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# 44 out of 60 is what percentage?
Welcome to Warren Institute! In the fascinating world of Mathematics education, we strive to unravel the mysteries behind numbers and percentages. Today, we delve into the question: What percentage is 44 out of 60? Join us as we explore the methods and concepts needed to understand and calculate percentages with precision and clarity. Whether you're a student, educator, or simply a math enthusiast, this article will provide valuable insights into this fundamental concept. Let's embark on this mathematical journey together!
Sure, here are the
## Understanding the Percentage Calculation
To understand what percentage 44 out of 60 is, we need to grasp the concept of percentages in mathematics. A percentage is a way of expressing a number as a fraction of 100. In this case, we are determining what portion of 60 44 represents.
## Calculating the Percentage
To calculate the percentage, we can use the formula: (part/whole) x 100. In this scenario, 44 is the part and 60 is the whole. Therefore, the calculation would be: (44/60) x 100. By performing this calculation, we can find the percentage that 44 represents out of 60.
## Interpreting the Result
After performing the calculation, we can interpret the result as the percentage that 44 represents out of 60. This will provide insight into how much 44 is in relation to the total quantity of 60. It's important to understand the context in which the percentage is being used to properly interpret its significance.
## Real-Life Applications
Understanding percentages is essential in various real-life scenarios, such as calculating discounts, understanding proportions, and interpreting statistics. By mastering the concept of percentages, students can apply their knowledge to everyday situations, making mathematics education practical and relevant to their lives.
### How can I calculate the percentage of 44 out of 60 in a Mathematics education setting?
You can calculate the percentage of 44 out of 60 by dividing 44 by 60 and then multiplying the result by 100. The formula is: (44 / 60) * 100 which equals 73.33%.
### What strategies can be used to teach students how to find the percentage of 44 out of 60 in Mathematics education?
One strategy to teach students how to find the percentage of 44 out of 60 in Mathematics education is to use the formula: percentage = (part/whole) x 100. This can be followed by providing examples and practice exercises to reinforce the concept.
### In Mathematics education, why is it important for students to understand how to determine the percentage of 44 out of 60?
In Mathematics education, it is important for students to understand how to determine the percentage of 44 out of 60 because it helps develop numerical fluency and problem-solving skills.
### How does understanding how to calculate the percentage of 44 out of 60 contribute to a student's mathematical literacy in education?
Understanding how to calculate the percentage of 44 out of 60 contributes to a student's mathematical literacy in education by developing their numerical reasoning and problem-solving skills.
### What real-world applications can be used to demonstrate the relevance of finding the percentage of 44 out of 60 in Mathematics education?
Calculating the percentage of 44 out of 60 can be demonstrated in financial literacy when calculating discounts or sales prices, in data analysis when interpreting survey results, and in science and engineering when determining proportions or concentrations.
In conclusion, understanding percentages is crucial in mathematics education. Knowing that 44 out of 60 represents approximately 73.33% is an important skill for students to develop. This knowledge not only enhances their mathematical proficiency but also helps in real-world applications. Mastering percentages is a fundamental aspect of mathematical literacy that empowers students to analyze and interpret data effectively.
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# What is 75 percent of 65?
WRITTEN BY: supportmymoto.com STAFF
#### Answer for What’s 75 % of 65:
75 % * 65 =
(75:100)* 65 =
(75* 65):100 =
4875:100 = 48.75
Now we now have: 75 % of 65 = 48.75
Query: What’s 75 % of 65?
Step 1: Our output worth is 65.
Step 2: We symbolize the unknown worth with {x}.
Step 3: From step 1 above,{ 65}={100%}.
Step 4: Equally, {x}={75%}.
Step 5: This leads to a pair of easy equations:
{ 65}={100%}(1).
{x}={75%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand facet) of each
equations have the identical unit (%); we now have
frac{ 65}{x}=frac{100%}{75%}
Step 7: Once more, the reciprocal of each side offers
frac{x}{ 65}=frac{75}{100}
Rightarrow{x} = {48.75}
Subsequently, {75%} of { 65} is {48.75}
#### Answer for What’s 65 % of 75:
65 % *75 =
( 65:100)*75 =
( 65*75):100 =
4875:100 = 48.75
Now we now have: 65 % of 75 = 48.75
Query: What’s 65 % of 75?
Step 1: Our output worth is 75.
Step 2: We symbolize the unknown worth with {x}.
Step 3: From step 1 above,{75}={100%}.
Step 4: Equally, {x}={ 65%}.
Step 5: This leads to a pair of easy equations:
{75}={100%}(1).
{x}={ 65%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand facet) of each
equations have the identical unit (%); we now have
frac{75}{x}=frac{100%}{ 65%}
Step 7: Once more, the reciprocal of each side offers
frac{x}{75}=frac{ 65}{100}
Rightarrow{x} = {48.75}
Subsequently, { 65%} of {75} is {48.75}
NOTE : Please do not copy - https://supportmymoto.com
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1. Mixed Questions: Help Needed
3 5/8 ½ =
1Ύ divided by 2 =
Write as a ratio in simplest form
a) 24:40 (would it be 3:5)
b) 220 cm to 1 m
Leah and Rachael both drive the same car, and use it basically at the same rate.
If Leah drives for 8 hours at 60 km/h to visit her best friend, the trip costs $63 in fuel. Rachael however drives to see her father but has to drive a further 256 km. How much will the trip cost in fuel? Three People contribute money to a raffle -$15, $20 and$25. If the winnings are distributed in same ratio as the contributions how much would each person get, if they won \$270?
2. Originally Posted by Quester
3 5/8 ½ =
1Ύ divided by 2 =
Write as a ratio in simplest form
a) 24:40 (would it be 3:5)
b) 220 cm to 1 m
For the first one convert the mixed numbers into improper fractions:
$3~\frac{5}{8} = \frac{29}{8}$
so
$3~\frac{5}{8} - \frac{1}{2} = \frac{29}{8} - \frac{1}{2}$
$= \frac{29}{8} - \frac{4}{8} = \frac{25}{8}$
If you really must have the answer as a mixed number, then
$\frac{25}{8} = 3 + \frac{1}{8} = 3~\frac{1}{8}$
The same thing goes for the other one.
For the ratios:
a) 24:40
$\frac{24}{40} = \frac{3 \cdot 8}{5 \cdot 8} = \frac{3}{5}$
and this is 3:5.
Again, the other one works the same way.
-Dan
3. Simplify (expand and collect like terms)
a)4q + 12q 3
b)4 (3a + 1) + 2a
c)Subtract 3x 5 from 8x 18
Factorise Following
a)3a + 15 ( is it: 3a + 15 = 3 x a + 3 x 5 = 3(a+5)
b)4 x xy + 3x2
Solve Following
a)2n + 3 = 17
b)3x 4/2 6 = 4
c)5(x 2) = 4 (x + 9)
But this particually, I cant get:
a) The adjacent sides of a rectangle are (3x 8) and 6 cm.
Given that the area of the rectangle is 96 cm^2, find the length of the rectangle and the value of x.
c)The Three sides of a triangle are 1/1 y + 2 and y = 4. The perimeter is 87 cm. What are the lengths of the three sides?
4. oh sorry when i wrote 1/2 i mean the fraction 1/2
5. Originally Posted by Quester
Simplify (expand and collect like terms)
a)4q + 12q – 3
b)4 (3a + 1) + 2a
c)Subtract 3x – 5 from 8x – 18
Factorise Following
a)3a + 15 ( is it: 3a + 15 = 3 x a + 3 x 5 = 3(a+5)
b)4 x – xy + 3x2
Solve Following
a)2n + 3 = 17
b)3x – 4/2 – 6 = 4
c)5(x – 2) = 4 (x + 9)
But this particually, I can’t get:
a) The adjacent sides of a rectangle are (3x – 8) and 6 cm.
Given that the area of the rectangle is 96 cm^2, find the length of the rectangle and the value of x.
c)The Three sides of a triangle are 1/1 y + 2 and y = 4. The perimeter is 87 cm. What are the lengths of the three sides?
1. Do yourself and do us a favor and start a new thread if you have new questions.
2. to #1.c.):
Translate the sentence into a mathematical operation:
$(8x-18)-(3x-5)~\buildrel {expand} \over \longrightarrow~8x-18-3x+5 = 5x-13$
to #2.a.): Yes
to 2.b.):
$4x - xy + 3x^2= x(4-y+3x)$
to #3.c.):
$5(x-2) = 4 (x + 9)~\iff~ 5x-10=4x+36~\iff~x=46$ ....... The last step is: add (-4x+10) on both sides of the equation - you only have to know why this is necessary!
to #4.a.):
You are suposed to know that the area of a rectangle is calculated by:
$area = length\ \cdot \ width$ ....... with
$length = 3x-8$ ....... and ....... $width = 6$
So you have to solve for x:
$(3x-8) \cdot 6 = 96 ~\iff~ 18x - 48 = 96~\iff~ 18x = 144 ~\iff~ x = \frac{144}{18} = 8$
Therefore the rectangle has the dimensions: $l = 16$ and $w = 6$
6. Originally Posted by earboth
So you have to solve for x:
$(3x-8) \cdot 6 = 96 ~\iff~ 18x - 48 = 96~\iff~ 18x = 144 ~\iff~ x = \frac{144}{18} = 8$
Therefore the rectangle has the dimensions: $l = 16$ and $w = 6$
how did you get the answer for (3x-8).6 = 96 what does the dot stand for?
7. Originally Posted by Quester
how did you get the answer for (3x-8).6 = 96 what does the dot stand for?
Multiplication
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### Free Educational Resources
4 Tutorials that teach Introduction to Sampling Distribution
# Introduction to Sampling Distribution
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##### Description:
This lesson will explain the sampling distribution of the sample means.
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Tutorial
This tutorial is going to introduce you to what a sampling distribution is. You’ll learn about:
1. Sampling Distribution
## 1. Sampling Distribution
Sampling Distribution of Sample Means
A distribution that shows the means from all possible samples of a given size.
Consider the spinner shown here:
Spin it four times to obtain an average. Imagine that you spin it for one set of four. You get a 2 the first time, a 4 the second time, a 3 the third time, and a 1 the fourth time. The mean is the average of 2, 4, 3, and 1: 2+4+3+1 = 10, 10/4 = 2.5. So your first mean is 2.5.
But your mean won't be 2.5 every time:
Sample Numbers Average 1 2, 4, 3, 1 2.5 2 1, 4, 3, 1 2.25 3 4, 2, 4, 4 3.5 4 2, 2, 3, 1 2 5 3, 1, 1, 1 1.5 6 1, 1, 1, 2 1.25
So how can we represent all these distributions?
1. First, these sample means and graph them. Draw out an axis, for this one it should go from 0 to 4 because this set can’t average anything higher than four or lower than a one.
2. Take the average value, the mean of 2.5, and put a dot at 2.5 on the x-axis, much like a dot plot.
3. Do the same for all the means.
You can keep doing this over and over and over again. Ideally, you would do this hundreds or thousands of times, to show the distribution of all possible samples that could be taken of size four.
4. Once you’ve enumerated every possible sample of size 4 from this spinner, then the sampling distribution looks like this:
Notice the lowest number you can get is one and the highest number you can get is four. On the far right of the graph is the point that represents a spin of four fours. On the far left is the point that represents a spin of four ones.
In the middle, notice that there are more dots because there were more ones on the spinner than there were fours. And because there were more ones, the average is pulled down a bit. The most frequent average is 2.25, not 2.5, which would be the exact middle between 1 and 4. So this distribution is skewed slightly to the right.
A sampling distribution is the distribution of all possible means that you could have for a given size. So in a sampling distribution where you graph all the possible means for samples of size 4, you take the sample, calculate the mean, and plot it with a dot. Then you take another sample, calculate the mean again, and plot that. So it can be a long and tedious process! In large populations, the sampling distribution consists of lots and lots and lots of points.
Thank you and good luck!
Source: THIS WORK IS ADAPTED FROM SOPHIA AUTHOR JONATHAN OSTERS
Terms to Know
Sampling Distribution of Sample Means
A distribution that shows the means from all possible samples of a given size.
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# 7.6: Calculating Centers of Mass and Moments of Inertia
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We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density.
## Center of Mass in Two Dimensions
The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. Figure $$\PageIndex{1}$$ shows a point $$P$$ as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.
To find the coordinates of the center of mass $$P(\bar{x},\bar{y})$$ of a lamina, we need to find the moment $$M_x$$ of the lamina about the $$x$$-axis and the moment $$M_y$$ about the $$y$$-axis. We also need to find the mass $$m$$ of the lamina. Then
$\bar{x} = \dfrac{M_y}{m}$
and
$\bar{y} = \dfrac{M_x}{m}.$
Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.
If we allow a constant density function, then $$\bar{x} = \dfrac{M_y}{m}$$ and $$\bar{y} = \dfrac{M_x}{m}$$ give the centroid of the lamina.
Suppose that the lamina occupies a region $$R$$ in the $$xy$$-plane and let $$\rho (x,y)$$ be its density (in units of mass per unit area) at any point $$(x,y)$$. Hence,
$\rho(x,y) = \lim_{\Delta A \rightarrow 0} \dfrac{\Delta m}{\Delta A}$
where $$\Delta m$$ and $$\Delta A$$ are the mass and area of a small rectangle containing the point $$(x,y)$$ and the limit is taken as the dimensions of the rectangle go to $$0$$ (see the following figure).
Just as before, we divide the region $$R$$ into tiny rectangles $$R_{ij}$$ with area $$\Delta A$$ and choose $$(x_{ij}^*, y_{ij}^*)$$ as sample points. Then the mass $$m_{ij}$$ of each $$R_{ij}$$ is equal to $$\rho (x_{ij}^*, y_{ij}^*) \Delta A$$ (Figure $$\PageIndex{2}$$). Let $$k$$ and $$l$$ be the number of subintervals in $$x$$ and $$y$$ respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.
Hence, the mass of the lamina is
$m =\lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R \rho(x,y) dA.$
Let’s see an example now of finding the total mass of a triangular lamina.
Example $$\PageIndex{1}$$: Finding the Total Mass of a Lamina
Consider a triangular lamina $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density $$\rho(x,y) = xy \, kg/m^2$$. Find the total mass.
Solution
A sketch of the region $$R$$ is always helpful, as shown in the following figure.
Using the expression developed for mass, we see that
$m = \iint_R \, dm = \iint_R \rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} xy \, dy \, dx = \int_{x=0}^{x=3} \left[ \left. x \dfrac{y^2}{2} \right|_{y=0}^{y=3} \right] \, dx = \int_{x=0}^{x=3} \dfrac{1}{2} x (3 - x)^2 dx = \left.\left[ \dfrac{9x^2}{4} - x^3 + \dfrac{x^4}{8} \right]\right|_{x=0}^{x=3} = \dfrac{27}{8}.$
The computation is straightforward, giving the answer $$m = \dfrac{27}{8} \, kg$$.
Exercise $$\PageIndex{1}$$
Consider the same region $$R$$ as in the previous example, and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the total mass.
$$\dfrac{9\pi}{8} \, kg$$
Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment $$M_z$$ about the $$x$$-axis for $$R$$ is the limit of the sums of moments of the regions $$R_{ij}$$ about the $$x$$-axis. Hence
$M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R y\rho (x,y) dA$
Similarly, the moment $$M_y$$ about the $$y$$-axis for $$R$$ is the limit of the sums of moments of the regions $$R_{ij}$$ about the $$y$$-axis. Hence
$M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R x\rho (x,y) dA$
Example $$\PageIndex{2}$$: Finding Moments
Consider the same triangular lamina $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density $$\rho (x,y) = xy$$. Find the moments $$M_x$$ and $$M_y$$.
Solution
Use double integrals for each moment and compute their values:
$M_x = \iint_R y\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x y^2 \, dy \, dx = \dfrac{81}{20},$
$M_y = \iint_R x\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x^2 y \, dy \, dx = \dfrac{81}{20},$
The computation is quite straightforward.
Exercise $$\PageIndex{2}$$
Consider the same lamina $$R$$ as above and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the moments $$M_x$$ and $$M_y$$.
$$M_x = \dfrac{81\pi}{64}$$ and $$M_y = \dfrac{81\pi}{64}$$
Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by $$\bar{x}$$ and the y-coordinate by $$\bar{y}$$. Specifically,
$\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA}$
and
$\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA}$
Example $$\PageIndex{3}$$: center of mass
Again consider the same triangular region $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density function $$\rho (x,y) = xy$$. Find the center of mass.
Solution
Using the formulas we developed, we have
$\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5},$
$\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5}.$
Therefore, the center of mass is the point $$\left(\dfrac{6}{5},\dfrac{6}{5}\right).$$
Analysis
If we choose the density $$\rho(x,y)$$ instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,
$x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1,$
$y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1.$
Notice that the center of mass $$\left(\dfrac{6}{5},\dfrac{6}{5}\right)$$ is not exactly the same as the centroid $$(1,1)$$ of the triangular region. This is due to the variable density of $$R$$ If the density is constant, then we just use $$\rho(x,y) = c$$ (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.
Exercise $$\PageIndex{3}$$
Again use the same region $$R$$ as above and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the center of mass.
$$\bar{x} = \dfrac{M_y}{m} = \dfrac{81\pi/64}{9\pi/8} = \dfrac{9}{8}$$ and $$\bar{y} = \dfrac{M_x}{m} = \dfrac{81\pi}{9\pi/8} = \dfrac{0}{8}$$.
Once again, based on the comments at the end of Example $$\PageIndex{3}$$, we have expressions for the centroid of a region on the plane:
$x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} \, \text{and} \, y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA}.$
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# The Ultimate Guide to the Fundamental Theorem of Calculus in AP® Calculus
Fundamental theorem of calculus (Part 1) | AP Calculus AB | Khan Academy
Fundamental theorem of calculus (Part 1) | AP Calculus AB | Khan Academy
## Introduction
The Fundamental Theorem of Calculus brings together two essential concepts in calculus: differentiation and integration. There are two parts to the Fundamental Theorem: the first justifies the procedure for evaluating definite integrals, and the second establishes the relationship between differentiation and integration. Here, we will focus on the first statement, which is referred to as the First Fundamental Theorem of Calculus.
This theorem establishes the procedure for computing a definite integral. Recall that a definite integral is an integral where you are given the limits of integration. That is, you are integrating over an interval whose endpoints you use to evaluate the integral.
## Indefinite vs. Definite Integrals
An indefinite integral is an integral without limits of integration; for example,
\int { { x }^{ 2 } } dx
When integrating this function, we are looking for a curve whose derivative is {x}^{2}. There are infinitely many curves with that derivative, including
\dfrac { 1 }{ 3 } { x }^{ 3 }
\dfrac { 1 }{ 3 } { x }^{ 3 }+1
\dfrac { 1 }{ 3 } { x }^{ 3 }+8
\dfrac { 1 }{ 3 } { x }^{ 3 }-3
\dfrac { 1 }{ 3 } { x }^{ 3 }-\sqrt { 5 }
While there are infinitely many options, we note that they differ only by a constant. As a result, when we integrate a function of this type, we add an arbitrary constant, C, to our solution, as follows:
\int { { x }^{ 2 } } =\dfrac { 1 }{ 3 } { x }^{ 3 }+C
In contrast, a definite integral is one in which there are limits of integration. As an example, consider \int _{ 1 }^{ 4 }{ { x }^{ 2 } dx } . Here, we are being asked to integrate the same function as before, but we are doing so over the interval 1\le x\le 4. To compute this integral, we rely on the First Fundamental Theorem of Calculus.
## The First Fundamental Theorem of Calculus
The theorem states that if a function f(x) has an indefinite integral of F(x)+C and is continuous over an interval [a,b], then \int _{ a }^{ b }{ f(x)dx=F(b)-F(a) }.
Let’s go back to our prior example:
\int _{ 1 }^{ 4 }{ { x }^{ 2 } dx }
First, we integrate { x }^{ 2 } to obtain \dfrac { 1 }{ 3 } { x }^{ 3 } . When evaluating definite integrals, we leave out the constant (so there is no +C here).
Next, we plug the limits of integration into the integral we just found. We first evaluate the upper limit of integration (F(b)), then subtract the value of the function at the lower limit of integration (F(a)). Putting this all together, we obtain the following expression:
\int _{ 1 }^{ 4 }{ { x }^{ 2 } dx } =\dfrac{x^3}{3} \int_{1}^{4}=\dfrac { 1 }{ 3 } { (4) }^{ 3 }-\dfrac { 1 }{ 3 } { (1) }^{ 3 }=\dfrac { 1 }{ 3 } (64-1)=\dfrac { 1 }{ 3 } (63)=21
## Why does it Work?
Let G(x)=\int _{ a }^{ x }{ f(t)dt } .
We can show that F(x) is an antiderivative for f on some interval [a,b]. This result is actually because of the Second Fundamental Theorem of Calculus. We can show that the second fundamental theorem works without using the first, but interestingly, proving the first part is easiest if we start with the second part.
The two anti-derivatives are continuous over the given interval. Also, { G }^{ ‘ }(x)={ F }^{ ‘ }(x)=f(x) for all x in the interval [a,b] . In other words, since G(x) and F(x) are anti-derivatives of f(x), their derivative is f(x).
You might remember that a function has multiple anti-derivatives and that these differ by a constant. This is why you add +C at the end of your answers when you evaluate an indefinite integral (one without limits of integration). As a result, we know there exists another anti-derivative of f on some interval [a,b]. Let us call this anti-derivative F(x), where G(x)=F(x)+C.
Recall that G(a)=\int _{ a }^{ a }{ f(t)dt=0 }. That is, the integral at a particular point within the interval is equal to zero. From this statement, we arrive at the following:
0=G(a)=F(a)+C
C=-F(a)
We noted before that G(x)=F(x)+C. Since we just found a value for C, we have
G(x)=F(b)-F(a).
If we let x=b, we arrive at G(b)=F(b)-F(a). Using the relationship between anti-derivatives and derivatives, we conclude that
\int _{ a }^{ b }{ f(t)dt=F(b)-F(a) }.
Example 1:
Find \int _{ 0 }^{ \pi /2 }{ cos(x)dx }.
In this straightforward example, we are asked to evaluate a definite integral. Recall that the derivative of sin { x } is cos { x }, so the anti-derivative of cos { x } is sin { x } + C. However, as this is a definite integral (i.e. we are given the limits of integration), we need not add the constant (C). We evaluate the integral by using the first fundamental theorem as follows:
\int _{ 0 }^{ \pi /2 }{ cos { x }dx } =sin{x} \int_{0}^{\pi/2}= sin { (\pi /2) } – sin { (0) } =1-0=1
Example 2: (1988 AP® Calculus Exam Multiple Choice Problem 10)
If \int _{ 0 }^{ k }{ (2kx-{ x }^{ 2 })dx=18 } what is the value of k?
Here, we have removed the multiple choice answers, but the question is identical to one that appeared on the AP® exam in all other respects.
We are asked to find k, which involves evaluating the given definite integral. Invoking the first fundamental theorem of calculus, we know that we must find the difference between the values of the anti-derivative at k and at 0.
Keep in mind that k is a constant (a number), as this will help us with the anti-derivative. Specifically, when finding the anti-derivative of 2kx, we treat k as we might the number 2. That is, we leave it alone! It is not a variable and so there should be no \dfrac { { k }^{ 2 } }{ 2 } anywhere in your work!
We find the anti-derivative as follows:
\int _{ 0 }^{ k }{ (2kx-{ x }^{ 2 })dx } =\dfrac{2kx^2}{2}-\dfrac{\pi^3}{3} \int_{0}^{k}=kx^2-\dfrac{x^3}{3} \int_{0}^{k}=\dfrac{3kx^2}{3}-\dfrac{x^3}{3} \int_{0}^{k}=\dfrac{3kx^2-x^3}{3} \int_{0}^{k}
Now, we apply the Fundamental Theorem of Calculus. This means that we evaluate the definite integral by using the limits of integration (first substituting k for x, then 0 for x) as follows.
\dfrac{3kx^2-x^3}{3} \int_{0}^{k}=\dfrac { 3k \cdot { k }^{ 2 }-{ k }^{ 3 } }{ 3 } -\dfrac { 3k{ (0) }^{ 2 }-{ (0) }^{ 3 } }{ 3 } =\dfrac { 3{ k }^{ 3 }-{ k }^{ 3 } }{ 3 } -0=\dfrac { 2{ k }^{ 3 } }{ 3 }
Recall that
\int _{ 0 }^{ k }{ (2kx-{ x }^{ 2 })dx=18 }. Thus, \dfrac { 2{ k }^{ 3 } }{ 3 } =18 and { k }^{ 3 }=\dfrac{\left(18\cdot3\right)}{2}=27
k=3
Example 3: (1988 AP® Calculus Exam Multiple Choice Problem 13)
If the function f has a continuous derivative on [0,C] then \int _{ 0 }^{ c }{ { f }^{ ‘ }(x)dx=? }
1. f(c)-f(0)
2. | f(c)-f(0) |
3. f(c)
4. f(x)+C
5. { f }^{ ” }(c)-{ f }^{ ” }(0)
This problem is testing us on our understanding of the Fundamental Theorem of Calculus. The anti-derivative of the derivative of f is f itself. Next, we evaluate the definite integral by using the limits of integration. The answer is (A).
What does \int _{ a }^{ b }{ f(t)dt=F(b)-F(a) } actually mean?
We can understand the integral \int _{ a }^{ b }{ f(t) } dt in a number of ways. One of these is to think of the integral as the total change in the curve over the interval from a to b. The total change is the definite integral of the rate of change of a function. Below are some questions from actual AP® Calculus exams that require the use of the First Fundamental Theorem of Calculus and the fact that the definite integral gives the total change of the function over the given integral.
Example: (2007 AP® Calculus AB Exam Question 2 part b)
We will focus on part b and note that we are asked for the total distance that the particle traveled between t=0 and t=3. We can integrate the given velocity function to arrive at the position function. However, the distance traveled is irrespective of direction, so we must remember to take the absolute value of the velocity first.
The total distance traveled can be found using a definite integral. Namely, \int_{0}^{3} sin(t^2)dt=1.702
Example 5: (2007 AP® Calculus AB Form B Exam Question 2 part a)
We are asked for the total water entering the tank between t=0 and t=7. The rate of water entering the tank is f(t) and the total amount of water entering the tank can be found using a definite integral:
\int _{ 0 }^{ 7 }{ f(t) } dt=8264 gallons.
## The Fundamental Theorem of Calculus
As you can see, the fundamental theorem of calculus establishes a procedure for calculating a definite integral. Now, this theorem on its own is already useful, but it also supplies us with the fact that this definite integral is equivalent to the total change over a particular interval, which comes in handy in a number of situations as we saw in the last two problems above.
### Looking for AP® Calculus practice?
Kickstart your AP® Calculus prep with Albert. Start your AP® exam prep today.
You are watching: The Ultimate Guide to the Fundamental Theorem of Calculus in AP® Calculus. Info created by THVinhTuy selection and synthesis along with other related topics.
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# Scientific notation is a way of expressing really big numbers or really small numbers. It is most often used in “scientific” calculations where the.
## Presentation on theme: " Scientific notation is a way of expressing really big numbers or really small numbers. It is most often used in “scientific” calculations where the."— Presentation transcript:
Scientific notation is a way of expressing really big numbers or really small numbers. It is most often used in “scientific” calculations where the analysis must be very precise.
For very large and very small numbers, these numbers can be converted into scientific notation to express them in a more concise form. Numbers expressed in scientific notation makes it easier to solve problems with big and small numbers.
A number between 1 and 10 A power of 10 N x 10 x Are the following in scientific notation?
Place the decimal point so that there is one non-zero digit to the left of the decimal point. Count the number of decimal places the decimal point has “moved” from the original number. This will be the exponent on the 10.
If the original number was less than 1, then the exponent is negative. If the original number was greater than 1, then the exponent is positive.
See if the original number is greater than or less than one. › If the number is greater than one, the exponent will be positive. 348943 = 3.489 x 10 5 › If the number is less than one, the exponent will be negative..0000000672 = 6.72 x 10 -8
Positive Exponent: 2.35 x 10 8 Negative Exponent: 3.97 x 10 -7
93,000,000
Move decimal left Leave only one number in front of decimal
Write number without zeros
Count how many places you moved decimal Make that your power of ten
The power of ten is 7 because the decimal moved 7 places.
93,000,000 --- Standard Form 9.3 x 10 7 --- Scientific Notation
1) 98,500,000 = 9.85 x 10 ? 2) 64,100,000,000 = 6.41 x 10 ? 3) 279,000,000 = 2.79 x 10 ? 4) 4,200,000 = 4.2 x 10 ? Write in scientific notation. Decide the power of ten. 9.85 x 10 7 6.41 x 10 10 2.79 x 10 8 4.2 x 10 6
1) 734,000,000 = ______ x 10 8 2) 870,000,000,000 = ______x 10 11 3) 90,000,000,000 = _____ x 10 10 On these, decide where the decimal will be moved. 1)7.34 x 10 8 2) 8.7 x 10 11 3) 9 x 10 10
1) 50,000 2) 7,200,000 3) 802,000,000,000 Write in scientific notation. 1) 5 x 10 4 2) 7.2 x 10 6 3) 8.02 x 10 11
Given: 289,800,000 Use: 2.898 (moved 8 places) Answer: 2.898 x 10 8
Given: 0.000567 Use: 5.67 (moved 4 places) Answer: 5.67 x 10 -4
#1 #2
1) 9872432 2).0000345 3).08376 4) 5673 9.872432 x 10 6 3.45 x 10 -5 8.376 x 10 2 5.673 x 10 3
Move the decimal point to the right for positive exponent 10. Move the decimal point to the left for negative exponent 10. (Use zeros to fill in places.)
If an exponent is positive, the number gets larger, so move the decimal to the right. If an exponent is negative, the number gets smaller, so move the decimal to the left.
Move the decimal to the right 3.4 x 10 5 in scientific notation 340,000 in standard form 3.40000 --- move the decimal
Given: 5.093 x 10 6 Answer: 5,093,000 (moved 6 places to the right)
Given: 1.976 x 10 -4 Answer: 0.0001976 (moved 4 places to the left)
6.27 x 10 6 9.01 x 10 4 6,270,000 90,100
1) 9.678 x 10 4 2) 7.4521 x 10 -3 3) 8.513904567 x 10 7 4) 4.09748 x 10 -5 96780.0074521 85139045.67.0000409748
#3 #4
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# A quantity which expresses a part of the whole is called a or an(a) fraction (b) prime number (c) integer (d) None of these
Last updated date: 17th Jun 2024
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Hint:
Here, we need to complete the statement by choosing the correct option. We will use the definitions of a fraction, prime numbers, and an integer, with examples to check which of the given options is the correct answer.
Complete step by step solution:
We will check the definitions of each of the options to find the correct answers.
A fraction is a number which shows or represents a part of a group. It can be written as $\dfrac{a}{b}$, where $a$ is called the numerator and $b$ is called the denominator. The group is divided into $b$ equal parts. The fraction $\dfrac{a}{b}$ shows $a$ out of $b$ equal parts of the group.
For example: 105 out of 200 children can be represented as the fraction $\dfrac{{105}}{{200}}$.
Thus, option (a) is the correct answer.
We will also check the other options.
A prime number is a number which is divisible only by 2 numbers, that is by 1 and the number itself.
For example: 5 is a prime number because it is divisible by only 2 numbers only, that is 1 and 5.
Therefore, a prime number does not represent a part of a whole.
Thus, option (b) is incorrect.
An integer is a rational number that is not expressed as a fraction.
For example: 1, $- 1$, 3, $- 7$, are integers.
Integers can be positive or negative. Negative integers are the numbers $- 1$, $- 5$, $- 92$, $- 41$, etc.
Integers do not represent a part of a whole.
Thus, option (c) is incorrect.
Note:
Fractions can be either proper, improper, or mixed fraction. A proper fraction is a fraction whose denominator is greater than its numerator. An improper fraction is a fraction whose denominator is smaller than its numerator. A mixed fraction is a fraction in the form $a\dfrac{b}{c}$. Every mixed fraction $a\dfrac{b}{c}$ can be converted to an improper fraction $\dfrac{{c \times a + b}}{c}$.
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# Multinomial Distributions
## Probability distributions for experiments with more than 2 outcomes : P= n!/ n1!n2!n3!...nk! x (p1^n1 x p2^n2 x p3^n3...pk^nk
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Multinomial Distributions
You're spinning a spinner that has three equal sections of red, green, and blue. If you spin the spinner 10 times, what is the probability that you will land on red 4 times, green 3 times, and blue 3 times? How would you calculate and express this probability?
### Watch This
First watch this video to learn about multinomial distributions.
Then watch this video to see some examples.
### Guidance
In later Concepts, we will learn more about multinomial distributions. However, we are now talking about probability distributions, and as such, we should at least see how the problems change for these distributions. We will briefly introduce the concept and its formula here, and then we will get into more detail in later Concepts. Let’s start with a problem involving a multinomial distribution.
#### Example A
You are given a bag of marbles. Inside the bag are 5 red marbles, 4 white marbles, and 3 blue marbles. Calculate the probability that with 6 trials, you choose 3 marbles that are red, 1 marble that is white, and 2 marbles that are blue, replacing each marble after it is chosen.
Notice that this is not a binomial experiment, since there are more than 2 possible outcomes. For binomial experiments, $k = 2$ (2 outcomes). Therefore, we use the binomial experiment formula for problems involving heads or tails, yes or no, or success or failure. In this problem, there are 3 possible outcomes: red, white, or blue. This type of experiment produces what we call a multinomial distribution . In order to solve this problem, we need to use one more formula:
$P & = \frac{n!}{n_1!n_2!n_3!\ldots n_k!} \times \left (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}} \right )$
where:
$n$ is the number of trials.
$p$ is the probability for each possible outcome.
$k$ is the number of possible outcomes.
Notice that in this example, $k$ equals 3. If we had only red marbles and white marbles, $k$ would be equal to 2, and we would have a binomial distribution.
The probability of choosing 3 red marbles, 1 white marble, and 2 blue marbles in exactly 6 picks is calculated as follows:
$n & = 6 \ (6 \ \text{picks})\\p_1 & = \frac{5}{12} = 0.416 \ (\text{probability of choosing a red marble})\\p_2 & = \frac{4}{12} = 0.333 \ (\text{probability of choosing a white marble})\\p_3 & = \frac{3}{12} = 0.25 \ (\text{probability of choosing a blue marble})\\n_1 & = 3 \ (3 \ \text{red marbles chosen})\\n_2 & = 1 \ (1 \ \text{white marble chosen})\\n_3 & = 2 \ (2 \ \text{blue marbles chosen})\\k & = 3 \ (3 \ \text{possibilities})\\P & = \frac{n!}{n_1!n_2!n_3! \ldots n_k!} \times (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}})\\P & = \frac{6!}{3!1!2!} \times (0.416^3 \times 0.333^1 \times 0.25^2)\\P & = 60 \times 0.0720\times 0.333\times 0.0625\\P & = 0.0899$
Therefore, the probability of choosing 3 red marbles, 1 white marble, and 2 blue marbles is 8.99%.
#### Example B
You are randomly drawing cards from an ordinary deck of cards. Every time you pick one, you place it back in the deck. You do this 5 times. What is the probability of drawing 1 heart, 1 spade, 1 club, and 2 diamonds?
$n & = 5 \ (5 \ \text{trials})\\p_1 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a heart})\\p_2 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a spade})\\p_3 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a club})\\p_4 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a diamond})\\n_1 & = 1 \ (1 \ \text{heart})\\n_2 & = 1 \ (1 \ \text{spade})\\n_3 & = 1 \ (1 \ \text{club})\\n_4 & = 2 \ (2 \ \text{diamonds})\\k & = 4 \ (\text{4 possibilities})\\P & = \frac{n!}{n_1!n_2!n_3! \ldots n_k!} \times (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}})\\P & = \frac{5!}{1!1!1!2!} \times (0.25^1 \times 0.25^1 \times 0.25^1 \times 0.25^2)\\P & = 60 \times 0.25 \times 0.25\times 0.25\times 0.0625\\P & = 0.0586$
Therefore, the probability of choosing 1 heart, 1 spade, 1 club, and 2 diamonds is 5.86%.
#### Example C
When spinning a spinner, there is an equal chance of landing on orange, green, yellow, red, and black. Suppose you spin the spinner 12 times. What is the probability of landing of orange 2 times, green 3 times, yellow 2 times, red 3 times, and black 2 times?
$n & = 12 \ (12 \ \text{trials})\\p_1 & = \frac{1}{5} = 0.2 \ (\text{probability of landing on orange})\\p_2 & = \frac{1}{5} = 0.2 \ (\text{probability of landing on green})\\p_3 & = \frac{1}{5} = 0.2 \ (\text{probability of landing on yellow})\\p_4 & = \frac{1}{5} = 0.2 \ (\text{probability of landing on red})\\p_5 & = \frac{1}{5} = 0.2 \ (\text{probability of landing on black})\\n_1 & = 2 \ (2 \ \text{oranges})\\n_2 & = 3 \ (3 \ \text{greens})\\n_3 & = 2 \ (2 \ \text{yellows})\\n_4 & = 3 \ (3 \ \text{reds})\\n_5 & = 2 \ (2 \ \text{blacks})\\k & = 5 \ (\text{5 possibilities})\\P & = \frac{n!}{n_1!n_2!n_3! \ldots n_k!} \times (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}})\\P & = \frac{12!}{2!3!2!3!2!} \times (0.2^2 \times 0.2^3 \times 0.2^2 \times 0.2^3 \times 0.2^2)\\P & = 1,663,200 \times 0.04 \times 0.008 \times 0.04 \times 0.008 \times 0.04\\P & = 0.0068$
Therefore, the probability of landing on orange 2 times, green 3 times, yellow 2 times, red 3 times, and black 2 times is 0.68%.
### Guided Practice
In Austria, 30% of the population has a blood type of O+, 33% has A+, 12% has B+, 6% has AB+, 7% has O-, 8% has A-, 3% has B-, and 1% has AB-. If 15 Austrian citizens are chosen at random, what is the probability that 3 have a blood type of O+, 2 have A+, 3 have B+, 2 have AB+, 1 has O-, 2 have A-, 1 has B-, and 1 has AB-?
$n & = 15 \ (15 \ \text{trials})\\p_1 & = 0.30 \ (\text{probability of O+})\\p_2 & = 0.33 \ (\text{probability of A+})\\p_3 & = 0.12 \ (\text{probability of B+})\\p_4 & = 0.06 \ (\text{probability of AB+})\\p_5 & = 0.07 \ (\text{probability of O-})\\p_6 & = 0.08 \ (\text{probability of A-})\\p_7 & = 0.03 \ (\text{probability of B-})\\p_8 & = 0.01 \ (\text{probability of AB-})\\n_1 & = 3 \ (3 \ \text{O+})\\n_2 & = 2 \ (2 \ \text{A+})\\n_3 & = 3 \ (3 \ \text{B+})\\n_4 & = 2 \ (2 \ \text{AB+})\\n_5 & = 1 \ (1 \ \text{O-})\\n_6 & = 2 \ (2 \ \text{A-})\\n_7 & = 1 \ (1 \ \text{B-})\\n_8 & = 1 \ (1 \ \text{AB-})\\k & = 8 \ (\text{8 possibilities})\\P & = \frac{n!}{n_1!n_2!n_3! \ldots n_k!} \times (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}})\\P & = \frac{15!}{3!2!3!2!1!2!1!1!} \times (0.30^3 \times 0.33^2 \times 0.12^3 \times 0.06^2 \times 0.07^1 \times 0.08^2 \times 0.03^1 \times 0.01^1)\\P & = 4,540,536,000 \times 0.027 \times 0.1089 \times 0.001728 \times 0.0036 \times 0.07 \times 0.0064 \times 0.03 \times 0.01\\P & = 0.000011$
Therefore, if 15 Austrian citizens are chosen at random, the probability that 3 have a blood type of O+, 2 have A+, 3 have B+, 2 have AB+, 1 has O-, 2 have A-, 1 has B-, and 1 has AB- is 0.0011%.
### Practice
1. You are randomly drawing cards from an ordinary deck of cards. Every time you pick one, you place it back in the deck. You do this 9 times. What is the probability of drawing 2 hearts, 2 spades, 2 clubs, and 3 diamonds?
2. In question 1, what is the probability of drawing 3 hearts, 1 spade, 1 club, and 4 diamonds?
3. A telephone survey measured the percentage of students in ABC town who watch channels NBX, FIX, MMA, and TSA, respectively. After the survey, analysis showed that 35% watch channel NBX, 40% watch channel FIX, 10% watch channel MMA, and 15% watch channel TSA. What is the probability that from 8 randomly selected students, 1 will be watching channel NBX, 2 will be watching channel FIX, 3 will be watching channel MMA, and 2 will be watching channel TSA?
4. In question 3, what is the probability that 2 students will be watching channel NBX, 1 will be watching channel FIX, 2 will be watching channel MMA, and 3 will be watching channel TSA?
5. When spinning a spinner, there is an equal chance of landing on pink, blue, purple, white, and brown. Suppose you spin the spinner 10 times. What is the probability of landing of pink 1 time, blue 2 times, purple 2 times, white 2 times, and brown 3 times?
6. In question 5, what is the probability of landing of pink 2 times, blue 2 times, purple 2 times, white 2 times, and brown 2 times?
7. In a recent poll, 23% of the respondents supported candidate A, 19% supported candidate B, 13% supported candidate C, and 45% were undecided. If 7 people are chosen at random, what is the probability that 2 people support candidate A, 2 support candidate B, 1 supports candidate C, and 2 are undecided?
8. In question 7, what is the probability that 2 people support candidate A, 1 supports candidate B, 1 supports candidate C, and 3 are undecided?
9. Suppose you roll a standard die 14 times. What is the probability of rolling a 1 three times, a 2 two times, a 3 one time, a 4 four times, a 5 two times, and a 6 two times?
10. In question 9, what is the probability of rolling a 1 two times, a 2 three times, a 3 three times, a 4 two times, a 5 two times, and a 6 two times?
### Vocabulary Language: English Spanish
multinomial distribution
multinomial distribution
A multinomial distribution is a distribution produced where the number of possible outcomes is greater than 2 and where each outcome has a specific probability.
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These are the top 5 things I learned this year in my Precalculus 11 class. I chose some of the most important topics from this semester that I didn’t know how to do in previous years of math. Below are the 5 topics with a definition of each, a few examples, and the reasons why I chose them as part of this blog post.
Mixed and entire radicals are both forms in which to write a radical (root of a number). Entire radicals are simply the root of any numbers, written normally. Mixed radicals are a simplified form of these, where a number is multiplied by a root. For example:
√3 √17 ∛44
2√7 3√5 2∛4
As you can see, the mixed radicals are usually smaller numbers, as they were originally large numbers that were distributed outside the root sign. To write an entire radical as a mixed radical, follow these steps. √45 – find multiples of the number, with at least one square (or cube) root. Since 9 x 5 is 45, and 9 can be square rooted, these numbers will work. (If no multiples are squares, then creating a mixed radical is not possible). Find the square root of 9, which is 3. Put the 3 outside the root symbol, leaving the remaining number inside, which was 5. It now looks like this: 3√5 To go back to the original entire radical, multiply the outside number by itself, since the symbol is a square root (and not a cube or fourth root). This would be 9. Put it back inside the symbol and multiply by the number there – √9×5 = √45
Here is one more example, using a cube root. ∛16 – find multiples that include a cube root. You could also make a factor tree for 16. Since 8×2 is 16, you can make this into a mixed radical. The cube root of 8 is 2, so put that outside the symbol: 2∛2 To go backwards, multiply the outside number by itself two times, because it is a cube root. This makes 8. Then put the 8 inside and solve: ∛8×2 = ∛16
I chose this topic as one of my top 5 because it was something unique I hadn’t learned about until grade 11. I found this interesting because if there is a question or answer with a root, you can simplify it to make it easier to read. Personally, I find math more straightforward if it is written in lowest or most simple terms, and this is one way to do that.
2: Rationalizing the Denominator
Rationalizing the denominator is helpful when there is a fraction with a root on the bottom. This technique puts the fraction into a simpler form, with an integer on the bottom instead. Here is an example of a non-rationalized fraction:
To rationalize the denominator, just multiply the top and bottom of the fraction by √3. Because you are multiplying the top and bottom by the same value, it is as if you are multiplying by 1. This doesn’t change the ratio of the fraction, it just makes it easier to read.
Here is a final example, where I have continued simplifying the fraction:
This skill was in my top 5 because it is a quick, easy technique that I find super useful. I still remember it today, even after learning it a few months ago. I also think it makes any answer look cleaner and easier to read.
3: Graphing a Parabola
This is something new that I only learned in precalc 11. A parabola is a u-shaped line that curves on a graph. I learned that every quadratic function or equation creates a graph of a parabola, with certain characteristics. Equations can come in standard, vertex, or factored form. Vertex form is the easiest to graph. It looks like this: y = (x – p)². This is an example of a simple equation in vertex form, and it’s resulting parabola. y = x² – 3
The middle point, at the highest or lowest spot, is called the vertex. This is either at the top or bottom of the parabola. A parabola is symmetrical, with the axis of symmetry being in the center. Based on the sign (positive or negative) of the x², the parabola will open upwards or downwards. Here are a few other examples:
For the first graph, the equation would be y = -x². The negative means it opens downward. The vertex, or highest point, is at the coordinates (0,0), and the axis of symmetry is x = 0. The second graph below shows the equation y = -(x – 3)² + 2. The vertex is (3,2), and the axis of symmetry is x = 2. As you can see the -3 turned into +3 when finding the vertex. Make sure you switch to the opposite sign when finding the number for the x-coordinate in vertex form.
Parabolas are a new and interesting way of graphing, making them one of my top 5 things I learned. When I could see the relation between the equation and a graph, it made it much easier for me to read and understand a question because there was a visual.
4: Multiplying & Dividing Rational Expressions
This was one skill that I found really fun once I learned to properly solve the expressions. When multiplying and dividing these expressions, you can cancel out like terms, or terms that are the exact same. These terms must be either in both the numerator and denominator of the same fraction, or in opposite fractions, with one on the top and the other on the bottom. Here is an example showing the usual way to multiply fractions, which can take a while. Below it is a more efficient way, where the 4 is cancelled out to find the same answer.
Sometimes you have to factor an expression first, to check for any terms that can be cancelled out. Remember that terms in brackets are connected, and cannot be cancelled out unless another term in brackets is the exact same. For example, (x + 3) and x are not like terms, but (x – 4) and (x – 4) are. When you have cancelled out all like terms, always check to make sure a fraction cannot be simplified any further. Don’t forget to include restrictions on your answer (values that would be impossible because they would make the denominator 0). With dividing, just flip the second fraction to it’s reciprocal, and continue. These examples show more complicated questions:
This technique turns a complicated equation into a simple fraction, without having to do lots of difficult calculations. That is why I chose it for one of the top things I learned, because it is so simple and fun!
5: The Cosine Law
The cosine law is a way to find the angles and side lengths or a non-right triangle. This law is a formula which works for any non-right triangle where there are 3 sides and no angles, or finding a third side when 2 sides and the angle between them are given. The cosine law looks like this when you need to find a side:
To find an angle instead, just switch the same formula to this:
Here are 2 examples using the cosine law to find a missing side/angle:
This was one of my top 5 because I didn’t realize trigonometry was possible with non-right triangles before math 11. The cosine law makes it easy to find missing values on a triangle by putting them into a simple formula.
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## How do you solve 13 divided by 5?
What is 13 Divided by 5?
1. 13 divided by 5 in decimal = 2.6.
2. 13 divided by 5 in fraction = 13/5.
3. 13 divided by 5 in proportion = 260%
What is 13 divided by 5 as a fraction?
Using a calculator, if you typed in 13 divided by 5, you’d get 2.6. You could also categorical 13/5 as a blended fraction: 2 3/5.
What quantity divided by 5 is 13?
Therefore, the solution to what divided by 5 equals 13 is 65. You can end up this by taking Sixty five and dividing it by 5, and you will see that the solution is 13.
### Can 13 be divided equally?
We have calculated all the numbers that 13 is calmly divisible by. The numbers that 13 is divisible by are 1 and 13. You can be interested to know that all the numbers that 13 is divisible by are often referred to as the factors of 13.
What is the remainder of 13 divided by 6?
The quotient (integer department) of 13/6 equals 2; the remainder (“left over”) is 1. 13 is the dividend, and six is the divisor.
What number is divisible by 13?
Divisibility regulations for numbers 1–30
Divisor Divisibility situation Examples
13 Form the alternating sum of blocks of 3 from proper to left. The end result should be divisible by 13. 2,911,272: 272 – 911 + 2 = -637
Add 4 occasions the ultimate digit to the remainder. The consequence will have to be divisible by 13. 637: 63 + 7 × 4 = 91, 9 + 1 × 4 = 13.
#### What divided by 6 gives you 5?
Therefore, the solution to what divided by 6 equals 5 is 30. You can turn out this by taking 30 and dividing it by 6, and you will see that the solution is 5.
What is the Disability Rule of 13?
Divisibility regulations for numbers 1–30
Divisor Divisibility condition
13 Subtract the remaining two digits from 4 occasions the rest. The end result will have to be divisible by 13.
Subtract Nine instances the last digit from the rest. The end result will have to be divisible by 13.
14 It is divisible by 2 and by 7.
Can 13 be divided by anything else?
Is 13 Divisible By Anything? The numbers that 13 is divisible by are 1 and 13. You may also be to know that all of the numbers that 13 is divisible by are sometimes called the standards of 13. Not best that, but the entire numbers that are divisible by 13 are the divisors of 13.
## What numbers have a remainder of 5?
So the imaginable remainders are 5 or 11. For example take numbers 11,17,23,29 gets a remainder of 5 persistently but when divide by 12 you get a remainder of either 5 or 11.
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# Moment of a force
Moment of a force about a pivot is defined as the product of the force and the perpendicular distance of its line of action from pivot.
Moment of a force about a pivot is defined as the product of the force and the perpendicular distance of its line of action from pivot.
Formula for moment
$M=\mathrm{Fd}$
M: Moment [Nm]
F: Force [N]
d: perpendicular distance [m]
- Perpendicular distance must be a distance from the pivot to the force.
- Perpendicular distance must be at right angle to the force.
Example
Calculate the moment of the force at the pivot.
$M=\mathrm{Fd}$
Given that
F = 3N
d = 2m
$M=\mathrm{3\left(2\right)}$
$M=\mathrm{6Nm}$
## Principal of moment
Law For a body to be in equilibrium (balanced), the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point.
Formula
Total clockwise moment = Total anticlockwise moment
$\mathrm{F1 d1}=\mathrm{F2 d2}$
Conclusion: If a body is balanced, then the total clockwise moment is equal to the total anticlockwise moment.
### Example 1
Calculate the force F if it is balance
Solution
$\mathrm{F1 d1}=\mathrm{F2 d2}$
Given that
F1 = F ?
d1 = 0.5m
F2 = 100N
d2 = 0.4m
$F×0.5=\mathrm{100}×0.4$
$F0.5=\mathrm{40}$
$\frac{\mathrm{F0.5}}{0.5}=\frac{\mathrm{40}}{0.5}$
Answer: $F=\mathrm{80N}$
### Example 2
Calculate the distance d if it is balanced.
Solution
$\mathrm{F1 d1}=\mathrm{F2 d2}$
Given that
F1 = 5N
d1 = d ?
F2 = 3N, 1N
d2 = 1m , 2m
Find the two moments on the right hand side of the moment and then add those moments
$M=\mathrm{Fd}$
$M=3×1$ = 3Nm
$M=1×2$ = 2Nm
Therefore moment on the right hand side is: $\mathrm{3Nm}+\mathrm{2Nm}=\mathrm{5Nm}$
Merge the moments to find the distance d
$\mathrm{F1 d1}=\mathrm{F2 d2}$
$\mathrm{5 d}=5$
$\frac{\mathrm{5 d}}{5}=\frac{5}{5}$
Answer: $d=\mathrm{1m}$
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Home > Algebra > Algebraic Expressions
# Algebraic Expressions
Imagine you have 4 cups, and each cup has 3 marbles. Altogether, we have 4 cups and 12 marbles. How can we represent this statement algebraically?
Let ‘x’ be the number of cups, and ‘y’ be the number of marbles. Then we can write the above statement as 4x + 12y. Here 4x and 12y are called terms and the expression 4x + 12y is called an Algebraic expression.
Let us look at some examples of algebraic expressions from words:
• Sum of 4 and c is 4 + c
• Nine times d is 9d
• Product of x and y is xy
• p divided by 10 is p/10, or
• 2p take away 4f is 2p – 4f
• Subtract 2p from 4f is expressed as 4f – 2p
• 2 more than 10 times p is 2 + 10p
• 7 times w minus q is 7w – q
There are some algebraic conventions (or rules) that we follow:
1. When multiplying expressions, we leave out the multiplication signs. For example, 3 x n is written as 3n.
2. When multiplying a pronumeral with 1, we leave out the 1. For example, 1 x m is simply m.
3. When dividing one expression by another, we write the division as a fraction. 10 divided by m is expressed as
4. When a pronumeral is multiplied by itself, we write the product in index form, such as mxm = m2, not mm
5. When a pronumeral is multiplied by a numeral, the numeral is written first followed by the pronumeral. For example, y x 4 is expressed as 4y, not y4. Note: The numeral before the pronumeral is called the coefficient of the pronumeral.
## Evaluating algebraic expressions
To solve (or evaluate) an algebraic expression, substitute a number in place of the pronumeral (or letter), and do the concerned operation. Let us look at a few examples:
1. If m = 3,
7 – m
= 7 – 3 = 4
And 5m = 5 x 3 = 15
2. Evaluate the expression 4m + 3n when m = 2 and n = 3.
Substituting the values of m and n in the given equation, we get
4m + 3n = 4 x 2 + 3 x 3
= 8 + 9
= 17
3. If the price of a movie ticket is \$10, write an algebraic expression to give the cost of movie tickets for n people.
If P is the price of the movie tickets, and n is the number of people, then the above question can be expressed as –
P = 10 x n.
If there are 4 people, the cost will be
P = 4 x 10 = \$40
## Like terms and unlike terms
When the terms in an algebraic expression have the same pronumeral representation, then we call them like terms, else they are called unlike terms.
Example: a, 2a, -2a, are all like terms, whereas a, a2, ab are all unlike terms.
We’ll find the use of like and unlike terms under various operators with algebraic expressions.
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## Generating Pascal's Triangle
Pascal's Triangle is a well known mathematical pattern. Although this pattern was actually discovered in China, it has been named after the first Westerner to study it. According to Yunze He, "Pascal's" triangle was first developed during the Song Dynasty by a mathematician named Hue Yang.
The simplest view of Pascal's Triangle is that it may be generated by affixing a one a either end of the new row and then generating all numbers in between by by adding together the two numbers above it. For example, 3 = 2+1, as shown The numbers may also be generated by using the idea of combinations found in probability theory. To do this assign a column and row number to each value. Then use the combinations formula to produce the value in question. For example : to find the 3rd number in the 5th row use: 5C3, "5 things combined 3 at a time." 5C3 = 5!/3!(5-3)! = 5*4*3*2*1 / 3*2*1*(2*1) = 10 (red circle).
The Question which drove me was, "Can Pascal's Triangle be turned into Pascal's Plane by finding the numbers that exist outside the triangle to fill the plane?" The goal was to find an array of numbers that fulfill the requirements of Pascal's Triangle yet fill the plane.
The Approach I took was to create a function similar to the factorial but based on addition. I called this function a "perfect" for perfect triangle based on an oversimplified view I had, at that time, of triangular numbers.
The Perfect worked as such: a first perfect was the value of the triangular number of those dimensions:
A Perfect of higher dimensions would require that each number generated be perfected before being added. Examples:
1h 1 = 1 = 1 2h1 = 2 + 1 = 3 3h1 = 3 + 2 + 1 = 6 4h1 = 4 + 3 + 2 + 1 = 10
Start 4h2 = step 1 4h1 + 3h1 + 2h1 + 1h1 = step 2 10 + 6 + 3 + 1 = step 3 20
Start 3h3 = step 1 3h2 + 2h2 + 1h2 = step 2 (3h1+2h1+1h1) + (2h1+1h1) + 1h1 step 3 (6+3+1) + (3+1) + 1 = step 4 15
As with the combination a perfect of order 0 is defined to be 1: Nh0 = 1 .
This method creates Pascal's triangle in the following arrangement:
This method never did successfully define the items outside the triangle. It suffered the same limitations a the combinations method - items outside the triangle are not easily defined this way.
It did, however raise many questions about what geometrical arrangement should be used to plot Pascal's Triangle. Notice how the angle defined by the 1's varies from about 60' above to exactly 90' by this method. So which angle would be best for defining the rest of the plane?
In the end, simply reversing the original steps would define much of the plane. Instead of adding the two numbers above to get the number below, subtract one of the numbers above from the number below to get the other number. From Pascal's triangle you can easily get back to this point (shown in spread sheet format)
There does not exist a single clear solution to what numbers must appear above the red line. This part of the plane may be chosen arbitrarily, by picking a starting number then filling in as required. Here are two possible solutions to Pascal's plane: The numbers in red were picked arbitrarily. The rest were calculated using the definition of Pascal's Triangle. The first solution retains the pattern of 1's but with negatives. The second retains the symmetry of the plane. Line of symmetry shown in blue.
Pascal's Triangle Links math forum math forum lessons patterns & relationships Serpenski JAVA Serpenski JAVA 2 Serpenski JAVA 3 algebra 12 days of Xmas history of pinball Fibonacci 1 Fibonacci 2 Fibonacci 3 binomial 1 binomial 2
Other Number Theory Pages at this Site: Goldbach's Conjecture the golden mean measuring compositeness factor patterns spatial number patterns binary lessons division by 0 lesson base 12 chart lesson
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## Precalculus (6th Edition) Blitzer
The function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function and is symmetric about the origin.
In order to check if the function is even, odd, or neither, substitute $x$ by – $x$ and evaluate the value of $f\left( -x \right)$. \begin{align} & f\left( -x \right)={{\left( -x \right)}^{3}}-5\left( -x \right) \\ & =\left( -x \right)\left( -x \right)\left( -x \right)-5\left( -x \right) \\ & =-{{x}^{3}}+5x \\ & =-\left( {{x}^{3}}-5x \right) \end{align} Since, $f\left( x \right)={{x}^{3}}-5\left( x \right)$, therefore $f\left( -x \right)=-f\left( x \right)$. Now as $f\left( -x \right)=-f\left( x \right)$, it gives that the function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function The function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function, therefore by definition of an odd function, $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is symmetric about the origin. Hence, the function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function and is symmetric about the origin.
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# Skip Counting by 1
Last updated date: 23rd May 2024
Total views: 150.6k
Views today: 3.50k
## How to Skip Count by 1?
Skip counting is the method of counting forward by numbers like 1, 2, 3, and so on. In other words, by doing this, we skip over a particular number. For example, when we skip count by 3, we start from 0 and skip the next number to go to the third number, i.e., 3. Therefore, skip counting by 1 means skipping through 1.
Skip counting
## What is Skip Counting by 1?
Skip counting by 1, as explained above, means skipping through 1. If we start from 0, the next number will be 1 itself. If we continue, the next number will be 1, 2, 3, 4, 5, and so on, as shown in the image below.
Skip counting by 1 chart
Skip counting is useful in multiplication tables because we can skip numbers to find the multiple of a number. For example, if you want to know the multiple of 6 in a multiplication table of 1, you can skip count by one 6 times. Thus, the result will be 6. We can use skip counting by 1 to 100 worksheets like the one given above for practice.
## Types of Skip Counting
Skip counting has two types: forward skip counting and backward skip counting.
• ### Forward Skip Counting
When we perform "forward skip counting," we count numbers in the forward direction. For example, if we had to perform forward skip counting by 1, and we began with 0, then the next number would be 1. If we continue forward counting, the number will be 1, 2, 3, 4, and so on.
• ### Backwards Skip Counting
When we perform backwards skip counting, we count numbers in the backward direction. For example, if you skip count backwards by 1 starting from the number 9, the next number would be 8. If we continue skip-counting backward, the numbers will be 8, 7, 6, and so on.
## Skip Counting by 1 Worksheet
1. Ravi has 20 marbles. How many times will he have to skip count by one?
Ans: Since the total number of marbles is 20, Ravi would have to skip count by 1, 20 times to get the total number of marbles. He can start by counting 1, 2, 3, 4,... 20.
2. Sneha has 10 candies. How many times will she need to perform a skip count by 1?
Ans: The total number of candies is 10. Sneha would need to skip count by 1, 10 times to get the total number of candies. She can skip count by counting 1, 2, 3, 4, 5, and so on, up to 10.
3. How many times will you need to skip count by 1 while climbing 15 stairs?
Ans: Since there are 15 stairs, while climbing the first one, you skip counting by 1. As you skip count by 1, the numbers will be 1, 2, 3, 4,... 15.
## Conclusion
So, today you have learned about skip counting and skip counting by 1 type, i.e., forward and backward skip counting, with some examples. Now you can count anything easily by using this method. Following similar rules, you can skip counting higher numbers like 2 and 3.
## FAQs on Skip Counting by 1
1. Where can we use skip counting?
Skip counting is very useful for children while learning multiplication tables. If a child wants to know the result of 5 x 6, they can skip the count by 5 or 6 times to get the result, which will be 30. As the child learns one-digit skip, we can introduce skip counting by higher numbers.
2. How do we introduce skip counting to children?
To teach a child to skip count, we start with 1. As the child learns, we progress to 2, and so on. This way, the child will skip count by 10 or 20. We can also provide them a skip counting by 1 worksheet.
3. How do you make skip-counting fun for kids?
To learn, you can perform some skip counting by one activity, like playing with the kids while climbing stairs, giving them candies, and introducing skip counting, or giving them a chart with numbers like a snake and ladder game and introducing skip counting where they move their token to the said number by skipping counting.
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# Dimensional Analysis
Dimensional Analysis. In which you will learn about: Conversion factors Standard dimensional analysis Dimensional analysis with exponential units. Dimensional Analysis. Imagine math class (don’t panic) Imagine multiplying two fractions
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## Dimensional Analysis
E N D
### Presentation Transcript
1. Dimensional Analysis In which you will learn about: Conversion factors Standard dimensional analysis Dimensional analysis with exponential units
2. Dimensional Analysis • Imagine math class (don’t panic) • Imagine multiplying two fractions • Imagine the numerator of one fraction matches the denominator of the second (3/7 x 2/3) • The numerator and denominator cancel! • In dimensional analysis, we use this idea to cancel UNITS of measurements.
3. Equalities State the same measurement in two different units length 10.0 in. 25.4 cm
4. Conversion Factors Fractions in which the numerator and denominator are EQUAL quantities expressed in different units Example: 1 in. = 2.54 cm Factors: 1 in. and 2.54 cm 2.54 cm 1 in.
5. How many minutes are in 2.5 hours? Conversion factor 2.5 hr x 60 min = 150 min 1 hr cancel By using dimensional analysis / factor-label method, the UNITS ensure that you have the conversion right side up, and the UNITS are calculated as well as the numbers!
6. Sample Problem • You have \$7.25 in your pocket in quarters. How many quarters do you have? 7.25 dollars 4 quarters 1 dollar = 29 quarters X
7. Learning Check A rattlesnake is 2.44 m long. How long is the snake in cm? a) 2440 cm b) 244 cm c) 24.4 cm
8. Solution A rattlesnake is 2.44 m long. How long is the snake in cm? b) 244 cm 2.44 m x 100 cm = 244 cm 1 m
9. Learning Check How many seconds are in 1.4 days? Unit plan: days hr min seconds 1.4 days x 24 hr x 1 day 60 min 1 hr x 60 s 1 min = 1.2 x 105 s
10. Wait a minute! What is wrongwith the following setup? 1.4 day x 1 day x 60 min x 60 sec 24 hr 1 hr 1 min
11. English and Metric Conversions • If you know ONE conversion for each type of measurement, you can convert anything! • I will provide these equalities, but you must be able to use them: • Mass: 454 grams = 1 pound • Length: 2.54 cm = 1 inch • Volume: 0.946 L = 1 quart
12. Steps to Problem Solving • Read problem • Identify data • Make a unit plan from the initial unit to the desired unit (good practice at beginning, not necessary as you get comfortable with this) • Select conversion factors • Change initial unit to desired unit • Cancel units and check • Do math on calculator • Give an answer using significant figures
13. Dealing with Two Units If your pace on a treadmill is 65 meters per minute, how many seconds will it take for you to walk a distance of 8450 feet? HINT: Always start with the simplest label. You’re looking for seconds, so you can’t start there. 65 m/min has two labels so that’s not very simple. Best STARTING place is 8450 feet!
14. What about Square and Cubic units? • Use the conversion factors you already know, but when you square or cube the unit, don’t forget to cube the number also! • Best way: Square or cube the ENTIRE conversion factor • Example: Convert 4.3 cm3 to mm3 () 4.3 cm3 10 mm 3 1 cm 4.3 cm3 103 mm3 13 cm3 = = 4300 mm3
15. Learning Check • A Nalgene water bottle holds 1000 cm3 of dihydrogen monoxide (DHMO). How many cubic decimeters is that?
16. Solution 1000 cm3 1 dm 3 10 cm ( ) = 1 dm3 So, a dm3 is the same as a Liter ! A cm3 is the same as a milliliter.
17. How do I round multiple step problems with the correct sig figs? • If the problem has only one “type” of math (adding/subtracting OR multiplying/dividing), round at the end of the problem • Dimensional analysis is all M/D! Round at the end. • If the problem has more than one type, you must follow the order of operations, round after each type is complete. • A good example is percent error. Round using adding rules after 0-E, then finish the calculation and round again using multiplying rules.
18. Speaking of Sig Figs… • Exact conversion factors, such as 100 cm in 1 m, do NOT count toward the number of sig figs! • Numbers that are part of a mathematical formula, such as x100 in percent error, do NOT count toward the number of sig figs!
More Related
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Simplifying Polynomials Worksheet - Page 2 | Problems & Solutions
# Simplifying Polynomials Worksheet - Page 2
Simplifying Polynomials Worksheet
• Page 2
11.
Find GCF of $a$4, 24$a$2, 12$a$.
a. $a$ b. 12 c. 12$a$ d. $a$2
#### Solution:
a4 = a × a × a × a
[Write the factors of a4.]
24a2 = 12 × 2 × a × a
[Write the factors of 24a2.]
12a = 12 × a
[Write the factors of 12a.]
The GCF of two or more numbers is the product of their common factors.
The GCF of a4, 24a2 and 12a is: a
12.
Factor $x$3 - 3$x$2 + 2$x$ completely.
a. $x$($x$ - 2)($x$ + 1) b. $x$($x$ - 2)($x$ - 1) c. $x$($x$ + 2)($x$ - 1) d. None of the above
#### Solution:
x3 - 3x2 + 2x
[Original expression.]
The GCF of x3, 3x2 and 2x is x.
= x(x2 - 3x + 2)
[Use GCF to factor.]
= x(x - 2)(x - 1)
[Factor the trinomial.]
13.
Factor 2$x$3 + 4$x$2 - 16$x$ completely.
a. 2$x$($x$ - 4)($x$ - 2) b. 2$x$($x$ + 4)($x$ - 2) c. 2$x$($x$ + 4)($x$ + 2) d. None of the above
#### Solution:
2x3 + 4x2 - 16x
[Original expression.]
= 2x(x2 + 2x - 8)
[Use GCF to factor.]
= 2x(x + 4)(x - 2)
[Factor the trinomial.]
14.
Factor 3$x$3 + 9$x$2 + 54$x$ completely.
a. -3$x$($x$ + 6)($x$ + 3) b. -3$x$($x$ - 6)($x$ + 3) c. -3$x$($x$ - 6)($x$ - 3) d. None of the above
#### Solution:
-3x3 + 9x2 + 54x
[Original expression.]
= -3x(x2 - 3x - 18)
[Use GCF to factor.]
= -3x(x - 6)(x + 3)
[Factor the trinomial.]
15.
Factor - 3$x$3 + 9$x$2 + 30$x$ completely.
a. -3$x$($x$ - 5)($x$ - 2) b. -3$x$($x$ + 5)($x$ - 2) c. -3$x$($x$ - 5)($x$ + 2) d. None of the above
#### Solution:
-3x3 + 9x2 + 30x
[Original expression.]
= -3x(x2 - 3x - 10)
[Factor out the GCF.]
The factors are 2 and -5
[2 + (-5) = -3 and 2 x (-5) = -10]
= -3x(x - 5)(x + 2)
[Factorize.]
16.
Factor - 2$x$2 - 6$x$ + 36 completely.
a. -2($x$ - 6)($x$ - 3) b. -2($x$ + 6)($x$ - 3) c. -2($x$ + 6)($x$ + 3) d. None of the above
#### Solution:
-2x2 - 6x + 36
[Original expression.]
= -2(x2 + 3x - 18)
[Use GCF to factor.]
= -2(x + 6)(x - 3)
[Factor the trinomial.]
17.
Factor 3$x$3 - 9$x$2 - 54$x$ completely.
a. 3$x$($x$ + 6)($x$ + 3) b. 3$x$($x$ - 6)($x$ + 3) c. 3$x$($x$ - 6)($x$ - 3) d. None of the above
#### Solution:
3x3 - 9x2 - 54x
[Original expression.]
= 3x(x2 - 3x - 18)
[Use GCF to factor.]
= 3x (x - 6) (x + 3)
[Factor the trinomial.]
18.
Factor: - 6$x$3 + 3$x$2 + 54$x$ - 27
a. (-6$x$ + 3)($x$ - 3)($x$ - 3) b. (-6$x$ - 3)($x$ + 3)($x$ - 3) c. (-6$x$ + 3)($x$ + 3)($x$ - 3) d. None of the above
#### Solution:
-6x3 + 3x2 + 54x - 27
[Original expression.]
= (-6x3 + 3x2) + (54x - 27)
[Group terms.]
= x2(-6x + 3) - 9(-6x + 3)
[Factor each group.]
= (-6x + 3) (x2 - 9)
[Use distributive property.]
= (-6x + 3)(x + 3)(x - 3)
[Use the formula a2 - b2 = (a + b)(a - b).]
19.
Factor: - 5$x$3 - 6$x$2 + 20$x$ + 24
a. (-5$x$ - 6)($x$ - 2)($x$ - 2) b. (-5$x$ + 6)($x$ + 2)($x$ - 2) c. (-5$x$ - 6)($x$ + 2)($x$ - 2) d. None of the above
#### Solution:
-5x3 - 6x2 + 20x + 24
[Original expression.]
= (-5x3 - 6x2) + (20x + 24)
[Group terms.]
= -x2(5x + 6) + 4(5x + 6)
[Factor each group.]
= (-5x - 6) (x2 - 4)
[Use distributive property.]
= (-5x - 6) (x + 2)(x - 2)
[Use the formula a2 - b2 = (a + b)(a - b).]
20.
Factor: 4$x$3 - 7$x$2 + 4$x$ - 7
a. None of the above b. (4$x$ - 7)($x$2 - 1) c. (4$x$ - 7)($x$2 + 1) d. (4$x$ + 7)($x$2 + 1)
#### Solution:
4x3 - 7x2 + 4x - 7
[Original expression.]
= (4x3 - 7x2) + (4x - 7)
[Group terms.]
= x2(4x - 7) + 1(4x - 7)
[Factor each group.]
= (4x - 7)(x2 + 1)
[Use distributive property.]
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# How do you solve the system of equations 2x - 5y = 3 and 3x - 6y = 9?
Apr 7, 2018
$x = 9$
$y = 3$
#### Explanation:
To solve this system of equations, we will first manipulate the first equation so that $y$ is alone on the left-hand side of the equation.
$2 x - 5 y = 3$,
$- 5 y = 3 - 2 x$,
$5 y = - 3 + 2 x$,
$5 y = 2 x - 3$,
$y = \frac{2}{5} x - \frac{3}{5}$.
We now plug this value of $y$ into the second equation.
$3 x - 6 y = 9$,
$3 x - 6 \left(\frac{2}{5} x - \frac{3}{5}\right) = 9$,
$3 x - \frac{12}{5} x + \frac{18}{5} = 9$.
Move all the $x$'s to one side of the equation and the constants to the other.
$3 x - \frac{12}{5} x + \frac{18}{5} = 9$,
$3 x - \frac{12}{5} x = 9 - \frac{18}{5}$.
Multiplying both sides of the equation by $5$ makes it easier to manipulate.
$5 \left(3 x - \frac{12}{5} x\right) = 5 \left(9 - \frac{18}{5}\right)$,
$15 x - 12 x = 45 - 18$,
$3 x = 27$,
$x = 9$.
Now plug this $x$ back into the first equation.
$2 x - 5 y = 3$,
$2 \left(9\right) - 5 y = 3$,
$18 - 5 y = 3$,
$18 = 3 + 5 y$,
$15 = 5 y$,
$3 = y$.
Thus, we have $x = 9$ and $y = 3$.
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# How Will Aaron Eckhart Fare Today? (03/27/2020)
How will Aaron Eckhart do on 03/27/2020 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not at all guaranteed – do not take this too seriously. I will first calculate the destiny number for Aaron Eckhart, and then something similar to the life path number, which we will calculate for today (03/27/2020). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology experts.
PATH NUMBER FOR 03/27/2020: We will take the month (03), the day (27) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. We’ll show you how it works now. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 27 we do 2 + 7 = 9. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 9 + 4 = 16. This still isn’t a single-digit number, so we will add its digits together again: 1 + 6 = 7. Now we have a single-digit number: 7 is the path number for 03/27/2020.
DESTINY NUMBER FOR Aaron Eckhart: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Aaron Eckhart we have the letters A (1), a (1), r (9), o (6), n (5), E (5), c (3), k (2), h (8), a (1), r (9) and t (2). Adding all of that up (yes, this can get tedious) gives 52. This still isn’t a single-digit number, so we will add its digits together again: 5 + 2 = 7. Now we have a single-digit number: 7 is the destiny number for Aaron Eckhart.
CONCLUSION: The difference between the path number for today (7) and destiny number for Aaron Eckhart (7) is 0. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is of questionable accuracy. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Скачать презентацию Multiplying and Dividing Greater Numbers 1 3 5
• Количество слайдов: 44
Multiplying and Dividing Greater Numbers 1 3 5 2 4
Using Place Value Patterns We can use multiplication patterns to help us multiply by multiples of 10, 100, and 1, 000. What patterns do you notice below? 5 5 x x 1=5 10 = 50 100 = 500 1, 000 = 5, 000 8 8 x x 1=8 10 = 80 100 = 800 1, 000 = 8, 000
Here is a multiplication trick! When one of the factors you are multiplying has zeros on the end, you can multiply the nonzero digits, and then add on the extra zeros. 9 x 100 Multiply the non-zero digits. 9 00 9 x 100 = 900 Add the extra zeros.
Let’s try another! When one of the factors you are multiplying has zeros on the end, you can multiply the nonzero digits, and then add on the extra zeros. 4 x 1000 Multiply the non-zero digits. 4 000 4 x 1000 = 4000 Add the extra zeros.
Let’s try another! When one of the factors you are multiplying has zeros on the end, you can multiply the nonzero digits, and then add on the extra zeros. 6 x 1000 Multiply the non-zero digits. 6 000 6 x 1000 = 6000 Add the extra zeros.
Try some on your own! Solve the following problems in your Math notebook. Use place value patterns to help you! 3 x 100 = ____ 5 x 1, 000 = _____ 6 x 10 = ____ 18 x 100= ____ 2 x 1, 000 =____ 9 x 10 = _____
Using Place Value Patterns We can use division patterns to help us multiply by 10, 100, and 1, 000. What patterns do you notice below? 6÷ 3=2 60 ÷ 3 = 20 600 ÷ 3 = 200 6, 000 ÷ 3 = 2, 000 8÷ 4=2 80 ÷ 4 = 20 800 ÷ 4 = 200 8, 000 ÷ 4 = 2, 000
Here is a division trick! When there are zeros at the end of the dividend, you can move them aside and use a basic division fact to divide the nonzero digits. 120 ÷ 4 3 0 120 ÷ 4 = 30 Divide the nonzero digits. Add the extra zeros.
Let’s see another example! When there are zeros at the end of the dividend, you can move them aside and use a basic division fact to divide the nonzero digits. 800 ÷ 4 Divide the nonzero digits. 2 00 800 ÷ 4 = 200 Add the extra zeros.
Let’s see another example! When there are zeros at the end of the dividend, you can move them aside and use a basic division fact to divide the nonzero digits. 800 ÷ 4 Divide the nonzero digits. 2 00 800 ÷ 4 = 200 Add the extra zeros.
Try some on your own! Solve the following problems in your Math notebook. Use place value patterns to help you! 3 x 100 = ____ 5 x 1, 000 = _____ 6 x 10 = ____ 18 x 100= ____ 2 x 1, 000 =____ 9 x 10 = _____
Write Out… How can using place value patterns help you multiply and divide by multiples of 10?
Let’s review! How does using place value patterns help you multiply and divide by multiples of 10? What does a hundred look like using base ten blocks? What does a ten look like using base ten blocks? How do we show ones using base ten blocks?
We can use arrays and base ten blocks to help us multiply and divide greater numbers! You can draw a picture of an array to show multiplication. REMEMBER: An array is an orderly arrangement of objects in a row! 3 x 10 This means 3 rows of 10.
Check out an example! 4 x 21 = 84 What You Show: What You Think: 4 rows of 2 tens = 8 tens 4 rows of 1 ones = 4 ones 8 tens 4 ones = 84 To find the product count the tens and ones, then add them together.
Let’s try another! 3 x 32 = 96 What You Show: What You Think: 3 rows of 3 tens = 9 tens 3 rows of 2 ones = 6 ones 9 tens 6 ones = 96 To find the product count the tens and ones, then add them together.
Let’s try a few problems on our own! Remember: You can draw pictures using base ten blocks to help you solve multiplication problems! Be prepared to share your problem solving strategies with the group!
Let’s review! We have learned new strategies for multiplying and dividing greater numbers. We learned that we can use place value patterns to help us! Yesterday we learned how to draw pictures to help us solve problems. Today we will learn another new strategy to make multiplication easier!
You can make multiplication easier by breaking larger numbers apart by place value. 4 x 23 20 + 3 First multiply the ones. Then multiply the tens. Add the products! You can use place value to break 23 apart. How would you write 23 in expanded form? 4 x 3 = 12 4 x 20 = 80 80 + 12 = 92
You can make multiplication easier by breaking larger numbers apart by place value. 4 x 36 30 + 6 First multiply the ones. Then multiply the tens. Add the products! You can use place value to break 36 apart. How would you write 36 in expanded form? 4 x 6 = 24 4 x 30 = 120 + 24 = 144
You can make multiplication easier by breaking larger numbers apart by place value. 2 x 62 60 + 2 First multiply the ones. Then multiply the tens. Add the products! You can use place value to break 62 apart. How would you write 62 in expanded form? 2 x 2=4 2 x 60 = 120 + 4 = 124
Solve this problem on your own! Remember: You can break numbers apart to help you! 5 x 42
Solve this problem on your own! Remember: You can break numbers apart to help you! 3 x 27
Solve this problem on your own! Remember: You can break numbers apart to help you! 6 x 18
Let’s review! We learned that we can use place value patterns to help us multiply! We also learned how to draw pictures and how to break apart numbers to help us solve problems. Today we will learn another strategy for multiplying greater numbers!
What’s going on today? Today we will learn the traditional method for multiplying 2 digit numbers by 1 digit numbers! REMEMBER: There is more than one way to do the same thing! You will be able to choose the method that works best for you.
There is not enough room for the tens digit so it gets stored in the “add”-ic “Add”-ic Add the digits in the addic. 9+2=11 Second Floor 2 First Floor 37 X 3 Basement 111 Multiply the tens. 3 x 3=9 Start by multiplying the ones! 3 x 7 = 21
There is not enough room for the tens digit so it gets stored in the “add”-ic “Add”-ic Add the digits in the addic. 4+3=7 Second Floor 3 First Floor 18 X 4 Basement 7 2 Multiply the tens. 4 x 1=4 Start by multiplying the ones! 4 x 8 = 32
There is not enough room for the tens digit so it gets stored in the “add”-ic “Add”-ic Add the digits in the addic. 4+1=5 Second Floor 1 First Floor 26 X 2 Basement 5 2 Multiply the tens. 2 x 2=4 Start by multiplying the ones! 6 x 2 = 12
There is not enough room for the tens digit so it gets stored in the “add”-ic “Add”-ic Add the digits in the addic. 15 + 4 =19 Second Floor 4 First Floor 38 X 5 Basement 190 Multiply the tens. 3 x 5 = 15 Start by multiplying the ones! 8 x 5 = 40
Let’s try one on our own! You can use the HOUSE model to help you! 34 x 7
Let’s try one on our own! You can use the HOUSE model to help you! 18 x 9
Let’s try one on our own! You can use the HOUSE model to help you! 33 x 4
Let’s try one on our own! You can use the HOUSE model to help you! 81 x 7
Let’s try one on our own! You can use the HOUSE model to help you! 15 x 6
Let’s review! We have learned different strategies for multiplying two digit numbers by one digit numbers. Yesterday we learned the traditional multiplication algorithm in a HOUSE to help us! Today we will practice using the HOUSE method to help us and apply the strategy to story problems!
There is not enough room for the tens digit so it gets stored in the “add”-ic “Add”-ic Add the digits in the addic. 5+2=7 Second Floor 2 First Floor 14 X 5 Basement 7 0 Multiply the tens. 5 x 1=5 Start by multiplying the ones! 5 x 4 = 20
There is not enough room for the tens digit so it gets stored in the “add”-ic “Add”-ic Add the digits in the addic. 6 + 1 =7 Second Floor 1 First Floor 26 X 3 Basement 7 8 Multiply the tens. 3 x 2=6 Start by multiplying the ones! 3 x 6 = 18
Let’s try one on our own! You can use the HOUSE model to help you! 14 x 7
Let’s try one on our own! You can use the HOUSE model to help you! 13 x 3
Let’s try one on our own! You can use the HOUSE model to help you! 15 x 9
Let’s solve a story problem! You can use the HOUSE model to help you! Four classrooms received 62 plants for a science project. How many plants do they have altogether?
Let’s solve a story problem! You can use the HOUSE model to help you! Twenty-three second graders have baseball card collections. Each second grader has 8 baseball cards. How many do they have in all?
Let’s solve a story problem! You can use the HOUSE model to help you! A baseball diamond has four sides. Each side is 90 feet long. How far will Joe run if he hits a homerun and runs completely around the baseball diamond?
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Parametric Conditions
by
Priscilla Alexander
This write-up is for students who are learning about the effects of integers on the graphs formed by parametric equations.
If given
What happens when a and b are the same positive integers
What happens when a is a positive integer and b is a negative integer (and vica versa)
What happens when a and b are the same negative integers
What happens when a and b are different integers, where
When a and b are the same a circle is formed.
When a and b are the same number but one is negative and the other is positive, a circle is still formed.
See Graph
In general when a and b are the same integers between -3 and 3 a circle will always appear that has a radius of one. This happens because the ratio of a/b in this case reduces to one.
When a and b are different integers, say a is two and b is one, then a rotated parabola is formed.
In general, when the ratio of a/b reduces to two a rotated parabola will always be formed. For example, if a equals 4 and b equals 2 then the rotated parabola will be formed.
When a and b are different integers, say a is one and b is two, then a bow tie looking image appears.
In general when the ratio of a/b reduces to 1/2 this image will always be formed.
When a and b are different integers, say a is equal to 2 and b is equal to 3, the image below is formed.
In general when the ratio of a/b reduces to 2/3 the image will always appear like the above.
When a and b are different integers, say a is equal to 1 and b is equal to 3, then an image that is reflected about the x-axis appears.
In general, when the ratio a/b reduces to 1/3, then the above image appears.
When a is equal to 3 and b is equal 1, then the graph is the same as a equals 1 and b equals 3, but is reflected about the y-axis.
In general this happens for all integers that are odd when a or b is one and the other interger is not one. For example, if a is equal to 5 and b is equal to 1, then the graph of a is equal to 1 and b is equal to five will be reflected on the opposite axis.
When a is an even integer and b is one, then the amplitude of the graph will always be
In conclusion, integers make the graphs of parametric equations act in many ways. Some of the graphs are reflected or rotated. The ratio of a and b also have an effort on the graph. In addition, odd and even integers make the graph behave in a different way then the rest.
return
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# Most Expected Types of Maths Questions in Agniveer Exam 2024
The Most Expected Agniveer Mathematics Questions Types for 2024 Exam: To score maximum marks in the Mathematics section of the Agniveer Exam, students should have a clear idea about the types or level of questions that can be asked. As per the analysis of Agniveer last year papers, we have found some most repetitive types of questions that have a very high probability of being in this year too. So, must prepare these important types of questions before the exam to score full marks in the Mathematics section.
These question types are prepared by our expert faculties that have solid conceptual knowledge in this particular field of Indian Defence Exams such as NDA, CDS, AFCAT, CAPF, Agniveer, etc.
Here, you will get more than 10 types of Agniveer Mathematics expected Questions, mentioned one by one with example questions and answers.
## Most Expected Types of Maths Questions in Agniveer Exam 2024 – Chapterwise
Type 01: LCM and HCF
1. The ratio of 2 numbers is 4 ∶ 5. Their LCM is 700. Find their HCF.
• (A) 35
• (B) 140
• (C) 175
• (D) 70
2. Find the HCF of 8/21 , 12/35 , 32/7
• (A) 4/105
• (B) 4/140
• (C) 4/175
• (D) 4/70
3. If the ratio of two numbers is 2 : 7 and the product of their L.C.M and H.C.F is 686. Find the greatest number.
• (A) 14
• (B) 49
• (C) 7
• (D) 21
4. The LCM of two numbers is 48. The numbers are in the ratio 2:3. The sum of the numbers is –
• (A) 28
• (B) 32
• (C) 40
• (D) 64
5. Two numbers are in ratio 14 : 9, if HCF and LCM of both numbers are 3 and 378, then find the greater number.
• (A) 21
• (B) 27
• (C) 28
• (D) 42
6. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
• (A) 4
• (B) 10
• (C) 15
• (D) 16
Type 02: Ration and Proportion
7. If a sum of Rs. 3540 is divided in threeparts in a ratio of 2 : 6 : 7, then the largestpart will be:
• (A) Rs. 1652
• (B) Rs. 1416
• (C) Rs. 944
• (D) Rs. 472
8. Three numbers are in the ratio 1/4 : 5/9 : 7/12 .The difference between the greatest and the smallest number is 180. Find the sum of all the three numbers.
• (A) 500
• (B) 650
• (C) 750
• (D) 800
9. Rs.10000 is divided among 3 persons. If ratio of amount between first and second person is 2 : 1 and between second and third person is 3 : 1, then find the amount get by third person.
• 1. 3000
• 2. 1000
• 3. 6000
• 4. 2000
10. In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
• a) 50
• b) 100
• c) 150
• d) 200
11. A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B’s contribution in the capital?
• a) 5000
• b) 1000
• c) 15000
• d) 9000
12. Three partners shared the profit in a business in the ratio 5:7:8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?
• a) 20:49:64
• b) 20:49:28
• c) 20:64:49
• d) 49:20:64
Type 03: Probability
13. A bag contains 5 black and 6 white balls; one balls are drawn at random. What is the probability that the balls drawn are black?
• (A) 6/11
• (B) 5/6
• (C) 5/11
• (D) 2/5
14. A bag contains 8 red ball and 12 orange ball. If a ball is drawn from the bag, then find the probability that the ball is red ball.
• (A) 1/5
• (B) 3/5
• (C) 2/5
• (D) 4/5
15. A bag contains 15 red balls and some green balls. If the probability of drawing a green ball is 1/6, then the number of green balls is
• (A) 5
• (B) 4
• (C) 3
• (D) 2
16. A bag contains 2 black, 3 yellow and 2 purple balls. Two balls are drawn at random. What is the probability that none of the balls drawn is purple?
• a) 10/21
• b) 11/21
• c) 4/3
• d) 6/5
17. What is the probability of drawing a Red Numbered Card from a deck of Cards ?
• (A) 10/52
• (B) 27/52
• (C) 26/52
• (D) 25/52
18. Two dice are thrown simultaneously. Find the probability of getting a sum of more than 7.
• (A) 7/12
• (B) 1/3
• (C) 5/12
• (D) 19/36
19. What is the probability of getting atleast 2 Heads when a 3 coins are tossed simultaneously.
• (A) 1/2
• (B) 1/4
• (C) 1/8
• (D) None
Also Check Out:-
Type 04: Trigonometry
20. Find the value of cot1°.cot2°.cot3° _______ cot89°.
• 1. 0
• 2. 1
• 3. 2
• 4. 1/2
21. What is the value of –
sin260° + tan245 ° + sec245 ° – cosec230 °?
• (A) 4
• (B) 1/4
• (C) -1/4
• (D) -4
22. Find the value of tan135°?
• (A) 1
• (B) -√3
• (C) -1
• (D) √3
23. If
then k =?
• a) cos𝜽
• b) sec𝜽
• c) sin𝜽
• d) cosec𝜽
24. Find the value of
cot25° × cot35° × cot45° × cot55° × cot65°
• (A) √3
• (B) 2
• (C) 1
• (D) 0
25. Find the value
• (a) 2
• (b) 1
• (c) 0
• (d) -1
26. If Cot A = 12/5, then the value of (SinA + CosA) × Cosec A is-
• (a) 13/5
• (b) 17/5
• (c) 14/5
• (d) 1
27. Find the value of x if
• 1. 1
• 2. 1/2
• 3. √𝟑/𝟐
• 4. 0
28. An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:
• a) 21.6 m
• b) 23.2 m
• c) 24.72 m
• d) None of these
Type 05: Mensuration
29. The area of a square is 144 cm2 . What is the perimeter of the square formed with the diagonal of the original square as its side?
• (A) 48 √𝟐 cm
• (B) 24 √𝟐 cm
• (C) 48 cm
• (D) 24 cm
30. If the side of the square is increased by 5 cm, then the area increases by 175 cm2. The side of the square is:
• (A) 15 cm
• (B) 20 cm
• (C) 18 cm
• (D) 17 cm
31. Area of square is 3/5th of area of rectangle. If sides of rectangle are 9 cm and 15 cm, then find the perimeter of the square.
• 1. 9 cm
• 2. 18 cm
• 3. 36 cm
• 4. 81 cm
32. The diagonal of the rectangle is 17 cm and the length is 15 cm. find the area of the rectangle.
• (A) 160 cm2
• (B) 240 cm2
• (C) 108 cm2
• (D) 120 cm2
33. A rectangular grassy plot is 112 m long and 78 m broad. It has a travel path 2.5 m wide all around it on the sides. Find the area of the path.
• a) 825 m
• b) 725 m
• c) 925 m
• d) 900 m
34. The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq. m. The length of the larger side of the rectangle is :
• a) 70 m
• b) 60 m
• c) 40 m
• d) 50 m
35. The perimeter and the area of a rectangular sheet are 42 m and 108 m2, respectively. The length of the diagonal is:
• a) 12 m
• b) 15 m
• c) 10 m
• d) 14 m
36. Find the area of the right-angle triangle, if its base is 8 cm and hypotenuse is 17 cm.
• (A) 52 cm2
• (B) 24 cm2
• (C) 60 cm2
• (D) 30 cm2
37. If diagonal of cube is 𝟏𝟐 cm, then its volume in cm3
• a) 18
• b) 9
• c) 6
• d) 8
38. How many 3 metre cubes can be cut from a cuboid measuring 18m×12m×9m ?
• a) 18
• b) 90
• c) 60
• d) 72
39. If the radius of a sphere is doubled, what is the ratio of the volume of original sphere to that of new sphere
• a) 1:8
• b) 8:1
• c) 1:2
• d) 2:1
40. The slant height of a cone is 13cm. the diameter of the base is 10cm. Find the total surface area of the cone.
• a) 80π
• b) 10π
• c) 90π
• d) 70π
41. The floor of a room is to be paved with tiles of length 60 cm and breadth 40 cm. If the floor of the room is 72 m × 48 m, find the number of tiles required.
• 1. 10000
• 2. 8000
• 3. 6400
• 4. 14400
42. Area of a circle is equal to the area of a rectangle. Perimeter of rectangle is 150 cm. If length is 9 cm more than the breadth, then what will be the radius of circle?
• (A) 21 cm
• (B) 10.5 cm
• (C) 42 cm
• (D) 28 cm
43. Find the perimeter of the following figure- (All lengths are in cm.)
• a) 50 cm
• b) 52 cm
• c) 54 cm
• d) 5.4 cm
Type 06: Speed, Distance and Time
44. A man walks at the rate of 5 km/hr for 6 hours and at 4 km/hr for 12 hours. The average speed of the man is:
• a) 4
• b) 4𝟏/𝟐
• c) 4𝟏/𝟑
• d) 4𝟐/𝟑
45. A car traveling at 40 km/hr completes a journey in 5 hours. At what speed will it have to cover the same distance in 8 hours?
• (A) 30 km/hr
• (B) 40 km/hr
• (C) 35 km/hr
• (D) 25 km/hr
46. A man travelled from a point A to B at the rate of 25 Kmph and walked back at the rate of 4 Kmph. If the whole journey took 5 hrs 48 minutes, the distance between A and B is
• a) 30 km
• b) 24 km
• c) 20 km
• d) 51.6 km
47. A 900 meter long train crosses a platform with the speed of 90 km/hr. It takes 1 minutes to cross the platform. Find the length of platform.
• (A) 800 meter
• (B) 1000 meter
• (C) 600 meter
• (D) 900 meter
48. The speeds of two trains A and B are in the ratio 3 : 4. If train B takes 2 hours less than train A to travel 480 km then find the speed of train A.
• 1. 45 km/h
• 2. 80 km/h
• 3. 30 km/h
• 4. 60 km/h
49. Ram has to cover a certain distance in a specified time. If he moves at the speed of 24 kmph, he will get late by 1 hour and if he moves at the speed of 30 kmph, he will reach 1 hour early. Find the distance.
• (A) 160
• (B) 240
• (C) 380
• (D) 480
50. How much time taken to cross a 1600 meter long platform by a 500 meter long train which is running at the speed of 60 km/hr?
• 1. 156 sec
• 2. 126 sec
• 3. 115 sec
• 4. 220 sec
51. A boat goes 18 km upstream and comes back to its starting point in 2 h and 24 min. If the speed of the current is 4 km/h, what will be the speed of the boat in still water?
• a) 18
• b) 14
• c) 16
• d) 20
52. A thief is noticed by a policeman from a distance of 500 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 12 km per hour respectively. What is the distance between them after 12 minutes?
• 1.150m
• 2. 180m
• 3. 100m
• 4. 130m
53. A 130 meter long train running at a speed of 45 km/h crosses a platform in 30 seconds. What is the length of the platform?
• (A) 245 m
• (B) 375 m
• (C) 275 m
• (D) 345 m
54. A 152.5 meter long train running at a speed of 57 km/h crosses a platform in 39 seconds. What is the length of the platform?
• (A) 378 m
• (B) 278 m
• (C) 356.5 m
• (D) 465 m
Type 07: Average
55. Find the average of numbers between 8 and 74, which is divided by 7.
• (A) 49
• (B) 42
• (C) 56
• (D) 35
56. The average of three numbers is 45. First number is twice the second number and second number is thrice the third number. Then find the smallest number?
• (A) 13.50
• (B) 14.75
• (C) 15
• (D) 17.50
57. The average marks scored by boys in a class is 72. The average marks of class is 77 and there are total 40 students in the class. If number of girls in class is 25 then find the average marks of girls in class.
• 1. 78
• 2. 80
• 3. 82
• 4. 84
58. Mira scored some runs in her continuous 9 innings. She scored 50 runs in her 10th innings and thus her average decreased by 4 runs. What was the average runs at the end of the 10th innings?
• (A) 80
• (B) 94
• (C) 90
• (D) 86
59. 10 years ago the average age of 4 members of a family was 24 years. The present average age of 6 members after the birth of two children is 24 years. If the difference between the ages of the two children is 2 years, then find the age of the younger child.
• 1. 3 years
• 2. 5 years
• 3. 6 years
• 4. 4 years
Type 08: Simplification
60. If [(216)4÷ 364] × 65= 6m, then m = ?
• a) 13
• b) 11
• c) 10
• d) 9
61. If 𝐱 −𝟏/x= 𝟓, 𝒕𝒉𝒆𝒏 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇
• a) 125
• b) 140
• c) 135
• d) 150
62. Find the value of –
• (A) 115
• (B) 770
• (C) 885
• (D) 1000
63. The value of
• a) 31/40
• b) 1/8
• c) 40/9
• d) 40/31
64. If 𝟑𝟒𝒙−𝟐 = 𝟕𝟐𝟗, then find the value of x.
• (A) 4
• (B) 3
• (C) 2
• (D) 5
65. Find the value of –
• (A) 3√12
• (B) 5√2
• (C) 5
• (D) 2
66. What will come in place of question mark‘?’ in the following question?
4 of 8 + [(6 × 3) + (2 – 5 × 3)] – 2 = ?
• 1. 20
• 2. 24
• 3. 30
• 4. 35
67. If 3x = 4y, then find the value of (x + y)/(x – y).
• 1. 7
• 2. 1/7
• 3. 4
• 4. 1/4
68. Find the value of ‘?’ in the following question.
2 of 3 ÷ 9 × 18 + 1 (6 + 3 ) = ?
• 1. 20
• 2. 21
• 3. 22
• 4. 24
69. Find the value of ‘?’ in the following question.
• a) 94
• b) 115
• c) 168
• d) 84
70. Find the value
• 1. 2.5
• 2. 5
• 3. 25
• 4. 1
71. If 0.25x = 0.75y, then find the value of (x + y)/(x – y).
• 1. 0
• 2. 1
• 3. 2
• 4. 4
72. Simplify the following:
• (A) 4/5
• (B) 16/25
• (C) 8/25
• (D) 13/8
Type 09: Time and Work
73. X and Y together can complete a piece of work in 12 days, Y and Z can do it 18 days and X and Z can complete the same work in 15 days. If X, Y and Z can together complete the work, approximately how many days will be required to complete the work?
• (A) 12
• (B) 18
• (C) 14
• (D) 10
74. To complete a certain work, A and B together take 10 days; B and C together take 14 days, and C and A together take 20 days. All the three worked together for 8 days, then A and B left. C alone will complete the remaining work in:
• (A) 10.66 days
• (B) 11.5 days
• (C) 13 days
• (D) 20 days
75. A and B alone can complete work in 9 days and 18 days respectively. They worked together; however 3 days before the completion of the work A left. In how many days was the work completed?
• 1. 5
• 2. 6
• 3. 7
• 4. 8
76. P and Q can do a piece of work in 15 days and 20 days respectively. They work for 5 days and P left then in how many days Q alone can finish the remaining work?
• (A) 𝟖𝟏/𝟑 Days
• (B) 7𝟏/𝟑 Days
• (C) 𝟖 𝟏/𝟐 Days
• (D) 8 Days
77. 6 men or 8 boys can do a piece of work in 18 days, then how many days will 3 men and 5 boys take to do the same work?
• (A) 16 days
• (B) 20 days
• (C) 18 days
• (D) 22 days
78. A can do a piece of work in 6 days. ‘B’ destroy it in 4 days. A does work for 4 days, During the last 2 days of which B has been destroying A’s work. How many more days must A work alone to complete the work?
• 1. 4
• 2. 7
• 3. 5.5
• 4. 5
79. 6 men and 8 women can complete a piece of work in 10 days while 13 men and 24 women can complete the same piece of work in 4 days. Find the ratio of work efficiency of a man and a woman?
• (A) 1 : 2
• (B) 2 : 1
• (C) 2 : 3
• (D) 3 : 2
Also Check Out:-
Type 10: Percentage and Profit-Loss
80. If 9 books are purchased at Rs.18 and sold at Rs 20, then the profit % is
• (A) 11
• (B) 11𝟏/𝟗
• (C) 12
• (D) 𝟏𝟐 𝟏/𝟗
81. In an election between two candidates, one got 56% of the total valid votes, 14% of the votes are invalid. If the total number of votes was 90000, what was the number of valid votes the other candidate got?
• a) 42990
• b) 43344
• c) 52345
• d) 34056
82. Amit’s salary first increased by 5% and then decreased by 5%. What is the percentage change in his salary?
• (A) 25%
• (B) 0.25%
• (C) 2.5%
• (D) 1.25%
83. The cost price of an article is 75% of the marked price. If a discount of 15% is allowed, then the profit or loss percentage is:
• (A) 12.44% loss
• (B) 15% profit
• (C) 15.55% loss
• (D) 13.33% profit
84. The cost price of 14 articles is equal to the selling price of 10 articles. Find the gain per cent.
• (A) 20%
• (B) 40%
• (C) 30%
• (D) 10%
85. A trader sells his goods at a discount of 8%. He still makes a profit of 15%. In order to make a profit of20%, how much percent discount should he allow?
• (A) 3
• (B) 4
• (C) 3.5
• (D) 5
86. An article is sold for Rs 950 after a loss of 5%. If it was sold for Rs.1040, find the profit percent.
• 1. 2%
• 2. 4%
• 3. 6%
• 4. 8%
87. Two successive discounts of 10% and 20% are equivalent to a single discount of?
• 1. 28%
• 2. 27%
• 3. 25%
• 4. 30%
88. In an exam the ratio of girls and boys is 7 : 5. If 40% of the girls and 60% of the boys passed, then what is the ratio of failed girls to failed boys?
• 1. 12 : 13
• 2. 16 : 11
• 3. 41 : 11
• 4. 21 : 10
89. If a product is sold at Rs. 564 then a loss of 6% occurs. Find the selling price if the same product is sold at 6% profit.
• (A) Rs 600
• (B) Rs 636
• (C) Rs 672
• (D) Rs 650
90. The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be.
• (A) 54000
• (B) 54060
• (C) 54080
• (D) 54800
91. In a town there are two schools A and B. In school A there are 75% of total students out of which 20% students are failed in exam. If 10% students of school B are failed in exam then find the overall passing percentage of the town.
• (A) 75%
• (B) 87.5%
• (C) 82.5%
• (D) 77.5%
92. Profit earned by selling at Rs. 1540 is same as the loss incurred by selling the article for Rs. 1140. What is the selling price of the article if it sold at 25% profit?
• (A) Rs. 1600
• (B) Rs. 1680
• (C) Rs. 1520
• (D) Rs. 1675
93. In an election, The winner gets 60% of the total votes and wins the election by 1200 votes then find the number of total voters in the election.
• (A) 6000
• (B) 8000
• (C) 10000
• (D) 12000
94. There are 55 kg of apples in a box. If 3 kg of these are rotten. The price of the apple box is Rs. 2750. If he sell at Rs. 55 per kg. What is the profit after selling all the apples?
• (A) Rs. 110
• (B) Rs. 120
• (C) Rs. 105
• (D) Rs. 95
95. In an examination, 40% of students failed in English, 20% failed in math. If 10% of the students failed in both these subjects, then what percentage of students passed in both the subject?
• (A) 50%
• (B) 40%
• (C) 30%
• (D) None of these
96. A dealer sold a camera for 20% profit on the selling price. Find the selling price if the cost price is Rs. 28,800.
• (A) Rs. 18,000
• (B) Rs. 36,000
• (C) Rs. 54,000
• (D) Rs. 72,000
97. 5 kg honey contains 60% sugar. If 3 kg more sugar is added to it then find new sugar percentage.
• (A) 70%
• (A) 75%
• (A) 80%
• (A) 90%
98. Ravi sold two television sets, one at a profit of 10%, and the other at a loss of 10%. Find the overall loss or gain percentage in the overall transaction.
• (A) Profit, 1%
• (B) Loss, 1%
• (C) Profit, 2%
• (D) No profit, no loss
99. By selling an article for Rs 480 a person lost 20%. For what should he sell it to make a profit of 20%?
• (A) Rs. 675
• (B) Rs. 720
• (C) Rs. 700
• (D) Rs. 750
100. If the cost of production of an article is 2/5 of its marked retail price. If it is sold at 10% discount then what will be profit?
• (A) 25%
• (B) 40%
• (C) 50%
• (D) 125%
101. A person bought some apples at the rate of Rs.3 per apple. He again bought the same number of apples at the rate of Rs.4 per apple. If he sold all the apples at the rate of Rs.3 per apple, then find his loss percent.
• (A) 16.67%
• (B) 12.5%
• (C) 14.28%
• (D) 10%
Thank you!
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Double Math
# Logarithmic Differentiation and its Examples with Solution
Logarithmic Differentiation: Let f(x)=u^v where both u and v are variables or function of x, the derivative of f(x) can be obtained by taking natural logarithms of both of sides and the differentiating .
Example
f(x)=x^x
lnf(x)=ln(x^x)
now by usning property of ln
lnf(x)=x.lnx
differentiate w.r.t x
\frac d{dx}\ln f(x)=\frac d{dx}x.\ln x
property of ln and product rule
\frac1{f(x)}\frac d{dx}f(x)=\ln x\frac d{dx}x+x\frac d{dx}\ln x
\frac{f'(x)}{f(x)}=\ln x(1)+x\frac1x
\frac{f'(x)}{f(x)}=\ln x+1
f'(x)=f(x).\left(\ln x+1\right)
\boxed{f'(x)=x^x.\left(\ln x+1\right)}
Example derivative of lnx and graph
f(x)=lnx
differentiate w.r.t x
\frac d{dx}f(x)=\frac d{dx}\ln x
using ln property
\boxed{f'(x)=\frac1x}
Example derivative of ln x power 2 and graph
f(x)=\ln x^2
differentiate w.r.t x
\frac d{dx}f(x)=\frac d{dx}\ln x^2
f'(x)=\frac1{x^2}\frac d{dx}x^2
f'(x)=\frac1{x^2}2x
\boxed{f'(x)=\frac2x}
This is required derivative of ln of x square.
Example derivative of ln(sinx) and graph
y=\ln(\sin x)
differentiate w.r.t x
\frac d{dx}y=\frac d{dx}\ln(\sin x)
\frac d{dx}y=\frac1{\sin x}.\frac d{dx}(\sin x)
\frac d{dx}y=\frac1{\sin x}.\cos x
\boxed{\frac d{dx}y=cotx}
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# “0.1 Repeating” In Binary Equals 1
In decimal, “0.9 repeating”, or 0.9, equals 1. In binary, a similar thing is true: “0.1 repeating”, or 0.1, equals 1. I’ll show you three ways to prove it, using the three bicimal to fraction conversion algorithms I described recently.
## Proof Using the Series Method
A repeating bicimal is really an infinite geometric series, so one way to prove 0.1 = 1 is to use the series method:
0.1 = 0.111…
= 0.1 + 0.01 + 0.001 + …
= 1/10 + 1/100 + 1/1000 + …
= 1/2 + 1/4 + 1/8 + … (converting to decimal numerals makes it easier)
= 1/21 + 1/22 + 1/23 + …
= (1/2)1 + (1/2)2 + (1/2)3 + …
As I showed when I described the series method, that summation resolves to the formula
,
where r = 1/2 in this case:
### What Did We Prove Exactly?
This proof convinces you only inasmuch as you accept the proof of the formula for infinite geometric series. It’s a tough thing to accept at first — infinity is a strange concept. You have to think of the series as a fixed sum, not as a never-ending process of adding tinier and tinier numbers that gets you ever so close to — but never equal to — 1.
## Proof Using the Subtraction Method
You can also prove 0.1 = 1 using the subtraction method:
b = 0.1
10b = 1.1
10bb = 1
b = 1
(All numerals are in binary.)
Look at that: b started out as 0.1 and ended up as 1; it’s 0.1 and 1 at the same time!
### What Did We Prove Exactly?
This looks like mathematical sleight of hand. We shifted the bicimal left, yet the fractional part remained unchanged. And we subtracted an infinite string of 1s from an infinite string of 1s — is that legal? In the land of infinity, this is OK.
## Proof Using the Direct Method
You can prove 0.1 = 1 using the direct method, which is really just a distillation of the series and subtraction methods. The direct method says to treat the repeating portion as an integer, and to place that integer over a denominator consisting of as many 1s as there are digits in the repeating portion. 0.1 is a pure repeating bicimal with a repeating cycle of one digit, so the fraction it converts to is 1/1; in other words, 1.
## Discussion
In binary, 0.1 is another name for 1. Similar dual naming exists for all bicimals: just as decimal 0.29 = 0.3 and 0.1589 = 0.159, binary 0.01 = 0.1 and 0.10101 = 0.1011. The trailing repeating 1s are the same as a 1 in the next higher place. You can show this using the same three methods above.
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# 16.7 Area Under a Curve. (Don’t write ) We have been emphasizing the connection between the derivative and slopes. There is another fundamental concept.
## Presentation on theme: "16.7 Area Under a Curve. (Don’t write ) We have been emphasizing the connection between the derivative and slopes. There is another fundamental concept."— Presentation transcript:
16.7 Area Under a Curve
(Don’t write ) We have been emphasizing the connection between the derivative and slopes. There is another fundamental concept in Calculus we will lay the foundation for today. The idea of area under a curve. It is easy to find the area under a straight line (could be a rectangle, or triangle, or trapezoid), but what if it is curved or has lots of curves? The underlying idea comes from estimating the area using skinny (and then even more skinny) rectangles.
(Write now) a b Find the area under this curve. We can estimate it! Say we divided it into 4 parts & find areas of RECTANGLES A = b · h (*We can do as many as we want, but our practice problems usually have 4) (difficult task!) Using “inscribed rectangles” - rectangles are completely “under” the curve - for this graph, the rectangle heights are obtained from the left side of each strip (referred to as LRAM) OR Using “circumscribed rectangles” - rectangles are somewhat outside the curve - for this graph, the rectangle heights are obtained from the right side of each strip (referred to as RRAM) Note: In Calculus, these rectangle widths will get smaller & smaller. We would get a limit of the sum of all the areas. left rectangle approx method
Area ≈ base height of 4 rectangles Area ≈ (0.5)(f (1.5)) + (0.5)(f (2)) + (0.5)(f (2.5)) + (0.5)(f (3)) OR ≈ (0.5)(f (1.5) + f (2) + f (2.5) + f (3)) ≈ (0.5)(4.5 + 8 + 12.5 + 18) ≈ (0.5)(43) ≈ 21.5 Let’s learn by doing! Ex 1) Find an approximation of the area of the region bounded by f (x) = 2x 2, y = 0, x = 1, and x = 3 by: a) circumscribed rectangles with a width of 0.5. draw sketch 1 3 2 1.52.5 This is an overestimation! Why? I II III IV
Same process as before... Area ≈ (0.5)(f (1) + f (1.5) + f (2) + f (2.5)) ≈ (0.5)(2 + 4.5 + 8 + 12.5) ≈ (0.5)(27) ≈ 13.5 Ex 1) Find an approximation of the area of the region bounded by f (x) = 2x 2, y = 0, x = 1, and x = 3 by: b) inscribed rectangles with a width of 0.5. draw sketch 1 3 2 1.52.5 This is an underestimation! Why? *Note: the actual area will be some value between 13.5 and 21.5
Note: When the graph was concave up, inscribed rectangles used the left sides & circumscribed rectangles used the right sides. What if it was concave down? Draw a sketch! *This is why the sketch is so important! Don’t memorize – SKETCH! inscribed uses right sides circumscribed uses left sides
Homework #1607 Pg 892 #7 – 11 all
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## About "How to graph a parabola in vertex form"
How to graph a parabola in vertex form :
The equation which is in the form
y = ± a (x - h)2 + k
known as vertex form of the parabola.
Here (h, k) stands for vertex.
To graph the above kinds of parabola, we need to find the following things.
(i) Find the axis of symmetry
(ii) Find the vertex of the parabola
(iii) x - intercept
(iv) y-intercept
Axis of symmetry :
The parabola is symmetric about the variable, which is not having square.
Vertex of parabola :
This point, where the parabola changes direction, is called the "vertex".
x -intercept :
The point where the parabola intersects the x-axis is known as x-intercept
To find the x-intercept, we have to put y = 0 and solve for x.
y -intercept :
The point where the parabola intersects the y-axis is known as x-intercept
To find the y-intercept, we have to put x = 0 and solve for y.
Let us look into some example problems to understand the above concept.
Example 1 :
Graph the following parabola
y = (x + 8)2 - 7
Solution :
Axis of symmetry :
In the above equation we don't have square for the variable y.
Hence the parabola is symmetric about y-axis.
The coefficient of (x + 8)2 is positive, the parabola opens upward.
Equation of axis ==> x = -8
Vertex :
y = (x + 8)2 - 7 ----(1)
y = a (x - h)2 + k -----(2)
By comparing the above equations, we get
(h ,k) ==> (-8, 7)
x - intercept :
Put y = 0
0 = (x + 8)2 - 7
(x - 8)2 = 7
x - 8 = √7
x = ±√7 + 8
x = √7 + 8 and x = -√7 + 8
y - intercept :
Put x = 0
y = (0 + 8)2 - 7
y = 64 - 7
y = 57
Example 2 :
Graph the following parabola
y = -2(x + 5)2 - 3
Solution :
Axis of symmetry :
In the above equation we don't have square for the variable y.
Hence the parabola is symmetric about y-axis.
The coefficient of (x + 5)is negative, the parabola opens downward.
Equation of axis ==> x = -5
Vertex :
y = -2(x + 5)2 - 3 ----(1)
y = -a (x - h)2 + k -----(2)
By comparing the above equations, we get
(h ,k) ==> (-5, -3)
x - intercept :
Put y = 0
y = -2(x + 5)2 - 3
0 = -2(x + 5)2 - 3
-2(x + 5)2 = 3
x + 5 = -3/2
x + 5 = ±√(-3/2) (undefined)
y - intercept :
Put x = 0
y = -2(0 + 5)2 - 3
y = -2(5)2 - 3
y = -50 - 3 ==> -53
After having gone through the stuff given above, we hope that the students would have understood "How to graph a parabola in vertex form".
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HCF and LCM word problems
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Word problems on quadratic equations
Algebra word problems
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Word problems on comparing rates
Converting customary units word problems
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Word problems on fractions
Word problems on mixed fractrions
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OTHER TOPICS
Profit and loss shortcuts
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L.C.M method to solve time and work problems
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Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# Video: Finding the Reciprocals of Complex Numbers in Exponential Form
Given that π§ = 3 (cos (11π)/6 + π sin (11π)/6), find 1/π§ in exponential form.
02:01
### Video Transcript
Given that π§ equals three multiplied by cos of 11π over six add π sin of 11π over 6, find one over π§ in exponential form.
Weβre used to writing complex numbers in their polar and trigonometric form. However, with Eulerβs formula, we can rewrite the polar and trigonometric form of a complex number into its exponential form. This can be particularly useful when finding the reciprocal of a complex number.
In this case, the multiplicative inverse can be given by the formula as shown. Letβs start then by defining the modulus π and the argument π. The modulus can be identified by comparing coefficients in the general polar form here and the complex number in our question. In this case, the modulus is equal to three. The value of its argument is a little trickier to obtain.
Now mathematical convention tells us that the argument must be between π and negative π. But the value of π in our question is outside this range at 11π over six. In the case of the unit circle, since a full turn is two π, we can subtract 11π over six from two π. Two π minus 11π over six is π over six. And because we are measuring this angle in a clockwise direction, the value of our argument is negative π over six.
Once we have finalized our values for the modulus and argument, we can simply substitute these into the formula for the reciprocal. One over π§ is equal to a third π to the power of π multiplied by negative negative π over six. Simplifying fully, we get the reciprocal to be a third π to the power of π over six multiplied by π.
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# Which Of The Following Is An Even Function
An even function is a mathematical function that satisfies the property f(x) = f(-x) for all x in its domain. In simpler terms, an even function is symmetric with respect to the y-axis. This article will discuss what even functions are and provide examples of even functions.
## Characteristics of an Even Function
An even function has the following characteristics:
• Symmetry: An even function is symmetric with respect to the y-axis. This means that if you reflect the graph of the function across the y-axis, the resulting graph will be identical.
• Property: The property of an even function is f(x) = f(-x) for all x in its domain. This property defines the symmetric nature of even functions.
• Examples: Examples of even functions include cosine function, exponential function, and parabola with its vertex on the y-axis.
## Examples of Even Functions
Now, let’s look at some examples of even functions:
• Cosine Function: The cosine function, denoted as cos(x), is an even function. It is symmetric with respect to the y-axis, and cos(x) = cos(-x) for all x.
• Exponential Function: The exponential function, denoted as exp(x) or e^x, is also an even function. It is symmetric with respect to the y-axis, and exp(x) = exp(-x) for all x.
• Parabola: A parabola with its vertex on the y-axis is an even function. The function f(x) = x^2 is an example of an even function.
## Testing for Even Functions
To determine if a function is even, you can apply the following steps:
1. Substitute -x for x: Replace x with -x in the function and simplify the expression.
2. Compare f(x) and f(-x): Check if the resulting expressions are equal. If f(x) = f(-x), then the function is even.
## Which Of The Following Is An Even Function?
Consider the following functions and determine which of them are even functions:
1. f(x) = x^4 – 3x^2 + 2:
2. To test if this function is even, we substitute -x for x:
f(-x) = (-x)^4 – 3(-x)^2 + 2
f(-x) = x^4 – 3x^2 + 2
Since f(x) = f(-x), the function f(x) = x^4 – 3x^2 + 2 is an even function.
3. g(x) = 2sin(x):
4. To test if this function is even, we substitute -x for x:
g(-x) = 2sin(-x)
g(-x) = -2sin(x)
The function g(x) = 2sin(x) is not even, as g(x) is not equal to g(-x).
## Conclusion
In conclusion, even functions are mathematical functions that exhibit symmetry with respect to the y-axis. They satisfy the property f(x) = f(-x) for all x in their domain. Examples of even functions include cosine function, exponential function, and parabolas with vertices on the y-axis. Testing for even functions involves substituting -x for x and comparing the resulting expressions. By understanding the characteristics and examples of even functions, you can easily identify them in various mathematical contexts.
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# Solving Quadratic Equations and Inequalities
## Overview
### Description
There are many ways to solve quadratic equations. The solutions of a quadratic equation can be graphed by determining the $x$-intercepts. If a quadratic function has no $x$-intercepts, then the quadratic equation has no real solutions.
If a quadratic equation has one side equal to zero and the other side can be factored, the factors can be used to determine the solutions of the equation. Completing the square and the quadratic formula can also be used to solve quadratic equations. The quadratic formula contains the discriminant, which determines how many real roots the quadratic equation has.
One way of solving a quadratic inequality is by writing and graphing the related quadratic function. The graph can be used to determine the values of the domain, if any, for which the function rule meets the condition in the inequality. These values form the solution set of the quadratic inequality.
### At A Glance
• The solutions of a quadratic equation are the x-intercepts of the graph of the related function.
• Quadratic equations of the form x2 + bx + c = 0 may be solved by factoring the quadratic expression and setting each factor equal to zero.
• Quadratic equations of the form ax2 + bx + c = 0 may be solved by factoring the quadratic expression and setting each factor equal to zero.
• Recognizing special patterns of quadratic expressions, such as perfect square trinomials and the difference of squares, can help with factoring.
• Equations that are not written in standard form of a quadratic equation may be solved using quadratic methods by substituting a temporary variable for part of the expression.
• All quadratic equations can be solved by completing the square.
• Completing the square for the standard form of a quadratic equation results in the quadratic formula.
• The quadratic formula can be applied to solve any quadratic equation. The discriminant of the related quadratic equation can be used to determine the number and type of roots of the quadratic function.
• More than one method may be appropriate for solving a quadratic equation based on the characteristics of the equation.
• A quadratic inequality may have zero, one, or infinitely many real solutions. The solution set can be identified by graphing the related function and determining which intervals of the domain satisfy the inequality.
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# 3 Systems of Linear Equations and Matrices
## Presentation on theme: "3 Systems of Linear Equations and Matrices"— Presentation transcript:
3 Systems of Linear Equations and Matrices
Systems of Two Equations in Two Unknowns
Suppose you have \$3 in your pocket to spend on snacks and a drink. If x represents the amount you’ll spend on snacks and y represents the amount you’ll spend on a drink, you can say that x + y = 3. On the other hand, if for some reason you want to spend \$1 more on snacks than on your drink, you can also say that x – y = 1. These are simple examples of linear equations in two unknowns.
Systems of Two Equations in Two Unknowns
Linear Equations in Two Unknowns A linear equation in two unknowns is an equation that can be written in the form ax + by = c with a, b, and c being real numbers. The number a is called the coefficient of x and b is called the coefficient of y. A solution of an equation consists of a pair of numbers: a value for x and a value for y that satisfy the equation.
Systems of Two Equations in Two Unknowns
Quick Example In the linear equation 3x – y = 15, the coefficients are a = 3 and b = –1. The point (x, y) = (5, 0) is a solution, because 3(5) – (0) = 15. In fact, a single linear equation such as 3x – y = 15 has infinitely many solutions: We could solve for y = 3x – 15 and then, for every value of x we choose, we can get the corresponding value of y, giving a solution (x, y). These solutions are the points on a straight line, the graph of the equation.
Systems of Two Equations in Two Unknowns
In this section we are concerned with pairs (x, y) that are solutions of two linear equations at the same time. For example, (2, 1) is a solution of both of the equations x + y = 3 and x – y = 1, because substituting x = 2 and y = 1 into these equations gives = 3 (true) and – 1 = 1 (also true), respectively. So, in the simple example we began with, you could spend \$2 on snacks and \$1 on a drink.
Systems of Two Equations in Two Unknowns
In the Example 1, you will see how to graphically and algebraically solve a system of two linear equations in two unknowns.
Find all solutions (x, y) of the following system of two equations:
Example 1 – Two Ways of Solving a System: Graphically and Algebraically Find all solutions (x, y) of the following system of two equations: x + y = 3 x – y = 1. Solution: We will see how to find the solution(s) in two ways: graphically and algebraically. Remember that a solution is a pair (x, y) that simultaneously satisfies both equations.
Example 1 – Solution Method 1: Graphical
cont’d Method 1: Graphical We already know that the solutions of a single linear equation are the points on its graph, which is a straight line. For a point to represent a solution of two linear equations, it must lie simultaneously on both of the corresponding lines. In other words, it must be a point where the two lines cross, or intersect. A look at Figure 1 should convince us that the lines cross only at the point (2, 1), so this is the only possible solution. Figure 1
Example 1 – Solution Method 2: Algebraic
cont’d Method 2: Algebraic In the algebraic approach, we try to combine the equations in such a way as to eliminate one variable. In this case, notice that if we add the left-hand sides of the equations, the terms with y are eliminated. So, we add the first equation to the second (that is, add the left-hand sides and add the right hand sides):
Example 1 – Solution 2x = 4 x = 2.
cont’d 2x = 4 x = 2. Now that we know that x has to be 2, we can substitute back into either equation to find y. Choosing the first equation (it doesn’t matter which we choose), we have 2 + y = 3 y = 3 − 2 = 1. We have found that the only possible solution is x = 2 and y = 1, or (x, y) = (2, 1).
Systems of Two Equations in Two Unknowns
Graphical Method for Solving a System of Two Equations in Two Unknowns Graph both equations on the same graph. (For example, solve each for y to find the slope and y-intercept.) A point of intersection gives the solution to the system. To find the point, you may need to adjust the range of x-values you use. To find the point accurately you may need to use a smaller range (or zoom in if using technology).
Systems of Two Equations in Two Unknowns
Algebraic Method for Solving a System of Two Equations in Two Unknowns Multiply each equation by a nonzero number so that the coefficients of x are the same in absolute value but opposite in sign. Add the two equations to eliminate x; this gives an equation in y that we can solve to find its value. Substitute this value of y into one of the original equations to find the value of x. (Note that we could eliminate y first instead of x if it’s more convenient.)
Systems of Two Equations in Two Unknowns
We summarize the three possible outcomes we have encountered. Possible Outcomes for a System of Two Linear Equations in Two Unknowns 1. A single (or unique) solution: This happens when the lines corresponding to the two equations are distinct and not parallel so that they intersect at a single point. 2. No solution: This happens when the two lines are parallel. We say that the system is inconsistent.
Systems of Two Equations in Two Unknowns
3. An infinite number of solutions: This occurs when the two equations represent the same straight line, and we say that such a system is redundant, or dependent. In this case, we can represent the solutions by choosing one variable arbitrarily and solving for the other. In cases 1 and 3, we say that the system of equations is consistent because it has at least one solution.
Applications
Example 5 – Blending Acme Baby Foods mixes two strengths of apple juice. One quart of Beginner’s juice is made from 30 fluid ounces of water and 2 fluid ounces of apple juice concentrate. One quart of Advanced juice is made from 20 fluid ounces of water and 12 fluid ounces of concentrate. Every day Acme has available 30,000 fluid ounces of water and 3,600 fluid ounces of concentrate. If the company wants to use all the water and concentrate, how many quarts of each type of juice should it mix?
Example 5 – Solution In all applications we follow the same general strategy. 1. Identify and label the unknowns. What are we asked to find? To answer this question, it is common to respond by saying, “The unknowns are Beginner’s juice and Advanced juice.” Quite frankly, this is a baffling statement. Just what is unknown about juice? We need to be more precise: The unknowns are (1) the number of quarts of Beginner’s juice and (2) the number of quarts of Advanced juice made each day.
Example 5 – Solution So, we label the unknowns as follows: Let
cont’d So, we label the unknowns as follows: Let x = number of quarts of Beginner’s juice made each day y = number of quarts of Advanced juice made each day. 2. Use the information given to set up equations in the unknowns. This step is trickier, and the strategy varies from problem to problem. Here, the amount of juice the company can make is constrained by the fact that they have limited amounts of water and concentrate.
Example 5 – Solution cont’d This example shows a kind of application we will often see, and it is helpful in these problems to use a table to record the amounts of the resources used. We can now set up an equation for each of the items listed in the left column of the table.
Example 5 – Solution cont’d Water: We read across the first row. If Acme mixes x quarts of Beginner’s juice, each quart using 30 fluid ounces of water, and y quarts of Advanced juice, each using 20 fluid ounces of water, it will use a total of 30x + 20y fluid ounces of water. But we are told that the total has to be 30,000 fluid ounces. Thus, 30x + 20y = 30,000. This is our first equation.
Example 5 – Solution cont’d Concentrate: We read across the second row. If Acme mixes x quarts of Beginner’s juice, each using 2 fluid ounces of concentrate, and y quarts of Advanced juice, each using 12 fluid ounces of concentrate, it will use a total of 2x + 12y fluid ounces of concentrate. But we are told that the total has to be 3,600 fluid ounces. Thus, 2x + 12y = 3,600. Now we have two equations: 30x + 20y = 30,000 2x + 12y = 3,600.
Example 5 – Solution cont’d To make the numbers easier to work with, let’s divide (both sides of) the first equation by 10 and the second by 2: 3x + 2y = 3,000 x + 6y = 1,800. We can now eliminate x by multiplying the second equation by –3 and adding:
Example 5 – Solution So, y = 2,400/16 = 150.
cont’d So, y = 2,400/16 = 150. Substituting this into the equation x + 6y = 1,800 gives x = 1,800, and so x = 900. The solution is (x, y) = (900, 150). In other words, the company should mix 900 quarts of Beginner’s juice and 150 quarts of Advanced juice.
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# Tips And Tricks To Do The Match Quickly - PowerPoint PPT Presentation
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### Tips And Tricks To Do The Match Quickly
Easily solve all Math problems.
Here at the tips and tricks that every student must do in his mind to get the results, whether there is adding fraction or the dividing fraction, it doesn’t make a difference at all. After these tips, you will be able to do all the things.
1. Addition Of Large Numbers his mind to get the results, whether there is adding fraction or the dividing fraction, it doesn’t make a difference at all. After these tips, you will be able to do all the things.
• Adding fractions that are large enough is difficult to accommodate into the mind. This is the method that is shown here that is simplifying the process by multiplying all the numbers with 10. Such as:
• 288 + 344
As these numbers are difficult to calculate in mind. So you can round these numbers, and this will make it more manageable. Such as if you make 288 to 290 and 344 to 350. Now add the 290 to 350 and the total will be 640.
Now to get the exact answer of the values that we used, you must remember the numbers that you used and subtracted the number from the original value. For examples:
290-288 = 2 and 350-344 = 6
Now add the two numbers, i.e. 2 + 6 = 8.
If you want to get the exact answer, you have to subtract this 8 from the combined value that is 640 and the actual answer will be:
640 – 8 = 632
So, the exact answer is 632.
2. Subtract From 1000 this 8 from the combined value that is 640 and the actual answer will be:
• This is the basic rule to subtract any number that is large enough to be difficult to be subtracted. The simple rule of subtraction is that you must subtract all the numbers except the last one from the 9 and subtract the last number from 10. Let’s suppose:
• 1000 – 544 =
• Subtract 5 from 9 = 4
• Subtract 4 from 9 = 5
• Subtract 4 from 10 = 6
• The answer is the 456.
Multiply A Number 5 Times this 8 from the combined value that is 640 and the actual answer will be:
• Here are the two conditions for the multiplication process. If you have the even number, then multiply it by 5. Here is the example: 5 * 8
Step 1 this 8 from the combined value that is 640 and the actual answer will be:
• The even number that has been selected for the multiplication with the five must be cut into two parts so, eight must be cut into two parts, and it will become the number 4.
Step 2 this 8 from the combined value that is 640 and the actual answer will be:
• Add one zero to this number, and you will get the exact answer. In this case, add 0 to the end of 4, and you will get 40.
When you are multiplying five by an odd number. this 8 from the combined value that is 640 and the actual answer will be:
Such as 5 *3
Step 1 this 8 from the combined value that is 640 and the actual answer will be:
• Subtract one from the odd number that is three will become 2.
Step 2 this 8 from the combined value that is 640 and the actual answer will be:
• Now half the resulting number and add five at the end that is two will be 1 and 15 is the answer.
Source this 8 from the combined value that is 640 and the actual answer will be:
• Phone: (512) 788-5675
• Fax: (512) 519-1805
• Email:[email protected]
• Website:www.quickmath.com
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# Difference between revisions of "2017 AIME I Problems/Problem 9"
## Problem 9
Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$.
## Solution 1
Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives $$a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10$$ Which simplifies to $$a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)$$ Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$, we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$, we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{45}$.
## Solution 2
$$a_n \equiv a_{n-1} + n \pmod {99}$$ By looking at the first few terms, we can see that $$a_n \equiv 10+11+12+ \dots + n \pmod {99}$$ This implies $$a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}$$ Since $a_n \equiv 0 \pmod {99}$, we can rewrite the equivalence, and simplify $$0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}$$ $$0 \equiv n(n+1) - 90 \pmod {99}$$ $$0 \equiv n^2+n+9 \pmod {99}$$ $$0 \equiv 4n^2+4n+36 \pmod {99}$$ $$0 \equiv (2n+1)^2+35 \pmod {99}$$ $$64 \equiv (2n+1)^2 \pmod {99}$$ The smallest squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$, so $$2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}$$ $2n+1 \equiv -8 \pmod {99}$ yields $n=45$ as the smallest integer solution.
$2n+1 \equiv 8 \pmod {99}$ yields $n=53$ as the smallest integer solution.
$2n+1 \equiv -19 \pmod {99}$ yields $n=89$ as the smallest integer solution.
$2n+1 \equiv -8 \pmod {99}$ yields $n=9$ as the smallest integer solution. However, $n$ must be greater than $10$.
The smallest positive integer solution greater than $10$ is $n=\boxed{045}$.
Solution by MathLearner01
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# Division to Solve Decimal Equations
## Solve one - step equations using multiplication.
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Division to Solve Decimal Equations
Have you ever tried to pole vault?
The pole vault is a track and field event where a student uses a pole to launch themselves over a bar. Then the student lands on a large mat underneath the pole. Each part of the pole vault event is very specific. The height of the bar is specific. The length of the pole is specific, and the dimensions of the mat are specific as well.
The track and field team at Harrison Middle school had a special visitor after practice. Jody a pole vaulter from the nearby college visited to share his experiences with the students. He brought some pictures of himself in different events and took a long time answering student questions.
“Even the mat has specific dimensions,” Jody explained. “They measure the length, width and height of the mat to be sure that it has an accurate volume. The mat that we are using has a volume of 9009 cubic feet. The length of the mat is 16.5 feet and the width is 21 feet.”
Justin and Kara were listening intently to Jody’s explanation of the pole vault event. Justin, who loves numbers, began jotting down the dimensions of the mat on a piece of paper.
9009 cubic feet
16.5 feet in length
21 feet in width
Justin knows that he is missing the height of the mat.
“How high is the mat?” Justin asks Kara showing her his notes.
“Who cares?” Kara whispered looking back at Jody.
“I do,” Justin said turning away.
Justin begins to complete the math. But he can’t remember how to work the equation and the division.
This is where you come in. Pay attention to this Concept. By the end of it, you will need to help Justin with his dilemma.
### Guidance
Sometimes division is called the inverse , or opposite, of multiplication. This means that division will “undo” multiplication.
$5 \times 6 &= 30\\30 \div 6 &= 5$
See how that works? You can multiply two factors to get a product. Then when you divide the product by one factor, you get the other factor.
The Inverse Property of Multiplication states that for every number $x, x\left (\frac{1}{x} \right )= 1$ .
In other words, when you multiply $x$ by its opposite, $\frac{1}{x}$ (also known as the multiplicative inverse ), the result is one.
Let’s put whole numbers in place of $x$ to make the property clear: $3 \times \left ( \frac{1}{3} \right ) = \frac{3}{3} = 1$ . The Inverse Property of Multiplication may seem obvious, but it has important implications for our ability to solve variable equations which aren’t easily solved using mental math—such as variable equations involving decimals.
How does this apply to our work with equations?
With equations, the two expressions on either side of the equal sign must be equal at all times. The Inverse Property of Multiplication lets us multiply or divide the same number to both sides of the equation without changing the solution to the equation. This technique is called an inverse operation and it lets us get the variable $x$ alone on one side of the equation so that we can find its value. Take a look at how it’s done.
$5x & = 15\\\frac{5x}{5} &= \frac{15}{5} \rightarrow \text{inverse operation } = \text{divide} \ 5 \ \text{from both sides}\\x &= 3$
Remember how we said that division can “undo” multiplication? Well, this is a situation where that has happened.
Notice how, on the left side of the equation, $\frac{5}{5} = 1$ , leaving the $x$ alone on the left side. This is very instinctual to us. Once you have been working with equations, it comes naturally that when you divide a number by itself the answer is 1 and that has the variable be by itself on that side of the equals. After a while, we don’t even think about it, but it is still important to point out.
Solve $2.7x = 3.78$
We need to find a value of $x$ that, when multiplied by 2.7, results in 3.78. Let’s begin by using inverse operations to get $x$ alone on the left side of the equation.
$2.7x & = 3.78\\\frac{2.7}{2.7} x&= \frac{3.78}{2.7} \rightarrow \text{Inverse Operations} = \text{divide both sides by} \ 2.7$
Now, to find the value of $x$ , we complete the decimal division. First we multiply by ten and move the decimal places accordingly. $2.7 \rightarrow 27$ and $3.78 \rightarrow 37.8$
$& \overset{ \quad \ \ 1.4}{27 \overline{ ) {37.8 \;}}}\\& \quad \ \underline{27\;\;}\\& \qquad 108\\& \quad \underline{- \; 108}\\& \qquad \quad 0$
Use the inverse operation to solve each problem.
#### Example A
$2.3x = 5.06$
Solution: $2.2$
#### Example B
$1.6x= 5.76$
Solution: $3.6$
#### Example C
$4.7x = 10.81$
Solution: $2.3$
Here is the original problem once again.
The pole vault is a track and field event where a student uses a pole to launch themselves over a bar. Then the student lands on a large mat underneath the pole. Each part of the pole vault event is very specific. The height of the bar is specific. The length of the pole is specific, and the dimensions of the mat are specific as well.
The track and field team at Harrison Middle school had a special visitor after practice. Jody a pole vaulter from the nearby college visited to share his experiences with the students. He brought some pictures of himself in different events and took a long time answering student questions.
“Even the mat has specific dimensions,” Jody explained. “They measure the length, width and height of the mat to be sure that it has an accurate volume. The mat that we are using has a volume of 9009 cubic feet. The length of the mat is 16.5 feet and the width is 21 feet.”
Justin and Kara were listening intently to Jody’s explanation of the pole vault event. Justin, who loves numbers, began jotting down the dimensions of the mat on a piece of paper.
9009 cubic feet
16.5 feet in length
21 feet in width
Justin knows that he is missing the height of the mat.
“How high is the mat?” Justin asks Kara showing her his notes.
“Who cares?” Kara whispered looking back at Jody.
“I do, some things are worth figuring out,” Justin said turning away.
Justin begins to complete the math. But he can’t remember how to work the equation and the division.
To solve this problem of height, we need to remember that the formula for volume is length times width times height. Justin has the measurements for the volume and for the length and the width. He is missing the height. Justin can write the following formula.
$V &= lwh\\9009 & = 16.5 \times 21 \times h$
Next, we multiply $16.5 \times 21$ to begin our equation.
$16.5 \times 21 = 346.5$
We can write this equation.
$9009 = 346.5h$
Now we divide 9009 by 346.5.
$h = 26 \ feet$
The height of the mat is equal to 26 feet.
### Vocabulary
Divisor
the number outside the division box. This is the number that is doing the dividing.
Dividend
the number being divided. It is the number inside the division box.
Quotient
the answer in a division problem.
Estimation
Inverse
the opposite
Inverse Property of Multiplication
when you multiply a value by its opposite, the answer is one.
Inverse Operation
the opposite operation. The opposite operation to division is multiplication.
### Guided Practice
Here is one for you to try on your own.
Divide.
$4.5x = 12.6$
To solve this problem, we simply divide 12.6 by 4.5.
$2.8$
### Practice
Directions: Solve the following problems using what you have learned about dividing decimals and equations. Write an equation when necessary.
1. Solve $3.7x = 7.77$
2. Solve $3.1x = 10.23$
3. Solve $7.2x = 29.52$
4. Solve $2.7x = 11.34$
5. Solve $1.2x = 6.72$
6. Solve $11x = 27.5$
7. Solve $6.7x = 42.21$
8. Solve $8.2x = 51.66$
9. Solve $1.9x = 12.92$
10. Solve $5.7x = 54.72$
11. Solve $.55x = .31955$
12. Solve $9.8x = 114.66$
13. In a week of track practice, Rose ran 3.12 times more than Jamie. If Rose ran 17.16 kilometers, how many kilometers did Jamie run? Write an equation and solve.
14. Ling’s flower bed has an area of $23.12 \ m^2$ and a width of 3.4 meters. What is the length of Ling’s flower bed? Write an equation and solve.
15. A jet airplane travels 6.5 times faster than a car. If the jet travels at 627.51 kilometers per hour, how fast is the car? Write an equation and solve.
### Vocabulary Language: English
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Estimation
Estimation
Estimation is the process of finding an approximate answer to a problem.
Inverse Operation
Inverse Operation
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Inverse Property of Multiplication
Inverse Property of Multiplication
The inverse property of multiplication states that the product of any real number and its multiplicative inverse (reciprocal) is one. If $a$ is a nonzero real number, then $a \times \left(\frac{1}{a}\right)=1$.
Quotient
Quotient
The quotient is the result after two amounts have been divided.
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# What is the vertex form of y= 4x^2 -12x + 9 ?
##### 1 Answer
Jun 17, 2017
$y = 4 {\left(x - \frac{3}{2}\right)}^{2}$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"vertex form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where ( h , k ) are the coordinates of the vertex and a is a constant.
$\text{for a parabola in standard form } y = a {x}^{2} + b x + c$
"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)
$y = 4 {x}^{2} - 12 x + 9 \text{ is in standard form}$
$\text{with } a = 4 , b = - 12 , c = 9$
$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 12}{8} = \frac{3}{2}$
$\text{substitute this value into function for y-coordinate}$
$y = 4 {\left(\frac{3}{2}\right)}^{2} - 12 \left(\frac{3}{2}\right) + 9 = 9 - 18 + 9 = 0$
$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{3}{2} , 0\right)$
$\Rightarrow y = 4 {\left(x - \frac{3}{2}\right)}^{2} \leftarrow \textcolor{red}{\text{ in vertex form}}$
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# Ken Ward's Mathematics Pages
## General Formula Using Summation
In previous pages we have looked at various ways to sum the powers of the natural numbers: powers of 1 and 2. Here we will generalise and find one (of many) formulae to give us the sums of powers with much less work.
Some of the techniques we examined, worked only for some of the powers. The technique of summation works for all powers.
In general, the sum of the (k+1) terms is the sum of the k terms plus the (k+1) term:
[1.1]
Expanding the left-hand side, using the Binomial Theorem, we get:
[1.2]
Replacing the left-hand side of 1.1 with the right-hand side of 1.2, we get:
[1,3]
As expected the km+1 terms will cancel.
Leaving the km sum where it is, we can move the rest to the right-hand side;
[1.4]
More generally, we can say:
[1.5]
This is a recursive formula, of course, so you can find the sum of the m powers only when you know the sum of the (m-1) powers. By setting m=0, 1, 2, 3...m, you can find the sum of any m power. .
Let us use the formula, by setting m=0.
Because the lower sum, r=2 exceeds the upper 1, then all that remains is (n+1) which is the sum of 1's from 0 to n. ∑n01=(n+1)!
Let m=3, to find the sum of the first n cubes. Substituting in the equation:
We find:
Giving us the formula:
The sum of the cubes of the first n natural numbers is the square of the formula for the first n natural numbers, so it is easy to remember!
## Examples of Powers of the Natural Numbers
The following table collects information about the coefficients of the sums of the first n powers of the natural numbers.
Power m+1 m m-1 m-2 m-3 m-4 m-5 1 1/2 1/2 2 1/3 1/2 1/6 3 1/4 1/2 1/4 4 1/5 1/2 1/3 -1/30 5 1/6 1/2 5/12 -1/12 6 1/7 1/2 1/2 -1/6 1/42 7 1/2 7/12 -7/24 1/12
Ken Ward's Mathematics Pages
# Faster Arithmetic - by Ken Ward
Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle:
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Close
two lines
## How many intersections can two lines have?
Introduction. Lines that are non-coincident and non-parallel intersect at a unique point. Lines are said to intersect each other if they cut each other at a point. By Euclid’s lemma two lines can have at most 1 1 1 point of intersection.
## How many times can a line intersect a hyperbola?
Since equation (iii) is a quadratic equation in x it can have at most two roots. This shows that the line (i) can intersect the hyperbola (ii) at two points maximum.
## What is hyperbola equation?
The hyperbola is the set of all points (x,y) such that the difference of the distances from (x,y) to the foci is constant. The standard form of an equation of a hyperbola centered at the origin with vertices (±a,0) ( ± a , 0 ) and co-vertices (0±b) ( 0 ± b ) is x2a2−y2b2=1 x 2 a 2 − y 2 b 2 = 1 .
## How do you find the point of intersection of a hyperbola?
Points of Intersection of a Hyperbola and a Line
1. Solve the equation of the line for x to obtain:
2. We now substitute x by 3 – 5y into the equation of the hyperbola to obtain.
3. Expand and group like terms and rewrite the equation as.
4. Solve the quadratic equation for y to obtain two solutions.
5. We now substitute the values of y already obtained into x = 3 – 5y to obtain.
## What is the maximum number of points at which two line segments can meet?
n=2: As one line can’t intersect with itself, it can only intersect with the old lines. As there is only one line, you can get at most one intersection. So the maximum number of intersections of two lines is 1.
## What is the maximum number of points in which three non parallel lines can cross each other?
(xv) The maximum number of points of intersection of three lines is three. (xvi) The minimum number of points of intersection of three lines is one.
## Can three lines intersect at more than three points explain?
Now each of these pairs of lines intersects each other at only a single point as lines are defined as straight, with the y-coordinate of each point on the line having only one corresponding x-coordinate value and vice versa. Therefore we cannot form more than 3 triangles with the points of intersection of 3 lines.
## What does it mean if two equations have no solution?
No solution would mean that there is no answer to the equation. It is impossible for the equation to be true no matter what value we assign to the variable. Infinite solutions would mean that any value for the variable would make the equation true. Note that we have variables on both sides of the equation.
## What is the solution of the system of equations?
A solution to a system of equations means the point must work in both equations in the system. So, we test the point in both equations. It must be a solution for both to be a solution to the system. Hope this helps.
## Is the point a solution to the equation?
And you used this same procedure to graph the equation. This points out an important fact: Every point on the graph was a solution to the equation, and any solution to the equation was a point on the graph. Since this point is on both lines, it thus solves both equations, so it solves the entire system of equation.
## What is true when a system of equations has no solution?
A system of equations is a set of equations in the same variables. The solution set to a system of equations is the set of all values of the variables that make all of the equations in the system true. This system has no solution, so we would say that it’s inconsistent.
## What is the solution to the system of equations y =- 5x 3 and Y 1?
Answer Expert Verified The correct result would be (0.4, 1).
2019-08-16
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Note:
• A quadratic equation is a polynomial equation of degree 2.
• The ''U'' shaped graph of a quadratic is called a parabola.
• A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions.
• There are several methods you can use to solve a quadratic equation:
1. Factoring
2. Completing the Square
4. Graphing
Solve for x in the following equation.
Example 1:
The equation is already set to zero.
If you have forgotten how to manipulate fractions, click on Fractions for a review.
Remove all the fractions by writing the equation in an equivalent form without fractional coefficients. In this problem, you can do it by multiplying both sides of the equation by 2.
Method 1: Factoring
The equation is not easily factored. Therefore, we will not use this method.
Method 2: Completing the square
Add 10 to both sides of the equation
Add to both sides of the equation:
Factor the left side and simplify the right side:
Take the square root of both sides of the equation :
Add 16 to both sides of the equation :
In the equation ,a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Substitute 1 for a, -32 for b, and -10 for c in the quadratic formula and simplify.
Method 4: Graphing
Graph the left side of the equation, and graph the right side of the equation, The graph of is nothing more than the x-axis. So what you will be looking for is where the graph of crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.
You can see from the graph that there are two x-intercepts, one at 32.309506 and one at -0.309506.
The answers are 32.309506 and These answers may or may not be solutions to the original equations. You must verify that these answers are solutions.
Check these answers in the original equation.
Check the solution x=32.309506 by substituting 32.309506 in the original equation for x. If the left side of the equation
equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 32.309506 for x, then x=32.309506 is a solution.
Check the solution x=-0.309506 by substituting -0.309506 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -0.309506 for x, then x= - 0.309506 is a solution.
The solutions to the equation are 32.309506 and - 0.309506.
Comment: You can use the exact solutions to factor the original equation.
Since
Since
The product
Since
then we could say
However the product of the first terms of the factors does not equal
Multiply
Let's check to see if
The factors of are , and
If you would like to work another example, click on Example
If you would like to test yourself by working some problems similar to this example, click on Problem
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Author: Nancy Marcus
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3.2 Dividing Polynomials 11/28/2012. Review: Quotient of Powers Ex. In general:
Presentation on theme: "3.2 Dividing Polynomials 11/28/2012. Review: Quotient of Powers Ex. In general:"— Presentation transcript:
3.2 Dividing Polynomials 11/28/2012
Review: Quotient of Powers Ex. In general:
Use Long Division Find the quotient 985 23. ÷ Divide 98 by 23. 985 23 -92 Subtract the product. 4 () 23 = 92 65 Bring down 5. Divide 65 by 23. 19 Remainder ANSWER The result is written as. 23 19 42 -46 Subtract the product. 2 () 23 = 46 42
Example 1 Use Polynomial Long Division x 3x 3 +4x 24x 2 Subtract the product. () 4x + x 2x 2 = x 3x 3 4x 24x 2 + – 6x6xx 2x 2 – Bring down - 6x. Divide –x 2 by x – 4x4xx 2x 2 – Subtract the product. () 4x + x = x 2x 2 4x4x ––– – 2x2x – 4 Bring down - 4. Divide -2x by x 4 Remainder x 3x 3 + – 6x6x3x 23x 2 – 4x+4 x 3x 3 ÷x = x 2x 2 ANSWER The result is written as. x 2x 2 –– x2 x+4 4 + – 2x2x – 8 Subtract the product () 4x + 2 = 2x2x8 –––. x2x2 -x -2 - +
Synthetic division: Is a method of dividing polynomials by an expression of the form x - k
Example 1 Using Synthetic division x – (-4) in x – k form -4Coefficients of powers of x 1 3 -6 -4 k 1 -4 4 multiply -2 8 4 add coefficients of the power of x in descending order, starting with the power that is one less than that of the dividend. ANSWER x 2x 2 –– x2 x+4 4 + remainder
k Isn’t this the remainder when we performed synthetic division? Remainder Thm:If a polynomials f(x) is divided by x – k, then the remainder is r = f(k)
Example 2 Using Synthetic division and Remainder Theorem 3 Coefficients of powers of x 2 -7 0 6 -14 k 2 6 -3 multiply -3 -9 -23 add remainder -3 -9 P(3)= -23
Example 3 Use Polynomial Long Division Can’t use synthetic division because it isn’t being divided by x-k - + - remainder
Homework: Worksheet 3.2 #1-5all, 11-19odd, 23-25all
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# AP Statistics Curriculum 2007 Distrib Dists
(Difference between revisions)
Revision as of 15:51, 7 May 2010 (view source)IvoDinov (Talk | contribs) (→Geometric)← Older edit Current revision as of 19:35, 23 June 2012 (view source)IvoDinov (Talk | contribs) (→Normal approximation to Negative Binomial distribution) (24 intermediate revisions not shown) Line 6: Line 6: *Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is $P(X = x) = (1 - p)^{x-1} \times p$, for x = 1, 2, 3, 4,.... *Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is $P(X = x) = (1 - p)^{x-1} \times p$, for x = 1, 2, 3, 4,.... - * Expectation: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar | Expected Value]] of a geometrically distributed random variable ''X'' is ${1\over p}.$ + * Expectation: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar | Expected Value]] of a geometrically distributed random variable ''X'' is ${1\over p}.$ This is because [http://en.wikipedia.org/wiki/Geometric_progression geometric series have this property]: + :$\sum_{k=0}^{n} p(1-p)^k = p(1-p)^0+p(1-p)^1+p(1-p)^2+p(1-p)^3+\cdots+p(1-p)^n.$ + : Let r=(1-p), then p=(1-r) and \sum_{k=0}^{n} p(1-p)^k = \begin{align} + (1-r) \sum_{k=0}^{n} r^k & = (1-r)(r^0 + r^1+r^2+r^3+\cdots+r^n) \\ + & = r^0 + r^1+r^2+r^3+\cdots+r^n \\ + & -( r^1+r^2+r^3+\cdots +r^n + r^{n+1}) \\ + & = r^0 - r^{n+1} = 1 - r^{n+1}. + \end{align} + : Thus: $\sum_{k=0}^{n} p(1-p)^k = \frac{p - pr^{n+1}}{1-r} = 1-pr^{n+1},$ which converges to 1, as $n\longrightarrow \infty,$, and hence the above geometric density is well defined. + + : Denote the geometric expectation by E = E(X) = $\sum_{k=0}^{\infty} kpr^k$, where r=1-p. Then $pE = E - (1-p)E = \sum_{k=0}^{\infty} kpr^k - (\sum_{k=0}^{\infty} kpr^{k+1})=$ $\sum_{k=0}^{\infty} pr^k = 1$. Therefore, $E = \frac{1}{p}$. *Variance: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar | Variance]] is ${1-p\over p^2}.$ *Variance: The [[AP_Statistics_Curriculum_2007_Distrib_MeanVar | Variance]] is ${1-p\over p^2}.$ Line 12: Line 22: *Example: See [[SOCR_EduMaterials_Activities_Binomial_Distributions | this SOCR Geometric distribution activity]]. *Example: See [[SOCR_EduMaterials_Activities_Binomial_Distributions | this SOCR Geometric distribution activity]]. - * The Geometric distribution gets its name because its probability mass function is a geometric progression. It is the discrete analogue of the Exponential distribution and also known as Furry distribution. + * The Geometric distribution gets its name because its probability mass function is a [http://en.wikipedia.org/wiki/Geometric_progression geometric progression]. It is the discrete analogue of the Exponential distribution and is also known as Furry distribution. ===HyperGeometric=== ===HyperGeometric=== Line 42: Line 52: ====Examples==== ====Examples==== - * SOCR Activity: The [[SOCR_EduMaterials_Activities_BallAndRunExperiment | SOCR Ball and Urn Experiment]] provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can be varied with scroll bars. + * SOCR Activity: The [[SOCR_EduMaterials_Activities_BallAndRunExperiment | SOCR Ball and Urn Experiment]] provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can vary with scroll bars.
[[Image:SOCR_Activities_BallAndUrnExperiment_SubTopic_Chui_050307_Fig2.JPG|500px]]
[[Image:SOCR_Activities_BallAndUrnExperiment_SubTopic_Chui_050307_Fig2.JPG|500px]]
Line 51: Line 61:
[[Image:SOCR_EBook_Dinov_RV_HyperGeom_013008_Fig9.jpg|500px]]
[[Image:SOCR_EBook_Dinov_RV_HyperGeom_013008_Fig9.jpg|500px]]
- * Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the sample-size (n) is < 5% of the population size(N), we can use [[AP_Statistics_Curriculum_2007_Limits_Bin2HyperG |binomial approximation to hypergeometric]]. Thus if the sample of n=100 fish had 5 tagged, the sample-proportion (estimate of the population proportion) will be $\hat{p}={5\over 100}=0.05$. Thus, we can estimate that $0.05=\hat{p}={200\over N}$, and $N\approx 4,000$, as shown on the figure below. + * Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish, we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the sample-size (n) is < 5% of the population size(N), we can use [[AP_Statistics_Curriculum_2007_Limits_Bin2HyperG |binomial approximation to hypergeometric]]. Thus if the sample of n=100 fish had 5 tagged, the sample-proportion (estimate of the population proportion) will be $\hat{p}={5\over 100}=0.05$. Thus, we can estimate that $0.05=\hat{p}={200\over N}$, and $N\approx 4,000$, as shown in the figure below.
[[Image:SOCR_EBook_Dinov_Prob_HyperG_041108_Fig9a.jpg|500px]]
[[Image:SOCR_EBook_Dinov_Prob_HyperG_041108_Fig9a.jpg|500px]]
Line 74: Line 84: ====Application==== ====Application==== - Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and 70% chance that she'll fail. + Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and a 70% chance that she'll fail. * ''What's the probability mass function of the number of failures (''k=n-r'') to get ''r=6'' successes''?'' * ''What's the probability mass function of the number of failures (''k=n-r'') to get ''r=6'' successes''?'' - : In other words, ''What's the probability mass function that the last 6th state she succeeds to secure all electoral votes happens to be the at the ''n''th state she campaigns in?'' + : In other words, ''what's the probability mass function that the last 6th state she succeeds to secure all electoral votes happens to be at the ''n''th state she campaigns in?'' NegBin(''r'', ''p'') distribution describes the probability of ''k'' failures and ''r'' successes in ''n''=''k''+''r'' Bernoulli(''p'') trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is ''n''=''k+6''. The random variable we are interested in is '''X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}'''. So, ''n'' = ''k+6'', and $X\sim NegBin(r=6, p=0.3)$. Thus, for $n \geq 6$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is: NegBin(''r'', ''p'') distribution describes the probability of ''k'' failures and ''r'' successes in ''n''=''k''+''r'' Bernoulli(''p'') trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is ''n''=''k+6''. The random variable we are interested in is '''X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}'''. So, ''n'' = ''k+6'', and $X\sim NegBin(r=6, p=0.3)$. Thus, for $n \geq 6$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is: Line 92: Line 102:
[[Image:SOCR_EBook_Dinov_RV_NegBinomial_013008_Fig5.jpg|500px]]
[[Image:SOCR_EBook_Dinov_RV_NegBinomial_013008_Fig5.jpg|500px]]
- * Suppose the success of getting all electoral votes within a state is reduced to only 10%, then '''X~NegBin(r=6, p=0.1)'''. Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below)! + * Suppose the success of getting all electoral votes within a state is reduced to only 10%, then '''X~NegBin(r=6, p=0.1)'''. Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below). : ''What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?'' : ''What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?'' :$P(X\geq 50) = 0.632391$ :$P(X\geq 50) = 0.632391$ Line 98: Line 108: * SOCR Activity: If you want to see an interactive Negative-Binomial Graphical calculator you can go to [http://socr.ucla.edu/htmls/SOCR_Experiments.html this applet (select Negative Binomial)] and see [[SOCR_EduMaterials_Activities_NegativeBinomial |this activity]]. * SOCR Activity: If you want to see an interactive Negative-Binomial Graphical calculator you can go to [http://socr.ucla.edu/htmls/SOCR_Experiments.html this applet (select Negative Binomial)] and see [[SOCR_EduMaterials_Activities_NegativeBinomial |this activity]]. + + ====Normal approximation to Negative Binomial distribution==== + The [[AP_Statistics_Curriculum_2007_Limits_CLT|central limit theorem]] provides the foundation for approximation of negative binomial distribution by [[AP_Statistics_Curriculum_2007_Normal_Std| Normal distribution]]. Each negative binomial random variable, $$V_k \sim NB(k,p)$$, may be expressed as a sum of '''k''' independent, identically distributed ([[AP_Statistics_Curriculum_2007_Distrib_Dists#Geometric|geometric]]) random variables $$\{X_i\}$$, i.e., $$V_k = \sum_{i=1}^k{X_i}$$, where [[AP_Statistics_Curriculum_2007_Distrib_Dists |$$X_i \sim Geometric(p)$$]]. In various scientific applications, given a large '''k''', the distribution of $$V_k$$ is approximately normal with mean and variance given by $$\mu=k\frac{1}{p}$$ and $$\sigma^2=k\frac{1-p}{p^2}$$, as $$k \longrightarrow \infty$$. Depending on the parameter '''p''', '''k''' may need to be rather large for the approximation to work well. Also, when using the normal approximation, we should remember to use the continuity correction, since the negative binomial and Normal distributions are discrete and continuous, respectively. + + In the above example, $$P(X\le 8)$$, $$V_k \sim NegBin(k=r=6, p=0.3)$$, the normal distribution approximation, $$N(\mu=\frac{k}{p}=20, \sigma=\sqrt{k\frac{1-p}{p^2}}=6.83)$$, is shown it the following image and table: + +
[[Image:SOCR_EBook_Dinov_RV_NegBinomial_013008_Fig4a.png|500px]]
+ + The probabilities of the real [http://socr.ucla.edu/htmls/dist/NegativeBinomial_Distribution.html Negative Binomial] and [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html approximate Normal] distributions (on the range [2:4]) are not identical but are sufficiently close. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |- + ! Summary|| [http://socr.ucla.edu/htmls/dist/NegativeBinomial_Distribution.html $$NegativeBinomial(k=6,p=0.3)$$ ] || [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html $$Normal(\mu=20, \sigma=6.83)$$ ] + |- + | Mean||20.0||20.0 + |- + | Median||19.0||20.0 + |- + | Variance||46.666667||46.6489 + |- + | Standard Deviation||6.831301||6.83 + |- + | Max Density|| 0.062439||0.058410 + |- + ! colspan=3|Probability Areas + |- + | $$\le 8$$|| .011292|| 0.039433 + |- + | >8|| .988708||0.960537 + |} +
===Negative Multinomial Distribution (NMD)=== ===Negative Multinomial Distribution (NMD)=== Line 104: Line 146: ::''X=Total # of experiments (n) to get r successes'' (and therefore n-r failures); ::''X=Total # of experiments (n) to get r successes'' (and therefore n-r failures); : $X \sim Negative Multinomial(k_0,\{p_0,p_1\})$, : $X \sim Negative Multinomial(k_0,\{p_0,p_1\})$, - :: ''X=Total # of experiments (n) to get $k_0$ (default variable) and $n-k_0$ outcomes of 1 other possible outcome ($X_1$)''. + :: ''X=Total # of experiments (n) to get $k_0$ (default variable, $X_o$) and $n-k_0$ outcomes of the other possible outcome ($X_1$)''. ====Negative Multinomial Summary==== ====Negative Multinomial Summary==== Line 115: Line 157: ====Cancer Example==== ====Cancer Example==== - The [[AP_Statistics_Curriculum_2007_Prob_Rules| Probability Theory Chapter]] of the [[EBook]] shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as in the Table below. + The [[AP_Statistics_Curriculum_2007_Prob_Rules| Probability Theory Chapter]] of the [[EBook]] shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as shown in the Table below.
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 140: Line 182: : $X=\{X_1, X_2, X_3\} \sim NMD(k_0,\{p_1,p_2,p_3\})$. : $X=\{X_1, X_2, X_3\} \sim NMD(k_0,\{p_1,p_2,p_3\})$. - Different columns (sites) are considered to be different instances of the random multinomially distributed vector, X. Then we have the following estimates: + Different columns (sites) are considered to be different instances of the random negative-multinomially distributed vector, X. Then we have the following estimates: * [[AP_Statistics_Curriculum_2007_Estim_MOM_MLE |MLE estimate]] of the Mean: is given by: * [[AP_Statistics_Curriculum_2007_Estim_MOM_MLE |MLE estimate]] of the Mean: is given by: Line 211: Line 253: : Therefore, the best model distribution for the observed sample $x=\{x_1=5,x_2=1,x_3=5\}$ is $X \sim NMD\left (2, \left \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\right\} \right ).$ : Therefore, the best model distribution for the observed sample $x=\{x_1=5,x_2=1,x_3=5\}$ is $X \sim NMD\left (2, \left \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\right\} \right ).$ - : Notice that in this calculation, we explicitely used the complete cancer data table, not only the sample $x=\{x_1=5,x_2=1,x_3=5\}$, as we need multiple samples (multiple sites or columns) to estimate the $k_0$ parameter. + : Notice that in this calculation, we explicitly used the complete cancer data table, not only the sample $x=\{x_1=5,x_2=1,x_3=5\}$, as we need multiple samples (multiple sites or columns) to estimate the $k_0$ parameter. ====SOCR Negative Multinomial Distribution Calculator==== ====SOCR Negative Multinomial Distribution Calculator====
## General Advance-Placement (AP) Statistics Curriculum - Geometric, HyperGeometric, Negative Binomial Random Variables and Experiments
### Geometric
• Definition: The Geometric Distribution is the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set {1, 2, 3, ...}. The name geometric is a direct derivative from the mathematical notion of geometric series.
• Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is $P(X = x) = (1 - p)^{x-1} \times p$, for x = 1, 2, 3, 4,....
$\sum_{k=0}^{n} p(1-p)^k = p(1-p)^0+p(1-p)^1+p(1-p)^2+p(1-p)^3+\cdots+p(1-p)^n.$
Let r=(1-p), then p=(1-r) and \sum_{k=0}^{n} p(1-p)^k = \begin{align} (1-r) \sum_{k=0}^{n} r^k & = (1-r)(r^0 + r^1+r^2+r^3+\cdots+r^n) \\ & = r^0 + r^1+r^2+r^3+\cdots+r^n \\ & -( r^1+r^2+r^3+\cdots +r^n + r^{n+1}) \\ & = r^0 - r^{n+1} = 1 - r^{n+1}. \end{align}
Thus: $\sum_{k=0}^{n} p(1-p)^k = \frac{p - pr^{n+1}}{1-r} = 1-pr^{n+1},$ which converges to 1, as $n\longrightarrow \infty,$, and hence the above geometric density is well defined.
Denote the geometric expectation by E = E(X) = $\sum_{k=0}^{\infty} kpr^k$, where r=1-p. Then $pE = E - (1-p)E = \sum_{k=0}^{\infty} kpr^k - (\sum_{k=0}^{\infty} kpr^{k+1})=$ $\sum_{k=0}^{\infty} pr^k = 1$. Therefore, $E = \frac{1}{p}$.
• Variance: The Variance is ${1-p\over p^2}.$
• The Geometric distribution gets its name because its probability mass function is a geometric progression. It is the discrete analogue of the Exponential distribution and is also known as Furry distribution.
### HyperGeometric
The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. An experimental design for using Hypergeometric distribution is illustrated in this table:
Type Drawn Not-Drawn Total Defective k m-k m Non-Defective n-k N+k-n-m N-m Total n N-n N
• Explanation: Suppose there is a shipment of N objects in which m are defective. The Hypergeometric Distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective.
• Mass function: The random variable X follows the Hypergeometric Distribution with parameters N, m and n, then the probability of getting exactly k successes is given by
$P(X=k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}.$
This formula for the Hypergeometric Mass Function may be interpreted as follows: There are ${{N}\choose{n}}$ possible samples (without replacement). There are ${{m}\choose{k}}$ ways to obtain k defective objects and there are ${{N-m}\choose{n-k}}$ ways to fill out the rest of the sample with non-defective objects.
The mean and variance of the hypergeometric distribution have the following closed forms:
Mean: $n \times m\over N$
Variance: ${ {nm\over N} ( 1-{m\over N} ) (N-n)\over N-1}$
#### Examples
• SOCR Activity: The SOCR Ball and Urn Experiment provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can vary with scroll bars.
• A lake contains 1,000 fish; 100 are randomly caught and tagged. Suppose that later we catch 20 fish. Use SOCR Hypergeometric Distribution to:
• Compute the probability mass function of the number of tagged fish in the sample of 20.
• Compute the expected value and the variance of the number of tagged fish in this sample.
• Compute the probability that this random sample contains more than 3 tagged fish.
• Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish, we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the sample-size (n) is < 5% of the population size(N), we can use binomial approximation to hypergeometric. Thus if the sample of n=100 fish had 5 tagged, the sample-proportion (estimate of the population proportion) will be $\hat{p}={5\over 100}=0.05$. Thus, we can estimate that $0.05=\hat{p}={200\over N}$, and $N\approx 4,000$, as shown in the figure below.
### Negative Binomial
The family of Negative Binomial Distributions is a two-parameter family; p and r with 0 < p < 1 and r > 0. There are two (identical) combinatorial interpretations of Negative Binomial processes (X or Y).
#### X=Trial index (n) of the rth success, or Total # of experiments (n) to get r successes
• Probability Mass Function: $P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} \!$, for n = r,r+1,r+2,.... (n=trial number of the rth success)
• Mean: $E(X)= {r \over p}$
• Variance: $Var(X)= {r(1-p) \over p^2}$
#### Y = Number of failures (k) to get r successes
• Probability Mass Function: $P(Y=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k \!$, for k = 0,1,2,.... (k=number of failures before the rth successes)
• $Y \sim NegBin(r, p)$, the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial.
• Mean: $E(Y)= {r(1-p) \over p}$.
• Variance: $Var(Y)= {r(1-p) \over p^2}$.
• Note that X = Y + r, and E(X) = E(Y) + r, whereas VAR(X)=VAR(Y).
#### Application
Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and a 70% chance that she'll fail.
• What's the probability mass function of the number of failures (k=n-r) to get r=6 successes?
In other words, what's the probability mass function that the last 6th state she succeeds to secure all electoral votes happens to be at the nth state she campaigns in?
NegBin(r, p) distribution describes the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is n=k+6. The random variable we are interested in is X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}. So, n = k+6, and $X\sim NegBin(r=6, p=0.3)$. Thus, for $n \geq 6$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:
$P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} = {n - 1 \choose r-1} \cdot 0.3^6 \cdot 0.7^{n-r}$
• What's the probability that Jane finishes her campaign in the 10th state?
Let $X\sim NegBin(r=6, p=0.3)$, then $P(X=10) = {10-1 \choose 6-1}\cdot 0.3^6 \cdot 0.7^{10-6} = 0.022054.$
• What's the probability that Jane finishes campaigning on or before reaching the 8th state?
$P(X\leq 8) = 0.011292$
• Suppose the success of getting all electoral votes within a state is reduced to only 10%, then X~NegBin(r=6, p=0.1). Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below).
What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?
$P(X\geq 50) = 0.632391$
#### Normal approximation to Negative Binomial distribution
The central limit theorem provides the foundation for approximation of negative binomial distribution by Normal distribution. Each negative binomial random variable, $$V_k \sim NB(k,p)$$, may be expressed as a sum of k independent, identically distributed (geometric) random variables $$\{X_i\}$$, i.e., $$V_k = \sum_{i=1}^k{X_i}$$, where $$X_i \sim Geometric(p)$$. In various scientific applications, given a large k, the distribution of $$V_k$$ is approximately normal with mean and variance given by $$\mu=k\frac{1}{p}$$ and $$\sigma^2=k\frac{1-p}{p^2}$$, as $$k \longrightarrow \infty$$. Depending on the parameter p, k may need to be rather large for the approximation to work well. Also, when using the normal approximation, we should remember to use the continuity correction, since the negative binomial and Normal distributions are discrete and continuous, respectively.
In the above example, $$P(X\le 8)$$, $$V_k \sim NegBin(k=r=6, p=0.3)$$, the normal distribution approximation, $$N(\mu=\frac{k}{p}=20, \sigma=\sqrt{k\frac{1-p}{p^2}}=6.83)$$, is shown it the following image and table:
The probabilities of the real Negative Binomial and approximate Normal distributions (on the range [2:4]) are not identical but are sufficiently close.
Summary $$NegativeBinomial(k=6,p=0.3)$$ $$Normal(\mu=20, \sigma=6.83)$$
Mean20.020.0
Median19.020.0
Variance46.66666746.6489
Standard Deviation6.8313016.83
Max Density 0.0624390.058410
Probability Areas
$$\le 8$$ .011292 0.039433
>8 .9887080.960537
### Negative Multinomial Distribution (NMD)
The Negative Multinomial Distribution is a generalization of the two-parameter Negative Binomial distribution (NB(r,p)) to $m\ge 1$ outcomes. Suppose we have an experiment that generates $m\ge 1$ possible outcomes, $\{X_0,\cdots,X_m\}$, each occurring with probability $\{p_0,\cdots,p_m\}$, respectively, where with 0 < pi < 1 and $\sum_{i=0}^m{p_i}=1$. That is, $p_0 = 1-\sum_{i=1}^m{p_i}$. If the experiment proceeds to generate independent outcomes until $\{X_0, X_1, \cdots, X_m\}$ occur exactly $\{k_0, k_1, \cdots, k_m\}$ times, then the distribution of the m-tuple $\{X_1, \cdots, X_m\}$ is Negative Multinomial with parameter vector $(k_0,\{p_1,\cdots,p_m\})$. Notice that the degree-of-freedom here is actually m, not (m+1). That is why we only have a probability parameter vector of size m, not (m+1), as all probabilities add up to 1 (so this introduces one relation). Contrast this with the combinatorial interpretation of Negative Binomial (special case with m=1):
X˜NegativeBinomial(NumberOfSuccesses = r,ProbOfSuccess = p),
X=Total # of experiments (n) to get r successes (and therefore n-r failures);
X˜NegativeMultinomial(k0,{p0,p1}),
X=Total # of experiments (n) to get k0 (default variable, Xo) and nk0 outcomes of the other possible outcome (X1).
#### Negative Multinomial Summary
• Probability Mass Function: $P(k_1, \cdots, k_m|k_0,\{p_1,\cdots,p_m\}) = \left (\sum_{i=0}^m{k_i}-1\right)!\frac{p_0^{k_0}}{(k_0-1)!} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, or equivalently:
$P(k_1, \cdots, k_m|k_0,\{p_1,\cdots,p_m\}) = \Gamma\left(\sum_{i=1}^m{k_i}\right)\frac{p_0^{k_0}}{\Gamma(k_0)} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, where Γ(x) is the Gamma function.
• Mean (vector): $\mu=E(X_1,\cdots,X_m)= (\mu_1=E(X_1), \cdots, \mu_m=E(X_m)) = \left ( \frac{k_0p_1}{p_0}, \cdots, \frac{k_0p_m}{p_0} \right)$.
• Variance-Covariance (matrix): Cov(Xi,Xj) = {cov[i,j]}, where
$cov[i,j] = \begin{cases} \frac{k_0 p_i p_j}{p_0^2},& i\not= j,\\ \frac{k_0 p_i (p_i + p_0)}{p_0^2},& i=j.\end{cases}$.
#### Cancer Example
The Probability Theory Chapter of the EBook shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as shown in the Table below.
Type Site Totals Head and Neck Trunk Extremities Hutchinson's melanomic freckle 22 2 10 34 Superficial 16 54 115 185 Nodular 19 33 73 125 Indeterminant 11 17 28 56 Column Totals 68 106 226 400
The sites (locations) of the cancer may be independent, but there may be positive dependencies of the type of cancer for a given location (site). For example, localized exposure to radiation implies that elevated level of one type of cancer (at a given location) may indicate higher level of another cancer type at the same location. We want to use the Negative Multinomial distribution to model the sites cancer rates and try to measure some of the cancer type dependencies within each location.
Let's denote by xi,j the cancer rates for each site ($0\leq i \leq 2$) and each type of cancer ($0\leq j \leq 3$). For each (fixed) site ($0\leq i \leq 2$), the cancer rates are independent Negative Multinomial distributed random variables. That is, for each column index (site) the column-vector X has the following distribution:
X = {X1,X2,X3NMD(k0,{p1,p2,p3}).
Different columns (sites) are considered to be different instances of the random negative-multinomially distributed vector, X. Then we have the following estimates:
$\hat{\mu}_{i,j} = \frac{x_{i,.}\times x_{.,j}}{x_{.,.}}$
$x_{i,.} = \sum_{j=0}^{3}{x_{i,j}}$
$x_{.,j} = \sum_{i=0}^{2}{x_{i,j}}$
$x_{.,.} = \sum_{i=0}^{2}\sum_{j=0}^{3}{{x_{i,j}}}$
Example: $\hat{\mu}_{1,1} = \frac{x_{1,.}\times x_{.,1}}{x_{.,.}}=\frac{34\times 68}{400}=5.78$
• Variance-Covariance: For a single column vector, X = {X1,X2,X3NMD(k0,{p1,p2,p3}), covariance between any pair of Negative Multinomial counts (Xi and Xj) is:
$cov[X_i,X_j] = \begin{cases} \frac{k_0 p_i p_j}{p_0^2},& i\not= j,\\ \frac{k_0 p_i (p_i + p_0)}{p_0^2},& i=j.\end{cases}$.
Example: For the first site (Head and Neck, j=0), suppose that $X=\left \{X_1=5, X_2=1, X_3=5\right \}$ and X˜NMD(k0 = 10,{p1 = 0.2,p2 = 0.1,p3 = 0.2}). Then:
$p_0 = 1 - \sum_{i=1}^3{p_i}=0.5$
NMD(X | k0,{p1,p2,p3}) = 0.00465585119998784
$cov[X_1,X_3] = \frac{10 \times 0.2 \times 0.2}{0.5^2}=1.6$
$\mu_2=\frac{k_0 p_2}{p_0} = \frac{10\times 0.1}{0.5}=2.0$
$\mu_3=\frac{k_0 p_3}{p_0} = \frac{10\times 0.2}{0.5}=4.0$
$corr[X_2,X_3] = \left (\frac{\mu_2 \times \mu_3}{(k_0+\mu_2)(k_0+\mu_3)} \right )^{\frac{1}{2}}$ and therefore, $corr[X_2,X_3] = \left (\frac{2 \times 4}{(10+2)(10+4)} \right )^{\frac{1}{2}} = 0.21821789023599242.$
You can also use the interactive SOCR negative multinomial distribution calculator to compute these quantities, as shown on the figure below.
• There is no MLE estimate for the NMD k0 parameter (see this reference). However, there are approximate protocols for estimating the k0 parameter, see the example below.
• Correlation: correlation between any pair of Negative Multinomial counts (Xi and Xj) is:
$Corr[X_i,X_j] = \begin{cases} \left (\frac{\mu_i \times \mu_j}{(k_0+\mu_i)(k_0+\mu_j)} \right )^{\frac{1}{2}} = \left (\frac{p_i p_j}{(p_0+p_i)(p_0+p_j)} \right )^{\frac{1}{2}}, & i\not= j, \\ 1, & i=j.\end{cases}$.
• The marginal distribution of each of the Xi variables is negative binomial, as the Xi count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of $X=\{X_1,\cdots,X_m\}$ is negative multinomial, i.e., $X \sim NMD(k_0,\{p_1,\cdots,p_m\})$ .
Notice that the pair-wise NMD correlations are always positive, where as the correlations between multinomail counts are always negative. Also note that as the parameter k0 increases, the paired correlations go to zero! Thus, for large k0, the Negative Multinomial counts Xi behave as independent Poisson random variables with respect to their means $\left ( \mu_i= k_0\frac{p_i}{p_0}\right )$.
#### Parameter estimation
• Estimation of the mean (expected) frequency counts (μj) of each outcome (Xj):
The MLE estimates of the NMD mean parameters μj are easy to compute.
If we have a single observation vector $\{x_1, \cdots,x_m\}$, then $\hat{\mu}_i=x_i.$
If we have several observation vectors, like in this case we have the cancer type frequencies for 3 different sites, then the MLE estimates of the mean counts are $\hat{\mu}_j=\frac{x_{j,.}}{I}$, where $0\leq j \leq J$ is the cancer-type index and the summation is over the number of observed (sampled) vectors (I).
For the cancer data above, we have the following MLE estimates for the expectations for the frequency counts:
Hutchinson's melanomic freckle type of cancer (X0) is $\hat{\mu}_0 = 34/3=11.33$.
Superficial type of cancer (X1) is $\hat{\mu}_1 = 185/3=61.67$.
Nodular type of cancer (X2) is $\hat{\mu}_2 = 125/3=41.67$.
Indeterminant type of cancer (X3) is $\hat{\mu}_3 = 56/3=18.67$.
• Estimation of the k0 (gamma) parameter:
There is no MLE for the k0 parameter; however, there is a protocol for estimating k0 using the chi-squared goodness of fit statistic. In the usual chi-squared statistic:
$\Chi^2 = \sum_i{\frac{(x_i-\mu_i)^2}{\mu_i}}$, we can replace the expected-means (μi) by their estimates, $\hat{\mu_i}$, and replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data:
$\Chi^2(k_0) = \sum_{i}{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$.
Now we can derive a simple method for estimating the k0 parameter by varying the values of k0 in the expression Χ2(k0) and matching the values of this statistic with the corresponding asymptotic chi-squared distribution. The following protocol summarizes these steps using the cancer data above:
df = (# rows – 1)(# columns – 1) = (3-1)*(4-1) = 6
• Mean Counts Estimates: The mean counts estimates (μj) for the 4 different cancer types are:
$\hat{\mu}_1 = 185/3=61.67$; $\hat{\mu}_2 = 125/3=41.67$; and $\hat{\mu}_3 = 56/3=18.67$.
• Thus, we can solve the equation above Χ2(k0) = 5.261948 for the single variable of interest -- the unknown parameter k0. Suppose we are using the same example as before, x = {x1 = 5,x2 = 1,x3 = 5}. Then the solution is an asymptotic chi-squared distribution driven estimate of the parameter k0.
$\Chi^2(k_0) = \sum_{i=1}^3{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$. $\Chi^2(k_0) = \frac{(5-61.67)^2}{61.67(1+61.67/k_0)}+\frac{(1-41.67)^2}{41.67(1+41.67/k_0)}+\frac{(5-18.67)^2}{18.67(1+18.67/k_0)}=5.261948.$ Solving this equation for k0 provides the desired estimate for the last parameter.
Mathematica provides 3 distinct (k0) solutions to this equation: {50.5466, -21.5204, 2.40461}. Since k0 > 0 there are 2 candidate solutions.
• Estimates of Probabilities: Assume k0 = 2 and $\frac{\mu_i}{k_0}p_0=p_i$, we have:
$\frac{61.67}{k_0}p_0=31p_0=p_1$
20p0 = p2
9p0 = p3
Hence, 1 − p0 = p1 + p2 + p3 = 60p0. Therefore, $p_0=\frac{1}{61}$, $p_1=\frac{31}{61}$, $p_2=\frac{20}{61}$ and $p_3=\frac{9}{61}$.
Therefore, the best model distribution for the observed sample x = {x1 = 5,x2 = 1,x3 = 5} is $X \sim NMD\left (2, \left \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\right\} \right ).$
Notice that in this calculation, we explicitly used the complete cancer data table, not only the sample x = {x1 = 5,x2 = 1,x3 = 5}, as we need multiple samples (multiple sites or columns) to estimate the k0 parameter.
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# Difference between revisions of "2008 AMC 12A Problems/Problem 22"
The following problem is from both the 2008 AMC 12A #22 and 2008 AMC 10A #25, so both problems redirect to this page.
## Problem
A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$[asy]unitsize(4mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$1$$",(-0.5,3.8),S);[/asy]$
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$
## Solution 1 (Trigonometry)
Let one of the mats be $ABCD$, and the center be $O$ as shown:
$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686));[/asy]$
Since there are $6$ mats, $\Delta BOC$ is equilateral. So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$.
By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$.
Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C$.
## Solution 2 (without trigonometry)
Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$
$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label("$$E$$", EE,SE); [/asy]$
As proved in the first solution, $\angle OCD = 150^\circ$. That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$
Since $\Delta ODE$ is a right triangle, $\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0$
Solving for $x$ gives $x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C$
## Solution 3 (simply Pythagorean Theorem)
$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$E$$",(0,4.17)); label("$$F$$",(0.75,4.15),W); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$
By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (or this can also be deduced from Pythagoras on $\triangle CED$). Thus $OC = \frac{3\sqrt{7}-\sqrt{3}}{2}$. As previous solutions noted, $\triangle BOC$ is equilateral, and thus the desired length is $x = OC \implies C$.
## Solution 3
$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.95,3),E); label("$$A$$",(-3.6,2.5513),E); label("$$C$$",(0.05,3.20),E); label("$$E$$",(0.40,-3.60),E); label("$$B$$",(-0.75,4.15),E); label("$$D$$",(-2.62,1.5),E); label("$$F$$",(-2.64,-1.43),E); label("$$G$$",(-0.2,-2.8),E); label("$$\sqrt{3}x$$",(-1.5,-0.5),E); label("$$M$$",(-2,-0.9),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-2.7,2.3),S); label("$$1$$",(0.1,-3.4),S); label("$$8$$",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$
Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$. $AE = 2 + \sqrt{3}x$
Use the Pythagorean theorem on triangle $ABE$, we get $$(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow C$$
~Zeric Hang
## Soultion 4 (coordinate bashing)
$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$1$$",(-0.5,3.8),S);[/asy]$
We will let $O(0,0)$ be the origin. This way the coordinates of C would be $(0,x)$. By 30-60-90, the coordinates of D would be $(\sqrt{-1}{2}, x + \frac{\sqrt{3}}{2})$. The distance $(x, y)$ is from the origin is just $\sqrt{x^2 + y^2}$. Therefore, the distance D is from the origin is both 4 and $\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16$. We get the equation mentioned in all the previous solution, using the quadratic formula, we get that $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C$
## Solution 5
$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$E$$",(0.3,4.15),E); label("$$F$$",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2));[/asy]$
Notice that $\overarc{AE}$ is one-sixth the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius, or $4$. $\angle AFE$ is a right angle, therefore $\triangle AFE$ is a right triangle. $\overline{AF}$ is half the length of $1$, or $\frac{1}{2}$. The length of $\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles, or $x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get
$4^2 = \left(\frac{1}{2}\right)^2 + \left(x+\frac{\sqrt{3}}{2}\right)^2$
Solving for $x$, we get $x =$
$\frac{3\sqrt{7}-\sqrt{3}}{2}$ , or $\boxed{\text{C}}$.
|
Bayes’ Theorem Formula
Bayes’ Theorem formula is a very important method for calculating conditional probabilities. It is used to calculate posterior probabilities under some already give a probability. This theorem describes the probability of an event, based on conditions that might be related to the event. For example, a patient is observed to have a certain symptom. Here the Bayes’ formula can be used to compute the probability that a diagnosis is correct, with the given observation. In this topic, we will discuss conditional probability and Bayes’ theorem Formula with examples. Let us learn the interesting topic.
Bayes’ Theorem formula
What Is Conditional Probability?
Conditional probabilities arise naturally in the investigation of experiments where some outcome of a trial may affect the outcomes of the trials subsequently. We may try to calculate the probability of the second event say event B given that the first event says event A has already happened. If the probability of second event changes while taking the first event into consideration. Then we can safely say that the probability of event B will be dependent on the occurrence of event A.
We can write the conditional probability as $$P(A | B)$$, the probability of the occurrence of event A given that B has already happened.
$$P(A | B) = \frac{P(A and B)}{P(B)} = \frac {Probability of the occurrence of both A and B}{ Probability of B}$$
Source: en.wikipedia.org
Bayes Theorem:
In statistics and probability theory, the Bayes’ theorem or Bayes’ rule is a mathematical formula used to determine the conditional probability of the events. Actually the Bayes’ theorem describes the probability of an event based on prior knowledge of the conditions, relevant to the event.
Bayes’ theorem is named after Thomas Bayes. He first provided an equation that allows new evidence to update beliefs. If we know the conditional probability $$P(B | A)$$ , we can use the Bayes rule to find out the reverse probabilities $$P(A | B)$$ as well. This theorem says that,
$$P(A | B) = P(B | A) \times \frac{P(A)}{P(B)}$$
We can represent the above statement as the general statement as below:
$$P(A_i | B) = \frac{P(B | A_i) \times P(A_i)} {\displaystyle\sum\limits_{i=1}^n (P(B |A_i) \times P(A_i))}$$
$$A_i$$ is the ith event with probability $$P(A_i)$$
Solved Examples for Bayes’ Theorem Formula
Q.1: We wish to find a person’s probability of having rheumatoid arthritis if they have hay fever. Having hay fever is the test for rheumatoid arthritis i.e. the event in this case. A is the event “patient has rheumatoid arthritis.” Data indicates 10 percent of patients in a clinic have this type of arthritis. B is the test “patient has hay fever.” Data indicates 5 percent of patients in a clinic have hay fever. The clinic’s records also show that of the patients with rheumatoid arthritis, 7 percent have hay fever. In other words, the probability that a patient has hay fever, given they have rheumatoid arthritis, is 7 percent.
Solution: Given terms in the problem are:
• P(A) = 0.10
• P(B) = 0.05
• $$P(B | A)$$ =0.07
Now, we use the Bays theorem formula:
$$P(A | B) = P(B | A) \times \frac{P(A)}{P(B)}$$
Thus,
$$P(A | B) = \frac {0.07 \times 0.10}{0.05}$$
$$P(A | B) = 0.14$$
Therefore, if a patient has hay fever, then chance of having rheumatoid arthritis is 14 percent.
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One response to “Equation Formula”
1. KUCKOO B says:
I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26
|
# 2018 AMC 10B Problems/Problem 8
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Sara makes a staircase out of toothpicks as shown:
$[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } } [/asy]$
This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircase that used 180 toothpicks?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30$
## Solutions
### Solution 1
A staircase with $n$ steps contains $4 + 6 + 8 + ... + 2n + 2$ toothpicks. This can be rewritten as $(n+1)(n+2) -2$.
So, $(n+1)(n+2) - 2 = 180$
So, $(n+1)(n+2) = 182.$
Inspection could tell us that $13 \cdot 14 = 182$, so the answer is $13 - 1 = \boxed {(C) 12}$
### Solution 2
Layer $1$: $4$ steps
Layer $1,2$: $10$ steps
Layer $1,2,3$: $18$ steps
Layer $1,2,3,4$: $28$ steps
From inspection, we can see that with each increase in layer the difference in toothpicks between the current layer and the previous increases by $2$. Using this pattern:
$4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180$
From this we see that the solution is $\boxed {(C) 12}$
By: Soccer_JAMS
### Solution 3
We can find a function that gives us the number of toothpicks for every layer. Using finite difference, we know that the degree must be $2$ and the leading coefficient is $1$. The function is $f(n)=n^2+3n$ where $n$ is the layer and $f(n)$ is the number of toothpicks.
We have to solve for $n$ when $n^2+3n=180\Rightarrow n^2+3n-180=0$. Factor to get $(n-12)(n+15)$. The roots are $12$ and $-15$. Clearly $-15$ is impossible so the answer is $\boxed {(C) 12}$.
~Zeric Hang
### Not a Solution! Just an observation.
Notice that the number of toothpicks can be found by adding all the horizontal and all the vertical toothpicks. We can see that for the case of 3 steps, there are $2(3+3+2+1)=18$ toothpicks. Thus, the equation is $2S + 2(1+2+3...+S)=180$ with $S$ being the number of steps. Solving, we get $S = 12$, or $\boxed {(C) 12}$. -liu4505
### Solution 5 General Formula
There are $\frac{n(n+1)}{2}$ squares. Each has $4$ toothpick sides. To remove overlap, note that there are $4n$ perimeter toothpicks. $\frac{\frac{n(n+1)}{2}\cdot 4-4n}{2}$ is the number of overlapped toothpicks Add $4n$ to get the perimeter (non-overlapping). Formula is $\text{number of toothpicks} = n^2+3n$ Then you can "guess" or factor (also guessing) to get the answer $\boxed{\text{(C) }12}$. ~bjc
### Not a solution! Just an observation.
If you are trying to look for a pattern, you can see that the first column is made of 4 toothpicks. The second one is made from 2 squares: 3 toothpicks for the first square and 4 for the second. The third one is made up of 3 squares: 3 toothpicks for the first and second one, and 4 for the third one. The pattern continues like that. So for the first one, you have 0 "3 toothpick squares" and 1 "4 toothpick squares". The second is 1 to 1. The third is 2:1. And the amount of three toothpick squares increase by one every column.
The list is as follow for the number of toothpicks used... 4,4+3,4+6,4+9, and so on. 4, 7, 10, 13, 16, 19, ...
- Flutterfly
## Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
|
<meta http-equiv="refresh" content="1; url=/nojavascript/">
# Algebra Expressions with Exponents
%
Progress
Practice Algebra Expressions with Exponents
Progress
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Algebra Expressions with Exponents
Remember the tent dilemma from the Whole Number Exponents?
Well, the hikers were given a specific tent with specific dimensions. They were given a Kelty Trail Dome 6 tent.
What if a different tent was used? What if many different tents were used?
The square footage of the floor would always have an exponent of 2, but a variable would be needed for the base because different size tents would be being used.
Here is how we could write this.
$a^2$
In this case, a is the length of one side of a square tent.
What if a tent with 8 feet on one side was being used?
What if a tent with 15 feet on one side was being used?
What would the square footage of each tent be?
This Concept will teach you how to evaluate powers with variable bases. Pay attention and you will know how to work through this at the end of the Concept.
### Guidance
When we are dealing with numbers, it is often easier to just simplify. It makes more sense to deal with 16 than with $4^2$ . Exponential notation really comes in handy when we’re dealing with variables. It is easier to write $y^{12}$ than it is to write $yyyyyyyyyyyy$ .
Yes, and we can simplify by using exponential form and we can also write out the variable expression by using expanded form.
Write the following in expanded form: $x^5$
To write this out, we simply write out each $x$ five times.
$x^5=xxxxx$
We can work the other way to by taking an variable expression in expanded form and write it in exponential form.
$aaaa$
Our answer is $a^4$ .
What about when we multiply two variable terms with exponents?
To do this, we are going to need to follow a few rules.
$(m^3)(m^2)$
The first thing to notice is that these terms have the same base. Both bases are m’s. Because of this, we can simplify the expression quite easily.
Let’s write it out in expanded form.
$mmm(mm)$
Here we have five $m$ ’s being multiplied our answer is $m^5$ .
Here is the rule.
Let’s apply this rule to the next one.
$(x^6)(x^3)$
The bases are the same, so we add the exponents.
$x^{6+3}= x^9$
We can also have an exponential term raised to a power. When this happens, one exponent is outside the parentheses. This means something different.
$(x^2)^3$
Let’s think about what this means. It means that we are multiplying $x$ squared by itself three times. We can write this out in expanded form.
$(x^2)(x^2)(x^2)$
Now we are multiplying three bases that are the same so we use Rule 1 and add the exponents.
Our answer is $x^6$ .
We could have multiplied the two exponents in the beginning.
$(x^2)^3= x^{2(3)} =x^6$
Here is Rule 2.
Simplify $x^0$
Our answer is $x^0 = 1$
Anything to the power of 0 equals 1.
Now it's time for you to try a few on your own.
#### Example A
Write the following in exponential form: $aaaaaaa$
Solution: $a^7$
#### Example B
Simplify: $(a^3)(a^8)$
Solution: $a^{11}$
#### Example C
Simplify: $(x^4)^2$
Solution: $x^8$
Remember the tent dilemma from the beginning of the Concept? Well let's take a look at it again.
The hikers were given a specific tent with specific dimensions. Remember, they were given a Kelty Trail Dome 6.
What if a different tent was used? What if many different tents were used?
The square footage of the floor would always have an exponent of 2, but a variable would be needed for the base because different size tents would be being used.
Here is how we could write this.
$a^2$
In this case, a is the length of one side of a square tent.
What if a tent with 8 feet on one side was being used?
What if a tent with 15 feet on one side was being used?
What would the square footage of each tent be?
Here is our solution.
$8^2 = 64$ square feet is the first tent.
$15^2 = 225$ square feet is the second tent.
### Guided Practice
Here is one for you to try on your own.
Simplify: $(x^6)(x^2)$
When we multiply variables with exponents, we add the exponents.
Our answer is $x^8$ .
### Explore More
Directions: Evaluate each expression.
1. $2^3$
2. $4^2$
3. $5^2$
4. $9^0$
5. $5^3$
6. $2^6$
7. $3^3$
8. $3^2+4^2$
9. $5^3+2^2$
10. $6^2+2^3$
11. $6^2-5^2$
12. $2^4-2^2$
13. $7^2+3^3+2^2$
Directions: Simplify the following variable expressions.
14. $(m^2)(m^5)$
15. $(x^3)(x^4)$
16. $(y^5 )(y^3)$
17. $(b^7 )(b^2)$
18. $(a^5 )(a^2)$
19. $(x^9 )(x^3)$
20. $(y^4 )(y^5)$
Directions: Simplify.
21. $(x^2 )^4$
22. $(y^5 )^3$
23. $(a^5 )^4$
24. $(x^2 )^8$
25. $(b^3 )^4$
### Vocabulary Language: English
Evaluate
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Expanded Form
Expanded Form
Expanded form refers to a base and an exponent written as repeated multiplication.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Integer
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Parentheses
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
Volume
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
|
Introductory Statistics 2e
# 13.3Facts About the F Distribution
Introductory Statistics 2e13.3 Facts About the F Distribution
Here are some facts about the F distribution.
1. The curve is not symmetrical but skewed to the right.
2. There is a different curve for each set of dfs.
3. The F statistic is greater than or equal to zero.
4. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.
5. Other uses for the F distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter.
Figure 13.3
## Example 13.2
### Problem
Let’s return to the slicing tomato exercise in Example 13.1. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.
## Try It 13.2
There are multiple variants of the virus that causes COVID-19. The length of hospital stays for patients afflicted with various strains of COVID-19 is shown in Table 13.6.
Delta StrainOmicron Strain Alpha Strain Gamma Strain Beta Strain
13.9 11.7 18.2 16.9 9.3
14.9 15.1 14.6 12.8 15.8
16.8 9.9 10.1 11.2 16.4
Table 13.6
Test whether the mean length of hospital stay is the same or different for the various strains of COVID-19. Construct the ANOVA table, find the p-value, and state your conclusion. Use a 5% significance level.
## Example 13.3
Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.7.
Sorority 1 Sorority 2 Sorority 3 Sorority 4
2.17 2.63 2.63 3.79
1.85 1.77 3.78 3.45
2.83 3.25 4.00 3.08
1.69 1.86 2.55 2.26
3.33 2.21 2.45 3.18
Table 13.7 MEAN GRADES FOR FOUR SORORITIES
### Problem
Using a significance level of 1%, is there a difference in mean grades among the sororities?
## Try It 13.3
Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.8.
3.6 2.1 4.0 2.0
2.9 2.6 2.0 3.6
2.5 3.9 2.6 3.9
3.3 3.1 3.2 2.7
3.8 3.4 3.2 2.5
Table 13.8 GPAs FOR FOUR SPORTS TEAMS
Use a significance level of 5%, and determine if there is a difference in GPA among the teams.
## Example 13.4
A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in Table 13.9.
Tommy's Plants Tara's Plants Nick's Plants
24 25 23
21 31 27
23 23 22
30 20 30
23 28 20
Table 13.9
### Problem
Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.
## Try It 13.4
Another fourth grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.4.
## Collaborative Exercise
From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States they have visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets in Appendix E Solution Sheets.
|
Important Questions: Algebra
# Class 6 Maths Chapter 11 Important Question Answers - Algebra
Q1: Deepak’s present age is one-third his mother’s present age. If the mother’s age was five times his age 6 years ago, what are their present ages?
Ans:
Q2: Form expressions using y, 2 and 7. Every expression must have y in it. use only two number operations. These should be different.
Ans:
The different expressions that can formed are: 2y + 7, 2y – 7, 7y + 2, 7y-2, (y/2) – 7, (y/7)-2, y – (7/2), y + (7/2).
Q3: Give expressions for the following
(b) 7 subtracted from
(c) p multiplied by
(d) p divided by
(e) 7 subtracted
(f) – p multiplied by
(g) – p divided by
(h) p multiplied by – 5.
Ans: (a)
p + 7
(b) p – 7
(c) 7p
(d) p/7
(e) – m – 7
(f) -5p
(g)
(h) – 5p.
Q4: Write which letters give us the same rule as that given by L.
Ans:
The other letters which give us the same rule as L are T, V and X because the number of matchsticks required to make each of them is 2.
Q5: Fill in the blanks:
(a) The value of 2x – 12 is zero, when x = ________.
(b) The product of 2 and x is being added to the product of 3 and y is expressed as ________.
(c) The numerical coefficient of the terms is _________.
(d) The no. of terms in the expression is ______.
Ans: (a)
6;
(b) 2x + 3y;
(c) 1/2;
(d) 4
Q6: The teacher distributes 4 pencils per student. Can you tell how many pencils are needed for given number of students? (Use s for the number of students.)
Ans:
Let the number of pencils be ‘s’.
As, the number of pencils distributed to each student = 4
Thus, No. of pencils for ‘s’ students = 4 x s = 4s.
Q7: The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?
Ans:
breadth of a rectangular hall = b meters
let length of a rectangular hall be ‘l’ meter
according to the question, l = 3 times the breadth – 4 = 3b – 4.
Q8: Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder. She has 10 dots in a row. How many dots will her Rangoli have for r rows?
Ans:
Let the total number of rows be ‘r’.
As, No. Of dots in a row = 10.
So, the dots needed for 10 rows = r x 10 = 10r.
Q9: Find the value of the expression 2x – 3y + 4z, if x = 10, y = -12 and z = 11.
Ans:
Given expression = 2x – 3y + 4z
If x = 10, y = -12 and z = 11,
The expression becomes, (2 × 10) – (3 × –12) + (4 × 11)
= 20 – (-36) + 44
= 20 + 36 + 44
= 100.
Q10: The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students ? (Use s for number of students.)
Ans:
Number of pencils to be distributed to each student = 5
And, let the number of students in class be ‘s’.
As per the logic, Number of pencils needed = (Number of students in the class) x (Number of pencils to be distributed to one student )
So, Number of pencils needed = 5 x s = 5s.
Q11: Rearrange the terms of the following expressions in ascending order of powers of x:
5x2, 2x, 4x4, 3x3, 7x5
Ans:
If the given terms are arranged in the ascending order of powers of x, we get, 2x, 5x2, 3x3, 4x4, 7x5.
Q12: State whether the following statements are true or false:
(a) The parts of an algebraic exponent which are connected by + or – sign are called its terms.
(b) 5 times x subtracted from 8 times y is 5x - 8y.
(c) A number having fixed value is called variable.
(d) The numerical coefficient of -2x2y is -2.
Ans: (a) True
(b) False
(c) False
(d) True
Q13: Match the following:
Ans:
Q14: The _______ of the variable in an equation which satisfies the equation is called a solution to the equation.
Ans:
value, It is correct because the value of the variable must satisfy the equation.
Q15: Which of the following is expression with one variable?
x + y + z, y + 1, 1, x + y – 5
Ans:
y + 1, The equation has one variable as “y” whose value is not known. therefore, the equation is in one variable.
The document Class 6 Maths Chapter 11 Important Question Answers - Algebra is a part of the Class 6 Course Mathematics (Maths) Class 6.
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## Mathematics (Maths) Class 6
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# WORD PROBLEMS ON RATIO AND PROPORTION 2
Problem 1 :
Two numbers are respectively 20% and 50% are more than a third number, Find the ratio of the two numbers.
Solution :
Let "x" be the third number.
Then, the first number is
= (100+20)% of x
= 120% of x
= 1.2x
The second number is
= (100+50)% of x
= 150% of x
= 1.5x
The ratio between the first number and second number is
= 1.2x : 1.5x
= 1.2 : 1.5
= 12 : 15
= 4 : 5
Hence, the ratio of two numbers is 4 : 5.
Problem 2 :
The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water ?
Solution :
Vessel A :
[4:3 ------> 4+3 = 7, M----> 4/7, W----> 3/7]
Let "x" be the quantity of mixture taken from vessel A to obtain a new mixture in vessel C.
Quantity of milk in "x" = (4/7)x = 4x/7
Quantity of water in "x" = (3/7)x = 3x/7
Vessel B :
[ 2:3 ------> 2+3 = 5, M----> 2/5, W----> 3/5 ]
Let "y" be the quantity of mixture taken from vessel B to obtain a new mixture in vessel C.
Quantity of milk in "y" = (2/5)y = 2y/5
Quantity of water in "y" = (3/5)y = 3y/5
Vessel A and B :
Quantity of milk from A and B is
= (4x/7) + (2y/5)
= (20x + 14y) / 35
Quantity of water from A and B is
= (3x/7) + (3y/5)
= (15x + 21y) / 35
According to the question, vessel C must consist half of the milk and half of the water.
That is, in vessel C, quantity of milk and water must be same.
There fore,
Quantity of milk in (A+B) = Quantity of water in (A+B)
(20x+14y) / 35 = (15x+21y) / 35
20x + 14y = 15x+21y
5x = 7y
x/y = 7/5
x : y = 7 : 5
Hence, the required ratio is 7 : 5.
Problem 3 :
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. From the vessel, 10 liters of the mixture is removed and replaced with an equal quantity of pure milk. Find the ratio of milk and water in the final mixture obtained.
Solution :
[ 3 : 2 ------> 3+2 = 5, M----> 3/5, W----> 2/5 ]
In 20 liters of mixture,
no. of liters of milk = 20 ⋅ 3/5 = 12
no. of liters of water = 20 ⋅ 2/5 = 8
Now, 10 liters of mixture removed.
In this 10 liters of mixture, milk and water will be in the ratio 3:2.
No. of liters of milk in this 10 liters = 10 ⋅ 3/5 = 6
No. of liters of water in this 10 liters = 10 ⋅ 2/5 = 4
After removing 10 liters (1st time),
No. of liters of milk in the vessel = 12 - 6 = 6
No. of liters of water in the vessel = 8 - 4 = 4
Now, we add 10 liters of pure milk in the vessel,
After adding 10 liters of pure milk in the vessel,
No. of liters of milk in the vessel = 6 + 10 = 16
No. of liters of water in the vessel = 4 + 0 = 4
After removing 10 liters of mixture and adding 10 liters of pure milk, the ratio of milk and water is
= 16 : 4
= 4 : 1
Hence, the required ratio is 4 : 1.
Problem 4 :
If \$782 is divided among three persons A, B and C in the ratio 1/2 : 2/3 : 3/4, then find the share of A.
Solution :
Given ratio ---> 1/2 : 2/3 : 3/4
First let us convert the terms of the ratio into integers.
L.C.M of denominators (2, 3, 4) = 12
When we multiply each term of the ratio by 12, we get
12 ⋅ 1/2 : 12 ⋅ 2/3 : 12 ⋅ 3/4 ------> 6 : 8 : 9
From the ratio 6 : 8 : 9,
Share of A = 6x
Share of B = 8x
Share of C = 9x
We know that,
Share of A + Share of B + Share of C = 782
6x + 8x + 9x = 782
23x = 782
x = 34
Share of A is
= 6x
= 6 ⋅ 34
= 204
Hence, the share of A = \$ 204.
Problem 5 :
An amount of money is to be divided among P, Q and R in the ratio 3 : 7 : 12. The difference between the shares of P and Q is \$2400. What will be the difference between the shares of Q and R ?
Solution :
From the given ratio 3 : 7 : 12,
Share of P = 3x
Share of Q = 7x
Share of R = 12x
Difference between the shares of P and Q is \$ 2400
That is,
Share of Q - Share of P = 2400
7x - 3x = 2400
4x = 2400
x = 600
Difference between the shares of Q and R is
= 12x - 7x
= 5x
= 5 ⋅ 600
= 3000
Hence, the difference between the shares of Q and R is \$3000.
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# I Would Like to be a Part of the Group!
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Print Lesson
## Objective
SWBAT read and write fractions that name a part of a group.
#### Big Idea
Students love working in groups, in this lesson students will be using two-color counters to discover how to read and write fractions that name a part of a group.
## Trying out for the group!
20 minutes
Materials: Two-color counters two color counters activity.docx
Fraction Pad
Vocabulary:
Fraction- A number that names a part of a whole or a part of a group.
Prior Talk:
Students love belonging to a group. Even more, students love getting their equal share, or playing their equal role when it relates to a part of a group! I tell students that today we will be exploring how to read and write fractions that name part of a group.
Connection: I bring cookies and have them setting on a table in front of me.
I ask students, "What would happen if you have three cookies and want to share them equally with two friends? What fraction of the group do you each get?" (We would get one third.) Can anyone tell me how you know if a group is made up of equal parts? Check to see if the parts are the same size.
In this lesson I want my students to understand how to correctly write a fraction without inverting the numerator and the denominator. I remind students of our previous lesson, and I tell them that the bottom number names the total number of equal parts that make up the group. The top number is the number of those equal parts that they are counting.
Ask and Discuss:
I draw five circles on the board, and shade in 4 of them.
What is a fraction? Most students will say that a fraction is a number that names part of a whole. I need them to understand that a fraction can also name part of a group. I model how to use the two-color counters to model a problem. As I model, I ask students to tell me what I was asked to find. The fraction that names the part of the group that is shaded. I guide students through by counting the number of counters in all and the number of shaded counters. What does the shaded number represent? What does the total number of counters represent? I discuss how to write 4 out of 5 as a fraction. I ask students to read the fraction for me? ( Four fifths) I repeat this using additional fractions until students are able to discuss and explain how to read and write fractions that name part of a whole. MP4- Model with math.
*Students share the cookies after we have visually demonstrated multiple problems.
Mathematical Practices:
MP.4. Model with mathematics.
MP.7. Look for and make use of structure
## Group Work!
20 minutes
In this portion of the lesson, I want my students to have additional time working with fractions. I want them to work together to better understand the difference in writing and reading fractions that name a part of a whole. I also want them to be able to read and write fractions that name a part of a group. Although the concept is somewhat familiar my students always forget that the shaded part names the amount being discussed or used. (When using visual models, some students wrongfully focus on the amount that is left.) Students need plenty of practice working with new concepts because it helps them discover and understand how to explain and solve better.
Before students began their group activity, I want to discuss key concepts that are essential for them to become fluent in reading and writing fractions that name part of a group. I ask students what they need to know in order to write a fraction that names part of a group. We need to know the number of items in the group or the number of groups, and the number of parts being counted. I ask if someone can tell me how to find the bottom number of a fraction? Response should be: I can count the number of equal parts that make up the group or the number of equal groups. What is the bottom number called? It is called the denominator. How do you find the top number of a fraction? I can find the top number of equal parts that I am counting.
Math Talk: Problem 1.docx
Can anyone explain how to solve the problem?
Students’ explanation: Count the number of counters in the set to find the bottom number of the fraction. Then, count the number of counters that are shaded to find the top number of the fraction.
I provide two-color counters for struggling students to help them develop a conceptual understanding of reading and writing fractions that name part of a group. I give students about 20 minutes or so to discuss and solve their problems in their assigned groups. As students are working, I circle the room to reinforce how to identify the numerator and the denominator of a fraction. I use student responses to determine if additional time should be spent on reviewing this objective. I ask student volunteers to share out with the rest of the groups. Students eagerly raise their hands to volunteer what they know!
MP7- Look for and make use of structure.
## How can I help the group?
20 minutes
Materials: students independent work.docx
In this portion of the lesson, I ask students to move back into their assigned seat. I tell students that they will be exploring a bit on their own to show me what they have learned so far. Students seem excited and eager to show what they know! I distribute two-color counters to students who are having a hard time understanding how to read and write fractions that name part of a group. I point out that they should use the counters to create a visual representation of the problem to help them better understand the steps we discussed in the group activity. I give students about 15 minutes to complete their independent assignment. As students are working, I circle the room to probe students and see what they are thinking. I take notes and plan to use the notes to determine if additional practice should be granted on reading and writing fractions that name a part of a group.
Probing Questions:
What do you need to know to write a fraction that names part of a group?
How do you find the bottom number of the fraction?
What is the bottom number called?
How do you know? Explain?
How do you find the top number of a fraction?
What is the top number called?
Can you explain how you solve your problem?
|
# NCERT Solutions For Class 10th Maths Chapter 6 : Triangles
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CBSE NCERT Solutions For Class 10th Maths Chapter 6 : Triangles. NCERT Solutions For Class 10 Mathematics. Exercise 6.1, Exercise 6.2, Exercise 6.3, Exercise 6.4, Exercise 6.5, Exercise 6.6.
## NCERT Solutions for Class X Maths Chapter 6 Triangles
Page No: 122
Exercise 6.1
1. Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
► Similar
(ii) All squares are __________. (similar, congruent)
► Similar
(iii) All __________ triangles are similar. (isosceles, equilateral)
► Equilateral
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
► (a) Equal, (b) Proportional
2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures
(i) Two twenty-rupee notes, Two two rupees coins.
(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note.
3. State whether the following quadrilaterals are similar or not:
The given two figures are not similar because their corresponding angles are not equal.
Page No: 128
Exercise 6.2
1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ 1.5/3 = 1/EC
⇒ Σ EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ AD = 1.8×7.2/5.4 = 18/10 × 72/10 × 10/54 = 24/10
2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
In ΔPQR, E and F are two points on side PQ and PR respectively.
(i) PE = 3.9 cm, EQ = 3 cm (Given)
PF = 3.6 cm, FR = 2,4 cm (Given)
∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]
And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm
∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]
And, PF/RF = 8/9
So, PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given)
Here, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 … (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55 … (ii)
∴ PE/EQ = PF/FR.
Hence, EF is parallel to QR.
3. In the fig 6.18, if LM || CB and LN || CD, prove thatAM/MB = AN/AD
In the given figure, LM || CB
By using basic proportionality theorem, we get,
AM/MB = AL/AC … (i)
Similarly, LN || CD
∴ AN/AD = AL/AC … (ii)
From (i) and (ii), we get
4. In the fig 6.19, DE||AC and DF||AE. Prove that
BF/FE = BE/EC
In ΔABC, DE || AC (Given)
∴ BD/DA = BE/EC …(i) [By using Basic Proportionality Theorem]
In ΔABC, DF || AE (Given)
∴ BD/DA = BF/FE …(ii) [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
BE/EC = BF/FE
5. In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.
In ΔPQO, DE || OQ (Given)
∴ PD/DO = PE/EQ …(i) [By using Basic Proportionality Theorem]
In ΔPQO, DE || OQ (Given)
∴ PD/DO = PF/FR …(ii) [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
PE/EQ = PF/FR
In ΔPQR, EF || QR. [By converse of Basic Proportionality Theorem]
6. In the fig 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
In ΔOPQ, AB || PQ (Given)
∴ OA/AP = OB/BQ …(i) [By using Basic Proportionality Theorem]
In ΔOPR, AC || PR (Given)
∴ OA/AP = OC/CR …(ii) [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
OB/BQ = OC/CR
In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem].
7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Given: ΔABC in which D is the mid point of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
To Prove: E is the mid point of AC.
Proof: D is the mid-point of AB.
⇒ AD/BD = 1 … (i)
In ΔABC, DE || BC,
Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]
⇒1 = AE/EC [From equation (i)]
∴ AE =EC
Hence, E is the mid point of AC.
8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Given: ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.
To Prove: DE || BC
Proof: D is the mid point of AB (Given)
⇒ AD/BD = 1 … (i)
Also, E is the mid-point of AC (Given)
∴ AE=EC
⇒AE/EC = 1 [From equation (i)]
From equation (i) and (ii), we get
Hence, DE || BC [By converse of Basic Proportionality Theorem]
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Given: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.
To Prove: AO/BO = CO/DO
Construction: Through O, draw EO || DC || AB
OE || DC (By Construction)
∴ AE/ED = AO/CO …(i) [By using Basic Proportionality Theorem]
In ΔABD, we have
OE || AB (By Construction)
∴ DE/EA = DO/BO …(ii) [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
AO/CO = BO/DO
⇒ AO/BO = CO/DO
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO/BO = CO/DO.
To Prove: ABCD is a trapezium
Construction: Through O, draw line EO, where EO || AB, which meets AD at E.
Proof: In ΔDAB, we have
EO || AB
∴ DE/EA = DO/OB …(i) [By using Basic Proportionality Theorem]
Also, AO/BO = CO/DO (Given)
⇒ AO/CO = BO/DO
⇒ CO/AO = BO/DO
⇒ DO/OB = CO/AO …(ii)
From equation (i) and (ii), we get
DE/EA = CO/AO
Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.
Page No: 138
Exercise 6.3
1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
(i) In ΔABC and ΔPQR, we have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)
(ii) In ΔABC and ΔPQR, we have
AB/QR = BC/RP = CA/PQ
∴ ΔABC ~ ΔQRP (SSS similarity criterion)
(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF= 2.7/5 = 27/50
Here, MP/DE = PL/DF ≠ LM/EF
Hence, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)
(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here, AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)
Page No: 139
2. In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125°
= 55°
In ΔDOC,
∠DCO + ∠ CDO + ∠ DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
It is given that ΔODC ~ ΔOBA.
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠ OAB = 55°
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD
In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]
∴ DO/BO = OC/OA [ Corresponding sides are proportional]
⇒ OA/OC = OB/OD
Page No: 140
4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR …(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP …(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
In ΔRPQ and ΔRST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)
6. In the fig 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] …(i)
And, AD = AE [By cpct] …(ii)
AD/AB = AE/AC [Dividing equation (ii) by (i)]
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]
7. In the fig 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iv) ΔPDC ~ ΔBEC
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (By AA similarity criterion)
9. In the fig 6.39, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
(i) In ΔABC and ΔAMP, we have
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90° (each 90°)
∴ ΔABC ~ ΔAMP (By AA similarity criterion)
(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)
If two triangles are similar then the corresponding sides are equal,
Hence, CA/PA = BC/MP
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
(i) It is given that ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ΔACD ~ ΔFGH (By AA similarity criterion)
⇒ CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ΔDCB ~ ΔHGE (By AA similarity criterion)
(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ΔDCA ~ ΔHGF (By AA similarity criterion)
Page No: 141
11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∴ ΔABD ~ ΔECF (By using AA similarity criterion)
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Given: ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e., AB/PQ = BC/QR = AD/PM
To Prove: ΔABC ~ ΔPQR
Proof: AB/PQ = BC/QR = AD/PM
⇒ AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR …(i)
∠ABC = ∠PQR …(ii)
From equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD
∠ACD = ∠BCA (Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB =CD/CA
⇒ CA2 = CB.CD.
14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM
To Prove: ΔABC ~ ΔPQR
Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
Proof: In ΔABD and ΔCDE, we have
BD = DC [∴ AP is the median]
and, ∠ADB = ∠CDE [Vertically opp. angles]
∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]
⇒ AB = CE [CPCT] …(i)
Also, in ΔPQM and ΔMNR, we have
PM = MN [By Construction]
QM = MR [∴ PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [By SAS criterion of congruence]
⇒ PQ = RN [CPCT] …(ii)
Now, AB/PQ = AC/PR = AD/PM
⇒ CE/RN = AC/PR = AD/PM …[From (i) and (ii)]
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [By SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P …(iii)
Now, In ΔABC and ΔPQR, we have
AB/PQ = AC/PR (Given)
∠A = ∠P [From (iii)]
∴ ΔABC ~ ΔPQR [By SAS similarity criterion]
15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Length of the vertical pole = 6m (Given)
Shadow of the pole = 4 m (Given)
Let Height of tower = h m
Length of shadow of the tower = 28 m (Given)
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (By AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒ h = 6×28/4
⇒ h = 6 × 7
⇒ = 42 m
Hence, the height of the tower is 42 m.
16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
It is given that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR …(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 …(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM …(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.
Page No: 143
Exercise 6.4
1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
It is given that,
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm
and, ΔABC ~ ΔDEF
∴ Area of ΔABC/Area of ΔDEF = AB2/DE2
= AC2/DF2 = BC2/EF2 …(i)
[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]
∴ 64/121 = BC2/EF2
⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]
Now, Area of (ΔAOB)/Area of (ΔCOD)
= AB2/CD2 [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]
= (2CD)2/CD2 [∴ AB = CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1
3. In the fig 6.53, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
Given: ABC and DBC are triangles on the same base BC. Ad intersects BC at O.
To Prove: area (ΔABC)/area (ΔDBC) = AO/DO.
Construction: Let us draw two perpendiculars AP and DM on line BC.
Proof: We know that area of a triangle = 1/2 × Base × Height
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each equals to 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (By AA similarity criterion)∴ AP/DM = AO/DO
⇒ area (ΔABC)/area (ΔDBC) = AO/DO.
4. If the areas of two similar triangles are equal, prove that they are congruent.
Given: ΔABC and ΔPQR are similar and equal in area.
To Prove: ΔABC ≅ ΔPQR
Proof: Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2
⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC2/QR2
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [BY SSS criterion of congruence]
5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
Given: D, E and F are the mid-points of the sides AB, BC and CA respectively of the ΔABC.
To Find: area(ΔDEF) and area(ΔABC)
Solution: In ΔABC, we have
F is the mid point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, we have
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent,then they are equal in area.
So, area(ΔFBD) = area(ΔDEF) …(i)
area(ΔAFE) = area(ΔDEF) …(ii)
and, area(ΔEDC) = area(ΔDEF) …(iii)
Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) …(iv)
area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)
⇒ area(ΔDEF) = 1/4area(ΔABC) [From (i)(ii) and (iii)]
⇒ area(ΔDEF)/area(ΔABC) = 1/4
Hence, area(ΔDEF):area(ΔABC) = 1:4
6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
To Prove: area(ΔABC)/area(ΔDEF) = AM2/DN2
Proof: ΔABC ~ ΔDEF (Given)
∴ area(ΔABC)/area(ΔDEF) = (AB2/DE2) …(i)
and, AB/DE = BC/EF = CA/FD …(ii)
In ΔABM and ΔDEN, we have
∠B = ∠E [Since ΔABC ~ ΔDEF]
AB/DE = BM/EN [Prove in (i)]
∴ ΔABC ~ ΔDEF [By SAS similarity criterion]
⇒ AB/DE = AM/DN …(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2
7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Given: ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
To Prove: area(ΔBQC) = 1/2area(ΔAPC)
Proof: ΔAPC and ΔBQC are both equilateral triangles (Given)
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Tick the correct answer and justify:
8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
ΔABC and ΔBDE are two equilateral triangle. D is the mid point of BC.
∴ BD = DC = 1/2BC
Let each side of triangle is 2a.
As, ΔABC ~ ΔBDE
∴ area(ΔABC)/area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 4/1 = 4:1
Hence, the correct option is (C).
9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Let ABC and DEF are two similarity triangles ΔABC ~ ΔDEF (Given)
and, AB/DE = AC/DF = BC/EF = 4/9 (Given)
∴ area(ΔABC)/area(ΔDEF) = AB2/DE[the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides]
∴ area(ΔABC)/area(ΔDEF) = (4/9)= 16/81 = 16:81
Hence, the correct option is (D).
Page No: 150
Exercise 6.5
1. Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will get 49, 576, and 625.
49 + 576 = 625
(7)2 + (24)2 = (25)2
The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm
(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 32 + 62 ≠ 82
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
(iii) Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Clearly, 144 +25 = 169
Or, 122 + 52 = 132
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Length of the hypotenuse of this triangle is 13 cm.
2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.
To prove: PM2 = QM × MR
Proof: In ΔPQM, we have
PQ2 = PM2 + QM2 [By Pythagoras theorem]
Or, PM2 = PQ2 – QM2 …(i)
In ΔPMR, we have
PR2 = PM2 + MR2 [By Pythagoras theorem]
Or, PM2 = PR2 – MR2 …(ii)
Adding (i) and (ii), we get
2PM2 = (PQ2 + PM2) – (QM2 + MR2)
= QR2 – QM2 – MR2 [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 – QM2 – MR2
= 2QM × MR
∴ PM2 = QM × MR
3. In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD
(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD
(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
= 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 = DC × BC
(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD
4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .
Given that ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (Given)
AB2 = AC2 + BC2 ([By using Pythagoras theorem]
= AC2 + AC2 [Since, AC = BC]
AB2 = 2AC2
5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Given that ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2
In ΔACB,
AC = BC (Given)
AB2 = 2AC2 (Given)
AB2 = AC+ AC2
= AC2 + BC[Since, AC = BC]
Hence, By Pythagoras theorem ΔABC is right angle triangle.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
ABC is an equilateral triangle of side 2a.
AB = AC [Given]
Hence, BD = DC [by CPCT]
⇒ AD2 = 4a2 – a2
7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
ABCD is a rhombus whose diagonals AC and BD intersect at O. [Given]
We have to prove that,
AB+ BC+ CD2 + AD= AC+ BD2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO+ BO… (i) [By Pythagoras]
Similarly,
DC2 = DO+ CO… (iii)
BC2 = CO+ BO… (iv)
Adding equations (i) + (ii) + (iii) + (iv) we get,
AB+ AD+ DC+ BC2 = 2(AO+ BO+ DO+ CO)
= 4AO+ 4BO[Since, AO = CO and BO =DO]
= (2AO)+ (2BO)2 = AC+ BD2
Page No: 151
8. In Fig. 6.54, O is a point in the interior of a triangle
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Join OA, OB and OC
(i) Applying Pythagoras theorem in ΔAOF, we have
OA2 = OF2 + AF2
Similarly, in ΔBOD
OB2 = OD2 + BD2
Similarly, in ΔCOE
OC2 = OE2 + EC2
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.
(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Let BA be the wall and Ac be the ladder,
Therefore, by Pythagoras theorem,we have
AC2 = AB2 + BC2
102 = 82 + BC2
BC= 100 – 64
BC= 36
BC = 6m
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC2 = AB2 + BC2
242 = 182 + BC2
BC= 576 – 324
BC= 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.
11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane due north in hours (OA) = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane due west in hours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, we have
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √2250000 + 3240000
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.
12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Let CD and AB be the poles of height 11 m and 6 m.
Therefore, CP = 11 – 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ΔAPC, we get
AP2 = PC2 + AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25)m2 = 169 m2
AC = 13m
Therefore, the distance between their tops is 13 m.
13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Applying Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2 ….(i)
Applying Pythagoras theorem in ΔBCD, we get
BC2 + CD2 = BD2 ….(ii)
Using equations (i) and (ii), we get
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …(iii)
Applying Pythagoras theorem in ΔCDE, we get
DE2 = CD2 + CE2
Applying Pythagoras theorem in ΔABC, we get
AB2 = AC2 + CB2
Putting these values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB2 = 2AC2+ BC2.
Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
AB2 = AD2 + BD2 …(i)
AC2 = AD2 + DC2 …(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 – DC2
= 9CD2 – CD2 [∴ BD = 3CD]
= 9CD2 = 8(BC/4)[Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given that, BD = 1/3BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
Applying Pythagoras theorem in ΔADE, we get
⇒ 9 AD2 = 7 AB2
16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
Applying Pythagoras theorem in ΔABE, we get
AB2 = AE2 + BE2
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
We can observe that
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct option is (C).
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# Prove inequality: $\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$
Let $a,b,c$ the sides of a triangle, $P$ its perimeter. Prove inequality: $$\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$$
### My attempt:
1) $P=a+b+c$. Then $\frac{P+2004a}{P-2a}=\frac{2005a+b+c}{b+c-a}$. Here $b+c-a>0 -$ triangle inequality.
2) $\sqrt[3]{xyz}\ge\frac{3}{\frac1x+\frac1y+\frac1z} \Rightarrow xyz\ge \left(\frac{3}{\frac1x+\frac1y+\frac1z}\right)^3$
• @SubhadeepDey: 2007 Jun 19, 2016 at 16:18
Let $$b+c-a=s,\quad c+a-b=t,\quad a+b-c=u$$ Then, we can have $$a=\frac{t+u}{2},\quad b=\frac{s+u}{2},\quad c=\frac{s+t}{2}$$
So, using AM-GM inequality and letting $d=1003$, \begin{align}&\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\\\\&=\left(1+d\frac ts+d\frac us\right)\left(1+d\frac st+d\frac ut\right)\left(1+d\frac su+d\frac tu\right)\\\\&=2d^3+3d^2+1+(d^3+d^2+d)\left(\frac su+\frac st+\frac tu+\frac ts+\frac us+\frac ut\right)+d^2\left(\frac{s^2}{tu}+\frac{t^2}{su}+\frac{u^2}{st}\right)\\\\&\ge 2d^3+3d^2+1+6(d^3+d^2+d)\sqrt[6]{1}+3d^2\sqrt[3]{1}\\\\&=(2d+1)^3\\\\&=2007^3\end{align}
I think it has to be $3\cdot 2007$ instead of $2007^3$. Since $a,b,c$ are sides of a triangle, there exist $x,y,z>0$ such that $a=y+z$, $b=z+x$ and $c=x+y$. Hence, we have to prove: $$\sum_{cyc}\frac{2005(x+y)+(y+z)+(z+x)}{2(x+y+z)-2(x+y)}≥3\cdot 2007\iff\\ \sum_{cyc}\frac{2006x+2006y+2z}{2z}≥3\cdot 2007\iff\\ 3+1003\sum_{sym}\frac{x}{y}≥3\cdot 2007$$ Now with AM-GM, we have $\sum_{sym}\frac{x}{y}≥6$ and thus: $$3+1003\sum_{sym}\frac{x}{y}≥3+6\cdot1003=3\cdot 2007$$ Withe equality iff $x=y=z\iff a=b=c$.
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# Important Questions for Maths Class 8 Chapter 3 Understanding Quadrilaterals
Important questions for Class 8 chapter 3 – Understanding quadrilaterals are given here. Students who are preparing for final exams can practice these questions to score good marks. All the questions presented here are as per NCERT curriculum or CBSE syllabus.
Understanding quadrilaterals chapter deals with different types of quadrilaterals and their properties. You will also learn to compute the measure of any missing angle of quadrilateral. Let us see some important questions with solutions here.
Students can also reach Important Questions for Class 8 Maths to get important questions for all the chapters here.
## Class 8 Chapter 3 Important Questions
Questions and answers are given here based on important topics of class 8 Maths Chapter 3.
Q.1: A quadrilateral has three acute angles, each measure 80°. What is the measure of the fourth angle?
Solution:
Let x be the measure of the fourth angle of a quadrilateral.
Sum of the four angles of a quadrilateral = 360°
80° + 80° + 80° + x = 360°
x = 360° – (80° + 80° + 80°)
x = 360° – 240°
x = 120°
Hence, the fourth angle is 120°.
Q,2: In a quadrilateral ABCD, the measure of the three angles A, B and C of the quadrilateral is 110°, 70° and 80°, respectively. Find the measure of the fourth angle.
Solution: Let,
∠A = 110°
∠B = 70°
∠C = 80°
∠D = x
We know that the sum of all internal angles of quadrilateral ABCD is 360°.
∠A + ∠B+ ∠C+∠D = 360°
110° + 70° + 80° + x = 360°
260° + x = 360°
x = 360° – 260°
x = 100°
Therefore, the fourth angle is 100°.
Q.3: In a quadrilateral ABCD, ∠D is equal to 150° and ∠A = ∠B = ∠C. Find ∠A, ∠B and ∠C.
Solution: Given,
∠D = 150°
Let ∠A = ∠B = ∠C = x
By angle sum property of quadrilateral,
∠A + ∠B + ∠C + ∠D = 360°
x + x +x+∠D = 360°
3x+∠D = 360°
3x = 360° – ∠D
30 = 360° – 150°
3x = 210°
x = 70°
Hence, ∠A = ∠B = ∠C = 70°.
Q.4: The angles of a quadrilateral are in the ratio of 1 : 2 : 3 : 4. What is the measure of the four angles?
Solution: Given,
The ratio of the angles of quadrilaterals = 1 : 2 : 3 : 4
Let the four angles of the quadrilateral be x, 2x, 3x, and 4x respectively.
The sum of four angles of a quadrilateral is 360°.
Therefore,
x + 2x + 3x + 4x = 360°
10x = 360°
x = 360°/10
x = 36°
Therefore,
First angle = x = 36°
Second angle = 2x = 2 × 36 = 72°
Third angle = 3x = 3 × 36 = 108°
Fourth angle = 4x = 4 × 36 = 144°
Hence, the measure of four angles is 36°, 72°, 108° and 144°.
(i) which of them have their diagonals bisecting each other?
(ii) which of them have their diagonal perpendicular to each other?
(iii) which of them have equal diagonals?
Solution:
(i) Diagonals bisect each other in:
• Parallelogram
• Rhombus
• Rectangle
• Square
• Kite
(ii) Diagonals are perpendicular in:
• Rhombus
• Square
• Kite
(iii) Diagonals are equal to each other in:
• Rectangle
• Square
Q. 6: Adjacent sides of a rectangle are in the ratio 5 : 12, if the perimeter of the rectangle is 34 cm, find the length of the diagonal.
Solution:
Given,
Ratio of the adjacent sides of the rectangle = 5 : 12
Let 5x and 12x be the two adjacent sides.
We know that the sum of all sides of a rectangle is equal to its perimeter.
Thus,
5x + 12x + 5x + 12x = 34 cm (given)
34x = 34
x = 34/34
x = 1 cm
Therefore, the adjacent sides are 5 cm and 12 cm respectively.
i.e. l = 12 cm, b = 5 cm
Length of the diagonal = √(l2 + b2)
= √(122 + 52)
= √(144 + 25)
= √169
= 13 cm
Hence, the length of the diagonal is 13 cm.
Q. 7: The opposite angles of a parallelogram are (3x + 5)° and (61 – x)°. Find the measure of four angles.
Solution:
Given,
(3x + 5)° and (61 – x)° are the opposite angles of a parallelogram.
We know that the opposite angles of a parallelogram are equal.
Therefore,
(3x + 5)° = (61 – x)°
3x + x = 61° – 5°
4x = 56°
x = 56°/4
x = 14°
⇒ 3x + 5 = 3(14) + 5 = 42 + 5 = 47
61 – x = 61 – 14 = 47
The measure of angles adjacent to the given angles = 180° – 47° = 133°
Hence, the measure of four angles of the parallelogram are 47°, 133°, 47°, and 133°.
Q. 8: ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C meet each other at O. Find the measure of the three angles of ΔBCO.
Solution:
Given,
∠A = 80°
We know that the opposite angles of a parallelogram are equal.
∠A = ∠C = 80°
And
∠OCB = (1/2) × ∠C
= (1/2) × 80°
= 40°
∠B = 180° – ∠A (the sum of interior angles on the same side of the transversal is 180)
= 180° – 80°
= 100°
Also,
∠CBO = (1/2) × ∠B
= (1/2) × 100°
= 50°
By the angle sum property of triangle BCO,
∠BOC + ∠OBC + ∠CBO = 180°
∠BOC = 180° – (∠OBC + CBO)
= 180° – (40° + 50°)
= 180° – 90°
= 90°
Hence, the measure of all the three angles of a triangle BCO is 40°, 50° and 90°.
Q. 9: Find the measure of all four angles of a parallelogram whose consecutive angles are in the ratio 1 : 3.
Solution:
Given,
The ratio of two consecutive angles of a parallelogram = 1 : 3
Let x and 3x be the two consecutive angles.
We know that the sum of interior angles on the same side of the transversal is 180°.
Therefore, x + 3x = 180°
4x = 180°
x = 180°/4
x = 45°
⇒ 3x = 3(45°) = 135°
Thus, the measure of two consecutive angles is 45° and 135°.
As we know, the opposite angles of a parallelogram are equal.
Hence, the measure of all the four angles is 45°, 135°, 45°, and 135°.
Q. 10: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus.
Solution:
Let ABCD be the rhombus.
Thus, AB = BC = CD = DA
Given that a side and diagonal are equal.
AB = BD (say)
Therefore, AB = BC = CD = DA = BD
Now, all the sides of a triangle ABD are equal.
Therefore, ΔABD is an equilateral triangle.
Similarly,
ΔBCD is also an equilateral triangle.
Thus, ∠A = ∠ABD = ∠ADB = ∠DBC = ∠C = ∠CDB = 60°
∠B = ∠ABD + ∠DBC = 60° + 60° = 120°
And
∠D = ∠ADB + ∠CDB = 60° + 60° = 120°
Hence, the angles of the rhombus are 60°, 120°, 60° and 120°.
### Extra Practice Questions For Class 8 Maths Chapter 3 Understanding Quadrilateral
1. How many sides do a regular polygon have, if the measure of an exterior angle is given as 24° ?
2. What is the measure of each exterior angle of a regular polygon of 15 sides?
(a) 30°
(b) 45°
(c) 60°
(d) 24°
3. When the sum of the internal angles of a polygon is 10 right angles, then how many sides does it have?
(a) 6
(b) 5
(c) 8
(d) 7
4. Select a false statement from those given below:
(a) A square is a rectangle that has equal adjacent sides
(b) A square is a rhombus whose one angle is a right angle
(c) The diagonals of a square bisect each other at right angles
(d) The diagonals of a square do not divide the whole square into four equal parts.
5. When one angle of a parallelogram is a right angle, then what is the name of the quadrilateral?
(a) kite
(b) rectangle
(c) rhombus
(d) square
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# 9.6: The Pythagorean Theorem
The Pythagorean Theorem And Simplest Radical Form
The Pythagorean Theorem And Simplest Radical Form
9.6: The Pythagorean Theorem
Page ID
19739
Pythagoras was a Greek mathematician and philosopher, born on the island of Samos (ca. 582 BC). He founded a number of schools, one in particular in a town in southern Italy called Crotone, whose members eventually became known as the Pythagoreans. The inner circle at the school, the Mathematikoi, lived at the school, rid themselves of all personal possessions, were vegetarians, and observed a strict vow of silence. They studied mathematics, philosophy, and music, and held the belief that numbers constitute the true nature of things, giving numbers a mystical or even spiritual quality.
Today, nothing is known of Pythagoras’s writings, perhaps due to the secrecy and silence of the Pythagorean society. However, one of the most famous theorems in all of mathematics does bear his name, the Pythagorean Theorem.
Let c represent the length of the hypotenuse, the side of a right triangle directly opposite the right angle (a right angle measures 90º) of the triangle. The remaining sides of the right triangle are called the legs of the right triangle, whose lengths are designated by the letters a and b.
The relationship involving the legs and hypotenuse of the right triangle, given by
$a^2 + b^2 = c^2 \label{1}$
is called the Pythagorean Theorem.
Note that the Pythagorean Theorem can only be applied to right triangles. Let’s look at a simple application of the Pythagorean Theorem (Equation \ref{1}).
Given that the length of one leg of a right triangle is 4 centimeters and the hypotenuse has length 8 centimeters, find the length of the second leg.
Let’s begin by sketching and labeling a right triangle with the given information. We will let x represent the length of the missing leg.
Here is an important piece of advice.
The hypotenuse is the longest side of the right triangle. It is located directly opposite the right angle of the triangle. Most importantly, it is the quantity that is isolated by itself in the Pythagorean Theorem (Equation \ref{1}). Always isolate the quantity representing the hypotenuse on one side of the equation. The “legs” go on the other side of the equation.
So, taking the tip to heart, and noting the lengths of the legs and hypotenuse in Figure 1, we write
$$4^2+x^2 = 8^2$$.
Square, then isolate x on one side of the equation.
$$16+x^2 = 64$$
$$x^2 = 48$$
Normally, we would take plus or minus the square root in solving this equation, but x represents the length of a leg, which must be a positive number. Hence, we take just the positive square root of 48.
$$x = \sqrt{48}$$
$$x = \sqrt{16}\sqrt{3}$$
$$x = 4\sqrt{3}$$
If need be, you can use your graphing calculator to approximate this length. To the nearest hundredth of a centimeter, $$x \approx 6.93$$ centimeters.
## Proof of the Pythagorean Theorem
It is not known whether Pythagoras was the first to provide a proof of the Pythagorean Theorem. Many mathematical historians think not. Indeed, it is not even known if Pythagoras crafted a proof of the theorem that bears his name, let alone was the first to provide a proof.
There is evidence that the ancient Babylonians were aware of the Pythagorean Theorem over a 1000 years before the time of Pythagoras. A clay tablet, now referred to as Plimpton 322 (see Figure 2), contains examples of Pythagorean Triples, sets of three numbers that satisfy the Pythagorean Theorem (such as 3, 4, 5).
One of the earliest recorded proofs of the Pythagorean Theorem dates from the Han dynasty (206 BC to AD 220), and is recorded in the Chou Pei Suan Ching (see Figure 3). You can see that this figure specifically addresses the case of the 3, 4, 5 right triangle. Mathematical historians are divided as to whether or not the image was meant to be part of a general proof or was just devised to address this specific case. There is also disagreement over whether the proof was provided by a more modern commentator or dates back further in time.
However, Figure 3 does suggest a path we might take on the road to a proof of the Pythagorean Theorem. Start with an arbitrary right triangle having legs of lengths a and b, and hypotenuse having length c, as shown in Figure 4(a).
Next, make four copies of the triangle shown in Figure 4(a), then rotate and translate them into place as shown in Figure 4(b). Note that this forms a big square that is c units on a side.
Further, the position of the triangles in Figure 4(b) allows for the formation of a smaller, unshaded square in the middle of the larger square. It is not hard to calculate the length of the side of this smaller square. Simply subtract the length of the smaller leg from the larger leg of the original triangle. Thus, the side of the smaller square has length b − a.
Now, we will calculate the area of the large square in Figure 4(b) in two separate ways.
• First, the large square in Figure 4(b) has a side of length c. Therefore, the area of the large square is
$$Area = c^2$$.
• Secondly, the large square in Figure 4(b) is made up of 4 triangles of the same size and one smaller square having a side of length b−a. We can calculate the area of the large square by summing the area of the 4 triangles and the smaller square.
1. The area of the smaller square is $$(b−a)^2$$.
2. The area of each triangle is $$\frac{ab}{2}$$. Hence, the area of four triangles of equal size is four times this number;
i.e., $$4(\frac{ab}{2})$$. Thus, the area of the large square is
Area = Area of small square + $$4 \cdot$$ Area of triangle
=$$(b−a)^2+4(\frac{ab}{2})$$.
We calculated the area of the larger square twice. The first time we got $$c^2$$; the second time we got $$(b−a)^2+4(\frac{ab}{2})$$. Therefore, these two quantities must be equal.
$$c^2 = (b−a)^2+4(\frac{ab}{2})$$.
Expand the binomial and simplify.
$$c^2 = b^2−2ab+a^2 +2ab$$
$$c^2 = b^2+a^2$$
That is,
$$a^2+b^2 = c^2$$,
and the Pythagorean Theorem is proven.
## Applications of the Pythagorean Theorem
In this section we will look at a few applications of the Pythagorean Theorem, one of the most applied theorems in all of mathematics. Just ask your local carpenter.
The ancient Egyptians would take a rope with 12 equally spaced knots like that shown in Figure 5, and use it to square corners of their buildings. The tool was instrumental in the construction of the pyramids.
The Pythagorean theorem is also useful in surveying, cartography, and navigation, to name a few possibilities.
Let’s look at a few examples of the Pythagorean Theorem in action.
One leg of a right triangle is 7 meters longer than the other leg. The length of the hypotenuse is 13 meters. Find the lengths of all sides of the right triangle.
Let x represent the length of one leg of the right triangle. Because the second leg is 7 meters longer than the first leg, the length of the second leg can be represented by the expression x + 7, as shown in Figure 6, where we’ve also labeled the length of the hypotenuse (13 meters).
Remember to isolate the length of the hypotenuse on one side of the equation representing the Pythagorean Theorem. That is,
$$x^2+(x+7)^2 = 13^2$$.
Note that the legs go on one side of the equation, the hypotenuse on the other. Square and simplify. Remember to use the squaring a binomial pattern.
$$x^2+x^2+14x+49 = 169$$
$$2x^2 +14x+49 = 169$$
This equation is nonlinear, so make one side zero by subtracting 169 from both sides of the equation.
$$2x^2+14x+49−169 = 0$$
$$2x^2 +14x−120 = 0$$
Note that each term on the left-hand side of the equation is divisible by 2. Divide both sides of the equation by 2.
$$x^2+7x−60 = 0$$
Let’s use the quadratic formula with a = 1, b = 7, and c = −60.
$$x = \frac{−b \pm \sqrt{b^2−4ac}}{2a} = \frac{−7 \pm \sqrt{7^2 −4(1)(−60)}}{2(1)}$$
Simplify.
$$x = \frac{−7 \pm \sqrt{289}}{2}$$
Note that 289 is a perfect square ($$17^2 = 289$$). Thus,
$$x = \frac{−7 \pm 17}{2}$$.
Thus, we have two solutions,
x = 5 or x = −12.
Because length must be a positive number, we eliminate −12 from consideration. Thus, the length of the first leg is x = 5 meters. The length of the second leg is x+7, or 12 meters.
Check. Checking is an easy matter. The legs are 5 and 12 meters, respectively, and the hypotenuse is 13 meters. Note that the second leg is 7 meters longer than the first. Also,
$$5^2+12^2 = 25+144 = 169$$,
which is the square of 13.
The integral sides of the triangle in the previous example, 5, 12, and 13, are an example of a Pythagorean Triple.
A set of positive integers a, b, and c, is called a Pythagorean Triple if they satisfy the Pythagorean Theorem; that is, if
$$a^2+b^2 = c^2$$.
If the greatest common factor of a, b, and c is 1, then the triple (a, b, c) is called a primitive Pythagorean Triple.
Thus, for example, the Pythagorean Triple (5, 12, 13) is primitive. Let’s look at another example.
If (a,b,c) is a Pythagorean Triple, show that any positive integral multiple is also a Pythagorean Triple.
Thus, if the positive integers (a, b, c) is a Pythagorean Triple, we must show that(ka, kb, kc), where k is a positive integer, is also a Pythagorean Triple.
However, we know that
$$a^2+b^2 = c^2$$.
Multiply both sides of this equation by $$k^2$$.
$$k^{2}a^2+k^{2}b^2 = k^{2}c^2$$
This last result can be written
$$(ka)^2 + (kb)^2 = (kc)^2$$.
Hence, (ka, kb, kc) is a Pythagorean Triple.
Hence, because (3, 4, 5) is a Pythagorean Triple, you can double everything to get another triple (6, 8, 10). Note that $$6^2 + 8^2 = 10^2$$ is easily checked. Similarly, tripling gives another triple (9, 12, 15), and so on.
In Example 5, we showed that (5, 12, 13) was a triple, so we can take multiples to generate other Pythagorean Triples, such as (10, 24, 26) or (15, 36, 39), and so on.
Formulae for generating Pythagorean Triples have been know since antiquity.
The following formula for generating Pythagorean Triples was published in Euclid’s (325–265 BC) Elements, one of the most successful textbooks in the history of mathematics. If m and n are positive integers with m > n, show
$$a = m^2−n^2$$,
b = 2mn, (7)
$$c = m^2+n^2$$,
generates Pythagorean Triples.
We need only show that the formulae for a, b, and c satisfy the Pythagorean Theorem. With that is mind, let’s first compute $$a^2+b^2$$.
$$a^2+b^2 = (m^2−n^2)^2+(2mn)^2$$
= $$m^4−2m^{2}n^{2}+n^4+4m^{2}n^2$$
= $$m^4+2m^{2}n^2+n^4$$
On the other hand,
$$c^2 = (m^2+n^2)^2$$
= $$m^4+2m^{2}n^2+n^4$$.
Hence, $$a^2+b^2 = c^2$$, and the expressions for a, b, and c form a Pythagorean Triple.
It is both interesting and fun to generate Pythagorean Triples with the formulae from Example 6. Choose m = 4 and n = 2, then
$$a = m^2−n^2 = (4)^2−(2)^2 = 12$$,
$$b = 2mn = 2(4)(2) = 16$$,
$$c = m^2+n^2 =(4)^2+(2)^2 = 20$$.
It is easy to check that the triple (12, 16, 20) will satisfy $$12^2+16^2 = 20^2$$. Indeed, note that this triple is a multiple of the basic (3, 4, 5) triple, so it must also be a Pythagorean Triple.
It can also be shown that if m and n are relatively prime, and are not both odd or both even, then the formulae in Example 6 will generate a primitive Pythagorean Triple. For example, choose m = 5 and n = 2. Note that the greatest common divisor of m = 5 and n = 2 is one, so m and n are relatively prime. Moreover, m is odd while n is even. These values of m and n generate
$$a = m^2−n^2 = (5)^2−(2)^2 = 21$$,
$$b = 2mn = 2(5)(2) = 20$$,
$$c = m^2+n^2 = (5)^2+(2)^2 = 29$$.
Note that
$$21^2+20^2 = 441+400 = 841 = 29^2$$.
Hence, (21, 20, 29) is a Pythagorean Triple. Moreover, the greatest common divisor of 21, 20, and 29 is one, so (21, 20, 29) is primitive.
The practical applications of the Pythagorean Theorem are numerous.
A painter leans a 20 foot ladder against the wall of a house. The base of the ladder is on level ground 5 feet from the wall of the house. How high up the wall of the house will the ladder reach?
Consider the triangle in Figure 7. The hypotenuse of the triangle represents the ladder and has length 20 feet. The base of the triangle represents the distance of the base of the ladder from the wall of the house and is 5 feet in length. The vertical leg of the triangle is the distance the ladder reaches up the wall and the quantity we wish to determine.
Applying the Pythagorean Theorem,
$$5^2+h^2 = 20^2$$.
Again, note that the square of the length of the hypotenuse is the quantity that is isolated on one side of the equation.
Next, square, then isolate the term containing h on one side of the equation by subtracting 25 from both sides of the resulting equation.
$$25+h^2 = 400$$
$$h^2 = 375$$
We need only extract the positive square root.
$$h = \sqrt{375}$$
We could place the solution in simple form, that is, $$h = 5\sqrt{15}$$, but the nature of the problem warrants a decimal approximation. Using a calculator and rounding to the nearest tenth of a foot,
$$h \approx 19.4$$.
## The Distance Formula
We often need to calculate the distance between two points P and Q in the plane. Indeed, this is such a frequently recurring need, we’d like to develop a formula that will quickly calculate the distance between the given points P and Q. Such a formula is the goal of this last section.
Let P(x1, y1) and Q(x2, y2) be two arbitrary points in the plane, as shown in Figure 8(a) and let d represent the distance between the two points.
To find the distance d, first draw the right triangle △PQR, with legs parallel to the axes, as shown in Figure 8(b). Next, we need to find the lengths of the legs of the right triangle △PQR.
• The distance between P and R is found by subtracting the x coordinate of P from the x-coordinate of R and taking the absolute value of the result. That is, the distance between P and R is $$|x_{2}−x_{1}|$$.
• The distance between R and Q is found by subtracting the y-coordinate of R from the y-coordinate of Q and taking the absolute value of the result. That is, the distance between R and Q is $$|y_{2}−y_{1}|$$.
We can now use the Pythagorean Theorem to calculate d. Thus,
$$d^2 = (|x_{2}−x_{1}|)^2+(|y_{2}−y_{1}|)^2$$.
However, for any real number a,
$$(|a|)^2 = |a|·|a| = |a^2| = a^2$$,
because a2 is nonnegative. Hence, $$(|x_{2} − x_{1}|)^2 = (x_{2} − x_{1})^2 and (|y_{2} − y_{1}|)^2 = (y_{2} − y_{1})^2$$ and we can write
$$d^2 = (x_{2}−x_{1})^2+(y_{2}−y_{1})^2$$.
Taking the positive square root leads to the Distance Formula.
Let P (x1, y1) and Q(x2, y2) be two arbitrary points in the plane. The distance d between the points P and Q is given by the formula
$$d = \sqrt{(x_{2}−x_{1})^2+(y_{2}−y_{1})^2}$$. (9)
The direction of subtraction is unimportant. Because you square the result of the subtraction, you get the same response regardless of the direction of subtraction (e.g.$$(5 − 2)^2 = (2 − 5)^2$$). Thus, it doesn’t matter which point you designate as the point P, nor does it matter which point you designate as the point Q. Simply subtract x- coordinates and square, subtract y-coordinates and square, add, then take the square root.
Let’s look at an example.
Find the distance between the points P (−4, −2) and Q (4, 4).
It helps the intuition if we draw a picture, as we have in Figure 9. One can now take a compass and open it to the distance between points P and Q. Then you can place your compass on the horizontal axis (or any horizontal gridline) to estimate the distance between the points P and Q. We did that on our graph paper and estimate the distance $$d \approx 10$$.
Let’s now use the distance formula to obtain an exact value for the distance d. With $$(x_{1}, y_{1})$$ = P (−4, −2) and $$(x_{2}, y_{2})$$ = Q (4, 4),
$$d = \sqrt{(x_{2}−x_{1})^2+(y_{2}−y_{1})^2}$$
= $$\sqrt{(4−(−4))^2+(4−(−2))^2}$$
= $$\sqrt{8^2+6^2}$$
= $$\sqrt{64+36}$$
= $$\sqrt{100}$$
= 10.
It’s not often that your exact result agrees with your approximation, so never worry if you’re off by just a little bit.
## Exercise
In Exercises 1-8, state whether or not the given triple is a Pythagorean Triple. Give a reason for your answer.
(8, 15, 17)
Yes, because $$8^2 + 15^2 = 17^2$$
(7, 24, 25)
(8, 9, 17)
No, because $$8^2+9^2 \ne 17^2$$
(4, 9, 13)
(12, 35, 37)
Yes, because $$12^2 + 35^2 = 37^2$$
(12, 17, 29)
(11, 17, 28)
No, because $$11^2 + 17^2 \ne 28^2$$
(11, 60, 61)
In Exercises 9-16, set up an equation to model the problem constraints and solve. Use your answer to find the missing side of the given right triangle. Include a sketch with your solution and check your result.
In Exercises 17-20, set up an equation that models the problem constraints. Solve the equation and use the result to answer the question. Look back and check your result.
The legs of a right triangle are consecutive positive integers. The hypotenuse has length 5. What are the lengths of the legs?
The legs have lengths 3 and 4.
The legs of a right triangle are consecutive even integers. The hypotenuse has length 10. What are the lengths of the legs?
One leg of a right triangle is 1 centimeter less than twice the length of the first leg. If the length of the hypotenuse is 17 centimeters, find the lengths of the legs.
The legs have lengths 8 and 15 centimeters.
One leg of a right triangle is 3 feet longer than 3 times the length of the first leg. The length of the hypotenuse is 25 feet. Find the lengths of the legs.
Pythagoras is credited with the following formulae that can be used to generate Pythagorean Triples.
a = m
$$b = \frac{m^2−1}{2}$$,
$$c = \frac{m^2+1}{2}$$
Use the technique of Example 6 to demonstrate that the formulae given above will generate Pythagorean Triples, provided that m is an odd positive integer larger than one. Secondly, generate at least 3 instances of Pythagorean Triples with Pythagoras’s formula.
(3, 4, 5), (5, 12, 13), and (7, 24, 25), with m = 3, 5, and 7, respectively.
Plato (380 BC) is credited with the following formulae that can be used to generate Pythagorean Triples.
$$a = 2m$$
$$b = m^2 − 1$$,
$$c = m^2 + 1$$
Use the technique of Example 6 to demonstrate that the formulae given above will generate Pythagorean Triples, provided that m is a positive integer larger than 1. Secondly, generate at least 3 instances of Pythagorean Triples with Plato’s formula.
In Exercises 23-28, set up an equation that models the problem constraints. Solve the equation and use the result to answer. Look back and check your answer.
Fritz and Greta are planting a 12-foot by 18-foot rectangular garden, and are laying it out using string. They would like to know the length of a diagonal to make sure that right angles are formed. Find the length of a diagonal. Approximate your answer to within 0.1 feet.
21.63 ft
Angelina and Markos are planting a 20-foot by 28-foot rectangular garden, and are laying it out using string. They would like to know the length of a diagonal to make sure that right angles are formed. Find the length of a diagonal. Approximate your answer to within 0.1 feet.
The base of a 36-foot long guy wire is located 16 feet from the base of the telephone pole that it is anchoring. How high up the pole does the guy wire reach? Approximate your answer to within 0.1.
32.25 ft
The base of a 35-foot long guy wire is located 10 feet from the base of the telephone pole that it is anchoring. How high up the pole does the guy wire reach? Approximate your answer to within 0.1 feet.
A stereo receiver is in a corner of a 13-foot by 16-foot rectangular room. Speaker wire will run under a rug, diagonally, to a speaker in the far corner. If 3 feet of slack is required on each end, how long a piece of wire should be purchased? Approximate your answer to within 0.1 feet.
26.62 ft
A stereo receiver is in a corner of a 10-foot by 15-foot rectangular room. Speaker wire will run under a rug, diagonally, to a speaker in the far corner. If 4 feet of slack is required on each end, how long a piece of wire should be purchased? Approximate your answer to within 0.1 feet.
In Exercises 29-38, use the distance formula to find the exact distance between the given points.
(−8, −9) and (6, −6)
$$\sqrt{205}$$
(1, 0) and (−9, −2)
(−9, 1) and (−8, 7)
$$\sqrt{37}$$
(0, 9) and (3, 1)
(6, −5) and (−9, −2)
$$\sqrt{234} = 3\sqrt{26}$$
(−5, 6) and (1, 4)
(−7, 7) and (−3, 6)
$$\sqrt{17}$$
(−7, −6) and (−2, −4)
(4, −3) and (−9, 6)
$$\sqrt{250} = 5\sqrt{10}$$
(−7, −1) and (4, −5)
In Exercises 39-42, set up an equation that models the problem constraints. Solve the equation and use the result to answer the question. Look back and check your result.
Find k so that the point (4, k) is $$2\sqrt{2}$$ units away from the point (2, 1).
k = 3, −1.
Find k so hat the point (k, 1) is $$2\sqrt{2}$$ units away from the point (0, −1)
Find k so hat the point (k, 1) is $$\sqrt{17}$$ units away from the point (2, −3)
k = 1, 3.
Find k so that the point (−1, k) is $$\sqrt{13}$$ units away from the point (−4, −3).
Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Plot the points P (0, 5) and Q (4, −3) on your coordinate system.
a) Plot several points that are equidistant from the points P and Q on your coordinate system. What graph do you get if you plot all points that are equidistant from the points P and Q? Determine the equation of the graph by examining the resulting image on your coordinate system.
b) Use the distance formula to find the equation of the graph of all points that are equidistant from the points P and Q. Hint: Let (x, y) represent an arbitrary point on the graph of all points equidistant from points P and Q. Calculate the distances from the point (x, y) to the points P and Q separately, then set them equal and simplify the resulting equation. Note that this analytical approach should provide an equation that matches that found by the graphical approach in part (a).
Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Plot the point P (0, 2) and label it with its coordinates. Draw the line y = −2 and label it with its equation.
a) Plot several points that are equidistant from the point P and the line y = −2 on your coordinate system. What graph do you get if you plot all points that are equidistant from the points P and the line y = −2.
b) Use the distance formula to find the equation of the graph of all points that are equidistant from the points P and the line y = −2. Hint: Let(x, y) represent an arbitrary point on the graph of all points equidistant from points P and the line y = −2. Calculate the distances from the point(x,y) to the points P and the line y = −2 separately, then set them equal and simplify the resulting equation.
Copy the following figure onto a sheet of graph paper. Cut the pieces of the first figure out with a pair of scissors, then rearrange them to form the second figure. Explain how this proves the Pythagorean Theorem.
Compare this image to the one that follows and explain how this proves the Pythagorean Theorem.
You are watching: 9.6: The Pythagorean Theorem. Info created by THVinhTuy selection and synthesis along with other related topics.
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# PERCENT DIFFERENCE CALCULATOR
## Calculate the percentage difference between two numbers
PERCENT DIFFERENCE CALCULATOR
Start value:
End value:
Our percent difference calculator uses this formula:
| (y2 - y1) / ((y1 + y2)/2) | *100 = percentage difference
(where y1=start value and y2=end value)
How to calculate the percentage difference?
The percentage difference is the difference in percent between two positive numbers. The percentage difference between the given numbers is the absolute value of the difference between these numbers divided by the average and multiplied by 100.
This is the formula to calculate the percent difference:
| (V1 - V2) / ((V1 + V2)/2) | * 100
Note: the "|" symbol indicates an absolute value, which means that negative values become positive.
Percent difference versus percent change
The percent difference and the percent change are two different ways of measuring the relationship between two values.
The percent difference is a measure of the absolute difference between two values. To calculate the percent difference, subtract one value from the other, divide the result by the average of the two values, and then multiply by 100. This calculation gives the percent difference that the two values are apart from one another.
The percent change is a measure of the relative difference between two values. To calculate the percent change, subtract one value from the other, divide the result by the original value, and then multiply by 100. This calculation gives the percent change of the two values compared to the starting value.
The percent difference and the percent change are useful for comparing values and understanding the magnitude of change between two points in time. The percent difference is typically used to compare absolute values, while the percent change is used to compare relative values.
Examples
1. Calculate the percentage difference between two numbers
How to figure the percentage difference between 90 and 115?
The order of the values does not matter, as we simply divide the difference between two values by the average to get the percentage difference.
Assuming V1 = 90 and V2 = 110, let's now enter the values into the formula for percentage difference.
| (V1 - V2) / ((V1 + V2)/2) | * 100 = | (90 - 115) / ((90 + 115)/2) | * 100 = 24.39 %
Thus, the percentage difference between 80 and 120 is exactly 24.39%.
Note that the percentage difference is an absolute value. In this case, the negative result becomes positive.
2. Find the percentage difference between two prices
John and Alex want to buy a new refrigerator. John's refrigerator costs 750 USD, while Alex's comes to 850 USD. How to determine the percentage difference between the two prices?
We now enter V1 = 550 and V2 = 690 into our formula. Remember that the order of the values does not matter.
| (V1 - V2) / ((V1 + V2)/2) | * 100 = | (550 - 690) / ((550 + 690)/2) | * 100 = 22.58 %
This results in a price difference of 22.58 % between the two refrigerators.
Other online calculators
Here you can compute the percent change between two values.
Here you can figure the percent increase between two numbers.
Here you can find the percent decrease between two numbers.
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# What Is 4/55 as a Decimal + Solution With Free Steps
The fraction 4/55 as a decimal is equal to 0.072.
The long division makes complex problems into simple steps. While performing long division, some steps need to be followed in order and precisely; otherwise, long division can go wrong and other complexes the problem.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 4/55.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 4
Divisor = 55
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 4 $\div$ 55
This is when we go through the Long Division solution to our problem.
      Figure 1
## 4/55 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 4 and 55, we can see how 4 is Smaller than 55, and to solve this division, we require that 4 be Bigger than 55.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 4, which after getting multiplied by 100 becomes 400.
We take this 400 and divide it by 55; this can be done as follows:
 400 $\div$ 55 $\approx$ 7
Where:
55 x 7 = 385
This will lead to the generation of a Remainder equal to 400 – 385 = 15. Now this means we have to repeat the process by Converting the 15 into 150 and solving for that:
150 $\div$ 55 $\approx$ 2Â
Where:
55 x 2 = 110
This, therefore, produces another Remainder which is equal to 150 – 110 = 40.
Finally, we have a Quotient generated after combining the three pieces of it as 0.072=z, with a Remainder equal to 40.
Images/mathematical drawings are created with GeoGebra.
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## Precalculus (6th Edition) Blitzer
The solution is, ${{x}_{1}}=2-37t,{{x}_{2}}=16t,{{x}_{3}}=1-7t,{{x}_{4}}=t$.
Consider the given system of equations \begin{align} & {{x}_{1}}+4{{x}_{2}}+3{{x}_{3}}-6{{x}_{4}}=5 \\ & {{x}_{1}}+3{{x}_{2}}+{{x}_{3}}-4{{x}_{4}}=3 \\ & 2{{x}_{1}}+8{{x}_{2}}+7{{x}_{3}}-5{{x}_{4}}=11 \\ & 2{{x}_{1}}+5{{x}_{2}}-6{{x}_{4}}=4 \end{align} Therefore, in matrix form the system of equations can be written as $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 \\ 1 & 3 & 1 & -4 \\ 2 & 8 & 7 & -5 \\ 2 & 5 & 0 & -6 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 5 \\ 3 \\ 11 \\ 4 \\ \end{array} \right];X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ {{x}_{4}} \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 & 5 \\ 1 & 3 & 1 & -4 & 3 \\ 2 & 8 & 7 & -5 & 11 \\ 2 & 5 & 0 & -6 & 4 \\ \end{array} \right]$ By applying elementary row operation on $A$ we will convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-2{{R}_{1}},{{{R}'}_{4}}={{R}_{4}}-2{{R}_{1}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 & 5 \\ 0 & -1 & -2 & 2 & -2 \\ 0 & 0 & 1 & 7 & 1 \\ 0 & -3 & -6 & 6 & -6 \\ \end{array} \right]$ Step 2: Apply the operation ${{{R}'}_{4}}={{R}_{4}}-3{{R}_{3}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 & 5 \\ 0 & -1 & -2 & 2 & -2 \\ 0 & 0 & 1 & 7 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]$ Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $A$ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=3$
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# Difference between revisions of "2005 Alabama ARML TST Problems/Problem 7"
## Problem
Find the sum of the infinite series:
$3+\frac{11}4+\frac 94 + \cdots + \frac{n^2+2n+3}{2^n}+\cdots$.
## Solution
$\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}=\sum_{n=1}^{\infty} \left(\frac{n^2}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)$
We can compute those sums:
$\begin{eqnarray} \sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\ =3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\ 2x=3\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\\ x=3(1)=3\\ \sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)=y\\ =\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\\ 2y=2+\frac{4}{2}+\frac{6}{4}+\frac{8}{8}+\cdots\\ y=2+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=4\\ \sum_{n=1}^{\infty} \left(\frac{n^2}{2^n}\right)=z\\ =\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\cdots\\ 2z=1+\frac{4}{2}+\frac{9}{4}+\frac{16}{8}+\cdots\\ z=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots\\ 2z=2+3+\frac{5}{2}+\frac{7}{4}+\frac{9}{8}+\cdots\\ z=4+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\cdots=6\\ 3+4+6=\boxed{13} \end{eqnarray}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)
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# Smartick - Math, one click away
Jul12
## Learn How to Calculate Factorials
Today we are going to learn how to calculate factorials. Calculating factorials is quite simple; let’s see what it’s all about:
##### What is the Factorial Function?
We represent the factorial function with the exclamation point “!”, placing it behind the number. This exclamation means that we need to multiply all of the positive whole numbers that fall between the number and 1.
For example:
We generally say “6 factorial”, although it can also be “factorial of 6”.
On your calculator, you’ll see a button with “n!” or “x!”. You can use this button to calculate the factorial of whatever number that you want to calculate.
##### A Few Examples of Factorials
We’re going to take a look at some more examples of factorials:
As you can see, 100! is a huge number…
And, what do we do with the smaller numbers? 1 factorial is, logically, 1, because it’s simply 1 x 1:
But, how can we calculate the 0 factorial? Well, when we apply the norms of multiplying all of the positive whole integers that fall between 0 and 1, it doesn’t make sense to calculate it because 0 x 1 is 0.
So, the solution is to equate the 0 factorial to 1. So, just remember that:
##### What do we use factorials for?
Above all, the factorial numbers are used in combinatorial analysis, in order to calculate combinations and permutations. In combinatorial analysis, factorials can also be used to calculate probabilities.
We are going to take a look at a simple problem where we can apply factorials.
Paula took out the 4 aces from a deck of cards. She is going to put them in a line on the table. How many different ways could she line them up?
In this problem, we have to solve a “permutation”, or in other words, we have to find out all of the possible ways that these 4 cards can be ordered.
If we start by making all of the possible lines that start off with the ace of diamonds, we can make 6 combinations:
We’ll also have 6 possible combinations with the ace of clovers, hearts and spades. In other words, 6 combinations starting with each one of the 4 aces: 4 x 6 = 24
She could order them in 24 different ways.
Using the factorial function, we could have solved the problem through a much simpler way:
• When we choose the first one, there are only 3 left to choose from
• When we choose the second one, there are only 2 left to choose from
• Then when we choose the third one, there is only 1 left to choose from
So as a result, all of the possible combinations are 4 x 3 x 2 x 1.
It’s the same thing as 4! = 24
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### High School Mathematics1.20 Linear Equations - Two Variables
Linear equation with one variable: has one solution. Example: x + 4 = 12 x + 2x + 6 = 9 simplifies to 3x + 6 = 9 Linear equation with two variables: A linear equation in two variables has infinitely many solutions. Example: x + y = 12 For x = 1, y = 11, x = 2, y = 10, x = 3, y = 9, x = 4, y = 8..... and so on Linear equation with three variables Example: x + y + z = 12 Note: You will need one equation to solve for one variable, two equations to solve two variable and three equations to solve three variable and so on...... Method 1: Elimination Method: Solving Linear Equations with two variables. Solve for the values of a and b for the following equations: Example 1: a + b = 12 -----equation 1 a - b = 4 ------- equation 2. Adding both equations we get a + b = 12 a - b = 4 ------------------The terms +b and -b cancels, and we get 2a = 16 a = 16/8 = 2 a = 2 substituting the value of a in equation 1 we have a + b = 12 2 + b = 12 b = 12 - 2 b = 10 The solution of the two equations is a = 2 and b = 10 Example 2: c + 5b = 16 -----equation 1 c + 2b = 7 ------- equation 2 . Multiplying the equation 2 with (-1) we get c + 5b = 16 -----equation 1 -1(c + 2b = 7) => -c - 2b = -7-------equation 2 c + 5b = 16 -c - 2b = -7 ---------------------The terms c and -c cancel each other and we get 3b = 9 b = 9/3 = 3 substituting b in the equation c + 5b = 16 we get c + 5(3) = 16 c + 15 = 16 - 3 c = 16 - 15 c = 1 Example 3: 2a + 4b = 12 -----equation 1 a - b = 3 ------- equation 2 First pick a variable either a or b. Lets pick b. Second find the common multiple of b so we can eliminate that variable and find the value of the other variable. so multiply both sides of the equation 2 by 4 we have 2a + 4b = 12 -----equation 1 4(a - b = 3) => 4a - 4b = 12------- equation 2 Now adding equation 1 and 2 we have 2a + 4b = 12 4a - 4b = 12 ----------------------The terms +4b in equation 1 and -4b in equation 2 cancels, and we get 6a = 24 a = 24/6 =4 substituting value of a in equation a - b = 3 we get 4 - b = 3 4- 3 = b b = 1 Therefore a = 4 and b = 1 Verification: To verify substitute the values of a and b in the equations above 2a + 4b = 12 2a + 4b = 2(4) + 4(1) = 8 + 4 = 12 NOTE: In the above question if you pick b 2a + 4b = 12 -----equation 1 a - b = 3 ------- equation 2 multiplying equation 2 by 2 we have 2(a - b = 3) ------- equation 2 2a - 2b = 6 multiplying equation by -1 we gave -1(2a - 2b = 6) we get -2a + 2b = -6 Solving both equations we have 2a + 4b = 12 -----equation 1 -2a + 2b = -6-----equation 2 ------------------------- 6b = 6 b = 1 Example 4: 2a + b = 4 -----equation 1 3a + 2b = 3 ------- equation 2. Multiplying the equation 1 with 3 and equation 2 with (-2) we get 3(2a + b = 4) -----equation 1 -2(3a + 2b = 3) ------- equation 2 6a + 3b = 12 -----equation 1 -6a - 4b = -6) ------- equation 2 -----------------------The terms 6a and -6a cancel each other and we get -b = 6 b = -6 substituting b in the equation 2a + b = 4 we get 2a + (-6) = 4 2a - 6 = 4 2a = 4 + 6 2a = 10 a = 10/2 = 5 Method 2: Substitution Method: Solve the equations: 3x + 5y = 26 y = 2x substitute the second equation in the first 3x + 5(2x) = 26 3x + 10x = 26 13x = 26 x = 2 y = 2x = 2x2 = 4 Solve the equations: x + y = 26 ------- equation 1 x - y = 4 -------equation 2 x - y = 4 -------equation 2 can be written as x = y + 4. Substituting x in the equation 1 we have: x + y = 26 ------- equation 1 y + 4 + y = 26 2y + 4 = 26 2y + 4 - 4 = 26 - 4 2y = 22 y = 11 x = y + 4 = 11 + 4 = 15 Directions: Solve for the variables using either Elimination or Substitution Method.
Q 1: Solve for the values of x and y for the following equations: x + y = 5 and x - y = 5x = 5 and y = 5x = 5 and y = 0x = 5 and y = 1 Q 2: Solve for the values of x and y for the following equations: 3x + 6y = 24 and x + 3y = 4x = 16 and y = -4x = -16 and y = 4x = 6 and y = 8 Q 3: Solve for the values of x and y for the following equations: 2x + y = 6 and 2x + 2y = 24x = -6 and y = 18x = 6 and y = -18x = 18 and y = 6 Q 4: Solve for the values of x and y for the following equations: y = 3x + 8 and y = -xx = 2 and y = -2x = -2 and y = 2x = 4 and y = -2 Q 5: Solve for the values of x and y for the following equations: x + 3y = 7 and 2x + 5y = 12x = 2 and y = 2x = 1 and y = 1x = 1 and y = 2 Q 6: Solve for the values of x and y for the following equations: 3x - y = 7 and 2x + 3y = 1.x = 2 and y = 1x = 2 and y = -1x = -2 and y = -1 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
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If the determinant of a matrix is 0 then the matrix is singular and it does not have an inverse. compared to the previous example. We cannot go any further! An online Matrix calculation. So, let us check to see what happens when we multiply the matrix by its inverse: And, hey!, we end up with the Identity Matrix! Remember it must be true that: A × A-1 = I. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. So I want to essentially find a inverse, and I want to do it just using a formula that it just applies to this matrix right here. We can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. The result should be the identity matrix I … See if you also get the Identity Matrix: Because with matrices we don't divide! A Matrix (This one has 2 Rows and 2 Columns) The determinant of that matrix is (calculations are explained later): problem and check your answer with the step-by-step explanations. So this is really any 2 by 2 matrix. Also note how the rows and columns are swapped over We know that the result is going to be a 2×2 matrix because the first matrix, A, has two rows and the second matrix, B, has two columns. To multiply matrix A by matrix B, we use the following formula: A x B =. When we multiply a matrix by its inverse we get the Identity Matrix (which is like "1" for matrices): We just mentioned the "Identity Matrix". Example: Determine the inverse of matrix … A11 * B12 + A12 * B22. 2x2 Sum of Two Determinants. And anyway 1/8 can also be written 8-1, When we multiply a number by its reciprocal we get 1. 02 Jul, 2015 . 2x2 Sum of Two Determinants. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. To find a 2×2 determinant we use a simple formula that uses the entries of the 2×2 matrix. Matrix Calculator. (1 × x) − (4 × −2) = 5 Formula 2*2 matrix is 2x2 Squared Matrix is given by, 3*3 matrix is 3x3 Squared Matrix is given by, X11 = a11*a11 + a12*a21 + a13*a31, X12 = a11*a12 + a12*a22 + a13*a32, It is like the inverse we got before, but Note: Not all square matrices have inverses. It can be done that way, but we must be careful how we set it up. Matrix2. We take the product of the elements … Determinant of 2×2 Matrix … Inverse of a Matrix Matrix Inverse Multiplicative Inverse of a Matrix For a square matrix A, the inverse is written A-1. So we select an area on the worksheet 2 cells wide by 2 cells high: Next, with the area still selected, type the array formula and select the arguments (I assigned names to the matrices A and B): Let us try an example: How do we know this is the … For those larger matrices there are three main methods to work out the inverse: Inverse of a Matrix using Elementary Row Operations (Gauss-Jordan), Inverse of a Matrix using Minors, Cofactors and Adjugate. The result should be the identity matrix I … So how do we solve this one? How to find the determinant of a matrix (2x2): formula, 1 example, and its solution. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses.. Matrix determinant 3x3 formula. If the determinant of a matrix is 0 then the matrix is singular and it does not have an inverse. To multiply a matrix by another matrix we need to follow the rule “DOT PRODUCT”. In Mathematics one matrix by another matrix. Search. AB = [c i j], where c i j = a i 1 b 1 j + a i 2 b 2 j + … + a in b n j. Matrix Determinant Calcualtor. 16. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Find the determinant of the remaining 2 x 2 matrix, multiply by the chosen element, and refer to a matrix sign chart to determine the sign. First, let us set up the matrices (be careful to get the rows and columns correct! In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Inverse Matrix Calculator (2X2) Enter the 4 values of a 2 x 2 matrix into the calculator. The first element of row one is occupied by the number 1 … My problem: I don't understand how to make c and b = 0 using elementary row operations. Given the matrix D we select any row or column. Please submit your feedback or enquiries via our Feedback page. Before we can find the inverse of a matrix, we need to first learn how to get the determinant of a matrix. Reference. So this is really any 2 by 2 matrix. The calculations are done by computer, but the people must understand the formulas. Note that if A ~ B, then ρ(A) = ρ(B) Try the given examples, or type in your own Diagonalizable Matrices vs Hermitian matrices. 5. Understanding the singular value decomposition (SVD) 1. Find the determinant of a larger matrix. 3x3 Sum of Determinants. What I want to do is use our technique for finding an inverse of this matrix to essentially find a formula for the inverse of a 2 by 2 matrix. BOOK FREE CLASS; COMPETITIVE ... Determinants occur throughout mathematics. This Matrix has no Inverse. To find a 2×2 determinant we use a simple formula that uses the entries of the 2×2 matrix. Determinant of a 2×2 Matrix It looks so neat! But we can take the reciprocal of 2 (which is 0.5), so we answer: The same thing can be done with matrices: Say we want to find matrix X, and we know matrix A and B: It would be nice to divide both sides by A (to get X=B/A), but remember we can't divide. Please read our Introduction to Matrices first. 02 Jul, 2015 . Multiplying a matrix by its inverse is the identity matrix. 3x3 Cramers Rule. 2x2 covariance matrix can be represented by an ellipse. So I'm going to keep it really general. You should check that this answer is correct by performing the matrix multiplication AA−1. Such a matrix is called "Singular", which only happens when the determinant is zero. Eigenvalues and eigenvectors - … 2x2 matrix inverse calculator The calculator given in this section can be used to find inverse of a 2x2 matrix. It is important to know how a matrix and its inverse are related by the result of their product. A good way to double check your work if you’re multiplying matrices by hand is to confirm your answers with a matrix calculator. OK, how do we calculate the inverse? For block matrix and its inverse, which generalizes this problem. Usefulness of Why Eigenvectors Corresponding to Distinct Eigenvalues of Symmetric Matrix are Orthogonal 0 Which $2\times 2$ matrices with entries from finite field are similar to upper triangular matrix? But what if we multiply both sides by A-1 ? If the determinant of a matrix is 0 then the matrix is singular and it does not have an inverse. It’s when you get into the depth that you discover both its power and flexibility. A21 * B11 + A22 * B21. For more details on matrix determinant follow the guidelines from Wikipedia. Its inverse in terms of A -1 or D -1 can be found in standard textbooks on linear algebra, e.g., [1-3]. Matrix determinant 3x3 formula. So then, If a 2×2 matrix A is invertible and is multiplied by its inverse (denoted by the symbol A−1 ), the resulting product is the Identity matrix which is denoted by. Search. In linear algebra, a rotation matrix is a transformation matrix that is used to perform a rotation in Euclidean space.For example, using the convention below, the matrix = [ − ] rotates points in the xy-plane counterclockwise through an angle θ with respect to the x axis about the origin of a two-dimensional Cartesian coordinate system. First of all, to have an inverse the matrix must be "square" (same number of rows and columns). 2Ã2 determinants 4. Why don't you have a go at multiplying these? AB is almost never equal to BA. So matrices are powerful things, but they do need to be set up correctly! 2x2 matrix inverse calculator The calculator given in this section can be used to find inverse of a 2x2 matrix. x + 8 = 5 The Calculator. What is the general formula for raising a square 2x2 matrix to a power such as 10 or 20? Determinant Formula, 2x2 determinant formula, 3x3 determinant formula, 4x4 determinant formula, matrices and determinants formulas. 2x2 Matrix. Parametric equations and formulas for radii + rotation are provided for covariance matrix shown below. 2x2 Cramers Rule. Do not assume that AB = BA, it is almost never true. This page introduces specific examples of cofactor matrix (2x2, 3x3, 4x4). But it is based on good mathematics. But also the determinant cannot be zero (or we end up dividing by zero). Note that if A ~ B, then ρ(A) = ρ(B) Solution Using the formula A−1 = 1 (3)(2)− (1)(4) 2 −1 −4 3! 2. Matrix inversion lemmas. Try the free Mathway calculator and How to find the determinant of a matrix (2x2): formula, 1 example, and its solution. They took the train back at $3.50 per child and$3.60 per adult for a total of $135.20. 3x3 Sum of Determinants. The remaining corresponding blocks are also equal. Eigenvalues and eigenvectors - … ... and someone asks "How do I share 10 apples with 2 people?". Determinant of a Matrix. A good way to double check your work if you’re multiplying matrices by hand is to confirm your answers with a matrix calculator. If A = [a i j] is an m × n matrix and B = [b i j] is an n × p matrix, the product AB is an m × p matrix. What I want to do is use our technique for finding an inverse of this matrix to essentially find a formula for the inverse of a 2 by 2 matrix. The Inverse matrix is also called as a invertible or nonsingular matrix. Matrix 2x2 Multiplication Calculator . Here 'I' refers to the identity matrix. I think I prefer it like this. If A and B are two equivalent matrices, we write A ~ B. Seriously, there is no concept of dividing by a matrix. The inverse formula (1.1) of a 2 x 2 block matrix appears frequently in many subjects and has long been studied. The multiplicative identity matrix obeys the following equation: IA = AI = A The multiplicative identity matrix for a 2x2 matrix is: It does not give only the inverse of a 2x2 matrix, and also it gives you the determinant and adjoint of the 2x2 matrix that you enter. Or if we could rewrite this as saying lambda is an eigenvalue of A if and only if-- I'll write it as if-- the determinant of lambda times the identity matrix minus A is equal to 0. Here 'I' refers to the identity matrix. While there are many matrix calculators online, the simplest one to use that I have come across is this one by Math is Fun. Eigenvalues and eigenvectors of similar matrices. 3x3 Matrix Determinants. The Inverse matrix is also called as a invertible or nonsingular matrix. While there are many matrix calculators online, the simplest one to use that I have come across is this one by Math is Fun. Matrix Inversion Formulas Next, comparing the upper-left blocks of (2) and (4), we see that [A BD 1C] 1 =A 1 +A 1B[D CA 1B] 1CA 1; (7) which is known as the Sherman–Morrison–Woodbury formula or sometimes just the Woodbury formula. Let us discuss how to multiply a matrix by another matrix, its algorithm, formula, 2×2 and 3×3 matrix multiplication. It is also a way to solve Systems of Linear Equations. Diagonalizable Matrices vs Hermitian matrices. The examples above illustrated how to multiply 2×2 matrices by hand. Search. First, the original matrix should be in the form below. Say that we are trying to find "X" in this case: This is different to the example above! This method helps visualize multivariate normal distributions and correlation matrices. That equals 0, and 1/0 is undefined. Students now can learn 10x faster and retain 98% of knowledge. 3x3 Inverse Matrix I. Embedded content, if any, are copyrights of their respective owners. First, the original matrix should be in the form below. There is also a general formula based on matrix conjugates and the determinant. 2. (Imagine in our bus and train example that the prices on the train were all exactly 50% higher than the bus: so now we can't figure out any differences between adults and children. For more details on matrix determinant follow the guidelines from Wikipedia. The determinant of matrix A is calculated as If you can’t see the pattern yet, this is how it looks when the elements of the matrix are color-coded. You should check that this answer is correct by performing the matrix multiplication AA−1. ... Inverse Matrix Formula. Selecting row 1 of this matrix will simplify the process because it contains a zero. 2x2 Sum of Determinants. Detailed Answer 2x2 Matrices Multiplication Formula. My problem: I don't understand how to make c and b = 0 using elementary row operations. Determinants are useful properties of square matrices, but can involve a lot of computation. This could be written as 1 −1 2 −2 3 2! can be used to find the area of a parallelogram and to determine invertibility of a 2Ã2 matrix. I'm supposed to find the inverse of the 2x2 matrix [a b] [c d] Now I don't want anyone to solve it for me, I would just like to know how to start finding the rref with elementary row operations, starting with making c and b = 0. 2x2 Sum of Determinants. 3x3 Matrix Determinants. When we multiply a matrix by a scalar value, then the process is known as scalar multiplication. Find the inverse of the matrix A = 3 1 4 2!. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. RE: singular matrix and eigenvectors. Thus, the rank of a matrix does not change by the application of any of the elementary row operations. Determinant of a 2×2 Matrix I'm supposed to find the inverse of the 2x2 matrix [a b] [c d] Now I don't want anyone to solve it for me, I would just like to know how to start finding the rref with elementary row operations, starting with making c and b = 0. Find the determinant of a larger matrix. A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. So I'm going to keep it really general. Matrix determinant 4x4 formula. In Mathematics one matrix by another matrix. RE: singular matrix and eigenvectors. We welcome your feedback, comments and questions about this site or page. Determining a 2x2 Inverse Matrix Using a Formula This video explains the formula used to determine the inverse of a 2x2 matrix, if one exists. 2x2 Cramers Rule. In that example we were very careful to get the multiplications correct, because with matrices the order of multiplication matters. 5. Below there are the formulas used to compute the determinant depending on the matrix ordin: Matrix determinant 2x2 formula. So I want to essentially find a inverse, and I want to do it just using a formula that it just applies to this matrix right here. Assuming using only your memory recall you can master and remember for one month how to compute multiplication of 2 x 2 matrix in 60 minutes. A 2Ã2 If A and B are two equivalent matrices, we write A ~ B. pka Elite Member. Joined Jan 29, 2005 Messages 10,712. If your matrix is 3 x 3 or larger, finding the determinant takes a bit more work: 3 x 3 matrix: Choose any element and cross out the row and column it belongs to. 3x3 Inverse Matrix Inverse of a Matrix Matrix Inverse Multiplicative Inverse of a Matrix For a square matrix A, the inverse is written A-1. Copyright © 2005, 2020 - OnlineMathLearning.com. And the determinant lets us know this fact. = 1 2 2 −1 −4 3! The Leibniz formula for the determinant of a 2 × 2 matrix is | | = −. The Inverse of a Matrix is the same idea but we write it A-1, Why not 1/A ? Determinant of a 2×2 Matrix Suppose we are given a square matrix with four elements: , , , and . Eigenvalues and eigenvectors of similar matrices. To multiply a matrix by another matrix we need to follow the rule “DOT PRODUCT”. [A | I]), and then do a row reduction until the matrix is of the form [I | B], and then B is the inverse of A. Solution Using the formula A−1 = 1 (3)(2)− (1)(4) 2 −1 −4 3! In the last video we were able to show that any lambda that satisfies this equation for some non-zero vectors, V, then the determinant of lambda times the identity matrix minus A, must be equal to 0. 2x2 Matrix Determinants. Matrix Calculator. In the last video we were able to show that any lambda that satisfies this equation for some non-zero vectors, V, then the determinant of lambda times the identity matrix minus A, must be equal to 0. The 2x2 matrix addition and subtraction calculator, formula, example calculation (work with steps), real world problems and practice problems would be very useful for grade school students (K-12 education) to understand the addition and subtraction of two or more matrices. X is now after A. Let the payoff matrix of a 2 x 2 game be characterized by the matrix All entries are positive real numbers. It is given by the property, I = A A-1 = A-1 A. The examples above illustrated how to multiply 2×2 matrices by hand. Using the same method, but put A-1 in front: Why don't we try our bus and train example, but with the data set up that way around. Or if we could rewrite this as saying lambda is an eigenvalue of A if and only if-- I'll write it as if-- the determinant of lambda times the identity matrix minus A is equal to 0. It does not give only the inverse of a 2x2 matrix, and also it gives you the determinant and adjoint of the 2x2 matrix that you enter. x = −3. Given the matrix D we select any row or column. 2x2 Matrix Game Formula | Math & Physics Problems Wikia | FANDOM powered by Wikia FANDOM The inverse of a 2x2 is easy ... compared to larger matrices (such as a 3x3, 4x4, etc). Selecting row 1 of this matrix will simplify the process because it contains a zero. Multiplying a matrix by its inverse is the identity matrix. The Leibniz formula for the determinant of a 2 × 2 matrix is | | = −. Find the inverse of the matrix A = 3 1 4 2!. ("Transposed") Below there are the formulas used to compute the determinant depending on the matrix ordin: Matrix determinant 2x2 formula. Given the matrix in the form: Let us discuss how to multiply a matrix by another matrix, its algorithm, formula, 2×2 and 3×3 matrix multiplication. It is "square" (has same number of rows as columns). Enter the numbers in this online 2x2 Matrix Inverse Calculator to find the inverse of the square matrix. Fast way to calculate Eigen of 2x2 matrix using a formula. We can remove I (for the same reason we can remove "1" from 1x = ab for numbers): And we have our answer (assuming we can calculate A-1). 2×2 determinants can be used to find the area of a parallelogram and to determine invertibility of a 2×2 matrix. This results in a 2×2 matrix. Thus, the rank of a matrix does not change by the application of any of the elementary row operations. A21 * B12 + A22 * B22. This page introduces specific examples of cofactor matrix (2x2, 3x3, 4x4). Matrix determinant 4x4 formula. Transposed (rows and columns swapped over). Matrix Determinant Calcualtor. B 22. 2x2 Matrix Multiply Formula & Calculation. We can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. The Woodbury formula is maybe one of the most ubiquitous trick in basic linear algebra: it starts with the explicit formula for the inverse of a block 2x2 matrix and results in identities that can be used in kernel theory, the Kalman filter, to combine multivariate normals etc. The determinant of matrix A is calculated as If you can’t see the pattern yet, this is how it looks when the elements of the matrix are color-coded. 2×2 determinants can be used to find the area of a parallelogram and to determine invertibility of a 2×2 matrix. find a 2Ã2 determinant we use a simple formula that uses the entries of the 2Ã2 matrix. Determinant of a 2×2 Matrix Suppose we are given a square matrix with four elements: , , , and . We take the product of the elements … Determinant of 2×2 Matrix … A11 * B11 + A12 * B21. In the following, DET is the determinant of the matrices at the left-hand side. It is given by the property, I = A A-1 = A-1 A. —Simon Trussler40 Mention “2 ×2 matrix” to someone in a business context, and more often than not, that person will think of the BCG Grid. And it makes sense ... look at the numbers: the second row is just double the first row, and does not add any new information. The determinant of a matrix is a special number that can be calculated from a square matrix.. A Matrix is an array of numbers:. 16. problem solver below to practice various math topics. The Calculator. A group took a trip on a bus, at$3 per child and $3.20 per adult for a total of$118.40. Matrix1. With matrices the order of multiplication usually changes the answer. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses.. … There needs to be something to set them apart.). Feb 4, 2008 #2 Re: Formula for matrix raised to power n. JohnfromTampa said: When we multiply a matrix by a scalar value, then the process is known as scalar multiplication. Fast way to calculate Eigen of 2x2 matrix using a formula. Calculations like that (but using much larger matrices) help Engineers design buildings, are used in video games and computer animations to make things look 3-dimensional, and many other places. Given the matrix in the form: = 1 2 2 −1 −4 3! Reference. Inverse Matrix Calculator (2X2) Enter the 4 values of a 2 x 2 matrix into the calculator. If your matrix is 3 x 3 or larger, finding the determinant takes a bit more work: 3 x 3 matrix: Choose any element and cross out the row and column it belongs to. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. In this case, (ad-bc) is also known as the magnitude of the original matrix. Determinant of a Matrix. Enter the numbers in this online 2x2 Matrix Inverse Calculator to find the inverse of the square matrix. The multiplicative identity matrix is so important it is usually called the identity matrix, and is usually denoted by a double lined 1, or an I, no matter what size the identity matrix is. So it must be right. The first element of row one is occupied by the number 1 … 4. In linear algebra, a rotation matrix is a transformation matrix that is used to perform a rotation in Euclidean space.For example, using the convention below, the matrix = [ − ] rotates points in the xy-plane counterclockwise through an angle θ with respect to the x axis about the origin of a two-dimensional Cartesian coordinate system. 3x3 Sum of Three Determinants. Formula 2*2 matrix is 2x2 Squared Matrix is given by, 3*3 matrix is 3x3 Squared Matrix is given by, X11 = a11*a11 + a12*a21 + a13*a31, X12 = a11*a12 + a12*a22 + a13*a32, The following formula is used to calculate the determinant of a 2×2 matrix. How about this: 24-24? In this case, (ad-bc) is also known as the magnitude of the original matrix. A Matrix (This one has 2 Rows and 2 Columns) The determinant of that matrix is (calculations are explained later): Search. ... Inverse Matrix Formula. The following examples illustrate how to multiply a 2×2 matrix with a 2×2 matrix using real numbers. But we can multiply by an inverse, which achieves the same thing. AB = [c i j], where c i j = a i 1 b 1 j + a i 2 b 2 j + … + a in b n j. If A = [a i j] is an m × n matrix and B = [b i j] is an n × p matrix, the product AB is an m × p matrix. determinant is much easier to compute than the determinants of larger matrices, like 3Ã3 matrices. Because we don't divide by a matrix! 3x3 Sum of Three Determinants. The determinant of a matrix is a special number that can be calculated from a square matrix.. A Matrix is an array of numbers:. 2x2 Matrix Determinants. The following formula is used to calculate the determinant of a 2×2 matrix. A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. Find the determinant of the remaining 2 x 2 matrix, multiply by the chosen element, and refer to a matrix sign chart to determine the sign. Understanding the singular value decomposition (SVD) 1. To BCG: Product Portfolio Matrix Bruce Hendersen The framework is simple on the surface, but has a lot of hidden depth. There are the formulas used to calculate the determinant depending on the matrix is called singular... 2 matrix be true that: a × A-1 = A-1 a like the inverse of the …... Be characterized by the matrix multiplication AA−1 also note how the rows and columns are over! The property, I = a A-1 = A-1 a about this site page! Appears frequently in many subjects and has long been studied on the matrix AA−1! First element of row one is occupied by the matrix ordin: matrix determinant follow the from! 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Via our feedback page below to practice various math topics is written.. 3 1 4 2! is like the inverse of a 2 x 2 matrix columns are swapped over.! Equivalent matrices, we write a ~ B 1 × x ) − ( 4 2... 2Ã2 determinants can be calculated by finding the determinants of larger matrices like! Take the product of the matrices ( be careful how we set up. As 1 −1 2 −2 3 2! 1 ) ( 2 ) − ( ×... What if we multiply a matrix by another matrix, we use a simple formula that uses the of... With four elements:,,,, and its inverse, which only happens the. Multiplication usually changes the answer back at \$ 3.50 per child and 3.60! A by matrix B, we need to be set up the matrices at the left-hand side by B... The example above provided for covariance matrix can be represented by an ellipse matrix appears in. Examples, or type in your own problem and check your answer with step-by-step! A go at multiplying these go at multiplying these it must be true that: a × A-1 A-1. 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matrix formula 2x2 2021
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# Simplification Quiz 9 | Quantitative Aptitude
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# Simplification Quiz 9 | Quantitative Aptitude
### Introduction
What is Quantitative Aptitude test? Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.
What is Simplification? Simplification is the most widely asked topic in almost every competitive exam. Simplification is based on basic math calculations and some other algebraic topics. Simplification is less time consuming and having higher accuracy.
Simplification Problems is based on BODMAS rule, where
B → Brackets,
O → Of,
D → Division,
M → Multiplication,
A → Addition, and
S → Subtraction.
BODMAS Rule: BODMAS is about simplifying an expression by firstly removing the brackets in the order i.e. (), {}, []. Removal of brackets is followed by addition, subtraction, multiplication, division, square roots, cube roots, powers, cancellation of numerator/ denominator and so on. The article Simplification Quiz 9 | Quantitative Aptitude provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc.
### Q1
The price of 80 apples is equal to that of 120 oranges. The price of 60 apples and 75 oranges together is Rs. 1320. The total price of 25 apples and 40 oranges is
A. 660 B. 620 C. 820 D. 780
B
80 apples ≡ 120 oranges
⇒ 2 apples ≡3 oranges
⇒60 apples ≡90 oranges
60 apples +75 oranges =1320
⇒ 90 oranges + 75 oranges = 1320
⇒ 165 oranges =1320
⇒1 orange = 8
1 apple = $\frac {3 \times 8}{2}$ = 12
25 apples + 40 oranges
= 25 × 12 + 40 × 8 = 620
### Q2
The price of 3 tables and 4 chairs is Rs. 3300. With the same money one can buy 2 tables and 10 chairs. If one wants to buy 1 table and 1 chair, how much does he need to pay?
A. 940 B. 1050 C. 1040 D. 950
B
3 tables + 4 chairs ≡ 2 tables + 10 chairs
⇒1 table ≡ 6 chairs
3 tables + 4 chairs =3300
⇒ 3(6 chairs) + 4 chairs =3300
⇒ 22 chairs =3300
⇒ 1 chair =150
1 table +1 chair
≡ 6 chairs + 1 chair
=7 chairs = 7 × 150 = 1050
### Q3
There are 6 working days in a regular week and for each day, the working hours are 10. A man earns Rs. 2.10/h for regular work and Rs. 4.20/h for overtime. If he earns Rs. 525 in 4 weeks, how many hours did he work?
A. 245 B. 285 C. 275 D. 255
A
Regular working hours in 44 weeks
= 4 × 6 × 10 = 240 hours
Amount earned by working in these regular working hours
= 240 × 2.10 = Rs. 504
= 525 − 504 = Rs. 21
Hours he worked overtime
= $\frac {21}{4.2}= \frac {210}{42}=5$hours
Total hours he worked
= 240 + 5 = 245 hours
### Q4
A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
A. 30 B. 22 C. 18 D. 38
A
B missed 27 times. Hence, B fired 27 × 2 = 54 shots.
Therefore, number of shots fired by A
= $\frac {54}{3} × 5 = 90$
Therefore, number of birds killed by A
= $90 \times \frac {1}{3 }= 30$
### Q5
A man has Rs. 312 in the denominations of one-rupee notes, five-rupee notes and twenty-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?
A. 36 B. 24 C. 28 D. 32
A
Let number of notes of each denomination =x. Then
x + 5x + 20x = 312
⇒ 26x = 312
⇒ x = $\frac { 312}{26}= 12$
Total number of notes = 3x = 3 × 12 = 36
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Convert native decimal to fraction. Convert 3.3 to Fraction. Decimal to portion chart and calculator. Writes any kind of decimal number together a fraction.
You are watching: What is 3.3 repeating as a fraction
## How to transform a Decimal come a fraction - Steps
Step 1: compose down the decimal together a fraction of one (decimal/1);Step 2: If the decimal is not a totality number, main point both top and bottom by 10 until you obtain an interger in ~ the numerator.
Learn much more reading the examples below or usage our self-explaining calculator above
## Convert decimal 0.05 come a fraction
0.05 = 1/20 together a fraction
### Step by step Solution
To convert the decimal 0.05 to a portion follow this steps:
Step 1: create down the number as a fraction of one:
0.05 = 0.05/1
Step 2: multiply both top and bottom by 10 because that every number ~ the decimal point:
As we have actually 2 numbers after the decimal point, we multiply both numerator and denominator through 100. So,
0.05/1 = (0.05 × 100)/(1 × 100) = 5/100.
Step 3: simplify (or reduce) the fraction:
5/100 = 1/20 when lessened to the simplest form.
## What is 0.45 as a fraction?
0.45 = 9/20 together a fraction
### Step by action Solution
To transform the decimal 0.45 to a portion follow these steps:
Step 1: write down the number as a fraction of one:
0.45 = 0.45/1
Step 2: main point both top and also bottom by 10 because that every number ~ the decimal point:
As we have 2 number after the decimal point, we multiply both numerator and also denominator by 100. So,
0.45/1 = (0.45 × 100)/(1 × 100) = 45/100.
Step 3: simplify (or reduce) the fraction:
45/100 = 9/20 when lessened to the simplest form.
## Equivalent fraction for 1.3 percent
1.3 = 13/10 = 13/10 together a fraction
### Step by action Solution
To convert the decimal 1.3 to a fraction follow these steps:
Step 1: compose down the number as a fraction of one:
1.3 = 1.3/1
Step 2: main point both top and also bottom by 10 for every number after ~ the decimal point:
As we have actually 1 numbers after the decimal point, us multiply both numerator and also denominator by 10. So,
1.3/1 = (1.3 × 10)/(1 × 10) = 13/10.
(This fraction is alread reduced, us can"t minimize it any further).
As the numerator is higher than the denominator, we have an not correct fraction, so we can also express it together a mixed NUMBER, therefore 13/10 is additionally equal to 1 3/10 when expressed as a blended number.
## Conversion table: portion to decimal inches and also millimeter equivalence
To transform fractions to decimals and also millimeters and also vice-versa use this formula: 1 inch = 25.4 mm exactly, therefore ...To transform from customs to millimeter main point inch value by 25.4.To convert from millimeter inch divide millimeter worth by 25.4.
an easier means to perform it is to use the table below. How?
### Example 1
Convert 1 1/32" come mm: discover 1 1/32 and read to the best under mm column! you will discover 26.1938.
### Example 2
Convert 0.875 decimal inches come inches (fraction form).Look under the decimal column until you discover 0.875, then check out to the left to find 7/8 inchesor move to the right shaft to find the mm value!
fractioninchesmm
1/640.01560.3969
1/320.03130.7938
3/640.04691.1906
1/160.06251.5875
5/640.07811.9844
3/320.09382.3813
7/640.10942.7781
1/80.12503.1750
9/640.14063.5719
5/320.15633.9688
11/640.17194.3656
3/160.18754.7625
13/640.20315.1594
7/320.21885.5563
15/640.23445.9531
1/40.25006.3500
17/640.26566.7469
9/320.28137.1438
19/640.29697.5406
5/160.31257.9375
21/640.32818.3344
11/320.34388.7313
23/640.35949.1281
3/80.37509.5250
25/640.39069.9219
13/320.406310.3188
27/640.421910.7156
7/160.437511.1125
29/640.453111.5094
15/320.468811.9063
31/640.484412.3031
1/20.500012.7000
33/640.515613.0969
17/320.531313.4938
35/640.546913.8906
9/160.562514.2875
37/640.578114.6844
19/320.593815.0813
39/640.609415.4781
5/80.625015.8750
41/640.640616.2719
21/320.656316.6688
43/640.671917.0656
11/160.687517.4625
45/640.703117.8594
23/320.718818.2563
47/640.734418.6531
3/40.750019.0500
49/640.765619.4469
25/320.781319.8438
51/640.796920.2406
13/160.812520.6375
53/640.828121.0344
27/320.843821.4313
55/640.859421.8281
7/80.875022.2250
57/640.890622.6219
29/320.906323.0188
59/640.921923.4156
15/160.937523.8125
61/640.953124.2094
31/320.968824.6063
63/640.984425.0031
11.000025.4000
fractioninchesmm
1 1/641.015625.7969
1 1/321.031326.1938
1 3/641.046926.5906
1 1/161.062526.9875
1 5/641.078127.3844
1 3/321.093827.7813
1 7/641.109428.1781
1 1/81.125028.5750
1 9/641.140628.9719
1 5/321.156329.3688
1 11/641.171929.7656
1 3/161.187530.1625
1 13/641.203130.5594
1 7/321.218830.9563
1 15/641.234431.3531
1 1/41.250031.7500
1 17/641.265632.1469
1 9/321.281332.5438
1 19/641.296932.9406
1 5/161.312533.3375
1 21/641.328133.7344
1 11/321.343834.1313
1 23/641.359434.5281
1 3/81.375034.9250
1 25/641.390635.3219
1 13/321.406335.7188
1 27/641.421936.1156
1 7/161.437536.5125
1 29/641.453136.9094
1 15/321.468837.3063
1 31/641.484437.7031
1 1/21.500038.1000
1 33/641.515638.4969
1 17/321.531338.8938
1 35/641.546939.2906
1 9/161.562539.6875
1 37/641.578140.0844
1 19/321.593840.4813
1 39/641.609440.8781
1 5/81.625041.2750
1 41/641.640641.6719
1 21/321.656342.0688
1 43/641.671942.4656
1 11/161.687542.8625
1 45/641.703143.2594
1 23/321.718843.6563
1 47/641.734444.0531
1 3/41.750044.4500
1 49/641.765644.8469
1 25/321.781345.2438
1 51/641.796945.6406
1 13/161.812546.0375
1 53/641.828146.4344
1 27/321.843846.8313
1 55/641.859447.2281
1 7/81.875047.6250
1 57/641.890648.0219
1 29/321.906348.4188
1 59/641.921948.8156
1 15/161.937549.2125
1 61/641.953149.6094
1 31/321.968850.0063
1 63/641.984450.4031
22.000050.8000
fractioninchesmm
2 1/642.015651.1969
2 1/322.031351.5938
2 3/642.046951.9906
2 1/162.062552.3875
2 5/642.078152.7844
2 3/322.093853.1813
2 7/642.109453.5781
2 1/82.125053.9750
2 9/642.140654.3719
2 5/322.156354.7688
2 11/642.171955.1656
2 3/162.187555.5625
2 13/642.203155.9594
2 7/322.218856.3563
2 15/642.234456.7531
2 1/42.250057.1500
2 17/642.265657.5469
2 9/322.281357.9438
2 19/642.296958.3406
2 5/162.312558.7375
2 21/642.328159.1344
2 11/322.343859.5313
2 23/642.359459.9281
2 3/82.375060.3250
2 25/642.390660.7219
2 13/322.406361.1188
2 27/642.421961.5156
2 7/162.437561.9125
2 29/642.453162.3094
2 15/322.468862.7063
2 31/642.484463.1031
2 1/22.500063.5000
2 33/642.515663.8969
2 17/322.531364.2938
2 35/642.546964.6906
2 9/162.562565.0875
2 37/642.578165.4844
2 19/322.593865.8813
2 39/642.609466.2781
2 5/82.625066.6750
2 41/642.640667.0719
2 21/322.656367.4688
2 43/642.671967.8656
2 11/162.687568.2625
2 45/642.703168.6594
2 23/322.718869.0563
2 47/642.734469.4531
2 3/42.750069.8500
2 49/642.765670.2469
2 25/322.781370.6438
2 51/642.796971.0406
2 13/162.812571.4375
2 53/642.828171.8344
2 27/322.843872.2313
2 55/642.859472.6281
2 7/82.875073.0250
2 57/642.890673.4219
2 29/322.906373.8188
2 59/642.921974.2156
2 15/162.937574.6125
2 61/642.953175.0094
2 31/322.968875.4063
2 63/642.984475.8031
33.000076.2000
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# Looking For Pythagoras: Homework Examples From ACE
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Looking for Pythagoras: Homework Examples from ACEInvestigation 1: Coordinate Grids, ACE #20, #37Investigation 2: Squaring Off, ACE #16, #44, #65Investigation 3: The Pythagorean Theorem, ACE #2, #9, #17Investigation 4: Using the Pythagorean Theorem: Understanding Real Numbers, ACE #6, #34Investigation 5: Using the Pythagorean Theorem: Analyzing Triangles and Circles, ACE #7Investigation 1: Coordinate GridsACE #2020. Find the area of each triangle. If necessary, copy the triangles onto dot paper.Students know that the area of a triangle can be found by using the formulaArea ½ ( base height ). (See Covering and Surrounding.) For this problem, thechallenge is that the base length or the height are not immediately clear. When studentsknow the Pythagorean Theorem, they will use it to find these lengths, but for now they needa different strategy.20. Students commonly use 2 different strategies to find areas: they subdivide the area intoshapes for which they know the area; or they surround the shape by a rectangle andsubtract areas from the rectangle.Subdividing the area as shown below may not be very helpful if the areas of the shadedtriangular shapers are not easy to find.Surrounding the triangle with a 4-by-3 rectangle (see below) and then subtracting theshaded areas (using the formulas for the area of a triangle) gives the followingcalculation:(Area of the un-shaded triangle) (area of 4-by-3 rectangle) – (A1 A2 A3) 12 – (2 2 3) 5 square units.
Investigation 1: Coordinate GridsACE #3737. Marcia finds the area of a figure on dot paper by dividing it into smaller shapes. Shefinds the area of each smaller shape and writes the sum of the areas as ½ (3) ½ ½ 1.a. What is the total area of the figure?b. On dot paper, draw a figure Marcia might have been looking at.37.a. ½ (3) ½ ½ 1 3.5 square unitsb. This problem makes students attend to the format of the expression. This developsstudent symbol sense.There seem to be 4 areas summed together in the expression ½ (3) ½ ½ 1 inwhich each term of the expression represents a geometric figure. The term “½ (3)”implies a triangle with the area ½ (base height ) where the base could be 3 and theheight could be 1 (see below). The terms “½ ” each imply a triangle with the area ½(base height ) where the base could be 1 and the height could be 1 (see below).The term “1” could be a square with side 1.Putting all these clues together, one of many possible figures could be:
Investigation 2: Squaring OffACE #16Tell whether each statement is true.16. 11 (101)Students learn in this Investigation that to find the area of a square they must multiply thelength of a side by itself, AND, to find the length of the side of a square from its area, theymust find the square root of the area.16. This question asks: if a square has area 101 square units then is its side 11 units long?112 121, so if a square had side length 11 units then its area is 121 square units.Therefore, 11 (101) is not true.Since 102 100, (100) 10; and since 112 121, (121) 11, then (101) must liebetween 10 and 11.
Investigation 2: Squaring OffACE #4444. Multiple Choice.Which line segment has a length of 17 units?44. We need to check each figure to find which segment is the side of a square with area 17square units.Figure F shows a slanting line segment, which can be thought of as the side of a square.If we build the rest of the square we see it has the area 20 square units.A1 A2 A3 A4 A5 4 4 4 4 4 20 square units.Therefore, the line segment is 20 units long.
Investigation 2: Squaring OffACE #6565.a. Which of the triangles below are right triangles?b. Find the area of each right triangle.At this stage students cannot use the Pythagorean theorem to show whether these trianglesare right angled or not. But they can use the strategy they have been using to draw a squareto decide if two sides of the triangle are perpendicular. In Investigation 1 of this unit, in orderto create a square on a slanting line segment, students discussed the use of slope (SeeMoving Straight Ahead). If the given line segment has slopeadjacent side of the square has slope–ba/b, then the slope of the/a. (Students may not have formalized this yet.)65.2a. For figure U we can see that the slopes of the bolded line segments are /1 andTherefore, these two line segments are perpendicular. This is a right triangle.-1/2.b. Area of figure U can be found by subdividing or surrounding (see ACE 20Investigation 1). Or it can be found by noticing that Figure U is half of a square witharea 5 square units (not drawn here). So area of figure U 2.5 square units.The other triangles can be investigated the same way.
Investigation 3: The Pythagorean TheoremACE #22. The triangle below is a right triangle. Show that this triangle satisfies the PythagoreanTheorem.Students might approach this problem either by showing that the triangle is right angled,in which case the Pythagorean Theorem applies. Or, they might find the areas of thesquares on the sides, and check that these fit the Pythagorean relationship. If they takethe first approach they must have a way to show that two sides of the triangle areperpendicular. The relationship between the slopes of perpendicular lines has not beenformally stated in any unit thus far (will be formalized in Shapes of Algebra), but someclasses may have drawn a conclusion about this relationship in this unit.The second approach is illustrated below.The area A3 can be found by subdividing the square.A3 1.5 1.5 1.5 1.5 4 10 square units.Likewise, A2 5, and A1 5 square units. So, A1 A2 A3.
Investigation 3: The Pythagorean TheoremACE #99. Find the flying distance in blocks between the two landmarks, Greenhouse and Stadium,without using a ruler.The segment joining the two landmarks can be thought of as the hypotenuse of a righttriangle, with legs of lengths 4 and 3 units.222GS 3 4 25. Therefore, GS 5 units. The distance between the greenhouse andthe stadium is 5 blocks.
Investigation 3: The Pythagorean TheoremACE #1717. The prism has a base that is a right trianglea. What is the length a?b. Do you need to know the length a to find the volume of the prism? Do you need toknow it to find the surface area? Explain.c. What is the volume?d. What is the surface area?17.222a. Applying the Pythagorean Theorem 2.5 6 a .2So, 42.25 a . So, a 6.5 cm.b. To find the volume you need to know the area of the base of the prism and the heightof the prism. The area of the base is 0.5 (2.5 x 6) 7.5 square units. You don’t needto know a to find this base area. To find the surface area you need to know the areasof all faces. The triangular base areas can be found as above. But the area of one ofthe rectangular faces is 4 x a square units. So we do need to know a to find surfacearea.3c. 30 cm ; 0.5 ( 6 2.5 ) 4 302d. 75 cm ; (2.5 4) 2 [ 0.5 (6 2.5) ] (6 4) (6.5 4) 10 15 24 26 75e. The net should show 3 rectangular faces and 2 triangular faces.
Investigation 4: Using the Pythagorean Theorem: Understanding Real NumbersACE #66. Write each fraction as a decimal. Tell whether the decimal is terminating or repeating. Ifthe decimal is repeating, tell which digits repeat.4996. In an earlier unit, Let’s Be Rational, students learned to think of a fraction in differentways; for example, a fraction might be thought of as parts out of a whole, or as a ratio, oras a division. The last interpretation helps to connect decimals to fractions.04040499 4.0000003964003964003964/99 0.040404.The decimal is repeating, and the digits that repeat are “04”.Note: Every fraction can be written as a decimal by dividing as above. Since there can bea finite number of choices for the remainders created by such a division we eventuallycome to a situation where the remainder is zero, in which case the decimal terminates, ora previous remainder repeats, in which case the decimal answer repeats. For example,when dividing by 99 we could theoretically have any remainder from 0 to 98. After allremainders have been used once one of them must repeat. In fact the only remaindercreated by the above division is 4, and so the division process repeats very quickly.
Investigation 4: Using the Pythagorean Theorem: Understanding Real NumbersACE #34Estimate the square root to one decimal place without using the key on your calculator.Then, tell whether the number is rational or irrational34. 1534. We can use the perfect squares that we know to find an estimate for 15. We knowthe perfect squares of 9 and 16, and that 9 15 16. So, 9 15 16. 16 4. 9 3.So, 3 15 4.We can see that 15 is closer to 16 than to 9. Therefore, we might try 3.9 as a firstapproximation.23.9 15.21.23.8 14.44.So, 3.9 appears to be a better approximation than 3.8.Since there is not exact decimal answer for 15 it is an irrational number (that is, thedecimal answer neither terminates nor repeats).
Investigation 5: Using the Pythagorean Theorem: Analyzing Triangles and CirclesACE #77. Find the perimeter of triangle KLM.7. Solution.In this Investigation students applied the Pythagorean theorem to a particular triangle,with angles 30, 60 and 90 degrees. By observing that this triangle is half of an equilateraltriangle they were able to conclude that the shortest side is always half of thehypotenuse; and by applying the Pythagorean Theorem they were able to conclude thatthe longer side is always 3 times the shortest side. These relationships apply to any 3060-90 triangle, because all such triangles are similar, or scale copies of each other. Ageneral 30-60-90 triangle is pictured below:Applying this to triangle KLN we have:(Continued on next page)
(Continued from prior page)Now look at triangle MLN. It is also a 30-60-90 triangle, and we know the shortest side is 3(3) units, or approximately 5.2MWe can deduce the length of hypotenuse LM and longer leg MN by using the length ofthe shortest side LN.(Not completed here)
Looking for Pythagoras: Homework Examples from ACE Investigation 1: Coordinate Grids, ACE #20, #37 Investigation 2: Squaring Off, ACE #16, #44, #65 Investigation 3: The Pythagorean Theorem, ACE #2, #9, #17 Investigation 4: Using the Pythagorean Theorem: Understanding Real Numbers, ACE #6, #34 Investigatio
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# How to Subtract Fractions in 5 Easy Steps
Updated on April 6, 2016
The subtraction of fractions is not as difficult as it seems. The key to subtracting fractions is to make them compatible. For example before you can begin to subtract 1/3 from 1/2 you have to convert the 2 fractions so that they have the same denominator (if you don't know what a denominator is, check the definitions, below).
Sometimes a common denominator just jumps out at you, but if not just multiplying the two denominators will always result in a common denominator. Once you have mastered the technique for finding a common denominator and converting fractions so they are compatible, you will be able tackle the subtraction (and addition - which uses the same technique) of fractions with confidence.
## Subtracting Fractions: Definitions
There are a few terms that are regularly used and you need to be familiar with them:
## Improper number = 9/4 - this is where the numerator (the top number, see below) is greater than the denominator (the bottom number, see below).
To define numerator and denominator, a picture paints a thousand words:-
## Example 2¼ - 3/7
Here are the 5 easy steps to the subtraction of fractions illustrated by an example:-
## Step 1 Convert Any Mixed Numbers to Improper Fractions
In this example 2¼ is a mixed number. To convert a mixed number into an improper fraction:-
1) Multiply the denominator of the fraction (in this case 4) by the whole number:-
2 x 4 = 8
2) Add the answer from 1) to the numerator of the fraction (in this case 1):-
8 + 1 = 9
3) Place the answer from 2) over the denominator of the fraction. Therefore 2¼ expressed as an improper fraction = 9/4
2¼ = 9/4
## Step 2: Find A Common Denominator
Remember our example is 2¼ - 3/7 = 9/4 - 3/7.
With simple examples a common denominator may just jump at you. But we need a straightforward method that will work with all fractions. All we have to do is multiply the 2 denominators
So 9/4 - 3/7 Common denominator = 4 x 7 = 28
## Step 3:
Use same amount you multiplied the denominators to multiply the numerators:-
### 63/28 - 12/28
You need to understand this step, which, is the most challenging step. This is best achieved by just practice. The illustration here will help you understand the logic.
## Subtract One Numerator From The Other
Now the fractions are compatible because the denominators are the same, we can simply subtract one numerator from the other:
First a reminder of the first 3 steps:-
1) Convert mixed numbers to improper fractions:
2¼ - 3/7 = 9/4 - 3/7
2) Find a common denominator:
9/4 - 3/7 Common denominator = 4 x 7 = 28
3) Convert the numerators
9/4 - 3/7 = 63/28 - 12/28
## Step 5: Simplify and Convert Improper Fractions to Mixed Numbers
This final step is just tidying up your answer. Although the answer from step 4 is technically correct, you may need to tidy up your final answer in order to satisfy your teacher, or much more importantly, your exam board!
## Summary
This hub defined the key terms;- improper fraction, mixed number, denominator and numerator. It then explained the five steps to subtract one fraction from another fraction:-
1. Convert any mixed numbers to improper fractions.
2. Find the common denominators.
3. Convert the numerators by the same factors you converted the denominators
4. Subtract one numerator from the other
5. Simplify and convert any improper fraction answer to a mixed number
## Did You Remember the Key Definitions?
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# What is the integral of int tan^4x dx?
##### 1 Answer
Mar 28, 2016
$\frac{{\tan}^{3} x}{3} - \tan x + x + C$
#### Explanation:
Solving trig antiderivatives usually involves breaking the integral down to apply Pythagorean Identities, and them using a $u$-substitution. That's exactly what we'll do here.
Begin by rewriting $\int {\tan}^{4} x \mathrm{dx}$ as $\int {\tan}^{2} x {\tan}^{2} x \mathrm{dx}$. Now we can apply the Pythagorean Identity ${\tan}^{2} x + 1 = {\sec}^{2} x$, or ${\tan}^{2} x = {\sec}^{2} x - 1$:
$\int {\tan}^{2} x {\tan}^{2} x \mathrm{dx} = \int \left({\sec}^{2} x - 1\right) {\tan}^{2} x \mathrm{dx}$
Distributing the ${\tan}^{2} x$:
$\textcolor{w h i t e}{X X} = \int {\sec}^{2} x {\tan}^{2} x - {\tan}^{2} x \mathrm{dx}$
Applying the sum rule:
$\textcolor{w h i t e}{X X} = \int {\sec}^{2} x {\tan}^{2} x \mathrm{dx} - \int {\tan}^{2} x \mathrm{dx}$
We'll evaluate these integrals one by one.
First Integral
This one is solved using a $u$-substitution:
Let $u = \tan x$
$\frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} x$
$\mathrm{du} = {\sec}^{2} x \mathrm{dx}$
Applying the substitution,
$\textcolor{w h i t e}{X X} \int {\sec}^{2} x {\tan}^{2} x \mathrm{dx} = \int {u}^{2} \mathrm{du}$
$\textcolor{w h i t e}{X X} = {u}^{3} / 3 + C$
Because $u = \tan x$,
$\int {\sec}^{2} x {\tan}^{2} x \mathrm{dx} = \frac{{\tan}^{3} x}{3} + C$
Second Integral
Since we don't really know what $\int {\tan}^{2} x \mathrm{dx}$ is by just looking at it, try applying the ${\tan}^{2} = {\sec}^{2} x - 1$ identity again:
$\int {\tan}^{2} x \mathrm{dx} = \int \left({\sec}^{2} x - 1\right) \mathrm{dx}$
Using the sum rule, the integral boils down to:
$\int {\sec}^{2} x \mathrm{dx} - \int 1 \mathrm{dx}$
The first of these, $\int {\sec}^{2} x \mathrm{dx}$, is just $\tan x + C$. The second one, the so-called "perfect integral", is simply $x + C$. Putting it all together, we can say:
$\int {\tan}^{2} x \mathrm{dx} = \tan x + C - x + C$
And because $C + C$ is just another arbitrary constant, we can combine it into a general constant $C$:
$\int {\tan}^{2} x \mathrm{dx} = \tan x - x + C$
Combining the two results, we have:
$\int {\tan}^{4} x \mathrm{dx} = \int {\sec}^{2} x {\tan}^{2} x \mathrm{dx} - \int {\tan}^{2} x \mathrm{dx} = \left(\frac{{\tan}^{3} x}{3} + C\right) - \left(\tan x - x + C\right) = \frac{{\tan}^{3} x}{3} - \tan x + x + C$
Again, because $C + C$ is a constant, we can join them into one $C$.
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# 5.7 Inverses and radical functions
Page 1 / 7
In this section, you will:
• Find the inverse of an invertible polynomial function.
• Restrict the domain to find the inverse of a polynomial function.
A mound of gravel is in the shape of a cone with the height equal to twice the radius.
The volume is found using a formula from elementary geometry.
$\begin{array}{ccc}V& \hfill =& \frac{1}{3}\pi {r}^{2}h\hfill \\ & =& \frac{1}{3}\pi {r}^{2}\left(2r\right)\hfill \\ & =& \frac{2}{3}\pi {r}^{3}\hfill \end{array}$
We have written the volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of the radius $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula
$r=\sqrt[3]{\frac{3V}{2\pi }}$
This function is the inverse of the formula for $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}r.$
In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.
## Finding the inverse of a polynomial function
Two functions $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are inverse functions if for every coordinate pair in $\text{\hspace{0.17em}}f,\left(a,b\right),\text{\hspace{0.17em}}$ there exists a corresponding coordinate pair in the inverse function, $\text{\hspace{0.17em}}g,\left(b,\text{\hspace{0.17em}}a\right).\text{\hspace{0.17em}}$ In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.
For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in [link] . We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.
Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ measured horizontally and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ measured vertically, with the origin at the vertex of the parabola. See [link] .
From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form $\text{\hspace{0.17em}}y\left(x\right)=a{x}^{2}.\text{\hspace{0.17em}}$ Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor $\text{\hspace{0.17em}}a.$
$\begin{array}{ccc}\hfill 18& =& a{6}^{2}\hfill \\ \hfill a& =& \frac{18}{36}\hfill \\ & =& \frac{1}{2}\hfill \end{array}$
Our parabolic cross section has the equation
$y\left(x\right)=\frac{1}{2}{x}^{2}$
We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ the width will be given by $\text{\hspace{0.17em}}2x,\text{\hspace{0.17em}}$ so we need to solve the equation above for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.
To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values. On this domain, we can find an inverse by solving for the input variable:
$\begin{array}{ccc}\hfill y& =& \frac{1}{2}{x}^{2}\hfill \\ \hfill 2y& =& {x}^{2}\hfill \\ \hfill x& =& ±\sqrt{2y}\hfill \end{array}$
This is not a function as written. We are limiting ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values, so we eliminate the negative solution, giving us the inverse function we’re looking for.
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
|
Talia wants to write the equation of the graphed line in point-slope form. These are the steps she plans to use: Step 1: Choose a point on the line, such as (2, 5). Step 2: Choose another point on the line, such as (1, 3). Step 3: Count units to determine the slope ratio. The line runs 1 unit to the right and rises 2 units up, so the slope is . Step 4: Substitute those values into the point-slope form. y – y1 = m(x – x1) y – 3 = (x – 1) Which of Talia’s steps is incorrect? Step 1 is incorrect because it uses a point that is not on the line. Step 2 is incorrect because it uses a point that is not on the line. Step 3 is incorrect because it shows an incorrect ratio for the slope. Step 4 is incorrect because it shows an incorrect substitution of (1, 3) into the point-slope form. - AssignmentGrade.com
# Talia wants to write the equation of the graphed line in point-slope form. These are the steps she plans to use: Step 1: Choose a point on the line, such as (2, 5). Step 2: Choose another point on the line, such as (1, 3). Step 3: Count units to determine the slope ratio. The line runs 1 unit to the right and rises 2 units up, so the slope is . Step 4: Substitute those values into the point-slope form. y – y1 = m(x – x1) y – 3 = (x – 1) Which of Talia’s steps is incorrect? Step 1 is incorrect because it uses a point that is not on the line. Step 2 is incorrect because it uses a point that is not on the line. Step 3 is incorrect because it shows an incorrect ratio for the slope. Step 4 is incorrect because it shows an incorrect substitution of (1, 3) into the point-slope form.
QUESTION POSTED AT 01/06/2020 - 03:45 PM
Steps 3 and 4 are incorrect because
(a) An incorrect value for slope was obtained in step 3.
(b) The incorrect value for the slope was used in step 4.
Explanation:
Let us evaluate Talia's steps.
Step 1: Select (2, 5) as a point on the line.
CORRECT
Step 2: Select another point (1, 3) on the line.
CORRECT
Step 3: Count units to the right and count units up, to determine the slope.
Units right = 2 - 1 = 1
Units up = 5 - 3 = 2
Slope = (Units up)/(Units right) = 2/1 = 2
INCORRECT because Talia did not obtain a slope of 2.
Step 4: Substitute obtained values in the point-slope form.
(y - y1) = m(x - x1). If we select the point (1, 3), then
y - 2 = 2(x - 1).
Talia's equation is INCORRECT because she used an incorrect value
for the slope.
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# If the letters of the word ASSASSINATION
Question:
If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
(a) Four S’s come consecutively in the word
(b) Two I’s and two N’s come together
(c) All A’s are not coming together
(d) No two A’s are coming together.
Solution:
Given word is ASSASSINATION
Total number of letters in ASSASSINATION is 13
In word ASSASSINATION, there are 3A’s, 4S’s, 2I’s, 2N’s, 1T’s and 1O’s
Total number of ways these letters can be arranged =
$\mathrm{n}(\mathrm{S})=\frac{13 !}{3 ! 4 ! 2 ! 2 !}$
(a) Four S's come consecutively in the word
If $4 S^{\prime}$ s come consecutively then word ASSASSINATION become.
$\therefore \mathrm{n}(\mathrm{E})=\frac{10 !}{3 ! 2 ! 2 !}$
$\therefore$ Required Probability $=\frac{\frac{10 !}{3 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
$=\frac{10 !}{3 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !}$
The above equation can be written as
$=\frac{10 ! \times 4 !}{13 \times 12 \times 11 \times 10 !}$
$=\frac{4 \times 3 \times 2 \times 1}{13 \times 12 \times 11}$
On simplifying we get
$=\frac{2}{143}$
(b) Two I's and two N's come together So, now numbers of letters is $1+9=10$
$\therefore \mathrm{n}(\mathrm{E})=\frac{4 !}{2 ! 2 !} \times \frac{10 !}{3 ! 4 !}$
$\therefore$ Required Probability $=\frac{\frac{4 ! 10 !}{3 ! 4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
The above equation can be written as
$=\frac{10 ! 4 !}{3 ! 4 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !}$
$=\frac{10 ! \times 4 !}{13 \times 12 \times 11 \times 10 !}$
$=\frac{4 \times 3 \times 2 \times 1}{13 \times 12 \times 11}$
On sımplityıng we get
$=\frac{2}{143}$
On simplifying we get
$=\frac{2}{143}$
(c) All A's are not coming together
Firstly, we find the probability that all A's are coming together If all A's are coming together then
$\therefore$ Number of words when all A'scome together $=\frac{11 !}{4 ! 2 ! 2 !}$
$\therefore$ Probability when all A's come together $=\frac{\frac{11 !}{4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
$=\frac{11 !}{4 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !}$
The above equation can be written as
$=\frac{11 ! \times 3 !}{13 \times 12 \times 11 !}$
$=\frac{3 \times 2 \times 1}{13 \times 12}$
On simplifying we get
$=\frac{1}{26}$
Now, $\mathrm{P}$ (all $\mathrm{A}^{\prime}$ s does not come together) $=1-\mathrm{P}$ (all $\mathrm{A}^{\prime}$ s come together)
$=1-\frac{1}{26}$
$=\frac{25}{26}$
(d) No two A's are coming together First we arrange the alphabets except A's
$\therefore$ Number ofways of arranging all alphabets except $\mathrm{A}^{\prime} \mathrm{s}=\frac{10 !}{4 ! 2 ! 2 !}$
There are 11 vacant places between these alphabets. Total A's in the word ASSASSINATION are 3
$\therefore 3$ A's can be placed in 11 place in ${ }^{11} C_{3}$ ways
$=\frac{11 !}{3 !(11-3) !}$
$=\frac{11 !}{3 ! 8 !}$
$\therefore$ Total number of words when no two A's together
$=\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !}$
$\therefore$ Required Probability $=\frac{\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
The above equation can be written as
$=\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !}$
$=\frac{11 ! \times 10 \times 9 \times 8 !}{8 ! \times 13 \times 12 \times 11 !}$
$=\frac{10 \times 9}{13 \times 12}$
On simplifying we get
$=\frac{15}{26}$
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How do you differentiate g(x) = (2x^2 + 4x - 3) ( 5x^3 + 2x + 2) using the product rule?
Mar 14, 2018
$\left(4 x + 4\right) \left(5 {x}^{3} + 2 x + 2\right) + \left(15 {x}^{2} + 2\right) \left(2 {x}^{2} + 4 x - 3\right)$
$50 {x}^{4} + 80 {x}^{3} - 33 {x}^{2} + 24 x + 2$
Explanation:
First the product rule is, $g \left(x\right) = f p r i m e \left(x\right) h \left(x\right) + h p r i m e \left(x\right) f \left(x\right)$
Where $f \left(x\right) = 2 {x}^{2} + 4 x - 3$
And $h \left(x\right) = 5 {x}^{3} + 2 x + 2$
Now take the derivative of both, this gives you...
$f p r i m e \left(x\right) = \left(4 x + 4\right)$
$h p r i m e \left(x\right) = \left(15 {x}^{2} + 2\right)$
So now plug into the product rule formula
$\left(4 x + 4\right) \left(5 {x}^{3} + 2 x + 2\right) + \left(15 {x}^{2} + 2\right) \left(2 {x}^{2} + 4 x - 3\right)$
After multiplying and adding like terms you get
$50 {x}^{4} + 80 {x}^{3} - 33 {x}^{2} + 24 x + 2$
|
# Solving Systems of Equations with Matrices
## Matrix -> Systems of Equations -> Solving Systems of Equations with Matrices
When working with systems of equations, matrices provide a powerful tool for solving them efficiently. In this tutorial, we will delve into the relationship between matrices and systems of equations, and learn how to solve systems of equations using matrices.
### Understanding Matrices
Let's start by understanding what matrices are. A matrix is a rectangular array of numbers, arranged in rows and columns. Each number in the matrix is called an element. For example, consider the following matrix:
``````[ 1 2 3 ]
[ 4 5 6 ]
[ 7 8 9 ]
``````
This is a 3x3 matrix, meaning it has 3 rows and 3 columns. Matrices can have any number of rows and columns, and they are often denoted by uppercase letters. In this tutorial, we will primarily work with square matrices, where the number of rows is equal to the number of columns.
### Systems of Equations
A system of equations is a set of equations that are to be solved simultaneously. Each equation in the system represents a relationship between variables. For example, consider the following system of equations:
``````2x + 3y = 8
4x - 2y = 2
``````
In this system, we have two equations with two variables, x and y. The goal is to find the values of x and y that satisfy both equations simultaneously.
### Matrix Representation of Systems of Equations
To solve systems of equations using matrices, we can represent the system in matrix form. Let's take the previous system of equations as an example:
``````[ 2 3 ] [ x ] [ 8 ]
[ 4 -2 ] * [ y ] = [ 2 ]
``````
In this representation, the coefficients of the variables are arranged in a matrix called the coefficient matrix. The variables are arranged in a column matrix, and the constants on the right-hand side of the equations are arranged in another column matrix. Multiplying the coefficient matrix by the column matrix of variables gives us the column matrix of constants.
### Solving Systems of Equations with Matrices
To solve the system of equations, we can use matrix operations. The key idea is to transform the coefficient matrix into an identity matrix, which will allow us to determine the values of the variables.
Let's continue with our example:
``````[ 2 3 ] [ x ] [ 8 ]
[ 4 -2 ] * [ y ] = [ 2 ]
``````
To transform the coefficient matrix into an identity matrix, we can perform row operations. The goal is to eliminate the coefficients below and above the main diagonal, starting from the top-left corner.
First, let's eliminate the coefficient below the main diagonal in the first column. We can achieve this by multiplying the first row by -2 and adding it to the second row:
``````[ 2 3 ] [ x ] [ 8 ]
[ 0 -8 ] * [ y ] = [-14]
``````
Next, let's eliminate the coefficient above the main diagonal in the second column. We can achieve this by multiplying the second row by 3 and adding it to the first row:
``````[ 2 0 ] [ x ] [ 2 ]
[ 0 -8 ] * [ y ] = [-14]
``````
Now, we have an upper triangular matrix. To transform it into an identity matrix, we can perform row operations again. This time, we will divide the second row by -8:
``````[ 2 0 ] [ x ] [ 2 ]
[ 0 1 ] * [ y ] = [ 7/4 ]
``````
Finally, we have an identity matrix on the left-hand side, which means we have determined the values of x and y. From the last equation, we can see that y is equal to 7/4. Substituting this value into the first equation, we can solve for x:
``````2x + 3(7/4) = 2
2x + 21/4 = 2
2x = 2 - 21/4
2x = 8/4 - 21/4
2x = -13/4
x = -13/8
``````
Therefore, the solution to the system of equations is x = -13/8 and y = 7/4.
### Conclusion
In this tutorial, we explored the relationship between matrices and systems of equations. We learned how to represent systems of equations using matrices and how to solve them using matrix operations. Matrices provide an efficient and systematic approach to solving systems of equations, making them a valuable tool for programmers and mathematicians alike.
Now that you have a solid understanding of solving systems of equations with matrices, you can apply this knowledge to various real-world problems that involve multiple equations and variables. Happy coding!
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b b 2 c b g c e b c 6 Title
Intermediate Algebra Tutorial 34: Complex Fractions
WTAMU > Virtual Math Lab > Intermediate Algebra
Learning Objectives
After completing this tutorial, you should be able to: Simplify complex fractions.
Introduction
Are fractions getting you down? If so, you are not alone. Well, you are in luck. We have one more tutorial devoted to rational expressions (fractions). In this tutorial we will be looking at simplifying complex fractions. If you need a review on simplifying and dividing rational expressions, feel free to go back to Tutorial 32: Multiplying and Dividing Rational Expressions. If you need a review on finding the LCD of fractions feel free to go back to Tutorial 33: Adding and Subtracting Rational Expressions. It is time to get started with this tutorial.
Tutorial
Complex Fraction
A complex fraction is a rational expression that has a fraction in its numerator, denominator or both. In other words, there is at least one small fraction within the overall fraction. Some examples of complex fractions are: and There are two ways that you can simplify complex fractions. We will call them method I and method II.
Method I Simplifying a Complex Fraction
In other words, you will be combining all the parts of the numerator to form one fraction and all of the parts of the denominator to form another fraction. If you need a review on adding and subtracting rational expressions, go to Tutorial 33: Adding and Subtracting Rational Expressions.
Step 2: Divide the numerator by the denominator by multiplying the numerator by the reciprocal of the denominator.
Step 3: If needed, simplify the rational expression.
If you need a review on simplifying rational expressions, go to Tutorial 32: Multiplying and Dividing Rational Expressions.
Example 1: Simplify .
Combining only the numerator we get:
*Rewrite fractions with LCD of 12
Combining only the denominator we get:
*Rewrite fractions with LCD of 8
Putting these back into the complex fraction we get:
*Write numerator over denominator
Step 2: Divide the numerator by the denominator by multiplying the numerator by the reciprocal of the denominator AND Step 3: If needed, simplify the rational expression.
*Rewrite div. as mult. of reciprocal *Divide out a common factor of 4
Example 2: Simplify .
Combining only the numerator we get:
*Rewrite fractions with LCD of ab
The denominator is already written as one fraction:
Putting these back into the complex fraction we get:
*Write numerator over denominator
Step 2: Divide the numerator by the denominator by multiplying the numerator by the reciprocal of the denominator AND Step 3: If needed, simplify the rational expression.
*Rewrite div. as mult. of reciprocal *Divide out common factors of a and b
Method II Simplifying a Complex Fraction
Step 1: Multiply the numerator and denominator of the overall complex fractions by the LCD of the smaller fractions.
Recall that when you multiply the exact same thing to the numerator and the denominator, the result is an equivalent fraction. If you need a review on finding the LCD, go back to Tutorial 33: Adding and Subtracting Rational Expressions.
Step 2: If needed, simplify the rational expression.
Example 3: Simplify .
Step 1: Multiply the numerator and denominator of the overall complex fractions by the LCD of the smaller fractions.
The denominator of the numerator’s fraction has the following two factors:
The denominator of the denominator’s fraction has the following factor:
Putting all the different factors together and using the highest exponent, we get the following LCD for all the small fractions:
Multiplying numerator and denominator by the LCD we get:
*Mult. num. and den. by (x + 5)(x - 5)
Step 2: If needed, simplify the rational expression.
*Divide out the common factor of (2x + 1)
Example 4: Simplify .
Step 1: Multiply the numerator and denominator of the overall complex fractions by the LCD of the smaller fractions.
The denominator of the numerator’s fraction has the following factor:
The denominator of the denominator’s fraction has the following factors:
y and
Putting all the different factors together and using the highest exponent, we get the following LCD for all the small fractions:
Multiplying numerator and denominator by the LCD we get:
*Mult. num. and den. by y squared
Step 2: If needed, simplify the rational expression.
*Num. factors as a difference of two squares *Den. factors as a perfect square trinomial. *Divide out the common factor of (3y + 1)
Example 5: Simplify .
At first glance, this does not look like a complex fraction. However, once you rewrite it with positive exponents you will see that we really do have a complex fraction.
*Rewrite with positive exponents
Step 1: Multiply the numerator and denominator of the overall complex fractions by the LCD of the smaller fractions.
The two denominators of the numerator’s fractions have the following factors:
a and b
The two denominators of the denominator’s fractions have the following factors:
and
Putting all the different factors together and using the highest exponent, we get the following LCD for all the small fractions:
Multiplying numerator and denominator by the LCD we get:
*Mult. num. and den. by a squared b squared
Step 2: If needed, simplify the rational expression.
*Factor out the GCF of ab in the num.
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1b: Simplify.
Need Extra Help on these Topics?
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on July 17, 2011 by Kim Seward.
|
# Solving Percent Problems and Estimating With Percents
### Popular Tutorials in Solving Percent Problems and Estimating With Percents
#### How Do You Solve a Word Problem Using a Percent Proportion?
Word problems allow you to see the real world uses of math! This tutorial shows you how to take a words problem and turn it into a percent proportion. Then see how to solve for the answer using the mean extremes property of proportions. Take a look!
#### How Do You Set Up a Percent Proportion from a Word Problem?
Sometimes the hardest part of a word problem is figuring out how to turn the words into an equation you can solve. This tutorial let's you see the steps to take in order to do just that! Take a look! You'll be glad you did!
#### How Do You Use a Proportion to Find What Percent a Part is of a Whole?
A part is some percent of a whole. Trying to calculate the percent? Use a percent proportion to solve! This tutorial will show you how!
#### How Do You Figure Out a Tip?
If you need to leave a tip at a restaurant, you can quickly estimate the amount in your head! This tutorial shows you how to use estimation and mental math to estimate a tip!
#### How Do You Estimate a Sale Price?
Sales are great, but how much are you really saving? This tutorial shows you how to estimate the sales price of an item.
#### How Do You Estimate a Percent?
When something is on sale, it's good to know how much you're saving! This tutorial shows you how to estimate a percent using an original price and a coupon!
#### How Do You Use Mental Math to Estimate a Percent?
Mental math is a powerful tool! See how to use mental math in this tutorial to estimate a percent!
#### How Do You Use a Proportion to Find a Whole?
Taking a percent of a number? Trying to figure out the result? Use a percent proportion to solve! This tutorial will show you how!
#### How Do You Use an Equation to Find a Whole?
Taking a percent of a number? Trying to figure out the result? Use a percent equation to solve! This tutorial will show you how!
#### How Do You Find Percents Mentally Using Fractions?
Some fractions are seen so often in math that it can be helpful to know the percent that goes with it. This tutorial shows you some common percent-fraction relationships!
#### How Do You Use Compatible Numbers to Estimate a Part of a Whole?
If you're trying to find the percent of a number, it may be helpful to use compatible numbers to find an estimated answer. Follow along with this tutorial to see how to use compatible numbers to estimate the percent of a number!
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# RELATION IN SET THEORY WORKSHEET
## About "Relation in Set Theory Worksheet"
Relation in Set Theory Worksheet :
Here we are going to see some practice questions on relations in set theory.
## Relation in Set Theory Worksheet - Practice questions
(1) Let A = {1, 2, 3, 7} and B = {3, 0, –1, 7}, which of the following are relation from A to B ?
(i) R1 = {(2, 1), (7, 1)}
(ii) R= {(–1,1)}
(iii) R3 = {(2, –1), (7, 7), (1, 3)}
(iv) R4= {(7,–1), (0,3), (3,3), (0,7)} Solution
(2) Let A = {1, 2, 3, 4,..., 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A x A. Also, find the domain and range of R. Solution
(3) A Relation R is given by the set {(x, y) /y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range. Solution
(4) Represent each of the given relations by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.
(i) {(x, y)| x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4} Solution
(ii) {(x, y) | y = x + 3, x, y are natural numbers < 10} Solution
(5) A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4and A5 were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers and E1, E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram. Solution
After having gone through the stuff given above, we hope that the students would have understood, "Relation in Set Theory Worksheet".
Apart from the stuff given in this section "Relation in Set Theory Worksheet"if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
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OTHER TOPICS
Profit and loss shortcuts
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Sum of all three digit numbers divisible by 6
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Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
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Sum of all three four digit numbers formed using 0, 1, 2, 3
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Applications of Trigonometry to Waves
Introduction
4.5
Waves and vibrations occur in many contexts. The water waves on the sea and the vibrations of
a stringed musical instrument are just two everyday examples. If the vibrations are simple ‘to and
fro’ oscillations they are referred to as ‘sinusoidal’ which implies that a knowledge of trigonometry,
particularly of the sine and cosine functions, is a necessary pre-requisite for dealing with their analysis.
In this Section we give a brief introduction to this topic.
Prerequisites
have a knowledge of the basics of trigonometry
be aware of the standard trigonometric identities
Before starting this Section you should
On completion you should be able to
Learning Outcomes
use simple trigonometric functions to describe waves
combine two waves of the same frequency as a single wave in amplitude-phase form
HELM (2005):
Section 4.5: Applications of Trigonometry to Waves
65
1. Applications of trigonometry to waves
Two-dimensional motion
Suppose that a wheel of radius R is rotating anticlockwise as shown in Figure 38.
y
B
Q
R
ωt
A
O
P
x
Figure 38
Assume that the wheel is rotating with an angular velocity ω radians per second about O so that, in
a time t seconds, a point (x, y) initially at position A on the rim of the wheel moves to a position B
such that angle AOB = ωt radians.
Then the coordinates (x, y) of B are given by
x = OP = R cos ωt
y = OQ = PB = R sin ωt
We know that both the standard sine and cosine functions have period 2π. Since the angular velocity
is ω radians per second the wheel will make one complete revolution in
2π
seconds.
ω
The time
2π
ω
(measured in seconds in this case) for one complete revolution is called the period of
rotation of the wheel. The number of complete revolutions per second is thus
1
T
= f say which is
called the frequency of revolution. Clearly f =
1
T =
ω
2π
relates the three quantities
introduced here. The angular velocity ω = 2πf is sometimes called the angular frequency.
One-dimensional motion
The situation we have just outlined is two-dimensional motion. More simply we might consider one-dimensional motion.
An example is the motion of the projection onto the x-axis of a point B which moves with uniform angular velocity ω round a circle of radius R (see Figure 39). As B moves round, its projection P moves to and fro across the diameter of the circle.
66
HELM (2005):
Workbook 4: Trigonometry
The position of P is given by
x = R cos ωt
y
B
R
ωt
A
O
x
P
Figure 39
x
(1)
Clearly, from the known properties of the cosine function, we can deduce the following:
1. x varies periodically with t with period T =
2π
ω
.
2. x will have maximum value +R and minimum value R.
(This quantity R is called the amplitude of the motion.)
Using (1) write down the values of x at the following times:
t = 0, t =
2ω , t = ω , t = 2ω , t = 2π
π
π
3π
ω
.
π π 3π 2π t 0 2ω ω 2ω ω x
π π 3π 2π t 0 2ω ω 2ω ω x R 0 −R 0 R
HELM (2005):
Section 4.5: Applications of Trigonometry to Waves
67
Using (1) this ’to and fro’ or ‘vibrational’ or ‘oscillatory’ motion between R and R continues indefinitely. The technical name for this motion is simple harmonic. To a good approximation it is the motion exhibited (i) by the end of a pendulum pulled through a small angle and then released (ii) by the end of a hanging spring pulled down and then released. See Figure 40 (in these cases damping of the pendulum or spring is ignored).
Figure 40
Using your knowledge of the cosine function and the results of the previous Task
sketch the graph of x against t where
x = R cos ωt
for t = 0 to t =
ω
x = R cos ωt
period
R
t
ω
ω
−R
This graph shows part of a cosine wave, specifically two periods of oscillation. The shape of the
graph suggests that the term wave is indeed an appropriate description.
68
HELM (2005):
Workbook 4: Trigonometry
We know that the shape of the cosine graph and the sine graph are identical but offset by radians horizontally. Bearing this in mind, attempt the following Task.
π
1.7
Write the equation of the wave x(t), part of which is shown in the following graph. You will need to find the period T and angular frequency ω .
x
5
4
8
t (secs)
− 5
Your solution Answer From the shape of the graph we have a sine wave rather than a cosine wave. The amplitude is 5. The period T = 4s so the angular frequency ω = 2π = π . Hence x = 5 sin πt . 4 2 2
The quantity x, a function of t, is referred to as the displacement of the wave.
Phase of a wave
θ π = sin θ which means that the graph of
2
but is shifted to the right by
π
2 .
We recall that cos
as that of
Suppose now that we consider the waves
x = cos θ
x 1 = R cos 2t
x 2 = R sin 2t
x = sin θ
is the same shape
Both have amplitude R, angular frequency ω = 2 rad s 1 . Also
x 2 = R cos 2t π = R cos 2 t π
2
4
The graphs of x 1 against t and of x 2 against t are said to have a phase difference of π . Specifically
x 1 is ahead of, or ‘leads’ x 2 by π radians.
4
4
More generally, consider the following two sine waves of the same amplitude and frequency:
x 1 (t) = R sin ωt
x 2 (t) = R sin(ωt α)
HELM (2005):
Section 4.5: Applications of Trigonometry to Waves
69
Now x 1 t α = R sin ω t α = R sin(ωt α) = x 2 (t)
ω
ω
so it is clear that the waves x 1 and x 2 are out of phase by α . Specifically x 1 leads x 2 by
ω
α
ω .
Calculate the phase difference between the waves
1 3 cos(10πt)
x
x 2
=
=
3 cos 10πt + π
4
where the time t is in seconds.
Note firstly that the waves have the same amplitude 3 and angular frequency 10π (corresponding
to a common period
2π
10π =
1
5 s)
Now cos 10πt + π = cos 10π t +
4
40
1
so x 1 t + 40 = x 2 (t).
1
In other words the phase difference is
1
40 s, the wave x 2 leads the wave x 1 by this amount.
Alternatively we could say that x 1 lags x 2 by
1
40 s.
70
HELM (2005):
Workbook 4: Trigonometry
Key Point 20
The equations
x = R cos ωt
x = R sin ωt
both represent waves of amplitude R and period 2π ω .
π
π
The phase difference between these waves is 2ω because cos ω t − 2ω = sin ωt.
Combining two wave equations
A situation that arises in some applications is the need to combine two trigonometric terms such as
A cos θ + B sin θ
where A and B are constants.
For example this sort of situation might arise if we wish to combine two waves of the same frequency
but not necessarily the same amplitude and with a phase difference. In particular we wish to be able
to deal with an expression of the form
R 1 cos ωt + R 2 sin ωt
where the individual waves have, as we have seen, a phase difference of
π
.
2ω
Consider an expression A cos θ + B sin θ. We seek to transform this into the single form
C cos(θ α) (or C sin(θ α)), where C and α have to be determined. The problem is easily solved
with the aid of trigonometric identities.
We know that
C cos(θ α) C(cos θ cos α + sin θ sin α)
Hence if A cos θ + B sin θ = C cos(θ α) then
A cos θ + B sin θ = (C cos α) cos θ + (C sin α) sin θ
For this to be an identity (true for all values of θ) we must be able to equate the coefficients of cos θ and sin θ on each side. Hence
A = C cos α and B = C sin α (2) HELM (2005): 71
Section 4.5: Applications of Trigonometry to Waves
By squaring and adding the Equations (2), obtain C in terms of A and B.
Your solution Answer A = C cos α and B = C sin α gives A 2 + B 2 = C 2 cos 2 α + C 2 sin 2 2 α = C (cos 2 α + sin 2 α) = C 2 . . . C = √ A 2 + B 2 (We take the positive square root.)
By eliminating C from Equations (2) and using the result of the previous Task,
obtain α in terms of A and B.
Your solution Answer C B A = tan α so α is obtained by solving tan α = = sin α B A . However, care must be By division, taken to obtain the correct quadrant for α. C cos α
Key Point 21
C = A 2 + B 2 and tan α = B A .
Note that the following cases arise for the location of α:
1. A > 0,B> 0 : 1st quadrant 3 A < 0,B< 0 : 3rd quadrant 2. A < 0,B> 0 : 2nd quadrant 4 A > 0,B< 0 : 4th quadrant
If
A cos θ + B sin θ =
C cos(θ α)
then
72
HELM (2005):
Workbook 4: Trigonometry
In terms of waves, using Key Point 21 we have
R 1 cos ωt + R 2 sin ωt = R cos(ωt α)
where R = R 2 + R
1
2
2
and
tan α = R 2 .
R
1
The form R cos(ωt α) is said to be the amplitude/phase form of the wave.
Example 5
Express in the form C cos(θ α) each of the following:
(a) 3 cos θ + 3 sin θ (b) −3 cos θ + 3 sin θ (c) −3 cos θ − 3 sin θ (d) 3 cos θ − 3 sin θ
Solution
In each case C =
√ A 2 2 = √ 9+9= √ 18 + B 3 = 3 B A = 1 gives α = 45
(a)
tan α =
(A and B are both positive so the first quadrant
is the correct one.) Hence 3 cos θ + 2 sin θ =
√ 18 cos(θ − 45 ) = ◦ √ 18 cos θ − π 4
(b)
The angle α must be in the second quadrant as A = 3 < 0, B = +3 > 0. By
calculator : tan α = 1 gives α = 45
that tan α has period π or 180
obtain the correct α value of 135 . Hence
to the calculator value to
but this is in the 4th quadrant. Remembering
we must therefore add 180
3 cos θ + 3 sin θ =
18 cos(θ 135 )
3 = 1 giving
α = 45 by calculator. Hence adding 180 to this tells us that
(d)
3 cos θ 3 sin θ = 18 cos(θ 225 )
Here A = 3 B = 3 so α is in the 4th quadrant. α = 45 so
3 cos θ 3 sin θ = 18 cos(θ + 45 ).
tan α = 1 gives us (correctly)
Note that in the amplitude/phase form the angle may be expressed in degrees or radians.
(c)
Here A = 3,
B = 3 so α must be in the 3rd quadrant.
tan α = 3
HELM (2005):
Section 4.5: Applications of Trigonometry to Waves
73
Write the wave form
the phase in radians to 3 d.p
x = 3 cos ωt +4 sin ωt
in amplitude/phase form. Express
Your solution Answer We have x = R cos(ωt − α) where R = √ 3 2 + 4 2 = 5 and tan α = 4 from which, using the 3 calculator in radian mode, α = 0.927 radians. This is in the first quadrant 0 <α< π 2 which is correct since A = 3 and B = 4 are both positive. Hence x = 5 cos(ωt − 0.927).
74
HELM (2005):
Workbook 4: Trigonometry
Exercises
1. Write down the amplitude and the period of
2.
y = 2 5 sin 2πt.
Write down the amplitude, frequency and phase of
(a)
y = 3 sin 2t π
3
(b) y = 15 cos 5t 3π
2
3.
The current in an a.c. circuit is i(t) = 30 sin 120πt amp where t is measured in seconds.
What is the maximum current and at what times does it occur?
4. The depth y of water at the entrance to a small harbour at time t is
y = a sin b t π +k
2
5.
where k is the average depth. If the tidal period is 12 hours, the depths at high tide and low
tide are 18 metres and 6 metres respectively, obtain a, b, k and sketch two cycles of the graph
of y.
The Fahrenheit temperature at a certain location over 1 complete day is modelled by
6.
7.
8.
9.
F (t) = 60 + 10 sin
π
12
(t 8)
0 t 24
where t is in the time in hours after midnight.
(a)
(b)
(c)
What are the temperatures at 8.00 am and 12.00 noon?
At what time is the temperature 60 F?
Obtain the maximum and minimum temperatures and the times at which they occur.
In each of the following write down expressions for shifted sine and shifted cosine functions
that satisfy the given conditions:
2π
3
(a) Amplitude 3, Period
(b)
, Phase shift π
3
Amplitude 0.7, Period 0.5, Phase shift 4.
Write the a.c. current i = 3 cos 5t + 4 sin 5t in the form i = C cos(5π α).
Show that if A cos ωt + B sin ωt = C sin(ωt + α) then
C
= A 2 + B 2 ,
cos α = B C ,
sin α = C A .
Using Exercise 8 express the following in the amplitude/phase form C sin(ωt + α)
(a) y = 3 sin 2t + cos 2t
(b) y = cos 2t + 3 sin 2t
10. The motion of a weight on a spring is given by
2
1
y = 3 cos 8t 6 sin 8t.
11.
Obtain C and α such that
Show that for the two a.c. currents
y = C sin(8t + α)
i 1 = sin ωt + π
3
and
i 2 = 3 cos ωt π then
6
HELM (2005):
Section 4.5: Applications of Trigonometry to Waves
i 1 + i 2 = 4 cos ωt π .
6
75
2
12. Show that the power P = v R in an electrical circuit where
P
2
V
0
= 2R (1 sin 2ωt)
13. Show that the product of the two signals
v = V 0 cos ωt + π
4
is
f 1 (t) = A 1 sin ωt
f 2 (t) = A 2 sin {ω(t + τ ) + φ}
is given by
f 1 (t)f 2 (t) =
A 1 A 2
2
{cos(ωτ + φ) cos(2ωt + ωτ + φ)}.
5
2
5
2
2π
2π
Check: y(t + 1) =
5
2
5
2
5
2
2π π (a) 2 = π. Writing y = 3 sin 2 t − 0
2. Amplitude 3, Period
phase shift of
π
6
radians in this wave compared with y = 3 sin 2t.
we see that there is a
2π 3π (b) Amplitude 15, Period . Clearly y = 15 cos 5 t − 5 10
3π
10
compared with y = 15 cos 5t.
so there is a phase shift of
3. Maximum current = 30 amps at a time t such that 120πt =
π
2
. i.e. t =
1
240
s.
1
240
+
n
60
2π
This maximum will occur again at
s,
n = 1, 2, 3,
4. y = a sin b t − π + h. The period is = 12 hr . . b = . π 2 b 6
Also since y
m, a
y
6 m. i.e. y = 6 sin π t π + 12.
= a + k
= a + k we have
2
max
min
6
=
π
5. = 60 + 10 sin 12 (t 8)
a + k = 18
a + k = 6
(a) At t = 8 : temp = 60 F.
At t = 12: temp = 60 + 10 sin π = 68.7 F
3
(b) F (t) = 60 when
π
12 (t 8)=0, π,
so k = 12 hours so
t = 8, 20, 32,
hours i.e. in 1 day at t = 8 (8.00 am) and t = 20 (8.00 pm)
(c) Maximum temperature is 70 F when
i.e. at t = 14 (2.00 pm).
Minimum temperature is 50 F when
i.e. at t = 26 (2.00 am).
1. y =
sin 2πt has amplitude
. The period is
= 1.
sin(2π(t + 1)) =
sin(2πt + 2π) =
sin 2πt = y(t)
hr
1
.
F (t)
0
t < 24
2π,
giving t
8=0, 12, 24,
12 (t = 8) = π
2
π
12 (t 8) = 3π
π
2
76
HELM (2005):
Workbook 4: Trigonometry
6. (a) y = 3 sin(3tπ)
y = 3 cos(3tπ)
(b) y = 0.7 sin(4πt16)
y = 0.7 cos(4πt16)
7.
C = 3 2 + 4 2 = 5 tan α = 4 and α must be in the first quadrant (since A = 3, B = 4 are
3
both positive.)
.
.
.
α = tan 1 4 = 0.9273 rad
3
.
.
.
i = 5 cos(5t 0.9273)
8. Since sin(ωt + α) = sin ωt cos α + cos ωt sin α then A = C sin α (coefficients of cos ωt)
B = C cos α (coefficients of sin ωt) from which C
2
2
A
C
B
C
so α is in the second quadrant,
5π 5π π α = . . y = 2 sin . 2t + (b) y = 2 sin 2t + 6 6 6
10. C 2 4 1 17 C = √ 17 − 1 6 1 sin α = 2 3 4 = + = so cos α = =
9 36 36 6 √ 17 − √ 17 √ 17 = √ 17 6
6
π
6
so α is in the second quadrant. α = 1.8158 radians.
π π π π x − sin ωt + = cos ωt + − = cos ωt − 2 3 3 2
11. Since sin x = cos
π 6 = π π 6 V 0 ωt − = √ 2 . i . . 1 + i 2 = cos ωt − + 3 cos ωt − 6 = 4 cos 12. ωt + π V 0 cos ωt cos π − sin ωt sin π 4 4 4
v = V 0 cos (cos ωt − sin ωt) 2 2 V 0 (cos 2 ωt + sin 2 ωt − 2 sin ωt cos ωt) = 2 V 0 (1 − sin 2ωt) . . . v =
2
2
P =
v
2
=
V
2
R 2R
13. Since the required answer involves the difference of two cosine functions we use the identity
cos A cos B
= 2 sin A + B sin B A
2
2
Hence with
A + B = ωt,
2
B A ωt + ωτ + φ.
2
We find, by adding these equations B = 2ωt+ωτ +φ and by subtracting A = ωτ φ.
Hence
1
sin(ωt) sin(ωt + ωτ + φ) = 2 {cos(ωτ + φ) cos(2ωt + ωτ + φ)}.
(Recall that cos(x) = cos x.) The required result then follows immediately.
2 = A
2
2
+ B , sin α =
, cos α =
3
1
9.
(a)
C =
3+1 = 2;
cos α =
sin α
and hence
0
(1 sin 2ωt.)
HELM (2005):
Section 4.5: Applications of Trigonometry to Waves
77
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## Arithmetical Operations – Rational Numbers
Rational numbers are numbers which can be represented in the form of p/q, where p and q are any two integers and q is not equal to zero(q ≠ 0).
A rational number p/q is said to be in its standard form if p and q do not have any common factors other than 1.
In p/q, p is a numerator and q is a denominator.
A rational number is represented by “Q”.
Example: (i) 3/4 is a rational number of form p/q where p = 3 and q = 4
(ii) 9/16 = 3/4 is a rational number of form p/q where p = 9 and q = 16.
Here p/q is in the lowest form, i.e. p and q have no common factors.
(iii) 5/7 is a rational number of form p/q where p = 5 and q = 7
(iv) 0/9 is a rational number of form p/q where p = 0 and q = 9
Arithmetical Operations on Rational Numbers
Rational numbers can also be used with the other arithmetic operations. The rational numbers can be used in mathematical operations like addition, subtraction, multiplication, and division much like real numbers.
The Basic arithmetic operations performed on rational numbers are,
Subtraction of Rational Numbers
Multiplication of Rational Numbers
Division of Rational Numbers
In the process of addition and subtraction of rational numbers there are two types of categories.
(1) Addition of Rational Numbers with Same Denominator
(2) Addition of Rational Numbers with Different Denominator
(1) Addition of Rational Numbers with Same Denominator
A rational number is a fraction, therefore the denominator play a important role in the operation.
Any two rational numbers in the form of p/q, where q ≠ 0 can be added just like two integers.
Addition of Rational Numbers with same denominator is like addition of like fractions and result is the sum of the numerators divided by their common denominators.
Example: Add two rational numbers 2/3 + 5/3
2/3 + 5/3
= (2 + 5)/3
= 7/3
we just have to add the numerators since denominator is common.
Therefore the result after adding the rational numbers as 7/3.
Addition of Rational Numbers with Different Denominator
Any two rational number with different denominators can be added by making the denominator same, that is taking out their LCM, to convert them into equivalent rational numbers with a common denominator.
First LCM of the denominators is to be carried out, so the LCM of 2 and 3 is 6.
Example: 5/3 + 4/2
= (5 x 2)+(4 x 3)/6
= (10 + 12)/6
= 22/6
= 11/3
Subtraction of Rational Numbers with same Denominators
Any two rational numbers in the form of p/q, where q ≠ 0 can be subtracted just like two integers.
Example: Subtract the rational numbers 7/3 – 5/3
7/3 – 5/3
= (7 – 5)/3
= 2/3
we just have to subtract the numerators since denominator is common.
Therefore the result after subtracting the rational numbers as 2/3.
Subtraction of Rational Numbers with Different Denominator
Any two rational number with different denominators can be subtracted by making the denominator same, that is taking out their LCM, to convert them into equivalent rational numbers with a common denominator.
Example: Subtract 5/3 – 3/2
First, LCM of the denominators is to be carried out, so the LCM of 3 and 2 is 6.
= {(5 x 2)-(3 x 3)}/6
= (10 – 9)/6
Next subtract the rational numbers.
= 1/6
Multiplication of Two Rational Numbers
Multiplication of Two Rational Numbers is just like multiplication of two integers. Product of two or more rational numbers is founded by multiplying the corresponding numerators and denominators of the numbers and writing them in the standard form.
Product of rational numbers = Product of Numerators/ Product of Denominators
Example: (1) 5/2 x 4/2 = (5 x 4)/(2 x 2) = 20/4 = 5/1
(ii) 3/4 x 2/5 = (3 x 2)/(4 x 5) = 6/20 = 3/10
Product of numerators and product of denominators is results.
Division of Two Rational Numbers
Any two rational numbers can be divided by the following method.
Rational numbers are written as fractions, therefore the to divide a given rational number by another rational number, we have to multiply the given rational number by the reciprocal of the second rational number.
Step 1: Take the reciprocal of the divisor.
Step 2: Find the product of the numerator and the denominator to get the result.
Example: 2/3 ÷ 5/4 becomes 2/3 x 4/5
= (2 x 4)/(3 x 5)
= 8/20
= 2/5
That is, division is simply divisor value can be reciprocated and multiplied by the numerator.
Remember
• A rational number is always written in the form of p/q, where q ≠ 0.
• A rational number is represented as Q.
• Rational numbers can be added, subtracted, multiplied and divided like integers.
• In addition of rational numbers with similar denominators, any two rational number in the form of p/q, where q ≠ 0, can be added like two integers.
• In addition of rational numbers with different denominators, any two rational number with different denominators can be added by making the denominator same, that is taking out their LCM, to convert them into equivalent rational numbers with a common denominator.
• In subtraction of rational numbers with similar denominators, any two rational number in the form of p/q, where q ≠ 0, can be subtracted like two integers.
• In subtraction of rational numbers with different denominators, any two rational number with different denominators in the form of p/q, where q ≠ 0, can be subtracted by making the denominator same, that is taking out their LCM, to convert them into equivalent rational numbers with a common denominator.
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# Geometric resolution of 2nd degree equations
In the first two grades of high school, second degree equations and the formula for their solution are presented, while limiting the treatment to real solutions.
Here we want to present a general method of geometric resolution of all second degree equations, even those with using an app made in geogebra. The aim is to show how usefully the two worlds of translation and rotation are complementary.
Click to open the Geogebra app
Let’s start from the classic second degree equation which admits two real solutions
Students are trained to transform these equations into parabolas and to recognize the intersections with the axes, the symmetries, the position of the vertex in analytic form. In this way, starting from algebraic geometry, they build those mathematical capabilities essential for scientific studies.
We could say – with a play on words – that geometric algebra complements this training by geometrizing mathematics.
For this exercise, in fact, we want to see that equation in a much more naive form. In very simple terms, it says, “let’s take an unknown quantity , square it and subtract one unit. If all of this comes out zero, then that quantity is exactly what we’re looking for.” We can imagine that this story takes place along the real axis: then we will have two quantities that add algebraically (we can use two vectors) and we will be satisfied when the sum coincides with the origin.
After setting up the app (check the “Re” box to indicate the real solutions and the correct coefficients) you can act on the slider that tests all the values of : you can verify that the value of the function will coincide with the origin when You can play freely, for example with the equation to find the solutions geometrically.
If we now represent the equation we realize that the point V never manages to reach the origin. We must access a further degree of freedom: that is, to be able to rotate the vectors in the plane, as well as scale them: this is the only way to reach the origin.
If we uncheck the “Re” checkbox, an extra slider will appear, which corresponds to the second parameter of our unknown: the angle of rotation. So now we will not denote the unknown with the expressing only a real number with a sign, but we will indicate or even better
It remains to understand how to represent the terms of the equation when the unknown is a rotation: if the known term is a simple real vector and the linear term is a vector scaled by and rotated by , the quadratic term will be scaled by a factor of and rotated by (angles sum up).
These three vectors add up and if the figure closes in a triangle, that is, in the origin, the rotation is a solution of the second degree equation.
Let’s try the equation
The triangle closes when and or because the two solutions are symmetric, if the coefficients of the equations are real (cf. fundamental theorem of algebra).
Of course this is only a visual demonstration: in secondary school you have all the tools to be able to proceed to a geometric-analytical resolution. In fact it is clear that the triangle is necessarily isosceles and therefore we immediately obtain furthermore the perpendicular to the linear vector divides it into two halves and therefore with the Pythagorean theorem we also find the angle .
For example for the case illustrated above, and it turns out that the equal angles are each worth and therefore the solutions are
If you really want the complex number, you can easily find that but at this point it will be evident that we consider complex numbers as rotations + dilations manages to enrich a purely algebraic problem with a geometric sense.
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What's commutative property of multiplication?
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
What is an example of the commutative property of multiplication?
Commutative property of multiplication: Changing the order of factors does not change the product. For example, 4 × 3 = 3 × 4 4 \times 3 = 3 \times 4 4×3=3×44, times, 3, equals, 3, times, 4.
What is an example of commutative property?
Wearing shoes, gloves or putting on socks are examples of Commutative Property, as the order in which you wear them is not important! Commutative property only applies to addition and multiplication. However, subtraction and division are not commutative.
How do you do commutative property of multiplication?
According to the commutative property of multiplication, changing the order of the numbers we are multiplying, does not change the product. Here's an example of how the product does NOT change, even if the order of the factors is changed.
What's commutative property in math?
This law simply states that with addition and multiplication of numbers, you can change the order of the numbers in the problem and it will not affect the answer. Subtraction and division are NOT commutative.
Commutative Property | Addition and Multiplication | Math Help with Mr. J
15 related questions found
What is the formula of commutative property?
The commutative property formula for multiplication is defined as the product of two or more numbers that remain the same, irrespective of the order of the operands. For multiplication, the commutative property formula is expressed as (A × B) = (B × A).
Which is commutative property?
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
What is commutative property of multiplication 3rd grade?
The commutative property says that when two numbers are multiplied together, they will always give the same product no matter how they are arranged.
Why is commutative property important?
1. The Commutative Property. The commutative property is the simplest of multiplication properties. It has an easily understandable rationale and impressive immediate application: it reduces the number of independent basic multiplication facts to be memorized.
What are commutative operations?
In mathematics, a binary operation is commutative if changing the order of the operands does not change the result. ... The idea that simple operations, such as the multiplication and addition of numbers, are commutative was for many years implicitly assumed.
What are the 4 types of properties?
The four main number properties are:
• Commutative Property.
• Associative Property.
• Identity Property.
• Distributive Property.
What's an example of the distributive property?
The distributive property of multiplication over addition can be used when you multiply a number by a sum. For example, suppose you want to multiply 3 by the sum of 10 + 2. 3(10 + 2) = ? According to this property, you can add the numbers and then multiply by 3.
What are 2 examples of commutative property?
Commutative property of addition: Changing the order of addends does not change the sum. For example, 4 + 2 = 2 + 4 4 + 2 = 2 + 4 4+2=2+44, plus, 2, equals, 2, plus, 4. Associative property of addition: Changing the grouping of addends does not change the sum.
What is commutative property sentence?
The commutative property says that the sum does not change when the order of the addends is changed. The Commutative Property. Click on the example of the commutative property. 14 × 9 = 9 × 14. 14 + 9 = 17 + 6.
How is commutative property useful to learners?
The Commutative Property is a great strategy to use when adding multi-digit numbers. When I taught first grade, I taught counting up as an addition and subtraction strategy. But if students know that they can switch the order of the addends and start adding with the greater number FIRST, it makes counting up easier.
What is commutative property of subtraction?
Commutative property or commutative law states that the result of a mathematical operation remains the same even when the order of the operands are reversed. ... So, subtraction and division operations do not satisfy the commutative law.
What is commutative property integers?
Commutative property for addition: Integers are commutative under addition when any two integers are added irrespective of their order, the sum remains the same. a+b =b+a. The sum of two integer numbers is always the same. This means that integer numbers follow the commutative property.
What is the distributive property of multiplication 3rd grade?
The distributive property says that when you multiply a factor by two addends, you can first multiply the factor with each addend, and then add the sum.
What is a commutative property in 3rd grade math?
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. The property holds for Addition and Multiplication, but not for subtraction and division.
How is understanding the commutative property of multiplication helpful?
Using the commutative property, you can switch the and the so that they are in a different order. Adding and is the same as subtracting from . The sum is . Rewrite in a different way, using the commutative property of multiplication.
What is commutative property Byjus?
The commutative property or commutative law, when two numbers are added or multiplied together, then change in their positions does not change the result.
What's the difference between symmetric and commutative property?
The only difference I can see between the two terms is that commutativity is a property of internal products X×X→X while symmetry is a property of general maps X×X→Y in which Y might differ from X.
What is commutative property and associative property?
The associative property of addition states that you can group the addends in different ways without changing the outcome. The commutative property of addition states that you can reorder the addends without changing the outcome.
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### Calculus
Topics within Calculus
## Practice Problems
Here are some problems to practice what you have learned! If you need a hint on any of them, there's a few for each problem.
1. $\displaystyle{\lim_{x\to 0} x+1 = }$
Remember that our ultimate goal is to substitute $x$ for $0$ when we can. Is there anything preventing us from computing $x+1$ when $x=0$?
We can plug in $x=0$ to the function, and see that $$\lim_{x\to 0} x+1=0+1=1$$.
2. $\displaystyle{\lim_{x \to 2} \frac{x^2 -4}{x-2}}=$
Always, when computing a limit, ask yourself: "Is there anything preventing me from just plugging in the value to the function?"
When you find what's preventing you from computing $\frac{x^2 -4}{x-2}$ when $x=2$, try to eliminate it.
We can't compute $\frac{x^2 -4}{x-2}$ when $x=2$ because then we would be dividing by $0$. However, remember that $$f(x)=\frac{x^2 -4}{x-2}= \frac{(x -2)(x+2)}{x-2}.$$ Remember that we are not finding $f(2)$, we are finding the limit of $f(x)$ as $x$ approaches $2$. So, we can divide the top and bottom of the fraction by $(x-2)$ because we know $(x-2)$ is not equal to 0, but rather very, very, close. Hence, our answer is $x+2=2+2=4$.
3. $\displaystyle{\lim_{x\to 0} \frac{1}{x}}=$
A limit, as you might remember, is a value that a function approaches as it nears the specified destination. Try graphing this function. What do you think the limit is?
Limits sometimes don't exist. For example, the limit of $g(x)=x$ as $x$ approaches $\infty$ doesn't exist, because there's really no way around not getting $\infty$ as your answer, which isn't an acceptable one. In this case, we say the limit doesn't exist. You may have heard the phrase in some video. Now you know what it means.
In the graph above, you see that the function just goes on and on and on... it doesn't stop! Who knows where it lands up at $0$?? Is it positive or negative? We say that the limit doesn't exist because as $x$ approaches $0$ from one side, it goes down, quite in disagreement with the other side. Additionally, there's nothing smart we can think of to get around it - at least I can't! Because the graph clearly tells us that the limit does not exist, this is what we live with ;). Note that this is a quite common case. Are there any other limits you can think of that don't exist?
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## Trig and hyperbolic expressions
Last updates: 17 June 2011
## Trig and hyperbolic expressions
The exponential expression is $ex = 1+x +x22! +x33! +x44! +x55! +x66! +x77! +⋯.$
The sine and cosine expressions are $sinx = eix -e-ix 2i and cosx = eix +e-ix 2 , with i2=-1.$ The tangent, cotangent, secant and cosecant expressions are $tanx = sinxcosx, cotx = 1tanx, secx = 1cosx, and cscx = 1sinx .$
The hyperbolic sine and hyperbolic cosine expressions are $sinhx= ex- e-x 2 and coshx= ex +e-x 2 .$ The hyperbolic tangent, hyperbolic cotangent, hyperbolic secant and hyperbolic cosecant functions are $tanhx= sinhxcoshx, cothx = 1tanhx, sechx = 1coshx, and cschx= 1sinhx.$
Example. Prove that ${e}^{ix}=\mathrm{cos}x+i\mathrm{sin}x$.
Proof. $cosx+isinx = eix +e-ix 2 +i( eix -e-ix 2i) = eix +e-ix +eix -e-ix 2 = eix.$ $\square$
Example. Prove that $sinx = x-x33! +x55! -x77! +x99! -x1111! +x1313! -⋯ and$ $cosx = 1-x22! +x44! -x66! +x88! -x1010! +x1212! -⋯ .$
Proof. Compare coefficients of $1$ and $i$ on each side of $cosx+isinx = eix = 1+ix+ (ix) 22! + (ix) 33! + (ix) 44! + (ix) 55! + (ix) 66! + (ix) 77! +⋯ = 1+ix +i2x2 2! + i3x3 3! + i4x4 4! + i5x5 5! + i6x6 6! + i7x7 7! +⋯ = 1+ix + i2x2 2! + i· i2x3 3! + (i2)2 x4 4! + i· (i2)2 x5 5! + (i2)3 x6 6! + i· (i2)3 x77! +⋯ = 1+ix + (-1) x2 2! + i·(-1) x3 3! + (-1)2 x44! + i· (-1)2 x55! + (-1)3 x6 6! + i· (-1)3 x7 7! +⋯ = 1+ix -x22! -ix33! +x44! +ix55! -x66! -ix77! +⋯ = ( 1- x22! +x44! -x66! +⋯) +i( x- x33! +x55! -x77! +⋯) .$ □
Example. Prove that $\mathrm{cos}\left(-x\right)=\mathrm{cos}x$ and $\mathrm{sin}\left(-x\right)=-\mathrm{sin}x$.
Proof. $cos(-x) = ei(-x) + e-i(-x) 2 = e-ix +eix 2 =cosx.$ and $sin(-x) = ei(-x) -e-i(-x) 2i = e-ix -eix 2i = -sinx.$ □
Example. Prove that ${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1$.
Proof. $1 = e0 = e ix+ (-ix) = eix e-ix = eix ei(-x) = (cosx+isinx) (cos(-x) +isin(-x)) = (cosx+isinx) (cosx-isinx) = cos2x -isinxcosx +isinxcosx -i2sin2x = cos2x -(-1)sin2x = cos2x +sin2x.$ □
Example. Prove that $cosx+y=cosxcosy-sinxsiny,andsinx+y=sinxcosy+cosxsiny.$
Proof. Comparing the coefficients of $1$ and $i$ on each side of $cos(x+y) +isin(x+y) = ei(x+y) = eix+iy = eix eiy = (cosx+isinx) (cosy+isiny) = cosxcosy +icosxsiny +isinxcosy +i2sinxsiny = cosxcosy +icosxsiny +isinxcosy -sinxsiny = (cosxcosy -sinxsiny) +i (cosxsiny +sinxcosy) .$ gives $cos(x+y) = cosxcosy -sinxsiny, and sin(x+y) = sinxcosy+cosxsiny .$ □
Example. Prove that ${e}^{x}=\mathrm{cosh}x+\mathrm{sinh}x$.
Proof. $ex = 1+x +x22! +x33! +x44! +x55! +x66! +x77! +⋯ = ( 1+x22! +x44! +x66! +⋯) +( x+x33! +x55! +x77! +⋯) = coshx+sinhx .$ □
Example. Prove that $\mathrm{cosh}\left(-x\right)=\mathrm{cosh}x$ and $\mathrm{sinh}\left(-x\right)=-\mathrm{sinh}x$.
Proof. $cosh(-x) = 1+ (-x)2 2! + (-x)4 4! + (-x)6 6! + (-x)8 8! + (-x)10 10! + (-x)12 12! +⋯ = 1+ x22! +x44! +x66! +x88! +x1010! +x1212! +⋯ = coshx$ and $sinh-x=-x+-x33!+-x55!+-x77!+-x99!+-x1111!+-x1313!+⋯=-x-x33!-x55!-x77!-x99!-x1111!-x1313!-⋯=-sinhx.$ □
Example. Prove that $sinhx=x+x33!+x55!+x77!+x99!+x1111!+x1313!+⋯ and$ $coshx=1+x22!+x44!+x66!+x88!+x1010!+x1212!+⋯.$
Proof. $coshx = 12ex+e-x =121+x+x22!+x33!+x44!+x55!+x66!+x77!+⋯+1+-x+-x22!+-x33!+-x44!+-x55!+-x66!+-x77!+⋯=121+x+x22!+x33!+x44!+x55!+x66!+x77!+⋯+1-x+x22!-x33!+x44!-x55!+x66!-x77!+⋯=1+x22!+x44!+x66!+x88!+⋯$ $sinhx= 12ex-e-x =121+x+x22!+x33!+x44!+x55!+x66!+x77!+⋯-1--x--x22!--x33!--x44!--x55!--x66!--x77!-⋯=121+x+x22!+x33!+x44!+x55!+x66!+x77!+⋯-1+x-x22!+x33!-x44!+x55!-x66!+x77!-⋯=x+x33!+x55!+x77!+x99!+⋯$ □
Example. Prove that ${\mathrm{cosh}}^{2}x-{\mathrm{sinh}}^{2}x=1$.
Proof. $1 = e0 = ex+(-x) = exe-x = (coshx+sinhx) (cosh(-x) + sinh(-x) ) = (coshx+sinhx) (coshx-sinhx) = cosh2x -sinhxcoshx +sinhxcoshx-sinh2x = cosh2x -sinh2x.$
Example. Prove that $cosh(x+y) = coshxcoshy +sinhxsinhy, and sinh(x+y) = sinhxcoshy +coshxsinhy.$
Proof. We have $coshxcoshy+sinhxsinhy=ex+e-x2ey+e-y2+ex-e-x2ey-e-y2=exey+e-xey+exe-y+e-xe-y4+exey-e-xey-exe-y+e-xe-y4=2exey+2e-xe-y4=exey+e-xe-y2=ex+y+e-x+y2=coshx+y$ and $sinhxcoshy+coshxsinhy=ex-e-x2ey+e-y2+ex+e-x2ey-e-y2=exey-e-xey+exe-y-e-xe-y4+exey+e-xey-exe-y-e-xe-y4=2exey-2e-xe-y4=exey-e-xe-y2=ex+y-e-x+y2=sinhx+y.$ □
## Notes and References
The student should learn to do these proofs at the same time that the trig expressions are introduced.
## References
[Bou] N. Bourbaki, Algebra II, Chapters 4–7 Translated from the 1981 French edition by P. M. Cohn and J. Howie, Reprint of the 1990 English edition, Springer-Verlag, Berlin, 2003. viii+461 pp. ISBN: 3-540-00706-7. MR1994218
[Mac] I.G. Macdonald, Symmetric functions and Hall polynomials, Second edition, Oxford University Press, 1995. ISBN: 0-19-853489-2 MR1354144
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# Addition and Subtraction Part 3: Facts Strategies KG-3rd
This is part three in a series of strategies regarding addition and subtraction strategies. This part will focus on a variety of strategies to help toward memorization of facts, meaning automatic computation. While children are learning their number bonds (building up to 5 in KG, to 10 in first grade, and to 20 in second grade), there are other facts which cross several number bonds that students can work towards. These strategies to build mental math automaticity are highlighted below. Get some freebies in the section on doubles / near doubles.
Identity (or Zero) Property:
• The value of the number does not change when zero is added or subtracted.
• 3 + 0 = 3
• 9 – 0 = 9
Subtracting All:
• The answer is always zero when you take away / subtract all.
• 9 – 9 = 0
• 50 – 50 = 0
• Adding 1 results in the next number in the counting sequence.
• Subtracting 1 means naming the number that comes right before it in the counting sequence.
• With manipulatives, lay out an amount for student to count. Slide one more and see if he/she can name the amount without recounting.
• Do the same as above, but take one away from the group to see if he/she can name the amount without recounting.
• Show this concept using a number line.
• 6 + 1 = 7; 26 + 1 = 27
• 7 – 1 = 6; 37 – 1 = 36
• After +1 or -1 strategies are in place, then go for +2 or -2 for automatic processing.
Next-Door Neighbor Numbers:
• If subtracting two sequential numbers (ie 7 subtract 6), the answer is always one because you are taking away almost all of the original amount.
• Help students identify these types of problems: 8-7; 10-9; 98-97; 158-157
• Guide students to writing these types of problems.
• Relate these to subtracting 1 problems. If 10-1 = 9; then 10 – 9 = 1.
• Show on a number line.
Doubles (with freebies):
Near Doubles:
• After learning the doubles facts, work on doubles + 1 or doubles – 1 facts.
• If 3 + 3 = 6; then 3 + 4 = 7 (six plus one more)
• If 4 + 4 = 8; then 3 + 4 = 7 (eight minus one)
• I can picture this with a ten frame.
• See some freebies in above section.
Adding 10, 20, 30 . . . to a one digit number
• Start with problems such as 10 + 1, 10 + 2, 10 + 3, etc. This shows a basic understanding of the teen numbers and place value. Use the “cover zero” method. For example, to add 10 + 3, hold up a 10 card in your right hand (facing students) and a 3 card in your left hand. Slide the 3 card over the zero of the 10 card so it looks like 13.
• Then move to other multiples of 10 such as 20 + 5 = 25, 30 + 8 = 38, 40 + 1 = 41, etc.
• Make sure students can also decompose these types of numbers. 14 = 10 and 4; 38 = 30 and 8
• Wait until their place value concepts are stronger to work on adding 10 to any 2 or 3 digit number or to subtract 10 from any 2 digit number.
Note: Most of us (me included) tend to write math problems in this format a + b = ____ which is the “result unknown” format. When we write it as a + ___ = c or __ + b = c you will most likely notice that students will add the two shown numbers together (because they see the plus sign) and put the sum in the blank, no matter where the blank appears. (So for this problem: 5 + ___ = 8, the child is likely to put 13 in the blank.) This is because they do not see the function of the equal sign. The equal sign does not mean “the sum goes here.” The equal sign means “the same as.” What is on the left of the equal sign must equal what is on the right of the equal sign. To help with this, alter your equations to reflect different positions of the missing addend. In next week’s post, I will address some strategies to deal with these types of problems (change unknown, start unknown) and how they relate to counting up and subtraction . . . so stay tuned!!
Would love to hear from some of you! Comments always welcomed . . . share your thoughts or your strategies.
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# Maths Essential formulas for competitive Exams
### ALGEBRA:
1. Sum of first n natural numbers = n(n+1)/2
2. Sum of the squares of first n natural numbers = n(n+1)(2n+1)/6
3. Sum of the cubes of first n natural numbers = [n(n+1)/2]^2
4. Sum of first n natural odd numbers = n^2
5. Average = (Sum of items)/Number of items
### Arithmetic Progression (A.P.):
An A.P. is of the form a, a+d, a+2d, a+3d…
Where a is called the ‘first term’ and d is called the ‘common difference’
1. nth term of an A.P. tn = a + (n-1)d
2. Sum of the first n terms of an A.P. Sn = n/2[2a+(n-1)d] or Sn = n/2(first
term + last term)
### Geometrical Progression (G.P.):
A G.P. is of the form a, ar, ar2, ar3…
Where a is called the ‘first term’ and r is called the ‘common ratio’.
1. nth term of a G.P. tn = arn-1
2. Sum of the first n terms in a G.P. Sn = a|1-rn|/|1-r|
### Tests of Divisibility:
1. A number is divisible by 2 if it is an even number.
2. A number is divisible by 3 if the sum of the digits is divisible by 3.
3. A number is divisible by 4 if the number formed by the last two digits is
divisible by 4.
4. A number is divisible by 5 if the unit digit is either 5 or 0.
5. A number is divisible by 6 if the number is divisible by both 2 and 3.
6. A number is divisible by 8 if the number formed by the last three digits is
divisible by 8.
7. A number is divisible by 9 if the sum of the digits is divisible by 9.
8. A number is divisible by 10 if the unit digit is 0.
9. A number is divisible by 11 if the difference of the sum of its digits at odd
places and the sum of its digits at even places, is divisible by 11.
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# Trapezoids
## Quadrilaterals with exactly one pair of parallel sides.
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Trapezoids
What if you were told that the polygon ABCD is an isoceles trapezoid and that one of its base angles measures $38^\circ$ ? What can you conclude about its other angles? After completing this Concept, you'll be able to find the value of a trapezoid's unknown angles and sides given your knowledge of the properties of trapezoids.
### Guidance
A trapezoid is a quadrilateral with exactly one pair of parallel sides. Examples look like:
An isosceles trapezoid is a trapezoid where the non-parallel sides are congruent. The third trapezoid above is an example of an isosceles trapezoid. Think of it as an isosceles triangle with the top cut off. Isosceles trapezoids also have parts that are labeled much like an isosceles triangle. Both parallel sides are called bases.
Recall that in an isosceles triangle, the two base angles are congruent. This property holds true for isosceles trapezoids.
Theorem: The base angles of an isosceles trapezoid are congruent.
The converse is also true: If a trapezoid has congruent base angles, then it is an isosceles trapezoid. Next, we will investigate the diagonals of an isosceles trapezoid. Recall, that the diagonals of a rectangle are congruent AND they bisect each other. The diagonals of an isosceles trapezoid are also congruent, but they do NOT bisect each other.
Isosceles Trapezoid Diagonals Theorem: The diagonals of an isosceles trapezoid are congruent.
The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides. There is only one midsegment in a trapezoid. It will be parallel to the bases because it is located halfway between them. Similar to the midsegment in a triangle, where it is half the length of the side it is parallel to, the midsegment of a trapezoid also has a link to the bases.
##### Investigation: Midsegment Property
Tools Needed: graph paper, pencil, ruler
1. Draw a trapezoid on your graph paper with vertices $A(-1, 5), \ B( 2, 5), \ C(6, 1)$ and $D(-3, 1)$ . Notice this is NOT an isosceles trapezoid.
2. Find the midpoint of the non-parallel sides either by using slopes or the midpoint formula. Label them $E$ and $F$ . Connect the midpoints to create the midsegment.
3. Find the lengths of $AB, \ EF$ , and $CD$ . Can you write a formula to find the midsegment?
Midsegment Theorem: The length of the midsegment of a trapezoid is the average of the lengths of the bases, or $EF=\frac{AB+CD}{2}$ .
#### Example A
Look at trapezoid $TRAP$ below. What is $m \angle A$ ?
$TRAP$ is an isosceles trapezoid. So, $m \angle R = 115^\circ$ . To find $m \angle A$ , set up an equation.
$115^\circ + 115^\circ + m \angle A + m \angle P & = 360^\circ\\230^\circ + 2m \angle A & = 360^\circ \rightarrow m \angle A = m \angle P\\2m \angle A & = 130^\circ\\m \angle A & = 65^\circ$
Notice that $m \angle R + m \angle A = 115^\circ + 65^\circ = 180^\circ$ . These angles will always be supplementary because of the Consecutive Interior Angles Theorem. Therefore, the two angles along the same leg (or non-parallel side) are always going to be supplementary. Only in isosceles trapezoids will opposite angles also be supplementary.
#### Example B
Write a two-column proof.
Given : Trapezoid $ZOID$ and parallelogram $ZOIM$
$\angle D \cong \angle I$
Prove : $\overline{ZD} \cong \overline{OI}$
Statement Reason
1. Trapezoid $ZOID$ and parallelogram $ZOIM, \ \angle D \cong \angle I$ Given
2. $\overline{ZM} \cong \overline{OI}$ Opposite Sides Theorem
3. $\angle I \cong \angle ZMD$ Corresponding Angles Postulate
4. $\angle D \cong \angle ZMD$ Transitive PoC
5. $\overline{ZM} \cong \overline{ZD}$ Base Angles Converse
6. $\overline{ZD} \cong \overline{OI}$ Transitive PoC
#### Example C
Find $x$ . All figures are trapezoids with the midsegment.
1. a) $x$ is the average of 12 and 26. $\frac{12+26}{2}=\frac{38}{2}=19$
b) 24 is the average of $x$ and 35.
$\frac{x+35}{2}&=24\\x+35&=48\\x&=13$
c) 20 is the average of $5x-15$ and $2x-8$ .
$\frac{5x-15+2x-8}{2}&=20\\7x-23& =40\\7x&=63\\x&=9$
Watch this video for help with the Examples above.
#### Concept Problem Revisited
Given an isosceles trapezoid with $m \angle B = 38^\circ$ , find the missing angles.
In an isosceles trapezoid, base angles are congruent.
$\angle B \cong \angle C$ and $\angle A \cong \angle D$
$m \angle B = m \angle C = 38^\circ\\38^\circ + 38^\circ + m \angle A + m \angle D = 360^\circ\\m \angle A = m \angle D = 142^\circ$
### Vocabulary
A trapezoid is a quadrilateral with exactly one pair of parallel sides. An isosceles trapezoid is a trapezoid where the non-parallel sides and base angles are congruent. The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides.
### Guided Practice
$TRAP$ an isosceles trapezoid.
Find:
1. $m \angle TPA$
2. $m \angle PTR$
3. $m \angle PZA$
4. $m \angle ZRA$
1. $\angle TPZ \cong \angle RAZ$ so $m\angle TPA=20^\circ + 35^\circ = 55^\circ$ .
2. $\angle TPA$ is supplementary with $\angle PTR$ , so $m\angle PTR=125^\circ$ .
3. By the Triangle Sum Theorem, $35^\circ + 35^\circ + m\angle PZA=180^\circ$ , so $m\angle PZA=110^\circ$ .
4. Since $m\angle PZA = 110^\circ$ , $m\angle RZA=70^\circ$ because they form a linear pair. By the Triangle Sum Theorem, $m\angle ZRA=90^\circ$ .
### Practice
1. Can the parallel sides of a trapezoid be congruent? Why or why not?
For questions 2-7, find the length of the midsegment or missing side.
Find the value of the missing variable(s).
Find the lengths of the diagonals of the trapezoids below to determine if it is isosceles.
1. $A(-3, 2), B(1, 3), C(3, -1), D(-4, -2)$
2. $A(-3, 3), B(2, -2), C(-6, -6), D(-7, 1)$
3. $A(1, 3), B(4, 0), C(2, -4), D(-3, 1)$
4. $A(1, 3), B(3, 1), C(2, -4), D(-3, 1)$
1. Write a two-column proof of the Isosceles Trapezoid Diagonals Theorem using congruent triangles.
Given : $TRAP$ is an isosceles trapezoid with $\overline{TR} \ || \ \overline{AP}$ .
Prove : $\overline{TA} \cong \overline{RP}$
1. How are the opposite angles in an isosceles trapezoid related?.
2. List all the properties of a trapezoid.
### Vocabulary Language: English
Diagonal
Diagonal
A diagonal is a line segment in a polygon that connects nonconsecutive vertices
midsegment
midsegment
A midsegment connects the midpoints of two sides of a triangle or the non-parallel sides of a trapezoid.
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# Complex Analysis
Complex analysis is known as one of the classical branches of mathematics and analyses complex numbers concurrently with their functions, limits, derivatives, manipulation, and other mathematical properties. Complex analysis is a potent tool with an abruptly immense number of practical applications to solve physical problems. Let’s understand various components of complex analysis one by one here.
## Complex numbers
A number of the form x + iy where x, y are real numbers and i2 = -1 is called a complex number.
In other words, z = x + iy is the complex number such that the real part of z is x and is denoted by Re(z), whereas the imaginary part of z is iy and is denoted by I(z).
### Modulus and Argument of a Complex Number
The modulus of a complex number z = x + iy is the real number √(x2 + y2) and is denoted by |z|.
The amplitude or argument of a complex number z = x + iy is given by:
arg(z) = θ = tan-1(y/x), where x, y ≠ 0.
Also, the arg(z) is called the principal argument when it satisfies the inequality -π < θ ≤ π, and it is denoted by Arg(z).
## Complex Functions
In complex analysis, a complex function is a function defined from complex numbers to complex numbers. Alternatively, it is a function that includes a subset of the complex numbers as a domain and the complex numbers as a codomain. Mathematically, we can represent the definition of complex functions as given below:
A function f : C → C is called a complex function that can be written as
w = f(z), where z ∈ C and w ∈ Z.
Also, z = x + iy and w = u + iv such that u = u(x, y) and v = v(x, y). That means u and v are functions of x and y.
## Limits of Complex Functions
Let w = f(z) be any function of z defined in a bounded closed domain D. Then the limit of f(z) as z approaches z0 is denoted by “l”, and is written as
$$\begin{array}{l}\displaystyle \lim_{z \to z_0}f(z)=l\end{array}$$
, i.e., for every ϵ > 0, there exists δ > 0 such that |f(z) – l| < ϵ whenever |z – z0| < δ where ϵ and δ are arbitary small positive real numbers. Here, l is the simultaneous limit of f(z) as z → z0.
## Continuity of Complex Functions
Let’s understand what is the continuity of complex functions in complex analysis.
A complex function w = f(z) defined in the bounded closed domain D, is said to be continuous at a point Z0, if f(z0) is defined
$$\begin{array}{l}\displaystyle \lim_{z \to z_0}f(z)\end{array}$$
exists and
$$\begin{array}{l}\displaystyle \lim_{z \to z_0}f(z)=f(z_0)\end{array}$$
.
## Complex Differentiation
Some of the standard results of complex differentiation are listed below:
• dc/dz = 0; where c is a complex constant
• d/dz (f ± g) = (df/dz) ± (dg/dz)
• d/dz [c.f(z)] = c . (df/dz)
• d/dz zn = nzn-1
• d/dz (f.g) = f (dg/dz) + g (df/dz)
• d/dz (f/g) = [g (df/dz) – f (dg/dz)]/ g2
All these formulas are used to solve various problems in complex analysis.
## Analytic Functions
A function f(z) is said to be analytic at a point z0 if f is differentiable not only at z, but also at every point in some neighbourhood of z0. Analytic functions are also called regular, holomorphic, or monogenic functions.
Also, check:Analytic functions
## Harmonic Function
A function u(x, y) is said to be a harmonic function if it satisfies the Laplace equation. Also, the real and imaginary parts of an analytic function are harmonic functions.
## Complex Integration
Suppose f(z) be a function of complex variable defined in a domain D and “c” be the closed curve in the domain D.
Let f(z) = u(x, y) + i v(x, y)
Here, z = x + iy
(or)
f(z) = u + iv and dz = dx + i dy
c f(z) dz = ∫c (u + iv) dz
= ∫c (u + iv) (dx + idy)
= ∫c (udx – vdy) +i ∫c (udy + vdx)
Here, ∫c f(z) dz is known as the contour integral.
## Cauchy’s Integral Theorem
If f(z) is analytic function in a simply-connected region R, then ∫c f(z) dz = 0 for every closed contour c contained in R.
Converse:
If a function f(z) is continuous throughout the simple connected domain D and if ∫c f(z) dz = 0 for every closed contour c in D, then f(z) is analytic in D.
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There are many ways to solve quadratic functions, but one of the most popular methods is known as the quadratic formula. This formula is based on the fact that any quadratic equation can be rewritten in the form of ax^2 + bx + c = 0. The quadratic formula then states that the roots of the equation are given by: x = (-b +/- sqrt(b^2 - 4ac)) / (2a). In other words, the roots of a quadratic equation are always symmetrical around the axis of symmetry, which is given by x = -b/(2a). To use the quadratic formula, simply plug in the values of a, b, and c into the formula and solve for x. Keep in mind that there may be more than one root, so be sure to check all possible values of x. If you're struggling to remember the quadratic formula, simply Google it or look it up in a math textbook. With a little practice, you'll be solvingquadratics like a pro!
A linear algebra solver can be used to find the solutions to systems of linear equations. Additionally, it can be used to find the inverse of a matrix, determinants, and eigenvectors. Linear algebra solvers are a valuable tool for mathematicians and engineers alike. Whether you're solving simple equations or working with more complex mathematical models, a linear algebra solver can be an invaluable resource.
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This a website that enables you to get detailed solutions to your math word problems. Just enter the problem in the text box and click on the "Solve" button. will then show you step-by-step how to solve the problem. You can also use the site to check your answers to make sure you are on the right track. is a great resource for students of all levels who are struggling with math word problems.
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# Question #fca5a
Jul 23, 2017
(3)
$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 3 \left(\vec{c} \times \vec{a}\right)$
#### Explanation:
Given that $\vec{a} + 2 \vec{b} + 3 \vec{c} = \vec{O}$
Now, Taking cross product or vector product with $\vec{a}$ on both sides:-
$\left(\vec{a} + 2 \vec{b} + 3 \vec{c}\right) \times \vec{a} = \vec{O} \times \vec{a}$
Since cross product is distributive and cross product with $\vec{O}$ OR null vector is $\vec{O}$ itself,
$\implies \vec{a} \times \vec{a} + 2 \left(\vec{b} \times \vec{a}\right) + 3 \left(\vec{c} \times \vec{a}\right) = \vec{O}$
$\because$ the cross product of a vector with itself is null vector or $\vec{O}$ $\therefore$ $\vec{a} \times \vec{a} = \vec{O}$
$\implies 2 \left(\vec{b} \times \vec{a}\right) + 3 \left(\vec{c} \times \vec{a}\right) = \vec{O}$ ------------------- 1.
Similarly, taking cross product with $\vec{b} \mathmr{and} \vec{c}$ on both sides:-
$\left(\vec{a} \times \vec{b}\right) + 3 \left(\vec{c} \times \vec{b}\right) = \vec{O}$ ------------ 2.
$\left(\vec{a} \times \vec{c}\right) + 2 \left(\vec{b} \times \vec{c}\right) = \vec{O}$ ------------ 3.
$2 \left(\vec{b} \times \vec{a}\right) + 3 \left(\vec{c} \times \vec{a}\right) + \left(\vec{a} \times \vec{b}\right) + 3 \left(\vec{c} \times \vec{b}\right) + \left(\vec{a} \times \vec{c}\right) + 2 \left(\vec{b} \times \vec{c}\right) = \vec{O}$
Now for any two vectors $\vec{x}$ and $\vec{y}$, $\textcolor{red}{\vec{x} \times \vec{y} = - \vec{y} \times \vec{x}}$.
Also, sum of any vector with null vector is the vector itself.
$\implies \left(\vec{a} \times \vec{b}\right) - 2 \left(\vec{a} \times \vec{b}\right) + 2 \left(\vec{b} \times \vec{c}\right) - 3 \left(\vec{b} \times \vec{c}\right) + 3 \left(\vec{c} \times \vec{a}\right) - \left(\vec{c} \times \vec{a}\right) = \vec{O}$
$\implies - \left(\vec{a} \times \vec{b}\right) - \left(\vec{b} \times \vec{c}\right) - \left(\vec{c} \times \vec{a}\right) + 3 \left(\vec{c} \times \vec{a}\right) = \vec{O}$
$\implies - \left(\vec{a} \times \vec{b}\right) - \left(\vec{b} \times \vec{c}\right) - \left(\vec{c} \times \vec{a}\right) = - 3 \left(\vec{c} \times \vec{a}\right)$
Taking product on both sides with $- 1$,
$\implies \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 3 \left(\vec{c} \times \vec{a}\right)$
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Resultant Vector, how to calculate a resultant using the parallelogram method and the head to tail method. A resultant is simply...
# Resultant Vector, Sum of Vectors
How to calculate the resultant vector
The resultant vector is the vector that 'results' from adding two or more vectors together. There are a two different ways to calculate the resultant vector.
Methods for calculating a Resultant Vector:
In the picture on the left, the black vector is the resultant of the two red vectors. To try to understand what a resultant is consider the following story.
If you drove from your house, centered at the origin. To your friends house, at the point (3, 4), imagine that you had to take two different roads these are the two red vectors. However, the resultant vector vector would be the straight line path from your home to your friend's house, and the black vector represents that path.
### Head to Tail Method
The head to tail method is way to find the resultant vector. The steps are quite straight forward. The head to tail method considers the head of a vector to be the end with the arrow, or the 'pointy end'. The tail of the vector is where the vector begins.
• Place the two vectors next to each other such that the head of the one vector is touching the tail of the other vector.
• Draw the resultant vector by starting where the tail of first vector is to the head of second vector.
### Calculate the magnitude resultant vector
To find the resultant vector's magnitude, use the pythagorean theorem.
### Practice Problems
##### Problem 1
How long is the vector that you drew?
### Parallelogram Method to Calculate Resultant
Before tackling the parallelogram method for solving resultant vectors, you should be comfortable with the following topics.
##### Problem 3
Step 1
Draw a parallelogram based on the two vectors that you already have. These vectors will be two sides of the parallelogram (not the opposite sides since they have the angle between them).
Step 2
We now have a parallelogram and know two angles (opposite angles of parallelograms are congruent). We can also figure out the other pair of angles since the other pair are congruent and all four angles must add up to 360.
Step 3
Draw the paprallelograms diagonal. This diagonal is the resultant vector.
Step 4
Use the law of cosines to determine the length of the resultant.
Use the law of cosines to calculate the resultant.
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# Breaking Down Barriers in Algebra with Monic Quadratics + –
## Transcript
Now again, it’s a bit different because now we have positive here and negative here, but don’t worry too much about this number, guys, because the the first thing you should consider is the factors of this one. Okay? Now it’s negative again. How do we make negative? Well, we need to have plus and minus, don’t we? Plus times negative makes negative. So again, it’s going to be one of one plus and one negative okay?
Again, I’m going to start with my x and x and then think about 56. I’m gonna try to use 7 and 8 again okay? Because that’s the one I can think of at the top of my head, and we want this plus one x, isn’t it? I know that the difference of these two numbers is one. So, this time, we need a positive one. How do we make a positive one with these numbers?
Well, I can say plus 8 minus 7 makes plus 1, doesn’t it?
So this time, I’m going to have minus 7 and plus 8. Draw your cross x times negative 7 is negative 7x. X times 8 is 8x. Negative 7x plus 8x is 1x or just x. Positive x, which is the same as what we have here, I’ll rub that off. Yeah, we’ve got the positive x. So the answer is x minus 7, x plus 8, okay?
Remember there’s a plus there isn’t it? 12, okay? Again! We’ve got a plus here and a negative here, but again, don’t worry too much about this one. Let’s find the right factors first and then see if we get that, so again, I’m going to put my x and x on my left. 48, what do you think? Well, I’m going to try and think of some factors, though again, 48 has a lot of factors, yeah? Like, I don’t know 4 and 12 and 6 and 8, but I’m going to actually use 6 and 8 because the difference is 2. That’s what I want, isn’t it? But I need to make, make sure this time this is negative. It’s going to be plus and minus in one of them, isn’t it?
So, How do I make plus 2 with these numbers? Minus 6 plus 8 would be positive, isn’t it? Ok! So plus 8 minus 6 makes plus 2. So that’s where I put the negative, that’s where I put the positive. If you do what if you did what I did before, you’d probably end up with a negative two, which is wrong. So I’m gonna put my negative 6 here and plus 8 there and then cross it. Negative 6, cross it 8x, negative 6x plus 8x is 2x which is exactly the same as what we have here.
Our answer is x minus 6, x plus 8, and that’s the right answer. This is your final answer.
okay? So when we multiply those two factors together.
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Problem 26 in penguin book of puzzles.
# A Walking Puzzle
A man set out at noon to walk from Appleminster to Boneyham, and a friend of his started at two p.m. on the same day to walk from Boneyham to Appleminster. They met on the road at five minutes past four o’clock and each man reached his destination at exactly the same time. Can you say at what time they both arrived?
What’s interesting and beautiful about this problem is the surprizing lack of information about speeds and distances; all you’re being told is the time.
# Solution 1 - Forming equations.
Such algebraic problems ask for an element of faith that when you simplify the relevant equations, the unknowns cancel out and leave you with a simple equation.
The crux to this is to note that they met at 4:05pm and ended at some time T. This duration, call it $t$, is fixed so we can form an equation here. Let $d_{xy}$ be the distance from point x to point y, and let m be the point which they met.
• $d_{mb}/s_a = t = d_{ma}/s_b$
• $s_a = d_{am}/(245\text{ min})$
• $s_b = d_{bm}/(125\text{ min})$
• Substituting them into the first equation gives $(d_{ma}/d_{mb})^2 = 245/125$
• Substituting $s_a$ into the first equation, we get $t = (245\text{ min}) * d_{mb}/d_{ma} = 175 \text{ min}$
# Solution 2 - Using distance time graph
My first attempt was to draw the time distance graph which looked a lot like the cross ladder problem mentioned in harmonic mean.
Putting it in the form, we get
$$\frac{1}{t-245} = \frac{1}{t} + \frac{1}{t-120}$$
I felt clever doing it, but it turns out this equation is quite a pain to solve by hand because of the algebraic manipulation and then solving the quadratic.
We can first apply a translation $y = x - 60$ to make the RHS somewhat symmetric which reduces the amount of manipulation, but we still end up having to solve a quadratic.
Faithfully solving the equation, or if we’re lazy just key it into wolfram alpha, we get $t = 420$.
# Thoughts
This question made me think about two things. First, equations involving reciprocals make the problem disproportionate tedious.
Second, what the general form of the solution to the equation looked like:
$$\frac{1}{x-c} = \frac{1}{x-a} + \frac{1}{x+a}$$ $$x^2 - a^2 = 2x (x-c)$$ $$c^2 - a^2 = (x - c)^2$$ $$x = c \pm \sqrt{c^2 - a^2}$$
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Q:
# How do you solve fraction problems using basic math?
A:
To solve fraction problems using basic math, understand the operations necessary to solve different fraction problems, such as simplifying fractions, understanding equivalent fractions, adding fractions, subtracting fractions and multiplying fractions. Also understand dividing fractions, comparing fractions, converting to tenths and hundredths, and equivelating fractions to tenths and hundredths.
## Keep Learning
Use the order of operations to work down multiple fractions, starting with two at a time. To multiply fractions, simply multiply the numerator by the numerator and the denominator by the denominator. The results that you get from multiplying these two numbers is the answer to the multiplication problem.
Dividing fractions is similar to multiplying fractions, except you flip the second fraction over to make the denominator's number the numerator and the numerator's number the denominator before you begin multiplying. Once the numbers are flipped and multiplied, the fraction you receive is the answer to the division problem.
To add and subtract fractions, first find the common denominator between the fractions. This is achieved by multiplying the two spaces in both fractions by each other's denominators and then simplifying both fractions to simplest terms. Once the denominators are the same and simplified, perform the subtraction and addition operations with the numerators, and keep the denominators the same.
Sources:
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i) What must be added to $3\dfrac{3}{4}$ to get $4\dfrac{2}{8}$? ii) What must be subtracted from $4\dfrac{1}{4}$ to get $2\dfrac{3}{{12}}$?
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Hint: Here we have to find the sum and difference between the given data. Here the data is in the form of a fraction. Since the given data is in fraction the value of the denominator is different so we find the LCM for both the denominators and then we add the numbers. Hence, we obtain the required solution for the given question.
Complete step by step solution:
I.We apply the arithmetic operations on the fractions. Here in this question, we add and subtract the two fractions. The LCM is the least common multiple. The LCM will be common for both the numbers.
Now we will consider the first question
What must be added to $3\dfrac{3}{4}$ to get $4\dfrac{2}{8}$
Let we consider the unknown number as x
The question is written in the form of algebraic equation as
$\Rightarrow 3\dfrac{3}{4} + x = 4\dfrac{2}{8}$
Convert the mixed fractions into improper fractions we have
$\Rightarrow \dfrac{{15}}{4} + x = \dfrac{{34}}{8}$
Take $\dfrac{{15}}{4}$ to RHS we have
$\Rightarrow x = \dfrac{{34}}{8} - \dfrac{{15}}{4}$
The LCM for the numbers 8 and 4 is 8.
So we have
$\Rightarrow x = \dfrac{{\dfrac{{34}}{8} \times 8 - \dfrac{{15}}{4} \times 8}}{8}$
On simplifying we have
$\Rightarrow x = \dfrac{{34 - 30}}{8}$
$\Rightarrow x = \dfrac{4}{8}$
On dividing the both numerator and the denominator by 4 we have
$\Rightarrow x = \dfrac{1}{2}$
So, the correct answer is “$x = \dfrac{1}{2}$”.
II.Now we will consider the second question
What must be subtracted from $4\dfrac{1}{4}$ to get $2\dfrac{3}{{12}}$
Let we consider the unknown number as x
The question is written in the form of algebraic equation as
$\Rightarrow 4\dfrac{1}{4} - x = 2\dfrac{3}{{12}}$
Convert the mixed fractions into improper fractions we have
$\Rightarrow \dfrac{{17}}{4} - x = \dfrac{{27}}{{12}}$
Take $\dfrac{{17}}{4}$ to RHS we have
$\Rightarrow - x = \dfrac{{27}}{{12}} - \dfrac{{17}}{4}$
The LCM for the numbers 12 and 4 is 12.
So we have
$\Rightarrow - x = \dfrac{{\dfrac{{27}}{{12}} \times 12 - \dfrac{{17}}{4} \times 12}}{{12}}$
On simplifying we have
$\Rightarrow - x = \dfrac{{27 - 51}}{{12}}$
$\Rightarrow - x = \dfrac{{ - 24}}{{12}}$
On dividing the both numerator and the denominator by 12 we have
$\Rightarrow - x = - 2$
On cancelling the minus we have
$\Rightarrow x = 2$
So, the correct answer is “ x = 2”.
Note: While adding the two fractions we need to check the values of the denominator, if both denominators are having the same value then we can add the numerators. Suppose if the fractions have different denominators, we have to take LCM for the denominators and we simplify for further.
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# artanh function
Martin McBride
2020-02-03
The artanh function is a hyperbolic function. It is the inverse of the tanh, and is also known as the inverse hyperbolic tangent function.
Why is it called the artanh rather than the arctanh? See here.
## Equation and graph
The artanh function is defined as the inverse of tanh, ie if:
$$x = \tanh y$$
then:
$$y = \operatorname{artanh} x$$
Here is a graph of the function:
The function is only valid for -1 < x < 1.
## artanh as inverse of tanh
This animation illustrates the relationship between the tanh function and the artanh function:
The first, blue, curve is the tanh function.
The grey dashed line is the line $y=x$.
The second, red, curve is the artanh function. As with any inverse function, it is identical to the original function reflected in the line $y=x$.
## Logarithm formula for artanh
There is a also a formula for finding artanh directly:
$$\operatorname{artanh} x =\frac12\ln\left(\frac{1+x}{1-x}\right)$$
Here is a proof of the logarithm formula for artanh. This follows similar lines to the proof for arsinh.
We will use:
$$u = \operatorname{artanh}{x}$$
The tanh of u will be x, because tanh is the inverse of artanh:
$$x = \tanh {u}$$
One form of the formula for tanh is:
$$\tanh{u} = \frac{e^{2u}-1}{e^{2u}+1}$$
This gives us:
$$x = \tanh{u} = \frac{e^{2u}-1}{e^{2u}+1}$$
Multiplying both sides by $(e^{2u}+1)$ gives:
$$({e^{2u}+1})x = e^{2u} - 1$$
Rearranging:
$$({e^{2u}+1})x - e^{2u} + 1 = 0$$
So:
$$e^{2u}(x - 1) + 1 + x = 0$$
$$e^{2u}(x - 1) = -(1 + x)$$
$$e^{2u} = -\frac{1 + x}{x - 1} = \frac{1 + x}{1-x}$$
Taking the log of both sides:
$$2u = \ln\left(\frac{1+x}{1-x}\right)$$
This gives the required result:
$$u = \operatorname{artanh} x =\frac12\ln\left(\frac{1+x}{1-x}\right)$$
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# Circle of Death
The basic problem here comes from the Josephus Problem. In Josephus problem, N people stands in a circle and every second person is killed until only one person is left in the circle. Then that man is free to go. The problem in Josephus Problem is to find the index of the free man.
We can find the solution of the Josephus problem by a recursive solution. The solution is follow:
J(1) = 1
J(2n) = 2J(n) - 1; when n >= 1 (even case)
J(2n+1) = 2J(n) + 1; when n >= 1 (odd case)
With some pattern finding, and mathematics, we can show that the final result of a Josephus circle will be a cyclic left shift of the binary of N.
If we take some examples:
J(1) = 1
J(2) = 1
J(3) = 3
J(4) = 1
J(5) = 3
J(6) = 5
J(7) = 7
We can see that, J(2m + l) = 2l + 1
Now if we look at the binary of the N. We have something like this.
N = (bm bm-1 ... b1 b0)2
l = (0m bm-1 ... b1 b0)2
2l = (bm-1 ... b1 b00)2
2l + 1= (bm-1 ... b1 b01)2
J(N) = (bm-1 ... b1 b01)2
So, the answer is the cyclic left shift of N
So, if 5 people were standing in a circle, the person to survive is the one with id 3. Example: 5 = (101)2. After one cyclic left shift, 5 becomes 3 = (11)2. So if we start again with 3 people in a circle, cyclic left shift of 3 is exactly 3, so we would stop playing the game. And 3 is thus our lucky number.
So, given N, the lucky number can be found by calculating the number of 1s in its binary form. Let the count be x. So the lucky number is 2x-1
To count how many integers are there, we only have to count how many integers have x 1s in its binary form. This can be done with combinatorics. Then the integers from 1 to 2K-1 has K bits in them.The answer is KCx.
### Statistics
66% Solution Ratio
ragibshahriarEarliest, Apr '20
JUNIORHMMC_CODFastest, 0.0s
JUNIORHMMC_CODLightest, 131 kB
bokaifShortest, 245B
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LINEAR FUNCTIONS
By: Carly Cantrell
LetŐs explore the two linear functions given below:
f(x) = x + 1
g(x) = x – 1
The above graph shows two linear functions. These functions are parallel because they have the same slope. The slope of f(x) and g(x) are both 1. These functions differ by a single translation. Compared to the parent function, f(x) is translated up one unit and g(x) is translated down one unit. The next graph shows the summative relationship of f(x) and g(x).
The sum of the two functions:
h(x) = f(x) + g(x)
h(x) = x + 1 + x – 1
h(x) = 2x
This function remains linear because two linear functions were added together, which results in a single linear function. This new function, h(x), has a slope greater than either of the original functions. The slope increases at a rate twice as fast because the original two functions both had the slope of 1. Now, this function has a slope of 2. Notice, h(x) goes through the origin. Neither, f(x) or g(x) went through the origin. The reason h(x) goes through the origin is because the original functions were equidistant away from the origin in opposite directions. The next graph will show the multiplicative relationship between f(x) and g(x).
The product of the two functions:
h(x) = f(x)*g(x)
h(x) = (x+1)(x-1)
h(x) = x2 -1
This function is no longer linear! In fact, this is now a quadratic function. The graph shown above is a parabola. This particular function has a minimum because the parabola opens upwards. The domain of quadratic functions will always be all real numbers, while the range depends on the specific function. This function has two roots, one at x = -1 and the second at x = 1. The minimum is located at the vertex: (0,-1). This is shown in the function h(x) = x2 – 1. The shift downwards justifies the location of the vertex and the reason for having two roots. Next, we will look at the quotient of the two original linear functions.
The quotient of the two functions:
h(x) =
h(x) =
This function becomes a rational function. First, the domain of a rational function consists of all real numbers except the zeroes of the polynomial in the denominator. The graph takes on the shape from above because this function contains both a vertical and horizontal asymptote. Vertical asymptotes are the imaginary vertical lines that form boundaries in the graph. The is where the function is undefined; the vertical asymptote for h(x) is x=1. This is shown in the graph above because the function gets very close to x=1, while never touching it. Horizontal asymptotes are very similar. Horizontal asymptotes are imaginary horizontal lines that the graph will approach as x increase or decreases to . The horizontal asymptotes for the function h(x) is y=1 because the degree of the leading coefficients are equal and both 1. When the degrees are equal you divide the coefficients, respectively, hence for h(x) it is 1 / 1 = 1.
The composition of the two functions:
h(x) = f(g(x))
h(x) = (x – 1) + 1
h(x) = x
This function remains a linear function because each of these original functions was linear. Will this always be the case?
LetŐs look at the same relationships using different linear functions:
LetŐs try two linear functions with different and opposite slopes:
f(x)= -2x+1
g(x)= x+3
Summative Relationship:
h(x)= f(x)+g(x)
Multiplicative Relationship:
h(x)= f(x)*g(x)
Quotient Relationship:
h(x)=
Composed Relationship:
h(x)= f(g(x))
In summary, the same relationship upheld when the two linear function were changed. When adding two linear functions, the result is a different linear function. When multiplying two linear functions, the result is a quadratic function. When dividing two linear functions, the result is a rational function. Lastly, when composing two linear functions, the result is a different linear function.
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## Trigonometry (11th Edition) Clone
RECALL: (1) The function $y=a\cdot \sec{x}$ has a period of $2\pi$. (2) Consecutive asymptotes of the secant function are $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ (3) The function $y=a \cdot \sec{[b(x-d)]}$ has a period of $\frac{2\pi}{|b|}$ and involves a phase shift of $|d|$ (to the right when $d\gt0$, to the left when $d\lt0$). Write the given equation in the form $y=a \cdot \sec{[b(x-d)]}$ by factoring out $2$ within the secant function to obtain: $y=\sec{[2(x+\frac{\pi}{6})]}$ Thus, the given function has $a=1, b=2,$and $d=-\frac{\pi}{6}$. This means that it has: period = $\frac{2\pi}{2}=\pi$ phase shift = $|-\frac{\pi}{6}|=\frac{\pi}{6}$, to the left The guide function for this secant is $y=\cos{[2(x+\frac{\pi}{6})]}$. One period of the function $y=\sec{(2x)}$ is in the interval $[0, 2\pi]$. This means that one period of the function $y=\sec{[2(x+\frac{\pi}{6})]}$, which involves a $\frac{\pi}{6}$ shift to the left, will be at $[-\frac{\pi}{6}, \frac{5\pi}{6}]$. Divide this interval into four equal parts to get the key x-values $-\frac{\pi}{6}, \frac{\pi}{12}, \frac{\pi}{3}, \frac{7\pi}{12}, \frac{5\pi}{6}$. Find the consecutive vertical asymptotes by equating $2x+\frac{\pi}{3}$ to $\frac{\pi}{2}$ and to $\frac{3\pi}{2}$, then solve each equation to obtain: \begin{array}{ccc} 2x+\frac{\pi}{3}&=\frac{\pi}{2} &\text{or} &2x+\frac{\pi}{3} = \frac{3\pi}{2} \\2x&=\frac{\pi}{6} &\text{or} &2x=\frac{7\pi}{6} \\x&=\frac{\pi}{12} &\text{or} &x=\frac{7\pi}{12} \end{array} To graph the given function, perform the following steps: (1) Create a table of values for the guide function $y=\cos{[2(x+\frac{\pi}{6}]}$ using the key x-values listed above. (Refer to the table below.) (2) Plot the points from the table of values and connect them using a dashed curve (as the curve will only serve as a guide). (3) Graph the consecutive vertical asymptotes $x=\frac{\pi}{12}$ and $x=\frac{7\pi}{12}$. (4) Sketch the graph of $y=\sec{[2(x+\frac{\pi}{6})]}$ by drawing (i) a U-shaped curve below the x-axis and between the consecutive vertical asymptotes. (ii) a half U-shaped curve from the point $(-\frac{\pi}{6}, 1)$ to the asymptote $x=\frac{\pi}{12}$. (iii) a half U-shaped curve from the asymptote $x=\frac{7\pi}{12}$ to the point $(\frac{5\pi}{6}, 1)$ . (Refer to the graph in the answer part above.)
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Learning Objectives
In this section students will:
3.3.1 – Using Rational Roots
Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.
Understanding nth Roots
Suppose we know that $\,{a}^{3}=8.\,$ We want to find what number raised to the 3rd power is equal to 8. Since $\,{2}^{3}=8,$ we say that 2 is the cube root of 8.
The nth root of $\,a\,$ is a number that, when raised to the nth power, gives $\,a.\,$ For example, $\,-3\,$ is the 5th root of $\,-243\,$ because $\,{\left(-3\right)}^{5}=-243.\,$ If $\,a\,$ is a real number with at least one nth root, then the principal nth root of $\,a\,$ is the number with the same sign as $\,a\,$ that, when raised to the nth power, equals $\,a.$
The principal nth root of $\,a\,$ is written as $\,\sqrt[n]{a},$ where $\,n\,$ is a positive integer greater than or equal to 2. In the radical expression, $\,n\,$ is called the index of the radical.
Principal nth Root
If $\,a\,$ is a real number with at least one nth root, then the principal nth root of $\,a,$ written as $\,\sqrt[n]{a},$ is the number with the same sign as $\,a\,$ that, when raised to the nth power, equals $\,a.\,$ The index of the radical is $\,n.$
Example 1 – Simplifying nth Roots
Simplify each of the following:
1. $\sqrt[5]{-32}$
2. $\sqrt[4]{4}\cdot \sqrt[4]{1,024}$
3. $-\sqrt[3]{\frac{8{x}^{6}}{125}}$
4. $8\sqrt[4]{3}-\sqrt[4]{48}$
1. $\sqrt[5]{-32}=-2\,$ because $\,{\left(-2\right)}^{5}=-32$
2. First, express the product as a single radical expression. $\,\sqrt[4]{4,096}=8\,$ because $\,{8}^{4}=4,096$
3. $\begin{array}{cc}\frac{-\sqrt[3]{8{x}^{6}}}{\sqrt[3]{125}}\hfill & \phantom{\rule{3em}{0ex}}\text{Write as quotient of two radical expressions}.\hfill \\ \frac{-2{x}^{2}}{5}\hfill & \phantom{\rule{3em}{0ex}}\text{Simplify}.\hfill \end{array}$
4. $\begin{array}{cc}8\sqrt[4]{3}-2\sqrt[4]{3}\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify to get equal radicands}.\hfill \\ 6\sqrt[4]{3} \hfill & \phantom{\rule{2em}{0ex}}\text{Add}.\hfill \end{array}$
Try It
Simplify.
1. $\sqrt[3]{-216}$
2. $\frac{3\sqrt[4]{80}}{\sqrt[4]{5}}$
3. $6\sqrt[3]{9,000}+7\sqrt[3]{576}$
1. $-6$
2. $6$
3. $88\sqrt[3]{9}$
Using Rational Exponents
Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index $\,n\,$ is even, then $\,a\,$ cannot be negative.
${a}^{\frac{1}{n}}=\sqrt[n]{a}$
We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an nth root. The numerator tells us the power and the denominator tells us the root.
${a}^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^{m}=\sqrt[n]{{a}^{m}}$
All of the properties of exponents that we learned for integer exponents also hold for rational exponents.
Rational Exponents
Rational exponents are another way to express principal nth roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is
${a}^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^{m}=\sqrt[n]{{a}^{m}}$
How To
Given an expression with a rational exponent, write the expression as a radical.
1. Determine the power by looking at the numerator of the exponent.
2. Determine the root by looking at the denominator of the exponent.
3. Using the base as the radicand, raise the radicand to the power and use the root as the index.
Example 2 – Writing Rational Exponents as Radicals
Write $\,{343}^{\frac{2}{3}}\,$ as a radical. Simplify.
The 2 tells us the power and the 3 tells us the root.
${343}^{\frac{2}{3}}={\left(\sqrt[3]{343}\right)}^{2}=\sqrt[3]{{343}^{2}}$
We know that $\,\sqrt[3]{343}=7\,$ because $\,{7}^{3}=343.\,$ Because the cube root is easy to find, it is easiest to find the cube root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power.
${343}^{\frac{2}{3}}={\left(\sqrt[3]{343}\right)}^{2}={7}^{2}=49$
Try It
Write $\,{9}^{\frac{5}{2}}\,$ as a radical. Simplify.
${\left(\sqrt{9}\right)}^{5}={3}^{5}=243$
Example 3 – Writing Radicals as Rational Exponents
Write $\,\frac{4}{\sqrt[7]{{a}^{2}}}\,$ using a rational exponent.
The power is 2 and the root is 7, so the rational exponent will be $\,\frac{2}{7}.\,$ We get $\,\frac{4}{{a}^{\frac{2}{7}}}.\,$ Using properties of exponents, we get $\,\frac{4}{\sqrt[7]{{a}^{2}}}=4{a}^{\frac{-2}{7}}.$
Try It
Write $\,x\sqrt{{\left(5y\right)}^{9}}\,$ using a rational exponent.
$x{\left(5y\right)}^{\frac{9}{2}}$
Example 4 – Simplifying Rational Exponents
Simplify:
1. $5\left(2{x}^{\frac{3}{4}}\right)\left(3{x}^{\frac{1}{5}}\right)$
2. ${\left(\frac{16}{9}\right)}^{-\frac{1}{2}}$
1. $\begin{array}{cc}30{x}^{\frac{3}{4}}{x}^{\frac{1}{5}}\hfill & \phantom{\rule{2.5em}{0ex}}\text{Multiply the coefficients}.\hfill \\ 30{x}^{\frac{3}{4}+\frac{1}{5}}\hfill & \phantom{\rule{2.5em}{0ex}}\text{Use properties of exponents}.\hfill \\ 30{x}^{\frac{19}{20}}\hfill & \phantom{\rule{2.5em}{0ex}}\text{Simplify}.\hfill \end{array}$
2. $\begin{array}{cc}{\left(\frac{9}{16}\right)}^{\frac{1}{2}}\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Use definition of negative exponents}.\hfill \\ \sqrt{\frac{9}{16}}\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Rewrite as a radical}.\hfill \\ \frac{\sqrt{9}}{\sqrt{16}}\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Use the quotient rule}.\hfill \\ \frac{3}{4}\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Simplify}.\hfill \end{array}$
Try It
Simplify $\,\left[{\left(8x\right)}^{\frac{1}{3}}\right]\left(14{x}^{\frac{6}{5}}\right).$
$28{x}^{\frac{23}{15}}$
Access these online resources for additional instruction and practice with radicals and rational exponents.
Key Concepts
• The principal nth root of $\,a\,$ is the number with the same sign as $\,a\,$ that when raised to the nth power equals $\,a.\,$ These roots have the same properties as square roots. See Example 1.
• Radicals can be rewritten as rational exponents and rational exponents can be rewritten as radicals. See Example 2 and Example 3.
• The properties of exponents apply to rational exponents. See Example 4.
Glossary
principal square root
the nonnegative square root of a number $\,a\,$ that, when multiplied by itself, equals $\,a$
principal nth root
the number with the same sign as $\,a\,$ that when raised to the nth power equals $\,a$
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The Complex Exponential Function Examples 1
# The Complex Exponential Function Examples 1
Recall from The Complex Exponential Function page that if $z = x + yi \in \mathbb{C}$ then we defined the complex exponential function to be:
(1)
\begin{align} \quad e^z = e^x ( \cos y + i \sin y) \end{align}
We will now look at some example problems regarding this function.
## Example 1
Write $e^{3 + \pi i}$ in the form $a + bi$.
We have that:
(2)
\begin{align} \quad e^{3 + \pi i} = e^3 (\cos \pi + i \sin \pi) = e^3 (-1) + e^3 (0)i = -e^3 + 0i \end{align}
## Example 2
Find all values $z \in \mathbb{C}$ for which $\overline{e^{iz}} = e^{i\overline{z}}$.
Let $z = x + yi$. Then $iz = -y + xi$, and so:
(3)
\begin{align} \quad \overline{e^{iz}} = \overline{e^{-y + xi}} = \overline{e^{-y} (\cos x + i \sin x)} = \overline{ e^{-y} \cos x + i e^{-y} \sin x} = e^{-y} \cos x + i (-e^{-y} \sin x) \quad (*) \end{align}
And we also have that:
(4)
\begin{align} \quad e^{i\overline{z}} = e^{i(x - yi)} = e^{y + xi} = e^y (\cos x + i \sin y) = e^y \cos x + i e^y \sin x \quad (**) \end{align}
Set $(*)$ equal to $(**)$. Then we must simultaneously solve the following:
(5)
\begin{align} \quad (1) & \: e^{-y} \cos x = e^y \cos x \\ \quad (2) & \: -e^{-y} \sin x = e^y \sin x \end{align}
First assume that $\cos x = 0$. Then $x = \frac{\pi}{2} + k\pi$ for some $k \in \mathbb{Z}$. If $x = ..., -\frac{3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2}, ...$ for some $m \in \mathbb{Z}$ then $\sin x = 1$, and so $-e^{-y} = e^y$ which has no solutions. If $x = ..., -\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}, ...$ then $\sin x = -1$ and so $e^{-y} = -e^y$ which has no solutions.
So we may assume that $\cos x \neq 0$. Dividing equation (1) by $\cos x$ on both sides gives us that $e^{-y} = e^y$ which implies that $y = 0$. Substituting this into equation (2) gives us that $- \sin x = \sin x$. So $2 \sin x = 0$ and $\sin x = 0$ which happens when $x = k\pi$ for some $k \in \mathbb{Z}$.
Therefore the values of $z = x + yi \in \mathbb{C}$ for which $\overline{e^{iz}} = e^{i\overline{z}}$ are:
(6)
\begin{align} \quad z = k\pi + 0i \:, \: k \in \mathbb{Z} \end{align}
## Example 3
Write $e^{z^2}$ in the form $a + bi$. What is $\mid e^{z^2} \mid$? What is $\arg (e^{z^2})$?
Let $z = x + yi$. Then:
(7)
\begin{align} \quad z^2 = (x + yi)(x + yi) = (x^2 - y^2) + 2xyi \end{align}
Therefore:
(8)
\begin{align} \quad e^{z^2} = e^{(x^2 - y^2) + 2xyi} = e^{x^2 - y^2} (\cos 2xy + i \sin 2xy) = (e^{x^2 - y^2}\cos 2xy) +i(e^{x^2 - y^2} \cos 2xy) \end{align}
We have that $\mid e^{z^2} \mid = e^{x^2 - y^2}$, and $\arg (e^{z^2}) = 2xy$.
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A+ » VCE » Further Maths U3 & 4 Master Notes » A1 Data Analysis » FM Centre
# FM Centre
## 1.8 Statistical Analysis of Numerical Distributions
### Guide to Analysing a Numerical Distribution
• Begin with some context: what does the data represent?
• Always mention the minimum, centre and maximum.
• Check for outliers and mention if there are any.
• Describe the shape of the distribution.
• If there are outliers, mention the values of the lower and upper fences.
Read More »1.8 Statistical Analysis of Numerical Distributions
## 1.7 Box Plots and the Five Number Summary
### The Box Plot
• The box plot is a graphical tool used to analyse the shape, spread and outliers of a numerical distribution.
• It consists of a box with the bottom drawn at the value of quartile 1 and the top at quartile 3, a line drawn through the box at the median and a line either end of the box drawn to the lower and upper fences.
• If the median line is in the middle of the box, the distribution is approximately symmetric, if it is drawn closer to the bottom of the box, it is positively skewed, if it is drawn closer to the top of the box, it is negatively skewed.
• If the distribution has any outliers, they are represented as dots or crosses at their respective value along the y-axis and placed parallel to box.
Read More »1.7 Box Plots and the Five Number Summary
## 1.6 Describing Numerical Distributions
### Shape
• The shape of a numerical distribution relies on two factors: symmetry and outliers.
• If you can draw a vertical line through some point in the distribution whereby the distribution to the left of the line looks similar to a mirror image of the distribution to the right of it, it is an approximately symmetrical distribution. If this is not the case, the distribution is asymmetric.
Note: in some cases, you may find situations where the distribution has perfect symmetry. In these situations, you can drop the “approximately” term and refer to it simply as symmetrical.
Read More »1.6 Describing Numerical Distributions
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# How to Divide Fractions
Math can be scary, but don't give up. Follow these simple steps, and soon you'll be dividing fractions like a pro.
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### Up next in Basic Math Skills (15 videos)
Do you and arithmetic add up to disaster? Sharpen your basic bath skills with the lessons in this Howcast video series.
#### You Will Need
• A pencil
• A sheet of paper
## Steps
1. Step 1
#### Write the problem
Write out the first fraction, a divided-by sign, and the second fraction. This will help you visualize the steps you need to take to divide the fractions.
2. Step 2
#### Invert the second fraction
Invert the second fraction by flipping it so that the bottom number is at the top. Rewrite the problem with the first fraction multiplied by the second, now inverted, fraction.
3. For example, rewrite one-half divided by three-fourths as one-half multiplied by four-thirds.
4. Step 3
#### Multiply the top two numbers
Multiply the top two numbers in the fractions to get the top number in your answer, which is known as the numerator.
5. Step 4
#### Multiply the bottom two numbers
Multiple the bottom two numbers in the fractions to get the bottom number in the answer, which is known as the denominator.
6. The result in the example is four-sixths.
7. Step 5
Reduce the answer by dividing both the numerator and the denominator by the greatest common factor, which is the highest number by which both can be divided.
8. In the example, divide the numerator and denominator by two, the greatest common factor. The final answer is two-thirds.
9. Step 6
Add an equal sign and your final, reduced answer at the end of the original problem. Keep practicing, and soon you may be solving tough math puzzles without even needing a pencil and paper.
10. The word fraction comes from the Latin frangere, which means, "to break."
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# Fundamental Identities
## Key Questions
• "The fundamental trigonometric identities" are the basic identities:
•The reciprocal identities
•The pythagorean identities
•The quotient identities
They are all shown in the following image:
When it comes down to simplifying with these identities, we must use combinations of these identities to reduce a much more complex expression to its simplest form.
Here are a few examples I have prepared:
a) Simplify: $\tan \frac{x}{\csc} x \times \sec x$
Apply the quotient identity $\tan \theta = \sin \frac{\theta}{\cos} \theta$ and the reciprocal identities $\csc \theta = \frac{1}{\sin} \theta$ and $\sec \theta = \frac{1}{\cos} \theta$.
$= \frac{\sin \frac{x}{\cos} x}{\frac{1}{\sin} x} \times \frac{1}{\cos} x$
$= \sin \frac{x}{\cos} x \times \sin \frac{x}{1} \times \frac{1}{\cos} x$
$= {\sin}^{2} \frac{x}{\cos} ^ 2 x$
Reapplying the quotient identity, in reverse form:
$= {\tan}^{2} x$
b) Simplify: $\frac{\csc \beta - \sin \beta}{\csc} \beta$
Apply the reciprocal identity $\csc \beta = \frac{1}{\sin} \beta$:
$= \frac{\frac{1}{\sin} \beta - \sin \beta}{\frac{1}{\sin} \beta}$
Put the denominator on a common denominator:
$= \frac{\frac{1}{\sin} \beta - {\sin}^{2} \frac{\beta}{\sin} \beta}{\frac{1}{\sin} \beta}$
Rearrange the pythagorean identity ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$, solving for ${\cos}^{2} \theta$:
${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$
$= \frac{{\cos}^{2} \frac{\beta}{\sin} \beta}{\frac{1}{\sin} \beta}$
$= {\cos}^{2} \frac{\beta}{\sin} \beta \times \sin \frac{\beta}{1}$
$= {\cos}^{2} \beta$
c) Simplify: $\sin \frac{x}{\cos} x + \cos \frac{x}{1 + \sin x}$:
Once again, put on a common denominator:
$= \frac{\sin x \left(1 + \sin x\right)}{\cos x \left(1 + \sin x\right)} + \frac{\cos x \left(\cos x\right)}{\cos x \left(1 + \sin x\right)}$
Multiply out:
$= \frac{\sin x + {\sin}^{2} x + {\cos}^{2} x}{\cos x \left(1 + \sin x\right)}$
Applying the pythagorean identity ${\cos}^{2} x + {\sin}^{2} x = 1$:
$= \frac{\sin x + 1}{\cos x \left(1 + \sin x\right)}$
Cancelling out the $\sin x + 1$ since it appears both in the numerator and in the denominator.
=cancel(sinx + 1)/(cosx(cancel(sinx + 1))
$= \frac{1}{\cos} x$
Applying the reciprocal identity $\frac{1}{\cos} \theta = \sec \theta$
$= \sec x$
Finally, on a last note, I know that here in Canada, British Columbia more specifically, these identities are given on a formula sheet, but I don't know what it's like elsewhere. In any event, many students, me included, memorize these identities because they're that important to mathematics. I would highly recommend memorization.
Practice exercises:
Simplify the following expressions:
a) $\cos \alpha + \tan \alpha \sin \alpha$
b) $\csc \frac{x}{\sin} x - \cot \frac{x}{\tan} x$
c) ${\sin}^{4} \theta - {\cos}^{4} \theta$
d) $\frac{\tan \beta + \cot \beta}{\csc} ^ 2 \beta$
Hopefully this helps, and good luck!
• Even & Odd Functions
A function $f \left(x\right)$ is said to be $\left\{\begin{matrix}\text{even if "f(-x)=f(x) \\ "odd if } f \left(- x\right) = - f \left(x\right)\end{matrix}\right.$
Note that the graph of an even function is symmetric about the $y$-axis, and the graph of an odd function is symmetric about the origin.
Examples
$f \left(x\right) = {x}^{4} + 3 {x}^{2} + 5$ is an even function since
$f \left(- x\right) = {\left(- x\right)}^{4} + {\left(- x\right)}^{2} + 5 = {x}^{4} + 3 {x}^{2} + 5 = f \left(x\right)$
$g \left(x\right) = {x}^{5} - {x}^{3} + 2 x$ is an odd function since
$g \left(- x\right) = {\left(- x\right)}^{5} - {\left(- x\right)}^{3} + 2 \left(- x\right) = - {x}^{5} + {x}^{3} - 2 x = - f \left(x\right)$
I hope that this was helpful.
• Divide the fundamental identity ${\sin}^{2} x + {\cos}^{2} x = 1$ by ${\sin}^{2} x$ or ${\cos}^{2} x$ to derive the other two:
${\sin}^{2} \frac{x}{\sin} ^ 2 x + {\cos}^{2} \frac{x}{\sin} ^ 2 x = \frac{1}{\sin} ^ 2 x$
$1 + {\cot}^{2} x = {\csc}^{2} x$
${\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$
${\tan}^{2} x + 1 = {\sec}^{2} x$
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# Multiplication Tips and Tricks
## Some Tips and Tricks
Everyone thinks differently, so just ignore any tricks that don't make sense to you.
First of all: Memory is your best friend!
With the multiplication table in your memory you simply know that 3×5=15, 6×8=48 etc.
Memory is fast, too.
Math Trainer - Multiplication is designed to
## The Best Trick
Every multiplication has a twin, which may be easier to remember.
For example if you forget 8×2, you might remember 2×8=16. This way, you only have to remember half the table.
## Tricks by Number
2
add the number to itself (in other words, double it)
Example 2×9 = 9+9 = 18
4
double, then double again
Example 4×9: double 9 is 18, double 18 is 36
5
Cut in half, then times 10
Example: 5x6: Cut 6 in half to get 3, then times 10 for 30
Or times 10 then cut in half
Example: 5x9: 9 times 10 is 90, then cut in half for 45
Also the last digit goes 5, 0, 5, 0, ... like this: 5, 10, 15, 20, ...
6
when you multiply 6 by an even number, they both end in the same digit.
Examples: 6×2=12, 6×4=24, 6×6=36, etc
7×8
Think "5,6,7,8": 56=7×8
8
Double, double, double!
Example: 8×6: double 6 is 12, double 12 is 24, double 24 is 48
9
is 10× the number minus the number.
Example: 9×6 = 10×66 = 60−6 = 54
the ones digit goes 9, 8, 7, 6, ...: 9, 18, 27, 36, 45, ...
the tens digit goes 0, 1, 2, 3, ...: 9, 18, 27, 36, 45, ...
subtract one to get the tens digit, and the tens and ones digit together make 9
Example: 9×5: tens digit is 4, 4 and 5 make 9, so 45
Example: 9×8: tens digit is 7, 7 and 2 make 9, so 72
your hands can help! Example: to multiply 9 by 8, hold your 8th finger down, and count "7" and "2", the answer is 72
10
put a zero after it
Example: 10×2 = 20
11
up to 11x9: just repeat the digit. Example: 11x4 = 44
for 11×10 to 11×18: write the sum of the digits between the digits
Example: 11×15 = 1(1+5)5 = 165
Note: this works for any two-digit number, but when the sum of the digits is more than 9, we need to"carry the one". Example: 11×75 = 7(7+5)5 = 7(12)5 = 825.
12
is 10× plus 2×
Example: 12×4 = 40+8 = 48
15
multiply by 10, then add half again
Example: 15×4 = 40+20 = 60
Example: 15×9 = 90+45 = 135
20
multiply by 10, then double
Example: 20×4 = 40+40 = 80
Example: 20×7 = 70+70 = 140
## Remembering Squares Can Help
This may not work for you, but it worked for me. I like remembering the squares (where you multiply a number by itself):
1×1 =1
2×2 =4
3×3 =9
4×4 =16
5×5 =25
6×6 =36
7×7 =49
8×8 =64
9×9 =81
10×10 =100
11×11 =121
12×12 =144
And this gives us one more trick. When the numbers we are multiplying are separated by 2 (example 7 and 5), then multiply the number in the middle by itself and subtract one. See this:
5×5 = 25 is just one bigger than 6×4 = 24 6×6 = 36 is just one bigger than 7×5 = 35 7×7 = 49 is just one bigger than 8×6 = 48 8×8 = 64 is just one bigger than 9×7 = 63 etc ...
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## Area Formulas
Now that students know the formula for the area of a triangle, it's time to build on this knowledge to develop a formula for the area of a parallelogram.
Materials: Overhead transparency, an activity sheet with six parallelograms on it with base and height measures given for four of the parallelograms, rulers for students to use
Preparation: Draw a parallelogram on an overhead transparency.
• Ask: Does anyone know what this figure is called?
Students should say that it is a parallelogram. If they don't, tell them what it is.
• Say: Let's list some of the properties of this figure on the board. Who can tell me something that is true about all parallelograms? Students will probably come up with several ideas such as the opposites sides are parallel and congruent, the opposite angles are congruent, all rectangles are parallelograms, a rhombus is a parallelogram, the sum of the angles of a parallelogram is 360 degrees, etc.
Draw one of the diagonals of the parallelogram as shown below.
• Ask: What can you tell me about the two triangles – triangle MNP and triangle OPN?
Students should say that the two triangles are congruent. If they don't, suggest that they are. If you need to, cut out a parallelogram then cut the parallelogram in half and show that the two triangles are congruent.
• Say: That's right. They are congruent. Now we know how to find the area of a triangle and since this parallelogram is made up of two congruent triangles, we could find the area of one of the triangles and double it.
• Ask: How would I find the area of triangle MNP?
Students should say you need to draw a line segment to show the height of the triangle. So, draw the line segment PQ in triangle MNP as shown below.
• Ask: What do I do now?
Students will say you need to find the measure of line segments MN and PQ and substitute them into the formula A = 1/2 (b x h). So measure the sides and find the area.
• Say: Now that I know the area of the triangle, how does that help me find the area of the parallelogram?
Students will say you need to double that value to find the area of the parallelogram.
• Say: That's right, so the formula for the area of a parallelogram is twice the area for a triangle, A = 1/2 (b x h) + A = 1/2 (b x h), or A = b x h.
Do another problem like this in which the measure for the height 8 feet and the base is 12 feet. Draw the picture of that parallelogram on the board. Have them find the area of the figure at their desks. Have someone come to the board to do it for the class to see.
Pass out an activity sheet for students to work on individually or in pairs.
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