text:
# Blog

Read the following SAT test question and then select the correct answer.

Always read the problem carefully and determine the bottom line, the question that you must answer.  Assess your options for solving the problem and choose the most efficient method to attack the problem.  When you have an answer, loop back to make sure that you completed all the necessary steps and solved for the bottom line.

If $\sqrt{x-a}=\sqrt{x+b}$ , which of the following must be true?

Bottom Line: Which of the following . . . ?

Assess your Options:  Many "Which of the following . . . " questions require you to look at the answer choices to solve the problem, but you should always check to see whether you can simplify the equation that you have been given.  Instead of jumping to the answer choices, work the equation into a form that is not as intimidating.

Attack the Problem:  The original equation has a square root on each side.  How do you get rid of these square root signs?  Square both sides of the equation, and the roots will cancel out.  You are left with:

xa = x + b

You just showed that when something is on both sides of the equation, you can cancel it out.  There is a positive x on both sides of the equation.  If you subtract it from one side, you must subtract it from the other, and the x is eliminated.  You are left with:

-a = b

This looks fairly simple, so glance down at your answer choices.  All of them are set equal to 0.  Set your equation equal to zero by adding an a to each side.

0 = b + a

Remember, it doesn’t matter what order you use when adding two variables.

Loop Back:  You put your answer in the same form as the answers on the test, so now all you have to do is match your answer to the correct one!

(A) a = 0
(B) b = 0
(C) a + b = 0
(D) a b = 0
(E) a² + b² = 0

On sat.collegeboard.org, 54% of the responses were correct.

For more help with SAT vocabulary, visit www.myknowsys.com!

# Circles

Anytime something new happens to something very old, the result is a rich current event that could be interpreted in many different ways.  The Catholic Church has chosen a new pope, and for the first time ever, the pope is from the Americas.  Look for broad themes in this article that would make it easy to relate this current event to an SAT essay topic.

## Geometry: Circles

Read the following SAT test question and then select the correct answer.

Read the question carefully and identify the bottom line.  Then assess your options and choose the most efficient method to attack the problem.  When you have an answer, loop back to make sure that your answer matches the bottom line.
In the figure above, a shaded circle with diameter  is tangent to a large semicircle with diameter  at points C and D. Point is the center of the semicircle, and  is perpendicular to . If the area of the large semicircle is 24, what is the area of the shaded circle?

Bottom Line: A sm =? (What is the area of the small, shaded circle?)

Assess your Options:  There are two good ways to approach this problem.  Both ways require you to know the formula for the area of a circle. On collegeboard.org you will find a method that is especially efficient for students who are good at writing equations.  The method used here will focus on geometry skills and estimation in order to avoid the mistakes that often come with working more abstract formulas.

Attack the Problem:  You know the most about the large circle, so start there.  A semicircle is just half of a whole circle.  Therefore, to find the area of the whole circle, you would simply double the 24.

24 × 2 = 48

If you know the area of the large circle, you can use the area formula to find out more information.  The area of a circle is $A=\Pi(r)^{2}$  Plug in the area you just found to find the radius.

Note: working backwards using the area formula for a circle is difficult, because using pi will always result in icky decimals.  If you glance at your answer choices, all of them are whole numbers.  You can estimate pi as 3 instead of 3.14 in order to keep this problem as easy as possible.

48 = 3r²
16 =
4 = r

You now have the radius for the big circle.  Now look back up at the diagram.  The radius for the big circle is also the diameter for the little circle!  If the diameter of the little circle is 4, the radius will be half of that.  Once you know that the radius of the little circle is 2, you are ready to find the area!

$A=\Pi(r)^{2}$
A = 3 × 2²

A = 3 × 4
A = 12

Loop Back:  You found the area of the small circle, so you are ready to look at your answer choices.

(A) 8
(B) 10
(C) 12
(D) 14
(E) It cannot be determined from the information given.

On sat.collegeboard.org, 56% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Sequence Problems

## Arithmetic: Sequence Problems

Always read the question carefully and identify the bottom line.  Then assess your options and use the most efficient method to attack the problem.  When you have an answer, loop back to make sure that you solved for the bottom line.

8, a, 14, b, 20, …
The first term of the sequence above is 8. Which of the following could be the formula for finding the nth term of this sequence for any positive integer n?

Bottom Line: You want a formula to describe this number sequence.

Assess your Options:  You could try to write a formula, but you will have a hard time doing that because you do not know the second and fourth terms in your pattern.  You also do not need to find numbers for the variables a and b in order to solve this problem.  Instead, use the answer choices to help you find an answer.

Attack the Problem: The first thing to do is realize that n is not a variable that you have to find algebraically; the nth term just describes the number of that term in the sequence, like the first, second, third, fourth, or fifth.  Therefore:
8,   a,  14,  b,   20, …
1,   2,   3,   4,     5

That means that when you plug in 1 to the formula, you should always get 8, when you plug in 3, you should always get 14, and when you plug in 5, you should always get 20.

(A) 2n + 6
(B) 3n + 5
(C) 5n + 3
(D) 6n + 2
(E) 6n + 5

You could start by plugging in 1 and finding out which of these equals 8, eliminate any that do not, and then try plugging in 3 and then 5 (this method is used on collegeboard.org).  However, just by looking at the numbers (a lot of 2s and 6s and a lot of 3s and 5s) you should be able to tell that a lot of these will equal 8.  To save time, start by plugging in the biggest term you know, the fifth, and see which answer choices equal 20.

(A) 2(5) + 6 = 16
(B) 3(5) + 5 = 20
(C) 5(5) + 3 = 28
(D) 6(5) + 2 = 32
(E) 6(5) + 5 = 35

Note: if you use logic, you do not even have to work out (C), (D), and (E) because the product of the first two numbers is larger than 20 before you even add to them.

Only one answer choice results in the correct 5th term of 20.  You don’t need to check any other numbers!

Loop back:  You found the only formula that will work for every number in the sequence, so select that answer.

(A) 2n + 6
(B) 3n + 5
(C) 5n + 3
(D) 6n + 2
(E) 6n + 5

On sat.collegeboard.org, 43% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Coordinate Geometry

## Geometry: Coordinate Geometry

Approach every question the same way to minimize mistakes.  Start by reading the question carefully and identifying the bottom line.  Then assess your options and choose the most efficient method to attack the problem.  When you have an answer, loop back to verify that it matches your bottom line.

In the xy-plane, line l passes through the points (0, 0) and (2, 5). Line m is perpendicular to line l. What is the slope of line m?

Bottom line: slope m = ?

Assess your Options:  You could draw out a graph and solve this visually, but that is a waste of time if you know the formula to find the slope of a line.

Attack the Problem:  You are given the most information about line l, so start with that line.  You should have the formula for slope memorized:

$slope=\frac{rise}{run}=\frac{\Delta y}{\Delta x}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

It is easiest just to think about slope as the change in y-values over the change in x-values.  If you look up at the original points that you have been given, from zero the y-values go up to 5 and the x-values go up to 2.  You now have 5 over 2.

The slope of line l is $\frac{5}{2}$.

At this point, some students will think they are finished and select answer (D).  However, your bottom line was the slope of line m!  The problem tells you that line m is perpendicular to line l.  In order to find a perpendicular line, you must take the opposite reciprocal of the first line; in essence you must flip the sign (negative or positive) and the numbers (a fraction or whole number).

The slope of line m is $-\frac{2}{5}$.

(A) $-\frac{5}{2}$
(B) $-\frac{2}{5}$
(C) $\frac{2}{5}$
(D) $\frac{5}{2}$
(E) 5

On sat.collegeboard.org, 47% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Equations

## Algebra: Equations

Read the following SAT test question and then select the correct answer.

Always read the question carefully and identify the bottom line so that you do not waste time finding something unrelated to the question.  Assess your options for solving the problem and choose the most efficient method to attack the problem.  When you have an answer, take a second or two to loop back and make sure that your answer matches the bottom line.

If a, b, and c are numbers such that $\frac{a}{b}=3$ and $\frac{b}{c}=7$, then $\frac{a+b}{b+c}$ is equal to which of the following?

Bottom line:  $\frac{a+b}{b+c}$

Assess your Options:  There are two ways that you can solve this equation, and both will arrive at the correct answer.  You can solve it algebraically by substituting information into the equation, or you can pick your own numbers for the variables.  Choose the method that is easier and faster for you.

Attack the problem:  If you are going to solve a problem algebraically, always look for ways to simplify the problem that you are given.  In this case, you will want to get rid of unnecessary fractions.  Look at the first piece of information that you are given.  If a divided by b is 3, you can get rid of the fraction by multiplying each side of the equation by b.

Now you have a = 3b.

Look at the numerator (the top part of the fraction) of your bottom line.  You can now make sure that there is only one variable in this portion of the equation.   Substitute 3b for a.  Now you have 3b + b, which will simplify to 4b

Here are the steps you just completed:

$\frac{a+b}{b+c}=\frac{3b+b}{b+c}=\frac{4b}{b+c}$

Look at the denominator of your equation.  How can you simplify b + c?  You might be tempted to substitute 7c for b, but remember your goal is to get to a number without a variable.  If you have the same variable in the top and bottom, the two variables cancel. Therefore, you need to find what c is equal to in terms of b

When you are given the information that b divided by c is 7, then you know that c divided by b is 1 over 7.  You flip both equations.  Solve for c by multiplying both sides of the equation by b.

$\frac{b}{c}=7$ so  $\frac{c}{b}=\frac{1}{7}$ so $c =\frac{1}{7}b$

Plug this information into your bottom line equation and combine like terms.

$\frac{4b}{b+c}=\frac{4b}{b+\frac{1}{7}b}=\frac{4b}{\frac{8}{7}b}$

A fraction over a fraction is ugly, but remember that dividing by a fraction is the same thing as multiplying by the reciprocal of that fraction.  In other words:

$\frac{4b}{\frac{8}{7}b}=4b(\frac{7}{8b})=4(\frac{7}{8})=\frac{28}{8}=\frac{7}{2}$

Notice that the variable b moves to the bottom of the second fraction and cancels out.  You solved the equation!

Alternatively:  If you dislike algebra, use the strategy of picking numbers to solve this problem.  You want to get rid of ugly fractions, and the best way to do that is to put a number over 1.  You cannot just put b = 1 because b affects two different equations and you might end up with numbers that are difficult to use in your other equation.   However, c is on the bottom of a fraction in one equation.  Pick c = 1.  Plug 1 into the second piece of information with c and solve for b.

$\frac{b}{c}=7$ so $\frac{b}{1}=7$ so b = 7.

The variable b must equal 7. Now plug that into the first piece of information that you were given.  If b is 7, then a must equal 21.

$\frac{a}{b}=3$ so $\frac{a}{7}=3$ so a = 21.

Now that you have numbers for a, b, and c, plug those into your bottom line equation:

$\frac{a+b}{b+c}=\frac{21+7}{7+1}=\frac{28}{8}=\frac{7}{2}$

Bottom Line:  As soon as you have a value to represent your bottom line, look down at your answer choices.

(A) $\frac{7}{2}$
(B) $\frac{7}{8}$
(C) $\frac{3}{7}$
(D) $\frac{1}{7}$
(E) 21

On sat.collegeboard.org, 42% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Sets

## Arithmetic: Sets

Read the following SAT test question and then select the correct answer.

Approach each math question on the SAT the same way.  Read the question carefully to be sure you take into account all of the information as you solve it, and be sure to identify and note the bottom line.  Assess your options for solving the problem, and then choose the most efficient method to attack the problem.  Never forget to loop back and make sure that your final answer solves for the bottom line, the question that you were asked.

If S is the set of positive integers that are multiples of 7, and if T is the set of positive integers that are multiples of 13, how many integers are in the intersection of S and T?

Bottom Line: # of intersections = ?

Assess your Options:  When you have a question that asks about number properties, ignore your answer choices!  If you look down and see a 0, you could think to yourself that both 7 and 13 are prime, so they have nothing in common.  Are you looking for factors?  No!  You are looking for multiples.  Think through all of the information that you are given before looking at the answer choices.

Attack the Problem:  A set is just a collection of data.  You are given two different sets and asked to find the intersections, the data that the two have in common.  The only restriction on both sets is that all of the numbers must be positive.

Now think about what multiples are.  Multiples are the product of a number and an integer.  So Set S contains 7, 14, 21, 28… and continues in this manner into infinity.  Set T contains 13, 26, 39, 52… and continues in this manner into infinity.

If you keep listing numbers in each set, it will take you forever to find the answer to this problem.  Instead, think logically about where you know you must have multiples that match.  For example, if you multiply 7 times 13, you will find a number that belongs in both sets.  If you multiply 14 times 13, you will find another intersection.  Notice that you can keep doing this because you will never reach infinity.  The answer to this problem is that there are an infinite number of intersections between S and T.

Loop Back:  You found your bottom line, so look down and see which answer choice it matches.

(A) None
(B) One
(C) Seven
(D) Thirteen
(E) More than thirteen

On sat.collegeboard.org, 40% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Coordinate Geometry

## Geometry: Coordinate Geometry

Read the following SAT test question and then select the correct answer.

Always be sure to read the question carefully and make a note of the bottom line.  Then assess your options and choose the most efficient method to attack the problem.  When you have an answer, loop back to verify that it matches the bottom line.

In the xy-plane, line l passes through the points (a, 0) and (0, 2a), where a > 1.  What is the slope of line l?

Bottom Line: slope of l = ?

Assess Your Options: You could select a number larger than 1, plug it in for the variable a, and then work the problem.  However, if you peek down at the answer choices, notice that some have a variable still in the problem.  It will take you longer to plug in a number than to work the problem using the variables.

Attack the Problem: Your bottom line is a slope, so use the formula for the slope of a line. The formula for slope of a line is:

$\frac{rise}{run}\: or\: \frac{\Delta y}{\Delta x}$

To find the change in y coordinates, subtract the first y-value from the second y-value.  Do the same with the x values:

$\frac{2a-0}{0-a}=\frac{2a}{-a}=-2$

The variable will cancel when you simplify the problem.  Your answer is -2.

Loop Back:  You found the slope of the line, so you are ready to look down at your answer choices.

(A) -2
(B)$-\frac{1}{2}$
(C) 2
(D) -2a
(E) 2a

On sat.collegeboard.org, 42% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Triangles

## Geometry: Triangles

Read the following SAT test question and then select the correct answer.

Always read the question carefully so that you don’t misapply any information.  Identify the bottom line and assess your options for solving the problem.  Choose the most efficient method to attack the problem.  When you think you have the answer, loop back to make sure that it matches the bottom line.

If triangle ABC above is congruent to triangle DEF (not shown), which of the following must be the length of one side of triangle DEF?

Bottom Line: side of DEF = ?

Assess your Options: Many students go straight to the Pythagorean Theorem whenever they see a right triangle.  This formula, a² + b² = c², will not help you in this case because you do not know a or b.  Instead, use your knowledge of special triangles to solve this problem.

Attack the problem:  As soon as you see that this is a 30° – 60° – 90° triangle, you should think about the sides that relate to this special triangle.  Those sides, which you should have memorized, are x - x√3 – 2x.  Remember that the longest side has to be across from the biggest angle, the 90° angle.  That is your 2x.  This triangle has a 12 in that position.  Solve for x.

2x = 12
x = 6

Now you know that the side across from the 30° angle, AB, must be 6.  Label it.  Look at the side across from the 60° angle.  AC must be x√3.  You know that x = 6, so this side must be 6√3.  Label it.  You now know all the sides of triangle ABC:

x - x√3 – 2x
6 -6√3 – 12

Your bottom line is a side on triangle DEF, not on triangle ABC.  However, the problem tells you that ABC is congruent to DEF.  Congruent triangles have the same shape and size; they are basically the same triangle with different labels.  That means that the side lengths from triangle DEF will match the lengths you already found for ABC.

Loop back: You took into account all of the information that you were given and solved for your bottom line.  Look down at your answer choices.  One of the three side lengths you found will be there.

(A) 18
(B) 24
(C) 3√6
(D) 6√3
(E) It cannot be determined from the information given.

On sat.collegeboard.org, 38% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Pattern Problems

How important is history?  Should people take more responsibility for solving problems that affect their communities or the nation in general?  That second question is a previous SAT essay question.  Before you answer it, read here and here about how the people of Mali reacted to a threat against manuscripts as old as the 13th century.  What important themes do you see in these articles that would be easy to write about as a current event example?  Write down details and facts that could help to support an opinion on a broad topic.

## Arithmetic: Pattern Problems

Read the following SAT test question and then select the correct answer.

Work each math problem by reading the question carefully and identifying the bottom line.  Then assess your options and choose the most efficient method to attack the problem.  When you have an answer, loop back to make sure that you finished all of the steps to match your bottom line.

If it is now 4:00 p.m. Saturday, in 253 hours from now, what time and day will it be? (Assume no daylight saving time changes in the period.)

Bottom Line: 253 hours from now = ?

Assess your Options: You could try to count the hours elapsed from the answer choices, but that will be time consuming if you don’t guess the right one first and end up working the problem five times.  Instead, use logic to methodically work through this problem.

Attack the Problem:  You are given 253 hours, but you know that there will also be a change in the day.  There are only 24 hours in a day.  Find out how many days there are in this time period by dividing.  Now, this problem is similar to the pattern problems in your Knowsys book.  You want to know how many days and hours have passed.  Instead of dividing with your calculator, use long division to find out how many days pass and how many hours remain. 253 ÷ 24 is 10 with a remainder of 13.  That means that there are 10 days and 13 hours that pass.

Continue to think about this logically.  If it is 4:00 p.m. on Saturday and a week passes, it will be the same day.  So 7 days will get you back to the same place.  Then you have 3 of your 10 days still to go.  Count 3 days from Saturday, (Sun, Mon, Tues), and you are now at 4:00 p.m. on Tuesday.

That accounts for all of the days that have passed, but you still have 13 hours.  If you add 12 hours to 4:00 p.m., it becomes 4:00 a.m. on the next day, Wednesday.  Add 1 more hour and you get 5:00 a.m. on Wednesday.

Loop Back:  You accounted for all of the 253 hours by counting out 10 days and 13 hours.  Look down at your answer choices.

(A) 5:00 a.m. Saturday
(B) 1:00 a.m. Sunday
(C) 5:00 p.m. Tuesday
(D) 1:00 a.m. Wednesday
(E) 5:00 a.m. Wednesday

On sat.collegeboard.org, 49% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Functions

## Algebra: Functions

Read the following SAT test question and then select the correct answer.

Always use the same process for math problems on the SAT.  Read carefully and make a note of the bottom line.  Then assess your options and choose the most efficient method to attack the problem.  When you have an answer, loop back to be sure it matches your bottom line.

If the function f is defined by , where 0 < a < b < c, for which of the following values of x is f undefined?

I. a
II. b
III. c

Bottom Line: For which value(s) of x is f undefined?

Assess your Options: You could pick numbers, but that will get confusing with three variables.  You could just start plugging in the variables a, b, and c for x and then simplify the function, but you will end up wasting time.  Time is precious on the SAT!  Start with the information that you are given and think about it logically.

Attack the Problem:  Always think about the information that you are given before you jump into the problem.  The inequality that you are given simply tells you that all of your variables are positive numbers.  A function or a fraction is undefined whenever it is divided by zero because you cannot divide by zero.

Think about it logically:  do you care what is on the top of the fraction?  No!  Focus on the bottom of the fraction.  How can you make x c = 0?  The variable that you are changing in this problem is x.  If you set x = to c, then cc = 0.

Note:  You do not know whether a or b is equal to c, so you cannot assume that ac or bc would equal 0.  If you plug those variables in, you still have a lot of variables on the bottom!

Loop Back:  You found the only answer that will work out of the three that you were given.  Look down at your answer choices.

(A) None
(B) I only
(C) III only
(D) I and II only
(E) I, II, and III

On sat.collegeboard.org, 53% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Probability

## Data Analysis: Probability

Read the following SAT test question and then select the correct answer.

Always read each question carefully and make a note of the bottom line (what you are trying to find).  Assess your options to find the best strategic method and use that method to attack the problem.  When you have an answer, loop back to verify that the answer matches the bottom line.

A jar contains only red marbles and green marbles. If a marble is selected at random from the jar, the probability that a red marble will be selected is $\frac{2}{3}$. If there are 36  green marbles in the jar, how many red marbles are there in the jar?

Bottom Line:  You want to know how many red marbles there are, so use r to represent red and just write r = ?

Assess your Options:  You could try to work backwards from the answer choices to find a number that, when combined with 36, makes the right fraction.  That won’t be any faster than just solving the problem.  Use the probability formula.

Attack the Problem:  The probability formula is:

$\frac{relevant\: outcomes}{total\: outcomes}$

In this problem, you know the red marbles are the relevant outcome, while the red and green marbles together are the total (all that is in the jar).  Use g for the green marbles.  There are 36 green marbles.

$\frac{r}{g + r}=\frac{r}{36+r}$

You have already been given the probability that a red marble will be selected.  Set the formula that you created equal to the probability that you were given.  Then solve for r with cross-multiplication.

$\frac{r}{36+r}=\frac{2}{3}$
3r = 2(36 + r)
3r = 72 + 2r
r = 72

Loop Back:  You solved for your bottom line, so you are ready to look at the answer choices.

(A) 18
(B) 24
(C) 54
(D) 72
(E) 108

On sat.collegeboard.org, 47% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Probability

## Data Analysis: Probability

Read the following SAT test question and then select the correct answer.

Read each question carefully to avoid making any mistakes. Identify the bottom line (what the question is asking) and assess your options for reaching it by asking yourself “What could I do?” and “What should I do?” Choose the most efficient method to attack the problem and find an answer. Last, loop back to make sure that your answer addresses the bottom line.

If a number is chosen at random from the set {-10, -5, 0, 5, 10}, what is the probability that it is a member of the solution set of both 3x – 2 <10 and x + 2?

Bottom Line: Prb = ?

Assess Your Options: You cannot solve for a probability until you know whether each number in the set meets the requirements that you are given. You could plug numbers from the set into each inequality and see if they work, but it is much faster to simplify the inequalities before you begin working with them.

Attack the Problem: Simplify the inequalities by solving both for x.

3x – 2 < 10
3x < 12
x < 4

x + 2 > -8
x > -10

You now know that x must be less than 4, but greater than -10. The question asked you to find a number that fits both of these solution sets. Look at the original set that you were given. The only two answers that are between -10 and 4 are -5 and 0 (-10 does not work because it cannot be equal to negative -10; it has to be greater than -10). You found 2 numbers out of 5 that you were given that work. To write this as a probability, you must set the number of relevant outcomes over the number of total possible outcomes. Your answer is .

Loop Back: You found a probability matching the restrictions you were given. Look down at your answer choices.

(A) 0

(B)

(C)

(D)

(E)

On sat.collegeboard.org, 50% of the responses were correct.

For more help with SAT math, visitwww.myknowsys.com!

# Rates

## Arithmetic: Rates

Read the following SAT test question and then select the correct answer.

Using the same method with every math problem to minimize mistakes.  Read the question carefully.  Identify the bottom line and assess your options for finding it.  Choose the most efficient method to attack the problem.  Once you have an answer, loop back to make sure it addresses the bottom line.

A woman drove to work at an average speed of 40 miles per hour and returned along the same route at 30 miles per hour. If her total traveling time was 1 hour, what was the total number of miles in the round trip?

Bottom Line: Td = ?  (Total distance)

Assess your Options: Remember that speed is really a rate.  There are 4 key rate scenarios: separation, overtake, round trip, and meet in between--this one is a round trip.  You can figure all of these out by using the distance formula (rate × time = distance), but it can be difficult to keep track of which scenario you have unless you treat all of them the same way.  Knowsys recommends that you use a chart to quickly organize your thoughts so that you can be sure that you accounted for all of the information in the problem. (Spoiler: many students make mistakes on these types of problems!  You do not get any extra points for ignoring the chart, so use it!)

Attack the Problem:  Here is the chart that you should use with all rate scenarios:

 1 2 Total Rate Time Distance

Start filling in the information that you know.  The first trip was at a rate of 40 miles per hour and the second trip was at a rate of 30 miles per hour.  The total time was 1 hour.

 Trip 1 Trip 2 Total Rate 40 30 Time 1 Distance

If you don’t know the time for the first trip, choose a variable to represent the unknown.  Put an “x” in that box.  You know that the time for the trips together must total 1 hour (x + ? = 1).  Therefore, the second trip is equal to 1 minus x

 Trip 1 Trip 2 Total Rate 40 30 Time x 1 – x 1 Distance

You already know that rate × time = distance, so multiply the two columns representing the trips.

 Trip 1 Trip 2 Total Rate 40 30 Time x 1 – x 1 Distance 40x 30(1 – x)

Before you start worrying about the total number of miles, remember that this person is using the same route each time.  That means the distance traveled each time is an equal length.  Set the distances equal to each other.

40x = 30(1 – x)
40x = 30 – 30x
70x = 30
$x=\frac{3}{7}$

If you know x, you can now find a number value for each part of your chart.  What was the bottom line?  You need to find the total number of hours.  You could plug x into both distances and add them up; however, there is an even faster method.  Take the first distance and multiply it by 2.  (Remember that the distances are the same.)

$2\times40\times \frac{3}{7}=Total\; distance$

$\frac{240}{7}=Total\; distance$

$34\frac{2}{7}=Total\; distance$

(A) 30
(B) $30\frac{1}{7}$
(C) $34\frac{2}{7}$
(D) 35
(E) 40

On sat.collegeboard.org, 38% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Lines

## Geometry: Lines and Angles

Read the following SAT test question and then select the correct answer.

Work all math problems the same way.  Read the problem carefully, identify the bottom line, and assess your options for solving the question.  Choose the most efficient method to attack the problem, and loop back to make sure that your answer matches the bottom line.

Four distinct lines lie in a plane, and exactly two of them are parallel. Which of the following could be the number of points where at least two of the lines intersect?

I. Three
II. Four
III. Five

Bottom Line: # intersections

Assess your Options: You could just start drawing any combination that you can think of, but try to think of the particular answer choices that you are given.  Examine options I, II, and III independently.

Attack the Problem: Think first about option I.  Go ahead and draw out two horizontal parallel lines.  The other lines cannot be parallel to these lines or to each other because the problem says that there are "exactly" two parallel lines.  How could you create three intersections?  One way is to make those next two lines into an “X” and put the middle of the X on one of the preexisting parallel lines.  If you extend the legs of the X out far enough (remember these are lines, not line segments), they will cross the other parallel line in two places.  You have created an image with 3 intersections while following all of the stipulations.

Now turn your attention to option II.  If you have two parallel lines, those lines will never cross.  You know that the other two lines cannot be parallel because this situation has "exactly two" parallel lines.  If you have two lines that are not parallel, no matter how close their slopes are, eventually they must cross.  That is why representing these lines as an X is a good idea.  Any combination of the lines other than the above combination will result in 1 intersection between the lines that are not parallel and 4 intersections where these lines cross the parallel lines for a total of 5 intersections.  There is no way to get only 4 intersections.

Look at option III.  You already thought about it conceptually while examining option II, but you can prove this possibility by drawing a picture.  Draw two horizontal parallel lines.  You can keep the other two lines as an X, but move the middle of the X off of the parallel lines.  Whether the middle of the X is between the parallel lines, above them, or below them, you will now have 5 intersections while following all of the stipulations. (This also confirms your thinking about option II.)

Loop Back:  You examined each option separately, so you are ready to look down at your answer choices.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

On sat.collegeboard.org, 29% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Writing Equations

New things can be exciting, but also scary.  Several years ago, Y2K (the year 2000) frightened many people.  Now people are worried about the end of the Mayan calendar on Dec 21, 2012.  Take a look at this article to see how people are reacting to rumors about the end of the world.  How could you use this current event on an SAT essay?  It would easily relate to questions about whether the world is getting better, how people understand themselves and those in authority, feelings and rationality, and many other topics.  Make sure to pick out specific details to mention in your essay if you choose this as one of your current event examples!

## Algebra: Writing Equations

Read the following SAT test question and then select the correct answer.

Always read math problems carefully so that you don’t miss an important piece of information.  Identify the bottom line, and assess your options for reaching it.  Choose the most efficient method to attack the problem.  Many problems have multiple steps, so be sure to loop back and make sure that you solved for the bottom line.

The stopping distance of a car is the number of feet that the car travels after the driver starts applying the brakes. The stopping distance of a certain car is directly proportional to the square of the speed of the car, in miles per hour, at the time the brakes are first applied. If the car’s stopping distance for an initial speed of 20 miles per hour is 17 feet, what is its stopping distance for an initial speed of 40 miles per hour?

Bottom Line: d (distance) = ?

Assess your Options:  You have to decide how to use the information in this problem; in other words, you need to write an equation.  Plugging in the answer choices will take a lot of guess work.  Instead, carefully work through each piece of information that you are given.

Attack the Problem:  You have probably worked with distance, rate, and time before.  One formula that is often used in Knowsys classes is distance = rate × time.  This problem is asking you to write a similar equation.  The problem says: “The stopping distance of a certain car is directly proportional to the square of the speed of the car, in miles per hour, at the time the brakes are first applied.”  In other words, you know that distance is (is means equals in math) directly proportional to something.  Now pay particular attention to the part that says “directly proportional.  This phrase just means that when the distance gets bigger, so does the other side of your equation.  For that to happen, you need another constant number on the other side of the equation.  Your distance is equal to some constant number times speed squared.  Your formula should look like this:

distance = constant number × speed²

Now that you have written an equation to show what is happening in this problem, you are ready to look at the next piece of information.  Plug in the first situation in which an initial speed of 20 miles per hour results in a distance of 17 feet.

d = c × s²
17 = c × 20²

Now you can solve for c by isolating the variable.  Use your calculator when it will be faster than mental math.

17 = c × 400  (divide each side by 400)
.0425 = c

Now you have enough information to find your bottom line. Plug in the second situation in which the car is going 40 miles per hour and solve for the distance.

d = c × s²
d = .0425 × 40²
d = .0425 × 1600
d = 68

Loop Back:  You solved for the stopping distance of a car traveling 40 mph, just as the question asked.  You are ready to look at your answer choices.

(A)  34 feet
(B)  51 feet
(C)  60 feet
(D)  68 feet
(E)  85 feet

On sat.collegeboard.org, 39% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Group Problems

## Arithmetic: Group Problems

Read the following SAT test question and then select the correct answer.

Work all math problems the same way so that you can approach even the most difficult problems with confidence.  Start by reading the question carefully.  Many problems have several steps, so you must identify the bottom line: what is the question asking?  Assess your options and choose the most efficient method to attack the problem.  Finally, loop back to make sure that your answer matches the bottom line.

In a community of 416 people, each person owns a dog or a cat or both. If there are 316 dog owners and 280 cat owners, how many of the dog owners own no cat?

Bottom Line: just dogs = ?
When you get to the step where you look at the answers, notice that (E) comes from not reading carefully.  Yes, there are 316 total dog owners, but some of them also own cats.  You must find how many own only dogs.

Assess your Options:  You could try to work backwards using the answer choices, but trying to think about the steps of a problem backwards often leads to mistakes.  You could also realize that this is a problem involving two overlapping groups and draw a Venn Diagram.  Forget those methods because the fastest method is to use the Group Formula.  Take a moment now to memorize this formula if you have not already done so: Total = Group 1 + Group 2 + Neither – Both.

Attack the Problem:  Plug all the information that you know into the formula.  How many total people are there? 416.  Then there are your two groups: Those who own dogs and those who own cats.  Plug in the numbers 316 and 280 to represent these groups.   Now, the problem tells you that “each person owns a dog or a cat or both,” so how many people own neither animal?  Zero.  The only thing that you are not given in the problem is how many people own both a dog and a cat.  Your formula should now look like this:

Total = Group 1 + Group 2 + Neither – Both
416 = 316 + 280 + 0 – B

That B represents the unknown Both, but you can now solve for it because it is the only variable left in your equation.  Start by simplifying the problem.

416 = 316 + 280 + 0 – B
416 = 596 – B   (add B to each side to make it positive)
416 + B = 596   (subtract 416 from each side)
B = 180

You have finished one step, but you have not yet reached your bottom line!  Do not look at the answer choices yet or you will be tempted to pick a wrong answer!

You just solved for the number of people who own both a cat and a dog.  How do you find the number of people who own just a dog?  Take the number who own both and subtract it from the total number of dog owners.  Remember that the total number of dog owners was given in the problem as 316.

316 – 180 = 136

(A) 36
(B) 100
(C) 136
(D) 180
(E) 316

On sat.collegeboard.org, 46% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Functions

How do medical breakthroughs happen?  How do you feel about animal testing?  A few dogs that were once paralyzed are now walking again with the aid of some cells from healthy dogs.  Scientists recognize that they are not ready to apply their findings to humans with spinal cord injuries, but they are hopeful about the future.  This development would make a great current event example for your SAT essay.  If you choose to use it as a current event, take notes detailing the facts in this article and how you could apply them to a wide variety of topics.

## Algebra: Functions

Read the following SAT test question and then select the correct answer.

Take the time to read math questions carefully because this will save you from wasting time as you solve the problem.  Start by identifying the bottom line and assessing your options for reaching it.  Choose the most efficient method to attack the problem.  When you think you have an answer, loop back to make sure that it matches the bottom line.
• f(2n) = 2f(n) for all integers n
• f(4) = 4
If f is a function defined for all positive integers n, and f satisfies the two conditions above, which of the following could be the definition of f?

Bottom line: f(n) = ?

Assess your options:  You could start by plugging in f(4) to see which of your answer choices results in the number 4, then check any that do against the first condition.  This is the method recommended by collegeboard.com.  However, this requires multiple steps as you compare each answer choice to both conditions.  Instead, take a moment to think logically about the two conditions that you are given.

Attack the problem:  Start with the second condition because it is already in a format that is easy to use.  If f(4) = 4 and you can use the same variable to represent numbers that are the same,  that is the same as saying that f(n) = n.  Now look at the first condition and think about it logically.  If you multiply the variable within the function by 2, that gives you the same number as multiplying the result of the function by 2.  In order for those two numbers to be the same, the final result of the function has to match the number that is plugged into the function.  In other words, f(n) = n.  Both conditions give you the same definition of the function.

Loop Back:  You found a simple way to define both of the conditions for f(n), so look down at your answer choices.

(A)  f(n) = n - 2
(B)  f(n) = n
(C)  f(n) = 2n
(D)  f(n) = 4
(E)  f(n) = 2n – 4

On sat.collegeboard.org, 46% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Triangles

## Geometry: Triangles

Read the following SAT test question and then select the correct answer.

You should start by reading the problem carefully and identifying your bottom line.  Then assess your options and choose the most efficient method to attack the problem.  Finally, loop back to make sure that the answer you found matches the bottom line that you set out to find.

Which of the following CANNOT be the lengths of the sides of a triangle?

Bottom Line: The word “CANNOT” tells you that you are looking for something that is not true.

Assess your Options:  Problems that ask you to find what cannot be true are often impossible to predict; you will have to look at the answer choices to determine whether they work.  Go ahead and look at the answers.

(A) 1, 1, 1
(B) 1, 2, 4
(C) 1, 75, 75
(D) 2, 3, 4
(E) 5, 6, 8

You could try to use logic and your experience with triangles to eliminate some choices.  For example, answer choice A is not the answer because you know there is such a thing as an equilateral triangle and the sides 1, 1, and 1 would create that kind of triangle.  However, after that point you would probably just be guessing.  If you know the Triangle Inequality Theorem, you can systematically check each answer choice.

Attack the problem:  When you are dealing with three sides of a triangle and you do not know that the triangle is a right triangle, you should always think of the Triangle Inequality Theorem.  This theorem states that for any triangle, side x is less than the sum and greater than the difference of the other two sides.  In other words, each side of the triangle must be less than the other two sides added together and greater than the difference of the other two sides.  If you have a triangle with sides x, y and z, you would write the theorem this way:
|yz| < x < y + z. For the subtraction part you can use absolute value or just always do the bigger side minus the smaller side; the result will be the same. The easiest way to think about the theorem is this: for any triangle, other sides subtracted < one side < other sides added.

Start by checking (E).  It has sides 5, 6, and 8.  Plug these sides into your formula by using the first side, 5, as your x.  Make sure that 5 is greater than the difference of the other two sides, but smaller than the other two sides added together.

|y – z| < x < y + z
8 - 6 < 5 < 6 + 8
2 < 5 < 14
This is true!  (E) works as a triangle.

Now check (D). It has sides 2, 3, and 4.  Plug those numbers into the Inequality Theorem.
|y – z| < x < y + z
4 – 3 < 2 < 3 + 4
1 < 2 < 7
This is true!  (D) works as a triangle.

Now check (C).  It has sides 1, 75, and 75.  Plug those numbers into the Inequality Theorem.
|y – z| < x < y + z
75 – 75 < 1 < 75 + 75
0 < 1  < 150
This is true!  (C) works as a triangle.

Now check (B).  It has sides 1, 2, and 4.  Plug those numbers into the Inequality Theorem.
|y – z| < x < y + z
4 – 2 < 1 < 2 + 4
2 < 1 < 6.
Is 2 less than 1?  No!  This is false.  You cannot have a triangle with these three side lengths.

Loop Back:  Your bottom line was to find an answer choice that cannot be a triangle, so you are finished!

On sat.collegeboard.org, 49% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

Read the following SAT test question and then select the correct answer.

Math questions should always be read carefully.  You will also avoid making errors by identifying the bottom line and assessing your options for solving the question.  Choose the most efficient method to attack the problem.  When you have finished, loop back to be sure that even if there were multiple steps, you reached the bottom line.

The length of a rectangle is increased by 20%, and the width of the rectangle is increased by 30%. By what percentage will the area of the rectangle be increased?

Bottom Line: % change = ?

Assess your Options:  You could work this problem without picking any numbers; however, picking easy numbers will allow you to think about the problem in a more concrete way and avoid errors.

Attack the Problem:  One of the easiest numbers to work with is one.  Think of your original rectangle as having a length of one and a width of one.  The formula for area of a rectangle is length times width.  If L × W = A, for your first rectangle you have 1 × 1 = 1.   The area of the original rectangle is one.

Then think about the changes that occur to that rectangle.  The length increases by 20%.  In order to find 20% of 1, all you have to do is move the decimal over twice to .2.  The new length is 1.2.  Use the same method to find the new width, and an increase of 30% becomes 1.3.  The area of the rectangle after the change is 1.2 × 1.3 = 1.56

The formula for percent change would require you to find the difference between these two areas and divide that by the original number.  You use the same formula whether you are looking for an increase or a decrease.  Notice that your original number is one, so dividing by one will not change your answer.  All you need to do is find the difference between the areas: 1.56 – 1 = .56.  What is .56 as a percent?  Your answer is 56%.

Loop Back:  You found the percent change, which was your bottom line.  Look down at your answer choices.

(A) 25%
(B) 36%
(C) 50%
(D) 56%
(E) 60%

On sat.collegeboard.org, 34% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

# Coordinate Geometry

## Geometry:  Coordinate Geometry

Read the following SAT test question and then select the correct answer.

Use the same method for all the math questions on the SAT.  First, read the question carefully to avoid making mistakes.  Identify the bottom line and assess your options for reaching it.  Next, choose an efficient method to attack the problem.  Finally, loop back to make sure that your answer addresses the bottom line.  Many problems have multiple steps.

If the graph of the function f in the xy-plane contains the points (0, -9), (1, -4), and (3, 0), which of the following CANNOT be true?

Bottom Line:  You are looking for something false.

Assess your Options:  You could try drawing an xy-plane and graphing the points to help you visualize the question, but your graph may be inaccurate without graph paper.  Instead, try to find the relationship between the three points.

Attack the problem:  To find the relationship between these points, you will need to find the slope of the line between each point.  The formula for slope is:

Then check the slope of the line between (1, -4) and (3, 0):

$\frac{0--4}{3-1}= \frac{0+4}{2}=2$

The function in this problem has a very steep slope between the first two points, but becomes less steep between the second two.  This is a “which of the following” question, so start with answer (E) as you work through your answer choices.

(A) The graph of f has a maximum value.
(B) y ≤ 0 for all points (x, y) on the graph of f.
(C) The graph of f is symmetric with respect to a line.
(D) The graph of f is a line.
(E) The graph of f is a parabola.

(E) The function could be a downward facing parabola if it continues to the right.  You are only given three points, but there could be many more points on this function.

(D)  In geometry, a line is always straight, without any curves.  Notice that there are different slopes connecting the three points.  You cannot draw one straight line through all three of these points, so this choice cannot be true.

Loop Back:  Your goal was to find an answer choice that was false.  You did so, so you are finished!  If you have extra time, you can check the other answer choices and see that they are all possible, depending on how you draw the rest of the function.  (E), (C), (B), and (A) could all describe a downward facing parabola with the equation y = -(x – 3)².

The correct answer choice is (D).

On sat.collegeboard.org, 42% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!


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# Ordinary linear regression vs penalized regression: predictive MSE

Does predicting on a test set with an ordinary linear regression model result in a smaller predictive MSE compared to a penalized regression model (LASSO or ridge)?

• if that were the case, why would we have Lasso or Ridge? – user75138 Oct 21 '15 at 1:56
• I was under the impression that using ridge reduces the variance of the estimates – K23 Oct 21 '15 at 2:02
• That is correct, but that will also increase the bias. The goal is to minimize the MSE by adding a little bias and reducing the variability substantially. – user75138 Oct 21 '15 at 2:37

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# Can Iron(III) oxide oxidize elemental iron in a nitrogen environment under room temperature without any catalysis? [closed]

Will there be electron transfer to allow this process to happen: $$\ce{2 Fe(III) + Fe(0) -> 3 Fe(II)}$$

## closed as unclear what you're asking by airhuff, Mithoron, M.A.R. ಠ_ಠ, Todd Minehardt, Mathew MahindaratneJun 3 at 2:19

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• In what situation? Also what speed? – Mithoron Jun 1 at 21:45
• What are the counter ions? – Karsten Theis Jun 2 at 11:02

Whether this qualifies as "without catalysis" is a matter of perspective, but it certainly does happen under ambient conditions. One mechanism of atmospheric rusting involves the $$\ce{Fe(III)}$$ bearing rust reacting with fresh iron to form $$\ce{Fe3O4}$$, which then oxidizes in air to form more rust. See Section 1.6 here.

We can capture the $$\ce{Fe3O4}$$ stage by merely placing the rusty material in water with a little electrolyte, such as rinse water after an acid pickling process, where air cannot easily penetrate. The rust, being converted to $$\ce{Fe3O4}$$, turns black. This emphasizes the fact that the iron reacts with the $$\ce{Fe(III)}$$ in the rust without oxygen, once the rust has first formed.

I've actually seen the $$\ce{Fe3O4}$$ formation happen on steel products during rinsing after pickling, giving black stains or stains bearing a black component. To prevent it, we had to design and maintain the rinsing (and drying) processes to prevent the rusting.

• Thanks for answering this question! I guess I should rephrase my question: if solid Fe2O3 and elemental iron were mixed in powder form, and stored under anhydrous environment, will there be electron transfer from iron(0) Fe2O3, leading to the formation of Fe(II) species? – Z-BO Jun 2 at 1:02

Reactions between solids are generally slow because diffusion only goes so fast. However, if you produced the two materials in nano size to increase surface area and contact, mixed thoroughly (in absence of air) and then perhaps under pressure (density of $$\ce{FeO}$$ is 5.745, density of $$\ce{Fe2O3}$$ is 5.25) with an increase in temperature ($$\ce{FeO}$$ is stable above 575 °C: https://en.wikipedia.org/wiki/Iron%28II%29_oxide) after some time (how long depends on temperature and pressure), you could get complete transformation of $$\ce{Fe2O3 + Fe}$$ to $$\ce{3 FeO}$$.

If magnetite ($$\ce{Fe3O4)}$$ forms as an intermediate product, its low density (5.175) suggests it would not survive.

But on the other hand, the heats of formation do not favor $$\ce{FeO}$$: for $$\ce{FeO}$$, 65 kcal/mol; for $$\ce{Fe2O3}$$, 200 kcal/mol; for $$\ce{Fe3O4}$$, 268 kcal/mol (CRC Handbook).

It's hard to say. The electrochemical data do not take oxygen or solid states into consideration. But if you squeeze the mixture hard enough, I'm sure it will proceed.

Theoretically, yes. Just look at the following redox reaction: \begin{align} \ce{Fe &<=> Fe^2+ + 2e-} &\quad \mathrm{E^\circ} &= \pu{0.447 V} \tag{Oxidation}\\ \ce{Fe^3+ + e- &<=> Fe^2+} &\quad \mathrm{E^\circ} &= \pu{0.771 V} \tag{Reduction}\\ \ce{2Fe^3+ + Fe &<=> 3Fe^2+ } &\quad \mathrm{E_{cell}^\circ} &= \pu{1.218 V} \tag{Total redox} \end{align}

Accordingly, $$\mathrm{E_{cell}^\circ}$$ is a positive value, and therefore, the reaction is spontaneous.

• Thanks! But I thought this is enthalpy calculation, am I wrong? I am not familiar with the notation E0 here. Do you have a reference article about this process? Or probably a book chapter I can read? – Z-BO Jun 2 at 1:09
• The values I used are from sites.chem.colostate.edu/diverdi/all_courses/…. Yet, some certain values you can find in any General Chemistry book. – Mathew Mahindaratne Jun 2 at 4:04


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# How do you simplify 2/3np^2(30p^2+9n^2p-12)?

May 27, 2017

$20 n {p}^{4} + 6 {n}^{3} {p}^{3} - 8 n {p}^{2}$

#### Explanation:

The first step is distributing the $\textcolor{red}{\frac{2}{3}} \textcolor{g r e e n}{n} \textcolor{b l u e}{{p}^{2}}$ across the parentheses.

$\left(\textcolor{red}{\frac{2}{3}} \textcolor{g r e e n}{n} \textcolor{b l u e}{{p}^{2}}\right) \left(\textcolor{red}{30} \textcolor{b l u e}{{p}^{2}}\right) + \left(\textcolor{red}{\frac{2}{3}} \textcolor{g r e e n}{n} \textcolor{b l u e}{{p}^{2}}\right) \left(\textcolor{red}{9} \textcolor{g r e e n}{{n}^{2}} \textcolor{b l u e}{p}\right) - \left(\textcolor{red}{\frac{2}{3}} \textcolor{g r e e n}{n} \textcolor{b l u e}{{p}^{2}}\right) \left(\textcolor{red}{12}\right)$

The colored variables/constants were added to differentiate variables from constants and to show clearly what is being multiplied.

Now we need to individually multiply these terms.

$\textcolor{red}{20} \textcolor{g r e e n}{n} \textcolor{b l u e}{{p}^{4}} + \textcolor{red}{6} \textcolor{g r e e n}{{n}^{3}} \textcolor{b l u e}{{p}^{3}} - \textcolor{red}{8} \textcolor{g r e e n}{n} \textcolor{b l u e}{{p}^{2}}$

And that's all there is to it.


text:
# browseURL

0th

Percentile

##### Load URL into an HTML Browser

Load a given URL into an HTML browser.

Keywords
file
##### Usage
browseURL(url, browser = getOption("browser"),
encodeIfNeeded = FALSE)
##### Arguments
url

a non-empty character string giving the URL to be loaded. Some platforms also accept file paths.

browser

a non-empty character string giving the name of the program to be used as the HTML browser. It should be in the PATH, or a full path specified. Alternatively, an R function to be called to invoke the browser.

Under Windows NULL is also allowed (and is the default), and implies that the file association mechanism will be used.

encodeIfNeeded

Should the URL be encoded by URLencode before passing to the browser? This is not needed (and might be harmful) if the browser program/function itself does encoding, and can be harmful for file:// URLs on some systems and for http:// URLs passed to some CGI applications. Fortunately, most URLs do not need encoding.

##### Details

The default browser is set by option "browser", in turn set by the environment variable R_BROWSER if that is set, otherwise to NULL. To suppress showing URLs altogether, use the value "false".

Some browsers have required : be replaced by | in file paths: others do not accept that. All seem to accept \ as a path separator even though the RFC1738 standard requires /.

To suppress showing URLs altogether, set browser = "false".

##### URL schemes

Which URL schemes are accepted is platform-specific: expect http://, https:// and ftp:// to work, but mailto: may or may not (and if it does may not use the user's preferred email client).

For the file:// scheme the format accepted (if any) can depend on both browser and OS.

• browseURL
##### Examples
library(utils) # NOT RUN { ## for KDE users who want to open files in a new tab options(browser = "kfmclient newTab") browseURL("https://www.r-project.org") # }
Documentation reproduced from package utils, version 3.4.3, License: Part of R 3.4.3

### Community examples

Looks like there are no examples yet.


text:
# Cycles with consecutive odd lengths

• Jie Ma
• Published 2016 in Eur. J. Comb.

#### Abstract

In this paper we prove that there exists an absolute constant c > 0 such that for every natural number k, every non-bipartite 2-connected graph with average degree at least ck contains k cycles with consecutive odd lengths. This implies the existence of the absolute constant d > 0 that every non-bipartite 2-connected graph with minimum degree at least dk contains cycles of all lengths modulo k, thus providing an answer (in a strong form) to a question of Thomassen in [14]. Both results are sharp up to the constant factors.

DOI: 10.1016/j.ejc.2015.09.002

### Cite this paper

@article{Ma2016CyclesWC, title={Cycles with consecutive odd lengths}, author={Jie Ma}, journal={Eur. J. Comb.}, year={2016}, volume={52}, pages={74-78} }


text:
# Contents

## Definition

Given a type $T$, the partial function classifier $T_\bot$ is inductively generated by

• a function $inj:T \to T_\bot$
• a term $\bot:T_\bot$
• a function $\Vee_{n:\mathbb{N}} (-)(n): (\mathbb{N} \to T) \to T$

and the partial order type family $\leq$ is simultaneously inductively generated by

• a family of dependent terms

$a:T, b:T \vdash \pi(a,b):isProp(a \leq b)$

representing that each type $a \leq b$ is a proposition.

• a family of dependent terms

$a:T \vdash \rho(a):a \leq a$

representing the reflexive property of $\leq$.

• a family of dependent terms

$a:T, b:T, c:T \vdash \tau(a, b, c):(a \leq b) \times (b \leq c) \to (a \leq c)$

representing the transitive property of $\leq$.

• a family of dependent terms

$a:T, b:T \vdash \sigma(a, b):(a \leq b) \times (b \leq a) \to (a = b)$

representing the anti-symmetric property of $\leq$.

• a family of dependent terms

$a:P \vdash t(a):\bot \leq a$

representing that $\bot$ is initial in the poset.

• a family of dependent terms

$n:\mathbb{N}, s:\mathbb{N} \to T \vdash i(n, s):s(n) \leq \Vee_{n:\mathbb{N}} s(n)$
• a family of dependent terms

$x:\Sigma, s:\mathbb{N} \to \Sigma \vdash i(s, x):\left(\prod_{n:\mathbb{N}}(s(n) \leq x)\right) \to \left(\Vee_{n:\mathbb{N}} s(n) \leq x\right)$

representing that denumerable/countable joins exist in the poset.


text:
# B is for Brilliant

In each cell of the following $$7\times7$$ square table, write one of the numbers $$1,2,3,4,5,6,7$$ such that the numbers in each row and each column are distinct. What is the greatest possible value of the sum of the numbers written in the blue squares?

×


text:
Asymmetric butterfly velocities in 2-local Hamiltonians

@article{Zhang2019AsymmetricBV,
title={Asymmetric butterfly velocities in 2-local Hamiltonians},
author={Yong-Liang Zhang and Vedika Khemani},
journal={arXiv: Statistical Mechanics},
year={2019}
}
• Published 2019
• Physics, Mathematics
• arXiv: Statistical Mechanics
The speed of information propagation is finite in quantum systems with local interactions. In many such systems, local operators spread ballistically in time and can be characterized by a "butterfly velocity", which can be measured via out-of-time-ordered correlation functions. In general, the butterfly velocity can depend asymmetrically on the direction of information propagation. In this work, we construct a family of simple 2-local Hamiltonians for understanding the asymmetric hydrodynamics… Expand
4 Citations

Figures from this paper

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• Medicine, Physics
• Physical review letters
• 2021
This study rigorously demonstrate that the OTOC shows a polynomial growth over time as long as α>D and the necessary scrambling time over a distance R is larger than t≳R^{[(2 α-2D)/(2α-D+1)]}. Expand
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• Physics
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Polynomial growth of out-of-time-order correlator in arbitrary realistic long-range interacting systems
• Physics
• 2020
In this study, we investigate out-of-time-order correlators (OTOCs) in systems with power-law decaying interactions such as $R^{-\alpha}$, where $R$ is the distance. In such systems, the fastExpand
Entanglement Membrane in Chaotic Many-Body Systems
• Physics
• 2019
In certain analytically-tractable quantum chaotic systems, the calculation of out-of-time-order correlation functions, entanglement entropies after a quench, and other related dynamical observables,Expand

References

SHOWING 1-10 OF 150 REFERENCES
Asymmetric Particle Transport and Light-Cone Dynamics Induced by Anyonic Statistics.
• Physics, Medicine
• Physical review letters
• 2018
The nonequilibrium dynamics of Abelian anyons in a one-dimensional system is studied and it is shown that anyonic statistics induces asymmetric spreading of quantum information, characterized by asymmetric light cones of out-of-time-ordered correlators. Expand
Velocity-dependent Lyapunov exponents in manybody quantum
• semiclassical, and classical chaos, Phys. Rev. B 98, 144304
• 2018
Phys
• Rev. Lett. 111, 127205
• 2013
Phys
• Rev. Lett. 106, 050405
• 2011
Phys
• Rev. B 98, 144304
• 2018
and D
• A. Huse, arXiv: 1802.05589
• 2018
Finite-Size Scaling of Out-of-Time-Ordered Correlators at Late Times.
• Mathematics, Physics
• Physical review letters
• 2019
It is shown that the residual value of the out-of-time-ordered correlator of generic traceless local operators can provide useful insights into the chaotic dynamics of quantum many-body systems. Expand
Information scrambling in chaotic systems with dissipation
• Physics, Mathematics
• Physical Review B
• 2019
Chaotic dynamics in closed local quantum systems scrambles quantum information, which is manifested quantitatively in the decay of the out-of-time-ordered correlators (OTOC) of local operators. HowExpand
Phys
• Rev. B 99, 014303
• 2019
Accessing scrambling using matrix product operators
• Physics, Mathematics
• 2018
Scrambling, a process in which quantum information spreads over a complex quantum system, becoming inaccessible to simple probes, occurs in generic chaotic quantum many-body systems, ranging fromExpand


text:
# Order 1.5 strong SDE integration methods for systems with diagonal additive noise

I'm looking into simple-to-implement and efficient order 1.5 strong SDE integration schemes for my system. My noise is diagonal and additive (possibly time-varying). Thus methods designed for either Itô or Stratonovich are fine.

Are there any explicit, derivative-free, order 1.5 methods that only require a single random variate per dimension per time-step (or that reduce to this in the diagonal additive noise case)? Based on a proof by Rümelin 1982, acccording to Burrage & Burrage 1996 (p. 83):

More general Runge-Kutta type schemes can be constructed but it is possible to show that a strong order of 1.5 cannot be surpassed if just the increments $\Delta W_n$ of the Wiener process are used.

In the same Burrage & Burrage paper, a method claimed to be order 1.5 strong is developed1 that indeed just uses one random variate. In simple tests, the Burrage & Burrage scheme performs slightly better than Euler-Maruyama (order 1.0 strong for additive noise), but I've yet to do anything formal. However, in a later paper (Burrage, et al. 2004), what looks to be the identical method (p. 388) is labelled as order 1.0 strong. Additionally, in my perusal of the recent literature, every explicit stochastic Runge-Kutta method of order 1.5 strong has required at least two random variates, e.g., those in Rößler 2010.

So, what's the deal? Is there such a scheme? And if so, can anyone point me to one? Or are there reasons why I might not be coming across such schemes in the literature (other than that they might be less exciting academically)? Perhaps order 1.5, being the limit for a single random variable, is hard to achieve in practice for real problems? Hence, why methods that claim to be order 1.5 generally use at least two variates in order achieve the desired order of convergence.

1 Let me know and I'll include the equations for the integration scheme if you think it necessary.


text:
Integration by Parts - Maple Programming Help

Integration by Parts

 Description Apply integration by parts to the integral thereby obtaining

Integration by Parts

Enter the integral :

 >
 ${∫}{x}{}{\mathrm{sin}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}$ (1)

Declare $u$:

 > ${x}$
 ${x}$ (2)

Execute integration by parts:

 > $\mathrm{IntegrationTools}\left[\mathrm{Parts}\right]\left(,\right)$
 ${-}{\mathrm{cos}}{}\left({x}\right){}{x}{-}\left({∫}\left({-}{\mathrm{cos}}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)$ (3)

 >
 Commands Used


text:
# Why hydrogen bonding in some acids make them a stronger acid when it is present even before deprotonation?

If we take the example of salicylic acid, hydrogen bonding is present in the acid as follows:

Even after deprotonation, it has intramolecular hydrogen bonding as follows:

My question:

p-hydroxybenzoic acid ($$\mathrm pK_\mathrm a = 4.54$$) is less acidic than o-hydroxybenzoic acid ($$\mathrm pK_\mathrm a = 2.97$$), the reason being hydrogen bonding in the latter after deprotonation.

If the same reason accounts for the acidity of salicylic acid (o-hydroxybenzoic acid), then why is the H-bonding before deprotonation not considered (which should decrease the acidity of the compound)?

Some of the conformers of salicylic acid are given below, all of which are planar in configuration.$$\mathrm{^{[1]}}$$

Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) is the conformer with the second lowest energy as described in the published paper as given below:

The molecule can adopt several conformations, although only one of them is highly populated and was assigned in IR experiments. The spectrum was recorded by Fiedler et al., $$\mathrm{^{[2]}}$$ in tetrachloride solution and indicated a strong intramolecular H-bond. The deviation of the OH −1 stretching frequency from that in phenol was 395 cm . The theoretically calculated deviation is 359 cm−1 at the B3LYP/6-311+G(d,p) level. The lowest-energy conformer is planar, the phenolic OH is a H-bond donor to the carbonyl oxygen of the syn carboxylic group. The =O...H distance and the =O...H–O bond angle were calculated at 176 pm and 145°, respectively. The second-most-stable conformer is higher in energy by 14.3 kJ/mol, where the phenolic OH is the H-bond donor to the syn carboxylic OH. Similar conclusions were drawn by Yahagi et al.,$$\mathrm{^{[3]}}$$ by interpreting the gas-phase IR frequencies of the phenolic OH.

Much more convincing is the fact that the lowest energy conformer has a fractional population distribution$$\mathrm{^*}$$ of nearly 99.7% which further supports the argument regarding the higher acidity of salicylic acid in comparison to its para isomer. The higher acidity of the compound thus is contributed by the first conformer. The lesser interaction of the carboxyl group’s acidic proton and the stability of the conjugate base due to hydrogen bonding thus increases the acidity of the compound.
Regarding detailed mechanism of acidity of salicylic acid you may refer this and this post in chemistry SE. You may further note that the other structures might have decreased the acidity of the compound if their population distributions would have been higher. The same has been mentioned in the published paper as given below:

In a former calculation by Nagy et al.$$\mathrm{^{[4]}}$$, the two conformers above were found also to be the most stable with MP2/6-31G*//HF/6-31G* energy separation of 13.7 kJ/mol and free energy difference of 12.1 kJ/mol at T = 298 K. All other conformers are much higher in free energy, supporting the estimate of Fiedler that the population of the lowest-energy form is 99.7%.

## Notes and References

1. Nagy, Peter. Competing Intramolecular vs. Intermolecular Hydrogen Bonds in Solution. Int. J. Mol. Sci. 2014, DOI: 10.3390/ijms151119562
2. Fiedler, P.; Böhm, S.; Kulhánek, J.; Exner, O. Acidity of ortho-substituted benzoic acids: An infrared and theoretical study of the intramolecular hydrogen bonds. Org. Biomol. Chem. 2006, 4, 2003–2011. DOI: https://doi.org/10.1039/B601875K
3. Yahagi, T.; Fujii, A.; Ebata, T.; Mikam, N. Infrared spectroscopy of the OH stretching vibrations of jet-cooled salicylic acid and its dimer in S0 and S1. J. Phys. Chem. A 2001, 105, 10673–10680. DOI: https://doi.org/10.1021/jp0126199
4. Nagy, P.I.; Dunn, W.J, III.; Alagona, G.; Ghio, C. Theoretical studies of the 2- and 4-hydroxybenzoic acids with competing hydrogen bonds in the gas phase and aqueous solution. J. Phys. Chem. 1993, 97, 4628–4642. DOI: https://doi.org/10.1039/B601875K

$$\mathrm{*}$$ Note that here fractional population distribution refers to the population of each conformer in the compound. You may find more information about it on the Wikipedia article.
• So, if I got this right, the answer here is experimental fact? Or did I miss the theoretical explanation? Jun 2, 2021 at 13:23
• @Buraian Yes here the answer which I have posted only contains experimental facts about the population distribution of the conformers while the mechanism of acidity that has been explained briefly is theoretical. Jun 2, 2021 at 14:23
• I don't understand your argument about acidity of the ortho-isomer (which is the main question of OP). What does lesser interaction of the acidic proton have to do with higher acidity of ortho vs. para? Because there is no interaction in para-isomer either. Then you have mentioned the stability of the conjugate base due to H-bonding: does this mean that the conjugate base is more stable than the acid? There is H-bond in the acid as well, as you show in the first conformer. Also, is this part from a paper, or are you concluding it from the paper? Jun 2, 2021 at 17:51
• @ShoubhikRMaiti (1)The question of OP is that at the end Why the hydrogen bonding is to be considered(or why not) even before the deprotonation? (not about ortho vs para)(2) The block quoted texts are from a openly accesssable review paper(ref 1).(3) By lesser interaction I mean it is not involved in Hydrogen Bonding as in conformer 2(Refer statement 1 of this comment) Jun 2, 2021 at 18:11
• @ShoubhikRMaiti I apologise for not reading your comment properly, I missed the last 2 questions in your comment. (1) ” does this mean that the conjugate base is more stable than the acid?” The first conformer’s high population results from the fact that it is the most stable conformer but I do not know whether the conjugate base is less stable than the acid but I think the data for that should be there in one of the other references which are not openly accessible. (Continued... Jun 4, 2021 at 16:03

The other answer has given the major conformer of the protonated form of the o-hydroxybenzoic acid, but I believe there is a part of your question that still remains to be answered. You asked why the hydrogen bond makes the ortho-isomer more acidic than the para-isomer, when the hydrogen bond is already present in the protonated form of the ortho-isomer.

For an explanation of this, let's consider the ionization equilibrium of the ortho-acid:

(I am considering only the major conformer as mentioned in the other answer)

Now, there is hydrogen bond in both the protonated and deprotonated form. But, consider the strength of the hydrogen bond in both cases. In the protonated acid, the H-bond donor is a neutral $$\ce{>C=O:}$$. In the deprotonated state, the H-bond donor is an oxygen of a carboxyl anion. The negative charge is delocalised over both of the oxygen, so the H-bond donor $$\ce{O}$$ in this case has some extra electron density compared to the neutral $$\ce{>C=O}$$. So, it can form a stronger H-bond.

This means the H-bond is stronger in the deprotonated state than in the protonated state of the o-hydroxybenzoic acid. So, the presence of H-bond still makes the ortho-isomer a stronger acid (than the para-isomer where there is no intramolecular hydrogen bond), even if the H-bond can exist in both protonated and deprotonated forms.

• This answer imo seems more easily understandable and simpler, answers should always strike a balance in accuracy and accessibility. (Not saying this is inaccurate) Jun 27, 2021 at 9:10
• It kind of would be fun to look at this with calculations, see if it actually get stronger. Well, "see", more like guess the electronic structure, do some QTAIM, and guess some more... anyway, I've mentioned the number I would look at here: How to identify hydrogen bonds and other non-covalent interactions from structure considerations? Feb 2 at 22:24


text:
Lesson 1

A Towering Sequence

The practice problem answers are available at one of our IM Certified Partners

Problem 1

Here is a rule to make a list of numbers: Each number is the sum of the previous two numbers. Start with the numbers 0 and 1, then follow the rule to build a sequence of 10 numbers.

Problem 2

A sequence starts $$\frac{1} {2}, \frac{1}{4}, \frac{1}{8}, \dots$$

1. Give a rule that the sequence could follow.
2. Follow your rule to write the next 3 terms in the sequence.

Problem 3

A sequence of numbers follows the rule: multiply the previous number by -2 and add 3. The fourth term in the sequence is -7.

1. Give the next 3 terms in the sequence.
2. Give the 3 terms that came before -7 in the sequence.

Problem 4

A sequence starts 0, 5, . . .

1. Give a rule the sequence could follow and the next 3 terms for that rule.
2. Give a different rule the sequence could follow and the next 3 terms for that rule.


text:
# IBPS Clerk :: English Language :: Test 18 IBPS Recruitment Latest Govt Jobs

## Home IBPS Clerk / English Language Test 18 Questions and Answers

1 . In the following questions, a sentence has been given with some of its parts in bold. To make the sentence grammatically correct, you have to replace the bold part with the correct alternative given below. If the sentence is correct as it is, give 5) as your answer (ie No correction required).
Raman Singh is aman of many asobriquets.
people of many
man of many
person of much
human of most of
No correction required
2 . In the following questions, a sentence has been given with some of its parts in bold. To make the sentence grammatically correct, you have to replace the bold part with the correct alternative given below. If the sentence is correct as it is, give 5) as your answer (ie No correction required).
Delhi's three-time chief minister was known forher ability to read the pulse of voters
has been known for
has known for
has been knowing for
No correction required
3 . In the following questions, a sentence has been given with some of its parts in bold. To make the sentence grammatically correct, you have to replace the bold part with the correct alternative given below. If the sentence is correct as it is, give 5) as your answer (ie No correction required).
Proteins are the primarybuilding brickof the human body
building fabric
construction slabs
building blocks
framework entity
No correction required
4 . In the following questions, a sentence has been given with some of its parts in bold. To make the sentence grammatically correct, you have to replace the bold part with the correct alternative given below. If the sentence is correct as it is, give 5) as your answer (ie No correction required).
Doing something for the first time can be difficult and frustrated
a difficulty and frustration
difficult and frustrating
difficult to frustration
difficult with frustration
No correction required
5 . Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph and then answer the questions given below.

(A) In economic terms, any product being capable of giving customer satisfaction needs to have 'place' and 'time' utility.

(B) For an effective conclusion of marketing, the functioning of other elements in the chain is as important as that of the producer.

(C) This is provided by the 'Distribution' function of marketing mix.

(D) One of the four Ps of marketing mix is 'place' or 'Distribution'.

(E) In the normal goods distribution process, entitiesother than the producers are also associated and there is a chain between the producer and the consumer.

(F) All endeavours of production will come to nought, if the products do not reach the consumers for consumption
Which of the following should be the FIRST sentence after rearrangement?
F
A
B
C
D
6 . Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph and then answer the questions given below.

(A) In economic terms, any product being capable of giving customer satisfaction needs to have 'place' and 'time' utility.

(B) For an effective conclusion of marketing, the functioning of other elements in the chain is as important as that of the producer.

(C) This is provided by the 'Distribution' function of marketing mix.

(D) One of the four Ps of marketing mix is 'place' or 'Distribution'.

(E) In the normal goods distribution process, entitiesother than the producers are also associated and there is a chain between the producer and the consumer.

(F) All endeavours of production will come to nought, if the products do not reach the consumers for consumption
Which of the following should be the SECOND sentence after rearrangement?
B
A
D
F
E
7 . Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph and then answer the questions given below.

(A) In economic terms, any product being capable of giving customer satisfaction needs to have 'place' and 'time' utility.

(B) For an effective conclusion of marketing, the functioning of other elements in the chain is as important as that of the producer.

(C) This is provided by the 'Distribution' function of marketing mix.

(D) One of the four Ps of marketing mix is 'place' or 'Distribution'.

(E) In the normal goods distribution process, entitiesother than the producers are also associated and there is a chain between the producer and the consumer.

(F) All endeavours of production will come to nought, if the products do not reach the consumers for consumption
Which of the following should be the THIRD sentence after rearrangement?
E
B
F
C
A
8 . Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph and then answer the questions given below.

(A) In economic terms, any product being capable of giving customer satisfaction needs to have 'place' and 'time' utility.

(B) For an effective conclusion of marketing, the functioning of other elements in the chain is as important as that of the producer.

(C) This is provided by the 'Distribution' function of marketing mix.

(D) One of the four Ps of marketing mix is 'place' or 'Distribution'.

(E) In the normal goods distribution process, entitiesother than the producers are also associated and there is a chain between the producer and the consumer.

(F) All endeavours of production will come to nought, if the products do not reach the consumers for consumption
Which of the following should be the LAST (SIXTH) sentence after rearrangement?
B
C
E
D
F
9 . Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph and then answer the questions given below.

(A) In economic terms, any product being capable of giving customer satisfaction needs to have 'place' and 'time' utility.

(B) For an effective conclusion of marketing, the functioning of other elements in the chain is as important as that of the producer.

(C) This is provided by the 'Distribution' function of marketing mix.

(D) One of the four Ps of marketing mix is 'place' or 'Distribution'.

(E) In the normal goods distribution process, entitiesother than the producers are also associated and there is a chain between the producer and the consumer.

(F) All endeavours of production will come to nought, if the products do not reach the consumers for consumption
Which of the following should be the FIFTH sentence after rearrangement?
A
F
E
B
C
10 . In the following passage, some of the words have been left out, each of which is indicated by a number. Find the suitable word from the options given against each number and fill up the blanks with appropriate words to make the paragraph meaningfully complete.

The framework released by the RBI for entry and expansion of foreign banks in India makes sense. If foreign banks do enter India, it would drive local banks to become $(1)$ efficient and encourage mergers and acquisitions.These banks will also bring $(2)$technology and expertise to foster financial inclusion. The RBI's final rules provide a foreign bank nearly the $(3)$ freedom as a private sector bank in opening branches if it takes the form of a local subsidiary. New or existing $(4)$ will have to set up whollyowned arms if they are systemically important foreign banks, or banks that have a complex $(5)$or those that do not have adequate disclosure requirement in their home jurisdictions or offer home residents preferential protection in case of a problem.

However, those present in India before August 2010 - Citi, HSBC and Standard Chartered - have the $(6)$to continue as branches. Rightly, the regulator must encourage their conversion into local subsidiaries, following Indian laws. It will ring-fence the capital and assets of the bank within India and $(7)$ effective risk control. Incentives by way of operational freedom could persuade some banks to $(8)$ into wholly-owned subsidiaries. India is a growing market and more branches would do no harm. However, technology has made physical branch networks much less $(9)$today. An example is the credit card $(10)$ of foreign banks that has far surpassed their branch-network size
$(1)$
less
further
more
also


text:
Drake
Drake C++ Documentation
Geometry Optimization

## Detailed Description

Provides an abstraction for reasoning about geometry in optimization problems, and using optimization problems to solve geometry problems.

Warning
This feature is considered to be experimental and may change or be removed at any time, without any deprecation notice ahead of time.

### Relationship to other components in Drake.

SceneGraph handles many types of geometry (see Shape). It is specialized to 3D and puts a strong emphasis on efficient implementation of a particular subset of geometry queries, like collision detection and signed-distance queries. SceneGraph also provides a lot of valuable tools for content management, including parsing geometries from file (via multibody::Parser) and Role.

MathematicalProgram has many relevant solvers::Cost / solvers::Constraint for reasoning about geometry (e.g. the LorentzCone or even LinearConstraint). The class and methods in this group add a level of modeling power above these individual constraints (there are many different types of constraints one would write given various optimization on these sets).

The geometry::optimization tools support:

• The ability to model sets in arbitrary dimensions, not just 3D.
• Many advanced queries (intersections, Minkowski sum, set containment, etc).
• The ability to use sets in mathematical programs
• The ability to use mathematical programs to perform operations on sets.
• The ability to include the geometry from a SceneGraph into a MathematicalProgram.
• The ability to use SceneGraph queries for some of the geometry operations (signed distance, collision), when the set is in 3D.
• The ability to use SceneGraph for rendering the geometries when they are in 3D.

## Classes

class  CartesianProduct
The Cartesian product of convex sets is a convex set: S = X₁ × X₂ × ⋯ × Xₙ = {(x₁, x₂, ..., xₙ) | x₁ ∈ X₁, x₂ ∈ X₂, ..., xₙ ∈ Xₙ} We currently require the sets X to be bounded. More...

class  ConvexSet
Abstract base class for defining a convex set. More...

class  GraphOfConvexSets
GraphOfConvexSets implements the design pattern and optimization problems first introduced in the paper "Shortest Paths in Graphs of Convex Sets". More...

class  HPolyhedron
Implements a polyhedral convex set using the half-space representation: {x| A x ≤ b}. More...

class  Hyperellipsoid
Implements an ellipsoidal convex set represented by the quadratic form {x | (x-center)ᵀAᵀA(x-center) ≤ 1}. More...

class  Intersection
A convex set that represents the intersection of multiple sets: S = X₁ ∩ X₂ ∩ ... More...

struct  IrisOptions
Configuration options for the IRIS algorithm. More...

class  MinkowskiSum
A convex set that represents the Minkowski sum of multiple sets: S = X₁ ⨁ X₂ ⨁ ... More...

class  Point
A convex set that contains exactly one element. More...

class  VPolytope
A polytope described using the vertex representation. More...

## Functions

HPolyhedron Iris (const ConvexSets &obstacles, const Eigen::Ref< const Eigen::VectorXd > &sample, const HPolyhedron &domain, const IrisOptions &options=IrisOptions())
The IRIS (Iterative Region Inflation by Semidefinite programming) algorithm, as described in. More...

ConvexSets MakeIrisObstacles (const QueryObject< double > &query_object, std::optional< FrameId > reference_frame=std::nullopt)
Constructs ConvexSet representations of obstacles for IRIS in 3D using the geometry from a SceneGraph QueryObject. More...

HPolyhedron IrisInConfigurationSpace (const multibody::MultibodyPlant< double > &plant, const systems::Context< double > &context, const IrisOptions &options=IrisOptions())
A variation of the Iris (Iterative Region Inflation by Semidefinite programming) algorithm which finds collision-free regions in the configuration space* of plant. More...

## ◆ Iris()

 HPolyhedron drake::geometry::optimization::Iris ( const ConvexSets & obstacles, const Eigen::Ref< const Eigen::VectorXd > & sample, const HPolyhedron & domain, const IrisOptions & options = IrisOptions() )

The IRIS (Iterative Region Inflation by Semidefinite programming) algorithm, as described in.

R. L. H. Deits and R. Tedrake, “Computing large convex regions of obstacle-free space through semidefinite programming,” Workshop on the Algorithmic Fundamentals of Robotics, Istanbul, Aug. 2014. http://groups.csail.mit.edu/robotics-center/public_papers/Deits14.pdf

This algorithm attempts to locally maximize the volume of a convex polytope representing obstacle-free space given a sample point and list of convex obstacles. Rather than compute the volume of the polytope directly, the algorithm maximizes the volume of an inscribed ellipsoid. It alternates between finding separating hyperplanes between the ellipsoid and the obstacles and then finding a new maximum-volume inscribed ellipsoid.

Parameters
 obstacles is a vector of convex sets representing the occupied space. sample provides a point in the space; the algorithm is initialized using a tiny sphere around this point. The algorithm is only guaranteed to succeed if this sample point is collision free (outside of all obstacles), but in practice the algorithm can often escape bad initialization (assuming the require_sample_point_is_contained option is false). domain describes the total region of interest; computed IRIS regions will be inside this domain. It must be bounded, and is typically a simple bounding box (e.g. from HPolyhedron::MakeBox).

The obstacles, sample, and the domain must describe elements in the same ambient dimension (but that dimension can be any positive integer).

## ◆ IrisInConfigurationSpace()

 HPolyhedron drake::geometry::optimization::IrisInConfigurationSpace ( const multibody::MultibodyPlant< double > & plant, const systems::Context< double > & context, const IrisOptions & options = IrisOptions() )

A variation of the Iris (Iterative Region Inflation by Semidefinite programming) algorithm which finds collision-free regions in the configuration space* of plant.

 plant describes the kinematics of configuration space. It must be connected to a SceneGraph in a systems::Diagram. context is a context of the plant. The context must have the positions of the plant set to the initialIRIS seed configuration. options provides additional configuration options. In particular, options.enabled_ibex may have a significant impact on the runtime of the algorithm.
 ConvexSets drake::geometry::optimization::MakeIrisObstacles ( const QueryObject< double > & query_object, std::optional< FrameId > reference_frame = std::nullopt )