge cracks criss-cross the region and geysers spew ice crystals hundreds of kilometers into the sky. Several huge gouges, more than 100 km (62 miles) long, run across the region of Enceladus's southern pole. These giant cracks have been nicknamed "tiger stripes". Scientists can tell that the area around the South Pole is geologically younger than the rest of the moon's surface. They have also discovered that the South Pole is by far the warmest place on the moon, especially near the tiger stripes. That wouldn't be so if the only source of heat on Enceladus is sunlight. The young surface, enormous "tiger stripe" cracks, and high temperatures all indicate that Enceladus (or at least the area around its South Pole) is geologically active. The Cassini spacecraft found evidence of geologic activity. It took pictures of ice geysers erupting from the southern polar region of Enceladus! There are only four bodies in our Solar System on which we have seen eruptions - Jupiter's moon Io, Neptune's moon Triton, Earth (of course!), and now Enceladus. Scientists are still trying to figure out how the "plumbing" of the geysers works and what supplies heat for the geysers. There might be liquid water beneath the surface on Enceladus. That is an idea that gets astrobiologists excited. Scientists have spotted several different plumes of ice from geysers. The ice crystals are flung hundreds of kilometers above Enceladus's surface. Some even leave the moon altogether, supplying icy material to one of Saturn's rings. Ice that falls back to the moon's surface covers it with a bright layer like fresh snow. This makes Enceladus the most reflective body in the Solar System. Shop Windows to the Universe Science Store! Our online store on science education, classroom activities in The Earth Scientist specimens, and educational games You might also be interested in: Saturn has // Call the moon count function defined in the document head print_moon_count('saturn'); moons. Many of those are tiny chunks of rock or ice only a few kilometers (miles) across. One of Saturn's...more A spacecraft named Cassini will study the planet Saturn for several years. Cassini blasted off from Earth in October 1997. After flying past Venus, Earth, and Jupiter, Cassini finally arrived at Saturn...more This picture of the Earth surface was taken from high above the planet in the International Space Station. In this view from above, we can see that there are lots of different things that cover the Earth....more There's a lot of strange and interesting stuff going on at both the North and South Poles of Saturn. Two of Saturn's moons also have interesting polar regions. Let's take a look! The atmosphere and clouds...more Titan is similar to the other icy moons, but Titan is the only icy moon to have a big atmosphere. It is natural to ask how is this possible? The nebula was colder near Saturn, than near Jupiter. The nebula...more Titan's atmosphere is a lot like the Earth's, except that it is very cold, from -330 degrees to -290 degrees! Like the Earth, there is a lot of Nitrogen and other complex molecules. There also may be an...more Pandora is a small moon of Saturn. It was discovered by S. Collins and others in 1980 from photos taken by the Voyager 1 spacecraft. Pandora's name comes from Greek mythology. Pandora was the first woman,...more

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# Question with acceleration vector 1. Feb 21, 2015 ### goonking 1. The problem statement, all variables and given/known data http://imgur.com/LravIr3 2. Relevant equations 3. The attempt at a solution We know A is wrong, because a object going in a circle has acceleration (i'm not sure why that is, maybe someone can explain) B is wrong because if a car in reverse is slowing down, it technically has positive acceleration, right? D is wrong because refer to my explanation for B. C and E, I have no idea what that formula is but V x T is distance but I have no idea what the formulas are implying. Anyone can shed some light on what C and E means? 2. Feb 21, 2015 ### Suraj M The formula $v(T + Δ T)$ refers to the velocity of the object at time$(T +Δ T)$. Now can you decide if C and E are right or wrong? 3. Feb 22, 2015 ### goonking how does velocity multiplying the time give you velocity again? 4. Feb 22, 2015 ### Suraj M Its not multiplication, It's a way of representing the velocity at a particular time, as the velocity varies with time! 5. Feb 22, 2015 ### goonking can you make up a word problem that uses v(T+ΔT)? 6. Feb 22, 2015 ### Suraj M Okay instead of a word problem with $V(T+ΔT)$ try this. Let $~~ V(t) = ƒ(t)$ and $ƒ(t) = u(0) + at$ here i used $u(0)$ itts actually $u(t=0)$ we often miss out the $t=$ so in your question it should have been- $v(t_2 = T+ΔT)$ and $v(t_1 = T)$ we often omit the t= to make things easier, Its represented like this because velocity is not a constant and is a function of time, skips the steps you'd have to involve to define $v_1 ~ and~ v_2$ its just like writing $V(initial) and V(final)$ see this Last edited: Feb 22, 2015 7. Feb 22, 2015 ### Staff: Mentor EDIT: (D) is not always true Last edited: Feb 22, 2015 8. Feb 22, 2015 ### Staff: Mentor Where the acceleration of a body is known to be constant, we can determine that acceleration by making two measurements of its velocity some time apart, then using the formula; a = Δv / Δt = ( v(T+ΔT) - v(T) ) / ΔT 9. Feb 22, 2015 ### goonking if I'm in a car, and i'm in reverse while increasing my speed. I'm accelerating while going backwards so my acceleration is negative. I then slow down a bit while reversing but I'm still moving backwards, my acceleration is now positive even though i'm still moving backwards. It is positive because I slowed down in reverse. How can D still be true? 10. Feb 22, 2015 ### Staff: Mentor You are right. I was not correct because in that case (D) is not true. I'll amend my earlier post. Thanks for the correction.

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## You’re carpeting 2 rooms of your house room 1 measurements are 12ft and 10ft. Room 2 is 10ft and 18ft. The carpet costs \$1.48 per square foo Question You’re carpeting 2 rooms of your house room 1 measurements are 12ft and 10ft. Room 2 is 10ft and 18ft. The carpet costs \$1.48 per square foot. How much does it cost to carpet the rooms? in progress 0 2 weeks 2021-11-21T15:28:50+00:00 1 Answer 0 Step-by-step explanation: Since the cost per square foot is \$ 1.48 , we need to calculate the square feet per  room in order to know the cost Room 1 : the square feet = 10 X 12 = 120 Cost = 120 x 1.48 = \$177.6 Room 2 : the square feet = 10 x 18 = 180 cost = 180 x 1.48 = \$266.4 The total cost = \$177.6 + \$ 266. 4 = \$ 444

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Climate disasters and extreme events The climate varies naturally on all timescales. Variations can be caused by external forces such as volcanic eruptions or changes in the sun's energy output. They can also result from the internal interactions of the climate system's various components – the atmosphere, oceans, biosphere, ice cover, and land surface. These internal interactions can cause fairly regular fluctuations, such as the El Nino/Southern Oscillation (ENSO) phenomenon, or apparently random changes in climate. Natural variability often leads to climate extremes. On timescales of days, months, and years, weather and climate variability can produce heat waves, frosts, floods, droughts, avalanches, and severe storms. Such extremes represent a significant departure from the average state of the climate system, irrespective of their actual impact on life or the earth's ecology. Record-breaking extremes occur from time to time in every region of the world. Growing human vulnerability is transforming more and more extreme events into climatic disasters. A climate extreme is called a climatic disaster when it has a major adverse impact on human welfare. In some parts of the world, climatic disasters occur so frequently that they may be considered part of the norm. Vulnerability to disasters is increasing as growing numbers of people are forced to live in exposed and marginal areas. Elsewhere, greater vulnerability is being caused by the development of more highvalue property in highrisk zones. Climate change is expected to increase the frequency and severity of heat waves. More hot weather will cause more deaths and illnesses among the elderly and urban poor. Together with increased summer drying, it will lead to greater heat stress for livestock and wildlife, more damage to crops, more forest fires, and more pressure on water supplies. Other likely impacts are a shift in tourist destinations and a boost in demand for energy. Meanwhile, fewer cold snaps should reduce cold-related risks to humans and agriculture and reduce the energy demand for heating while extending the range and activity of some pests and diseases. More intense rainfall events may lead to greater flooding in some regions. Global warming is expected to accelerate the hydrological cycle and thus raise the percentage of precipitation that falls in violent bursts. In addition to floods, this could contribute to more landslides, avalanches, and soil erosion. Greater flood runoff could decrease the amount of surface water captured for irrigation and other purposes, although it could help to recharge some floodplain aquifers. The intensity of tropical cyclones is likely to worsen over some areas. The risks include direct threats to human life, epidemics and other health risks, damage to infrastructure and buildings, coastal erosion, and destruction of ecosystems such as coral reefs and mangroves. Major climate patterns could shift. Although centered in the Southern Pacific, the El Nino/Southern Oscillation (ENSO) phenomenon affects the weather and climate in much of the tropics. Climate change could intensify the droughts and floods that are associated with El Niño events in these regions. Similarly, new patterns could emerge for the Asian summer monsoon, which affects large areas of temperate and tropical Asia. Likely impacts would include a greater annual variability in the monsoon’s precipitation levels, leading to more intense floods and droughts. It is difficult to predict local and regional trends for extreme events. For example, a warming of the tropical oceans would by itself be expected to increase the frequency, and perhaps the severity, of tropical cyclones. But other factors, such as changing winds or storm tracks, might offset this effect at the local level. Another example: because climate models are poor at representing small-scale events, they tend to disagree on whether or not the intensity of mid-latitude storms will change. While extreme events are inherently abrupt and random, the risks they pose can be reduced. Improved preparedness planning is urgently needed in many parts of the world, with or without climate change. Better information, stronger institutions, and new technologies can minimize human and material losses. For example, new buildings can be designed and located in ways that minimize damage from floods and tropical cyclones, while sophisticated irrigation techniques can protect farmers and their crops from droughts. Climate change also has the potential to cause large-scale singular events. Unlike most extreme events, singular events would have broad regional or global implications and be essentially irreversible. Examples of such calamities would include a significant slowing of the ocean’s transport of warm water to the North Atlantic (which is responsible for Europe’s relatively benign climate), a major shrinking of the Greenland or West Antarctic ice sheets (which would raise sea levels by three metres each over the next 1,000 years), and an accelerated warming due to carbon cycle feedbacks in the terrestrial biosphere, the release of carbon from melting permafrost, or the emission of methane from coastal sediments. Such risks have not yet been reliably quantified, but fortunately they are expected to be quite low.

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Artificial nighttime lighting – from the dim glow of your iProduct to the alley streetlight – could be detrimental to your mental health. Ohio State University Medical Center neuroscientists found that hamsters who slept 8 hours a night with a dim nightlight no longer enjoyed their old hamster games. The nightlight had altered the brain structure of these depressed rodents. Many studies in the past, including some conducted by this team, have linked nighttime light pollution with health problems, including mood disorders, breast cancer, and significant weight gain. At night, a gland in your brain secretes the hormone melatonin, which regulates light-related functions and biological rhythms – such as letting your body know when it's nighttime or what season it is. Exposure to light at night suppresses melatonin secretion, disrupting the body’s natural cycles. The researchers kept one group of female Siberian hamsters (Phodopus sungorus) in a cycle of 16 hours of light followed by 8 hours of complete darkness. Another group had the same 16 hours of light, but slept 8 hours in dim lamplight – about as illuminated as a dark room with the TV on. After several weeks, the researchers analyzed the hamsters for symptoms of depression, utilizing the same assessments pharmaceutical companies use to test antidepressive drugs in animals before testing on humans. When the ‘dim’ hamsters were forced to swim, they spent less time performing an escape-directed swim, suggesting that they gave up sooner than the ‘dark’ hamsters. Dim hamsters were also less interested in drinking their sugar water treat; they were no longer getting the same "pleasurable and rewarding feeling” said co-author Tracy Bedrosian. “Even dim light at night is sufficient to provoke depressive-like behaviors in hamsters,” Bedrosian concluded. After processing the hamsters’ brains, the researchers found changes to the brain structure and chemistry of the depressed hamsters. A part of their brain called the hippocampus had less dense networks of dendritic spines – the branchlike growths on brain cells responsible for transmitting messages between the cells. The hippocampus plays a key role in depressive disorders. Nighttime low level lighting, according to the authors, may be a significant contributor to the increasing prevalence of mood disorders in North America and Europe. Additionally, the potential dangers of disrupted sleep would affect not only people who work at night, but also those who sleep with the lights on, fly frequently on red-eyes, and spend nights in hospitals. If this link applies to humans, “people might want to think about getting dark shades," Bedrosian said, "and making sure to give themselves darkness when they go to sleep.” Co-author Randy Nelson added: “You would expect to see an impact if we were blasting these hamsters with bright lights, but this was a very low level, something that most people could easily encounter every night.” Image: Sleeping hamster by ASL via Flickr This post was originally published on Smartplanet.com

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This ghostly image from NASA's Spitzer Space Telescope shows the disembodied remains of a dying star, called a planetary nebula. Planetary nebulas are a late stage in a sun-like star's life, when its outer layers have sloughed off and are lit up by ultraviolet light from the central star. The Ghost of Jupiter, also known as NGC 3242, is located roughly 1,400 light-years away in the constellation Hydra. Spitzer's infrared view shows off the cooler outer halo of the dying star, colored here in red. Also evident are concentric rings around the object, the result of material being periodically tossed out in the star's final death throes. In this image, infrared light at wavelengths of 3.6 microns is rendered in blue, 4.5 microns in green, and 8.0 microns in red. > Read more NASA's Jet Propulsion Laboratory, Pasadena, Calif., manages the Spitzer Space Telescope mission for NASA's Science Mission Directorate, Washington. Science operations are conducted at the Spitzer Science Center at the California Institute of Technology in Pasadena. Data are archived at the Infrared Science Archive housed at the Infrared Processing and Analysis Center at Caltech. Caltech manages JPL for NASA. Image Credit: NASA/JPL-Caltech/Harvard-Smithsonian CfA Page Last Updated: October 30th, 2013 Page Editor: Sarah Loff

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Astronomers have discovered that the previously discovered exoplanet WASP-12b almost does not reflect light. The discovery was made by an international team of scientists who analyzed the data obtained by the STIS spectrometer of the Hubble Space Telescope. The researchers found that the WASP-12b albedo is 0.064, which is 2 times less than that of the Moon, whose albedo is 0.12. Thus, this exoplanet was darker than fresh asphalt. WASP-12b revolves around the sun-like star WASP-12A, located about 1.4 thousand light-years from the Earth. Exoplanet was discovered in 2008. Its radius is almost twice the radius of Jupiter. One revolution around the star exoplanet commits in one day. WASP-12b is classified as a hot jupiter. Since the planet experiences a strong gravitational influence of its star, it has an ovoid shape. The temperature on the star-lit side is 2.6 thousand degrees Celsius. Scientists believe that a small albedo is due to the high temperature of the planet. If, in the case of other dark planets, it could be assumed that they are enveloped in clouds and alkali metals that contribute to the absorption of light, then in the case of WASP-12b, the situation is different, because this planet is too hot to form clouds. And with alkali metals there would be ionization. Scientists believe that the unusual reflectivity of WASP-12b is due to the fact that the atmosphere of this planet is more like the atmosphere of a small star than the atmosphere of the planet. The atmosphere of WASP-12b probably consists of atomic hydrogen and helium. WASP-12b – the second planet, which managed to measure the albedo. The first was another hot Jupiter HD 189733b.

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# Function Notation In Russia Materials Needs Pencil Notebook • Slides: 27 Function Notation (In Russia!) Materials Needs • Pencil • Notebook • Scissors • Glue • Passport Table of Contents Pg Title 4 LI/SC, Warm Up, Guided Practice Pg Title 5 Function Notation Notes Warm Up Are the following points solutions to the equation? 1. 3 x – 2 y = 11 Point (5, 2) 2. y = -2 x – 3 Point (-2, 1) 3. x – 7 y = 3 Point (-4, -1) 4. y = 3 x – 1 Point (3, 5) Warm Up Are the following points solutions to the equation? 1. 3 x – 2 y = 11 True Point (5, 2) 2. y = -2 x – 3 Point (-2, 1)True 3. x – 7 y = 3 Point (-4, -1)True 4. y = 3 x – 1 Point (3, 5)False Golden Ring, Northeast Moscow, Russia Learning Intention/Success Criteria LI: We are learning to solve equations while using function notation SC: I know how to… 1. use the order of operations 2. use function notation to evaluate functions 3. use function notation to solve problems using tables 4. use function notation to solve problems using graphs 5. write solutions to equations as coordinates or in a table Today’s Vocabulary Function Notation: The way an equation is written. Used to evaluate functions. In Context: What the answer means in terms of the problem Table: Data arranged in rows and columns Think, Pair, Share f( x) = x 2 + 3 x – 1 f(\$) = \$2 + 3\$ – 1 Think, Pair, Share f(x) = |x 3 – 2 x| – 2 f( )=| 3 – 2 |– 2 Think, Pair, Share f( x) = 3 x + 1 f( a) = 3 a + 1 Think, Pair, Share f( x) = ½ x + 9 f(4) = ½ (4) + 9 Think, Pair, Share f( x) = 2 x 2 + x – 1 f(3) = 2(3)2 + (3) – 1 Example 1: Evaluate f(x) = 3 x – 8, when x = 2 f(x) = 3 x – 8 SC 7 f(2) = 3(2) – 8 SC 2 f(2) = 6 – 8 SC 1 f(2) = – 2 When x is 2, f(x) is -2. This makes the point SC 5 (2, -2) Guided Practice 1 Evaluate f(x) = 2 x + 10, when x = 7 A) f(-7) = -4 B) f(-7) = 4 C) f(-7) = -140 Guided Practice 1 Solution Evaluate f(x) = 2 x + 10, when x = 7 f(x) = 2 x +10 f(-7) = 2(-7) +10 f(-7) = -14 +10 f(-7) = -4 Guided Practice 2 Evaluate g(x) = x + 9, when x = 5 A) g(5) = 4 B) g(5) = 14 C) g(5) = 45 D) g(5) = -4 Guided Practice 2 Solution Evaluate g(x) = x + 9, when x = 5 g( x) = x + 9 g(5) = 5 + 9 g(5) = 14 Guided Practice 3 Evaluate h(x) = ½x – 6, when x = 0 A) h(0) = 6 B) h(0) = -6 C) h(0) = 0 D) h(0) = 2 Guided Practice 3 Solution Evaluate h(x) = ½ x – 6 , when x = 0 h( x) = ½ x – 6 h(0) = ½ (0) – 6 h(0) = 0 – 6 h(0) = – 6 Think, Pair, Share What is Example 2 a: changing and M(w) represents how much? the amount of --Amount of \$\$ money in your --Goes up \$2/wk bank account each How much week. You start money did you with \$6 in your start with? account and add \$2 --Started with \$6 each week, where w is the number of M( t ) = 2 w + 6 weeks. Write the SC 6 Example 2 b: Fill in the table defined by the function: M(w)=2 w + 6 w 12 3 4 5 6 M(w) 8 10 12 14 16 18 M(w) = 2 w + 6 SC 2, SC 3 M(1) = 2(1) + 6 SC 7 M(1) = 2 + 6 SC 1 M(1) = 8 SC 5 Amount of money, M(w) Example 2 c: Plot the points from the equation, found in the table. SC 4 20 16 12 8 4 1 2 3 4 5 6 Number of weeks, w What is the value of M(4)? What does M(4) represent in context? For which values of t does M(w) = 10? M(4) = 14 SC 3, SC 4, SC 6 M(4) represents the amount of money in your account, after 4 SC 3, SC 4, weeks. SC 6 M(w) = 10, when w = 2 SC 3, SC 4, SC 6 Quiz Evaluate the following expressions, given the function: 1. g(x) = -3 x + 1, when x = 10 2. f(m) = m 2 + 7, when m = 3 3. j(k) = 2 k + 9, when k = 7 Passport Stamp Grade yourself as a passport stamp at the Golden Ring. Are you a gold, silver, or bronze medal? Represent your learning as an Olympic Medal. Include a short reason why you chose that medal.

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An ultraviolet-visible spectrophotometer is a device used to analyze the degree to which a given sample absorbs different spectra of light - as the name suggests, the machine functions in the visible and ultraviolet wavelengths. In other words, they tell you what color a sample is, and to what degree, out to a large number of decimal points. The only type of UV-VIS machines I have experience with are diode arrays. These work by shining light through the sample - which is in solution - at different wavelengths, and measuring the amount picked up by a collector on the other side of the sample. It generates a series of data points relating wavelength to absorption number; the absorption number is a relative unit; all the units put into the calculation cancel. UV-VIS spectrophotometry is useful for things such as titrations, because the machine can identify a much smaller change in the color of the pH indicator than is visible to the human eye. As an example, consider a lab from my biochemistry class: we were studying enzyme kinetics, and we were using a yeast hexokinase enzyme that catalyzed the phosphorylation of six-membered sugars. The byproduct of this reaction is the release of a proton, one per sugar molecule phosphorylated. The change in the proton (hydronium ion, actually, to be pedantic) concentration can be observed using a pH indicator such as cresol red, whose color change is measured with the UV-VIS by tracking the change in absorption at a constant wavelength (at physiological pH ranges, 560 nm seemed to work well for cresol red). The results are calibrated to the color change produced by adding a known amount of HCl or another strong acid to a solution of cresol red in the same concentration. They are also useful for measuring the UV absorbance of a sample, which we would be unable to see with our naked eyes anyway; this can be useful for determining the concentrations of protein vs. DNA in solution.

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