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# The position of an object moving along a line is given by p(t) = 2t^3 - 2t^2 +1. What is the speed of the object at t = 3 ?
Aug 11, 2017
$v = 42 \text{ units}$
#### Explanation:
The speed of an object is the magnitude of the object's velocity , which is the derivative of displacement (magnitude is position).
To find the speed of the object at time $t$, we can begin by taking the derivation of the provided equation for position:
$p \left(t\right) = 2 {t}^{3} - 2 {t}^{2} + 1$
$\implies p ' \left(t\right) = v \left(t\right) = 6 {t}^{2} - 4 t$
At $t = 3$, we have:
$v \left(t\right) = 6 {\left(3\right)}^{2} - 4 \left(3\right)$
$= 54 - 12$
$= 42$
$\therefore$ At $t = 3$, the object has a speed of $42 \text{ units.}$ |
# What are tessellations in math
Tessellations
A Tessellation (or Tiling) is when we cover a surface with a pattern of flat shapes so that there are no overlaps or gaps. A tessellation is a pattern of shapes repeated to fill a plane. The shapes do not overlap and there are no gaps. The figure above composed of squares is a tessellation since the are no gaps or overlaps between any 2 squares. The figure above composed of regular pentagons is not a tessellation since there are gaps between the tessellations in grey.
A tessellation is a pattern of shapes repeated to fill a plane. The shapes do not overlap and there are no gaps. Tessellations are something we often see in quilts, carpets, floors, and more. In the figure below are three examples. The shape that is repeated, or tessellated, is called a motif. The following are the motifs for the tessellations above.
This is true for any vertex in the tessellation. A regular tessellation is made up of regular congruent polygons. There are only three tessellations that are composed entirely of regular, congruent polygons. A semi-regular tessellation is a tessellation that is composed of two or more regular polygons such that the arrangement of the polygons is the what does the word superior mean for each vertex in the tessellation.
The pattern around each vertex is identical. We call this pattern the order of the vertex of the tessellation, and name it based on the number of sides of each regular polygon surrounding the vertex. The order of the semi-regular tessellation composed of equilateral triangles, squares, and regular hexagons shown above is Start with the polygon with the fewest number of sides first, then rotate clockwise or counterclockwise and count the number of sides for the successive polygons to complete the order.
A non-regular tessellation is a tessellation that is composed of other shapes that may or may not be polygons. The figure above composed of squares is a tessellation since the are no gaps or overlaps between any 2 squares. The figure above composed of regular pentagons is not a tessellation since there are gaps between the tessellations in grey.
The figure above composed of regular octagons is not a tessellation since there is overlap between consecutive octagons in grey. Triangular tessellation.
Each polygon is a non-overlapping equilateral triangle. What are tessellations in math hexagons and equilateral triangles tessellate around each vertex in the order of Regular hexagons, equilateral triangles, and squares tessellate around each vertex in the order of Regular octagons and squares tessellate around each vertex in the order of Regular square and equilateral triangles tessellate around each vertex in the order of Regular dodecagonshexagons, and squares tessellate around each vertex in the order of Regular dodecagons and equilateral triangles tessellate around each vertex in the order of Equilateral triangles and squares tessellate around each vertex in the order of
Did you find this activity helpful?
Illustrated definition of Tessellation: A pattern made of one or more shapes: the shapes must fit together without any gaps the shapes. A tessellation is the tiling of a plane using one or more geometric shapes such that there are no overlaps or gaps. In other words, a tessellation is a never-ending pattern on a flat 2-D surface (such as a piece of paper) where all of the shapes fit together perfectly like . Apr 11, · A tessellation is a regular pattern made up of flat shapes repeated and joined together without any gaps or overlaps. These shapes do not all need to be the same, but the pattern should repeat. Another word for tessellation is tiling. Here are some tessellation examples.
A tessellation is a regular pattern made up of flat shapes repeated and joined together without any gaps or overlaps. These shapes do not all need to be the same, but the pattern should repeat. Another word for tessellation is tiling. The word tessellation is derived from the Greek "tesseres", which means "four" and refers to the four sides of a square, the first shape to be tiled.
A regular tessellation is a pattern made by repeating a regular polygon. A regular polygon is one having all its sides equal and all it's interior angles equal. So there are only 3 kinds of regular tessellations - ones made from squares, equilateral triangles and hexagons. Where the shapes join together, the corner point, we call that the vertex. By looking at the vertex and counting the sides of all the shapes that meet at the vertex you are able to name a tessellation.
Choose a vertex and count the sides of the polygons that touch it. In the example above of a regular tessellation of hexagons, next to the vertex are three polygons and each has six sides, so this tessellation is called " 6. A semi-regular tessellation is made of two or more regular polygons e.
The pattern at each vertex should be the same. There are also demi-regular tessellations or polymorph tessellations , but they are difficult to define. Some have described them as a tilings of the 3 regular and 8 semi-regular tessellations, but this is not a very precise definition.
Here are some tessellation examples Regular Tessellations A regular tessellation is a pattern made by repeating a regular polygon. For example Semi-Regular Tessellations A semi-regular tessellation is made of two or more regular polygons e. There are only 8 semi-regular tessellations Other Tessellations There are also demi-regular tessellations or polymorph tessellations , but they are difficult to define.
And some people allow for tessellations of curved shapes. The above exampe of the tessellated lizards was done by famous artist MC Escher. All rights reserved.
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# How do you solve -4y+16=4?
Aug 16, 2016
y = 3
#### Explanation:
-4y = 4 - 16 (subtract 16 from both sides)
-4y = -12 (simplify RHS)
y = 3 (divide both sides by -4)
Aug 16, 2016
$y = 3$
#### Explanation:
To solve this equation, isolate the - 4y term on the left of the equation while having numeric values on the right.
To do this, subtract 16 from both sides.
$\Rightarrow - 4 y \cancel{+ 16} \cancel{- 16} = 4 - 16$
we are left with $- 4 y = - 12$
now divide both sides by - 4.
$\frac{{\cancel{- 4}}^{1} y}{\cancel{- 4}} ^ 1 = \frac{- 12}{- 4} \Rightarrow y = 3 \text{ is the solution}$ |
# 1.6: Subtract Whole Numbers (Part 2)
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## Translate Word Phrases to Math Notation
As with addition, word phrases can tell us to operate on two numbers using subtraction. To translate from a word phrase to math notation, we look for key words that indicate subtraction. Some of the words that indicate subtraction are listed in Table $$\PageIndex{1}$$.
Table $$\PageIndex{1}$$
Operation Word Phrase Example Expression
Subtraction minus 5 minus 1 5 - 1
difference the difference of 9 and 4 9 - 4
decreased by 7 decreased by 3 7 - 3
less than 5 less than 8 8 - 5
subtracted from 1 subtracted from 6 6 - 1
##### Example $$\PageIndex{11}$$: translate
Translate and then simplify:
1. the difference of $$13$$ and $$8$$
2. subtract $$24$$ from $$43$$
Solution
1. The word difference tells us to subtract the two numbers. The numbers stay in the same order as in the phrase.
the difference of 13 and 8 Translate. 13 - 8 Simplify. 5
1. The words subtract from tells us to take the second number away from the first. We must be careful to get the order correct.
subtract 24 from 43 Translate. 43 - 24 Simplify. 19
##### Exercise $$\PageIndex{21}$$
Translate and simplify:
1. the difference of $$14$$ and $$9$$
2. subtract $$21$$ from $$37$$
$$14-9=5$$
$$37-21=16$$
##### Exercise $$\PageIndex{22}$$
Translate and simplify:
1. $$11$$ decreased by $$6$$
2. $$18$$ less than $$67$$
$$11-6=5$$
$$67-18=49$$
## Subtract Whole Numbers in Applications
To solve applications with subtraction, we will use the same plan that we used with addition. First, we need to determine what we are asked to find. Then we write a phrase that gives the information to find it. We translate the phrase into math notation and then simplify to get the answer. Finally, we write a sentence to answer the question, using the appropriate units.
##### Example $$\PageIndex{12}$$: difference
The temperature in Chicago one morning was $$73$$ degrees Fahrenheit. A cold front arrived and by noon the temperature was $$27$$ degrees Fahrenheit. What was the difference between the temperature in the morning and the temperature at noon?
Solution
We are asked to find the difference between the morning temperature and the noon temperature.
Write a phrase. the difference of 73 and 27 Translate to math notation. Difference tells us to subtract. 73 - 27 Then we do the subtraction. Write a sentence to answer the question. The difference in temperatures was 46 degrees Fahrenheit.
##### Exercise $$\PageIndex{23}$$
The high temperature on June $$1^{st}$$ in Boston was $$77$$ degrees Fahrenheit, and the low temperature was $$58$$ degrees Fahrenheit. What was the difference between the high and low temperatures?
The difference is $$19$$ degrees Fahrenheit.
##### Exercise $$\PageIndex{24}$$
The weather forecast for June $$2^{nd}$$ in St Louis predicts a high temperature of $$90$$ degrees Fahrenheit and a low of $$73$$ degrees Fahrenheit. What is the difference between the predicted high and low temperatures?
The difference is $$17$$ degrees Fahrenheit.
##### Example $$\PageIndex{13}$$: difference
A washing machine is on sale for $$399$$. Its regular price is $$588$$. What is the difference between the regular price and the sale price?
Solution
We are asked to find the difference between the regular price and the sale price.
## Everyday Math
1. Road trip Noah was driving from Philadelphia to Cincinnati, a distance of 502 miles. He drove 115 miles, stopped for gas, and then drove another 230 miles before lunch. How many more miles did he have to travel?
2. Test Scores Sara needs 350 points to pass her course. She scored 75, 50, 70, and 80 on her first four tests. How many more points does Sara need to pass the course?
## Writing Exercises
1. Explain how subtraction and addition are related.
## Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section
(b) What does this checklist tell you about your mastery of this section? What steps will you take to improve? |
# Triangle A has an area of 15 and two sides of lengths 4 and 9 . Triangle B is similar to triangle A and has a side of length 12 . What are the maximum and minimum possible areas of triangle B?
##### 1 Answer
Jul 17, 2016
135 and $\approx 15.8$, respectively.
#### Explanation:
The tricky thing in this problem is that we do not know which of the tree sides of the original triangle corresponds to the one of length 12 in the similar triangle.
We know that the area of a triangle can be calculated from Heron's formula
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - x\right)}$
For our triangle we have $a = 4$ and $b = 9$ and so $s = \frac{13 + c}{2}$, $s - a = \frac{5 + c}{2}$, $s - b = \frac{c - 5}{2}$ and $s - c = \frac{13 - c}{2}$. Thus
${15}^{2} = \frac{13 + c}{2} \times \frac{5 + c}{2} \times \frac{c - 5}{2} \times \frac{13 - c}{2}$
This leads to a quadratic equation in ${c}^{2}$ :
${c}^{4} - 194 {c}^{2} + 7825 = 0$
which leads to either $c \approx 11.7$ or $c \approx 7.5$
So the maximum and minimum possible value for the sides of our original triangle are 11.7 and 4, respectively. Thus the maximum and minimum possible value of the scaling factor are $\frac{12}{4} = 3$ and $\frac{12}{11.7} \approx 1.03$. Since area scales as square of length, the maximum and minimum possible values of the area of the similar triangle are $15 \times {3}^{2} = 135$ and $15 \times {1.03}^{2} \approx 15.8$, respectively. |
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# The plot represents the flow of current through a wire at three different times. The ratio of charge flowing through the wire at three different times is (A) $2:1:2$(B) $1:3:3$(C) $1:1:1$(D) $2:3:4$
Last updated date: 13th Jun 2024
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Hint In a current v/s time plot, the charge is given by the area under the curve. So we can calculate the area under the curve at three different times and then by finding their ratio, we can find the charge flowing through the wire at three different times.
Formula Used: In this solution we will be using the following formula,
$\Rightarrow I = \dfrac{{dq}}{{dt}}$
where $I$ is the current in the wire,
$q$ is the charge flowing and $t$ is the time.
In the question we are given the plot of the current flowing through a wire. Now the current in a wire and the charge flowing is related by the formula,
$\Rightarrow I = \dfrac{{dq}}{{dt}}$
Now taking the $dt$ from the denominator of the RHS to the LHS to get,
$\Rightarrow dq = Idt$
Therefore, in a plot of current and time, the charge will be given by the area under the curve. So let us name the 3 curves as 1, 2 and 3.
Now for the first plot, we can see that it is a rectangle.
According to the figure, the sides of the rectangle are $2A$ along the Y-axis and $1s$along the X-axis.
So we have the area as,
$\Rightarrow {A_1} = 2 \times 1 = 2C$
Now for the second plot, we can see that it is a rectangle.
According to the figure, the sides of the rectangle are $1A$ along the Y-axis and $2s$along the X-axis.
So we have the area as,
$\Rightarrow {A_2} = 1 \times 2 = 2C$
Now for the third plot, we can see that it is a triangle.
According to the figure, the height of the triangle is $2A$ along the Y-axis and the base is $2s$ along the X-axis.
So we have the area as,
$\Rightarrow {A_2} = \dfrac{1}{2} \times 2 \times 2 = 2C$
Therefore the ratio of the charges will be the ratio of the areas under the plot,
$\Rightarrow {A_1}:{A_2}:{A_3} = 2:2:2$
So the ratio is $1:1:1$
Hence the correct answer is option C.
Note
When there are charges moving through a wire, then a current is generated in the wire. Hence the rate of flow of charge through a wire per unit time is called the current in the wire. The unit of current in the wire is coulomb and it is denoted by the letter C. |
# Perimeter and Circumference Worksheets
What is the Difference Between Perimeter & Circumference? The word perimeter is derived from the Greek word peri meaning around and metron, which means to measure. Hence, in geometry, the perimeter is defined as the boundary or the path that surrounds a geometrical shape or object. In simpler terms, the perimeter is also defined as the outline of a geometrical shape. When we find the perimeter of the shape, we typically find the total length of sides of any two-dimensional shape. Initially, students are taught to use rulers and thread to measure the perimeter of a shape. But as they start learning geometry, they are taught different formulas to find out the perimeter of the shape. For example, if you are given a rectangle of length 3 inches by 8 inches, to find the perimeter. We all know that the opposite sides a rectangle are of the same lengths so that we will have two sides of measurement 3 inches and two sides of measurements 8 inches. We will add these four lengths to calculate the perimeter of a rectangle. We will have 3 + 3 + 8 + 8 = 22 inches. We use the same approach to find out the perimeter of most polygons. But finding the perimeter of a circular figure is different. For circles, we find the circumference, which is the complete distance around a circle. We can find the circumference by using the formula: C = 2πr, Where π = 3.14116 and r is the radius of the circle.
• ### Basic Lesson
Guides students through finding the perimeter of a square and the circumference of a circle.
• ### Intermediate Lesson
Demonstrates how to find the perimeter of a parallelogram and an odd shaped figure.
• ### Independent Practice 1
A really great activity for allowing students to reinforce the concept of Perimeter & Circumference.
• ### Independent Practice 2
Students draw and Perimeter & Circumference in assorted problems. The answers can be found below.
• ### Independent Practice 3
An in-depth review of Perimeter & Circumference are found on this worksheet.
• ### Independent Practice 4
Students draw on past knowledge to solve this set of Perimeter & Circumference problems. The answers can be found below.
• ### Homework Worksheet
Students are provided with problems to achieve the concepts of Perimeter & Circumference.
• ### Skill Quiz
This tests the students ability to master Perimeter & Circumference.
Answers for the homework and quiz.
• ### Answer Key Part 2
Answers for lessons and both practice sheets.
• ### Basic Lesson
Guides students through finding the area of a rectangle and the circumference of a circle.
• ### Intermediate Lesson
Demonstrates how to use this skill in a word problem.
• ### Independent Practice 1
A really great activity for allowing students to understand the concepts of the Perimeter of Polygons & Circumference of Circles.
• ### Independent Practice 2
Students use Perimeter of Polygons & Circumference of Circles in 20 assorted problems. The answers can be found below.
• ### Homework Worksheet
Students are provided with 12 problems to achieve the concepts of Perimeter of Polygons & Circumference of Circles.
• ### Skill Quiz
This tests the students ability to understand Perimeter of Polygons & Circumference of Circles. |
# How do you differentiate f(x)=sqrt(tane^(4x) using the chain rule.?
Feb 8, 2016
f'(x) =( 2e^(4x)sec^2e^(4x))/sqrt(tane^(4x)
#### Explanation:
making use of the chain rule
d/dx(f(g(x)) = f'(g(x)).g'(x)
and knowing $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x \textcolor{b l a c k}{\text{ and }} \frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$
first step is to rewrite $\sqrt{\tan {e}^{4 x}} = {\left(\tan {e}^{4 x}\right)}^{\frac{1}{2}}$
$f ' \left(x\right) = \frac{1}{2} {\left(\tan {e}^{4 x}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(\tan {e}^{4 x}\right)$
now $\frac{d}{\mathrm{dx}} \left(\tan {e}^{4 x}\right) = {\sec}^{2} {e}^{4 x} \frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) \ldots \ldots \ldots \ldots . \left(1\right)$
and $\frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) = {e}^{4 x} \frac{d}{\mathrm{dx}} \left(4 x\right) = 4 {e}^{4 x} \ldots \ldots \ldots \ldots \ldots . \left(2\right)$
replacing the results of (1) and (2) back into f'(x)
$f ' \left(x\right) = \frac{1}{2} {\left(\tan {e}^{4 x}\right)}^{- \frac{1}{2}} . {\sec}^{2} {e}^{4 x} . 4 {e}^{4 x}$
and rewriting negative exponent as radical
rArr f'(x) =( 2e^(4x)sec^2e^(4x))/sqrt(tane^(4x) |
## Basics of Statistics
Statistics is the analysis of data or information. In order for you to solve statistics, you have to know the basic information or data.
#### What is data?
Data is a group of different facts. It is divided into two types i.e. the qualitative and the quantitative fact.
Qualitative fact is a fact that describes. This type of fact is subjective and broad. For example, “it was funny”.
On the other hand, quantitative fact, from the name itself, represents the quantity of the data. You will see numbers in this data. And since quantitative data shows values and numbers, we can use this to solve for problems that require quantity and figures.
Data can be presented in many ways. Some of these examples are as follows:
1. Bar graph
This shows data in using bars.
2. Pie chart
Imagine a pie being divided into parts and what you get is a pie chart. Each part of the pie represents the data you present.
3. Line graph
This presents data that shows relationship such as changes in time, performance in time, among others.
4. Pictograms
This presentation of data can be used for presenting data to children since it shows the picture of the data collected and their relationship with each other.
In statistics, there are terms that we always encounter. These are mean, median and mode.
#### Mean
Mean number is the average of all the data you have presented.
To solve for the mean value of the data presented, all you have to do is to add up all the numbers and then divide the product by how many numbers you have added.
#### Example:
Find the mean of the following data
1. 5, 8, -2, 6, 7, 10
= 5 + 8 + (-2) + 6 + 7 + 10
= 34/6
= 5.67
2. 12, -8, -2, 5, 8, 14
= 12 + (-8) + (-2) + 5 + 8 + 14
= 29/6
= 4.83
3. 21, 22, 21, -9, 16, 15
= 21 + 22 + 21 + (-9) + 16 + 15
= 86/6
= 14.3
4. 32, 20, 18, 11, 10, -2
= 32 + 20 + 18 + 11 + 10 + (-2)
= 89/6
= 14.83
5. 16, 12, -8, 12, 32, 30, 5
= 16 + 12 + (-8) + 12 + 32 + 30 + 5
= 99/7
= 14.14
6. 10, -12, 25, 15
= 10 + (-12) + 25 + 15
= 38/4
= 9.5
7. 6, 8, 12, 10, 5
= 6 + 8 + 12 + 10 + 5
= 41/5
= 8.2
8. 13, 12, 8, -25
= 13 + 12 + 8 + (-25)
= 8/4
= 2
#### Median
When we say median, we mean the middle.
To find the median of a presented data, all you have to do is arrange the given numbers in the data presented and then find the middle number. In the event that you have two median values, all you have to do is to add the two medians and find its mean.
#### Example:
Find the median of the following:
1. 1, 2, 5, 16, 25
= 1, 2, 5, 16, 25
= 5
2. 36, 102, 75, 18, 3, 5
= 3, 5, 18, 36, 75, 102
= 18 + 36
2
= 27
3. 5, 2, 1, 32, 75, 18
= 1, 2, 5, 18, 32, 75
= 5 + 18
2
= 11.5
4. 1, 3, 14, 15, 8, 9
= 3, 8, 9, 14, 15
= 9
5. 75, 76, 101, 32, 10
= 10, 32, 75, 76, 101
= 75
#### Mode
Mode or the modal value is the number which appears the most.
To find it, you just have to look at the data presented, tally it and then find the number which appears the most. In the event that you find two modes in the given data, that is called “bimodal”. It could also have more than to modes, and that is called “multimodal”.
#### Examples:
1. 13, 25, 6, 25, 25, 11, 8, 9
= 25
2. 26, 3, 3, 3, 3, 2, 2, 2, 2, 5, 7, 28
= 3 and 2
3. 1, 2, 5, 3, 8, 8, 8, 9, 10, 21
= 8
4. 5, 3,16, 17, 17, 17, 3, 58
= 17
5. 16, 9, 9, 9, 9, 12, 1, 8, 8,8, 8
=9 and 8 |
Last updated:
# Multiplicative Inverse Modulo Calculator
What is the multiplicative inverse in modular arithmetic?When does the multiplicative modular inverse exist?How to use this multiplicative inverse modulo calculator?Finding the multiplicative modulo inverse with Bézout's identityFAQs
The multiplicative inverse modulo calculator is of immeasurable value whenever you need to quickly find the multiplicative inverse modulo for some m, be it for a math assignment, a programming project, or any other scientific endeavor you deal with.
In the brief article below, we'll explain how to find the multiplicative inverse modulo — both by Bézout's identity and by brute force (depending on how much you care about mathematical subtlety). And to spare you useless work, we'll also tell you how to check if the multiplicative modular inverse exists in the first place.
🙋 We assume you're already familiar with the modulo operation in math. If this is not the case (or you feel you need a refresher), check out Omni's modulo calculator.
## What is the multiplicative inverse in modular arithmetic?
Let a and x be integers. We say that x is the modular multiplicative inverse of a (modulo m) if
a × x ≡ 1 (mod m).
That is, when a × x and 1 are congruent modulo m, i.e., when (a × x) mod m = 1.
In even simpler words, the remainder from the division of a × x by m must equal 1.
❗ It may happen that the multiplicative inverse modulo does not exist! For instance, the multiplicative modular inverse of 3 modulo 6 does not exist: you may check yourself that for every number x ∊ {1, 2, 3, 4, 5}, the result of (3 × x) mod 6 is different from 1.
## When does the multiplicative modular inverse exist?
The modular multiplicative inverse of a modulo m exists if and only if a and m are coprime (a.k.a. relatively prime), i.e., if their greatest common factor equals one: gcd(a,m) = 1.
If m is prime, then the multiplicative modular inverse modulo m exists for every non-zero integer a that is not a multiple of m.
🙋 To quickly determine the greatest common divisor of two integers, use Omni's GCF calculator.
As you can see, it's easy to verify if the multiplicative modular inverse exists, but computing it is quite a different story. The fastest method is to use our multiplicative inverse modulo calculator!
## How to use this multiplicative inverse modulo calculator?
Using this multiplicative inverse modulo calculator is really simple:
1. Enter a positive integer m: the number with which we calculate the modulo.
2. Enter an integer a: the number whose multiplicative inverse modulo m we look for.
3. Our calculator returns the answer immediately. A short explanation is provided as well.
Are you curious how our tool can solve this modulo problem so quickly? In the next section, we explain the method implemented in our calculator.
## Finding the multiplicative modulo inverse with Bézout's identity
Let a and m be integers. Bézout's identity says that there exist two integers x and y such that:
a×x + m×y = gcd(a, m).
That is, we can represent gcd(a, m) as a linear combination of a and m with coefficients x and y.
It turns out we can use this representation to find the multiplicative inverse of a modulo m. Recall that a and m must be coprime, so gcd(a,m) = 1 — Bézout's identity says that there exist integers x and y such that:
a×x + m×y = 1
Next, we apply the mod m operation to both sides:
(a×x + m×y) mod m = 1 mod m
⇒ a×x + m×y ≡ 1 (mod m)
The second form is just short-hand for the first form — they mean the same. The left-hand side simplifies to a×x because m×y is divisible by m. That is, we get:
a×x ≡ 1 (mod m)
This means that we have the solution right in front of our eyes: x is the multiplicative inverse of a modulo m!
🔎 Finding x and y for Bézout's identity is not trivial. The best method is to use the .
FAQs
### How do I find the multiplicative modulo inverse by hand?
To find the multiplicative inverse of a modulo m by brute force:
1. Take any number x from the set {0, 1, ..., m − 1} and calculate a × x.
2. Find the remainder from the division of a × x by m.
3. If this remainder is 1, you've found the solution.
4. If not, repeat Steps 1–3 for a different number x.
5. If all numbers from {0, 1, ..., m − 1} fail, then the multiplicative inverse of a modulo m does not exists.
### Is the multiplicative inverse modulo m unique?
No — if x is a multiplicative inverse of a modulo m, then every number of the form x + (k×m) (where k is an integer) is a multiplicative inverse of a modulo m as well. However, the solution is unique in the set {1, ..., m − 1}.
### Does 142 have a multiplicative inverse modulo 76?
No, 142 does not have a multiplicative inverse modulo 76. The reason is that these numbers are not relatively prime, i.e., their greatest common factor exceeds 1. Indeed, we immediately see that both 142 and 76 are both divisible by 2.
### What numbers have multiplicative inverses modulo 11?
Every integer that is not a multiple of 11 has a multiplicative inverse modulo 11. This is because 11 is a prime number, and so it is not coprime only with its multiplicities. Calculating the multiplicative inverses may be tiresome, so don't hesitate to use an online multiplicative inverse modulo calculator.
We find x such that a × x ≡ 1 (mod m) |
# Factorize each of the following polynomials:
Question:
Factorize each of the following polynomials:
1. $x^{3}+13 x^{2}+31 x-45$ given that $x+9$ is a factor
2. $4 x^{3}+20 x^{2}+33 x+18$ given that $2 x+3$ is a factor
Solution:
1. $x^{3}+13 x^{2}+31 x-45$ given that $x+9$ is a factor
let, $f(x)=x^{3}+13 x^{2}+31 x-45$
given that (x + 9) is the factor
divide f(x) with (x + 9) to get other factors
by , long division
$x^{2}+4 x-5$
$x+9 x^{3}+13 x^{2}+31 x-45$
$x^{3}+9 x^{2}$
(-) (-)
$4 x^{2}+31 x$
$4 x^{2}+36 x$
(-) (-)
-5x – 45
-5x – 45
(+) (+)
0
$=x^{3}+13 x^{2}+31 x-45=(x+9)\left(x^{2}+4 x-5\right)$
Now,
$x^{2}+4 x-5=x^{2}+5 x-x-5$
= x(x + 5) -1(x + 5)
= (x + 5) (x – 1) are the factors
Hence, $x^{3}+13 x^{2}+31 x-45=(x+9)(x+5)(x-1)$
2. $4 x^{3}+20 x^{2}+33 x+18$ given that $2 x+3$ is a factor
let, $f(x)=4 x^{3}+20 x^{2}+33 x+18$
given that 2x + 3 is a factor
divide f(x) with (2x + 3) to get other factors
by, long division
$2 x^{2}+7 x+6$
$2 x+3,4 x^{3}+20 x^{2}+33 x+18$
$4 x^{3}+6 x^{2}$
(-) (-)
$14 x^{2}-33 x$
$14 x^{2}-21 x$
(-) (+)
12x + 18
12x + 18
(-) (-)
0
$=>4 x^{3}+20 x^{2}+33 x+18=(2 x+3)\left(2 x^{2}+7 x+6\right)$
Now,
$2 x^{2}+7 x+6=2 x^{2}+4 x+3 x+6$
= 2x(x + 2) + 3(x + 2)
= (2x + 3)(x + 2) are the factors
Hence, $4 x^{3}+20 x^{2}+33 x+18=(2 x+3)(2 x+3)(x+2)$ |
# How do you factor and simplify sin^4x-cos^4x?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
34
Nghi N Share
Apr 18, 2017
= - cos 2x
#### Explanation:
${\sin}^{4} x - {\cos}^{4} x = \left({\sin}^{2} x + {\cos}^{2} x\right) \left({\sin}^{2} x - {\cos}^{2} x\right)$
Reminder:
${\sin}^{2} x + {\cos}^{2} x = 1$, and
${\cos}^{2} x - {\sin}^{2} x = \cos 2 x$
Therefore:
${\sin}^{4} x - {\cos}^{4} x = - \cos 2 x$
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
24
Apr 18, 2017
$\left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$
#### Explanation:
Factorizing this algebraic expression is based on this property:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
Taking ${\sin}^{2} x = a$ and ${\cos}^{2} x = b$ we have :
${\sin}^{4} x - {\cos}^{4} x = {\left({\sin}^{2} x\right)}^{2} - {\left({\cos}^{2} x\right)}^{2} = {a}^{2} - {b}^{2}$
Applying the above property we have:
${\left({\sin}^{2} x\right)}^{2} - {\left({\cos}^{2} x\right)}^{2} = \left({\sin}^{2} x - {\cos}^{2} x\right) \left({\sin}^{2} x + {\cos}^{2} x\right)$
Applying the same property on${\sin}^{2} x - {\cos}^{2} x$
thus,
${\left({\sin}^{2} x\right)}^{2} - {\left({\cos}^{2} x\right)}^{2}$
$= \left(\sin x - C o s x\right) \left(\sin x + \cos x\right) \left({\sin}^{2} x + {\cos}^{2} x\right)$
Knowing the Pythagorean identity, ${\sin}^{2} x + {\cos}^{2} x = 1$ we simplify the expression so,
${\left({\sin}^{2} x\right)}^{2} - {\left({\cos}^{2} x\right)}^{2}$
$= \left(\sin x - C o s x\right) \left(\sin x + \cos x\right) \left({\sin}^{2} x + {\cos}^{2} x\right)$
$= \left(\sin x - \cos x\right) \left(\sin x + \cos x\right) \left(1\right)$
$= \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$
Therefore,
${\sin}^{4} x - {\cos}^{4} x = \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$
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Home | Math | 10 Examples of Systems of Equations
# 10 Examples of Systems of Equations
October 16, 2023
written by Rida Mirza
In mathematics, system of equations is a collection of two or more equations that share the same set of variables.
In this article, we will discuss ten examples of systems of equations in mathematics.
## Examples of Systems of Equations
These are 10 examples of system of equations.
### 1: Linear Equations in Two Variables
Classic example of a system of equations involves two linear equations in two variables.
For example,
2x + 3y = 7
4x – y = 6
### 2: Three Equations with Three Unknowns
Systems can have more equations and variables. A system with three equations in three variables is,
x + y – z = 5
2x – 3y + z = 1
3x + y + 2z = 8
### 3: Distance, Rate, and Time Problems
Systems of equations are useful for solving distance, rate, and time problems.
For example, If you travel 300 miles at a certain speed, it takes 5 hours. If you increase your speed by 10 mph, you can cover the same distance in 4 hours. Find the original speed.
### 4: Mixing Problems
Mixing problems involve combining solutions with different concentrations. A system of equations help determine the quantities of each solution.
### 5: Investment and Interest Problems
When calculating investments with different interest rates, a system of equations help determine the amount to invest at each rate to achieve a specific total return.
### 6: Physics Problems
Physics often involves systems of equations.
For example, equations related to velocity, acceleration, and time are used to solve problems involving motion.
### 7: Chemical Reactions
Balancing chemical equations is a classic application of systems of equations in chemistry.
### 8: Economics and Market Equilibrium
Systems of equations are used to model supply and demand in economics, helping to find equilibrium prices and quantities.
### 9: Circuit Analysis
Electrical circuits are analyzed using systems of equations. The Kirchhoff’s circuit laws are expressed through a system of equations.
### 10: Optimization Problems
In optimization, we seek to maximize or minimize a particular quantity. Systems of equations are used to formulate constraints and objectives in these problems.
File Under: |
### Hallway Borders
What are the possible dimensions of a rectangular hallway if the number of tiles around the perimeter is exactly half the total number of tiles?
### Not a Polite Question
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?
##### Age 11 to 14 ShortChallenge Level
Answer: $30$
Using algebra and adding the sums together
Let $39=a+b$, $48=a+c$ and $51=b+c$.
$c$ is the largest number because $a+b=39$ is the smallest sum.
Adding together the sums containing $c$, \begin{align}48+51&=a+c+b+c\\ \Rightarrow 99&=\underbrace{a+b}_{39}+2c\\ 99&=39+2c\\ \Rightarrow60&=2c\\ \Rightarrow30&=c\end{align}
Using algebra and elimination
Let $39=a+b$, $48=a+c$ and $51=b+c$.
$c$ is the largest number because $a+b=39$ is the smallest sum.
Subtracting $39=a+b$ from $48=a+c$ we can eliminate $a$:
\begin{align}48-39&=a+c-(a+b)\\ \Rightarrow 9&=c-b\end{align}
Adding $9=c-b$ and $51=b+c$ we can eliminate $b$:
\begin{align}9+51&=c-b+c+b\\ \Rightarrow60&=2c\\ \Rightarrow30&=c\end{align}
Trying out numbers
51 = largest number added to middle number.
51 = 25 + 26, so the largest number must be at least 26.
largest number middle number (=51-largest number) smallest number (=39- middle number) smallest number + largest number (should be 48)
26 25 14 40 - much too small
32 19 20 - too large 52 - too large
29 22 17 46 - too small
30 21 18 48
So the largest number is 30.
Finding the differences between the numbers
39 is the smallest number so it must be the sum of the smallest and middle numbers.
51 is the largest number so it must be the the sum of the largest and middle numbers.
48 must be the sum of the largest and smallest numbers.
So when the smallest number is added to the middle and largest numbers, the sums are 39 and 48 respectively. So the difference between 39 and 48 must be the same as the difference between the middle and largest numbers. So the largest number must be 9 more than the middle number.
So 51 is the sum of two numbers whose difference is 9.
Ways to make 51
25 + 26 = 51 difference: 1
23 + 28 = 51 difference: 5
21 + 30 = 51 difference: 9
So the largest number is 30.
39 is the smallest number so it must be the sum of the smallest and middle numbers.
51 is the largest number so it must be the the sum of the largest and middle numbers.
48 must be the sum of the largest and smallest numbers.
We are interested in the largest number. Adding together the sums containing the largest number,
the largest number + the middle number + the largest number + the smallest number = 51 + 48 = 99.
But that is the largest number twice, plus the sum of the smallest and middle numbers. But we know that the sum of the smallest and middle numbers is 39! So the largest number twice is 99 $-$ 39 = 60.
So the largest number is 30.
You can find more short problems, arranged by curriculum topic, in our short problems collection. |
AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.2
AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.2 Textbook Questions and Answers.
AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.2
Question 1.
In the figure, if AB = CD and ∠AOB = 90° find ∠COD.
Solution:
‘O’ is the centre of the circle.
AB = CD (equal chords from the figure)
∴ ∠AOB = ∠COD
[ ∵ equal chords make equal angles at the centre]
∴ ∠COD = 90° [ ∵ ∠AOB = 90° given]
Question 2.
In the figure, PQ = RS and ∠ORS = 48°.
Find ∠OPQ and ∠ROS.
Solution:
‘O’ is the centre of the circle.
PQ = RS [given, equal chords]
∴∠POQ = ∠ROS [ ∵ equal chords make equal angles at the centre]
∴ In ΔROS
∠ORS + ∠OSR + ∠ROS = 180°
[angle sum property]
∴ 48° + 48° + ∠ROS = 180°
[ ∵ OR = OS(radii); ΔORS is isosceles]
∴ ∠ROS = 180° – 96° = 84°
Also ∠POQ = ∠ROS = 84°
∴ ∠OPQ = ∠OQP
[∵ OP = OQ; radii]
= $$\frac { 1 }{ 2 }$$[180°-84°] = 48°
Question 3.
In the figure, PR and QS are two diameters. Is PQ = RS ?
Solution:
‘O’ is the centre of the circle.
[ ∵ PR, QS are diameters]
OP = OR (∵ radii) .
OQ = OS (∵ radii)
∠POQ = ∠ROS [vertically opp. angles]
∴ ΔOPQ ≅ ΔORS [SAS congruence]
∴ PQ = RS [CPCT] |
## What is the greatest common factor mean?
The greatest common factor (GCF) of a set of numbers is the largest factor that all the numbers share. For example, 12, 20, and 24 have two common factors: 2 and 4. GCF is often used to find common denominators.
## How do you find the greatest common factor?
To find the GCF of two numbers:
1. List the prime factors of each number.
2. Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1.
## What is the GCF of 36 and 54?
1, 2, 3, 6, 9, 18, 27, and 54. Although the numbers in bold are all common factors of both 36 and 54, 18 is the greatest common factor. The second method to find the greatest common factor is to list the prime factors, then multiply the common prime factors.
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## What is the greatest common factor of 12 and 18?
Example 1: 6 is the greatest common factor of 12 and 18.
## What is the GCF of 18 and 24?
18 and 24 share one 2 and one 3 in common. We multiply them to get the GCF, so 2 * 3 = 6 is the GCF of 18 and 24.
## What is the GCF of 12 and 30?
Earlier we found that the Common Factors of 12 and 30 are 1, 2, 3 and 6, and so the Greatest Common Factor is 6. The Greatest Common Factor of 12 and 30 is 6.
## What is the GCF of 27 and 30?
Answer: GCF of 27 and 30 is 3
GCF of 27 and 30 is the highest number that divides 27 and 30 exactly leaving the remainder 0.
## What is the GCF of 27 and 63?
GCF of 27 and 63 is 9.
## What is the GCF of 20 24 and 40?
The greatest common factor of 20, 24, and 40 is 4.
## What is the GCF of 18 and 27?
The factors of 27 are 1, 3, 9, 27. The common factors of 18 and 27 are 1, 3 and 9. The greatest common factor of 18 and 27 is 9.
## What is the GCF of 30?
The product of these factors is called the Greatest Common Factor, or GCF for short. Example: 30 = (5)(3)(2)
## What is the GCF for 36 and 48?
Answer: GCF of 36 and 48 is 12.
## What is the GCF of 35 and 30?
What is the Greatest Common Factor of 30 and 35? Greatest common factor (GCF) of 30 and 35 is 5. We will now calculate the prime factors of 30 and 35, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 30 and 35.
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## What is the HCF of 45 and 30?
Answer: HCF of 45 and 30 is 15
Highest Common Factor (HCF) of t45 and 30 is the highest possible number which divides both the numbers exactly without any remainder.
## What are the common factors of 18 and 30?
Greatest common factor (GCF) of 18 and 30 is 6. |
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# RD Sharma Solutions - Ex-23.3, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev
## Class 7: RD Sharma Solutions - Ex-23.3, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev
The document RD Sharma Solutions - Ex-23.3, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7
#### Question 1:
Find the median of the following data (1-8)
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Arranging the data in ascending order, we have:
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Here, the number of observations, n = 10 (Even).
Hence, the median of the given data is 49.5.
#### Question 2:
Find the median of the following data (1-8)
133, 73, 89, 108, 94, 104, 94, 85, 100, 120
Arranging the data in ascending order, we have:
73, 85, 89, 94, 94, 100, 104, 108, 120, 133.
Here, the number of observations n = 10 (Even).
Hence, the median of the given data is 97.
#### Question 3:
Find the median of the following data (1-8)
31, 38, 27, 28, 36, 25, 35, 40
Arranging the data in ascending order, we have:
25,27, 28, 31, 35, 36, 38, 40
Here, the number of observations n = 8 (Even).
Hence, the median of the given data is 33.
#### Question 4:
Find the median of the following data (1-8)
15, 6, 16, 8, 22, 21, 9, 18, 25
Arranging the data in ascending order, we have:
6, 8, 9, 15,16,18, 21, 22, 25
Here, the number of observations n = 9 (Odd).
⇒Median=Value of observati5on i.e , value of the 5th observation = 16
Hence, the median of the given data is 16
#### Question 5:
Find the median of the following data (1-8)
41, 43, 127, 99, 71, 92, 71, 58, 57
Arranging the given data in ascending order, we have:
41, 43, 57, 58, 71,71, 92, 99, 127
Here, n = 9, which is odd.
∴ Median = Value of observation, i.e., the 5th observation = 71.
#### Question 6:
Find the median of the following data (1-8)
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Arranging the given data in ascending order, we have:
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Here, n = 10, which is even.
Hence, the median is 27.5 for the given data.
#### Question 7:
Find the median of the following data (1-8)
12, 17, 3, 14, 5, 8, 7, 15
Arranging the given data in ascending order, we have:
3,5,7,8,12,14,15,17
Here, n = 8, which is even.
Hence, the median of the given data is 10.
#### Question 8:
Find the median of the following data (1-8)
92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Arranging the given data in ascending order, we have:
35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Here, n = 10, which is even.
Hence, the median of the given data is 68.
#### Question 9:
Numbers 50, 42, 35, 2x + 10, 2x − 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.
Here, the number of observations n is 9. Since n is odd , the median is the observation, i.e. the 5th observation.
As the numbers are arranged in the descending order, we therefore observe from the last.
Median = 5th observation.
⇒ 25 = 2x -8
⇒ 2x = 25 +8
⇒ 2x = 33
x = 33/2
x = 16.5
Hence, x = 16.5.
#### Question 10:
Find the median of the following observations : 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Arranging the given data in ascending order, we have:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Here, the number of observations n is 11 (odd).
Since the number of observations is odd, therefore,
Median = Value of observation = Value of the 6th observation = 58.
Hence, median = 58.
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99.
∴ New median = Value of the 6th observation = 58.
#### Question 11:
Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?
Arranging the given data in ascending order, we have:
41, 43, 57, 58, 61, 71, 92, 99,127
Here, the number of observations, n, is 9(odd).
∴ Median = Value of observation = Value of the 5th observation = 61.
Hence, the median = 61.
If 58 is replaced by 85 , then the new observations arranged in ascending order are:
41, 43, 57, 61, 71, 85, 92, 99,12 .
∴ New median = Value of the 5th observation = 71.
#### Question 12:
The weights (in kg) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Arranging the given data in ascending order, we have:
27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45.
Here, the number of observations n is 15(odd).
Since the number of observations is odd, therefore,
Median = Value of observation = Value of the 8th observation = 35.
Hence, median = 35 kg.
If 44 kg is replaced by 46 kg and 27 kg by 25 kg , then the new observations arranged in ascending order are:
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46.
∴ New median = Value of the 8th observation = 35 kg.
#### Question 13:
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x :
29, 32, 48, 50, xx + 2, 72, 78, 84, 95
Here, the number of observations n is 10. Since n is even,
⇒63 = x+1
⇒x = 63−1 = 62.
Hence, x = 62.
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# How to Calculate Acres in a Circle
••• YUCELOZBER/iStock/GettyImages
Print
Once upon a time, the acre was loosely defined as the amount of land that could be plowed in one day by one man working with one ox. While that's not a terribly useful means of measuring land today, the acre itself lingers as a measurement of land area in U.S. customary units of measure and the U.K.'s Imperial measurements. For the sake of visualization, an acre measures about three-quarters the size of a football field. With that said, if you want to calculate the number of acres in a circular area, you're going to have to do some measuring and calculating in feet first.
### First Feet, Then Acres
If you already know how many square feet are in your circular area, you can skip straight to converting that measurement into acres. But if you don't know the circle's area in square feet yet, you'll have to start by measuring its radius or diameter in feet. Here's why: you need that linear (or straight line) measurement in order to calculate the circle's area. And you can't take a linear measurement in acres because, by its very definition acres involves two dimensions (length and width), while a linear measurement has only one dimension (its length).
Measure the radius of the circle or, if it's easier, measure the diameter and then divide that by two to get the radius. The radius of the circle is the straight-line distance from its center point to any point on the circle; the diameter is the straight-line distance from any point on the circle, through the center point of the circle, and then on to the far side of the circle.
So if you're measuring an enormous circle that has a diameter of 200 feet, you can divide this by 2 to get the circle's radius:
200 \text{ feet} ÷ 2 = 100 \text{ feet}
Calculate the area of the circle in square feet, using the formula A = πr2, where A is the circle's area, r is the length of the circle's radius in feet and π is usually abbreviated as 3.14. This gives you:
A = 3.14 × (100 \text{ ft})^2
Which simplifies to:
A = 31400 \text{ ft}^2
As your final step, divide the result by 43560 to convert from square feet to acres. (One acre equals 43,560 square feet.) This gives you:
\frac{31400 \text{ ft}^2}{43560} = 0.72 \text{ acres}
#### Tips
• Note that because an acre is very large, it's not unusual to find yourself dealing with less than one acre, as in the example just given. In fact, in 2015 the average lot size for a newly constructed single-family home was just less than 1/5 or 0.2 acre.
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# Linear Equations in Real Life by JohnQuinn
VIEWS: 2,738 PAGES: 13
• pg 1
``` Real World
Applications of
Linear Equations
Yes, there really are uses for them!!
Math 208 ~ College Mathematics I
Objective: Apply the concepts of slope and intercept to real-life situations
Today’s Objectives
To define the slope of a
linear equation a real world
context
To find a meaning for an
equation’s x & y-intercepts
Brief Review of Graphing
Solve the equation for y 2x + y = 3
y = – 2x + 3
Make a table and find x y = –2x + 3 y
–1
some ordered pair y=2+3 5
0 Y=0+3 3
values 1 Y = –2 + 3 1
2 Y = –4 + 3 –1
Graph the Equation
Graph the ordered
pairs
(-1, 5)
Draw the line (0, 3)
(1, 1)
(shows all the (2, -1)
possible solutions)
First Problem to follow along.
Use the handout
What do we know?
A pillow company buys ribbon in 100 foot
rolls.
2.5 feet of ribbon are needed to decorate
each pillow.
What do we need to do?
Write an equation. Use y for the amount
ribbon left on the roll and x for the number
Fill in the table and draw the a graph.
Write Equation & Fill in the Table
# Pillows # of Feet of Ribbon
Made y = 100 – 2.5x Remaining on Roll
0 100 – 0 100
1 100 – 2.5 97.5
2 100 – 5 95
3 100 – 7.5 92.5
4 100 – 10 90
Draw the graph
100
90
80
70
60
50
40
30
20
10
1 2 3 4 5 6 7 8 9 10
Questions for
100
90
80
Discussion
70
60
50
40
30
y = 100 – 2.5x
20
1. What is the slope?
10
1 2 3 4 5 6 7 8 9 10
What does it represent?
2. Why is the slope negative?
3. What is the y-intercept? What does
it represent?
4. What is the x-intercept? What does
it represent?
Second Problem
What do we know?
A swimming pool holds 450 gallons of water.
The pool currently contains 100 gallons of
water.
A hose deposits 50 gallons of water in the
pool each minute.
What do we need to do?
Write an equation. Use y for the amount of
water in the pool and x for the number of
minutes the hose runs.
Fill in the table and graph the information.
Fill in the Table
# of Minutes Hose # of Liters in
is in Pool y = 100 + 50x The Pool
0 100 + 0 100
1 100 + 50 150
2 100 + 100 200
3 100 + 150 250
4 100 + 200 300
Draw the Graph
500
450
400
350
300
250
200
150
100
50
1 2 3 4 5 6 7 8 9 10
500
450
Questions for 400
350
Discussion
300
250
200
150
y = 100 + 50x
100
1. What is the slope?
50
1 2 3 4 5 6 7 8 9 10
2. What does it represent?
3. What is the y-intercept? What
does it represent?
4. Why does this graph contain a
line segment (have an end)?
Conclusions
What kinds of real life applications are linear?
(What do these situations have in common?)
They both have a Constant Rate of Change (SLOPE)
What does the slope represent?
The slope represents the rate of change
What does the y=intercept represent?
The y-intercept represents the y value before the
rate begins. x = 0
```
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Fraction of a Number
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CCSS.Math.Content.3.NF.A.1CCSS.Math.Content.4.NF.A.1
What is
1
2
of 6?
To answer this question, let's first draw 6 stars.
Since our denominator is 2, we need to split up these stars into 2 even groups:
And since our numerator is 1, we need to look at one of the groups.
So, 3 is
1
2
of 6 .
Let's take a look at a few examples of fractions of numbers:
Example 1
What is
1
3
of 12?
Let's draw 12 stars and split them into 3 groups (because 3 is the denominator).
Since the numerator is 1, we'll circle 1 group and count up those stars:
There are 4 stars in 1 group.
So,
1
3
of 12 is 4.
Example 2
What is
3
4
of 8?
We need to draw 8 stars and split them up into 4 groups (because 4 is the denominator).
Since the numerator is 3, we'll circle 3 groups and count up those stars:
6 stars are circled, so our answer is 6.
3
4
of 8 is 6.
Example 3
What is
2
5
of 10?
Start with 10 stars and divide them into 5 groups (because the denominator is 5).
Then, circle 2 of these groups (because the numerator is 2).
So,
2
5
of 10 is equal to 4.
Summary
• Fractions show pieces of a whole and consists of a numerator and a denominator.
• The denominator (the bottom number) tells how many total pieces there are and the numerator (the top number) tells how many of those pieces we are talking about.
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# Practical Geometry Class 7 Extra Questions Maths Chapter 10
Practical Geometry Class 7 Extra Questions Maths Chapter 10
### Practical Geometry Class 7 Extra Questions Maths Chapter 10
Extra Questions for Class 7 Maths Chapter 10 Practical Geometry
### Practical Geometry Class 7 Extra Questions Very Short Answer Type
Question 1.
State whether the triangle is possible to construct if
(a) In ΔABC, m∠A = 80°, m∠B = 60°, AB = 5.5 cm
(b) In ΔPQR, PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
Solution:
(a) m∠A = 80°, m∠B = 60°
m∠A + m∠B = 80° + 60° = 140° < 180°
So, ΔABC can be possible to construct.
(b) PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
PQ + QR = 5 cm + 3 cm = 8 cm < 8.8 cm
or PQ + QR < PR
So, the ΔPQR can not be constructed.
Question 2.
Draw an equilateral triangle whose each side is 4.5 cm.
Solution:
Steps of construction:
(i) Draw AB = 4.5 cm.
(ii) Draw two arcs with centres A and B and same radius of 4.5 cm to meet each other at C.
(iii) Join CA and CB.
(iv) ΔCAB is the required triangle.
Question 3.
Draw a ΔPQR, in which QR = 3.5 cm, m∠Q = 40°, m∠R = 60°.
Solution:
Steps of construction:
(i) Draw QR = 3.5 cm.
(ii) Draw ∠Q = 40°, ∠R = 60° which meet each other at P.
(iii) ΔPQR is the required triangle.
Question 4.
There are four options, out of which one is correct. Choose the correct one:
(i) A triangle can be constructed with the given measurement.
(a) 1.5 cm, 3.5 cm, 4.5 cm
(b) 6.5 cm, 7.5 cm, 15 cm
(c) 3.2 cm, 2.3 cm, 5.5 cm
(d) 2 cm, 3 cm, 6 cm
(ii) (a) m∠P = 40°, m∠Q = 60°, AQ = 4 cm
(b) m∠B = 90°, m∠C = 120° , AC = 6.5 cm
(c) m∠L = 150°, m∠N = 70°, MN = 3.5 cm
(d) m∠P = 105°, m∠Q = 80°, PQ = 3 cm
Solution:
(i) Option (a) is possible to construct.
1.5 cm + 3.5 cm > 4.5 cm
(ii) Option (a) is correct.
m∠P + m∠Q = 40° + 60° = 100° < 180°
Question 5.
What will be the other angles of a right-angled isosceles triangle?
Solution:
In right angled isosceles triangle ABC, ∠B = 90°
∠A + ∠C = 180° – 90° = 90°
But ∠A = ∠B
∠A = ∠C = 90/2 = 45°
Hence the required angles are ∠A = ∠C = 45°
Question 6.
What is the measure of an exterior angle of an equilateral triangle?
Solution:
We know that the measure of each interior angle = 60°
Exterior angle = 180° – 60° = 120°
Question 7.
In ΔABC, ∠A = ∠B = 50°. Name the pair of sides which are equal.
Solution:
∠A = ∠B = 50°
AC = BC [∵ Sides opposite to equal angles are equal]
Hence, the required sides are AC and BC.
Question 8.
If one of the other angles of a right-angled triangle is obtuse, whether the triangle is possible to construct.
Solution:
We know that the angles other than right angle of a right-angled triangle are acute angles.
So, such a triangle is not possible to construct.
Question 9.
State whether the given pair of triangles are congruent.
Solution:
Here, AB = PQ = 3.5 cm
AC = PR = 5.2 cm
∠BAC = ∠QPR = 70°
ΔABC = ΔPQR [By SAS rule]
### Practical Geometry Class 7 Extra Questions Short Answer Type
Question 10.
Draw a ΔABC in which BC = 5 cm, AB = 4 cm and m∠B = 50°.
Solution:
Steps of construction:
(i) Draw BC = 5 cm.
(ii) Draw ∠B = 50° and cut AB = 4 cm.
(iii) Join AC.
(iv) ΔABC is the required triangle.
Question 11.
Draw ΔPQR in which QR = 5.4 cm, ∠Q = 40° and PR = 6.2 cm.
Solution:
Steps of construction:
(i) Draw QR = 5.4 cm.
(ii) Draw ∠Q = 40°.
(iii) Take R as the centre and with radius 6.2 cm, draw an arc to meet the former angle line at P.
(iv) Join PR.
(v) ΔPQR is the required triangle.
Question 12.
Construct a ΔPQR in which m∠P = 60° and m∠Q = 30°, QR = 4.8 cm.
Solution:
m∠Q = 30°, m∠P = 60°
m∠Q + m∠P + m∠R = 180° (Angle sum property of triangle)
30° + 60° + m∠R = 180°
90° + m∠R = 180°
m∠R = 180° – 90°
m∠R = 90°
Steps of construction:
(i) Draw QR = 4.8 cm.
(ii) Draw ∠Q = 30°.
(iii) Draw ∠R = 90° which meets the former angle line at P.
(iv) ∠P = 180° – (30° + 90°) = 60°
(v) ΔPQR is the required triangle.
### Practical Geometry Class 7 Extra Questions Higher Order Thinking Skills [HOTS] Type
Question 13.
Draw an isosceles right-angled triangle whose hypotenuse is 5.8 cm.
Solution:
Right angled triangle is an isosceles triangle
Each of its acute angles = 90/2 = 45°
Steps of construction:
(i) Draw AB = 5.8 cm.
(ii) Construct ∠A = 45° and ∠B = 45° to meet each other at C.
(iii) ∠C = 180° – (45° + 45°) = 90°
(iv) ΔACB is the required isosceles right angle triangle.
Question 14.
Construct a ΔABC such that AB = 6.5 cm, AC = 5 cm and the altitude AP to BC is 4 cm.
Solution:
Steps of construction:
(i) Draw a line l and take any point P on it.
(ii) Construct a perpendicular to l at P.
(iii) Cut AP = 4 cm.
(iv) Draw two arcs with centre A and radii 6.5 cm and 5 cm to cut the line l at B and C respectively.
(v) Join AB and AC.
(vi) ΔABC is the required triangle.
Question 15.
Construct an equilateral triangle whose altitude is 4.5 cm.
Solution:
Steps of construction:
(i) Draw any line l and take a point D on it.
(ii) Construct a perpendicular to l at D and cut AD = 4.5 cm.
(iii) Draw the angle of 30° at on either side of AD to meet the line l at B and C.
(iv) ΔABC is the required equilateral triangle.
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Prealgebra
1.5Divide Whole Numbers
Prealgebra1.5 Divide Whole Numbers
Learning Objectives
By the end of this section, you will be able to:
• Use division notation
• Model division of whole numbers
• Divide whole numbers
• Translate word phrases to math notation
• Divide whole numbers in applications
Be Prepared 1.4
Before you get started, take this readiness quiz.
1. Multiply: $27·3.27·3.$
If you missed this problem, review Example 1.44.
2. Subtract: $43−26.43−26.$
If you missed this problem, review Example 1.32
3. Multiply: $62(87).62(87).$
If you missed this problem, review Example 1.45.
Use Division Notation
So far we have explored addition, subtraction, and multiplication. Now let’s consider division. Suppose you have the $1212$ cookies in Figure 1.13 and want to package them in bags with $44$ cookies in each bag. How many bags would we need?
Figure 1.13
You might put $44$ cookies in first bag, $44$ in the second bag, and so on until you run out of cookies. Doing it this way, you would fill $33$ bags.
In other words, starting with the $1212$ cookies, you would take away, or subtract, $44$ cookies at a time. Division is a way to represent repeated subtraction just as multiplication represents repeated addition.
Instead of subtracting $44$ repeatedly, we can write
$12÷412÷4$
We read this as twelve divided by four and the result is the quotient of $1212$ and $4.4.$ The quotient is $33$ because we can subtract $44$ from $1212$ exactly $33$ times. We call the number being divided the dividend and the number dividing it the divisor. In this case, the dividend is $1212$ and the divisor is $4.4.$
In the past you may have used the notation $412412$, but this division also can be written as $12÷4,12/4,124.12÷4,12/4,124.$ In each case the $1212$ is the dividend and the $44$ is the divisor.
Operation Symbols for Division
To represent and describe division, we can use symbols and words.
Operation Notation Expression Read as Result
$DivisionDivision$ $÷÷$
$abab$
$baba$
$a/ba/b$
$12÷412÷4$
$124124$
$412412$
$12/412/4$
$Twelve divided by fourTwelve divided by four$ $the quotient of 12 and 4the quotient of 12 and 4$
Division is performed on two numbers at a time. When translating from math notation to English words, or English words to math notation, look for the words of and and to identify the numbers.
Example 1.56
Translate from math notation to words.
$64÷864÷8$ $427427$ $428428$
Try It 1.111
Translate from math notation to words:
$84÷784÷7$ $186186$ $824824$
Try It 1.112
Translate from math notation to words:
$72÷972÷9$ $213213$ $654654$
Model Division of Whole Numbers
As we did with multiplication, we will model division using counters. The operation of division helps us organize items into equal groups as we start with the number of items in the dividend and subtract the number in the divisor repeatedly.
Manipulative Mathematics
Doing the Manipulative Mathematics activity Model Division of Whole Numbers will help you develop a better understanding of dividing whole numbers.
Example 1.57
Model the division: $24÷8.24÷8.$
Try It 1.113
Model: $24÷6.24÷6.$
Try It 1.114
Model: $42÷7.42÷7.$
Divide Whole Numbers
We said that addition and subtraction are inverse operations because one undoes the other. Similarly, division is the inverse operation of multiplication. We know $12÷4=312÷4=3$ because $3·4=12.3·4=12.$ Knowing all the multiplication number facts is very important when doing division.
We check our answer to division by multiplying the quotient by the divisor to determine if it equals the dividend. In Example 1.57, we know $24÷8=324÷8=3$ is correct because $3·8=24.3·8=24.$
Example 1.58
Divide. Then check by multiplying. $42÷642÷6$ $729729$ $763763$
Try It 1.115
Divide. Then check by multiplying:
$54÷654÷6$ $279279$
Try It 1.116
Divide. Then check by multiplying:
$369369$ $840840$
What is the quotient when you divide a number by itself?
$1515=1because1·15=151515=1because1·15=15$
Dividing any number $(except 0)(except 0)$ by itself produces a quotient of $1.1.$ Also, any number divided by $11$ produces a quotient of the number. These two ideas are stated in the Division Properties of One.
Division Properties of One
Any number (except 0) divided by itself is one. $a÷a=1a÷a=1$ Any number divided by one is the same number. $a÷1=aa÷1=a$
Table 1.6
Example 1.59
Divide. Then check by multiplying:
1. $11÷1111÷11$
2. $191191$
3. $1717$
Try It 1.117
Divide. Then check by multiplying:
$14÷1414÷14$ $271271$
Try It 1.118
Divide. Then check by multiplying:
$161161$ $1414$
Suppose we have $0,0,$ and want to divide it among $33$ people. How much would each person get? Each person would get $0.0.$ Zero divided by any number is $0.0.$
Now suppose that we want to divide $1010$ by $0.0.$ That means we would want to find a number that we multiply by $00$ to get $10.10.$ This cannot happen because $00$ times any number is $0.0.$ Division by zero is said to be undefined.
These two ideas make up the Division Properties of Zero.
Division Properties of Zero
Zero divided by any number is 0. $0÷a=00÷a=0$ Dividing a number by zero is undefined. $a÷0a÷0$ undefined
Table 1.7
Another way to explain why division by zero is undefined is to remember that division is really repeated subtraction. How many times can we take away $00$ from $10?10?$ Because subtracting $00$ will never change the total, we will never get an answer. So we cannot divide a number by $0.0.$
Example 1.60
Divide. Check by multiplying: $0÷30÷3$ $10/0.10/0.$
Try It 1.119
Divide. Then check by multiplying:
$0÷20÷2$ $17/017/0$
Try It 1.120
Divide. Then check by multiplying:
$0÷60÷6$ $13/013/0$
When the divisor or the dividend has more than one digit, it is usually easier to use the $412412$ notation. This process is called long division. Let’s work through the process by dividing $7878$ by $3.3.$
Divide the first digit of the dividend, 7, by the divisor, 3. The divisor 3 can go into 7 two times since $2×3=62×3=6$. Write the 2 above the 7 in the quotient. Multiply the 2 in the quotient by 2 and write the product, 6, under the 7. Subtract that product from the first digit in the dividend. Subtract $7−67−6$. Write the difference, 1, under the first digit in the dividend. Bring down the next digit of the dividend. Bring down the 8. Divide 18 by the divisor, 3. The divisor 3 goes into 18 six times. Write 6 in the quotient above the 8. Multiply the 6 in the quotient by the divisor and write the product, 18, under the dividend. Subtract 18 from 18.
We would repeat the process until there are no more digits in the dividend to bring down. In this problem, there are no more digits to bring down, so the division is finished.
$So78÷3=26.So78÷3=26.$
Check by multiplying the quotient times the divisor to get the dividend. Multiply $26×326×3$ to make sure that product equals the dividend, $78.78.$
$216×3___78✓216×3___78✓$
It does, so our answer is correct.
How To
Divide whole numbers.
1. Step 1. Divide the first digit of the dividend by the divisor.
If the divisor is larger than the first digit of the dividend, divide the first two digits of the dividend by the divisor, and so on.
2. Step 2. Write the quotient above the dividend.
3. Step 3. Multiply the quotient by the divisor and write the product under the dividend.
4. Step 4. Subtract that product from the dividend.
5. Step 5. Bring down the next digit of the dividend.
6. Step 6. Repeat from Step 1 until there are no more digits in the dividend to bring down.
7. Step 7. Check by multiplying the quotient times the divisor.
Example 1.61
Divide $2,596÷4.2,596÷4.$ Check by multiplying:
Try It 1.121
Divide. Then check by multiplying: $2,636÷42,636÷4$
Try It 1.122
Divide. Then check by multiplying: $2,716÷42,716÷4$
Example 1.62
Divide $4,506÷6.4,506÷6.$ Check by multiplying:
Try It 1.123
Divide. Then check by multiplying: $4,305÷5.4,305÷5.$
Try It 1.124
Divide. Then check by multiplying: $3,906÷6.3,906÷6.$
Example 1.63
Divide $7,263÷9.7,263÷9.$ Check by multiplying.
Try It 1.125
Divide. Then check by multiplying: $4,928÷7.4,928÷7.$
Try It 1.126
Divide. Then check by multiplying: $5,663÷7.5,663÷7.$
So far all the division problems have worked out evenly. For example, if we had $2424$ cookies and wanted to make bags of $88$ cookies, we would have $33$ bags. But what if there were $2828$ cookies and we wanted to make bags of $8?8?$ Start with the $2828$ cookies as shown in Figure 1.14.
Figure 1.14
Try to put the cookies in groups of eight as in Figure 1.15.
Figure 1.15
There are $33$ groups of eight cookies, and $44$ cookies left over. We call the $44$ cookies that are left over the remainder and show it by writing R4 next to the $3.3.$ (The R stands for remainder.)
To check this division we multiply $33$ times $88$ to get $24,24,$ and then add the remainder of $4.4.$
$3×8___24+4___283×8___24+4___28$
Example 1.64
Divide $1,439÷4.1,439÷4.$ Check by multiplying.
Try It 1.127
Divide. Then check by multiplying: $3,812÷8.3,812÷8.$
Try It 1.128
Divide. Then check by multiplying: $4,319÷8.4,319÷8.$
Example 1.65
Divide and then check by multiplying: $1,461÷13.1,461÷13.$
Try It 1.129
Divide. Then check by multiplying: $1,493÷13.1,493÷13.$
Try It 1.130
Divide. Then check by multiplying: $1,461÷12.1,461÷12.$
Example 1.66
Divide and check by multiplying: $74,521÷241.74,521÷241.$
Try It 1.131
Divide. Then check by multiplying: $78,641÷256.78,641÷256.$
Try It 1.132
Divide. Then check by multiplying: $76,461÷248.76,461÷248.$
Translate Word Phrases to Math Notation
Earlier in this section, we translated math notation for division into words. Now we’ll translate word phrases into math notation. Some of the words that indicate division are given in Table 1.8.
Operation Word Phrase Example Expression
Division divided by
quotient of
divided into
$1212$ divided by $44$
the quotient of $1212$ and $44$
$44$ divided into $1212$
$12÷412÷4$
$124124$
$12/412/4$
$412412$
Table 1.8
Example 1.67
Translate and simplify: the quotient of $5151$ and $17.17.$
Try It 1.133
Translate and simplify: the quotient of $9191$ and $13.13.$
Try It 1.134
Translate and simplify: the quotient of $5252$ and $13.13.$
Divide Whole Numbers in Applications
We will use the same strategy we used in previous sections to solve applications. First, we determine what we are looking for. Then we write a phrase that gives the information to find it. We then translate the phrase into math notation and simplify it to get the answer. Finally, we write a sentence to answer the question.
Example 1.68
Cecelia bought a $160-ounce160-ounce$ box of oatmeal at the big box store. She wants to divide the $160160$ ounces of oatmeal into $8-ounce8-ounce$ servings. She will put each serving into a plastic bag so she can take one bag to work each day. How many servings will she get from the big box?
Try It 1.135
Marcus is setting out animal crackers for snacks at the preschool. He wants to put $99$ crackers in each cup. One box of animal crackers contains $135135$ crackers. How many cups can he fill from one box of crackers?
Try It 1.136
Andrea is making bows for the girls in her dance class to wear at the recital. Each bow takes $44$ feet of ribbon, and $3636$ feet of ribbon are on one spool. How many bows can Andrea make from one spool of ribbon?
Section 1.5 Exercises
Practice Makes Perfect
Use Division Notation
In the following exercises, translate from math notation to words.
343.
$54 ÷ 9 54 ÷ 9$
344.
$56 7 56 7$
345.
$32 8 32 8$
346.
$6 42 6 42$
347.
$48 ÷ 6 48 ÷ 6$
348.
$63 9 63 9$
349.
$7 63 7 63$
350.
$72 ÷ 8 72 ÷ 8$
Model Division of Whole Numbers
In the following exercises, model the division.
351.
$15 ÷ 5 15 ÷ 5$
352.
$10 ÷ 5 10 ÷ 5$
353.
$14 7 14 7$
354.
$18 6 18 6$
355.
$4 20 4 20$
356.
$3 15 3 15$
357.
$24 ÷ 6 24 ÷ 6$
358.
$16 ÷ 4 16 ÷ 4$
Divide Whole Numbers
In the following exercises, divide. Then check by multiplying.
359.
$18 ÷ 2 18 ÷ 2$
360.
$14 ÷ 2 14 ÷ 2$
361.
$27 3 27 3$
362.
$30 3 30 3$
363.
$4 28 4 28$
364.
$4 36 4 36$
365.
$45 5 45 5$
366.
$35 5 35 5$
367.
$72 / 8 72 / 8$
368.
$8 64 8 64$
369.
$35 7 35 7$
370.
$42 ÷ 7 42 ÷ 7$
371.
$15 15 15 15$
372.
$12 12 12 12$
373.
$43 ÷ 43 43 ÷ 43$
374.
$37 ÷ 37 37 ÷ 37$
375.
$23 1 23 1$
376.
$29 1 29 1$
377.
$19 ÷ 1 19 ÷ 1$
378.
$17 ÷ 1 17 ÷ 1$
379.
$0 ÷ 4 0 ÷ 4$
380.
$0 ÷ 8 0 ÷ 8$
381.
$5 0 5 0$
382.
$9 0 9 0$
383.
$26 0 26 0$
384.
$32 0 32 0$
385.
$12 0 12 0$
386.
$16 0 16 0$
387.
$72 ÷ 3 72 ÷ 3$
388.
$57 ÷ 3 57 ÷ 3$
389.
$96 8 96 8$
390.
$78 6 78 6$
391.
$5 465 5 465$
392.
$4 528 4 528$
393.
$924 ÷ 7 924 ÷ 7$
394.
$861 ÷ 7 861 ÷ 7$
395.
$5,226 6 5,226 6$
396.
$3,776 8 3,776 8$
397.
$4 31,324 4 31,324$
398.
$5 46,855 5 46,855$
399.
$7,209 ÷ 3 7,209 ÷ 3$
400.
$4,806 ÷ 3 4,806 ÷ 3$
401.
$5,406 ÷ 6 5,406 ÷ 6$
402.
$3,208 ÷ 4 3,208 ÷ 4$
403.
$4 2,816 4 2,816$
404.
$6 3,624 6 3,624$
405.
$91,881 9 91,881 9$
406.
$83,256 8 83,256 8$
407.
$2,470 ÷ 7 2,470 ÷ 7$
408.
$3,741 ÷ 7 3,741 ÷ 7$
409.
$8 55,305 8 55,305$
410.
$9 51,492 9 51,492$
411.
$431,174 5 431,174 5$
412.
$297,277 4 297,277 4$
413.
$130,016 ÷ 3 130,016 ÷ 3$
414.
$105,609 ÷ 2 105,609 ÷ 2$
415.
$15 5,735 15 5,735$
416.
$4,933 21 4,933 21$
417.
$56,883 ÷ 67 56,883 ÷ 67$
418.
$43,725 / 75 43,725 / 75$
419.
$30,144 314 30,144 314$
420.
$26,145 ÷ 415 26,145 ÷ 415$
421.
$273 542,195 273 542,195$
422.
$816,243 ÷ 462 816,243 ÷ 462$
Mixed Practice
In the following exercises, simplify.
423.
$15 ( 204 ) 15 ( 204 )$
424.
$74 · 391 74 · 391$
425.
$256 − 184 256 − 184$
426.
$305 − 262 305 − 262$
427.
$719 + 341 719 + 341$
428.
$647 + 528 647 + 528$
429.
$25 875 25 875$
430.
$1104 ÷ 23 1104 ÷ 23$
Translate Word Phrases to Algebraic Expressions
In the following exercises, translate and simplify.
431.
the quotient of $4545$ and $1515$
432.
the quotient of $6464$ and $1616$
433.
the quotient of $288288$ and $2424$
434.
the quotient of $256256$ and $3232$
Divide Whole Numbers in Applications
In the following exercises, solve.
435.
Trail mix Ric bought $6464$ ounces of trail mix. He wants to divide it into small bags, with $22$ ounces of trail mix in each bag. How many bags can Ric fill?
436.
Crackers Evie bought a $4242$ ounce box of crackers. She wants to divide it into bags with $33$ ounces of crackers in each bag. How many bags can Evie fill?
437.
Astronomy class There are $125125$ students in an astronomy class. The professor assigns them into groups of $5.5.$ How many groups of students are there?
438.
Flower shop Melissa’s flower shop got a shipment of $152152$ roses. She wants to make bouquets of $88$ roses each. How many bouquets can Melissa make?
439.
Baking One roll of plastic wrap is $4848$ feet long. Marta uses $33$ feet of plastic wrap to wrap each cake she bakes. How many cakes can she wrap from one roll?
440.
Dental floss One package of dental floss is $5454$ feet long. Brian uses $22$ feet of dental floss every day. How many days will one package of dental floss last Brian?
Mixed Practice
In the following exercises, solve.
441.
Miles per gallon Susana’s hybrid car gets $4545$ miles per gallon. Her son’s truck gets $1717$ miles per gallon. What is the difference in miles per gallon between Susana’s car and her son’s truck?
442.
Distance Mayra lives $5353$ miles from her mother’s house and $7171$ miles from her mother-in-law’s house. How much farther is Mayra from her mother-in-law’s house than from her mother’s house?
443.
Field trip The $4545$ students in a Geology class will go on a field trip, using the college’s vans. Each van can hold $99$ students. How many vans will they need for the field trip?
444.
Potting soil Aki bought a $128128$ ounce bag of potting soil. How many $44$ ounce pots can he fill from the bag?
445.
Hiking Bill hiked $88$ miles on the first day of his backpacking trip, $1414$ miles the second day, $1111$ miles the third day, and $1717$ miles the fourth day. What is the total number of miles Bill hiked?
446.
Reading Last night Emily read $66$ pages in her Business textbook, $2626$ pages in her History text, $1515$ pages in her Psychology text, and $99$ pages in her math text. What is the total number of pages Emily read?
447.
Patients LaVonne treats $1212$ patients each day in her dental office. Last week she worked $44$ days. How many patients did she treat last week?
448.
Scouts There are $1414$ boys in Dave’s scout troop. At summer camp, each boy earned $55$ merit badges. What was the total number of merit badges earned by Dave’s scout troop at summer camp?
Writing Exercises
449.
Explain how you use the multiplication facts to help with division.
450.
Oswaldo divided $300300$ by $88$ and said his answer was $3737$ with a remainder of $4.4.$ How can you check to make sure he is correct?
Everyday Math
451.
Contact lenses Jenna puts in a new pair of contact lenses every $1414$ days. How many pairs of contact lenses does she need for $365365$ days?
452.
Cat food One bag of cat food feeds Lara’s cat for $2525$ days. How many bags of cat food does Lara need for $365365$ days?
Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?
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# Percentage change
In this lesson, we will learn about percentage change. We will investigate how to increase or decrease an amount by a given percentage.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What is 73% of 85?
1/5
Q2.What is 71.69% of 92.4?
2/5
Q3.100% of any number is simply the number itself.
3/5
Q4.What is 730.4% of 730.4?
4/5
Q5.30% of 90 is the same as 90% of 30.
5/5
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What is 73% of 85?
1/5
Q2.What is 71.69% of 92.4?
2/5
Q3.100% of any number is simply the number itself.
3/5
Q4.What is 730.4% of 730.4?
4/5
Q5.30% of 90 is the same as 90% of 30.
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Percentage change
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.If the amount of flour in a bag increased by 20%, it would always mean we add 20g on to the amount.
1/5
Q2.If I increase 200kg by 30%, then the new amount I have is...
2/5
Q3.The decimal multiplier to increase an amount by 53% would be...
3/5
Q4.It is possible to increase an amount by more than 100%
4/5
Q5.If you decreased an amount by 60%, the decimal multiplier would be 1.60.
5/5
Quiz:
# Percentage change
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.If the amount of flour in a bag increased by 20%, it would always mean we add 20g on to the amount.
1/5
Q2.If I increase 200kg by 30%, then the new amount I have is...
2/5
Q3.The decimal multiplier to increase an amount by 53% would be...
3/5
Q4.It is possible to increase an amount by more than 100%
4/5
Q5.If you decreased an amount by 60%, the decimal multiplier would be 1.60.
5/5
# Lesson summary: Percentage change
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Jog
On the spot:
Dance |
# Difference between revisions of "2002 AMC 8 Problems/Problem 14"
## Problem 14
A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices and claims that the final price of these items is $50\%$ off the original price. The total discount is
$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$
## Solution #1
Let's assume that each item is $100. First we take off $30\%$ off of$100. $100\cdot0.70=$$70
Next, we take off the extra$(Error compiling LaTeX. )20\%$as asked by the problem. $70\cdot0.80=$56
So the final price of an item is $56. We have to do$(Error compiling LaTeX. )100-56$because$56$was the final price and we wanted the discount.$100-56=44$so the final discount was$44\%$$(Error compiling LaTeX. ) \text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$
## Solution #2
Assume the price was \$100. We can just do $100\cdot0.7\cdot0.8=56$ and then do $100-56=44$ That is the discount percentage wise.
$\text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$ |
# Relative frequency distribution.
A relative frequency distribution expresses the frequency distribution of a variable in percentages or proportions, either on a table or graph. In other words, it shows how often an event happens, compared to other events. This is useful for observing and comparing the frequencies of different values or classes of values.
For example, comparing the frequency of different weather events can be done with a relative frequency distribution.
## How to create a relative frequency distribution
First, to create a relative frequency distribution, we need to calculate the frequency of each value or class. The frequency refers to the number of observations that are within each category. The following table shows a frequency distribution with counts (“frequency”) showing how often an adverse weather event happens in a certain city in a calendar year:
After calculating the frequencies, we can then compute the relative frequencies by dividing each frequency by the total number of observations. Percentages and proportions are two ways to express relative frequencies. Percentages are denoted as numbers out of 100, whereas proportions are denoted as numbers out of 1. The following relative frequency distribution shows the same information on adverse weather events, except that this time there is a third “relative frequency” column. These are calculated as follows:
• Count the total number of items. In this table the total is 67.
• Divide the count (the frequency) by the total number of items. For example, hurricanes were seen 12 times, divided by the total (67) gives 12/67 = 0.179.
• Sum up the relative frequencies to get the total. This should equal 1 in decimal or 100% in percentages.
One way to visualize a data distribution is using a frequency distribution histogram. This displays the relative frequency of each event on the y-axis.
## Why use a relative frequency distribution?
Relative frequency distributions are an important tool for understanding the distribution of a variable. They can be used to compare the frequencies of different values or classes of values, and to answer a variety of questions, such as:
• What is the most common type of adverse weather event? From the graph above, that’s rain, which has the tallest bar.
• What percentage of events are hurricanes? From the above table, hurricanes happen at a relative frequency of 0.179 or 17.9%.
• What percentage of adverse weather events happen less than 10 times per year? From the above table, only one event happens less than 10 times per year: snow.
Use relative frequency distributions when you need to quickly and efficiently analyze the trends of a data set.
## Cumulative Relative Frequency
To obtain the cumulative relative frequency, follow the above-mentioned steps to generate a relative frequency distribution table. Then, in an additional column labeled “cumulative relative frequency,” add up the relative frequencies as you go down the column.
• The first entry in the column is the same as the first entry in the relative frequency column (.179).
• Add the first and second entries to get 0.179 + 0.224 = 0.403.
• Add the first, second and third entries to get 0.179 + 0.224 + 0.433 = 0.836.
• Add the first, second, third and fourth entries to get 0.179 + 0.224 + 0.433 + 0.15 = .851.
• Add the first, second, third, fourth and fifth entries to get 0.179 + 0.224 + 0.433 + 0.15 + .149 = 1.00.
Scroll to Top |
Browse Questions
# Find the particular solution satisfying the given condition $x^2dy+(xy+y^2)\;dx=0;y=\;1;\;when\;x=\;1$
$\begin{array}{1 1} (A)\;y + 2x = 3x^2 \\(B)y - 2x = 3x^2 \\ (C)\;y + x = 3x^2 \\(D)\;y + 4x = 3x^2 \end{array}$
Toolbox:
• To solve homogenous differnetial equation put $y = vx$ and $\large\frac{dy}{dx}$$= v+x\large\frac{dv}{dx} Step 1: Let us rearrange and write the equation as x^2dy = - (xy+y^2)dx Or \large\frac{dy}{dx}=-\large\frac{(xy+y^2)}{x^2} Using the information in the tool box, v + x\large\frac{dv}{dx }= -\frac{ (x.vx+v^2x^2)}{x^2} Taking x^2 as the common factor and cancelling in the RHS we get v+x\large\frac{dv}{dx}$$=- (v+v^2)$
Bringing v from the LHS to RHS we get
$x\large\frac{dv}{dx }$$= -v(v+2) Seperating the variables we get, \large\frac{dv}{v(v+2) }= -\large\frac{ dx}{x } Step 2: Integrating on both sides we get, \int \large\frac{dv}{v(v+2)} = - \int\large\frac{dx}{x} Integration of \large\frac{dv}{v(v+2)} can be done by the method of partial fraction \large\frac{A}{v }+\frac{ B}{(v+2)} = \frac{1}{v(v+2)} A(v+2) + B(v) = 1 Equating the coefficients of like terms we get, A+B = 0(equating for v) 2A = 1 (equating for constant term) or A = 1/2 and B = -1/2 Substituting this for A and B we get, \int [\large\frac{1}{2(v)} -\frac{ 1}{2(v+2)}] = - \int\large\frac{ dx}{x} \bigg[\large\frac{1}{2}\bigg]$$\log v - \bigg[\large\frac{1}{2}\bigg]$$\log (v-2) = \log x + log C \large\frac{1}{2}$$\log \large\frac{v}{(v+2) }=$$\log Cx \log\large\frac{v}{(v+2)}$$ = 2\log Cx$
$\log\bigg[\large\frac{v}{(v+2)}\bigg] =$$\log c^2x^2 \large\frac{v}{(v+2)} = c^2x^2 Substituting for v =\large\frac{ y}{x} we get, \Large\frac{(\Large\frac{y}{x})}{[(\Large\frac{y}{x}) + 2] }=$$C^2x^2$
$\large\frac{x^2y}{(y+2x)}$$= C^2 Step 3: Given y = 1 and x = 1 substituting for x and y to evaluate the value of C we get \large\frac{1}{3 }=$$ C^2$
Substituting for $C^2$ we get,
$\large\frac{x^2}{(y+2x)} =\frac{ 1}{3}$
$y + 2x = 3x^2$
This is the required equation.
edited Aug 1, 2013 |
# Subtraction of Mass
In subtraction of mass we will learn how to find the difference between the units of mass or weight. While subtracting we need to follow that the units of mass i.e., kilogram and gram are converted into grams before subtraction and then follow the simple subtraction process.
We can subtract units of mass like ordinary numbers.
We will learn two different methods to solve subtraction using the standard unit and smaller unit of mass. Students can practice both the methods.
(i) Subtracting units with conversion into gram
(ii) Subtracting units without conversion into gram
Worked-out examples on subtraction of mass:
1. Subtract 11 kg 460 g from 25 kg 765 g
Solution:
Method I: (with conversion into gram):
We know, 1 kg = 1000 grams
Now kg and g are converted into grams before doing subtraction and then we need to follow the simple subtraction process.
11 kg 460 g = (11 × 1000) g + 460 g = 11000 g + 460 g = 11460 grams
25 kg 765 g = (25 × 1000) g + 765 g = 25000 g + 765 g = 25765 grams
Now sum,
2 5 7 6 5 g
- 1 1 4 6 0 g
1 4 3 0 5 g
= 14 kg 305 g
Therefore, 25 kg 765 g - 11 kg 460 g = 11 kg 305 g
Method II: (without conversion into gram):
Here kg and g are arranged in different columns and then added like ordinary numbers.
(i) kg and g are arranged in columns
(ii) 765 g - 460 g = 305 g
(iii) 25 kg - 11 kg = 14 kg
kg g
25 765
- 11 460
14 305
= 14 kg 305 g
Therefore, difference of 11 kg 460 g from 25 kg 765 g = 14 kg 305 g
2. Subtract 24 kg 565 g from 45 kg 225 g
Solution:
Method I: (with conversion into gram):
We know, 1 kg = 1000 grams
Now kg and g are converted into grams before doing subtraction and then we need to follow the simple subtraction process.
24 kg 565 g = (24 × 1000) g + 565 g = 24000 g + 565 g = 24565 grams
45 kg 225 g = (45 × 1000) g + 225 g = 45000 g + 225 g = 45225 grams
Now sum,
45225 g
- 24565
g
20660 g
= 20 kg 660 g
Therefore, 45 kg 225 g - 24 kg 565 g = 20 kg 660 g
Method II: (without conversion into gram):
Here kg and g are arranged in different columns and then added like ordinary numbers.
(i) kg and g are arranged in columns
(ii) 225 g < 565 g, then 225 g - 565 g, so 1 kg from 45 kg is borrowed and added to 225 g
1 kg + 225 g = 1000 g + 225 g = 1225 g
1225 g - 565 g = 660 ml
(iii) 45 kg reduce into 44 kg
44 kg - 24 kg = 20 kg
kg g
45 225
- 24 565
20 660
= 20 kg 660 g
Therefore, difference of 24 kg 565 g from 45 kg 225 g = 20 kg 660 g
3. Subtract 21 kg 370 g from 37 kg 675 g without conversion into gram.
Solution:
Here kg and g are arranged in different columns and then added like ordinary numbers (without conversion into gram).
kg g
37 675
- 21 370
16 305
= 16 kg 305 g
Therefore, difference of 21 kg 370 g from 37 kg 675 g = 16 kg 305 g
4. Subtract 44 kg 900 g from 105 kg 60 g.
Solution:
Arrange the numbers vertically.First, subtract the grams.As 900 g > 60 g, we cannot subtract.We borrow 1 kg from 105 kg. So, 105 kg becomes 104 kg and 60 grams becomes 1000 + 60 = 1060 g.Subtract 900 g from 1060 g.1060 – 900 = 160Write 160 under grams column.
Subtract kilograms. 104 – 44 = 60 kg
Write 60 under kilograms column.
To subtract weights, write the number of kg and gin separate columns the subtract like ordinary numbers, starting from right.
5. Subtract 22 kg 70 g from 46 kg 90 g.
Solution:
kg g
4 6 9 0
- 2 2 7 0
2 4 2 0
6. Subtract 682 kg 541 g from 325 kg 193 g.
Solution:
kg g
7 12 4 13 11
6 8 2 5 4 1
- 3 2 5 1 9 3
3 5 7 3 4 8
7. Subtract 30 kg 664 g from 65 kg 282 g
Solution:
30 kg 664 from 65 kg 282 g65 kg 282 g - 30 kg 664First subtract g here 282 < 664So, we borrow 1 kg from 65 kg,leaving behind 64 kg.Also, 1 kg = 1000 gThus, 65 kg 282 g becomes 64 kg 1282 gSubtract 664 from 1282.i.e., 1282 - 664 = 618 gNow subtract kg = (64 kg - 30 kg) = 34 kg
Therefore, the difference of the mass = 34 kg 618 g.
The above problems on subtraction of mass will help the students to practice the worksheet on subtracting the different units with conversion or without conversion.
Questions and Answers on Subtraction of Mass:
1. Subtract the given weights:
(i) 76 kg 142 g – 24 kg 031 g
(ii) 90 g 622 mg – 48 g 503 mg
(iii) 62 kg 579 g – 51 kg 560 g
(iv) 60 g 222 mg – 34 g 083 mg
(v) 80 kg 885 g – 47 kg 000 g
(vi) 100 kg 529 g – 36 kg 610 g
(vii) 27 g 021 mg – 9 g 300 mg
(viii) 321 kg 450 g – 50 kg 290 g
(ix) 560 kg 000 g – 110 kg 850 g
1. (i) 52 kg 111 g
(ii) 42 g 119 mg
(iii) 11 kg 19 g
(iv) 26 g 139 mg
(v) 33 kg 885 g
(vi) 63 kg 919 g
(vii) 17 g 721 mg
(viii) 271 kg 160 g
(ix) 449 kg 150 g
2. Subtract the following:
(i) kg g 68 97 - 12 67 _____________ (ii) kg g 49 70 - 26 50 _____________
(iii) kg g 17 74 - 15 30 _____________ (iv) kg g 24 60 - 12 23 _____________
(v) kg g 52 800 - 19 485 ______________ (vi) kg g 73 423 - 38 365 ______________
(vii) kg g 62 125 - 47 496 ______________ (viii) kg g 403 320 - 304 743 ______________
(ix) kg g 247 400 - 153 575 ______________ (x) kg g 490 200 - 296 150 ______________
(xi) kg g 631 189 - 518 976 ______________ (xii) kg g 350 875 - 114 109 ______________
(xiii) kg g 761 02 - 253 37 ______________ (xiv) kg g 80 500 - 60 495 _____________
(xv) kg g 518 210 - 109 625 ______________ (xvi) kg g 610 582 - 108 644 ______________
2. (i) 56 kg 30 g
(ii) 23 kg 20 g
(iii) 2 kg 44 g
(iv) 12 kg 37 g
(v) 33 kg 315 g
(vi) 35 kg 58 g
(vii) 14 kg 629 g
(viii) 98 kg 577 g
(ix) 93 kg 825 g
(x) 194 kg 50 g
(xi) 112 kg 213 g
(xii) 236 kg 766 g
(xiii) 507 kg 965 g
(xiv) 20 kg 5 g
(xv) 408 kg 585 g
(xvi) 501 kg 938 g
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# Minimizing Expected Value
## Problem
Suppose $$X$$ is a random variable with $$E[X^2] < \infty$$. What is the constant $$c$$ that minimizes $$E[(X - c)^2]$$?
## Solution
As with most optimization problems, we will begin by taking the derivative of $$E[(X - c)^2]$$ with respect to $$c$$. To do so, we replace the expectation with its definition, letting $$f(x)$$ represent the probability density function of $$X$$.
$\begin{split} \dfrac{d}{dc}E\left[(X - c)^2\right] &= \dfrac{d}{dc} \int_{-\infty}^{\infty} (x-c)^2f(x)dx \\ &= -2\int_{-\infty}^{\infty}(x-c)f(x)dx \\ &= -2\left[ \int_{-\infty}^{\infty}xf(x)dx - \int_{-\infty}^{\infty}cf(x)dx \right] \\ &= -2\left( E[X] - c \right) \\ \end{split}$
Next, we set the derivative equal to $$0$$ and solve for $$c$$:
$-2\left( E[X] - c \right) = 0$ $c = E[X]$
Thus, $$E[(X - c)^2]$$ is minimized by $$c = E[X]$$. We confirm this by noting $$\dfrac{d^2}{dc}E[(X - c)^2] = 2$$, indicating upward concavity.
## Alternative Approach
There is another, cleaner approach to finding the solution. We begin by utilizing the linearity of expectation property:
$E[(X - c)^2] = E[X^2 - 2cX + c^2] = E[X^2] - 2cE[X] +c^2$
Then, we take the derivative of this new formulation and set the result to $$0$$:
$\dfrac{d}{dc} \left[ E[X^2] - 2cE[X] +c^2 \right] = 0$ $-2E[X] + 2c = 0$ $c = E[X]$ |
# 3.4 - Small Population Example
3.4 - Small Population Example
## Example 3-4: Wheat Production
( Reference Section 6.4 of the text)
unit (Farm) i 1 2 3 Selection Prob, $$p_i$$ 0.3 0.2 0.5 Wheat produced 11 6 25
N = 3 farms, n = 2 sample with replacement.
s p(s) ys 1, 1 0.3(0.3) = 0.09 (11, 11) 2, 2 0.2(0.2) = 0.04 (6, 6) 3, 3 0.5(0.5) = 0.25 (25, 25) 1, 2 0.3(0.2) = 0.06 (11, 6) 2, 1 0.2(0.3) = 0.06 (6, 11) 1, 3 0.3(0.5) = 0.15 (11, 25) 3, 1 0.5(0.3) = 0.15 (25, 11) 2, 3 0.2(0.5) = 0.10 (6, 25) 3, 2 0.5(0.2) = 0.10 (25, 6)
Question: Compute the Hansen-Hurwitz estimator.
When (1,1) is sampled, the Hansen-Hurwitz estimator is:
$$\hat{\tau}_p=\dfrac{1}{2}\left(\dfrac{y_1}{p_1}+\dfrac{y_1}{p_1}\right)=\dfrac{1}{2}\left(\dfrac{11}{0.3}+\dfrac{11}{0.3}\right)=36.67$$
When (1,2) is sampled, the Hansen-Hurwitz estimator is:
$$\hat{\tau}_p=\dfrac{1}{2}\left(\dfrac{y_1}{p_1}+\dfrac{y_2}{p_2}\right)=\dfrac{1}{2}\left(\dfrac{11}{0.3}+\dfrac{6}{0.2}\right)=33.33$$
Similarly, we can fill out the table and get the Hansen-Hurwitz estimators as shown below:
s p(s) ys $$\hat{\tau}_p$$ 1, 1 0.3(0.3) = 0.09 (11, 11) 36.67 2, 2 0.2(0.2) = 0.04 (6, 6) 30.00 3, 3 0.5(0.5) = 0.25 (25, 25) 50.00 1, 2 0.3(0.2) = 0.06 (11, 6) 33.33 2, 1 0.2(0.3) = 0.06 (6, 11) 33.33 1, 3 0.3(0.5) = 0.15 (11, 25) 43.33 3, 1 0.5(0.3) = 0.15 (25, 11) 43.33 2, 3 0.2(0.5) = 0.10 (6, 25) 40.00 3, 2 0.5(0.2) = 0.10 (25, 6) 40.00
Question: Compute the Horvitz-Thompson estimator.
$$\pi_1=0.09+0.06+0.06+0.15+0.15=0.51$$
$$\pi_2=0.04+0.06+0.06+0.10+0.10=0.36$$
$$\pi_3=0.25+0.15+0.15+0.10+0.10=0.75$$
When (1,1) is sampled, the Horvitz-Thompson estimator is:
$$\hat{\tau}_\pi=\left(\dfrac{11}{0.51}\right)=21.57$$
When (1,2) is sampled, the Horvitz-Thompson estimator is:
$$\hat{\tau}_\pi=\left(\dfrac{11}{0.51}+\dfrac{6}{0.36}\right)=38.24$$
Similarly, we can fill out the table and get the Horvitz-Thompson estimators as shown below:
s p(s) ys $$\hat{\tau}_p$$ $$\hat{\tau}_\pi$$ 1, 1 0.3(0.3) = 0.09 (11, 11) 36.67 21.57 2, 2 0.2(0.2) = 0.04 (6, 6) 30.00 16.67 3, 3 0.5(0.5) = 0.25 (25, 25) 50.00 33.33 1, 2 0.3(0.2) = 0.06 (11, 6) 33.33 38.24 2, 1 0.2(0.3) = 0.06 (6, 11) 33.33 38.24 1, 3 0.3(0.5) = 0.15 (11, 25) 43.33 54.90 3, 1 0.5(0.3) = 0.15 (25, 11) 43.33 54.90 2, 3 0.2(0.5) = 0.10 (6, 25) 40.00 50.00 3, 2 0.5(0.2) = 0.10 (25, 6) 40.00 50.00 Mean 42 42 Variance 34.67 146.46
$$\text{Mean of }\hat{\tau}_p = E(\hat{\tau}_p)=\sum{p(s) \cdot \hat{\tau}_p (s)}$$
$$= 0.09\times36.67+0.04\times30.00+0.25\times50.00+0.06\times33.33+0.06\times33.33$$
$$+0.15\times43.33+0.15\times43.33+0.10\times40.00+0.10\times40.00$$
$$= 42$$
From the table above we can see that both $$\hat{\tau}_p$$ and $$\hat{\tau}_\pi$$are unbiased. This example is a small population example to illustrate conceptually the properties of these estimators. We can compute the variance for $$\hat{\tau}_p$$and the variance for $$\hat{\tau}_\pi$$ directly from the definition of variance.
Since mean of $$\hat{\tau}_p$$ is 42 and E(g) = $$\sum p(s)*g(s)$$, we can compute the variance of $$\hat{\tau}_p$$ as:
$$Var(\hat{\tau}_p) = E[\hat{\tau}_p-\text{mean of }\hat{\tau}_p]^2$$
$$= 0.09\times(36.67-42)^2+0.04\times(30.00-42)^2+0.25\times(50.00-42)^2+0.06\times(33.33-42)^2$$
$$+0.06\times(33.33-42)^2+0.15\times(43.33-42)^2+0.15\times(43.33-42)^2+0.10\times(40.00-42)^2$$ $$+0.10\times(40.00-42)^2$$
$$=34.67$$
Similarly, we can compute that the variance of $$\hat{\tau}_\pi$$is 146.46.
Now, how do we compute the MSE of $$\hat{\tau}_p$$?
By definition, $$MSE (\hat{\tau}_p) = E(\hat{\tau}_p-\tau)^2$$, in this case, since $$\hat{\tau}_p$$is unbiased and $$E(\hat{\tau}_p) =\tau$$ ,$$MSE (\hat{\tau}_p)$$is the same as $$Var(\hat{\tau}_p)$$.
Remark 1. The above demonstration is just a teaching tool. In reality we will not know the population and will not come across small population problems like this other than in exams and homeworks. What we know are:
Unit 1 2 3 Selection probability 0.3 0.2 0.5
And, we draw a sample. If the sample we draw is (1,2) then $$\hat{\tau}_p$$ = 33.33 and$$\hat{\tau}_\pi=38.24$$ .
We will not be able to find the real population total nor the real variance of the estimator. However, we will be able to estimate them.
Remark 2. Now, should we use $$\hat{\tau}_p$$or should we use $$\hat{\tau}_\pi$$?
There are no clear answers. Both estimators ar acceptable when $$y_i$$ and $$p_i$$ are proportional.
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Question Video: Simplifying the Sum of Two Rational Functions and Determining Its Domain | Nagwa Question Video: Simplifying the Sum of Two Rational Functions and Determining Its Domain | Nagwa
# Question Video: Simplifying the Sum of Two Rational Functions and Determining Its Domain Mathematics • Third Year of Preparatory School
## Join Nagwa Classes
Simplify the function 𝑛(𝑥) = 3𝑥/(𝑥 + 8) + 6/(𝑥 + 8), and determine its domain.
03:13
### Video Transcript
Simplify the function 𝑛 of 𝑥 equals three 𝑥 over 𝑥 plus eight plus six over 𝑥 plus eight, and determine its domain.
The first thing that we can notice here is that we’re trying to add two fractions together. And to add fractions, we must have a common denominator. This is true when we’re working with whole numbers. It’s also true when we’re working with polynomials or with variables in the denominator. In our case, our two fractions already have a common denominator. This means that we’re able to add the numerators. Three 𝑥 plus six just equals three 𝑥 plus six over 𝑥 plus eight.
Our question does want us to try and simplify this problem. So there’s one other thing we can do. We can notice that the three and the six both have a common factor of three. If we take out that common factor, we can rewrite three 𝑥 plus six to say three times 𝑥 plus two, all over 𝑥 plus eight. Since there’s nothing else that cancels out or can be simplified, this is the simplest form of our function.
Now we’ll need to determine its domain. Remember that our domain is all the possible 𝑥-values. We want to ask the question here, “Is there any value for 𝑥 that would not yield a valid result?” We should notice that we’re dealing with a fraction and we have a variable in our denominator. No fraction can ever have a denominator value of zero because we can’t divide by zero. We want to know for what value of 𝑥 with the denominator of this fraction be equal to zero. To isolate 𝑥, we’ll subtract eight from both sides, and then we’ll have 𝑥 equal to negative eight. What that means is we cannot plug in negative eight into our equation. Here’s what would happen. We would end up with three times negative six over zero, and that’s impossible. We cannot divide by zero.
What we can say about our domain is this: our domain can be all real numbers with the exception of negative eight, or all reals minus negative eight which would be the case for our function 𝑛 of 𝑥 equals three times 𝑥 plus two over 𝑥 plus eight.
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# If ${{\cos }^{-1}}x-{{\cos }^{-1}}\left( \dfrac{y}{2} \right)=\alpha$ then $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$ is equal to?\begin{align} & \left( A \right)4{{\sin }^{2}}\alpha \\ & \left( B \right)-4{{\sin }^{2}}\alpha \\ & \left( C \right)2\sin 2\alpha \\ & \left( D \right)4 \\ \end{align}
Last updated date: 13th Jun 2024
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Hint: We first apply the formula of ${{\cos }^{-1}}x+{{\cos }^{-1}}y$ on the left hand side of the given equation. We then simplify the equation, and square it. Rearranging the terms and applying some basic trigonometric formulae, we get $\left( A \right)$ as the correct option.
The given equation is
${{\cos }^{-1}}x+{{\cos }^{-1}}\left( \dfrac{y}{2} \right)=\alpha$
We know the formula that ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)$ . Thus, applying this formula in the above equation the equation thus becomes,
$\Rightarrow {{\cos }^{-1}}\left( x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)} \right)=\alpha$
Taking $\text{cosine}$ on both sides on the above equation, we get,
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)} \right) \right)=\cos \alpha$
We know the simple formula that $\cos \left( {{\cos }^{-1}}x \right)=\cos x$ . So, applying this in the above equation, the equation thus becomes,
$\Rightarrow x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)}=\cos \alpha$
Simplifying the above equation, we get,
$\Rightarrow \dfrac{xy}{2}+\sqrt{\left( 1-{{x}^{2}} \right)\left( \dfrac{4-{{y}^{2}}}{4} \right)}=\cos \alpha$
Further simplifying the above equation, the equation thus becomes,
$\Rightarrow \dfrac{xy}{2}+\dfrac{\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}}{2}=\cos \alpha$
Multiplying both sides of the above equation by $2$ , we get,
$\Rightarrow xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}=2\cos \alpha$
Taking $\cos \alpha$ to the left hand side of the above equation and $\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}$ to the right hand side of the above equation, we get,
$\Rightarrow xy-2\cos \alpha =-\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}$
Squaring both sides of the above equation, the equation thus becomes,
$\Rightarrow {{\left( xy-2\cos \alpha \right)}^{2}}={{\left( -\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)} \right)}^{2}}$
Evaluating the above equation, we get
$\Rightarrow {{x}^{2}}{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha =\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)$
Opening the brackets in the above equation, we get,
$\Rightarrow {{x}^{2}}{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha =4-4{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}$
Subtracting ${{x}^{2}}{{y}^{2}}$ from both sides of the above equation, we get,
$\Rightarrow -4xy\cos \alpha +4{{\cos }^{2}}\alpha =4-4{{x}^{2}}-{{y}^{2}}$
Bringing the terms $4{{x}^{2}},{{y}^{2}}$ to the left hand side of the above equation and the term $4{{\cos }^{2}}\alpha$ to the right hand side of the above equation, we get,
$\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4-4{{\cos }^{2}}\alpha$
Taking $4$ common in the right hand side of the above equation, we get,
$\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4\left( 1-{{\cos }^{2}}\alpha \right)$
We know that $1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha$ . Thus, applying this formula in the above equation, we get,
$\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4{{\sin }^{2}}\alpha$
Rearranging the terms of the above equation, we get,
$\Rightarrow 4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha$
This is nothing but the thing that we have to prove. Therefore, we can conclude that $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$ is equal to $4{{\sin }^{2}}\alpha$ which is option $\left( A \right)$ .
Note: We must be very careful while carrying out the square as this expression deals with a little complex terms and students are prone to make mistakes here. This problem can also be solved by taking some values for $x,y,\alpha$ and find out which of the following options gives the correct answer. Let’s take $x=0,y=1$ . Then, $\alpha$ becomes ${{\cos }^{-1}}0-{{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{2}-\dfrac{\pi }{3}=\dfrac{\pi }{6}$ . The expression becomes $4{{\left( 0 \right)}^{2}}-4\left( 0 \right)\left( 1 \right)\cos \left( \dfrac{\pi }{6} \right)+{{\left( 1 \right)}^{2}}=1$ . Out of the following options, only $\left( A \right)$ satisfies by putting $\alpha =\dfrac{\pi }{6}$. |
# DAV Class 6 Maths Chapter 15 Worksheet 2 Solutions
The DAV Class 6 Maths Book Solutions and DAV Class 6 Maths Chapter 15 Worksheet 2 Solutions of Perimeter and Area offer comprehensive answers to textbook questions.
## DAV Class 6 Maths Ch 15 WS 2 Solutions
Question 1.
The area of a plastic sheet is 300 sq cm and its length is 30 cm. What is the breadth of the plastic sheet?
Area of the plastic sheet = 300 sq. cm
Its length = 30 cm
Breadth of the plastic sheet = $$\frac{\text { Area }}{\text { Length }}$$
= $$\frac{300}{30}$$
= 10 cm
Hence the required breadth = 10 cm.
Question 2.
The area of a football ground is 2900 m2. If its breadth is 29 m, find the length of the ground.
The area of the football ground
= 2900 m2
Length of the ground = $$\frac{\text { Area }}{\text { Breadth }}$$
= $$\frac{2900}{29}$$
= 100 m
Hence the required length = 100 m.
Question 3.
The perimeter of a square lawn is 72 m. Find the length of the side of the square lawn.
Perimeter of the square lawn = 72 m
Side = $$\frac{1}{4}$$ × Perimeter
= $$\frac{1}{4}$$ × 72 m
= 18 m
Hence the required length of side = 18 m.
Question 4.
A rectangular garden is 230 m long and its area is 13800 m2. Find the width of the garden.
Area of the rectangular garden = 13800 m2
Its length = 230 m
The width of the garden = $$\frac{\text { Area }}{\text { Length }}$$
= $$\frac{13800}{230}$$
= 60 m
Hence the required width = 60 m.
Question 5.
Neha covered 95.52 m by running around a square garden. Find the length of each side of the garden.
Perimeter of a square garden = 95.52 m
Side = $$\frac{\text { Perimeter }}{4}$$
= $$\frac{95.52}{4}$$
= 23.88 m
Hence the side of the square garden – 23.88 m
### DAV Class 6 Maths Chapter 15 Value Based Questions
Question 1.
On the occasion of Republic Day, a school organised a poster making competition for the students of Class VI. The topic was ‘National Integration’. Each child was given a square sheet of side 60 cm. They had to make a border of width 2 cm using the coloured tape and the remaining portion had to be divided into four equal portions to showT main teachings of different religions.
(a) Find the length of the coloured tape needed for the border.
Side of square sheet = 60 cm
The coloured taped needed for the border = Perimeter of the square
= 4 × side
= 4 × 60
= 240 cm
(b) What will be the side of the square into which the sheet has been divided?
The side of the square (ABCD) into which the sheet has been divided
= $$\frac{60-2 \times 2}{2}=\frac{60-4}{2}=\frac{56}{2}$$ = 28 cm
(c) What do you learn from the given situation?
To inculcate team work among children.
Question 2.
There was a deserted land near Rajesh’s house. The colony people use to throw garbage on that land. Rajesh along with his friends contacted the people of the colony and they all decided to develop the land into a park with a walking track where the children can play and elders can walk.
(a) Find the cost for levelling the park including the track at the rate of ₹ 15 per square metre. |
# ACT Math : How to find the solution to a binomial problem
## Example Questions
### Example Question #1 : How To Find The Solution To A Binomial Problem
Solve for .
Explanation:
Even though there are three terms in the equation, there are technically only two because it's a matter of collecting like terms.
The main objective in this problem is to solve for x. In order to do so, we need to get x by itself. The first step to accomplish this is to subtract 27 from the left side of the equation and do the same to the right. This follows the idea of "what you do to one side of an equation, you must do to the other." From here, you can collect like terms between -137 and -27 because they're both constants.
The goal of getting x by itself on one side has been almost achieved. We still have that coefficient "5" that's being multiplied by x. In order to make 5x just x, 5x needs to be divided by 5 and therefore so does -164. Again, this follows the concept of mimicking actions on both sides of an equation.
### Example Question #2 : How To Find The Solution To A Binomial Problem
Choose the answer below that is one of the solutions to the following equation:
Explanation:
The equation presented in the problem is:
First, you need to get everything on one side of the equation:
Then, you can reduce by dividing everything by :
Then, you factor:
Therefore, or .
### Example Question #3 : How To Find The Solution To A Binomial Problem
Choose the answer that is a potential solution to the binomial equation below:
Explanation:
The equation presented in the problem is:
To solve, first get all terms on the same side of the equation:
Now, you've eliminated your third term, which makes factoring difficult. If it makes things easier, you can throw a zero back in the equation:
Factor:
Therefore:
or
### Example Question #4 : How To Find The Solution To A Binomial Problem
Choose the answer below that is a solution to the following binomial equation:
Explanation:
First, get everything on one side of the equation:
Then, to make things easier, you can reduce by dividing everything by :
Next, factor:
Therefore:
or
### Example Question #2 : How To Find The Solution To A Binomial Problem
Choose the answer which is a solution to the following binomial equation:
Explanation:
To solve, first pull everything over to one side of the equation:
Then, reduce by dividing by the greatest common denominator—in this case, :
Now, factor:
Therefore:
or
### Example Question #3 : How To Find The Solution To A Binomial Problem
Choose the answer below that is a possible solution to the following binomial equation:
Explanation:
The equation presented in the problem is:
To solve this type of equation, you need to factor. First, get all of the terms of the equation on one side:
Then, you need to find two factors that will give you the equation in its current form:
Therefore, and , so or .
is listed as an answer, and must therefore be correct. |
# Slope Formula
• Last Updated : 15 May, 2022
A slope formula is used to determine the steepness or inclination of a line. The x and y coordinates of the points lying on the line are used to calculate the slope of a line. The change in the “y” coordinate with respect to the change in the “x” coordinates is called the slope of a line, and it is usually depicted by the letter “m”. Using the slope formula, we can determine whether two lines are perpendicular, parallel, or collinear.
### Slope Formula
In mathematics, the slope of a line is used to determine how much the line has inclined, i.e., the steepness of a line. To determine the slope of a line, we need the x and y coordinates of the points lying on the line. The slope formula is the net change in the “y” coordinate divided by the net change in the “x” coordinate. Δy is the change in the “y” coordinates, and Δx is the change in the “x” coordinates. Hence, the ratio of the change in “y” coordinates with respect to the change in the “x” coordinates is given by,
Slope (m) = change in y/change in x = Δy/Δx
m= (y2 – y1)/(x2 – x1)
Where x1 and x2 are the coordinates of the X-axis and y1 and y2 are the coordinates of the Y-axis.
• We know that tan θ is also a slope of the line hence the slope of a line can also be represented as,
Slope (m) = tan θ = Δy/Δx
Where θ is the angle made by the line w.r.t to the positive X-axis,
Δy = change in the “y” coordinates,
Δx = change in the “x” coordinates.
• We can also define the slope of a line as the ratio of rising with respect to run.
Slope (m) = Rise/Run
• Let ax + by + c = 0 be the general equation of a line. Now, the formula for the slope of the line is given by,
Slope (m) = – coefficient of x/coefficient of y = -a/b
• The slope-intercept form of a line using the line equation is given as,
y = mx + c
Where m is the slope of the line and c is the y-intercept of the line.
### Sample Problems
Problem 1: Find the slope of a line whose coordinates are (3, 7) and (5, 8).
Solution:
Given, (x1, y1) = (3,7) and (x2, y2) = (5,8)
Slope formula (m) = (y2 – y1)/(x2 – x1)
⇒ m = (8 – 7)/(5 – 3) = 1/2
Hence, the slope of the given line is 1/2.
Problem 2: Determine the slope of a line whose coordinates are (7, -5) and (2, -3).
Solution:
Given, (x1, y1) = (7, -5) and (x2, y2) = (2, -3)
Slope formula (m) = (y2 – y1)/(x2 – x1)
⇒ m = (-3 – (-5))/(2 – 7) = -2/5
Hence, the slope of the given line is -2/5
Problem 3: Find the value of a, if the slope of a line passing through the points (-4, a) and (2, 5) is 3.
Solution:
Given, (x1, y1) = (4,a) and (x2, y2) = (2, 5) and slope (m) = 3
We know that slope (m) = (y2 – y1)/(x2 – x1)
⇒ 3 = (5 – a)/(2 – 4)
⇒ 3 = (5 – a)/(-2)
⇒ -6 = 5 – a ⇒ a = 5 + 6 = 11
Hence, the value of a = 11
Problem 4: If a line makes an angle of 60° with the positive Y-axis, then what is the value of the slope of the line?
Solution:
Given data, Angle made by a line with the positive y-axis = 60°
We know that if the line makes an angle of 60° from the positive y-axis, then it makes an angle of (90° – 60° = 30°) with the x-axis.
Therefore, the value of the slope of the line (m) = tan 30° = 1/√3
Hence, the value of the slope of the line = 1/√3.
Problem 5: Sheela was checking a graph, she noticed that the raise was 12 units and the run was 4 units. Now calculate the slope of a line?
Solution:
Given data, rise = 12 units and run = 4 units
We know that slope (m) = rise/run
⇒ m = 12/4 = 3
Hence, the slope of the given line is 3
Problem 6: Find the slope of the line 3x – 7y + 8 = 0.
Solution:
Given data, The equation of the line = 3x – 7y + 8 = 0
Now, compare the given with the general equation of the line i.e., ax + by + c = 0
Therefore, a = 3, b = -7 and c = 8
We know that Slope (m) = – coefficient of x/coefficient of y = -a/b
⇒ m = -3/(-7) = 3/7
Hence, the slope of the given line is 3/7.
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# What is 10% off?
## What is 10% off?
This means the cost of the item to you is \$9. You will pay \$9 for a item with original price of \$10 when discounted 10%. In this example, if you buy an item at \$10 with 10% discount, you will pay 10 – 1 = 9 dollars.
### How do you work out 10% of a price?
How do I calculate a 10% discount?
1. Take the original price.
2. Divide the original price by 100 and times it by 10.
3. Alternatively, move the decimal one place to the left.
4. Minus this new number from the original one.
5. This will give you the discounted value.
6. Spend the money you’ve saved!
#### What is 10% off 10 pounds?
In this example, if you buy an item at £10 with 10% discount, you will pay 10 – 1 = 9 pounds.
What is 10% of a dollar 95?
You will pay \$85.5 for a item with original price of \$95 when discounted 10%. In this example, if you buy an item at \$95 with 10% discount, you will pay 95 – 9.5 = 85.5 dollars.
How do you take 10 off a number?
Taking 10 percent off a number changes depending on the original number:
1. Divide your number by 10.
2. Subtract this new number from your original number.
3. You’ve taken 10 percent off!
## How do you find 10% of a number?
To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent.
### What is 10% of an amount?
While 10 percent of any amount is the amount multiplied by 0.1, an easier way to calculate 10 percent is to divide the amount by 10. So, 10 percent of \$18.40, divided by 10, equates to \$1.84.
#### How do you add 10% to a price on a calculator?
How do I calculate a 10% increase? Divide the number you are adding the increase to by 10. Alternatively multiply the value by 0.1.
How do you calculate percentage off?
How to calculate percent off?
1. Divide the number by 100 (move the decimal place two places to the left).
2. Multiply this new number by the percentage you want to take off.
3. Subtract the number from step 2 from the original number. This is your percent off number.
What number is 10% of 90?
Percentage Calculator: What is 10 percent of 90? = 9.
## What number is 10 percent of 20?
Answer: 10% of 20 is 2.
### How do you calculate 10 percent marks?
To find the percentage of the marks, divide the marks obtained in the examination with the maximum marks and multiply the result with 100. Example 1: If 1156 is the total score obtained in the examination out of 1200 marks, then divide 1156 by 1200, and then multiply it by 100. |
# How do you factor 4(x+3)^2-9(x-1)^2?
Feb 14, 2016
This can be factored as a difference of squares.
#### Explanation:
Differences of squares are of the form ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$. We can find the factors by finding the square roots of each part of the expression:
$\sqrt{4 {\left(x + 3\right)}^{2}} = 2 \left(x + 3\right)$
$\sqrt{9 {\left(x - 1\right)}^{2}} = 3 \left(x - 1\right)$
Thus, the expression can be factored into:
$= \left(2 \left(x + 3\right) - 3 \left(x - 1\right)\right) \left(2 \left(x + 3\right) + 3 \left(x - 1\right)\right)$
$= \left(2 x + 6 - 3 x + 3\right) \left(2 x + 6 + 3 x - 3\right)$
$= \left(9 - x\right) \left(3 + 5 x\right)$
Hopefully this helps! |
# Maths – Short Cut Tricks for Square
One of the ardent follower and most helpful person of BankersAdda, TempleRun aka VT shared this trick for the Short Cut Trick for Squares of the number. We are sharing this with everybody, so that everybody can make use of it!!
Square of a number
From 1 to 30 (this is damn easy) You need to remember. Based on this I am going to explain square of a number from 31 to 100.
31 to 100 is divided into 3 parts
• 31 to 50
• 51 to 80
• 81 to 100
From 31 to 50:
Find the square of 38.
First Step :
122 =50-12 (haan bhai haan fir se 50 )
Second Step :
Last 2 Digit : 122 = 144 =44 (1 Carry)
First 2 Digit : 25-12+1 =14
25 hamesha ki tarah Constant ..You need to add
12 comes from (50-12)
1 upar wala carry
Final answer =1444 (First 2 Digit Last 2 Digit)
382=1444
Find the Square of 47.
First Step :
47 =50-03
Second Step :
Last 2 Digit : 09 ( kaise ? 3 ka Sqaure ) Carry ? Zero
First 2 Digit : 25 -03+0 =22
03 comes from 50-03
Carry to zero hi aya tha
So final Answer is 2209 (First 2 Digit Last 2 Digit)
472= 2209
Rapid Fire Round !
1)Find the Square of 33.
33 =50-17
Last 2 Digit =289 =89 ( 2 carry)
First 2 Digit =25-17+2 =10
2) Find the square of 42.
42 =50-08
Last 2 Digit =64 ( Carry Zero)
First 2 Digit =25-8+0 =17
3) Find the Square of 45.
Last 2 Digit =25
First 2 Digit =4*(4+1) =20
Let’s see 51 to 80 :
Find the Square of 62
First Step – 50+12 ( Yes, every time You have to think like this 50 + )
{If the number is 72 We will write 50+22}
Second Step – _ _ _ _ — 4 Digit number is the answer ( From 32 to 99 Every Number Square has 4 Digits)
Last 2 Digit : 122 =144 = 44 ( 1 will be carry )
First 2 Digit : 25+12+1 =38
25 is constant number for every number 51-80 You have to add this!!
12 comes from (50+12)
So our final result is 3844 (See First 2 Digit and Last 2 Digit above)
62= 3844
Thoda to jarur samjh me aaya hoga ? Solve another question
Find the Square of 77
50 +27 ( Every time You have to think like this)
_ _ _ _ — First 2 Digit Last 2 Digit
Last 2 Digit : 272 =729 = 29 ( 7 is carry )
First 2 Digit : 25+27+ 7 =59
25 is constant as I said before You have to add every time this number!!
27 comes from (50+27)
(Thought process will go like this 25+(25+2)+7=(25+25+9) =50+9=59)
So Final Answer is .. 5929
Hmm ..You can do it fast …just You need practice …Ghar me Practice karoge to Exam me without putting pen on the Paper You can do it.
Remember : The more you sweat in the peace ..the less you bleed
Lets Solve few examples in Simple way
1)Find the Square of 53 ?
Last 2 Digit =09 ( No Carry here )
First 2 Digit =25+3+0 =28
2)Find the Square of 66 ?
Last 2 Digit =256=56 ( 2 Carry )
First 2 Digit =25+16+2 =43
3)Find the Square of 75 ?
Don’t use above method
First 2 Digit =7(7+1) =56
Now, let’s see for 81 to 100 :
Find the square of 88
First Step :
88=100 -12 ( Yes 100-12 Only , not 50+38 or 80+8 )
{If the number is 92 then we have to think like 100-8}
Second Step :
_ _ _ _ 4 Digit number is the answer.As I said before Square of any number b/w 32 and 99 has 4 digits.
Last 2 Digit : 122 = 144 = 44 (1 carry )
First 2 Digit : 88 – 12 + 1 =77
88 is the given number jiska hum square find kar rhe he
12 comes from 100-12
1 is as usual carry ( 100-12 doesn’t mean that we have to minus 1 from
the number..here also we add )
So our final Answer is 7744 (First 2 Digit Last 2 Digit)
88= 7744
(Btw this number is Special,No other number square has this form XXYY)
Let’s do another sum
Find the Square of 92 .
First Step :
92 =100-8 ( Yes 100-8 only ,don’t confuse with 50+ )
Second Step :
Last 2 Digit : 82 = 64 ( It’s only the last 2 Digit …No carry)
First 2 Digit : 92-8+0 =84
92 is given number
8 is from 100-8
O is carry ( Carry to Anda hi aaya tha )
Final Answer is 8464 ( First 2 Digit Last 2 Digit )
922 =8464
Rapid Fire Round !
1)Find the square of 96.
96=100-4
Last 2 Digit =16 (How come 16 ? 4 ka Square ) Carry Zero
First 2 Digit =96-4+0 =92
2) Find the Square of 81.
81 =100-19
Last 2 Digit = 192= 361 =61 (Carry 3)
First 2 Digit =81-19+3 =65 |
# Strict inequality
A relation that expresses the comparison between the unequal quantities is called strict inequality.
## Introduction
In mathematics, A quantity is compared with another quantity to understand how different a quantity is from another quantity. If one quantity is different to another quantity, then the two quantities are said the quantities are not equal. The mathematical relation between them is called strict inequality.
Let’s learn the concept of strict inequality from the following cases with understandable examples.
### Greater Than
If one quantity is greater than another quantity in comparison, then a greater than symbol $(>)$ is written between big and small quantities as follows.
1. $3 \,>\, 2$
2. $0.1 \,>\, 0.025$
3. $4 \,>\, \sqrt{2}$
This type of inequality between two quantities is written algebraically as $a > b$ in mathematics.
### Less Than
If one quantity is less than another quantity in comparison, then a less than symbol $(<)$ is written between small and big quantities in the following way.
1. $2 \,<\, 5$
2. $-1 \,<\, 1$
3. $\sqrt{3} \,<\, 7$
This type of inequality between the quantities is written algebraically as $a < b$ in mathematics.
### Not Equal
If a quantity is not equal to another quantity in comparison, then a not equal symbol $(\ne)$ is written between the quantities as follows.
1. $3 \,\ne\, 4$
2. $2 \,\ne\, -9$
3. $\sqrt{2} \,\ne\, \sqrt{5}$
This type of inequality between two quantities is written as $a \ne b$ algebraically in mathematics.
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In the diagram shown above, because the lines AB and CD are parallel and EF is transversal, ∠FOB and ∠OHD are corresponding angles and they are congruent. arcsin [7/9] = 51.06°. Definitions: Complementary angles are two angles with a sum of 90º. Two lines are intersect each other and form four angles in which, the angles that are opposite to each other are verticle angles. Vertical angles are congruent, so set the angles equal to each other and solve for \begin {align*}x\end {align*}. Since vertical angles are congruent or equal, 5x = 4x + 30. m∠DEB = (x + 15)° = (40 + 15)° = 55°. Divide each side by 2. Use the vertical angles theorem to find the measures of the two vertical angles. Vertical angles are always congruent. 85° + 70 ° + d = 180°d = 180° - 155 °d = 25° The triangle in the middle is isosceles so the angles on the base are equal and together with angle f, add up to 180°. Subtract 4x from each side of the equation. The second pair is 2 and 4, so I can say that the measure of angle 2 must be congruent to the measure of angle 4. It means they add up to 180 degrees. You have a 1-in-90 chance of randomly getting supplementary, vertical angles from randomly tossing … Angles in your transversal drawing that share the same vertex are called vertical angles. 6. So I could say the measure of angle 1 is congruent to the measure of angle 3, they're on, they share this vertex and they're on opposite sides of it. The angles opposite each other when two lines cross. 5x = 4x + 30. For example, in the figure above, m ∠ JQL + m ∠ LQK = 180°. Vertical Angles: Vertically opposite angles are angles that are placed opposite to each other. Corresponding Angles. Using the vertical angles theorem to solve a problem. After you have solved for the variable, plug that answer back into one of the expressions for the vertical angles to find the measure of the angle itself. Vertical AnglesVertical Angles are the angles opposite each other when two lines cross.They are called "Vertical" because they share the same Vertex. Toggle Angles. 5. The formula: tangent of (angle measurement) X rise (the length you marked on the tongue side) = equals the run (on the blade). Big Ideas: Vertical angles are opposite angles that share the same vertex and measurement. We examine three types: complementary, supplementary, and vertical angles. Introduction: Some angles can be classified according to their positions or measurements in relation to other angles. Now we know c = 85° we can find angle d since the three angles in the triangle add up to 180°. Thus one may have an … Well the vertical angles one pair would be 1 and 3. Vertical angles are two angles whose sides form two pairs of opposite rays. They are always equal. For a rough approximation, use a protractor to estimate the angle by holding the protractor in front of you as you view the side of the house. The angles that have a common arm and vertex are called adjacent angles. Vertical Angle A Zenith angle is measured from the upper end of the vertical line continuously all the way around, Figure F-3. In the diagram shown below, if the lines AB and CD are parallel and EF is transversal, find the value of 'x'. Formula : Two lines intersect each other and form four angles in which the angles that are opposite to each other are vertical angles. Then go back to find the measure of each angle. This becomes obvious when you realize the opposite, congruent vertical angles, call them a a must solve this simple algebra equation: 2a = 180° 2 a = 180 °. Find m∠2, m∠3, and m∠4. Introduce and define linear pair angles. Improve your math knowledge with free questions in "Find measures of complementary, supplementary, vertical, and adjacent angles" and thousands of other math skills. Provide practice examples that demonstrate how to identify angle relationships, as well as examples that solve for unknown variables and angles (ex. The triangle angle calculator finds the missing angles in triangle. Because the vertical angles are congruent, the result is reasonable. Subtract 20 from each side. Students also solve two-column proofs involving vertical angles. Click and drag around the points below to explore and discover the rule for vertical angles on your own. We help you determine the exact lessons you need. ∠1 and ∠3 are vertical angles. a = 90° a = 90 °. ∠1 and ∠2 are supplementary. Example. These opposite angles (vertical angles ) will be equal. arcsin [14 in * sin (30°) / 9 in] =. When two lines intersect each other at one point and the angles opposite to each other are formed with the help of that two intersected lines, then the angles are called vertically opposite angles. Vertical Angles: Theorem and Proof. Vertical and adjacent angles can be used to find the measures of unknown angles. 120 Why? This forms an equation that can be solved using algebra. Their measures are equal, so m∠3 = 90. A vertical angle is made by an inclined line of sight with the horizontal. The line of sight may be inclined upwards or downwards from the horizontal. β = arcsin [b * sin (α) / a] =. Explore the relationship and rule for vertical angles. You have four pairs of vertical angles: ∠ Q a n d ∠ U ∠ S a n d ∠ T ∠ V a n d ∠ Z ∠ Y a n d ∠ X. Students learn the definition of vertical angles and the vertical angle theorem, and are asked to find the measures of vertical angles using Algebra. Given, A= 40 deg. Solution The diagram shows that m∠1 = 90. They’re a special angle pair because their measures are always equal to one another, which means that vertical angles are congruent angles. m∠1 + m∠2 = 180 Definition of supplementary angles 90 + m∠2 = 180 Substitute 90 for m∠1. Supplementary angles are two angles with a sum of 180º. Vertical angles are angles in opposite corners of intersecting lines. A o = C o B o = D o. Read more about types of angles at Vedantu.com To solve for the value of two congruent angles when they are expressions with variables, simply set them equal to one another. Two angles that are opposite each other as D and B in the figure above are called vertical angles. m∠CEB = (4y - 15)° = (4 • 35 - 15)° = 125°. For a pair of opposite angles the following theorem, known as vertical angle theorem holds true. Divide the horizontal measurement by the vertical measurement, which gives you the tangent of the angle you want. Students learn the definition of vertical angles and the vertical angle theorem, and are asked to find the measures of vertical angles using Algebra. So, the angle measures are 125°, 55°, 55°, and 125°. Why? Vertical angles are formed by two intersecting lines. Using Vertical Angles. Example: If the angle A is 40 degree, then find the other three angles. Introduce vertical angles and how they are formed by two intersecting lines. Acute Draw a vertical line connecting the 2 rays of the angle. Try and solve the missing angles. How To: Find an inscribed angle w/ corresponding arc degree How To: Use the A-A Property to determine 2 similar triangles How To: Find an angle using alternate interior angles How To: Find a central angle with a radius and a tangent How To: Use the vertical line test In the figure above, an angle from each pair of vertical angles are adjacent angles and are supplementary (add to 180°). They have a … The real-world setups where angles are utilized consist of; railway crossing sign, letter “X,” open scissors pliers, etc. Theorem of Vertical Angles- The Vertical Angles Theorem states that vertical angles, angles which are opposite to each other and are formed by … Angles from each pair of vertical angles are known as adjacent angles and are supplementary (the angles sum up to 180 degrees). In this example a° and b° are vertical angles. m∠AEC = ( y + 20)° = (35 + 20)° = 55°. As in this case where the adjacent angles are formed by two lines intersecting we will get two pairs of adjacent angles (G + F and H + E) that are both supplementary. Both pairs of vertical angles (four angles altogether) always sum to a full angle (360°). The intersections of two lines will form a set of angles, which is known as vertical angles. Theorem: In a pair of intersecting lines the vertically opposite angles are equal. Using the example measurements: … Determine the measurement of the angles without using a protractor. Vertical angles are pair angles created when two lines intersect. In some cases, angles are referred to as vertically opposite angles because the angles are opposite per other. Do not confuse this use of "vertical" with the idea of straight up and down. omplementary and supplementary angles are types of special angles. For the exact angle, measure the horizontal run of the roof and its vertical rise. "Vertical" refers to the vertex (where they cross), NOT up/down. Vertical Angles are Congruent/equivalent. To determine the number of degrees in … \begin {align*}4x+10&=5x+2\\ x&=8\end {align*} So, \begin {align*}m\angle ABC = m\angle DBF= (4 (8)+10)^\circ =42^\circ\end {align*} These opposite angles (verticle angles ) will be equal. It ranges from 0° directly upward (zenith) to 90° on the horizontal to 180° directly downward (nadir) to 270° on the opposite horizontal to 360° back at the zenith. Another pair of special angles are vertical angles. 5x - 4x = 4x - 4x + 30. 60 60 Why? Adjacent angles share the same side and vertex. So vertical angles always share the same vertex, or corner point of the angle. Note: A vertical angle and its adjacent angle is supplementary to each other. From the theorem about sum of angles in a triangle, we calculate that γ = 180°- α - β = 180°- 30° - 51.06° = 98.94°. Examples, videos, worksheets, stories, and solutions to help Grade 6 students learn about vertical angles. The rule for vertical angles and are supplementary ( the angles opposite how to find vertical angles and! The missing angles in your transversal drawing that share the same vertex, worksheets, stories, and.... Always sum to a full angle ( 360° ) whose sides form two of! 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how to find vertical angles 2021 |
VIC Natural logarithm. Natural exponential.
VI.C. The Natural Logarithm and Exponential Functions
In the previous sections we have considered independently the functions that solved two differential equations. The "natural logarithm" function, ln , was the solution to the differential equation L'(t) = 1/t with L(1) = 0 , while the "natural exponential" function, exp or e, was the solution to the differential equation P'(t) = P(t) with P(0) = 1. You probably have seen these two functions before in your pre-calculus course work where they were closely related as the next theorem shows.
Theorem VI.C.1: (a) For all t > 0, exp(ln(t)) = t. (b) For all x, ln(exp(x)) = x.
Proof: (a) Let F(t) = exp(ln(t))/t for t > 0.
Then using the quotient and the chain rules we see that
F'(t) = [t .exp(ln(t)) .1/t - exp(ln(t)) .1]/t 2 = 0 for all t > 0.
Thus F(t) is a constant and since F(1)= exp(ln(1))/1 = exp(0)/1 = 1 we have
exp(ln(t))/t = 1 or exp(ln(t)) = t for all t > 0.
(b) Let F(x) = ln(exp(x)) - x for all x .
Then using the chain rule we see that F'(x)= 1/exp(x) . exp(x)-1= 0 for all x .
Thus F(x) is a constant. Since F(0) = ln(exp(0)) - 0= ln(1)= 0 we have ln(exp(x)) - x = 0 for all x or ln(exp(x)) = x for all x .EOP
We can visualize the relationship described in the theorem in several ways. Using transformation figures as in Figure VI.C.i , we see that the arrows used to visualize the ln function as a transformation reverse the arrows used to visualize the exp function and vice versa.
In functional terms, the composition of ln with exp is the identity function, as is also the composition of exp with ln, i.e.,
ln ° exp = id = exp ° ln,
when the domain is suitably chosen. For both these reasons we say that the natural logarithm and natural exponential function are inversely related as functions.
Another graphical way to view this inverse relation is to note that if b = exp(a) then ln(b) = ln(exp(a)) = a . Thus if (a,b) is a point on the graph of the natural exponential function, then (b,a) is a point on the graph of the natural logarithm function. Similarly if a = ln(b) , b > 0 then exp(a) = exp(ln(b)) = b which shows that if (b,a) is a point on the graph of the natural logarithm function then (a,b) is a point on the graph of the natural exponential function. See Figure VI.C.ii. We summarize this by saying exp(a) = b if and only if b = ln(a) , or using the notation e a = exp(a),
e a = b if and only if b = ln(a).
This last statement justifies calling the natural logarithm "the logarithm with the base e" and writing in more conventional terms ln(a) = log e(a). For example, e 0 = 1 so, ln(1) = 0 and e1=e so ln(e) = 1. In general ln(e a) = a , which merely restates part (b) of the theorem.
Applications: The inverse relation of these two functions is most convenient for finding ways to express the solutions to differential equations that are in exponential form as the following examples illustrate.
Example VI.C.1: Suppose f(x) = A e kx . Find A and k when f(0) = 20 and f(1) = 40. Find t so that f(t) = 100.
Solution: First we merely substitute the given values into the formula for f: 20 = f(0) = Ae k(0) = A so A=20, while 40=f(1)=A e k(1) =20 e k. So 2 = e k.
Using the inverse relation between ln and exp in this equation means precisely that k = ln(2). Thus f(x) = 20 e (ln(2) x).
To find when f(t) = 100 we need to solve 100 = 20 e (ln(2) t), or 5 = e (ln(2) t). Using the inverse properties once again, we have ln(5) = t ln(2) so
t = ln(5) / ln(2) » 2.32.
Example VI.C.2: Suppose f(x) = Ae kx and f(3) = 2f(0). If f(1) = 40, find f(5).
Solution: f(0) = A so we have been given 2 A = A e k3 in the statement f(3) = 2f(0). Thus 2 = e 3k , so by the inverse relation of ln and exp we have that 3k = ln(2) or k = (1/3) ln(2).
The second equation f(1) = 40 becomes 40 = Ae k. At this point we can solve for A theoretically by dividing both sides of the last equation by e k, giving A = 40 / e k. We continue with this form for A to find f(5) = A e k5 = (40 / e k) e 5k = 40 e 4k = 40 e 4/3 ln(2).
Recognizing this result is not the easiest number to compute, let's notice that
e 4/3 ln(2) = (eln(2))4/3 = 2 4/3
because of the properties of exponents and the inverse relation of ln and exp.
So f(5) = 40 (2) 4/3.
The connection of the natural logarithm to other core exponential and logarithmic functions.
The natural logarithm function can be defined as a definite integral: `ln(x) = int _1^x 1/t dt` . Using this definition, the natural logarithm can provide the mathematical foundation for all other logarithmic and exponential functions. The following chart demonstrates the major connections of these functions in one "Big Picture."
Def'n.: `ln(x) = int_1^x 1/t dt` [ x>0] Def'n.: exp (x) = y `-=` ln(y) = x exp(1) `-=` e [so ln(e) = 1]. exp(x) `-=` ex Def'n.: For b >0, bx`-=`e xln(b). Note: ln(bx) = x ln(b) Def'n.: For b >0, log b (x) = y `-=` x= by ln(1) = 0 ln(x) > 0 for x >1 ln(x) < 0 for 0< x <1 exp(0) = e0=1 ex> 1 for x > 0 0< ex < 1 for x < 0 b0 = 1 For b > 1: b x > 1 for x > 0 0< bx < 1 for x < 0 For 0< b < 1: b x > 1 for x < 0 0< bx < 1 for x > 0 log b (1) = 0 For b > 1: log b (x)>0 for x >1 log b (x)<0 for 0< x <1 For 0< b < 1: log b (x)<0 for x >1 log b (x)>0 for 0< x <1 ln(A*B) = ln(A) + ln(B) eAeB =eA+B bAbB = bA+B log b (A*B) = log b (A) + log b (B) ln(A/B) = ln(A) - ln(B) eA / eB = eA-B bA /bB = bA-B log b (A/B) = log b (A ) - log b (B) ln(Ap/q) =p/q ln(A) (ex) p/q = e(p/q)*x (bx) p/q = b(p/q)*x log b (Ap/q )= p/q log b (A) ln'(x) = D ln(x) = 1/x [So ln is continuous and increasing for x >0] exp'(x) = D(ex) = ex D(bx) = ln(b) bx log b'(x) = D log b (x) = 1/( x ln(b)) `int 1/u du = ln|u| + C` `int e^u du = e^u + C` ` int b^u du = b^u / ln(b) + C` Not relevant! As `x -> oo , ln (x) -> oo`. As `x -> oo, e^x -> oo`. b >0: As `x -> oo, b^x -> oo` b <0:As `x -> oo, b^x -> 0` For b > 1:As `x -> oo, log_b (x) -> oo`. For 0< b < 1:As `x -> oo, log_b (x) -> - oo`. As `x -> 0^+ , ln(x) -> - oo` As `x -> - oo , e^x -> 0` b>0: As `x -> - oo , b^x -> 0` b< 0:As `x ->- oo, b^x -> oo` For b > 1:As `x -> 0^+, log_b (x) -> - oo`. For 0< b < 1: As `x -> 0^+, log_b (x) -> oo`
Exercises VI.C
Find the first derivative for the functions in problems 1-4 using logarithmic differentiation:
1. f(x) = 7 x
2. f(x) = 3 sin(x)
3. y = sin(x) 5 x
4. `g(t) = t^pi`
5. The numbers `e^pi` and `pi^e` have been of interest to mathematicians for many years. Consider the function f(x) = e x - x e.
6. Use calculus to determine the local extreme values of f and when f is increasing or decreasing. With this information determine whether `f(pi)` is positive or negative.
Which is larger, `e^pi` or `pi^e`?
Find the first derivative for the functions in problems 6-9.
7. f(x) = log 5(x)
8. f(x) = log 3(sin(x))
9. y = sin(x) log 5(x)
10. g(t) = log t(2) for t > 1.
11. Find `int 5^x dx`.
12. Find `int 5^{sin(x)}cos(x) dx`.
13. Find `int_e^{e^2} (ln(x))/x dx`.
14. Find `int_e^{e^2} 5/(ln(x)x) dx`. |
Student Question
# `int cos(5theta)cos(3theta) d theta` Find the indefinite integral
Indefinite integrals are written in the form of `int f(x) dx = F(x) +C`
where:` f(x)` as the integrand
`F(x)` as the anti-derivative function
`C` as the arbitrary constant known as constant of integration
For the given problem `int cos(5theta)cos(3theta) d theta` has an integrand in a form of a trigonometric function. To evaluate this, we apply the identity:
`cos(X)cos(Y) =[cos(X+Y) +cos(X-Y)]/2`
The integral becomes:
`int cos(5theta)cos(3theta) d theta = int[cos(5theta+3theta) + cos(5theta -3theta)]/2 d theta`
Apply the basic properties of integration:` int c*f(x) dx= c int f(x) dx` .
`int[cos(5theta+3theta) + cos(5theta -3theta)]/2d theta = 1/2int[cos(5theta+3theta) + cos(5theta -3theta)] d theta`
Apply the basic integration property: `int (u+v) dx = int (u) dx + int (v) dx` .
`1/2 *[int cos(5theta +3theta)d theta+cos(5theta -3theta)d theta]`
Then apply u-substitution to be able to apply integration formula for cosine function:` int cos(u) du= sin(u) +C` .
For the integral: `int cos(5theta +3theta)d theta` , we let ` u =5theta +3theta =8theta` then `du= 8 d theta` or `(du)/8 =d theta` .
`int cos(5theta +3theta)d theta=int cos(8theta)d theta`
`=intcos(u) *(du)/8`
`= 1/8 int cos(u)du`
`= 1/8 sin(u) +C`
Plug-in `u =8theta` on `1/8 sin(u) +C` , we get:
`int cos(5theta +3theta)d theta=1/8 sin(8theta) +C`
For the integral: `intcos(5theta -3theta)d theta` , we let `u =5theta -3theta =2theta` then `du= 2 d theta` or `(du)/2 =d theta` .
`int cos(5theta -3theta)d theta = intcos(2theta) d theta`
`=intcos(u) *(du)/2`
`= 1/2 int cos(u)du`
`= 1/2 sin(u) +C`
Plug-in `u =2 theta` on `1/2 sin(u) +C` , we get:
`intcos(5theta -3theta)d theta =1/2 sin(2theta) +C`
Combing the results, we get the indefinite integral as:
`int cos(5theta)cos(3theta)d theta = 1/2*[1/8 sin(8theta) +1/2 sin(2theta)] +C`
or `1/16 sin(8theta) +1/4 sin(2theta) +C`
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# How do you simplify ln(e^2lne^3)?
Nov 27, 2016
$= 2 + \ln 3$
#### Explanation:
Simplifying the expression is determined by applying the
$\text{ }$
following properties:
$\text{ }$
$\textcolor{b l u e}{\ln \left(a \times b\right) = \ln a + \ln b}$
$\text{ }$
$\textcolor{red}{\ln {e}^{a} = a}$
$\text{ }$
$\ln \left({e}^{2} \ln {e}^{3}\right)$
$\text{ }$
$= \ln \left({e}^{2} \times \textcolor{red}{3}\right)$
$\text{ }$
$= \ln \left({e}^{2} \times 3\right)$
$\text{ }$
=color(blue)(lne^2 + ln3
$\text{ }$
$= \textcolor{red}{2} + \ln 3$
$\text{ }$ |
# Difference between revisions of "2015 AMC 8 Problems/Problem 7"
Each of two boxes contains three chips numbered $1$, $2$, $3$. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?
$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}$
## Contents
### Solution
We can instead calculate the probability that their product is odd, and subtract this from $1$. In order to get an odd product, we have to draw an odd number from each box. We have a $\frac{2}{3}$ probability of drawing an odd number from one box, so there is a $\left ( \frac{2}{3} \right )^2=\frac{4}{9}$ probability of having an odd product. Thus, there is a $1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}$ probability of having an even product.
### Solution 2
You can also make this problem into a spinner problem. You have the first spinner with $3$ equally divided
sections, $1, 2$ and $3.$ You make a second spinner that is identical to the first, with $3$ equal sections of
$1$,$2$, and $3$. If the first spinner lands on $1$, to be even, it must land on two. You write down the first
combination of numbers $(1,2)$. Next, if the spinner lands on $2$, it can land on any number on the second
spinner. We now have the combinations of $(1,2) ,(2,1), (2,2), (2,3)$. Finally, if the first spinner ends on $3$, we
have $(3,2).$ Since there are $3*3=9$ possible combinations, and we have $5$ evens, the final answer is
$\boxed{\textbf{(E) }\frac{5}{9}}$.
### Solution 3
We can also list out the numbers. Hat A has chips $1$, $2$, and $3$, and Hat B also has chips $1$, $2$, and $3$. Chip $1$(from Hat A) could be with 3 partners from Hat B. This is also the same for chips $2$ and $3$ from Hat A. $3+3+3=9$ total sums. Chip $1$ could be added with 2 other chips to make an even sum, just like chip $3$. Chip $2$ can only add with 1 chip. $2+2+1=5$. The answer is $\boxed{\textbf{(E) }\frac{5}{9}}$. |
# Chapter 4 - Force System Resultants
Moment of a Force 1 Objectives 1. Introduce the concept of the moment of a force and show how to calculate it in 2 dimensions. 2. Define the moment of a couple.
3. Present methods for determining the resultants force systems. 2 Force F tends to rotate the beam clockwise about A with moment MA = FdA Force F tends to rotate the beam counterclockwise
about B with moment MB = FdB Hence support at A prevents the rotation 3 Magnitude For magnitude of MO,
MO = Fd where d = moment arm or perpendicular distance from the axis at point O to its line of action of the force Units for moment is N.m 4 Direction
FMD 5 Example1 For each case, find the moment of the force about the point O
6 A MO = 7 B
MO = 8 C m N
m m MO = 9 D m
m m N MO = 10
E MO = 11 Example2 Determine the
moment of the 800 N force about points A, B, C, and D 12 Resultant Moment cw
M Fd A 13 Principles of Moments Also known as Varignons Theorem Moment of a force about a point is equal to the sum of the moments of the
forces components about the point 14 Example 3 Determine the resultant moment
of the four forces. 15 Example4 Determine the moment of the force about A. 16 17
Example5 Determine the moment of the force about 0. 18 19 Moment of a Couple
A couple is two parallel forces having the same magnitude and opposite directions separated by a distance d. 20
Moment of a Couple Resultant Force is zero. Effect of couple is a moment 21
Moment of a Couple A Couple consists of two parallel forces, equal magnitude, opposite directions, and separated a distant d apart. A Couple Moment about any point O equals the sum of the moments of both forces. 22
Example6 150N 600N 100N 250N A 4.80m long beam is subjected to the forces shown. Reduce the given system of forces to
a) an equivalent force couple system at A b) single force or resultant 23 Important Points 1. The moment of a force indicates the tendency of the body to cause rotation about a point.
2. Magnitude M=Fd d is distance from point O to line of action of F 3. Principle of Moments 24 QUIZ 1. What is the moment of the 10 N force about point A (MA)?
F = 12 N A) 3 Nm B) 36 Nm C) 12 Nm D) (12/3) Nm E) 7 Nm d=3m A 25
QUIZ 2. Using the CW direction as positive, the net moment of the two forces about point P is A) 10 N m B) 20 N m C) - 20 N m D) 40 N m E) - 40 N m 10 N
3m P 2m 5N
26 QUIZ 3. If F1 = 1 N, x1 = 1 m, F2 = 2 N and x2 = 2 m, what is the location of FR, i.e., the distance x. A) 1 m B) 1.33 m C) 1.5 m D) 1.67 m E) 2 m
F1 x2 F2 x FR
x1 27
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# How do you find the equation of a parabola when given two points (2,10) and (4,10)?
Jul 26, 2015
There's not enough information here to uniquely determine the parabola, but enough to find:
$y = a {x}^{2} - 6 a x + \left(8 a + 10\right)$ for some constant $a \in \mathbb{R}$
#### Explanation:
Since the $y$ coordinate of both points is the same, the axis must be midway between the two points at $x = 3$.
So the equation of the parabola takes the form:
$y = a {\left(x - 3\right)}^{2} + k$
If we substitute $x = 4$ and $y = 10$ into this equation we get:
$10 = a {\left(4 - 3\right)}^{2} + k = a + k$
So $k = 10 - a$ and we can write the equation as:
$y = a {\left(x - 3\right)}^{2} + \left(10 - a\right)$
$= a {x}^{2} - 6 a x + 9 a + 10 - a$
$= a {x}^{2} - 6 a x + \left(8 a + 10\right)$
The constant $a$ can be given any value in $\mathbb{R}$, except that when $a = 0$ the 'parabola' is the straight line $y = 10$.
Here are the parabolas for $a = \pm 1$ and $a = \pm 2$
graph{(y-x^2+6x-18)(y+x^2-6x-2)(y-2x^2+12x-26)(y+2x^2-12x+6)=0 [-2.585, 7.415, 7.22, 12.22]} |
# Video: CBSE Class X • Pack 1 • 2018 • Question 5
CBSE Class X • Pack 1 • 2018 • Question 5
02:20
### Video Transcript
Find the distance of a point 𝑃 𝑥, 𝑦 from the origin.
The distance formula is a useful formula that’s derived from the Pythagorean theorem that helps us find the distance between two points on a pair of axes. The distance formula says that, for two points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the distance between these points is the square root of the square of the differences between the 𝑥-coordinates plus the square of the difference between the 𝑦-coordinates. It’s the square root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared.
For this question, the first ordered pair is the one at 𝑃. It’s 𝑥, 𝑦. The second ordered pair is the one that corresponds to the origin. That’s zero, zero. So actually, we’re trying to find the distance between the points 𝑥, 𝑦 and zero, zero. Now it doesn’t matter which of these ordered pairs we choose to be 𝑥 one, 𝑦 one and which we choose to be 𝑥 two, 𝑦 two. The formula works either way round as long as we’re consistent.
Let’s see what this looks like. First, let’s choose the ordered pair 𝑥 and 𝑦 to the 𝑥 one, 𝑦 one. Then the origin becomes 𝑥 two, 𝑦 two. Substituting these values into the distance formula, we get that the distance between them is the square root of zero minus 𝑥 all squared plus zero minus 𝑦 all squared. That simplifies to the square root of negative 𝑥 squared plus negative 𝑦 squared.
Remember, when we square a negative number, we always get a positive result. So the distance between the point 𝑃 and the origin is given by the square root of 𝑥 squared plus 𝑦 squared.
Now let’s just check what this would’ve looked like had we chosen our coordinates the other way round. Substituting them into the formula gives us the square root of 𝑥 minus zero all squared plus 𝑦 minus zero all squared. Once again, that simplifies to the square root of 𝑥 squared plus 𝑦 squared. The distance of the point 𝑃 from the origin is the square root of 𝑥 squared plus 𝑦 squared. |
### About Collinear Points Distance Formula Method:
In some cases, we need to determine if three points are collinear, which means they lie on the same line. There are many methods that can be used for this procedure. Here we will use a method that relies on our distance formula. Essentially, if three points are collinear, then the sum of the distances between two pairs of points will be equal to the distance between the remaining point. We can use our distance formula to find the three distances involved and then check to see if the two smaller distances sum to the largest distance.
Test Objectives
• Demonstrate an understanding of the Pythagorean Formula
• Demonstrate an understanding of the distance formula
• Demonstrate the ability to determine if three points are collinear
Collinear Points Distance Formula Method Practice Test:
#1:
Instructions: determine whether the points are collinear.
$$a)\hspace{.2em}(-1,-2), (1,8), (-2,-7)$$
#2:
Instructions: determine whether the points are collinear.
$$a)\hspace{.2em}(3,1), (-2,-1), (-8,8)$$
#3:
Instructions: determine whether the points are collinear.
$$a)\hspace{.2em}(4,0), (-1,5), (8,-4)$$
#4:
Instructions: determine whether the points are collinear.
$$a)\hspace{.2em}(-2,-1), (-3,-3), \left(\frac{3}{2},6\right)$$
#5:
Instructions: determine whether the points are collinear.
$$a)\hspace{.2em}(-5,-5), (-2,0), (0,2)$$
Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}Collinear$$
#2:
Solutions:
$$a)\hspace{.2em}Not \hspace{.2em}Collinear$$
#3:
Solutions:
$$a)\hspace{.2em}Collinear$$
#4:
Solutions:
$$a)\hspace{.2em}Collinear$$
#5:
Solutions:
$$a)\hspace{.2em}Not \hspace{.2em}Collinear$$ |
# Solving ASA Triangles
"ASA" means "Angle, Side, Angle"
This means we are given two angles and a side between the angles.
To solve an ASA Triangle find the third angle using the three angles add to 180° then use The Law of Sines to find each of the other two sides.
### Example 1
In this triangle we know:
• angle A = 76°
• angle B = 34°
• and c = 9
It's easy to find angle C by using 'angles of a triangle add to 180°':
So C = 180° - 76° - 34° = 70°
We can now find side a by using The Law of Sines:
a/sinA = c/sin C
a/sin76° = 9/sin70°
a = (9 × sin76°)/sin70° = 9.29 to 2 decimal places.
Similarly we can find side b by using The Law of Sines:
b/sinB = c/sin C
b/sin34° = 9/sin70°
b = (9 × sin34°)/sin70° = 5.36 to 2 decimal places.
Now we have completely solved the triangle i.e. we have found all its angles and sides.
### Example 2
This is also an ASA triangle.
First find angle X by using 'angles of a triangle add to 180°':
X = 180° - 87° - 42° = 51°
Now find side y by using The Law of Sines:
y/sinY = x/sin X
So y/sin(87°) = 18.9/sin(51°)
So y = (18.9 × sin(87°))/sin(51°) = 24.29 to 2 decimal places.
Similarly we can find z by using The Law of Sines:
z/sinZ = x/sin X
So z/sin(42°) = 18.9/sin(51°)
So a = (18.9 × sin(42°))/sin(51°) = 16.27 to 2 decimal places. |
# Probability Distribution of a Random Variable
## Probability Distribution of a Random Variable:
In statistics, we had sufficient discussions about frequency distribution, which were based on observations. In a frequency distribution, the frequencies for different values of the variable under consideration are based on actual observation. Thus, if an unbiased coin is tossed 20 times, we may get head 13 times, though theoretically, we shall expect head 10 times. But 13 is the observed frequency here.
In this section, we shall discuss probability distribution based on theoretical considerations. A symbol that can assume any of the prescribed set of values is known as a variable. We consider S to be the Sample Space of some given random experiment. The outcomes, i.e., the points of the sample space are not always numbers. But we may assign a real number to each sample point according to some definite rule, which gives us a function defined on the sample space S. This real-valued function defined on the sample space of an experiment is known as a random variable. The values of a random variable are real numbers connected with the outcomes of an experiment.
In the random experiment of tossing 3 coins, the sample space is given by-
S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }
If x be the random variable denoting the ‘number of heads’ in this case, then we assign a number to each sample point as follows:
One can define many other random variables on the same sample space. For example, if x denotes the random variable defined as the ‘square of the number of tails’ in the above experiment, then we have
In the first example, the random variable x can assume only four discrete values 0, 1, 2, and 3 and in the second example, x can assume the values 0, 1, 4, and 9. The random variables defined in these two cases are called discrete random variables. We can define a discrete random variable as the one which can assume only finite number of values. If, however, the random variable assumes any value between certain limits, then it is known as a continuous random variable.
The required probability distribution is-
The required probability distribution is-
The required probability distribution is-
The required probability distribution is-
The required probability distribution is- |
Theory:
Let us learn a few results on the similarity of triangles.
1.
A perpendicular line drawn from the vertex of a right angled triangle divides the triangle into two triangles similar to each other and the original triangle.
Here, $$\triangle ADB \sim \triangle ADC$$, $$\triangle BAC \sim \triangle BDA$$, $$\triangle BAC \sim \triangle ADC$$.
2.
If two triangles are similar, then the ratio of the corresponding sides are equal to the ratio of their corresponding altitudes.
That is, $$\triangle ABC \sim \triangle PQR$$, then $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{AD}{PS} = \frac{BE}{QT} = \frac{CF}{RU}$$
3.
If two triangles are similar, then the ratio of the corresponding sides are equal to the ratio of the corresponding perimeters.
That is, $$\triangle ABC \sim \triangle XYZ$$, then $$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX} = \frac{AB + BC + CA}{XY + YZ + ZX}$$
4.
The ratio of the area of two similar triangles are equal to the ratio of the squares of their corresponding sides.
That is, $$\frac{ar(\triangle ABC)}{ar(\triangle XYZ)} = \frac{AB^2}{XY^2} = \frac{BC^2}{YZ^2} = \frac{CA^2}{ZX^2}$$
5.
If two triangles have a common vertex and their bases are on the same straight line, the ratio between their areas is equal to the ratio between the length of their bases.
That is, $$\frac{ar(\triangle ABD)}{ar(\triangle ADC)} = \frac{BD}{DC}$$ |
# How do you factor 12x^2 + 4xy + 9x + 3y?
Jul 11, 2016
$\left(3 x + y\right) \left(4 x + 3\right) .$
#### Explanation:
Given Exp. $= 12 {x}^{2} + 4 x y + 9 x + 3 y$
We notice that $4 x$ is common in first & second term, so, taking that common , we are left with $\left(3 x + y\right)$.
Similarly, from third & fourth term, we can take $3$ common, then there remains $\left(3 x + y\right) .$
This way, in turn, we'll have (3x+y) as a common factor. So, the
Given Exp. $= 4 x \left(3 x + y\right) + 3 \left(3 x + y\right)$
$= \left(3 x + y\right) \left(4 x + 3\right) .$
Enjoyed it?! |
# Partners Make Ten
8 teachers like this lesson
Print Lesson
## Objective
SWBAT increase their automaticity with number facts that make 10.
#### Big Idea
This lesson builds a foundation for adding larger numbers which is supported by the Common Core standards for second grade..
## Warming Up
10 minutes
I begin with a word problem on the board. I ask students to solve the problem in their math journal notebooks.
There are 6 fish in the pond. 4 more fish swim into the pond. How many fish are there altogether?
I give students work time (2 - 3 minutes) and then ask for students to explain how they solved the problem. I take 2 to 3 different students to explain how they solved the problem.
I put a second problem on the board.
There are 8 apples in the basket. I pick 2 more. How many apples do I have now?
I give students work time and then ask for volunteers to tell how they solved the problem. I again ask several students to share their solutions.
I ask students if they notice anything about the two problems. I take a variety of suggestions and reinforce the idea that both pairs of numbers add to 10.
I ask if they can think of any other pairs that add up to 10. I record them on the board. I tell students that these are the tens partners and that they should get to know these partners.
We look at the other partners of ten, especially those that are reverse of each other, such as 8+ 2 =10 and 2 + 8 =10. What do they notice about these? (Students were introduced to fact families in first grade and may recognize that these are partners of 10. If they don't, I will remind them of the term and see if they think these represent a fact family.)
We also look at the order of the numbers. Does order in numbers make a difference? 8+2 =10 and 2 + 8 =10 we can count and see are both true. What if we try the subtraction facts for the family? 10 - 2 =8, but can I say that 8 - 2 = 10? Why not? I try to let students discover that in subtraction the larger number always has to be first or the answer will be less than zero.
I also ask if we are representing the same amount (10) using different combinations of numbers? Can you always use different numbers to show the same amount? I ask for a few examples.
## Teaching the Lesson
40 minutes
Today I introduce a rhyme to go with the ten's partners. I present the rhyme on the Smart Board, but it could be projected on an overhead, or written on a chart.
We read the rhyme together several times. I ask if anyone could write a number sentence next to each part of the rhyme. Volunteers come up and write in the number sentences.
I tell students that in one of their centers today they will add the number sentences to the rhyme and illustrate each part. This will help them to remember the rhyme and the ten's partners.
I tell students to hold up 10 fingers. I show them that the ten's partners are all on their hands. If I call out 5, how many more fingers do they need to hold up to get to 10? If I hold up 8, how many? We practice this for a minute or two.
I encourage students to see that their fingers are a useful tool, especially for partners of ten. The visual for 10 is right in front of them at all times. They need only look down and they see the partner of ten. Seeing it and not having to count it is one step towards automaticity with number facts.
It is easy for adults to count by tens. If children can be taught to look for partners of tens in lists of numbers, or see things that are almost partners of ten, they are developing strategies that allow them to "just know" (which is an expression students use a lot when they know a fact automatically) a fact rather than have to count it.
Now I will individualize this lesson to meet the needs of students. The other activities they do will be in small groups. The reason for this is that some children need to really work with manipulatives to understand how the ten's partners work. Other children have a good grasp of this and can work with larger numbers.
One group of students who are still not sure of the concept of numbers that make ten (a first grade Common Core skill) will work with an adult to make tens partners. They will use snap blocks in many colors. They will all take 10 of one color and make a rod. Next they will start with 1 of one color and then make a rod of ten (the same length as the first rod) with a new color. What number partners do they see (9 +1). They will take that apart and continue with other number combinations.
A second group who have mastered this first grade skill will work with bundles of tens to make 100 (the second grade Common Core expectation). These students will use base 10 blocks for tens and hundreds. They will lay out the 100 block and then one ten. They will then add groups of ten until they have 100. How many bundles of ten did they need to add? (9 bundles) How much is 9 bundles of 10? (90). Students will continue to do number pairs of bundles of 10s that equal 100.
A group who is secure with bundles of tens for 100 could do bundles of 100 to make 1,000.
The groups will work for 15 minutes each and then work on their rhyme paper, or work on the rhyme paper first and then come to the table to do the activity.
## Closing
10 minutes
It is important for students to come together at the end of the lesson to reinforce their learning of partners of ten and one hundred. I ask students to clean up their materials and return to their desks. I ask for students to tell me what a partner of ten is. I take several definitions from students. I repeat them so that everyone is hearing them.
Next I put the problem, "There were 8 snowflakes on my mitten. Now there are 10. How many more snowflakes did I catch?" I ask students to raise their hands if they know the answer and to show me with fingers. I ask children how many of them used a partner of ten to solve the problem?
Even though a child may copy peers in this show of hands activity, they are also noticing how many of their peers used partners of ten and this may be an incentive for them to try it next time. While I want to make sure some children are using the partners of ten, I also know that this is a beginning of using the partners of ten, and I want students to become familiar so I am not as concerned with children copying their peers on this process. |
# What is angle subtended by an arc of circle?
## What is angle subtended by an arc of circle?
In geometry, an angle is subtended by an arc, line segment or any other section of a curve when its two rays pass through the endpoints of that arc, line segment or curve section. For example, one may speak of the angle subtended by an arc of a circle when the angle’s vertex is the centre of the circle.
## What will be the measure of the angle subtended by the arc of a circle at any point in the remaining part of the circle?
Since the angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the circle. So, 2 ∠PSQ = ∠POQ. ⇒ ∠PSQ = 80°.
## How do you find the angle of a circle with an arc?
Arc length = 2πr (θ/360) θ = the angle (in degrees) subtended by an arc at the center of the circle. 360 = the angle of one complete rotation.
## What does same arc mean?
The angles at the circumference subtended by the same arc are equal. More simply, angles in the same segment are equal.
## Is the arc double the inscribed angle?
A summary of what we did. We set out to prove that the measure of a central angle is double the measure of an inscribed angle when both angles intercept the same arc. In Case A, we spotted an isosceles triangle and a straight angle.
## What is arc in circle?
An arc of a circle is any part of the circumference. The angle subtended by an arc at any point is the angle formed between the two line segments joining that point to the end-points of the arc. For example, in the circle shown below, OP is the arc of the circle with center Q.
## What is angle subtended by diameter?
A special case of the theorem is Thales’ theorem, which states that the angle subtended by a diameter is always 90°, i.e., a right angle. As a consequence of the theorem, opposite angles of cyclic quadrilaterals sum to 180°; conversely, any quadrilateral for which this is true can be inscribed in a circle.
## What is the angle between a tangent and a chord?
The angle between a tangent and a chord is equal to the angle in the alternate segment.
## What are the 9 circle theorems?
Circle Theorem 1 – Angle at the Centre.
• Circle Theorem 2 – Angles in a Semicircle.
• Circle Theorem 3 – Angles in the Same Segment.
• Circle Theorem 4 – Cyclic Quadrilateral.
• Circle Theorem 5 – Radius to a Tangent.
• Circle Theorem 6 – Tangents from a Point to a Circle.
• Circle Theorem 7 – Tangents from a Point to a Circle II. |
# LCM of 60 and 96
The lcm of 60 and 96 is the smallest positive integer that divides the numbers 60 and 96 without a remainder. Spelled out, it is the least common multiple of 60 and 96. Here you can find the lcm of 60 and 96, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the lcm of 60 and 96, but also that of three or more integers including sixty and ninety-six for example. Keep reading to learn everything about the lcm (60,96) and the terms related to it.
## What is the LCM of 60 and 96
If you just want to know what is the least common multiple of 60 and 96, it is 480. Usually, this is written as
lcm(60,96) = 480
The lcm of 60 and 96 can be obtained like this:
• The multiples of 60 are …, 420, 480, 540, ….
• The multiples of 96 are …, 384, 480, 576, …
• The common multiples of 60 and 96 are n x 480, intersecting the two sets above, $\hspace{3px}n \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$.
• In the intersection multiples of 60 ∩ multiples of 96 the least positive element is 480.
• Therefore, the least common multiple of 60 and 96 is 480.
Taking the above into account you also know how to find all the common multiples of 60 and 96, not just the smallest. In the next section we show you how to calculate the lcm of sixty and ninety-six by means of two more methods.
## How to find the LCM of 60 and 96
The least common multiple of 60 and 96 can be computed by using the greatest common factor aka gcf of 60 and 96. This is the easiest approach:
lcm (60,96) = $\frac{60 \times 96}{gcf(60,96)} = \frac{5760}{12}$ = 480
Alternatively, the lcm of 60 and 96 can be found using the prime factorization of 60 and 96:
• The prime factorization of 60 is: 2 x 2 x 3 x 5
• The prime factorization of 96 is: 2 x 2 x 2 x 2 x 2 x 3
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(60,60) = 480
In any case, the easiest way to compute the lcm of two numbers like 60 and 96 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 60,96. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 60 and 96
What is the least common multiple of 60 and 96 used for? Answer: It is helpful for adding and subtracting fractions like 1/60 and 1/96. Just multiply the dividends and divisors by 8 and 5, respectively, such that the divisors have the value of 480, the lcm of 60 and 96.
$\frac{1}{60} + \frac{1}{96} = \frac{8}{480} + \frac{5}{480} = \frac{13}{480}$. $\hspace{30px}\frac{1}{60} – \frac{1}{96} = \frac{8}{480} – \frac{5}{480} = \frac{3}{480}$.
## Properties of LCM of 60 and 96
The most important properties of the lcm(60,96) are:
• Commutative property: lcm(60,96) = lcm(96,60)
• Associative property: lcm(60,96,n) = lcm(lcm(96,60),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the lcm of three or more numbers; our calculator makes use of it.
To sum up, the lcm of 60 and 96 is 480. In common notation: lcm (60,96) = 480.
If you have been searching for lcm 60 and 96 or lcm 60 96 then you have come to the correct page, too. The same is the true if you typed lcm for 60 and 96 in your favorite search engine.
Note that you can find the least common multiple of many integer pairs including sixty / ninety-six by using the search form in the sidebar of this page.
Questions and comments related to the lcm of 60 and 96 are really appreciated. Use the form below or send us a mail to get in touch.
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# What is the axis of symmetry and vertex for the graph y=x^2+8x+12?
May 18, 2018
$x = - 4 \text{ and vertex } = \left(- 4 , - 4\right)$
#### Explanation:
"given a parabola in standard form "color(white)(x);ax^2+bx+c
$\text{then the x-coordinate of the vertex which is also the }$
$\text{equation of the axis of symmetry is}$
•color(white)(x)x_(color(red)"vertex")=-b/(2a)
$y = {x}^{2} + 8 x + 12 \text{ is in standard form}$
$\text{with "a=1,b=8" and } c = 12$
$\Rightarrow {x}_{\text{vertex}} = - \frac{8}{2} = - 4$
$\text{substitute this value into the equation for y}$
${y}_{\text{vertex}} = {\left(- 4\right)}^{2} + 8 \left(- 4\right) + 12 = - 4$
$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 4 , - 4\right)$
$\text{axis of symmetry is } x = - 4$
graph{(y-x^2-8x-12)(y-1000x-4000)=0 [-10, 10, -5, 5]} |
# What is 71/81 as a decimal?
## Solution and how to convert 71 / 81 into a decimal
71 / 81 = 0.877
Fraction conversions explained:
• 71 divided by 81
• Numerator: 71
• Denominator: 81
• Decimal: 0.877
• Percentage: 0.877%
To convert 71/81 into 0.877, a student must understand why and how. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. If we need to convert a fraction quickly, let's find out how and when we should.
71 / 81 as a percentage 71 / 81 as a fraction 71 / 81 as a decimal
0.877% - Convert percentages 71 / 81 71 / 81 = 0.877
## 71/81 is 71 divided by 81
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(71) / denominator (81) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is our equation:
### Numerator: 71
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Any value greater than fifty will be more difficult to covert to a decimal. 71 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Let's take a look below the vinculum at 81.
### Denominator: 81
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. Larger values over fifty like 81 makes conversion to decimals tougher. But 81 is an odd number. Having an odd denominator like 81 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. Next, let's go over how to convert a 71/81 to 0.877.
## How to convert 71/81 to 0.877
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 81 \enclose{longdiv}{ 71 }$$
Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 81 \enclose{longdiv}{ 71.0 }$$
Uh oh. 81 cannot be divided into 71. Place a decimal point in your answer and add a zero. Now 81 will be able to divide into 710.
### Step 3: Solve for how many whole groups you can divide 81 into 710
$$\require{enclose} 00.8 \\ 81 \enclose{longdiv}{ 71.0 }$$
We can now pull 648 whole groups from the equation. Multiple this number by our furthest left number, 81, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.8 \\ 81 \enclose{longdiv}{ 71.0 } \\ \underline{ 648 \phantom{00} } \\ 62 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. This is also true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are examples of when we should use each.
### When you should convert 71/81 into a decimal
Dining - We don't give a tip of 71/81 of the bill (technically we do, but that sounds weird doesn't it?). We give a 87% tip or 0.877 of the entire bill.
### When to convert 0.877 to 71/81 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 71/81 = 0.877 what would it be as a percentage?
• What is 1 + 71/81 in decimal form?
• What is 1 - 71/81 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.877 + 1/2?
### Convert more fractions to decimals
From 71 Numerator From 81 Denominator What is 71/71 as a decimal? What is 61/81 as a decimal? What is 71/72 as a decimal? What is 62/81 as a decimal? What is 71/73 as a decimal? What is 63/81 as a decimal? What is 71/74 as a decimal? What is 64/81 as a decimal? What is 71/75 as a decimal? What is 65/81 as a decimal? What is 71/76 as a decimal? What is 66/81 as a decimal? What is 71/77 as a decimal? What is 67/81 as a decimal? What is 71/78 as a decimal? What is 68/81 as a decimal? What is 71/79 as a decimal? What is 69/81 as a decimal? What is 71/80 as a decimal? What is 70/81 as a decimal? What is 71/81 as a decimal? What is 71/81 as a decimal? What is 71/82 as a decimal? What is 72/81 as a decimal? What is 71/83 as a decimal? What is 73/81 as a decimal? What is 71/84 as a decimal? What is 74/81 as a decimal? What is 71/85 as a decimal? What is 75/81 as a decimal? What is 71/86 as a decimal? What is 76/81 as a decimal? What is 71/87 as a decimal? What is 77/81 as a decimal? What is 71/88 as a decimal? What is 78/81 as a decimal? What is 71/89 as a decimal? What is 79/81 as a decimal? What is 71/90 as a decimal? What is 80/81 as a decimal? What is 71/91 as a decimal? What is 81/81 as a decimal?
### Convert similar fractions to percentages
From 71 Numerator From 81 Denominator 72/81 as a percentage 71/82 as a percentage 73/81 as a percentage 71/83 as a percentage 74/81 as a percentage 71/84 as a percentage 75/81 as a percentage 71/85 as a percentage 76/81 as a percentage 71/86 as a percentage 77/81 as a percentage 71/87 as a percentage 78/81 as a percentage 71/88 as a percentage 79/81 as a percentage 71/89 as a percentage 80/81 as a percentage 71/90 as a percentage 81/81 as a percentage 71/91 as a percentage |
# How do you solve 4 - |3k+1| <2?
Apr 26, 2017
$k < - 1$
$k > \frac{1}{3}$
#### Explanation:
To make the absolute positive multiply everything by (-1). Note that this turns the inequality sign round the other way. So now we have:
$- 4 + | 3 k + 1 | > - 2$
$| 3 k + 1 | > + 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There are what I will call 'trigger' values that this condition relates to.
Lets determine these 'trigger' values.
Set $| 3 k + 1 | = 2$ this is the same as $| \pm 2 | = 2$
So the 'trigger' values are such that
$3 k + 1 = - 2 \text{ "=>" } k = - \frac{3}{3} = - 1$
$3 k + 1 = + 2 \text{ "=>" } k = \frac{1}{3}$
so we have $| \text{increasingly negative} | > 2 \implies k < - 1$
and we have $| \text{increasingly positive} | > 2 \implies k > \frac{1}{3}$ |
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# 14.2b: Double Integrals Over General Regions
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Learning Objectives
• Recognize when a function of two variables is integrable over a general region.
• Evaluate a double integral by computing an iterated integral over a region bounded by two vertical lines and two functions of $$x$$, or two horizontal lines and two functions of $$y$$.
• Simplify the calculation of an iterated integral by changing the order of integration.
• Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region.
• Solve problems involving double improper integrals.
Previously, we studied the concept of double integrals and examined the tools needed to compute them. We learned techniques and properties to integrate functions of two variables over rectangular regions. We also discussed several applications, such as finding the volume bounded above by a function over a rectangular region, finding area by integration, and calculating the average value of a function of two variables.
In this section we consider double integrals of functions defined over a general bounded region $$D$$ on the plane. Most of the previous results hold in this situation as well, but some techniques need to be extended to cover this more general case.
## General Regions of Integration
An example of a general bounded region $$D$$ on a plane is shown in Figure $$\PageIndex{1}$$. Since $$D$$ is bounded on the plane, there must exist a rectangular region $$R$$ on the same plane that encloses the region $$D$$ that is, a rectangular region $$R$$ exists such that $$D$$ is a subset of $$R (D \subseteq R)$$.
Suppose $$z = f(x,y)$$ is defined on a general planar bounded region $$D$$ as in Figure $$\PageIndex{1}$$. In order to develop double integrals of $$f$$ over $$D$$ we extend the definition of the function to include all points on the rectangular region $$R$$ and then use the concepts and tools from the preceding section. But how do we extend the definition of $$f$$ to include all the points on $$R$$? We do this by defining a new function $$g(x,y)$$ on $$R$$ as follows:
$g(x,y) = \begin{cases} f(x,y) \text{if} \; (x,y) \; \text{is in}\; D \\[4pt] 0 \text{if} \;(x,y) \; \text{is in} \; R \;\text{but not in}\; D \end{cases}$
Note that we might have some technical difficulties if the boundary of $$D$$ is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in the section on Double Integrals over Rectangular Regions used an integrable function $$f(x,y)$$ we must be careful about $$g(x,y)$$ and verify that $$g(x,y)$$ is an integrable function over the rectangular region $$R$$. This happens as long as the region $$D$$ is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.
We consider two types of planar bounded regions.
Definition: Type I and Type II regions
A region $$D$$ in the $$(x,y)$$-plane is of Type I if it lies between two vertical lines and the graphs of two continuous functions $$g_1(x)$$ and $$g_2(x)$$. That is (Figure $$\PageIndex{2}$$),
$D = \big\{(x,y)\,|\, a \leq x \leq b, \space g_1(x) \leq y \leq g_2(x) \big\}.$
A region $$D$$ in the $$xy$$-plane is of Type II if it lies between two horizontal lines and the graphs of two continuous functions $$h_1(y)$$ and $$h_2(y)$$. That is (Figure $$\PageIndex{3}$$),
$D = \big\{(x,y)\,| \, c \leq y \leq d, \space h_1(y) \leq x \leq h_2(y) \big\}.$
Example $$\PageIndex{1}$$: Describing a Region as Type I and Also as Type II
Consider the region in the first quadrant between the functions $$y = \sqrt{x}$$ and $$y = x^3$$ (Figure $$\PageIndex{4}$$). Describe the region first as Type I and then as Type II.
When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region $$D$$ is bounded above by $$y = \sqrt{x}$$ and below by $$y = x^3$$ in the interval for $$x$$ in $$[0,1]$$. Hence, as Type I, $$D$$ is described as the set $$\big\{(x,y)\,| \, 0 \leq x \leq 1, \space x^3 \leq y \leq \sqrt[3]{x}\big\}$$.
However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region $$D$$ is bounded on the left by $$x = y^2$$ and on the right by $$x = \sqrt[3]{y}$$ in the interval for y in $$[0,1]$$. Hence, as Type II, D is described as the set $$\big\{(x,y) \,| \, 0 \leq y \leq 1, \space y^2 \leq x \leq \sqrt[3]{y}\big\}$$.
Exercise $$\PageIndex{1}$$
Consider the region in the first quadrant between the functions $$y = 2x$$ and $$y = x^2$$. Describe the region first as Type I and then as Type II..
Hint
Graph the functions, and draw vertical and horizontal lines.
Type I and Type II are expressed as $$\big\{(x,y) \,|\, 0 \leq x \leq 2, \space x^2 \leq y \leq 2x\big\}$$ and $$\big\{(x,y)|\, 0 \leq y \leq 4, \space \frac{1}{2} y \leq x \leq \sqrt{y}\big\}$$, respectively.
## Double Integrals over Non-rectangular Regions
To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.
Theorem: Double Integrals over Nonrectangular Regions
Suppose $$g(x,y)$$ is the extension to the rectangle $$R$$ of the function $$f(x,y)$$ defined on the regions $$D$$ and $$R$$ as shown in Figure $$\PageIndex{1}$$ inside $$R$$. Then $$g(x,y)$$ is integrable and we define the double integral of $$f(x,y)$$ over $$D$$ by
$\iint\limits_D f(x,y) \,dA = \iint\limits_R g(x,y) \,dA.$
The right-hand side of this equation is what we have seen before, so this theorem is reasonable because $$R$$ is a rectangle and $$\iint\limits_R g(x,y)dA$$ has been discussed in the preceding section. Also, the equality works because the values of $$g(x,y)$$ are $$0$$ for any point $$(x,y)$$ that lies outside $$D$$ and hence these points do not add anything to the integral. However, it is important that the rectangle $$R$$ contains the region $$D$$.
As a matter of fact, if the region $$D$$ is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle $$R$$ containing the region.
Theorem: Fubini’s Theorem (Strong Form)
For a function $$f(x,y)$$ that is continuous on a region $$D$$ of Type I, we have
$\iint\limits_D f(x,y)\,dA = \iint\limits_D f(x,y)\,dy \space dx = \int_a^b \left[\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy \right] dx.$
Similarly, for a function $$f(x,y)$$ that is continuous on a region $$D$$ of Type II, we have
$\iint\limits_D f(x,y)\,dA = \iint\limits_D f(x,y)\,dx \space dy = \int_c^d \left[\int_{h_1(y)}^{h_2(y)} f(x,y)\,dy \right] dy.$
The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate $$f(x,y)$$ with $$x$$ being held constant and the limits of integration being $$g_1(x)$$ and $$g_2(x)$$. In the inner integral in the second expression, we integrate $$f(x,y)$$ with $$y$$ being held constant and the limits of integration are $$h_1(x)$$ and $$h_2(x)$$.
Example $$\PageIndex{2}$$: Evaluating an Iterated Integral over a Type I Region
Evaluate the integral $$\displaystyle \iint \limits _D x^2 e^{xy} \,dA$$ where $$D$$ is shown in Figure $$\PageIndex{5}$$.
Solution
First construct the region as a Type I region (Figure $$\PageIndex{5}$$). Here $$D = \big\{(x,y) \,|\, 0 \leq x \leq 2, \space \frac{1}{2} x \leq y \leq 1\big\}$$. Then we have
$\iint \limits _D x^2e^{xy} \,dA = \int_{x=0}^{x=2} \int_{y=1/2x}^{y=1} x^2e^{xy}\,dy\,dx. \nonumber$
Therefore, we have
\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}\\ &=\int_{x=0}^{x=2} \left.\left[ x^2 \frac{e^{xy}}{x} \right] \right|_{y=1/2x}^{y=1}\,dx & \text{Integrate with respect to y}\\ &= \int_{x=0}^{x=2} \left[xe^x - xe^{x^2/2}\right]dx &\text{Integrate with respect to x} \\ &=\left[xe^x - e^x - e^{\frac{1}{2}x^2} \right] \Big|_{x=0}^{x=2} = 2. \end{align*}
In Example $$\PageIndex{2}$$, we could have looked at the region in another way, such as $$D = \big\{(x,y)\,|\,0 \leq y \leq 1, \space 0 \leq x \leq 2y\big\}$$ (Figure $$\PageIndex{6}$$).
This is a Type II region and the integral would then look like
$\iint \limits _D x^2e^{xy}\,dA = \int_{y=0}^{y=1} \int_{x=0}^{x=2y} x^2 e^{xy}\,dx \space dy.$
However, if we integrate first with respect to $$x$$ this integral is lengthy to compute because we have to use integration by parts twice.
Example $$\PageIndex{3}$$: Evaluating an Iterated Integral over a Type II Region
Evaluate the integral
$\iint \limits _D (3x^2 + y^2) \,dA \nonumber$
where $$D = \big\{(x,y)\,| \, -2 \leq y \leq 3, \space y^2 - 3 \leq x \leq y + 3\big\}$$.
Solution
Notice that $$D$$ can be seen as either a Type I or a Type II region, as shown in Figure $$\PageIndex{7}$$. However, in this case describing $$D$$ as Type I is more complicated than describing it as Type II. Therefore, we use $$D$$ as a Type II region for the integration.
Choosing this order of integration, we have
\begin{align*} \iint \limits _D (3x^2 + y^2)\,dA &= \int_{y=-2}^{y=3} \int_{x=y^2-3}^{x=y+3} (3x^2 + y^2) \,dx \space dy \\ &=\int_{y=-2}^{y=3} \left. (x^3 + xy^2) \right|_{y^2-3}^{y+3} \,dy & \text{Iterated integral, Type II region}\\ &=\int_{y=-2}^{y=3} \left((y + 3)^3 + (y + 3)y^2 - (y^2 - 3)y^2\right)\,dy \\ &=\int_{-2}^3 (54 + 27y - 12y^2 + 2y^3 + 8y^4 - y^6)\,dy &\text{Integrate with respect to x.} \\ &= \left[ 54y + \frac{27y^2}{2} - 4y^3 + \frac{y^4}{2} + \frac{8y^5}{5} - \frac{y^7}{7} \right]_{-2}^3 \\ &=\frac{2375}{7}. \end{align*}
Exercise $$\PageIndex{2}$$
Sketch the region $$D$$ and evaluate the iterated integral $\iint \limits _D xy \space dy \space dx$ where $$D$$ is the region bounded by the curves $$y = \cos \space x$$ and $$y = \sin \space x$$ in the interval $$[-3\pi/4, \space \pi/4]$$.
Hint
Express $$D$$ as a Type I region, and integrate with respect to $$y$$ first.
$$\frac{\pi}{4}$$
Recall from Double Integrals over Rectangular Regions the properties of double integrals. As we have seen from the examples here, all these properties are also valid for a function defined on a non-rectangular bounded region on a plane. In particular, property 3 states:
If $$R = S \cup T$$ and $$S \cap T = 0$$ except at their boundaries, then
$\iint \limits _R f(x,y)\,dA = \iint\limits _S f(x,y)\,dA + \iint_T f(x,y) \,dA.$
Similarly, we have the following property of double integrals over a non-rectangular bounded region on a plane.
Theorem: Decomposing Regions into Smaller Regions
Suppose the region $$D$$ can be expressed as $$D = D_1 \cup D_2$$ where $$D_1$$ and $$D_2$$ do not overlap except at their boundaries. Then
$\iint \limits _D f(x,y) \,dA = \iint \limits _{D_1} f(x,y) \,dA + \iint \limits _{D_2} f(x,y) \,dA.$
This theorem is particularly useful for non-rectangular regions because it allows us to split a region into a union of regions of Type I and Type II. Then we can compute the double integral on each piece in a convenient way, as in the next example.
Example $$\PageIndex{4}$$: Decomposing Regions
Express the region $$D$$ shown in Figure $$\PageIndex{8}$$ as a union of regions of Type I or Type II, and evaluate the integral
$\iint \limits _D (2x + 5y)\,dA. \nonumber$
Solution
The region $$D$$ is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions $$D_1$$, $$D_2$$, and $$D_3$$ where, $$D_1 = \big\{(x,y)\,| \, -2 \leq x \leq 0, \space 0 \leq y \leq (x + 2)^2 \big\}$$, $$D_2 = \big\{(x,y)\,| \, 0 \leq y \leq 4, \space 0 \leq x \leq \big(y - \frac{1}{16} y^3 \big) \big\}$$, and $$D_3 = \big\{(x,y)\,| \, -4 \leq y \leq 0, \space -2 \leq x \leq \big(y - \frac{1}{16} y^3 \big) \big\}$$. These regions are illustrated more clearly in Figure $$\PageIndex{9}$$.
Here $$D_1$$ is Type I and $$D_2$$ and $$D_3$$ are both of Type II. Hence,
\begin{align*} \iint\limits_D (2x + 5y)\,dA &= \iint\limits_{D_1} (2x + 5y)\,dA + \iint\limits_{D_2} (2x + 5y)\,dA + \iint\limits_{D_3} (2x + 5y)\,dA \\ &= \int_{x=-2}^{x=0} \int_{y=0}^{y=(x+2)^2} (2x + 5y) \,dy \space dx + \int_{y=0}^{y=4} \int_{x=0}^{x=y-(1/16)y^3} (2 + 5y)\,dx \space dy + \int_{y=-4}^{y=0} \int_{x=-2}^{x=y-(1/16)y^3} (2x + 5y)\,dx \space dy \\ &= \int_{x=-2}^{x=0} \left[\frac{1}{2}(2 + x)^2 (20 + 24x + 5x^2)\right]\,dx + \int_{y=0}^{y=4} \left[\frac{1}{256}y^6 - \frac{7}{16}y^4 + 6y^2 \right]\,dy +\int_{y=-4}^{y=0} \left[\frac{1}{256}y^6 - \frac{7}{16}y^4 + 6y^2 + 10y - 4\right] \,dy\\ &= \frac{40}{3} + \frac{1664}{35} - \frac{1696}{35} = \frac{1304}{105}.\end{align*}
Now we could redo this example using a union of two Type II regions (see the Checkpoint).
Exercise $$\PageIndex{3}$$
Consider the region bounded by the curves $$y = \ln x$$ and $$y = e^x$$ in the interval $$[1,2]$$. Decompose the region into smaller regions of Type II.
Hint
Sketch the region, and split it into three regions to set it up.
$\big\{(x,y)\,| \, 0 \leq y \leq 1, \space 1 \leq x \leq e^y \big\} \cup \big\{(x,y)\,| \, 1 \leq y \leq e, \space 1 \leq x \leq 2 \big\} \cup \big\{(x,y)\,| \, e \leq y \leq e^2, \space \ln y \leq x \leq 2 \big\} \nonumber$
Exercise $$\PageIndex{4}$$
Redo Example $$\PageIndex{4}$$ using a union of two Type II regions.
Hint
$\big\{(x,y)\,| \, 0 \leq y \leq 4, \space 2 + \sqrt{y} \leq x \leq \big(y - \frac{1}{16}y^3\big) \big\} \cup \big\{(x,y)\,| \, - 4 \leq y \leq 0, \space -2 \leq x \leq \big(y - \frac{1}{16}y^{13}\big) \big\} \nonumber$
Same as in the example shown.
## Changing the Order of Integration
As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.
Example $$\PageIndex{5}$$: Changing the Order of Integration
Reverse the order of integration in the iterated integral
$\int_{x=0}^{x=\sqrt{2}} \int_{y=0}^{y=2-x^2} xe^{x^2} \,dy \space dx. \nonumber$
Then evaluate the new iterated integral.
Solution
The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to Figure $$\PageIndex{10}$$.
We can see from the limits of integration that the region is bounded above by $$y = 2 - x^2$$ and below by $$y = 0$$ where $$x$$ is in the interval $$[0, \sqrt{2}]$$. By reversing the order, we have the region bounded on the left by $$x = 0$$ and on the right by $$x = \sqrt{2 - y}$$ where $$y$$ is in the interval $$[0, 2]$$. We solved $$y = 2 - x^2$$ in terms of $$x$$ to obtain $$x = \sqrt{2 - y}$$.
Hence
\begin{align*} \int_0^{\sqrt{2}} \int_0^{2-x^2} xe^{x^2} dy \space dx &= \int_0^2 \int_0^{\sqrt{2-y}} xe^{x^2}\,dx \space dy &\text{Reverse the order of integration then use substitution.} \\[4pt] &= \int_0^2 \left[\left.\frac{1}{2}e^{x^2}\right|_0^{\sqrt{2-y}}\right] dy = \int_0^2\frac{1}{2}(e^{2-y} - 1)\,dy \\[4pt] &= -\left.\frac{1}{2}(e^{2-y} + y)\right|_0^2 = \frac{1}{2}(e^2 - 3). \end{align*}
Example $$\PageIndex{6}$$: Evaluating an Iterated Integral by Reversing the Order of Integration
Consider the iterated integral
$\iint\limits_R f(x,y)\,dx \space dy$
where $$z = f(x,y) = x - 2y$$ over a triangular region $$R$$ that has sides on $$x = 0, \space y = 0$$, and the line $$x + y = 1$$. Sketch the region, and then evaluate the iterated integral by
1. integrating first with respect to $$y$$ and then
2. integrating first with respect to $$x$$.
Solution
A sketch of the region appears in Figure $$\PageIndex{11}$$.
We can complete this integration in two different ways.
a. One way to look at it is by first integrating $$y$$ from $$y = 0$$ to $$y = 1 - x$$ vertically and then integrating $$x$$ from $$x = 0$$ to $$x = 1$$:
\begin{align*} \iint\limits_R f(x,y) \,dx \space dy &= \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} (x - 2y) \,dy \space dx = \int_{x=0}^{x=1}\left[xy - 2y^2\right]_{y=0}^{y=1-x} dx \\[4pt] \int_{x=0}^{x=1} \left[ x(1 - x) - (1 - x)^2\right] \,dx = \int_{x=0}^{x=1} [ -1 + 3x - 2x^2] dx = \left[ -x + \frac{3}{2}x^2 - \frac{2}{3} x^3 \right]_{x=0}^{x=1} = -\frac{1}{6}. \end{align*}
b. The other way to do this problem is by first integrating $$x$$ from $$x = 0$$ to $$x = 1 - y$$ horizontally and then integrating $$y$$ from $$y = 0$$ to $$y = 1$$:
\begin{align*} \iint \limits _D (3x^2 + y^2)\,dA &= \int_{y=-2}^{y=3} \int_{x=y^2-3}^{x=y+3} (3x^2 + y^2) \,dx \space dy \\[4pt] &=\int_{y=-2}^{y=3} \left. (x^3 + xy^2) \right|_{y^2-3}^{y+3} \,dy & \text{Iterated integral, Type II region}\\[4pt] &=\int_{y=-2}^{y=3} \left((y + 3)^3 + (y + 3)y^2 - (y^2 - 3)y^2\right)\,dy \\[4pt] &=\int_{-2}^3 (54 + 27y - 12y^2 + 2y^3 + 8y^4 - y^6)\,dy &\text{Integrate with respect to x.} \\ &= \left[ 54y + \frac{27y^2}{2} - 4y^3 + \frac{y^4}{2} + \frac{8y^5}{5} - \frac{y^7}{7} \right]_{-2}^3 \\ &=\frac{2375}{7}. \end{align*}
Exercise $$\PageIndex{5}$$
Evaluate the iterated integral $$\iint\limits_D (x^2 + y^2)\,dA$$ over the region $$D$$ in the first quadrant between the functions $$y = 2x$$ and $$y = x^2$$. Evaluate the iterated integral by integrating first with respect to $$y$$ and then integrating first with resect to $$x$$.
Hint
Sketch the region and follow Example $$\PageIndex{6}$$.
$$\frac{216}{35}$$
## Calculating Volumes, Areas, and Average Values
We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in Double Integrals over Rectangular Regions, but without the restriction to a rectangular region, we can now solve a wider variety of problems.
Example $$\PageIndex{7}$$: Finding the Volume of a Tetrahedron
Find the volume of the solid bounded by the planes $$x = 0, \space y = 0, \space z = 0$$, and $$2x + 3y + z = 6$$.
Solution
The solid is a tetrahedron with the base on the $$xy$$-plane and a height $$z = 6 - 2x - 3y$$. The base is the region $$D$$ bounded by the lines, $$x = 0$$, $$y = 0$$ and $$2x + 3y = 6$$ where $$z = 0$$ (Figure $$\PageIndex{12}$$). Note that we can consider the region $$D$$ as Type I or as Type II, and we can integrate in both ways.
First, consider $$D$$ as a Type I region, and hence $$D = \big\{(x,y)\,| \, 0 \leq x \leq 3, \space 0 \leq y \leq 2 - \frac{2}{3} x \big\}$$.
Therefore, the volume is
\begin{align*} V &= \int_{x=0}^{x=3} \int_{y=0}^{y=2-(2x/3)} (6 - 2x - 3y) \,dy \space dx = \int_{x=0}^{x=3} \left[ \left.\left( 6y - 2xy - \frac{3}{2}y^2\right)\right|_{y=0}^{y=2-(2x/3)} \right] \,dx\\[4pt] &= \int_{x=0}^{x=3} \left[\frac{2}{3} (x - 3)^2 \right] \,dx = 6. \end{align*}
Now consider $$D$$ as a Type II region, so $$D = \big\{(x,y)\,| \, 0 \leq y \leq 2, \space 0 \leq x \leq 3 - \frac{3}{2}y \big\}$$. In this calculation, the volume is
\begin{align*} V &= \int_{y=0}^{y=2} \int_{x=0}^{x=3-(3y/2)} (6 - 2x - 3y)\,dx \space dy = \int_{y=0}^{y=2} \left[\left.(6x - x^2 - 3xy)\right|_{x=0}^{x=3-(3y/2)} \right] \,dy \\[4pt] &= \int_{y=0}^{y=2} \left[\frac{9}{4}(y - 2)^2 \right] \,dy = 6.\end{align*}
Therefore, the volume is 6 cubic units.
Exercise $$\PageIndex{6}$$
Find the volume of the solid bounded above by $$f(x,y) = 10 - 2x + y$$ over the region enclosed by the curves $$y = 0$$ and $$y = e^x$$ where $$x$$ is in the interval $$[0,1]$$.
Hint
Sketch the region, and describe it as Type I.
$$\frac{e^2}{4} + 10e - \frac{49}{4}$$ cubic units
Finding the area of a rectangular region is easy, but finding the area of a non-rectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general non-rectangular region, as stated in the next definition.
Definition: Double Integrals
The area of a plane-bounded region $$D$$ is defined as the double integral
$\iint\limits_D 1\,dA.$
We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter.
Example $$\PageIndex{8}$$: Finding the Area of a Region
Find the area of the region bounded below by the curve $$y = x^2$$ and above by the line $$y = 2x$$ in the first quadrant (Figure $$\PageIndex{13}$$).
Solution
We just have to integrate the constant function $$f(x,y) = 1$$ over the region. Thus, the area $$A$$ of the bounded region is $$\int_{x=0}^{x=2} \int_{y=x^2}^{y=2x} dy \space dx \space or \space \int_{y=0}^{y=4} \int_{x=y/2}^{x=\sqrt{y}} dx \space dy:$$
\begin{align*} A &= \iint\limits_D 1\,dx \space dy \\[4pt] &= \int_{x=0}^{x=2} \int_{y=x^2}^{y=2x} 1\,dy \space dx \\[4pt] &= \int_{x=0}^{x=2} \left[\left.y\right|_{y=x^2}^{y=2x} \right] \,dx \\[4pt] &= \int_{x=0}^{x=2} (2x - x^2)\,dx \\[4pt] &= \left.x^2 - \frac{x^3}{3} \right|_0^2 = \frac{4}{3}. \end{align*}
Exercise $$\PageIndex{7}$$
Find the area of a region bounded above by the curve $$y = x^3$$ and below by $$y = 0$$ over the interval $$[0,3]$$.
Hint
Sketch the region.
$$\frac{81}{4}$$ square units
We can also use a double integral to find the average value of a function over a general region. The definition is a direct extension of the earlier formula.
Definition: The Average Value of a Function
If $$f (x,y)$$ is integrable over a plane-bounded region $$D$$ with positive area $$A(D)$$, then the average value of the function is
$f_{ave} = \frac{1}{A(D)} \iint\limits_D f(x,y) \,dA.$
Note that the area is $$A(D) = \iint\limits_D 1\,dA$$.
Example $$\PageIndex{9}$$: Finding an Average Value
Find the average value of the function $$f(x,y) = 7xy^2$$ on the region bounded by the line $$x = y$$ and the curve $$x = \sqrt{y}$$ (Figure $$\PageIndex{14}$$).
Solution
First find the area $$A(D)$$ where the region $$D$$ is given by the figure. We have
$A(D) = \iint\limits_D 1\,dA = \int_{y=0}^{y=1} \int_{x=y}^{x=\sqrt{y}} 1\,dx \space dy = \int_{y=0}^{y=1} \left[\left. x \right|_{x=y}^{x=\sqrt{y}} \right] \,dy = \int_{y=0}^{y=1} (\sqrt{y} - y) \,dy = \frac{2}{3}\left. y^{2/3} - \frac{y^2}{2} \right|_0^1 = \frac{1}{6}$
Then the average value of the given function over this region is
\begin{align*} f_{ave} = \frac{1}{A(D)} \iint\limits_D f(x,y) \,dA = \frac{1}{A(D)} \int_{y=0}^{y=1}\int_{x=y}^{x=\sqrt{y}} 7xy^2 \,dx \space dy = \frac{1}{1/6} \int_{y=0}^{y=1} \left[ \left. \frac{7}{2} x^2y^2 \right|_{x=y}^{x=\sqrt{y}} \right] \,dy \\ = 6 \int_{y=0}^{y=1} \left[ \frac{7}{2} y^2 (y - y^2)\right] \,dy = 6\int_{y=0}^{y=1} \left[ \frac{7}{2} (y^3 -y^4) \right] \,dy = \frac{42}{2} \left. \left( \frac{y^4}{4} - \frac{y^5}{5}\right) \right|_0^1 = \frac{42}{40} = \frac{21}{20}. \end{align*}
Exercise $$\PageIndex{8}$$
Find the average value of the function $$f(x,y) = xy$$ over the triangle with vertices $$(0,0), \space (1,0)$$ and $$(1,3)$$.
Hint
Express the line joining $$(0,0)$$ and $$(1,3)$$ as a function $$y = g(x)$$.
$$\frac{3}{4}$$
## Improper Double Integrals
An improper double integral is an integral $$\displaystyle \iint\limits_D f \,dA$$ where either $$D$$ is an unbounded region or $$f$$ is an unbounded function. For example, $$D = \big\{(x,y) \,|\,|x - y| \geq 2\big\}$$ is an unbounded region, and the function $$f(x,y) = 1/(1 - x^2 - 2y^2)$$ over the ellipse $$x^2 + 3y^2 \geq 1$$ is an unbounded function. Hence, both of the following integrals are improper integrals:
1. $\iint\limits_D xy \space dA \space \text{where} \space D = \big\{(x,y)| | \, x - y| \geq 2 \big\};$
2. $\iint\limits_D \frac{1}{1 - x^2 -2y^2}\,dA \space \text{where} \space D = \big\{(x,y)| \, x^2 + 3y^2 \leq 1 \big\}.$
In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that has only finitely many discontinuities. Not all such improper integrals can be evaluated; however, a form of Fubini’s theorem does apply for some types of improper integrals.
Theorem: Fubini’s Theorem for Improper Integrals
If $$D$$ is a bounded rectangle or simple region in the plane defined by
$$\big\{(x,y)\,: a \leq x \leq b, \space g(x) \leq y \leq h(x) \big\}$$ and also by
$$\big\{(x,y)\,: c \leq y \leq d, \space j(y) \leq x \leq k(y)\big\}$$ and $$f$$ is a nonnegative function on $$D$$ with finitely many discontinuities in the interior of $$D$$ then
$\iint\limits_D f \space dA = \int_{x=a}^{x=b} \int_{y=g(x)}^{y=h(x)} f(x,y) \,dy \space dx = \int_{y=c}^{y=d} \int_{x=j(y)}^{x=k(y)} f(x,y) \,dx \space dy$
It is very important to note that we required that the function be nonnegative on $$D$$ for the theorem to work. We consider only the case where the function has finitely many discontinuities inside $$D$$.
Example $$\PageIndex{10}$$: Evaluating a Double Improper Integral
Consider the function $$f(x,y) = \frac{e^y}{y}$$ over the region $$D = \big\{(x,y)\,: 0 \leq x \leq 1, \space x \leq y \leq \sqrt{x}\big\}.$$
Notice that the function is nonnegative and continuous at all points on $$D$$ except $$(0,0)$$. Use Fubini’s theorem to evaluate the improper integral.
Solution
First we plot the region $$D$$ (Figure $$\PageIndex{15}$$); then we express it in another way.
The other way to express the same region $$D$$ is
$D = \big\{(x,y)\,: \, 0 \leq y \leq 1, \space y^2 \leq x \leq y \big\}. \nonumber$
Thus we can use Fubini’s theorem for improper integrals and evaluate the integral as
$\int_{y=0}^{y=1} \int_{x=y^2}^{x=y} \frac{e^y}{y} \,dx \space dy. \nonumber$
Therefore, we have
$\int_{y=0}^{y=1} \int_{x=y^2}^{x=y} \frac{e^y}{y} \,dx \space dy = \int_{y=0}^{y=1} \left. \frac{e^y}{y}x\right|_{x=y^2}^{x=y} \,dy = \int_{y=0}^{y=1} \frac{e^y}{y} (y - y^2) \,dy = \int_0^1 (e^y - ye^y)\,dy = e - 2.$
As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function $$f$$ is continuous in an unbounded rectangle $$R$$.
Theorem: Improper Integrals on an Unbounded Region
If $$R$$ is an unbounded rectangle such as $$R = \big\{(x,y)\,: \, a \leq x \leq \infty, \space c \leq y \leq \infty \big\}$$, then when the limit exists, we have
$\iint\limits_R f(x,y) \,dA = \lim_{(b,d) \rightarrow (\infty, \infty)} \int_a^b \left(\int_c^d f (x,y) \,dy \right) dx = \lim_{(b,d) \rightarrow (\infty, \infty)} \int_c^d \left(\int_a^b f(x,y) \,dx \right) dy.$
The following example shows how this theorem can be used in certain cases of improper integrals.
Example $$\PageIndex{11}$$
Evaluate the integral $$\iint\limits_R xye^{-x^2-y^2}\,dA$$ where $$R$$ is the first quadrant of the plane.
Solution
The region $$R$$ is the first quadrant of the plane, which is unbounded. So
\begin{align*} \iint\limits_R xye^{-x^2-y^2} \,dA &= \lim_{(b,d) \rightarrow (\infty, \infty)} \int_{x=0}^{x=b} \left(\int_{y=0}^{y=d} xye^{-x^2-y^2} dy\right) \,dx \\ &= \lim_{(b,d) \rightarrow (\infty, \infty)} \int_{y=0}^{x=b} xye^{-x^2-y^2} \,dy \\ &= \lim_{(b,d) \rightarrow (\infty, \infty)} \frac{1}{4} \left(1 - e^{-b^2}\right) \left( 1 - e^{-d^2}\right) = \frac{1}{4} \end{align*}
Thus,
$\iint\limits_R xye^{-x^2-y^2}\,dA \nonumber$
is convergent and the value is $$\frac{1}{4}$$.
Exercise $$\PageIndex{9}$$
$\iint\limits_D \frac{y}{\sqrt{1 - x^2 - y^2}}dA$ where $$D = \big\{(x,y)\,: \, x \geq 0, \space y \geq 0, \space x^2 + y^2 \leq 1 \big\}$$.
Hint
Notice that the integral is nonnegative and discontinuous on $$x^2 + y^2 = 1$$. Express the region $$D$$ as $$D = \big\{(x,y)\,: \, 0 \leq x \leq 1, \space 0 \leq y \leq \sqrt{1 - x^2} \big\}$$ and integrate using the method of substitution.
$$\frac{\pi}{4}$$
In some situations in probability theory, we can gain insight into a problem when we are able to use double integrals over general regions. Before we go over an example with a double integral, we need to set a few definitions and become familiar with some important properties.
Definition: Joint Density Function
Consider a pair of continuous random variables $$X$$ and $$Y$$ such as the birthdays of two people or the number of sunny and rainy days in a month. The joint density function $$f$$ of $$X$$ and $$Y$$ satisfies the probability that $$(X,Y)$$ lies in a certain region $$D$$:
$P((X,Y) \in D) = \iint\limits_D f(x,y) \,dA.$
Since the probabilities can never be negative and must lie between 0 and 1 the joint density function satisfies the following inequality and equation:
$f(x,y) \geq 0 \space and \space \iint\limits_R f(x,y) \,dA = 1.$
Definition: Independent Random Variables
The variables $$X$$ and $$Y$$ are said to be independent random variables if their joint density function is the product of their individual density functions:
$f(x,y) = f_1(x) f_2(y).$
Example $$\PageIndex{12}$$: Application to Probability
At Sydney’s Restaurant, customers must wait an average of 15 minutes for a table. From the time they are seated until they have finished their meal requires an additional 40 minutes, on average. What is the probability that a customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the meal are independent events?
Solution
Waiting times are mathematically modeled by exponential density functions, with $$m$$ being the average waiting time, as
$f(t) = \begin{cases} 0 \text{if}\; t<0 \\ \dfrac{1}{m}e^{-t/m} \text{if} \; t\geq 0.\end{cases} \nonumber$
if $$X$$ and $$Y$$ are random variables for ‘waiting for a table’ and ‘completing the meal,’ then the probability density functions are, respectively,
$f_1(x) = \begin{cases} 0 \text{if}\; x<0. \\ \dfrac{1}{15} e^{-x/15} \text{if} \; x\geq 0. \end{cases} \quad \text{and} \quad f_2(y) = \begin{cases} 0 \text{if}\; y<0 \\ \dfrac{1}{40} e^{-y/40} \text{if}\; y\geq 0. \end{cases}$
Clearly, the events are independent and hence the joint density function is the product of the individual functions
$f(x,y) = f_1(x)f_2(y) = \begin{cases} 0 \text{if} \; x<0 \; \text{or} \; y<0, \\ \dfrac{1}{600} e^{-x/15} \text{if} \; x,y\geq 0 \end{cases} \nonumber$
We want to find the probability that the combined time $$X + Y$$ is less than 90 minutes. In terms of geometry, it means that the region $$D$$ is in the first quadrant bounded by the line $$x + y = 90$$ (Figure $$\PageIndex{16}$$).
Hence, the probability that $$(X,Y)$$ is in the region $$D$$ is
$P(X + Y \leq 90) = P((X,Y) \in D) = \iint\limits_D f(x,y) \,dA = \iint\limits_D \frac{1}{600}e^{-x/15} e^{-y/40} \,dA.$
Since $$x + y = 90$$ is the same as $$y = 90 - x$$, we have a region of Type I, so
\begin{align} D = \big\{(x,y)\,|\,0 \leq x \leq 90, \space 0 \leq y \leq 90 - x\big\}, \\ P(X + Y \leq 90) = \frac{1}{600} \int_{x=0}^{x=90} \int_{y=0}^{y=90-x} e^{-(/15}e^{-y/40}dx \space dy = \frac{1}{600} \int_{x=0}^{x=90} \int_{y=0}^{y=90-x}e^{-x/15}e^{-y/40} dx \space dy \\ = \frac{1}{600} \int_{x=0}^{x=90} \int_{y=0}^{y=90-x} e^{-(x/15+y/40)}dx \space dy = 0.8328 \end{align}
Thus, there is an $$83.2\%$$ chance that a customer spends less than an hour and a half at the restaurant.
Another important application in probability that can involve improper double integrals is the calculation of expected values. First we define this concept and then show an example of a calculation.
Definition: Expected Values
In probability theory, we denote the expected values $$E(X)$$ and $$E(Y)$$ respectively, as the most likely outcomes of the events. The expected values $$E(X)$$ and $$E(Y)$$ are given by
$E(X) = \iint\limits_S x\,f(x,y) \,dA \space and \space E(Y) = \iint\limits_S y\,f (x,y) \,dA,$
where $$S$$ is the sample space of the random variables $$X$$ and $$Y$$.
Example $$\PageIndex{13}$$: Finding Expected Value
Find the expected time for the events ‘waiting for a table’ and ‘completing the meal’ in Example $$\PageIndex{12}$$.
Solution
Using the first quadrant of the rectangular coordinate plane as the sample space, we have improper integrals for $$E(X)$$ and $$E(Y)$$. The expected time for a table is
\begin{align} E(X) = \iint\limits_S x\frac{1}{600} e^{-x/15}e^{-y/40} dA = \frac{1}{600} \int_{x=0}^{x=\infty} \int_{y=0}^{y=\infty} xe^{-x/15} e^{-y/40}dA\\ = \frac{1}{600} \lim_{(a,b) \rightarrow (\infty, \infty)} \int_{x=0}^{x=a} \int_{y=0}^{y=b} xe^{-x/15} e^{-y/40} dx \space dy \\ = \frac{1}{600} \left(\lim_{a\rightarrow \infty}\int_{x=0}^{x=a} xe^{-x/15} dx \right) \left( \lim_{b\rightarrow \infty} \int_{y=0}^{y=b} e^{-y/40} dy \right) \\ \frac{1}{600}\left(\left. (\lim_{a\rightarrow \infty} (-15e^{-x/15}(x + 15)))\right|_{x=0}^{x=a} \right) \left( \left. ( \lim_{b\rightarrow \infty}(-40e^{-y/40})) \right|_{y=0}^{y=b}\right) \\ =\frac{1}{600} \left( \lim_{a \rightarrow \infty} (-15e^{-a/15} (x + 15) + 225) \right)\left(\lim_{b\rightarrow \infty} (-40e^{-b/40} + 40) \right) \\ = \frac{1}{600} (225)(40) = 15. \end{align}
A similar calculation shows that $$E(Y) = 40$$. This means that the expected values of the two random events are the average waiting time and the average dining time, respectively.
Exercise $$\PageIndex{10}$$
The joint density function for two random variables $$X$$ and $$Y$$ is given by
$f(x,y) =\begin{cases}\frac{1}{600} (x^2 + y^2),\; & \text{if} \; \leq x \leq 15, \; 0 \leq y \leq 10 \\ 0, & \text{otherwise} \end{cases} \nonumber$
Find the probability that $$X$$ is at most 10 and $$Y$$ is at least 5.
Hint
Compute the probability
$P(X \leq 10, \space Y \geq 5) = \int_{x=-\infty}^{10} \int_{y=5}^{y=10} \frac{1}{6000} (x^2 + y^2) dy \space dx. \nonumber$
$$\frac{55}{72} \approx 0.7638$$
## Key Concepts
• A general bounded region $$D$$ on the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
• To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
• We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
• We can use Fubini’s theorem for improper integrals to evaluate some types of improper integrals.
## Key Equations
• Iterated integral over a Type I region
$\iint\limits_D f(x,y) \,dA = \iint\limits_D f(x,y) \,dy \space dx = \int_a^b \left[\int_{g_1(x)}^{g_2(x)} f(x,y) \,dy \right] dx$
• Iterated integral over a Type II region
$\iint\limits_D f(x,y) \,dA = \iint\limits_D (x,y) \,dx \space dy = \int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) \,dx \right] dy$
## Glossary
improper double integral
a double integral over an unbounded region or of an unbounded function
Type I
a region $$D$$ in the $$xy$$- plane is Type I if it lies between two vertical lines and the graphs of two continuous functions $$g_1(x)$$ and $$g_2(x)$$
Type II
a region $$D$$ in the $$xy$$-plane is Type II if it lies between two horizontal lines and the graphs of two continuous functions $$h_1(y)$$ and $$h_2(h)$$ |
# Engineering Mathematics Algebra: Linear and Polynomial Equations
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## 6 Questions
### What is a linear equation?
A mathematical equation representing a straight line in a two- or-3-d graph.
### How are linear equations solved?
By performing operations on the rows of the augmented matrix.
### What is a system of linear equations?
A set of simultaneous linear equations that have the form of a matrix equation.
### What method is used to solve systems of linear equations?
Gaussian Elimination.
### What is a polynomial equation?
A mathematical equation that involves variables and coefficients.
### What is involved in solving polynomial equations?
Involves variables and coefficients.
## Study Notes
Engineering mathematics algebra is a fundamental subject that is used in a wide spectrum of mathematical techniques and algorithms, both symbolic and numerical. It is the study of systems of polynomial equations in many variables, which requires a good understanding of linear equations, quadratic equations, polynomials, systems of equations, and inequalities. In this article, we will discuss these subtopics in the field of engineering mathematics algebra.
1. Linear Equations: A linear equation is a mathematical equation that represents a straight line in a two- or-3-d graph. It is usually written in the form of y = f(x), where f(x) is a linear polynomial in x. Linear equations are solved by performing operations on the rows of the augmented matrix, which simplify it to a form from which the solution of the system can be ascertained by inspection.
2. Linear Systems of Equations: A system of linear equations is a set of simultaneous linear equations that have the form of a matrix equation. Gaussian Elimination is a method used to solve systems of linear equations. It involves performing certain operations on the rows of the augmented matrix that simplify it to a form from which the solution of the system can be ascertained by inspection
3. Polynomial Equations: A polynomial equation is a mathematical equation that involves variables and coefficients. It is usually written in the form of f(x) = 0, where f(x) is a polynomial in x. The Fundamental Theorem of Algebra is used to solve polynomial equations. It ascertained that a polynomial of degree d in one variable x can be written as a linear equation in the variable x - d, and the solutions of the polynomial equation can be found by solving the linear equation
4. Systems of Polynomial Equations: A system of polynomial equations is a set of simultaneous polynomial equations that have the form of a matrix equation. Solving systems of polynomial equations is a classical problem in mathematics. There are many methods for solving polynomial systems, such as continuation methods, numerical methods, and symbolic methods
Explore the fundamental concepts of linear equations, linear systems of equations, polynomial equations, and systems of polynomial equations in the field of engineering mathematics algebra. Learn about solving methods and practical applications of these subtopics.
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Polynomials are algebraic terms which are two or more in numbers. They are also known as expressions and consist of the variable, constants, and coefficient. Besides this, they are of two types i.e. linear and quadratic polynomials plus they can be high degree and low degree. For solving them, you should be used to factoring methods and basic of algebra. So, let’s start our lessons.
How to solve linear polynomial?
At first, you should know how a linear polynomial looks like? In it, you will see variable with exponents one for example 3x+2. Let’s look at the steps for solving it:
• Sat first you have to set the polynomials in which you have to insert an equal sign in front of it like this 3x+2=0.
• Many people suggest you isolate terms but why to do that when you can give the answer by just looking at the expression.
• What you have to do is to separate like terms like shift 2 to the RHS and 3x to LHS. In other change their sides like this, 3x=0-2. Sort out like terms, and with this simple trick, you can obtain the answer quickly.
• Now, 3x= -2 and now bring three under two as like this x=-2/3.
You should take care of the similar terms, sides, and sign. Guys keep in mind that signs changes when you change the sides of the expression or constants.
How to solve a quadratic equation?
In it, the polynomial is of second degree which means that variables have exponent 2.
Let’ solve.
• First thing guys, always make sure that the polynomial must be in right order. In other words, higher degree expression should come first or their degree must be in order.
• Now, guys find out the factors of 20 plus you should use these factors in a way that it will give (-8) and will also give 20 when multiplied together.
• You have to multiply one with 20 as the coefficient of is one plus for finding factors, it is an important to step.
• 2, 2 and 5 are its factors so you should simplify it in a way mentioned above. Let’s multiply 5 and 2 to get 10 and leave two alone. It is also known as a grouping of factors.
• Now, write four expressions like this
. Now, divided them from the middle and formed two groups. Do like this: x(x+10)-2(x+10)=0 . Here we have taken x and two as common terms.
• Also, while splinting into expressions, make sure you write signs right. The two group of binomials are visible to you i.e. ( x+10) and (x-2).
• Now, you have to solve it like linear polynomials and get an answer(x=-10,x=2). Again, while using zero in the polynomial, change sings
Algebra is a very easy topic, and polynomials are like basics for other topics like linear or quadratic equations. You can solve another example of polynomials by the steps describe above plus understand properly how to group factors in it? |
## Engage NY Eureka Math 4th Grade Module 6 Lesson 6 Answer Key
### Eureka Math Grade 4 Module 6 Lesson 6 Problem Set Answer Key
Question 1.
Shade the area models to represent the number, drawing horizontal lines to make hundredths as needed. Locate the corresponding point on the number line. Label with a point, and record the mixed number as a decimal.
a. 1$$\frac{15}{100}$$ = __1__ . _15___
1 x 15/100 = 1.15.
Explanation:
In the above-given question,
given that,
1 x 15/100.
115/100.
1.15.
b. 2$$\frac{47}{100}$$ = __2__ . __47__
2 x 47/100 = 2.47.
Explanation:
In the above-given question,
given that,
2 x 47/100.
2 x 100 + 47.
247/100.
2.47.
Question 2.
Estimate to locate the points on the number lines.
a. 2$$\frac{95}{100}$$
2 x 95/100 = 2.95.
Explanation:
In the above-given question,
given that,
2 x 95/100.
295/100.
2.95.
b. 7$$\frac{52}{100}$$
7 x 52/100 = 7.52.
Explanation:
In the above-given question,
given that,
7 x 52/100.
752/100.
7.52.
Question 3.
Write the equivalent fraction and decimal for each of the following numbers.
a. 1 one 2 hundredths
b. 1 one 17 hundredths
c. 2 ones 8 hundredths
d. 2 ones 27 hundredths
e. 4 ones 58 hundredths
f. 7 ones 70 hundredths
a.1$$\frac{2}{100}$$.
Explanation:
In the above-given question,
given that,
1 one 2 hundredths.
1 x 2/100.
202/100.
2.02.
b.1$$\frac{17}{100}$$.
Explanation:
In the above-given question,
given that,
1 one 17 hundredths.
1 x 17/100.
117/100.
1.17.
c.2$$\frac{8}{100}$$.
Explanation:
In the above-given question,
given that,
2 one 8 hundredths.
2 x 8/100.
208/100.
2.08.
d.2$$\frac{27}{100}$$.
Explanation:
In the above-given question,
given that,
2 one 27 hundredths.
2 x 27/100.
227/100.
2.27.
e.4$$\frac{58}{100}$$.
Explanation:
In the above-given question,
given that,
4 one 58 hundredths.
4 x 58/100.
458/100.
4.58.
f.7$$\frac{70}{100}$$.
Explanation:
In the above-given question,
given that,
7 one 70 hundredths.
7 x 70/100.
770/100.
7.70.
Question 4.
Draw lines from dot to dot to match the decimal form to both the unit form and fraction form. All unit forms and fractions have at least one match, and some have more than one match.
a.7 ones 13 hundredths.
Explanation:
In the above-given question,
given that,
7 one 13 hundredths.
7 x 13/100.
713/100.
7.13.
b.7 ones 3 hundredths.
Explanation:
In the above-given question,
given that,
7 one 3 hundredths.
7 x 3/100.
73/100.
0.73.
c.7 ones 3 tenths.
Explanation:
In the above-given question,
given that,
7 one 3 tenths.
7 x 3/10.
73/10.
7.3.
d.7 tens 3 ones.
Explanation:
In the above-given question,
given that,
7 tens 3 ones.
7/10 x 3.
0.7 + 3.
### Eureka Math Grade 4 Module 6 Lesson 6 Exit Ticket Answer Key
Question 1.
Estimate to locate the points on the number lines. Mark the point, and label it as a decimal.
a. 7$$\frac{20}{100}$$
a.7 ones 20 hundredths.
Explanation:
In the above-given question,
given that,
7 one 20 hundredths.
7 x 20/100.
720/100.
7.20.
b. 1$$\frac{75}{100}$$
a.1 ones 75 hundredths.
Explanation:
In the above-given question,
given that,
1 one 75 hundredths.
1 x 75/100.
175/100.
1.75.
Question 2.
Write the equivalent fraction and decimal for each number.
a. 8 ones 24 hundredths
b. 2 ones 6 hundredths
a.8$$\frac{24}{100}$$.
Explanation:
In the above-given question,
given that,
8 one 24 hundredths.
8 x 24/100.
824/100.
8.24.
b.2$$\frac{6}{100}$$.
Explanation:
In the above-given question,
given that,
2 one 6 hundredths.
2 x 6/100.
206/100.
2.06.
### Eureka Math Grade 4 Module 6 Lesson 6 Homework Answer Key
Question 1.
Shade the area models to represent the number, drawing horizontal lines to make hundredths as needed. Locate the corresponding point on the number line. Label with a point, and record the mixed number as a decimal.
a. 2$$\frac{35}{100}$$ = __2_ . __35__
2 x 35/100 = 2.35.
Explanation:
In the above-given question,
given that,
2 x 35/100.
2 x 100 + 35.
235/100.
2.35.
b. 3$$\frac{17}{100}$$ = _3___ . __17__
3 x 17/100 = 3.17.
Explanation:
In the above-given question,
given that,
3 x 17/100.
3 x 100 + 17.
317/100.
3.17.
Question 2.
Estimate to locate the points on the number lines.
a. 5$$\frac{90}{100}$$
b. 3$$\frac{25}{100}$$
3 x 25/100 = 2.47.
Explanation:
In the above-given question,
given that,
3 x 25/100.
3 x 100 + 25.
325/100.
3.25.
Question 3.
Write the equivalent fraction and decimal for each of the following numbers.
a. 2 ones 2 hundredths
b. 2 ones 16 hundredths
c. 3 ones 7 hundredths
d. 1 one 18 hundredths
e. 9 ones 62 hundredths
f. 6 ones 20 hundredths
a.2$$\frac{2}{100}$$.
Explanation:
In the above-given question,
given that,
2 one 2 hundredths.
2 x 2/100.
202/100.
2.02.
b.2$$\frac{16}{100}$$.
Explanation:
In the above-given question,
given that,
2 one 16 hundredths.
2 x 16/100.
216/100.
2.16.
c.3$$\frac{7}{100}$$.
Explanation:
In the above-given question,
given that,
3 one 7 hundredths.
3 x 7/100.
307/100.
3.07.
d.1$$\frac{18}{100}$$.
Explanation:
In the above-given question,
given that,
1 one 18 hundredths.
1 x 18/100.
118/100.
1.18.
e.9$$\frac{62}{100}$$.
Explanation:
In the above-given question,
given that,
9 one 62 hundredths.
9 x 62/100.
962/100.
9.62.
f.6$$\frac{20}{100}$$.
Explanation:
In the above-given question,
given that,
6 one 20 hundredths.
6 x 20/100.
620/100.
6.20.
Question 4.
Draw lines from dot to dot to match the decimal form to both the unit form and fraction form. All unit forms and fractions have at least one match, and some have more than one match.
a.4 ones 18 hundredths.
Explanation:
In the above-given question,
given that,
4 one 18 hundredths.
4 x 18/100.
418/100.
4.18.
b.4 ones 8 hundredths.
Explanation:
In the above-given question,
given that,
4 one 8 hundredths.
4 x 8/100.
408/100.
4.08.
c.4 ones 8 tenths.
Explanation:
In the above-given question,
given that,
4 one 8 tenths.
4 x 8/10.
48/10.
4.8. |
You are on page 1of 9
# PROJECT
CHAPTER 3 SUMMARY
BY SAM WARREN
11/15/13
CLASS
70:56:81:af:eb:a5
DATE
Warren, Samuel
MATH 2
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
Properties
The Distributive Property is when you split up a two-digit number into two numbers then multiply both numbers to the constant and then add the sums. 5(48)= (5x40) + (5x8)= 200+40= 240 The Associative Property is when you switch the parenthesis around in the equation. 5(4 ! 3)= (5 ! 4)3, 7+(3+9)=(7+3)+9 The Commutative Property is when you change the order of the digits in the equation. 7+3= 3+7, 9!5= 5!9
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
Simplification
The rst step you take to simplify the expression 4(7x-3)+2x is to make 7x 9x by adding 2x to it. Then after you do that, you use the distributive property. You do 4 times 9x which equals 36x and 4 times 3 which equals 12. The nal equation is 36x- 12
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
Solving Equations #1
The rst step you do to solve the equation 7x-10-9x=-13 is to make 7x -2x by subtracting 9x from it. Then, you add 10 to -13 which equals -3 so -2x equals -3. Then, you divide -3 by -2 which equals 1.5 so x =1.5
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
Solving Equations #2
The rst step to solving the equation 5x/6+x/3 -1/2=11/6 is multiplying everything by 6 so that makes the equation 5x+2x-3=11. The next step is combining 5x and 2x and making that 7x. Then. you add 3 to both sides that makes the equation 7x=14. Then, you divide 14 by 7 and that makes x=2
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
Solving Inequalities #1
The rst step on solving the inequality 2 < x/-5 is multiply both sides by -5 and then switch the variable to the left side which would make the inequality x < -10. You also have to switch the sings because of multiplying with negative numbers. When graphing the inequality, you would make the point lled in because -10 is part of the inequality. -10 0
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
Solving Inequalities #2
The rst step in solving the inequality -4x+4 > -8 is subtracting 4 from both sides which would make the inequality -4x > -12. Then, you divide -12 by -4 which makes the inequality x < 3. You switch the signs because of dividing with negative numbers. When graphing the inequality, the point is lled in because 3 is part of the inequality. 0 3
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
## Open and Closed Points
An open point on a graph is when the number is not part of the solution set. This means that if the equation was x > 4, the point would be open. A closed point on a graph is when the number is part of the solution set. This means that if the equation was x > 4, the point would be closed.
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5
Word Problems
Chris would have to save for 14 weeks. 14b > 160.
0 160
Marks brother would spend more than an hour and 15 minutes. m > 75 0 75 x/3 - 36 < 22 x/3 < 58 x < 174
+ 36 + 36 !3 !3
Warren, Samuel
## Monday, November 18, 2013 9:52:28 AM Pacic Standard Time
70:56:81:af:eb:a5 |
# Factorize expression with sum of cubes
In this chapter we will try to factorize the algebraic expression given in the form of \mathtt{a^{3} +b^{3}}
At the end of the chapter, some solved problems are also given for your practice.
## How to factorize \mathtt{a^{3} +b^{3}} ?
To factorize the expression \mathtt{a^{3} +b^{3}} , use the following formula;
\mathtt{a^{3} +b^{3} =\ ( a+b)\left( a^{2} -ab+b^{2}\right)}
This formula is very important.
Please memorize it as it the questions related to it are directly asked in math exams.
### Problems on \mathtt{a^{3} +b^{3}}
Example 01
Factorize the below expression;
\mathtt{\Longrightarrow \ x^{3} +\ 1331}
Solution
The above expression can be expressed as;
\mathtt{\Longrightarrow \ x^{3} +( 11)^{3}}
Referring the formula;
\mathtt{a^{3} +b^{3} =\ ( a+b)\left( a^{2} -ab+b^{2}\right)}
Where;
a = x
b = 11
Putting the values in the formula, we get;
\mathtt{\Longrightarrow \ ( x\ +11)\left( x^{2} -x.11\ +( 11)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x+11)\left( x^{2} -11x\ +121\right)}
Hence, the above expression is the solution.
Example 02
Factorize the below expression
\mathtt{\Longrightarrow 343\ x^{3} +\ 216y^{3}}
Solution
The above expression can be written as;
\mathtt{\Longrightarrow \ ( 7x)^{3} +( 6y)^{3}}
Referring the formula;
\mathtt{a^{3} +b^{3} =\ ( a+b)\left( a^{2} -ab+b^{2}\right)}
Where;
a = 7x
b = 6y
Putting the values in formula;
\mathtt{\Longrightarrow \ ( 7x\ +6y)\left(( 7x)^{2} -( 7x)( 6y) \ +( 6y)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( 7x+6y)\left( 49x^{2} -42xy\ +36y^{2}\right)}
Hence, the above expression is the solution.
Example 03
Factorize the below expression.
\mathtt{\Longrightarrow 3x^{3} +\ 81}
Solution
The expression can be written as;
\mathtt{\Longrightarrow \ 3\ \left( x^{3} +27\right)}\\\ \\ \mathtt{\Longrightarrow \ 3\ \left( x^{3} +\ 3^{3}\right)}
Referring the formula;
\mathtt{a^{3} +b^{3} =\ ( a+b)\left( a^{2} -ab+b^{2}\right)}
Where;
a = x
b = 3
Using the formula;
\mathtt{\Longrightarrow \ 3\ ( x+3) \ \left( x^{2} +x.3\ +3^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 3\ ( x+3)\left( x^{2} +3x+9\right)}
Hence, the above expression is the solution.
Example 04
Factorize the below expression.
\mathtt{\Longrightarrow \ x^{6} +y^{6}}
Solution
The above expression can be written as;
\mathtt{\Longrightarrow \ \left( x^{2}\right)^{3} +\left( y^{2}\right)^{3}}
Using the sum of cubes formula;
\mathtt{\Longrightarrow \left( x^{2} +y^{2}\right)\left(\left( x^{2}\right)^{2} -\left( x^{2}\right)\left( y^{2}\right) \ +\left( y^{2}\right)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} +y^{2}\right)\left( x^{4} -x^{2} .y^{2} \ +y^{4}\right)}
Hence, the above expression is the solution.
Example 05
Solve the below expression
\mathtt{\Longrightarrow \ 15x^{9} +120y{^{21}}}
Solution
The above expression can be expressed as;
\mathtt{\Longrightarrow \ 15\left(\left( x^{3}\right)^{3} +8\left( y^{7}\right)^{3}\right)}\\\ \\ \mathtt{\Longrightarrow \ 15\left(\left( x^{3}\right)^{3} +\left( 2y^{7}\right)^{3}\right)}
Referring the formula;
\mathtt{a^{3} +b^{3} =\ ( a+b)\left( a^{2} -ab+b^{2}\right)}
Where;
a = \mathtt{x^{3}}
b = \mathtt{2y^{7}}
Using the formula, we get;
\mathtt{\Longrightarrow \ 15\ \left( x^{3} +2y^{7}\right)\left(\left( x^{3}\right)^{2} +x^{3} .2y^{7} +\left( 2y^{7}\right)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 15\left( x^{2} +2y^{7}\right)\left( x^{6} +2x^{3} y^{7} +4y^{14}\right)}
Hence, the above expression is the solution.
Example 06
Find the value of below expression using sum of cube formula.
\mathtt{\Longrightarrow \ 12^{3} +13{^{3}}}
Solution
Applying sum of cube formula;
\mathtt{\Longrightarrow \ ( 12\ +\ 13)\left( 12^{2} -12.13+13^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( 25) .\ ( 144\ -\ 156\ +\ 169)}\\\ \\ \mathtt{\Longrightarrow \ ( 25) .\ ( 157)}\\\ \\ \mathtt{\Longrightarrow \ 3925}
Hence, 3925 is the solution
Next chapter : Problems on factorization of algebraic expression |
Home » What Equals 160 In Multiplication? Update
# What Equals 160 In Multiplication? Update
Let’s discuss the question: what equals 160 in multiplication. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## What equals in multiply?
The values you are multiplying are called the factors. The answer in a multiplication problem is called the product. You find the product when you multiply two or any number of factors.
## What multiply equals 150?
What are the factors of 150? The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, and 150.
### A number when increased by 12 equals 160 times its reciprocal. find the number
A number when increased by 12 equals 160 times its reciprocal. find the number
A number when increased by 12 equals 160 times its reciprocal. find the number
## What can equal 15 multiplication?
1 x 15 = 15. 3 x 5 = 15. 5 x 3 = 15. 15 x 1 = 15.
## What can equal 180 in multiplication?
5 x 36 = 180. 9 x 20 = 180. 10 x 18 = 180. 12 x 15 = 180.
## What can you multiply to get 6?
List of Factor Pairs for 6
• 1 x 6 = 6.
• 2 x 3 = 6.
• 3 x 2 = 6.
• 6 x 1 = 6.
## What can u multiply to get 140?
The pairs of numbers that when multiplied together make 140 are the factors of 140. 1 × 140, 2 × 70 , 4 × 35 , 5 × 28 , 7 × 20, 10 × 14 are the pairs that make 140 and they are all the factors of 140.
## What times what gives you 392?
4 x 98 = 392. 7 x 56 = 392. 8 x 49 = 392. 14 x 28 = 392.
## What equals 112 when multiplied?
Factor pairs of 112 are the pairs of factors which, when multiplied together, result in a product equal to the number 112. (1 × 112), (2 × 56), (4 × 28), (7 × 16), (8 ×14) make 112.
## What can equal 48?
The factor pairs of the number 48 are: 1 x 48, 2 x 24, 3 x 16, 4 x 12, and 6 x 8.
## What do you multiply to get 24?
1 × 24 = 24. 2 × 12 = 24. 3 × 8 = 24. 4 × 6 = 24.
## What do you multiply to get 54?
1 x 54 = 54. 2 x 27 = 54. 3 x 18 = 54.
### Prime factorization of 160 and 198
Prime factorization of 160 and 198
Prime factorization of 160 and 198
## What times what can equal 56?
List of Factor Pairs for 56
• 1 x 56 = 56.
• 2 x 28 = 56.
• 4 x 14 = 56.
• 7 x 8 = 56.
• 8 x 7 = 56.
• 14 x 4 = 56.
• 28 x 2 = 56.
• 56 x 1 = 56.
## What multiplication makes 162?
Factor pairs are the pairs of two numbers that, when multiplied, give the number. Factor pairs of 162 are (1,162) (2,81) (3,54) (6,27) and (9,18).
## What multiplied equals 224?
32 × 7 = 224. 28 × 8 = 224. 14 × 16 = 224.
## What equals 270 when multiplied?
5 x 54 = 270. 10 x 27 = 270. 15 x 18 = 270. 18 x 15 = 270.
## What can you multiply to get 7?
Number 7 has infinite multiples as it can be multiplied with any whole number and we have infinite whole numbers. A multiple can be the common multiple of two or more numbers. Example: 20 is the common multiple of 2, 4, 5, 10,and 20.
First 20 Multiples of 7.
Multiplication Multiples of 7
7 × 20 140
## What gives you 8 in multiplication?
The multiples of 8 are the numbers that are generated when 8 is multiplied by any natural number.
8, 16, 24, 32,….., 72, 80, 88,….
8 × 6 = 48 8 multiplied by 6 to get 48
8 × 7 = 56 8 multiplied by 7 to get 56
8 × 8 = 64 8 multiplied by 8 to get 64
8 × 12 = 96 8 multiplied by 12 to get 96
17 thg 12, 2020
## What are the factors of 160?
The factors of 160 are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 and 160.
## What numbers can be multiplied to get 120?
As you can see, all the factor pairs of 120 are equal to the number 120 if you multiply them together.
• 1 x 120 = 120.
• 2 x 60 = 120.
• 3 x 40 = 120.
• 4 x 30 = 120.
• 5 x 24 = 120.
• 6 x 20 = 120.
• 8 x 15 = 120.
• 10 x 12 = 120.
## Which number is a factor of 200?
The factors of 200 are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200.
## What can equal 98?
Thus we have found that the 6 factors of 98 are 1, 2, 7, 14, 49, and 98.
### 3 A natural number, when increased by 12, equals 160 times its reciprocal Find the number
3 A natural number, when increased by 12, equals 160 times its reciprocal Find the number
3 A natural number, when increased by 12, equals 160 times its reciprocal Find the number
## What can go into 196?
The factors of 196 are 1, 2, 4, 7, 14, 28, 49, 98, and 196.
## What times what gives you 192?
Factors of 192 in Pairs
3 × 64 = 192. 4 × 48 = 192. 6 × 32 = 192. 8 × 24 = 192.
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# 4.1.1: Introduction to Probability
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CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Understand and use the terminology of probability.
• Determine whether two events are mutually exclusive and whether two events are independent.
• Calculate probabilities using the Addition Rules and Multiplication Rules.
• Construct and interpret Contingency Tables.
• Construct and interpret Venn Diagrams.
• Construct and interpret Tree Diagrams.
It is often necessary to "guess" about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.
You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.
Collaborative Exercise
Your instructor will survey your class. Count the number of students in the class today.
• Raise your hand if you have any change in your pocket or purse. Record the number of raised hands.
• Raise your hand if you rode a bus within the past month. Record the number of raised hands.
• Raise your hand if you answered "yes" to BOTH of the first two questions. Record the number of raised hands.
Use the class data as estimates of the following probabilities. $$P(\text{change})$$ means the probability that a randomly chosen person in your class has change in his/her pocket or purse. $$P(\text{bus})$$ means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.
• Find $$P(\text{change})$$.
• Find $$P(\text{bus})$$.
• Find $$P(\text{change AND bus})$$. Find the probability that a randomly chosen student in your class has change in his/her pocket or purse and rode a bus within the last month.
• Find $$P(\text{change|bus})$$. Find the probability that a randomly chosen student has change given that he or she rode a bus within the last month. Count all the students that rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus.
This page titled 4.1.1: Introduction to Probability is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. |
# Search by Topic
#### Resources tagged with Practical Activity similar to Can You Make 100?:
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### There are 60 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Practical Activity
### Triangles to Tetrahedra
##### Age 11 to 14 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### Tower of Hanoi
##### Age 11 to 14 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### Sociable Cards
##### Age 11 to 14 Challenge Level:
Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
### Cunning Card Trick
##### Age 11 to 14 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Getting an Angle
##### Age 11 to 14 Challenge Level:
How can you make an angle of 60 degrees by folding a sheet of paper twice?
### Making Maths: Double-sided Magic Square
##### Age 7 to 14 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Making Maths: Snake Pits
##### Age 5 to 14 Challenge Level:
A game to make and play based on the number line.
### Sea Defences
##### Age 7 to 14 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Tangram Pictures
##### Age 5 to 14 Challenge Level:
Use the tangram pieces to make our pictures, or to design some of your own!
### Rolling Triangle
##### Age 11 to 14 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Factors and Multiples Puzzle
##### Age 11 to 14 Challenge Level:
Using your knowledge of the properties of numbers, can you fill all the squares on the board?
### Which Solids Can We Make?
##### Age 11 to 14 Challenge Level:
Interior angles can help us to work out which polygons will tessellate. Can we use similar ideas to predict which polygons combine to create semi-regular solids?
### Constructing Triangles
##### Age 11 to 14 Challenge Level:
Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
### Back to the Practical?
##### Age 7 to 14
In this article for teachers, Bernard uses some problems to suggest that once a numerical pattern has been spotted from a practical starting point, going back to the practical can help explain. . . .
### More Marbles
##### Age 11 to 14 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
### Factors and Multiples Game
##### Age 7 to 16 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### Marbles
##### Age 11 to 14 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### Nine Colours
##### Age 11 to 16 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Factors and Multiples Game for Two
##### Age 7 to 14 Challenge Level:
Factors and Multiples game for an adult and child. How can you make sure you win this game?
### Making Maths: String and Circles
##### Age 7 to 14 Challenge Level:
You could use just coloured pencils and paper to create this design, but it will be more eye-catching if you can get hold of hammer, nails and string.
### Making Maths: Archimedes' Spiral
##### Age 7 to 14 Challenge Level:
Make a spiral mobile.
### Fractions Jigsaw
##### Age 11 to 14 Challenge Level:
A jigsaw where pieces only go together if the fractions are equivalent.
### Well Balanced
##### Age 5 to 18 Challenge Level:
Exploring balance and centres of mass can be great fun. The resulting structures can seem impossible. Here are some images to encourage you to experiment with non-breakable objects of your own.
### First Forward Into Logo 5: Pen Up, Pen Down
##### Age 7 to 16 Challenge Level:
Learn about Pen Up and Pen Down in Logo
### Attractive Rotations
##### Age 11 to 14 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### First Forward Into Logo 12: Puzzling Sums
##### Age 11 to 18 Challenge Level:
Can you puzzle out what sequences these Logo programs will give? Then write your own Logo programs to generate sequences.
### Celtic Knotwork Patterns
##### Age 7 to 14
This article for pupils gives an introduction to Celtic knotwork patterns and a feel for how you can draw them.
### The Best Card Trick?
##### Age 11 to 16 Challenge Level:
Time for a little mathemagic! Choose any five cards from a pack and show four of them to your partner. How can they work out the fifth?
### Making Maths: Walking Through a Playing Card?
##### Age 7 to 14 Challenge Level:
It might seem impossible but it is possible. How can you cut a playing card to make a hole big enough to walk through?
### Cubic Conundrum
##### Age 7 to 16 Challenge Level:
Which of the following cubes can be made from these nets?
### Making Maths: Equilateral Triangle Folding
##### Age 7 to 14 Challenge Level:
Make an equilateral triangle by folding paper and use it to make patterns of your own.
### Amazing Card Trick
##### Age 11 to 14 Challenge Level:
How is it possible to predict the card?
### Making Maths: Clinometer
##### Age 11 to 14 Challenge Level:
Make a clinometer and use it to help you estimate the heights of tall objects.
### Making Maths: Celtic Knot Tiles
##### Age 7 to 16 Challenge Level:
Make some celtic knot patterns using tiling techniques
### First Forward Into Logo 10: Count up - Count Down
##### Age 11 to 18 Challenge Level:
What happens when a procedure calls itself?
### Paper Folding - Models of the Platonic Solids
##### Age 11 to 16
A description of how to make the five Platonic solids out of paper.
### Making Maths: Make a Pendulum
##### Age 7 to 14 Challenge Level:
Galileo, a famous inventor who lived about 400 years ago, came up with an idea similar to this for making a time measuring instrument. Can you turn your pendulum into an accurate minute timer?
### Making Rectangles, Making Squares
##### Age 11 to 14 Challenge Level:
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
### First Forward Into Logo 2: Polygons
##### Age 7 to 16 Challenge Level:
This is the second in a twelve part introduction to Logo for beginners. In this part you learn to draw polygons.
### Straw Scaffold
##### Age 11 to 14 Challenge Level:
Build a scaffold out of drinking-straws to support a cup of water
### Gym Bag
##### Age 11 to 16 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Cool as Ice
##### Age 11 to 16 Challenge Level:
Design and construct a prototype intercooler which will satisfy agreed quality control constraints.
### Notes on a Triangle
##### Age 11 to 14 Challenge Level:
Can you describe what happens in this film?
### Witch's Hat
##### Age 11 to 16 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Make Your Own Pencil Case
##### Age 11 to 14 Challenge Level:
What shape would fit your pens and pencils best? How can you make it?
##### Age 7 to 14 Challenge Level:
What shape and size of drinks mat is best for flipping and catching?
### Observing the Sun and the Moon
##### Age 7 to 14 Challenge Level:
How does the time of dawn and dusk vary? What about the Moon, how does that change from night to night? Is the Sun always the same? Gather data to help you explore these questions. |
# Telling the time to 5 minutes
## Home learning focus
Learn to read the time to the nearest 5 minutes.
This lesson includes:
• a learning summary
• a video
• two activities
# Learn
Today we're going to look at how to tell the time to the nearest 5 minutes.
But first, watch the video below to recap on some key words you need to tell the time.
There are many more Teacher Talks covering different Maths topics on BBC iPlayer.
## Telling the time to 5 minutes
When the minute hand on a clock moves from one number to the next, it has moved 5 minutes.
There are 12 numbers around the clock face, and each interval is 5 minutes, so there are 12 × 5 = 60 minutes in an hour.
Let's use this knowledge to tell the time on the clocks below.
## Example 1:
This clock shows 5 minutes past 2 o'clock, or '5 past 2'.
This clock shows 20 minutes past 1 o'clock, or '20 past 1'.
If the number of minutes past the hour is more than 30, we say how many minutes to the next hour.
So instead of saying '50 minutes past 3 o’clock' we say ‘10 to 4’.
## Top tip
Check the minute hand to see if it shows minutes past or minutes to the hour.
## Example 2:
How many minutes is it from 5 past 7 until 25 past 7?
Between these two times, the minute hand moves from 1 to 5, which is 4 numbers. Each of these number represents 5 minutes. So:
4 × 5 = 20
Therefore, there are 20 minutes between 5 past 7 and 25 past 7.
How many minutes is it from 10 past 3 until 20 to 4?
The minute hand moves from 2 to 8, which is 6 numbers.
6 × 5 = 30 minutes
## Remember
When the minute hand is on 3, we say 'quarter past', and when it is on 9 we say 'quarter to'.
# Practise
## Activity 1
What time is it now?
Practise reading the time during the day.
Notice when the minute hand is on a number and work out how many minutes past or minutes to the hour.
What time is it now? How many minutes is it until the next hour?
## Activity 2
Test your knowledge with this quiz!
# There's more to learn
Have a look at these other resources around the BBC and the web. |
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Found in: Page 317
### College Physics (Urone)
Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000
# To get up on the roof, a person (mass $$70.0\;{\rm{kg}}$$) places a $$6.00\;{\rm{m}}$$aluminum ladder (mass $$10.0\;{\rm{kg}}$$) against the house on a concrete pad with the base of the ladder $$2.00\;{\rm{m}}$$from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is $$2\;{\rm{m}}$$from the bottom. The person is standing $$3\;{\rm{m}}$$ from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?
The magnitude of the force at the top is $$133\;{\rm{N}}$$ and at the bottom is $$784\;{\rm{N}}$$.
See the step by step solution
## Step 1: Given Data
Given:
The mass of the person is $${m_1} = 70.0\;{\rm{kg}}$$.
The mass of the ladder is $${m_2} = 10.0\;{\rm{kg}}$$.
The length of the ladder is $$l = 6\;{\rm{m}}$$.
The distance between the ladder and the concrete pad is $$2.0\;{\rm{m}}$$.
The distance between the bottom of the ladder and the person is $$3\;{\rm{m}}$$.
## Step 2: Calculation of the force
The diagram of the ladder can be shown as below,
Applying the condition of equilibrium under forces we get,
\begin{align}n - {m_1}g - mg &= 0\\n = {m_1}g + mg\\n &= 70.0 \times 9.8 + 10.0 \times 9.8\\n &= 784\;{\rm{N}}\end{align}
Applying the condition of equilibrium under torque, we get,
\begin{align}mg{l_1}\cos \theta - mg{l_2}\cos \theta + f\sqrt {{l^2} - {d^2}} - nl\cos \theta &= 0\\f &= \frac{{nl\cos \theta - mg{l_1}\cos \theta + mg{l_2}\cos \theta }}{{\sqrt {{l^2} - {d^2}} }}\\f &= \frac{{784 \times 2 - 10 \times 9.8 \times 4 \times \frac{2}{6} - 70 \times 9.8 \times 3 \times \frac{2}{6}}}{{\sqrt {{6^2} - {2^2}} }}\\ &= 133\;{\rm{N}}\end{align} |
# Ncert Solutions For Class 12 Maths Ex 13.2
## NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 – Free PDF Download
Exercise 13.2 of NCERT Solutions for Class 12 Maths Chapter 13 – Probability is based on the following topics:
1. Multiplication Theorem on Probability
2. Independent Events: Two events such that the probability of occurrence of one of them is not affected by the occurrence of the other are called independent events.
The second exercise of this chapter contains problems based on the above mentioned topics. Solving questions related to this help understand the application of probability in real life.
## Download PDF of NCERT Solutions for Class 12 Maths Chapter 13- Probability Exercise 13.2
### Access other exercises of Class 12 Maths Chapter 13
Exercise 13.1 Solutions 17 Questions
Exercise 13.3 Solutions 14 Questions
Exercise 13.4 Solutions 17 Questions
Exercise 13.5 Solutions 15 Questions
Miscellaneous Exercise On Chapter 13 Solutions 10 Questions
#### Access Answers of Maths NCERT Class 12 Chapter 13.2
1. If P (A) = 3/5 and P (B) = 1/5, find P (A ∩ B) if A and B are independent events.
Solution:
Given P (A) = 3/5 and P (B) = 1/5
As A and B are independent events.
⇒ P (A ∩ B) = P (A).P (B)
= 3/5 × 1/5 = 3/25
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Solution:
Given a pack of 52 cards.
As we know there are 26 cards in total which are black. Let A and B denotes respectively the events that the first and second drawn cards are black.
Now, P (A) = P (black card in first draw) = 26/52 = ½
Because the second card is drawn without replacement so, now the total number of black card will be 25 and total cards will be 51 that is the conditional probability of B given that A has already occurred.
Now, P (B/A)Â = P (black card in second draw) =Â 25/51
Thus the probability that both the cards are black
⇒ P (A ∩ B) = ½ × 25/51 = 25/102
Hence, the probability that both the cards are black = 25/102.
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
Given a box of oranges.
Let A, B and C denotes respectively the events that the first, second and third drawn orange is good.
Now, P (A) = P (good orange in first draw) =Â 12/15
Because the second orange is drawn without replacement so, now the total number of good oranges will be 11 and total oranges will be 14 that is the conditional probability of B given that A has already occurred.
Now, P (B/A) = P (good orange in second draw) = 11/14
Because the third orange is drawn without replacement so, now the total number of good oranges will be 10 and total orangs will be 13 that is the conditional probability of C given that A and B has already occurred.
Now, P (C/AB) = P (good orange in third draw) =Â 10/13
Thus the probability that all the oranges are good
⇒ P (A ∩ B ∩ C) = 12/15 × 11/14 × 10/13 = 44/91
Hence, the probability that a box will be approved for sale = 44/91
4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
Solution:
Given a fair coin and an unbiased die are tossed.
We know that the sample space S:
S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Let A be the event head appears on the coin:
⇒ A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
⇒ P (A) = 6/12 = ½
Now, Let B be the event 3 on the die
⇒ B = {(H, 3), (T, 3)}
⇒ P (B) = 2/12 = 1/6
As, A ∩ B = {(H, 3)}
⇒ P (A ∩ B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A ∩ B) = P (A). P (B)
Therefore, A and B are independent events.
5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?
Solution:
The sample space for the dice will be
S = {1, 2, 3, 4, 5, 6}
Let A be the event, the number is even:
⇒ A = {2, 4, 6}
⇒ P (A) = 3/6 = ½
Now, Let B be the event, the number is red:
⇒ B = {1, 2, 3}
⇒ P (B) = 3/6 = 1/2
As, A ∩ B = {2}
⇒ P (A ∩ B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A ∩ B) ≠P (A). P (B)
Therefore, A and B are not independent events.
6. Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ≠1/5
⇒ P (E ∩ F) ≠ P (E). P (F)
Therefore, E and F are not independent events.
7. Given that the events A and B are such that P (A) = 1/2, P (A ∪ B) = 3/5 and P (B) = p. Find p if they are (i) mutually exclusive (ii) independent.
Solution:
Given P (A) = ½, P (A ∪ B) = 1/5 and P (B) = p
(i) Mutually exclusive
When A and B are mutually exclusive.
Then (A ∩ B) = ϕ
⇒ P (A ∩ B) = 0
As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 3/5 = ½ + p -0
⇒ P = 3/5 – ½ = 1/10
(ii) Independent
When A and B are independent.
⇒ P (A ∩ B) = P (A). P (B)
⇒ P (A ∩ B) = ½ p
As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 3/5 = ½ + 2 – p/2
⇒ p/2 = 3/5 – ½
⇒ p = 2 × 1/10 = 1/5
8. Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find
(ii) P (A ∪ B)
(iii) P (A|B)
(iv) P (B|A)
Solution:
Given P (A) = 0.3 and P(B) = 0.4
When A and B are independent.
⇒ P (A ∩ B) = P (A). P (B)
⇒ P (A ∩ B) = 0.3 × 0.4
⇒ P (A ∩ B) = 0.12
(ii) P (A ∪ B)
As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ P (A ∪ B) = 0.3 + 0.4 – 0.12
⇒ P (A ∪ B) = 0.58
(iv) P (B|A)
9. If A and B are two events such that P (A) = 1/4 , P (B) = 1/2 and P (A ∩ B) = 1/8, find P (not A and not B).
Solution:
Given P (A) = ¼, P (B) = ½ and P (A ∩ B) = 1/8
P (not A and not B) = P (A’ ∩ B’)
As, { A’ ∩ B’ = (A ∪ B)’}
⇒ P (not A and not B) = P ((A ∪ B)’)
= 1 – P (A ∪ B)
= 1- [P (A) + P (B) – P (A ∩ B)]
10. Events A and B are such that P (A) = 1/2, P (B) = 7/12 and P (not A or not B) = 1/4. State whether A and B are independent?
Solution:
Given P (A) = ½, P (B) =7/12 and P (not A or not B) = 1/4
⇒ P (A’∪ B’) = 1/4
⇒ P (A ∩ B)’ = 1/4
⇒ 1 – P (A ∩ B) = 1/4
⇒ P (A ∩ B) = 1 – 1/4
⇒ P (A ∩ B) = 3/4…….. (1)
And P (A). P (B) = ½ × 7/12 = 7/24 …. (2)
From (1) and (2) P (A ∩ B) ≠ P (A). P (B)
Therefore, A and B are not independent events.
11. Given two independent events A and B such that P (A) = 0.3, P(B) = 0.6.
Find
(i) P (A and B)
(ii) P (A and not B)
(iii) P (A or B)
(iv) P (neither A nor B)
Solution:
Given P (A) = 0.3, P (B) = 0.6.
(i) P (A and B)
As A and B are independent events.
⇒ P (A and B) = P (A ∩ B) = P (A). P (B)
= 0.3 × 0.6
= 0.18
(ii) P (A and not B)
⇒ P (A and not B) = P (A ∩ B’) = P (A) – P (A ∩ B)
= 0.3 – 0.18
= 0.12
(iii) P (A or B)
⇒ P (A or B) = P (A ∪ B)
As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ P (A ∪ B) = 0.3 + 0.6 – 0.18
⇒ P (A ∪ B) = 0.72
(iv) P (neither A nor B)
P (neither A nor B) = P (A’ ∩ B’)
As, { A’ ∩ B’ = (A ∪ B)’}
⇒ P (neither A nor B) = P ((A ∪ B)’)
= 1 – P (A ∪ B)
= 1 – 0.72
= 0.28
12. A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
Given a die is tossed thrice.
Then the sample space S = {1, 2, 3, 4, 5, 6}
Let P (A) = probability of getting an odd number in first throw.
⇒ P (A) = 3/6 = ½.
Let P (B) = probability of getting an even number.
⇒ P (B) =3/6 = ½.
Now, probability of getting an even number in three times = ½ × ½ × ½ = 1/8
So, probability of getting an odd number at least once
= 1 – probability of getting an odd number in no throw
= 1 – probability of getting an even number in three times
= 1 – 1/8
∴ Probability of getting an odd number at least once = 7/8
13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) First ball is black and second is red.
(iii) One of them is black and other is red.
Solution:
Given A box containing 10 black and 8 red balls.
Total number of balls in box = 18
(i) Both balls are red.
Probability of getting a red ball in first draw =Â 8/18 = 4/9
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw = 8/18 = 4/9
Now, Probability of getting both balls red = 4/9 × 4/9 = 16/81
(ii) First ball is black and second is red.
Probability of getting a black ball in first draw = 10/18 = 5/9
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw = 8/18 = 4/9
Now, Probability of getting first ball is black and second is red = 5/9 × 5/9 = 20/81
(iii) One of them is black and other is red.
Probability of getting a black ball in first draw = 10/18 = 5/9
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw = 8/18 = 4/9
Now, Probability of getting first ball is black and second is red = 5/9 × 4/9 = 20/81
Probability of getting a red ball in first draw =Â 8/18 = 4/9
As the ball is replaced after first throw,
Hence, Probability of getting a black ball in second draw = 10/18 = 5/9
Now, Probability of getting first ball is red and second is black = 5/9 × 4/9 = 20/81
Therefore, Probability of getting one of them is black and other is red:
= Probability of getting first ball is black and second is red + Probability of getting first ball is red and second is black
= 20/81 + 20 /81 = 40/81
14. Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that
(i) The problem is solved
(ii) Exactly one of them solves the problem.
Solution:
Given,
P (A) = Probability of solving the problem by A = 1/2
P (B) = Probability of solving the problem by B = 1/3
Because A and B both are independent.
⇒ P (A ∩ B) = P (A). P (B)
⇒ P (A ∩ B) = ½ × 1/3 = 1/6
P (A’) = 1 – P (A) = 1 – 1/2 = 1/2
P (B’) = 1 – P (B) = 1 – 1/3 = 2/3
(i) The problem is solved
The problem is solved, i.e. it is either solved by A or it is solved by B.
= P (A ∪ B)
As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ P (A ∪ B) = ½ + 1/3 – 1/6 = 4/6
⇒ P (A ∪ B) = 2/3
(ii) Exactly one of them solves the problem
That is either problem is solved by A but not by B or vice versa
That is P (A).P (B’) + P (A’).P (B)
= ½ (2/3) + ½ (1/3)
= 1/3 + 1/6 = 3/6
⇒ P (A).P (B’) + P (A’).P (B) = ½
15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’ F: ‘the card drawn is an ace’
(ii) E: ‘the card drawn is black’ F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen’ F: ‘the card drawn is a queen or jack’.
Solution:
Given: A deck of 52 cards.
(i) In a deck of 52 cards, 13 cards are spade and 4 cards are ace and only one card is there which is spade and ace both.
Hence, P (E) = the card drawn is a spade =Â 13/52 = 1/4
P (F) = the card drawn is an ace =Â 4/52 = 1/13
P (E ∩ F) = the card drawn is a spade and ace both = 1/52….. (1)
And P (E). P (F)
= ¼ × 1/13 = 1/52…. (2)
From (1) and (2)
⇒ P (E ∩ F) = P (E). P (F)
Hence, E and F are independent events.
(ii) In a deck of 52 cards, 26 cards are black and 4 cards are king and only two card are there which are black and king both.
Hence, P (E) = the card drawn is of black = 26/52 = ½
P (F) = the card drawn is a king =Â 4/52 = 1/13
P (E ∩ F) = the card drawn is a black and king both = 2/52 = 1/26…. (1)
And P (E). P (F)
= ½ × 1/13 = 1/26…. (2)
From (1) and (2)
⇒ P (E ∩ F) = P (E). P (F)
Hence, E and F are independent events.
(iii) In a deck of 52 cards, 4 cards are queen, 4 cards are king and 4 cards are jack.
Hence, P (E) = the card drawn is either king or queen = 8/52 = 2/13
P (F) = the card drawn is either queen or jack = 8/52 = 2/13
There are 4 cards which are either king or queen and either queen or jack.
P (E ∩ F) = the card drawn is either king or queen and either queen or jack = 4/52 = 1/13 … (1)
And P (E). P (F)
= 2/13 × 2/13 = 4/169…. (2)
From (1) and (2)
⇒ P (E ∩ F) ≠P (E). P (F)
Hence, E and F are not independent events.
16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper
Solution:
Given:
Let H and E denote the number of students who read Hindi and English newspaper respectively.
Hence, P (H) = Probability of students who read Hindi newspaper=Â 60/100 = 3/5
P (E) = Probability of students who read English newspaper =Â 40/100 = 2/5
P (H ∩ E) = Probability of students who read Hindi and English both newspaper = 20/100 = 1/5
(a) Find the probability that she reads neither Hindi nor English newspapers.
P (neither H nor E)
P (neither H nor E) = P (H’ ∩ E’)
As, { H’ ∩ E’ = (H ∪ E)’}
⇒ P (neither A nor B) = P ((H ∪ E)’)
= 1 – P (H ∪ E)
= 1- [P (H) + P (E) – P (H ∩ E)]
17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
A. 0
B. 1/3
C. 1/12
D. 1/36
Solution:
D. 1/36
Explanation:
Given A pair of dice is rolled.
Hence the number of outcomes = 36
Let P (E) be the probability to get an even prime number on each die.
As we know the only even prime number is 2.
So, E = {2, 2}
⇒ P € = 1/36
18. Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P (A′B′) = [1 – P (A)] [1 – P (B)] (C) P (A) = P (B)
(D) P (A) + P (B) = 1
Solution:
(B) P (A′B′) = [1 – P (A)] [1 – P (B)]
Explanation:
P (A′B′) = [1 – P (A)] [1 – P (B)]
⇒ P (A′∩ B′) = 1 – P (A) – P (B) + P (A) P (B)
⇒ 1 – P (A ∪ B) =1 – P (A) – P (B) + P (A) P (B)
= – [P (A) + P (B) – P (A ∩ B)] = – P (A) – P (B) + P (A) P (B)
= – P (A) – P (B) + P (A ∩ B) = – P (A) – P (B) + P (A) P (B)
⇒ P (A ∩ B) = P (A). P (B)
Hence, it shows A and B are Independent events. |
# How do you integrate int x^3sinx by parts?
Jan 6, 2018
$I = - {x}^{3} \cos x + 3 {x}^{2} \sin x + 6 x \cos x - 6 \sin x + \text{C}$
#### Explanation:
The formula for Integration by Parts (IBP):$\int$ $u$ $\mathrm{dv} = u v - \int$ $v$ $\mathrm{du}$
Let color(red)(u_1=x^3;dv_1=sinx
Thus, color(red)(du_1=3x^2dx;v_1=-cosx
$I = \textcolor{red}{\left[{x}^{3}\right] \left[- \cos x\right]} - \int \left[- \cos x\right] \left[3 {x}^{2} \mathrm{dx}\right]$
$I = \textcolor{red}{- {x}^{3} \cos x} + \int 3 {x}^{2} \cos x \mathrm{dx}$
Apply IBP again:
Let color(blue)(u_2=3x^2;dv_2=cosx
Thus, color(blue)(du_2=6xdx;v_2=sinx
$I = \textcolor{red}{- {x}^{3} \cos x} + \left\{\textcolor{b l u e}{\left[3 {x}^{2}\right] \left[\sin x\right]} - \int \left[\sin x\right] \left[6 x \mathrm{dx}\right]\right\}$
$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \int 6 x \sin x \mathrm{dx}$
Apply IBP once more:
Let color(green)(u_3=6x;dv_3=sinx
Thus, color(green)(du_3=6dx;v_3=-cosx
$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \left\{\textcolor{g r e e n}{\left[6 x\right] \left[- \cos x\right]} - \int \left[- \cos x\right] \left[6 \mathrm{dx}\right]\right\}$
$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \left\{\textcolor{g r e e n}{- 6 x \cos x} + 6 \int \cos x \mathrm{dx}\right\}$
Since intcosxdx=color(purple)(sinx+"C"
$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \left\{\textcolor{g r e e n}{- 6 x \cos x} + \textcolor{p u r p \le}{6 \left[\sin x\right]}\right\} + \text{C}$
$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} \textcolor{g r e e n}{+ 6 x \cos x} \textcolor{p u r p \le}{- 6 \sin x + \text{C}}$ |
Image credit: Desmos
Before us begin, let’s review what an equilateral triangle is — a triangle with three equal side lengths and three equal internal angles of 60° each. Now, let’s acquire one thing straight: The area of an it is intended triangle is not the perimeter that an equilateral triangle. It"s the total an are of the triangle’s surface.
You are watching: Function for the area of an equilateral triangle
As you know, over there are numerous different species of triangles: best triangles, scalene triangles, and also isosceles triangles. Again, in an it is intended triangle, the length of the political parties of an it is intended triangle room equal.
To recognize the area that an it is intended triangle, girlfriend must know its next lengths. So, prior to diving into the equilateral triangle area formula, let"s look at at just how to discover the side lengths.
## How to find the side Lengths of an it is provided Triangle
When you recognize the elevation of the triangle, you can determine the side lengths. When you"ve discovered the next length, you deserve to then identify the area of an it is intended triangle.
When you divide an it is intended triangle into two right triangles, you view the elevation of an it is intended triangle. This is excellent by slicing an equilateral triangle in half from the pointer of a vertex come the midpoint the one side to type an edge bisector.
Image credit: Desmos
The perpendicular bisector, the right line that creates two 90° angles, to represent the elevation of the it is provided triangle, as significant by height h. By producing this bisector, we’ve divided this equilateral right into two ideal triangles. To find the height, you can use the Pythagorean theorem:
Since all sides the an it is intended triangle are the same, side A=side C. And also since the base of the best triangle is half the side length of the it is provided triangle, next A=side C/2. Currently let’s plugin the height, base, and side size of C because that the hypotenuse to isolation the worth of h:
If girlfriend only know the height of an it is intended triangle’s perpendicular bisector, you deserve to use this formula to identify the length of each equal side:
Let’s use this formula to a triangle in i m sorry h = 9 to uncover the next lengths:
Now that us know exactly how to use the height of an equilateral triangle to identify its absent side length, let"s learn just how to resolve for the area.
See more: How To Change Oil For Ez Go Golf Carts, How Much Oil Does A Ezgo Golf Cart Take
## Formula for the Area of it is provided Triangles
Image credit: Desmos
The figure above is an it is intended triangle. The following formula is offered to recognize the area of the triangle:
Let"s use this formula to identify the area that the triangle above:
## Mastering the Area of it is provided Triangles
Equilaterals are triangles with three equal sides and also angles the all measure 60°. As soon as you create a perpendicular bisector line through the peak of an equilateral, you type two ideal triangles. You have the right to use the Pythagorean theorem and also height of the ideal triangles in ~ the equilateral to identify the absent side lengths that an it is provided triangle.
Then, you have the right to use the formula A = √3/4 (a²) to identify the area the an it is intended triangle. Knowing how to uncover the height and area that a triangle v equal sides provides it therefore much simpler to discover other trigonometry formulas. |
Simple Probability
The ratio of the number of outcomes favourable for the event to the total number of possible outcomes is termed as probability. In other words, a measure of the likelihood of an event (or measure of chance) is called probability. The basic terms involved in probability are listed below:
Experiment is one of several possible outcomes that are obtained from any process.
Sample space is the possible outcomes of the experiment.
Events are the subsets of the sample space.
Example 1:
Consider an experiment of playing cards.
The shapes in the deck of cards are listed below:
Obtain the sample space for a deck of cards.
Each shape in the deck of cards has numbers from 2 to 10, Ace (A), Jack (J), Queen (Q), and King (K). Therefore, the number of cards in each type is 13. Thus, the total number of cards is 52(=13×4).
Sample events are as follows:
N : Drawn card is a 5.
S : Drawn card is a spade.
Probability line:
The value of probability lies between 0 and 1.
Formula:
Where,
P(E) : Probability of incident of event
n(E) : Number of outcomes favourable for the event
n(S) : Total number of possible outcomes
Example 2:
Consider an experiment of rolling a die.
If a single die is rolled, then the probability of getting a number less than 4 is obtained below:
The sample space for rolling one die is, S= {1,2,3,4,5,6}.
Thus, n(S )= 6.
Thus, the probability of getting a number less than 4 is,
See more Prealgebra topics
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# Prime Factorization
Understand concepts involved in topic prime factorization
Last updated : 30 April 2024, Tuesday
Verified for accuracy
## Introduction
Prime factorization is a fundamental concept in mathematics that involves breaking down a number into its prime components. It serves as the foundation for various mathematical and computational algorithms, making it an essential skill for students and enthusiasts. In this article, we will delve into the world of prime factorization, exploring its significance, methods, and real-world applications. Prime factorization involves expressing a given integer as a product of its prime factors. This process not only provides a unique representation of the number but also allows for a deeper understanding of its mathematical properties.
## Factors and Multiples
A factor is a term used in mathematics to describe a number or expression that can be evenly divided into another number or expression without leaving a remainder.
A multiple is a number that can be evenly divided by another number. In other words, when one number is a multiple of another, it means that the first number can be obtained by multiplying the second number by an integer (a whole number).
## Methods of prime factorisation
1. Trial Division Method
• The trial division method involves dividing the number by progressively larger prime numbers.
• Start with the smallest prime number (2) and continue to divide until the number is no longer divisible by the current prime.
• Record each prime factor and continue with the quotient until it becomes 1.
For example, let’s find the prime factors of 36 using trial division:
• Continue with 2: 18 ÷ 2 = 9
• Continue with 3: 9 ÷ 3 = 3
• Continue with 3: 3 ÷ 3 = 1
The prime factors of 36 are 2 and 3
1. Prime Factorisation Tree
• Another method involves constructing a factorization tree, which visually represents the prime factors of a number.
• Start by dividing the number by its smallest prime factor and write down the quotient.
• Continue this process for each quotient until you reach 1, and then collect all the prime factors from the tree.
## Applications
Prime factorization has numerous practical applications in mathematics, computer science, and cryptography. Here are a few areas where it plays a crucial role:
1. Cryptography: Prime factorization is at the heart of encryption techniques like RSA (Rivest-Shamir-Adleman). The security of these systems relies on the difficulty of factoring the product of two large prime numbers.
2. Number Theory: Prime factorization is an essential tool for solving various problems in number theory, such as finding the greatest common divisor (GCD) and the least common multiple (LCM) of numbers.
3. Algorithms: Prime factorization algorithms are used in computer science for tasks like simplifying fractions, reducing fractions to their lowest terms, and solving diophantine equations.
4. Computational Mathematics: In computational mathematics, prime factorization is used in tasks like integer factorization, which has applications in solving complex mathematical problems and simulating physical systems.
## Quiz
Prime Factorization
Questions 1-10 have single answer correct
Questions 11-20 have one or more answers correct
1 / 20
What is the prime factorization of 196?
2 / 20
Which of the following is the prime factorization of 72?
3 / 20
What is the prime factorization of 60?
4 / 20
Which of the following is a prime factor of 91?
5 / 20
What is the prime factorization of 81?
6 / 20
The prime factorization of 210 is:
7 / 20
Find the prime factorization of 1024.
8 / 20
Which of the following numbers is a prime factor of 45?
9 / 20
What is the prime factorization of 125?
10 / 20
If a number is divisible by 6 and 15, which prime number must be a factor of it?
11 / 20
Choose all correct prime factorizations of 72.
12 / 20
Select all prime factors of 48
13 / 20
Which of these sets of numbers are the prime factors of 100?
14 / 20
Identify all prime factors of 56.
15 / 20
Which of the following numbers are prime factors of 210?
16 / 20
What is the prime factorization of 60?
17 / 20
Select all prime factors of 126
18 / 20
Which of the following are prime factors of 84?
19 / 20
Identify the prime factors of 45.
20 / 20
Choose all the prime factors of 132. |
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# How to Find a Sum of Fractions
In this post, we are going to learn how to find a sum of fractions.
Before you begin to add fractions, it is recommended that you know how to calculate the least common multiple (LCM) of two or more numbers.
To calculate a sum of fractions, the important thing is that the fractions have the same denominator.
###### Sum of fractions with the same denominator:
To add fractions with the same denominator you have to add the numerators and leave the same denominator.
For example:
Since the 2 fractions have the same denominator, what we have to do is keep the same denominator, which is 4, and add the numerators.
3 + 2 = 5
And the result of the sum of fractions is:
###### Sum of fractions with different denominators:
To add fractions with different denominators, the first thing that you have to do is find a common denominator: this is the least common multiple of the denominators that you have. Then we multiply each numerator by the number that we have multiplied the denominator by. Finally, we add the numerators that we have obtained and keep the same denominator.
For example,
The first thing to do is find a common denominator between 3 and 5. To do this, we calculate the least common multiple between both numbers.
LCM(3,5) = 15
So 15 is the common denominator of the two fractions.
Now we have to multiply each numerator by the number that we have multiplied the denominator by. To do this, we divide the LCM by the initial denominator and multiply the result by the numerator of that fraction.
For the first fraction:
15 / 3 = 5
5 x 2 = 10
So 10 is the numerator of the first fraction.
For the second fraction:
15 / 5 = 3
3 x 4 =12
So 12 is the numerator of the second fraction.
Now, all we have left to do is add the numerators:
10 + 12 = 22
And the result of the sum of fractions is:
I hope that you have learned with this post how to to find a sum of fractions.
Do not hesitate to leave your comments!
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## 6.1. Integrating a Function that Blows Up
When we integrate a function that blows up to infinity at some point in the interval we're integrating, the result may be either finite or infinite.
#### Example 75
◊ Integrate the function $$y = 1/\sqrt{x}$$ from x=0 to x=1.
◊ The function blows up to infinity at one end of the region of integration, but let's just try evaluating it, and see what happens.
\begin{align} \int_0^1 x^{-1/2}dx &= \left.2x^{1/2}\right|_0^1 \\ &= 2 \end{align}
The result turns out to be finite. Intuitively, the reason for this is that the spike at x=0 is very skinny, and gets skinny fast as we go higher and higher up.
Figure A. The integral $$\int_0^1 dx/\sqrt{x}$$ is finite.
#### Example 76
◊ Integrate the function y = 1/x2 from x=0 to x=1.
\begin{align} \int_0^1 x^{-2}dx &= \left.-x^{-1}\right|_0^1 \\ &= -1+\frac{1}{0} \end{align}
Division by zero is undefined, so the result is undefined.
Another way of putting it, using the hyperreal number system, is that if we were to integrate from ε to 1, where ε was an infinitesimal number, then the result would be -1+1/ε, which is infinite. The smaller we make ε, the bigger the infinite result we get out.
Intuitively, the reason that this integral comes out infinite is that the spike at x=0 is fat, and doesn't get skinny fast enough.
Figure B. The integral $$\int_0^1 dx/x^2$$ is infinite.
These two examples were examples of improper integrals. |
# Number and codes in digital systems
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1 Number and codes in digital systems Decimal Numbers You are familiar with the decimal number system because you use them everyday. But their weighted structure is not understood. In the decimal number system each of the ten digits, 0 through 9, represents a certain quantity (or weight ). These are base TEN numbers. Consider x x x x 1 5x10 + 6x10 + 7x10 + 8x10 and for x10 + 8x1 + 2 x x x10 + 8x10 + 2x10 + 5x10 Problem : Express the following as the sum of values of each digit = = = Binary Numbers The binary number system is another way of counting and it is simpler than the decimal system, since it has only two digits( 0 and 1). These are base TWO numbers. Consider 1011 (binary) = and for (binary) = x8 + 0x4 + 1x2 + 1x1 1x2 + 0x2 + 1x2 + 1x2 B431 Principles of Digital Systems : Hassan Parchizadeh Page 1
2 1x2 + 0x1 + 0x x x x2 + 0x2 + 0x2 + 0x2 + 1x2 Problem : Express the following as the sum of values of each digit = = = Decimal To Binary Conversion We can convert a decimal number into a binary equivalent by dividing the decimal number successively by 2 and noting the remainders. The equvalent number is then found by writing these remainders in the REVERSE order. Example : Convert 37 decimal into binary. 37 / 2 = 18 remainder 1 LSB 18 / 2 = 9 remainder 0! 9 / 2 = 4 remainder 1!! 4 / 2 = 2 remainder 0!!! 2 / 2 = 1 remainder 0!!!! 1 / 2 = 0 remainder 1!!!!! MSB!!!!!! Binary equivalent of 37 Example : Convert decimal into binary x 2 = Carry 0 MSB x 2 = 1.25 Carry 1! 0.25 x 2 = 0.50 Carry 0!! 0.5 x 2 = 1.00 Carry 1!!! LSB!!!! equivalent of Problem : Convert the following decimals to binary form. : i. 23 ii. 49 iii) B431 Principles of Digital Systems : Hassan Parchizadeh Page 2
3 Binary Arithmetic All the digital computers and microprocessors work on a binary basis, processing numbers and instructions coded in 1 and 0 patterns. Binary arithmetic is very similar to normal decimal arithmetic, perhaps even easir, since we are only dealing with 0's and 1's. The four operations are shown below :- Addition Substraction Multiplication Division = = 0 0 x 0 = = = 1 C=1 0 x 1 = 0 0 / 1 = = = 1 1 x 0 = = 0 C=1 1-1 = 0 1 x 1 = 1 1 / 1 = 1 C = Carry / Borrow Example : Perform binary i. Addition ii. Subtraction iii. Multiplication and iv. Division on the 110 and 10 binary numbers. 11 i) ii) iii) 110 x iv) Signed Numbers Digital Systems, such as the computer, must be able to handle both positive and negative numbers. A signed binary number consists of both sign and magnitude information. There are three ways in which signed numbers can be represented. Sign-magnitude numbers : = - 10 In the sign-magnitude system, the negative and positive numbers have the same magnitude, but the sign bit(msb) is 1 for negative numbers. 1'S Complements of Binary Numbers = - 10 B431 Principles of Digital Systems : Hassan Parchizadeh Page 3
4 positive numbers are represented in the same way as sign-magnitude numbers, but the negative numbers are 1's complement of the positive number.. The 1'S complement is found by simply changing all 1s to 0s and all 0s to 1s. 2'S Complements of Binary Numbers 's complement = -10 2's complement The 2'S complement is found by adding 1 to 1'S complement. In computers, the 2's complement system is the most widely used for handling signed numbers. Problem :Determine the 2'S complement for each 8-bit binary number. i ii iii Hexadecimal Numbers The hexadecimal system has a base of 16, that is composed of 16 digits (0,1,2,3,4,5,6,7,8,9,A,B,C,D,E and F) where A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15. Hexadecimal to Decimal conversion Consider 1A34 1x x16 + 3x16 + 4x Problem : Express i and ii. ABCD as the sum of values of each digit. Hexadecimal to Binary conversion Consider 1A Problem : Determine the binary numbers for i and ii. CDEF 16. B431 Principles of Digital Systems : Hassan Parchizadeh Page 4
5 Binary to Hexadecimal conversion Consider 100,1110,0110, E 6 4 4E64 16 Problem : Determine the hexadecimal numbers for the following. i ii Decimal To Hexadecimal Conversion We can convert a decimal number into a hexadecimal equivalent by dividing the decimal number succesively by 16 and noting the remainders. The equvalent number is then found by writing these remainders in the REVERSE order. Example : Convert 37 decimal into binary. 650 / 16 = 40 remainder = A 16 LSB 40 / 16 = 2 remainder 8 10 = / 16= 0 remainder 2 10 = 2 16 MSB 28A 16 hexadecimal equivalent of Problem : Determine the hexadecimal numbers for i and ii Note to Students : Also look at BCD (Binary Coded Decimal) and Gray Codes. B431 Principles of Digital Systems : Hassan Parchizadeh Page 5
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## Profit & Loss Quiz-3
Maths is an important subject in CLAT,DULLB & Other Law Exams. In any of law exam, Maths carries weightage of 20 -25 % of questions. With focused practice good marks can be fetched from this section. These questions are very important in achieving your success in CLAT, DULLB and Other Law Exams..
Q1.Raghu bought 4 dozen oranges at Rs 12 per dozen and 2 dozen oranges at Rs 16 per dozen. He sold them all to earn % profit. At what price per dozen did he sell the oranges?
• Rs 14.40
• Rs 16
• Rs 16.80
• Rs 19.20
Solution
CP = (4 x 12 + 2 x 16) / (4 + 2) = 80/6
SP = 1.2CP = 1.2 x 80/6 = 16
Q2. A single discount equivalent to a discount series of 40% and 20% is:
• 50%
• 52%
• 55%
• 60%
Solution
SP = (1 - 0.4) (1 - 0.2)CP = 0.6 x 0.8CP = 0.48
CP = (1 - 0.52) CP
Q3. Rajesh buys an article with 25% discount on its marked price. He makes a profit of 10% by selling it at Rs 660 the marked price is:
• Rs 600
• Rs 685
• Rs 700
• None of these
Solution
SP = 660 = 1.1 CP
CP = 600
CP = (1 - 0.25)MP = 0.75MP
MP =
Q4.What price should a shopkeeper mark on an article, costing him Rs 153, to gain 20% after allowing a discount of 15%?
• Rs 224
• Rs 216
• Rs 184
• Rs 162
Solution
0.85 MP = SP = 1.2CP
MP = C P = x 153 = 216
Q5.If the SP of Rs 24 results in a discount on list price, what SP would result in a 30% discount on list price?
• Rs 27
• Rs 21
• Rs 20
Solution
We need discount on the current price
Q6. If the cost price of 12 books is same as the selling price of 16 books, the loss percentage is
• 15%
• 20%
• 25%
• 30%
CP of 12 books = SP of 16 books
= CP of 16 books + (P/2) of 16 books
CP of 4 books = Loss of 16 books
CP of 1 book = loss of 4 books
Loss = 1/4 CP = 25% CP
Alternate Method:
Let the CP of 1 book = Re 1
CP of 16 books = Rs 16
SP of 16 Books = Rs 12
Loss % = (4/16) x 10 = 25%
Q7. A man loses the selling price of 4 apples on selling 36 apples. His loss percentage is
• 12.5%
• 11.1%
• 10%
• None of these
Solution
Let SP of 1 apple = x; SP of 36 apples = 36x
Loss = 4x
CP = 40x
Therefore, Loss %age = (4/40) x 100 = 40%
Q8.A man sells two cows for Rs. 4000 each, neither gaining nor losing in the deal. If he sold one cow at a gain of 25%, then the other cow is sold at a loss of
• 16.66%
• 18.22%
• 25%
• None of these
Solution
4000 = 1.25 CP1
CP1 = 3200, Profit = 800
CP2 = 4000 + 800 = 4800
Loss = = 16.66%
Q9. The cost price of an article is 40% of the selling price. The percentage that the selling price is of cost price is
• 250%
• 240%
• 60%
• 40%
Solution
CP = 40% of SP = 0.4 SP
SP1 = 2.5 CP = 250%
Q10. Two horses were sold for Rs.12000 each, one at a loss of 20% and the other at a gain of 20%. The entire transaction resulted in
• no loss, no gain
• loss of 4%
• gain of 4%
• gain of 8%
Solution
1.2 CP = 12000
CP = 10000
P = 2000
0.8 CP = 12000
CP = 15000
L = 3000
Loss = 3000 - 2000 = 1000
Alternate Method:
Using direct formula:
Loss % =
= (20/10)2 = 4% loss
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• Slides: 28
Addition Relate addition to combining two groups and counting on and record in a number sentence using + and = signs. 13 + 24 = 37 Record addition by: model and images +
Your turn • 23 + 26 =
When the children are comfortable with this we move onto adding on a number line. 13 + 24 = 37 +10 13 23 +1 34 +1 35 +1 36 37
Your turn • 32 + 25 =
From this we move onto using partitioning both numbers. 35 + 23 30 + 20 = 50 5+ 3=8 50 + 8 = 58
Your turn 43 + 56 =
Develop methods for adding two digit and three digit numbers by partitioning second number only. 246 + 37= 246 + 30 = 276 + 7 = 282
Your turn • 145 + 54 =
Then we develop written methods for addition of two and three digit numbers using expanded methods of recording. 375 + 67 300 + 70 60 300 130 5 7 12 = 442
How can we apply this knowledge?
Problem solving T= Target I= Information N= Number sentence A=Answer
Mina and Ben play a game. Mina scores 70 points. Ben scores 42 points. How many points did they score in total? T= How many points did they score in total? I = 70 42 + N= 70 + 42 = A=
Your turn Mina and Ben play a game. Mina scores 35 points. Ben scores 63 points. How many points did they score in total? T= I= N= A=
There are 265 children at Hill School. 102 children have a packed lunch. 27 children go home for lunch. The other children have a school lunch. How many children have school lunch?
Subtraction Relate subtraction to taking away by counting back and as counting on and record in a number sentence using the – and = signs.
Subtraction not crossing through the tens boundary. 46 – 25 = - 25 = 21 In a subtraction calculation we always write the digits of the second number, we don’t draw the image!
Subtraction crossing the tens boundary. 53 – 27= - 27 = 26
Your turn Have a go, using models and images at these two calculations. 1. ) 57 - 32 = 2. ) 42 – 16 =
Once the children are comfortable with this strategy we move them onto subtraction on a number line. 49 – 14 = -1 35 -1 36 -1 37 - 10 -1 38 39 49
Your turn 56 – 38 =
Subtraction by partitioning not crossing the tens boundary. 48 – 32 = 40 – 30 = 10 8 - 2 = 6 10 + 6 = 16
Subtraction crossing the tens boundary using partitioning. 53 – 37 = 50 – 30 = 20 10 3 - 7= We can’ t do this so we take one of the above tens. 13 – 7 = 6 10 + 6 = 16
Partitioning only the second number. 78 – 43 = 78 – 40= 38 38 - 3= 35
This is the same process even when crossing through the tens barrier. 63 - 28 = 63 - 20 = 43 43 - 8= 35
Begin to record calculations in preparation for an efficient standard method- expanded decomposition. 81 - 57 70 and 11 50 and 7 20 and 4 = 24
Problem Solving Sita had 87 p. She spent 35 p. How much money did she have left? T= How much money did she have left? I = 87 p 35 p left - N= 87 p – 35 p= A=
There are 104 children at Delton School. 48 children are girls. How many are boys? T= How many are boys? I= 104 48 - N=104 - 48 A= |
Ferry Boat Problem
Two ferry boats serve the same route on a river, but travel at different speeds. They depart from opposite ends of the river at the same time, meeting at a point 720 yards from the nearest shore.
When each boat reaches the other side, it takes 10 minutes to unload and load passengers, then begins the return trip. This time, the boats meet at a point 400 yards from the other shore.
How wide is the river?
The ferry boat problem is created by well-known puzzle author Sam Loyd.
Solution
At the first meeting point, the total distance traveled by both boats is the width of the river. When the boats meet again, they’ve each traveled the full width of the river individually, plus another width of the river combined – so they’ve traveled 3 widths of the river combined at that point, thus each traveling 3 times as far as when they first met.
So at this second meeting point, one boat has traveled 3 x 720 yards, leaving it 400 yards away from shore. So the width of the river is the distance traveled by that boat at the second meeting point minus the extra distance it is from the shore: 3 x 720 – 400 = 1,760.
Alternatively, you can set up a proper system of 4 equations that gives you the same result. You may notice that the 10 min loading time is a complete red herring, it does not factor into the calculation at all.
1. Michele Baruffi
Boat A Boat B
first trip 720 width of river (wr -720)
second trip (wr-720) +400 720 + (wr-400)
3wr = 720 + (wr-720) + (wr-720) + 400 +720 + (wr-400)
How do you get 3×720 -400?
• Michele Baruffi
Boat A first trip 720 Boat B first trip=width of river (wr -720) Boat A second trip= (wr-720) +400 Boat B second trip=720 + (wr-400) So 3wr = 720 + (wr-720) + (wr-720) + 400 +720 + (wr-400) How do you get 3×720 -400?
• BrainEaser
Unfortunately the equation you set up doesn’t tell us anything – if you simplify it, you get 0 = 0.
To get 3×720 – 400:
1. When the boats first meet, they have traveled WR combined (that’s how they met) and Boat A traveled 720.
2. When the boats meet again, they have traveled 3WR combined. The key is, this means each boat traveled 3 times as far as when they met the first time. So Boat A has traveled 3×720 total.
3. Also, just by following Boat A, we know it has traveled WR + 400 by the 2nd time the boats meet. So:
WR + 400 = 3×720
WR = 3×720 – 400
2. Jay
This solution assumes that the slower boat made it to the other shore and turned around. It is possible, however, that it was too slow to do this.
3. Baibhab
If its 3WR then how each boat covers 3 times the initial distance
• BrainEaser
What do you mean? The boats travel at constant speed. When they meet the 1st time, they have traveled some initial distance individually, but they have traveled 1WR combined. When they meet the 2nd time, they have traveled 3WR combined, so if their speed is constant, they must have also traveled 3 times the initial distance, individually. |
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# A piece of wire is $\dfrac{7}{8}\,m$ long and is broken into two pieces. One piece was $\dfrac{1}{4}\,m$ long. How long is the other piece?
Last updated date: 09th Aug 2024
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Hint: The problem can be solved using simple mathematics. Since the wire is broken into two pieces and the original and length of one of its pieces is given, we can simply calculate the length of this piece from the total length to get the length of another piece. So, simply subtraction would help us solve the question.
It is written in the question that the total length of the piece of wire is $\dfrac{7}{8}\,m$
Further it is broken into two pieces. The length of one piece is given as $\dfrac{1}{4}\,m$
$\dfrac{1}{4}\,m$$+ Length of second piece =$$\dfrac{7}{8}\,m$
Length of second piece$= \dfrac{7}{8} - \dfrac{1}{4} = \dfrac{{7 - 2}}{8} = \dfrac{5}{8}$
So Length of second piece is $\dfrac{5}{8}$ meters
Suppose the length of wire was $=$$\dfrac{7}{8}\,m$ and it got broken into two equal pieces. So now the length of one of both the pieces can be obtained by dividing original length by $2$
So the length of each piece $= \dfrac{7}{{8 \times 2}} = \dfrac{7}{{16}}$ meters. |
# Find the equation of tangent to the y=F(x) at x=1, where F(x)=∫x3xdt√1+t2
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## To find the equation of the tangent to the curve y=F(x) at x=1, where F(x)=∫x3xdt√1+t2,we will follow these steps:Step 1: Calculate F(1)Substituting x=1:F(1)=∫131dt√1+t2=∫11dt√1+t2.Since the upper and lower limits of the integral are the same, we have:F(1)=0.Step 2: Find the derivative F′(x)To find the derivative F′(x), we use the Leibniz rule for differentiation under the integral sign:F′(x)=ddx(∫x3xdt√1+t2)=ddx(∫x3dt√1+t2−∫xdt√1+t2).Using the Fundamental Theorem of Calculus:F′(x)=ddx(∫x3dt√1+t2)−ddx(∫xdt√1+t2).This gives:F′(x)=ddx(x3√1+(x3)2⋅ddx(x3))−ddx(x√1+x2⋅ddx(x)).Calculating these derivatives, we have:F′(x)=3x2√1+x6−1√1+x2.Step 3: Evaluate F′(1)Now, substituting x=1:F′(1)=3(12)√1+16−1√1+12=3√2−1√2=3−1√2=2√2=√2.Step 4: Write the equation of the tangent lineThe equation of the tangent line at the point (x1,y1)=(1,0) with slope m=F′(1)=√2 is given by:y−y1=m(x−x1).Substituting the values:y−0=√2(x−1).This simplifies to:y=√2x−√2.Final Equation of the TangentThus, the equation of the tangent to the curve y=F(x) at x=1 is:y=√2x−√2.---
|
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# Interactive Mathematics
Square root of 15
Square root of 15
Irrational Numbers
By Murray Bourne, 10 Aug 2010
## Challenge
Can you construct the length √7 just by using a measuring stick, a pencil and a set-square? (A set-square is a triangular device used for producing right angles.)
Before we answer this question. First, some background on several very important ancient mathematicians and how irrational numbers came to be.
## Length of the Hypotenuse
In the 5th century B.C., mathematicians were fascinated – yet exasperated – with irrational numbers. They believed the only meaningful numbers were the natural numbers (1,2,3,…) and any ratios involving these numbers (like 5/2, 7/9, etc). So they only accepted rational numbers and any other numbers were “unmeasureable”.
Pythagoras (or someone in his metaphysical school of mathematicians) had shown the famous result that for a right angled triangle, the area of the square on the hypotenuse (in green below) equals the sum of the areas of the squares on the other 2 sides (the 2 light red squares).
You normally see Pythagoras’ Theorem written as follows, where c is the hypotenuse and a and b are the lengths of the other 2 sides:
a2 + b2 = c2
In my diagram above, we have the common 3-4-5 triangle. Each number is an integer and for the ancient Greek mathematicians, this presented no problem.
However, since they believed irrational numbers did not exist there was a problem when they extended the Pythagorean formula to other values.
The length of the hypotenuse involved a square root:
Depending on the values of a and b, we could easily get irrational values for c. How could they measure these distances if they didn’t actually exist?
## Theodorus of Cyrene
Theodorus of Cyrene was a 5th century B.C. mathematician and was born around 100 years after Pythagoras. (Cyrene is now called Shahhat, in Libya.)
He apparently proved that the square roots of 2, 3, 5, 6 and so on up to 17 were all irrational, except the perfect squares 4, 9, 16. (Unfortunately we no longer have the proofs.) He also went on to construct these supposedly non-existent distances.
He proceeded as follows.
Start with a right triangle with equal sides 1, giving a hypotenuse of √2 (which of course was a problem, because this distance didn’t officially exist):
Then, extend a line with length 1 unit (using your 1-unit measuring stick) at right angles to the first hypotenuse as follows. This gives us the length √3 after we apply Pythagoras’ Theorem to the new triangle.
Do it again, and you now get the length √4 = 2. Theodorus had discovered one hypotenuse with a rational number length.
He kept going and found that the next one to have a “rational” length was √9 = 3.
He continued on to √16 = 4, constructed one more, √17, then stopped.
And so now you know how to construct the square root of any number using a straight edge, a pencil and a set-square.
## Eudoxus
It was one hundred years later when the Greek astronomer Eudoxus (around 370 B.C.) concluded that because we can measure irrational distances (as we did above), then irrational numbers must exist. Problem solved.
## Conclusion
It’s interesting that throughout history, people have yelled “Impossible!” when some new type of number was proposed. But subsequently, mathematicians have shown that not only are many of those numbers possible, but they have proved to be very useful.
Apart from irrational numbers as we discussed above, people originally did not believe in the existence of the number zero, imaginary numbers and “infinitesimals” in calculus. But in each case, they have been accepted as true numbers and used in many real applications. |
Home | | Maths 7th Std | Median
Median
We have discussed the situations where arithmetic mean and mode are the representative values of the given data. Let us think of any other alternative representative value or measures of central tendency. For this let us consider the following situation.
Median
We have discussed the situations where arithmetic mean and mode are the representative values of the given data. Let us think of any other alternative representative value or measures of central tendency. For this let us consider the following situation.
Rajam an old student of the school wanted to provide financial support to a group of 15 students, who are selected for track events. She wanted to support them on the basis of their family income. The monthly income of those 15 families are given below.
₹ 3300, ₹ 5000, ₹ 4000, ₹ 4200, ₹ 3500, ₹ 4500, ₹ 3200, ₹ 3200, ₹ 4100, ₹ 4000, ₹ 4300, ₹ 3000, ₹ 3200, ₹ 4500, ₹ 4100.
Rajam would like to give them an amount to their family.
If we find the mean, we get
Arithmetic mean, A.M = sum of all values / 15
= [ 3300 + 5000 + 4000 + 4200 + 3500 + 4500 + 3200 + 4100 + 4000 + 4300 + 3000 + 3200 + 4500 + 4100] / 15
= 58100/15 = 3873.3
Can the amount of ₹ 3873.3 be given to all of them irrespective of their salary? Is ₹ 3873.3 is the suitable representative here? No, this is not suitable here because a student with family income ₹ 3000 and a student with family income ₹ 5000 will receive the same amount. Because the representative measure used here is not sutable for the above data, let us find the mode for this data.
Here mode is 3200 which means there are more number of students with a family income of ₹ 3200. But this does not suite our purpose.
Hence, mode is also not suitable. Is there any other representative measures that can be used here? Yes.
Let us look at another representative value which divides the data into two halves exactly. First, let us arrange the data in ascending order.
That is, ₹ 3000, ₹ 3200, ₹ 3200, ₹ 3200, ₹ 3300, ₹ 3500, ₹ 4000, ₹ 4000, ₹ 4100, ₹ 4100, ₹ 4200, ₹ 4300, ₹ 4500, ₹ 4500, ₹ 5000.
After arranging the income in ascending order, Rajam finds 8th value ( 4000) which divides the data into two halves. It helps her to decide the amount of financial support that can be given to each of the students. Note that ₹ 4000 is the middle most value.
This kind of representative value which is obtained by choosing the middle item is known as Median.
Thus in a given data, arranged in ascending or descending order, the median gives us the middle value.
Consider another example, where the data contains even number of terms 13, 14, 15, 16, 17 and 18. How to find the middle term for this example? Here the number of terms is 6, that is an even number. So we get, two middle terms namely 3rd and 4th term. Then, we take the average of the two terms (3rd and 4th term) and the value we get is the median.
That is, Median = 1/2 {3rd term + 4th term}
=1/2 {15 + 16}
= [ 15+ 16 ] /2 = 31/2 = 15.5
Here, to find median we arrange the values of the given data either in ascending or descending order, then find the average of the two middle values.
So we conclude that, to find median,
(i) arrange the data in ascending or descending order.
(ii) If the number of terms (n) is odd, then ([n + 1]/2)th term is the median
(iii) If the number of terms (n) is even, then average of ( n/2)th and ([n/2] + 1)th terms is the median.
Try these
1.Find the median of 3, 8, 7, 8, 4, 5, 6.
Solution:
Arranging in ascending order: 3, 4, 5, 6, 7, 8.8.
Here n = 7, which is odd.
Hence the median is 6.
2. Find the median of 11, 14, 10, 9, 14, 11, 12, 6, 7, 7.
Solution:
Arranging in ascending order: 6, 7, 7, 9, 10, 11, 11, 12, 14, 14
Here is n = 10, which is even.
Median
Median = 10.5
Activity
Create a group of 6 to 7 students and collect the data of weight of the students in your class. In each group find mean, median and mode. Also, compare the averages among groups. Are they same for all the groups?
Also find all the three averages for entire class. Now, compare the results with the average of each of the groups.
Example 5.10
Find the median of the following golf scores. 68, 79, 78, 65, 75, 70, 73.
Solution:
Arranging the golf scores in ascending order, we have, 65, 68, 70, 73, 75, 78, 79
Here n = 7 , which is odd.
Therefore, Median = ([n+1] / 2)th term
= ([7+1] / 2 )th term.
= (8 /2)th term
= 4th term = 73
Hence, the Median is 73.
Example 5.11
The weights of 10 students (in kg) are 35, 42, 40, 38, 25, 32, 29, 45, 20, 24 Find the median of their weight?
Solution:
Arranging the weights in ascending order, we have, 20, 24, 25, 29, 32, 35, 38, 40, 42, 45
Here, n = 10 , which is even.
Hence, Median is 33.5 kg.
Example 5.12
Create a collection of 12 observations with median 16.
Solution:
As the number of observations is even, there are two middle values.
The average of those values must be 16.
We will now find any pair of numbers whose average is 16. Say 14 and 18.
Now an example of data with median 16 can be 2, 4, 7, 9, 12, 14, 18, 24, 28, 30, 45, 62.
Note
We can find more than one answer for this question
Example 5.13
The lifetime (in days) of 11 types of LED bulbs is given in days. 365, 547, 730, 1095, 547, 912, 365, 1460, 1825, 1500, 2000. Find the median life time of the LED bulbs.
Solution:
Arranging the data in ascending order, we have, 365, 365, 547, 547, 730, 912, 1095, 1460, 1500, 1825, 2000.
The number of observations are 11, which is odd.
Median ={ [n+1] / 2} th term
= {[11+1] /2}th term
= 6th term = 912
Therefore, the median is 912.
Hence, the median lifetime of the LED bulb is 912 days.
Example 5.14
Find the Median of the following data.
12, 7, 23, 14, 19, 10, 5, 26
Solution:
Arranging the data in ascending order, we have
5, 7, 10, 12, 14, 19, 23, 26.
Here, n = 8 , which is even.
Therefore, the Median is 13.
Think
Complete the table given below and observe it to answer the following questions.
(i) Which are all the series having common mean and median?
Solution: A, B and C.
(ii) Why median is same for all the 4 series?
Solution: Since the middle value is 100.
(iii) How mean is unchanged in the series A, B and C.
Solution: The difference between the given numbers are equal.
(iv) What change is to be made in the data, so that mean and median of ‘D’ series is equal to other series?
Solution: If 99 becomes 0 or 200 becomes 101 then mean becomes 100.
Tags : Statistics | Term 3 Chapter 5 | 7th Maths , 7th Maths : Term 3 Unit 5 : Statistics
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 3 Unit 5 : Statistics : Median | Statistics | Term 3 Chapter 5 | 7th Maths |
# Video: Systems of Equations
A cellphone factory produces 𝑥 cellphones with cost 𝐶(𝑥) = 150𝑥 + 10000 and revenue 𝑅(𝑥) = 200𝑥. What is the breakeven point?
02:14
### Video Transcript
A cellphone factory produces 𝑥 cellphones with cost 𝐶 of 𝑥 equals 150𝑥 plus 10000 and revenue 𝑅 of 𝑥 equals 200𝑥. What is the break-even point?
Well, if we look at what the words mean in this question, we’ve got revenue. Well, revenue is the money taken. So, it’s the money taken by the factory. Then, we next got another word which might be unfamiliar to you, and that is breakeven or the break-even point. This is the point where the cost and the revenue are the same. So, the amount the company has paid out and the amount that the company is taking in are equal. And because of this relationship, what we can do is set our two equations equal to each other.
So, we can say that 150𝑥 plus 10000 is equal to 200𝑥. And we can do that because that’s means that if we find out what 𝑥 is, then we can work out how many phones the factory will need to sell in order to break even because 𝑥 is the number of phones. So, to solve the equation, the first thing we do is subtract 150𝑥 from both sides. When we do that, we get 10000 is equal to 50𝑥. And then, in order to work out what 𝑥 is, we divide both sides of the equation by 50. So, we get that 200 is equal to 𝑥.
And if we think about how we could work that out, well, if we have 10000 over 50, well then the first thing we can do is divide both the numerator and denominator by 10. So, we’re gonna knock off a zero. So then, we’ve got 1000 divided by five. And therefore, if we’ve got 1000 divided by five, we can think of it as fives into 10 goes twice. And then, we’ve got two zeros left, so that’s gonna be 200. So therefore, we can say that 1000 divided by five will be 200. So therefore, 10000 divided by 50 will be 200. So therefore, we can say that the break-even point is gonna be when 𝑥 is equal to 200. So, what this means is the break-even point will be when the factory sells 200 cellphones. |
# Lesson 4 Build Fractions from Unit Fractions
• Let’s build other fractions from unit fractions.
## Warm-up Number Talk: 3 and Another Factor
Find the value of each expression mentally.
## Activity 1 Introduce Secret Fractions
The goal of the game is to be the first to build 2 secret fractions with unit fractions.
1. Make two stacks: one for secret fractions and one for unit fractions. Place all cards face down.
2. Each player draws 2 secret fraction cards. These are the fractions you are trying to make with your unit fractions.
3. On your turn, you can make one of these moves:
• Pick up 1 unit fraction card.
• Trade both of your secret fractions for 2 new secret fractions from the stack.
4. When you have enough unit fractions to make one of your secret fractions, shade your gameboard to represent your secret fraction. Then, pick a new secret fraction.
5. The first player to make 2 secret fractions wins.
## Activity 2 Represent Fraction Situations
Here are four situations about playing Pilolo and four diagrams. Each diagram represents the length of a street where the game is played.
Represent each situation on a diagram. Be prepared to explain your reasoning.
1. A student walks the length of the street and hides a rock.
2. A student walks the length of the street and hides a penny.
3. A student walks the length of the street and hides a stick.
4. A student walks the length of the street and hides a penny.
5. This diagram represents the location of a hidden stick.
About what fraction of the length of the street did the student walk to hide it? Be prepared to explain how you know.
## Problem 1
Jada walks across the street at a stoplight of her way from home to school. Represent the situation on the fraction strip. Explain your reasoning. |
# Ratios and Unit Rates Lesson 6-1 & 6-2.
## Presentation on theme: "Ratios and Unit Rates Lesson 6-1 & 6-2."— Presentation transcript:
Ratios and Unit Rates Lesson 6-1 & 6-2
Ratios A ratio is a comparison between two numbers by division. It can be written in three different ways: 5 to 2 5 : 2 5 2
Equal Ratios When two ratios name the same number, they are equal. It’s like writing an equivalent fraction. 20 : 30 Equal Ratios: 80 : 120 10 : 15 2 : 3
Ratios in Simplest Form
Ratios can be written in simplest form. Divide both terms in the ratio by their GCF. Example: 12 to 8 3 to 2
Vocabulary A rate is a ratio that compares two quantities measured in different units. The unit rate is the rate for one unit of a given quantity. Unit rates have a denominator of 1.
Vocabulary A rate is a ratio that compares two quantities measured in different units. The unit rate is the rate for one unit of a given quantity. Unit rates have a denominator of 1.
Examples Rate: 150 heartbeats 2 minutes Unit Rate (Divide to get it):
150 ÷ 2 = 75 heartbeats per minute.
Find the Unit Rate Amy can read 88 pages in 4 hours. What is the unit rate? (How many pages can she read per hour?) 88 pages 22 pages / hour 4 hours
Using Unit Rates You can find the missing terms of equal ratios.
Use the unit rate, and set it equal to another ratio. Solve for what is missing by dividing or multiplying.
Example Joe’s car goes 25 miles per gallon of gasoline. How far can it go on 8 gallons of gasoline? x 8 25 miles = Unit Rate 1 gallon 8 gallons x 8 25 x 8 = Joe’s car can go 200 miles on 8 gallons of gas.
Comparing Unit Prices Use division to find the unit prices of the two products in question. The unit rate that is smaller (costs less) is the better value.
Example Juice is sold in two different sizes. A 48-fluid ounce bottle costs \$ A 32-fluid ounce bottle costs \$ Which is the better buy? \$2.07 \$0.04 per fl.oz. 48 fl.oz. \$1.64 \$0.05 per fl.oz. 32 fl.oz. The 48 fl.oz. bottle is the better value.
Classwork: Check out this interactive for ratios and unit rates
Classwork: Check out this interactive for ratios and unit rates. You can click on lessons, interactives, or applications. Try it at BRAINCAMP Homework Time: 6-1 and 6-2 Handout |
# Figures on the Same Base and between the Same Parallels
• Last Updated : 24 Jun, 2022
A triangle is a three-sided polygon and a parallelogram is a four-sided polygon or simply a quadrilateral that has parallel opposite sides. We encounter these two polynomials almost everywhere in our everyday lives. For example: Let’s say a farmer has a piece of land that is in the shape of a parallelogram. He wants to divide this land into two parts for his daughters. Now when divided into two parts, it will generate two triangles. So, in this case, it becomes essential for us to know their areas. Simple formulas to calculate the areas of triangles and parallelograms can be sometimes cumbersome. So, we use some theorems and properties to simplify our calculations in such scenarios. Let’s look at these properties in detail.
### Parallelograms and Triangles
Triangles have three sides and parallelograms are the quadrilaterals that have four sides while the opposite sides are parallel to each other. The figure below shows a parallelogram and a triangle. Let’s say “b2” and “h2″ is the length of the base and height of the triangle respectively and “b1″ and “h1” are the base and height of the parallelogram.
Area of triangle =
Area of parallelogram =
### Congruent Figures and Their Areas
We know that two figures are called congruent if they have the same size and shape. So if two figures are congruent, you can superimpose them over each other and both will completely cover the other figure. That means that if any two figures are congruent, their areas must be equal. But notice that the converse of this statement is not true, if two figures have equal areas, it is not necessary that both are congruent. In the figure below, we can see two pairs of figures which are congruent and have the same areas, but the other pair of figures have the same areas, but they are not congruent.
Two Congruent figures with the same area
In the above figure, let’s compare the areas of two quadrilaterals.
Ar(PQRS) = 9 × 4 = 36
Ar(TUVW) = 62 = 36
Notice that both have the same area, but they are not congruent. This verifies that the converse of our statement that we gave earlier is not true. So, now the definition of the area can be summarized as follows,
Area of a figure is the measure of the plane that is enclosed by that figure. It has following two properties:
1. Let’s say we have two figures X and Y. If X and Y are congruent figures, then ar(X) = ar(Y).
2. If both figure combine without overlapping to make another figure T. Area of figure T will be given by, ar(T) = ar(X) + ar(Y).
### Parallelograms on the same Base and Between the Same Parallels
Our goal is to learn about the relation between the areas of two parallelograms when they have the same base and are between the same parallels. The figure below shows two parallelograms with a common base and between the same parallel lines. Let’s prove the relation between these areas with some theorems.
### Theorem: Parallelograms with the same base and between the same parallels have the same area.
Proof:
Let’s assume we have two parallelograms PQRS and RSTU as shown in the figure below. Both have the same base RS and are between same parallels. The objective is to prove that ar(PQRS) = ar(RSTU).
In the figure, RSTQ is common to both the parallelograms. Now, if we can prove that ar(PST) = ar(QRU). We can prove that areas of both parallelograms are equal.
Let’s look at the triangle PST and QRU.
∠SPT = ∠RQU (Corresponding angles)
∠PTS = ∠QUR (Corresponding angles)
Now since two angles of triangles are equal, the third angle will also be equal due to angle sum property.
∠PST = ∠QRU
Now both of these triangles are congruent
ΔPST≅ ΔQRU
Thus, ar(PST) = ar(QRU)
Now we know that,
Ar(PQRS) = ar(RSTQ) + ar(PST)
Ar(RSTU) = ar(RSTQ) + ar(QRU)
Since ar(RSTQ) is common and ar(PST) = ar(QRU).
Thus, ar(PQRS) = ar(RSTU)
### Triangles on the same Base and Between the Same Parallels
The figure below represents two triangles that are on the same base and are between the same parallels. Our goal is to find the relation between the areas of these two triangles. Let’s see theorems related to
### Theorem: Two triangles on the same base and between the same parallels have equal area.
Proof:
We know that area of triangle is given by
So, two triangles will have same area if they have same base and height.
Our triangles have a common base. Now since they are between two parallels, they must have the same height. Thus, both the triangles have same area.
### Sample Questions
Question 1: Find the area of the triangle and parallelogram given in the figures below.
Solution:
We know,
Area of triangle =
Area of parallelogram = b2 × h2
= 3 × 5
= 15
Question 2: State three properties of a parallelogram.
Three properties of parallelogram:
1. Opposite sides are parallel and equal.
2. Opposite angles are equal.
3. Adjacent Angles sum up to 180°.
Question 3: In the triangle ΔPQR given in the figure below, PS is the median. Prove that ar(PSR) = ar(PQS).
Now if we draw a perpendicular from the vertex P to the base QR. We’ll see that this perpendicular is common to both the triangle. Thus, both of them have same base and same height. So, they must have the same area.
Question 4: In the figure given below, we have a rectangle RSTU and a parallelogram PQRS. It is given that, PL is perpendicular to RS. Now, prove:
1. Ar(RSTU) = ar(PQRS)
2. Ar(PQRS) = RS x AL.
Solution:
1. We know that rectangles are also parallelograms. So, the theorem we studied also apply on the rectangle. Both of these figures have same base and lie between same parallels. Thus, both should have the same area.
Ar(RSTU) = ar(PQRS)
2. The area of parallelogram = base x height.
Ar(PQRS) = RU x RS
Since, PURL is also a rectangle, RU = AL. Thus,
Ar(PQRS) = RS x AL.
Question 5: A farmer has a field that is in the shape of a parallelogram PQRS. The farmer takes a point on RS and joins it to P, Q. Answer the following questions:
1. How many portions does he have now in the field?
2. He wants to sow corn and sugarcane, which portions he should use so that both are grown in the same area.
Solution:
1. Let the point farmer chose to be X. Join X to P and Q. This divides the field into three parts.
2. Now, we know that the if a triangle and a parallelogram have same base and are in between same parallels. Then, area of triangle is half of the area of the parallelogram.
So, in this case, ar(XPQ) = 1/2(ar(PQRS)
So, the remaining two portions must make the other half of the area. That means,
Ar(XPQ) = ar(XPS) + ar(XQR)
So, he should grow one crop in XPQ and other crop in both XPS and XQR.
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# 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-1 Graphs and Functions Chapter 3.
## Presentation on theme: "1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-1 Graphs and Functions Chapter 3."— Presentation transcript:
1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-1 Graphs and Functions Chapter 3
2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-2 3.1 – Graphs 3.2 – Functions 3.3 – Linear Functions: Graphs and Applications 3.4 – The Slope-Intercept Form of a Linear Equation 3.5 – The Point-Slope Form of a Linear Equation 3.6 – The Algebra of Functions 3.7 – Graphing Linear Inequalities Chapter Sections
3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-3 § 3.3 Linear Functions: Graphs and Applications
4 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-4 Graph Linear Functions Linear Function A linear function is a function of the form f(x) = ax + b The graph of any linear equation is a straight line. The domain of any linear function is all real numbers. If a ≠ 0, then the range of any linear function is all real numbers.
5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-5 Graph Linear Functions Using Intercepts Standard Form of Linear Equation The standard form of a linear equation is ax + by = c where a, b, and c are real numbers, and a and b are not both 0.
6 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-6 Graph Linear Functions Using Intercepts x- and y-intercepts The x-intercept is the point at which a graph crosses the x-axis. The x- intercept will always be of the form (x, 0). The y-intercept is the point at which a graph crosses the y-axis. The y-intercept will always be of the form (0, y).
7 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-7 Graph Linear Functions Using Intercepts To Graph Linear Equations Using the x- and y- intercepts 1.Find the y-intercept. Set x equal to 0 and find the corresponding value for y. 2.Find the x-intercept. Set y equal to 0 and find the corresponding value for x. 3.Plot the intercepts. 4.Draw the line. Using a straightedge, draw a line through the points. Draw an arrowhead at both ends of the line.
8 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-8 Graph Using Intercepts This gives us the y-intercept (0, 2). Example: Graph 5x = 10y – 20 using the x- and y- intercepts. Solution To find the y-intercept, set x = 0 and solve for y.
9 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-9 Graph Using Intercepts This gives us the x-intercept (-4, 0). Use the intercepts (0, 2) and (-4, 0) to graph the linear function. To find the x-intercept, set y = 0 and solve for x.
10 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-10 Graph Using Intercepts Plot the points and draw the line.
11 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-11 Graph Equations of the Form x = a and y = b Equations of a Horizontal Line The graph of any equation of the form y = b will always be a horizontal line for any real number b. Equations of a Vertical Line The graph of any equation of the form x = a will always be a vertical line for any real number a. |
# Powers and Roots: Law of Exponents – II
We began looking at exponent rules in Part I. We’ll continue our discussion here. So enjoy the video and transcript while you absorb these important concepts.
# Let’s Continue Looking at Exponent Rules
Now we can expand the laws of exponents a little bit further. Back in the arithmetic module, we learned about the Distributive Law, and really, the Distributive Law is one of the big ones. It’s really one of the big mathematical ideas.
And of course the Distributive Law says that P to M–plus or minus N. What we can do is just multiply the P separately times each one of those terms, that is the Distributive Law.
Multiplication distributes over addition and subtraction. As it turns out, division also distributes over addition and subtraction. Much in the same way, exponents distribute over multiplication and division. If I have (ab) to the n, or (a/b) to the n, I can distribute the exponent to each factor. So (ab) to the n equals a to the n times b to the n, a divided by b, that fraction to the n = a to the n divided by b to the n.
## Exponent Rules: Example
So, according to the exponent rules we can distribute an exponent across multiplication or division.
Here’s a very quick numerical example. Suppose we have 18 to the 8th. Well, we know that we could write 18 as a product. We could write it as its prime factorization. Of course, the prime factorization of 18 is 2 times 3 squared.
So 18 to the 8th = 2 times 3 to the 8th. Well, we can distribute that exponent to each one of those factors. We’ll get a 2 to the 8th and then we’ll get a 3 squared to the 8th. And for 3 squared to the 8th of course we’ll use the rule for the product of for a power to a power, which means multiply the exponents and we’ll get 2 to the 8th x 3 to the 16th.
So notice it’s very easy to go from the prime factorization of the individual number to the prime factorization of one of its powers.
## Exponent Rules: Practice Problem
Image by triocean
Okay. In the numerator, all we’re gonna do is multiply out that 4. We’re just going to distribute it in each one of those terms. And for each one of those terms, we’re going to have a power to a power which means multiply the exponents. So we’re gonna wind up with x to the 8th, y to the 12th, then we have to deal with the division.
Well, x to the 8th divided by x to the 5th–we subtract. The exponent has to be x cubed, y to the 12th divided by y to the- 5th. That’s gonna be 12- -5, which is 12 + 5, which is 17. And that’s why we get x cubed y to the 17th. It’s important to beware of a very common and tempting trap, because it’s close to what is true.
## Look Out for Traps
Image by solar22,/sub>
Now we’re gonna talk about a trap. First of all it’s legal to distribute multiplication over addition and subtraction. That’s 100% legal. It’s legal to distribute exponents over multiplication and division. That’s 100% legal.
But it’s illegal to distribute an exponent over addition and subtraction. So that line, that’s just the distributive law. That’s 100% legal. That’s one of the fundamental patterns in mathematics. This is, it’s also a version of the distributive law. We’re distributing the exponent over multiplication and division.
That’s also 100% legal. The thing that is illegal, is distributing the exponent over addition or subtraction. That is always illegal. In fact, M plus or minus N to the p means that we’re taking that what’s in the parentheses M plus or minus N and multiplying it by itself p times. So these were variables, we have to foil out several times.
So you’ll never actually have to do that. But it’s just important to keep in mind that that’s what it would be, not multiplying, not raising the individual terms of these powers. And I will say this is a very tricky one because even when you understand that this third line is illegal, the human brain’s inborn pattern-matching software is tempted to make that mistake again–especially when you are under pressure.
You really need to know this cold. Then even when you walk into the test and you’re stressed in the middle of this test, you don’t accidentally fall into making this mistake again–because it is a very tempting trap.
## Distributive Law with Powers
Again let’s look at all of this with numbers. Here is just the ordinary distributive law with numbers. Multiplication distributing over addition.
Here is the distributive law with powers. So that exponent distributing over multiplication and division. But it would be illegal if we had 8 plus or minus 5 to the 3rd. That would not be 8 to the 3rd plus or minus 5 to the 3rd. And one way to see this is to just think, let’s just take the subtraction case. If we look at (8- 5) to the 3rd, well what is that?
Of course that is 3 to the 3rd, which is 27. Whereas if we looked at something different, 8 cubed minus 5 cubed, well 8 cubed as we’ve mentioned in other videos is 512. 5 cubed is 125, and we subtract them, we get 387. And those two are not equal.
## Distributing, or Factoring Out P
In other words, we get two different numerical answers. And that’s why we can’t set those things equal. We can do some legal math with sums or differences of power. First we need to go back to that most impressive pattern in the distributive law. Now this is very tricky, when we read this equation from left to right, we say that we are distributing P.
When we read this equation from right to left, we say that we are factoring out P. So distributing and factoring out are two sides of the same coin. It’s just a matter whether we were reading this equation going from left to right or from right to left, but it’s the same fundamental pattern. It’s also important to remember that any higher power of a base. Is divisible by any lower power of that same base.
Thus, in the sum of a higher power and a lower power of the same base, the greatest common factor of the two terms is the lower power, and this can be factored out because a lower power is always a factor of a higher power. So for example, 17 to the 30th + 17 to the 20th. Well, first of all we know that 17 to the 30th has to be divisible by 17 to the 20th.
We know one is a factor of the other. And so 17 to the 20th is the greatest common factor of these two terms. So I’m gonna factor that out, 17 to the 30th, I can write that as 17 to 20th times 17 to the 10 by the multiplication of powers law, I can write it that way. And of course 17 to the 20th, I can write that as 17 to the 20th times 1.
I factor out 17 to the 20th and I get 17 to the 20th times parentheses 17 to the 10th plus 1. And that is a factored out form of those powers. Obviously the powers here are too large to simplify any of these resultant terms, but if the two powers in the sum are closer sometimes such simplification is easy. So I will say, pause the video and see if you can simplify this.
Okay. 3 to the 32nd- 3 to the 28th. 3 to the 28th is a factor of 3 to the 32nd. In fact 3 to the 28th is the greatest common factor of these two terms. So we’re gonna express both of them as products involving 3 to the 28th.
So 3 to the 32, we can write that as 3 to the 28th times 3 to the 4th. And of course 3 to the 28th we can write that as 3 to the 28th times 1. Factor out 3 to the 28th we get 3 to the 1/4- 1. Now 3 to the 4th, that’s something we can calculate. 3 to the fourth, so one way to think about that is if you have memorized 3 to the 4th is 81.
Also 3 to the 4th is 3 squared squared. Well, 3 squared is 9, and 9 squared is 81. So that simplifies to 81. I do 81- 1, that’s 80. And so this is 80 times 3 to the 28th. We don’t often have to solve for something in the exponent, because this usually involves much more advanced ideas on exponent rules than are found on the test.
The test will expect us to know that if bases are the same we have b to the s equals b to the t, then it must mean that the exponents are equal. That will be very important in the lesson equations with exponents, which we’ll get to much later in the module.
## Summary
In summary, exponents distribute over multiplication and division, and those are the patterns.
Exponents do not distribute over addition or subtraction. Those are very tempting mistake patterns. And we can simplify the sum or difference of powers by factoring out the lower power. And finally, if we have bases are equal and we have a to the m = a to the n, we can equate the exponents. |
# Compound Interest as Repeated Simple Interest
We will learn how to calculate compound interest as repeated simple interest.
If the compound interest of any particular year is $z; then the compound interest for the next year on the same sum and at the same rate =$ z + Interest for one year on $z. Thus the compound interest on a principal P for two years = (Simple interest SI on the principal for 1 year) + (simple interest SI' on the new principal (P + SI), that is, the amount at the end of first year, for one year) In the same way, if the amount at compound interest in a particular year is$ z; then the amount for the next year, on the same sum and the same rate = $x + Interest of$ z for one year.
Thus, the compound interest on a principal P for three years = (Simple interest SI on the principal for 1 year) + (simple interest SI' on the new principal (P + SI), that is, the amount at the end of first year, for one year) + (simple interest SI'' on the new principal (P + SI + SI'), that is, the amount at the end of second years, for one year)
This method of calculating compound interest is known as the method of repeated simple interest computation with a growing principal.
In case of simple interest the principal remains the same for the whole period but in case of compound interest the principal changes every year.
Clearly, the compound interest on a principal P for 1 year =simple interest on a principal for 1 year, when the interest is calculated yearly.
The compound interest on a principal for 2 years > the simple interest on the same principal for 2 years.
Remember, if the principal = P, amount at the end of the period = A and compound interest = CI, CI = A - P
Solved examples on Compound Interest as Repeated Simple Interest:
1. Find the compound interest on $14000 at the rate of interest 5% per annum. Solution: Interest for the first year = $$\frac{14000 × 5 × 1}{100}$$ =$700
Amount at the end of first year = $14000 +$700
= $14700 Principal for the second year =$14700
Interest for the second year = $$\frac{14700 × 5 × 1}{100}$$
= $735 Amount at the end of second year =$14700 + $735 =$15435
Therefore, compound interest = A – P
= final amount – original principal
= $15435 -$14000
= $1435 2. Find the compound interest on$30000 for 3 years at the rate of interest 4% per annum.
Solution:
Interest for the first year = $$\frac{30000 × 4 × 1}{100}$$
= $1200 Amount at the end of first year =$30000 + $1200 =$31200
Principal for the second year = $31200 Interest for the second year = $$\frac{31200 × 4 × 1}{100}$$ =$1248
Amount at the end of second year = $31200 +$1248
= $32448 Principal for the third year =$32448
Interest for the third year = $$\frac{32448 × 4 × 1}{100}$$
= $1297.92 Amount at the end of third year =$32448 + $1297.92 =$33745.92
Therefore, compound interest = A – P
= final amount – original principal
= $33745.92 -$30000
= $3745.92 3. Calculate the amount and compound interest on$10000 for 3 years at 9% p.a.
Solution:
Interest for the first year = $$\frac{10000 × 9 × 1}{100}$$
= $900 Amount at the end of first year =$10000 + $900 =$10900
Principal for the second year = $10900 Interest for the second year = $$\frac{10900 × 9 × 1}{100}$$ =$981
Amount at the end of second year = $10900 +$981
= $11881 Principal for the third year =$11881
Interest for the third year = $$\frac{11881 × 9 × 1}{100}$$
= $1069.29 Amount at the end of third year =$11881 + $1069.29 =$12950.29
Therefore, the required amount = $12950.29 Therefore, compound interest = A – P = final amount – original principal =$12950.29 - $10000 =$2950.29
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# How do you slove these simultaneous equations? Please help.1. 2y-x=-4 and x+3y=-5 2. 5y-2x=4 and 6x+y=4
malkaam | Student
1. 2y-x=-4 and x+3y=-5
Using ELIMINATION METHOD:
2y-x=-4
-x+2y=-4 -(1)
x+3y=-5 -(2)
Hence, by using the addition method we join both equations:
-x+2y+x+3y=-4-5
-x+x+2y+3y=-9
5y=-9
y=-9/5
Putting the value of y in equation (1):
-x+2(-9/5)=-4
-x-18/5=-4
-5x-18 /5 = -4
-5x-18=-4*5
-5x-18=-20
-5x=-20+18
-5x=-2
x=-2/-5
x=2/5
2. 5y-2x=4 and 6x+y=4
Using Substitution Method:
Considering equation (2):
6x+y=4
y=4-6x
Putting value of y in equation (1):
5(4-6x) -2x = 4
20-30x-2x=4
20-32x=4
-32x=4-20
-32x=-16
x=-16/-32
x=16/32
x=1/2
Now, since y=4-6x,
Therefore, putting derived value of x into equation of y:
y=4-6(0.5)
y=4-3
y=1
Wiggin42 | Student
For the first problem use the method of elimination. By adding the two equations together, you will cancel out the x term. Then its simple to solve for y as you normally would. Then plug this y value back into one of the original equations and solve for x.
A similar approach can be used for the second problem, but you will have to manipulate one of the equations before you add or subtract them from each other to ensure that a term cancels. For instance, you could multiply the first equation by 3 which will cancel out the x term.
krishna-agrawala | Student
1.
The two simultaneous equations are
2y - x = -4 ... (1)
x + 3y = -5 ... (2)
Adding equation (1) to equation (2) we get
x - x + 3y + 2y = - 5 - 4
Simplifying this we get:
5y = -9
Therefor y = -9/5 = -2.25
Substituting this value of y in equation (1) we get:
2*(-2.25) - x = -4
Simplifying this equation we get:
x = 0.5
Answer: x = 0.5 and y = -2.25
2.
The two simultaneous equations are:
5y - 2x = 4 ... (1)
6x + y = 4 ... (2)
Multiplying equation (1) by 3 we get:
-6x + 15y = 12 ... (3)
Adding equation (2) and (30 we get:
6x - 6x + y + 15y = 4 + 12
Simplifying this equation we get:
16y = 16
Therefore:
y = 1
Substituting this value of y in equation (1) we get:
5*1 - 2x = 4
Simplifying this equation we get:
-2x = -1
Therefor x = 1/2 = 0.5
Answer: x = 0.5 and y = 1
neela | Student
1)
2y-x=4..........(1)
x+3y =-5........(2)
If we add the equations, x's get cancelled and the result is:
(1)+(2) : 2y-x+x+3y = 4-5 = -1.Or
5y = -1 Or y = -1/5 = -0.2
From (1), 2(-1/5) -x = 4. Or
-x = 4+2/5 Or
x = -(4+2/5) = -4.4
2)
5y-2x=4 ..........(i)and
6x+y=4............(ii)
From the equation (ii), y = 4-6x, we replace y's in equation (i). So from (i), 5(4-6x)-2x=4. Or
20-30x-2x =4.Or
20-4 = 32x . Or x = 16/32 = 1/2 = 0.5.
From the (ii), 6(0.5)+y =4. Or y = 4-3 = 1 |
# Using The Socratic Method...
Using the Socratic Method
_________________________________________________________________________________________________
The Socratic Method (also known as Socratic teaching or questioning) is probably the most ancient of all teaching methodologies. In its original form, the master tutor questioned the student until the student actually became confused about what he knew regarding a particular subject. Then slowly the master tutor asked open-ended questions that not only reintroduced subject knowledge to the student, but also raised the student's level of thinking to a higher level of synthesis and creativity.
This is a dialogue I had with a GED Mathematics student using The Socratic Method...
Triangles
1 What are triangles?
A triangle is a plane (flat figure) with three sides.
2 Could you elaborate further? Could you give me an example?
Has three angles that the sum equal to 180 degrees, example is a triangle with angles of 90 degrees, 30 degrees and 60 degrees. Each of the three points is called a vertax (sp.).
3 How can we determine if that is true? How can we verify your statements?
I don't know.
You could reference the book...
4 Could you be more specific? Could you provide more details?
I don't know.
The angles of a triangle are easy to understand if you start with a square.
5 What are the different types of triangles?
Scalene Triangle/Equilateral Triangle/Isosceles Triangle/Right Triangle
6 What are some of the complexities of this question? What factors need to be considered?
One complexity is that the third side of an isosceles triangle maybe longer or shorter than either of the two equal sides. You have to consider how many equal sides, how many equal angles, if there is a 90 degree angle.
7 What is a vertex?
Each of the three points where the sides of a triangle meet is called a vertex.
8 Compare and Contrast similarity and congruence.
In Geometry, similar figures has the same shape but different sizes. Congruent has the same shape and same size.
9 Define corresponding.
Corresponding refers to matching sides of similar figures, corresponding sides can be written as a proportion.
10 How can we tell if triangles are congruent?
Triangles are congruent if they meet any one of the three conditions; Angle Side Angle,Side Angle Side, Side Side Side.
\$35p/h
Amy P. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts - Honors Go to the latest version.
# 2.2: Equations with Variables on Both Sides
Difficulty Level: Advanced Created by: CK-12
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Practice Equations with Variables on Both Sides
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Thomas has $50 and Jack has$100. Thomas is saving $10 per week for his new bike. Jack is saving$5 a week for his new bike. Can you represent this situation with an equation? How long will it be before the two boys have the same amount of money?
### Guidance
The methods used for solving equations with variables on both sides of the equation are the same as the methods used to solve equations with variables on one side of the equation. What differs is that there is the added step of combining like terms with the variables before isolating the variable to find the solution.
Combining like terms means that you are putting all of the variables that match on the same side of the equation. A like term is one in which the degrees match and the variables match. So, for example, 3x\begin{align*}3x\end{align*} and 4x\begin{align*}4x\end{align*} are like terms, 3x\begin{align*}3x\end{align*} and 4z\begin{align*}4z\end{align*} are not. Three apples and four apples are like terms, three apples and four oranges are not.
Remember that your goal for solving any equation is to get the variables on one side and the constants on the other side. You do this by adding and subtracting terms from both sides of the equals sign. Then you isolate the variables by multiplying or dividing. You must remember in these problems, as with any equation, whatever operation (addition, subtraction, multiplication, or division) you do to one side of the equals sign, you must do to the other side. This is a big rule to remember in order for equations to remain equal or to remain in balance.
#### Example A
x+4=2x6\begin{align*}x+4=2x-6\end{align*}
We will solve this problem using the balance method.
You could first try to get the variables all on one side of the equation. You do this by subtracting x\begin{align*}x\end{align*} from both sides of the equation.
Next, isolate the x\begin{align*}x\end{align*} variable by adding 6 to both sides.
Therefore x=10\begin{align*}x = 10\end{align*}.
Check:x+4(10)+41414=2x6=2(10)6=206=14
#### Example B
143y=4y\begin{align*}14-3y=4y\end{align*}
We will solve this problem using algebra tiles.
We first have to combine our variables (x)\begin{align*}(x)\end{align*} tiles onto the same side of the equation. We do this by adding 3x\begin{align*}3 x\end{align*} tiles to both sides of the equals sign. In this way the 3y\begin{align*}-3y\end{align*} will be eliminated from the left hand side of the equation.
By isolating the variable (y)\begin{align*}(y)\end{align*} we are left with these algebra tiles.
Rearranging we will get the following. {Note: Remember that rearranging is not necessary, it simply allows you to quickly see what the value for the variable is.}
Check:143y143(2)1468=4y=4(2)=8=8
Therefore y=2\begin{align*}y = 2\end{align*}.
#### Example C
We can use these same methods for any of the equations involving variables. Sometimes, however, numbers are so large that one method is more valuable than the other. Let’s look at the following problem.
53a99=42a\begin{align*}53a-99=42a\end{align*}
To solve this problem, we would need to have a large number of algebra tiles! It might be more efficient to use the balance method to solve this problem.
Check:53a9953(9)47799378=42a=42(9)=378=378
Therefore, a=9\begin{align*}a = 9\end{align*}.
#### Concept Problem Revisited
Thomas has $50 and Jack has$100. Thomas is saving $10 per week for his new bike. Jack is saving$5 a week for his new bike.
If we let w\begin{align*}w\end{align*} be the number of weeks, we can write the following equation.
We can solve the equation now by first combining like terms.
We can now solve for \begin{align*}x\end{align*} to find the number of weeks until the boys have the same amount of money.
Therefore, in 10 weeks Jack and Thomas will each have the same amount of money.
### Vocabulary
Degree
The degree is the exponent on the variable in a term. For example, in the term \begin{align*}4x\end{align*}, the exponent is 1 so the degree is 1.
Like Terms
Like terms refer to terms in which the degrees match and the variables match. For example \begin{align*}3x\end{align*} and \begin{align*}4x\end{align*} are like terms.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is sometimes referred to as the literal coefficient.
### Guided Practice
1. Use algebra tiles to solve for the variable in the problem \begin{align*}6x+4=5x-5\end{align*}.
2. Use the balance (seesaw) method to solve for the variable in the problem \begin{align*}7r-4=3+8r\end{align*}.
3. Determine the most efficient method to solve for the variable in the problem \begin{align*}10b-22=29-7b\end{align*}. Explain your choice of method for solving this problem.
Answers:
1. \begin{align*}6x+4=5x-5\end{align*}
Therefore \begin{align*}x = -9\end{align*}.
2. \begin{align*}7r-4=3+8r\end{align*}
You can begin by combining the \begin{align*}r\end{align*} terms. Subtract \begin{align*}8r\end{align*} from both sides of the equation.
You next have to isolate the variable. To do this, add 4 to both sides of the equation.
But there is still a negative sign with the \begin{align*}r\end{align*} term. You now have to divide both sides by -1 to finally isolate the variable.
Therefore \begin{align*}r = -7\end{align*}.
3. \begin{align*}10b-22=29-7b\end{align*}
You could choose either method but there are larger numbers in this equation. With larger numbers, the use of algebra tiles is not an efficient manipulative. You should solve the problem using the balance (or seesaw) method. Work through the steps to see if you can follow them.
Therefore \begin{align*}b = 17\end{align*}.
### Practice
Use the balance method to find the solution for the variable in each of the following problems.
1. \begin{align*}5p+3=-3p-5\end{align*}
2. \begin{align*}6b-13=2b+3\end{align*}
3. \begin{align*}2x-5=x+6\end{align*}
4. \begin{align*}3x-2x=-4x+4\end{align*}
5. \begin{align*}4t-5t+9=5t-9\end{align*}
Use algebra tiles to find the solution for the variable in each of the following problems.
1. \begin{align*}6-2d=15-d\end{align*}
2. \begin{align*}8-s=s-6\end{align*}
3. \begin{align*}5x+5=2x-7\end{align*}
4. \begin{align*}3x-2x=-4x+4\end{align*}
5. \begin{align*}8+t=2t+2\end{align*}
Use the methods that you have learned for solving equations with variables on both sides to solve for the variables in each of the following problems. Remember to choose an efficient method to solve for the variable.
1. \begin{align*}4p-7=21-3p\end{align*}
2. \begin{align*}75-6x=4x-15\end{align*}
3. \begin{align*}3t+7=15-t\end{align*}
4. \begin{align*}5+h=11-2x\end{align*}
5. \begin{align*}9-2e=3-e\end{align*}
For each of the following models, write a problem to represent the model and then find the variable for the problem.
### Vocabulary Language: English
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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### Application Of Differential Equation
Learn how to solve logistic differential equation & orthogonal trajectories differential equations. Practice to find the orthogonal trajectory of the family of curve & population of a certain area is modeled by the logistic differential equation.
# Application to Electric Circuit (Differential Equation)
Inductor-Resistor Circuit :
Consider a simple electric circuit as shown.
Let E be the battery which causes the current $$I(t)$$ to flow in the circuit. The current is the function of time.
• By Kirchhoff's rule
$$E-\displaystyle\underbrace {RI}_{\text{potential drop across resistor}}-\underbrace {L\;dI/dt}_{\text{potential drop across inductor}}=0$$
$$\Rightarrow \,L\dfrac{dI}{dt}+RI=E$$
$$\Rightarrow\,\dfrac{dI}{dt}+\dfrac{RI}{L}=\dfrac{E}{L}$$
which is a linear differential equation in which independent variable is $$t$$(time) and dependent variable is $$I$$ (current).
• Solving this will give the current in the circuit as a function of time.
• First we find the integrating factor and then write the solution.
• For equation of the form $$\to \dfrac{dy}{dx}+y\,P(x)=Q(x)$$
$$I.F.=e^{\int P(x)dx}$$
and solution is $$y×I.F.=\int (Q×I.F.)dx$$
#### Consider the figure as shown. The value of resistance $$R$$ is $$10\Omega$$, and the inductance $$L$$ is $$5H$$. If battery given a constant voltage of $$70V$$, switch is closed when $$t=0,\,I(0)=0$$ i.e. no current in the beginning find the current in the circuit after 1 sec. $$[e^{-2}=0.1353]$$
A 15 A
B 6.053 A
C 10 A
D 2.31 A
×
By Kirchhoff law the differential equation
$$\dfrac{dI}{dt}+\dfrac{RI}{L}=\dfrac{E}{L}$$ will be followed.
Putting the values given in the problem
$$R=10,\,L=5,\,E=70$$ we get
$$\dfrac{dI}{dt}+\dfrac{10I}{5}=\dfrac{70}{5}$$
$$\Rightarrow\,\dfrac{dI}{dt}+2I=14$$ (which is linear differential equation)
$$I.F.=e^{\int P(t)dt}$$
$$=e^{\int 2dt}$$
$$=e^{2t}$$
$$\therefore$$ Solution is $$I(t)×e^{2t}=\int14e^{2t}dt$$
$$\Rightarrow\,I(t)e^{2t}=\dfrac{14e^{2t}}{2}+c$$
$$\Rightarrow\,I(t)×e^{2t}=7e^{2t}+c$$
$$\Rightarrow\,I(t)=7+ce^{-2t}$$
Use the initial condition $$\to$$ when $$t=0,\,I=0$$
$$\Rightarrow\,0=7+c$$
$$\Rightarrow\,c=-7$$
$$\Rightarrow \,I(t)=7-7e^{-2t}$$
$$\Rightarrow \,I(t)=7(1-e^{-2t})$$
$$\therefore\,I(1)$$= value asked = $$7(1-e^{-2})=7(1-0.1353)=7×0.865$$
$$=6.053\, Amp.$$
### Consider the figure as shown. The value of resistance $$R$$ is $$10\Omega$$, and the inductance $$L$$ is $$5H$$. If battery given a constant voltage of $$70V$$, switch is closed when $$t=0,\,I(0)=0$$ i.e. no current in the beginning find the current in the circuit after 1 sec. $$[e^{-2}=0.1353]$$
A
15 A
.
B
6.053 A
C
10 A
D
2.31 A
Option B is Correct
# Models of Population Growth
## The law of natural growth
The elementary model for growth of population is based on the assumption that population grows at rate proportional to the size of present population.
P = number of individuals in the population (dependent variable).
t = time (the independent variable)
then $$\dfrac{dP}{dt}=kP$$, where k is the position proportionality constant.
• We say this is the differential equation of population growth. It is a simple variable separable equation.
$$\dfrac{dP}{dt}=kP$$
$$\Rightarrow \dfrac{dP}{P}=k\,dt$$
Integrating both sides we get
$$\displaystyle\int\dfrac{dP}{P}=\int k\,dt$$
$$\Rightarrow\,ln\,P=kt+C$$
$$\Rightarrow\,P(t)=e^{kt+C}$$
$$\Rightarrow\,P(t)=Ce^{kt}$$
If initially i.e. at $$t=0$$, the population is P0 then
$$P_0=Ce^0\Rightarrow\,C=P_0$$
$$\Rightarrow\,P(t)=P_0e^{kt}$$
$$\therefore\,P(t)=P_0e^{kt}$$
#### The population of a certain place is known to increase at the rate proportional to the number of people presently living in that place. If after two years the population has doubled and after three years the population in 20000. Find the population today. [ $$ln2=0.6931$$ , $$e^{1.041}= 2.832$$ ]
A 15181
B 7062
C 56
D 25000
×
Apply the natural growth model
$$\dfrac{dP}{dt}=kP$$ whose solution is
$$P=P_0e^{kt}$$ ...(1)
Now when $$t=2,\,P=2P_0\to$$ put in equation ...(1)
$$\therefore\,2P_0=P_0e^{2k}$$
$$\Rightarrow\,e^{2k}=2$$
$$\Rightarrow\,2k=ln2$$
$$\Rightarrow\,k=\dfrac{1}{2}ln2=\dfrac{1}{2}(0.6931)=0.347$$
When $$t=3,\,P=20000\to$$ put in equation ...(1)
$$20000=P_0e^{3×0.347}$$
$$\Rightarrow\,P_0=\dfrac{20000}{e^{3×0.347}}$$
$$\Rightarrow\,P_0=\dfrac{20000}{e^{0.041}}$$
$$\Rightarrow\,P_0=\dfrac{20000}{2.832}=7062$$
### The population of a certain place is known to increase at the rate proportional to the number of people presently living in that place. If after two years the population has doubled and after three years the population in 20000. Find the population today. [ $$ln2=0.6931$$ , $$e^{1.041}= 2.832$$ ]
A
15181
.
B
7062
C
56
D
25000
Option B is Correct
#### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population increasing ?
A $$P<15000$$
B $$P<6000$$
C $$P>20000$$
D $$P>2$$
×
population is increasing when $$\dfrac{dP}{dt}>0$$
In this case
$$\dfrac{dP}{dt}=1.5P\,\left(1–\dfrac{P}{6000}\right)$$
$$\therefore\,\dfrac{dP}{dt}>0$$
$$\Rightarrow\,1.5P\left(1–\dfrac{P}{6000}\right)>0$$
$$\Rightarrow\,1-\dfrac{P}{6000}>0$$
$$\Rightarrow\,P<6000$$
### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population increasing ?
A
$$P<15000$$
.
B
$$P<6000$$
C
$$P>20000$$
D
$$P>2$$
Option B is Correct
#### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population decreasing ?
A $$P<500$$
B $$P>6000$$
C $$P>5000$$
D $$P<20000$$
×
Population is decreasing when $$\dfrac{dP}{dt}<0$$
In this case
$$\dfrac{dP}{dt}=1.5P\,\left(1–\dfrac{P}{6000}\right)$$
$$\therefore\,\dfrac{dP}{dt}<0$$
$$\Rightarrow\,1.5P\left(1–\dfrac{P}{6000}\right)<0$$
$$\Rightarrow\,1-\dfrac{P}{6000}<0$$
$$\Rightarrow\,P>6000$$
### The population of a certain area is modeled by the differential equation $$\dfrac{dP}{dt}=1.5P\,\left(1-\dfrac{P}{6000}\right)$$ For when values of $$P$$ is the population decreasing ?
A
$$P<500$$
.
B
$$P>6000$$
C
$$P>5000$$
D
$$P<20000$$
Option B is Correct
# Models of Population Growth (Logistics Model)
• Logistic model
In the natural growth model
$$\dfrac{dP}{dt}=kP$$ when P is the population at any time t.
• Initially the population increases exponentially (according to natural growth model) but it levels off eventually because of limited resources. So $$\dfrac{dP}{dt}\cong kP$$ if P is small.
• As P increases and approaches M which is the carrying capacity, (the maximum population the environment is capable of sustaining) the rate of change of population does not follow the natural growth model.
• $$\therefore$$ we have logistics model
$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$
• If P > M then $$\dfrac{dP}{dt}<0$$ and the population will decrease.
• As long as P < M and $$\dfrac{dP}{dt}>0$$,the population will increase.
• Consider the logistics equation
$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$
$$\Rightarrow\,\dfrac{dP}{dt}=\dfrac{kP(M-P)}{M}$$
$$\dfrac{M}{P(M-P)}dP=k\,dt$$ (separate the variable)
Integrate both sides
$$\displaystyle\int \dfrac{MdP}{P(M-P)}=kt+C$$
$$\displaystyle\,\Rightarrow\,\int\left(\dfrac{1}{P}+\dfrac{1}{M-P}\right)dP=kt+C$$ (partial fraction method)
$$\Rightarrow\,lnP+ln|M-P|×-1=kt+C$$
$$\Rightarrow\,ln\left|\dfrac{P}{M-P}\right|=kt+C$$
$$\Rightarrow\,ln\left|\dfrac{M-P}{P}\right|=-kt-C$$
$$\Rightarrow\,\left|\dfrac{M-P}{P}\right|=e^{-kt-C}$$
$$\Rightarrow\,\left|\dfrac{M-P}{P}\right|=Ae^{-kt}(e^{-C}=A)$$
$$\Rightarrow\,\dfrac{M}{P}-1=Ae^{-kt}$$
$$\Rightarrow\,\dfrac{M}{P}=1+Ae^{-kt}$$
$$\Rightarrow \,P=\dfrac{M}{1+Ae^{-kt}}$$
Now when $$t=0$$ (Initially), $$P=P_0$$ So.
$$P_0=\dfrac{M}{1+A}$$
$$\Rightarrow\,A=\dfrac{M-P_0}{P_0}$$
$$\therefore$$ solution to logistic model is
$$P(t)=\dfrac{M}{1+Ae^{-kt}}$$ when $$A=\dfrac{M-P_0}{P_0}$$
Graph showing various values of $${P_0}$$ and population eventually reaching M.
#### Suppose population at a certain place develops according to logistic model. $$\dfrac{dP}{dt}=0.07P-0.0035P^2$$ where t is measured in weeks. What is the carrying capacity M
A $$M=2$$
B $$M=20$$
C $$M=500$$
D $$M=36$$
×
In logistic model, the standard form of differential equation is
$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$ ...(1)
Put the given equation in this form.
$$\therefore\,\dfrac{dP}{dt}=0.07P-0.0035P^2$$
$$\Rightarrow\,\dfrac{dP}{dt}=0.07P(1-0.05P)$$
$$\Rightarrow\,\dfrac{dP}{dt}=0.07P\left(1-\dfrac{P}{20}\right)$$
$$\therefore$$ comparing with (1)
$$k=0.07,\,M=20$$
### Suppose population at a certain place develops according to logistic model. $$\dfrac{dP}{dt}=0.07P-0.0035P^2$$ where t is measured in weeks. What is the carrying capacity M
A
$$M=2$$
.
B
$$M=20$$
C
$$M=500$$
D
$$M=36$$
Option B is Correct
#### Find the population $$P(50)$$ for the logistics model differential equation. $$\dfrac{dP}{dt}=0.05P(1-\dfrac{P}{10000})$$, $$P(0)=1000$$ where $$t$$ is in weeks i.e. population after 50 weeks. [ $$e^{-2.5}=0.082$$ ]
A 1250
B 5750
C 15120
D 7805
×
The solution to the logistics model
$$\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)$$
is given by $$P(t)=\dfrac{M}{1+Ae^{-kt}}$$ where $$A=\dfrac{M-P_0}{P_0}$$
In this case $$M=10000,\,P_0=1000$$
$$\therefore\,A=\dfrac{10000–1000}{1000}=9$$ and $$k=0.05$$
$$\therefore$$ $$P(t)=\dfrac{10000}{1+9e^{-kt}}$$
put $$k=0.05$$
$$\Rightarrow \,P(t)=\dfrac{10000}{1+9e^{-0.05t}}$$
$$\therefore\,P(50)=\dfrac{10000}{1+9e^{-0.05×50}}$$
$$=\dfrac{10000}{1+9e^{-2.5}}$$
$$\because\,e^{-2.5}=0.082$$
$$=\dfrac{10000}{1.739}=5750$$
### Find the population $$P(50)$$ for the logistics model differential equation. $$\dfrac{dP}{dt}=0.05P(1-\dfrac{P}{10000})$$, $$P(0)=1000$$ where $$t$$ is in weeks i.e. population after 50 weeks. [ $$e^{-2.5}=0.082$$ ]
A
1250
.
B
5750
C
15120
D
7805
Option B is Correct
# Orthogonal Trajectories
The orthogonal trajectory of family of curves is a curve that intersects each curve of the family orthogonally that is at right angles.
e.g $$y=2x$$ line is an orthogonal trajectory to the family of curves $$x^2+y^2=r^2$$
• In general, a family of curves is called orthogonal trajectories of another family of curves if each member of one family is orthogonal to every member of the other.
• Each line of the family $$y=mx$$ is orthogonal to each member of the circle family $$x^2+y^2=r^2$$.
i.e. the two families of curves are orthogonal trajectories of each other.
• To find orthogonal trajectories of a family of curves
1. First, find the differential equation of the given family of curves by eliminating 'c'.
2. In this differential equation replace $$\dfrac{dy}{dx}$$ by $$\dfrac{–1}{\dfrac{dy}{dx}}$$.
3. Solve the new differential equation obtained, this will be the orthogonal trajectory.
#### Find the orthogonal trajectory of the family of curve $$y^2=4(x–a)$$ where 'a' is a parameter.
A $$y^2=4Cx+7$$
B $$y=Cx^2$$
C $$y=C\,sinx+cosx$$
D $$y=Ce^{-x/2}$$
×
Find the differential equation of family of curves which is given $$y^2=4(x-a)$$
$$\Rightarrow\,2yy'=4$$
$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{2}{y}$$
Replace $$\dfrac{dy}{dx}$$ by $$\dfrac{\dfrac{–1}{dy}}{dx}$$
$$\Rightarrow\,\dfrac{\dfrac{–1}{dy}}{dx}=\dfrac{2}{y}$$
$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{-y}{2}$$ (differential equation required family)
Now solve this equation
$$\dfrac{dy}{dx}=\dfrac{-y}{2}$$
$$\Rightarrow\,\dfrac{dy}{y}=\dfrac{–1}{2}dx$$ (variable separable)
$$\Rightarrow\,\ln y=\dfrac{–1}{2}x+C$$
$$\Rightarrow\,y=e^{-x/2+C}$$
$$\Rightarrow\,y=e^{\dfrac{-1}{2}{x}}×e^C$$
$$y=Ce^{-x/2}$$
### Find the orthogonal trajectory of the family of curve $$y^2=4(x–a)$$ where 'a' is a parameter.
A
$$y^2=4Cx+7$$
.
B
$$y=Cx^2$$
C
$$y=C\,sinx+cosx$$
D
$$y=Ce^{-x/2}$$
Option D is Correct
# Mixing Problem
A mixing problem involves a tank of fixed capacity (volume) filled with mixed solution of some substance such as salt.
Now a solution of given concentration enters the tank at a fixed rate and the mixture thoroughly stirred leaves at a fixed rate, which is usually different from the entering rate.
If $$y(t)$$ is the amount of substances in the tank at time 't' then $$y'(t)$$ is the rate at which substance is being added minus the rate at which it is removal.
$$y'(t)$$ = (rate of supply) – (rate of removal)
• This will lead to a differential equation of variable separable category and using initial condition. We can find a particular solution.
#### A tank contains 50 kg of salt dissolved in $$10000 \,\ell$$ of water. Brine which contains 0.05 kg of salt per liter of water enters the tank at the rate of $$25\,l/min.$$ The solution is thoroughly mixed and drain from the tank at some rate. Find the amount of salt in the tank after 20 min. [ $$e^{-1/20}=0.9512$$ ]
A 54.83 kg
B 71.96 kg
C 2 kg
D 78 kg
×
$$y'(t)$$ = rate of supply – rate of removal
Rate of supply of salt = $$\dfrac{0.05kg}{l}×\dfrac{25\,l}{min}$$
$$=1.25\,kg/min$$
Rate of removal of salt = $$\underbrace { \dfrac{y(t)}{10000}}_{\text{concentration at time }'t'}×\dfrac{25\,l}{min}$$
$$\dfrac{dy}{dt}=\dfrac{y(t)}{400}kg/min$$
$$\therefore\,y'(t)=\dfrac{dy}{dt}$$
$$\Rightarrow\,\dfrac{dy}{dt}=1.25–\dfrac{y(t)}{400}$$
$$\Rightarrow\,\dfrac{dy}{dt}=\dfrac{500–y(t)}{400}$$
which is variable separable.
$$\therefore\,\dfrac{dy}{dt}=\dfrac{500–y}{400}$$
$$\Rightarrow\,\dfrac{1}{500–y}dy=\dfrac{1}{400}dt$$
Integrate both side
$$\displaystyle\int\dfrac{1}{500–y}\,dy=\int\dfrac{1}{400}dt$$
$$\Rightarrow\,–ln|500–y|=\dfrac{t}{400}+c$$
Now $$y(0)=50$$
$$\Rightarrow\,–ln(500–50)=\dfrac{t}{400}+c$$
$$\Rightarrow\,c=-\ln450$$
$$\therefore\,–ln|500–y|=\dfrac{t}{400}–ln(450)$$
$$\Rightarrow\,ln\left[\dfrac{500–y}{450}\right]=\dfrac{–t}{400}$$
$$|500–y|=450e^{–t/400}$$ (both sides one position)
$$\Rightarrow\,y=500–450e^{-t/400}$$
$$\Rightarrow\,y(t)=500–450e^{–t/400}$$
$$\therefore\,$$ The amount of salt after 20 min.
$$=y(20)=500–450×e^{–20/400}$$
$$=500–450×e^{–1/20}$$
$$=500–450×0.9512$$
$$=71.96\,kg$$
### A tank contains 50 kg of salt dissolved in $$10000 \,\ell$$ of water. Brine which contains 0.05 kg of salt per liter of water enters the tank at the rate of $$25\,l/min.$$ The solution is thoroughly mixed and drain from the tank at some rate. Find the amount of salt in the tank after 20 min. [ $$e^{-1/20}=0.9512$$ ]
A
54.83 kg
.
B
71.96 kg
C
2 kg
D
78 kg
Option B is Correct |
# Difference between revisions of "Ceva's Theorem"
Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.
## Statement
Let $ABC$ be a triangle, and let $D, E, F$ be points on lines $BC, CA, AB$, respectively. Lines $AD, BE, CF$ concur iff (if and only if)
$\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1$,
where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of $1$ is $1$.
(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
The proof using Routh's Theorem is extremely trivial, so we will not include it.
## Proof
We will use the notation $[ABC]$ to denote the area of a triangle with vertices $A,B,C$.
First, suppose $AD, BE, CF$ meet at a point $X$. We note that triangles $ABD, ADC$ have the same altitude to line $BC$, but bases $BD$ and $DC$. It follows that $\frac {BD}{DC} = \frac{[ABD]}{[ADC]}$. The same is true for triangles $XBD, XDC$, so
$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}$.
Similarly, $\frac{CE}{EA} = \frac{[BCX]}{[BXA]}$ and $\frac{AF}{FB} = \frac{[CAX]}{[CXB]}$, so
$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1$.
Now, suppose $D, E,F$ satisfy Ceva's criterion, and suppose $AD, BE$ intersect at $X$. Suppose the line $CX$ intersects line $AB$ at $F'$. We have proven that $F'$ must satisfy Ceva's criterion. This means that
$\frac{AF'}{F'B} = \frac{AF}{FB}$,
so
$F' = F$,
and line $CF$ concurrs with $AD$ and $BE$.
## Proof by Barycentric Coordinates
Since $D\in BC$, we can write its coordinates as $(0,d,1-d)$. The equation of line $AD$ is then $z=\frac{1-d}{d}y$.
Similarly, since $E=(1-e,0,e)$, and $F=(f,1-f,0)$, we can see that the equations of $BE$ and $CF$ respectively are $x=\frac{1-e}{e}z$ and $y=\frac{1-f}{f}x$
Multiplying the three together yields the solution to the equation:
$xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y$
Since the coordinates are homogeneous, we can resize $x,y,$and$z$ to have $xyz=1$. This yields:
$1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}$, which is equivalent to Ceva's theorem
QED
## Trigonometric Form
The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians $AD,BE,CF$ concur if and only if
$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.$
### Proof
First, suppose $AD, BE, CF$ concur at a point $X$. We note that
$\frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC}$,
and similarly,
$\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$.
It follows that
$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$
$\qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1$.
Here, sign is irrelevant, as we may interpret the sines of directed angles mod $\pi$ to be either positive or negative.
The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem.
## Problems
### Introductory
• Suppose $AB, AC$, and $BC$ have lengths $13, 14$, and $15$, respectively. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$, find $BD$ and $DC$. (Source) |
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### Using Matrices to Solve Systems of Linear Equations
Note: The ideas in this lesson can be rather difficult to follow with just words. The video will help explain this a lot, as it has a lot of visual diagrams to show what's going on step-by-step.
• We can represent an entire linear system with an an augmented matrix:
x
+ y
+ z
=
3
2x
−2y
+3z
=
−4
−x
−z
=
0
⇒
1
1
1
3
2
−2
3
−4
−1
0
−1
0
• Each row represents an equation of the system,
• Each left-side column gives a variable's coefficients,
• If a variable does not appear, it has coefficient 0,
• The vertical line represents `=',
• The right-side column gives the constant terms.
• There are three row operations that we can perform on an augmented matrix:
1. Interchange the locations of two rows.
2. Multiply (or divide) a row by a nonzero number.
3. Add (or subtract) a multiple of one row to another.
While row operations are a simple idea, working with them involves a lot of arithmetic and steps. With that much calculation going on, it's easy to make mistakes. Counter this by noting each step you take and what you did. This will make it easier to avoid making mistakes and will help you find any that manage to creep through.
• We can use row operations to put a matrix in reduced row-echelon form. While it has a formal definition, it will be enough for us to think of it as a diagonal of 1's starting at the top-left with 0's above and below (entries on the far right can be any number). If we can use row operations to put an augmented matrix in reduced row-echelon form, we will have solved its associated linear system.
• Gauss-Jordan elimination is a method we can follow to produce reduced row-echelon form matrices through row operations and thus solve linear systems.
1. Write the linear system as an augmented matrix.
2. Use row operations to attain a 1 in the top left and zeros below. Then move on to creating the next 1 diagonally down, with zeros below. Repeat until end.
3. Now, work from the bottom right of the diagonal, canceling out everything above the 1's. Continue up the diagonal until you have only 0's above as well.
4. The matrix should now be in reduced row-echelon form, giving you the solutions to the system. [Note: If there is no solution or infinitely many, you will not be able to achieve reduced row-echelon form.]
• We can also find the solutions to a linear system by using determinants and Cramer's Rule. Let A be the coefficient matrix for our linear system. [The coefficient matrix is the left-hand side of an augmented matrix. It is all the coefficients from the linear system arranged in a matrix.] Let Ai be the same as A except the ith column is replaced with the column of constants from the linear system (the right side of the equations). Then, if det(A) ≠ 0, the ith variable xi is
xi = det Ai
det A
.
If det(A) = 0, then the system has either no solutions or infinitely many solutions.
• Using matrix multiplication, we can write a linear system as an equation with matrices:
AX = B, where
• A is the coefficient matrix of the linear system,
• X is a single-column matrix of the variables,
• B is a single-column matrix of the constants.
Notice that if we could get X alone on one side of the equation, we will have solved for the variables.
• If A is invertible, then there exists some A−1 we can multiply by that will cancel out A above, allowing us to get X alone. By computing A−1 and A−1B, we will have solved the system. (Notice that we have to multiply A−1 from the left on both sides of the equation because matrix multiplication is affected by direction.) If A is not invertible (det(A) = 0), then the system either has no solutions or infinitely many solutions.
• While all of these methods work great for solving linear systems, they all have the downside of being tedious: they take lots and lots of arithmetic. Good news! Almost all graphing calculators have the ability to do matrix (and vector) operations. You can enter matrices, then multiply them, take determinants, find inverses, or put them in reduced-row echelon form. Even if you don't have a graphing calculator, there are many websites where you can do these things for free.
### Using Matrices to Solve Systems of Linear Equations
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
• Introduction 0:12
• Augmented Matrix 1:44
• We Can Represent the Entire Linear System With an Augmented Matrix
• Row Operations 3:22
• Interchange the Locations of Two Rows
• Multiply (or Divide) a Row by a Nonzero Number
• Add (or Subtract) a Multiple of One Row to Another
• Row Operations - Keep Notes! 5:50
• Suggested Symbols
• Gauss-Jordan Elimination - Idea 8:04
• Gauss-Jordan Elimination - Idea, cont.
• Reduced Row-Echelon Form
• Gauss-Jordan Elimination - Method 11:36
• Begin by Writing the System As An Augmented Matrix
• Gauss-Jordan Elimination - Method, cont.
• Cramer's Rule - 2 x 2 Matrices 17:08
• Cramer's Rule - n x n Matrices 19:24
• Solving with Inverse Matrices 21:10
• Solving Inverse Matrices, cont.
• The Mighty (Graphing) Calculator 26:38
• Example 1 29:56
• Example 2 33:56
• Example 3 37:00
• Example 3, cont.
• Example 4 51:28
### Transcription: Using Matrices to Solve Systems of Linear Equations
Hi--welcome back to Educator.com.0000
Today, we are finally going to see why we have been studying matrices--just how powerful they are.0002
We are going to use matrices to solve systems of linear equations.0006
Consider the following system of linear equations: x + y + z = 3; 2x - 2y + 3z = -4; and -x - z = 0.0010
Notice that, as long as we keep the variables in the same order for each equation (we can't swap it0019
to y + x + z; we keep it in xyz order every time), we could write these coefficients to all of the variables as a coefficient matrix.0023
What we have in front of this x is just a 1; in front of this y is just a 1; and in front of this z is just a 1.0031
So, we can write a first row of 1, 1, 1, all of these coefficients that are on that row right there of the equation.0037
For the next one, we have 2, -2, 3; so we put them all down here; so we have done that equation as the coefficients in that row, showing up there.0045
And since we can always trust that we are going to have the x, the y, and the z here,0055
because we are always staying in this order of x, y, z (that is why we have to keep the variables in the same order each time),0060
we can create this coefficient matrix; and finally, we have -1 here and -1 here.0068
And why do we have a 0? Well, if y doesn't show up, it must be because we have a 0y; so that is why we get a 0.0074
So, we have done all 3 equations, the coefficients to the variables in all three equations.0081
This idea of converting the information in a linear system into a matrix0085
will allow us to explore ways that we can have linear systems interact with matrices and vice versa.0089
How can a matrix allow us to solve a linear system of equations?0094
Our first idea is the augmented matrix; we can take this idea of the coefficient matrix and expand on it.0098
Instead of just representing coefficients for the equations, we can represent an entire linear system,0104
the solutions included, with an augmented matrix.0109
So, previously we didn't have the constants for the equations, what was on the right side of the equals sign.0112
So now, we have that show up on the right side over here.0118
So now, we have the coefficients, and we have what each of those equations is equal to.0121
So, each row represents an equation of the system: x + y + z = 3 is 1, 1, 1, 3, because that first 10126
represents the coefficient on x; the first one on the y; the first one on the z; and it all comes together to equal 3,0136
because we know that just addition is what is going on between all of those coefficients, because it is a linear system.0142
Each left-side column gives a variable's coefficients: all of the coefficients to x show up here.0147
We have 1, 2, and -1 on the x, and we have 1, 2, and -1 in that column, as well, on the left side.0153
Where the variable does not appear, it has a coefficient of 0.0160
The 0 is to show that we have nothing for y here, because if you have nothing, we can think of it as just 0 times y.0163
The vertical line represents equalities; since this vertical line here is representing each of these, there is an equals sign here.0171
And finally, the right-side column, this column here, gives us...0179
This column of our matrix is the constant terms from our equations.0183
So, we have a way of converting that entire set of equations into a single matrix.0187
So now, we can look at how we can play around with this matrix to have it give up what those variables are equal to.0192
Since we can represent a linear system as an augmented matrix, we can do operations on the matrix the same way we interact with a linear system.0199
Anything that would make sense to do to a linear system, to an equation in a linear system,0206
should make sense--there should be a way to do it over on the matrix version,0209
because since we can do it to a linear system, and our augmented matrix is just showing us a linear system--0213
it is a way to portray a linear system--then if we can figure out a way to interact with our matrix0220
that is the same as interacting with a linear system, we know that it is just fine.0224
So, that gives us the idea of 3 row operations.0228
The first one is to interchange the locations of two rows.0231
If I have row 1 and row 2, I can swap their places.0235
Our next idea is to multiply or divide a row by a non-zero number.0239
I can multiply an entire row by 2 or by 5 or by -10--whatever I want to--0243
or divide by 2, because that is just the same as multiplying by 1/2.0249
And then finally, add or subtract a multiple of one row to another.0253
If I have row 2 here and row 5 here, I can have row 5 subtract twice on itself; so it is row 2 - 2(row 5).0257
I can have a multiple of one row subtract from another one.0267
So, why does this make sense--how is this like a linear system?0270
Well, all of these operations are completely reasonable, if we had a linear system.0274
If we have an equation here and an equation here, it is totally meaningless for us to swap the order that the equations come in.0278
The location of the equation isn't an important thing when we are working with a linear system.0284
We just have to look at all of them, so it doesn't matter which came first and which came last.0289
So, we can move them around, and it doesn't matter.0292
That means that we can move our rows around, and it doesn't matter, because it is just representing a linear system.0294
Similarly, multiplying both sides of an equation is just algebra.0298
If I have an equation, I can multiply 2 on the left side and 2 on the right side, and that is just fine.0302
You can think of that as multiplying each of the numbers inside of the equation.0308
If we multiply each of the numbers inside of the equation, that is the same thing as just multiplying an entire row of our augmented matrix.0313
Finally, adding a multiple of an equation is elimination.0318
Remember: when we were talking about linear systems at first, elimination was when we can add a multiple of one equation to another equation.0322
Well, that is the same thing as adding a multiple of one row to another row over in the augmented matrix version.0327
So, everything here has a perfect parallel between the two ideas.0333
Each one of our row operations makes total sense, if we were just working with a linear system.0336
And since our augmented matrix is just representing a linear system, they make sense over here with the augmented matrices, as well.0341
While row operations are a simple idea (each one of these is pretty simple--swap; multiply; or add a multiple--not that crazy)--0348
working with them will involve a lot of arithmetic and steps.0356
You are going to have a lot of calculations going on, and while none of them will probably be very hard calculations,0360
you are going to be doing so many that it is easy to make mistakes.0364
So, when you are working on row operations, when you are working on doing this stuff,0368
I want you to be careful with what you are doing, and counter the fact that you are likely to make mistakes0372
by noting each step: note what you just did for each step.0376
In addition to this being a good way to keep you from making mistakes, some teachers will simply require it,0380
and won't give you credit if you don't do it; and that is pretty reasonable,0384
because you will definitely end up making mistakes sooner or later if you don't do this sort of thing.0386
So, take a note of what you did on each step; show what you did between the first one, and then the second one,0391
and then the third one, by writing on the side what you just did.0397
This will make it easier to avoid making mistakes, and it will help you find any that manage to creep through.0401
If you get to the end, and you see that this doesn't make sense--something must have gone wrong--0405
you can go up and carefully analyze each of your steps, and figure out where you made a mistake.0408
Or maybe things actually do come out to be weird for some reason.0413
So, here are some suggested symbols for each of the three row operations.0416
You don't have to use them; use whatever makes sense to you (and if your teacher cares about it, whatever makes sense to your teacher).0420
But these work well for me, and I think they make sense, pretty clearly.0424
If we are interchanging row i and row j (we are swapping their locations), I like just a little arrow, left-right, between them.0428
We show some row by using a capital R to talk about a row, and then the number of it.0435
For example, if we want to talk about the second row, we would just talk about it as R2.0440
If we wanted to talk about the ninth row, it would be R9; and so on, and so on.0444
So, if we are swapping row i and row j, we have Ri, little arrow going back and forth, Rj.0450
If we want to multiply row i by the number k (we are multiplying by k), we just have k times row i--as simple as that.0455
And finally, if we are adding a multiple of row i to row j, then we have kRi (k times row i,0463
the thing that we are adding the multiple of) to what it is being added to.0472
We will see these pretty soon, when we are actually starting to see how this stuff gets done.0477
Gauss-Jordan elimination: here is an idea: since all of our row operations make sense for solving a linear system--0482
they all make sense for a way to do it--we can apply them to find the value of each variable.0489
If we manage to get our augmented matrix in a form like this form right here, we would immediately know what each variable is; why?0494
Well, notice: this first row here has 1, 0, 0; well, 1 here would correspond to the x here.0505
And then, this one here would be non-existent, because there are two 0's there; so there would be no y; there would be no z.0511
And we know that it is equal to -17.0516
The exact same thing is going on here; we know that y, because that is the y column, must be equal to 8 (nothing else shows up),0520
and that z (since that is the z column) must be equal to 47.0528
We have managed to solve the system by just moving stuff around in this augmented matrix.0531
We know that that has to be true, because that augmented matrix must be equivalent to the linear system,0536
because we turned our linear system into an augmented matrix, and then we had all of these row operations0541
that are just the same as working with a linear system.0546
So, what we have here is still the same as our original linear system.0548
So, we can convert back to the linear system and see what our answers are.0551
We call this format for a matrix reduced row echelon form; it is a mouthful, and it is kind of confusing at first.0556
Echelon has something to do with a triangle shape; it is "row echelon" because the rows are kind of arranged in a triangle,0563
and "reduced" because they all start with 1's, and there are 0's...0569
Honestly, don't really worry about it; just know reduced row echelon form, that really long name, and you will be fine.0572
While it has a formal definition (there is a way to formally define it), it is going to be enough0577
for us to just think of it in this casual way, where it is a diagonal of 1's.0581
That main diagonal will have all 1's on it; it will start at the top left, and it will continue down diagonally.0585
And it will have 0's above and below.0591
So, we have 1's along the main diagonal, and above and below our 1's, there will be 0's.0593
So, we have a 0 here above the 1's and a 0 here below the 1's.0603
We have 0's here above the 1's and a 0 here below the 1's.0607
Finally, the numbers on the far right can be any number.0611
Over here, -5, 8...it doesn't matter; over here, we manage to have two columns,0616
because the 1 diagonal's part of the reduced row echelon form is just an identity matrix.0620
So, it can't get any farther than whatever the square portion of it would be.0628
So, we can end up having multiple columns, as well.0632
But in our case, when we are working on this elimination to solve linear systems, we will always only have a single column on the far right.0634
Anyway, the point is that we have this diagonal of 1's with 0's above and below; and the stuff on the far right can be any number.0641
Now, from what we have just discussed, if we can use row operations to put an augmented matrix in reduced row echelon form,0648
like this one right here, we will solve its associated linear system,0653
because we will know that, since there is only one of a variable in that row, it must be equal to the constant0657
on the other side of the augmented matrix--one the right side of the augmented matrix, past that vertical line that shows equality.0662
We will have solved its associated linear system.0669
Gauss-Jordan elimination is just named after the people who created it.0672
It is a method we can follow to produce reduced row echelon form matrices through row operations to solve linear systems.0676
It is just a simple method of being able to get 0's to show up and 1's to show up on the diagonal, and then we are done.0683
So, it is just a method that we can follow through that will always end up resulting in a reduced row echelon form.0689
All right, so let's see how it is done.0695
The very first thing you do for Gauss-Jordan elimination is: you take your linear system, and you write it in augmented matrix form.0697
So, we have our augmented matrix form over here.0705
We look at the coefficients; we convert them over; the line to show the equals signs is here, and then our constants are on the far right side.0707
All right, the next thing: you use row operations to attain a 1 in the top left0715
(we start in the top left here), and then 0's below.0722
We get a 1 here; and then we wanted 0's below it, because it is a 1, and 0's are above and below.0726
So, we start working with 1's, creating 1's...well, sorry, from your point of view...creating 1's and creating 0's underneath them.0731
So, we first get a 1 up here; and then we create the 0's underneath it.0740
Once we have done that, we move on to creating the next one diagonally down and doing 0's below that.0746
And we just keep repeating until we have made it all the way down the diagonal.0751
So, you just keep going until you are all the way down the diagonal, creating 1's and creating 0's below.0754
First, we started with a 1 up here; we already have the first thing done.0760
So, our next step is...how do we get this stuff to turn into 0's?0764
We do row operations: we want to get rid of this 2, so since row 1 has a 1 there, we subtract by 2(row 1).0767
2 row 1 gets us -2 here, -2 here; this becomes a 0; -2 times 1 on -2 gets us -4; 1 times -2, added to 3, gets us 1; and -2 times 3, added to -4, gets us -10.0776
Next, we are adding row 1, because we need to get rid of this -1 here.0795
So, we add row 1, because it is positive 1; add +1 to -1; we get 0; add +1 to 0; we get 1; add +1 to -1; we get 0; add +3,0800
because it is just a multiple--how many times we are adding whatever is on that first row, because it was just 1 of row 1-- 3 on 0 gets us 3.0813
All right, at this point, we have 0's below; great.0821
So now, we are ready to move on to the next step in the diagonal.0824
All right, our next step in the diagonal is going to be this -4 here.0828
We want to get that to turn into a 1.0832
We could do this by manipulating it with canceling things out, or by dividing both sides of that entire row by -4,0834
or multiplying the entire row by -1/4; but we might notice that we already have a 1 here.0844
Well, let's just use that: if we swap these two rows, we will manage to have a 1 in our next location; great--we do that instead.0850
So, we make row 2 swap with row 3, because up here, they used to be row 2 and row 3.0858
And now, they take their new spots; they swap locations; and we have our new matrix right here.0864
The next step: we see that we have the 1 here, so our next step is...we need to turn everything below the 1's to 0's on this first portion.0871
So, how do we do that? Well, we can add +4 times row 2, because we have a 1 here.0882
So, we add 4 times that, and that will cancel out the -4.0887
4 times 1, added to this, gets us 0; 0 times 4, added to 0--we still have 0.0891
0 times 4, added to this--we just have 1 still; 3 times 4, added to -10, gets us 2.0897
So, at this point, notice that we have nothing but 1's on our main diagonal, and we have 0's all below it.0903
However, we don't have 0's above it yet; there are things other than that; so that is the next step.0910
Once you have 1's all along the main diagonal and 0's all along below that main diagonal,0917
the next step is to cancel out the stuff above it.0927
Our first thing is: we work from the bottom right of the diagonal, and we cancel out above the 1's.0930
You work your way up: so our first step is to cancel out everything here and here.0940
But we notice that, because this row is 0, 1, 0, we can actually do it in one step,0944
where we can subtract one of row 3 and one of row 2 (it is getting kind of hard to see, with all of those colors there).0948
So, subtracting one of row 3: we have 1 here, minus 1; so now we subtract 2 on this, so we have 3 - 2 so far,0954
for what is going to show up here; let's also subtract by row 2; row 2...minus here...0964
1 - 1 comes out to be 0; since it is zeroes everywhere else on that row, we don't have to worry about them interfering,0971
except for over here; we have it subtracting by another 3; so 3 - 2 - 3 comes out to be equal to -2, and we get -2 here.0976
At this point, we have reduced row echelon form.0986
We have 1's on the main diagonal and 0's above and below, so we can convert this.0990
Our x here becomes -2; the representative -1 of y here becomes 3, so we have y = 3.0995
And the representative 1z becomes z = 2; so now we have solved the thing.1002
And I want you to know: if there is no solution, or infinitely many--if our linear system can't be solved,1007
or it has infinitely many solutions--this method will end up not working.1013
You will not be able to achieve reduced row echelon form if there is no solution or infinitely many solutions.1017
So, that is something to keep in mind.1023
All right, a new way to do this: we are going to look at a total of 3 different ways to solve linear systems, each using matrices.1026
Another way to do this is through Cramer's Rule.1033
We can find the solutions to a linear system by this rule.1035
Given a two-variable system (we will start with 2 x 2, until we get a good understanding of what is going on),1037
where we have the a's (each of our a's here is just a constant), a11x, a12y, a21x, a22y,1042
is equal to constants on the right side (b1 and b2 are also constants),1050
we can create a normal coefficient matrix (A is the normal coefficient matrix).1055
All of our coefficients, a11, a12, a21, a22...1063
show up just like they would normally in a coefficient matrix.1068
Now, Ax...what it is going to do is take this column here on a, and it is going to replace it with the constants to the equation.1071
It is going to replace a1, a21 with this; and so, we have b1, b2 for that column.1083
And then, the rest of A is like normal.1089
Similarly, for y, we are going to swap out the y constants from A.1092
And so, we are going to have A like normal, except we swap out the constant column for where the y variables were occupying.1098
The y column gets swapped to the constant column.1105
So, the constant column goes in there.1108
Notice that Ax is just like A, except that it has this constant column replacing the x column.1111
Similarly, Ay has the constant column replacing the y column from our normal coefficient matrix.1119
All right, that is the idea; now if the system has a single solution, if it comes out to be just one solution--1125
it isn't infinitely many; it isn't no solutions at all; if the system has a single solution,1131
then x will be equal to the determinant of Ax, over the determinant of A,1137
the determinant of its special matrix, divided by the determinant of the general coefficient matrix.1143
Similarly, y is going to be equal to the determinant of its special matrix, divided by the determinant of the general coefficient matrix.1150
OK, this method can be generalized to any linear system with n variables.1160
Let A be the coefficient matrix for this n-variable linear system.1165
Let Ai be the same as A, except the ith column; the column that represents1169
the variable we are currently working with--the variable that we want to solve for--that column1177
will be replaced with the column of constants.1182
So, we replace the column for the variable we are interested in solving for with the column of constants.1185
And that makes A for whatever variable we were looking for: Ai in this case,1191
if we are looking for the ith variable, up until we are looking for the ith one.1195
We replace it with the column of constants from the linear system, just the right side of the equations,1202
the b1, b2 on our previous 2 x 2 example.1206
OK, with that idea in mind, if the determinant of A, the determinant of our normal coefficient matrix, right up here,1210
is not equal to 0, then the ith variable, xi, this variable we are trying to solve for,1216
is equal to the determinant of its special matrix that has that column replacing it,1223
divided by the determinant of the normal coefficient matrix; that is what we have right here.1227
It is just like in the 2 x 2 form, except we can do it on a larger scale, as well.1232
You swap out this one column; you take the determinant of that special matrix; you divide it by the determinant of the normal coefficient matrix.1236
And that gives you the variable for whatever column you had swapped.1242
We will get the chance to see this done on a more confusing scale (which is the sort of thing1247
that we want to be able to understand--this on a larger scale) in Example 3.1251
And we will just see this get applied normally in Example 2 for a 2 x 2 matrix.1255
Also notice: if the determinant of A is equal to 0, then the system will have either no solutions or infinitely many solutions.1259
All right, the final method to do this: we can solve with inverse matrices.1266
This one is my personal favorite for understanding how this stuff works; I think it is the easiest to understand.1271
But that is maybe just me.1276
Using matrix multiplication, we can write a linear system as an equation with matrices.1278
How can we do this as an equation?1282
It made sense with an augmented matrix, because we talked about the special thing.1284
But how is 1, 1, 1, 2, -2, 3, -1, 0, -1--notice that that is just our normal coefficient matrix A showing up here--1287
if we multiply it by the column matrix x, y, z equals our coefficient column here, that ends up being just the same--1298
it is completely equivalent; these two ideas here are completely equivalent--multiplying the matrices1313
versus the linear system, and the linear system versus the matrices being multiplied together; they are completely the same.1319
Let's see why; let's just do some basic matrix multiplication on this.1323
What is going to come out of this? We have a 3 x 3 (3 rows by 3 columns), 3 rows by 1 column;1327
so yes, they match up, so they can multiply; that is going to produce a 3 x 1, 3 row by 1 column, matrix in the end.1335
So, let's see what is going to get made out of this.1342
We are going to have a 3 x 3; our first row times the only column is 1, 1, 1, 1; so I'll make this a little bit larger,1346
so we can see the full size of what is going to go in...1 times x + 1 times y + 1 times z is x + y + z.1356
Next, 2, -2, 3 on x, y, z gets us 2x - 2y + 3z.1369
Finally, -1, 0, -1 on x, y, z gets us -x + 0y (so let's just leave it blank) - z.1378
Now, if we know that that is equal to our coefficient matrix, because we said it from the beginning,1389
then all we are saying...3, -4, 0...well, for two matrices to be the same thing, for them to be equal to each other,1395
every entry in the two matrices has to be equal to its entry in the same location.1403
So, the top one, x + y + z, equals 3; that is just the exact same thing as this.1409
2x - 2y + 3z has to be equal to -4; well, that is the same thing as saying 2x - 2y + 3z = -4.1416
And the same thing: -x - z is saying it is equal to 0 through the matrices; and that is the same thing it was saying by the linear system.1424
So, the linear system, taken as a whole, is just the same thing as taking the coefficient matrix,1430
multiplying it by this coefficient column matrix...and that is going to come out to be equal to our constant matrix,1434
which was the constants for the equations; that is the idea that is going to really be the driving force behind using inverse matrices.1444
All right, we can symbolically write this whole thing as Ax = B; A times x equals B,1451
where A is the coefficient matrix right here; then X is a single-column matrix of the variables;1463
that is our x, y, z; whatever variables we end up using are going to go like that; and then finally,1476
B is a single-column matrix of the constants (3, -4, 0 gets the same thing right here); OK.1484
Notice: if we could somehow get X alone, if we could get our variable matrix, our variable column, alone on one side,1492
whatever it was on the other side, if it equals numbers on the other side in a matrix,1499
then we would have solved for it, because we would say that x is equal to whatever the corresponding location is on the other side;1503
y is equal to whatever its corresponding location is on the other side; z is equal to whatever its corresponding location is on the other side.1508
We would have solved for this.1514
So, if we can somehow get X alone, we will be done; we will have figured out what x, y, and z are equal to.1515
How can we do that, though--how can we get rid of A? Through inverse matrices!1521
That is no surprise, since this thing is titled Inverse Matrices.1525
We cancel out A; if A is invertible, then there exists some A-1 that we can multiply that by that will cancel out A.1527
So, we started with Ax = B; we can multiply by A-1 on the left side on both sides.1537
Remember: if you multiply by the left and the right, for matrix equations that doesn't work.1542
You have to always multiply both from the left or both from the right.1548
You are not allowed to do them on opposite sides; they have to both be coming from the same side when you multiply.1553
We multiply by A-1 on the left side on both cases; the A-1 here and the A cancel out, and we are left with just X = A-1B.1558
So, if we can compute A-1, and then we can compute what is A-1 times B, we will have solved our system.1566
We will have what our system is equal to.1573
Just make sure that you multiply from the same side for your inverse on both sides of the equation.1575
You have to multiply both from the left; otherwise it won't work out.1579
Finally, if A is not invertible (if the determinant of A is equal to 0, then you can't invert it),1582
then that means that the system has either no solutions or infinitely many solutions.1588
All right, let's try putting these things to use.1593
Oh, sorry; before we get into using them, the mighty graphing calculator:1596
all of these methods work great for solving linear systems: augmented matrices with Gauss-Jordan elimination,1601
Cramer's Rule, inverse matrices--they are all great ways to solve linear systems.1606
But they all have the downside of being really tedious; they take so much arithmetic to use.1611
We can work through it; we can see that we can do this stuff; but it is going to take us forever to actually work through this stuff by hand.1617
I have great news; it turns out that, if you have a graphing calculator, you can already do this right now, really fast, really quickly, and really easily.1623
Almost all graphing calculators have the ability to do matrix and vector operations.1631
You can enter matrices into your calculator, and then you can multiply them; you can take determinants of the matrices;1636
you can find inverses; or you can put them in reduced row echelon form.1641
Look on your graphing calculator, if you have a graphing calculator, for something that talks about where...1645
just look for a button about matrices; look for something like that.1649
And it will probably have more information about how to create a matrix, and then how to do things with it.1652
Inverse is probably just raising the matrix to the -1, and it will give out the value.1657
Each graphing calculator will end up being a little bit different for how it handles inputting the matrices.1661
But they will almost all have this ability, for sure.1665
If you have real difficulty figuring out how to do it on your calculator, just do a quick Internet search for "[name of your calculator] put in matrices."1669
Use "matrices," and you will be able to figure out an easy way to do it very quickly.1675
Someone has a guide up somewhere.1678
Also, if you don't have a graphing calculator, don't despair; it is still possible to get this stuff done really easily and really quickly.1680
There are a lot of websites out there where you can do these things for free.1687
Just try doing a quick Internet search for matrix calculator; just simply search the words matrix and calculator,1690
and the first 5 hits or so will all be matrix calculators, where you can plug in matrices,1698
and you can normally multiply them, or you can take their determinants, or you can get their inverses.1704
Or you can do other things that you don't even know you can do with matrices yet.1708
But just look for the things that you are looking for; there are lots of things you can do with it.1711
Do a quick Internet search for the words matrix and calculator, and you will be able to find all sorts of stuff for those.1714
So, even if you don't have a graphing calculator, there are lots of things out there.1721
If you are watching this video right now, you can go and find websites that will let you do this for free.1724
Finally, while this is great that we can do all of this stuff with a calculator,1729
and the calculator will do the work for us, I still want to point out that it is important to be able to do this stuff without a calculator.1732
So, it makes it so much easier to be able to use a calculator; but we still have to understand what is going on underneath the hood.1738
We don't have to constantly be using it, but we have to have some sense of what is going on under the hood1744
if we are going to be able to understand more, higher, complex-level stuff in later classes.1749
So, you want to be able to understand this stuff, just because you want to be able to understand things,1754
if you are going to be able to make sense of things that come later.1757
And also, you usually need to show your work on your tests.1759
Your teacher is not going to be very happy if you are taking a test, and you just say, "My calculator said it!"1762
You are not going to get any points for that.1767
So, you can't just get away with it all the time.1769
That said, it can be a great help for checking your work, so you can work through the thing by hand,1771
and then just do a quick check on your calculator to tell you that you got the problem right; that is really useful on tests.1775
Or if you are dealing with really huge matrices, where it is 4 x 4, 5 x 5, 6 x 6, or even larger,1780
where you can't reasonably be able to do that by hand, you just use a calculator, and that is perfectly fine.1786
All right, now let's go on to the examples.1791
The first example: Using Gauss-Jordan elimination, solve 2x + 5y = -3, 4x + 7y = 3.1793
Our very first thing to do is: we need to convert it into an augmented matrix.1799
We have 2 as the first coefficient on the x, and then 5 as the first coefficient on the y, and that equals -3.1803
So, there is our bar there; 4, 7, 3; we have converted it into an augmented matrix.1809
Our coefficients are on the left part of the matrix, and our constant terms are on the right part of the matrix.1818
All right, at this point, we just start working through it.1825
The very first thing that we need to do is to get that to turn into a 1--get the top left corner to turn into a 1.1828
So, we will do that by multiplying the first row: 1/2 times row 1.1834
All right, that is what we will do there: 2 times 1/2 becomes 1; 5 times 1/2 becomes 5/2; -3 times 1/2 becomes -3/2.1839
4, 7, 3; the bottom row didn't get touched, so it just stays there.1849
The next thing to do: we want to get this to turn into a 0.1853
So, we will subtract the top row; the top row is not going to end up doing anything on this step,1858
but we will subtract the top row 4 times, because we have 4 here; so - 4R1 + our second row.1866
-4 times 1 plus 4 gets us 0; -4 times 5/2 gets us -10; -10 + 7 gets us -3; -3/2 times -4 gets us +6; we got +6 out of that, so that gets us +9.1877
Our next step: we want to get this to turn into a 1; we will bring this whole thing up here.1900
The next step is to get the second row to turn into a 1: 1, 5/2, -3/2.1908
We multiply the bottom part by -1/3 times row 2; so 0 times -1/3 is still 0; -3 times -1/3 becomes +1; 9 times -1/3 becomes -3.1916
At this point, we can now turn this into a 0; we don't need to do anything to our bottom row; it is still 0, 1, -3.1930
But we will add -5/2 of row 2 to row 1; so 0 times anything is still going to be 0; so added there...it is still 1 there.1940
Then, -5/2 on 1 + 5/2 becomes 0; -5/2 times -3 becomes +15/2, and then still -3/2.1953
Let's simplify that: we have 1, 0, 0, 1...15/2 - 3/2 becomes 12/2; 12/2 is 6...6, -3.1963
So, at this point, we can convert that into answers; x = 6; y = -3.1975
There are our answers; however, we did have to do a whole lot of calculation to get to this point.1983
And it could be even more if we were working on a larger augmented matrix; they get big really fast.1987
So, it might be a good idea to do a quick check; let's just check our work and make sure it is correct.1992
Let's plug it into the first one: 2 times 6, plus 5 times -3; what does that come out to be?1998
We hope it will come out to be -3: 12 + -15 = -3; indeed, that is true.2004
We could check it again with this equation, as well, if we want to be really, really extra careful.2011
4 times 6, plus 7 times -3, equals positive 3; 24 - 21 = 3; that is true.2015
So, both of our checks worked out; we know that x = 6, y = -3; that is definitely a solution--great.2025
All right, the second example: let's see Cramer's Rule in action.2031
The first thing we want to do, if we are going to use Cramer's Rule, is: we need to get a coefficient matrix going.2034
A =...what are our coefficients here? We have a 2 x 2: 2, 5, 4, 7.2039
There are the coefficients; our next step is...we want Ax--what is Ax going to be?2048
Here is our x column; we are going to swap that out for the constants here.2055
-3 and 3 replaces what had been our x column here.2062
And then, the rest of it is just like normal; so we replace that one column, but everything else is just the same.2067
Ay: what will Ay be?2073
The same sort of thing, except now we are replacing the y column--what is the y column going to turn into?2075
It is going to also become -3, 3; so 2, 4 is just as it was before; the first column is still the same, because that is the x column.2081
But now we are swapping out the y column, so it becomes the constants, -3 and 3.2088
All right, so we were told that Cramer's Rule says that x is equal to the determinant of its special, swapped-out matrix, Ax,2093
divided by the determinant of the normal coefficient matrix.2103
The determinant of -5, 3, 5, 7, divided by the determinant of 2, 5, 4, 7:2107
-3 times 7 gets us -21; minus 3 times 5 (is 15), over 2 times 7 (is 14), minus 2 times 5 (is 20);2118
we have -36/-6 =...that comes out to be positive 6, so we now have x = +6.2130
That checks out with what we just did in the previous one.2140
If you didn't notice, these equations are the same as what they were in the previous example,2142
so we are just seeing two different ways to do the same problem2146
(well, at least the part where we are trying to solve for x and y).2149
So, that checks out, because we checked it in the previous problem.2153
Now, what about y? y is going to be the same thing; y is equal to the determinant...same structure, at least...2155
of Ay, its special matrix, divided by the determinant of A, once again.2162
We could calculate the determinant of A, but we already calculated the determinant of A.2168
We figured out that it comes out to be -6, so we just drop in -6 here.2173
You only have to do it once; it is not going to change--the determinant of A will stay the determinant of A, as long as A doesn't change.2177
Then, the determinant of Ay: we don't know what Ay, that special matrix, is, yet.2183
2, -3, 4, 3: 2 times 3 is 6, minus 4 times -3 (-12); that cancels out; so we have 6 + 12,2187
divided by -6; 18/-6 equals -3; so y comes out to be -3.2197
Once again, that is the same answer as we had on the previous example, so we know that this checks out.2205
If you had just done this for the very first time, I would recommend doing a check,2210
because once again, you have to do a lot of arithmetic to get to this point.2213
And it is going to be even more if you are doing a larger Cramer's Rule, like, say, this one.2216
All right, so if we are working on this one, we only have to solve for the value of y.2220
There is one slight downside to only solving for one variable.2224
It means you can't check your work, because we can't plug in y and be sure that it works out to be true.2227
But we can at least get out what it should be.2231
All right, using Cramer's Rule, the first thing we need to do is figure out what our coefficient matrix A is.2234
A =...all of our first variables is w, so 2w, 3w...there is no w there at all, so it must be 0w; -2w.2240
Next, our x's: there are no x's in the first equation, so it is 0 x's; -2, -3, +5.2250
Next, 4y, 1y, 2y, 0y, -1z, 4z, 0z, 1z.2258
Great; now, we are looking to figure out what y is going to be.2270
We figure out that Ay is going to be the same thing, except it is going to have its y column swapped.2274
Notice that the y column is the third column in; so we are going to swap out the third column for the constants.2281
Other than that, it is going to look like our normal coefficient one.2287
So, we can copy over what we had in the previous one, except for that one column: 2, 3, 0, -2, 0, -2, -3, 5.2289
Now, this is the third column--this one right here--so we are swapping out for the column of constants.2301
That is 5, 16, 0, 17; and then back to copying the rest of it: -1, 4, 0, 1.2309
So, at this point, we have Ay; we have A; so we are going to need to figure out determinants.2317
First, let's figure out what the determinant of A is.2322
The determinant of A: if we want to figure out this here, remember: if we are going to be figuring out determinants,2324
we are going to be using cofactor expansion; so the very first thing we want to do is make a little +/- field: + - + -, - + - +, + - + -, - + - +.2331
We can use that as a reference point.2346
Which one would be the best one if we are looking to get the determinant of A?2348
If we are looking to get the determinant of A, which would be the best row or column to expand on?2354
I see two 0's on this column, so let's work off of that one.2358
The first 0 just disappears, because it is 0 times its cofactor; blow out that cofactor, because it is 0.2362
The next one is -3; so we are on the third row, second column, so that corresponds to that symbol, a negative.2368
It is negative, and then -3, what we have for what we are expanding around; negative -3 is minus -3;2376
and then times...what happens if we cut out everything on a line with that -3?2384
We have 2, 3, -2, 4, 1, 0, -1, 4, 1.2390
We have to keep going; now we are on the 2; let's swap to a new color.2400
2 here; we are on a + now, so it is +2 times...cut out what is on a line with that 2; so we have 2, 3, -2, 0, -2, 5, -1, 4, 1.2406
All right, at this point, we want to figure out what are the easiest rows or columns to expand on for these two matrices.2426
I notice that there is a 0 here and a 0 here; I personally find it easier to do expanding2433
based on a row than based on a column, so I will just choose to do rows.2440
- -3; these cancel to +3; so +3 times...expand on -2 first, so...2445
oh, and we are on a 3 x 3 now, so we are on that + there; so it is still positive...2452
it is 3 times...now we are figuring out the determinant of that matrix: -2 times...2457
cross out what is on a line with that: 4, -1, 1, 4; minus 0...but - 0 cancels out, so what is next after that?2462
Another plus: + 1 times...cross out what is on a line with that 1 here; 2, 4, 3, 1.2471
All right, let's work on our other half, the other determinant.2485
+2, times whatever the determinant is inside of this matrix: 2 here; 2 is a positive here,2489
because it is in the top left: so 2 times...whatever is on a line with that gets cancelled out.2495
So, we are left with -2, 4, 5, 1.2504
Then, 0 next: the 0 here we don't have to worry about.2508
And then, we are finally onto a +; but it is a negative 1, so + -1 times...cross out what is on a line with the -1; we are left with 3, -2, -2, 5.2511
OK, at this point, we just have a lot of arithmetic to work through.2525
3, -2...take the determinant of this matrix; we have 4 times 4; that is 16; 16 - -1 gets us +17.2530
Plus 1 times...forget about the 1...2 times 1 is 2, minus 3 times 4 is 12; so 2 - 12 is -10.2540
Plus 2 times...-2 times 1 is -2; minus 5 times 4 (is 20); so -2 - 20 is -22.2550
Plus -1...let's make it a negative...times...3 times 5 is 15; minus...-2 times -2 is +4; so 3 times 5 is 15, plus 4 is 19.2560
If you have difficulty doing that in your head, just write out the 2 x 2, as well.2577
Keep working through this: 3 times...-2 times 17 comes out to be -34; -34 - 10 + 2 times -22 (is -44), minus 19;2582
3 times -44 + 2(-44) - 19 becomes...oops, a mistake was made...oh, here it is.2602
I just caught my mistake--see how easy it is to make mistakes here?2622
That should be an important point: be really, really careful with this; it is really easy to make mistakes.2625
3 times 5 is 15, minus...-2 times -2 (let's work this one out carefully)...3 times 5, minus -2 times -2...2629
well, these cancel out, and we are left with +4; so 15 - 4 becomes 11.2640
So, this shouldn't be 19; it should be 11.2644
This shouldn't be a 19 here, either; it should be 11.2649
So, -44 - 11 becomes -55; see how easy it is to make mistakes?2652
I make mistakes; it is really easy to make mistakes; be very careful with this sort of stuff.2658
It is really, really a sad way to end up missing things, when you understand what is going on, but it is just one little, tiny arithmetic error.2662
All right, let's finish this one out.2670
3 times -44 becomes -132; plus 2 times -55 becomes -110; we combine those together, and we get -242.2671
-242 is the determinant of A; it is equal to -242; it takes a while to work through, doesn't it?2691
All right, the next one: We figured out the determinant of A; that comes out to be -242.2701
So, to use Cramer's Rule, we know that y is going to be equal to the determinant of Ay, over the determinant of A.2706
We now need to figure out what Ay comes out to be.2716
The determinant of Ay...let's figure this out.2720
We work through this one; I notice this nice row right here--we have three 0's on it.2727
That is going to make it easy to work through; if we are doing a cofactor expansion, we want to make our sign table.2733
OK, we can work along with that.2744
Our first one is a + on 0, but that doesn't matter; the next one is a - on -3.2745
- -3 on...we cut out what is on a line with that -3; we have 2, 3, -2, 5, 16, 17, -1, 4, 1.2752
Next is 0; once again, we don't have to worry about that.2772
Next is 0; once again, we don't have to worry about that.2774
All right, so we see that these cancel out, and we have 3 times...now we need to choose what we are going to expand along--which row or column.2777
Personally, I like the top row; I like expanding along rows, and 2, 5, -1 does at least have some kind of small numbers.2786
So, I will expand across that, just because I feel like it.2793
2, 5, -1: we will do it in three different colors here.2797
2 corresponds to this, so it is a positive 2, times what cuts along this...we are left with 16, 17, 4, 1.2801
The next one (do it with green): that corresponds to a negative there, so that is minus 5;2814
what does it cut out? We are left with 3, -2, 4, 1.2822
And then finally, go back to red; -1 corresponds to a positive, so + -1 times...what does it cross out? We have 3, -2, 16, 17 left.2831
All right, let's work this out: we have 3 times all of this stuff; 2 times...16 times 1 is 16; minus 17 times 4...2850
17 times 4 comes out to be -68; the next one: minus 5 times...3 times 1 is 32860
(after that mistake last time, let's be careful) minus...-2 times 4 is -8; so that will cancel out to plus.2874
Finally, we turn this to a minus, since it was times -1; 3 times 17 comes to 51; minus -2 times 16 becomes -32.2882
OK, keep working this out: 3 times 2 times 16 - 68 is -52,2899
minus 5 times 3 + 8 is 11, minus 51 - -32 becomes 51 + 32; 51 + 32 is - 83.2911
So, 3 times...2 times -52 becomes -104; minus 5 times 11 becomes -55; and still, minus 83.2929
We combine all of those together, and that gets us 3 times -242.2941
Now, you could go through and multiply this together, and you would get a number out of it.2948
But notice: we have -242 here, and later on, in just a few moments, we are about to divide by that previous detA at -242.2953
So, why don't we just leave this as 3 times -242; that is equal to the determinant of Ay, our special matrix for Ay.2962
At this point, we know from Cramer's Rule that y equals (cut out a little space for it)2976
the determinant of its special matrix, Ay, divided by the determinant of the coefficient matrix.2983
We figured out that the determinant of our special matrix is 3(-242), so 3(-242) divided by the determinant2991
of our coefficient matrix--that is also -242; -242/-242--those parts cancel out, and we are left with 3; so y = 3.3002
Sadly, there is no good way to check it at this point, if we are going to have to work through the whole thing,3013
because we would have to solve for each one of them, w and x and z.3018
On the bright side, solving for w, x, and z is only having to figure out the determinant of Aw, Ax, and Az,3023
because we have already figured out the determinant of A.3030
But still, it clearly takes some effort to take the determinants of even just a size 4 x 4, so it is pretty difficult.3031
However, if you have a graphing calculator, it would be pretty easy to go through and enter the matrix,3036
and then enter an augmented matrix, including the constants, and then get the reduced row echelon form3042
and see if y = 3 pops out as the answer that you would have from it.3049
It would be the case that that is what you would get out of it.3052
Or you could use Cramer's Rule and do determinants: figure out what Ax is; figure out what Aw is;3055
figure out what Az is; and then, be able to plug them all in and check afterwards.3061
Or you could also go through and do it with inverse matrices and see if y comes out to be 3, once you have figured out that on your calculator.3066
You can do this stuff by hand if you have to do it on a test;3073
but then, you can also, if you are allowed to just use the calculator (you just have to show your work)...3075
you can check your work in a second, different way to make sure that your work did come out to be true.3079
So, you can definitely get the problem right.3083
All right, the final example: do you remember that monster from solving systems of linear equations?3086
It is back, and we are going to solve it: we are going to knock out this thing that was way too difficult for us then.3090
It is going to be really easy for us now, because we have access to how inverse matrices work.3095
We can use calculators to be able to calculate an inverse matrix very quickly; this thing is going to be easy.3099
Our plan: remember, the idea was that we have the coefficient matrix A, times the column of the variables, is equal to the column of the constants.3104
So, AX = B: if we can figure out what A-1 is, we can multiply by A-1 on the left side on both cases.3117
A-1 cancels out there, and we are left with X = A-1B.3125
We already know what B is: B is this thing right here, so that part is pretty easy.3134
Can we figure out what A-1 is?3139
Well, this is A; I am assuming that we have access to a graphing calculator or some way to do matrix calculations.3141
Once again, matrix calculations are easy to do, but really tedious.3149
They take all of this time; it is easy to make a mistake, because just doing 100 calculations, you tend to make a mistake somewhere.3153
But that is what calculators and computers are for; that is why humans invented those sorts of things--3159
to be able to make tedious calculations like that go away, where we can trust the calculator3164
to do the number-crunching part, and we can trust us to do the thinking part (hopefully).3168
We figure what A is; it is going to be a big one: our u's first: 1u, -4u, 1u, -2, 1/5u, 2u;3174
next, our v's: 2v, 2v, 1v, 1/2v, -3v, +4v; 7w, 1w, 0w (because it didn't show up), 3w, -1w, -1w;3188
-3x, 1/3x, 0x, 0x, 2x, -3x; 4y, 2y, 1y, 2y, -1y, 5y; 2z, 1z, 1z, 4z, 4z, 0z.3207
So, what you do is: you take A, and you enter that into your graphing calculator.3225
You put that into a graphing calculator; you put that into some sort of matrix calculator.3230
You enter this into a calculator, or a computer, or something that is able to work with matrices.3233
Lots of programs are, because matrices are very useful.3243
Once again, we aren't even beginning to scratch the surface of how useful they are; we are just getting some sense with this one problem.3245
So, we enter this whole thing into a calculator; then you tell the calculator to take the inverse.3251
So, we do that; and I want to point out, before we actually go on to talk about the inverse:3257
you tell the calculator to take the inverse; before you do that, double-check that you entered the matrix correctly.3261
If you entered this wrong--if you entered this A, 6 x 6...that is 36 numbers that you just put into your calculator.3268
Chances are that you might have accidentally entered one of them wrong.3274
If you enter one of them wrong, your entire answer is going to be wrong.3277
Chances are it will end up being this awful decimal number, so you will think,3280
"Well, my teacher probably didn't give me something that would come out to be an awful decimal number."3283
But if you are working with something like physics, where you don't already know what the answer is going to be,3286
it is up to you to make sure that you get it in correctly the first time.3290
So, double-check: if you are entering a very large matrix, make certain that you entered that matrix correctly.3294
We have the entire matrix set up in our calculator, and we have double-checked that it is correct.3300
Now, we punch out A-1: on most calculators, that is going to end up being: take the matrix and raise it to the -1.3304
What does it come out to be; it comes out to be really ugly--it is awful.3310
For example, the very first term is going to be 1780/14131; the first row, second column, would be 45/14131; the third...this is awful.3315
So, what are we going to end up doing?3337
Do we have to write the whole thing down?3339
No, we don't have to write the whole thing down--it is in our calculator.3340
We just tell the calculator A-1, and then we don't have to worry about A-1 at all.3342
We don't have to figure it out and write the whole thing down on paper; there is no need for it.3348
The calculator will keep track of what the numbers for A-1 are,3352
because all we are concerned about is taking A-1 and applying it against B.3356
We leave it in the calculator; we know that X is going to be equal to A-1 times B.3362
So, we have, in our calculator, that A-1 is in there.3374
We have it in the calculator; we don't have to actually see what the whole thing is, because it is already there.3378
What is our B? We enter in the column matrix, 41, 39, 4, 23, -30, 44; we make sure that our A-1 is multiplying from the left side.3383
Otherwise, it won't work at all.3396
And what does this end up coming out to be?3398
This comes out to be the deliciously simple -5, 4, 1, -3, 6, -1.3400
So, we just figured out that our X (all of our variables at once) is equal to...what were all of our variables?3410
It was u, and then we put in v, and then we put in w, and then we put in x, y, z.3420
So, they go in that order in our column: u, v, w, x, y, z = this thing that we just punched out, -5, 4, 1, -3, 6, -1.3425
So, u = -5; v = 4; w = 1; x = -3; y = 6; z = -1.3443
If we really wanted to at this point, we could check it; we could plug each one of these into any one of these equations;3450
and if it came out right, chances are that we probably got the entire thing right.3455
So, it might not be a bad idea to check at that point.3458
But also, as long as we were really careful with entering in our A, and careful with entering in our column of constants, our B,3460
everything should have worked out fine there; otherwise there is some other error that cropped up.3467
So, it becomes really, really easy, with just a little bit of thinking, and this calculator3470
(to take care of the awful manual work of the numbers, of just having to work through that many numbers)--3475
as long as we have the calculator to be able to do that part, so that it is quick and easy,3482
and we can trust that it came out right, and we are able to do the thought of what is going on,3485
we see that A-1, our coefficient matrix inverted, times what the equations come out to be,3489
our constant column matrix, just comes out to be the answers for each one of them.3496
It is really cool, really fast, and really easy; any time you have a large linear system,3500
or even a small linear system, and you just want to check it, you can have it done like that,3506 |
Posted by: Dave Richeson | June 1, 2012
## Angle trisection using origami
It is well known that it is impossible to trisect an arbitrary angle using only a compass and straightedge. However, as we will see in this post, it is possible to trisect an angle using origami. The technique shown here dates back to the 1970s and is due to Hisashi Abe.
Assume, as in the figure below, that we begin with an acute angle ${\theta}$ formed by the bottom edge of the square of origami paper and a line (a fold, presumably), ${l_{1}}$, meeting at the lower left corner of the square. Create an arbitrary horizontal fold to form the line ${l_{2}}$, then fold the bottom edge up to ${l_{2}}$ to form the line ${l_{3}}$. Let ${B}$ be the lower left corner of the square and ${A}$ be the left endpoint of ${l_{2}}$. Fold the square so that ${A}$ and ${B}$ meet the lines ${l_{1}}$ and ${l_{3}}$, respectively. (Note: this is the non-Euclidean move—this fold line cannot, in general, be drawn using compass and straightedge.) With the paper still folded, refold along ${l_{3}}$ to create a new fold ${l_{4}}$. Open the paper and fold it to extend ${l_{4}}$ to a full fold (this fold will extend to the corner of the square, ${B}$). Finally, fold the lower edge of the square up to ${l_{4}}$ to create the line ${l_{5}}$. Having accomplished this, the lines ${l_{4}}$ and ${l_{5}}$ trisect the angle ${\theta}$.
Let us see why this is true. Consider the diagram below. We have drawn in ${CD}$, which is the location of the segment ${AB}$ after it is folded, ${AC}$, the fourth side of the isosceles trapezoid ${ABDC}$, and ${AD}$, the second diagonal of ${ABDC}$. We must show that ${\theta=3\alpha}$, where ${\alpha=\angle DBE}$ and ${\theta=\angle CBE}$.
Because ${BE}$ and ${DF}$ are parallel, ${\angle DBE=\angle BDF}$, and because ${DF}$ is the altitude of the isosceles triangle ${ABD}$, ${\angle BDF=\angle ADF}$. Thus ${\alpha=\angle DBE=\angle BDF=\angle ADF}$. Now, ${ABDC}$ is an isosceles trapezoid and ${ABD}$ is an isosceles triangle, so ${ABD}$ and ${BCD}$ are congruent isosceles triangles. Thus ${\angle CBD=\angle ADB= \angle BDF+\angle ADF}$. It follows that ${\theta=\angle CBE=\angle DBE+\angle CBD=\angle DBE+\angle BDF+\angle ADF=3\alpha}$.
The geometric properties of origami constructions are quite interesting. Every point that is constructible using a compass and straightedge is constructible using origami. But more is constructible. As we’ve seen, it is possible to trisect any angle using origami (I’ll leave the obtuse angles as an exercise). It is possible to double a cube. It is possible to construct regular heptagons and nonagons. In fact, where the constructability of ${n}$-gons is related to Fermat primes, the origami-constructibility of ${n}$-gons is related to Pierpont primes. While the field of constructible numbers is the smallest subfield of ${\mathbb{R}}$ that is closed under square roots, the field of origami-constructible numbers is the smallest subfield that is closed under square roots and cube roots. In fact, it is possible to solve any linear, quadratic, cubic, or quartic equation using origami!
There are quite a few places to read about geometric constructions using origami, but a good starting point is this online article (pdf) by Robert Lang.
Posted by: Dave Richeson | April 18, 2012
## An interesting multivariable calculus example
Earlier this semester in my Multivariable Calculus course we were discussing the second derivative test. Recall the pesky condition that if ${(a,b) }$ is a critical point and ${D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^{2}=0}$, then the test fails.
A student emailed me after class and asked the following question. Suppose a function ${f}$ has a critical point at ${(0,0)}$ and ${D(0,0)=0}$. Moreover, suppose that as we approach ${(0,0)}$ along ${x=0}$ we have ${f_{xx}(0,y)>0}$ when ${y<0}$ and ${f_{xx}(0,y)<0}$ when ${y>0}$. Is that enough to say that the critical point is a not a maximum or a minimum? His thought process was that if we look at slices ${y=k}$ we get curves that are concave up when ${k<0}$ and curves that are concave down when ${k>0}$—surely that could not happen at a maximum or minimum.
I understood his intuition, but I was skeptical. Indeed, after a little playing around I came up with the following counterexample. The function is
${\displaystyle f(x,y)=\begin{cases} x^{4}+y^{2}e^{-x^{2}} & y\ge 0\\ x^{4}+x^{2}y^{2}+y^{2} & y<0. \end{cases}}$
The first partial derivatives are
${\displaystyle f_{x}(x,y)=\begin{cases} 4x^{3}-2xy^{2}e^{-x^{2}} & y\ge 0\\ 4x^{3}+2xy^{2} & y<0, \end{cases}}$
${\displaystyle f_{y}(x,y)=\begin{cases} 2ye^{-x^{2}} & y\ge 0\\ 2x^{2}y+2y & y<0. \end{cases}}$
Clearly ${(0,0)}$ is a critical point. The second partial derivatives are
${\displaystyle f_{xx}(x,y)=\begin{cases} 12x^{2}-2y^{2}e^{-x^{2}}+4x^{2}y^{2}e^{-x^{2}} & y\ge 0\\ 12x^{2}+2y^{2} & y<0, \end{cases}}$
${\displaystyle f_{yy}(x,y)=\begin{cases} 2e^{-x^{2}} & y\ge 0\\ 2x^{2}+2 & y<0, \end{cases}}$
${\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=\begin{cases} -4xye^{-x^{2}} & y\ge 0\\ 4xy & y<0. \end{cases}}$
Thus ${D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^{2}=0\cdot 2-0^{2}=0}$. So the second derivative test fails. But observe that when ${x=0}$ we have
${\displaystyle f_{xx}(0,y)=\begin{cases} -2y^{2} & y\ge 0\\ 2y^{2} & y<0. \end{cases}}$
So ${f_{xx}(0,y)>0}$ when ${y<0}$ and ${f_{xx}(0,y)<0}$ when ${y>0}$. Yet it is easy to see that ${(0,0)}$ is a minimum: ${f(0,0)=0}$ and ${f(x,y)>0}$ for all ${(x,y)\ne (0,0)}$. A graph of the function is shown below. You can see the concave down cross sections for $x=0$ and $y>0$.
Posted by: Dave Richeson | February 21, 2012
## Parametric curve project for multivariable calculus
I’m teaching two sections of Multivariable Calculus this semester. Each class has 3 hours of lecture and a 1 hour 20 minute lab each week. Last week the students were learning about parametric equations. So in lab I wanted to give them some hands-on experience with 2-dimensional parametric curves. Their assignment was to create a work of art using Grapher (Apple’s free graphing program) and parametric curves. They worked on the projects in lab for about an hour (in pairs, mostly). Some finished in that time, but others finished outside of class. The results were fantastic, so I thought I’d share them here.
Here is a pdf of the lab assignment.
This lab was not my idea. It was written by my colleague Lorelei Koss. It was also classroom-tested by my colleague Jen Schaefer. This semester I took their feedback, changed the lab a little, and used it in my class. Lorelei said that she got the idea from Judy Holdener and Keith Howard at Kenyon College and Tommy Ratliff at Wheaton College. I’ve also found some similar ideas in the mathematical literature: there’s an article by Barry Tesman (also a colleague of mine) and Marc Sanders, “MATH and other four-letter words,” College Math Journal, Nov. 1998, and “Painting by Parametric Curves and Van Gogh’s Starry Night,” by Stephen Lovett, Matthew Arildsen, Jon Jones, Anna Larson and Rebecca Russ, Math Horizons, Nov. 2010.
Here are their amazing works of art (click to see a slide show):
Posted by: Dave Richeson | October 28, 2011
## Cantor set applet
I made this Cantor set applet for my Real Analysis class. It is nothing fancy, but it saves me from drawing it on the board.
Posted by: Dave Richeson | October 28, 2011
## Applet to illustrate the epsilon-delta definition of limit
Here’s a GeoGebra applet that I made for my Real Analysis class. It can be used to explore the definition of limit:
Definition. The limit of $f(x)$ as $x$ approaches $c$ is $L$, or equivalently $\displaystyle \lim_{x\to c}f(x)=L,$ if for any $\varepsilon>0$ there exists $\delta>0$ such that whenever $0<|x-c|<\delta$, it follows that $|f(x)-L|<\varepsilon$.
Posted by: Dave Richeson | October 7, 2011
## The danger of false positives
As I mentioned earlier, I’m teaching a first-year seminar this semester called “Science or Nonsense?” On Monday and Wednesday this week we discussed some math/stats/numeracy topics. We talked about the Sally Clark murder trial, the prosecutor’s fallacy, the use of DNA testing in law enforcement, Simpson’s paradox, the danger of false positives, and the 2009 mammogram screening recommendations.
I made a GeoGebra applet to illustrate the dangers of false positives. So I thought I’d share that here. Here’s the statement of the problem.
Suppose Lenny Oiler visits his doctor for a routine checkup. The doctor says that he must test all patients (regardless of whether they have symptoms) for rare disease called analysisitis. (This horrible illness can lead to severe pain in a patient’s epsylawns and del-tahs. It should not to be confused with analysis situs.) The doctor says that the test is 99% effective when given to people who are ill (the probability the test will come back positive) and it is 95% effective when given to people who are healthy (it will come back negative).
Two days later the doctor informs Lenny that the test came back positive.
Should Lenny be worried?
Surprisingly, we do not have enough information to answer the question, and Lenny (being pretty good at math) realizes this. After a little investigating he finds out that approximately 1 in every 2000 people have analysisitis (about 0.05% of the population).
Now should Lenny be worried?
Obviously he should take notice because he tested positive. But he should not be too worried. It turns out that there is less than a 1% chance that he has analysisitis.
Notice that there are four possible outcomes for a person in Lenny’s position. A person is either ill or healthy and the test may come back positive or negative. The four outcomes are shown in the chart below.
Test result Ill Healthy Positive true positive false positive Negative false negative true negative
Obviously, the two red boxes are the ones to worry about because the test is giving the incorrect result. But in this case, because the test came back positive, we’re interested in the top row.
For simplicity, suppose the city that is being screened has a population of 1 million. Then approximately (1000000)(0.0005)=500 people have the illness. Of these (500)(.99)=495 will test positive and (500)(0.01)=5 will test negative. Of the 999,500 healthy people (999500)(.05)=49975 will test positive and (999500)(.95)=949525 will test negative. This is summarized in the following chart.
Test result Ill Healthy Positive 495 49975 Negative 5 949525
Thus, 495+49975=50470 people test positive, and of these only 495 are ill. So the chance that a recipient of positive test result is sick is 495/50470=0.0098=0.98%. That should seem shockingly low! I wonder how many physicians are aware of this phenomenon.
You can try out this or other examples using this GeoGebra applet that I made.
Posted by: Dave Richeson | September 14, 2011
## A neat probability rule-of-thumb
Disclaimer: I am NOT a probabilist. Not only have I never taught probability, the last time I took a course in probability was in my sophomore year of college. So if this is well known (or totally wrong), forgive me.
A non-mathematician friend of mine shared this link with me. It compares the lifetime risk of dying by various means—cancer, heart disease, shark attack, etc. There are many problems with the analysis presented on this web page (for example, you are not equally likely to die from the flu in each of your 77.6 years (the average lifespan), conditional probability would be a more useful measure of risk for some of these, etc.), but I will ignore all of that. I want to focus on the last line. It says:
Lifetime risk is calculated by dividing 2003 population (290,850,005) by the number of deaths, divided by 77.6, the life expectancy of a person born in 2003.
For example, for drowning the risk is 1 in $290850005/(3306\cdot 77.6)=1133.7$
Stated another way, they are claiming that if $D$ people die each year from a given cause, the total population is $P$, and the life expectancy is $L$, then the probability of dying from the given cause is $DL/P$. I saw this and I thought, “Surely this is wrong. Why would that formula give the probability?”
So I tried to calculate it myself. Here is my back-of-the-envelope calculation. The chance of dying from this cause in one year is $D/P$. The chance of not dying from this cause in one year is $1-D/P$, the chance of not dying from this cause for $L$ years is $(1-D/P)^L$, and so the chance of dying from the cause in $L$ years is $1-(1-D/P)^L$. (Of course, this leaves open the possibility of dying several times in those $L$ years, but we’ll ignore that.)
Let’s use this formula with the drowning example. I get $1-(1-3306/290850005)^{77.6}=0.000881671\ldots$, or 1 in $1134.2$.
What?!?! I was shocked to see an answer almost identical to the one using the “wrong” technique. There must be more to this than I first thought. Let’s look a little closer.
First, notice that $1-(1-D/P)^L=1-((1+(-D)/P)^P)^{L/P}$. Sitting inside this expression is a sub-expression that looks a lot like the limit definition of $e^x$. In particular, because $P$ is a large number, this expression is very nearly $1-(e^{-D})^{L/P}=1-e^{-DL/P}$. Aha! There’s the $DL/P$ term! But we still don’t quite have what we want.
What we’ve shown is that if the probability someone dies of a given cause in one year is $x$, then the probability that they will die from it in $L$ years is approximately $1-e^{-Lx}$. Now suppose the probability $x$ is small (like the probability of dying by drowning). We will compute the linear approximation to this function at $x=0$. We see that $d(1-e^{-Lx})/dx=Le^{-Lx}$. At $x=0$, that derivative is $L$. So the linear approximation at $x=0$ is simply $Lx$. In particular, if we evaluate it at our specific annual probability value $D/P$, we obtain $DL/P$. And there it is! [Update: thank you to the commenters for pointing out that the introduction of the exponential function, while fine, is unnecessary. Quicker: just use the linear approximation for $1-(1-x)^L$ at $x=0$.]
Again, I’ve never seen this before. Perhaps it is well known. For example, maybe it is a good rule-of-thumb that all good actuaries know.
I’d be happy to hear people’s thoughts about this formula and my reasoning. Maybe there’s another, different way to see this.
[I’d like to thank my colleague Jeff Forrester for talking through this with me.]
Posted by: Dave Richeson | August 30, 2011
## Advice for new college students
I’m teaching a first year seminar this semester. This isn’t a math course. (The title of my course is “Science or Nonsense?” We will look at a wide range of topics including the paranormal, evolution, climate change, the vaccine/autism controversy, alternative medicines, etc.) We are required to focus on academic writing, library research, oral communication, etc. I will also be the academic advisor to the students in my class until they declare a major. With this last role in mind, I decided to write up some advice for these new students. Here’s my list. I gave them only statements in bold, the plain text is what I told them as we went through the list.
1. Get to class on time.
3. Spend a summer on campus. Work for a professor, be a tour guide, do research, etc.
4. Use proper grammar and capitalization in the email messages to your professors. The email shorthand that may be appropriate between friends is not appropriate when corresponding with your professor (e.g., “hey, prof. when r u going 2 b in yr office?”).
5. Call your teachers “Professor —” not “Mr. —” or “Mrs. —.” Almost all of your professors have the highest degree in their field (usually a PhD). (Addressing them as “Dr. —” is appropriate too, although it isn’t common at our school.)
6. Get to know your professors and let them get to know you. They’re nice people. Ask your professors about their research, their family, their schooling, etc. Tell them about your summer research projects, your internships, etc. Down the road you may want to ask them for a letter of recommendation and they will be able to write you a much better letter if they know you. Besides, they are human beings, if you are rude to them, they will be less enthusiastic about helping you.
7. Don’t skip class. Either you won’t be able to learn the material that you missed or the “free hour” that you gained will be lost several times over trying to catch up. If you do skip class, DON’T ask the professor what you missed—get notes from a classmate.
8. Take classes outside of your comfort zone.
10. Don’t sell your books back, especially for classes in your major.
11. Don’t be a member of a clique. For many of you college will be the most diverse living experience of your life. Get to know as many people as possible and not just those with the same background as you.
12. Be organized, use a calendar, and pay attention to due dates.
13. Find a good distraction-free place to study.
14. Learn to write well. I’ve seen far too many mathematics and science students avoid writing courses. They are under the impression that it won’t be relevant to them. Writing is an extremely important skill that is a prerequisite for almost all careers. You will be amazed at how much you will need to write.
16. Do the assigned work. And the related…
17. Don’t ask for extra credit. I don’t give extra credit and neither do most other college professors; if they do, they would give it to the entire class not just to you individually. Extra credit is great for the strong students—it can boost their grades from an A to an A+. Weaker students who need a grade booster should spend their time doing the assigned work (which they often haven’t done—that’s why their grade is in trouble in the first place). Doing the assigned work is the best preparation for the exams in the class—it gives the best “bang for the buck.”
18. Start assignments early and start studying early. Related: don’t email the professor late the night before (or worse, the day of) an exam or the due date for an assignment asking for help.
19. Admit when you are wrong. It may be difficult, painful, or embarrassing, but it is liberating. Living with a lie or a guilty conscience is worse than coming clean.
20. If you choose to drink alcohol, do so in moderation. Not all college students drinks alcohol. According to a survey given here last year at fall break, approximately one fourth of the first year students had not consumed any alcohol in the past year.
21. Stay healthy: eat well, exercise, and get enough sleep.
22. Take the courses you want to take, not the ones your parents want you to take.
23. Beware of technology such as video games, movies, social media, etc. They can be unhealthy, addictive time sinks.
24. Don’t read or send text messages in class.
25. Try new things (clubs, sports, volunteering, etc.), but don’t spread yourself too thin.
26. Call home, but not too often.
27. Get off campus and explore the area. Eat in restaurants, go for a hike, see a movie, visit a museum, etc.
30. Show up for appointments and be punctual.
31. Don’t let your parents fight your battles. Professors cannot speak with your parents anyway (without a FERPA release).
32. If things start going wrong, see a counselor. Each year the counseling center is used by 15–20% of the student body. The service is completely confidential; they won’t notify your parents, your professors, your friends, or your insurance company.
33. Let go of your high school anxieties. Your classmates didn’t know you in high school. Make new friends, wear new clothes, listen to different music, and try new things.
34. Don’t lie to your professors; they’ve heard them all (otherwise known as the “dead grandmother rule”). (A retired professor I know used to send a condolence card to the student’s parents every time a student informed him of a death in the family.)
35. Be considerate of the neighbors. Not everyone in town is a college student. Keep this in mind when you are returning from a party at 2:00 AM.
36. Be a good roommate.
37. Don’t cheat. The penalties are steep if you are caught. If you are not caught you will have to contend with a guilty conscience. Cheating will produce a short-term gain and a long-term loss. Besides, it is a slippery slope—this is not the way you want to conduct the rest of your life.
38. Become a novice. You’ll learn more and get more out of college if you don’t hold onto the attitude that you know everything already.
39. Go on a road trip.
40. Look at your final exam schedule before scheduling your flight home.
Posted by: Dave Richeson | August 17, 2011
## A quick guide to LaTeX
This semester I’ll be teaching real analysis. I am going to have the students type their homework in LaTeX. To make this as easy for them as possible, I will give them a template that is all ready for them to enter their solutions. They shouldn’t have to worry about headers, packages, font sizes, margins, etc. Furthermore, I decided that I should give them a LaTeX cheat sheet—a single document that has all the LaTeX information that they will need. I’ve created LaTeX cheat sheets like this before—but one was for real analysis, one was for topology, one was for linear algebra, and one was for discrete math. Each cheat sheet had different symbols.
So, I decided to bring them all together into a one-size-fits-all LaTeX cheat sheet. I kept it to two pages, so it can be printed (double-sided) on one piece of paper. (I have also posted the LaTeX code. Feel free to take it, edit it, and use it.)
It doesn’t have everything. As I said, I’ve left out all information about headers, etc. Also, since these students will probably not be using figures or tables, I’ve left them out. [Update: I added information about figures. I also added links to some online resources.]
Please let me know in the comments if there is anything that you think I should add. I still have a week and a half to tinker with it before classes begin. Also, please let me know if you find any errors. Thanks!
[Note: When I began this project I intended to modify this cheat sheet by Winston Chang to suit my needs. But in the end, I wiped it clean and started from scratch. (I did use his very nice three-column format though.)]
Posted by: Dave Richeson | August 12, 2011
## Highlights from MathFest 2011
Last weekend I was in Lexingon, Kentucky for MathFest 2011. I had a very nice time and saw some very good talks. I thought, just for fun, that I’d share a couple of juicy mathematical tidbits I learned.
Fibonacci numbers and the golden ratio
Ed Burger of Williams College gave a talk entitled “Planting your roots in the natural numbers: A rational and irrational look at 1, 2, 3, 4,…” From his talk I learned the following interesting facts.
In 1939 Edouard Zeckendorf proved that every natural number can be decomposed uniquely into a sum of Fibonacci numbers in such a way that no two of the Fibonacci numbers are consecutive. Recall, of course, that the Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,… In particular, they satisfy the relation $F_1=1$, $F_2=1$, and $F_{k+1}=F_k+F_{k-1}$.
For example:
1=1
2=2
3=1+2
4=1+3
5=5
6=1+5
7=2+5
$\vdots$
30=1+8+21
$\vdots$
48=1+13+34
$\vdots$
Then, in 1957 G. Bergman proved that every natural number can be written uniquely as the sum of distinct nonconsecutive integer powers of $\varphi$ (where $\varphi$ the “golden ratio$(1+\sqrt{5})/2$). For example:
$1=\varphi^0$
$2=\varphi^{-2}+\varphi^1$
$3=\varphi^{-2}+\varphi^2$
$4=\varphi^{-2}+\varphi^0+\varphi^{2}$
$5=\varphi^{-4}+\varphi^{-1}+\varphi^{3}$
$6=\varphi^{-4}+\varphi^1+\varphi^{3}$ (check it here if you don’t believe it)
Then, in 2008 Dale Gerdemann noticed that these facts are related.
First of all, the fact that $\varphi=1+\varphi^{-1}$ implies that $\varphi^{k+1}=\varphi^{k}+\varphi^{k-1}$, which is a very Fibonacci-like relation.
Moreover, notice that $30=6\cdot 5=6\cdot F_5$ and that $30=1+8+21=F_1+F_6+F_8$.
Similarly, $48=6\cdot 8=6\cdot F_6$ and $48=1+13+34=F_2+F_7+F_9$.
$6=\varphi^{-4}+\varphi^1+\varphi^{3}$
$6\cdot F_5=F_{5-4}+F_{5+1}+F_{5+3}$
$6\cdot F_6=F_{6-4}+F_{6+1}+F_{6+3}$
Indeed, Gerdemann proved that $n=\varphi^{k_1}+\cdots+\varphi^{k_n}$ if and only if $nF_m=F_{m+k_1}+\cdots+F_{m+k_n}$ (for $m$ sufficiently large).
So, for example, $7\cdot F_6=7\cdot 8= 56=1+55=F_2+F_{10}=F_{6-4}+F_{6+4}$. So from this we can conclude that $7=\varphi^{-4}+\varphi^{4}$, which it is. Isn’t that cool?
Burger went on to describe some work he did with his REU students to extend these results to other sequences and other irrational numbers.
Beyond the Pythagorean theorem
Roger Nelson gave an excellent talk entitled “Math Icons.” It is base on material in his new book (with Claudi Alsina) Icons of Mathematics. They look at the mathematics behind several famous images (icons) in mathematics.
He started by talking about the “bride’s chair.” This is the famous image which gives the geometric interpretation of the Pythagorean theorem. Rather than our usual algebraic $a^2+b^2=c^2$, it shows that the sum of the areas of the squares on sides $a$ and $b$ is equal to the area of the square on the side $c$.
He went on to point out, for instance, that the figures on the sides of the triangle need not be squares. Any similar shapes will do. For example, in the figure below we see that the area sum property holds for semicircles as well. (This is in Euclid’s Elements, VI.31: In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle.)
He also discussed various properties of the so-called Vecten configuration. This is the same as the brides’ chair, but for triangles that aren’t right.
One property that I thought was particular nice is that if we take a Vecten configuration and draw in the three “flanks” (the red triangles below), then the area of each of the three flanks is the same as the area of the original (blue) triangle.
Finally, we turn to a Vecten-type configuration, but with equilateral triangles on each face. In this case, if we join the midpoints of each of the equilateral triangles, we obtain a new equilateral triangle (the red triangle below). This is now known as Napoleon’s theorem (yes, that Napoleon, and no, although he was interested in mathematics, we don’t believe that he discovered or proved this theorem).
This entire talk was fascinating. There was a lot more great material in it. I’ll have to check out their book!
How to draw a towel on a beach
Annalisa Crannell gave an amazing talk called “In the shadow of Desargues” on math, art, and perspective drawing. The main focus of her talk was Desargues’s theorem and using it to draw a towel on a beach. I couldn’t do the topic justice here, so you’ll have to check out her new book (with Marc Franz) called Viewpoints: Mathematical Perspective and Fractal Geometry in Art. I’m excited to read it.
MAA: The Musical
Finally, I was honored to be asked to participate in MAA: The Musical, which was performed during the opening banquet. I was happy to be asked and even happier not to be asked to sing in the production. I was enlisted as tech support (running the slide-show that went along with their songs). That was right up my alley. The MAA players were Alissa Crans, Annalisa Crannell, Art Benjamin, Bud Brown (musical director), Dan Kalman, David Bressoud, Francis Su, Frank Farris, Jennifer Beineke, Jenny Quinn, Matthew DeLong, Norm Richert, Paul Zorn, Talithia Williams. They did an amazing job (at least one song is now on YouTube).
[Update: Francis Su recorded the entire performance on his phone. It is now available online (audio only). Enjoy!]
All-in-all, it was a great conference. |
# 6. The Hyperbola
Cooling towers for a nuclear power plant have a hyperbolic cross-section.
[Image source: Flickr.]
A hyperbola is a pair of symmetrical open curves. It is what we get when we slice a pair of vertical joined cones with a vertical plane.
### How do we create a hyperbola?
Take 2 fixed points A and B and let them be 4a units apart. Now, take half of that distance (i.e. 2a units).
Now, move along a curve such that from any point on the curve,
(distance to A) − (distance to B) = 2a units.
The curve that results is called a hyperbola. There are two parts to the curve.
Let's see how this works with some examples.
The Equation of a Hyperbola
East-West Opening Hyperbola
Definition of a Hyperbola
More Forms of the Equation of a Hyperbola
Even More Forms of the Equation of a Hyperbola
### Applications of Hyperbolas
• Navigation: Ship's navigators can plot their position by comparing GPS signals from different satellites. The technique involves hyperbolas.
• Physics: The movement of objects in space and of subatomic particles trace out hyperbolas in certain situations.
• Sundials: Historically, sundials made use of hyperbolas. Place a stick in the ground and trace out the path made by the shadow of the tip, and you'll get a hyperbola.
• Construction: Nuclear power plant smoke stacks have a hyperbolic cross section as illustrated above. Such 3-dimensional objects are called hyperboloids.
### Example 1
Let the distance between our points A and B be 4 cm. For convenience in our first example, let's place our fixed points A and B on the number line at (0, 2) and (0, −2), so they are 4 units apart. In this case, a = 1 cm and 2a = 2 cm.
Now we start tracing out a curve such that P is a point on the curve, and:
distance PB − distance PA = 2 cm.
We start at (0, 1).
Shown below is one of the points P, such that PB − PA = 2.
If we continue, we obtain the blue curve:
Now, continuing our curve on the left side of the axis gives us the following:
We also have another part of the hyperbola on the opposite side of the x-axis, this time using:
distance PA − distance PB = 2.
Once again a typical point P is shown, and we can see from the lengths given that PA − PB = 2.
We observe that the curves become almost straight near the extremities. In fact, the lines y=x/sqrt3 and y=-x/sqrt3 (the red dotted lines below) are asymptotes:
[An asymptote is a line that forms a "barrier" to a curve. The curve gets closer and closer to an asymptote, but does not touch it.]
In Example 1, the points (0, 1) and (0, -1) are called the vertices of the hyperbola, while the points (0, 2) and (0, -2) are the foci (or focuses) of the hyperbola.
### The equation of our hyperbola
For the hyperbola with a = 1 that we graphed above in Example 1, the equation is given by:
y^2-x^2/3=1
Notice that it is not a function, since for each x-value, there are two y-values.
We call this example a "north-south" opening hyperbola.
### Where did this hyperbola equation come from?
The equation follows from the distance formula and the requirement (in this example) that distance PB − distance PA = 2.
Here's the proof.
Proof
For any point P(x, y) on the hyperbola,
text(distance)\ PB=sqrt(x^2+(y+2)^2
text(distance)\ PA=sqrt(x^2+(y-2)^2
Since PBPA = 2 in our example, then:
sqrt(x^2+(y+2)^2)-sqrt(x^2+(y-2))=2
Rearrange:
sqrt(x^2+(y+2)^2)=sqrt(x^2+(y-2))+2
Square both sides:
x^2+(y+2)^2 =[x^2+(y-2)^2] +4sqrt(x^2+(y-2)^2)+4
Expand brackets and simplify:
2y-1=sqrt(x^2+(y-2)^2
Square both sides again:
4y2 − 4y + 1 = x2 + y2 − 4y + 4
Simplifying gives the equation of our hyperbola:
y^2-x^2/3=1
The asymptotes (the red dotted boundary lines for the curve) are obtained by setting the above equation equal to 0, rather than 1.
y^2-x^2/3=0
This gives us the 2 lines:
y=-x/sqrt3, and y=x/sqrt3
### Interactive graph
See an interactive graph of this example, and the others on this page, here:
### General Equation of North-South Hyperbola
For the hyperbola with focal distance 4a (distance between the 2 foci), and passing through the y-axis at (0, c) and (0, −c), we define
b2 = c2a2
Applying the distance formula for the general case, in a similar fashion to the above example, we obtain the general form for a north-south hyperbola:
y^2/a^2-x^2/b^2=1
Continues below
### Example 2
Here's another example of a "north-south" hyperbola.
It's equation is:
y2x2 = 1
"North-South" hyperbola (in green) with its asymptotes (in magenta color).
Similar to Example 1, this hyperbola passes through 1 and −1 on the y-axis, but it has a different equation and a slightly different shape (and different asymptotes). Where are the 2 foci for this hyperbola? We need to find the value of c.
By inspection (of the equation of this hyperbola), we can see a = 1 and b = 1. Using the formula given above, we have:
b2 = c2a2
So
12 = c2 − 12
c2 = 2
c = ±√2
So the points A and B (the foci) for this hyperbola are at A (0, √2) and B (0, −√2).
## East-West Opening Hyperbola
By reversing the x- and y-variables in our second example above, we obtain the following equation.
### Example 3
x2y2 = 1
This gives us an "East-West" opening hyperbola, as follows. Our curve passes through -1 and 1 on the x-axis and once again, the asymptotes are the lines y = x and y = −x.
"East-West" hyperbola (in green) with its asymptotes (in magenta color).
The general formula for an East-West hyperbola is given by:
x^2/a^2-y^2/b^2=1
Note the x and y are reversed, compared the formula for the North-South hyperbola.
Don't miss the interactive graph of this example, and the others on this page, here:
## Technical Definition of a Hyperbola
A hyperbola is the locus of points where the difference in the distance to two fixed foci is constant.
This technical definition is one way of describing what we were doing in Example 1, above.
### Hyperbolas in Nature
Throw 2 stones in a pond. The resulting concentric ripples meet in a hyperbola shape.
## More Forms of the Equation of a Hyperbola
There are a few different formulas for a hyperbola.
Considering the hyperbola with centre (0, 0), the equation is either:
1. For a north-south opening hyperbola:
y^2/a^2-x^2/b^2=1
The slopes of the asymptotes are given by:
+-a/b
2. For an east-west opening hyperbola:
x^2/a^2-y^2/b^2=1
The slopes of the asymptotes are given by:
+-b/a
In Examples 2 & 3 given above, both a and b were equal to 1, so the slopes of the asymptotes were simply ± 1 and our asymptotes were the lines y = x and y = −x.
What effect does it have if we change a and b?
### Example 4
Sketch the hyperbola
y^2/25-x^2/4=1
First, we recognise that it is a north-south opening hyperbola, with a = 5 and b = 2 . It will look similar to Example 1 above, which was also a north-south opening hyperbola.
We need to find:
• The y-intercepts (there are no x-intercepts for this example)
• The asymptotes
y-intercepts: Simply let x = 0 in the equation given in the question:
y^2/25-x^2/4=1
We have:
y^2/25=1
Solving gives us 2 values (as expected):
y = -5 and y = 5
Alternatively, we note that the vertices of the hyperbola are a units from the centre of the hyperbola. In this example, it means our vertices will be at x = 0 and y = -5 and y = 5.
Aymptotes: We have a north-south opening hyperbola, so the slopes of the asymptotes will be given by
+-a/b
In this example, a = 5 and b = 2. So the slopes of the asymptotes will be simply:
-5/2 and 5/2.
The equations for the asymptotes, since they pass through (0, 0), are given by:
y = -(5x)/2 and y = (5x)/2
So we are ready to include the above information on our graph:
Asymptotes and y-intercepts (& vertices in this case).
All that remains is to complete the arms of the hyperbola, making sure that they get closer and closer to the asymptotes, as follows:
"North-South" hyperbola (in green) with its asymptotes (in magenta color).
### Interactive graph
See an interactive graph of this example, and the others on this page, here:
## Even More Forms of the Equation of a Hyperbola
(1) Possibly the simplest equation of a hyperbola is given in the following example.
### Example 5 - Equilateral Hyperbola
xy = 1
This is known as the equilateral or rectangular hyperbola.
Rectangular hyperbola. Asymptotes are the x- and y-axes.
Notice that this hyperbola is a "north-east, south-west" opening hyperbola. Compared to the other hyperbolas we have seen so far, the axes of the hyperbola have been rotated by 45°. Also, the asymptotes are the x- and y-axes.
### Hyperbola with axis not at the Origin
(2) Our hyperbola may not be centred on (0, 0). In this case, we use the following formulas:
For a "north-south" opening hyperbola with centre (h, k), we have:
((y-k)^2)/a^2-((x-h)^2)/b^2=1
For an "east-west" opening hyperbola with centre (h, k), we have:
((x-h)^2)/a^2-((y-k)^2)/b^2=1
### Example 6 - Hyperbola with Axes Shifted
Sketch the hyperbola
((x-2)^2)/36-((y+3)^2)/64=1
We note that this is an "east-west opening" hyperbola, with a = 6 and b = 8.
The center of this hyperbola will be at (2, -3), since h = 2 and k = -3 in this example.
The best approach is to ignore the shifting (for now) and figure out the other parameters for the hyperbola.
So I'm assuming (for this part) that the hyperbola is actually X^2/36+Y^2/64=1, and I'll use upper case X and Y.
The vertices of the parabola are found when Y = 0. This gives us X = -6 or X = 6.
The asymptotes will have slope -8/6 = -4/3, OR 8/6 = 4/3.
Now we can sketch the asymptotes and the vertices, remembering to shift everything so that the center is (2, -3):
Asymptotes, and vertices (-4,-3) and (8,-3).
Now for the hyperbola:
"East-West" hyperbola (in green) with its asymptotes (in magenta color).
3. We could expand our equations for the hyperbola into the following form:
Ax^2+ Bxy + Cy^2+ Dx + Ey + F = 0 (such that B^2>4AC)
In the earlier examples on this page, there was no xy-term involved. As we saw in Example 5, if we do have an xy-term, it has the effect of rotating the axes. We no longer have "north-south" or "east-west" opening arms - they could open in any direction.
### Example 7 - Hyperbola with Shifted and Rotated Axes
The graph of the hyperbola x2 + 5xy − 2y2 + 3x + 2y + 1 = 0 is as follows:
Shifted and rotated hyperbola.
We see that the axes of the hyperbola have been rotated and have been shifted from (0, 0).
[Further analysis is beyond the scope of this section. ]
### Exercise
Sketch the hyperbola
x^2/9-y^2/16=1
This is an east-west opening hyperbola, with a = 3 and b = 4. It will look similar to the east-west opening Example 3, given above.
x-intercepts: Letting y = 0 in the equation given in the question, and we have:
x^2/9=1
Solving gives us:
x = -3 and x = 3
Aymptotes: We have an east-west opening hyperbola, so the slopes of the asymptotes will be given by
+-b/a
In this example, a = 3 and b = 4. So the slopes of the asymptotes will be simply:
-4/3 and 4/3.
The equations for the asymptotes, since they pass through (0, 0), are given by:
y = -(4x)/3 and y = (4x)/3
Including the above information on our graph:
Asymptotes (in magenta color) and vertices.
Completing the hyperbola:
"East-West" hyperbola (in green) with its asymptotes (in magenta color).
### Conic section: Hyperbola
How can we obtain a hyperbola from slicing a cone?
We start with a double cone (2 right circular cones placed apex to apex):
When we slice the 2 cones vertically, we get a hyperbola, as shown.
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# Definitions, simple applications, and graphs of trigonometric (Page 6/7)
Page 6 / 7
$\theta$ 0 ${}^{\circ }$ 30 ${}^{\circ }$ 60 ${}^{\circ }$ 90 ${}^{\circ }$ 120 ${}^{\circ }$ 150 ${}^{\circ }$ $cos\theta$ $\theta$ 180 ${}^{\circ }$ 210 ${}^{\circ }$ 240 ${}^{\circ }$ 270 ${}^{\circ }$ 300 ${}^{\circ }$ 330 ${}^{\circ }$ 360 ${}^{\circ }$ $cos\theta$
Let us look back at our values for $cos\theta$
$\theta$ ${0}^{\circ }$ ${30}^{\circ }$ ${45}^{\circ }$ ${60}^{\circ }$ ${90}^{\circ }$ ${180}^{\circ }$ $cos\theta$ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ 0 $-1$
If you look carefully, you will notice that the cosine of an angle $\theta$ is the same as the sine of the angle ${90}^{\circ }-\theta$ . Take for example,
$cos{60}^{\circ }=\frac{1}{2}=sin{30}^{\circ }=sin\left({90}^{\circ }-{60}^{\circ }\right)$
This tells us that in order to create the cosine graph, all we need to do is to shift the sine graph ${90}^{\circ }$ to the left. The graph of $cos\theta$ is shown in [link] . As the cosine graph is simply a shifted sine graph, it will have the same period and amplitude as the sine graph.
## Functions of the form $y=acos\left(x\right)+q$
In the equation, $y=acos\left(x\right)+q$ , $a$ and $q$ are constants and have different effects on the graph of the function. The general shape of the graph of functions of this form is shown in [link] for the function $f\left(\theta \right)=2cos\theta +3$ .
## Functions of the form $y=acos\left(\theta \right)+q$ :
1. On the same set of axes, plot the following graphs:
1. $a\left(\theta \right)=cos\theta -2$
2. $b\left(\theta \right)=cos\theta -1$
3. $c\left(\theta \right)=cos\theta$
4. $d\left(\theta \right)=cos\theta +1$
5. $e\left(\theta \right)=cos\theta +2$
Use your results to deduce the effect of $q$ .
2. On the same set of axes, plot the following graphs:
1. $f\left(\theta \right)=-2·cos\theta$
2. $g\left(\theta \right)=-1·cos\theta$
3. $h\left(\theta \right)=0·cos\theta$
4. $j\left(\theta \right)=1·cos\theta$
5. $k\left(\theta \right)=2·cos\theta$
Use your results to deduce the effect of $a$ .
You should have found that the value of $a$ affects the amplitude of the cosine graph in the same way it did for the sine graph.
You should have also found that the value of $q$ shifts the cosine graph in the same way as it did the sine graph.
These different properties are summarised in [link] .
$a>0$ $a<0$ $q>0$ $q<0$
## Domain and range
For $f\left(\theta \right)=acos\left(\theta \right)+q$ , the domain is $\left\{\theta :\theta \in \mathbb{R}\right\}$ because there is no value of $\theta \in \mathbb{R}$ for which $f\left(\theta \right)$ is undefined.
It is easy to see that the range of $f\left(\theta \right)$ will be the same as the range of $asin\left(\theta \right)+q$ . This is because the maximum and minimum values of $acos\left(\theta \right)+q$ will be the same as the maximum and minimum values of $asin\left(\theta \right)+q$ .
## Intercepts
The $y$ -intercept of $f\left(\theta \right)=acos\left(x\right)+q$ is calculated in the same way as for sine.
$\begin{array}{ccc}\hfill {y}_{int}& =& f\left({0}^{\circ }\right)\hfill \\ & =& acos\left({0}^{\circ }\right)+q\hfill \\ & =& a\left(1\right)+q\hfill \\ & =& a+q\hfill \end{array}$
## Comparison of graphs of $sin\theta$ And $cos\theta$
Notice that the two graphs look very similar. Both oscillate up and down around the $x$ -axis as you move along the axis. The distances between the peaks of the two graphs is the same and is constant along each graph. The height of the peaks and the depths of the troughs are the same.
The only difference is that the $sin$ graph is shifted a little to the right of the $cos$ graph by 90 ${}^{\circ }$ . That means that if you shift the whole $cos$ graph to the right by 90 ${}^{\circ }$ it will overlap perfectly with the $sin$ graph. You could also move the $sin$ graph by 90 ${}^{\circ }$ to the left and it would overlap perfectly with the $cos$ graph. This means that:
$\begin{array}{ccc}\hfill sin\theta & =& cos\left(\theta -90\right)\phantom{\rule{1.em}{0ex}}\left(\mathrm{shift}\phantom{\rule{2.pt}{0ex}}\mathrm{the}cos\mathrm{graph}\phantom{\rule{2.pt}{0ex}}\mathrm{to}\phantom{\rule{2.pt}{0ex}}\mathrm{the}\phantom{\rule{2.pt}{0ex}}\mathrm{right}\right)\hfill \\ & \mathbf{a}\mathrm{nd}& \\ \hfill cos\theta & =& sin\left(\theta +90\right)\phantom{\rule{1.em}{0ex}}\left(\mathrm{shift}\phantom{\rule{2.pt}{0ex}}\mathrm{the}sin\mathrm{graph}\phantom{\rule{2.pt}{0ex}}\mathrm{to}\phantom{\rule{2.pt}{0ex}}\mathrm{the}\phantom{\rule{2.pt}{0ex}}\mathrm{left}\right)\hfill \end{array}$
## Graph of $tan\theta$
Complete the following table, using your calculator to calculate the values correct to 1 decimal place. Then plot the values with $tan\theta$ on the $y$ -axis and $\theta$ on the $x$ -axis.
$\theta$ 0 ${}^{\circ }$ 30 ${}^{\circ }$ 60 ${}^{\circ }$ 90 ${}^{\circ }$ 120 ${}^{\circ }$ 150 ${}^{\circ }$ $tan\theta$ $\theta$ 180 ${}^{\circ }$ 210 ${}^{\circ }$ 240 ${}^{\circ }$ 270 ${}^{\circ }$ 300 ${}^{\circ }$ 330 ${}^{\circ }$ 360 ${}^{\circ }$ $tan\theta$
#### Questions & Answers
where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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# How do you simplify ((x^-3)^4x^4)/(2x^-3) and write it using only positive exponents?
Feb 5, 2017
See the entire simplification process below:
#### Explanation:
First, use the rule of exponents to simplify the term in parenthesis in the numerator: ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
$\frac{{\left({x}^{\textcolor{red}{- 3}}\right)}^{\textcolor{b l u e}{4}} {x}^{4}}{2 {x}^{-} 3} \to \frac{{x}^{\textcolor{red}{- 3} \times \textcolor{b l u e}{4}} {x}^{4}}{2 {x}^{-} 3} \to \frac{{x}^{-} 12 {x}^{4}}{2 {x}^{-} 3}$
Next, use this rule for exponents to simplify the numerator:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
$\frac{{x}^{\textcolor{red}{- 12}} {x}^{\textcolor{b l u e}{4}}}{2 {x}^{-} 3} \to {x}^{\textcolor{red}{- 12} + \textcolor{b l u e}{4}} / \left(2 {x}^{-} 3\right) \to {x}^{-} \frac{8}{2 {x}^{-} 3}$
Now, use this rule of exponents to complete the simplification using only positive exponents: ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$
${x}^{\textcolor{red}{- 8}} / \left(2 {x}^{\textcolor{b l u e}{- 3}}\right) \to \frac{1}{2 {x}^{\textcolor{b l u e}{- 3} - \textcolor{red}{- 8}}} \to \frac{1}{2 {x}^{\textcolor{b l u e}{- 3} + \textcolor{red}{8}}} \to \frac{1}{2 {x}^{5}}$
Feb 5, 2017
$\frac{1}{2 {x}^{5}}$
#### Explanation:
$\frac{{\left({x}^{-} 3\right)}^{4} {x}^{4}}{2 {x}^{-} 3} = \frac{{x}^{-} 12 \cdot {x}^{4}}{2 {x}^{-} 3} = {x}^{-} \frac{8}{2 {x}^{-} 3} = \frac{{x}^{-} 5 \cdot \cancel{{x}^{-} 3}}{2 \cdot \cancel{{x}^{-} 3}} = {x}^{-} \frac{5}{2} = \frac{1}{2 {x}^{5}}$ [Ans] |
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