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# mill weight of balls

## mill weight of balls

### The grinding balls bulk weight in fully unloaded mill

1kg of 20mm white ceramic grinding balls. 20mm diameter alumina (Al2O3) ceramic balls, very hard wearing and ideal for ball milling. For volume calculations, 1kg will take up roughly half a litre. 1kg of 20mm balls is roughly 55 balls, but the exact quantity can vary.

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### (PDF) A comparison of wear rates of ball mill grinding media

Calculate and Select Ball Mill Ball Size for Optimum Grinding. In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do.

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### New and Used Ball Mills for Sale | Ball Mill Supplier ...

· As we have written it before, there are two methods for determining the grinding balls bulk weight in a mill: Method with complete grinding media discharge from the mill inner drum. Method without unloading grinding balls. In the previous article we considered the 1-st method for determining the bulk weight of grinding media. This article analyzes the second method.

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### Calculate and Select Ball Mill Ball Size for Optimum Grinding

In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do. Often, the ball used in ball mills is oversize “just in case”.

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### ball mill weight of balls marble grinding roller

SILICON CARBIDE BALLS Silicon Carbide Balls are very high-cost grinding media that are used for milling same materials (silicon carbide ball to mill silicon carbide materials) to avoid contamination. They are only available in 5mm,10mm, 15mm and 20mm sizes. They are a special order item.

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### White Ceramic Grinding Balls - Size: 20mm - Weight: 1kg ...

Ball Weight and Density How much will a ball of a given diameter in a certain material weigh? The answer is calculated by multiplying the volume of the ball by the density of the material. $\text"Weight" = \text"Volume" ⋅ \text"Density"$ For example, calculate the weight of a two inch diameter lead ball:

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### Grinding Media & Grinding Balls | Union Process, Inc.

China Ball Mill manufacturers - Select 2020 high quality Ball Mill products in best price from certified Chinese Plastic Machinery manufacturers, Milling Machine suppliers, wholesalers and factory on Made-in …

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### Fitness Equipment and Smart Tech | Les Mills

Ball Mill for Sale. APPROXIMATE CAPACITIES. All ball mills operate on the same principles. One of these principle is that the total weight of the charge in the mill-the sum of the weight of the grinding media, the weight of the material to be ground, and any water in the mill—is a function of the percentage of the volume of the mill it occupies.

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### Ball Mill - RETSCH - powerful grinding and homogenization

The ball mill is the key equipment to crush the material after being crushed. This type of mill is equipped with a number of steel balls in its cylinder as grinding medium. It is widely used in cement, silicate products, new building materials, refractories, fertilizers, black and non-ferrous metal beneficiation and glass and ceramics industries.

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### What You Ought To Know About Ball Screws | Machine Design

ball weight effect in ball mill - tga-crest. Ball mill Wikipedia. A ball mill, a type of grinder, is a cylindrical dev used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and mills rotate around a horizontal axis, partially filled with the material to be ground plus the grinding medium.

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### Solid Precision Plastic Balls, Beads & Shapes - Oakland ...

Best way to determine the ball-to-powder ratio in ball-milling? ... Total weight of ball = number of ball X weight of single ball. ... The maximum power draw in ball mill is when ball bed is 35-40 ...

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### Ball Mill Jars 1 Liter - The Ceramic Shop

· In this article, we will consider the technique for determining the grinding balls bulk weight in fully unloaded mill. This method used in the mills repair (armor plates replacement). The grinding balls unload from the mill into a special pit (needs to open hatches and pour the grinding balls from the drum during mill scroll).

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### Calculating Ball Weight In A Ball Mill

Ball Mill. Raymond Mill. MW Series Micro Powder Mill. T130X Superfine Grinding Mill. MTW Trapezium Mill. LM Vertical Mill. 5X Pendulum Roller Mill. VSI6X Series Vertical Crusher. Impact Crusher. European Type Jaw Crusher. Jaw Crusher. HPT Hydraulic Cone Crusher. CS Series Cone Crusher.

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### Ball Mill Loading - Wet Milling - Paul O. Abbe

The Les Mills SMARTBAND™ is a durable and functional resistance band. This new innovation in resistance band technology is scientifically-engineered to deliver a consistent force increase throughout the extension range.

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### Joyal-Ball Mill,Ball Mill For Sales,Ball Mill Manufacturer

steel balls in a ball mill, or large lumps of ore in an autogenous mill or a mixture in a semi-autogenous mill, as well as the slurry that makes up the operating charge. Let Jt be the fraction of the mill volume that is occupied by the total charge, Jb the fraction of the mill volume that is occupied by steel balls and E the voidage of the balls

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### Ball Mill - an overview | ScienceDirect Topics

In all ore dressing and milling Operations, including flotation, cyanidation, gravity concentration, and amalgamation, the Working Principle is to crush and grind, often with rob mill & ball mills, the ore in order to liberate the minerals. In the chemical and process industries, grinding is an important step in preparing raw materials for subsequent treatment.

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### Balls for Bearings | Precision Balls - Abbott Ball Company

McMaster-Carr is the complete source for your plant with over 580,000 products. 98% of products ordered ship from stock and deliver same or next day.

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### Fitness Equipment & SMART TECH – Les Mills

Mandala® Yarn Mill End Ball. We often have a lot of sample and mill-end yarn in the development process. This yarn is slightly over or under in weight or yardage and the colors may not exactly match. ***All Mill End Balls are sold as-is. Colors will be chosen automatically and no replacements are allowed.

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### How can one select ball size in ball milling and how much ...

Abbott Ball manufactures balls in a variety of sizes and alloys, including stainless steel, carbon steel, aluminum, gold, platimum, brass, titanium and plastic balls for bearings and other applications.

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### Bal-tec - Ball Weight and Density

A ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paints.Ball mills rotate around a horizontal axis, partially filled with the material to be ground plus the grinding medium. Different materials are used as media, including ceramic balls, flint pebbles and stainless steel balls.

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### Balls | McMaster-Carr

Note: Ball must be larger than interior pipe diameter for proper cleaning. All sizes available in soft, medium, and hard styles. Frequently referred to as Concrete Pump Clean Out Balls, Clean-Out Balls, Cleanout Balls, Wiper Balls, Cleaning Balls, and Blow-Out Balls …

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### How Can I calculate new ball size and weight desing for ...

New and Used Ball Mills for Sale Savona Equipment is a new and used Ball Mill supplier worldwide.A ball mill is a type of grinder used to grind materials into extremely fine powder for use in mineral dressing processes, paints, pyrotechnics, ceramics and selective laser sintering.

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### Calculating Ball Weight In A Ball Mill- SPECIAL Mining machine

Multiple parameters are explored including mass of starting material, ball size, rotational speed of the mill, number of balls employed, and material used for the milling container.

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### Ball mill_Henan Hongji Mine Machinery Co., Ltd.

1. Introduction. The ball size is one of the critical factors for determining the mill performance of ball mills. It is well known that larger balls are needed for the effective breakage of large-size particles, whereas smaller balls are more effective for the breakage of fine particles.

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### Grinding mills - for mining and minerals processing -

More balls with small size results in fine powder. As a thumb rule powder to be milled should be taken as 25% of total ball weight. If the quantity of charge is very less then milling balls will ...

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The viability of the multinational corporate system depends upon the degree to which people will tolerate the unevenness it creates. It is well to remember that the ‘New Imperialism’ which began after $1870$ in a spirit of Capitalism Triumphant, soon became seriously troubled and after $1914$ was characterized by war, depression, breakdown of the international economic system and war again, rather than Free Trade, Pax Britannica and material Improvement. A major reason was Britain’s inability to cope with the by-products of its own rapid accumulation of capital; i.e., a class-conscious labour force at home; a middle class in the hinterland; and rival centres of capital on the Continent and in America. Britain’s policy tended to be atavistic and defensive rather than progressive – more concerned with warding off new threats than creating new areas of expansion. Ironically, Edwardian England received the paraphernalia of the landed aristocracy it had just destroyed. Instead of embarking on a ‘big push’ to develop the vast hinterland of the Empire, colonial administrators often adopted policies to arrest the development of either a native capitalist class or a native proletariat which could overthrow them.

As time went on, the centre had to devote an increasing share of government activity to military and other unproductive expenditures; they had to rely on alliances with an inefficient class of landlords, officials and soldiers in the hinterland to maintain stability at the cost of development. A great part of the surplus extracted from the population was thus wasted locally.

The New Mercantilism (as the Multinational Corporate System of special alliances and privileges, aid and tariff concessions is sometimes called) faces similar problems of internal and external division. The centre is troubled; excluded groups revolt and even some of the affluent are dissatisfied with the roles. Nationalistic rivalry between major capitalist countries remains an important devisive factor. Finally, there is the threat presented by the middle classes and the excluded groups of the undeveloped countries. The national middle classes in the underdeveloped countries came to power when the centre weakened but could not, through their policy of import substitution manufacturing, establish a viable basis for sustained growth. They now face a foreign exchange crisis and an unemployment (or population) crisis – the first indicating their inability to function in the international economy and the second indicating their alienation from the people they are supposed to lead. IN the immediate future, these national middle classes will gain a new lease of life as they take advantage of the spaces created by the rivalry between American and non-American oligopolists striving to establish global market positions.

The native capitalists will again become the champions of national independence as they bargain with multinational corporations. But the conflict at this level is more apparent than real, for in the end the fervent nationalism of the middle class asks only for promotion within the corporate structure and not for a break with that structure. In the last analysis their power derives from the metropolis and they cannot easily afford to challenge the international system. They do not command the loyalty of their own population and cannot really complete with the large, powerful, aggregate capitals from the centre.

The main threat from the excluded groups. It is not unusual in underdeveloped countries for the top $5$ per cent to obtain between $30$ and $40$ per cent of the total national income, and for the top one-third to obtain anywhere from $60$ to $70$ per cent. At most, one-third of the population can be said to benefit in some sense from the dualistic growth that characterizes development in the hinterland. The remaining two-thirds, who together get only one-third of the income, are outsiders, not because they do not contribute to the economy, but because they do not share in the benefits. They provide a source of cheap labour which helps keep exports-to the developed world at a low price and which has financed the urban-biased growth of recent years. In fact, it is difficult to see how the system in most underdeveloped countries could survive without cheap labour since removing it (e.g. diverting it to public works projects as is done in socialist countries) would raise consumption costs to capitalists and professional elites.

Under New Mercantilism, the fervent nationalism of the native middle classes does not create conflict with the multinational corporations because they (the middle classes)

1. negotiate with the multinational corporate.
2. are dependent on the international system for their continued prosperity
3. are not in a position to challenge the status quo
4. do not enjoy popular support

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NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### AIEEE 2007

In a geometric progression consisting of positive terms, each term equals the sum of the next two terns. Then the common ratio of its progression is equals
A
$${\sqrt 5 }$$
B
$$\,{1 \over 2}\left( {\sqrt 5 - 1} \right)$$
C
$${1 \over 2}\left( {1 - \sqrt 5 } \right)$$
D
$${1 \over 2}\sqrt 5$$.

## Explanation

Let the series $$a,ar,$$ $$a{r^2},........$$ are in geometric progression.

given, $$a = ar + a{r^2}$$

$$\Rightarrow 1 = r + {r^2}$$

$$\Rightarrow {r^2} + r - 1 = 0$$

$$\Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}$$

$$\Rightarrow r = {{ - 1 \pm \sqrt 5 } \over 2}$$

$$\Rightarrow r = {{\sqrt 5 - 1} \over 2}$$

[ As terms of $$G.P.$$ are positive

$$\therefore$$ $$r$$ should be positive]
2

### AIEEE 2007

The sum of series $${1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - .......$$ upto infinity is
A
$${e^{ - {1 \over 2}}}$$
B
$${e^{ + {1 \over 2}}}$$
C
$${e^{ - 2}}$$
D
$${e^{ - 1}}$$

## Explanation

We know that $${e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty$$

Put $$x=-1$$

$$\therefore$$ $${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty$$

$$\therefore$$ $${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {1 \over {5!}}........\infty$$
3

### AIEEE 2006

Let $${a_1}$$, $${a_2}$$, $${a_3}$$.....be terms on A.P. If $${{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p \ne q,\,then\,{{{a_6}} \over {{a_{21}}}}\,$$ equals
A
$${{41} \over {11}}$$
B
$${7 \over 2}$$
C
$${2 \over 7}$$
D
$${{11} \over {41}}$$

## Explanation

$${{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}$$

$$\Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}$$

$${{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}$$

For $${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$$

$$\Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}$$
4

### AIEEE 2006

If $${{a_1},{a_2},....{a_n}}$$ are in H.P., then the expression $${{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$$ is equal to
A
$$n({a_1}\, - {a_n})$$
B
$$(n - 1)({a_1}\, - {a_n})$$
C
$$n{a_1}{a_n}$$
D
$$(n - 1)\,\,{a_1}{a_n}$$

## Explanation

$${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)

Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}$$

$$\therefore$$ $${a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}$$

$$= {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}$$

$$= {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}$$

Also, $${1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d$$

$$\Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d$$

$$\Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}$$

Which is the required result.

### Joint Entrance Examination

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### Graduate Aptitude Test in Engineering

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Class 12


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# The Structure and Signals of Neutron Stars, from Birth to Death

Mar 24 – 28, 2014
Europe/Rome timezone

## Statistical ages and the cooling rate of X-ray dim isolated neutron stars

Mar 27, 2014, 2:50 PM
20m

### Speaker

Dr Ramandeep Gill (CITA (University of Toronto))

### Description

The cooling theory of neutron stars is corroborated by its comparison with observations of thermally emitting isolated neutron stars and accreting neutron stars in binary systems. An important ingredient for such an analysis is the age of the object, which, typically, is obtained from the spin-down history. This age is highly uncertain if the object's magnetic field varies appreciably over time. Other age estimators, such as supernova remnant ages and kinematic ages, only apply to few handful of neutron stars. We conduct a population synthesis study of the nearby isolated thermal emitters and obtain their ages statistically from the observed luminosity function of these objects. We argue that a more sensitive blind scan of the galactic disk with the upcoming space telescopes can help to constrain the ages to higher accuracy.

### Primary author

Dr Ramandeep Gill (CITA (University of Toronto))

 Slides


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[HFSS] Simulation of S21 parameter of CPW line

Status
Not open for further replies.

mys ruby

Newbie level 3
Hello,

I would like to simulate a simple CPW line on a large band of frequency 0-20 GHz. I have created the excitation waveport and the boundary as the same as the CPW tutorial.
https://obrazki.elektroda.pl/8201766200_1432220823.jpg
The E-field on waveport is ok.
https://obrazki.elektroda.pl/5336836600_1432220824.jpg
The S11 parameter have a good agreement between the measurement and the simulation.
https://obrazki.elektroda.pl/7835222900_1432220825.jpg
But with S21 parameter, the level of S21 simulation is inferior to the S21 measurement (while S21 simulation level must be superior to measurement because of connector losses in calibration OSLT). S21 Momentum simulation is very good even though it doesn't include metal losses. (Measurement: red, simulation HFSS: blue, simulation Momentum: green)
https://obrazki.elektroda.pl/3860663900_1432220826.jpg
I would like to find my errors in HFSS simulation, could you help me, please. Thanks you very much.

volker@muehlhaus

simulation level must be superior to measurement because of connector losses in calibration OSLT

From experience, I do not agree with this part. Contact resistance exists in calibration standards and in the device under test (DUT). It can be the same for CAL and DUT, or larger in the DUT resulting in too high measured DUT resistance, or larger in the CAL resulting in too small measured DT resistance. You really don't know.

mys ruby

Newbie level 3
From experience, I do not agree with this part. Contact resistance exists in calibration standards and in the device under test (DUT). It can be the same for CAL and DUT, or larger in the DUT resulting in too high measured DUT resistance, or larger in the CAL resulting in too small measured DT resistance. You really don't know.

Dear volker@muehlhaus,

Could you explain in detail, I really don't know this part. I have measured with an OSLT calibration that includes the connectors losses due to the transitions between the input-output coaxial cables and the cpw line. I think S21 simulation level must be slightly superior to measurement even if I can't estimate exactly these connectors losses and the S21 Momentum simulation seems more reasonable than HFSS. I don't know where are my errors in HFSS.

volker@muehlhaus

I have measured with an OSLT calibration that includes the connectors losses due to the transitions between the input-output coaxial cables and the cpw line. I think S21 simulation level must be slightly superior to measurement even if I can't estimate exactly these connectors losses and the S21 Momentum simulation seems more reasonable than HFSS. I don't know where are my errors in HFSS.

Ok, if you have calibrated with coax cal standards you are right. I expected that you did wafer prober measurement with coplanar cal standards.

For HFSS, you should describe your model in more detail. How did you model the conductors (surface impedance or solve inside)?

mys ruby

Newbie level 3
I have assigned the conductors (Copper adhesive tape) as Finite Conductivity Boundary with the conductivity 2e7 S/m. The substrate is synthetic paper and there is a adhesive layer between the substrate and the conductors. The waveport is renormalized to 50 ohm.

volker@muehlhaus

conductivity 2e7 S/m.

2e7 S/m is very pessimistic for copper. The usual value is 5.8e7 S/m.

mys ruby

Newbie level 3
2e7 S/m is very pessimistic for copper. The usual value is 5.8e7 S/m.

I have tried with the usual value 5.8e7 S/m: The level is a little superior, but it is also inferior to the measurement. I don't think it is a principal reason.

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#### Authors

Mark Plutowski, Shinichi Sakata, Halbert White

#### Abstract

Integrated Mean Squared Error (IMSE) is a version of the usual mean squared error criterion, averaged over all possible training If it could be observed, it could be used sets of a given size. to determine optimal network complexity or optimal data sub(cid:173) sets for efficient training. We show that two common methods of cross-validating average squared error deliver unbiased estimates of IMSE, converging to IMSE with probability one. These esti(cid:173) mates thus make possible approximate IMSE-based choice of net(cid:173) work complexity. We also show that two variants of cross validation measure provide unbiased IMSE-based estimates potentially useful for selecting optimal data subsets.

1 Summary To begin, assume we are given a fixed network architecture. (We dispense with this assumption later.) Let zN denote a given set of N training examples. Let QN(zN) denote the expected squared error (the expectation taken over all possible examples) of the network after being trained on zN. This measures the quality of fit afforded by training on a given set of N examples. Let IMSEN denote the Integrated Mean Squared Error for training sets of size N. Given reasonable assumptions, it is straightforward to show that IMSEN = E[Q N(ZN)] - 0"2, where the expectation is now over all training sets of size N, ZN is a random training set of size N, and 0"2 is the noise variance. Let CN = CN(zN) denote the "delete-one cross-validation" squared error measure for a network trained on zN. CN is obtained by training networks on each of the N training sets of size N -1 obtained by deleting a single example; the measure follows


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1. ## Combinations problem

I've been challenged to come up with the total possible number of combinations for a given set of ingredients. For the purposes of this challenge, AB == BA. In other words, given ingredients 'A', 'B', and 'C', the possible combinations are as follows:

A, B, C, AB, AC, BC, ABC

BA,CA,CB,ACB,BAC,BCA,CAB, and CBA are not counted because for our purposes order is irrelevant. Also, each item may only be used once. It's been a LONG time since I got out of school, but I can't think of an equation that satisfies this set of rules. Am I missing something obvious? Is there a more complex formula that is required? Any assistance would be greatly appreciated.

2. ## Re: Combinations problem

The number of nonempty subsets of a set with n elements is $2^n-1$ (see Wikipedia).


text:
# Proof of Liouville`s Theorem

Consider the autonomous ODE $$y'(t)=f(y(t))$$ where $f \in C^{\infty}(\mathbb{R}^n, \mathbb{R}^n)$. The flow $F(t,y_0)$ associated with this equation is defined as follows. Let $y(t)$ denote the (unique) solution to this equation with initial condition $y(0) = y_0$ and $t \in I_x$, where $I_x$ is the maximal possible interval about $0$ where the solution exists. Then $F(t,y_0) : = y(t)$ for $t \in I_x$. Notice that $x \mapsto F(t,x)$ is smooth and injective (Why?). For a bounded open set $U \subset \mathbb{R}^n$, set $$V(t) = \text{Vol}\,(F(t, U)) : = \int_{F(t,U)}1 \,dx$$

Theorem: If $U\subset \mathbb{R}^n$ is a bounded open set, then $$\frac{d\, V}{dt} = \int_{F(t,U)} \operatorname{div} f(x) \, dx$$

Proof: Notice that $F(t_0 + h, U) = F(h, F(t_0,U))$ provided that the right hand side is defined, which will be the case if $|h|$ is sufficiently small. By a change of variables

$\begin{matrix} V(t_0+h) & = & \int_{F(t_0+h,U)} 1\, dx \\ & = & \int_{F(h, F(t_0, U))} 1\, dx\\ & = & \int_{F(t_o,U)} \operatorname{det}\left(\frac{\partial F}{\partial x}(h,x) \right) dx. \end{matrix}$

We claim that for $x \in F(t_0, U)$, $$\frac{\partial F}{\partial x}(t,x) = I + tDf(x) + o(t)$$

if $|t|$ is sufficiently small. Indeed, note that $$F(t,x) = x + \int_{0}^t f(F(s,x))ds,$$ and after taking partial derivatives with respect to $x$ we get $$\frac{\partial F}{\partial x}(t,x) = I + \int_{0}^t \frac{\partial}{\partial x}f(F(s,x)) ds,$$ so that $$\frac{\partial}{\partial t} \frac{\partial F}{\partial x}(t,x) = \frac{\partial}{\partial x}f(F(t,x))$$ Evaluating at $t=0$ and using the chain rule we get $${\frac{\partial}{\partial t} \frac{\partial F}{\partial x}(t,x)}\Big|_{t=0} = Df(F(0,x))\cdot \frac{\partial F}{\partial x}(0,x) = Df(x)$$ The claim follows from this and from the definition of the derivative.

From the claim we deduce that $\operatorname{det} \left(\frac{\partial F}{\partial x}(t,x) \right) = \operatorname{det} (I + t\,Df(x)) + o(t)$. In the expansion of the determinant, the only summand that does not contribute an $o(t)$ term is $$\prod_{i=1}^n(1 + t \frac{\partial f_{i}}{\partial x_{i}})= 1 + t \sum_{i=1}^n \frac{\partial f_{i}}{\partial x_{i}} + o(t)= 1 + t \operatorname{div} f(x) + o(t)$$

It follows that $$V(t_0+h) = \int_{F(t_0,U)} 1 + h\operatorname{div} f(x) + o(h) dx = V(t_0) + h \int_{F(t_0,U)} \operatorname{div} f(x) \,dx + o(h)$$ and this establishes the theorem from the definition of the derivative.

Corollary: The flow of a Hamiltonian system preserves volume since the corresponding $f = (-H_q, H_p)$ satisfies $\operatorname{div} f \equiv 0$.

Reference: Teschl, Gerald: Ordinary Differential Equations and Dynamical Systems, 2004. [1]


text:
INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

Eric Smith

Linear Equations in Two Variables

Slide Duration:

Section 1: Properties of Real Numbers
Basic Types of Numbers

30m 41s

Intro
0:00
Objectives
0:07
Basic Types of Numbers
0:36
Natural Numbers
1:02
Whole Numbers
1:29
Integers
2:04
Rational Numbers
2:38
Irrational Numbers
5:06
Imaginary Numbers
6:48
Basic Types of Numbers Cont.
8:09
The Big Picture
8:10
Real vs. Imaginary Numbers
8:30
Rational vs. Irrational Numbers
8:48
Basic Types of Numbers Cont.
10:55
Number Line
11:06
Absolute Value
11:44
Inequalities
12:39
Example 1
13:16
Example 2
17:30
Example 3
21:56
Example 4
24:27
Example 5
27:48
Operations on Numbers

19m 26s

Intro
0:00
Objectives
0:06
Operations on Numbers
0:25
0:53
Subtraction
1:33
Multiplication & Division
2:19
Exponents
3:24
Bases
4:04
Square Roots
4:59
Principle Square Roots
5:09
Perfect Squares
6:32
Simplifying and Combining Roots
6:52
Example 1
8:16
Example 2
12:30
Example 3
14:02
Example 4
16:27
Order of Operations

12m 6s

Intro
0:00
Objectives
0:06
The Order of Operations
0:25
Work Inside Parentheses
0:42
Simplify Exponents
0:52
Multiplication & Division from Left to Right
0:57
Addition & Subtraction from Left to Right
1:11
Remember PEMDAS
1:21
The Order of Operations Cont.
2:27
Example
2:43
Example 1
3:55
Example 2
5:36
Example 3
7:35
Example 4
8:56
Properties of Real Numbers

18m 52s

Intro
0:00
Objectives
0:07
The Properties of Real Numbers
0:23
Commutative Property of Addition and Multiplication
0:44
Associative Property of Addition and Multiplication
1:50
Distributive Property of Multiplication Over Addition
3:20
Division Property of Zero
4:46
Division Property of One
5:23
Multiplication Property of Zero
5:56
Multiplication Property of One
6:17
6:29
Why Are These Properties Important?
6:53
Example 1
9:16
Example 2
13:04
Example 3
14:30
Example 4
16:57
Section 2: Linear Equations
The Vocabulary of Linear Equations

12m 22s

Intro
0:00
Objectives
0:09
The Vocabulary of Linear Equations
0:44
Variables
0:52
Terms
1:09
Coefficients
1:40
Like Terms
2:18
Examples of Like Terms
2:37
Expressions
4:01
Equations
4:26
Linear Equations
5:04
Solutions
5:55
Example 1
6:16
Example 2
7:16
Example 3
8:45
Example 4
10:20
Solving Linear Equations in One Variable

28m 52s

Intro
0:00
Objectives
0:08
Solving Linear Equations in One Variable
0:34
Conditional Cases
0:51
Identity Cases
1:09
1:30
Solving Linear Equations in One Variable Cont.
2:00
2:10
Multiplication Property of Equality
2:43
Steps to Solve Linear Equations
3:14
Example 1
4:22
Example 2
8:21
Example 3
12:32
Example 4
14:19
Example 5
17:25
Example 6
22:17
Solving Formulas

12m 2s

Intro
0:00
Objectives
0:06
Solving Formulas
0:18
Formulas
0:26
Use the Same Properties as Solving Linear Equations
1:36
1:55
Multiplication Property of Equality
1:58
Steps to Solve Formulas
2:43
Example 1
3:56
Example 2
6:09
Example 3
8:39
Applications of Linear Equations

28m 41s

Intro
0:00
Objectives
0:10
Applications of Linear Equations
0:43
The Six-Step Method to Solving Word Problems
0:55
Common Terms
3:12
Example 1
5:03
Example 2
9:40
Example 3
13:48
Example 4
17:58
Example 5
23:28
Applications of Linear Equations, Motion & Mixtures

24m 26s

Intro
0:00
Objectives
0:21
Motion and Mixtures
0:46
Motion Problems: Distance, Rate, and Time
1:06
Mixture Problems: Amount, Percent, and Total
1:27
The Table Method
1:58
The Beaker Method
3:38
Example 1
5:05
Example 2
9:44
Example 3
14:20
Example 4
19:13
Section 3: Graphing
Rectangular Coordinate System

22m 55s

Intro
0:00
Objectives
0:11
The Rectangular Coordinate System
0:39
The Cartesian Coordinate System
0:40
X-Axis
0:54
Y-Axis
1:04
Origin
1:11
1:26
Ordered Pairs
2:10
Example 1
2:55
The Rectangular Coordinate System Cont.
6:09
X-Intercept
6:45
Y-Intercept
6:55
Relation of X-Values and Y-Values
7:30
Example 2
11:03
Example 3
12:13
Example 4
14:10
Example 5
18:38
Slope & Graphing

27m 58s

Intro
0:00
Objectives
0:11
Slope and Graphing
0:48
Standard Form
1:14
Example 1
2:24
Slope and Graphing Cont.
4:58
Slope, m
5:07
Slope is Rise over Run
6:11
Don't Mix Up the Coordinates
8:20
Example 2
9:39
Slope and Graphing Cont.
14:26
Slope-Intercept Form
14:34
Example 3
16:55
Example 4
18:00
Slope and Graphing Cont.
19:00
Rewriting an Equation in Slope-Intercept Form
19:39
Rewriting an Equation in Standard Form
20:09
Slopes of Vertical & Horizontal Lines
20:56
Example 5
22:49
Example 6
24:09
Example 7
25:59
Example 8
26:57
Linear Equations in Two Variables

20m 36s

Intro
0:00
Objectives
0:13
Linear Equations in Two Variables
0:36
Point-Slope Form
1:07
Substitute in the Point and the Slope
2:21
Parallel Lines: Two Lines with the Same Slope
4:05
Perpendicular Lines: Slopes are Negative Reciprocals of Each Other
4:39
Perpendicular Lines: Product of Slopes is -1
5:24
Example 1
6:02
Example 2
7:50
Example 3
10:49
Example 4
13:26
Example 5
15:30
Example 6
17:43
Section 4: Functions
Introduction to Functions

21m 24s

Intro
0:00
Objectives
0:07
Introduction to Functions
0:58
Relations
1:03
Functions
1:37
Independent Variables
2:00
Dependent Variables
2:11
Function Notation
2:21
Function
3:43
Input and Output
3:53
Introduction to Functions Cont.
4:45
Domain
4:46
Range
4:55
Functions Represented by a Diagram
6:41
Natural Domain
9:11
Evaluating Functions
12:02
Example 1
13:13
Example 2
15:03
Example 3
16:18
Example 4
19:54
Graphing Functions

16m 12s

Intro
0:00
Objectives
0:09
Graphing Functions
0:54
Using Slope-Intercept Form
1:56
Vertical Line Test
2:58
Determining the Domain
4:20
Determining the Range
5:43
Example 1
6:06
Example 2
7:18
Example 3
8:31
Example 4
11:04
Section 5: Systems of Linear Equations
Systems of Linear Equations

25m 54s

Intro
0:00
Objectives
0:13
Systems of Linear Equations
0:46
System of Equations
0:51
System of Linear Equations
1:15
Solutions
1:35
Points as Solutions
1:53
Finding Solutions Graphically
5:13
Example 1
6:37
Example 2
12:07
Systems of Linear Equations Cont.
17:01
One Solution, No Solution, or Infinite Solutions
17:10
Example 3
18:31
Example 4
22:37
Solving a System Using Substitution

20m 1s

Intro
0:00
Objectives
0:09
Solving a System Using Substitution
0:32
Substitution Method
1:24
Substitution Example
2:35
One Solution, No Solution, or Infinite Solutions
7:50
Example 1
9:45
Example 2
12:48
Example 3
15:01
Example 4
17:30
Solving a System Using Elimination

19m 40s

Intro
0:00
Objectives
0:09
Solving a System Using Elimination
0:27
Elimination Method
0:42
Elimination Example
2:01
One Solution, No Solution, or Infinite Solutions
7:05
Example 1
8:53
Example 2
11:46
Example 3
15:37
Example 4
17:45
Applications of Systems of Equations

24m 34s

Intro
0:00
Objectives
0:12
Applications of Systems of Equations
0:30
Word Problems
1:31
Example 1
2:17
Example 2
7:55
Example 3
13:07
Example 4
17:15
Section 6: Inequalities
Solving Linear Inequalities in One Variable

17m 13s

Intro
0:00
Objectives
0:08
Solving Linear Inequalities in One Variable
0:37
Inequality Expressions
0:46
Linear Inequality Solution Notations
3:40
Inequalities
3:51
Interval Notation
4:04
Number Lines
4:43
Set Builder Notation
5:24
Use Same Techniques as Solving Equations
6:59
'Flip' the Sign when Multiplying or Dividing by a Negative Number
7:12
'Flip' Example
7:50
Example 1
8:54
Example 2
11:40
Example 3
14:01
Compound Inequalities

16m 13s

Intro
0:00
Objectives
0:07
Compound Inequalities
0:37
'And' vs. 'Or'
0:44
'And'
3:24
'Or'
3:35
'And' Symbol, or Intersection
3:51
'Or' Symbol, or Union
4:13
Inequalities
4:41
Example 1
6:22
Example 2
9:30
Example 3
11:27
Example 4
13:49
Solving Equations with Absolute Values

14m 12s

Intro
0:00
Objectives
0:08
Solve Equations with Absolute Values
0:18
Solve Equations with Absolute Values Cont.
1:11
Steps to Solving Equations with Absolute Values
2:21
Example 1
3:23
Example 2
6:34
Example 3
10:12
Inequalities with Absolute Values

17m 7s

Intro
0:00
Objectives
0:07
Inequalities with Absolute Values
0:23
Recall…
2:08
Example 1
3:39
Example 2
6:06
Example 3
8:14
Example 4
10:29
Example 5
13:29
Graphing Inequalities in Two Variables

15m 33s

Intro
0:00
Objectives
0:07
Graphing Inequalities in Two Variables
0:32
Split Graph into Two Regions
1:53
Graphing Inequalities
5:44
Test Points
6:20
Example 1
7:11
Example 2
10:17
Example 3
13:06
Systems of Inequalities

21m 13s

Intro
0:00
Objectives
0:08
Systems of Inequalities
0:24
Test Points
1:10
Steps to Solve Systems of Inequalities
1:25
Example 1
2:23
Example 2
7:28
Example 3
12:51
Section 7: Polynomials
Integer Exponents

44m 51s

Intro
0:00
Objectives
0:09
Integer Exponents
0:42
Exponents 'Package' Multiplication
1:25
Example 1
2:00
Example 2
3:13
Integer Exponents Cont.
4:50
Product Rule for Exponents
4:51
Example 3
7:16
Example 4
10:15
Integer Exponents Cont.
13:13
Power Rule for Exponents
13:14
Power Rule with Multiplication and Division
15:33
Example 5
16:18
Integer Exponents Cont.
20:04
Example 6
20:41
Integer Exponents Cont.
25:52
Zero Exponent Rule
25:53
Quotient Rule
28:24
Negative Exponents
30:14
Negative Exponent Rule
32:27
Example 7
34:05
Example 8
36:15
Example 9
39:33
Example 10
43:16

18m 33s

Intro
0:00
Objectives
0:07
0:25
Terms
0:33
Coefficients
0:51
1:13
Like Terms
1:29
Polynomials
2:21
Monomials, Binomials, Trinomials, and Polynomials
5:41
Degrees
7:00
Evaluating Polynomials
8:12
9:25
Example 1
11:48
Example 2
13:00
Example 3
14:41
Example 4
16:15
Multiplying Polynomials

25m 7s

Intro
0:00
Objectives
0:06
Multiplying Polynomials
0:41
Distributive Property
1:00
Example 1
2:49
Multiplying Polynomials Cont.
8:22
Organize Terms with a Table
8:23
Example 2
13:40
Multiplying Polynomials Cont.
16:33
Multiplying Binomials with FOIL
16:48
Example 3
18:49
Example 4
20:04
Example 5
21:42
Dividing Polynomials

44m 56s

Intro
0:00
Objectives
0:07
Dividing Polynomials
0:29
Dividing Polynomials by Monomials
2:10
Dividing Polynomials by Polynomials
2:59
Dividing Numbers
4:09
Dividing Polynomials Example
8:39
Example 1
12:35
Example 2
14:40
Example 3
16:45
Example 4
21:13
Example 5
24:33
Example 6
29:02
Dividing Polynomials with Synthetic Division Method
33:36
Example 7
38:43
Example 8
42:24
Section 8: Factoring Polynomials
Greatest Common Factor & Factor by Grouping

28m 27s

Intro
0:00
Objectives
0:09
Greatest Common Factor
0:31
Factoring
0:40
Greatest Common Factor (GCF)
1:48
GCF for Polynomials
3:28
Factoring Polynomials
6:45
Prime
8:21
Example 1
9:14
Factor by Grouping
14:30
Steps to Factor by Grouping
17:03
Example 2
17:43
Example 3
19:20
Example 4
20:41
Example 5
22:29
Example 6
26:11
Factoring Trinomials

21m 44s

Intro
0:00
Objectives
0:06
Factoring Trinomials
0:25
Recall FOIL
0:26
Factor a Trinomial by Reversing FOIL
1:52
Tips when Using Reverse FOIL
5:31
Example 1
7:04
Example 2
9:09
Example 3
11:15
Example 4
13:41
Factoring Trinomials Cont.
15:50
Example 5
18:42
Factoring Trinomials Using the AC Method

30m 9s

Intro
0:00
Objectives
0:08
Factoring Trinomials Using the AC Method
0:27
Factoring when Leading Term has Coefficient Other Than 1
1:07
Reversing FOIL
1:18
Example 1
1:46
Example 2
4:28
Factoring Trinomials Using the AC Method Cont.
7:45
The AC Method
8:03
Steps to Using the AC Method
8:19
Tips on Using the AC Method
9:29
Example 3
10:45
Example 4
16:50
Example 5
21:08
Example 6
24:58
Special Factoring Techniques

30m 14s

Intro
0:00
Objectives
0:07
Special Factoring Techniques
0:26
Difference of Squares
1:46
Perfect Square Trinomials
2:38
No Sum of Squares
3:32
Special Factoring Techniques Cont.
4:03
Difference of Squares Example
4:04
Perfect Square Trinomials Example
5:29
Example 1
7:31
Example 2
9:59
Example 3
11:47
Example 4
15:09
Special Factoring Techniques Cont.
19:07
Sum of Cubes and Difference of Cubes
19:08
Example 5
23:13
Example 6
26:12

23m 38s

Intro
0:00
Objectives
0:08
0:19
0:20
Zero Factor Property
1:39
Zero Factor Property Example
2:34
Example 1
4:00
Solving Quadratic Equations by Factoring Cont.
5:54
Example 2
7:28
Example 3
11:09
Example 4
14:22
Solving Quadratic Equations by Factoring Cont.
18:17
Higher Degree Polynomial Equations
18:18
Example 5
20:22

29m 27s

Intro
0:00
Objectives
0:12
0:29
Linear Factors
0:38
1:22
Principle of Square Roots
3:36
Completing the Square
4:50
Steps for Using Completing the Square
5:15
Completing the Square Works on All Quadratic Equations
6:41
7:28
Discriminants
8:25
10:11
Example 1
11:54
Example 2
13:03
Example 3
16:30
Example 4
21:29
Example 5
25:07

16m 47s

Intro
0:00
Objectives
0:08
0:24
Using a Substitution
0:53
U-Substitution
1:26
Example 1
2:07
Example 2
5:36
Example 3
8:31
Example 4
11:14

29m 4s

Intro
0:00
Objectives
0:09
0:35
Squared Variable
0:40
Principle of Square Roots
0:51
Example 1
1:09
Example 2
2:04
3:34
Example 3
4:42
Example 4
13:33
Example 5
20:50

26m 53s

Intro
0:00
Objectives
0:06
0:39
Axis of Symmetry
1:46
Vertex
2:12
Transformations
2:57
3:23
Example 1
5:06
Example 2
6:02
Example 3
9:07
11:26
Completing the Square
12:02
Vertex Shortcut
12:16
Example 4
13:49
Example 5
17:25
Example 6
20:07
Example 7
23:43
Polynomial Inequalities

21m 42s

Intro
0:00
Objectives
0:07
Polynomial Inequalities
0:30
Solving Polynomial Inequalities
1:20
Example 1
2:45
Polynomial Inequalities Cont.
5:12
Larger Polynomials
5:13
Positive or Negative Intervals
7:16
Example 2
9:01
Example 3
13:53
Section 10: Rational Equations
Multiply & Divide Rational Expressions

26m 41s

Intro
0:00
Objectives
0:09
Multiply and Divide Rational Expressions
0:44
Rational Numbers
0:55
Dividing by Zero
1:45
Canceling Extra Factors
2:43
Negative Signs in Fractions
4:52
Multiplying Fractions
6:26
Dividing Fractions
7:17
Example 1
8:04
Example 2
14:01
Example 3
16:23
Example 4
18:56
Example 5
22:43

20m 24s

Intro
0:00
Objectives
0:07
0:41
Common Denominators
0:52
Common Denominator Examples
1:14
Steps to Adding and Subtracting Rational Expressions
2:39
Example 1
3:34
Example 2
5:27
Adding and Subtracting Rational Expressions Cont.
6:57
Least Common Denominators
6:58
Transitioning from Fractions to Rational Expressions
9:08
Identifying Least Common Denominators for Rational Expressions
9:56
10:41
Example 3
11:19
Example 4
12:36
Example 5
15:08
Example 6
16:46
Complex Fractions

18m 23s

Intro
0:00
Objectives
0:09
Complex Fractions
0:37
Dividing to Simplify Complex Fractions
1:10
Example 1
2:03
Example 2
3:58
Complex Fractions Cont.
9:15
Using the Least Common Denominator to Simplify Complex Fractions
9:16
10:07
Example 3
10:42
Example 4
14:28
Solving Rational Equations

16m 24s

Intro
0:00
Objectives
0:07
Solving Rational Equations
0:23
Isolate the Specified Variable
1:23
Example 1
1:58
Example 2
5:00
Example 3
8:23
Example 4
13:25
Rational Inequalities

18m 54s

Intro
0:00
Objectives
0:06
Rational Inequalities
0:18
Testing Intervals for Rational Inequalities
0:38
Steps to Solving Rational Inequalities
1:05
Tips to Solving Rational Inequalities
2:27
Example 1
3:33
Example 2
12:21
Applications of Rational Expressions

20m 20s

Intro
0:00
Objectives
0:07
Applications of Rational Expressions
0:27
Work Problems
1:05
Example 1
2:58
Example 2
6:45
Example 3
13:17
Example 4
16:37
Variation & Proportion

27m 4s

Intro
0:00
Objectives
0:10
Variation and Proportion
0:34
Variation
0:35
Inverse Variation
1:01
Direct Variation
1:10
Setting Up Proportions
1:31
Example 1
2:27
Example 2
5:36
Variation and Proportion Cont.
8:29
Inverse Variation
8:30
Example 3
9:20
Variation and Proportion Cont.
12:41
Constant of Proportionality
12:42
Example 4
13:59
Variation and Proportion Cont.
16:17
Varies Directly as the nth Power
16:30
Varies Inversely as the nth Power
16:53
Varies Jointly
17:09
Combining Variation Models
17:36
Example 5
19:09
Example 6
22:10
Rational Exponents

14m 32s

Intro
0:00
Objectives
0:07
Rational Exponents
0:32
Power on Top, Root on Bottom
1:05
Example 1
1:37
Rational Exponents Cont.
4:04
Using Rules from Exponents for Radicals as Exponents
4:05
Combining Terms Under a Single Root
4:50
Example 2
5:21
Example 3
7:39
Example 4
11:23
Example 5
13:14
Simplify Rational Exponents

15m 12s

Intro
0:00
Objectives
0:07
Simplify Rational Exponents
0:25
0:26
Product Rule to Simplify Square Roots
1:11
1:42
Applications of Product and Quotient Rules
2:17
Higher Roots
2:48
Example 1
3:39
Example 2
6:35
Example 3
8:41
Example 4
11:09

17m 22s

Intro
0:00
Objectives
0:07
0:33
Like Terms
1:29
Bases and Exponents May be Different
2:02
Bases and Powers Must be Same when Adding and Subtracting
2:42
3:55
Example 1
4:47
Example 2
6:00
7:10
Simplify the Bases to Look the Same
7:25
Example 3
8:23
Example 4
11:45
Example 5
15:10

19m 24s

Intro
0:00
Objectives
0:08
0:25
0:26
1:11
Don’t Distribute Powers
2:54
4:25
Rationalizing Denominators
6:40
Example 1
7:22
Example 2
8:32
9:23
Rationalizing Denominators with Higher Roots
9:25
Example 3
10:51
Example 4
11:53
13:13
Rationalizing Denominators with Conjugates
13:14
Example 5
15:52
Example 6
17:25

15m 5s

Intro
0:00
Objectives
0:07
0:17
0:18
Isolate the Roots and Raise to Power
0:34
Example 1
1:13
Example 2
3:09
7:04
7:05
Example 3
7:54
Example 4
13:07
Complex Numbers

29m 16s

Intro
0:00
Objectives
0:06
Complex Numbers
1:05
Imaginary Numbers
1:08
Complex Numbers
2:27
Real Parts
2:48
Imaginary Parts
2:51
Commutative, Associative, and Distributive Properties
3:35
4:04
Multiplying Complex Numbers
6:16
Dividing Complex Numbers
8:59
Complex Conjugate
9:07
Simplifying Powers of i
14:34
Shortcut for Simplifying Powers of i
18:33
Example 1
21:14
Example 2
22:15
Example 3
23:38
Example 4
26:33
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• ## Related Books

### Linear Equations in Two Variables

• To build the equation of a line you can use point-slope form ( y – y1 = m(x-x1) ). To use this you must know the slope of the line and one point on the line.
• Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals of one another.
• You can test if two lines are perpendicular by multiplying their slopes together. If you get -1 then you know they are perpendicular.
• You may have to use other formulas such as the slope formula, before using the point-slope form.

### Linear Equations in Two Variables

A line has a slope of - 4 and passes through the point ( - 4, - 3).
Find the equation of this line in point slope form.
• y − y1 = m(x − x1)
• y − ( − 3) = m[x − ( − 3)]
• y + 3 = m(x + 3)
y + 3 = − 4(x + 3)
A line has a slope of - 7 and passes through the point (11, - 1).
Find the equation of this line in point slope form.
• y − y1 = m(x − x1)
• y − ( − 1) = − 7(x − 11)
y + 1 = − 7(x − 11)
A line has a slope of - 8 and passes through the point ( - 14,10).
Find the equation of this line in point slope form.
• y − y1 = m(x − x1)
• y − 10 = − 8[x − ( − 14)]
y − 10 = − 8(x + 14)
A horizontal line passes through the point (5,3). Find the equation of this line in point slope form.
• slope = 0 for a horizontal line
m = 0
• y − y1 = m(x − x1)
y − 3 = 0(x − 5)
A horizontal line passes through the point ( - 12, - 13). Find the equation of this line in point slope form.
• slope = 0 for a horizontal line
m = 0
• y − y1 = m(x − x1)
• y − ( − 13) = 0[x − ( − 12)]
y + 13 = 0(x + 12)
The equation of a line is
y − 5 = [1/5](x + 3)
Find the equation of this line in slope intercept form.
• y − 5 = [1/5](x + 3)
• 5(y − 5) = [ [1/5](x + 3) ]5
• 5y − 25 = 1(x + 3)
• 5y − 25 = x + 3
• 5y − 25 + 25 = x + 3 + 25
• 5y = x + 28
• y = [(x + 28)/5]
y = [1/5]x + [28/5]
The equation of a line is
y + 10 = [7/30](x − 2)
Find the equation of this line in slope intercept form.
• y + 10 = [7/30](x − 2)
• 30(y + 10) = [ [7/30](x − 2) ]30
• 30y + 300 = 7(x − 2)
• 30y + 300 = 7x − 14
• 30y = 7x − 314
• y = [(7x − 314)/30]
• y = [7/30]x − [314/30]
y = [7/30]x − [157/15]
The equation of a line is y + 12 = [5/6](x + 9)
Find the equation of this line in slope intercept form.
• y + 12 = [5/6](x + 9)
• 6(y + 12) = [ [5/6](x + 9)]6
• 6y + 72 = 5(x + 9)
• 6y + 72 = 5x + 45
• 6y = 5x − 27
y = [(5x − 27)/6]
A line passes through the points (7,9) and (2,5). Find the equation of this line in point slope form.
• m = [(y2 − y1)/(x2 − x1)]
• m = [(5 − 9)/(2 − 7)]
• m = [( − 4)/( − 5)]
• m = [4/5]
• y − y1 = m(x − x1)
y − 9 = [4/5](x − 7)
A line passes through the points ( - 1,3) and ( - 7, - 11). Find the equation of this line in point slope form.
• m = [(y2 − y1)/(x2 − x1)]
• m = [( − 11 − 3)/( − 7 − ( − 1))]
• m = [( − 14)/( − 6)]
• m = [14/6] = [7/3]
• y − y1 = m(x − x1)
• y − 3 = [7/3][x − ( − 1)]
y − 3 = [7/3](x + 1)
A line is parallel to the line whose equation is
y = − [1/4]x − 3
This line also passes through the point ( - 5,7)
Find the equation of this line in slop intercept form.
• y = mx + b
parallel lines have the same slope
m = − [1/4]b = ?
• 7 = − [1/4]( − 5) + b
• 7 = [5/4] + b
• 7 = 1[1/4] + b
• 5[3/4] = b
y = − [1/4]x + 5[3/4]
A line is parallel to the line whose equation is
y = − [3/7]x + 6
This line also passes through the point ( - 4,2)
Find the equation of this line in slop intercept form.
• y = mx + b
parallel lines have the same slope
m = − [3/7]b = ?
• 2 = − [3/7]( − 4) + b
• 2 = [12/7] + b
• 2 = 1[5/7] + b
• [2/7] = b
y = - [3/7]x + [2/7]
Are the lines determined by the equations
3x − 6y = 18
4x − 2y = 12
parallel, perpendicular, or neither?
• 3x − 6y = 18
• − 6y = 18 − 3x
• y = [(18 − 3x)/( − 6)]
• y = [18/( − 6)] − [3x/( − 6)]
• y = − 3 + [x/2]m = [1/2]
• 4x − 2y = 12
• − 2y = − 4x + 12
• y = [( − 4x + 12)/( − 2)]
• y = 2x − 6m = 2
• ([1/2]) ≠ − 1
• ([1/2]) ≠ (2)
neither
Are the lines determined by the equations
5x − 2y = 10
2x + 12y = 24
parallel, perpendicular, or neither?
• 5x − 2y = 10
• − 2y = − 5x + 10
• y = [( − 5x)/( − 2)] + [10/( − 2)]
• y = [5/2]x − 5m = [5/2]
• 2x + 12y = 24
• 12y = − 2x + 24
• y = [( − 2x)/12] + [24/12]
• y = − [x/6] + 2m = − [1/6]
• (−[1/6])([5/2]) ≠ − 1
• (−[1/6]) ≠ ([5/2])
neither
A line passes through the point ( - 4, - 3) and is perpendicular to the line whose equation is y = 1/5 x − 6. Find the equation of this line in slope intercept form.
• y = mx + b
m = [1/5]
negative reciprocal = - 5
• − 3 = − 5( − 4) + b
• − 3 = 20 + b
• − 23 = b
y = − 5x − 23
A line passes through the point (5,4) and is perpendicular to the line whose equation is y = − 7/10 x + 3. Find the equation of this line in slope intercept form.
• y = mx + b
m = − [7/10]
negative reciprocal = [10/7]
• 4 = [10/7](5) + b
• 4 = [50/7] + b
• 4 = 7[1/7] + b
• - 3[1/7] = b
y = [10/7]x − 3[1/7]
A line is perpendicular to the line whose equation is 2x − 4y = 8. This line also passes through the y - intercept of the graph of x − 5y = 20. Find the equation of this line in slope intercept form.
• 2x − 4y = 8
• − 4y = − 2x + 8
• y = [( − 2x + 8)/( − 4)]
• y = [1/2]x − 2
• m = [1/2]
slope of perpendicular line = - 2
• x − 5y = 20
• − 5y = − x + 20
• y = [( − x + 20)/( − 5)]
• y = [1/5]x − 4b = − 4
y = − 2x + 4
A line is perpendicular to the line whose equation is 3x − 7y = 42. This line also passes through the y - intercept of the graph of 6x + 3y = 30. Find the equation of this line in slope intercept form.
• 3x − 7y = 42
• − 7y = − 3x + 42
• y = [( − 3x + 42)/( − 7)]
• y = [3/7]x − 6
negative reciprocal = − [7/3]
• 6x + 3y = 30
• 3y = − 6x + 30
• y = [( − 6x + 30)/3]
• y = − 2x + 10
b = 10
y = − [7/3]x + 10
A line is parallel to the line whose equation is
y = − [2/5]x − 6
This line also passes through the point ( - 10,12)
Find the equation of this line in slop intercept form.
• y = mx + b
parallel lines have the same slope
m = − [2/5]b = ?
• 12 = − [2/5]( − 10) + b
• 12 = [20/5] + b
• 12 = 4 + b
• 16 = b
y = − [2/5]x + 16
A line passes through the point ( - 6, - 4) and is perpendicular to the line whose equation is y = − 2/3 x + 8. Find the equation of this line in slope intercept form.
• m = − [2/3]
negative reciprocal = [3/2]
• − 4 = [3/2]( − 6) + b
• − 4 = [( − 18)/2] + b
• − 4 = − 9 + b
• 5 = b
y = [3/2]x + 5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Linear Equations in Two Variables

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:13
• Linear Equations in Two Variables 0:36
• Point-Slope Form
• Substitute in the Point and the Slope
• Parallel Lines: Two Lines with the Same Slope
• Perpendicular Lines: Slopes are Negative Reciprocals of Each Other
• Perpendicular Lines: Product of Slopes is -1
• Example 1 6:02
• Example 2 7:50
• Example 3 10:49
• Example 4 13:26
• Example 5 15:30
• Example 6 17:43

### Transcription: Linear Equations in Two Variables

Welcome back to www.educator.com.0000

In this lesson we are going to continue on with our linear equations and look at how we can build these linear equations for ourselves.0002

Specifically, some of things that we will look at is we will look at the point slope form on a line.0014

It will help us when building those equations.0020

For some of our examples, we will also have to know a little bit more about parallel and perpendicular lines.0023

Watch out for my explanation on what these are and how you can tell if two lines are parallel or perpendicular.0029

Earlier, we looked at two forms of a line.0038

We looked at standard form and we looked at slope intercept form.0041

Remember both of these forms are important because they were good for graphing.0045

When we want to build our own line, sometimes we can do this as long as we have enough information.0057

To form that we use for building our line is known as point slope form.0064

The reason why it is called that is because those are the two bits of information that we need.0068

We need to know at least one point on a line and we need to know what the slope of that line is.0072

Once we have both those bits of information and we can drop it into point slope form.0077

This form here requires a little bit of explanation.0086

You will see that there is an x and y that has some subscripts on it and that is where the point goes in terms of its x and y.0090

The slope is the n here so we can just put that in and then we also have a couple of other x’s and y’s which do not have subscripts.0098

With the ones that do not have any subscripts whatsoever, you will not be putting any values in for those.0108

Those simply stay in our equation.0114

One last thing to note is even though we will use point slope to actually build the equation on a line.0116

Usually you take point slope and end up converting it into slope intercept form because you are looking to do some other things without a line,0122

other than just than build it or maybe you want to eventually graph it.0129

It is good to go ahead and move it in to one of those other forms.0132

To build lines using point slope form, we want to substitute in our point and we will go ahead and substitute in our slope.0143

I already have point slope form, I will put it over here and I have a point and I have a slope.0151

Let us go ahead and just substitute these in and see how it works.0157

I will have y - -3 then the m represent the slope, I will put in ¾, x – xy.0161

Notice how the x and y which did not have any subscripts are still in there.0176

Once we have everything substitute in here, we want to go ahead and start rewriting this0181

either into standard form or slope intercept form so we can do some other stuff with it.0186

I’m going to go ahead and put this into slope intercept form by just getting y all by itself and maybe cleaning up a few other terms.0190

I'm subtracting a negative on the left side and that will be the same as adding 3.0198

I will go ahead and distribute my ¾ here.0203

I have ¾ x and -3, looking much better.0206

I will go ahead and subtract 3 from both sides.0216

Y = 3/4x – 3 – 3 is 6.0219

Our point slope gets our foot in the door so we can actually build the equation and then put it into a different form, maybe a slope intercept down here.0227

We can go ahead and graph it and do some other things with it.0241

Sometimes we will be given information about another line in order to build the one we want.0247

And information about that other line, we may know that it is parallel or perpendicular to the one we want.0253

What exactly does that mean?0258

We can say that two lines are parallel if they had exactly the same slope.0261

In my little picture here, you can see what that does.0266

You will end up with two lines and they usually have a little bit of a gap or space in between them, but they have exactly the same slope.0269

They are going in the same direction.0275

You might also have lines that are perpendicular.0279

With this one, they end up meeting at a right angle.0282

We do not talk much about angles in this course, another way that you can say two lines or perpendicular is0286

if their slopes are negative reciprocals of one another.0291

Let me show you what that means.0295

I suppose this blue line had a slope of 4/5, if the red one was perpendicular then I wanted to be the negative reciprocal of the blue line.0296

I made it negative and I flipped it over.0312

With perpendicular lines, they will always be different in sign, one will be positive and one will be negative and they will be reciprocal when flipped over.0314

Another way that you can test if two lines are perpendicular or not is actually you take both of their slopes and just multiply them together.0325

Two things could happen if you multiply them together and their slope is negative or their combined value is -10333

then you will know for sure that they are perpendicular.0342

If you multiply them together and you get anything else other than -1 then you will know that they are not perpendicular.0346

It is a nice and easy task you can use to know how the two lines are related.0352

We will definitely know this as we get into more of those examples.0358

Let us start off with example 1, in this one we want to write the equation on a line using the given point and the slope.0363

Since the two bits of information that I need for point slope, I will just go ahead and nearly drop it into the formula.0372

y -y1 equals slope x - x1.0378

Let us put in the y value first, y - 3 equals my slope is -2/3x - -6.0386

You will know the formula has a minus sign in there since the x value is a negative, go ahead and put that negative sign in there as well.0399

Now all we got to do is clean it up a little bit and maybe turned it into a different form.0408

I will see what we can do.0412

y – 3 = -2/3, when I subtract the negative that is the same as addition.0413

x + 6 and I think I will go ahead and distribute this -2/3 in there.0422

-2/3x – 12 ÷ 3 and a -12 ÷ 3 = -4, it looks pretty good.0430

Let us add 3 to both sides and now we have that same line written into slope intercept form.0449

Again we use point slope form to go ahead and create the equation of the line and then probably put it into some other form.0462

This one is a little different, in this one we simply want to determine whether the two lines are parallel0472

or maybe the perpendicular or maybe they do not fall into either of those two cases.0477

With these ones, notice how many of them are written not in one of the forms that we have covered.0482

In other words they are not in slope intercept form but the second one is in standard form but we have to be able to figure out what their slope is.0489

A way to determine what the slope of the line is to go ahead and rewrite it into our slope intercept form and we can just read it out of the equation.0501

Let us do that first before we actually compare what these slopes are.0509

Starting with this first one here, I can move the 2 to the other side by subtracting 2 so y =5x – 2.0512

Let us see what the second one, let us go ahead and move the x to the other side, -x – 15 and then we will divide it by 5.0521

Here is our first line, here is our second one.0539

Now they are both in slope intercept form we can say that the first one has a slope of 50545

since it is right next to x and the other one has a slope of -1/5.0550

Looking at the two slopes, they are definitely not the same, they are not parallel.0555

It looks like one is positive and one is negative and they are reciprocals.0562

I think these two are perpendicular.0567

If you want to test out feel free to just take the two slopes and multiply them together and see that when you do this you get a -1.0570

You will know that these two lines are perpendicular.0579

Let us try this process with the next pair of lines.0590

I will begin by just putting this into slope intercept form.0595

This one is almost in slope intercept form.0599

I just have to reverse the y and putting it first.0602

This one I will go ahead and move the 3x to the other side, so now I have my two lines right here.0606

y = 2x + 1 and y = -3x + 4, in this form, I can read off what both of their slopes are.0616

Looking at their slopes I can see that they are not the same, they are definitely not parallel.0627

One is positive and one is negative, but they are not reciprocal so they are not perpendicular either.0633

They are not parallel or not perpendicular, I will put this in the neither categories.0638

They are just two lines hanging around on the graph.0643

Let us try that same process with another pair of lines.0650

Let us see if we can find a couple of that which actually are parallel.0652

We will begin by putting these into a better form so we can read off that slope.0656

I'm moving the 4x to the other side, this one the y is still not completely all by itself.0662

Let us go ahead and divide everything through by 2, now we have one of our lines.0668

The second one let us go ahead and rearrange things.0678

The y is on the left and let us get it completely isolated by multiplying everything through by -1.0683

Now we have both of our lines.0693

Now that they are in this form, the slope of the first one is -2 and the slope of the second one is -2.0702

I can see that they are exactly the same and we will call these pair of lines parallel.0709

Onto one more pair of lines, let us see what they are, parallel, perpendicular, or neither, we will find out.0719

I need to get my y isolated I will start off by moving the 4x to the other side.0726

I will divide both sides by 3, y =4/3x +2.0734

The second one, let us go ahead and move the 2 over.0745

I have 3x + 2 and I will divide everything by the 4, ¾ x + ½.0749

Here is equation 1 and here is equation 2.0760

If you look at them in this form, we can pick out what each of their slopes are, I have 4/3 and 3/4.0767

These ones are pretty close, they are definitely reciprocals of one another since you would take 4/3 and flip it over and get ¾.0775

When those have been both positive so we can not say that these are perpendicular.0783

Even though they are reciprocals they are not negative reciprocals of one another.0788

They are definitely so they are not parallel.0792

Even though they are close, they do end up in the neither category.0795

They are not parallel or not perpendicular, they are simply neither.0799

They are just two lines.0802

With this one we want to write the equation of a line that happens to be parallel to the given line and also goes through the given point.0809

This one is a little different, notice how we need to know what the slope of the line is0818

and we need to know what the point is but they have not quite given us what the slope is.0822

We are going to have to extract that out of this other line that they have given us by knowing that it is parallel to the one we want to build.0827

Let us hunt down what the slope of this other line is.0834

We will do that by dividing everything through by 4.0837

It looks good so y = ¼ x + 5.0845

Let us see, I know that the slope is ¼.0850

Now that I have a point, I have a slope, I can use point slope form and build our line.0855

Y - -3 = ¼ x – xy and now let us clean it up and see what line we have.0861

y + 3 = ¼ x, ¼ of 2 would be -1/2.0872

Now we will subtract 3 from both sides.0890

y = ¼ x – ½ - 3, get a common denominator over there.0893

Our line is y = ¼ x – 7/2.0908

The way we built this line, I know for sure that it goes through the given point 2, -3.0917

I already explored what the slope of the other line was so I know it has a slope of 1/4.0923

Let us try this one, write the equation of a line which is perpendicular to the given line and it goes through the given point.0932

Similar to the other one, but it have not given us the slope directly.0939

It just told us it is perpendicular to this other line over here.0943

Let us figure out what slope is, y equals, I will move the 8 to the other side, -8x + 3.0948

The slope of this line is -8 which is good but that is not exactly the slope we want to use.0958

Our line is perpendicular to this line, so we need to use the slope -1, 1/8.0966

That way it is the negative reciprocal of the other.0979

Now we will drop it into our formula.0984

Let us go ahead and put in our y value.0994

We have our slope and our x value.0997

We will clean it up and see what line we have.1002

y - 8 = 1/8 - 4/8, my x in there.1005

y – 8, 1/8x – ½ and we will go ahead and 8 to both sides, find a common denominator and combine the last of our like terms.1017

1/8x + 15/2.1044

We can see that its slope is the negative reciprocal of the other line.1053

I know it is perpendicular and the way we build that it, for sure it goes through the point 4/8.1057

This last one is a little bit more of the word problem but you can see how we can still use these techniques in order to build the equation of the line.1066

We want to build a line that represents the following problem.1074

It cost a $20 flat fee to rent a drill +$2 every day starting with the first day.1077

Let x represent the number of days that we rent this drill and y represent the charge to the users.1085

How much will we end up willing to get from them?1091

When we are all done, let us see if we can write this line in a slope intercept form.1095

Using the techniques that we know about so far, we need to figure out what our slope is and what point this will go through.1100

Let us see, one thing that we can think of is the slope being what changes or your variable costs.1108

The $20 that you have to pay up front is not a variable.1120 It is going to be there no matter what, what does change is a$2 every single day.1123

The variable cost, we can think of that as our slope.1134

Let us see what we can do with that.1144

What happens with that flat cost?1146

We will be charged of that no matter what, that is like our y intercept.1151

If this one, we can just drop in our variable cost and we can drop in our flat fee.1158

What we are left here is the equation of the line but it represents how much we end up charging the person1177

and we can see this by substituting some values.1183

Suppose they rent this drill for one day, it cost $2 for that day +$20 flat fee, it cost them $22.1186 If they rent it for two days, we can put that in for x,$4 for the variable + $20 flat fee,$24.1196

You can go on and on figuring out how much it cost for each of the different days.1206

But in the end, this formula right here would represent how much we need to charge them.1211

Just simply plug in the number of days for x and y would give you the cost.1215

You can see that building a line is not so bad and using point slope form is handy in that entire process.1221

If you do use point slope form, you need to know the slope of the line and you need to know a point on that line.1228

Thanks for watching www.educator.com.1234

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text:
Showing results for
Do you mean
Solved!

How do I add a text note to a dimension ? I have the standard dimension and am trying to add a note after the dimension i.e. 200mm (text note).

Regards,

Anthony

SE ST8

2 REPLIES
Solution
Solution
Accepted by topic author agoodwin
02-29-2016 05:06 AM

Betreff: Add text note to dimension

RMB on dimension and dimension prefix will give YOu the option to add prefix, suffix super or subfix

regards
Wolfgang

Betreff: Add text note to dimension

Perfect, thank you


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E. Choosing The Commander
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.

Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.

Each warrior is represented by his personality — an integer number pi. Each commander has two characteristics — his personality pj and leadership lj (both are integer numbers). Warrior i respects commander j only if ( is the bitwise excluding OR of x and y).

Initially Vova's army is empty. There are three different types of events that can happen with the army:

• pi — one warrior with personality pi joins Vova's army;
• pi — one warrior with personality pi leaves Vova's army;
• pi li — Vova tries to hire a commander with personality pi and leadership li.

For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.

Input

The first line contains one integer q (1 ≤ q ≤ 100000) — the number of events.

Then q lines follow. Each line describes the event:

• pi (1 ≤ pi ≤ 108) — one warrior with personality pi joins Vova's army;
• pi (1 ≤ pi ≤ 108) — one warrior with personality pi leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment);
• pi li (1 ≤ pi, li ≤ 108) — Vova tries to hire a commander with personality pi and leadership li. There is at least one event of this type.
Output

For each event of the third type print one integer — the number of warriors who respect the commander Vova tries to hire in the event.

Example
Input
51 31 43 6 32 43 6 3
Output
10
Note

In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (, and 2 < 3, but , and 5 ≥ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him.