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# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
For the homogeneous differential equations, we use the substitution method and we reduce the equation to the variable separable. Having an exercise in which you have to solve the differential equation, you firstly have to figure out what kind of differential equation is the equation, so you know what method it's better to use.
Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that:
A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together.
f to fn are the functions of x.
In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral.
## Solving differential equations video lesson
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RD Sharma Solutions Class 12 Area Bounded Regions Exercise 21.4
RD Sharma Solutions for Class 12 Maths Exercise 21.4 Chapter 21 Areas of Bounded Regions is the most preferred study material due to its unique description of the concepts. In this RD Sharma Solutions for Class 12 Maths Chapter 21, a distinctive attempt is made to build an understanding of the problems. Pursuing this chapter would ensure that you develop a piece of in-depth knowledge about the steps and methods of solving problems.
Download PDF of RD Sharma Solutions for Class 12 Maths Chapter 21 Exercise 4
Access RD Sharma Solutions for Class 12 Maths Chapter 21 Exercise 4
EXERCISE 21.4
Question. 1
Solution:
From the question it is given that, parabola x = 4y – y2 and the line x = 2y – 3,
As shown in the figure,
x1 = 4y – y2
x2 = 2y – 3
So,
2y – 3 = 4y – y2
y2 + 2y – 4y – 3 = 0
y2 – 2y – 3 = 0
y2 – 3y + y – 3 = 0
y(y – 3) + 1(y – 3) = 0
(y – 3) (y + 1) = 0
y = -1, 3
Now, we have to find the area of the bounded region,
Applying limits, we get,
= [- (33/3) + 2(32)/2 + 3(3)] – [- ((-1)3/3) + 2(-12)/2 + 3(-1)]
= [- 32 + 32 + 9] – [(1/3) + 1 – 3]
= [9] – [(1/3) – 1 + 3]
= 9 – (1/3) + 2
= 11 – (1/3)
= (33 – 1)/3
= 32/3 square units
Therefore, the required area is 32/3 square units.
Question. 2
Solution:
From the question it is given that, parabola x = 8 + 2y – y2 and the line y = – 1, y = 3
As shown in the figure,
Applying limits, we get,
= [8(3) + (32) – (3)3/3] – [8(-1) + (-12) – (-1)3/3]
= [24 + 9 – 9] – [-8 + 1 + (1/3)]
= [24] – [-7 + 1/3]
= 24 + 7 – (1/3)
= 31 – (1/3)
= (93 – 1)/3
= 92/3 square units
Therefore, the required area is 92/3 square units.
Question. 3
Solution:
From the question it is given that, parabola y2 = 4x and the line y = 2x – 4,
As shown in the figure,
So,
Now, we have to find the points of intersection,
2x – 4 = √(4x)
Squaring on both side,
(2x – 4)2 = (√(4x))2
4x2 + 16 – 16x = 4x
4x2 + 16 – 16x – 4x = 0
4x2 + 16 – 20x = 0
Dividing both side by 4 we get,
x2 – 5x + 4 = 0
x2 – 4x – x + 4 = 0
x(x – 4) – 1(x – 4) = 0
(x – 4) (x – 1) = 0
x = 4, 1
Applying limits, we get,
= [(42/4) + 2(4) – (43/12)] – [((-22)/4) + 2(-2) – ((-2)3/12)]
= [4 + 8 – (64/12)] – [1 – 4 + (8/12)]
= [12 – (16/3)] – [-3 + (2/3)]
= 12 – (16/3) + 3 – (2/3)
= 15 – 18/3
= 15 – 6
= 9 square units
Therefore, the required area is 9 square units.
Question. 4
Solution:
From the question it is given that, parabola y2 = 2x and the line x – y = 4,
As shown in the figure,
y2 = 2x … [equation (i)]
x = y + 4 … [equation (ii)]
Now, we have to find the points of intersection,
So,
y2 = 2(y + 4)
y2= 2y + 8
Transposing we get,
y2 – 2y – 8 = 0
y2 – 4y + 2y – 8 = 0
y (y – 4) + 2(y – 4) = 0
(y – 4) (y + 2) = 0
y = 4, -2
Applying limits, we get,
= 4(4 – (-2)) + ½ (42 – (-2)2) – (1/6) (43 + 23)
= 4(4 + 2) + ½ (16 – 4) – (1/6) (64 + 8)
= 4(6) + ½ (12) – 1/6 (72)
= 24 + 6 – 12
= 30 – 12
= 18 square units
Therefore, the required area is 18 square units.
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Simultaneous equations can be thought of as being two equations in two unknowns, say x and y. Note that the word simultaneous means ‘at the same time’. It follows that for the values of x and y found both equations must be true at the same time. Sometimes it is easy to inspect the equations and guess the answers. However, when one of the equations is quadratic this becomes less likely. The answers could be surds, in which case, this is very difficult to guess.
Note that a question may ask you to solve simultaneous equations explicitly. In others, it will be implied and you must deduce that it is simultaneous equations to solve. For example, you could be asked to find out which points two curves have in common. See Example 2 below.
## Methods for solving Simultaneous Equations
There are three methods for solving simultaneous equations:
1. Elimination – this is where you multiply both equations through by different coefficient in order to eliminate one of the unknowns. This page will focus on substitution since it works for more complicated simultaneous equations. For example, when one of the equations is a quadratic. Click here to see an example using elimination.
2. Substitution – one of the equations can be quadratic, in which case, substitution is the method to use. You will need to know how to solve quadratics. By making x or y the subject of one of the equations, it can be substituted into the other. See the Worked Example and Example 1 below.
3. Graphical method – the solution of simultaneous equations can be interpreted as the intersection of their graphs. This plot shows the graphs of $y=2x-3$ in red and $4x+5y=6$ in blue. Their intersection lies on the x-axis and has coordinates (1.5,0). This is the solution when solving simultaneously. Also see Example 2 below.
## Simultaneous Equations Worked Example
Solve the simultaneous equations
$x^2+y^2=10$
and
$x+2y=5$.
This example requires solution via substitution, i.e. make either x or y the subject of one equation and insert it into the other. The obvious choice would be to make x the subject of the second equation – it is the quickest, least complicated choice. The second equation tells us that $x=5-2y$. We can insert this into the first equation: $(5-2y)^2+y^2=10$. By multiplying out the brackets and simplifying we see that this is a quadratic equation in y:
$(5-2y)^2+y^2=10$
Write out the brackets: $(5-2y)(5-2y)+y^2=10$
Expand the brackets: $25-10y-10y+4y^2+y^2=10$
Simplify: $5y^2-20y+15=0$
Divide both by sides by 5: $y^2-4y+3=0$
Factorise: $(y-3)(y-1)=0$
This tells us that y has to be either 3 or 1. If $y=3$, then $x=5-2\times 3=-1$ (from the second equation rearranged) and if $y=1$ then $x=5-2\times 1=3$.
We obtain the solutions $(x_1,y_1)=(-1,3)$ and $(x_2,y_2)=(3,1)$.
### Example 1
Solve the simultaneous equations:
$y=x-4$
$2x^2-xy=8$
### Example 2
Sketch the graphs of $x^2+y^2=10$ and $x+2y=5$ on the same plot. Determine the coordinates of the intersection points.
Click here to find Questions by Topic all scroll down to all past SIMULTANEOUS EQUATIONS questions to practice some more questions.
Are you ready to test your Pure Maths knowledge? If so, visit our Practice Papers page and take StudyWell’s own Pure Maths tests. Alternatively, try the
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APPLICATIONS OF ANTIDERIVATIVES; DIFFERENTIAL EQUATIONS - Antidifferentiation - Calculus AB and Calculus BC
CHAPTER 5 Antidifferentiation
E. APPLICATIONS OF ANTIDERIVATIVES; DIFFERENTIAL EQUATIONS
The following examples show how we use given conditions to determine constants of integration.
EXAMPLE 48
Find f (x) if f (x) = 3x2 and f (1) = 6.
SOLUTION:
Since f (1) = 6, 13 + C must equal 6; so C must equal 6 − 1 or 5, and f (x) = x3 + 5.
EXAMPLE 49
Find a curve whose slope at each point (x, y) equals the reciprocal of the x-value if the curve contains the point (e, −3).
SOLUTION: We are given that and that y = −3 when x = e. This equation is also solved by integration. Since
Thus, y = ln x + C. We now use the given condition, by substituting the point (e, −3), to determine C. Since −3 = ln e + C, we have −3 = 1 + C, and C = −4. Then, the solution of the given equation subject to the given condition is
y = ln x − 4.
DIFFERENTIAL EQUATIONS: MOTION PROBLEMS.
An equation involving a derivative is called a differential equation. In Examples 48 and 49, we solved two simple differential equations. In each one we were given the derivative of a function and the value of the function at a particular point. The problem of finding the function is called aninitial-value problem and the given condition is called the initial condition.
In Examples 50 and 51, we use the velocity (or the acceleration) of a particle moving on a line to find the position of the particle. Note especially how the initial conditions are used to evaluate constants of integration.
EXAMPLE 50
The velocity of a particle moving along a line is given by v(t) = 4t3 − 3t2 at time t. If the particle is initially at x = 3 on the line, find its position when t = 2.
SOLUTION: Since
Since x(0) = 04 − 03 + C = 3, we see that C = 3, and that the position function is x(t) = t4 t3 + 3. When t = 2, we see that
x(2) = 24 − 23 + 3 = 16 − 8 + 3 = 11.
EXAMPLE 51
Suppose that a(t), the acceleration of a particle at time t, is given by a(t) = 4t − 3, that v(1) = 6, and that f (2) = 5, where f (t) is the position function.
(a) Find v(t) and f (t).
(b) Find the position of the particle when t = 1.
SOLUTIONS:
Using v(1) = 6, we get 6 = 2(1)2 − 3(1) + C1, and C1 = 7, from which it follows that v(t) = 2t2 − 3t + 7. Since
Using f (2) = 5, we get + 14 + C2, so Thus,
For more examples of motion along a line, see Chapter 8, Further Applications of Integration, and Chapter 9, Differential Equations.
Chapter Summary
In this chapter, we have reviewed basic skills for finding indefinite integrals. We’ve looked at the antiderivative formulas for all of the basic functions and reviewed techniques for finding antiderivatives of other functions.
We’ve also reviewed the more advanced techniques of integration by partial fractions and integration by parts, both topics only for the BC Calculus course.
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# Into Math Grade 2 Module 12 Lesson 3 Answer Key Represent and Record Two-Digit Addition
We included HMH Into Math Grade 2 Answer Key PDF Module 12 Lesson 3 Represent and Record Two-Digit Addition to make students experts in learning maths.
## HMH Into Math Grade 2 Module 12 Lesson 3 Answer Key Represent and Record Two-Digit Addition
I Can represent and record two-digit addition with and without regrouping.
How can you represent Brianna’s cat and dog books? How many books about cats or dogs does she have?
Brianna has _________ cat or dog books.
Read the following: Brianna has 12 books about cats. She has 11 books about dogs. How many books about cats or dogs does she have?
Given that,
The total number of books about cats near Brianna is 12
The total number of books about dogs near Brianna is 11
Therefore 12 + 11 = 23
There are 23 books she has.
Build Understanding
Question 1.
Kurt has 57¢. His friend gives him 35¢. How much money does Kurt hove now?
A. How can you use tools to show the two addends for this problem? Draw to show what you did.
Given that
Kurt has money = 57 cents.
Her friend given = 35 cents.
The total money near Kurt = 57 + 35 = 92
Kurt has 92 cents.
B. Are there 10 ones to regroup?
Yes, there are 10 ones to regroup.
Adding 57 + 35 in this case you need to regroup the numbers.
when you add the ones place digits 7 + 5, you get 12 which means 1 ten and 2 ones.
Know to regroup the tens into the tens place and leave the ones. Then 57 + 35 = 92.
C. Regroup 10 ones as 1 ten. Write a 1 in the tens column to show the regrouped ten.
D. How many ones are left after regrouping? Write the number of ones left over in the ones place.
After regrouping the number of ones left over in the one place is 2.
E. How many tens are there in all? Write the number of tens ¡n the tens place.
The number of tens in the tens place is 9.
F. How much money does Kurt have now?
________ ¢
Given that
Kurt has money = 57 cents.
Her friend given = 35 cents.
The total money near Kurt = 57 + 35 = 92
Kurt has 92 cents.
Question 2.
Mateo and his friends make a list of two-digit numbers. He chooses two of the numbers to add.
A. How can you draw quick pictures to help you find the sum of 26 and 46?
B. How can you add the ones? Regroup if you need to. Show your work in the chart.
26 + 46 = 72
Adding 26 + 46 in this case you need to regroup the numbers.
when you add the ones place digits 6 + 6, you get 12 which means 1 ten and 2 ones.
Know to regroup the tens into the tens place and leave the ones. Then 26 + 46 = 72.
C. How can you odd the tens? Show your work in the chart.
D. What is the sum?
26 + 46 = 72
Adding 26 with 46 then we get 72.
Turn and Talk Are there two numbers from that Mateo could add without regrouping?
52, 11, 25 and 74
Any two numbers can add without regrouping. Because the addition of one’s place digit is less than the 10.
Step It Out
Question 1.
Add 47 and 37.
A. Find How many ones in all. Regroup if you need to. Write a I in the tens column to show the regrouped ten.
Adding 47 + 37 in this case you need to regroup the numbers.
when you add the ones place digits 7 + 7, you get 14 which means 1 ten and 4 ones.
Know to regroup the tens into the tens place and leave the ones. Then 47 + 37 = 84.
B. Write the number of ones left over in the ones place.
Number of ones left over in the ones place is 4.
C. Write the number of tens in the tens place.
Number of tens in the tens place is 8.
D. Write the sum.
47 + 37 = 84
Adding 47 with 37 then we get 84.
Check Understanding
Question 1.
There are 65 apples on a tree. There are 28 apples on another tree. How many apples are on the trees? Draw to show the addition.
________ apples
Given that,
The total number of apples on the tree = 65
The total number of apple on the another tree = 28
The total number of apples = 65 + 28 = 93.
Question 2.
Attend to Precision Mrs. Meyers plants 34 flowers. Mrs. Owens plants 42 flowers. How many flowers do they plant? Draw to show the addition.
_________ flowers
Given that,
Mrs. Meyers plants 34 flowers.
Mrs. Owens plants 42 flowers.
The total number of flowers = 34 + 42 = 76.
Question 3.
Reason Did you need to regroup 10 ones as 1 ten in Problem 2? Explain.
No need to regroup 10 ones as 1 ten. Because the addition of one’s place digits is less than 10.
So, there is no need to regroup.
Question 4.
Open Ended Rewrite Problem 2 with different numbers so that you need to regroup when you odd. Then solve.
Mrs. Meyers plants 36 flowers. Mrs. Owens plants 45 flowers. How many flowers do they plant? Draw to show the addition.
36 + 45 = 81
Adding 36 + 45 in this case you need to regroup the numbers.
when you add the ones place digits 6 + 5, you get 11 which means 1 ten and 1 ones.
Know to regroup the tens into the tens place and leave the ones.
Question 5.
Use Structure There are 25 big dogs and 19 small dogs at the dog park. How many dogs are at the park?
_________ dogs
Given that,
The total number of big dogs = 25.
The total number of small dogs = 19.
The total number of dogs = 25 + 19 = 44.
Question 6.
Add 16 and 23.
16 + 23 = 39
There is no need of regrouping.
Question 7.
Add 44 + 49
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# 6.06 Sales tax and tip
Lesson
Two ways that percentages are commonly used in the U.S. are when calculating the amount of sales tax we will need to pay for purchasing an item and when calculating the proper tip to leave the waitstaff at a restaurant. Interestingly, we will find that in different parts of the United States the amount of tax that we will pay for the purchase of goods or services can vary. In addition, while tipping for good service is a common practice in the U.S., there are other countries that do not engage in the practice of tipping at all.
### Tips
This image below, from mint.com, displays the tipping customs of many countries around the world. You can see that tips vary from $0$0 to $20%$20%
Tipping is an amount of money left for the staff, in addition to paying the bill, as a sign that we appreciate good service. Tips are common in the service industry, but in other sectors like government receiving a tip can be considered illegal. So, it is important to know the customary amount to tip for different services and who we should not offer a tip to.
#### Worked examples
##### Question 1
David is paying for a meal with lots of friends. They received great service, so he is giving a $20%$20% tip. The meal came to $\$182.30$$182.30. How much will he leave as a tip? Think: I need to work out 20%20% of the total meal charge. 20%20% as a fraction is \frac{20}{100}20100. Do: 20%20% of \182.30$$182.30
$20%$20% of $\$182.30$$182.30 == \frac{20}{100}\times\182.3020100×182.30 20%20% is \frac{20}{100}20100 and of in mathematics means multiplication. == \frac{20\times182.30}{100}20×182.30100 == \36.46$$36.46
### Outcomes
#### 7.RP.3
Use proportional relationships to solve multi-step ratio, rate, and percent problems. Examples: simple interest, tax, price increases and discounts, gratuities and commissions, fees, percent increase and decrease, percent error.
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# Painting a Wall
Alignments to Content Standards: 5.NF.B
Nicolas is helping to paint a wall at a park near his house as part of a community service project. He had painted half of the wall yellow when the park director walked by and said,
This wall is supposed to be painted red.
Nicolas immediately started painting over the yellow portion of the wall. By the end of the day, he had repainted $\frac56$ of the yellow portion red.
What fraction of the entire wall is painted red at the end of the day?
## IM Commentary
The purpose of this task is for students to find the answer to a question in context that can be represented by fraction multiplication. This task is appropriate for either instruction or assessment depending on how it is used and where students are in their understanding of fraction multiplication. If used in instruction, it can provide a lead-in to the meaning of fraction multiplication. If used for assessment, it can help teachers see whether students readily see that this is can be solved by multiplying $\frac56\times \frac12$ or not, which can help diagnose their comfort level with the meaning of fraction multiplication.
The teacher might need to emphasize that the task is asking for what portion of the total wall is red, it is not asking what portion of the yellow has been repainted.
## Solutions
Solution: Solution 1
In order to see what fraction of the wall is red we need to find out what $\frac56$ of $\frac12$ is. To do this we can multiply the fractions together like so:
$\frac56 \times \frac12 = \frac{5 \times 1}{6 \times 2} = \frac{5}{12}$
So we can see that $\frac{5}{12}$ of the wall is red.
Solution: Solution 2
The solution can also be represented with pictures. Here we see the wall right before the park director walks by:
And now we can break up the yellow portion into 6 equally sized parts:
Now we can show what the wall looked like at the end of the day by shading 5 out of those 6 parts red.
And finally, we can see that if we had broken up the wall into 12 equally sized pieces from the beginning, that finding the fraction of the wall that is red would be just a matter of counting the number of red pieces and comparing them to the total.
And so, since 5 pieces of the total 12 are red, we can see that $\frac{5}{12}$ of the wall is red at the end of the day.
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# 2003 AMC 10B Problems/Problem 4
The following problem is from both the 2003 AMC 12B #3 and 2003 AMC 10B #4, so both problems redirect to this page.
## Problem
Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $1$ each, begonias $1.50$ each, cannas $2$ each, dahlias $2.50$ each, and Easter lilies $3$ each. What is the least possible cost, in dollars, for her garden?
$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((6,0)--(0,0)--(0,1)--(6,1)); draw((0,1)--(0,6)--(4,6)--(4,1)); draw((4,6)--(11,6)--(11,3)--(4,3)); draw((11,3)--(11,0)--(6,0)--(6,3)); label("1",(0,0.5),W); label("5",(0,3.5),W); label("3",(11,1.5),E); label("3",(11,4.5),E); label("4",(2,6),N); label("7",(7.5,6),N); label("6",(3,0),S); label("5",(8.5,0),S);[/asy]$
$\textbf{(A) } 108 \qquad\textbf{(B) } 115 \qquad\textbf{(C) } 132 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156$
## Solution
The areas of the five regions from greatest to least are $21,20,15,6$ and $4$.
If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4$, which simplifies to $108$. Therefore the answer is $\boxed{\textbf{(A) } 108}$.
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# Limits of a Function: Indeterminate Forms – Calculus
by | Sep 27, 2021 | Math Learning
While studying calculus or other branches of mathematics we may need to find the limits of a function, a sequence, or an expression, and in doing so we stumble on a situation where we cannot determine the limits, in this article we will learn about the different indeterminate forms and how to work around them in order to find the limits we are looking for.
## Indeterminate Forms
We call an indeterminate form, when computing limits the case when we get an expression that we cannot determine the limit. In total there is seven indeterminate forms, here they are:
Here are some examples to illustrate each of these indeterminate cases:
Indeterminate form
Indeterminate form
Indeterminate form
Indeterminate form
Indeterminate form
Indeterminate form
Indeterminate form
## L’Hôpital’s rule and how to solve indeterminate forms
L’Hôpital’s rule is a method used to evaluate limits when we have the case of a quotient of two functions giving us the indeterminate form of the type or .
The L’Hôpital rule states the following:
Theorem: L’Hôpital’s Rule:
To determine the limit of
where is a real number or infinity, and if we have one of the following cases:
Then we calculate the limit of the derivatives of the quotient of and , i.e.,
Examples:
Case of :
Case of :
In this case, after we get the derivatives of the quotient, we still get the indeterminate form of the type so we apply L’Hôpital’s Rule again, and therefore we get:
For other Indeterminate forms, we have to do some transformation on the expression to bring it to one of the two forms that L’Hôpital’s rule solves. Let’s see some examples of how to do that!!!
L’Hôpital’s rule with the form :
Let’s compute
Here we have the indeterminate form , to use L’Hôpital’s rule we re-write the expression as follow:
Now by computing the limit we have the form , therefore we can apply L’Hôpital’s rule and we get:
L’Hôpital’s rule with the form :
Let’s compute
This limit gives us the form , to apply the L’Hôpital’s rule we need to take a few steps as follow:
Let’s be:
By applying the natural logarithm, we get:
And now we compute the limit:
And since we know that:
Therefore, we can write the limit as:
And from what we got before; we can solve the problem as follow:
L’Hôpital’s rule with the form :
Let’s compute
This limit gives us the form , to apply the L’Hôpital’s rule we need to re-write the expression, in this case, all we need to do is combine the two fractions as follow:
Now the limit of the expression gives us the form . Now by applying the L’Hôpital’s rule twice (because we get the indeterminate form after the first time) we get:
L’Hôpital’s rule with the form :
Let’s compute
This limit gives us the form , to avoid it and be able to apply L’Hôpital’s rule we need to re-write the expression as follow:
Let
Then
Using L’Hôpital’s rule we get:
And therefore, we get:
L’Hôpital’s rule with the form :
Let’s compute
This limit gives us the indeterminate form , to use the L’Hôpital’s rule we need to re-write the expression as follow:
Now we calculate the limit of the exponent using L’Hôpital’s rule:
Therefore,
## Limits of a composite function
Theorem:
Let , and represent real numbers or or , and let , , and be functions that verify .
If the limit of the function when tends to is , and the limit of the function when tend to is then the limit of the function when tends to is .
Meaning: if and if then
Example:
Let’s consider the function defined on the domain as
and we want to determine the limit of the function when tends to , i.e.,
We notice that the function is a composite of two functions, precisely is a composite of the functions and in this order (), where
and
Since
And
Therefore
## Limits with comparisons
Theorem 1:
Suppose , , and three functions, and a real number; if we have and and if for big enough we have then .
Example:
Let’s consider the function defined on as
We know that for every from , we have
And therefore, for every in , we have
And since we conclude that
Theorem 2:
Suppose and two functions and a real number; if we have , and if for big enough we have then .
Theorem 3:
Suppose and two functions and a real number; if we have , and if for big enough we have then .
Remarque: these three theorems can be extended to the two cases for the limit when tends to or a real number.
Example:
Let’s consider the function defined on as
We know that for every from , we have , and then for every from , we have
Therefore:
Since
Then
And since
Then
## Conclusion
In this article, we discovered the different indeterminate forms and how to avoid them and calculate the limits using L’Hôpital’s rule, with examples of the various cases. Also, we learned about how to determine the limits of composite function and how to determine limits with comparison. Don’t miss the previous articles about the idea of limits, their properties, and the arithmetic operations on them.
Also, if you want to learn more fun subjects, check the post about Functions and some of their properties, or the one about How to solve polynomial equations of first, second, and third degrees!!!!!
And don’t forget to join us on our Facebook page for any new articles and a lot more!!!!!
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# NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4
These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.4
Question 1.
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$
Write
A = IA
Question 2.
$$\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$
Write
A = IA
Question 3.
$$\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$$
Write
A = IA
Question 4.
$$\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$$
Write
A = IA
Question 5.
$$\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$$
Write
A = IA
Question 6.
$$\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$$
Write
A = IA
Question 7.
$$\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$$
Write
A = IA
Question 8.
$$\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}$$
Write
A = IA
Question 9.
$$\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$$
Write
A = IA
Question 10.
$$\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$$
Write
A = IA
Question 11.
$$\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$$
Write
A = IA
Question 12.
$$\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$$
To use column transformation write A = AI
$$\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$$ = A$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Applying C1 → C1 + 2C2
$$\left[\begin{array}{ll} 0 & -3 \\ 0 & 1 \end{array}\right]$$ = A$$\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]$$
Question 13.
$$\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$$
Write
A = IA
Question 14.
$$\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$$
Solution:
Let $$A=\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$$
Write
A = IA
Question 15.
$$\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right]$$
Solution:
Question 16.
$$\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 2 \end{matrix} \right]$$
Solution:
Question 17.
$$\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right]$$
Solution:
Row transformation
Let $$A=\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right]$$
Question 18.
Choose the correct answer in the following question:
Matrices A and B will be inverse of each other only if
(a) AB = BA
(b) AB = BA = 0
(c) AB = 0, BA = 1
(d) AB = BA = I
Solution:
Choice (d) is correct
i.e., AB = BA = I
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# Cardinal Numbers - Definition with Examples
The Complete K-5 Math Learning Program Built for Your Child
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## Cardinal Numbers
Cardinal numbers are counting numbers. The numbers that we use for counting are called cardinal numbers.
Cardinal numbers tell us “How many?”
For example:
How many dogs are there in all?
To know the total number of dogs, we need to count the dogs given in the image.10804
Therefore, there are 8 dogs in all.
Example: How many kites are there in all?
Count the kites to know the total number of kites.
On counting, we get:
Thus, from the above examples, we see that to know “how many?” of something is there, we need to use cardinal numbers.
From where do cardinal numbers start?
Cardinal numbers or counting numbers start from
How many cardinal numbers are there?
Cardinal numbers can go on and on and on. That means there are infinite counting numbers.
(Imagine, how many numbers you’ll need to count the number of stars in the sky or the number of sand grains in a desert?)
How are cardinal numbers formed?
The digits 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0 are used to form several other cardinal numbers.
For example:
Is zero (0) a cardinal number?
No, zero (0) is not a cardinal number. To know “how many” there should be something. Since 0 means nothing; it is not a cardinal number.
We can write cardinal numbers in numerals as 1, 2, 3, 4, and so on as well as in words like one, two, three, four, and so on.
The chart shows the cardinal numbers in figures as well in words.
Cardinality
The cardinality of a group (set) tells how many objects or terms are there in that set or group.
Example: What is the cardinality of the flowers in the vase?
Here, there are 5 flowers in the vase. Therefore, the cardinality of flowers is 5.
Cardinal numbers start from 1. Fractions and decimals represent a part (less than one) of a whole or a group. Therefore, fractions and decimals are not cardinal numbers.
Fun Facts Cardinal numbers are also called natural numbers. Natural Numbers (Cardinal numbers) along with 0 form a set of whole numbers.
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Mathematics
# Calculus the Basics
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Calculus is basically the use of differentiation and integration on a given polynomial. At the AS level Mathematics in the UK calculus is introduced for the first time and is a rather different, though simple concept for students to grasp.
Differentiation
This is the process of taking each part of any polynomial (equation containing any powers of x) and multiplying it by the power of x then decreasing the power of x by one. For example:
4x^2 -> 2 x (4x^2) -> 2 x 4x -> 8x
Step one: Take the number.
Step two: Multiply by power, in this case 2.
Step Three: Take one from the power, in this case taking it down to 1.
The basic uses of differentiation are quite helpful in many things in mathematics. Given the equation of a line you can use the first derivative (differentiated once) to find the gradient of a curve by substituting in a point on the curve. The first derivative is known as 'dy/dx'.
Differentiation can also be used to find stationary points on graphs, ie. where the gradient is equal to zero. This is done by taking the equation of the graph, finding the first derivative of it and setting that equal to zero, this will find the points on the graph that are equal to zero.
The second derivative (differentiated twice) can be used then to find if this is a maximum point, the stationary point is at the top of the curve; a minimum point, the stationary point is at the bottom of the curve; or a point of inflexion where the graph goes level in the middle of a graph.
Here are some very simple diagrams of what I mean:
Maximum point:
_
/
/
Minimum point:
/
_/
Point of inflexion:
/
__/
/
/
That just about covers the basic uses for differentiation, there are further uses for it which are more advanced.
Integration
This is basically the opposite of differentiation. If you took the second derivative of an equation and integrated it you would get the first derivative. When anything is integrated, there is always an unknown, which is commonly referred to as 'c'. This can be worked out if there is a point given.
To integrate something you firstly take each part seperately then add one to the power of x then divide it al by that power. For example:
8x -> 8x^2 -> (8x^2)/2 -> 4x^2
Step One: Take the part of your equation you want to integrate.
Step Two: Raise the power by one.
Step Three: Divide by the new power.
Step Four: Simplify the outcome.
Integration, like differentiation has many uses. The main use in basic calculus of integration is finding the area under a curve between two points. It can also find the area between two curves, between two points.
This is done by first taking the equation of the line and integrating it. Let us say the equtation of the line is y = 3x^2 + 4x + 2, not too difficult. When this equation is fully integrated it comes out as: x^3 + 2x^2 + 2x + c.
Right, though I said earlier that all integrations bring out an unknown, 'c', this is different. Though the integration does make an unknown 'c' it is not needed for the equation. Say in this we want to find the area between x points 1 and 3. We need to substitute in 3 and 1 into the equation:
3^3 + 2(3^2) + 2(3) = 27 + 18 + 6 = 51
1^3 + 2(1^2) + 2(1) = 1 + 2 + 1 = 4
After this we substitute the lesser x value, in this case 1, from the higher value, in this case 3. This substitution would have cancelled out the 'c' in the equation therfore it was not necessary to work it out.
51 - 4 = 47
This is the area under the curve between points 1 and three and above the x axis.
Calculus can take you far in maths and it is a handy basic tool to know about. If you are still in secondary education, getting the grasp of this early if you intend to go on into further education will give you a great advantage. If this didn't help then look it up elsewhere. A useful tool for anyone.
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## How is the graph of Tangent different from sine and cosine?
The sine, cosine and tangent functions are said to be periodic. This means that they repeat themselves in the horizontal direction after a certain interval called a period. The sine and cosine functions have a period of 2π radians and the tangent function has a period of π radians.
## How do you graph tangent?
How to Graph a Tangent Function
1. Find the vertical asymptotes so you can find the domain. These steps use x instead of theta because the graph is on the x–y plane.
2. Determine values for the range.
3. Calculate the graph’s x-intercepts.
4. Figure out what’s happening to the graph between the intercepts and the asymptotes.
What is tangent in graph?
In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that “just touches” the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve. The word “tangent” comes from the Latin tangere, “to touch”.
How do you tell if a graph is cosine or sine?
The graph of the cosine is the darker curve; note how it’s shifted to the left of the sine curve. The graphs of y = sin x and y = cos x on the same axes. The graphs of the sine and cosine functions illustrate a property that exists for several pairings of the different trig functions.
### How do you go from sin to csc?
The secant of x is 1 divided by the cosine of x: sec x = 1 cos x , and the cosecant of x is defined to be 1 divided by the sine of x: csc x = 1 sin x .
### What is the difference between sine and cosine?
Key Difference: Sine and cosine waves are signal waveforms which are identical to each other. The main difference between the two is that cosine wave leads the sine wave by an amount of 90 degrees. A sine wave depicts a reoccurring change or motion.
What is the tangent of a graph?
A tangent of a curve is a line that touches the curve at one point. It has the same slope as the curve at that point. A vertical tangent touches the curve at a point where the gradient (slope) of the curve is infinite and undefined. On a graph, it runs parallel to the y-axis.
How do you calculate sine?
The trigonometric function sine, like the cosine and the tangent, is based on a right-angled triangle. In mathematics, you can find the sine of an angle by dividing the length of the side opposite the angle by the length of the hypotenuse.
## What is the equation for a sine graph?
The general equation of a sine graph is y = A sin(B(x – D)) + C. The general equation of a cosine graph is y = A cos(B(x – D)) + C. Example: Given a transformed graph of sine or cosine, determine a possible equation.
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tgt
## Saturday, 2 August 2014
### CHAPTER 6- Worked Out Examples
Example: 1
If ${x^2} + {y^2} = t + \dfrac{1}{t}\,\,$ and $\,{x^4} + {y^4} = {t^2} + \dfrac{1}{{{t^2}}}$, then prove that $\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$
Solution: 1
We first try to use the two given relations to get rid of the parameter $t$, so that we obtain a (implicit) relation between $x$ and $y$.
${x^2} + {y^2} = t + \dfrac{1}{t}$
Squaring, we get
${x^4} + {y^4} + 2{x^2}{y^2} = {t^2} + \dfrac{1}{{{t^2}}} + 2$ $\ldots(i)$
Using the second relation in ($i$), we get
$2{x^2}{y^2} = 2$ $\Rightarrow \,\, {y^2} = \dfrac{1}{{{x^2}}}$
Differentiating both sides w.r.t $x$, we get
$2y\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{{x^3}}}$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$
Example: 2
If ${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$, then prove that
$\dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$
Solution: 2
The final relation that we need to obtain is independent of $\sin x$ and $\cos x$; this gives us a hint that using the given relation, we must first get rid of $\sin x$and $\cos x$:
${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$ $= \dfrac{1}{2}\left\{ {{a^2}\left( {2{{\cos }^2}x} \right) + {b^2}\left( {2{{\sin }^2}x} \right)} \right\}$ $= \dfrac{1}{2}\left\{ {{a^2}\left( {1 + \cos 2x} \right) + {b^2}\left( {1 - \cos 2x} \right)} \right\}$ $= \dfrac{1}{2}\left\{ {\left( {{a^2} + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\cos 2x} \right\}$ $\Rightarrow \, 2{y^2} - \left( {{a^2} + {b^2}} \right) = \left( {{a^2} - {b^2}} \right)\cos 2x$ $\ldots(i)$
Differentiating both sides of ($i$) w.r.t $x$, we get
$4y\dfrac{{dy}}{{dx}} = - 2\left( {{a^2} - {b^2}} \right)\sin 2x$ $\Rightarrow \,\, - 2y\dfrac{{dy}}{{dx}} = \left( {{a^2} - {b^2}} \right)\sin 2x$ $\ldots(ii)$
We see now that squaring ($i$) and ($ii$) and adding them will lead to an expression independent of the trig. terms:
${\left( {2{y^2} - \left( {{a^2} + {b^2}} \right)} \right)^2} + 4{y^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {{a^2} - {b^2}} \right)^2}$
A slight rearrangement gives:
${\left( {\dfrac{{dy}}{{dx}}} \right)^2} + {y^2} - \left( {{a^2} + {b^2}} \right) = - \dfrac{{{a^2}{b^2}}}{{{y^2}}}$ $\ldots(iii)$
Differentiating both sides of ($iii$) w.r.t $x$:
$2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + 2y\dfrac{{dy}}{{dx}} = \dfrac{{2{a^2}{b^2}}}{{{y^3}}}\dfrac{{dy}}{{dx}}$ $\Rightarrow \,\, \dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$
Example:3
If the derivatives of $f(x)$ and $g(x)$ are known, find the derivative of $y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$
Solution: 3
We cannot directly differentiate the given relation since no rule tells us how to differentiate a term ${p^q}$ where both $p$ and $q$ are variables.
What we can instead do is take the logarithm of both sides of the given relation:
$y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$ $\Rightarrow \,\, \ln y = g\left( x \right)\ln \left( {f\left( x \right)} \right)$
Now we differentiate both sides w.r.t $x$:
$\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = g\left( x \right) \cdot \dfrac{1}{{f\left( x \right)}} \cdot f'\left( x \right) + \ln \left( {f\left( x \right)} \right) \cdot g'\left( x \right)$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right.$ $= {\left( {f\left( x \right)} \right)^{g\left( x \right)}}\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right\}$
As a simple example, suppose we have to differentiate $y = {x^x}$:
$\ln y = x\ln x$ $\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \cdot \dfrac{1}{x} + \ln x \cdot 1$ $= 1 + \ln x$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left( {1 + \ln x} \right)$ $= {x^x}\left( {1 + \ln x} \right)$
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Courses
RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Class 10 : RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
The document RD Sharma Solutions (Part 1): Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
Exercise 3.1
Q.1. Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically. Sol: The pair of equations formed is:
Solution.
The pair of equations formed is:
i.e., x - 2y = 0 ....(1)
3x + 4y = 20 ....(2)
Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in Table
Recall from Class IX that there are infinitely many solutions of each linear equation. So each of you choose any two values, which may not be the ones we have chosen. Can you guess why we have chosen x =O in the first equation and in the second equation? When one of the variables is zero, the equation reduces to a linear equation is one variable, which can be solved easily. For instance, putting x =O in Equation (2), we get 4y = 20 i.e.,
y = 5. Similarly, putting y =O in Equation (2), we get 3x = 20 ..,But asis not an integer, it will not be easy to plot exactly on the graph paper. So, we choose y = 2 which gives x = 4, an integral value.
Plot the points A (O,O) , B (2,1) and P (O,5) , Q (412) , corresponding to the draw the lines AB and PQ, representing the equations x - 2 y = O and 3x + 4y= 20, as shown in figure
In fig., observe that the two lines representing the two equations are intersecting at the point (4,2),
Q.2. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Is not this interesting? Represent this situation algebraically and graphically.
Sol: Let the present age of Aftab and his daughter be x and y respectively. Seven years ago.
Age of Ahab = x - 7
Age of his daughter y - 7
According to the given condition.
(x - 7) = 7(y - 7)
⇒ x - 7 = 7y - 49
⇒ x - 7y = -42
Three years hence
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,
(x + 3) = 3 (y + 3)
⇒ x+3 = 3y +9
⇒ x - 3y = 6
Thus, the given condition can be algebraically represented as
x - 7y = - 42
x - 3y = 6
x - 7y = - 42 ⇒ x = -42 + 7y
Three solution of this equation can be written in a table as follows:
x - 3y = 6 ⇒ x = 6+3y
Three solution of this equation can be written in a table as follows:
The graphical representation is as follows:
Concept insight In order to represent a given situation mathematically, first see what we need to find out in the problem. Here. Aftab and his daughters present age needs to be found so, so the ages will be represented by variables z and y. The problem talks about their ages seven years ago and three years from now. Here, the words ’seven years ago’ means we have to subtract 7 from their present ages. and ‘three years from now’ or three years hence means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Q.3. The path of a train A is given by the equation 3x + 4y - 12 = 0 and the path of another train B is given by the equation 6x + 8y - 48 = 0. Represent this situation graphically.
Sol:
The paths of two trains are giver by the following pair of linear equations.
3x + 4 y -12 = 0 ...(1)
6x + 8 y - 48 = 0 ... (2)
In order to represent the above pair of linear equations graphically. We need two points on the line representing each equation. That is, we find two solutions of each equation as given below:
We have,
3x + 4 y -12 = 0
Putting y = 0, we get
3x + 4 x 0 - 12 = 0
⇒ 3x = 12
Putting x = 0, we get
3 x 0 + 4 y -12 = 0
⇒ 4y = 12
Thus, two solution of equation 3x + 4y - 12 = 0 are ( 0, 3) and ( 4, 0 )
We have,
6x + 8y -48 = 0
Putting x = 0, we get
6 x 0 + 8 y - 48 = 0
⇒ 8y = 48
⇒ y = 6
Putting y = 0, we get
6x + 8 x 0 = 48 = 0
⇒ 6x = 48
Thus, two solution of equation 6 x + 8y - 48= 0 are ( 0, 6 ) and (8, 0 )
Clearly, two lines intersect at ( -1, 2 )
Hence, x = -1,y = 2 is the solution of the given system of equations.
Q.4. Gloria is walking along the path joining (— 2, 3) and (2, — 2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
Sol:
It is given that Gloria is walking along the path Joining (-2,3) and (2, -2), while Suresh is walking along the path joining (0,5) and (4,0).
We observe that the lines are parallel and they do not intersect anywhere.
Q.5. On comparing the ratios and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide:
(i) 5x- 4y + 8 = 0
7x + 6y - 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x - 3y + 10 = 0
2x - y + 9 = 0
Sol:
We have,
5x - 4 y + 8 = 0
7 x + 6 y - 9 = 0
Here,
a= 5, b1 = -4, c1 = 8
a2 = 7, b2 = 6, c2 = -9
We have,
∴ Two lines are intersecting with each other at a point.
We have,
9 x + 3 y +12 = 0
18 + 6 y + 24 = 0
Here,
a1 = 9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
Now,
And
∴ Both the lines coincide.
We have,
6 x - 3 y +10 = 0
2 x - y + 9 = 0
Here,
a1 = 6, b= -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
Now,
And
∴ The lines are parallel
Q.6. Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Sol:
We have,
2x + 3 y - 8 = 0
Let another equation of line is
4x + 9 y - 4 = 0
Here,
a1 = 2, b1 = 3, c1 = -8
a= 4, b2 = 9, c2 = -4
Now,
And
∴ 2x + 3 y - 8 = 0 and 4 x + 9 y - 4 = 0 intersect each other at one point.
Hence, required equation of line is 4 x + 9y - 4 = 0
We have,
2x + 3y -8 = 0
Let another equation of line is:
4x +6y -4 = 0
Here,
a1 = 2, b1 = 3, c1 = -8
a2 = 4, b2 = 6, c2 = -4
Now,
And
∴ Lines are parallel to each other.
Hence, required equation of line is 4 x + 6y - 4 = 0.
Q.7. The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4kg of apples and 2kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
Sol:
Let the cost of 1 kg of apples and 1 kg grapes be Rs x and Rs y.
The given conditions can be algebraically represented as:
2 x + y = 160 4 x + 2 y = 300
2x + y = 160 ⇒ y = 160 - 2x
Three solutions of this equation cab be written in a table as follows:
4x + 2y = 300 ⇒ y =
Three solutions of this equation cab be written in a table as follows:
The graphical representation is as follows:
Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1kg grapes will be taken as the variables from the given condition of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then In order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are Large so take the suitable scale.
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Mathematics (Maths) Class 10
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# What Is 17/37 as a Decimal + Solution With Free Steps
The fraction 17/37 as a decimal is equal to 0.459.
The division of two numbers is usually shown as p $\boldsymbol\div$ q, where p is the dividend and q is the divisor. This is mathematically equivalent to the numeral p/q, called a fraction. In fractions, though, the dividend is called the numerator and the divisor is called the denominator.
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 17/37.
## 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 = 17
Divisor = 37
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 = 17 $\div$ 37
This is when we go through the Long Division solution to our problem.
Figure 1
## 17/37 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 17 and 37, we can see how 17 is Smaller than 37, and to solve this division, we require that 17 be Bigger than 37.
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 17, which after getting multiplied by 10 becomes 170.
We take this 170 and divide it by 37; this can be done as follows:
 170 $\div$ 37 $\approx$ 4
Where:
37 x 4 = 148
This will lead to the generation of a Remainder equal to 170 – 148 = 22. Now this means we have to repeat the process by Converting the 22 into 220 and solving for that:
220 $\div$ 37 $\approx$ 5Â
Where:
37 x 5 = 185
This, therefore, produces another Remainder which is equal to 220 – 185 = 35. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 350.
350 $\div$ 37 $\approx$ 9Â
Where:
37 x 9 = 333
Finally, we have a Quotient generated after combining the three pieces of it as 0.459, with a Remainder equal to 17.
Images/mathematical drawings are created with GeoGebra.
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# An aqueous solution of 3.47 M silver nitrate, AgNO_3, has a density of 1.47 g/mL. What is percent by mass of AgNO_3 in the solution?
Nov 16, 2015
40.1%
#### Explanation:
Here's your strategy for this problem - you need to pick a sample volume of this solution, use the given density to find its mass, then the number of moles of silver nitrate to get mass of silver nitrate it contains.
So, to make the calculations easier, let's take a $\text{1.00-L}$ sample of this solution. Use the given density to determine its mass
1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.47 g"/(1color(red)(cancel(color(black)("mL")))) = "1470 g"
Now, this $\text{1.00-L}$ solution will contain
$c = \frac{n}{V} \implies n = c \cdot V$
$n = 3.47 \text{moles"/color(red)(cancel(color(black)("L"))) * 1.00color(red)(cancel(color(black)("L"))) = "3.47 moles}$
Use silver nitrate's molar mass to help you determine how many grams of silver nitrate would contain this many moles
3.47color(red)(cancel(color(black)("moles AgNO"_3))) * "169.87 g"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "589.4 g AgNO"_3
Now, the solution's percent concentration by mass is defined as the mass of the solute, in your case silver nitrate, divided by the mass of the solution, and multiplied by $100$.
$\textcolor{b l u e}{\text{%w/w" = "mass of solute"/"mass of solution} \times 100}$
Plug in your values to get
"%w/w" = (589.4color(red)(cancel(color(black)("g"))))/(1470color(red)(cancel(color(black)("g")))) xx 100 = color(green)("40.1%")
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# Active Calculus
## Section2.6Derivatives of Inverse Functions
Much of mathematics centers on the notion of function. Indeed, throughout our study of calculus, we are investigating the behavior of functions, with particular emphasis on how fast the output of the function changes in response to changes in the input. Because each function represents a process, a natural question to ask is whether or not the particular process can be reversed. That is, if we know the output that results from the function, can we determine the input that led to it? And if we know how fast a particular process is changing, can we determine how fast the inverse process is changing?
One of the most important functions in all of mathematics is the natural exponential function $$f(x) = e^x\text{.}$$ Its inverse, the natural logarithm, $$g(x) = \ln(x)\text{,}$$ is similarly important. One of our goals in this section is to learn how to differentiate the logarithm function. First, we review some of the basic concepts surrounding functions and their inverses.
### Preview Activity2.6.1.
The equation $$y = \frac{5}{9}(x - 32)$$ relates a temperature given in $$x$$ degrees Fahrenheit to the corresponding temperature $$y$$ measured in degrees Celcius.
a. Solve the equation $$y = \frac{5}{9} (x - 32)$$ for $$x$$ to write $$x$$ (Fahrenheit temperature) in terms of $$y$$ (Celcius temperature).
$$x =$$ .
b. Now let $$C(x) = \frac{5}{9} (x - 32)$$ be the function that takes a Fahrenheit temperature as input and produces the Celcius temperature as output. In addition, let $$F(y)$$ be the function that converts a temperature given in $$y$$ degrees Celcius to the temperature $$F(y)$$ measured in degrees Fahrenheit. Use your work above to write a formula for $$F(y)\text{.}$$
$$F(y) =$$ .
c. Next consider the new function defined by $$p(x) = F(C(x))\text{.}$$ Use the formulas for $$F$$ and $$C$$ to determine an expression for $$p(x)$$ and simplify this expression as much as possible. What do you observe?
• $$\displaystyle p(x)=x$$
• $$\displaystyle p(x)=0$$
• $$\displaystyle p(x)=1$$
• $$\displaystyle p(x)=5x+9$$
• $$\displaystyle p(x)=9x+5$$
d. Now, let $$r(y) = C(F(y))\text{.}$$ Use the formulas for $$F$$ and $$C$$ to determine an expression for $$r(y)$$ and simplify this expression as much as possible. What do you observe?
• $$\displaystyle r(y)=y$$
• $$\displaystyle r(y)=1$$
• $$\displaystyle r(y)=9y+5$$
• $$\displaystyle r(y)=5y+9$$
• $$\displaystyle r(y)=0$$
e. What is the value of $$C'(x)\text{?}$$
$$C'(x) =$$
What is the value of $$F'(y)\text{?}$$
$$F'(y) =$$
How do $$C'(x)$$ and $$F'(y)$$ appear to be related?
• They add up to 1
• They are equal
• They are reciprocals
• They are opposites
### Subsection2.6.1Basic facts about inverse functions
A function $$f : A \to B$$ is a rule that associates each element in the set $$A$$ to one and only one element in the set $$B\text{.}$$ We call $$A$$ the domain of $$f$$ and $$B$$ the codomain of $$f\text{.}$$ If there exists a function $$g : B \to A$$ such that $$g(f(a)) = a$$ for every possible choice of $$a$$ in the set $$A$$ and $$f(g(b)) = b$$ for every $$b$$ in the set $$B\text{,}$$ then we say that $$g$$ is the inverse of $$f\text{.}$$
We often use the notation $$f^{-1}$$ (read “$$f$$-inverse”) to denote the inverse of $$f\text{.}$$ The inverse function undoes the work of $$f\text{.}$$ Indeed, if $$y = f(x)\text{,}$$ then
\begin{equation*} f^{-1}(y) = f^{-1}(f(x)) = x\text{.} \end{equation*}
Thus, the equations $$y = f(x)$$ and $$x = f^{-1}(y)$$ say the same thing. The only difference between the two equations is one of perspective — one is solved for $$x\text{,}$$ while the other is solved for $$y\text{.}$$
Here we briefly remind ourselves of some key facts about inverse functions.
#### Note2.6.1.
For a function $$f : A \to B\text{,}$$
• $$f$$ has an inverse if and only if $$f$$ is one-to-one
1
A function $$f$$ is one-to-one provided that no two distinct inputs lead to the same output.
and onto
2
A function $$f$$ is onto provided that every possible element of the codomain can be realized as an output of the function for some choice of input from the domain.
;
• provided $$f^{-1}$$ exists, the domain of $$f^{-1}$$ is the codomain of $$f\text{,}$$ and the codomain of $$f^{-1}$$ is the domain of $$f\text{;}$$
• $$f^{-1}(f(x)) = x$$ for every $$x$$ in the domain of $$f$$ and $$f(f^{-1}(y)) = y$$ for every $$y$$ in the codomain of $$f\text{;}$$
• $$y = f(x)$$ if and only if $$x = f^{-1}(y)\text{.}$$
The last fact reveals a special relationship between the graphs of $$f$$ and $$f^{-1}\text{.}$$ If a point $$(x,y)$$ that lies on the graph of $$y = f(x)\text{,}$$ then it is also true that $$x = f^{-1}(y)\text{,}$$ which means that the point $$(y,x)$$ lies on the graph of $$f^{-1}\text{.}$$ This shows us that the graphs of $$f$$ and $$f^{-1}$$ are the reflections of each other across the line $$y = x\text{,}$$ because this reflection is precisely the geometric action that swaps the coordinates in an ordered pair. In Figure 2.6.2, we see this illustrated by the function $$y = f(x) = 2^x$$ and its inverse, with the points $$(-1,\frac{1}{2})$$ and $$(\frac{1}{2},-1)$$ highlighting the reflection of the curves across $$y = x\text{.}$$
To close our review of important facts about inverses, we recall that the natural exponential function $$y = f(x) = e^x$$ has an inverse function, namely the natural logarithm, $$x = f^{-1}(y) = \ln(y)\text{.}$$ Thus, writing $$y = e^x$$ is interchangeable with $$x = \ln(y)\text{,}$$ plus $$\ln(e^x) = x$$ for every real number $$x$$ and $$e^{\ln(y)} = y$$ for every positive real number $$y\text{.}$$
### Subsection2.6.2The derivative of the natural logarithm function
In what follows, we find a formula for the derivative of $$g(x) = \ln(x)\text{.}$$ To do so, we take advantage of the fact that we know the derivative of the natural exponential function, the inverse of $$g\text{.}$$ In particular, we know that writing $$g(x) = \ln(x)$$ is equivalent to writing $$e^{g(x)} = x\text{.}$$ Now we differentiate both sides of this equation and observe that
\begin{equation*} \frac{d}{dx}\left[e^{g(x)}\right] = \frac{d}{dx}[x]\text{.} \end{equation*}
The righthand side is simply $$1\text{;}$$ by applying the chain rule to the left side, we find that
\begin{equation*} e^{g(x)} g'(x) = 1\text{.} \end{equation*}
Next we solve for $$g'(x)\text{,}$$ to get
\begin{equation*} g'(x) = \frac{1}{e^{g(x)}}\text{.} \end{equation*}
Finally, we recall that $$g(x) = \ln(x)\text{,}$$ so $$e^{g(x)} = e^{\ln(x)} = x\text{,}$$ and thus
\begin{equation*} g'(x) = \frac{1}{x}\text{.} \end{equation*}
#### Natural Logarithm.
For all positive real numbers $$x\text{,}$$ $$\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$$
This rule for the natural logarithm function now joins our list of basic derivative rules. Note that this rule applies only to positive values of $$x\text{,}$$ as these are the only values for which $$\ln(x)$$ is defined.
Also notice that for the first time in our work, differentiating a basic function of a particular type has led to a function of a very different nature: the derivative of the natural logarithm is not another logarithm, nor even an exponential function, but rather a rational one.
Derivatives of logarithms may now be computed in concert with all of the rules known to date. For instance, if $$f(t) = \ln(t^2 + 1)\text{,}$$ then by the chain rule, $$f'(t) = \frac{1}{t^2 + 1} \cdot 2t\text{.}$$
There are interesting connections between the graphs of $$f(x) = e^x$$ and $$f^{-1}(x) = \ln(x)\text{.}$$
In Figure 2.6.3, we are reminded that since the natural exponential function has the property that its derivative is itself, the slope of the tangent to $$y = e^x$$ is equal to the height of the curve at that point. For instance, at the point $$A = (\ln(0.5), 0.5)\text{,}$$ the slope of the tangent line is $$m_A = 0.5\text{,}$$ and at $$B = (\ln(5), 5)\text{,}$$ the tangent line’s slope is $$m_B = 5\text{.}$$
At the corresponding points $$A'$$ and $$B'$$ on the graph of the natural logarithm function (which come from reflecting $$A$$ and $$B$$ across the line $$y = x$$), we know that the slope of the tangent line is the reciprocal of the $$x$$-coordinate of the point (since $$\frac{d}{dx}[\ln(x)] = \frac{1}{x}$$). Thus, at $$A' = (0.5, \ln(0.5))\text{,}$$ we have $$m_{A'} = \frac{1}{0.5} = 2\text{,}$$ and at $$B' = (5, \ln(5))\text{,}$$ $$m_{B'} = \frac{1}{5}\text{.}$$
In particular, we observe that $$m_{A'} = \frac{1}{m_A}$$ and $$m_{B'} = \frac{1}{m_B}\text{.}$$ This is not a coincidence, but in fact holds for any curve $$y = f(x)$$ and its inverse, provided the inverse exists. This is due to the reflection across $$y = x\text{.}$$ It changes the roles of $$x$$ and $$y\text{,}$$ thus reversing the rise and run, so the slope of the inverse function at the reflected point is the reciprocal of the slope of the original function.
#### Activity2.6.2.
For each function given below, find its derivative.
1. $$\displaystyle h(x) = x^2\ln(x)$$
2. $$\displaystyle p(t) = \frac{\ln(t)}{e^t + 1}$$
3. $$\displaystyle s(y) = \ln(\cos(y) + 2)$$
4. $$\displaystyle z(x) = \tan(\ln(x))$$
5. $$\displaystyle m(z) = \ln(\ln(z))$$
### Subsection2.6.3Inverse trigonometric functions and their derivatives
Trigonometric functions are periodic, so they fail to be one-to-one, and thus do not have inverse functions. However, we can restrict the domain of each trigonometric function so that it is one-to-one on that domain.
For instance, consider the sine function on the domain $$[-\frac{\pi}{2}, \frac{\pi}{2}]\text{.}$$ Because no output of the sine function is repeated on this interval, the function is one-to-one and thus has an inverse. Thus, the function $$f(x) = \sin(x)$$ with $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$ and codomain $$[-1,1]$$ has an inverse function $$f^{-1}$$ such that
\begin{equation*} f^{-1} : [-1,1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]\text{.} \end{equation*}
We call $$f^{-1}$$ the arcsine (or inverse sine) function and write $$f^{-1}(y) = \arcsin(y)\text{.}$$ It is especially important to remember that
\begin{equation*} y = \sin(x) \ \ \text{and} \ \ x = \arcsin(y) \end{equation*}
say the same thing. “The arcsine of $$y$$” means “the angle whose sine is $$y\text{.}$$” For example, $$\arcsin(\frac{1}{2}) = \frac{\pi}{6}$$ means that $$\frac{\pi}{6}$$ is the angle whose sine is $$\frac{1}{2}\text{,}$$ which is equivalent to writing $$\sin(\frac{\pi}{6}) = \frac{1}{2}\text{.}$$
Next, we determine the derivative of the arcsine function. Letting $$h(x) = \arcsin(x)\text{,}$$ our goal is to find $$h'(x)\text{.}$$ Since $$h(x)$$ is the angle whose sine is $$x\text{,}$$ it is equivalent to write
\begin{equation*} \sin(h(x)) = x\text{.} \end{equation*}
Differentiating both sides of the previous equation, we have
\begin{equation*} \frac{d}{dx}[\sin(h(x))] = \frac{d}{dx}[x]\text{.} \end{equation*}
The righthand side is simply $$1\text{,}$$ and by applying the chain rule applied to the left side,
\begin{equation*} \cos(h(x)) h'(x) = 1\text{.} \end{equation*}
Solving for $$h'(x)\text{,}$$ it follows that
\begin{equation*} h'(x) = \frac{1}{\cos(h(x))}\text{.} \end{equation*}
Finally, we recall that $$h(x) = \arcsin(x)\text{,}$$ so the denominator of $$h'(x)$$ is the function $$\cos(\arcsin(x))\text{,}$$ or in other words, “the cosine of the angle whose sine is $$x\text{.}$$” A bit of right triangle trigonometry allows us to simplify this expression considerably.
Let’s say that $$\theta = \arcsin(x)\text{,}$$ so that $$\theta$$ is the angle whose sine is $$x\text{.}$$ We can picture $$\theta$$ as an angle in a right triangle with hypotenuse $$1$$ and a vertical leg of length $$x\text{,}$$ as shown in Figure 2.6.5. The horizontal leg must be $$\sqrt{1-x^2}\text{,}$$ by the Pythagorean Theorem.
Now, because $$\theta = \arcsin(x)\text{,}$$ the expression for $$\cos(\arcsin(x))$$ is equivalent to $$\cos(\theta)\text{.}$$ From the figure, $$\cos(\arcsin(x)) = \cos(\theta) = \sqrt{1-x^2}\text{.}$$
Substituting this expression into our formula, $$h'(x) = \frac{1}{\cos(\arcsin(x))}\text{,}$$ we have now shown that
\begin{equation*} h'(x) = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
#### Inverse sine.
For all real numbers $$x$$ such that $$-1 \lt x \lt 1\text{,}$$
\begin{equation*} \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}
#### Activity2.6.3.
The following prompts in this activity will lead you to develop the derivative of the inverse tangent function.
1. Let $$r(x) = \arctan(x)\text{.}$$ Use the relationship between the arctangent and tangent functions to rewrite this equation using only the tangent function.
2. Differentiate both sides of the equation you found in (a). Solve the resulting equation for $$r'(x)\text{,}$$ writing $$r'(x)$$ as simply as possible in terms of a trigonometric function evaluated at $$r(x)\text{.}$$
3. Recall that $$r(x) = \arctan(x)\text{.}$$ Update your expression for $$r'(x)$$ so that it only involves trigonometric functions and the independent variable $$x\text{.}$$
4. Introduce a right triangle with angle $$\theta$$ so that $$\theta = \arctan(x)\text{.}$$ What are the three sides of the triangle?
5. In terms of only $$x$$ and $$1\text{,}$$ what is the value of $$\cos(\arctan(x))\text{?}$$
6. Use the results of your work above to find an expression involving only $$1$$ and $$x$$ for $$r'(x)\text{.}$$
While derivatives for other inverse trigonometric functions can be established similarly, for now we limit ourselves to the arcsine and arctangent functions.
#### Activity2.6.4.
Determine the derivative of each of the following functions.
1. $$\displaystyle \displaystyle f(x) = x^3 \arctan(x) + e^x \ln(x)$$
2. $$\displaystyle \displaystyle p(t) = 2^{t\arcsin(t)}$$
3. $$\displaystyle \displaystyle h(z) = (\arcsin(5z) + \arctan(4-z))^{27}$$
4. $$\displaystyle \displaystyle s(y) = \cot(\arctan(y))$$
5. $$\displaystyle \displaystyle m(v) = \ln(\sin^2(v)+1)$$
6. $$\displaystyle \displaystyle g(w) = \arctan\left( \frac{\ln(w)}{1+w^2} \right)$$
### Subsection2.6.4The link between the derivative of a function and the derivative of its inverse
In Figure 2.6.3, we saw an interesting relationship between the slopes of tangent lines to the natural exponential and natural logarithm functions at points reflected across the line $$y = x\text{.}$$ In particular, we observed that at the point $$(\ln(2), 2)$$ on the graph of $$f(x) = e^x\text{,}$$ the slope of the tangent line is $$f'(\ln(2)) = 2\text{,}$$ while at the corresponding point $$(2, \ln(2))$$ on the graph of $$f^{-1}(x) = \ln(x)\text{,}$$ the slope of the tangent line is $$(f^{-1})'(2) = \frac{1}{2}\text{,}$$ which is the reciprocal of $$f'(\ln(2))\text{.}$$
That the two corresponding tangent lines have reciprocal slopes is not a coincidence. If $$f$$ and $$g$$ are differentiable inverse functions, then $$y = f(x)$$ if and only if $$x = g(y)\text{,}$$ then$$f(g(x)) = x$$ for every $$x$$ in the domain of $$f^{-1}\text{.}$$ Differentiating both sides of this equation, we have
\begin{equation*} \frac{d}{dx} [f(g(x))] = \frac{d}{dx} [x]\text{,} \end{equation*}
and by the chain rule,
\begin{equation*} f'(g(x)) g'(x) = 1\text{.} \end{equation*}
Solving for $$g'(x)\text{,}$$ we have $$g'(x) = \frac{1}{f'(g(x))}\text{.}$$ Here we see that the slope of the tangent line to the inverse function $$g$$ at the point $$(x,g(x))$$ is precisely the reciprocal of the slope of the tangent line to the original function $$f$$ at the point $$(g(x),f(g(x))) = (g(x),x)\text{.}$$
To see this more clearly, consider the graph of the function $$y = f(x)$$ shown in Figure 2.6.6, along with its inverse $$y = g(x)\text{.}$$ Given a point $$(a,b)$$ that lies on the graph of $$f\text{,}$$ we know that $$(b,a)$$ lies on the graph of $$g\text{;}$$ because $$f(a) = b$$ and $$g(b) = a\text{.}$$ Now, applying the rule that $$g'(x) = 1/f'(g(x))$$ to the value $$x = b\text{,}$$ we have
\begin{equation*} g'(b) = \frac{1}{f'(g(b))} = \frac{1}{f'(a)}\text{,} \end{equation*}
which is precisely what we see in the figure: the slope of the tangent line to $$g$$ at $$(b,a)$$ is the reciprocal of the slope of the tangent line to $$f$$ at $$(a,b)\text{,}$$ since these two lines are reflections of one another across the line $$y = x\text{.}$$
#### Derivative of an inverse function.
Suppose that $$f$$ is a differentiable function with inverse $$g$$ and that $$(a,b)$$ is a point that lies on the graph of $$f$$ at which $$f'(a) \ne 0\text{.}$$ Then
\begin{equation*} g'(b) = \frac{1}{f'(a)}\text{.} \end{equation*}
More generally, for any $$x$$ in the domain of $$g'\text{,}$$ we have $$g'(x) = 1/f'(g(x))\text{.}$$
The rules we derived for $$\ln(x)\text{,}$$ $$\arcsin(x)\text{,}$$ and $$\arctan(x)$$ are all just specific examples of this general property of the derivative of an inverse function. For example, with $$g(x) = \ln(x)$$ and $$f(x) = e^x\text{,}$$ it follows that
\begin{equation*} g'(x) = \frac{1}{f'(g(x))} = \frac{1}{e^{\ln(x)}} = \frac{1}{x}\text{.} \end{equation*}
### Subsection2.6.5Summary
• For all positive real numbers $$x\text{,}$$ $$\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$$
• For all real numbers $$x$$ such that $$-1 \lt x \lt 1\text{,}$$ $$\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.}$$ In addition, for all real numbers $$x\text{,}$$ $$\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}\text{.}$$
• If $$g$$ is the inverse of a differentiable function $$f\text{,}$$ then for any point $$x$$ in the domain of $$g'\text{,}$$ $$g'(x) = \frac{1}{f'(g(x))}\text{.}$$
### Exercises2.6.6Exercises
#### 1.
Let $$(x_0, y_0) = (1, 5)$$ and $$(x_1, y_1) = (1.1, 5.4)\text{.}$$ Use the following graph of the function $$f$$ to find the indicated derivatives.
If $$h(x)=(f(x))^{2}\text{,}$$ then
$$h'(1) =$$
If $$g(x)=f^{-1}(x)\text{,}$$ then
$$g'(5) =$$
#### 2.
Find the derivative of the function $$f(t)\text{,}$$ below.
$$f(t)=\ln(t^{4}+3)$$
$$f'(t) =$$
#### 3.
Let
\begin{equation*} f(x) = 4\sin^{-1}\mathopen{}\left(x^{4}\right) \end{equation*}
$$f'( x ) =$$
NOTE: The webwork system will accept $$\arcsin(x)$$ or $$\sin^{-1}(x)$$ as the inverse of $$\sin (x)\text{.}$$
#### 4.
If $$f(x) = 6 x^{3}\arctan(6 x^{4})\text{,}$$ find $$f' ( x ).$$
$$f' (x)$$ =
#### 5.
For each of the given functions $$f(x)\text{,}$$ find the derivative $$\left(f^{-1}\right)'(c)$$ at the given point $$c\text{,}$$ first finding $$a=f^{-1}(c)\text{.}$$
a) $$f(x)= 5 x + 7 x^{21}\text{;}$$ $$c = -12$$
$$a$$ =
$$\left(f^{-1}\right)'(c)$$ =
b) $$f(x)= x^2 - 12 x + 51$$ on the interval $$[6,\infty)\text{;}$$ $$c = 16$$
$$a$$ =
$$\left(f^{-1}\right)'(c)$$ =
#### 6.
Given that $$f(x)=2x+\cos(x)$$ is one-to-one, use the formula
\begin{equation*} \displaystyle (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \end{equation*}
to find $$(f^{-1})'(1)\text{.}$$
$$(f^{-1})'(1) =$$
#### 7.
Determine the derivative of each of the following functions. Use proper notation and clearly identify the derivative rules you use.
1. $$\displaystyle f(x) = \ln(2\arctan(x) + 3\arcsin(x) + 5)$$
2. $$\displaystyle r(z) = \arctan(\ln(\arcsin(z)))$$
3. $$\displaystyle q(t) = \arctan^2(3t) \arcsin^4(7t)$$
4. $$\displaystyle g(v) = \ln\left( \frac{\arctan(v)}{\arcsin(v) + v^2} \right)$$
#### 8.
Consider the graph of $$y = f(x)$$ provided in Figure 2.6.7 and use it to answer the following questions.
1. Use the provided graph to estimate the value of $$f'(1)\text{.}$$
2. Sketch an approximate graph of $$y = f^{-1}(x)\text{.}$$ Label at least three distinct points on the graph that correspond to three points on the graph of $$f\text{.}$$
3. Based on your work in (a), what is the value of $$(f^{-1})'(-1)\text{?}$$ Why?
#### 9.
Let $$f(x) = \frac{1}{4}x^3 + 4\text{.}$$
1. Sketch a graph of $$y = f(x)$$ and explain why $$f$$ is an invertible function.
2. Let $$g$$ be the inverse of $$f$$ and determine a formula for $$g\text{.}$$
3. Compute $$f'(x)\text{,}$$ $$g'(x)\text{,}$$ $$f'(2)\text{,}$$ and $$g'(6)\text{.}$$ What is the special relationship between $$f'(2)$$ and $$g'(6)\text{?}$$ Why?
#### 10.
Let $$h(x) = x + \sin(x)\text{.}$$
1. Sketch a graph of $$y = h(x)$$ and explain why $$h$$ must be invertible.
2. Explain why it does not appear to be algebraically possible to determine a formula for $$h^{-1}\text{.}$$
3. Observe that the point $$(\frac{\pi}{2}, \frac{\pi}{2} + 1)$$ lies on the graph of $$y = h(x)\text{.}$$ Determine the value of $$(h^{-1})'(\frac{\pi}{2} + 1)\text{.}$$
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# How to Find the Height of a Triangle
Author Info
Updated: September 6, 2019
To calculate the area of a triangle you need to know its height. To find the height follow these instructions. You must at least have a base to find the height.
### Method 1 of 3: Using Base and Area to Find Height
1. 1
Recall the formula for the area of a triangle. The formula for the area of a triangle is
A=1/2bh.
[1]
• A = Area of the triangle
• b = Length of the base of the triangle
• h = Height of the base of the triangle
2. 2
Look at your triangle and determine which variables you know. You already know the area, so assign that value to A. You should also know the value of one side length; assign that value to "'b'".
Any side of a triangle can be the base,
regardless of how the triangle is drawn. To visualize this, just imagine rotating the triangle until the known side length is at the bottom.
Example
If you know that the area of a triangle is 20, and one side is 4, then:
A = 20 and b = 4.
3. 3
Plug your values into the equation A=1/2bh and do the math. First multiply the base (b) by 1/2, then divide the area (A) by the product. The resulting value will be the height of your triangle!
Example
20 = 1/2(4)h Plug the numbers into the equation.
20 = 2h Multiply 4 by 1/2.
10 = h Divide by 2 to find the value for height.
### Method 2 of 3: Finding an Equilateral Triangle's Height
1. 1
Recall the properties of an equilateral triangle. An equilateral triangle has three equal sides, and three equal angles that are each 60 degrees. If you
cut an equilateral triangle in half, you will end up with two congruent right triangles.
[2]
• In this example, we will be using an equilateral triangle with side lengths of 8.
2. 2
Recall the Pythagorean Theorem. The Pythagorean Theorem states that for any right triangle with sides of length a and b, and hypotenuse of length c:
a2 + b2 = c2.
We can use this theorem to find the height of our equilateral triangle![3]
3. 3
Break the equilateral triangle in half, and assign values to variables a, b, and c. The hypotenuse c will be equal to the original side length. Side a will be equal to 1/2 the side length, and side b is the height of the triangle that we need to solve.
• Using our example equilateral triangle with sides of 8, c = 8 and a = 4.
4. 4
Plug the values into the Pythagorean Theorem and solve for b2. First square c and a by multiplying each number by itself. Then subtract a2 from c2.
Example
42 + b2 = 82 Plug in the values for a and c.
16 + b2 = 64 Square a and c.
b2 = 48 Subtract a2 from c2.
5. 5
Find the square root of b2 to get the height of your triangle! Use the square root function on your calculator to find Sqrt(2. The answer is the height of your equilateral triangle!
• b = Sqrt (48) = 6.93
### Method 3 of 3: Determining Height With Angles and Sides
1. 1
Determine what variables you know. The height of a triangle can be found if you have 2 sides and the angle in between them, or all three sides. We'll call the sides of the triangle a, b, and c, and the angles, A, B, and C.
• If you have all three sides, you'll use
Heron's formula
, and the formula for the area of a triangle.
• If you have two sides and an angle, you'll use the formula for the area given two angles and a side.
A = 1/2ab(sin C).[4]
2. 2
Use Heron's formula if you have all three sides. Heron's formula has two parts. First, you must find the variable
s, which is equal to half of the perimeter of the triangle.
This is done with this formula:
s = (a+b+c)/2.[5]
Heron's Formula Example
For a triangle with sides a = 4, b = 3, and c = 5:
s = (4+3+5)/2
s = (12)/2
s = 6
Then use the second part of Heron's formula, Area = sqr(s(s-a)(s-b)(s-c). Replace Area in the equation with its equivalent in the area formula: 1/2bh (or 1/2ah or 1/2ch).
Solve for h. For our example triangle this looks like:
1/2(3)h = sqr(6(6-4)(6-3)(6-5).
3/2h = sqr(6(2)(3)(1)
3/2h = sqr(36)
Use a calculator to calculate the square root, which in this case makes it 3/2h = 6.
Therefore, height is equal to 4, using side b as the base.
3. 3
Use the area given two sides and an angle formula if you have a side and an angle. Replace area in the formula with its equivalent in the area of a triangle formula: 1/2bh. This gives you a formula that looks like 1/2bh = 1/2ab(sin C). This can be simplified to
h = a(sin C)
, thereby eliminating one of the side variables.[6]
Finding Height with 1 Side and 1 Angle Example
For example, with a = 3, and C = 40 degrees, the equation looks like this:
h = 3(sin 40)
Use your calculator to finish the equation, which makes h roughly 1.928.
## Community Q&A
Search
• Question
How do I find the area of an equilateral triangle when only the height is given?
H = height, S = side, A = area, B = base. You know that each angle is 60 degrees because it is an equilateral triangle. If you look at one of the triangle halves, H/S = sin 60 degrees because S is the longest side (the hypotenuse) and H is across from the 60 degree angle, so now you can find S. The base of the triangle is S because all the sides are the same, so B = S. Using A = (1/2)*BH, you get A = (1/2)*SH, which you can now find.
• Question
How do I calculate the height of a right triangle, given only the length of the base and the interior angle at the base?
Donagan
Look up the tangent of the angle in a trigonometry table. Multiply the tangent by the length of the base.
• Question
How do I determine the height of a triangle when I know the length of all three sides?
You already know the base, so calculate the area by Heron's formula. Then, substitute the values you know in the formula. Area=1/2 * base * height or height=2 * Area/base and find your answer.
• Question
How do I find the height of a triangle?
Donagan
You need to know both the length of the base of the triangle and its area. Divide the base into the area, then double that.
• Question
What is height of a triangle if the base is 5 and the two sides are 3?
Donagan
This is a trigonometry question. Draw the height from the obtuse angle to the "5" side. This forms two right triangles inside the main triangle, each of whose hypotenuses are "3". The cosine of either of the original acute angles equals 2½÷3, or 0.833. Look up that angle in a trig table. Find the sine of that angle, and multiply that by 3 to get the height.
• Question
How do I find the height of a triangle with the length of the three sides?
Follow Method 3. "If you have all three sides, you'll use Heron's formula, and the formula for the area of a triangle." Use Heron's formula to determine the area of the triangle. Set the result equal to 1/2bh, and solve for h.
• Question
How can I determine the height of an isosceles triangle?
Since the two opposite sides on an isosceles triangle are equal, you can use trigonometry to figure out the height. You can find it by having a known angle and using SohCahToa. For example, say you had an angle connecting a side and a base that was 30 degrees and the sides of the triangle are 3 inches long and 5.196 for the base side. In order to find the height, you would need to set it up as this: S=o/h, S=sine, o=opposite (the height), h=hypotenuse (the side), S30=o/3 because it's the opposite divided by the three, you can multiply by the reciprocal on both sides so the three gets cancelled on one side and the other is multiplied by 3, 3sin30=o and 3sin30=1.5=height.
• Question
How do I find the height if the area and base of the quadrilateral are given?
Donagan
If it's a regular quadrilateral, divide the area by the base.
• Question
Can I determine the base of a triangle if I know its height is eight feet?
Donagan
You would also need to know other information, such as the area or some of the sides or angles.
• Question
What is the formula for determining the area of a triangle?
Donagan
One-half base times height.
• How do I find the height of a right angle triangle if I know the base length and the two remaining angles?
• How do I calculate the distance of a height if two sides are known in a triangle?
• If I only have the base measurement (5 cm) and possibly the angles, how do I find the height?
• Is there a simpler version of Heron's Formula?
• How do I find the angle of a triangle when I know the base and the height?
200 characters left
## Video.By using this service, some information may be shared with YouTube.
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 36 people, some anonymous, worked to edit and improve it over time. Together, they cited 6 references. This article has also been viewed 1,718,650 times.
Co-authors: 36
Updated: September 6, 2019
Views: 1,718,650
Categories: Geometry
Article SummaryX
If you know the base and area of the triangle, you can divide the base by 2, then divide that by the area to find the height. To find the height of an equilateral triangle, use the Pythagorean Theorem, a^2 + b^2 = c^2. Cut the triangle in half down the middle, so that c is equal to the original side length, a equals half of the original side length, and b is the height. Plug a and c into the equation, squaring both of them. Then subtract a^2 from c^2 and take the square root of the difference to find the height. If you want to learn how to calculate the area if you only know the angles and sides, keep reading!
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# General Form of the Equation of a Circle – Definition, Formula, Examples | How to find the General Form Equation of a Circle?
Know the definition of a circle, the general form of the equation of a circle. Get the various terms involved in the general and standard form of a circle, formulae, and definition, etc. Refer to solved examples of a circle, standard equation of a circle. For your reference, we have included the solved examples on how to find the general form of an equation of the circle, conversion from standard form to general form and vice versa, etc.
Also, Read: Circumference and Area of Circle
## Circle Definition
The circle is defined as the locus of a point that moves in a plane such that its distance from a fixed point in that plane is always constant. The center of the circle is the fixed point. The set of points in the plane at a fixed distance is called the radius of the circle.
### General Form of the Equation of a Circle
To find the general form of the equation of a circle, we use the below-given graph. Each circle form has its own advantages. Here, we can take an example of a standard form which is great for determining the radius and center just with a glance at the above equation. The general form of a circle is good at substituting ordered pairs and testing them. We use both of these forms. So this gives us an idea that we should interchange between these forms. Firstly, we will transform the standard form to the general form.
General form of equation is (x-h)2Â + (y-k)2Â = r2
where r is defined as the radius of the circle
h, k is defined as the center coordinates
#### Standard Form to General Form
Here, we will take an example that gives us an idea to transform an equation from a Standard form to a general form
Eg: Transform (x – 3)2 + (y + 5)2 = 64 to general form.
(x – 3)2 + (y + 5)2 = 64
Now, all the binomial should be multiplied and rearranged till we get the general form.
(x – 3) (x – 3) + (y + 5) (y + 5) = 64
(x2 – 3x – 3x + 9) + (y2 + 5y + 5y + 25) = 64
x2 – 6x + 9 + y2 + 10y + 25 = 64
x2 + y2 – 6x + 10y + 9 + 25 – 64 = 0
(x2) + (y2) – 6(x) + 10(y) – 30 = 0
x2+y2–6x+10y–30 = 0
This is the general form of the equation as transformed from Standard from.
#### General to Standard Form
To transform an equation to standard form from a general form, we must first complete the equation balanced and complete the square. Here, completing the square implies creating Perfect Square Trinomials(PST’s).
To give you an idea about Perfect Square Trinomials, here are some examples
Example 1:
x2 + 2x + 1
When we factor PSTs, we get two identical binomial factors.
x2 + 2x + 1 = (x + 1)(x + 1) = (x + 1)2
Example 2:
x2 – 4x + 4
When we factor PSTs, we get two identical binomial factors.
x2 – 4x + 4= (x – 2)(x – 2) = (x – 2)2
We can observe that the sign for the middle term can either be positive or negative.
We have a relationship between the last term and the coefficient of the middle term
(b/2)2
Now, we see a few examples of circle equation that include the transformation of the equation from a standard form to the general form
### General Form of the Equation of a Circle Examples
Problem 1:
The circle equation is: x2 + y2 – 8x + 4y + 11 = 0. Find the centre and radius?
Solution:
To find the centre and radius of the circle, we first need to transform the equation from general form to standard form
x2 + y2 – 8x + 4y + 11 = 0
x2 – 8x + y2 + 4y + 11 = 0
(x2 – 8x + ) + (y2 + 4y + ) = -11
We are leaving the spaces empty for PST’s.
We must complete the square of the PST’ds by adding appropriate values
To maintain balance on the above equation, we must add same values on the right side which we add on the left side of the equation to keep the equation equal on both the sides
(x2 – 8x + 16) + (y2 + 4y + 4) = -11 + 16 + 4
(x – 4)2 + (y + 2)2 = 9
By comparing the above equation with the standard form of the circle, we observe that
Centre =(4,-2)
Radius = 3
Problem 2:
Find the standard form of the equation of a circle of radius 4 whose centre is (-3,2). Convert the equation into general form
Solution:
As given in the question,
radius = 4
h = -3
k = 2
General form of equation is (x-h)2Â + (y-k)2Â = r2
(x-(-3))2 + (y-2)2 = 42
(x+3)2 + (y-2)2 = 16
x2Â + 6x + 9Â + y2 -4y + 4 = 16
x2Â + y2Â + 6x – 4y – 3 = 0
Therefore, the general solution is x2Â + y2Â + 6x – 4y – 3 = 0
Problem 3:
Write the equation in the general form given the radius and centre
r = 3, centre = (1,2)
Solution:
As given in the question,
r = 3
h = 1
k = 2
General form of equation is (x-h)2Â + (y-k)2Â = r2
(x-1)2 + (y-2)2 = 32
x2 – 2x + 1Â + y2 -4y + 4 = 9
x2 + y2 – 2x – 4y – 4 = 0
Therefore, the general solution is x2 + y2 – 2x – 4y – 4 = 0
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# Perpendicular Bisector Theorem
Written by
Malcolm McKinsey
Fact-checked by
Paul Mazzola
## Perpendicular Bisector Theorem (Proof, Converse, & Examples)
### Perpendicular
All good learning begins with vocabulary, so we will focus on the two important words of the theorem. Perpendicular means two line segments, rays, lines or any combination of those that meet at right angles. A line is perpendicular if it intersects another line and creates right angles.
### Bisector
bisector is an object (a line, a ray, or line segment) that cuts another object (an angle, a line segment) into two equal parts. A bisector cannot bisect a line, because by definition a line is infinite.
### Perpendicular bisector
Putting the two meanings together, we get the concept of a perpendicular bisector, a line, ray or line segment that bisects an angle or line segment at a right angle.
Before you get all bothered about it being a perpendicular bisector of an angle, consider: what is the measure of a straight angle? 180°180°; that means a line dividing that angle into two equal parts and forming two right angles is a perpendicular bisector of the angle.
## Perpendicular bisector theorem
Okay, we laid the groundwork. So putting everything together, what does the Perpendicular Bisector Theorem say?
### How does it work?
Suppose you have a big, square plot of land, 1,000 meters on a side. You built a humdinger of a radio tower, 300 meters high, right smack in the middle of your land. You plan to broadcast rock music day and night.
Anyway, that location for your radio tower means you have 500 meters of land to the left, and 500 meters of land to the right. Your radio tower is a perpendicular bisector of the length of your land.
You need to reinforce the tower with wires to keep it from tipping over in high winds. Those are called guy wires. How long should a guy wire from the top down to the land be, on each side?
Because you constructed a perpendicular bisector, you do not need to measure on each side. One measurement, which you can calculate using geometry, is enough. Use the Pythagorean Theorem for right triangles:
Your tower is 300 meters. You can go out 500 meters to anchor the wire's end. The tower meets your land at 90°. So:
You need guy wires a whopping 583.095 meters long to run from the top of the tower to the edge of your land. You repeat the operation at the 200 meter height, and the 100 meter height.
For every height you choose, you will cut guy wires of identical lengths for the left and right side of your radio tower, because the tower is the perpendicular bisector of your land.
## Proving the perpendicular bisector theorem
Behold the awesome power of the two words, "perpendicular bisector," because with only a line segment, HM, and its perpendicular bisector, WA, we can prove this theorem.
We are given line segment HM and we have bisected it (divided it exactly in two) by a line WA. That line bisected HM at 90° because it is a given. This means, if we run a line segment from Point W to Point H, we can create right triangle WHA, and another line segment WM creates right triangle WAM.
What do we have now? We have two right triangles, WHA and WAM, sharing side WA, with all these congruences:
1. WA ≅ WA (by the reflexive property)
2. ∠WAH ≅ ∠WAM (90° angles; given)
3. HA ≅ AM (bisector; given)
What does that look like? We hope you said Side Angle Side, because that is exactly what it is.
That means sides WH and WM are congruent, because CPCTC (corresponding parts of congruent triangles are congruent). WHAM! Proven!
## Practice proof
You can tackle the theorem yourself now. You will either sink or swim on this one. Here is a line segment, WM. We construct a perpendicular bisector, SI.
How can you prove that SW ≅ SM? Do you know what to do?
1. Construct line segments SW and SM.
2. You now have what? Two right triangles, SWI and SIM. They have right angles, ∠SIW and ∠SIM.
3. Identify WI and IM as congruent, because they are the two parts of line segment WM that were bisected by SI.
4. Identify SI as congruent to itself (by the reflexive property).
What does that give you? Two congruent sides and an included angle, which is what postulate? The SAS Postulate, of course! Therefore, line segment SW ≅ SM.
So, did you sink or SWIM?
## Converse of the perpendicular bisector theorem
Notice that the theorem is constructed as an "if, then" statement. That immediately suggests you can write the converse of it, by switching the parts:
We can show this, too. Construct a line segment HD. Place a random point above it (but still somewhere between Points H and D) and call it Point T.
If Point T is the same distance from Points H and D, this converse statement says it must lie on the perpendicular bisector of HD.
You can prove or disprove this by dropping a perpendicular line from Point T through line segment HD. Where your perpendicular line crosses HD, call it Point U.
If Point T is the same distance from Points H and D, then HU ≅ UD. If Point T is not the same distance from Points H and D, then HU ≆ UD.
You can go through the steps of creating two right triangles, △THU and △TUD and proving angles and sides congruent (or not congruent), the same as with the original theorem.
You would identify the right angles, the congruent sides along the original line segment HD, and the reflexive congruent side TU. When you got to a pair of corresponding sides that were not congruent, then you would know Point T was not on the perpendicular bisector.
Only points lying on the perpendicular bisector will be equidistant from the endpoints of the line segment. Everything else lands with a THUD.
## Lesson summary
After you worked your way through all the angles, proofs and multimedia, you are now able to recall the Perpendicular Bisector Theorem and test the converse of the Theorem. You also got a refresher in what "perpendicular," "bisector," and "converse" mean.
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# Intercept Theorem
Reviewed by:
Last updated date: 13th Sep 2024
Total views: 207.6k
Views today: 4.07k
## An Overview of the Theorem
The Intercept Theorem is a fundamental tool of Euclidean Geometry. The concept of parallel lines and transversal is of great importance in our day-to-day life. And, the Intercept theorem extends our understanding of parallel lines and transversal and we can apply these concepts in our day-to-day life.
A Transversal
In the above figure, we can see that there are 3 parallel lines \${L}_{1}\$,\${L}_{2}\$, \${L}_{3}\$ and then there is a transversal \$PR\$ which is intersecting all the 3 parallel lines at an equal distance. The intercept theorem, also known as Thale’s theorem, Basic Proportionality Theorem, or side splitter theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels.
### History of the Mathematician
Euclid
• Year of Birth: 325 BC
• Year of Death: 270 BC
• Contribution: He contributed significantly in the field of Mathematics and Physics by discovering the intercept theorem.
## Statement of the Theorem
If there are three or more parallel lines and the intercepts made by them on one transversal are equal, the corresponding intercepts of any transversal are also equal.
## Proof of the Theorem
Two Parallel Lines
Given:
\$l\$, \$m\$, \$n\$ are three parallel lines.
\$P\$ is a transversal intersecting the parallel lines such that \$AB=BC\$.
The transversal \$Q\$ has the intercepts \$DE\$ and \$HE\$ by the parallel lines \$l\$, \$m\$, \$n\$.
To prove:
\$DE=EF\$
Proof:
Draw a line \$E\$ parallel to the line \$P\$ which intersects the line \$n\$ at \$H\$ and line \$l\$ at \$G\$.
\$AG||BE\$ (Given)
\$GE||AB\$ (By construction)
From the information above, we can say that \$AGBE\$ is a parallelogram.
According to the properties of a parallelogram:
\$AB=GE\$ - (1)
Similarly, we can say that \$BEHC\$ is a parallelogram.
\$BC=HE\$ - (2)
From the given information, we know that \$AB=BC\$.
So, from equations (1) and (2), we can say that \$GE=HE\$.
In \$\Delta GED\$ and \$\Delta HEF\$,
\$GE=HE\$(Proved)
\$\angle GED=\angle FEH\$(Vertically Opposite Angles)
\$\angle DGE=\angle FHE\$(Alternate Interior Angles)
Hence, \$\Delta GED\cong \Delta HEF\$
As\$\Delta GED\cong \Delta HEF\$, the sides\$DE=EF\$.
Hence proved.
## Applications of the Theorem
The intercept theorem can be used to prove that a certain construction yields parallel line segments:
• If the midpoints of two triangle sides are connected, then the resulting line segment is parallel to the third triangle side (Mid point theorem of triangles).
• If the midpoints of the two non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid.
## Limitations of the Theorem
• The intercept theorem is not able to help us in finding the midpoint of the sides of the triangle.
• The basic proportionality theorem is an advanced version of the intercept theorem and it gives us a lot of information on the sides of the triangles.
## Solved Examples
1. In a \[\Delta ABC\], sides \[AB\] and \[AC\] are intersected by a line at \[D\] and \[E\], respectively, which is parallel to side \[BC\]. Prove that \[\dfrac{AD}{AB}=\dfrac{AE}{AC}\].
Ans:
Scalene Triangle
\[DE||BC\] (Given)
Interchanging the ratios,
Interchanging the ratios again,
Hence proved.
2. Find DE
Basic Proportionality Theorem
Ans: According to the basic proportionality theorem,
\[\dfrac{AE}{DE}=\dfrac{BE}{CE}\]
\[\dfrac{4}{DE}=\dfrac{6}{8.5}\]
\[\dfrac{4*8.5}{6}=DE\]
\[DE=5.66\]
So, \[DE=5.66\]
3. In \[\Delta ABC\], \[D\] and \[E\] are points on the sides \[AB\] and \[AC\], respectively, such that \[DE||BC\]. If \[\dfrac{AD}{DB}=\dfrac{3}{4}\] and \[AC=15cm\], find \[AE\].
Intercept Theorem
Ans:\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\] (According to the intercept theorem)
Let \[AE=x\] and \[EC=15-x\]
So, \[\dfrac{3}{4}=\dfrac{x}{15-x}\]
\[3(15-x)=4x\]
\[45=7x\]
\[x=\dfrac{45}{7}\]
\[x=6.4cm\]
So, \[x=6.4cm\]
## Important Points
• The intercept theorem can only be applied when the lies are parallel, if the transversal is cutting lines that are not parallel, then the intercept theorem is not valid.
• The basic proportionality theorem and mid-point theorem are all applications of the intercept theorem but they are not the same theorems.
## Conclusion
In the above article, we have discussed the Equal intercept Theorem and its proof. We have also discussed the applications of the theorem. So, we can conclude that Intercept Theorem is a fundamental tool of Geometry and is based on applications of parallel lines and transversal and reduces our computational work based on its application as we have seen in the examples based on the theorem.
Competitive Exams after 12th Science
## FAQs on Intercept Theorem
1. Are the basic proportionality theorem, mid-point theorem, and intercept theorem the same?
No, basic proportionality theorem, intercept theorem, and mid-point theorem are 3 different kinds of theorems. The intercept theorem is a very broad theorem and it tells about how the properties of a transversal change when they interact with parallel lines. The basic proportionality theorem is covering about the interaction of lines parallel to one side of the triangle with the other two sides and the basic proportionality theorem is derived from the intercept theorem. The midpoint theorem talks about when a line bisects the other two sides of a triangle, then it is parallel to the third side, and the midpoint theorem is derived from the basic proportionality theorem.
2. What is the mid-point theorem?
The line segment of a triangle connecting the midpoint of two sides of the triangle is said to be parallel to its third side and is also half the length of the third side, according to the midpoint theorem. The mid-point theorem is derived from the basic proportionality theorem and it is a very useful theorem in solving the questions of geometry as it directly gives the value of the sides of a triangle and also the parallel line helps to get the values of angles.
3. Why can’t the intercept theorem be used for non-parallel lines?
The intercept theorem cannot be used for non-parallel lines because there are a lot of things in the derivation of the intercept theorem which we won’t get if the lines are not parallel. Like we won’t be able to use the angles such as the alternate interior angles and the vertically opposite angles and much more. The construction that we make in the derivation is only made because the lines are parallel, hence if the lines are not parallel the intercept theorem is not valid.
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## Chapter 5 : Derivatives and limits
### By Lund University
This chapter is entirely devoted to the derivative of a function of one variable. The derivative of function is defined as a limit of a specific ratio (the Newton quotient) and we begin the chapter with a brief introduction to limits. In the rest of the chapter we will learn how to differentiate various functions. To our help we will have a bunch of rules such as the chain rule. We will also need higher order derivatives. For example, we can sometimes use the second order derivative to distinguish between a maximum point and a minimum point. This chapter is concluded with a few more advanced topics such as implicit differentiation.
## Limits and continuity
The main objective of this chapter is to study the derivative of a function. However, in order to understand the definition of a derivative we must look at limits. The limit of a function is the value that a function takes when x gets close to, but is not exactly equal to, a given value. Limits are closely related to another concept called continuity. Informally, a function is continuous if its graph is “connected”.
## Limits and continuity: Problems
Exercises on limits and continuity
## Basic derivatives
This section introduces derivatives. It begins by defining the tangent, a straight line that just touches the graph of the function. The slope of this tangent is precisely the derivative of the function at the touching point. From this, the formal definition of a derivative is presented as the limit of the Newton quotient. We then look at rules which will help us finding the derivative of a function. Finally we look at the relationship between derivatives and whether the function is increasing or decreasing.
## Basic derivatives: Problems
Exercises on derivatives
## Chain rule
This section is devoted to the chain rule. More complex functions can be written as a composition of simpler functions. Such functions can be differentiated using the chain rule where we only need to differentiate the simpler functions.
## Higher order derivatives
By differentiating the derivative of a function we get what is called the second derivative. The same idea can be extended to higher order derivatives. Second derivatives will be important in the next chapter. In this section we also look at the relationship between second derivatives and whether the function is convex or concave.
## Implicit differentiation and the derivative of the inverse
This section contains two topics that are a bit more advanced. First, we look at implicit differentiation. This is a method that allows us to find a derivative when we only have an implicit relationship between two variables. Second, we look at a method which allows us to find the derivative of the inverse of the function without actually finding the inverse function.
## Implicit differentiation and the derivative of the inverse: Problems
Exercises on implicit differentiation and the derivative of the inverse
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How to Calculate Percentage
Percentage
The word percentage means per hundred. For instance, if a person saves 15% of his salary, he is said to save 15 parts out of 100 parts. Which can also be written as (15 / 100).
Information conveyed in percentage does not give the exact but gives as an approximation. The Percentage is commonly used to present data in graphs and tables.
The definition and utility of percentage can be made clear with the help of examples and further discussion in the following paragraphs.
How to convert any Fraction to percentage and vice versa
To convert any faction a / b to rate percentage, multiply it by 100 and put % sign.
Alternatively, to convert a rate percent to a fraction, divide it by 100 and remove the percentage sign.
Example 01
Quantity of water in milk constitutes 5 parts of every 20 parts of mixture. What is the percentage of water in the mixture?
Solution:
$\text{Percentage of water in the mixture}=\left( \frac{5}{20}\times 100 \right)\%=25\%$
Percentage Increase & Percentage Decrease
Increase or decrease is always in absolute terms whereas percentage increase/decrease is expressed in percentage terms
Percentage increase/decrease is calculated with respect to the base (Previous) value unless mentioned otherwise. $\text{Percentage Increase}=\frac{\text{Increase}}{\text{Base value}}\times 100$$\text{Percentage Decrease}=\frac{\text{Decrease}}{\text{Base value}}\times 100$
1. If a quantity increased by r %, then final quantity (after increase) is obtained by $\text{Final Quantity}=\text{Original Quantity}\times \left( \frac{100+r}{100} \right)$2. Likewise, if a quantity is decreased by r %, the final quantity (after decrease) is obtained by $\text{Final Quantity}=\text{Original Quantity}\times \left( \frac{100-r}{100} \right)$
Example 02
If A’s income is 20% more than that of B, then how much percent is B’s income less than that of A?
Solution:
Let the income of B be Rs. 100, then income of A = Rs. 120.
In the question B’s income is being compared with that of A and hence base value to find the percentage decrease will be the income of A,
$\text{Percentage Decrease}=\frac{\text{Decrease}}{\text{Base value}}\times 100$
$=\frac{\left( 120-100 \right)}{120}\times 100=\frac{20}{120}\times 100$
$=\frac{50}{3}\%=16\frac{2}{3}\%$
Successive Increase/ Decrease Percentage
In the case of successive changes appears. All successive changes in percentage (increase or decrease) can be represented as a single percentage.
The resultant percentage can be obtained by \left[ a+b+\frac{ab}{100} \right]\% where a and b show the first and second percentage changes.
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# Square Root And Cube Root
#### Video Lesson on Square Root And Cube Root
Loading...
Exponents, also called powers, are a way of expressing a number multiplied by itself by a certain number of times.
## Important points to remember:
• Square root: If a2 = b, we say that the square root of b is a It is written as √ b = a
• 2) Cube root: Cube root of a is denoted as 3√ a
• 3) √ab = √a × √b
• 5) Number ending in 8 can never be a perfect square.
• 6) Remember the squares and cubes of 2 to 10. This will help in easily solving the problems.
### Quick Tips and Tricks
1) Finding square root of 5, 4 and 3 digit numbers
How to find the square root of 5 digit number ?
• Step 1: Ignore last two digits 24 and just consider first three digits.
• Step 2: Find a number near or less than 174.169 is a number near to 174 which is perfect square of 13. Hence, the number in ten’s place is 13.
• Step 3: Find the number in units’s place; 4 is the number in unit’s place. Hence, remember the table, 2 and 8 numbers have unit’s digit 4.
• Step 4: now we have to find the correct number among 2 and 8.
• Step 5 : Multiply 13 (First number) with next higher number (14) i.e 13 × 14 = 182. Number 182 is greater than the first two digits, hence consider the smallest number among 2 and 8 i.e 2. Therefore, second number is 2.
Square root of 17424 = 132
• ### How to find the square root of 4 digit number?
• Step 1: Ignore last two digits 24 and just consider first three digits.
• Step 2: Find a number near or less than 60.49 is a number near to 60 which is perfect square of 7. Hence, the number in ten’s place is 7.
• Step 3: Find the number in unit’s place: 4 is the number in unit’s place. Hence, remember the table, 2 and 8 numbers have unit’s digit 4.
• Step 4: Now we have to find the correct number among 2 and 8.
• Step 5: Multiply 7 with next higher number (7+1) = 8 i .e 7×8 = 56. Number 56 is less than the first two digits, hence consider the largest number among 2 and 8 i.e 8. Therefore, second number is 8.
Square root of 6084 = 78
### How to find the square root of 3 digit number?
• Step 1: Ignore last two digits 24 and just consider first three digits.
• Step 2: Find a number near or less than 7.4 is a number near to 7 which is perfect square of 2. Hence, the number in ten’s place is 2.
• Step 3: Find the number in unit’s place: 4 is the number in unit’s place. Hence, remember the table, 2 and 8 numbers have unit’s digit 4.
• Step 4: Now we have to find the correct number among 2 and 8.
• Step 5: Multiply 7 with next higher number (2+1 = 8) i .e 2×3 = 6. Number 6 is less than the first digits, hence consider the largest number among 2 and 8 i.e 8. Therefore, second number is 8.
Square root of 784 = 28
### 2) Finding the square of large numbers
Example: 472 = 2209
Square of 47 can be easily determined by following the steps shown below:
Step 1: Split the number 47 as 4 and 7.
Step 2: Use the formula: (a + b)2 = a2 + 2ab + b2
Here, (4 + 7)2 = 42 + 2 × 4 × 7 + 72
Without considering the plus sign, write the numbers as shown below:
[16] [56] [49]
• Step 1: Write down 9 from 49 and carry 4 to 56. [-----9]
• Step 2: After adding 4 to 6, we get 10. Therefore, write down zero and carry 1 (5 + 1 = 6) to 16. [----09]
• Step 3: 6 + 6 = 12, write down 2 and carry one. [---209]
• Step 4: Finally write the answer along with (1 + 1 = 2). [2209]
### 3) Finding the cube root of 6 digit number?
Note: Cube roots of 6, 5, 4 or 3 digit numbers can be easily found out by using the same trick as used to find the square root of larger digits.
Example: 3√132651
Remember: The last 3 numbers are to cut off and the nearby cube of first remaining numbers is to be found out.
• Step 1: Split the number 132 and 651
• Step 2: 125 is the cube of 5, which is the closest number to 132. Hence, first number i.e. the number in ten’s place is 5.
• Step 3: 1 is the digit in unit’s place. Hence, the digit in unit’s place is 1.
Hence, the cube root of 132651 is 51.
• ### 4) How to find a number to be added or subtracted to make a number a perfect square ?
For easy understanding, let’s take an example. Example: 8888
• Step 1: Divide 8888 by 9. We get remainder 7.
• Step 2: Add Divisor and Quotient [9 + 9 = 18]
• Step 3: Now the next divisor will be (18 and number x) which will divide the next dividend. In this case, 4 is the number x and now the divisor becomes 184 × 4 =736.
• Step 1: This step is to be followed depend the number of digits in the dividend.
Case 1: If we have to find a number to be added to make a number perfect square, then Consider a number greater than the quotient. Her quotient is 94, hence consider 95.
942 < 8888 < 952
8836 < 8888 < 9025
Number to be added = Greater number – Given number
Number to be added = 9025 – 8888 = 137
Case 2: If we have to find a number to be subtracted to make a number perfect square, then
942 < 8888 < 952
8836 < 8888 < 9025
Number to be subtracted = Given number - Smaller number
Number to be added = 8888 – 8836 = 52
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# Mathematics for Technology I (Math 1131)
Durham College, Mathematics
Free
• 55 lessons
• 1 quizzes
• 10 week duration
• ##### Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.
• ##### Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.
• ##### Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.
• ##### Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.
• ##### Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
## Mathematics for Technology I (Math 1131)
When a general form quadratic has an a coefficient greater than 1, the trial-and-error method no longer works. Take, for example, the equation:
y = 3x² + 5x + 6
You can’t choose 3 and 2 as factors that multiply to 6 and add to 5 – it doesn’t work that way.
Arguably you could common factor the 3, leaving x² with a coefficient of 1:
y = 3 ( x² + 5/3x + 2 )
But then you’re left with finding two factors of 2 that add to 5/3!
To factor quadratics whose a > 1, we use a technique known as factoring by decomposition, which involving breaking up the middle term – hence the name.
Let’s see a few examples of this technique in action.
To summarize, factoring by decomposition involves finding two integers whose product is a × c and whose sum is b. Then, break up the middle term and factor by grouping.
Interestingly, referring back to the initial equation:
y = 3x² + 5x + 6
If you try factoring by decomposition here, it still won’t yield a factored-form quadratic. In that case, you’d have to use the quadratic formula to find the roots (more on this to come). Therefore, not all quadratic expressions of the form ax² + bx + c can be factored over the integers. The trinomial factorability test is shown below:
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## Wednesday, October 08, 2014
### Finding the Least Common Denominator of Three Fractions
The least common denominator of two or more fractions is the smallest number that can be divided evenly by each of the fractions' denominators. You can determine the LCM (least common multiple) by finding multiples of the denominators of the fractions.
Find the least common denominator of the following fractions: 5/12, 7/36, and 3/8.
8, 16, 24, 36
12, 24, 36
36
The least common denominator is 36.
## Tuesday, October 07, 2014
### Least Common Multiple
Find the least common denominator of 6, 8, 12.
6, 12, 18, 24
8, 16, 24
12, 24
The least common multiple is 24.
## Monday, October 06, 2014
### Help With GED Math Problems: Finding Lowest Common Denominator for Fractions
Building the LCD or lowest common denominators for two or more fractions can be challenging. But it is an important skill for knowing how to add and subtract fractions and one that anyone studying their GED math test will need to know.
First step: Take each denominator and factor to product of prime numbers.
Second step: Build the lowest common denominator by using each factor with the greatest exponent.
What is the lowest common denominator for the following fractions: 7/12, 7/15, 19/30? Use the product of prime factor method.
12 = 2 x 2 x 3 or 2^2 x 3
15 = 3 x 5
30 = 2 x 3 x 5
Build the lowest common denominator by using each factor (i.e. 2^2) with the greatest exponents.
If I were demonstrating the concept of building lowest common denominators to students, it would go something like this, " Let's start with the denominator twelve. The denominator 12 needs at least two twos and a three. The denominator fifteen needs a three, but because we have one from the twelve... we do not need to write another one. However, the denominator twelve needs a five, so we need to add a five. The denominator thirty needs a two... which we have so we do not need to add one. It also needs a three and a five, but because we already have both, again we do not need to add. We have now build our LCD and all we need to do is multiply the factors together. So 2 x 2 x 3 x 5 = 60. The LCD of 12, 15, and 30 is 60.
LCD = 2 x 2 x 3 x 5 = 60
## Friday, October 03, 2014
### Lowest Common Denominator
Find the lowest common denominator for the following fractions: 1/2, 1/4, 1/5
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
4, 8, 12, 16, 20
5, 10, 15, 20
Because 20 is the first common multiple of 2, 4, and 5..... it is the lowest common denominator or LCD.
## Wednesday, July 30, 2014
### GED Math Test Prep: Area of Rectangle
GED Skill: Area of rectangles
You have decided to put carpet in your 10ft by 15 ft. living room. What is the area of carpet needed?
Answer: 10ft x 15ft = 150 cubic feet
### GED Math Test Prep: Simplify the equation 3x + 7y - 2z + 3 - 6x - 5z +15
Simplify the following equation.
3x + 7y - 2z + 3 - 6x - 5z +15
Step 1: Using the associative property, rearrange the terms of the equation so that "like" terms are next to each other.
3x - 6x - 2z - 5z + 7y + 3 + 15
Step 2: Combine like terms.
-3x - 7z + 7y + 18
## Monday, May 12, 2014
### Using the Product Rule with Exponents
When you multiply constants (variables) that have the same base, you add the exponents... but keep the base unchanged.
For example:
x^2c · x^3 = x^(2+3) = x^5
(x · x) (x · x · x) = x^5
"X" squared times "X" cubed equals "X" to the fifth power.
Try a few more.
1) p^5 · p^4 =
2) 2t^2 · 3t^4
3) r^2 · 2^3 · r^5
4) 3x^2 · 2x^5 · x^4
5) (p^2)(3p^4)(3p^2)
1) p^9
2) 6t^6
3) 2r^10
4) 6r^11
5) 9p^8
## Thursday, May 08, 2014
### Simplify and Solve Using the Addition Principal of Equality
4 ( 8 - 15) + (-10) = x - 7
4 ( 8 - 15) + (-10) = x - 7
32 - 60 + (-10) = x - 7
-28 + (-10) = x - 7
-38 = x - 7
-38 + 7 = x -7 + 7
-31 = x + 0
-31 = x
## Wednesday, May 07, 2014
### Solving Equations Using the Addition Principle of Equality
Can you find the error in the following problem?
5² + (4 - 8) = x + 15
25 + 4 = x + 15
29 = x + 15
29 + (-15) = x + 15 + (-15)
14 = x + 0
14 = x
## Tuesday, May 06, 2014
### Practice Solving Simple Equations Using the Addition Property of Equality
It is important to practice the addition property of equality. See below and solve five simple equations using the addition property of equality.
Practice Problem #1
x - 11 = 41
Practice Problem #2
x - 17 = -35
Practice Problem #3
84 = 40 + x
Practice Problem #4
45 = -15 + x
Practice Problem #5
-21 = -52 + x
Practice Problem #1
x - 11 = 41
x - 11 + 11 = 41 + 11
x + 0 = 52
x = 52
check
52 - 11 = 41
41 = 41
Practice Problem #2
x - 17 = -35
x - 17 + 17 = -35 + 17
x + 0 = -18
x = -18
check
-18 - 17 = -35
-35 = -35
Practice Problem #3
84 = 40 + x
84 + ( - 40) = 40 + (-40) + x
44 = 0 + x
44 = x
check
84 = 40 + 44
84 = 84
Practice Problem #4
45 = -15 + x
45 + 15 = -15 + 15 + x
60 = 0 + x
60 = x
check
45 = -15 + 60
45 = 45
Practice Problem #5
-21 = -52 + x
-21 + 52 = -52 + 52 + x
31 = 0 + x
31 = x
check
-21 = -52 + 31
-21 = -21
## Monday, May 05, 2014
### Solving Equations Using the Addition Property of Equality
The addition principle of equality states that if a = b, then a + c = b + c.
When you solve equations using this addition principle of equality, you need to use the additive inverse property. In other words, you must add the same number to both sides of an equation.
Example #1:
x - 5 = 10
x - 5 + 5 = 10 + 5 We add the opposite of (-5) to both sides of the equation.
x + 0 = 15 We simplify -5 + 5 = 0.
x = 15 The solution is x = 15
To check the answer, simply substitute 15 in for x, in the original equation and solve.
15 - 5 = 10
10 = 10
Example #2:
x + 12 = -5
x + 12 + (- 12) = -5 + (- 12) We add the opposite of (+12) to both sides of the equation.
x + 0 = -17 We simplify +12 - 12 = 0.
x = -17 The solution is x = -17
(-17) + 12 = -5
-5 = -5
## Friday, April 11, 2014
### Practice Percent Word Problem
A car which is normally priced at \$25,437 is marked down 10%. How much would Karen save if she purchased the car at the sale price?
(Spanish translation coming soon...)
## Tuesday, March 11, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty-one more than a number is 51. What is the number?
Veinte y uno más que el número es 51. ¿Cuál es el número?
2. Thirty-seven less than a number is 45. Find the number.
Treinta y siete menos que el número es 45. Encuentre el número.
1. 30
2. 82
## Monday, March 10, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 50 is 73. Find the number.
La suma del número y 50 es 73. Encuentre el número.
2. Thirty-one more than a number is 69. What is the number?
Treinta y uno más que el número es 69. ¿Cuál es el número?
3. A number decreased by 46 is 20. Find the number.
El número que está reducido por 46 es 20. Encuentre el número.
1. 23
2. 38
3. 66
## Friday, March 07, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 28 is 74. Find the number.
La suma del número y 28 es 74. Encuentre el número.
2. Thirty-nine more than a number is 72. What is the number?
Treinta y nueve más que el número es 72. ¿Cuál es el número?
3. Eighteen less than a number is 48. Find the number.
Dieciocho menos que el número es 48. Encuentre el número.
1. 46
2. 33
3. 66
## Thursday, March 06, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by 21 is 52. Find the number.
El número que está aumentado por 21 es 52. Encuentre el número.
2. Twenty-five more than a number is 68. What is the number?
Veinte y cinco más que el número es 68. ¿Cuál es el número?
3. Forty-two more than a number is 58. What is the number?
Cuarenta y dos más que el número es 58. ¿Cuál es el número?
1. 31
2. 43
3. 16
## Wednesday, March 05, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty more than a number is 42. What is the number?
Veinte más que el número es 42. ¿Cuál es el número?
2. Forty-three more than a number is 85. What is the number?
Cuarenta y tres más que el número es 85. ¿Cuál es el número?
3. Twenty-two more than a number is 62. What is the number?
Veinte y dos más que el número es 62. ¿Cuál es el número?
1. 22
2. 42
3. 40
## Tuesday, March 04, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 26 is 42. Find the number.
La suma del número y 26 es 42. Encuentre el número.
2. Thirty more than a number is 51. What is the number?
Treinta más que el número es 51. ¿Cuál es el número?
3. Fifteen more than a number is 47. What is the number?
Quince más que el número es 47. ¿Cuál es el número?
1. 16
2. 21
3. 32
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. One-half of a number is 13. Find the number.
Una media de un número es 13. Encuentre el número.
2. A number decreased by 29 is 39. Find the number.
Un número que está reducido por 29 es 39. Encuentre el número.
3. The sum of a number and 39 is 56. Find the number.
La suma del número y 39 es 56. Encuentre el número.
1. 26
2. 68
317
## Tuesday, January 28, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by eight is 14. Find the number.
El número que aumenta por ocho es 14. Encuentre el número.
2. Three less than a number is 2. Find the number.
Tres menos que el número es dos. Encuentre el número.
1. 6
2. 5
## Monday, January 27, 2014
### Translating Words Into Algebraic Expressions Examples: (Spanish & English)
1. Six less than a number is 9. Find the number.
Seis menos que el número es nueve. Encuentre el número.
2. Ten less than a number is 9. Find the number.
Diez menos que el número es nueve. Encuentre el número.
3. A number increased by seven is 12. Find the number.
El número que aumenta por siete es 12. Encuentre el número.
1. 15
2. 19
3. 5
## Friday, January 24, 2014
### Easy Tanslating Algebra Word Problems: (Spanish & English)
1. Seven more than a number is 11. What is the number?
Siete más que el número es 11. Encuentre el número.
2. The sum of a number and six is 16. Find the number.
La Suma del número y seis es 16. Encuentre el número.
3. A number diminished by 9 is 3. Find the number.
El número que reduce por nueve es tres. Encuentre el número.
1. 4
2. 10
3. 12
## Thursday, January 23, 2014
### Translating Words into Algebraic Expressions Simple: (Spanish & English)
1. A number diminished by 2 is 7. Find the number.
El número que reduce por dos es siete. Encuentre el número.
2. A number decreased by 7 is 8. Find the number.
El número que reduce por siete es ocho. Encuentre el número.
3. A number increased by three is 13. Find the number.
El número que aumenta por tres es 13. Encuentre el número.
1. 9
2. 15
3. 10
## Wednesday, January 22, 2014
### Translating Simple Number Word Problems: (Spanish & English)
1. Six less than a number is 5. Find the number.
Seis menos que el número es cinco. Encuentre el número.
2. Six less than a number is 7. Find the number.
Seis menos que el número es siete. Encuentre el número.
3. The sum of a number and three is 11. Find the number.
La suma del número y tres es 11. Encuentre el número.
1. 11
2. 13
3. 8
## Tuesday, January 21, 2014
### Translating Word Problems Simple: (Spanish & English)
1. One-third of a number is 1. Find the number.
Un tercer del número es uno. Encuentre el número.
2. A number increased by five is 13. Find the number.
El número que aumenta por cinco es 13. Encuentre el número.
3. One-third of a number is 2. Find the number.
Un tercer del número es dos. Encuentre el número.
1. 3
2. 8
3. 6
## Monday, January 20, 2014
### Translating Simple Algebra Word Problems: (Spanish & English)
1. Two more than a number is 8. What is the number?
Dos más que el número es ocho. ¿Cuál es el número?
2. Three more than a number is 5. What is the number?
Tres más que el número es cinco. ¿Cuál es el número?
3. A number decreased by 2 is 5. Find the number.
El número que reduce por dos es cinco. Encuentre el número.
1. 10
2. 8
3. 7
## Tuesday, January 14, 2014
### Algebra Word Problem: Setting up Problem (Spanish & English)
A total of r players came to a basketball practice. The coach divides them into four groups of t players each, but two players are left over. Which expression shows the relationship between the number of players out for basketball and the number of players in each group?
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The Normal Distribution
# The Standard Normal Distribution
The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:
x = μ + (z)(σ) = 5 + (3)(2) = 11
The z-score is three.
The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = produces the distribution Z ~ N(0, 1). The value x in the given equation comes from a normal distribution with mean μ and standard deviation σ.
### Z-Scores
If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is:
The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.
Suppose X ~ N(5, 6). This says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then:
This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5.
Notice that: 5 + (2)(6) = 17 (The pattern is μ + = x)
Now suppose x = 1. Then: z = = = –0.67 (rounded to two decimal places)
This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)
Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative x is less than μ.
Try It
What is the z-score of x, when x = 1 and X ~ N(12,3)?
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.
a. Suppose a person lost ten pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.
b. Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.
b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean.
c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z?
c. z = = = 2 where µ = 2 and σ = 1.
The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two (of their own) standard deviations to the right of their respective means.
The z-score allows us to compare data that are scaled differently. To understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.
Try It
Fill in the blanks.
Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16,4). Suppose Jerome scores ten points in a game. The z–score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).
The Empirical RuleIf X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following:
• About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean).
• About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean).
• About 99.7% of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean.
• The z-scores for +1σ and –1σ are +1 and –1, respectively.
• The z-scores for +2σ and –2σ are +2 and –2, respectively.
• The z-scores for +3σ and –3σ are +3 and –3 respectively.
The empirical rule is also known as the 68-95-99.7 rule.
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).
a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
a. –0.32, 0.32, left, 170
b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
b. 177.98 cm, 1.27, right
Try It
Use the information in (Figure) to answer the following questions.
1. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males from 1984 to 1985. Then Y ~ N(172.36, 6.34).
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).
Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations?
The z-score for x = -160.58 is z = –1.5.
The z-score for y = 162.85 is z = –1.5.
Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.
Try It
In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then X ~ N(496, 114).
Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21 as they compare to their respective means and standard deviations?
Suppose x has a normal distribution with mean 50 and standard deviation 6.
• About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively.
• About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively.
• About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 95% of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 from the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively.
Try It
Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie?
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in 1984 to 1985. Then Y ~ N(172.36, 6.34).
1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
3. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
1. About 68% of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1.
2. About 95% of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2.
3. About 99.7% of the values lie between 153.34 cm and 191.38 cm. The z-scores are –3 and 3.
Try It
The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points.
1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
3. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
### References
“Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013).
“The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013).
“2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013).
“Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013).
Data from the San Jose Mercury News.
Data from The World Almanac and Book of Facts.
“List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013).
Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013).
### Chapter Review
A z-score is a standardized value. Its distribution is the standard normal, Z ~ N(0, 1). The mean of the z-scores is zero and the standard deviation is one. If z is the z-score for a value x from the normal distribution N(µ, σ) then z tells you how many standard deviations x is above (greater than) or below (less than) µ.
### Formula Review
z = a standardized value (z-score)
mean = 0; standard deviation = 1
To find the kth percentile of X when the z-scores is known:
k = μ + (z)σ
z-score: z =
Z = the random variable for z-scores
A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________.
ounces of water in a bottle
A normal distribution has a mean of 61 and a standard deviation of 15. What is the median?
<!– <solution id=”fs-idm63161360″> 61 –>
X ~ N(1, 2)
σ = _______
2
A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________.
<!– <solution id=”fs-idp15866016″> diameter of a rubber ball –>
X ~ N(–4, 1)
What is the median?
–4
X ~ N(3, 5)
σ = _______
<!– <solution id=”fs-idm120454704″> 5 –>
X ~ N(–2, 1)
μ = _______
–2
What does a z-score measure?
<!– <solution id=”fs-idm57247936″> The number of standard deviations a value is from the mean. –>
What does standardizing a normal distribution do to the mean?
The mean becomes zero.
Is X ~ N(0, 1) a standardized normal distribution? Why or why not?
<!– <solution id=”fs-idp48272224″> Yes because the mean is zero, and the standard deviation is one. –>
What is the z-score of x = 12, if it is two standard deviations to the right of the mean?
z = 2
What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean?
<!– <solution id=”fs-idp6573504″> z = –1.5 –>
What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean?
z = 2.78
What is the z-score of x = 7, if it is 0.133 standard deviations to the left of the mean?
<!– <solution id=”fs-idp38051616″> z = –0.133 –>
Suppose X ~ N(2, 6). What value of x has a z-score of three?
x = 20
Suppose X ~ N(8, 1). What value of x has a z-score of –2.25?
<!– <solution id=”fs-idp19875360″> x = 5.75 –>
Suppose X ~ N(9, 5). What value of x has a z-score of –0.5?
x = 6.5
Suppose X ~ N(2, 3). What value of x has a z-score of –0.67?
<!– <solution id=”fs-idp31923120″> x = –0.01 –>
Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?
x = 1
Suppose X ~ N(4, 2). What value of x is two standard deviations to the right of the mean?
<!– <solution id=”fs-idm47755696″> x = 8 –>
Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?
x = 1.97
Suppose X ~ N(–1, 2). What is the z-score of x = 2?
<!– <solution id=”fs-idm119642576″> z = 1.5 –>
Suppose X ~ N(12, 6). What is the z-score of x = 2?
z = –1.67
Suppose X ~ N(9, 3). What is the z-score of x = 9?
<!– <solution id=”fs-idm56843792″> z = 0 –>
Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5?
z ≈ –0.33
In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.
<!– <solution id=”fs-idm62539280″> 1.25, left –>
In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean.
0.67, right
In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean.
<!– <solution id=”fs-idm102069296″> six, right –>
In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean.
3.14, left
In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean.
<!– <solution id=”fs-idm98531584″> 1.7, left –>
About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution?
About what percent of the x values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution?
<!– <solution id=”fs-idm47813520″> about 95.45% –>
About what percent of x values lie between the second and third standard deviations (both sides)?
Suppose X ~ N(15, 3). Between what x values does 68.27% of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15).
<!– <solution id=”fs-idm33513040″> between 12 and 18 –>
Suppose X ~ N(–3, 1). Between what x values does 95.45% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3).
between –5 and –1
Suppose X ~ N(–3, 1). Between what x values does 34.14% of the data lie?
<!– <solution id=”fs-idm68579024″> between –4 and –3 or between –3 and –2 –>
About what percent of x values lie between the mean and three standard deviations?
About what percent of x values lie between the mean and one standard deviation?
<!– <solution id=”fs-idp3556992″> about 34.14% –>
About what percent of x values lie between the first and second standard deviations from the mean (both sides)?
About what percent of x values lie betwween the first and third standard deviations(both sides)?
<!– <solution id=”fs-idp56159920″> about 34.46% –>
Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.
Define the random variable X in words. X = _______________.
The lifetime of a Sunshine CD player measured in years.
X ~ _____(_____,_____)
<!– <solution id=”fs-idp89883200″> X ~ N(4.1, 1.3) –>
### Homework
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the median recovery time?
1. 2.7
2. 5.3
3. 7.4
4. 2.1
<!– <solution id=”fs-idm68359392″> b –>
What is the z-score for a patient who takes ten days to recover?
1. 1.5
2. 0.2
3. 2.2
4. 7.3
c
The length of time to find it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true?
1. The data cannot follow the uniform distribution.
2. The data cannot follow the exponential distribution..
3. The data cannot follow the normal distribution.
1. I only
2. II only
3. III only
4. I, II, and III
<!– <solution id=”fs-idp16013392″> b –>
The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences.
1. 77 inches
2. 85 inches
3. If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer.
1. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.
2. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.
3. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely.
The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution.
1. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters.
2. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him?
<!– <solution id=”eip-774″> Use the z-score formula. 100 – 125 14 ≈ –1.8 and 100 – 125 14 ≈ 1.8 I would tell him that 2.5 standard deviations below the mean would give him a blood pressure reading of 90, which is below the range of 100 to 150. –>
Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score then X ~ N (125, 14).
1. Kyle’s systolic blood pressure is 175.
2. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age.
3. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age.
4. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.
2. Calculate Kyle’s blood pressure.
1. iv
2. Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5.
Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them.
1. 11 kg
2. 7.9 kg
3. 12.2 kg
<!– <solution id=”eip-182″> 11 – 10.2 0.8 = 1 A child who weighs 11 kg is one standard deviation above the mean of 10.2 kg. 7.9 – 10.2 0.8 = –2.875 A child who weighs 7.9 kg is 2.875 standard deviations below the mean of 10.2 kg. 12.2 – 10.2 0.8 = 2.5 A child who weighs 12.2 kg is 2.5 standard deviation above the mean of 10.2 kg. –>
In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115.
1. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence.
2. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score?
3. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?
Let X = an SAT math score and Y = an ACT math score.
1. X = 720 = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520.
2. z = 1.5
The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.
3. = ≈ 1.59, the z-score for the SAT. = ≈ 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).
<!– <para id=”fs-idp207361600″>Use the following information to answer the next three exercises: X ~ U(3, 13) –>
### Glossary
Standard Normal Distribution
a continuous random variable (RV) X ~ N(0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N(0, 1).
z-score
the linear transformation of the form z = ; if this transformation is applied to any normal distribution X ~ N(μ, σ) the result is the standard normal distribution Z ~ N(0,1). If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ, the result is called the z-score of x. The z-score allows us to compare data that are normally distributed but scaled differently.
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# Calculations With the Gamma Function
The gamma function is defined by the following complicated looking formula:
Γ ( z ) = ∫0e - ttz-1dt
One question that people have when they first encounter this confusing equation is, “How do you use this formula to calculate values of the gamma function?” This is an important question as it is difficult to know what this function even means and what all of the symbols stand for.
One way to answer this question is by looking at several sample calculations with the gamma function. Before we do this, there are a few things from calculus that we must know, such as how to integrate a type I improper integral, and that e is a mathematical constant
## Motivation
Before doing any calculations, we examine the motivation behind these calculations. Many times the gamma functions show up behind the scenes. Several probability density functions are stated in terms of the gamma function. Examples of these include the gamma distribution and students t-distribution, The importance of the gamma function cannot be overstated.
## Γ ( 1 )
The first example calculation that we will study is finding the value of the gamma function for Γ ( 1 ). This is found by setting z = 1 in the above formula:
0e - tdt
We calculate the above integral in two steps:
• The indefinite integral ∫e - tdt= -e - t + C
• This is an improper integral, so we have ∫0e - tdt = limb → ∞ -e - b + e 0 = 1
## Γ ( 2 )
The next example calculation that we will consider is similar to the last example, but we increase the value of z by 1. We now calculate the value of the gamma function for Γ ( 2 ) by setting z = 2 in the above formula. The steps are the same as above:
Γ ( 2 ) = ∫0e - tt dt
The indefinite integral ∫te - tdt=- te - t -e - t + C. Although we have only increased the value of z by 1, it takes more work to calculate this integral. In order to find this integral, we must use a technique from calculus known as integration by parts. We now use the limits of integration just as above and need to calculate:
limb → ∞ - be - b -e - b -0e 0 + e 0.
A result from calculus known as L’Hospital’s rule allows us to calculate the limit limb → ∞ - be - b = 0. This means that the value of our integral above is 1.
## Γ (z +1 ) =zΓ (z )
Another feature of the gamma function and one which connects it to the factorial is the formula Γ (z +1 ) =zΓ (z ) for z any complex number with a positive real part. The reason why this is true is a direct result of the formula for the gamma function. By using integration by parts we can establish this property of the gamma function.
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Here, we will show you how to work with Radical equation solver with steps. How to solve using substitution is best explained with an example. Let's say you have the equation 4x + 2y = 12. To solve this equation using substitution, you would first need to isolate one of the variables. In this case, let's isolate y by subtracting 4x from both sides of the equation. This gives us: y = (1/2)(12 - 4x). Now that we have isolated y, we can substitute it back into the original equation in place of y. This gives us: 4x + 2((1/2)(12 - 4x)) = 12. We can now solve for x by multiplying both sides of the equation by 2 and then simplifying. This gives us: 8x + 12 - 8x = 24, which simplifies to: 12 = 24, and therefore x = 2. Finally, we can substitute x = 2 back into our original equation to solve for y. This gives us: 4(2) + 2y = 12, which simplifies to 8 + 2y = 12 and therefore y = 2. So the solution to the equation 4x + 2y = 12 is x = 2 and y = 2.
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# Find the square root of: $12\sqrt 5 + 2\sqrt {55}$ A. $\left( {\sqrt {11} + 1} \right)\sqrt[4]{5}$ B. $\sqrt[4]{5}\left( {1 + \sqrt 5 } \right)$ C. $\sqrt[4]{5}\left( {\sqrt {11} + \sqrt 5 } \right)$ D. $\sqrt 5 \left( {\sqrt {11} + 1} \right)$
Last updated date: 19th Jun 2024
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Hint: In this question, we will use factorization, and expansion of algebraic identities. For this problem, we will use the algebraic identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$ .
Complete step by step solution:
Now, in this question, we have to find the square root of $12\sqrt 5 + 2\sqrt {55}$.
So it will become: $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, which on simplification will become:
$\sqrt {12\sqrt 5 + 2\sqrt {55} } \\ = \sqrt {12\sqrt 5 + 2\sqrt {5 \times 11} } \\ = \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\$
Now, to solve $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, we will take $\sqrt 5$ common within the under root and get:
$= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\$
Now we will change the term $(12 + 2\sqrt {11} )$ inside the under root sign to express it in terms of ${(a + b)^2}$ .
Now we can write $(12 + 2\sqrt {11} )$ as:
$(1 + 11 + 2\sqrt {11} )$ which can we reframed as:
$({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$, comparing it with the RHS of the expansion of the algebraic identity ${(a + b)^2}$ which is given as:
${a^2} + 2ab + {b^2}$
We will get $({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$= ${a^2} + 2ab + {b^2}$
So that,
${a^2} = {1^2}, \\ 2ab = 2.1.\sqrt {11} \\ {b^2} = {(\sqrt {11} )^2} \\$
Such that we get :
$a = 1, \\ 2ab = 2.1.\sqrt {11} \\ b = \sqrt {11} \\$
Now, since
${a^2} + 2ab + {b^2} = {(a + b)^2}$
Then putting the values obtained above:
$({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} ) = {(1 + \sqrt {11} )^2}$
Therefore $\sqrt {12\sqrt 5 + 2\sqrt {55} }$ will now become:
$= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 ({1^2} + {{(\sqrt {11} )}^2} + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 {{(1 + \sqrt {11} )}^2}} \\ = \sqrt {\sqrt 5 } (1 + \sqrt {11} ) \\ = \sqrt[4]{5}(1 + \sqrt {11} ) \\$
So, finally we can say that :
Square root of $12\sqrt 5 + 2\sqrt {55}$
$= \sqrt[4]{5}(\sqrt {11} + 1)$
Hence, the correct answer is option A.
Note: We cannot afford to forget the square root operation throughout the solution of this problem. For such problems, which require us to find the square root of another square root, we need to identify the algebraic expansion accurately so that we can get the correct corresponding algebraic identity to simplify and evaluate the square root.
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# उपसूत्र ६. यावदूनं तावदुनं
The Upasūtra: yāvadūnaṃ tāvadūnaṃ (Lessen by the deficiency) is used for cubing (x3) a number, that is close to a power of ten (10n). The technique followed by this Upasūtra uses Rekhanks & Vinculum Numbers (discussed here »).
As an illustration, let us use this Upasūtra for:
963
Steps 963 1. Consider the nearest power of 10 as the Base In this case, the Base, closest to 96, is 100. 2. Find the deficiency, and represent as Rekhank for negative deficiency. In this case, the Deficiency =100 - 96 = 4 3. Subtract the twice the deficiency from that number In this case, 96 - (2 × 4) = 88 4. Set-up twice of as many Zeroes, as the Base In this case, 100 has two Zeroes. So, we get 88,0000 5. Multiply thrice the Deficiency, with the Deficiency - and set-up as many Zeroes as the Base. In this case, (3 × 4) × 4 = 12 × 4 = 48 And set-up with two Zeroes, we get 4800 6. Cube the Deficiency, and subtract from the sum previous numbers to get the answer. In this case, 43 = 64 And, 88,00,00 + 48,00 - 64 = 884736, which is the answer!
Let us take another example, for something like:
1033
The Base is 100 and, the Deficiency is: 100 - 103 = -3 = 3 And, 33 = 27 Also, (3 × 3) × 3 = 27 Now, 103 - (2 × 3) = 103 + 6 = 109 Also, 109,00,00 + 2700 = 1092700 And, 1092700 - 27 = 1092700 + 27 = 1092727, which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
9913
Clearly, the Deficiency is 9And, 93 = 729 Also, (3 × 9) × 9 = 243 Now, 991 - (2 × 9) = 991 - 18 = 973 Also, 973,000,000 + 243,000 = 973,243,000 And, 973,243,000 - 729 = 973242271 Thus, 9913 = 973,242,271
Again, for something like:
100063
Clearly, the Deficiency is 6And, 63 = 216 Also, (3 × 6) × 6 = 108 Now, 10006 - (2 × 6) = 10006 + 12 = 10018 Also, 10018,0000,0000 + 108,0000 = 10018,0108,0000 And, 10018,0108,0000 - 216 = 10018,0108,0000 + 216 = 10018,0108,0216 Thus, 100063 = 1,001,801,080,216
Similarly, one take take any (n-digit) number and execute the steps above to obtain the desired calculated value.
But, why does it work? For the above Upasūtra (yāvadūnaṃ tāvadūnaṃ), let us consider the following:
Assuming N is a number close (and less) to a power of 10, then N = a - b 'a' being the power of 10, and 'b' being the Deficiency Now, N3 = (a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - a2b - 2a2b + 3ab2 - b3 = a2(a - b - 2b) + a(3b × b) + b3 But, N = a - b. So, substituting a = N + b N3 = a2(N + b - b - 2b) + a(3b × b) + b3 = a2(N - 2b) + a(3b × b) + b3 This is exactly what this Upasūtra makes us do.
Lastly, please remember that, as in any other form of mathematics, the mastery of Vedic Mathematics require practice and the judgement of applying the optimal method for a given scenario - a guideline of which, is presented in Applications »
« Upasūtra 5 Upasūtra 7 »
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maths > commercial-arithmetics
Proportions
what you'll learn...
overview
Two ratios are said to be in proportion, if the ratios are equivalent. For example 2:4$2 : 4$ and 3:6$3 : 6$ are equivalent. Such equivalent ratios are formally represented as a proportion. The representation is 2:4::3:6$2 : 4 : : 3 : 6$.
illustrative example
A ratio of two quantities helps: to understand and to use the comparative measure of quantities.
Let us consider making dough for chappati or pizza. The recipe gives that for 400$400$ gram of flour, 150$150$ml water is used.
One person has only 200$200$ gram of flour. In that case, the person can reduce the water to $75$ml.
For $400$ gram flour $150$ml water is used. This is in $400 : 150$ ratio.
For $200$ gram flour $75$ml water is used. This is in $200 : 75$ ratio.
These two ratios can be simplified to $8 : 3$.
To denote that two ratios are identical, they are said to be in same proportion. That is $400 : 150$ is in the same proportion as $200 : 75$. This is given as $400 : 150 : : 200 : 75$.
proportion
The word "proportion" means: comparative measurement of quantities".
Pro-portion was from root word meaning "person's portion or share".
Proportion : Two ratios are said to be in proportion if the corresponding terms of ratio are identical in the simplified form.
Consider the example $2 : 3 : : 4 : 6$ proportion.
• The numbers in the proportion are called first term, second term, third term, and fourth term in the order.
• The first and fourth terms are called the extremes of the proportion.
• The second and third terms are called the means of the proportion.
The word "extreme" means: farthest from the center.
The root word is from "exter" meaning outer.
The word "mean" means: average. The word is derived from a root word meaning "middle".
illustrative example
• Fruit basket $A$ has $4$ apples and $16$ oranges. Ratio of apples to oranges is $4 : 16$ which is $1 : 4$.
• Fruit basket $B$ has $20$ apples and $80$ oranges. Ratio of apples to oranges is $20 : 80$ which is $1 : 4$.
The proportion of apples to oranges in the two baskets is $4 : 16 : : 20 : 80$.
The proportion of apples to oranges in basket $A$ and basket $B$ is $4 : 16 : : 20 : 80$.
The word "Proportion" is also used to specify the simplified ratio, as in the following.
The proportion of apples to oranges in basket $A$ and basket $B$ is $1 : 4$.
Students are reminded to note the context in which the word "proportion" is used.
The proportion of count in basket A to count in basket B for apples and oranges is $4 : 20 : : 16 : 80$.
This proportion is given as
apples of basket $A$ to basket $B$ is in the same proportion as oranges of basket $A$ to basket $B$
• Fruit basket $A$ has $4$ apples and $16$ oranges.
• Fruit basket $B$ has $20$ apples and $80$ oranges.
The proportion of apples to oranges in basket $A$ and basket $B$ is $4 : 16 : : 20 : 80$.
The proportion of apples and oranges in basket $A$ to basket $B$ is $4 : 20 : : 16 : 80$.
Students are reminded to note the context in which proportion is defined.
proportion to fractions
We learned that "Ratio can be equivalently represented as a fraction.".
A basket has $3$ apples and $4$ oranges.
• The ratio of the number of apples to number of oranges is $3 : 4$.
• Number of apples are $\frac{3}{4}$ of the number of oranges.
The number of apples to number of oranges in basket A and B is in proportion $3 : 4 : : 6 : 8$.The following are all true.
the number of apples is $\frac{3}{4}$ of the number of oranges in basket A
the number of apples is $\frac{6}{8}$ of the number of oranges in basket B
the number of apples is $\frac{3}{4}$ of the number of oranges -- in both basket A and basket B
Note that in a proportion, the two fractions are equivalent fractions.
$\frac{6}{8}$, when simplified, is $\frac{3}{4}$.
Given the proportion $3 : 4 : : 6 : 8$,
$\frac{3}{4} = \frac{6}{8}$, that is, the two ratios given as fractions are always equal.
formula
Given a proportion, $a : b : : c : d$, it is understood that $\frac{a}{b} = \frac{c}{d}$. In that case,
$a \times d = b \times c$.
Given a proportion, $a : b : : c : d$, it is understood that $\frac{a}{b} = \frac{c}{d}$.
Given that
$\frac{a}{b} = \frac{c}{d}$
Note that $a , b , c , d$ are numbers. As per the properties of numbers, if two numbers are equal, then the numbers multiplied by another number are equal. (eg: if $4 = 2 \times 2$, then multiplying by $5$ we get $4 \times 5 = 2 \times 2 \times 5$.)
$\frac{a}{b}$ and $\frac{c}{d}$ are two numbers that are equal. On multiplying these numbers by $b d$, we get the two numbers $\frac{a}{b} \times b d$ and $\frac{c}{d} \times b d$. Simplifying these two numbers we get, $a d$ and $b c$. As per the property, these two numbers are equal.
$a d = b c$.
That is product of extremes and product of means are equal.
examples
Given the proportion $3 : 4 : : x : : 12$ find the value of $x$.
The answer is "$9$". Products of extremes equals product of means.
$3 \times 12 = 4 \times x$
$x = \frac{36}{4}$
$x = 9$
summary
Proportion : Two ratios are said to be in proportion if the corresponding terms of ratio are identical in the simplified form.
Mean-Extreme Property of Proportions : product of extremes = product of means
If $a : b : : c : d$ is a proportion, then $a d = b c$
Outline
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# Expand Logarithms Using The Product Property Worksheet
Assignment will be available soon
Expanding logarithms using the product property means turning one log of a multiplication into several logs added together. For example, $$\log_b(xy)$$ becomes $$\log_b(x) + \log_b(y)$$. This makes tricky log problems easier to handle and solve because we break them into simpler parts. It's like taking a big math problem and splitting it into smaller, more manageable pieces.
Example: Expand $$\log_2(8x)$$ using the product property of logarithms.
Algebra 2
Logarithms
## How Will This Worksheet on "Expand Logarithms Using the Product Property" Benefit Your Student's Learning?
• Splitting logarithms using the product property makes tough math problems easier to handle.
• Helps solve log equations by changing multiplications into easier additions.
• Learning this technique aids in understanding advanced math, such as calculus.
• Reduces errors and simplifies complex problems.
• Enhances logical thinking and problem-solving skills by manipulating mathematical expressions.
## How to Expand Logarithms Using the Product Property?
• Start with a logarithm of a product, such as $$\log_b(xy)$$.
• Use the product property of logarithms, which states $$\log_b(xy) = \log_b(x) + \log_b(y)$$.
• Separate the logarithm into the sum of logarithms of each factor involved in the product.
• Expand the logarithmic expression by writing it as a sum of simpler logarithmic terms, facilitating easier calculation and manipulation.
## Solved Example
Q. Expand the logarithm. Assume all expressions exist and are well-defined. Write your answer as a sum or difference of common logarithms or multiples of common logarithms. The inside of each logarithm must be a distinct constant or variable. $\log uv$
Solution:
1. Identify Property: Identify the property of logarithm used to expand $\log uv$. Product property is used to expand a product within a logarithm.
2. Apply Property: Apply the product property to expand $\log uv$. Product Property: $\log_b (PQ) = \log_b P + \log_b Q$$\newline$ $\log uv = \log(u) + \log(v)$
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# Scientific Notation and Significant Figures
Document Sample
``` Bell Ringer: Oct. 4, 2010:
Complete the table below. Place X in the appropriate
box to indicate the type of each measurement unit.
Reference: Physical Science, page 16
Measurement SI Unit Derived Unit
Gram per centimeter
cubed (g/cm3)
Decimeter (dm)
Liter (L)
Meter cubed (m3)
Kilogram (kg)
Glenn C. Soltes
Integrated Science
Biology
2010-2011
Objectives:
Define Scientific Notation and Significant Figures.
Identify the rules in writing scientific notation and
significant figures.
Use scientific notation and significant figures in
problem solving.
Identify the significant figures in calculations.
Scientific Notation
A short-hand way of writing large
numbers without writing all of the zeros.
A number is expressed in scientific
notation when it is in the form
a x 10n
where a is between 1 and 10
and n is an integer
The Distance From the Sun to the
Earth
93,000,000 miles
Write the width of the universe in
scientific notation.
210,000,000,000,000,000,000,000 miles
Where is the decimal point now?
After the last zero.
Where would you put the decimal to make
this number be between 1 and 10?
Between the 2 and the 1
2.10,000,000,000,000,000,000,000.
How many decimal places did you move
the decimal?
23
When the original number is more than
1, the exponent is positive.
The answer in scientific notation is
2.1 x 1023
Write 28750.9 in scientific notation.
1. 2.87509 x 10-5
2. 2.87509 x 10-4
3. 2.87509 x 104
4. 2.87509 x 105
2) Express 1.8 x 10-4 in decimal
notation.
0.00018
3) Express 4.58 x 106 in decimal
notation.
4,580,000
On the graphing calculator, scientific
notation is done with the button.
4.58 x 106 is typed 4.58 6
Practice Problem
Write in scientific notation.
Decide the power of ten.
1) 98,500,000 =
2) 64,100,000,000 =
3) 279,000,000 =
4) 4,200,000 =
5) .000567 =
A prescribed decimal place that determines the
amount of rounding off to be done based on the
precision of the measurement.
There are 2 kinds of
numbers:
Exact: the amount of
Known with certainty.
Approximate: weight,
height—anything
MEASURED.
No measurement is
perfect.
When a measurement
is recorded only those
digits that are
dependable are
written down.
If you measured the
width of a paper with
record 21.7cm.
To a mathematician
21.70, or 21.700 is the
same.
But, to a scientist 21.7cm and
21.70cm is NOT the same
21.700cm to a scientist
means the
measurement is
accurate to within one
thousandth of a cm.
If you used an ordinary
ruler, the smallest
marking is the mm, so
to be recorded as 21.7cm.
Rule: All digits are
significant starting
with the first non-zero
digit on the left.
Exception to rule: In
whole numbers that
end in zero, the zeros
at the end are not
significant.
How many significant
figures?
7 1
40 1
0.5 1
0.00003 1
7 x 105 1
7,000,000 1
2ndException to rule:
If zeros are
sandwiched between
non-zero digits, the
zeros become
significant.
How many significant figures
here?
1.2 2
2100 2
56.76 4
4.00 3
0.0792 3
7,083,000,000 4
How many sig figs here?
3401 4
2100 2
2100.0 5
5.00 3
0.00412 3
8,000,050,000 6
Practice: Count the number of
significant figures.
1. 80000
2. 0.0015
3. 8 002 000
4. 1.12
5. 1.oo5
subtracting measured
have no more places after
the decimal than the LEAST
of the measured numbers.
2.45cm + 1.2cm = 3.65cm,
Round off to = 3.7cm
7.432cm + 2cm = 9.432
round to 9cm
Multiplication and Division
Rule: When
multiplying or
dividing, the result
can have no more
significant figures
than the least reliable
measurement.
A couple of examples
56.78 cm x 2.45cm = 139.111 cm2
Round to 139cm2
75.8cm x 9.6cm = ?
Perform the following calculations, and write the answer
with the correct number of significant figures.
a. 12.65 cm x 42.1 cm
b. 3.02 cm x 6.3 cm x 8.225 cm
c. 3.7 g ÷ 1.o83 cm3
Credit:
Holt, Physical Science 2006
```
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# Geometric Probability
## Use geometric properties to evaluate probability
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Basic Geometric Probabilities
A rectangular dartboard that measures 12 inches by 24 inches has a 2-inch by 2-inch red square painted at its center. What is the probability that a dart that hits the dartboard will land in the red square?
### Geometric Probabilities
Sometimes we need to use our knowledge of geometry to determine the likelihood of an event occurring. We may use areas, volumes, angles, polygons or circles.
Let's solve the following problems.
1. A game of pin-the-tale-on-the-donkey has a rectangular poster that is 2ft by 2ft. The area in which the tale should be pinned is shown as a circle with radius 1 inch. Assuming that the pinning of the tale is completely random and that it will be pinned on the poster (or the player gets another try), find the probability of pinning the tale in the circle?
This probability can be found by dividing the area of the circle target by the area of the poster. We must have the same units of measure for each area so we will convert the feet to inches.
\begin{align*}\frac{1^2 \pi}{24^2} \thickapprox 0.005454 \ or \ \text{about} \ 0.5 \% \ \text{chance}.\end{align*}
1. In a game of chance, a pebble is dropped onto the board shown below. If the radius of each of blue circle is 1 cm, find the probability that the pebble will land in a blue circle.
The area of the square is \begin{align*}16 \ cm^2\end{align*}. The area of each of the 16 circles is \begin{align*}1^2 \pi=\pi\end{align*}. The probability of the pebble landing in a circle is the sum of the areas of the circles divided by the area of the square.
\begin{align*}P(\text{blue circle}) = \frac{16 \pi}{64} \thickapprox 0.785\end{align*}
1. What is the probability that a randomly thrown dart will land in a red area on the dart board shown? What is the probability that exactly two of three shots will land in the red? The radius of the inner circle is 1 unit and the radius of each annulus is 1 unit as well.
First we need to determine the probability of landing in the red. There are four rings of width 1 and the radius of the center circle is also 1 so the total radius is 5 units. The area of the whole target is thus \begin{align*}25 \pi\end{align*} square units. Now, we need to find the areas of the two red rings and the red circular center. The center circle area is \begin{align*}\pi\end{align*} square units. The outside ring area can be found by subtracting the area inside from the entire circle’s area. The inside circle will have a radius of 4 units, the area of the outer ring is \begin{align*}25 \pi -16 \pi=9 \pi\end{align*} square units. This smaller red ring’s area can be found similarly. The circle with this red ring on the outside has a radius of 3 and the circle inside has a radius of 2 so, \begin{align*}9 \pi -4 \pi=5 \pi\end{align*} square units. Finally, we can add them together to get the total red area and divide by the area of the entire target. \begin{align*}\frac{9 \pi+5 \pi+ \pi}{25 \pi}=\frac{15 \pi}{25 \pi}=\frac{3}{5}\end{align*}. So the probability of hitting the red area is \begin{align*}\frac{3}{5}\end{align*} or 60%.
For the second part of the problem we will use a binomial probability. There are 3 trials, 2 successes and the probability of a success is 0.6: \begin{align*}\dbinom{3}{2}(0.6)^2(0.4)=0.432\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the probability that a dart that hits the dartboard will land in the red square.
This probability can be found by dividing the area of the red square by the area of the dartboard. The area of the dartboard is \begin{align*}12\times24=288\end{align*}. The area of the red square is \begin{align*}2\times2=4\end{align*}. Therefore, the probability of the dart landing in the red square is
\begin{align*}\frac{4}{288}\\ \frac{1}{72}\\ \approx 0.0139\end{align*}
Therefore, there is about a 1.39% chance the dart will hit the red square.
#### Example 2
Consider the picture below. If a “circle” is randomly chosen, what is the probability that it will be:
1. red
\begin{align*}\frac{29}{225}\end{align*}
1. yellow
\begin{align*}\frac{69}{225}\end{align*}
1. blue or green
\begin{align*}\frac{84}{225}\end{align*}
1. not orange
\begin{align*}\frac{182}{225}\end{align*}
#### Example 3
If a dart is randomly thrown at the target below, find the probability of the dart hitting in each of the regions and show that the sum of these probabilities is 1. The diameter of the center circle is 4 cm and the diameter of the outer circle is 10 cm. What is the probability that in 5 shots, at least two will land in the 4 region?
\begin{align*}P(1)&=\frac{2^2 \pi}{5^2 \pi}=\frac{4}{25} \\ P(2)&=\frac{120}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{3} \times \frac{21 \pi}{25 \pi}=\frac{7}{25} \\ P(3)&=\frac{90}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{4} \times \frac{21 \pi}{25 \pi}=\frac{21}{100} \\ P(4)&=\frac{150}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{5}{12} \times \frac{21 \pi}{25 \pi}=\frac{35}{100}\end{align*}
\begin{align*}& P(1) + P(2) + P(3) + P(4) = \\ &=\frac{4}{25}+\frac{7}{25}+\frac{21}{100}+\frac{35}{100} \\ &=\frac{16}{100}+\frac{28}{100}+\frac{21}{100}+\frac{35}{100} \\ &=\frac{100}{100}=1\end{align*}
The probability of landing in region 4 at least twice in five shots is equivalent to \begin{align*}1 - \left [ P(0) + P(1) \right ]\end{align*}.
Use binomial probability to determine these probabilities:
\begin{align*}&1-\left [ \dbinom{5}{0} \left(\frac{35}{100}\right)^0 \left(\frac{65}{100}\right)^5+ \dbinom{5}{1} \left(\frac{35}{100}\right)^1 \left(\frac{65}{100}\right)4\right ] \\ &=1-(0.116029+0.062477) \\ & \thickapprox 0.821\end{align*}
### Review
Use the diagram below to find the probability that a randomly dropped object would land in a triangle of a particular color.
1. yellow
2. green
3. plum
4. not yellow
5. not yellow or light blue
The dart board to the right has a red center circle (bull’s eye) with area \begin{align*}\pi \ cm^2\end{align*}. Each ring surrounding this bull’s eye has a width of 2 cm. Use this information to answer the following questions.
1. Given a random throw of a dart, what is the probability that it will land in a white ring?
2. What is the probability of a bull’s eye?
3. What is the probability that in 10 throws, exactly 6 land in the black regions?
4. What is the probability that in 10 throws, at least one will land in the bull’s eye?
5. How many darts must be thrown to have a 95% chance of making a bull’s eye?
To see the Review answers, open this PDF file and look for section 12.11.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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# 4th Grade Multiplication Worksheet PDF | Multiplication Activities for Grade 4
Students of Grade 4 spend more time-solving multiplication topics such as mental math multiplication, multi-digit multiplication. Use our comprehensive collection of 4th Grade Math Multiplication Worksheets PDF for free and ace your preparation. Math Worksheets on Multiplication for Class 4 provided will have questions ranging from easy to difficult ones. Multiplying Worksheets with Answers makes it easy for you to develop mental math multiplication skills right in your head, multiply in columns, etc.
Do Check:
I. Multiply the numbers given below
1. Multiply the numbers 132 and 24
Solution:
1. Arrange the number 132 one on top as the multiplicand and 24 at the bottom. Line up the place values in columns.
2. Starting with the one’s digit of the bottom number, i.e. 4 × 2=8. Write the answer below the equals line
3. Multiply the one digit of the bottom number to the next digit to the left in the top number i.e. i.e. 4 × 3=12. Here the answer is greater than nine, write2 in the one’s place as the answer and carry the tens digit 1 Proceed right to left.
4. Multiply 4 with 1.i.e. 4 × 1=4. Add the carry forward digit 1 to 4.Write 5 as the answer.
5. Move to the tens digit in the bottom number i.e. 2. Multiply as above, but this time write your answers in a new row, shifted one digit place to the left.
6. Multiply 2 with 2 i.e. 2 × 2=4.
Multiply 2 with 3 i.e. 2 × 3=6
Multiply 2 with 1 i.e. 2 × 1=2
7. After multiplying, draw another answer line below the last row of answer numbers.
8. Use long addition to add the number columns from right to left, carrying as you normally do for long addition.
Therefore, the product of 132 and 24 is 3168.
2. Multiply the numbers 543 and 12
Solution:
1. Arrange the number 543 one on top as the multiplicand and 12 at the bottom. Line up the place values in columns.
2. Starting with the one’s digit of the bottom number, i.e. 2 × 3=6. Write the answer below the equals line
3. Multiply the one digit of the bottom number to the next digit to the left in the top number i.e. i.e. 2 × 4=8. Write 8 as the answer.
4. Move to the tens digit in the bottom number i.e. 1. Multiply as above, but this time write your answers in a new row, shifted one digit place to the left.
6. Multiply 1 with 3 i.e. 1 × 3=3.
Multiply 1 with 4 i.e. 1 × 4=4
Multiply 1 with 5 i.e. 1 × 5=5
7. After multiplying, draw another answer line below the last row of answer numbers.
8. Use long addition to add the number columns from right to left, carrying as you normally do for long addition.
Therefore, the product of 543 and 12 is 6516.
3. Multiply the numbers 2158 and 35
Solution:
1. Arrange the number 2158Â on top as the multiplicand and 35 at the bottom. Line up the place values in columns.
2. Starting with the one’s digit of the bottom number, i.e. 5 × 8=40. Write 0 as the answer and carry forward 4.
3. Multiply the one digit of the bottom number to the next digit to the left in the top number i.e. i.e. 5× 5=25. Add the carry forward 4 to the 25 i.e. 25+4=29. write 9 in the one’s place as the answer and carry the tens digit 2 Proceed right to left.
4. Multiply 5 with 1. i.e. 5 × 1=5. Add the carry forward digit 2 to 5.Write 7 as the answer.
5. Multiply 5 with 2. i.e. 5 × 2=10. Write 10 as the answer.
5. Move to the tens digit in the bottom number i.e. 3. Multiply as above, but this time write your answers in a new row, shifted one digit place to the left.
6. Multiply 3 with 8 i.e. 3 × 8=24. Write 4 as answer and carryforward 2.
Multiply 3 with 5 i.e. 3 × 5=15. Add the carry forward 2 to 15 i.e.15+2=17. Write 7 as the answer and carry forward 1.
Multiply 3 with 1 i.e. 3 × 1=3. Add the carry forward 1 to 3 i.e. 3+1=4. Write 4 as the answer.
Multiply 3 with 2 i.e. 3 × 2=6. Write 4 as the answer.
7. After multiplying, draw another answer line below the last row of answer numbers.
8. Use long addition to add the number columns from right to left, carrying as you normally do for long addition.
Therefore, the product of 2158 and 35 is 75530.
II.
1. Find the number which is twice of 3478?
Solution:
1.Start multiplying with one’s place digits. i.e. 2 × 8=16. Write 6 as the answer and carry forward 1.
2. Multiply tens. i.e. 2 × 4=8. Add carry forward 1 to 8 i.e. 8+1=9. Write 6 as the answer.
3. Multiply Hundreds. i.e. 2 × 7=14. Write 4 as the answer and carry forward 1.
4. Multiply thousands. i.e. 2 × 3=6. Add carry forward 1to 6 i.e. 6+1=7. Write 7 as the answer.
Therefore, twice as of 3478 is 7496.
2. Find the number which is thrice of 124?
Solution:
1.Start multiplying with one’s place digits. i.e. 3 × 4=12. Write 2 as the answer and carry forward 1.
2. Multiply tens. i.e. 3 × 2=6. Add carry forward 1 to 6 i.e. 6+1=7. Write 7 as the answer.
3. Multiply Hundreds. i.e. 3 × 1=3. Write 3 as the answer.
Therefore, thrice as of 124 is 372.
III. Estimate the product by rounding the given numbers
1.
S.No Question Rounded numbers Estimated product Exact product Which is Bigger
1 128 × 23 Nearest 10’s
2 250 ×  35 Nearest 10’s
3 570 × 48 Nearest 10’s
Solution:
S.No Question Rounded numbers Estimated product Exact product Which is Bigger
1 128 × 23 Nearest 10’s 442000 447496 Exact product
2 250 ×  35 Nearest 10’s 10000 8750 Estimated product
3 570 × 48 Nearest 10’s 28500 27360 estimated product
IV. Sandhya bought 8 chocolate packets to distribute on her birthday. Each packet has 110 chocolates. Find how many chocolates were there in 8 packets?
Solution:
Given,
No. of chocolate packets Sandhya bought=8
No. of chocolates in each packet=110
Total no. of chocolates in each packet=110 × 8=880
Therefore, No. of chocolates in 8 packets is 880.
V. Arjun’s monthly salary is Rs 35000. Arun’s monthly salary is three times that of Arjun. Find how much salary does Arun earns?
Solution:
Given,
Arjun’s monthly salary is =Rs 35000
Arun’s monthly salary is three times that of Arjun=3 × Rs 35000=Rs105000
Therefore, Aruns monthly salary is Rs 1,05,000.
VI. In a class there are 20 students. Each student has 15 books. Find out the total no. of books do students have?
Solution:
Given,
No. of students in the class=20
No. of books each student have=15
Total no. of books students have=20 × 15=300
Therefore, students have a total of 300 books.
VII. There are 250 papers in a bundle. How many papers were there in 6 bundles?
Solution:
Given,
No. of papers in a bundle=250
No. of papers in 6 bundles=250 × 6=1500.
Therefore, the total no. of papers in 6 bundles is 1500.
VIII. Raju does typing work and earns money of Rs 530 every day. Find out how much money will Raju earn if he works for 30 days?
Solution:
Given,
Raju earns money every day=Rs 530
Money earned by Raju if he works for 30 days=Rs 530 × 30=Rs 15900
Therefore, the total money earned by Raju in 30 days is Rs 15900.
1. 638 × 16=_____
a. 10250
b. 10208Â Â Â b
c. 10100
2. 848 × 24=______
a. 21524
b. 20245
c. 20352Â Â c
3. The number to be multiplied by another is called ______
a. Product
b. multiplier
c. Multiplicand  c
4. The number that you are multiplying by is called ______
a. Multiplier
b. Product
c. Multiplicand
5. 25 × 3 × 0=______
a. 0
b. 75
c. 85
6. Thrice of 65 =_____
a. 165
b. 195
c. 155
7. The product of 6 and 8 is same as the product of 4 and _____
a. 8
b. 12
c. 14
Solution:
1. b
2. c
3. c
4. a
5. a
6. b
7. b
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# An object 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high:(i) What is the focal length of the mirror?(ii) What is the position of image?(iii) Draw a ray-diagram to show the formation of image.
(i) Given:
Distance of the object from the mirror $u$ = $-$8 cm
Height of the object, $h_{1}$ = 3 cm
Height of the image $h_{2}$ = 4.5 cm
To find: The focal length of the mirror $f$, and the distance of the image $v$ from the mirror $v$.
Solution:
From the magnification formula, we know that-
$m=\frac{{h}_{2}}{{h}_{1}}=-\frac{v}{u}$
Substituting the given values in the magnification formula we get-
$\frac{4.5}{3}=-\frac{v}{(-8)}$
$\frac{4.5}{3}=\frac{v}{8}$
$3v={4.5}\times {8}$
$v=\frac{{4.5}\times {8}}{3}$
$v={1.5}\times {8}$
$v=+12cm$
Thus, the distance of the image $v$ is 12 cm, and the positive sign implies that the image forms behind the mirror (on the right).
Now, from the mirror formula, we know that-
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Substituting the given values in the mirror formula we get-
$\frac{1}{f}=\frac{1}{12}+\frac{1}{(-8)}$
$\frac{1}{f}=\frac{1}{12}-\frac{1}{8}$
$\frac{1}{f}=\frac{2-3}{24}$
$\frac{1}{f}=\frac{-1}{24}$
$f=-24cm$
Thus, the focal length of the concave mirror $f$ is 24 cm, and the negative sign implies that the focus is in front of the concave mirror (on the left).
(ii) $v=12$ (calculated above).
Hence, the position of the image is 12 cm behind the mirror.
(iii) Ray diagram to show the formation of the image.
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Updated on: 10-Oct-2022
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# A Fraction as a Percent
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The authors
Chris S.
## Basics on the topicA Fraction as a Percent
After this lesson you will be able to convert fractions to percents and back again, using visual models and double number lines.
The lesson begins with a review of the definition of percent, which leads to a method for writing any percent as a fraction over 100. It continues by applying this method to solve real world problems, with the help of visual models and your knowledge of equivalent fractions.
Learn how to convert fractions to percents and back again, by helping Olivia battle wildfires!
This video includes key concepts, notation, and vocabulary such as: percent (a quantity out of 100); equivalent fractions (two fractions which represent the same proportional relationship); and double number line (a visual representation of a ratio, in which the units are represented by two simultaneous number lines).
Before watching this video, you should already be familiar with the relationship between percents and part-to-whole ratios, and rates where the whole is 100.
After watching this video, you will be prepared to solve problems by finding the percent of a quantity.
Common Core Standard(s) in focus: 6.RP.A.3.c A video intended for math students in the 6th grade Recommended for students who are 11-12 years old
### TranscriptA Fraction as a Percent
Olivia has spent the last months training herself physically and mentally to be ready to battle wildfires. She’s passed every test so far and now she’s reached the last challenge. The firefighters have set a controlled burn in a field at their training center and she must extinguish 85% of the flames, by herself, before sundown. In order to track her progress and accomplish the mission, Olivia will need to convert fractions to percents, and back again. Time is ticking but before grabbing a firehose, Olivia needs to perform some calculations. The whole field is divided into 20 squares. If she needs to extinguish 85% of the field, how many squares IS that? Remember, a percent is a quantity out of 100. That means 85% is the same as the fraction 85 over 100. We can use equivalent fractions to rewrite 85 over 100 to what we need. For this problem, we want to find the equivalent fraction that has 20 in the denominator. 85 over 100 equals WHAT over 20? In order to change the denominator from 100 to 20, we need to divide by 5. But since we want an equivalent fraction, we also have to divide the numerator by 5. Dividing 85 by 5 gives us 17 which tells us 85 percent is equivalent to the fraction seventeen twentieths. Olivia needs to extinguish 17 out of the 20 squares. After a few hours of hard work, she takes a short break giving us time to calculate her progress so far. Olivia has extinguished the fire in 8 squares out of 20. What percentage has she extinguished so far? Let's take a look at the grid. Here are the eight squares where the fire has already been extinguished. On the right side we can write the number of squares in each row: 4, 8, 12, 16 and 20. Hmm, this reminds us of a double number line, so let's put the strip right here and turn it on it's side. The whole is always equal to 100 percent, so we can write that 20 squares is equal to 100 percent. We need to divide the 100 percent into 5 parts. Since 100 divided by 5 is 20, each part is 20 percent. That gives us 20 percent, 40 percent, 60 percent, and 80 percent. Looking at the double number line, that shows us that 8 squares represents 40 percent. That's 40 percent of the fire already extinguished. Olivia just remembered that she can earn a special commendation if she extinguishes at least 90 percent of the field before noon and now that seems like a real possibility. It looks like Olivia has extinguished all the squares but one! Has she earned the special commendation? What is the fraction of the fire that is still burning? One twentieth. What percent is that? Again, a percent is a quantity out of 100. That means we have to find an equivalent fraction with 100 in the denominator. What can we multiply 20 by to get 100? 100 divided by 20 is 5, so we need to multiply the denominator by 5. But since we want to find an equivalent fraction, we also have to multiply the numerator by 5. That gives us 5 over 100, which is equal to 5%. If only 5% is still burning, what percent did Olivia extinguish? 100%, which represents one whole, minus 5% gives us 95%. She did it! While Olivia takes a well-earned rest, let's review. Because percents are a quantity out of 100, we can convert fractions to percents, and vice versa. To convert a percent to a fraction, first write the percent as a fraction with 100 in the denominator. Then multiply or divide the numerator and denominator by the same number to find the equivalent fraction with the denominator you want. To convert a fraction to a percent, we first find what number we need to multiply or divide the denominator by in order to get 100. Then we multiply or divided both the numerator and denominator by that value to get a fraction out of 100. The numerator shows us the percent. This allows us to compare numbers even when they are given in different forms. Congratulations Olivia! With careful conversions of fractions to percents and back, you'll always be able to track your firefighting progress. But what’s this? Hmm, I’m not sure fractions and percents will help here. Olivia has to rescue 100% of the cat!
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## Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions)
### Locus and Its Constructions Exercise 16A – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC.
Prove— Q is equidistant from A and B.
Solution:
Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90 )
PQ = PQ (Common)
By Side-Angle-Side criterian of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.
Question 2.
Given— CP is bisector of angle C of ∆ ABC.
Prove— P is equidistant from AC and BC.
Solution:
Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.
Prove—
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:
Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than $$\frac{1}{2} \mathrm{BC}$$ draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Hence, D is equidistant from B and C.
Question 5.
In each of the given figures: PA = PH and QA = QB.
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
Construction: Join PQ which meets AB in D.
Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.
Question 6.
Construct a right angled triangle PQR, in which ∠ Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
∆PQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Hence, T is equidistant from PQ and QR.
Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Hence P is equidistant from AC and BC.
Solution:
Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Solution:
Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that –
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:
Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M
Question 10.
Use ruler and compasses only for this question:
(i) construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Solution:
Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60o. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
⇒ BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
PB=3.6 cm
Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E
Prove that:
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:
Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC …. (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD …. (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.
Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:
Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.
Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B.
Solution:
Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.
Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.
Solution:
Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.
Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Solution:
Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.
Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Solution:
Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.
Question 19.
On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.
Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.
### Locus and Its Constructions Exercise 16B – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Describe the locus of a point at a distance of 3 cm from a fixed point.
Solution:
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle.
Question 2.
Describe the locus of a point at a distance of 2 cm from a fixed line.
Solution:
The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.
Question 3.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:
The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.
Question 4.
Describe the locus of the moving end of the minute hand of a clock.
Solution:
The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.
Question 5.
Describe the locus of a stone dropped from the top of a tower.
Solution:
The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.
Question 6.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:
The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.
Question 7.
Describe the locus of the door handle as the door opens.
Solution:
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.
Question 8.
Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.
Question 9.
Describe the locus of the centers of all circles passing through two fixed points.
Solution:
The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.
Question 10.
Describe the locus of vertices of all isosceles triangles having a common base.
Solution:
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.
Question 11.
Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.
Question 12.
Describe the locus of a point P so that:
AB2 = AP2 + BP2, where A and B are two fixed points.
Solution:
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2 + BP2.
Question 13.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
i) AB and BC
ii) B and D.
Solution:
i)
The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.
ii)
The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.
Question 14.
The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
Solution:
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.
Question 15.
Describe:
i) The locus of points at distances less than 3 cm from a given point.
ii) The locus of points at distances greater than 4 cm from a given point.
iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
iv) The locus of points at distances greater than or equal to 35 mm from a given point.
v)The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.
vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.
Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.
Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Solution:
Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively.
Hence, a, b, c and d are the required four points.
Question PQ.
By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Solution:
Steps of construction:
i) Draw a linen.
ii) Take a point Lonn and draw a perpendicular to n.
iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
iv) At M, draw a line m making an angle of 90°.
v) Produce LM and mark a point P such that PM = 2 cm.
vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.
A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.
Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
i) Draw an angle of 60° with AB = BC = 8 cm
ii) Draw the angle bisector BX of ∠ABC
iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
iv) Join MP, NP
BMPN is a rhombus since MP = BM = NB = NP = 4 cm
Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:
Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.
Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠ A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:
Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.
Question PQ.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
P moves along AB, and Qmoves in such a way that PQ is always equal to OP.
But Pis the mid-point of OQ
Now in ∆OQQ’
P’and P” are the mid-points of OQ’ and OQ”
Therefore, AB||Q’Q”
Therefore, Locus of Q is a line CD which is parallel to AB.
Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:
Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.
Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:
Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.
Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:
Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.
Question 25.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.
Question 26.
State the locus of a point in a rhombus ABCD, which is equidistant
(ii) from the vertices A and C.
Solution:
Question 27.
Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Steps of Construction:
i) Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm
Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Solution:
Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.
Question PQ.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Solution:
Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at Mand meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof:
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore. Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.
Question 29.
Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Solution:
Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A.
Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Solution:
i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
ii) On measuring AB = 5.7 cm
Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
i. Steps of construction:
1. Draw BC = 6.5 cm using a ruler.
2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
3. With Q as center and same radius, cut the previous arc at P.
4. Join BP and extend it.
5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
6. Join AC to obtain ΔABC.
ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.
iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.
iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)
More Resources for Selina Concise Class 10 ICSE Solutions
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Integration by parts is one of the two important integration methods to learn in C4 (the other is substitution ((“What about function-derivative!?” That’s a special case of substitution.)) ). In this article, I want to run through when you do it, how you do it, and why it works, just in case you’re interested.
### When you use integration by parts
Integration by parts is most useful:
• When substitution doesn’t look promising; or
• When integrating two things multiplied together; or
• When you have a logarithm knocking around; or
• When you can’t think what else to do.
The last one of those doesn’t always work, but it can be worth a shot in an exam if you’re completely out of other ideas.
### How you integrate by parts
To integrate by parts, you always start with a product: two things multiplied together. A typical example:
$\int xe^x\d x$
There are two things there: an $x$ and and $e^x$. You’re going to call these $u$ and $v^\prime$ (($\d v/\d x$ is more ideal, but saving space beats proper notation here)) - the question is, which way around?
The rule is, you usually want $u$ to be something that gets simpler when you differentiate it. In C4, that means:
• If you have a $\ln$ anywhere, that will be $u$; otherwise
• If you have an $x$ or an $x^n$ anywhere, that will be $u$.
Here, we don’t have a $\ln$, but we have an $x$, so we’ll let $u = x$ and $v^\prime = e^x$.
We need $u^\prime$, which we get by differentiating $u$, so $u^\prime = 1$; we get $v$ by integrating $v^\prime$ with respect to $x$, getting $v^\prime = e^x$. (You don’t need a constant of integration right here).
If you look in the formula book, you’ll see something like $\int u \frac{\d v}{\d x}~\d x = uv - \int v \frac{\d u}{\d x}~\d x$, which is your template for how to proceed. You started with what’s on the left hand side, so now you need to figure out the right.
$uv$ is easy enough - that’s $xe^x$. The other term is harder: $\int vu^\prime~\d x = \int (e^x)(1)~\d x$… oh, wait, it’s not actually that hard. It’s just $e^x + c$. That gives a final answer:
$\int xe^x~\d x = xe^x - (e^x + c)$
You can tidy that up a bit - there’s a factor of $e^x$ in the first two terms, and you can finagle the constant to be nicer, so you end up with $e^x(x-1) + C$. Lovely!
(If you’re not sure, you can differentiate it back using the product rule to check it gives the original integrand $xe^x$. It does.)
Another example, in less excruciating detail: let’s find $\int 4x \ln(x)~\d x$
Here, we have a $\ln$, so that’s going to be $u$; that makes $u^\prime = \frac 1x$.
$v^\prime$ is clearly $4x$, making $v = 2x^2$.
Now apply the parts formula: $\int u \frac{\d v}{\d x}~\d x = uv - \int v \frac{\d u}{\d x}~\d x$
The terms on the right hand side are $uv = 2x^2 \ln (x)$ and $\int vu^\prime~\d x = \int 2x^2 \times \frac 1x~\d x$, or - more reasonably - $\int 2x~\d x = x^2 + c$.
Putting it together, $\int x\ln(x)~\d x = 2x^2 \ln(x) - (x^2 + c)$, or (with some tidying up) $x^2(2 \ln(x) - 1) + C$. Again, you can check it using the product rule.
### But where does the integration by parts formula come from?
You notice how you checked both of these with the product rule? That’s not a coincidence. Let’s start there:
$\frac{\d}{\d x}(uv) = u \frac{\d v}{\d x} + v\frac{\d u}{\d x}$
If you integrate both sides with respect to $x$, you get:
$uv = \int u \frac{\d v}{\d x}~\d x + \int v \frac{\d u}{\d x}~\d x$
See where it’s going? Take the second term on the right over to the left, and you have:
$uv - \int v \frac{\d u}{\d x}~\d x = \int u \frac{\d v}{\d x}~\d x$, or
$\int uv^\prime~\d x = uv - \int vu^\prime~\d x$, as required.
See? It all fits together quite neatly.
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# Given $y=-2f\left( 1-2x \right)+3$. How do you describe the transformation?
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Hint: Assume $f\left( x \right)$ as the reference function. Now, compare the given function $y=-2f\left( 1-2x \right)+3$ with the general form $y=af\left( b\left( x+c \right) \right)+d$. Find the corresponding values of a, b, c and d. Using these values find the vertical compression or stretch (a), horizontal compression or stretch (b), horizontal shift of the function (c) and vertical shift of the function (d).
Complete step-by-step solution:
Here, we have been provided with the function $y=-2f\left( 1-2x \right)+3$ and we are asked to explain the transformations of the function. Generally, we assume $y=f\left( x \right)$ as our reference function.
Now, the general form of the transform of a function is given as $y=af\left( b\left( x+c \right) \right)+d$. Here, a, b, c and d have their own meanings. For this general function the transform is described as: -
1. If $\left| a \right|>1$ then vertical stretch takes place and if 0 < a < 1 then vertical compression takes place.
2. If ‘a’ is negative then the function is reflected about the x – axis.
3. If $\left| b \right|>1$ then horizontal stretch takes place and if 0 < b < 1 then horizontal compression takes place.
4. If ‘b’ is negative then the function is reflected about y – axis.
5. If ‘c’ is negative then the function is shifted $\left| c \right|$ units to the right and if ‘c’ is positive then the function is shifted $\left| c \right|$ units to the left.
6. If ‘d’ is negative then the function is shifted $\left| k \right|$ units down and if ‘d’ is positive then the function is shifted $\left| k \right|$ units up.
Now, let us write our given function $y=-2f\left( 1-2x \right)+3$ in the general form and compare the values of a, b, c and d. So, we have,
$y=-2f\left( -2\left( x-\dfrac{1}{2} \right) \right)+3$
On comparing the values of a, b, c and d, we have,
$\Rightarrow$ a = -2, b = -2, $c=\dfrac{-1}{2}$ and d = +3
So, the description of the transform of the function can be given as: -
1. The function will have a vertical stretch of 2 units and it will be reflected about x – axis.
2. The function will have a horizontal stretch of 2 units and it will be reflected about y – axis.
3. The function will have a horizontal shift of $\dfrac{1}{2}$ units to the right.
4. The function will have a vertical shift of 3 units in the upward direction.
Note: One may note that you can also understand the situations by taking the examples of some trigonometric functions like: - sine or cosine function. Assume $y=\sin x$ and transform it into the function $y=A\sin \left( B\left( x+C \right) \right)+D$. Draw the graph of the two functions and compare the phase shifts, vertical and horizontal shifts etc. You must remember the six rules that we have stated in the solution. These transform rules are very useful in drawing the graph of functions which are provided to us in the chapter ‘area under curve’ in the topic ‘Integration’.
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# 7.5: Indirect Measurement
Difficulty Level: At Grade Created by: CK-12
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What if you wanted to measure the height of a flagpole using your friend George? He is 6 feet tall and his shadow is 10 feet long. At the same time, the shadow of the flagpole was 85 feet long. How tall is the flagpole? After completing this Concept, you'll be able to use indirect measurement to help you answer this question.
### Guidance
An application of similar triangles is to measure lengths indirectly. The length to be measured would be some feature that was not easily accessible to a person, such as the width of a river or canyon and the height of a tall object. To measure something indirectly, you need to set up a pair of similar triangles.
#### Example A
A tree outside Ellie’s building casts a 125 foot shadow. At the same time of day, Ellie casts a 5.5 foot shadow. If Ellie is 4 feet 10 inches tall, how tall is the tree?
Draw a picture. From the picture to the right, we see that the tree and Ellie are parallel, therefore the two triangles are similar to each other. Write a proportion.
4ft,10inxft=5.5ft125ft\begin{align*}\frac{4ft, 10in}{xft}=\frac{5.5ft}{125ft}\end{align*}
Notice that our measurements are not all in the same units. Change both numerators to inches and then we can cross multiply.
58inxft=66in125ft58(125)7250x=66(x)=66x109.85 ft\begin{align*}\frac{58in}{xft}=\frac{66in}{125ft} \longrightarrow 58(125) &= 66(x)\\ 7250 &= 66x\\ x & \approx 109.85 \ ft\end{align*}
#### Example B
Cameron is 5 ft tall and casts a 12 ft shadow. At the same time of day, a nearby building casts a 78 ft shadow. How tall is the building?
To solve, set up a proportion that compares height to shadow length for Cameron and the building. Then solve the equation to find the height of the building. Let x\begin{align*}x\end{align*} represent the height of the building.
5ft12ft12xx=x78ft=390=32.5ft\begin{align*} \frac{5 ft}{12 ft}&=\frac{x}{78 ft} \\ 12x&=390 \\ x&=32.5 ft\end{align*}
The building is 32.5\begin{align*}32.5\end{align*} feet tall.
#### Example C
The Empire State Building is 1250 ft. tall. At 3:00, Pablo stands next to the building and has an 8 ft. shadow. If he is 6 ft tall, how long is the Empire State Building’s shadow at 3:00?
Similar to Example B, solve by setting up a proportion that compares height to shadow length. Then solve the equation to find the length of the shadow. Let x\begin{align*}x\end{align*} represent the length of the shadow.
6ft8ft6xx=1250ftx=10000=1666.67ft\begin{align*}\frac{6 ft}{8 ft}&=\frac{1250 ft}{x}\\ 6x&=10000\\ x&=1666.67 ft\end{align*}
The shadow is approximately 1666.67\begin{align*}1666.67\end{align*} feet long.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
It is safe to assume that George and the flagpole stand vertically, making right angles with the ground. Also, the angle where the sun’s rays hit the ground is the same for both. The two trianglesare similar. Set up a proportion.
1085=6x10xx=510=51 ft.\begin{align*}\frac{10}{85} = \frac{6}{x} \longrightarrow 10x &= 510\\ x &= 51 \ ft.\end{align*}
The height of the flagpole is 51 feet.
### Vocabulary
Two triangles are similar if all their corresponding angles are congruent (exactly the same) and their corresponding sides are proportional (in the same ratio). Solve proportions by cross-multiplying.
### Guided Practice
In order to estimate the width of a river, the following technique can be used. Use the diagram.
Place three markers, O,C,\begin{align*}O, C,\end{align*} and E\begin{align*}E\end{align*} on the upper bank of the river. E\begin{align*}E\end{align*} is on the edge of the river and OC¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯\begin{align*}\overline {OC} \perp \overline{CE}\end{align*}. Go across the river and place a marker, N\begin{align*}N\end{align*} so that it is collinear with C\begin{align*}C\end{align*} and E\begin{align*}E\end{align*}. Then, walk along the lower bank of the river and place marker A\begin{align*}A\end{align*}, so that CN¯¯¯¯¯¯¯¯NA¯¯¯¯¯¯¯¯\begin{align*}\overline{CN} \perp \overline{NA}\end{align*}. OC=50\begin{align*}OC = 50\end{align*} feet, CE=30\begin{align*}CE = 30\end{align*} feet, NA=80\begin{align*}NA = 80\end{align*} feet.
1. Is OCEANE?\begin{align*}\triangle OCE \sim \triangle ANE?\end{align*} How do you know?
2. Is OC¯¯¯¯¯¯¯¯NA¯¯¯¯¯¯¯¯?\begin{align*}\overline {OC} \| \overline {NA}?\end{align*} How do you know?
3. What is the width of the river? Find EN\begin{align*}EN\end{align*}.
1. Yes. CN\begin{align*}\angle{C} \cong \angle{N}\end{align*} because they are both right angles. OECAEN\begin{align*}\angle{OEC} \cong \angle{AEN}\end{align*} because they are vertical angles. This means OCEANE\begin{align*}\triangle OCE \sim \triangle ANE\end{align*} by the AA Similarity Postulate.
2. Since the two triangles are similar, we must have EOCEAN\begin{align*}\angle{EOC} \cong \angle{EAN}\end{align*}. These are alternate interior angles. When alternate interior angles are congruent then lines are parallel, so OC¯¯¯¯¯¯¯¯NA¯¯¯¯¯¯¯¯\begin{align*}\overline {OC} \| \overline {NA}\end{align*}.
3. Set up a proportion and solve by cross-multiplying.
30ftEN50(EN)EN=50ft80ft=2400=48\begin{align*}\frac{30 ft}{EN}&=\frac{50 ft}{80 ft}\\ 50(EN)&=2400 \\ EN&=48\end{align*}
The river is 48\begin{align*}48\end{align*} feet wide.
### Practice
The technique from the guided practice section was used to measure the distance across the Grand Canyon. Use the picture below and OC=72 ft,CE=65 ft\begin{align*}OC = 72 \ ft , CE = 65 \ ft\end{align*}, and NA=14,400 ft\begin{align*}NA = 14,400 \ ft\end{align*} for problems 1 - 3.
1. Find EN\begin{align*}EN\end{align*} (the distance across the Grand Canyon).
2. Find OE\begin{align*}OE\end{align*}.
3. Find EA\begin{align*}EA\end{align*}.
1. Janet wants to measure the height of her apartment building. She places a pocket mirror on the ground 20 ft from the building and steps backwards until she can see the top of the build in the mirror. She is 18 in from the mirror and her eyes are 5 ft 3 in above the ground. The angle formed by her line of sight and the ground is congruent to the angle formed by the reflection of the building and the ground. You may wish to draw a diagram to illustrate this problem. How tall is the building?
2. Sebastian is curious to know how tall the announcer’s box is on his school’s football field. On a sunny day he measures the shadow of the box to be 45 ft and his own shadow is 9 ft. Sebastian is 5 ft 10 in tall. Find the height of the box.
3. Juanita wonders how tall the mast of a ship she spots in the harbor is. The deck of the ship is the same height as the pier on which she is standing. The shadow of the mast is on the pier and she measures it to be 18 ft long. Juanita is 5 ft 4 in tall and her shadow is 4 ft long. How tall is the ship’s mast?
4. Evan is 6 ft tall and casts a 15 ft shadow. A the same time of day, a nearby building casts a 30 ft shadow. How tall is the building?
5. Priya and Meera are standing next to each other. Priya casts a 10 ft shadow and Meera casts an 8 ft shadow. Who is taller? How do you know?
6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow?
7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother?
8. Ray is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow?
9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree?
10. Jason, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building?
11. Alexandra, who is 5 ft 8 in tall is casting a 12 ft shadow. A nearby lamppost is casting a 20 ft shadow. How tall is the lamppost?
12. Use shadows or a mirror to measure the height of an object in your yard or on the school grounds. Draw a picture to illustrate your method.
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### Vocabulary Language: English
TermDefinition
AA Similarity Postulate If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar.
Proportion A proportion is an equation that shows two equivalent ratios.
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Question Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point | Nagwa Question Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point | Nagwa
# Question Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point Mathematics • Second Year of Secondary School
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Given that lim_(π₯ β β2) (π(π₯))/(3π₯Β²) = β3, determine lim_(π₯ β β2) π(π₯).
02:04
### Video Transcript
Given that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to negative three, determine the limit as π₯ tends to negative two of π of π₯.
In this question, weβ²ve been given the limit as π₯ tends to negative two of π of π₯ over three π₯ squared. We can break this limit down using the properties of limits. We have the property for limits of quotients of functions, which tells us that the limit as π₯ tends to π of π of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as π₯ tends to π of π of π₯. For our limit, weβre taking the limit as π₯ tends to negative two. Therefore, π is equal to negative two. And we have a quotient of functions. In the numerator, we have π of π₯. And in the denominator, we have three π₯ squared. Applying this rule for limits of quotients of functions, we obtain that our limit is equal to the limit as π₯ tends to negative two of π of π₯ over the limit as π₯ tends to negative two of three π₯ squared.
Letβ²s now consider the limit in the denominator of the fraction. Thatβ²s the limit as π₯ tends to negative two of three π₯ squared. We can apply direct substitution to this limit, giving us that the limit as π₯ tends to negative two of three π₯ squared is equal to three times negative two squared. Negative two squared is equal to four. We then simplify to obtain that this limit is equal to 12. We can substitute this value of 12 back in to the denominator of our fraction, giving us that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to the limit as π₯ tends to negative two of π of π₯ all over 12.
However, weβ²ve been given in the question that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to negative three. And since this is on the left-hand side of our equation, we can set our equation equal to negative three. So we now have that the limit as π₯ tends to negative two of π of π₯ over 12 is equal to negative three. We simply multiply both sides of the equation by 12. Here, we reach our solution which is that the limit as π₯ tends to negative two of π of π₯ is equal to negative 36.
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Modular Arithmetic/What is a Modulus?
Modular Arithmetic What is a Modulus? Modular Arithmetic →
In modular arithmetic, 38 can equal 14 — what??
You might be wondering how so! (Or you might already know how so, but we will assume that you don't.) Well, modular arithmetic works as follows:
Think about military time which ranges from 0000 to 2359. For the sake of this lesson, we will only consider the hour portion, so let us consider how hourly time works from 0 to 23. After a 24 hour period, the time restarts at 0 and builds again to 23. So, using this reasoning, the 27th hour would be equivalent to the third hour in military time, as the remainder when 27 is divided by 24 is 3 (make sure you understand why this is true as it is vital to understanding everything that follows in this book). So to get any hour into military time we just find the number's remainder when divided by 24. What we really are doing is finding that number modulo 24.
To write the earlier example mathematically, we would write:
${\displaystyle 27\equiv 3{\pmod {24}}}$
which reads as "27 is congruent to 3 modulo 24". Let us try one more example before setting you loose: what is 34 congruent to modulo 4, which can also be written as "34 (mod 4)" which is read as "what is 34 congruent to modulo 4?". We find the remainder when 34 is divided by 4; the remainder is 2. So, ${\displaystyle 34\equiv 2{\pmod {4}}}$.
Now, try some on your own:
Exercises: Definition of Modular Arithmetic Replace the '?' by the correct value. ${\displaystyle 25\equiv {\text{?}}{\pmod {7}}}$ ${\displaystyle 10^{10}\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {729}}}$
Challenge: Definition of Modular Arithmetic These two are a lot harder. Simplify: ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {11}}}$ You do not need to work out what (936 + 3) is to solve these, but they will require some calculation.
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# 11th Class Mental Ability Geometry Notes - Geometry
Notes - Geometry
Category : 11th Class
# Geometry
Learning Objectives
• Geometry
Geometry
Geometry is the visual study of shapes, sizes, patterns, and positions. It occurred in all cultures, through at least one of these five strands of human activities: The following formulas and relationships are important in solving geometry problems.
Angle Relationships
1. The base angles of an isosceles triangle are equal
2. The sum of the measures of the interior angles of any n-sided polygon is 180(n - 2) degrees.
3. The sum of the measures of the exterior angles of any n-sided polygon is 360°.
4. If two parallel lines are cut by a transversal, the alternate interior angles are equal, and the corresponding angles are equal.
Angle Measurement Theorems
1. A central angle of a circle is measured by its intercepted arc.
2. An inscribed angle in a circle is measured by one-half of its intercepted arc,
3. An angle formed by two chords intersecting within a circle is measured by one-half the sum of the opposite intercepted arcs.
4. An angle formed by a tangent and a chord is measured by one-half its intercepted arc.
5. An angle formed by two secants, or by two tangents, or by a tangent and a secant, is measured by one-half the difference of the intercepted arcs.
Proportion Relationships
1. A line parallel to one side of triangle divides the other two sides proportionally.
2. In two similar triangles, corresponding sides, medians, altitudes, and angle bisectors are proportional.
3. If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other.
4. If a tangent and a secant are drawn to a circle from an outside point, the tangent is the mean proportional between the secant and the external segment.
5. In similar polygons the perimeters have the same ratio as any pair of corresponding sides.
Triangle Relationships
1. If an altitude is drawn to the hypotenuse of a right triangle, it is the mean proportional between the segments of the hypotenuse, and either leg is the mean proportional between the hypotenuse and the segment adjacent to that leg.
2. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. (Remember the Pythagorean triples: 3, 4, 5; 5, 12, 13.)
3. In a 30°- 60° right triangle, the leg opposite the 30° angle is one-half the hypotenuse, and the leg opposite the 60° angle is one-half the hypotenuse times $\sqrt{3}\,.$
4. In a right isosceles triangle the hypotenuse is equal to either leg times $\sqrt{2}\,.$
5. In an equilateral triangle of sides, the altitude equals $\frac{s}{2}\sqrt{3}\,.$
If the sides of a triangle are produced then the sum of the exterior angles i.e.
$\angle DAB+\angle EBC+\angle FCA$is equal to:
(a) 180° (b) 270°
(c) 360° (d) 240°
(e) None of these
Ans. (c)
Explanation: Sum of exterior angles = 360°
2. In a $\Delta \mathbf{ABC},\text{ }\angle \mathbf{BAC>90{}^\circ },$then $\angle \mathbf{ABC}$and $\angle \mathbf{ACB}$must be:
(a) acute (b) obtuse
(c) one acute and one obtuse (d) can't be determined
(e) None of these
Ans. (a)
Explanation: $\angle ABC+\angle ACB<90{}^\circ$
1. If the angles of a triangle are in the ratio 1:4:7, then the value of the largest is:
(a) 135° (b) 84°
(c) 105° (d) 110°
(e) None of these
Ans. (c)
Explanation: $x+4x+7x=180{}^\circ$
$\Rightarrow$ $x=15{}^\circ$
$\therefore$ $7x=105{}^\circ$
In the adjoining figure of $\Delta \mathbf{ABC}=\mathbf{12}{{\mathbf{0}}^{\mathbf{{}^\circ }}}$ and $\mathbf{AB=c,}\,\mathbf{BC=a,}\,\,\mathbf{AC=b}$then:
(a) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}+ba$ (b) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-ba$
(c) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ba$ (d) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}+2ba$
(e) None of these
2. What is the ratio of side and height of and equilateral triangle?
(a) $2:1$ (b) $1:1$
(c) $2:\sqrt{\text{3}}$ (d) $\sqrt{3}:2$
(e) None of these
In the$\Delta \mathbf{ABC}$, BD bisects$\angle \mathbf{B}$, and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, then find the value of x and y:
(a) 6, 12 (b) 10, 12
(c) 16, 8 (d) 8, 15
(e) None of these
In the given diagram AB||CD, then which one of the following is true?
(a) $\frac{AB}{AC}=\frac{AO}{OC}$ (b) $\frac{AB}{CD}=\frac{BO}{OD}$
(c) $\Delta AOB-\Delta COD$ (d) Ail of these
(e) None of these
In the figure BC||AD. Find the value of x:
(a) 9, 10 (b) 7, 8
(c) 10, 12 (d) 8, 9
(e) None of these
1. Find the maximum area that can be enclosed in a triangle of perimetre 24 cm:
(a) $32c{{m}^{2}}$
(b) $16\text{ }\sqrt{3}\text{ }c{{m}^{2}}$
(c) $16\text{ }\sqrt{\text{2}}\,\,c{{m}^{2}}$
(d) $27\text{ }c{{m}^{2}}$
(e) None of these
1. If one of the interior angles of a regular polygon is equal to 5/6 times of one of the interior angles of a regular pentagon, then the number of sides of the polygon is:
(a) 3 (b) 4
(c) 6 (d) 8
(e) None of these
1. If each interior angle of a regular polygon is 3 times its exterior angle, the number of sides of the polygon is:
(a) 4 (b) 5
(c) 6 (d) 8
(e) None of these
In the adjoining figure, O is the centre of circle and diametre AC = 26 cm. If chord AB = 10 cm, then the distance between chord AB and centre O of the circle is:
(a) 24cm (b) 16cm
(c) 12cm (d) 11cm
(e) None of these
1. A polygon has 54 diagonals. The number of sides in the polygon is :
(a) 7 (b) 9
(c) 12 (d) 11
(e) None of these
#### Other Topics
##### Notes - Geometry
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2013 AMC 10A Problems/Problem 18"
## Problem
Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$?
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$
## Solution
First, various area formulas (shoelace, splitting, etc) allow us to find that $[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.
Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$.
Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \frac{27}{8}$.
From this, we know that $E = (\frac{27}{8}, \frac{15}{8})$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$
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# David Arquette And The Stars (12/22/2019)
How will David Arquette get by on 12/22/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is of questionable accuracy – don’t get too worked up about the result. I will first calculate the destiny number for David Arquette, and then something similar to the life path number, which we will calculate for today (12/22/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology experts.
PATH NUMBER FOR 12/22/2019: We will consider the month (12), the day (22) and the year (2019), 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 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 22 we do 2 + 2 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 4 + 12 = 19. This still isn’t a single-digit number, so we will add its digits together again: 1 + 9 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 12/22/2019.
DESTINY NUMBER FOR David Arquette: 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 David Arquette we have the letters D (4), a (1), v (4), i (9), d (4), A (1), r (9), q (8), u (3), e (5), t (2), t (2) and e (5). Adding all of that up (yes, this can get tiring) gives 57. This still isn’t a single-digit number, so we will add its digits together again: 5 + 7 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the destiny number for David Arquette.
CONCLUSION: The difference between the path number for today (1) and destiny number for David Arquette (3) is 2. That is less 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 not at all guaranteed. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### 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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# Finding area of largest rectangle between the axes and a line
The question is as follows: Find the area of the largest rectangle that has sides parallel to the coordinate axes, one corner at the origin and the opposite corner on the line 3x+2y=12 in the first quadrant.
I get that the equation I have to maximize is in the form of A=bh but I don't know how to eliminate one of the variables to continue.
• Hint: If the opposite corner has coordinates $(h,k)$, then $3h+2k=12$ and the area is given by $A=hk$. Oct 8, 2018 at 19:10
Since the bottom left corner of the rectangle is at the origin, then the $$(x,\,y)$$ coordinates of the top right corner will be the base and height (draw a figure to help visualize). We know that this point is on the line $$3x+2y=12$$, so that $$y=-\frac{3}{2}x+6$$. Plugging this in gives $$A=x(-\frac{3}{2}x+6)$$, which has only one variable.
Suppose that your rectangle has vertices $$(0, 0)$$, $$(x, 0)$$, $$(0, y)$$, and $$(x, y)$$, where $$x > 0$$, $$y > 0$$, and $$3x + 2y = 12. \tag{1}$$ Then the area of your rectangle is given by $$A = xy. \tag{2}$$ But (1) implies that $$y = 6 - \frac{3x}{2}. \tag{3}$$ Putting the value of $$y$$ from (3) into the formula in (2), we obtain $$A = A(x) = x \left( 6 - \frac{3x}{2} \right) = 6x - \frac{3x^2}{2}. \tag{4}$$ Now (4) gives area $$A$$ as a function of $$x$$ for $$x > 0$$.
Differentiating both sides of (4) w.r.t. $$x$$ we obtain $$A^\prime(x) = 6 - 3x.$$ Thus we see that $$A^\prime(x) \ \begin{cases} > 0 \ & \ \mbox{ for } x < 2, \\ = 0 \ & \ \mbox{ for } x = 2, \\ < 0 \ & \ \mbox{ for } x > 2. \end{cases}$$ Thus the area attains its (relative) maximum value at $$x = 2$$, and since this is the only relative extreme value of $$A$$, this is in fact the absolute maximum value of $$A$$.
Therefore the largest possible area is given by $$A(2) = 12 - 6 = 6.$$
Any point on the line can be represented as (x,(12-3x)/2)
Now we have to find the maximum area of a rectangle given one point as (0,0) and opposite point as $$(x,(12-3x)/2$$)
Area=$$(x)*(12-3x)/2$$
We have to maximize this area taking derivative and equating to $$0$$ we get value of $$x=2$$
Substituting this value in area we get area=6 which should be maximum
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## How do you write the standard form of the equation of a circle?
The standard form of a circle’s equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius. To convert an equation to standard form, you can always complete the square separately in x and y.
## What is standard form of a circle?
The equation of a circle comes in two forms: 1) The standard form: (x – h)2 + (y-k)2 = r2. 2) The general form : x2 + y2 + Dx + Ey + F = 0, where D, E, F are constants. If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r .
## How do you find the equation of a circle?
The formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.
## What’s a standard form?
Standard form is a way of writing down very large or very small numbers easily. So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
## Is circle a function?
A circle is a curve. It can be generated by functions, but it’s not a function itself. Something to careful about is that defining a circle with a relation from x to y is NOT a function as there is multiple points with a given x-value, but it can be defined with a function parametrically.
## How do you plot a circle?
follow these steps:Realize that the circle is centered at the origin (no h and v) and place this point there.Calculate the radius by solving for r. Set r-squared = 16. Plot the radius points on the coordinate plane. Connect the dots to graph the circle using a smooth, round curve.
You might be interested: Q10 equation
## How do you find the standard form of a circle with the center and a point?
1 AnswerThe equation of a circle with center (h,k) and radius r is: r2=(x−h)2+(y−k)2.We are told that the center of this circle is (2,4) , so. r2=(x−2)2+(y−4)2.The radius of the circle is 5 , which means the equation of the circle is: 25=(x−2)2+(y−4)2.
## How do I calculate the area of a circle?
The area of a circle is pi times the radius squared (A = π r²).
## How do you convert standard form to standard form?
Step 1: Subtract how many decimal places there are from the power. Step 2: Remove the decimal point, add as many zeros as the number in Step 1. Step 1: Subtract how many decimal places there are from the power. There are 3 decimal places.
## What is Standard Form calculator?
Standard Form Calculator is a free online tool that displays the number in the standard form. The number can be in either integer form or the decimal form.
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#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]
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# Properties of a Parallelogram
In Geometry, a parallelogram is a type of quadrilateral. It is a two-dimensional figure with four sides. The most important properties of a parallelogram are that the opposite sides are parallel and congruent and the opposite angles are also equal. In this article, let us discuss all the properties of a parallelogram with a complete explanation and many solved examples.
Table of Contents:
## Important Properties of a Parallelogram
A parallelogram is a closed four-sided two-dimensional figure in which the opposite sides are parallel and equal in length. Also, the opposite angles are also equal. Learning the properties of a parallelogram is useful in finding the angles and sides of a parallelogram. The four most important properties of a parallelogram are:
• The opposite sides of a parallelogram are equal in measurement and they are parallel to each other.
• The opposite angles of a parallelogram are equal.
• The sum of interior angles of a parallelogram is equal to 360°.
• The consecutive angles of a parallelogram should be supplementary (180°).
## Theorems on Properties of a Parallelogram
The 7 important theorems on properties of a parallelogram are given below:
### Theorem 1:
A diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Proof:
Assume that ABCD is a parallelogram and AC is a diagonal of a parallelogram.
The diagonal AC divides the parallelogram into two congruent triangles, such as ∆ABC and ∆CDA. Now, we need to prove that the triangles ∆ABC and ∆CDA are congruent.
From the triangles, ∆ABC and ∆CDA, AD || BC, and AC is a transversal.
Hence, ∠ BCA = ∠ DAC (By using the property of pair of alternate angles)
Also, we can say that AB || DC and line AC is transversal.
Thus, ∠ BAC = ∠ DCA (Using the pair of alternate angles)
Also, AC = CA (Common side)
By using the Angle side Angle rule (ASA rule), we can conclude that ∆ABC is congruent to ∆CDA.
(i.e) ∆ABC ≅ ∆CDA.
Thus, the diagonal AC divides a parallelogram ABCD into two congruent triangles ABC and CDA.
Hence, proved.
### Theorem 2:
The opposite sides of a parallelogram are equal.
Proof:
From theorem 1, it is proved that the diagonals of a parallelogram divide it into two congruent triangles. When you measure the opposite sides of a parallelogram, it is observed that the opposite sides are equal.
Hence, we conclude that the sides AB = DC and AD = BC.
### Theorem 3:
If each pair of opposite sides of a quadrilateral is equal, then the quadrilateral is a parallelogram.
Proof:
Assume that the sides AB and CD of the quadrilateral ABCD are equal and also AD = BC
Now, draw the diagonal AC.
Clearly, we can say that ∆ ABC ≅ ∆ CDA (From theorem 1)
Therefore, ∠ BAC = ∠ DCA and ∠ BCA = ∠ DAC.
From this result, we can say that the quadrilateral ABCD is a parallelogram because each pair of opposite sides is equal in measurement.
Thus, conversely, we can say that if each pair of opposite sides of a quadrilateral is equal, then the quadrilateral is a parallelogram.
Hence, proved.
### Theorem 4:
The opposite angles are equal in a parallelogram.
Proof:
Using Theorem 3, we can conclude that the pairs of opposite angles are equal.
(i.e) ∠A = ∠C and ∠B = ∠D
Thus, each pair of opposite angles is equal, a quadrilateral is a parallelogram.
### Theorem 5:
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof:
We can say that Theorem 5 is the converse of Theorem 4.
### Theorem 6:
The diagonals of a parallelogram bisect each other.
Proof:
Consider a parallelogram ABCD and draw both the diagonals and it intersects at the point O.
Now, measure the lengths, such as OA, OB, OC and OD.
You will observe that OA = OC and OB = OD
So, we can say that “O” is the midpoint of both the diagonals.
Hence, proved.
### Theorem 7:
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Proof:
This theorem is the converse of Theorem 6.
Now, consider a parallelogram ABCD which is given below:
From the figure, we can observe that OA = OC and OB = OD.
Hence, we can say ∆ AOB ≅ ∆ COD.
∠ ABO = ∠ CDO
Thus, we can conclude that AB || CD and BC || AD.
Hence, ABCD is a parallelogram.
Also, read:
### Properties of a Parallelogram Example
Question:
Prove that the bisectors of angles of a parallelogram form a rectangle.
Solution:
Assume that ABCD is a parallelogram. Let P, Q, R, S be the point of intersection of bisectors of
∠A and ∠B, ∠B and ∠C, ∠C and ∠D, and ∠D and ∠A, respectively.
From the triangle, ASD, we can observe that DS bisects ∠D and AS bisects ∠A.
Therefore,
∠ADS + ∠DAS = (½)∠D + (½)∠A
∠ADS + ∠DAS = ½ (∠A + ∠D)
∠ADS + ∠DAS = ½ (180°) [Since ∠A and ∠D are the interior angles on the same side of the transversal]
Therefore, ∠ADS + ∠DAS = 90°.
Using the angle sum property of a triangle, we can write:
∠ DAS + ∠ ADS + ∠ DSA = 180°
Now, substitute ∠ADS + ∠DAS = 90° in the above equation, we get
90° + ∠DSA = 180°
Therefore, ∠DSA = 180° – 90°
∠DSA = 90°.
Since, ∠PSR is being vertically opposite to ∠DSA,
We can say ∠PSR = 90°
Likewise, we can be shown that ∠ APB = 90° or ∠ SPQ = 90°
Similarly, ∠ PQR = 90° and ∠ SRQ = 90°.
Since all the angles are at right angles, we can say that PQRS is a quadrilateral.
Now, we need to conclude that the quadrilateral is a rectangle.
We have proved that ∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°.
As, both the pairs of opposite angles are equal to 90°, we can conclude that PQRS is a rectangle.
Hence, proved.
### Practice Problems
1. A quadrilateral ABCD is a parallelogram where AP and CQ are perpendiculars from vertices A and C on diagonal BD as shown in the figure.
2. Prove that (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ
3. If the diagonals of a parallelogram are equal, prove that it is a rectangle.
Stay tuned with BYJU’S – The Learning App and learn all the class-wise concepts easily by exploring more videos.
## Frequently Asked Questions on Properties of a Parallelogram
### What type of polygon is a parallelogram?
A parallelogram is a quadrilateral.
### What are the properties of the parallelogram?
The properties of the parallelogram are:
The opposite sides of a parallelogram are parallel and congruent
The consecutive angles of a parallelogram are supplementary
The opposite angles are equal
A diagonal bisect the parallelogram into two congruent triangles
Diagonals bisect each other
### What are the two special types of a parallelogram?
The two special types of a parallelogram are square and rectangle.
### What is the order of rotational symmetry of a parallelogram?
The order of rotational symmetry of a parallelogram is 2.
### Does a parallelogram have reflectional symmetry?
No, a parallelogram does not have reflectional symmetry.
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# Which Is The Graph Of F(X) = X2 – 2x + 3?
A parabola is a quadratic equation, which is a polynomial equation of degree two. F(X) = X2 – 2x + 3 is a parabola equation, which has a graph that is a parabola. Understanding the graph of F(X) = X2 – 2x + 3 is an important part of calculus and algebra. In this article, we will explain the graph of F(X) = X2 – 2x + 3 and its properties.
## Understanding F(X)=X^2 – 2X + 3
F(X) = X2 – 2x + 3 is a quadratic equation. It is a polynomial equation of degree two, which means that it has two terms, one being X2 and the other being -2x + 3. The graph of this equation is a parabola, which is a symmetrical curve that opens upwards or downwards. The graph of F(X) = X2 – 2x + 3 is an upward-opening parabola.
## Analyzing the Graph of F(X).
The graph of F(X) = X2 – 2x + 3 is a parabola that opens upwards. The parabola has a vertex, which is the highest or lowest point of the parabola. The vertex of the graph of F(X) = X2 – 2x + 3 is (1,2). This means that the parabola has a maximum value at the point (1,2).
The graph also has a y-intercept, which is the point where the graph crosses the y-axis. The y-intercept of the graph of F(X) = X2 – 2x + 3 is (0,3). This means that the graph crosses the y-axis at the point (0,3).
The graph of F(X) = X2 – 2x + 3 is symmetrical. This means that the graph is the same on both sides of the vertex. The graph is also continuous, which means that it does not have any gaps or breaks.
In conclusion, the graph of F(X) = X2 – 2x + 3 is a parabola that opens upwards. The graph has a vertex at (1,2) and a y-intercept at (0,3). The graph is symmetrical and continuous. Understanding the graph of F(X) = X
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Back
Introduction to Unit Conversions
> ... Math > Unit Conversions > Introduction to Unit Conversions
The following video will show how to do unit conversions and explains why it works.
Introduction to Unit Conversions
### Introduction to Unit Conversions
Unit conversion = taking the measurement of something in one set of units and changing it to an equivalent measurement in another set of units.
The keys to doing unit conversions are the concepts that anything divided by itself is equal to 1 and that anything multiplied by 1 is equal to itself.
$${\dfrac {x}{x}} = 1$$
and
$$1x = x$$
### Conversion Factors
Conversion factors are fractions where the item in the numerator is equal to the item in the denominator, essentially making the fraction equal to 1.
Examples:
1 inch = 2.54 cm, therefore:
$${\dfrac {1 \ inch}{2.54 \ cm}} = 1$$
This is the conversion factor between inches and centimeters.
Another example is that 60 minutes = 1 hour, therefore:
$${\dfrac {60 \ minutes}{1 \ hour}} = 1$$
This is the conversion factor between minutes and hours.
### Example 1: Inches to Centimeters
How do we convert 3 inches into the equivalent length in centimeters?
2. Determine what we want to get in the end: centimeters.
3. Determine the conversion factor to use: 1 inch = 2.54 cm.
4. Multiply by 1 in the form of the conversion factor that cancels out the unwanted units. In this case, $${\frac {2.54 \ cm}{1 \ in}}$$.
$$\dfrac{\color{DarkGreen}3 {\cancel{\text{ inches}}}}{1}\left ( \dfrac{{\color{DarkOrange} 2.54\:\text{cm}}}{{\color{DarkGreen} 1\:\cancel{\text{ inch}} }} \right ) = \dfrac{\left ( {\color{DarkGreen} 3} \right )\left ( {\color{DarkOrange} 2.54\:\text{cm}}\right )}{1} = {\color{Black} 7.62\:\text{cm}}$$
Answer: 3 in = 7.62 cm
### Example 2: Minutes to Hours
How do we convert 14 minutes into the equivalent time in hours?
$$14 \ minutes = ? \ hours$$
2. Determine what we want to get in the end: hours.
3. Determine the conversion factor to use: 60 minutes = 1 hour.
4. Multiply by 1 in the form of the conversion factor that cancels out the unwanted units. In this case, $${\frac {1 \ hour}{60 \ minutes}}$$.
$${14\color{Magenta}\cancel{\text{ minutes}}}\times \dfrac{{1\color{green} \text{ hour}}}{{60\color{Magenta} \cancel{\text{ minutes}}}}=\dfrac{(14\times 1){\color{green} \text{hours}}}{60}$$
$$=0.23{\color{green}\text{ hours}}$$
Answer: 14 minutes = 0.23 hours
### Practice Problems
Video Solutions for Problems 1 - 5
1. The length of a table measures 250 centimeters. There are 100 centimeters in 1 meter. Convert the length of the table to meters. Round to the nearest tenth. (
Solution
Solution:
$$2.5$$ meters
)
2. Margaret has a pitcher filled with 2 liters of water. One liter is equal to approximately $$4.23$$ cups. Convert the amount of water in Margaret’s pitcher to cups. Round to the nearest hundredth. (
Solution
Solution:
$$8.46$$ cups
)
3.Kim walked $$2.5$$ miles to the grocery store. There are 1760 yards in one mile. Convert the distance Kim walked to yards. Round to the nearest whole number. (
Solution
Solution:
4400 yd
Details:
We know that Kim walked $$2.5$$ miles to the grocery store, but we want to know how far that is in yards. To figure that out we need 3 things:
1. The original amount: $$2.5$$ miles
2. The desired units: yards
3. The conversion rate: 1760 yards = 1 mile
The first step is to write $$2.5$$ miles as a fraction. The equivalent fraction is the following:
$$\dfrac{2.5\text{ miles}}{1}$$
Now we need to figure out the conversion factor that we need to use. We need it to cancel out the miles and leave us with yards. Using our conversion rate, 1760 yards = 1 mile, we know it will be one of the following:
$$\dfrac{1760\text{ yards}}{1\text{ mile}}$$ or $$\dfrac{1\text{ mile}}{1760\text{ yards}}$$
Since miles is at the top of our original fraction, we must use the conversion factor with miles in the bottom of the fraction so that miles will cancel out when we multiply. We set it up like this:
$$\displaystyle\frac{2.5\:\text{miles}}{1}\cdot\frac{1760\:\text{yards}}{1\:\text{mile}}$$
The first step is to cancel out the miles:
$$\displaystyle\frac{2.5\:{\color{DarkOrange} \cancel{\text{miles}}}}{1}\cdot \frac{1760\:\text{yards}}{1\:{\color{DarkOrange} \cancel{\text{mile}}}}$$
Then we multiply straight across:
$$\displaystyle\frac{2.5\:\cdot \: 1760 \text{ yards}}{1\:\cdot\: 1}$$
Then simplify both the top and bottom of the fraction:
$$= \dfrac{4400\text{ yards}}{1}$$
Which equals to the following:
= 4400 yards
So $$2.5$$ miles is the same distance as 4400 yards.
)
4. Brent has $$£35$$ (British pounds) that he would like to convert to US dollars ($$US$$). Assume the current conversion rate is $$£1 = 1.27$$. How much money will he have in dollars rounded to the nearest hundredth? (
Solution
Solution:
$$44.45$$
Details:
Brent has $$£35$$ and needs to know how much it is in US dollars. To figure that out we need three things:
1. The original amount: $$£35$$
2. The desired units: US dollars ($$US$$)
3. The conversion rate: $$£1 = 1.27$$
The first step is to write our original amount as a fraction:
$$\dfrac{{\text{£}}35}{1}$$
Using the conversion rate, $$£1 = 1.27$$, we can figure out our conversion factors. The two options are the following:
$$\dfrac{£1}{\1.27}$$ or $$\dfrac{\1.27}{£1}$$
We are changing British pounds ($$£$$) to US dollars () and £ is in the top (or numerator) of our original amount so we must use the conversion rate with $$£$$ in the bottom of the fraction so that we can cancel out the $$£$$.
$$\dfrac{\1.27}{£1}$$
Next, we multiply our original amount by our conversion factor:
$$\displaystyle\frac{£35}{1}\times\frac{\1.27}{£1}$$
Now we cancel out the $$£$$ since there is $$£$$ in the top and $$£$$ in the bottom:
$$\displaystyle\frac{\cancel{{\color{Magenta} £}}35}{1}\times\frac{\1.27}{\cancel{{\color{Magenta} £}}1}$$
Next, we multiply straight across:
$$\dfrac{35\:\cdot\:\1.27}{1\:\cdot\:1}$$
Then simplify top and bottom:
$$=\dfrac{\44.45}{1}$$
Which gives us the following:
$$= 44.45$$
This means that $$£35$$ is the same amount of money as $$44.45$$.
)
5. Max ran a half marathon which is $$13.1$$ miles in length. There are approximately $$0.62$$ miles in 1 kilometer. Convert the length of the half marathon to kilometers. Round to the nearest whole number. (
Solution
$$21$$ km
)
Video Solution for Problems 6 - 7
6. A piano weighs 325 pounds. There are approximately $$2.2$$ pounds in 1 kilogram. Convert the weight of the piano to kilograms. Round to the nearest whole number. (
Solution
$$148$$ kg
)
7. Zeezrom tried to tempt Amulek by offering him six onties of silver, which are of great worth, if he would deny the existence of a Supreme Being. (See Alma 11:21–25.) Of course, Amulek refuses the money. In the Nephite market, one onti of silver is the amount of money a judge earns after seven working days (Alma 11:3, 6–13.) How many days would a judge have to work to earn six onties of silver? (
Solution
$$42$$ days
)
## Need More Help?
1. Study other Math Lessons in the Resource Center.
2. Visit the Online Tutoring Resources in the Resource Center.
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By accessing our 180 Days of Math for Fifth Grade Answers Key Day 174 regularly, students can get better problem-solving skills.
Directions: Solve each problem.
Question 1.
Calculate the difference between 348 and 96.
Subtraction is one of the four basic arithmetic operations in mathematics. We can observe the applications of subtraction in our everyday life in different situations. For example, when we purchase fruits and vegetables for a certain amount of money say Rs. 200 and we have given an Rs. 500 note to the vendor. Now, the vendor returns the excess amount by performing subtraction such as 500 – 200 = 300. Then, the vendor will return Rs. 300.
Now we need to calculate the above-given question:
we need to subtract 348 from 96
348 = Minuend; 96 = Subtrahend; 252 = Difference
Question 2.
In mathematics, multiplication is a method of finding the product of two or more numbers. It is one of the basic arithmetic operations, that we use in everyday life. The major application we can see in multiplication tables.
In arithmetic, the multiplication of two numbers represents the repeated addition of one number with respect to another. These numbers can be whole numbers, natural numbers, integers, fractions, etc. If m is multiplied by n, then it means either m is added to itself ‘n’ number of times or vice versa.
The formula for multiplication:
The multiplication formula is given by:
Multiplier × Multiplicand = Product
– The multiplicand is the total number of objects in each group
– A multiplier is the number of equal groups
– Product is the result of multiplication of multiplier and multiplicand
Question 3.
Is 129 evenly divisible by 9?
No, it is not divisible by 9.
The division is breaking a number into an equal number of parts. The division is an arithmetic operation used in Maths. It splits a given number of items into different groups.
There are a number of signs that people may use to indicate division. The most common one is ÷, but the backslash / and a horizontal line (-) is also used in the form of Fraction, where a Numerator is written on the top and a Denominator on the bottom.
The division formula is:
Dividend ÷ Divisor = Quotient (or) Dividend/Divisor=quotient
14 is the quotient; 3 is the remainder.
Question 4.
What digit is in the thousands place in the number 95,387?
Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on.
The above-given number is 95,387
There are five digits in the number 95,387.
7 is in the unit’s place.
8 is in the tens place
3 is in the hundreds place
5 is in the thousands place
9 is in the ten thousand place
Question 5.
Write $$\frac{5}{2}$$ as a mixed number.
Step 1: Find the whole number
Calculate how many times the denominator goes into the numerator. To do that, divide 5 by 2 and keep only what is to the left of the decimal point:
5 / 2 = 2.5000 = 2
Step 2: Find a new numerator
Multiply the answer from Step 1 by the denominator and deduct that from the original numerator.
5-(2×2)
=5-4
=1
Step 3: Get a solution
Keep the original denominator and use the answers from Step 1 and Step 2 to get the answer. 5/2 as a mixed number is:
2 1/2
Question 6.
5 × 5 – 3 × 5 = ____________
The above equation can be solved as:
5 × 5 = 25
3 × 5 = 15
Now subtract the numbers 25 and 15
=25 – 15
=10
Therefore, 5 × 5 – 3 × 5 = 10.
Question 7.
165 – b = 87
b = ___________
We need to find out the value of b
The above-given equation is: 165 – b = 87
If we ‘b’ to right-hand side then it becomes ‘+b’
Now the equation will be:
165=87+b
Now get 87 to the left-hand side then it will subtract. Here the equation will be:
165-87=b
78=b
Therefore, the value of b is 78
Now verify the answer. Put the number 78 in the place of b in the above-given equation.
165 – b = 87
165-78=87
87=87
Question 8.
2$$\frac{1}{2}$$ hours = _________ minutes
2$$\frac{1}{2}$$ hours can be written as: 2 1/2 hours
1 hour is equal to 60 minutes
for 2 hours:
2*60=120 minutes.
1/2 (half an hour)=1/2*60
=30 minutes.
Now add 120 and 30 then we get:
120+30=150
Therefore, 2 1/2 hours = 150 minutes.
Question 9.
Which 3-dimensional figure has only square faces?
Three-dimensional shapes are those figures that have three dimensions, such as height, width, and length, like any object in the real world. It is also known as 3D.
Polyhedral or polyhedrons are straight-sided solid shapes. They are based on polygons, two- dimensional plane shapes with straight lines. Polyhedrons are defined as having straight edges, corners called vertices, and flat sides called faces. The polyhedrons are often defined by the number of edges, faces and vertices, as well as their faces are all the same size and same shape. Like polygons, polyhedrons can be regular and irregular based on regular and irregular polygons. Polyhedrons can be convex or concave. The cube is the most familiar and basic polyhedron. It has 6 square faces, 12 edges and eight vertices.
Question 10.
What is the mean of these numbers?
528, 455, 537
Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean=sum of observations/number of observations
The above-given numbers are 528, 455, 537
528+455+537=1520
The number of observations is 3
Now divide by 3 (total number of observations)
Mean=1520/3
Mean=506.66667
Therefore, the mean is 506.6
Question 11.
If the probability that someone knows how to swim is $$\frac{5}{6}$$, how many people in a group of 100 will likely not know how to swim?
P(S)=5/6=0.83
The probability of knowing how to swim is 0.83 that is nothing but 83%
The number of people is there in a group is 100
We know the probability of swimming. Now we need to find out the number will likely not know how to swim. Assume it as X
Now subtract the total group percentage and the probability we know how to swim
X=100-83
X=17
therefore, 17 people do not know how to swim.
Question 12.
Quadruple 46, then divide by 2.
First, we need to know what is quadruple.
to become four times as big, or to multiply a number or amount by four.
This means we need to multiply 46 with 4
46 * 4 = 184
And asked we need to divide by 2.
Therefore, quotient is 92 and remainder is 0
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# Limit
A limit is the behavior of a function as its inputs approach arbitrarily close to a given value.
## Intuitive Definition
We can examine the limit of a simple continuous function, $f(x)=x^2 \,$.
The limit at x=0, would simply be f(x)=0, since as the function is continuous and differentiable at every point, $f(0)=0$.
Indeed for this function, we define the limit as $\lim_{x \to a}f(x) = f(a) \,$. But this is a special case of limit, in the majority of limits cannot be solved in this manner.
For example, given
$f(x)=\left\{\begin{matrix} {x}^2, & \mbox{if }x\ne 0 \\ \\ 1, & \mbox{if }x=0. \end{matrix}\right.$
The limit of $f(x)\,$ as x approaches 0 is 0 (just as in $f(x)\,$), but $\lim_{x\to 0}f(x)\neq f(0)$; $f(x)$ is not continuous at $x = 0$ (as shown on the right).
In other cases a limit can fail to exist, as approaching the limit from different sides produces conflicting values.
For cases in general, the expression below satisfies our needs. $\lim_{x \to a}f(x) = L$
The expression above states that when $x\,$ approaches arbitrarily close to $a\,$, the function $f(x)\,$ becomes arbitrarily close to the value $L\,$, which is called the limit.
## Rigorous Definition of a Limit
This definition is more appropriate for 2nd year calculus students and higher.
The original statement $\lim_{x \to a}f(x) = L$ now means that given any $\varepsilon > 0$, a $\delta > 0 \,$ exists such that if $0 < |x-a| < \delta \,$, then $|f(x)-L|< \varepsilon$.
We use the definition as follows to prove for the function $f(x) = 3x-1 \,$, $\lim_{x \to 2}f(x) = 5$
Given any $\delta\,$, we choose a $\varepsilon\,$ such that $\delta \leq \varepsilon/3$
If $|x-2|<\delta\,$, we can derive $|f(x)-5|<3\delta\,$.
$3|x-2|<3\delta\,$
$|3x-6|<3\delta\,$
$|f(x)-5|<3\delta\,$
$3\delta \leq 3*\varepsilon/3 = \varepsilon\,$
Thus $\lim_{x \to 2}f(x) = 5$
## Properties of a Limit
The limit of a sum of two functions is equal to the sum of the limits of the functions.
$\lim (f(x)+g(x)) = \lim f(x) + \lim g(x)$
The limit of a difference between two functions is equal to the difference between the limits of the functions.
$\lim (f(x)-g(x)) = \lim f(x) - \lim g(x)$
The limit of a product of two functions is equal to the product of the limits of the functions.
$\lim (f(x)*g(x)) = \lim f(x) * \lim g(x)$
The limit of a quotient of two functions is equal to the quotient of the limits of the functions (assuming a non-zero denominator).
$\lim (f(x)/g(x)) = \lim f(x) / \lim g(x)$
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PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)
Section1.2Sets and Equivalence Relations
Subsection1.2.1Set Theory
A set is a well-defined collection of objects; that is, it is defined in such a manner that we can determine for any given object $$x$$ whether or not $$x$$ belongs to the set. The objects that belong to a set are called its elements or members. We will denote sets by capital letters, such as $$A$$ or $$X\text{;}$$ if $$a$$ is an element of the set $$A\text{,}$$ we write $$a \in A\text{.}$$
A set is usually specified either by listing all of its elements inside a pair of braces or by stating the property that determines whether or not an object $$x$$ belongs to the set. We might write
\begin{equation*} X = \{ x_1, x_2, \ldots, x_n \} \end{equation*}
for a set containing elements $$x_1, x_2, \ldots, x_n$$ or
\begin{equation*} X = \{ x :x \text{ satisfies }{\mathcal P}\} \end{equation*}
if each $$x$$ in $$X$$ satisfies a certain property $${\mathcal P}\text{.}$$ For example, if $$E$$ is the set of even positive integers, we can describe $$E$$ by writing either
\begin{equation*} E = \{2, 4, 6, \ldots \} \quad \text{or} \quad E = \{ x : x \text{ is an even integer and } x \gt 0 \}\text{.} \end{equation*}
We write $$2 \in E$$ when we want to say that 2 is in the set $$E\text{,}$$ and $$-3 \notin E$$ to say that $$-3$$ is not in the set $$E\text{.}$$
Some of the more important sets that we will consider are the following:
\begin{align*} {\mathbb N} &= \{n: n \text{ is a natural number}\} = \{1, 2, 3, \ldots \}\\ {\mathbb Z} &= \{n : n \text{ is an integer} \} = \{\ldots, -1, 0, 1, 2, \ldots \}\\ {\mathbb Q} &= \{r : r \text{ is a rational number}\} = \{p/q : p, q \in {\mathbb Z} \text{ where } q \neq 0\}\\ {\mathbb R} &= \{ x : x \text{ is a real number} \}\\ {\mathbb C} &= \{z : z \text{ is a complex number}\}\text{.} \end{align*}
We can find various relations between sets as well as perform operations on sets. A set $$A$$ is a subset of $$B\text{,}$$ written $$A \subset B$$ or $$B \supset A\text{,}$$ if every element of $$A$$ is also an element of $$B\text{.}$$ For example,
\begin{equation*} \{4,5,8\} \subset \{2, 3, 4, 5, 6, 7, 8, 9 \} \end{equation*}
and
\begin{equation*} {\mathbb N} \subset {\mathbb Z} \subset {\mathbb Q} \subset {\mathbb R} \subset {\mathbb C}\text{.} \end{equation*}
Trivially, every set is a subset of itself. A set $$B$$ is a proper subset of a set $$A$$ if $$B \subset A$$ but $$B \neq A\text{.}$$ If $$A$$ is not a subset of $$B\text{,}$$ we write $$A \notsubset B\text{;}$$ for example, $$\{4, 7, 9\} \notsubset \{2, 4, 5, 8, 9 \}\text{.}$$ Two sets are equal, written $$A = B\text{,}$$ if we can show that $$A \subset B$$ and $$B \subset A\text{.}$$
It is convenient to have a set with no elements in it. This set is called the empty set and is denoted by $$\emptyset\text{.}$$ Note that the empty set is a subset of every set.
To construct new sets out of old sets, we can perform certain operations: the union $$A \cup B$$ of two sets $$A$$ and $$B$$ is defined as
\begin{equation*} A \cup B = \{x : x \in A \text{ or } x \in B \} \end{equation*}
and the intersection of $$A$$ and $$B$$ is defined by
\begin{equation*} A \cap B = \{x : x \in A \text{ and } x \in B \}\text{.} \end{equation*}
If $$A = \{1, 3, 5\}$$ and $$B = \{ 1, 2, 3, 9 \}\text{,}$$ then
\begin{equation*} A \cup B = \{1, 2, 3, 5, 9 \} \quad \text{and} \quad A \cap B = \{ 1, 3 \}\text{.} \end{equation*}
We can consider the union and the intersection of more than two sets. In this case we write
\begin{equation*} \bigcup_{i = 1}^{n} A_{i} = A_{1} \cup \ldots \cup A_n \end{equation*}
and
\begin{equation*} \bigcap_{i = 1}^{n} A_{i} = A_{1} \cap \ldots \cap A_n \end{equation*}
for the union and intersection, respectively, of the sets $$A_1, \ldots, A_n\text{.}$$
When two sets have no elements in common, they are said to be disjoint; for example, if $$E$$ is the set of even integers and $$O$$ is the set of odd integers, then $$E$$ and $$O$$ are disjoint. Two sets $$A$$ and $$B$$ are disjoint exactly when $$A \cap B = \emptyset\text{.}$$
Sometimes we will work within one fixed set $$U\text{,}$$ called the universal set. For any set $$A \subset U\text{,}$$ we define the complement of $$A\text{,}$$ denoted by $$A'\text{,}$$ to be the set
\begin{equation*} A' = \{ x : x \in U \text{ and } x \notin A \}\text{.} \end{equation*}
We define the difference of two sets $$A$$ and $$B$$ to be
\begin{equation*} A \setminus B = A \cap B' = \{ x : x \in A \text{ and } x \notin B \}\text{.} \end{equation*}
Example1.2.1.Set Operations.
Let $${\mathbb R}$$ be the universal set and suppose that
\begin{equation*} A = \{ x \in {\mathbb R} : 0 \lt x \leq 3 \} \quad \text{and} \quad B = \{ x \in {\mathbb R} : 2 \leq x \lt 4 \}\text{.} \end{equation*}
Then
\begin{align*} A \cap B & = \{ x \in {\mathbb R} : 2 \leq x \leq 3 \}\\ A \cup B & = \{ x \in {\mathbb R} : 0 \lt x \lt 4 \}\\ A \setminus B & = \{ x \in {\mathbb R} : 0 \lt x \lt 2 \}\\ A' & = \{ x \in {\mathbb R} : x \leq 0 \text{ or } x \gt 3 \}\text{.} \end{align*}
Proof.
We will prove (1) and (3) and leave the remaining results to be proven in the exercises.
(1) Observe that
\begin{align*} A \cup A & = \{ x : x \in A \text{ or } x \in A \}\\ & = \{ x : x \in A \}\\ & = A\\ \end{align*}
and
\begin{align*} A \cap A & = \{ x : x \in A \text{ and } x \in A \}\\ & = \{ x : x \in A \}\\ & = A\text{.} \end{align*}
Also, $$A \setminus A = A \cap A' = \emptyset\text{.}$$
(3) For sets $$A\text{,}$$ $$B\text{,}$$ and $$C\text{,}$$
\begin{align*} A \cup (B \cup C) & = A \cup \{ x : x \in B \text{ or } x \in C \}\\ & = \{ x : x \in A \text{ or } x \in B, \text{ or } x \in C \}\\ & = \{ x : x \in A \text{ or } x \in B \} \cup C\\ & = (A \cup B) \cup C. \end{align*}
A similar argument proves that $$A \cap (B \cap C) = (A \cap B) \cap C\text{.}$$
Proof.
(1) We must show that $$(A \cup B)' \subset A' \cap B'$$ and $$(A \cup B)' \supset A' \cap B'\text{.}$$ Let $$x \in (A \cup B)'\text{.}$$ Then $$x \notin A \cup B\text{.}$$ So $$x$$ is neither in $$A$$ nor in $$B\text{,}$$ by the definition of the union of sets. By the definition of the complement, $$x \in A'$$ and $$x \in B'\text{.}$$ Therefore, $$x \in A' \cap B'$$ and we have $$(A \cup B)' \subset A' \cap B'\text{.}$$
To show the reverse inclusion, suppose that $$x \in A' \cap B'\text{.}$$ Then $$x \in A'$$ and $$x \in B'\text{,}$$ and so $$x \notin A$$ and $$x \notin B\text{.}$$ Thus $$x \notin A \cup B$$ and so $$x \in (A \cup B)'\text{.}$$ Hence, $$(A \cup B)' \supset A' \cap B'$$ and so $$(A \cup B)' = A' \cap B'\text{.}$$
The proof of (2) is left as an exercise.
Example1.2.4.Other Relations on Sets.
Other relations between sets often hold true. For example,
\begin{equation*} ( A \setminus B) \cap (B \setminus A) = \emptyset\text{.} \end{equation*}
To see that this is true, observe that
\begin{align*} ( A \setminus B) \cap (B \setminus A) & = ( A \cap B') \cap (B \cap A')\\ & = A \cap A' \cap B \cap B'\\ & = \emptyset\text{.} \end{align*}
Subsection1.2.2Cartesian Products and Mappings
Given sets $$A$$ and $$B\text{,}$$ we can define a new set $$A \times B\text{,}$$ called the Cartesian product of $$A$$ and $$B\text{,}$$ as a set of ordered pairs. That is,
\begin{equation*} A \times B = \{ (a,b) : a \in A \text{ and } b \in B \}\text{.} \end{equation*}
Example1.2.5.Cartesian Products.
If $$A = \{ x, y \}\text{,}$$ $$B = \{ 1, 2, 3 \}\text{,}$$ and $$C = \emptyset\text{,}$$ then $$A \times B$$ is the set
\begin{equation*} \{ (x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3) \} \end{equation*}
and
\begin{equation*} A \times C = \emptyset\text{.} \end{equation*}
We define the Cartesian product of $$n$$ sets to be
\begin{equation*} A_1 \times \cdots \times A_n = \{ (a_1, \ldots, a_n): a_i \in A_i \text{ for } i = 1, \ldots, n \}\text{.} \end{equation*}
If $$A = A_1 = A_2 = \cdots = A_n\text{,}$$ we often write $$A^n$$ for $$A \times \cdots \times A$$ (where $$A$$ would be written $$n$$ times). For example, the set $${\mathbb R}^3$$ consists of all of 3-tuples of real numbers.
Subsets of $$A \times B$$ are called relations. We will define a mapping or function $$f \subset A \times B$$ from a set $$A$$ to a set $$B$$ to be the special type of relation where $$(a, b) \in f$$ if for every element $$a \in A$$ there exists a unique element $$b \in B\text{.}$$ Another way of saying this is that for every element in $$A\text{,}$$ $$f$$ assigns a unique element in $$B\text{.}$$ We usually write $$f:A \rightarrow B$$ or $$A \stackrel{f}{\rightarrow} B\text{.}$$ Instead of writing down ordered pairs $$(a,b) \in A \times B\text{,}$$ we write $$f(a) = b$$ or $$f : a \mapsto b\text{.}$$ The set $$A$$ is called the domain of $$f$$ and
\begin{equation*} f(A) = \{ f(a) : a \in A \} \subset B \end{equation*}
is called the range or image of $$f\text{.}$$ We can think of the elements in the function’s domain as input values and the elements in the function’s range as output values.
Example1.2.7.Mappings.
Suppose $$A = \{1, 2, 3 \}$$ and $$B = \{a, b, c \}\text{.}$$ In Figure 1.2.6 we define relations $$f$$ and $$g$$ from $$A$$ to $$B\text{.}$$ The relation $$f$$ is a mapping, but $$g$$ is not because $$1 \in A$$ is not assigned to a unique element in $$B\text{;}$$ that is, $$g(1) = a$$ and $$g(1) = b\text{.}$$
Given a function $$f : A \rightarrow B\text{,}$$ it is often possible to write a list describing what the function does to each specific element in the domain. However, not all functions can be described in this manner. For example, the function $$f: {\mathbb R} \rightarrow {\mathbb R}$$ that sends each real number to its cube is a mapping that must be described by writing $$f(x) = x^3$$ or $$f:x \mapsto x^3\text{.}$$
Consider the relation $$f : {\mathbb Q} \rightarrow {\mathbb Z}$$ given by $$f(p/q) = p\text{.}$$ We know that $$1/2 = 2/4\text{,}$$ but is $$f(1/2) = 1$$ or 2? This relation cannot be a mapping because it is not well-defined. A relation is well-defined if each element in the domain is assigned to a unique element in the range.
If $$f:A \rightarrow B$$ is a map and the image of $$f$$ is $$B\text{,}$$ i.e., $$f(A) = B\text{,}$$ then $$f$$ is said to be onto or surjective. In other words, if there exists an $$a \in A$$ for each $$b \in B$$ such that $$f(a) = b\text{,}$$ then $$f$$ is onto. A map is one-to-one or injective if $$a_1 \neq a_2$$ implies $$f(a_1) \neq f(a_2)\text{.}$$ Equivalently, a function is one-to-one if $$f(a_1) = f(a_2)$$ implies $$a_1 = a_2\text{.}$$ A map that is both one-to-one and onto is called bijective.
Example1.2.8.One-to-One and Onto Mappings.
Let $$f:{\mathbb Z} \rightarrow {\mathbb Q}$$ be defined by $$f(n) = n/1\text{.}$$ Then $$f$$ is one-to-one but not onto. Define $$g : {\mathbb Q} \rightarrow {\mathbb Z}$$ by $$g(p/q) = p$$ where $$p/q$$ is a rational number expressed in its lowest terms with a positive denominator. The function $$g$$ is onto but not one-to-one.
Given two functions, we can construct a new function by using the range of the first function as the domain of the second function. Let $$f : A \rightarrow B$$ and $$g : B \rightarrow C$$ be mappings. Define a new map, the composition of $$f$$ and $$g$$ from $$A$$ to $$C\text{,}$$ by $$(g \circ f)(x) = g(f(x))\text{.}$$
Example1.2.10.Composition of Mappings.
Consider the functions $$f: A \rightarrow B$$ and $$g: B \rightarrow C$$ that are defined in Figure 1.2.9 (top). The composition of these functions, $$g \circ f: A \rightarrow C\text{,}$$ is defined in Figure 1.2.9 (bottom).
Example1.2.11.Composition is not Commutative.
Let $$f(x) = x^2$$ and $$g(x) = 2x + 5\text{.}$$ Then
\begin{equation*} (f \circ g)(x) = f(g(x)) = (2x + 5)^2 = 4x^2 + 20x + 25 \end{equation*}
and
\begin{equation*} (g \circ f)(x) = g(f(x)) = 2x^2 + 5\text{.} \end{equation*}
In general, order makes a difference; that is, in most cases $$f \circ g \neq g \circ f\text{.}$$
Example1.2.12.Some Mappings Commute.
Sometimes it is the case that $$f \circ g= g \circ f\text{.}$$ Let $$f(x) = x^3$$ and $$g(x) = \sqrt[3]{x}\text{.}$$ Then
\begin{equation*} (f \circ g )(x) = f(g(x)) = f( \sqrt[3]{x}\, ) = (\sqrt[3]{x}\, )^3 = x \end{equation*}
and
\begin{equation*} (g \circ f )(x) = g(f(x)) = g( x^3) = \sqrt[3]{ x^3} = x\text{.} \end{equation*}
Example1.2.13.A Linear Map.
Given a $$2 \times 2$$ matrix
\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\text{,} \end{equation*}
we can define a map $$T_A : {\mathbb R}^2 \rightarrow {\mathbb R}^2$$ by
\begin{equation*} T_A (x,y) = (ax + by, cx +dy) \end{equation*}
for $$(x,y)$$ in $${\mathbb R}^2\text{.}$$ This is actually matrix multiplication; that is,
\begin{equation*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx +dy \end{pmatrix}\text{.} \end{equation*}
Maps from $${\mathbb R}^n$$ to $${\mathbb R}^m$$ given by matrices are called linear maps or linear transformations.
Example1.2.14.A Permutation.
Suppose that $$S = \{ 1,2,3 \}\text{.}$$ Define a map $$\pi :S\rightarrow S$$ by
\begin{equation*} \pi( 1 ) = 2, \qquad \pi( 2 ) = 1, \qquad \pi( 3 ) = 3\text{.} \end{equation*}
This is a bijective map. An alternative way to write $$\pi$$ is
\begin{equation*} \begin{pmatrix} 1 & 2 & 3 \\ \pi(1) & \pi(2) & \pi(3) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}\text{.} \end{equation*}
For any set $$S\text{,}$$ a one-to-one and onto mapping $$\pi : S \rightarrow S$$ is called a permutation of $$S\text{.}$$
Proof.
We will prove (1) and (3). Part (2) is left as an exercise. Part (4) follows directly from (2) and (3).
(1) We must show that
\begin{equation*} h \circ (g \circ f) = (h \circ g) \circ f\text{.} \end{equation*}
For $$a \in A$$ we have
\begin{align*} (h \circ (g \circ f))(a) & = h((g \circ f)(a))\\ & = h(g(f(a))) \\ & = (h \circ g)(f(a))\\ & = ((h \circ g) \circ f)(a)\text{.} \end{align*}
(3) Assume that $$f$$ and $$g$$ are both onto functions. Given $$c \in C\text{,}$$ we must show that there exists an $$a \in A$$ such that $$(g \circ f)(a) = g(f(a)) = c\text{.}$$ However, since $$g$$ is onto, there is an element $$b \in B$$ such that $$g(b) = c\text{.}$$ Similarly, there is an $$a \in A$$ such that $$f(a) = b\text{.}$$ Accordingly,
\begin{equation*} (g \circ f)(a) = g(f(a)) = g(b) = c\text{.} \end{equation*}
If $$S$$ is any set, we will use $$id_S$$ or $$id$$ to denote the identity mapping from $$S$$ to itself. Define this map by $$id(s) = s$$ for all $$s \in S\text{.}$$ A map $$g: B \rightarrow A$$ is an inverse mapping of $$f: A \rightarrow B$$ if $$g \circ f = id_A$$ and $$f \circ g = id_B\text{;}$$ in other words, the inverse function of a function simply “undoes” the function. A map is said to be invertible if it has an inverse. We usually write $$f^{-1}$$ for the inverse of $$f\text{.}$$
Example1.2.16.An Inverse Function.
The function $$f(x) = x^3$$ has inverse $$f^{-1}(x) = \sqrt[3]{x}$$ by Example 1.2.12.
Example1.2.17.Exponential and Logarithmic Functions are Inverses.
The natural logarithm and the exponential functions, $$f(x) = \ln x$$ and $$f^{-1}(x) = e^x\text{,}$$ are inverses of each other provided that we are careful about choosing domains. Observe that
\begin{equation*} f(f^{-1}(x)) = f(e^x) = \ln e^x = x \end{equation*}
and
\begin{equation*} f^{-1}(f(x)) = f^{-1}(\ln x) = e^{\ln x} = x \end{equation*}
whenever composition makes sense.
Example1.2.18.A Matrix Inverse Yields an Inverse of a Linear Map.
Suppose that
\begin{equation*} A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\text{.} \end{equation*}
Then $$A$$ defines a map from $${\mathbb R}^2$$ to $${\mathbb R}^2$$ by
\begin{equation*} T_A (x,y) = (3x + y, 5x + 2y)\text{.} \end{equation*}
We can find an inverse map of $$T_A$$ by simply inverting the matrix $$A\text{;}$$ that is, $$T_A^{-1} = T_{A^{-1}}\text{.}$$ In this example,
\begin{equation*} A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \end{equation*}
and hence, the inverse map is given by
\begin{equation*} T_A^{-1} (x,y) = (2x - y, -5x + 3y)\text{.} \end{equation*}
It is easy to check that
\begin{equation*} T^{-1}_A \circ T_A (x,y) = T_A \circ T_A^{-1} (x,y) = (x,y)\text{.} \end{equation*}
Not every map has an inverse. If we consider the map
\begin{equation*} T_B (x,y) = (3x , 0 ) \end{equation*}
given by the matrix
\begin{equation*} B = \begin{pmatrix} 3 & 0 \\ 0 & 0 \end{pmatrix}\text{,} \end{equation*}
then an inverse map would have to be of the form
\begin{equation*} T_B^{-1} (x,y) = (ax + by, cx +dy) \end{equation*}
and
\begin{equation*} (x,y) = T \circ T_B^{-1} (x,y) = (3ax + 3by, 0) \end{equation*}
for all $$x$$ and $$y\text{.}$$ Clearly this is impossible because $$y$$ might not be 0.
Example1.2.19.An Inverse Permutation.
Given the permutation
\begin{equation*} \pi = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \end{equation*}
on $$S = \{ 1,2,3 \}\text{,}$$ it is easy to see that the permutation defined by
\begin{equation*} \pi^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \end{equation*}
is the inverse of $$\pi\text{.}$$ In fact, any bijective mapping possesses an inverse, as we will see in the next theorem.
Proof.
Suppose first that $$f:A \rightarrow B$$ is invertible with inverse $$g: B \rightarrow A\text{.}$$ Then $$g \circ f = id_A$$ is the identity map; that is, $$g(f(a)) = a\text{.}$$ If $$a_1, a_2 \in A$$ with $$f(a_1) = f(a_2)\text{,}$$ then $$a_1 = g(f(a_1)) = g(f(a_2)) = a_2\text{.}$$ Consequently, $$f$$ is one-to-one. Now suppose that $$b \in B\text{.}$$ To show that $$f$$ is onto, it is necessary to find an $$a \in A$$ such that $$f(a) = b\text{,}$$ but $$f(g(b)) = b$$ with $$g(b) \in A\text{.}$$ Let $$a = g(b)\text{.}$$
Conversely, let $$f$$ be bijective and let $$b \in B\text{.}$$ Since $$f$$ is onto, there exists an $$a \in A$$ such that $$f(a) = b\text{.}$$ Because $$f$$ is one-to-one, $$a$$ must be unique. Define $$g$$ by letting $$g(b) = a\text{.}$$ We have now constructed the inverse of $$f\text{.}$$
Subsection1.2.3Equivalence Relations and Partitions
A fundamental notion in mathematics is that of equality. We can generalize equality with equivalence relations and equivalence classes. An equivalence relation on a set $$X$$ is a relation $$R \subset X \times X$$ such that
• $$(x, x) \in R$$ for all $$x \in X$$ (reflexive property);
• $$(x, y) \in R$$ implies $$(y, x) \in R$$ (symmetric property);
• $$(x, y)$$ and $$(y, z) \in R$$ imply $$(x, z) \in R$$ (transitive property).
Given an equivalence relation $$R$$ on a set $$X\text{,}$$ we usually write $$x \sim y$$ instead of $$(x, y) \in R\text{.}$$ If the equivalence relation already has an associated notation such as $$=\text{,}$$ $$\equiv\text{,}$$ or $$\cong\text{,}$$ we will use that notation.
Example1.2.21.Equivalent Fractions.
Let $$p\text{,}$$ $$q\text{,}$$ $$r\text{,}$$ and $$s$$ be integers, where $$q$$ and $$s$$ are nonzero. Define $$p/q \sim r/s$$ if $$ps = qr\text{.}$$ Clearly $$\sim$$ is reflexive and symmetric. To show that it is also transitive, suppose that $$p/q \sim r/s$$ and $$r/s \sim t/u\text{,}$$ with $$q\text{,}$$ $$s\text{,}$$ and $$u$$ all nonzero. Then $$ps = qr$$ and $$ru = st\text{.}$$ Therefore,
\begin{equation*} psu = qru = qst\text{.} \end{equation*}
Since $$s \neq 0\text{,}$$ $$pu = qt\text{.}$$ Consequently, $$p/q \sim t/u\text{.}$$
Example1.2.22.An Equivalence Relation From Derivatives.
Suppose that $$f$$ and $$g$$ are differentiable functions on $${\mathbb R}\text{.}$$ We can define an equivalence relation on such functions by letting $$f(x) \sim g(x)$$ if $$f'(x) = g'(x)\text{.}$$ It is clear that $$\sim$$ is both reflexive and symmetric. To demonstrate transitivity, suppose that $$f(x) \sim g(x)$$ and $$g(x) \sim h(x)\text{.}$$ From calculus we know that $$f(x) - g(x) = c_1$$ and $$g(x)- h(x) = c_2\text{,}$$ where $$c_1$$ and $$c_2$$ are both constants. Hence,
\begin{equation*} f(x) - h(x) = ( f(x) - g(x)) + ( g(x)- h(x)) = c_1 - c_2 \end{equation*}
and $$f'(x) - h'(x) = 0\text{.}$$ Therefore, $$f(x) \sim h(x)\text{.}$$
Example1.2.23.Equivalent Circles.
For $$(x_1, y_1 )$$ and $$(x_2, y_2)$$ in $${\mathbb R}^2\text{,}$$ define $$(x_1, y_1 ) \sim (x_2, y_2)$$ if $$x_1^2 + y_1^2 = x_2^2 + y_2^2\text{.}$$ Then $$\sim$$ is an equivalence relation on $${\mathbb R}^2\text{.}$$
Example1.2.24.Equivalent Matrices.
Let $$A$$ and $$B$$ be $$2 \times 2$$ matrices with entries in the real numbers. We can define an equivalence relation on the set of $$2 \times 2$$ matrices, by saying $$A \sim B$$ if there exists an invertible matrix $$P$$ such that $$PAP^{-1} = B\text{.}$$ For example, if
\begin{equation*} A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} -18 & 33 \\ -11 & 20 \end{pmatrix}\text{,} \end{equation*}
then $$A \sim B$$ since $$PAP^{-1} = B$$ for
\begin{equation*} P = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\text{.} \end{equation*}
Let $$I$$ be the $$2 \times 2$$ identity matrix; that is,
\begin{equation*} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\text{.} \end{equation*}
Then $$IAI^{-1} = IAI = A\text{;}$$ therefore, the relation is reflexive. To show symmetry, suppose that $$A \sim B\text{.}$$ Then there exists an invertible matrix $$P$$ such that $$PAP^{-1} = B\text{.}$$ So
\begin{equation*} A = P^{-1} B P = P^{-1} B (P^{-1})^{-1}\text{.} \end{equation*}
Finally, suppose that $$A \sim B$$ and $$B \sim C\text{.}$$ Then there exist invertible matrices $$P$$ and $$Q$$ such that $$PAP^{-1} = B$$ and $$QBQ^{-1} = C\text{.}$$ Since
\begin{equation*} C = QBQ^{-1} = QPAP^{-1} Q^{-1} = (QP)A(QP)^{-1}\text{,} \end{equation*}
the relation is transitive. Two matrices that are equivalent in this manner are said to be similar.
A partition $${\mathcal P}$$ of a set $$X$$ is a collection of nonempty sets $$X_1, X_2, \ldots$$ such that $$X_i \cap X_j = \emptyset$$ for $$i \neq j$$ and $$\bigcup_k X_k = X\text{.}$$ Let $$\sim$$ be an equivalence relation on a set $$X$$ and let $$x \in X\text{.}$$ Then $$[x] = \{ y \in X : y \sim x \}$$ is called the equivalence class of $$x\text{.}$$ We will see that an equivalence relation gives rise to a partition via equivalence classes. Also, whenever a partition of a set exists, there is some natural underlying equivalence relation, as the following theorem demonstrates.
Proof.
Suppose there exists an equivalence relation $$\sim$$ on the set $$X\text{.}$$ For any $$x \in X\text{,}$$ the reflexive property shows that $$x \in [x]$$ and so $$[x]$$ is nonempty. Clearly $$X = \bigcup_{x \in X} [x]\text{.}$$ Now let $$x, y \in X\text{.}$$ We need to show that either $$[x] = [y]$$ or $$[x] \cap [y] = \emptyset\text{.}$$ Suppose that the intersection of $$[x]$$ and $$[y]$$ is not empty and that $$z \in [x] \cap [y]\text{.}$$ Then $$z \sim x$$ and $$z \sim y\text{.}$$ By symmetry and transitivity $$x \sim y\text{;}$$ hence, $$[x] \subset [y]\text{.}$$ Similarly, $$[y] \subset [x]$$ and so $$[x] = [y]\text{.}$$ Therefore, any two equivalence classes are either disjoint or exactly the same.
Conversely, suppose that $${\mathcal P} = \{X_i\}$$ is a partition of a set $$X\text{.}$$ Let two elements be equivalent if they are in the same partition. Clearly, the relation is reflexive. If $$x$$ is in the same partition as $$y\text{,}$$ then $$y$$ is in the same partition as $$x\text{,}$$ so $$x \sim y$$ implies $$y \sim x\text{.}$$ Finally, if $$x$$ is in the same partition as $$y$$ and $$y$$ is in the same partition as $$z\text{,}$$ then $$x$$ must be in the same partition as $$z\text{,}$$ and transitivity holds.
Let us examine some of the partitions given by the equivalence classes in the last set of examples.
Example1.2.27.A Partition of Fractions.
In the equivalence relation in Example 1.2.21, two pairs of integers, $$(p,q)$$ and $$(r,s)\text{,}$$ are in the same equivalence class when they reduce to the same fraction in its lowest terms.
Example1.2.28.A Partition of Functions.
In the equivalence relation in Example 1.2.22, two functions $$f(x)$$ and $$g(x)$$ are in the same partition when they differ by a constant.
Example1.2.29.A Partition of Circles.
We defined an equivalence class on $${\mathbb R}^2$$ by $$(x_1, y_1 ) \sim (x_2, y_2)$$ if $$x_1^2 + y_1^2 = x_2^2 + y_2^2\text{.}$$ Two pairs of real numbers are in the same partition when they lie on the same circle about the origin.
Example1.2.30.A Partition of Integers.
Let $$r$$ and $$s$$ be two integers and suppose that $$n \in {\mathbb N}\text{.}$$ We say that $$r$$ is congruent to $$s$$ modulo $$n\text{,}$$ or $$r$$ is congruent to $$s$$ mod $$n\text{,}$$ if $$r - s$$ is evenly divisible by $$n\text{;}$$ that is, $$r - s = nk$$ for some $$k \in {\mathbb Z}\text{.}$$ In this case we write $$r \equiv s \pmod{n}\text{.}$$ For example, $$41 \equiv 17 \pmod{ 8}$$ since $$41 - 17=24$$ is divisible by 8. We claim that congruence modulo $$n$$ forms an equivalence relation of $${\mathbb Z}\text{.}$$ Certainly any integer $$r$$ is equivalent to itself since $$r - r = 0$$ is divisible by $$n\text{.}$$ We will now show that the relation is symmetric. If $$r \equiv s \pmod{ n}\text{,}$$ then $$r - s = -(s -r)$$ is divisible by $$n\text{.}$$ So $$s - r$$ is divisible by $$n$$ and $$s \equiv r \pmod{ n}\text{.}$$ Now suppose that $$r \equiv s \pmod{ n}$$ and $$s \equiv t \pmod{ n}\text{.}$$ Then there exist integers $$k$$ and $$l$$ such that $$r -s = kn$$ and $$s - t = ln\text{.}$$ To show transitivity, it is necessary to prove that $$r - t$$ is divisible by $$n\text{.}$$ However,
\begin{equation*} r - t = r - s + s - t = kn + ln = (k + l)n\text{,} \end{equation*}
and so $$r - t$$ is divisible by $$n\text{.}$$
If we consider the equivalence relation established by the integers modulo 3, then
\begin{align*} {[0]} & = \{ \ldots, -3, 0, 3, 6, \ldots \}\\ {[1]} & = \{ \ldots, -2, 1, 4, 7, \ldots \}\\ {[2]} & = \{ \ldots, -1, 2, 5, 8, \ldots \}\text{.} \end{align*}
Notice that $$[0] \cup [1] \cup [2] = {\mathbb Z}$$ and also that the sets are disjoint. The sets $$[0]\text{,}$$ $$[1]\text{,}$$ and $$[2]$$ form a partition of the integers.
The integers modulo $$n$$ are a very important example in the study of abstract algebra and will become quite useful in our investigation of various algebraic structures such as groups and rings. In our discussion of the integers modulo $$n$$ we have actually assumed a result known as the division algorithm, which will be stated and proved in Chapter 2.
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## Divide Decimals
Division of Decimals
The division algorithm for dividing a decimal by a whole number is similar to the division algorithm for whole numbers. An example of dividing money can be used to illustrate this.
Divide \$56.34 equally among 6 individuals. How much does each person receive? Think of the \$56.34 as five \$10 bills, six \$1 bills, three dimes, and four pennies.
If the divisor and dividend are multiplied by the same number, the quotient remains the same. If you are dividing apples equally among a group of children and triple both the number of apples and the number of children, then the number of apples per child stays the same. This is the principle used when dividing by a decimal. Use multiples of 10 to convert the division by a decimal into a problem of dividing by a whole number.
23 ÷ 3.2 = 230 ÷ 32
452.893 ÷ 0.23 = 45,289.3 ÷ 23
Multiply by 10.
Multiply by 100.
Such conversions are usually shown as
The quotient of two decimals can always be written as a quotient of whole numbers. For example, the quotient 452.893 ÷ 0.23 = 452,893 ÷ 230 (multiply both the dividend and the divisor by 1,000).
Repeating and Terminating Decimals
All fractions represent division and thus can be written as decimals. However, not all division ends, or terminates. A terminating decimal occurs when the division has a remainder of 0. An example is 1 ÷ 2 = 0.5. Fractions like cannot be represented by a terminating decimal. The decimal equivalent for is 0.333…. This means the threes continue on and on indefinitely since there is always a remainder of 1. A complete explanation of why = 0.333… requires an understanding of infinite sums and is beyond most students in elementary school. However, students do understand patterns and can see the patterns that occur in repeating decimals. A repeating decimal occurs when a pattern of numbers repeats indefinitely after the decimal point and the numbers are not all zeros. In writing repeating decimals, a bar is placed over the pattern of numbers that repeats. Thus,
Teaching Model 14.7: Divide a Decimal by a Decimal
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FUNDAMENTAL PRINCIPLES OF INTEGRATION - General Methods of Integration - The Calculus Primer
## The Calculus Primer (2011)
### Chapter 45. FUNDAMENTAL PRINCIPLES OF INTEGRATION
12—3. Basic Relationships. From what has already been said, it will be seen that many integrations can be performed “by inspection.” If the reader has thoroughly mastered the fundamental formulas for derivatives, he will, with practice, be able to integrate many expressions at sight. However, we would wish to impress him with the need for having the fundamental formulas for differentiating, especially in differential notation, “at fingers’ ends.” Ready command of these formulas will go a long way toward success in integrating. Facility in integrating can be achieved only by diligent practice.
One of the basic principles for integrating is, of course, that
dx = x + C, [1]
which follows at once from the definitions of the symbols and the meaning of the constant of integration.
Another principle is the following:
Since
or d(av) = a dv;
therefore a dv = a dv. [2]
Or, expressing the role played by a constant multiplier in the integrand in another way, we have,
av dx = a v dx.
In words, a constant factor may be removed from one side of the integration sign to the other without affecting the value of the integral. This is a very useful procedure, but the reader should be warned that this can be done only with a constant, and never with a variable. For example:
2x3 dx = 2 x3 dx.
πa2x2 dx = πa2 x2 dx.
A third basic principle is the following:
(du + dv − dw) = du + dv − dw, [3]
or, in different symbolism,
{f(x) + ø(x) − F(x)} dx = f(x) dx + ø(x) dx − F(x) dx.
In words, the integral of the sum of any number of functions is equal to the sum of the integrals of the several functions. This follows, of course, from the fact that
EXAMPLE 1. Find (ax2 + bx + c) dx.
Solution.
(ax2 + bx + c) dx = ax2 dx + bx dx + c dx
= a x2 dx + b x dx + c dx
EXAMPLE 2. Find
Solution.
12—4. The Process of Integration. It is important to point out that the result of an integration does not always lead to a simple function. That is to say, performing an inverse operation in mathematics sometimes yields unusual results. Thus in arithmetic, when adding positive numbers only, the inverse operation of subtraction may lead to negative numbers, which differ in kind from the sort of numbers operated upon in arithmetic in the first place. Again, in algebra, when raising real numbers to the second power, the inverse operation of taking the square root may lead to imaginaries, a different kind of number. So in calculus: when we differentiate various so-called elementary functions (i.e., algebraic expressions resulting from the operations of addition, subtraction, multiplication, division, involution, evolution; exponential and logarithmic functions; and trigonometric functions and their inverses) the result of the differentiation contains only elementary functions; but the inverse operation of integration may yield a new kind of function, not an elementary function at all. However, in this book we shall limit ourselves to integrals which are elementary functions.
12—5. The Form Un dU. The integral of a power of a variable, when the power is not −1, is given by the following:
This may be readily verified by reversing the procedure:
Thus, to integrate a variable to a given power, we increase the exponent by that amount and also divide the expression by the amount of the new exponent.
EXAMPLE 1. Find x6 dx,
Solution.
EXAMPLE 2. Find ax dx.
Solution.
EXAMPLE 3. Find
Solution.
EXAMPLE 4. Find x¾ dx.
Solution.
EXAMPLE 5. Find
Solution.
EXAMPLE 6. Find (2a − 3x)2 dx.
Solution. Multiplying out first:
(2a − 3x)2 dx = (4a2 − 12ax + 9x2) dx
= 4a2 dx − 12a x dx + 9 x2 dx
= 4a2x − 6ax2 + 3x3 + C.
EXERCISE 12—1
Integrate each of the following; check your result by differentiating:
1. 8x3 dx
2. 10ax4 dx
3. y4 dy
4. x¾dx
5. 2 dx
6. x dx
7. z dz
8. 5y dy
9. xm/n dx
15. kz dz
16. 2x−2 dx
23. (x + 3)2 dx
24. (3a − x2)2 dx
25. 5(x + 2)3 dx
12—6. Variations of the Form Un dU. At this point it might be well to suggest to the reader that there is no general or infallible method of finding the integral of a given function, since it is impossible always to retrace the steps of a differentiation. What we must do is this: depending upon how familiar we are with the derivative formulas, we try to recognize the given function as the derivative of some known function, or to change it to a form which can so be recognized. Every process of integration involves bringing the integrand to a form to which some standard, known formula applies. Considerable ingenuity is often required to do this.
EXAMPLE 1. Find .
Solution.
Now, we may write dx = d(a + x).
Hence, the given integral may be written:
(a + x)½ dx = (a + x)½ d(a + x);
here, if we consider u = a + x, and n = , then our integral can be found by means of equation [4]:
EXAMPLE 2. Find .
EXAMPLE 3. Find (b2a2x2)½x dx.
Solution. We note that
d(b2 a2x2) = −2a2x dx.
Multiplying inside the integral sign by −2a2, and dividing outside by −2a2, leaves the integral unchanged in value, but transforms it to the form un du, where u = (b2a2x2), and n = ;
hence,
EXAMPLE 4. Find (x3 + 1)2x2 dx.
Solution. We can first multiply out:
(x3 + 1)2x2 dx = (x6 + 2x3 + 1)x2 dx
= (x8 + 2x5 + x2)dx
Or, an alternative, more elegant solution. Since d(x3 + 1) = 3x2 dx, we may replace the “x2 dx” in the original problem by “d(x3 + 1)”; thus
We leave it to the reader to verify the fact that the right-hand members of (1) and (2), respectively, will both check when differentiated.
EXERCISE 12—2
Integrate; check by differentiating:
14. (m + x)(a + x) dx
15. Integrate (x4 1)2x3 dx in two ways, explain why the results are not identical, and show that both are correct.
12—7. The Form . The integral of the reciprocal of a variable is the logarithm of the variable; or
This, of course, is readily verified by differentiation; thus,
NOTE. Strictly speaking, we should write = log |u| + C, since, although is defined for all values of u ≠ 0, log u is defined only for values of u > 0.
EXAMPLE 1. Find .
Solution.
EXAMPLE 2. Find
Solution. Since, from the definition of a differential, we may write, if we care to, dx = d(x + 1).
Hence
where we are regarding u as equal to x + 1, and
= 1, or du = 1(dx) = dx.
By differentiating the answer, the result is seen to check; thus
In fact, whenever the numerator of an integrand is (or can be made) the differential of the denominator, the integral is the logarithm of the denominator.
EXAMPLE 3. Find .
Solution. d(x2 + 1) = 2x.
Rewrite the given example:
now, considering u as x2 + 1, and as 2x, or du = 2x dx, we have:
EXAMPLE 4. Find
Solution. First rewrite the integral as follows:
EXERCISE 12—3
Find the following integrals; check by differentiation:
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## Which applies the power of a power rule property to simplify the expression?
The power of a power rule says, when a number with an exponent is raised to another exponent, we can simplify the exponent by keeping the base and multiplying the exponents. The general form of the rule is (am)n=am·n. For example, to find the power of the power of the expression, (x2)7=x2·7=x14.
## What is the rule for power of product?
The power of a product rule tells us that we can simplify a power of a power by multiplying the exponents and keeping the same base.
## How do you rewrite an expression as a power of a product?
To find a power of a product, find the power of each factor and then multiply. In general, (ab)m=am⋅bm. am⋅bm=(ab)m. In other words, you can keep the exponent the same and multiply the bases.
## What is the difference between the product rule and the power rule?
The power rule to follow when finding the derivative of a variable base, raised to a fixed power. … How the product rule allows us to find the derivative of a function that is defined as the product of another two (or more) functions.
## What is the product of powers rule for exponents?
Lesson Summary
When you are multiplying like terms with exponents, use the product of powers rule as a shortcut to finding the answer. It states that when you are multiplying two terms that have the same base, just add their exponents to find your answer.
## What is the power rule in exponents?
The Power Rule for Exponents: (am)n = am*n. To raise a number with an exponent to a power, multiply the exponent times the power. Negative Exponent Rule: xn = 1/xn. Invert the base to change a negative exponent into a positive.
## Which rule is used to simplify 34 2?
The power rule says that if we have an exponent raised to another exponent, you can just multiply the exponents together. For example, suppose we wanted to simplify (34)2. Our rule tells us: (34)2=34⋅2=38 And to prove this we can write out the multiplication.
## How do you simplify?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## Why can’t you use the product of powers rule to simplify this expression explain 34 28?
Why can’t you use the product of powers rule to simplify this expression? Explain. They are not the same number. to do that it would have to be 3 and a 3 or 2 and a 2 can not combine unlike terms.
## Which rules of exponents will be used to evaluate the expression?
Divide the coefficients and subtract the exponents of matching variables. All of these rules of exponents—the Product Rule, the Power Rule, and the Quotient Rule—are helpful when evaluating expressions with common bases.
## Which is the value of this expression when a =- 2 and B =- 3?
IF your expression “a-2 b-3” is meant to be “(a-2)(b-3)” then Bill is correct, and the answer is -5 when a is 3 and b is -2. However, if you mean”a – 2b -3″ then, substituting the values you gave for a=3 and b=-2, that would be “3 – 2(-2) -3” which is the same as “3 +4 -3” and the expression evaluates to 4.
## How does the use of exponents simplify the way we write expressions?
Any non-zero number or variable raised to a power of 0 is equal to 1. When dividing two terms with the same base, subtract the exponent in the denominator from the exponent in the numerator: Power of a Power: To raise a power to a power, multiply the exponents.
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# Finding the Inverse of an Exponential Function
I will go over three examples in this tutorial showing how to determine algebraically the inverse of an exponential function. But before you take a look at the worked examples, I suggest that you review the suggested steps below first in order to have a good grasp of the general procedure.
## Steps to Find the Inverse of an Exponential Function
STEP 1: Change $f\left( x \right)$ to $y$.
$\large{f\left( x \right) \to y}$
STEP 2: Interchange $\color{blue}x$ and $\color{red}y$ in the equation.
$\large{x \to y}$
$\large{y \to x}$
STEP 3: Isolate the exponential expression on one side (left or right) of the equation.
The exponential expression shown below is a generic form where $b$ is the base, while $N$ is the exponent.
STEP 4: Eliminate the base $b$ of the exponential expression by taking the logarithms of both sides of the equation.
• To make the simplification much easier, take the logarithm of both sides using the base of the exponential expression itself.
• Using the log rule,
STEP 5: Solve the exponential equation for $\color{red}y$ to get the inverse. Finally, replace $\color{red}y$ with the inverse notation ${f^{ – 1}}\left( x \right)$ to write the final answer.
Replace $y$ with ${f^{ – 1}}\left( x \right)$
Let’s apply the suggested steps above to solve some problems.
### Examples of How to Find the Inverse of an Exponential Function
Example 1: Find the inverse of the exponential function below.
This should be an easy problem because the exponential expression on the right side of the equation is already isolated for us.
Start by replacing the function notation $f\left( x \right)$ by $y$.
The next step is to switch the variables $\color{red}x$ and $\color{red}y$ in the equation.
Since the exponential expression is by itself on one side of the equation, we can now take the logarithms of both sides. When we get the logarithms of both sides, we will use the base of $\color{blue}2$ because this is the base of the given exponential expression.
Apply the Log of Exponent Rule which is ${\log _b}\left( {{b^k}} \right) = k$ as part of the simplification process. The rule states that the logarithm of an exponential number where its base is the same as the base of the log is equal to the exponent.
We are almost done! Solve for $y$ by adding both sides by $5$ then divide the equation by the coefficient of $y$ which is $3$. Don’t forget to replace $y$ to ${f^{ – 1}}\left( x \right)$. This means that we have found the inverse function.
If we graph the original exponential function and its inverse on the same $XY-$ plane, they must be symmetrical along the line $\large{\color{blue}y=x}$. Which they are!
Example 2: Find the inverse of the exponential function below.
The only difference of this problem from the previous one is that the exponential expression has a denominator of $2$. Other than that, the steps will be the same.
We change the function notation $f\left( x \right)$ to $y$, followed by interchanging the roles of $\color{red}x$ and $\color{red}y$ variables.
At this point, we can’t perform the step of taking the logarithms of both sides just yet. The reason is that the exponential expression on the right side is not fully by itself. We first have to get rid of the denominator $2$.
We can accomplish that by multiplying both sides of the equation by $2$. The left side becomes $2x$ and the denominator on the right side is gone!
By isolating the exponential expression on one side, it is now possible to get the logs of both sides. When you do this, always make sure to use the base of the exponential expression as the base of the logarithmic operations.
In this case, the base of the exponential expression is $5$. Therefore, we apply log operations on both sides using the base of $5$.
Using this log rule, ${\log _b}\left( {{b^k}} \right) = k$ , the fives will cancel out leaving the exponent $\color{blue}4x+1$ on the right side of the equation after simplification. This is great since the log portion of the equation is gone.
We can now finish this up by solving for the variable $y$, then replacing that by ${f^{ – 1}}\left( x \right)$ to denote that we have obtained the inverse function.
As you can see, the graphs of the exponential function and its inverse are symmetrical about the line $\large{\color{green}y=x}$.
Example 3: Find the inverse of the exponential function below.
I see that we have an exponential expression being divided by another. The good thing is that the exponential expressions have the same base of $3$. We should be able to simplify this using the Division Rule of Exponent. To divide exponential expressions having equal bases, copy the common base and then subtract their exponents. Below is the rule. The assumption is that $b \ne 0$.
Observe how the original problem has been greatly simplified after applying the Division Rule of Exponent.
At this point, we can proceed as usual in solving for the inverse. Rewrite $f\left( x \right)$ as $y$, followed by interchanging the variables $\color{red}x$ and $\color{red}y$.
Before we can get the logs of both sides, isolate the exponential portion of the equation by adding both sides by $4$.
Since the exponential expression is using base $3$, we take the logs of both sides of the equation with base $3$ as well! By doing so, the exponent $\color{blue}2y-1$ on the right side will drop, so we can continue on solving for $y$ which is the required inverse function.
It verifies that our answer is correct because the graph of the given exponential functions and its inverse (logarithmic function) are symmetrical along the line $\large{y=x}$.
You may also be interested in these related math lessons or tutorials:
Inverse of a 2×2 Matrix
Inverse of Absolute Value Function
Inverse of Constant Function
Inverse of Linear Function
Inverse of Logarithmic Function
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# FINDING THE LIKELIHOOD OF AN EVENT
## About "Finding the likelihood of an event"
Finding the likelihood of an event :
Let us consider the two experiments given below.
Each time we roll a number cube, a number from 1 to 6 lands face up.
In the same way, each time we toss a coin, the face will turn up either head or tail.
Getting a number from 1 to 6 in the first experiment is called an event.
And getting head or tail in the second experiment is called an event.
## Finding the likelihood of an event (Rolling a cube)
Let us roll a cube where the faces numbered from 1 to 6.
Now, order the below mentioned events and write their possibilities.
1. Rolling a number less than 7.
{1, 2, 3, 4, 5, 6 }
6 of 6 possible rolls
The likelihood of the event = 6/6 = 1
2. Rolling an 8.
There is no 8 in the six faced cube.
0 of 6 possible rolls
The likelihood of the event = 0/6 = 0.
3. Rolling a number greater than 4.
{5, 6}
2 of 6 possible rolls
The likelihood of the event = 2/6 = 1/3.
4. Rolling a 5.
{5}
1 of 6 possible rolls
The likelihood of the event = 1/6.
5. Rolling a number other than 6.
{1, 2, 3, 4, 5}
5 of 6 possible rolls
The likelihood of the event = 5/6.
6. Rolling an even number.
{2, 4, 6}
3 of 6 possible rolls
The likelihood of the event = 3/6 = 1/2.
7. Rolling a number less than 5.
{1, 2, 3, 4}
4 of 6 possible rolls
The likelihood of the event = 4/6 = 2/3.
8. Rolling an odd number.
{1, 3, 5}
3 of 6 possible rolls
The likelihood of the event = 3/6 = 1/2.
9. Rolling a number divisible by 3.
{3, 6}
2 of 6 possible rolls
The likelihood of the event = 2/6 = 1/3.
10. Rolling a number divisible by 2.
{2, 4, 6}
3 of 6 possible rolls
The likelihood of the event = 3/6 = 1/2.
## Finding the likelihood of an event (Tossing a coin)
Let us toss a coin.
Now, order the below mentioned events and write their possibilities.
1. Getting head.
{H}
1 of 2 possible outcomes
The likelihood of the event = 1/2
2. Getting tail.
{T}
1 of 2 possible outcomes
The likelihood of the event = 1/2
3. Getting head or tail.
{H, T}
2 of 2 possible outcomes
The likelihood of the event = 2/2 = 1.
4. Getting a number divisible by 2.
A coin will turn up either head or tail and it will never turn up a number.
0 of 2 possible outcomes
The likelihood of the event = 0/2 = 0.
After having gone through the stuff given above, we hope that the students would have understood "How to find the likelihood of an event".
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# Matrix equation solver
Looking for Matrix equation solver? Look no further! We can solve math word problems.
## The Best Matrix equation solver
This Matrix equation solver helps to quickly and easily solve any math problems. distance = sqrt((x2-x1)^2 + (y2-y1)^2) When using the distance formula, you are trying to find the length of a line segment between two points. The first step is to identify the coordinates of the two points. Next, plug those coordinates into the distance formula and simplify. The last step is to take the square root of the simplify equation to find the distance. Let's try an example. Find the distance between the points (3,4) and (-1,2). First, we identify the coordinates of our two points. They are (3,4) and (-1,2). Next, we plug those coordinates into our distance formula: distance = sqrt((x2-x1)^2 + (y2-y1)^2)= sqrt((-1-3)^2 + (2-4)^2)= sqrt(16+4)= sqrt(20)= 4.47 Therefore, the distance between the points (3,4) and (-1,2) is 4.47 units.
It is usually written with an equals sign (=) like this: 4 + 5 = 9. This equation says that the answer to 4 + 5 (9) is equal to 9. So, an equation is like a puzzle, and solving it means finding the value of the missing piece. In the above example, the missing piece is the number 4 (because 4 + 5 = 9). To solve an equation, you need to figure out what goes in the blank space. In other words, you need to find the value of the variable. In algebra, variables are often represented by letters like x or y. So, an equation like 2x + 3 = 7 can be read as "two times x plus three equals seven." To solve this equation, you would need to figure out what number multiplied by 2 and added to 3 would give you 7. In this case, it would be x = 2 because 2 * 2 + 3 = 7. Of course, there are many different types of equations, and some can be quite challenging to solve. But with a little practice, you'll be solving equations like a pro in no time!
In math, a function is a set of ordered pairs in which each element in the set corresponds to a unique output. In other words, a function takes an input and produces an output. College algebra deals with the study of functions and their properties. There are many different types of functions that can be studied, and each has its own set of characteristics. College algebra students learn how to identify, graph, and manipulate different types of functions. They also learn how to solve problems involving functions. By understanding functions, college algebra students are better prepared to tackle advanced math concepts.
Math questions and answers can be very helpful when it comes to studying for a math test or exam. There are many websites that offer free math questions and answers, as well as online forums where students can ask questions and get help from other students. Math question and answer collections can also be found in most libraries. In addition, there are many textbook companies that offer supplements with practice questions and answers. Math questions and answers can be a great way to review material, identify areas of weakness, and get extra practice. With so many resources available, there is no excuse for not getting the math help you need!
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# Unit Fraction
In mathematics, when we divide a complete part into some equal parts then each part shows the fraction of whole part. Like pizza, pizza is a circular in shape, if we divide it in 8 equal parts then each part is the fraction of whole part.
Fractions are written in the form of ratio of two numbers. The upper number is called as numerator and the number below is called as denominator.
For example: a/b is the fraction, in which a is numerator and b is the denominator.
## Unit fractions:
Unit fractions are the fractions in which numerator is always one and denominator is any whole number except zero.
For example:
a/b is the unit fraction only if a= 1 and b≠0.
1/5, 1/8, 1/12 all are the unit fractions, because here numerator is one.
Here, numerator is 1 and denominator is 8≠0.
If a complete circle is divided into 4 equal parts then each part will be the fraction of whole part. And each part takes the value as ¼ and it is the unit fraction.
• Here, complete part / total equal parts = 1/4
• Here, each part contributes ¼ th part of the whole part and it is in the unit fraction form.
• And total part of circle = ¼ + ¼ + ¼ + ¼ = 1
• Similarly, if we divide a complete circle into 8 equal parts then each part will contribute the 1/8th part of the complete part and it is in the form of proper fraction as numerator is less than denominator.
• Here, each fraction takes value 1/8 because, complete part/ number of equal parts = 1/8
• And hence, 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 = 1
The following figures shows the different fraction on the basis of the portion selected out of complete part.
Here, in figure there is a complete circle hence it takes Value 1
• In fig. 2 we divided the complete circle into 2 equal parts hence each part takes the value ½.
Since, complete part/ number of equal parts = ½, and it is the proper fraction.
• In fig. 3 the whole circle is divided into 3 equal parts and hence each part takes value 1/3 and it is the proper fraction.
• In fig. 4 the whole circle is divided into 4 equal parts hence each part takes the value ¼ and it is the proper fraction.
• In fig. 5, the whole circle is divided into 5 equal parts and each part takes the value 1/5, and it is the proper fraction.
• Also in fig. 6, the whole circle is divided into 6 equal parts hence, each part takes the value 1/6 and it is the proper fraction.
Note:
As the numerator is less than denominator in proper fraction, the decimal value of proper fraction is always less than 1.
Updated: July 15, 2021 — 12:06 am
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# Graphing Polynomial Functions Examples
Here we look at how to approach graphing polynomial functions examples, where we want to try to make as accurate a sketch as we can of a polynomial graph on an appropriate axis.
Some examples of graphing quadratic graphs which are polynomials of degree 2 can be seen here.
When a polynomial is of larger degree though, sometimes we need to do a bit more work for some more information on what the shape of the graph should be.
Here are two examples of what a polynomial graph can look like.
Polynomial graphs are smooth graphs and have no holes or breaks in them.
### Turning Points:
The curves on polynomial graphs are called the ‘turning points’.
A polynomial of degree n,
has at most n − 1 turning points on its graph.
So the graph of a polynomial such as f(x) = x^3 + 5x^2 \space {\text{–}} \space 2x + 1,
will have at most 2 turning points.
With learning how to deal with graphing polynomial functions examples it’s important to be clear about the leading term and leading coefficient is in a polynomial.
For a polynomial, P(x) = a_{n}x^n + a_{n \space {\text{–}} \space 1}x^{n \space {\text{–}} \space 1} + … + a_{1}x + a_0.
So with g(x) = 2x^3 + x^2 \space {\text{–}} \space 2x + 5,
Something called the ‘leading coefficient test’ can help us establish the behaviour of the polynomial graph at each end.
As stated, for a polynomial P(x) = a_{n}x^n + a_{n \space {\text{–}} \space 1}x^{n \space {\text{–}} \space 1} + … + a_{1}x + a_0,
There are 4 cases we can have with the leading term, and what it can tell us.
1) a_{n} > 0 \space\space , \space\space n even. The graph will increase with no bound to the left and the right.
2) a_{n} > 0 \space\space , \space\space n odd. The graph will decrease to the left, and increase on the right with no bound.
3) a_{n} < 0 \space\space , \space\space n even. The graph will decrease with no bound to the left and the right.
4) a_{n} < 0 \space\space , \space\space n odd. The graph will increase on the left, and decrease to the right with no bound.
### Multiplicity of a Zero:
The multiplicity of a zero is the number of times it appears in the factored polynomial form.
For (x \space {\text{–}} \space 5)^3(x + 1)^2 = 0.
The zeros are 5 and {\text{-}}1.
Of which, 5 has multiplicity 3, and {\text{-}}1 has multiplicity 2.
So:
When (x \space {\text{–}} \space t)^m is a factor, t is a zero.
If m is even, the x-intercept at x = t will only touch the x-axis, but not cross it.
If m is odd, the x-intercept at x = t will cross the x-axis.
Also if m is greater than 1, the curve of the polynomial graph will flatten out at the zero.
## Graphing Polynomial Functions ExamplesSteps
So in light of what we’ve seen so far on this page, to make a sketch of the graph of a polynomial function there are some steps to take.
A) Find the zeros of the polynomial and their multiplicity.
B) Use the leading coefficient test to establish the graph end behaviour.
C) Find the y-intercept, (0,P(0)).
D) Establish the maximum number of turning points there can be, n \space {\text{–}} \space 1.
E) Plot some extra points for some more information of graph shape between the zeros.
Part E) will become more clear in an example.
Example
(1.1)
Sketch the graph of the polynomial f(x) = x^4 \space {\text{–}} \space 3x^3 \space {\text{–}} \space 9x^2 + 23x \space {\text{–}} \space 12.
Solution
Often we are required to find the factors and zeros, but as to concentrate on sketching a graph the polynomial here, we will state them.
f(x) = x^4 \space {\text{–}} \space 3x^3 \space {\text{–}} \space 9x^2 + 23x \space {\text{–}} \space 12 \space = \space (x \space {\text{–}} \space 1)^{\tt{2}}(x \space {\text{–}} \space 4)(x + 3)
(x \space {\text{–}} \space 1)^{\tt{2}}(x \space {\text{–}} \space 4)(x + 3) \space = \space 0 => x = 1 \space , \space x = 4 \space , \space x = {\text{-}}3
Of the zeros, 1 has multiplicity 2, while 4 and {\text{-}}3 have multiplicity 1.
So the graph will cross the x-axis at –3 and 4, but only touch at 1.
The leading coefficient is 1, which is > 0, and n = 4.
So the graph will increase with no bound past both the left and right ends.
For the y-intercept. f(0) = {\text{-}}12 => (0,-12)
As n = 4, there can be at most 3 turning points, if our graph has any more we have done something wrong somewhere.
With what we know thus far, we can make a start at a sketch of the polynomial graph, on a suitable axis.
Now we look to establish some extra information about graph behaviour between the zeros or other point son the x-axis.
Between 1 and 4. f(3) = {\text{-}}24 \space , \space f(2) = {\text{-}}10
So there is a turning point between 1 and 4.
With n \space {\text{–}} \space 1 = 3, there can only be one more turning point.
Between -3 and 0. f({\text{-}}2) = {\text{-}}54 \space , \space f({\text{-}}\frac{3}{2}) = {\text{-}}51.56
There is a turning point between -3 and 0.
In calculus we can work out more exact turning points and make more exact drawings of polynomial graphs when dealing with graphing polynomial functions examples.
But for a rough sketch of the graph, what we have above gives us a good idea and is fairly accurate.
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# 10.5 Graphing quadratic equations (Page 7/15)
Page 7 / 15
$y=5{x}^{2}+2$
$y=2{x}^{2}-4x+1$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,1\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(1.7,0\right),\left(0.3,0\right);$
axis: $x=1;\phantom{\rule{0.2em}{0ex}}\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,-1\right)$
$y=3{x}^{2}-6x-1$
$y=2{x}^{2}-4x+2$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,2\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,0\right);$
axis: $x=1;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(1,0\right)$
$y=-4{x}^{2}-6x-2$
$y=\text{−}{x}^{2}-4x+2$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,2\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4.4,0\right),\left(0.4,0\right);$
axis: $x=-2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-2,6\right)$
$y={x}^{2}+6x+8$
$y=5{x}^{2}-10x+8$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=1;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,3\right)$
$y=-16{x}^{2}+24x-9$
$y=3{x}^{2}+18x+20$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,20\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4.5,0\right),\left(-1.5,0\right);$
axis: $x=-3;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-3,-7\right)$
$y=-2{x}^{2}+8x-10$
Solve Maximum and Minimum Applications
In the following exercises, find the maximum or minimum value.
$y=2{x}^{2}+x-1$
The minimum value is $-\frac{9}{8}$ when $x=-\frac{1}{4}$ .
$y=-4{x}^{2}+12x-5$
$y={x}^{2}-6x+15$
The minimum value is 6 when $x=3$ .
$y=\text{−}{x}^{2}+4x-5$
$y=-9{x}^{2}+16$
The maximum value is 16 when $x=0$ .
$y=4{x}^{2}-49$
In the following exercises, solve. Round answers to the nearest tenth.
An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic equation $h=-16{t}^{2}+168t+45$ to find how long it will take the arrow to reach its maximum height, and then find the maximum height.
In 5.3 sec the arrow will reach maximum height of 486 ft.
A stone is thrown vertically upward from a platform that is 20 feet high at a rate of 160 ft/sec. Use the quadratic equation $h=-16{t}^{2}+160t+20$ to find how long it will take the stone to reach its maximum height, and then find the maximum height.
A computer store owner estimates that by charging $x$ dollars each for a certain computer, he can sell $40-x$ computers each week. The quadratic equation $R=\text{−}{x}^{2}+40x$ is used to find the revenue, $R$ , received when the selling price of a computer is $x$ . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
20 computers will give the maximum of \$400 in receipts.
A retailer who sells backpacks estimates that, by selling them for $x$ dollars each, he will be able to sell $100-x$ backpacks a month. The quadratic equation $R=\text{−}{x}^{2}+100x$ is used to find the $R$ received when the selling price of a backpack is $x$ . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation $A=x\left(240-2x\right)$ gives the area of the corral, $A$ , for the length, $x$ , of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.
The length of the side along the river of the corral is 120 feet and the maximum area is 7,200 sq ft.
A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic equation $A=x\left(100-2x\right)$ gives the area, $A$ , of the dog run for the length, $x$ , of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.
## Everyday math
In the previous set of exercises, you worked with the quadratic equation $R=\text{−}{x}^{2}+40x$ that modeled the revenue received from selling computers at a price of $x$ dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
Graph the equation $R=\text{−}{x}^{2}+40x$ . Find the values of the x -intercepts.
1. $\left(0,0\right),\left(40,0\right)$
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How do you know if the sequence 1, -2, 4, -8, ... is arithmetic or geometric?
Nov 7, 2016
Because the sequence has a common ratio $r = - 2$, it is geometric.
Explanation:
Each term of an arithmetic sequence is generated by adding or subtracting a number to get the next term. The number is called the common difference $d$.
Each term of a geometric sequence is generated by multiplying or dividing by a number to get the next term. The number is called the common ratio $r$.
Let's look at the given sequence: $1 , - 2 , 4 , - 8. . .$
If you subtract the first term from the second, $- 2 - 1 = 3$. and the second term from the third, $4 - - 2 = 6$, you can see that there is no common difference between the terms. The difference between these terms is not the same. The sequence is not arithmetic.
If you divide the second term by the first, $\frac{- 2}{1} = - 2$, and the third term by the second, $\frac{4}{-} 2 = - 2$, you get the same quotient. In fact, if you divide the fourth term by the third, $\frac{- 8}{4} = - 2$, the pattern continues. The common ratio is $r = - 2$.
Because there is a common ratio, the sequence is geometric
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# Multiples of 37
Created by: Team Maths - Examples.com, Last Updated: May 28, 2024
## Multiples of 37
In mathematics, multiples of 37 are integers that result from multiplying 37 by any whole number. These numbers are part of a sequence where each term increases by a fixed amount, specifically 37 in this case. Understanding multiples is crucial as they help in identifying divisors and factors of larger numbers. Multiples of 37 play a significant role in multiplication problems and number theory. Recognizing these patterns aids in simplifying calculations and solving various mathematical problems efficiently.
## What are Multiples of 37?
Multiples of 37 are the numbers obtained by multiplying 37 with any integer. The sequence starts as 37, 74, 111, 148, 185, and continues infinitely. These multiples can be expressed as 37n, where n is an integer.
Prime Factorization of 37: 1 x 37 First 10 Multiples of 37 are 37, 74, 111, 148, 185, 222, 259, 296, 333, 370.
Table of 37
## Important Notes
• Definition: Multiples of 37 are the products obtained when 37 is multiplied by any integer (e.g., 37, 74, 111, etc.). They follow the form 37n, where n is an integer.
• Properties: Every multiple of 37 is a result of adding 37 repeatedly. For example, the first five multiples are 37, 74, 111, 148, and 185.
• Divisibility: Any multiple of 37 is divisible by 37 without leaving a remainder. This makes 37 a factor of all its multiples.
• Sequence: The sequence of multiples of 37 increases linearly, with each subsequent multiple being 37 units greater than the previous one.
• Applications: Multiples of 37 are used in various mathematical problems and concepts, including finding common multiples, solving equations involving divisors, and understanding number patterns.
## Examples on Multiples of 37
• 1×37 = 37
• 2×37 = 74
• 3×37 = 111
• 4×37 = 148
• 5×37 = 185
• 10×37 = 370
• 20×37 = 740
• 30×37 = 1110
• 40×37 = 1480
• 50×37 = 1850
### Real-Life Examples
• Time: If a task takes 37 minutes and you do it 3 times, it takes 3×37 = 11 minutes.
• Money: If an item costs \$37 and you buy 4, it costs 4×37 = 148 dollars.
• Distance: A marathon is 42.195 kilometers, and if a runner runs 37 kilometers per day, in 3 days they run 3×37 = 111 kilometers.
• Storage: If one box can hold 37 items, and you have 5 boxes, you can store 5×37 = 185 items.
• Construction: If each floor of a building is 37 feet high, a 10-floor building is 10×37 = 370 feet high.
## Practical Examples of Multiples of 37
### What is the 15th multiple of 37?
The 15th multiple of 37 is 37×15 = 55.
### Is 481 a multiple of 37?
Yes, 481 is a multiple of 37 because 37×13 = 481.
### What is the next multiple of 37 after 444?
The next multiple of 37 after 444 is 444+37 = 4814.
### Find the 25th multiple of 37.
The 25th multiple of 37 is 37×25 = 925.
### Is 950 a multiple of 37?
No, 950 is not a multiple of 37 because it cannot be expressed as 37×n for any integer n.
## What are multiples of 37?
Multiples of 37 are numbers obtained by multiplying 37 by any integer, such as 37, 74, 111, and so on.
## How do you find the nth multiple of 37?
To find the nth multiple of 37, simply multiply 37 by n (e.g., the 5th multiple is 37 × 5 = 185).
## Are all multiples of 37 odd?
No, multiples of 37 alternate between odd and even. For example, 37 (odd), 74 (even), 111 (odd), etc.
## What is the smallest multiple of 37?
The smallest multiple of 37 is 37 itself, as it is 37 multiplied by 1.
## Is 370 a multiple of 37?
Yes, 370 is a multiple of 37 because 370 ÷ 37 equals 10, an integer.
## Can multiples of 37 be negative?
Yes, multiples of 37 can be negative if you multiply 37 by a negative integer (e.g., -37, -74, -111).
## What are common multiples of 37 and 74?
Since 74 is itself a multiple of 37 (74 = 37 × 2), any multiple of 74 will also be a multiple of 37 (e.g., 74, 148, 222).
## How do multiples of 37 relate to factors?
Multiples of 37 are numbers that 37 divides evenly, making 37 a factor of these numbers.
## Are there any shortcuts to identify multiples of 37?
No specific shortcuts exist for identifying multiples of 37; you multiply 37 by integers and check for divisibility by 37.
## Why are multiples of 37 important?
Multiples of 37 are important in various mathematical contexts, including solving equations, finding common denominators, and understanding number properties.
Text prompt
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# Difference between revisions of "2012 AMC 10A Problems/Problem 24"
The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.
## Problem
Let $a$, $b$, and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$.
What is $a$?
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$
## Solution
Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$.
Now, this can be rearranged and factored.
$$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$$
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$, $b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that $14 = 9 + 4 + 1$.
$(a-c)^2 = 9 \rightarrow a-c = 3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$, or whether $b-c = 1$ or $2$.
We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$.
$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$, after some algebra, simplifies to $7a = 2021$. $2021$ is not divisible by $7$, so $a$ is not an integer.
The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$, which simplifies to $8a = 2024$. Thus, $a = 253$ and the answer is $\boxed{\textbf{(E)}\ 253}$.
~dolphin7
## See Also
2012 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2012 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Solving Equations By: Marinda, Ashley, Trevor, and Ti’Shawn Mrs. Sawyer 1 st block.
## Presentation on theme: "Solving Equations By: Marinda, Ashley, Trevor, and Ti’Shawn Mrs. Sawyer 1 st block."— Presentation transcript:
Solving Equations By: Marinda, Ashley, Trevor, and Ti’Shawn Mrs. Sawyer 1 st block
Essential Question Why is it important to be fair? Unit Questions How do you justify the steps necessary to solve equations? How does the understanding of balances and equality relate to solving equations? Content Questions How do you define a variable? What are the steps for solving equations?
1.02 Use formulas and algebraic expressions, including iterative and recursive forms, to model and solve problems. Student Objectives/Learning Outcomes The learner will: properly use equality formulas to evaluate for the given variable solve equations using addition, subtraction, multiplication and division solve two-step equations using addition, subtraction, multiplication and division use the distributive property to combine like terms and solve problems solve equations with variables on both sides identify equations as identity or no solution model distance-rate-time problems use a graphing calculator to check solutions of equations by graphing
Addition Property of Equality a + c = b + c Substitution Property of Equality a – c = b – c Multiplication Property of Equality a ∙ c = b ∙ c Division Property of Equality a = b c c Distributive Property a(b + c) = ab + ac a(b – c) = ab – ac a = b
Addition Property of Equality a + c = b + c If 10 = 6 + 4, then 10 + 2 = 6 + 4 + 2 Substitution Property of Equality a – c = b – c If 10 = 6 + 4, then 10 – 2 = 6 + 4 – 2 Multiplication Property of Equality a ∙ c = b ∙ c If 2 ∙ 6 = 12, then 2 ∙ 6 ∙ 5 = 12 ∙ 5 Division Property of Equality a = b c c If 5 + 1 = 6, then 5 + 1 = 6 2 2 Distributive Property a(b + c) = ab + ac a(b – c) = ab – ac If 3(4 + 1), then 3(4) + 3(1)
Example: Solve each equation. Check your answer. ONE STEP a. x – 8 = 0 x – 8 + 8 = 0 + 8 x = 8 b. 4c = -96 4c = -96 4 c = -24 n / 6 = 5 n / 6 (6) = 5(6) n = 30 3 / 4 x = 9 ( 4 / 3 ) 3 / 4 x = 9( 4 / 3 ) x = 12
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In mathematics, there are two types of shapes that we learn about: isosceles triangles and right triangles. The isosceles triangle and the right triangle are special triangles.
Since they are special triangles, they have their own characteristics. By learning what characteristics they have, we will be able to calculate angles and prove shapes.
Isosceles triangles and right triangles must be studied together. This is because these figures are often given as mixed problems. In other words, by understanding the properties of both isosceles and right triangles, we will be able to finally solve problems.
Therefore, after explaining the properties of isosceles triangles and right triangles, we will explain the theorems for proving that they are congruent.
## Isosceles and Right Triangles Are Special Triangles
Of all the shapes, the most frequently asked geometry problem is the triangle. However, there are different types of triangles. If a triangle satisfies certain conditions, it is called by another name.
Such triangles include the following.
• Isosceles triangle
• Right-angled triangles
Another typical example of a special triangle is the equilateral triangle. A triangle whose side lengths and angles are all the same is an equilateral triangle. As for equilateral triangles, they have simple properties.
On the other hand, isosceles and right triangles have more properties to remember than equilateral triangles. Also, isosceles triangles and right triangles are often given as mixed problems, and it is often impossible to solve them unless you understand the properties of both.
In addition, right triangles have a congruence condition that is available only for right triangles. You must remember this congruence theorem.
### Definition, Properties, and Theorems of Isosceles Triangles
It is important to understand the definition of special shapes. In the case of isosceles triangles, what is the definition? Any triangle that satisfies the following conditions is an isosceles triangle.
• Triangles with two equal sides
Whenever two sides are equal, it is an isosceles triangle. Conversely, if all sides are not equal in length, it is not an isosceles triangle. Since it is a definition, the following isosceles triangle will always have AB=AC.
If you are given an isosceles triangle in a math problem, the two sides have the same length.
Also, isosceles triangles have a property (theorem) derived from their definition. One of these theorems is that the base angles are equal. In an isosceles triangle, the two sides are equal, and the two angles at the base are also equal.
Properties derived from definitions are called theorems. One of the theorems of an isosceles triangle is that the base angles are equal.
### Proof That Base Angles of Isosceles Triangles Are Equal
Why are the base angles of an isosceles triangle equal? In mathematics, we study proofs, so let’s prove why the base angles are equal. Consider the following isosceles triangle where point D is the midpoint of BC.
The proof is as follows.
• In △ABD and △ACD
• AB=AC: Definition of an isosceles triangle – (1)
• BD = CD: Point D is the midpoint of BC – (2)
• From (1), (2), and (3), since Side – Side – Side (SSS), △ABD≅△ACD
• Since △ABD≅△ACD, ∠B=∠C
By proving that the triangles are congruent, we can prove that the base angles of an isosceles triangle are equal. In an isosceles triangle, the two sides are equal and the base angles are always equal.
## Properties of Right Triangles and Hypotenuse
Another special triangle that we need to learn at the same time as the properties of isosceles triangles is the right triangle. If a triangle has an angle of 90°, it is called a right triangle.
Right triangles have hypotenuse. General triangles do not have a hypotenuse. However, right triangles are special triangles, and we will use the hypotenuse to explain the properties of right triangles. The hypotenuse of a right triangle refers to the following part.
A triangle with an angle of 90° is the definition of a right triangle. Right triangles also have two acute angles in addition to the hypotenuse; any angle smaller than 90° is called an acute angle. In a right triangle, two angles that are not 90° are always acute angles.
### Congruence Theorem for Right Angle Triangles: HL
It is important to note that right triangles have their own congruence conditions in addition to the triangle congruence theorems. The four congruence theorems for triangles are as follows.
• Side – Side – Side (SSS) Congruence Postulate
• Side – Angle – Side (SAS) Congruence Postulate
• Angle – Side – Angle (ASA) Congruence Postulate
• Angle – Angle – Side (AAS) Congruence Postulate
In addition to these, there is a congruence theorem that exists only for right triangles. The following is a right triangle congruence theorem.
• Hypotenuse – Leg (HL); Sides other than the hypotenuse are called legs.
There are a total of five congruence theorems for triangles. In addition to the triangle congruence theorems, try to remember the right triangle congruence condition.
-It’s Not Enough That Two Angles Are Equal
Some people consider the congruence condition of right triangles when the two angles are equal. In this case, however, the two right triangles are not necessarily congruent. There are cases where they have different shapes, as shown below.
Just because the angles are equal does not mean that they are congruent. Triangles are not congruent if they do not satisfy the congruence condition.
## Exercises: Proof of Shapes
Q1: Prove the following figure.
For the following figure, △ABC is an isosceles triangle with AB = AC. If BD⊥AC and CE⊥AB, prove that △BCE≅△CBD.
Mixed problems of isosceles triangles and right triangles are frequently asked. So, try to understand the properties of the two shapes.
• In △BCE and △CBD
• ∠BEC = ∠CDB = 90°: Given – (1)
• BC = CB: Common line – (2)
• ∠EBC = ∠DCB: Base angles of an isosceles triangle are equal – (3)
• From (1), (2), and (3), since Angle – Angle – Side (AAS), △BCE≅△CBD
Q2: Prove the following figure.
For the figure below, △ABC is a right triangle with ∠ABC = 90°. From point B, draw a line BD perpendicular to side AC. Also, draw a bisector of ∠BAC, and let the intersection points be E and F, respectively, as shown below. Prove that BE=BF.
There are several possible ways to prove that the lengths of the sides are the same. The most common way is to find congruent triangles. However, looking at the figure, it seems that there are no two triangles that are congruent.
So, let’s think of another way to prove that the side lengths are the same. The property of an isosceles triangle is that the two sides are equal, and the base angles are equal. Therefore, if we can prove that the two angles are the same, then we know that the triangle is an isosceles triangle and the two side lengths are equal.
First, let’s focus on △ABF. ABF is a right triangle, and ∠AFB is as follows.
• $∠AFB=180°-90°-∠BAF$
• $∠AFB=90°-∠BAF$ – (1)
On the other hand, what about ∠AED? △AED is a right triangle, and the angle AED is as follows.
• $∠AED=180°-90°-∠EAD$
• $∠AED=90°-∠EAD$ – (2)
Also, since the vertical angles are equal, ∠AED = ∠BEF. Therefore, for (2), we can replace ∠AED with ∠BEF to get the following.
• $∠BEF=90°-∠EAD$ – (3)
Since ∠BAC is bisected, the angles of ∠EAD and ∠BAF are the same. Therefore, we have the following.
1. $∠AFB=90°-∠BAF$ – (1)
2. $∠AFB=90°-∠EAD$ – Substitute (3)
3. $∠AFB=90°-(90°-∠BEF)$
4. $∠AFB=∠BEF$
Thus, we know that △BEF is an isosceles triangle because the angles of ∠AFB and ∠BEF are equal. Also, since it is an isosceles triangle, BE=BF.
## Use the Properties of Special Triangles and Prove Them
Problems involving the calculation of angles and the proof of figures are frequently asked in mathematics. In these problems, special triangles are often used. Special triangles include isosceles triangles and right triangles.
There are two characteristics of isosceles triangles. The first is that the two sides are equal. The second is that each base angle is equal. On the other hand, right triangles have a congruence theorem. There is a congruence theorem available only for right triangles, so try to remember it.
It is important to remember that isosceles triangles and right triangles are sometimes asked as mixed problems. You often need to understand the properties of both isosceles triangles and right triangles to be able to solve the problem. Try to answer using the characteristics of both isosceles and right triangles.
As special triangles, isosceles triangles and right triangles are frequently asked. They are important figures in mathematics, so remember their definitions, properties, and congruence theorems before solving them.
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# Quadratic Expressions, Equations, and Functions
## Overview
Quadratic Expressions, Equations, and Functions are found throughout the SAT, so it’s very important to be comfortable with all the key aspects of this topic. In particular, make sure you know these concepts:
• Standard Form
• Vertex Form
• Factored Form
• Properties of Parabolas
• Graphical Solutions
• Discriminant
Let’s look at each of these in more detail:
Quadratic expressions include a variable whose exponent is 2, although the product of two linear variables (for example, xy) is also quadratic. Here are some examples of quadratic expressions:
Be sure you know how to translate a verbal expression into a quadratic expression like the ones shown above. Here are some examples:
“A number times itself.” “A number multiplied by one more than the number.” “The product of a number and one less than the number.”
Expressions aren’t equations but they are an important component of quadratic equations. Also, you can add and subtract quadratic expressions and still have a quadratic expression. Multiplying or dividing quadratic expressions will result in a non-quadratic expression. Here are some examples:
Some of the questions you’ll be asked will involve rewriting quadratic expressions to equivalent forms. To simplify this task familiarize yourself with these quadratic identities:
Binomial Expansion Binomial Squared Difference of Squares
Another important skill involves factoring quadratic expressions. This is a skill that also comes into play when solving quadratic equations. Here are some examples.
Factoring a constant. Factoring a linear term Factoring a quadratic into a binomial squared Factoring a quadratic into the product of two linear expressions.
To learn more about factoring quadratics, click on this link. Now look at some SAT-style questions that focus on quadratic expressions.
SAT Skill: Working with Quadratic Expressions Example 1 Simplify the following expression. Combine like terms and simplify. Keep track of how the subtraction symbol will change the signs of the terms in the expression on the right. Example 2 Simplify the following expression. Expand the term on the left (use a calculator for the squared decimal terms): Combine like terms and simplify. Keep track of how the subtraction sign changes the signs of the terms on the right. Example 3 In the equation below, what is a possible value for k? You should recognize that the term on the left is a difference of squares, which can be rewritten this way: In factored form, you can find the equivalent values for k and a: Example 4 What quadratic expression is equivalent to this expression? Factor the numerator, which is a difference of squares, then simplify: Example 5 What is this expression in expanded form? Use the binomial squared identity: Example 6 Look at the following equation. If a + b = 10, what are two possible values for c? Expand the term on the left to a quadratic in standard form. Compare the two terms in standard form: From this we can conclude the following: We also know that a + b = 10. We can substitute to create this equation: Factor this quadratic to find the values for b, and corresponding values for a: Now we can find the possible values for c:
Before studying what a quadratic function is, make sure you are comfortable with the following concepts, which we will also review:
• What a function is
• Independent variable
• Dependent variable
• Domain
• Range
• Different representations of functions
#### Brief Reviewof Functions
What Is a Function? A function is a one-to-one mapping of input values (the independent variable) to output values (the dependent variable). Click on this link to see a quick tutorial on what a function is. This slide show goes over the following key points:
• For every input value (x), there is a unique output value, f(x).
• Functions can be represented as equations, tables, and graphs.
• A function machine is a useful visual representation of the input/output nature of functions.
Dependent/Independent Variables. When one variable depends on another, then it is the dependent variable. For example, the faster your speed, the farther you travel. Suppose that speed is represented by the variable s and the distance traveled is represented by the variable d
Here’s how to describe the relationship between s and d:
The faster the speed, the more distance traveled.
Distance is dependent on speed.
Distance is a function of speed.
d = f(s)
When studying functions, make sure you are comfortable telling the difference between the independent variable and dependent variable. Get comfortable using function notation. To learn more about function notation, click on this link.
Domain and Range. A function shows the relationship between two variables, the independent variable and the dependent variable. The domain is the allowed values for the independent variable. The range is the allowed values for the dependent variable. The domain and range influence what the graph of the function looks like.
For a detailed review of what domain and range are, click on this link to learn more. You’ll see definitions of the terms domain and range, as well as examples of how to find the domain and range for given functions.
Multiple Representations of Functions. We mentioned previously that functions can be represented in different ways. In fact, any function can be represented by an equation, usually f(x) equal to some expression; a table; or a graph. For a detailed review of multiple representations of functions, click on this link, to see a slide show that includes examples of these multiple representations.
#### Quadratic Functions in Standard Form
The most common form of a quadratic function is the standard form.
The standard form is also the most common form used to solve a quadratic equation.
To see examples of graphing quadratic functions in standard form, click on this link. This slide show also includes a video tutorial.
#### Quadratic Functions in Vertex Form
Another way that a quadratic function can be written is in vertex form
To see examples of quadratic functions in vertex form, click on this link. This slide show tutorial walks you through the difference between standard form and vertex form. It also includes examples and two Desmos activities where you can graph these two types of equations.
#### Quadratic Functions in Factored Form
Another way that a quadratic function can be written is in factored form. From your work with quadratic expressions, you saw some techniques for factoring a quadratic. If you can write a quadratic function as the product of linear terms, then it is much easier to solve the corresponding the quadratic equation.
To see examples of quadratic functions in factored form, click on this link
#### Graphs of Quadratic Functions
The graph of a quadratic function is known as a parabola
To learn more about the properties of parabolas and their graphs, click on this link.
#### Special Case of Quadratics: Equation of a Circle
The equation of a circle is not a quadratic function, but it is a quadratic relation. This is the equation of a circle:
The coordinates of the center of the circle are (h, k) and the radius of the circle is r.
Quadratic equations are usually written as a quadratic expression in standard form equal to zero.
A quadratic equation can have two, one, or zero real number solutions. There are several ways to solve a quadratic. These are the methods we’ll be looking at:
• The Quadratic Formula
• Factoring
• Graphing the Quadratic Function.
Let’s look at the first method, which will work for any quadratic equation.
#### The Quadratic Formula
When a quadratic equation is written in standard form, like the one shown below, then you can use the quadratic formula to find the solutions to the equation.
Use the a, b, and c values from the quadratic equation and plug them into the quadratic formula:
To learn more about using the quadratic formula to solve quadratic equations, click on this link. This slide show includes a video overview of the quadratic formula and a number of detailed math examples.
Before using the quadratic formula, calculate the discriminant, which is the term under the square root sign of the quadratic formula.
To see examples of using the discriminant, click on this link.
#### Factoring
You’ve already seen how to factor quadratic expressions into the product of linear terms. That same idea can be used to factor certain quadratic expressions in order to find the solutions to the equation.
A factored quadratic equation will look something like this:
The solutions to this equation are x = a and x = b
A more simplified version of a factored quadratic can look like this:
The solutions to this equation are x = 0 and x = a
The previous two examples both had two solutions. There is a factored form that has one solution:
This is the case of the binomial squared. In this case the solution to the equation is x = a
The simplest example of the binomial squared is this:
The solution to this is x = 0.
If a quadratic cannot be easily factored, then you should use the quadratic formula or graph the quadratic.
To see examples of using factoring to solve a quadratic equation, click on this link.
#### Solving by Graphing
A visual approach to solving quadratic equations is to graph the parabola. There are three cases to look at.
Case 1: Two solutions. If the graph of the parabola intersects the x-axis twice, then there are two solutions.
Suppose you are solving this quadratic equation:
To find the solution graphically, then graph the corresponding quadratic function.
Notice that this parabola intersects the x-axis at x = 2 and x = 4. Those are the solutions to the quadratic equation. In fact, you can rewrite the quadratic in factored form:
Case 2: One solution. If the graph of the parabola intersects the x-axis once, then there is only one real number solution.
Suppose you are solving this quadratic equation:
To find the solution graphically, then graph the corresponding quadratic function.
Notice that this parabola intersects the x-axis at x = -2. This is the solution to the quadratic equation. In fact, you can rewrite the quadratic as a binomial squared:
Case 3: No real solutions. If the graph of the parabola doesn’t intersect the x-axis, then there are no real solutions to the quadratic equation.
Suppose you are solving this quadratic equation:
To find the solution graphically, then graph the corresponding quadratic function.
Notice that this parabola doesn’t intersect the x-axis. When this happens, the quadratic equation doesn’t have real number solutions. It does, however, have complex number solutions, which you can find using the quadratic formula.
Summary of Solving by Graphing. When a parabola intersects the x-axis, then the parabola has at least one real number solution. These intersection points are also referred to as:
• x-intercepts
• Zeros of the Quadratic Function
• Roots of the Quadratic Equation
To learn more about solving a quadratic equation graphically, click on the following link. This includes a video tutorial and several worked-out math examples.
SAT Skill: Solving Quadratic Equations Example 1 The parabola with the equation shown below intersects the line with equation y = 16 at two points, A and B. What is the length of segment AB? For a question of this type, where references are made to graphs and intersection points, it’s best to draw a diagram to get a better understanding of the problem. The equation shown is of a parabola in vertex form that intersects y = 16. Sketch that. We know the y coordinates for A and B; in both cases it’s 16. To find the corresponding x-coordinates, solve this equation: The x-coordinate for point A is x = 4 and the for B it’s x = 12. So the distance from A to B is the difference, or 8. Example 2 In the quadratic equation below, a is a nonzero constant. The vertex of the parabola has coordinates (c, d). Which of the following is equal to d? When a quadratic is written in standard form, the vertex has these coordinates: Write the function in standard form. Now find the corresponding x and y coordinates with this equation. Example 3 The parabola whose equation is shown below intersects the graph of y = x at (0, 0) and (a, a). What is the value of a? To get a better understanding of this problem, draw a diagram. To find the value of a, solve the following equation:
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# Solving equations symbolically
When Solving equations symbolically, there are often multiple ways to approach it. We can solve math word problems.
## Solve equations symbolically
We will also give you a few tips on how to choose the right app for Solving equations symbolically. The disparities between minority groups and the majority is a major problem in the United States. Exact statistics on how many minorities are unemployed and how many people of lower income are living in poverty are hard to track, but it’s clear that there is still much to be done. One way that the inequality gap can be closed is by encouraging more minorities to go into STEM fields. This will not only help them to earn more money, but it will also give them more recognition in the workplace and make it easier for them to get raises and promotions. Another way that inequality can be closed is by improving access to education. If more minorities have access to quality education, they will be less likely to end up stuck in low-paying jobs or trapped in poverty.
The intercept is the value that represents the y value of each data point when plotted on a graph. Sometimes it is useful to know the value of x at which y = 0. This is called the x-intercept and it can be used to estimate where y will be when x = 0. There are two main ways to determine the intercept: 1) The easiest way is to use a line of best fit. The line shows that when x increases, y increases by the same amount. Therefore, if you know x, you can calculate y based on that value and then plot the resulting line on your graph (see figure 1 below). If there is more than one data point, you can select the one that has the highest y value and plot that point on your graph (see figure 2 below). When you do this for all data points, you get an approximation of where the line of best fit crosses zero. This is called the x-intercept and it is equal to x minus y/2 (see figure 3). 2) Another way to find x-intercept involves using the equation y = mx + b. The left side is equation 1 and the right side is equation 2. When solving for b, remember that b depends on both m and x, so make sure to factor in your other values as well (for example, if you have both
If you want to calculate an individual’s natural log, then you need to measure their height and multiply it by three. The basic idea behind natural log is that trees grow in all directions, so if you take the total diameter of a tree and divide it by its height, you will get 1, 2 or 3. The more branches there are on a tree and the longer they are, the higher the log will be. The thicker a tree trunk is, the more logs it has. The larger a tree grows in diameter, the more logs it has, but only up to a certain point as it would have to have more branches and trunks to offset the increased surface area of each branch. There are two main ways to get around this problem: 1) Take out one branch in order to get less branches and increase your natural log. A common example of this is grafting where one sapling is grafted onto another sapling that has fewer branches. 2) Grow multiple trunks from one original trunk so that each new trunk has equal or
Linear equations are very common in every grade. They are used to show the relationship between two numbers or values. There are a few different ways to solve linear equations by graphing. You can graph the equation on a coordinate grid, plot points on a coordinate grid, or plot points on an axes grid. When graphing, always follow the order of operations. To graph an equation, start with an ordered pair (x, y). Then put points in between the coordinates that indicate how you want your equation to look. For example, if x = 2 and y = -8, then your graphed equation would look like this: (2,-8). Starting from the left and working from one point to the next will help you visualize how you want your graph to look.
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Marlee Gonzales
The best math app I have seen so far, definitely recommend it to others. The photo feature is more than amazing and the step-by-step detailed explanation is quite on point. Gave it a try never regretted.
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# Finding out the Basics Of Graphing Step-wise Formula Together with Cost-free Math concepts Worksheets
First, the Fundamentals! The x axis of a graph refers to the horizontal line while the y axis refers to the vertical line. Jointly these strains sort a cross and the level in which they the two meet up with is known as the origin. The value of the origin is constantly . So if you move your pencil from the origin to the proper, you are drawing a line throughout the constructive values of the x axis, i.e., one, two, 3 and so on. From the origin to the left, you happen to be shifting across the adverse values of the x axis, i.e., -one, -2, -3 and so on. If you go up from the origin, you are covering the optimistic values of the y axis. Likely down from the origin, will consider you to the negative values of the y axis.
How Do You Find Factors In A Graph? This set of figures (2, 3) is an illustration of an ordered pair. The initial amount refers to the benefit of x even though the 2nd amount stands for the price of y. When ordered pairs are utilised to uncover points on the grid, they are named the coordinates of the point. In over example, the x coordinate is two even though the y coordinate is three. With each other, they empower you to locate the position (2, three) on the grid. What is the point of all this? Nicely, ever wondered how ships explain exactly in which they are in the vastness of the ocean? To be able to track down places, individuals have to attract a grid more than the map and explain points with the assist of x and y coordinates. Why never you give it a try out? Envision remaining aspect wall of your room to be y axis and the wall at your back to be the x axis. The corner that connects them equally will be your origin. Evaluate equally in Daffynition decoder answers . If I say stand on coordinates (3, 2), would you know where to go? That indicates from the corner (origin) you ought to transfer 3 toes to the right and two feet forward.
Graphing Linear Equations. If you have many sets of x and y coordinates, you can now draw strains on a graph. Totally free math worksheets can drill you on plotting x and y coordinates although graphing linear equation. A linear equation when drawn on a line graph often yields a straight line. Consider “y = 2x + 1” for example – a linear equation. Assign any three numbers to x, and then resolve for y. No matter what figures you assign to x and whatsoever y arrives out to be, you will conclude up with a straight line. Remember to follow on easier math worksheets very first just before transferring on to writing a linear equation or to the programs of linear equations. Very good luck!
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# 3.1 Functions and function notation (Page 7/21)
Page 7 / 21
Using [link] , solve $\text{\hspace{0.17em}}f\left(x\right)=1.$
$x=0\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$
## Determining whether a function is one-to-one
Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in [link] at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of$1000.
However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in [link] .
A 4.0
B 3.0
C 2.0
D 1.0
This grading system represents a one-to-one function because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter.
To visualize this concept, let’s look again at the two simple functions sketched in [link] (a) and [link] (b) . The function in part (a) shows a relationship that is not a one-to-one function because inputs $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ both give output $\text{\hspace{0.17em}}n.\text{\hspace{0.17em}}$ The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.
## One-to-one function
A one-to-one function is a function in which each output value corresponds to exactly one input value. There are no repeated x - or y -values.
## Determining whether a relationship is a one-to-one function
Is the area of a circle a function of its radius? If yes, is the function one-to-one?
A circle of radius $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ has a unique area measure given by $\text{\hspace{0.17em}}A=\pi {r}^{2},$ so for any input, $\text{\hspace{0.17em}}r,\text{\hspace{0.17em}}$ there is only one output, $A.$ The area is a function of radius $\text{\hspace{0.17em}}r.$
If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is given by the formula $\text{\hspace{0.17em}}A=\pi {r}^{2}.\text{\hspace{0.17em}}$ Because areas and radii are positive numbers, there is exactly one solution: $\sqrt{\frac{A}{\pi }}.$ So the area of a circle is a one-to-one function of the circle’s radius.
1. Is a balance a function of the bank account number?
2. Is a bank account number a function of the balance?
3. Is a balance a one-to-one function of the bank account number?
a. yes, because each bank account has a single balance at any given time; b. no, because several bank account numbers may have the same balance; c. no, because the same output may correspond to more than one input.
1. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade?
2. If so, is the function one-to-one?
2. No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.
## Using the vertical line test
As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis.
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
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# What is the vertex of y=x^2+12x+26?
Apr 12, 2017
The vertex is at $\left(- 6 , - 10\right)$
#### Explanation:
You can find the vertex (turning point) by first finding the line that is the axis of symmetry.
$x = \frac{- b}{2 a} = \frac{- 12}{2 \left(1\right)} = - 6 \text{ } \leftarrow$ This is the $x$-value of the vertex.
Now find $y$.
$y = {x}^{2} + 12 x + 26$
$y = {\left(- 6\right)}^{2} + 12 \left(- 6\right) + 26$
$y = 36 - 72 + 26$
$y = - 10 \text{ } \leftarrow$ This is the $y$-value of the vertex.
The vertex is at $\left(- 6 , - 10\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You can also find the vertex by completing the square to get the equation in vertex form: $y = a {\left(x + b\right)}^{2} + c$
$y = {x}^{2} + 12 x + 26$
$y = {x}^{2} + 12 x \textcolor{red}{+ {6}^{2}} \textcolor{red}{- {6}^{2}} + 26 \text{ } \textcolor{red}{{\left(\frac{b}{2}\right)}^{2} = {\left(\frac{12}{2}\right)}^{2}}$
$y = {\left(x + 6\right)}^{2} - 10$
Vertex is at $\left(- b , c\right) \text{ } \rightarrow \left(- 6 , - 10\right)$
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# Functional Programming and Intelligent Algorithms
## Tutorial 1.3bis: More on Lists and Tuples
(Last update: 12 January 2015)
### Overview
• Reading: Simon Thompson, Chapter 5-6 (skip 5.3)
This is an optional tutorial, building on Tutorial 3.
### Problem 3: Positioned Pictures (Optional)
See Thompson Chapter 6.6
### Problem 4: Credit card numbers (optional)
This exercise is based on a an exercise given by Richard Eisenberg for the cis194 module at University of Pennsylvania. He in turn adapted it from a practicum assigned at the University of Utrecht functional programming course taught by Doaitse Swierstra, 2008-2009.
Working with long numbers, such as national insurance numbers, account numbers, ISBN numbers or credit card numbers, it is easy to make mistakes. Therefore such numbers are designed using an error correcting codes. The most probable mistakes lead to an invalid number, so that it is easily seen that an error has occurred.
In this exercise, we shall explore credit card numbers and write a functional program to validate such numbers. The validation comprises the following steps:
1. Double the value of every second digit beginning from the right. That is, the last digit is unchanged; the second-to-last digit is doubled; the third-to-last digit is unchanged; and so on. For example, [1,3,8,6] becomes [2,3,16,6].
2. Calculate the digit sum of each number in the list. For the one-digit numbers, this is just the number itself. For two-digit numbers, it is the sum of the two digits. For example, [2,3,16,6] becomes [2,3,7,6].
3. Add all the digit sums together. For example, [2,3,7,6] becomes 18.
4. Calculate the remainder when the sum is divided by 10. For example, 18 becomes 8.
5. If the result equals 0, then the number is valid. In the running example, we got 8, which means the number is invalid.
We will implement the validation through a series of small exercises.
#### Exercise 1
We first need to be able to split a number into its last digit and the rest of the number. Write these functions:
``` lastDigit :: Integer -> Integer dropLastDigit :: Integer -> Integer ```
Haskell has built-in functions or operators for integer division and remainder (modulo).
Examples:
``` lastDigit 123 == 3 lastDigit 9 == 9 dropLastDigit 123 == 12 dropLastDigit 9 == 0 ```
Task: Implement and test the two functions described above.
#### Exercise 2
Now, we male a function to break a number into a list of digits. The method is the standard approach to writing a number to base b (with b=10 in our case). You will need the functions from the previous exercise and apply them with recursion. The function should have the following signature,
``` toDigits :: Integer -> [Integer] ```
Positive numbers must be converted to a list of digits. For 0 or negative inputs, `toDigits` should return the empty list.
Examples:
``` toDigits 1234 == [1,2,3,4] toDigits 0 == [] toDigits (-17) == [] ```
Task: Implement and test the `toDigits` function.
#### Exercise 3
Once we have the digits in a list, we need to double every other one. Define a function ``` doubleEveryOther :: [Integer] -> [Integer] ```
Remember that doubleEveryOther doubles every other number starting from the right, i.e. the second-to-last, fourth-to-last, and so on. This is easiest to do if the the list is put reverse order, so that one can count from the left instead. Check out the builtin `reverse` function.
Examples:
``` doubleEveryOther [8,7,6,5] == [16,7,12,5] doubleEveryOther [1,2,3] == [1,4,3] ```
Task: Implement and test the `doubleEveryOther` function.
#### Exercise 4
We need to write a function to calculate the digit sum of an arbitrary integer. The digit sum is the sum of the digits, e.g. 7, 16, 25, and 34 each have 7 as the digit sum. Thus we are looking for a function with the following signature:
``` sumDigits :: Integer -> Integer ```
You probably want to use `toDigits` as an auxiliary.
Task: Implement and test the `digitSum` function.
#### Exercise 5
The output of doubleEveryOther has a mix of one-digit and two-digit numbers. Define the function ``` sumDigits :: [Integer] -> Integer ``` to calculate the sum of all digits.
Examples:
``` sumDigits [16,7,12,5] = 1 + 6 + 7 + 1 + 2 + 5 = 22 ```
The `digitSum` can be used as an auxiliary here.
Task: Implement and test the `sumDigits` function.
#### Exercise 6
Define the function ``` validate :: Integer -> Bool ``` that indicates whether an Integer could be a valid credit card number. This will use all functions defined in the previous exercises.
Examples:
``` validate 4012888888881881 = True validate 4012888888881882 = False ``` Task: Implement and test the `validate` function.
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# Power of a product rule
##### Intros
###### Lessons
1. What are exponent rules?
##### Examples
###### Lessons
1. Simplify the following:
1. $(-4xy)^4$
##### Practice
###### Topic Notes
We use the power of a product rule when there are more than one variables being multiplied together and raised to a power. The power of a product rule tells us that we can simplify a power of a power by multiplying the exponents and keeping the same base.
## Introduction to the Power of a Product Rule
The power of a product rule is a fundamental concept in exponent laws, essential for mastering algebra and higher mathematics. Our introduction video serves as a crucial starting point, offering a clear and concise explanation of this rule and its applications. By watching this video, students gain a solid foundation in understanding how exponents work when multiplying terms with the same base. Rather than simply memorizing formulas, the video emphasizes the importance of grasping the underlying principles of exponents. This approach enables learners to apply the power of a product rule confidently across various mathematical scenarios. Understanding this rule is key to unlocking more complex exponent laws and algebraic expressions. By focusing on the logic behind the rule, students develop a deeper, more intuitive understanding of exponents, setting the stage for advanced mathematical concepts and problem-solving skills.
## Understanding the Basics of Exponents
Exponents are a fundamental concept in mathematics that represent repeated multiplication. This powerful notation allows us to express large numbers concisely and perform complex calculations efficiently. At its core, an exponent indicates how many times a number, called the base, is multiplied by itself.
Let's start with a simple example to illustrate the concept of exponents. Consider the expression 2³. This means we multiply 2 by itself three times: 2 × 2 × 2 = 8. In this case, 2 is the base, and 3 is the exponent. We read this as "2 to the power of 3" or "2 cubed."
Here are a few more examples of positive integer exponents:
• 3² = 3 × 3 = 9
• 5 = 5 × 5 × 5 × 5 = 625
• 10³ = 10 × 10 × 10 = 1,000
As we can see, exponents provide a shorthand way to express repeated multiplication. This notation becomes especially useful when dealing with larger numbers or variables.
Speaking of variables, exponents work the same way with algebraic expressions with exponents. For instance:
• x² = x × x
• y = y × y × y × y × y
Understanding exponents as repeated multiplication naturally leads us to one of the most important rules in exponent arithmetic: the product of powers rule. This rule states that when multiplying expressions with the same base, we keep the base and add the exponents.
Mathematically, we express the product of powers rule as:
x^a × x^b = x^(a+b)
This rule makes sense when we think about exponents as repeated multiplication. Let's break it down with an example:
2³ × 2 = (2 × 2 × 2) × (2 × 2 × 2 × 2) = 2
We can see that we're simply combining all the 2s being multiplied, resulting in 2 to the power of 3 + 4 = 7.
The product rule for exponents works with any base, including variables:
• x² × x³ = x
• y × y² = y
• 5³ × 5² = 5
This rule is incredibly useful in simplifying algebraic expressions and solving complex mathematical problems. It allows us to quickly combine like terms and reduce expressions to their simplest form.
As we delve deeper into the world of exponents, we'll encounter more rules and properties that build upon this fundamental understanding. The product of powers rule is just the beginning, but it serves as a crucial foundation for mastering exponent operations.
In conclusion, exponents represent repeated multiplication, providing a concise way to express large numbers and repeated operations. The product of powers rule naturally extends from this concept, allowing us to simplify expressions by adding exponents when multiplying terms with the same base. This understanding of exponents and their properties is essential for advancing in algebra, calculus, and many other areas of mathematics and science.
## The Power of a Product Rule Explained
The power of a product rule, also known as the product of powers law, is a fundamental concept in algebra that simplifies the process of raising a product to a power. This rule states that when multiplying two or more factors raised to the same power, we can simply multiply the bases and keep the exponent the same. Mathematically, it can be expressed as (ab)^n = a^n * b^n, where a and b are the bases and n is the exponent.
Let's explore this rule with numerical examples first. Consider (2 * 3)^4. Without the rule, we would need to multiply 2 and 3, then raise the result to the fourth power: 6^4 = 1296. Using the power of a product rule, we can simplify this process: (2 * 3)^4 = 2^4 * 3^4 = 16 * 81 = 1296. This method is often more efficient, especially with larger numbers or variables.
The rule also applies to algebraic expressions simplification. For instance, (xy)^3 = x^3 * y^3. This simplification is particularly useful when dealing with complex algebraic expressions simplification. Consider (2ab)^5. Using the rule, we can expand this to 2^5 * a^5 * b^5 = 32a^5b^5, which is much easier to work with in further calculations.
To understand why this rule works, let's expand an expression like (ab)^3:
(ab)^3 = (ab)(ab)(ab) = a*a*a * b*b*b = a^3 * b^3
This expansion demonstrates that each base is indeed multiplied by itself as many times as the exponent indicates, validating the rule.
The versatility of the power of a product rule becomes evident when we apply it to different bases and exponents. For example:
(3x^2y)^4 = 3^4 * (x^2)^4 * y^4 = 81x^8y^4
Here, we see the rule applied to a numerical coefficient (3), a variable with its own exponent (x^2), and another variable (y).
The rule also works with fractional and negative exponents in algebra. For instance:
(2ab)^(1/2) = 2^(1/2) * a^(1/2) * b^(1/2) = 2 * a * b
(xy)^(-3) = x^(-3) * y^(-3) = 1/(x^3 * y^3)
These examples showcase how the rule simplifies expressions with various types of exponents, making it a powerful tool in algebra.
It's important to note that the power of a product rule is closely related to other exponent rules interaction, such as the power of a power rule ((a^m)^n = a^(mn)) and the power of a quotient rule ((a/b)^n = a^n / b^n). Understanding how these rules interact can greatly enhance one's ability to manipulate and simplify complex exponent rules interaction.
In practical applications, the power of a product rule is invaluable in fields such as physics, engineering, and computer science, where complex calculations involving powers are common. For instance, in calculating compound interest or analyzing exponential growth, this rule can significantly streamline computations.
To further illustrate the rule's applicability, consider a problem in physics where we need to calculate the force of gravity between two objects. The formula F = G(m1m2)/r^2 involves a product in the numerator. If we needed to cube this entire expression, the power of a product rule would allow us to distribute the exponent: (G(m1m2)/r^2)^3 = G^3 * m1^3 * m2^3 / r^6, greatly simplifying the calculation process.
In conclusion, the power of a product rule
## Applications and Examples of the Power of a Product Rule
The power of a product rule is a fundamental concept in algebra that simplifies expressions involving exponents. This rule states that when raising a product to a power, we can raise each factor to that power and then multiply the results. Let's explore various examples and applications of this rule, ranging from simple to complex scenarios.
Simple Examples:
1. (xy)² = x²y²
This basic example shows how the rule applies to a product of two variables.
2. (3a)³ = 3³a³ = 27a³
Here, we see how the rule works with a coefficient and a variable.
3. (2xy) = 2xy = 16xy
This example demonstrates the rule applied to a product with a coefficient and two variables.
Complex Examples:
4. (4abc)³ = 4³a³b³c³ = 64a³b³c³
The rule extends easily to products with multiple variables.
5. (x²y³z) = (x²)(y³)z = xy¹²z
This example shows how the rule works with variables already raised to powers.
6. ((a²b)(cd³))² = (a²b)²(cd³)² = ab²c²d
Here, we apply the rule to a more complex expression with grouped terms.
Negative Exponents:
7. (xy)² = x²y²
The rule applies similarly to negative exponents in algebra, simplifying reciprocal expressions.
8. (2ab¹)³ = 2³a³(b¹)³ = a³b³
This example combines negative exponents in algebra with the power of a product rule.
Applications in Different Scenarios:
9. Area of a rectangle: If the length and width of a rectangle are doubled, the new area is (2l)(2w) = 4lw, which is four times the original area.
10. Volume of a cube with tripled side length: If the side length of a cube is tripled, the new volume is (3s)³ = 27s³, which is 27 times the original volume.
11. Compound interest growth factor: In finance, (1 + r) represents the growth factor for compound interest, where r is the interest rate and n is the number of compounding periods.
12. Scientific notation multiplication: (5.2 × 10)(3.1 × 10²) = (5.2 × 3.1)(10 × 10²) = 16.12 × 10²
13. Simplifying algebraic fractions with exponents: (x²y³)/(xy²) = x²y³/(xy) = x²y
14. Physics equations: In kinetic energy (KE = ½mv²), doubling the velocity results in (½m(2v)²) = ½m(4v²) = 2mv², quadrupling the energy.
15. Probability calculations: If the probability of an event occurring twice independently is (0.3)², the probability of it not occurring twice is (1 - 0.3)² = 0.7² = 0.49.
These examples demonstrate the versatility and power of the product rule
## Common Mistakes and How to Avoid Them
When applying the power of a product rule, students often encounter several common mistakes that can lead to incorrect solutions. Understanding these errors and learning how to avoid them is crucial for mastering this important mathematical concept.
One of the most frequent mistakes is misapplying the rule by distributing the exponent to each factor individually. For example, students might incorrectly write (xy)² as x²y². This error stems from confusing the power of a product rule with the product rule for exponents. To avoid this, always remember that (xy)² means (xy)(xy), not x²y².
Another common error is forgetting to include all factors when raising a product to a power. For instance, (xyz)³ should be expanded to (xyz)(xyz)(xyz), not just x³y³. Students often overlook one or more factors, leading to incomplete or incorrect answers. To prevent this, carefully count the number of factors and ensure each is included in the expansion.
Students also frequently struggle with negative exponents in product expressions. For example, (xy)² is sometimes mistakenly written as x²y². The correct application would be 1/(xy)², which can be further simplified to 1/(x²y²). To avoid this error, remember that negative exponents indicate reciprocals, and the entire product should be treated as a single unit.
Confusion often arises when dealing with fractional exponents in product expressions. For instance, (xy)½ is not equal to x½y½. The correct approach is to treat the entire product as a single term and apply the fractional exponent to it, resulting in the square root of xy. To prevent this mistake, always consider the product as a whole when applying fractional exponents.
Another pitfall is incorrectly applying the rule to sums or differences. The power of a product rule does not apply to expressions like (x+y)². Students sometimes erroneously write this as x²+y². To avoid this, recognize that the rule only applies to products, not sums or differences. For expressions involving sums or differences, use the binomial theorem or FOIL method instead.
To master the power of a product rule, it's essential to understand its underlying principle rather than blindly applying a formula. Visualize the repeated multiplication of the entire product, and practice expanding various expressions step-by-step. For example, expand (ab)³ as (ab)(ab)(ab) = a³b³, reinforcing the correct application of the rule.
Regular practice with diverse examples, including those with multiple factors, negative exponents, and fractional powers, will help solidify understanding and reduce errors. Additionally, always double-check your work by expanding the expression manually to verify the result. By focusing on comprehension and careful application, students can significantly improve their accuracy when using the power of a product rule in mathematical problem-solving.
## Related Exponent Rules and Their Connections
When exploring exponent rules connections, it's essential to understand that they are interconnected and build upon one another. While the power of a product rule is fundamental, other rules like the quotient rule exponent and the power of a power rule are equally important in mastering exponents. By grasping these rules and their relationships, students can tackle complex problems with confidence.
The quotient rule exponent is closely related to the power of a product rule. It states that when dividing expressions same base, we subtract the exponents. Mathematically, this is expressed as (x^a) / (x^b) = x^(a-b). This rule complements the power of a product rule, which involves addition of exponents when multiplying expressions with the same base.
The power of a power rule, on the other hand, deals with exponents raised to another power. It states that (x^a)^b = x^(ab). This rule is particularly useful when simplifying nested exponents and can be seen as an extension of the power of a product rule. Understanding the power of a power rule helps in breaking down complex exponential expressions into simpler forms.
These exponent rules connections are interconnected in several ways. For instance, the quotient rule exponent can be derived from the power of a product rule by considering division as multiplication by the reciprocal. Similarly, the power of a power rule can be understood as repeated application of the power of a product rule. By recognizing these connections, students can develop a more intuitive understanding of exponents and their properties.
Applying multiple exponent rules to solve complex problems is a crucial skill. For example, consider the expression ((x^3)^2 * (x^4)) / (x^5). To simplify this, we can start by applying the power of a power rule to (x^3)^2, giving us x^6. Then, we can use the power of a product rule to combine x^6 and x^4, resulting in x^10. Finally, we apply the quotient rule exponent to divide x^10 by x^5, yielding x^5 as the simplified result.
Another example that combines multiple rules is (y^-2 * y^5)^3 / y^4. Here, we first use the power of a product rule inside the parentheses to get y^3. Then, we apply the power of a power rule to (y^3)^3, resulting in y^9. Lastly, we use the quotient rule exponent to divide y^9 by y^4, giving us y^5 as the final answer.
Understanding these interconnections helps in developing problem-solving strategies. When faced with a complex exponential expression, students can break it down into smaller parts and apply the appropriate rules step by step. This approach not only simplifies the problem-solving process but also reinforces the relationships between different exponent rules.
Moreover, recognizing the patterns in exponent rules can lead to a deeper understanding of mathematical concepts. For instance, the quotient rule exponent can be extended to negative exponents explanation, explaining why x^-n is equivalent to 1/(x^n). Similarly, the power of a power rule helps in understanding the concept of roots, as (x^(1/n))^n = x.
In conclusion, the quotient rule exponent, power of a power rule, and other exponent rules connections are closely interconnected. By understanding these relationships and practicing with diverse problems, students can enhance their mathematical skills and tackle complex exponential expressions with ease. The key is to recognize the patterns, apply the rules systematically, and always be mindful of how different rules interact with each other in various mathematical contexts.
## Practical Applications in Algebra and Beyond
The power of a product rule in algebra is a fundamental concept with wide-ranging practical applications across various mathematical fields and real-world scenarios. This rule states that when raising a product to a power, we can raise each factor to that power and then multiply the results. In algebra, this rule is crucial for solving equations and simplifying complex expressions efficiently.
In solving polynomial equations, the power of a product rule allows mathematicians to break down complicated terms into more manageable components. For instance, when dealing with equations involving variables raised to powers, this rule enables us to distribute the exponent across multiple factors, making it easier to isolate variables and find solutions. This technique is particularly useful in polynomial equations and exponential functions, which are common in scientific and engineering calculations.
The rule's application extends beyond basic algebra into more advanced mathematical concepts. In calculus, it plays a vital role in differentiating and integrating complex functions. When working with derivatives, the power rule combines with the product rule to simplify the process of finding rates of change for intricate expressions. In integral calculus, this rule aids in breaking down complex integrands, making integration more approachable.
Real-world applications of the power of a product rule are abundant. In physics, it's used to calculate work done by varying forces or to determine the kinetic energy calculations of objects in motion. Engineers apply this rule when designing structures, calculating stress and strain on materials, or optimizing energy systems. In finance, the rule is crucial for compound interest modeling, helping investors and economists model growth over time.
In computer science and cryptography, the power of a product rule is fundamental to many algorithms, particularly in public-key encryption systems. These systems rely on the difficulty of factoring large numbers, a process that involves extensive use of exponents and products. The rule also finds applications in probability theory, where it's used to calculate the likelihood of multiple independent events occurring simultaneously.
Environmental scientists use this rule when modeling population growth or decay, considering factors like birth rates, death rates, and environmental carrying capacities. In chemistry, it's applied in reaction kinetics in chemistry to understand how the concentration of reactants affects reaction rates. The versatility of this algebraic principle demonstrates its importance across diverse fields, making it a cornerstone of mathematical problem-solving in both theoretical and applied contexts.
## Conclusion
In summary, the product rule is a powerful tool in exponent laws, allowing us to simplify expressions by adding exponents when multiplying terms with the same base. Understanding the principle behind this rule is crucial, as it enables you to apply it confidently across various mathematical scenarios. Rather than memorizing formulas, focus on grasping the underlying concept. We encourage you to practice applying the product rule regularly, as this will reinforce your understanding and improve your problem-solving skills. For those seeking to deepen their knowledge, explore further resources on exponent laws and related topics. Remember, the introduction video serves as an excellent foundation for mastering exponent laws, including the product rule. By building on this knowledge, you'll develop a strong mathematical toolkit that will serve you well in future studies and real-world applications. Keep practicing, stay curious, and don't hesitate to revisit the video for a refresher on these fundamental concepts.
### Example:
Simplify the following: $(-4xy)^4$
#### Step 1: Understand the Power of a Product Rule
The Power of a Product Rule states that when you have a product raised to a power, you can distribute the exponent to each factor in the product. In mathematical terms, this means: $(ab)^n = a^n \cdot b^n$ In this example, we have $(-4xy)^4$. According to the Power of a Product Rule, we need to distribute the exponent 4 to each factor inside the parentheses.
#### Step 2: Separate the Terms
First, let's separate the terms inside the parentheses. We have three factors: -4, x, and y. So, we can rewrite the expression as: $(-4xy)^4 = (-4)^4 \cdot (x)^4 \cdot (y)^4$ This step helps us to handle each factor individually.
#### Step 3: Simplify the Negative Base
Next, we need to simplify $(-4)^4$. It's important to note that raising a negative number to an even power results in a positive number. This is because multiplying an even number of negative factors results in a positive product. Therefore: $(-4)^4 = 4^4$ Now, we only need to calculate $4^4$.
#### Step 4: Calculate the Power of 4
Now, let's calculate $4^4$. This means multiplying 4 by itself four times: $4^4 = 4 \cdot 4 \cdot 4 \cdot 4 = 256$ So, $(-4)^4$ simplifies to 256.
### FAQs
Here are some frequently asked questions about the power of a product rule:
#### What is the power of a product rule?
The power of a product rule states that when raising a product to a power, you can raise each factor to that power and then multiply the results. Mathematically, it's expressed as (ab)^n = a^n * b^n, where a and b are the factors and n is the exponent.
#### What is an example of the power of a product rule?
A simple example is (2x)^3 = 2^3 * x^3 = 8x^3. Here, we raise both 2 and x to the power of 3 separately and then multiply the results.
#### How do you apply the power of a product rule to simplify expressions?
To simplify expressions using this rule, identify the product within parentheses and the power it's raised to. Then, apply the power to each factor individually. For example, (3ab)^4 simplifies to 3^4 * a^4 * b^4 = 81a^4b^4.
#### Does the power of a product rule work with negative exponents?
Yes, the rule works with negative exponents. For instance, (xy)^-2 = x^-2 * y^-2 = 1/(x^2 * y^2). Remember that a negative exponent means the reciprocal of the positive exponent.
#### How is the power of a product rule different from the product rule for exponents?
The power of a product rule deals with raising a product to a power, while the product rule for exponents involves multiplying terms with the same base and adding their exponents. For example, the power of a product rule is (ab)^n = a^n * b^n, while the product rule for exponents is x^a * x^b = x^(a+b).
### Prerequisite Topics for Understanding the Power of a Product Rule
Mastering the power of a product rule in mathematics requires a solid foundation in several key areas. One of the most crucial prerequisites is combining the exponent rules. Understanding how exponents work and how to manipulate them is essential when dealing with products raised to powers.
Equally important is the ability to simplify rational expressions and understand their restrictions. This skill helps in breaking down complex expressions that often arise when applying the power of a product rule. Additionally, familiarity with the negative exponent rule is crucial, as it allows for the proper handling of expressions with negative powers.
When working with the power of a product rule, you'll often encounter situations that require solving polynomial equations. This prerequisite topic provides the tools needed to manipulate and solve equations that result from applying the rule. Moreover, understanding scientific notation is beneficial, especially when dealing with very large or small numbers in product expressions.
The power of a product rule has practical applications in various fields. For instance, in finance, it's used in compound interest calculations. Understanding this connection can provide real-world context and motivation for mastering the rule. Similarly, in chemistry, the rule is applied in reaction kinetics, demonstrating its importance beyond pure mathematics.
By building a strong foundation in these prerequisite topics, students can approach the power of a product rule with confidence. Each of these areas contributes to a deeper understanding of how products and exponents interact, making it easier to grasp and apply the rule in various contexts. Whether you're solving complex algebraic problems or applying the concept in scientific or financial scenarios, a solid grasp of these prerequisites will significantly enhance your ability to work with the power of a product rule effectively.
Remember, mathematics is a cumulative subject. Each new concept builds upon previous knowledge. By taking the time to review and master these prerequisite topics, you're not just preparing for understanding the power of a product rule, but also laying the groundwork for more advanced mathematical concepts that you'll encounter in your future studies.
$({a^mb^n} {)^p} = {a^{mp}b^{np}}$
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# Your Ultimate Guide to the Sankalana Vyavakanabhyam Sutra
The Sankalana Vyavakanabhyam is a popular Vedic Maths sutra used to solve linear equations in 2 variables. Literally meaning ‘addition and subtraction’, this 7th Vedic Maths sutra will help enthusiastic learners to breeze through linear equations. This article is dedicated to this wonderful method which will help you speed up your calculation and give accurate results. Check out other Vedic Maths lessons on the Podium Blog.
Only specific types of problems are solved with this method. For generic lengthy multiplication, look up the Nikhilam sutra and the Anurupyena sutra on the Podium blog.
Suppose we have a linear equation as:
ax+by= c(1)
bx+ay=c (2)
The linear equation should be such that, the coefficients are interchanged i.e. where x is a coefficient of a and y is a coefficient of b in equation (1), it is vice-versa in equation (2).
## Let us try with some examples
EXAMPLE 1
5x – 4y = 34
4x – 5y =11
(5x-4y) + (4x-5y) = 34 +11
5x – 4y + 4y- 5y = 45
9x + 9y = 45
x + y = 45…..(i)
Subtraction –
(5x-4y) – (4x-5y) = 23
5x – 4y – 4x + 5y= 23
x + y = 23…..(ii)
x + y +x – y = 28 / x – y – (x-y) =-18
x = 14 -2y = -18
y = 18
Therefore, x = 14 and y = 18 and we have our equation solved!
EXAMPLE 2
45x + 23y = 113
23x + 45y = 91
(45x + 23y) + (23x + 45y) = 204
68x + 68y = 204
x + y = 204/68
x + y = 3…….(i)
Subtraction –
45x + 23y – (23x + 45y) = 22
45x + 23y – 23x – 45y = 22
22x – 22y = 22
x – y = 1……(ii)
Taking (i) and (ii) together, we will repeat the addition and subtraction procedures till we get the desired values of x and y, respectively.
x + y + x – y = 4 (ADDITION)
Which implies to –
2x = 4
Therefore, x = 2
x + y – (x – y) = 2 (SUBTRACTION)
x + y – x + y =2
2y = 2
y = 1
### More Practice
This Sutra is all about practice in solving linear equation in two variables. The more we practice it, the easier it gets and the more it allows and helps us to do quick math.
EXAMPLE 3
9x – 7y = 40
7x – 9y = 24
9x – 7y + 7x – 9y = 64
16x – 16y = 64
x + y = 4…..(i)
Subtraction –
9x – 7y – (7x – 9y) = 16
9x – 7y – 7x + 9y = 16
2x – 2y = 16
(Here we will divide both sides by 2 since it is the common co-coefficient on LHS and RHS)
Which leaves us with x– y =8….(ii)
Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.
x + y + x – y = 4 (ADDITION)
Which gives us-
2x = 4
Therefore, x = 2
x + y – (x – y) = 8 (SUBTRACTION)
x + y – x + y = 8
2y = 8
y = 4
### Practicing Sankalana Vyavakanabhyamwith More Problems
EXAMPLE 4
69y – 5z = 42
5y – 69z = 32
69y – 5z + 5y – 69z = 74
74y – 74z = 74
x – y = 1…..(i)
Subtraction –
69y – 5z – (5y – 69z) = 74
69y – 5z – 5y + 69z = 74
y + z = 5/28……(ii)
Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.
y = 1 + z = 5/28 – z
1 + z = 5/28 – z
1 + 2z = 5/28
2z = -23/28
z = (-23/28)/2
z = -23/56 Therefore, y = 5/28 – -23/56
y = 10/56 + 23/56
y = 33/56
EXAMPLE 5
1955x + 476y = 2482
476x + 1955y = – 4913
1955x + 476y + 476x + 1955y = -2431
2431 x + 2431y = -2431
x + y = -1….(i)
Subtraction –
1955x + 476y – (476x +1955y) = -2431
1955x + 476y – 476x – 1955y = -2431
1479x + 1479y = 7395
x + y =5
Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.
x + y + x – y = 6 (ADDITION)
2x = 6
which gives us, x = 3
x + y – (x – y) = 4 (SUBTRACTION)
x + y – x + y = 4
2y = 4
y = 2
### Our Last Practice Problem!
EXAMPLE 6
3x + 2y = 18
2x + 3y = 17
3x + 2y + 2x + 3y = 35
5x + 5y = 35
x + y = 7…..(i)
Subtraction –
3x + 2y – (2x + 3y) = 1
3x + 2y – 2x + 3y= 1
x – y = 1……(ii)
Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.
x + y + x – y = 6 (ADDITION)
2x = 6
x = 3
x + y – (x – y) = (SUBTRACTION)
2y = 4
y = 2
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Discrete Distributions
# Discrete Distributions
## Discrete Distributions
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##### Presentation Transcript
1. Discrete Distributions
2. What is the binomial distribution? The binomial distribution is a discrete probability distribution. It is a distribution that governs the random variable, X, which is the number of successes that occur in "n" trials. The binomial probability distribution gives us the probability that a success will occur x times in the n trials, for x = 0, 1, 2, …, n. Thus, there are only two possible outcomes. It is conventional to apply the generic labels "success" and "failure" to the two possible outcomes. Discrete Distributions
3. A success can be defined as anything! "the axle failed" could be the definition of success in an experiment testing the strength of truck axles. Examples: 1.A coin flip can be either heads or tails 2.A product is either good or defective Binomial experiments of interest usually involve several repetitions or trials of the same basic experiments. These trials must satisfy the conditions outlined below: Discrete Distributions
4. When can we use it? Condition for use: Each repetition of the experiment (trial) can result in only one of two possible outcomes, a success or failure. See example BD1. The probability of a success, p, and failure (1-p) is constant from trial to trial. All trials are statistically independent; i.e. No trial outcome has any effect on any other trial outcome. The number of trials, n, is specified constant (stated before the experiment begins). Discrete Distributions
5. Example 1: binomial distribution: A coin flip results in a heads or tails A product is defective or not A customer is male or female Example 4: binomial distribution: Say we perform an experiment – flip a coin 10 times and observe the result. A successful flip is designated as heads. Assuming the coin is fair, the probability of success is .5 for each of the 10 trials, thus each trial is independent. We want to know the number of successes (heads) in 10 trials. The random variable that records the number of successes is called the binomial random variable. Random variable, x, the number of successes that occur in the n = 10 trials. Discrete Distributions
6. Do the 4 conditions of use hold? We are not concerned with sequence with the binomial. We could have several successes or failures in a row. Since each experiment is independent, sequence is not important. The binomial random variable counts the number of successes in n trials of the binomial experiment. By definition, this is a discrete random variable. Binomial Random Variable
7. Calculating the Binomial Probability • Rather than working out the binomial probabilities from scratch each time, we can use a general formula. • Say random variable "X" is the number of successes that occur in "n" trials. • Say p = probability of success in any trial • Say q = probability failure in any trial where q = (1 – p) • In general, The binomial probability is calculated by: Where x = 0, 1, 2, …, n
8. Example: For n = 3 Calculating the Binomial Probability
9. Discrete Distributions Each pair of values (n, p) determines a distinct binomial distribution. Two parameters: n and p where a parameter is: Any symbol defined in the functions basic mathematical form such that the user of that function may specify the value of the parameter.
10. Developing the Binomial Probability Distribution P(S2|S1 P(SSS)=p3 S3 P(S3|S2,S1) P(S3)=p S2 P(S2)=p P(S2|S1) S1 P(F3)=1-p F3 P(SSF)=p2(1-p) P(F3|S2,S1) P(S3|F2,S1) S3 P(S3)=p P(SFS)=p(1-p)p P(F2|S1) P(S1)=p P(F2)=1-p P(F3)=1-p Since the outcome of each trial is independent of the previous outcomes, we can replace the conditional probabilities with the marginal probabilities. F2 P(F3|F2,S1) F3 P(SFF)=p(1-p)2 S3 P(S3|S2,F1) P(FSS)=(1-p)p2 P(S3)=p S2 P(S2)=p P(F1)=1-p P(S2|F1) P(F3)=1-p P(F3|S2,F1) F3 P(FSF)=(1-p)P(1-p) S3 P(S3|F2,F1) P(FFS)=(1-p)2p F1 P(S3)=p P(F2|F1) P(F2)=1-p F2 P(F3)=1-p P(F3|F2,F1) F3 P(FFF)=(1-p)3
11. Let X be the number of successes in three trials. Then, P(SSS)=p3 SSS P(SSF)=p2(1-p) SS P(X = 3) = p3 X = 3 X =2 X = 1 X = 0 S S P(SFS)=p(1-p)p P(X = 2) = 3p2(1-p) P(SFF)=p(1-p)2 P(X = 1) = 3p(1-p)2 P(FSS)=(1-p)p2 SS P(X = 0) = (1- p)3 P(FSF)=(1-p)P(1-p) P(FFS)=(1-p)2p This multiplier is calculated in the following formula P(FFF)=(1-p)3
12. 5% of a catalytic converter production run is defective. A sample of 3 converter s is drawn. Find the probability distribution of the number of defectives. Solution A converter can be either defective or good. There is a fixed finite number of trials (n=3) We assume the converter state is independent on one another. The probability of a converter being defective does not change from converter to converter (p=.05). Binomial Example The conditions required for the binomial experiment are met
13. Let X be the binomial random variable indicating the number of defectives. Define a “success” as “a converter is found to be defective”. X P(X) 0 .8574 1 .1354 2 .0071 3 .0001
14. Discrete Distributions Example: The quality control department of a manufacturer tested the most recent batch of 1000 catalytic converters produced and found that 50 of them to be defective. Subsequently, an employee unwittingly mixed the defective converters in with the nondefective ones. Of a sample of 3 converters is randomly selected from the mixed batch, what is the probability distribution of the number of defective converters in the sample? Does this situation satisfy the requirements of a binomial experiment? n = 3 trials with 2 possible outcomes (defective or nondefective). Does the probability remain the same for each trial? Why or why not? The probability p of selecting a defective converter does not remain constant for each trial because the probability depends on the results of the previous trial. Thus the trials are not independent.
15. Discrete Distributions The probability of selecting a defective converter on the first trial is 50/1000 = .05. If a defective converter is selected on the first trial, then the probability changes to 49/999 = .049. In practical terms, this violation of the conditions of a binomial experiment is often considered negligible. The difference would be more noticeable if we considered 5 defectives out of a batch of 100.
16. Discrete Distributions If we assume the conditions for a binomial experiment hold, then consider p = .5 for each trial. Let X be the binomial random variable indicating the number o defective converters in the sample of 3. P(X = 0) = p(0) = [3!/0!3!](.05)0(.95)3 = .8574 P(X = 1) = p(1) = [3!/1!2!](.05)1(.95)2 = .1354 P(X = 2) = p(2) = [3!/2!1!](.05)2(.95)1 = .0071 P(X = 3) = p(3) = [3!/3!0!](.05)3(.95)0 = .0001 The resulting probability distribution of the number of defective converters in the sample of 3, is as follows:
17. Discrete Distributions xp(x) 0 .8574 1 .1354 2 .0071 3 .0001
18. Cumulative Binomial Distribution: F(x)= S from k=0 to x: nCx * p k q (n-k) Another way to look at things = cummulative probabilities Say we have a binomial with n = 3 and p = .05 x p(x) 0 .8574 1 .1354 2 .0071 3 .0001
19. Cumulative Binomial Distribution: this could be written in cumulative form: from x = 0 to x = k: x p(x) 0 .8574 1 .9928 2 .9999 3 1.000
20. Cumulative Binomial Distribution: What is the advantage of cummulative? It allows us to find the probability that X will assume some value within a range of values. Example 1: Cumulative: p(2) = p(x<2) – p(x<1) = .9999 - .9928 = .0071 Example 2: Cumulative: Find the probability of at most 3 successes in n=5 trials of a binomial experiment with p = .2. We locate the entry corresponding to k = 3 and p = .2 P(X < 3) = SUM p(x) = p(0) + p(1) + p(2) + p(3) = .993
21. E(X) = µ = np V(X) = s2 = np(1-p) Example 6.10 Records show that 30% of the customers in a shoe store make their payments using a credit card. This morning 20 customers purchased shoes. Use the Cummulative Binomial Distribution Table (A.1 of Appendix) to answer some questions stated in the next slide. Mean and Variance of Binomial Random Variable
22. What is the probability that at least 12 customers used a credit card? This is a binomial experiment with n=20 and p=.30. p k .01……….. 30 0 . . 11 P(At least 12 used credit card) = P(X>=12)=1-P(X<=11) = 1-.995 = .005 .995
23. What is the probability that at least 3 but not more than 6 customers used a credit card? p k .01……….. 30 0 2 . 6 P(3<=X<=6)= P(X=3 or 4 or 5 or 6) =P(X<=6) -P(X<=2) =.608 - .035 = .573 .035 .608
24. What is the expected number of customers who used a credit card? E(X) = np = 20(.30) = 6 Find the probability that exactly 14 customers did not use a credit card. Let Y be the number of customers who did not use a credit card.P(Y=14) = P(X=6) = P(X<=6) - P(x<=5) = .608 - .416 = .192 Find the probability that at least 9 customers did not use a credit card. Let Y be the number of customers who did not use a credit card.P(Y>=9) = P(X<=11) = .995
25. Poisson Distribution • What if we want to know the number of events during a specific time interval or a specified region? • Use the Poisson Distribution. • Examples of Poisson: • Counting the number of phone calls received in a specific period of time • Counting the number of arrivals at a service location in a specific period of time – how many people arrive at a bank • The number of errors a typist makes per page • The number of customers entering a service station per hour
26. Poisson Distribution • Conditions for use: • The number of successes that occur in any interval is independent of the number of successes that occur in any other interval. • The probability that a success will occur in an interval is the same for all intervals of equal size and is proportional to the size of the interval. • The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller. • Example: • The arrival of individual dinners to a restaurant would not fit the Poisson model because dinners usually arrive with companions, violating the independence condition.
27. The Poisson variable indicates the number of successes that occur during a given time interval or in a specific region in a Poisson experiment Probability Distribution of the Poisson Random Variable: Poisson Random Variable • = average number of successes occurring in a specific interval • Must determine an estimate for from historical data (or other source) • No limit to the number of values a Poisson random Variable can assume
28. Poisson Probability Distribution With m = 1 The X axis in Excel Starts with x=1!! 0 1 2 3 4 5
29. Poisson probability distribution with m =2 0 1 2 3 4 5 6 Poisson probability distribution with m =5 0 1 2 3 4 5 6 7 8 9 10 Poisson probability distribution with m =7 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
30. Cars arrive at a tollbooth at a rate of 360 cars per hour. What is the probability that only two cars will arrive during a specified one-minute period? (Use the formula) The probability distribution of arriving cars for any one-minute period is Poisson with µ = 360/60 = 6 cars per minute. Let X denote the number of arrivals during a one-minute period. Poisson Example
31. What is the probability that only two cars will arrive during a specified one-minute period? (Use table 2, Appendix B.) P(X = 2) = P(X<=2) - P(X<=1) = .062 - .017 = .045
32. What is the probability that at least four cars will arrive during a one-minute period? Use Cummulative Poisson Table (Table A.2 , Appendix) P(X>=4) = 1 - P(X<=3) = 1 - .151 = .849
33. When n is very large, binomial probability table may not be available. If p is very small (p< .05), we can approximate the binomial probabilities using the Poisson distribution. Use = np and make the following approximation: Poisson Approximation of the Binomial With parameters n and p With m = np
34. Example of Poisson Example: Poisson Approximation of the Binomial A warehouse engages in acceptance sampling to determine if it will accept or reject incoming lots of designer sunglasses, some of which invariably are defective. Specifically, the warehouse has a policy of examining a sample of 50 sunglasses from each lot and accepting the lot only if the sample contains no more than 2 defective pairs. What is the probability of a lot's being accepted if, in fact, 2% of the sunglasses in the lot are defective? This is a binomial experiment with n = 50 and p = .02. Our binomial tables include n values up to 25, but since p < .05 and the expected number of defective sunglasses in the sample is np = 50(.02) = 1, the required probability can be approximated by using the Poisson distribution with μ = 1. From Table A.1, we find that the probability that a sample contains at most 2 defective pairs o sunglasses is .920.
35. What is the probability of a lot being accepted if, in fact, 2% of the sunglasses are defective? Solution This is a binomial experiment with n = 50, p = .02. Tables for n = 50 are not available; p<.05; thus, a Poisson approximation is appropriate [ = (50)(.02) =1] P(Xpoisson<=2) = .920 (true binomial probability = .922) Poisson Example
36. Example of Poisson So how well does the Poisson approximate the Binomial? Consider the following table: x Binomial (n = 50, p = .02) Poisson (μ = np = 1) 0 .364 .368 1 .372 .368 2 .186 .184 3 .061 .061 4 .014 .015 5 .003 .003 6 .000 .001
37. Example of Poisson • A tollbooth operator has observed that cars arrive randomly at an average rate of 360 cars per hour. • Using the formula, calculate the probability that only 2 cars will arrive during a specified 1 minute period. • Using Table A.2 on page 360, find the probability that only 2 cars will arrive during a specified 1 minute period. • Using Table A.2 on page 360, find the probability that at least 4 cars will arrive during a specified 1 minute period. • P(X=2) = [(e-6)(62)] / 2! = (.00248)(36) / 2 = .0446 • P(X=2) = P(X < 2) - P(X < 1) = .062 = .017 = .045 • P(X > 4) = 1 - P(X < 3) = 1 - .151 = .849
38. Example of Poisson What if we wanted to know the probability of a small number of occurrences in a large number of trials and a very small probability of success? We use Poisson as a good approximation of the answer. When trying to decide between the binomial and the Poisson, use the following guidelines: n > 20 n > 100 or p < .05 or np < 10
39. Hypergeometric Distribution What about sampling without replacement? What is likely to happen to the probability of success? Probability of success is not constant from trial to trial. We have a finite set on N things, and "a" of them possess a property of interest. Thus, there are "a" successes in N things. Let X be the number of successes that occur in a sample, without replacement of "n" things from a total of N things. This is the hypergeometric distribution: P(x) = (aCx)(N-a C n-x) x = 0,1,2…. NCn
40. Binomial Approximation of Hypergeometric Distribution If N is large, then the probability of success will remain approximately constant from one trial to another. When can we use the binomial distribution as an approximation of the hypergeometric distribution when: N/10 > n
41. What if we want to perform a single experiment and there are only 2 possible outcomes? We use a special case of the binomial distribution where n=1: P(x)= 1Cx * px * (1-p) 1-x x =0,1 which yields p(0)= 1-p = px * (1-p) 1-x x=0,1 p(1)= p In this form, the binomial is referred to as the Bernoulli distribution. Bernoulli Distribution
42. Now, instead of being concerned with the number of successes in n Bernoulli trials, let’s consider the number of Bernoulli trial failures that would have to be performed prior to achieving the 1st success. In this case, we use the geometric distribution, where X is the random variable representing the number of failures before the 1st success. Mathematical form of geometric distribution: P(x)=p * (1-p)x x = 0,1,2… Geometric Distribution
43. What if we wanted to know the number of failures, x, that occur before the rth success (r = 1,2….)? In this case, we use the negative binomial distribution. The number of statistically independent trials that will be performed before the r success = X + r The previous r – 1 successes and the X failures can occur in any order during the X + r – 1 trials. Negative binomial distribution – mathematical form: P(x) = r+x-1 Cx * pr * (1-p)x x= 0,1,2….. Negative Binomial Distribution
44. Problem Solving • A Suggestion for Solving Problems Involving Discrete Random Variables • An approach: • Understand the random variable under consideration and determine if the random variable fits the description and satisfies the assumptions associated with any of the 6 random variables presented in Table 3.3. • If you find a match in Table 3.3 use the software for that distribution. • If none of the 6 random variables in table 3.3 match the random variable associated with your problem, use the sample space method
45. Discrete Bivariate Probability Distribution Functions • The first part of the chapter considered only univariate probability distribution functions.~ One variable is allowed to change. • What if two or more variables change? • When considering situations where two or more variables change, the definitions of • sample space • numerically valued functions • random variable • still apply.
46. To consider the relationship between two random variables, the bivariate (or joint) distribution is needed. Bivariate probability distribution The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x, Y = y) Bivariate Distributions
47. Discrete Bivariate Probability Distribution Functions Example Consider the following real estate data: We want to know how the size of the house varies with the cost.
48. Discrete Bivariate Probability Distribution Functions
49. Discrete Bivariate Probability Distribution Functions What is the next step? Construct frequency distributions: Frequency distribution of house size:
50. Discrete Bivariate Probability Distribution Functions Frequency distribution of selling price:
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# Inverse 3×3 matrix online
Calcula la matriz inversa 3x3 online step by step with our calculator that will allow you to find the inverse of a matrix instantly.
Sólo tienes que rellenar correctamente todos los valores que componen la matriz 3x3 y pulsar sobre el botón de calcular para obtener el resultado paso a paso cómo se ha calculado su matriz inversa a partir de la fórmula que veremos más adelante. Descubre cómo se calcula la inversa de una matriz 3x3.
## Fórmula para calcular la matriz inversa 3x3 y nxn
Para calcular la inversa de una matriz 3x3, 4x4 o del tipo nxn, sólo tienes que aplicar la fórmula que tienes encima de estas líneas.
In the formula to calculate the inverse matrix we can see that we have to calculate:
• The determinant of the starting matrix nxn
• Transpose the attached matrix
Cuando tengamos ambos datos, hacemos la división correspondiente y obtendremos la matriz inversa 3x3 o la del tipo nxn que busquemos.
## Ejemplos de matriz inversa 3x3
Let's take a practical example cómo calculamos la matriz inversa 3x3 that we have just put.
### 1 - Calcular determinante de matriz 3x3
As we have said in the previous point, the first thing we must do in order to sacar la matriz inversa 3x3 is calculate the determinant of the matrix.
When we have the result of the determinant, two things can happen:
• Si el valor del determinante 3x3 is equal to zero, then there is no inverse matrix.
• If the determinant has a non-zero heat, we can calculate its inverse.
### 2 - Calcular matriz adjunta
Now let's calculate the attached matrix and to do so, we must eliminate the row and the column in which each element of the matrix is located. With the remaining elements we will form a determinante 2x2 to be solved. This must be done with each and every one of the elements that form the matrix.
In addition, you must remember that the attached matrix has a specific order of symbols associated with each of the positions of the matrix as shown in the following figure:
Based on our example, it remains that the step by step to calculate the attached matrix is as follows:
Once we group the results obtained by calculating the adjoint determinant of each element of the matrix, we are left with the following matriz adjunta 3x3 is as follows:
### 3 - Transponemos la matriz adjunta
This step is very fast, because in order to make the transpose of the adjoining matrixIn the table below, we only have to exchange rows for columns. This leaves us with the following result:
### 4 - Aplicamos la fórmula para sacar la matriz inversa 3x3
The fourth and final step is to apply the fórmula para calcular la matriz inversa 3x3 we saw above.
We copy it, compile the results obtained in the previous operations and solve.
## Cómo sacar la matriz inversa 3x3 en Excel
If you want to make your own calculadora de matriz inversa 3x3 con ExcelYou can get it in a very simple way if you follow these steps:
1. Abre una nueva hoja de cálculo y escribe en ella la matriz 3x3 para la cual quieres calculate its inverse.
2. Cuando la tengas escriba, busca un rango 3x3 de celdas vacías y selecciónalo. Cuando lo tengas, escribe la siguiente función de Excel que te permitirá calcular la matriz inversa 3x3 (recuerda que entre los paréntesis irá el rango de celdas 3x3 en los que has escrito tu matriz):
=MINVERSE()
1. Now press simultaneously the CTRL + Shift keys on your keyboard and without releasing, press the ENTER key to accept the formula. If you have done it right, Excel calculará la matriz inversa 3x3 automatically.
If you have any doubts about the procedure to follow, we recommend you to watch the video and you will find out. If you still have any doubts, leave us a comment and we will try to help you as soon as possible.
### 1 thought on “Matriz inversa 3×3 online”
1. WILFRIDO RAMIREZ
The video on how to calculate the inverse matrix in excel is missing or not available.
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Question Video: Comparing the Distance and Displacement of a Body Moving in a Given Path | Nagwa Question Video: Comparing the Distance and Displacement of a Body Moving in a Given Path | Nagwa
# Question Video: Comparing the Distance and Displacement of a Body Moving in a Given Path Mathematics
According to the figure, a body moved from π΄ to π΅ along the line segment π΄π΅, and then it moved to πΆ along π΅πΆ. Finally, it moved to π· along πΆπ· and stopped there. Find the distance covered by the body πβ and the magnitude of its displacement πβ.
02:46
### Video Transcript
According to the figure, a body moved from π΄ to π΅ along the line segment π΄π΅, and then it moved to πΆ along π΅πΆ. Finally, it moved to π· along πΆπ· and stopped there. Find the distance covered by the body π one and the magnitude of its displacement π two.
Looking at the figure, weβre told that a body begins at point π΄, right here, and then follows the line segment π΄π΅ to point π΅ then moves to point πΆ following this path, finally moving along line segment πΆπ· to end up at point π·. Given this motion, we want to calculate the distance the body has covered, we call that π one, and the magnitude of its displacement π two.
Clearing some space, letβs first work on solving for the distance our body travels π one. We can recall that distance in general is equal to the total path length followed by some body as it moves from one location to another. In our case, our body moved from point π΄ to point π· along the path shown in orange. We see that that involves 6.6 centimeters of travel from point π΄ to point π΅, 8.8 centimeters from π΅ to πΈ, 16.4 centimeters from πΈ to πΆ, and on the last leg of the journey 12.3 centimeters. π one then equals the sum of these four distances. Adding them all up gives a result of 44.1 centimeters. This is the distance our body traveled.
And now letβs consider its displacement magnitude π two. Displacement is different from distance in that displacement only takes into account the start point and end point for some body. In our situation, our body begins at point π΄ and it ends up at point π·. So this straight pink line connecting these two points represents the displacement magnitude π two. We see that the length of this line segment can be divided up into two parts: one part right here and then the second part here. Each of these is a hypotenuse of a right triangle. This means we can use the Pythagorean theorem to solve for these lengths.
If we call the length of the first hypotenuse π one and that of the second π two, then we can say that the displacement magnitude π two is equal to their sum and that π one and π two are defined this way. π one equals the square root of 6.6 centimeters quantity squared plus 8.8 centimeters quantity squared. And π two equals the square root of 16.4 centimeters quantity squared plus 12.3 centimeters quantity squared. When we enter these expressions on our calculator, we find that π one is equal to exactly 11 centimeters, while π two is 20.5 centimeters. Adding these together gives us a result of 31.5 centimeters. This is the displacement magnitude of our body.
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# Finding the Median Value
The median is the middle value of a list of data values. To find the median of a list of data values:
1. Place the numbers in increasing or decreasing order by value.
2. Identify the number that appears in the middle position in the sorted list.
## Example 1
Find the median of the list
8, 2, 15, 3 and 6
Solution: Start by sorting these scores into increasing order. The sorted list is:
2, 3, 6, 8, 15.
As there are five values in the list, the middle value is the third one in the list. So, the median is 6.
## Example 2
Find the median of the list
15, 7, 8, 11, 23, 25, 17, 12, 15, 16, 19, 24, 21, 22, 31, 36, 47
Solution: Start by sorting these scores into increasing order. The sorted list is:
7, 8, 11, 12, 15, 15, 16, 17, 19, 21, 22, 23, 24, 25, 31, 36, 47
Now identify the middle value. There are 17 numbers in this list, so the median is the 9th value, or 19.
The repetition of values in the list (e.g. 15) does not affect the way we calculate the median. We simply include the repeated values in our sorted list.
## What About Lists With Even Numbers of Elements?
If a list has an even number of elements, we can still find the median, even though we can't identify the middle element exactly.
When a list has an even number of elements, it has two middle numbers. The median is defined to be the average of these values (add the two numbers together and divide by 2).
### Example 3
Find the median of the following list of values:
15, 7, 8, 11, 23, 25, 17, 12, 15, 16, 19, 24, 21, 22, 31, 36, 47, 54
This list has an even number (18) of entries, so it has a pair of middle values. Let's write the list in increasing order:
7, 8, 11, 12, 15, 15, 16, 17, 19, 21, 22, 23, 24, 25, 31, 36, 47, 54
The two middle entries are the 9th and 10th entries (19 and 21). So, to find the median of this list, we add them together and divide by 2 to give
$\dfrac{19 + 21}{2} = \dfrac{40}{2} = 20$.
The median of our list is $20$.
## Finding the Median of Longer Lists
Often, the most challenging part of finding the median is identifying the middle value in the list. However, there's an easy trick to get you out of trouble. Simply count the number of elements in your list, add one, and then divide the result by 2.
For example, if a list has $253$ elements, the middle element will be the $\dfrac{253 + 1}{2} = \dfrac{254}{2} = 127$th element in the list after it has been sorted into increasing order.
But, what if the list has an even number of elements? That's OK. Let's suppose we have a list with $366$ elements. When we apply our little trick, we end up with the middle element being element number $\dfrac{366 + 1}{2} = \dfrac{367}{2} = 183.5$, which doesn't exist. However, 183.5 is halfway between 183 and 184, and we know that we have to find the average of the corresponding list elements to find our median. So, to find the median of the numbers in this list, we sort the list into increasing or decreasing order, identify the 183rd and 184th elements, add them together and divide the result by 2.
## When is the Median Useful?
Mrs Fitzsimmons asks Sam's class to evaluate her teaching of Shakespeare. She receives an average score of 5.36 out of 10 from her 11 students, which makes her pretty happy until she decides to look at the data more closely. Then she sees that the 11 scores were 1,3,3,3,3,3,5,9,9,10,10. The median score of this list is a not very impressive 3 out of 10. This tells her that at least half of her students were not very happy with her Shakespeare teaching. The average hides this fact because the unexpected 9s and 10s (possibly from students who were trying to get a higher mark on their exam, or maybe they really liked Shakespeare) skewed the data, making the average higher than it should have been. In this case, the median is a much better choice for data analysis.
If you've ever looked at the property pages of a newspaper, you might have noticed that median house prices are more commonly reported than average house prices. This is because there are always a few house prices that are completely over the top, and which could over-inflate the apparent value of property if the mean was used as the reported measure.
The median is a better choice than the mean when you are looking for a more representative value for data sets that contain extreme values.
### Description
This chapter series is on Data and is suitable for Year 10 or higher students, topics include
• Accuracy and Precision
• Calculating Means From Frequency Tables
• Correlation
• Cumulative Tables and Graphs
• Discrete and Continuous Data
• Finding the Mean
• Finding the Median
• FindingtheMode
• Formulas for Standard Deviation
• Grouped Frequency Distribution
• Normal Distribution
• Outliers
• Quartiles
• Quincunx
• Quincunx Explained
• Range (Statistics)
• Skewed Data
• Standard Deviation and Variance
• Standard Normal Table
• Univariate and Bivariate Data
• What is Data
### Audience
Year 10 or higher students, some chapters suitable for students in Year 8 or higher
### Learning Objectives
Learn about topics related to "Data"
Author: Subject Coach
You must be logged in as Student to ask a Question.
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# Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.1 Questions and Answers.
## Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1
Question 1.
Verify whether the following sequences are G.P. If so, write tn.
(i) 2, 6, 18, 54, ……
(ii) 1, -5, 25, -125, …….
(iii) $$\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots$$
(iv) 3, 4, 5, 6, ……
(v) 7, 14, 21, 28, …..
Solution:
(i) 2, 6, 18, 54, …….
t1 = 2, t2 = 6, t3 = 18, t4 = 54, …..
Here, $$\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=3$$
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 2, r = 3
tn= arn-1
∴ tn = 2(3n-1)
(ii) 1, -5, 25, -125, ……
t1 = 1, t2 = -5, t3 = 25, t4 = -125, …..
Here, $$\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=-5$$
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 1, r = -5
tn = arn-1
∴ tn = (-5)n-1
(iii) $$\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots$$
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
(iv) 3, 4, 5, 6,……
t1 = 3, t2 = 4, t3 = 5, t4 = 6, …..
Here, $$\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{4}{3}, \frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}=\frac{5}{4}, \frac{\mathrm{t}_{4}}{\mathrm{t}_{3}}=\frac{6}{5}$$
Since, $$\frac{t_{2}}{t_{1}} \neq \frac{t_{3}}{t_{2}} \neq \frac{t_{4}}{t_{3}}$$
∴ the given sequence is not a geometric progression.
(v) 7, 14, 21, 28, …..
t1 = 7, t2 = 14, t3 = 21, t4 = 28, …..
Here, $$\frac{t_{2}}{t_{1}}=2, \frac{t_{3}}{t_{2}}=\frac{3}{2}, \frac{t_{4}}{t_{3}}=\frac{4}{3}$$
Since, $$\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}} \neq \frac{\mathrm{t}_{3}}{\mathrm{t}_{2}} \neq \frac{\mathrm{t}_{4}}{\mathrm{t}_{3}}$$
∴ the given sequence is not a geometric progression.
Question 2.
For the G.P.,
(i) if r = $$\frac{1}{3}$$, a = 9, find t7.
(ii) if a = $$\frac{7}{243}$$, r = $$\frac{1}{3}$$, find t3.
(iii) if a = 7, r = -3, find t6.
(iv) if a = $$\frac{2}{3}$$, t6 = 162, find r.
Solution:
Question 3.
Which term of the G. P. 5, 25, 125, 625, ….. is 510?
Solution:
Question 4.
For what values of x, $$\frac{4}{3}$$, x, $$\frac{4}{27}$$ are in G. P.?
Solution:
Question 5.
If for a sequence, $$t_{n}=\frac{5^{n-3}}{2^{n-3}}$$, show that the sequence is a G. P. Find its first term and the common ratio.
Solution:
The sequence (tn) is a G.P., if $$\frac{t_{n}}{t_{n-1}}$$ = constant, for all n ∈ N
∴ the sequence is a G. P. with common ratio $$\frac{5}{2}$$
First term, t1 = $$\frac{5^{\mathrm{l}-3}}{2^{1-3}}=\frac{2^{2}}{5^{2}}=\frac{4}{25}$$
Question 6.
Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.
Solution:
Let the three numbers in G. P. be $$\frac{a}{\mathrm{r}}$$, a, ar.
According to the first condition,
∴ the three numbers are 12, 6, 3 or 3, 6, 12.
Check:
First condition:
12, 6, 3 are in G.P. with r = $$\frac{1}{2}$$
12 + 6 + 3 = 21
Second condition:
122 + 62 + 32 = 144 + 36 + 9 = 189
Thus, both the conditions are satisfied.
Question 7.
Find four numbers in G. P. such that sum of the middle two numbers is $$\frac{10}{3}$$ and their product is 1.
Solution:
Let the four numbers in G.P. be $$\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$$.
According to the second condition,
$$\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1$$
∴ a4 = 1
∴ a = 1
According to the first condition,
Question 8.
Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
Solution:
Let the five numbers in G. P. be
$$\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}$$
According to the given conditions,
When a = 4, r = -2
$$\frac{a}{r^{2}}$$ = 1, $$\frac{a}{r}$$ = -2, a = 4, ar = -8, ar2 = 16
∴ the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.
Question 9.
The fifth term of a G. P. is x, eighth term of the G. P. is y and eleventh term of the G. P. is z. Verify whether y2 = xz.
Solution:
Question 10.
If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G.P.
Solution:
p, q, r, s are in G.P.
∴ p + q, q + r, r + s are in G.P.
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# 3.06 Unit rate
Lesson
### What is unit rate?
A rate is a measure of how quickly one measurement changes with respect to another. Some commonly used rates in our everyday lives are speed, which measures distance per time, and the price of food, which is often measured in dollars per pound.
Notice how both of these examples combines two different units into a single compound unit. We can write these compound units using a slash ( / ) between the different units, so "meters per second" becomes "m/s" and "dollars per pound" becomes "$/lb". This compound unit represents the division of one measurement by another to get a rate. When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates Let's have a look at an example. #### Exploration Consider an Olympic sprinter who runs$100$100 meters in$10$10 seconds. How fast does he run? We can find how far the sprinter runs in$1$1 second by dividing the$100$100 meters evenly between the$10$10 seconds. This calculation tells us that the sprinter runs$10$10 meters in one second. We can write this as a unit rate for the sprinter's speed in meters per second using the compound unit m/s to give us: Sprinter's speed$=$=$10$10 m/s Now let's try a more direct method to finding the sprinter's speed. Since the sprinter runs$100$100 meters in$10$10 seconds we can say that he runs at a rate of$100$100 meters per$10$10 seconds. Writing this as a fraction gives us: Sprinter's speed$=$=$100$100m/$10$10s$=$=$\frac{100}{10}$10010 m/s$=$=$10$10 m/s After some simplifying we find that the speed of the sprinter matches that from the previous method. Notice that we were able to separate the numbers and the units into separate fractions, this is the core concept we use for turning fractions of measurements into rates. Did you know? Not all compound units are written using a slash and instead use the letter "p" to represent "per". For example, "beats per minute" uses the compound unit bpm and "frames per second" uses fps. ### Finding unit rates Rates have two components, the numeric value and the compound unit. The compound unit tells us which units are being measured and the numeric value tells us how quickly the numerator unit changes with respect to the denominator unit. When constructing a rate we usually start with just a fraction of measurements. For example, let's find the speed of a car that travels$180$180 miles in$3$3 hours. We start by setting up the fraction as distance per time, written: Speed$=$=$180$180mi/$3$3hr Then we can separate the fraction into its numeric value and its compound unit. This gives us: Numeric value Compound unit$=$=$\frac{180}{3}$1803$=$=$60$60$=$= mi/hr We can then combine them again to get the unit rate which is: Speed$=$=$60$60 mi/hr Whenever we can, simplify the fraction to get the unit rate. This is much nicer to work with as we can now say that the car travels$60$60 miles per$1$1 hour, rather than$180$180 miles per$3$3 hours. ### Applying unit rate Now that we know how to make unit rates, it's time to use them. Rates are very similar to ratios in that we can use them to calculate how much one measurement changes based on the change in another. #### Worked examples ##### question 1 Returning to our sprinter, we found that they could run at a speed of$10$10 m/s. Assuming that they can maintain this speed, how far will the sprinter run in$15$15 seconds? Think: One way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by$15$15. This will give us: Do: Speed$=$=$10$10 m/s$\times$×$\frac{15}{15}$1515$=$=$150$150m/$15$15s By turning the rate back into a fraction we can see that the sprinter will run$150$150 meters in$15$15 seconds. Reflect: Another way to solve this problem is to apply the rate directly to the question. We can do this by multiplying the time by the rate. This gives us: Distance$=$=$15$15$\times$×$10$10m/s$=$=$\left(15\times10\right)$(15×10) m$\times$×s/s$=$=$150$150m Notice that the units for seconds from the time canceled with the units for second in the compound unit leaving only meters as the unit for distance. ##### question 2 We can also ask the similar question, how long will it take for the sprinter to run$220$220 meters? Think: Again, one way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by$22$22. This gives us: Do: Speed$=$=$10$10 m/s$\times$×$\frac{22}{22}$2222$=$=$220$220m/$22$22s By turning the rate back into a fraction we can see that the sprinter will take$22$22 seconds to run$220$220 meters. Reflect: Another way to solve this problem is to apply the rate directly to the question. We can do this by dividing the distance by the rate. This gives us: Time$=$=$220$220$\div$÷$10$10 m/s$=$=$\frac{220}{10}$22010 s$\times$×m/m =$22$22s This time we divided by the rate so that the compound unit would be flipped and the meters units would cancel out to leave only seconds as the unit for time. Careful! A rate of$10$10 meters per second ($10$10 m/s) is not the same as a rate of$10$10 seconds per meter ($10$10 s/m). In fact,$10$10 m/s$=$=$\frac{1}{10}$110 s/m. When we flip the compound unit we also need to take the reciprocal of the numeric value. ### Converting units When applying rates it's important to make sure that we are applying the right one. #### Worked example ##### question 3 Consider the car from before that traveled at a speed of$60$60 mi/hr. How many miles will the car travel in$7$7 minutes? Think: Before we use one of the methods we learned for applying rates we should first notice that the units in the question don't quite match up with our unit rate. Specifically, the question is asking for minutes as the units for time instead of hours. We can fix this by converting hours into minutes for our unit rate. Using the fact that$1$1 hour$=$=$60$60 minutes we can convert our speed from mi/hr to mi/min like so: Do: Speed$=$=$60$60mi/hr$=$=$60$60mi/$60$60min$=$=$\frac{60}{60}$6060 mi/min$=$=$1$1 mi/min Now that we have a speed with the appropriate units we can apply the unit rate to the question to find how far the car will travel in$7$7 minutes: Distance$=$=$7$7min$\times$×$1$1 mi/min$=$=$\left(7\times1\right)$(7×1) mi$\times$×min/min$=$=$7$7mi Now that we have some experience with this type of question you can try one yourself. #### Practice question ##### question 4 A car travels$320$320 km in$4$4 hours. 1. Complete the table of values. Time taken (hours) Distance traveled (kilometers)$4$4$2$2$1$1$320$320$\editable{}\editable{}$2. What is the speed of the car in kilometers per hour? ##### question 5 Adam really likes apples and eats them at a rate of$4$4 per day. 1. How many apples does Adam eat in one week? 2. If Adam buys$44$44 apples how many days will this last him? ##### question 6 A cyclist travels$70$70 meters per$10$10 seconds. 1. How many groups of$10$10 seconds are in$1\$1 minute?
2. What is the speed of the cyclist in meters per minute?
### Outcomes
#### 6.12b
Determine the unit rate of a proportional relationship and use it to find a missing value in a ratio table
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# SAT Math : How to multiply exponents
## Example Questions
← Previous 1 3 4 5 6 7
### Example Question #1 : Exponents
If (300)(400) = 12 * 10n, n =
12
2
3
4
7
4
Explanation:
(300)(400) = 120,000 or 12 * 104.
### Example Question #2 : Exponents
(2x103) x (2x106) x (2x1012) = ?
6x1021
6x1023
8x1021
8x1023
8x1021
Explanation:
The three two multiply to become 8 and the powers of ten can be added to become 1021.
### Example Question #1 : Exponents
If 3x = 27, then 22x = ?
64
32
3
8
9
64
Explanation:
1. Solve for x in 3x = 27. x = 3 because 3 * 3 * 3 = 27.
2. Since x = 3, one can substitute x for 3 in 22xÂ
3. Now, the expression is 22*3
4. This expression can be interpreted as 22 * 22 * 22. Since 22 = 4, the expression can be simplified to become 4 * 4 * 4 = 64.
5. You can also multiply the powers to simplify the expression. When you multiply the powers, you get 26, or 2 * 2 * 2 * 2 * 2 * 2
6. 26Â = 64.
### Example Question #4 : Exponents
Find the value of x such that:
8x-3Â = 164-x
4
11/3
19/4
7/2
25/7
25/7
Explanation:
In order to solve this equation, we first need to find a common base for the exponents. We know that 23 = 8 and 24 = 16, so it makes sense to use 2 as a common base, and then rewrite each side of the equation as a power of 2.
8x-3 = (23)x-3
We need to remember our property of exponents which says that (ab)c = abc.
Thus (23)x-3 = 23(x-3) = 23x - 9.
We can do the same thing with 164-x.
164-x = (24)4-x = 24(4-x)Â = 216-4x.
So now our equation becomes
23x - 9Â =Â 216-4x
In order to solve this equation, the exponents have to be equal.Â
3x - 9 = 16 - 4x
7x - 9 = 16
7x = 25
Divide by 7.
x = 25/7.
### Example Question #5 : Exponents
Which of the following is equal to 410 + 410 + 410 + 410 + 411?
240
223
250
215
260
223
Explanation:
We can start by rewriting 411 as 4 * 410. This will allow us to collect the like terms 410 into a single term.
410Â + 410Â + 410Â + 410Â + 411
= 410Â + 410Â + 410Â + 410Â + 4 * 410
= 8 * 410
Because the answer choices are written with a base of 2, we need to rewrite 8 and 4 using bases of two. Remember that 8 = 23, and 4 =Â 22.
8 * 410
= (23)(22)10
We also need to use the property of exponents that (ab)c = abc. We can rewrite (22)10 as 22x10 = 220.
(23)(22)10
= (23)(220)
Finally, we must use the property of exponents that ab * ac = ab+c.
(23)(220) = 223
### Example Question #1 : Exponents
If 3Â + 3n+3 = 81, what is 3n+2Â ?
3
81
9
26
18
26
Explanation:
3Â + 3n+3Â = 81
In this equation, there is a common factor of 3, which can be factored out.
Thus, 3(1 + 3n+2) = 81
Note: when 3 is factored out of 3n+3, the result is 3n+2 because (3n+3 = 31 * 3n+2). Remember that exponents are added when common bases are multiplied.  Also remember that 3 = 31.
3(1 + 3n+2) = 81
(1 + 3n+2) = 27
3n+2Â = 26
Note: do not solve for n individually. Â But rather seek to solve what the problem asks for, namely 3n+2. Â
### Example Question #7 : Exponents
If f(x) = (2 – x)(x/3), and 4n = f(10), then what is the value of n?
0
–2
5
2
–5
5
Explanation:
First, let us use the definiton of f(x) to find f(10).
f(x) = (2 – x)(x/3)
f(10) = (2 – 10)(10/3)
= (–8)(10/3)
In order to evaluate the above expression, we can make use of the property of exponents that states that abc = (ab)c = (ac)b.
(–8)(10/3) = (–8)10(1/3) = ((–8)(1/3))10.
(–8)(1/3) requires us to take the cube root of –8. The cube root of –8 is –2, because (–2)3 = –8.
Let's go back to simplifying ((–8)(1/3))10.
((–8)(1/3))10 = (–2)10 = f(10)
We are asked to find n such that 4n = (–2)10. Let's rewrite 4n with a base of –2, because (–2)2 = 4.
4n = ((–2)2)n = (–2)2n = (–2)10
In order to (–2)2n = (–2)10, 2n must equal 10.
2n = 10
Divide both sides by 2.
n = 5.
### Example Question #8 : Exponents
What is the value of n that satisfies the following equation?
2n·4n·8n·16 = 2-n·4-n·8-n
0
–1/3
–2/3
2/3
1/3
–1/3
Explanation:
In order to solve this equation, we are going to need to use a common base. Because 2, 4, 8, and 16 are all powers of 2, we can rewrite both sides of the equation using 2 as a base. Since 22 = 4, 23 = 8, and 24 = 16, we can rewrite the original equation as follows:
2n * 4n * 8n * 16 = 2–n * 4–n * 8–n
2n(22)n(23)n(24) = 2–n(22)–n(23)–n
Now, we will make use of the property of exponents which states that (ab)c = abc.
2n(22n)(23n)(24) = 2–n(2–2n)(2–3n)
Everything is now written as a power of 2. We can next apply the property of exponents which states that abac = ab+c.
2(n+2n+3n+4) = 2(–n + –2n + –3n)
We can now set the exponents equal and solve for n.
n + 2n + 3n + 4 = –n + –2n + –3n
Let's combine the n's on both sides.
6n + 4 = –6n
12n + 4 = 0
Subtract 4 from both sides.
12n = –4
Divide both sides by 12.
n = –4/12 = –1/3
### Example Question #9 : Exponents
If 1252x–4 = 6257–x, then what is the largest prime factor of x?
3
5
2
11
7
2
Explanation:
First, we need to solve 1252x–4 = 6257–x . When solving equations with exponents, we usually want to get a common base. Notice that 125 and 625 both end in five. This means they are divisible by 5, and they could be both be powers of 5. Let's check by writing the first few powers of 5.
51 = 5
52 = 25
53 = 125
54 = 625
We can now see that 125 and 625 are both powers of 5, so let's replace 125 with 53 and 625 with 54.Â
(53)2x–4 = (54)7–x
Next, we need to apply the rule of exponents which states that (ab)c = abc .
53(2x–4) = 54(7–x)
We now have a common base expressed with one exponent on each side. We must set the exponents equal to one another to solve for x.
3(2x – 4) = 4(7 – x)
Distribute the 3 on the left and the 4 on the right.
6x – 12 = 28 – 4x
10x – 12 = 28
10x = 40
Divide by 10 on both sides.
x = 4
However, the question asks us for the largest prime factor of x. The only factors of 4 are 1, 2, and 4. And of these, the only prime factor is 2.Â
### Example Question #10 : Exponents
(x3)2 * x–2 =Â
x–4
x
x6
x4
x2
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Before learning how to add fractions, make sure you first truly understand fractions. Please visit the rules of fractions main page if necessary. Also, you might get hungry during this lesson, so feel free to order pizza now.
Adding fractions is easy when the pieces are the same size. For example...
Because the pieces are the same size, it is fair to say that... 1 piece + 1 piece = 2 pieces. We now know how many pieces we have (thi is the top or "numerator"), we just need to know the total number of pieces that our pizza was cut into (that is the bottom or "denominator").
Adding fractions gets more difficult when the pieces are different sizes:
Because the pieces are different sizes, it is not fair to say that we have 2 pieces this time. Wouldn't you rather have these 2 pieces than the 2 pieces above?
To add fractions, we must first make the pieces the same size! Take a look at how we made that second, bigger piece into 2 smaller pieces so that all the slices are equal...
Because we made the pieces the same size (like in the first example) we can simply add them up. 1 piece + 2 pieces = 3 pieces.
How to Add Fractions: The #1 Rule
You MUST have all the pieces the same size before you add them.
If they are not the same size, make them the same size! If your pieces are not the same size, then somebody is getting ripped off (especially if you are dealing with money instead of pizza)! For more information on how to "make pieces the same size," please view our equivalent fractions page and then practice the skill with our equivalent fractions worksheets.
Determining the Number of Pieces
The left image shows our original pizza, before and after we sliced it up.
The right images shows the pizza sliced into 8 equal pieces.
Remember, the total number of pieces is the denominator (bottom) of the fraction. The number of pieces that you have is the numerator (top) of the fraction.
Case #1: Same Denominators
Case #2: Different (Unlike) Denominators
If the denominators are different, then you must first change the entire fraction into a fraction with the same denominator. That's the key to learning how to add fractions!
Example #1: 1/8 + 2/4 --> change to 1/8 + 4/8 = 5/8
Example #2: 1/2 + 2/8 --> change to 4/8 + 2/8 = 6/8
Return from How to Add Fractions to All Rules of Fractions.
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## Truth Tables
Because complex Boolean statements can get tricky to think about, we can create a truth table to keep track of what truth values for the simple statements make the complex statement true and false
### Truth Table
A table showing what the resulting truth value of a complex statement is for all the possible truth values for the simple statements.
### Example 1
Suppose you’re picking out a new couch, and your significant other says “get a sectional or something with a chaise.”
This is a complex statement made of two simpler conditions: “is a sectional,” and “has a chaise.” For simplicity, let’s use S to designate “is a sectional,” and C to designate “has a chaise.” The condition S is true if the couch is a sectional.
A truth table for this would look like this:
S C S or C
T T T
T F T
F T T
F F F
In the table, T is used for true, and F for false. In the first row, if S is true and C is also true, then the complex statement “S or C” is true. This would be a sectional that also has a chaise, which meets our desire.
Remember also that or in logic is not exclusive; if the couch has both features, it does meet the condition.
To shorthand our notation further, we’re going to introduce some symbols that are commonly used for and, or, and not.
### Symbols
The symbol ⋀ is used for and: A and B is notated AB.
The symbol ⋁ is used for or: A or B is notated A ⋁ B
The symbol ~ is used for not: not A is notated ~A
You can remember the first two symbols by relating them to the shapes for the union and intersection. A B would be the elements that exist in both sets, in A ⋂ B. Likewise, A B would be the elements that exist in either set, in A ⋃ B.
In the previous example, the truth table was really just summarizing what we already know about how the or statement work. The truth tables for the basic and, or, and not statements are shown below.
### Basic Truth Tables
A B A B
T T T
T F F
F T F
F F F
A B A B
T T T
T F T
F T T
F F F
A ~A
T F
F T
Truth tables really become useful when analyzing more complex Boolean statements.
### Example 2
Create a truth table for the statement A ⋀ ~(BC)
It helps to work from the inside out when creating truth tables, and create tables for intermediate operations. We start by listing all the possible truth value combinations for A, B, and C. Notice how the first column contains 4 Ts followed by 4 Fs, the second column contains 2 Ts, 2 Fs, then repeats, and the last column alternates. This pattern ensures that all combinations are considered. Along with those initial values, we’ll list the truth values for the innermost expression, BC.
A B C B ⋁ C T T T T T T F T T F T T T F F F F T T T F T F T F F T T F F F F
Next we can find the negation of BC, working off the BC column we just created.
A B C B ⋁ C ~(B ⋁ C) T T T T F T T F T F T F T T F T F F F T F T T T F F T F T F F F T T F F F F F T
Finally, we find the values of A and ~(BC)
A B C B ⋁ C ~(B ⋁ C) A ⋀ ~(B ⋁ C) T T T T F F T T F T F F T F T T F F T F F F T T F T T T F F F T F T F F F F T T F F F F F F T F
It turns out that this complex expression is only true in one case: if A is true, B is false, and C is false.
When we discussed conditions earlier, we discussed the type where we take an action based on the value of the condition. We are now going to talk about a more general version of a conditional, sometimes called an implication.
### Implications
Implications are logical conditional sentences stating that a statement p, called the antecedent, implies a consequence q.
Implications are commonly written as pq
Implications are similar to the conditional statements we looked at earlier; p → q is typically written as “if p then q,” or “p therefore q.” The difference between implications and conditionals is that conditionals we discussed earlier suggest an action—if the condition is true, then we take some action as a result. Implications are a logical statement that suggest that the consequence must logically follow if the antecedent is true.
### Example 3
The English statement “If it is raining, then there are clouds in the sky” is a logical implication. It is a valid argument because if the antecedent “it is raining” is true, then the consequence “there are clouds in the sky” must also be true.
Notice that the statement tells us nothing of what to expect if it is not raining. If the antecedent is false, then the implication becomes irrelevant.
### Example 4
A friend tells you that “if you upload that picture to Facebook, you’ll lose your job.” There are four possible outcomes:
There is only one possible case where your friend was lying—the first option where you upload the picture and keep your job. In the last two cases, your friend didn’t say anything about what would happen if you didn’t upload the picture, so you can’t conclude their statement is invalid, even if you didn’t upload the picture and still lost your job.
In traditional logic, an implication is considered valid (true) as long as there are no cases in which the antecedent is true and the consequence is false. It is important to keep in mind that symbolic logic cannot capture all the intricacies of the English language.
### Truth Values for Implications
p q p → q T T T T F F F T T F F T
### Example 5
Construct a truth table for the statement (m ⋀ ~p) → r
We start by constructing a truth table for the antecedent.
m p ~p m ⋀ ~p T T F F T F T T F T F F F F T F
Now we can build the truth table for the implication
m p ~p m ⋀ ~p r (m ⋀ ~p) → r T T F F T T T F T T T T F T F F T T F F T F T T T T F F F T T F T T F F F T F F F T F F T F F T
In this case, when m is true, p is false, and r is false, then the antecedent m ⋀ ~p will be true but the consequence false, resulting in a invalid implication; every other case gives a valid implication.
For any implication, there are three related statements, the converse, the inverse, and the contrapositive.
### Related Statements
The original implication is “if p then q”: p q
The converse is “if q then p”: qp
The inverse is “if not p then not q”: ~p → ~q
The contrapositive is “if not q then not p”: ~q → ~p
### Example 6
Consider again the valid implication “If it is raining, then there are clouds in the sky.”
The converse would be “If there are clouds in the sky, it is raining.” This is certainly not always true.
The inverse would be “If it is not raining, then there are not clouds in the sky.” Likewise, this is not always true.
The contrapositive would be “If there are not clouds in the sky, then it is not raining.” This statement is valid, and is equivalent to the original implication.
Looking at truth tables, we can see that the original conditional and the contrapositive are logically equivalent, and that the converse and inverse are logically equivalent.
Implication Converse Inverse Contrapositive
p q pq qp ~p → ~q ~q → ~p
T T T T T T
T F F T T F
F T T F F T
F F T T T T
### Equivalence
A conditional statement and its contrapositive are logically equivalent.
The converse and inverse of a statement are logically equivalent.
## Arguments
A logical argument is a claim that a set of premises support a conclusion. There are two general types of arguments: inductive and deductive arguments.
### Argument types
An inductive argument uses a collection of specific examples as its premises and uses them to propose a general conclusion.
A deductive argument uses a collection of general statements as its premises and uses them to propose a specific situation as the conclusion.
### Example 7
The argument “when I went to the store last week I forgot my purse, and when I went today I forgot my purse. I always forget my purse when I go the store” is an inductive argument.
The premises are:
I forgot my purse last week
I forgot my purse today
The conclusion is:
I always forget my purse
Notice that the premises are specific situations, while the conclusion is a general statement. In this case, this is a fairly weak argument, since it is based on only two instances.
### Example 8
The argument “every day for the past year, a plane flies over my house at 2pm. A plane will fly over my house every day at 2pm” is a stronger inductive argument, since it is based on a larger set of evidence.
### Evaluating inductive arguments
An inductive argument is never able to prove the conclusion true, but it can provide either weak or strong evidence to suggest it may be true.
Many scientific theories, such as the big bang theory, can never be proven. Instead, they are inductive arguments supported by a wide variety of evidence. Usually in science, an idea is considered a hypothesis until it has been well tested, at which point it graduates to being considered a theory. The commonly known scientific theories, like Newton’s theory of gravity, have all stood up to years of testing and evidence, though sometimes they need to be adjusted based on new evidence. For gravity, this happened when Einstein proposed the theory of general relativity.
A deductive argument is more clearly valid or not, which makes them easier to evaluate.
### Evaluating deductive arguments
A deductive argument is considered valid if all the premises are true, and the conclusion follows logically from those premises. In other words, the premises are true, and the conclusion follows necessarily from those premises.
### Example 9
The argument “All cats are mammals and a tiger is a cat, so a tiger is a mammal” is a valid deductive argument.
The premises are:
All cats are mammals
A tiger is a cat
The conclusion is:
A tiger is a mammal
Both the premises are true. To see that the premises must logically lead to the conclusion, one approach would be use a Venn diagram. From the first premise, we can conclude that the set of cats is a subset of the set of mammals. From the second premise, we are told that a tiger lies within the set of cats. From that, we can see in the Venn diagram that the tiger also lies inside the set of mammals, so the conclusion is valid.
### Analyzing Arguments with Venn Diagrams[1]
To analyze an argument with a Venn diagram
1. Draw a Venn diagram based on the premises of the argument
2. If the premises are insufficient to determine what determine the location of an element, indicate that.
3. The argument is valid if it is clear that the conclusion must be true
### Example 10
Premise: All firefighters know CPR
Premise: Jill knows CPR
Conclusion: Jill is a firefighter
From the first premise, we know that firefighters all lie inside the set of those who know CPR. From the second premise, we know that Jill is a member of that larger set, but we do not have enough information to know if she also is a member of the smaller subset that is firefighters.
Since the conclusion does not necessarily follow from the premises, this is an invalid argument, regardless of whether Jill actually is a firefighter.
It is important to note that whether or not Jill is actually a firefighter is not important in evaluating the validity of the argument; we are only concerned with whether the premises are enough to prove the conclusion.
In addition to these categorical style premises of the form “all ___,” “some ____,” and “no ____,” it is also common to see premises that are implications.
### Example 11
Premise: If you live in Seattle, you live in Washington.
Premise: Marcus does not live in Seattle
Conclusion: Marcus does not live in Washington
From the first premise, we know that the set of people who live in Seattle is inside the set of those who live in Washington. From the second premise, we know that Marcus does not lie in the Seattle set, but we have insufficient information to know whether or not Marcus lives in Washington or not. This is an invalid argument.
### Example 12
Consider the argument “You are a married man, so you must have a wife.”
This is an invalid argument, since there are, at least in parts of the world, men who are married to other men, so the premise not insufficient to imply the conclusion.
Some arguments are better analyzed using truth tables.
### Example 13
Consider the argument:
Premise: If you bought bread, then you went to the store
Conclusion: You went to the store
While this example is hopefully fairly obviously a valid argument, we can analyze it using a truth table by representing each of the premises symbolically. We can then look at the implication that the premises together imply the conclusion. If the truth table is a tautology (always true), then the argument is valid.
We’ll get B represent “you bought bread” and S represent “you went to the store”. Then the argument becomes:
Premise: BS
Premise: B
Conclusion: S
To test the validity, we look at whether the combination of both premises implies the conclusion; is it true that [(BS) ⋀ B] → S ?
B S B → S (B→S) ⋀ B [(B→S) ⋀ B] → S T T T T T T F F F T F T T F T F F T F T
Since the truth table for [(BS) ⋀ B] → S is always true, this is a valid argument.
### Analyzing arguments using truth tables
To analyze an argument with a truth table:
1. Represent each of the premises symbolically
2. Create a conditional statement, joining all the premises with and to form the antecedent, and using the conclusion as the consequent.
3. Create a truth table for that statement. If it is always true, then the argument is valid.
### Example 14
Premise: If I go to the mall, then I’ll buy new jeans
Premise: If I buy new jeans, I’ll buy a shirt to go with it
Conclusion: If I got to the mall, I’ll buy a shirt.
Let M = I go to the mall, J = I buy jeans, and S = I buy a shirt.
The premises and conclusion can be stated as:
Premise: MJ
Premise: JS
Conclusion: MS
We can construct a truth table for [(MJ) ⋀ (JS)] → (MS)
M J S M → J J → S (M→J) ⋀ (J→S) M → S [(M→J) ⋀ (J→S)] → (M→S) T T T T T T T T T T F T F F F T T F T F T F T T T F F F T F F T F T T T T T T T F T F T F F T T F F T T T T T T F F F T T T T T
From the truth table, we can see this is a valid argument.
1. Technically, these are Euler circles or Euler diagrams, not Venn diagrams, but for the sake of simplicity we’ll continue to call them Venn diagrams.
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# Percentage Problems for IBPS PO Exam - Part III
The quantitative aptitude section for IBPS PO is sure to have a few questions from the topic percentage. We had earlier discussed what percentage is and what the basic concepts of percentage are.
The most common percentage related questions in bank exams have been classified into 18 easy models. Sometimes, the questions are straight forward where we need to only apply one of these models to find the solution. However, more often than not, we are required to apply two or more of these models to solve the questions based on data interpretation. Which is why understanding each of these concepts and being aware of solving these in “Smart” methods play such an important role when preparing for any bank exam.
The first 5 models have been discussed here and here. We are now going to understand the next 2 models.
### Model 6: How to find out the percentage change between two values
Percentage change can be either an increase or a decrease. A negative answer implies a decrease and a positive solution implies an increase in the percentage.
Example 1: The profit made by a company in the present year is 15 lakhs. Two years ago, the profit made by the same company was 24 lakhs. What is the percentage change in the profit made by the company?
Solution:
The percentage change in the profit made by this company is -37.5%, which is nothing but a decrease of 37.5% in profit.
### Model 7: Net Percentage change or the Effective Percentage change when there is a percentage change between two other variables
Where a and b are the percentage changes in two variables.
This is an important concept which is also used in other topics such as Simple Interest and Compound Interest, Profit and Loss and others.
Example 1: The length of a rectangle increased by 40% and its breadth increased by 20%. What will be the percentage change in the area of the rectangle?
Solution:
By applying the percent change in the formula, we can say that:
Applying the percentage change formula:
And since this is a positive answer, we can say that the change is an increment of 68%.
SMART METHOD:
The same question can be solved in a ‘smart’ method that saves time in competitive exams such as SBI PO exam and IBPS PO exam since time is of the essence here.
As the length and the breadth increase, so will the area of the rectangle.
If the percentage change in length = a%
And percentage change in breadth = b%
There is a 68% increase in the area.
Similarly, the process would be reversed if the question was altered to:
The length of a rectangle decreased by 40% and its breadth decreased by 20%. What will be the percentage change in the area of the rectangle?
Only here, percentage change in length = -a%
And percentage change in breadth = -b%
Since there is a negative change in both length and breadth:
Since the final result is in negative, it is obvious that the area has decreased by 52%.
Example 2: The revenue of a shop in the month of March was ₨ 40,000. In the month of April, the shopkeeper announced a discount of 20% and hence, his sales went up by 20%. What will be the revenue in the month of April?
Solution:
Hence, the revenue in the month of April has decreased by 4%, when compared to the revenue in the month of March.
Example 3: The price of sugar increased by 25%. What should be the percentage decrease in the consumption of sugar by a family, such that their expenditure on sugar remains the same?
Solution:
Since there is no change in the expenditure, the change between the old expenditure and the new expenditure is 0.
Consumption of sugar must decrease by 20% to keep the expenditure same as before.
Hence, it is easy for us to solve the percentage word problems in bank exams using these smart methods. There are remaining 11 models on percentage. To learn all about them and to understand how to solve each of them in Smart methods, check out our online programs.
For more tips and tricks on calculating percentages for bank exams, sign up for the free mock tests and never stop practicing.
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Calculus Volume 2
# 5.1Sequences
Calculus Volume 25.1 Sequences
## Learning Objectives
• 5.1.1 Find the formula for the general term of a sequence.
• 5.1.2 Calculate the limit of a sequence if it exists.
• 5.1.3 Determine the convergence or divergence of a given sequence.
In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.
## Terminology of Sequences
To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form
$a1,a2,a3,…,an,….a1,a2,a3,…,an,….$
Each of the numbers in the sequence is called a term. The symbol $nn$ is called the index variable for the sequence. We use the notation
${an}n=1∞,or simply{an},{an}n=1∞,or simply{an},$
to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number $anan$ exists for each positive integer $n,n,$ we can also define a sequence as a function whose domain is the set of positive integers.
Let’s consider the infinite, ordered list
$2,4,8,16,32,….2,4,8,16,32,….$
This is a sequence in which the first, second, and third terms are given by $a1=2,a1=2,$ $a2=4,a2=4,$ and $a3=8.a3=8.$ You can probably see that the terms in this sequence have the following pattern:
$a1=21,a2=22,a3=23,a4=24,anda5=25.a1=21,a2=22,a3=23,a4=24,anda5=25.$
Assuming this pattern continues, we can write the $nthnth$ term in the sequence by the explicit formula $an=2n.an=2n.$ Using this notation, we can write this sequence as
${2n}n=1∞or{2n}.{2n}n=1∞or{2n}.$
Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the $nthnth$ term $anan$ in terms of the previous term $an−1.an−1.$ In particular, we can define this sequence as the sequence ${an}{an}$ where $a1=2a1=2$ and for all $n≥2,n≥2,$ each term $anan$ is defined by the recurrence relation$an=2an−1.an=2an−1.$
## Definition
An infinite sequence${an}{an}$ is an ordered list of numbers of the form
$a1,a2,…,an,….a1,a2,…,an,….$
The subscript $nn$ is called the index variable of the sequence. Each number $anan$ is a term of the sequence. Sometimes sequences are defined by explicit formulas, in which case $an=f(n)an=f(n)$ for some function $f(n)f(n)$ defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.
Note that the index does not have to start at $n=1n=1$ but could start with other integers. For example, a sequence given by the explicit formula $an=f(n)an=f(n)$ could start at $n=0,n=0,$ in which case the sequence would be
$a0,a1,a2,….a0,a1,a2,….$
Similarly, for a sequence defined by a recurrence relation, the term $a0a0$ may be given explicitly, and the terms $anan$ for $n≥1n≥1$ may be defined in terms of $an−1.an−1.$ Since a sequence ${an}{an}$ has exactly one value for each positive integer $n,n,$ it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence ${an}{an}$ consists of all points $(n,an)(n,an)$ for all positive integers $n.n.$ Figure 5.2 shows the graph of ${2n}.{2n}.$
Figure 5.2 The plotted points are a graph of the sequence ${2n}.{2n}.$
Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence
$3,7,11,15,19,….3,7,11,15,19,….$
You can see that the difference between every consecutive pair of terms is $4.4.$ Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation
${a1=3an=an−1+4forn≥2.{a1=3an=an−1+4forn≥2.$
Note that
$a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.$
Thus the sequence can also be described using the explicit formula
$an=3+4(n−1)=4n−1.an=3+4(n−1)=4n−1.$
In general, an arithmetic sequence is any sequence of the form $an=cn+b.an=cn+b.$
In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence
$2,−23,29,−227,281,….2,−23,29,−227,281,….$
We see that the ratio of any term to the preceding term is $−13.−13.$ Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as
$a1=2an=−13·an−1forn≥2.a1=2an=−13·an−1forn≥2.$
Alternatively, since
$a2=−13·2a3=(−13)(−13)(2)=(−13)2·2a4=(−13)(−13)(−13)(2)=(−13)3·2,a2=−13·2a3=(−13)(−13)(2)=(−13)2·2a4=(−13)(−13)(−13)(2)=(−13)3·2,$
we see that the sequence can be described by using the explicit formula
$an=2(−13)n−1.an=2(−13)n−1.$
The sequence ${2n}{2n}$ that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2.2.$ In general, a geometric sequence is any sequence of the form $an=crn.an=crn.$
## Example 5.1
### Finding Explicit Formulas
For each of the following sequences, find an explicit formula for the $nthnth$ term of the sequence.
1. $−12,23,−34,45,−56,…−12,23,−34,45,−56,…$
2. $34,97,2710,8113,24316,…34,97,2710,8113,24316,…$
## Checkpoint5.1
Find an explicit formula for the $nthnth$ term of the sequence ${15,−17,19,−111,…}.{15,−17,19,−111,…}.$
## Example 5.2
### Defined by Recurrence Relations
For each of the following recursively defined sequences, find an explicit formula for the sequence.
1. $a1=2,a1=2,$ $an=−3an−1an=−3an−1$ for $n≥2n≥2$
2. $a1=12,a1=12,$ $an=an−1+(12)nan=an−1+(12)n$ for $n≥2n≥2$
## Checkpoint5.2
Find an explicit formula for the sequence defined recursively such that $a1=−4a1=−4$ and $an=an−1+6.an=an−1+6.$
## Limit of a Sequence
A fundamental question that arises regarding infinite sequences is the behavior of the terms as $nn$ gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as $n→∞.n→∞.$ For example, consider the following four sequences and their different behaviors as $n→∞n→∞$ (see Figure 5.3):
1. ${1+3n}={4,7,10,13,…}.{1+3n}={4,7,10,13,…}.$ The terms $1+3n1+3n$ become arbitrarily large as $n→∞.n→∞.$ In this case, we say that $1+3n→∞1+3n→∞$ as $n→∞.n→∞.$
2. ${1−(12)n}={12,34,78,1516,…}.{1−(12)n}={12,34,78,1516,…}.$ The terms $1−(12)n→11−(12)n→1$ as $n→∞.n→∞.$
3. ${(−1)n}={−1,1,−1,1,…}.{(−1)n}={−1,1,−1,1,…}.$ The terms alternate but do not approach one single value as $n→∞.n→∞.$
4. ${(−1)nn}={−1,12,−13,14,…}.{(−1)nn}={−1,12,−13,14,…}.$ The terms alternate for this sequence as well, but $(−1)nn→0(−1)nn→0$ as $n→∞.n→∞.$
Figure 5.3 (a) The terms in the sequence become arbitrarily large as $n→∞.n→∞.$ (b) The terms in the sequence approach $11$ as $n→∞.n→∞.$ (c) The terms in the sequence alternate between $11$ and $−1−1$ as $n→∞.n→∞.$ (d) The terms in the sequence alternate between positive and negative values but approach $00$ as $n→∞.n→∞.$
From these examples, we see several possibilities for the behavior of the terms of a sequence as $n→∞.n→∞.$ In two of the sequences, the terms approach a finite number as $n→∞.n→∞.$ In the other two sequences, the terms do not. If the terms of a sequence approach a finite number $LL$ as $n→∞,n→∞,$ we say that the sequence is a convergent sequence and the real number $LL$ is the limit of the sequence. We can give an informal definition here.
## Definition
Given a sequence ${an},{an},$ if the terms $anan$ become arbitrarily close to a finite number $LL$ as $nn$ becomes sufficiently large, we say ${an}{an}$ is a convergent sequence and $LL$ is the limit of the sequence. In this case, we write
$limn→∞an=L.limn→∞an=L.$
If a sequence ${an}{an}$ is not convergent, we say it is a divergent sequence.
From Figure 5.3, we see that the terms in the sequence ${1−(12)n}{1−(12)n}$ are becoming arbitrarily close to $11$ as $nn$ becomes very large. We conclude that ${1−(12)n}{1−(12)n}$ is a convergent sequence and its limit is $1.1.$ In contrast, from Figure 5.3, we see that the terms in the sequence $1+3n1+3n$ are not approaching a finite number as $nn$ becomes larger. We say that ${1+3n}{1+3n}$ is a divergent sequence.
In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.
## Definition
A sequence ${an}{an}$ converges to a real number $LL$ if for all $ε>0,ε>0,$ there exists an integer $NN$ such that $|an−L|<ε|an−L|<ε$ if $n≥N.n≥N.$ The number $LL$ is the limit of the sequence and we write
$limn→∞an=Loran→L.limn→∞an=Loran→L.$
In this case, we say the sequence ${an}{an}$ is a convergent sequence. If a sequence does not converge, it is a divergent sequence, and we say the limit does not exist.
We remark that the convergence or divergence of a sequence ${an}{an}$ depends only on what happens to the terms $anan$ as $n→∞.n→∞.$ Therefore, if a finite number of terms $b1,b2,…,bNb1,b2,…,bN$ are placed before $a1a1$ to create a new sequence
$b1,b2,…,bN,a1,a2,…,b1,b2,…,bN,a1,a2,…,$
this new sequence will converge if ${an}{an}$ converges and diverge if ${an}{an}$ diverges. Further, if the sequence ${an}{an}$ converges to $L,L,$ this new sequence will also converge to $L.L.$
Figure 5.4 As $nn$ increases, the terms $anan$ become closer to $L.L.$ For values of $n≥N,n≥N,$ the distance between each point $(n,an)(n,an)$ and the line $y=Ly=L$ is less than $ε.ε.$
As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences ${1+3n}{1+3n}$ and ${(−1)n}{(−1)n}$ shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The sequence ${(−1)n}{(−1)n}$ diverges because the terms alternate between $11$ and $−1,−1,$ but do not approach one value as $n→∞.n→∞.$ On the other hand, the sequence ${1+3n}{1+3n}$ diverges because the terms $1+3n→∞1+3n→∞$ as $n→∞.n→∞.$ We say the sequence ${1+3n}{1+3n}$ diverges to infinity and write $limn→∞(1+3n)=∞.limn→∞(1+3n)=∞.$ It is important to recognize that this notation does not imply the limit of the sequence ${1+3n}{1+3n}$ exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence ${−5n+2}{−5n+2}$ diverges to negative infinity because $−5n+2→−∞−5n+2→−∞$ as $n→∞.n→∞.$ We write this as $limn→∞(−5n+2)=→−∞.limn→∞(−5n+2)=→−∞.$
Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence ${an}{an}$ and a related function $ff$ defined on all positive real numbers such that $f(n)=anf(n)=an$ for all integers $n≥1.n≥1.$ Since the domain of the sequence is a subset of the domain of $f,f,$ if $limx→∞f(x)limx→∞f(x)$ exists, then the sequence converges and has the same limit. For example, consider the sequence ${1n}{1n}$ and the related function $f(x)=1x.f(x)=1x.$ Since the function $ff$ defined on all real numbers $x>0x>0$ satisfies $f(x)=1x→0f(x)=1x→0$ as $x→∞,x→∞,$ the sequence ${1n}{1n}$ must satisfy $1n→01n→0$ as $n→∞.n→∞.$
## Theorem5.1
### Limit of a Sequence Defined by a Function
Consider a sequence ${an}{an}$ such that $an=f(n)an=f(n)$ for all $n≥1.n≥1.$ If there exists a real number $LL$ such that
$limx→∞f(x)=L,limx→∞f(x)=L,$
then ${an}{an}$ converges and
$limn→∞an=L.limn→∞an=L.$
We can use this theorem to evaluate $limn→∞rnlimn→∞rn$ for $0≤r≤1.0≤r≤1.$ For example, consider the sequence ${(1/2)n}{(1/2)n}$ and the related exponential function $f(x)=(1/2)x.f(x)=(1/2)x.$ Since $limx→∞(1/2)x=0,limx→∞(1/2)x=0,$ we conclude that the sequence ${(1/2)n}{(1/2)n}$ converges and its limit is $0.0.$ Similarly, for any real number $rr$ such that $0≤r<1,0≤r<1,$ $limx→∞rx=0,limx→∞rx=0,$ and therefore the sequence ${rn}{rn}$ converges. On the other hand, if $r=1,r=1,$ then $limx→∞rx=1,limx→∞rx=1,$ and therefore the limit of the sequence ${1n}{1n}$ is $1.1.$ If $r>1,r>1,$ $limx→∞rx=∞,limx→∞rx=∞,$ and therefore we cannot apply this theorem. However, in this case, just as the function $rxrx$ grows without bound as $n→∞,n→∞,$ the terms $rnrn$ in the sequence become arbitrarily large as $n→∞,n→∞,$ and we conclude that the sequence ${rn}{rn}$ diverges to infinity if $r>1.r>1.$
We summarize these results regarding the geometric sequence ${rn}:{rn}:$
$rn→0if01.rn→0if01.$
Later in this section we consider the case when $r<0.r<0.$
We now consider slightly more complicated sequences. For example, consider the sequence ${(2/3)n+(1/4)n}.{(2/3)n+(1/4)n}.$ The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences ${(2/3)n}{(2/3)n}$ and ${(1/4)n}.{(1/4)n}.$ As we describe in the following algebraic limit laws, since ${(2/3)n}{(2/3)n}$ and ${1/4)n}{1/4)n}$ both converge to $0,0,$ the sequence ${(2/3)n+(1/4)n}{(2/3)n+(1/4)n}$ converges to $0+0=0.0+0=0.$ Just as we were able to evaluate a limit involving an algebraic combination of functions $ff$ and $gg$ by looking at the limits of $ff$ and $gg$ (see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of $anan$ and $bnbn$ by evaluating the limits of ${an}{an}$ and ${bn}.{bn}.$
## Theorem5.2
### Algebraic Limit Laws
Given sequences ${an}{an}$ and ${bn}{bn}$ and any real number $c,c,$ if there exist constants $AA$ and $BB$ such that $limn→∞an=Alimn→∞an=A$ and $limn→∞bn=B,limn→∞bn=B,$ then
1. $limn→∞c=climn→∞c=c$
2. $limn→∞can=climn→∞an=cAlimn→∞can=climn→∞an=cA$
3. $limn→∞(an±bn)=limn→∞an±limn→∞bn=A±Blimn→∞(an±bn)=limn→∞an±limn→∞bn=A±B$
4. $limn→∞(an·bn)=(limn→∞an)·(limn→∞bn)=A·Blimn→∞(an·bn)=(limn→∞an)·(limn→∞bn)=A·B$
5. $limn→∞(anbn)=limn→∞anlimn→∞bn=AB,limn→∞(anbn)=limn→∞anlimn→∞bn=AB,$ provided $B≠0B≠0$ and each $bn≠0.bn≠0.$
### Proof
We prove part iii.
Let $ϵ>0.ϵ>0.$ Since $limn→∞an=A,limn→∞an=A,$ there exists a constant positive integer $N1N1$ such that $|an-A|<ε2|an-A|<ε2$ for all $n≥N1.n≥N1.$ Since $limn→∞bn=B,limn→∞bn=B,$ there exists a constant $N2N2$ such that $|bn−B|<ε/2|bn−B|<ε/2$ for all $n≥N2.n≥N2.$ Let $NN$ be the larger of $N1N1$ and $N2.N2.$ Therefore, for all $n≥N,n≥N,$
$|(an+bn)−(A+B)|≤|an−A|+|bn−B|<ε2+ε2=ε.|(an+bn)−(A+B)|≤|an−A|+|bn−B|<ε2+ε2=ε.$
The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence ${1n2}.{1n2}.$ As shown earlier, $limn→∞1/n=0.limn→∞1/n=0.$ Similarly, for any positive integer $k,k,$ we can conclude that
$limn→∞1nk=0.limn→∞1nk=0.$
In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.
## Example 5.3
### Determining Convergence and Finding Limits
For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.
1. ${5−3n2}{5−3n2}$
2. ${3n4−7n2+56−4n4}{3n4−7n2+56−4n4}$
3. ${2nn2}{2nn2}$
4. ${(1+4n)n}{(1+4n)n}$
## Checkpoint5.3
Consider the sequence ${(5n2+1)/en}.{(5n2+1)/en}.$ Determine whether or not the sequence converges. If it converges, find its limit.
Recall that if $ff$ is a continuous function at a value $L,L,$ then $f(x)→f(L)f(x)→f(L)$ as $x→L.x→L.$ This idea applies to sequences as well. Suppose a sequence $an→L,an→L,$ and a function $ff$ is continuous at $L.L.$ Then $f(an)→f(L).f(an)→f(L).$ This property often enables us to find limits for complicated sequences. For example, consider the sequence $5−3n2.5−3n2.$ From Example 5.3a. we know the sequence $5−3n2→5.5−3n2→5.$ Since $xx$ is a continuous function at $x=5,x=5,$
$limn→∞5−3n2=limn→∞(5−3n2)=5.limn→∞5−3n2=limn→∞(5−3n2)=5.$
## Theorem5.3
### Continuous Functions Defined on Convergent Sequences
Consider a sequence ${an}{an}$ and suppose there exists a real number $LL$ such that the sequence ${an}{an}$ converges to $L.L.$ Suppose $ff$ is a continuous function at $L.L.$ Then there exists an integer $NN$ such that $ff$ is defined at all values $anan$ for $n≥N,n≥N,$ and the sequence ${f(an)}{f(an)}$ converges to $f(L)f(L)$ (Figure 5.5).
### Proof
Let $ϵ>0.ϵ>0.$ Since $ff$ is continuous at $L,L,$ there exists $δ>0δ>0$ such that $|f(x)−f(L)|<ε|f(x)−f(L)|<ε$ if $|x−L|<δ.|x−L|<δ.$ Since the sequence ${an}{an}$ converges to $L,L,$ there exists $NN$ such that $|an−L|<δ|an−L|<δ$ for all $n≥N.n≥N.$ Therefore, for all $n≥N,n≥N,$ $|an−L|<δ,|an−L|<δ,$ which implies $|f(an)−f(L)|<ε.|f(an)−f(L)|<ε.$ We conclude that the sequence ${f(an)}{f(an)}$ converges to $f(L).f(L).$
Figure 5.5 Because $ff$ is a continuous function as the inputs $a1,a2,a3,…a1,a2,a3,…$ approach $L,L,$ the outputs $f(a1),f(a2),f(a3),…f(a1),f(a2),f(a3),…$ approach $f(L).f(L).$
## Example 5.4
### Limits Involving Continuous Functions Defined on Convergent Sequences
Determine whether the sequence ${cos(3/n2)}{cos(3/n2)}$ converges. If it converges, find its limit.
## Checkpoint5.4
Determine if the sequence ${2n+13n+5}{2n+13n+5}$ converges. If it converges, find its limit.
Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.
## Theorem5.4
### Squeeze Theorem for Sequences
Consider sequences ${an},{an},$ ${bn},{bn},$ and ${cn}.{cn}.$ Suppose there exists an integer $NN$ such that
$an≤bn≤cnfor alln≥N.an≤bn≤cnfor alln≥N.$
If there exists a real number $LL$ such that
$limn→∞an=L=limn→∞cn,limn→∞an=L=limn→∞cn,$
then ${bn}{bn}$ converges and $limn→∞bn=Llimn→∞bn=L$ (Figure 5.6).
### Proof
Let $ε>0.ε>0.$ Since the sequence ${an}{an}$ converges to $L,L,$ there exists an integer $N1N1$ such that $|an−L|<ε|an−L|<ε$ for all $n≥N1.n≥N1.$ Similarly, since ${cn}{cn}$ converges to $L,L,$ there exists an integer $N2N2$ such that $|cn−L|<ε|cn−L|<ε$ for all $n≥N2.n≥N2.$ By assumption, there exists an integer $NN$ such that $an≤bn≤cnan≤bn≤cn$ for all $n≥N.n≥N.$ Let $MM$ be the largest of $N1,N2,N1,N2,$ and $N.N.$ We must show that $|bn−L|<ε|bn−L|<ε$ for all $n≥M.n≥M.$ For all $n≥M,n≥M,$
$−ε<−|an−L|≤an−L≤bn−L≤cn−L≤|cn−L|<ε.−ε<−|an−L|≤an−L≤bn−L≤cn−L≤|cn−L|<ε.$
Therefore, $−ε and we conclude that $|bn−L|<ε|bn−L|<ε$ for all $n≥M,n≥M,$ and we conclude that the sequence ${bn}{bn}$ converges to $L.L.$
Figure 5.6 Each term $bnbn$ satisfies $an≤bn≤cnan≤bn≤cn$ and the sequences ${an}{an}$ and ${cn}{cn}$ converge to the same limit, so the sequence ${bn}{bn}$ must converge to the same limit as well.
## Example 5.5
### Using the Squeeze Theorem
Use the Squeeze Theorem to find the limit of each of the following sequences.
1. ${cosnn2}{cosnn2}$
2. ${(−12)n}{(−12)n}$
## Checkpoint5.5
Find $limn→∞2n−sinnn.limn→∞2n−sinnn.$
Using the idea from Example 5.5b. we conclude that $rn→0rn→0$ for any real number $rr$ such that $−1 If $r<−1,r<−1,$ the sequence ${rn}{rn}$ diverges because the terms oscillate and become arbitrarily large in magnitude. If $r=−1,r=−1,$ the sequence ${rn}={(−1)n}{rn}={(−1)n}$ diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.
$rn→0if|r|<1rn→0if|r|<1$
(5.1)
$rn→1ifr=1rn→1ifr=1$
(5.2)
$rn→∞ifr>1rn→∞ifr>1$
(5.3)
${rn}diverges ifr≤−1{rn}diverges ifr≤−1$
(5.4)
## Bounded Sequences
We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.
## Definition
A sequence ${an}{an}$ is bounded above if there exists a real number $MM$ such that
$an≤Man≤M$
for all positive integers $n.n.$
A sequence ${an}{an}$ is bounded below if there exists a real number $MM$ such that
$M≤anM≤an$
for all positive integers $n.n.$
A sequence ${an}{an}$ is a bounded sequence if it is bounded above and bounded below.
If a sequence is not bounded, it is an unbounded sequence.
For example, the sequence ${1/n}{1/n}$ is bounded above because $1/n≤11/n≤1$ for all positive integers $n.n.$ It is also bounded below because $1/n≥01/n≥0$ for all positive integers n. Therefore, ${1/n}{1/n}$ is a bounded sequence. On the other hand, consider the sequence ${2n}.{2n}.$ Because $2n≥22n≥2$ for all $n≥1,n≥1,$ the sequence is bounded below. However, the sequence is not bounded above. Therefore, ${2n}{2n}$ is an unbounded sequence.
We now discuss the relationship between boundedness and convergence. Suppose a sequence ${an}{an}$ is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms $anan$ that are arbitrarily large in magnitude as $nn$ gets larger. As a result, the sequence ${an}{an}$ cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.
## Theorem5.5
### Convergent Sequences Are Bounded
If a sequence ${an}{an}$ converges, then it is bounded.
Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence ${(−1)n}{(−1)n}$ is bounded, but the sequence diverges because the sequence oscillates between $11$ and $−1−1$ and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.
Consider a bounded sequence ${an}.{an}.$ Suppose the sequence ${an}{an}$ is increasing. That is, $a1≤a2≤a3….a1≤a2≤a3….$ Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that ${an}{an}$ converges. For example, consider the sequence
${12,23,34,45,…}.{12,23,34,45,…}.$
Since this sequence is increasing and bounded above, it converges. Next, consider the sequence
${2,0,3,0,4,0,1,−12,−13,−14,…}.{2,0,3,0,4,0,1,−12,−13,−14,…}.$
Even though the sequence is not increasing for all values of $n,n,$ we see that $−1/2<−1/3<−1/4<⋯.−1/2<−1/3<−1/4<⋯.$ Therefore, starting with the eighth term, $a8=−1/2,a8=−1/2,$ the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.
## Definition
A sequence ${an}{an}$ is increasing for all $n≥n0n≥n0$ if
$an≤an+1for alln≥n0.an≤an+1for alln≥n0.$
A sequence ${an}{an}$ is decreasing for all $n≥n0n≥n0$ if
$an≥an+1for alln≥n0.an≥an+1for alln≥n0.$
A sequence ${an}{an}$ is a monotone sequence for all $n≥n0n≥n0$ if it is increasing for all $n≥n0n≥n0$ or decreasing for all $n≥n0.n≥n0.$
We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.
## Theorem5.6
### Monotone Convergence Theorem
If ${an}{an}$ is a bounded sequence and there exists a positive integer $n0n0$ such that ${an}{an}$ is monotone for all $n≥n0,n≥n0,$ then ${an}{an}$ converges.
The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 5.7).
Figure 5.7 Since the sequence ${an}{an}$ is increasing and bounded above, it must converge.
In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.
## Example 5.6
### Using the Monotone Convergence Theorem
For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.
1. ${4nn!}{4nn!}$
2. ${an}{an}$ defined recursively such that
$a1=2andan+1=an2+12anfor alln≥2.a1=2andan+1=an2+12anfor alln≥2.$
## Checkpoint5.6
Consider the sequence ${an}{an}$ defined recursively such that $a1=1,a1=1,$ $an=an−1/2.an=an−1/2.$ Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.
## Student Project
### Fibonacci Numbers
The Fibonacci numbers are defined recursively by the sequence ${Fn}{Fn}$ where $F0=0,F0=0,$ $F1=1F1=1$ and for $n≥2,n≥2,$
$Fn=Fn−1+Fn−2.Fn=Fn−1+Fn−2.$
Here we look at properties of the Fibonacci numbers.
1. Write out the first twenty Fibonacci numbers.
2. Find a closed formula for the Fibonacci sequence by using the following steps.
1. Consider the recursively defined sequence ${xn}{xn}$ where $xo=cxo=c$ and $xn+1=axn.xn+1=axn.$ Show that this sequence can be described by the closed formula $xn=canxn=can$ for all $n≥0.n≥0.$
2. Using the result from part a. as motivation, look for a solution of the equation
$Fn=Fn−1+Fn−2Fn=Fn−1+Fn−2$
of the form $Fn=cλn.Fn=cλn.$ Determine what two values for $λλ$ will allow $FnFn$ to satisfy this equation.
3. Consider the two solutions from part b.: $λ1λ1$ and $λ2.λ2.$ Let $Fn=c1λ1n+c2λ2n.Fn=c1λ1n+c2λ2n.$ Use the initial conditions $F0F0$ and $F1F1$ to determine the values for the constants $c1c1$ and $c2c2$ and write the closed formula $Fn.Fn.$
3. Use the answer in 2 c. to show that
$limn→∞Fn+1Fn=1+52.limn→∞Fn+1Fn=1+52.$
The number $ϕ=(1+5)/2ϕ=(1+5)/2$ is known as the golden ratio (Figure 5.8 and Figure 5.9).
Figure 5.8 The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)
Figure 5.9 The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon appears to have these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)
## Section 5.1 Exercises
Find the first six terms of each of the following sequences, starting with $n=1.n=1.$
1.
$an=1+(−1)nan=1+(−1)n$ for $n≥1n≥1$
2.
$an=n2−1an=n2−1$ for $n≥1n≥1$
3.
$a1=1a1=1$ and $an=an−1+nan=an−1+n$ for $n≥2n≥2$
4.
$a1=1,a1=1,$ $a2=1a2=1$ and $an+2=an+an+1an+2=an+an+1$ for $n≥1n≥1$
5.
Find an explicit formula for $anan$ where $a1=1a1=1$ and $an=an−1+nan=an−1+n$ for $n≥2.n≥2.$
6.
Find a formula $anan$ for the $nthnth$ term of the arithmetic sequence whose first term is $a1=1a1=1$ such that $an+1−an=17an+1−an=17$ for $n≥1.n≥1.$
7.
Find a formula $anan$ for the $nthnth$ term of the arithmetic sequence whose first term is $a1=−3a1=−3$ such that $an+1−an=4an+1−an=4$ for $n≥1.n≥1.$
8.
Find a formula $anan$ for the $nthnth$ term of the geometric sequence whose first term is $a1=1a1=1$ such that $an+1an=10an+1an=10$ for $n≥1.n≥1.$
9.
Find a formula $anan$ for the $nthnth$ term of the geometric sequence whose first term is $a1=3a1=3$ such that $an+1an=1/10an+1an=1/10$ for $n≥1.n≥1.$
10.
Find an explicit formula for the $nthnth$ term of the sequence whose first several terms are ${0,3,8,15,24,35,48,63,80,99,…}.{0,3,8,15,24,35,48,63,80,99,…}.$ (Hint: First add one to each term.)
11.
Find an explicit formula for the $nthnth$ term of the sequence satisfying $a1=0a1=0$ and $an=2an−1+1an=2an−1+1$ for $n≥2.n≥2.$
Find a formula for the general term $anan$ of each of the following sequences.
12.
${1,0,−1,0,1,0,−1,0,…}{1,0,−1,0,1,0,−1,0,…}$ (Hint: Find where $sinxsinx$ takes these values)
13.
${ 1 , − 1 / 3 , 1 / 5 , − 1 / 7 ,… } { 1 , − 1 / 3 , 1 / 5 , − 1 / 7 ,… }$
Find a function $f(n)f(n)$ that identifies the $nthnth$ term $anan$ of the following recursively defined sequences, as $an=f(n).an=f(n).$
14.
$a1=1a1=1$ and $an+1=−anan+1=−an$ for $n≥1n≥1$
15.
$a1=2a1=2$ and $an+1=2anan+1=2an$ for $n≥1n≥1$
16.
$a1=1a1=1$ and $an+1=(n+1)anan+1=(n+1)an$ for $n≥1n≥1$
17.
$a1=2a1=2$ and $an+1=(n+1)an/2an+1=(n+1)an/2$ for $n≥1n≥1$
18.
$a1=1a1=1$ and $an+1=an/2nan+1=an/2n$ for $n≥1n≥1$
Plot the first $NN$ terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.
19.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ and for $n≥2,n≥2,$ $an=12(an−1+an−2);an=12(an−1+an−2);$ $N=30N=30$
20.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ $a3=3a3=3$ and for $n≥4,n≥4,$ $an=13(an−1+an−2+an−3),an=13(an−1+an−2+an−3),$ $N=30N=30$
21.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ and for $n≥3,n≥3,$ $an=an−1an−2;an=an−1an−2;$ $N=30N=30$
22.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ $a3=3,a3=3,$ and for $n≥4,n≥4,$ $an=an−1an−2an−3;an=an−1an−2an−3;$ $N=30N=30$
Suppose that $limn→∞an=1,limn→∞an=1,$ $limn→∞bn=−1,limn→∞bn=−1,$ and $0<−bn for all $n.n.$ Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.
23.
$lim n → ∞ ( 3 a n − 4 b n ) lim n → ∞ ( 3 a n − 4 b n )$
24.
$lim n → ∞ ( 1 2 b n − 1 2 a n ) lim n → ∞ ( 1 2 b n − 1 2 a n )$
25.
$lim n → ∞ a n + b n a n − b n lim n → ∞ a n + b n a n − b n$
26.
$lim n → ∞ a n − b n a n + b n lim n → ∞ a n − b n a n + b n$
Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.
27.
$n 2 2 n n 2 2 n$
28.
$( n − 1 ) 2 ( n + 1 ) 2 ( n − 1 ) 2 ( n + 1 ) 2$
29.
$n n + 1 n n + 1$
30.
$n1/nn1/n$ (Hint: $n1/n=e1nlnn)n1/n=e1nlnn)$
For each of the following sequences, whose $nthnth$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.
31.
$n/2n,n/2n,$ $n≥2n≥2$
32.
$ln ( 1 + 1 n ) ln ( 1 + 1 n )$
33.
$sin n sin n$
34.
$cos ( n 2 ) cos ( n 2 )$
35.
$n1/n,n1/n,$ $n≥3n≥3$
36.
$n−1/n,n−1/n,$ $n≥3n≥3$
37.
$tan n tan n$
38.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
$a1=2,a1=2,$ $a2=22,a2=22,$ $a3=222a3=222$ etc.
39.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
$a1=3,a1=3,$ $an=2an−1,an=2an−1,$ $n=2,3,….n=2,3,….$
Use the Squeeze Theorem to find the limit of each of the following sequences.
40.
$n sin ( 1 / n ) n sin ( 1 / n )$
41.
$cos ( 1 / n ) − 1 1 / n cos ( 1 / n ) − 1 1 / n$
42.
$a n = n ! n n a n = n ! n n$
43.
$a n = sin n sin ( 1 / n ) a n = sin n sin ( 1 / n )$
For the following sequences, plot the first $2525$ terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.
44.
[T] $an=sinnan=sinn$
45.
[T] $an=cosnan=cosn$
Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.
46.
$a n = tan −1 ( n 2 ) a n = tan −1 ( n 2 )$
47.
$a n = ( 2 n ) 1 / n − n 1 / n a n = ( 2 n ) 1 / n − n 1 / n$
48.
$a n = ln ( n 2 ) ln ( 2 n ) a n = ln ( n 2 ) ln ( 2 n )$
49.
$a n = ( 1 − 2 n ) n a n = ( 1 − 2 n ) n$
50.
$a n = ln ( n + 2 n 2 − 3 ) a n = ln ( n + 2 n 2 − 3 )$
51.
$a n = 2 n + 3 n 4 n a n = 2 n + 3 n 4 n$
52.
$a n = ( 1000 ) n n ! a n = ( 1000 ) n n !$
53.
$a n = ( n ! ) 2 ( 2 n ) ! a n = ( n ! ) 2 ( 2 n ) !$
Newton’s method seeks to approximate a solution $f(x)=0f(x)=0$ that starts with an initial approximation $x0x0$ and successively defines a sequence $xn+1=xn−f(xn)f′(xn).xn+1=xn−f(xn)f′(xn).$ For the given choice of $ff$ and $x0,x0,$ write out the formula for $xn+1.xn+1.$ If the sequence appears to converge, give an exact formula for the solution $x,x,$ then identify the limit $xx$ accurate to four decimal places and the smallest $nn$ such that $xnxn$ agrees with $xx$ up to four decimal places.
54.
[T] $f(x)=x2−2,f(x)=x2−2,$ $x0=1x0=1$
55.
[T] $f(x)=(x−1)2−2,f(x)=(x−1)2−2,$ $x0=2x0=2$
56.
[T] $f(x)=ex−2,f(x)=ex−2,$ $x0=1x0=1$
57.
[T] $f(x)=lnx−1,f(x)=lnx−1,$ $x0=2x0=2$
58.
[T] Suppose you start with one liter of vinegar and repeatedly remove $0.1L,0.1L,$ replace with water, mix, and repeat.
1. Find a formula for the concentration after $nn$ steps.
2. After how many steps does the mixture contain less than $10%10%$ vinegar?
59.
[T] A lake initially contains $20002000$ fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by $6%6%$ each month. However, factoring in all causes, $150150$ fish are lost each month.
1. Explain why the fish population after $nn$ months is modeled by $Pn=1.06Pn−1−150Pn=1.06Pn−1−150$ with $P0=2000.P0=2000.$
2. How many fish will be in the pond after one year?
60.
[T] A bank account earns $5%5%$ interest compounded monthly. Suppose that $10001000$ is initially deposited into the account, but that $1010$ is withdrawn each month.
1. Show that the amount in the account after $nn$ months is $An=(1+.05/12)An−1−10;An=(1+.05/12)An−1−10;$ $A0=1000.A0=1000.$
2. How much money will be in the account after $11$ year?
3. Is the amount increasing or decreasing?
4. Suppose that instead of $10,10,$ a fixed amount $dd$ dollars is withdrawn each month. Find a value of $dd$ such that the amount in the account after each month remains $1000.1000.$
5. What happens if $dd$ is greater than this amount?
61.
[T] A student takes out a college loan of $10,00010,000$ at an annual percentage rate of $6%,6%,$ compounded monthly.
1. If the student makes payments of $100100$ per month, how much does the student owe after $1212$ months?
2. After how many months will the loan be paid off?
62.
[T] Consider a series combining geometric growth and arithmetic decrease. Let $a1=1.a1=1.$ Fix $a>1a>1$ and $0 Set $an+1=a.an−b.an+1=a.an−b.$ Find a formula for $an+1an+1$ in terms of $an,an,$ $a,a,$ and $bb$ and a relationship between $aa$ and $bb$ such that $anan$ converges.
63.
[T] The binary representation $x=0.b1b2b3...x=0.b1b2b3...$ of a number $xx$ between $00$ and $11$ can be defined as follows. Let $b1=0b1=0$ if $x<1/2x<1/2$ and $b1=1b1=1$ if $1/2≤x<1.1/2≤x<1.$ Let $x1=2x−b1.x1=2x−b1.$ Let $b2=0b2=0$ if $x1<1/2x1<1/2$ and $b2=1b2=1$ if $1/2≤x<1.1/2≤x<1.$ Let $x2=2x1−b2x2=2x1−b2$ and in general, $xn=2xn−1−bnxn=2xn−1−bn$ and $bn−1=0bn−1=0$ if $xn<1/2xn<1/2$ and $bn−1=1bn−1=1$ if $1/2≤xn<1.1/2≤xn<1.$ Find the binary expansion of $1/3.1/3.$
64.
[T] To find an approximation for $π,π,$ set $a0=2+1,a0=2+1,$ $a1=2+a0,a1=2+a0,$ and, in general, $an+1=2+an.an+1=2+an.$ Finally, set $pn=3.2n+12−an.pn=3.2n+12−an.$ Find the first ten terms of $pnpn$ and compare the values to $π.π.$
For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of $NN$ integers $a1,a2,…,aNa1,a2,…,aN$ by fixing two special integers $KK$ and $MM$ and letting $an+1an+1$ be the remainder after dividing $K.anK.an$ into $M,M,$ then creates a bit sequence of zeros and ones whose $nthnth$ term $bnbn$ is equal to one if $anan$ is odd and equal to zero if $anan$ is even. If the bits $bnbn$ are pseudorandom, then the behavior of their average $(b1+b2+⋯+bN)/N(b1+b2+⋯+bN)/N$ should be similar to behavior of averages of truly randomly generated bits.
65.
[T] Starting with $K=16,807K=16,807$ and $M=2,147,483,647,M=2,147,483,647,$ using ten different starting values of $a1,a1,$ compute sequences of bits $bnbn$ up to $n=1000,n=1000,$ and compare their averages to ten such sequences generated by a random bit generator.
66.
[T] Find the first $10001000$ digits of $ππ$ using either a computer program or Internet resource. Create a bit sequence $bnbn$ by letting $bn=1bn=1$ if the $nthnth$ digit of $ππ$ is odd and $bn=0bn=0$ if the $nthnth$ digit of $ππ$ is even. Compute the average value of $bnbn$ and the average value of $dn=|bn+1−bn|,dn=|bn+1−bn|,$ $n=1,...,999.n=1,...,999.$ Does the sequence $bnbn$ appear random? Do the differences between successive elements of $bnbn$ appear random?
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# How do you simplify (sqrt5-x)(sqrt5+x)?
Oct 4, 2017
Use the FOIL method of multiplication, and the result is $5 - {x}^{2}$
#### Explanation:
The FOIL method, usually thought of as applying to binomial expressions can be used for this radical expression as well.
It tells us to multiply the First terms in each bracket, the Outside terms, the Inside terms and the Last terms.
Thus,
$\left(\sqrt{5} - x\right) \cdot \left(\sqrt{5} + x\right) = \left[\left(\sqrt{5} \cdot \sqrt{5}\right) + \sqrt{5} \cdot x + \sqrt{5} \cdot \left(- x\right) + x \cdot \left(- x\right)\right]$
Simplifying:
$\left(\sqrt{5} - x\right) \cdot \left(\sqrt{5} + x\right) = \left(5 + \sqrt{5} \cdot x - \sqrt{5} \cdot x - {x}^{2}\right)$
Noting that the second and third terms add to zero, we get the final result:
$\left(\sqrt{5} - x\right) \cdot \left(\sqrt{5} + x\right) = 5 - {x}^{2}$
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Multiplying fractions by an integer
Probably the best way to address an explanation of the correct way to resolve fractional and integer multiplication is to start by doing a theoretical review, in which some crucial definitions for understand this operation within its precise mathematical context.
Fundamental definitions
In this regard, it may also be prudent to delimit this revision to two specific notions: the very definitions of fractions and whole numbers, in order to be aware of the nature of the expressions or numerical elements involved in the Multiplication operation. Here’s each one:
Fractions
In this sense, it will begin to say that the different sources have been given the task of pointing out fractions as a type of mathematical expression, which is used to express fractional numbers, that is, that fractions will be representations of numbers or amounts that are not accurate or not whole.
Likewise, fractions are generally understood as an expression composed of two elements, each of which can be defined in turn as follows:
• Numerator: First, you will find the Numerator, who will fulfill the task of indicating what part of the whole has been taken or that represents the fraction. This element will be located at the top of the expression.
• Denominator: The Denominator will occupy the bottom of the expression. Its mission is to indicate in how many parts the whole is divided, of which the numerator represents a part.
Integers
In contrast, Whole Numbers will be elements that will serve to account for exact quantities. These numbers are made up of natural numbers, their negative inverses and zero, elements these which in turn constitute the Z Numeric Set. Unlike fractions, they will not have a numerator or denominator, and they are then made up of a single element.
Multiplying fractions with integers
Once these definitions have been revised, it will be much easier to understand the operation known as Fraction and Integer Multiplication, and that as the name that calls it indicates, it is an operation by which a product is sought to obtain between a fractional number, expressed as a fraction, and an integer.
However, because this is an operation that combines mixed elements, or of different mathematical nature, it will be necessary to specify the correct way to solve such operations, and that will be based on these simple steps:
• The first thing to remember is that every whole number, mathematically speaking, has a denominator equivalent to one.
• Assuming this, that is, the denominator of the whole number corresponds to the unit, the multiplication between two fractions can be expressed.
• Conselus with what the Mathematics dictates, it is necessary to proceed to multiply the numbers that serve as a numerator, thus obtaining the numerator of the product.
• Likewise, the names will be carried out in order to determine what is the product of the final fraction.
• If there is any chance of simplifying the fraction, it must be done.
The way to solve this type of operation can be expressed mathematically as follows:
Example of Multiplication of Fractions and Integers
However, the best way to complete an explanation of the correct way to multiply fractions and integers may be to use some examples, which allow us to see in a practical way how every whole number can be expressed as a fraction, assuming as its denominator the unit, thus allowing the solve of the Fraction Multiplication, as seen below:
Resolve the following operation:
Picture: pixabay.com
Multiplying fractions by an integer
Source: Education
September 26, 2019
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# Quiz on word problem fractions for children.
Word problems involving fractions can be challenging for students, as they require not only a strong understanding of fractions but also the ability to translate word problems into mathematical equations. Here are some tips and strategies to help students solve word problems involving fractions:
1. Read the problem carefully: It is important to understand what the problem is asking before trying to solve it. Be sure to read the problem at least twice, and highlight or underline any important information or key words.
2. Identify the relevant information: Look for information in the problem that will be useful in solving the problem. This might include the fractions being used, the operation being performed (e.g. addition, subtraction, multiplication, division), and any given quantities or values.
3. Translate the problem into a mathematical equation: Once you have identified the relevant information, you can use it to create a mathematical equation. For example, if the problem asks you to add two fractions together, you can write an equation using the fraction symbols and the appropriate operation.
4. Solve the equation: Use your knowledge of fractions to solve the equation. This might involve reducing fractions to their simplest form, finding a common denominator, or using the appropriate operation to combine the fractions.
5. Check your work: Once you have found a solution, it is important to check your work to ensure that it is correct. You can do this by substituting your solution back into the original equation to see if it works. You can also try solving the problem in a different way to check your work.
Here are a few examples of word problems involving fractions:
Example 1:
Samantha has 1/4 of a pie and Jessica has 1/6 of a pie. How much pie do they have in total?
To solve this problem, we can create an equation using the information given: 1/4 + 1/6 = x, where x represents the total amount of pie. We can then solve for x by adding the fractions together: 1/4 + 1/6 = 3/12 + 2/12 = 5/12. Therefore, Samantha and Jessica have a total of 5/12 of a pie.
Example 2:
There are 3/4 of a cup of sugar in a bag. If a recipe calls for 1/2 cup of sugar, how many bags of sugar will you need?
To solve this problem, we can create an equation using the information given: 3/4 / 1/2 = x, where x represents the number of bags of sugar needed. We can then solve for x by dividing the fractions: 3/4 / 1/2 = (32)/(41) = 6/4 = 3/2. Therefore, you will need 3/2 bags of sugar, which is equivalent to 1 and 1/2 bags.
By following these steps and practicing with various word problems involving fractions, students can improve their skills in solving these types of problems.
Also feel free to check other relevant educational content for your young ones Here.
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# {FREE} Multiply & Divide by Powers of Ten Game: Includes Exponents
Help your students better understand how to multiply & divide by powers of ten using this hands-on math game. Or use the game pieces as manipulatives for students to learn & practice.
We work in a base ten number system. Therefore, helping students understand place value & powers of ten in this number system is essential for developing deep number sense. But far too often, students are told facts or ‘tricks’ to memorize rather than being given time to play with numbers and make sense of this number system on their own.
The goal of this math center game is to aid 5th grade students in their understanding of powers of ten and the impact multiplying and dividing by powers of ten has on numbers.
In this game, students will also see powers of ten written with exponents. This allows them to begin to understand exponents as repeated multiplication, while working with only powers of ten.
## Multiply & Divide by Powers of Ten on a Place Value Mat:
In this game, each student works with their number using a place value mat. There are 2 different place value mats included:
• Whole Numbers Place Value Mat to Billions
• Decimal Place Value Mat: Millions to Thousandths
If your students are not yet ready to work with decimals, you can print the whole number mat for them.
## Why Use a Place Value Mat?
The main goal of this game is to help students see the connection between multiplying and dividing by powers of ten and place value within our base ten system.
Oftentimes, when kids learn this skill they are told to just ‘add zeros’ to the number or ‘move the decimal point.’
But this doesn’t help them understand WHY. Using the place value mat, students can see that it’s actually the DIGITS of a number that shift with each power of ten rather than just adding zeros or moving the decimal (the decimal doesn’t move).
Before allowing students to play the game, spend some time working on basic multiplication & division problems with the place value mat so students can see & think about what’s happening.
Start with a simple number like 5. Then ask students, what happens when we multiply 5 x 10?
They may know right away that the answer is 50. But remind them that 5 x 10 means 5 groups of 10. And because we work in a base ten system, that means we now have 5 tens instead of 5 ones.
Then show then how the 5 shifts from the ones place to the tens place.
Do this with a few more examples, increasing in difficulty and also thinking through division examples.
## Powers of Ten with Exponents:
Once students are comfortable using the place value mat to multiply & divide by 10, you can introduce them (if you haven’t already) to exponents.
You may want to start with a review of multiplication as repeated addition. Then explain that just as we use multiplication to show repeated addition, we use exponents to show repeated multiplication.
So if we want to multiply a number by 10 four times, meaning 10x10x10x10, we can write it as 10^4 (10 to the power of 4).
Do some work with powers of ten written with exponents to be sure students are comfortable & familiar with the notation before playing the game.
## Powers of Ten Math Center Game Set Up:
This game does require a bit of prep. But hopefully the prep will be worth it when it gives students a more solid understanding of multiplying & dividing by powers of ten.
Plus, you can prep most of the materials once to use again and again!
In addition to printing the materials in the download, you will need a single die and some game markers for each player.
First, print the place value mat of your choice on card stock paper. There is one for whole numbers and one that includes decimals. You will need a mat for each student.
Cut the place value mat on the dotted line and tape together so the place values go in order.
Then print the ‘digit’ cards and game directions on card stock paper of a different color and laminate for durability, then cut out all the cards.
You will notice there’s an additional sheet of ‘zero’ digits. This is because as students multiply, they may need additional zeros as place holders on their place value mat. Keep this stack face up next to the game board to use as needed.
Cut out the additional digit cards, shuffle and stack face down. This will be the pile they draw from. Leave a copy of the game directions for students to reference.
Print the game board on card stock paper of another color and laminate. You only need one game board for each small group.
Lastly, print score cards for each student. You can print these on regular paper for students to record their thinking and scores as they play, or you can print on card stock, laminate and provide dry erase markers. At the end of the game, students can then erase the score cards to be used again another day.
I recommend 3-4 students per game. If you will have multiple groups playing at the same time, complete the prep above for each game set.
Here’s a summary of what you need to print & prep:
• One game board per small group
• One set of ‘Digit’ cards & game directions per small group, cut out & shuffled (separate the ‘zero’ digits for students to use as needed)
• One place value mat per student, cut & taped together
• One score card per student
Provide a die & a set of game markers and you’re all set!
## Multiply & Divide by Powers of Ten: How to Play the Game
The goal of the game is to be the player with the largest number at the end of the game.
To begin, each player draws 2 digits from the deck. They then start with any 2-digit number, placing the digits on the tens and ones place, or the ones and tenths place.
Players then take turns rolling a die and moving their game piece along the board. They then follow the directions depending on where they land:
• Power of ten: multiply the number on your place value mat if the number you rolled was even or divide the number on your place value mat if the roll was odd.
• New Digit: draw a card from the ‘digit deck’ to add to the beginning or end of the number on your place value mat. This will be your new score and value used on your next turn.
• Digit Swap: player MUST swap one of the digits on their place value mat. They can either swap 2 of their own to form a new number or they can swap one of their digits with another player.
Once all players have made it to the finish, the player with the largest number on their place value mat wins. Each player then writes the final values in order from least to greatest at the bottom of their score card.
A quick note:students will keep a running tally on their score card, meaning they always start with their current value and multiply/divide from there. They do not start over with their original 2-digit number on each subsequent turn.
And that’s it! I hope this provides a fun way for students to make sense of place value, large numbers, decimals, powers of ten & exponents.
Love this idea? This game was originally created for Math Geek Mama+ members. If you want instant access to games like this and so much more, check it out today!
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# Lesson-1 -THE-CONICS
```Lesson No.
1
Topic:
The Conics
I. Objectives: At the end of the lessons, the learners shall be able to:
1. Illustrate the different types of conic sections – circle, parabola, ellipse, and hyperbola.
2. Determine the type of conic section defined by a given 2nd degree equation in x and y.
3. Appreciate the importance of the conic sections to different real life situations.
II. Key Concepts
The Conics
A conic section is the intersection of a plane and a right circular cone or two right circular cones
with a common vertex and parallel bases.
Consider the two – napped right circular cone in Figure 1. A line on the cone is called a
generator and the only point where the generators intersect is called the vertex. The vertical
line passing through the vertex is called the axis of the cone.
The angular orientation of the plane relative to the cone defines whether the conic section is a
circle, parabola, ellipse or hyperbola.
The circle and the ellipse are formed when the intersection of cone and plane is abounded
curve.
The circle is a special case of the ellipse in which the plane is perpendicular to the axis of the
cone.
The ellipse is formed when the tilted plane intersects only one cone to form a bounded curve.
If the plane is parallel to a generator line of the cone, the conic is a parabola.
Lastly, if the intersection is an unbounded curve and the plane is not parallel to a generator line
of the cone and the plane intersects both halves of the cone, the conic is a hyperbola. These
are illustrated in Figure 2.
The graph of the second – degree equation of the form 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
is determined by the values of 𝐵2 − 4𝐴𝐶.
Table 1 below provides the type of conic section given the values of A, B and C.
Conic sections can also be defined in terms of eccentricity e.
Table 1: Quadratic Relations and Conic Section
Conic Section
Circle
Parabola
Ellipse
Hyperbola
Value of 𝐵2 − 4𝐴𝐶
𝐵2 − 4𝐴𝐶 < 0, 𝐵 = 0 𝑜𝑟 𝐴 = 𝐶
𝐵2 − 4𝐴𝐶 = 0
𝐵2 − 4𝐴𝐶 < 0, 𝐵 ≠ 0 𝑜𝑟 𝐴 ≠ 𝐶
𝐵2 − 4𝐴𝐶 > 0
Eccentricity
𝑒=0
𝑒=1
0<𝑒<1
𝑒>1
Example : Determine the type of conic section that each general equation will produce.
Procedure:
Collect first all the values of A, B and C and solve for the value of 𝐵2 − 4𝐴𝐶. Then interpret the
result based on Table 1.
a.
𝟗𝒙𝟐 + 𝟗𝒚𝟐 − 𝟔𝒙 + 𝟏𝟖𝒚 + 𝟏𝟏 = 𝟎
𝐴 = 9, 𝐵 = 0, 𝐶 = 9
𝐵2 − 4𝐴𝐶 = 02 − 4(9)(9) = 0 − 324
= −324
𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 − 324 < 0
Note also that B = 0 and A = C.
Thus, the conic section is a circle.
b. 𝟐𝒙𝟐 + 𝟒𝒚𝟐 + 𝟖𝒙 + 𝟐𝟒𝒚 + 𝟒𝟒 = 𝟎
𝐴 = 2, 𝐵 = 0, 𝐶 = 4
𝐵2 − 4𝐴𝐶 = 02 − 4(2)(4) = 0 − 32
= −32
𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 − 32 < 0
Note also that B ≠ 0 and A ≠ C.
Thus, the conic section is an ellipse.
c.
𝟒𝒙𝟐 + 𝟒𝒙𝒚 + 𝒚𝟐 + 𝟖𝒙 + 𝟐𝟒𝒚 + 𝟑𝟔 = 𝟎
𝐴 = 4, 𝐵 = 4, 𝐶 = 1
𝐵2 − 4𝐴𝐶 = 42 − 4(4)(1) = 16 − 16
=0
Thus, the conic section is a parabola.
d. 𝟒𝒙𝟐 + 𝟔𝒙𝒚 + 𝟐𝒚𝟐 − 𝟒𝒙 − 𝟐𝒚 + 𝟒𝟑 = 𝟎
𝐴 = 4, 𝐵 = 6, 𝐶 = 2
𝐵2 − 4𝐴𝐶 = 62 − 4(4)(2) = 36 − 32
=4
Note that 4 0
Thus, the conic section is a hyperbola.
Conics are very important in our daily life for some of the following reasons:
1. The world is round.
2. There are situations that are cyclic in nature.
3. The paths of the planets around the sun are ellipses with the sun at one focus.
4. The parabola is used in the design of car headlights and in spotlights because it aids
in concentrating the light beam.
5. The ideas of ellipse and hyperbola serve as the student’s foundation in the elliptical
and hyperbolic turns and motion. Several applications in the fields of astronomy
and design use the ellipse and the hyperbola as the serving model.
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## Geometric Sequences and Series
Introduction
We have discussed about arithmetic sequences, its characteristics and its connection to linear functions. In this post, we will discuss another type of sequence.
The sequence of numbers 2, 6, 18, 54, 162, … is an example of an geometric sequence. The first term 2 is multiplied by 3 to get the second term, the second term is multiplied by 3 to get the third term, the third term is multiplied by 3 to get the fourth term, and so on. The same number that we multiplied to each term is called the common ratio. Expressing the sequence above in terms of the first term and the common ratio, we have 2, 2(3), 2(32), 2(33), …. Hence, a geometric sequence, also known as a geometric progression, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.
The Sierpinski triangle below is an example of a geometric representation of a geometric sequence. The number of blue triangles, the number of white triangles, their areas, and their side lengths form different geometric sequences. It is left to the reader, as an exercise, to find the rules of these geometric sequences.
Figure 1 - The Seriepinski Triangles.
To generalize, if a1 is its first term and the common ratio is r, then the general form of a geometric sequence is a1, a1r, a1r2, a1r3,…, and the nth term of the sequence is a1rn-1.
A geometric series, on the other hand, is the sum of the terms of a geometric sequence. Given a geometric sequence with terms a1r, a1r2, a1r3,…, the sum Sn of the geometric sequence with n terms is the geometric series a1 + a1r + a1r2, a1r3 + … + arn-1. Multiplying Sn by -r and adding it to Sn, we have
Hence, the sum of a geometric series with n terms, and $r \neq 1 = \displaystyle\frac{a_1(1-r^n)}{1-r}$.
Sum of Infinite Geometric Series and a Little Bit of Calculus
Note: This portion is for those who have already taken elementary calculus.
The infinite geometric series $\{a_n\}$ is the the symbol $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots$. From above, the sum of a finite geometric series with $n$ terms is $\displaystyle \sum_{k=1}^n \frac{a_1(1-r^n)}{1-r}$. Hence, to get the sum of the infinite geometric series, we need to get the sum of $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r}$. However, $\displaystyle \sum_{k=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n\to \infty} \frac{a_1(1-r^n)}{1-r}$.
Also, that if $|r| < 1$, $r^n$ approaches $0$ (try $(\frac{2}{3})^n$ or any other proper fraction and increase the value of $n$), thus, $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r} = \frac{a_1}{1-r}$. Therefore, sum of the infinite series $\displaystyle a_1 + a_2r + a_2r^2 + \cdots = \frac{a_1}{1-r}$.
One very common infinite series is $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$, or the sum of the areas of the partitions of the square with side length 1 unit shown below. Using the formula above,
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{a_1}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$.
Figure 2 - A representation of an infinite geometric series.
This is evident in the diagram because the sum of all the partitions is equal to the area of a square. We say that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n}$ converges to 1.
## Is 0.999… really equal to 1?
Introduction
Yes it is. 0.999… is equal to 1.
Before we begin our discussion, let me make a remark that the symbol “…” in the decimal 0.999… means that the there are infinitely many 9’s, or putting it in plain language, the decimal number has no end.
For non-math persons, you will probably disagree with the equality, but there are many elementary proofs that could show it, some of which, I have shown below. A proof is a series of valid, logical and relevant arguments (see Introduction to Mathematical Proofs for details), that shows the truth or falsity of a statement.
Proof 1
$\frac{1}{3} = 0.333 \cdots$
$\frac{2}{3} = 0.666 \cdots$
$\frac{1}{3} + \frac{2}{3} = 0.333 \cdots + 0.666 \cdots$
$\frac{3}{3} =0.999 \cdots$
But $\frac{3}{3} = 1$, therefore $1 =0.999 \cdots$
Proof 2
$\frac{1}{9} = 0.111 \cdots$
Multiplying both sides by 9 we have
$1 = 0.999 \cdots$
Proof 3
Let $x = 0.999 \cdots$
$10x = 9.999 \cdots$
$10x - x = 9.999 \cdots - 0.9999 \cdots$
$9x = 9$
$x = 1$
Hence, $0.999 \cdots = 1$
Still in doubt?
Many will probably be reluctant in accepting the equality $1 = 0.999 \cdots$ because the representation is a bit counterintuitive. The said equality requires the notion of the real number system, a good grasp of the concept of limits, and knowledge on infinitesimals or calculus in general. If, for instance,you have already taken sequences (in calculus), you may think of the $0.999 \cdots$ as a sequence of real numbers $(0.9, 0.99, 0.999,\cdots)$. Note that the sequence gets closer and closer to 1, and therefore, its limit is 1.
Infinite Geometric Sequence
My final attempt to convince you that $0.999 \cdots$ is indeed equal to $1$ is by the infinite geometric sequence. For the sake of brevity, in the remaining part of this article, we will simply use the term “infinite sequence” to refer to an infinite geometric sequence. We will use the concept of the sum of an infinite sequence, which is known as an infinite series, to show that $0.999 \cdots = 1$.
One example of an infinite series is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$. If you add its infinite number of terms, the answer is equal to 1. Again, this is counterintuitive.
How can addition of numbers with infinite number of terms have an exact (or a finite) answer?
There is a formula to get the sum of an infinite geometric sequence, but before we discuss the formula, let me give the geometric interpretation of the sum above. The sum $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ can be represented geometrically using a 1 unit by 1 unit square as shown below. If we divide the square into two, then we will have two rectangles, each of which has area $\frac{1}{2}$ square units. Dividing the other half into two, then we have three rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{4}$ square units. Dividing the one of the smaller rectangle into two, then we have four rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{8}$. Again, dividing one of the smallest rectangle into two, we have five rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$, and $\frac{1}{16}$ Since this process can go on forever, the sum of all the areas of all the rectangles will equal to 1, which is the area of the original square.
Now that we have seen that an infinite series can have a finite sum, we will now show that $0.999 \cdots$ can be expressed as a finite sum by expressing it as an infinite series. The number $0.999 \cdots$ can be expressed as an infinite series $0.9 + 0.09 + 0.009 + \cdots$. Converting it in fractional form, we have $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$.
We have learned that the sum of the infinite series with first term $\displaystyle a_1$ and ratio $r$ is described by $\displaystyle\frac{a_1}{1-r}$. Applying the formula to our series above, we have
$\displaystyle\frac{\frac{9}{10}}{1-\frac{1}{10}} = 1$
Therefore, the sum our infinite series is 1.
Implication
This implication of the equality $0.999 \cdots =1$ means that any rational number that is a non-repeating decimal can be expressed as a repeating decimal. Since $0.999 \cdots =1$, it follows that $0.0999 \cdots =0.1, 0.00999 \cdots=0.01$ and so on. Hence, any decimal number maybe expressed as number + 0.00…01. For example, the decimal $4.7$, can be expressed as $4.6 + 0.1 = 4.6 + 0.0999 \cdots = 4.6999 \cdots$. The number $0.874$ can also be expressed as $0.873 + 0.001 = 0.873 + 0.000999 \cdots = 0.873999 \cdots$
Conclusion
Any of the four proofs above is actually sufficient to show that $0.999 \cdots = 1$. Although this concept is quite hard to accept, we should remember that in mathematics, as long as the steps of operations or reasoning performed are valid and logical, the conclusion will be unquestionably valid.
There are many counterintuitive concepts in mathematics and the equality $0.999 \cdots = 1$ is only one of the many. In my post, Counting the Uncountable: A Glimpse at the Infinite, we have also encountered one: that the number of integers (negative, 0, positive) is equal to the number of counting numbers (positive integers) and we have shown it by one-to-one pairing. We have also shown that the number of counting numbers is the same as the number of rational numbers. Thus, we have shown that a subset can have the same element as the “supposed” bigger set. I guess that is what makes mathematics unique; intuitively, some concepts do not make sense, but by valid and logical reasoning, they perfectly do.
Notes:
1. You can find discussions about 0.999… = 1 here and here.
2. There is another good post about it here and here.
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# Resolve the following into partial fractions. 2x−1(1−x−x2)(x+2)
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## To resolve the expression 2x−1(1−x−x2)(x+2) into partial fractions, we will follow these steps:Step 1: Set up the partial fraction decompositionWe express 2x−1(1−x−x2)(x+2) as a sum of partial fractions. Since 1−x−x2 is a quadratic polynomial, we can write:2x−1(1−x−x2)(x+2)=Ax+2+Bx+C1−x−x2where A, B, and C are constants to be determined.Step 2: Combine the fractionsTo combine the fractions on the right side, we find a common denominator:A(1−x−x2)+(Bx+C)(x+2)(1−x−x2)(x+2)=2x−1(1−x−x2)(x+2)Step 3: Eliminate the denominatorsSince the denominators are the same, we can equate the numerators:2x−1=A(1−x−x2)+(Bx+C)(x+2)Step 4: Expand the right sideNow we expand the right-hand side:A(1−x−x2)=A−Ax−Ax2(Bx+C)(x+2)=Bx2+2Bx+Cx+2C=(B)x2+(2B+C)x+2CCombining these gives:2x−1=(B−A)x2+(2B+C−A)x+(A+2C)Step 5: Compare coefficientsNow we compare coefficients from both sides:1. Coefficient of x2: B−A=0 (1)2. Coefficient of x: 2B+C−A=2 (2)3. Constant term: A+2C=−1 (3)Step 6: Solve the system of equationsFrom equation (1), we have:B=ASubstituting B=A into equation (2):2A+C−A=2⟹A+C=2(4)Substituting B=A into equation (3):A+2C=−1(5)Now we solve equations (4) and (5):From (4): C=2−ASubstituting into (5):A+2(2−A)=−1A+4−2A=−1−A+4=−1−A=−5⟹A=5Now substituting A=5 back into (4):5+C=2⟹C=2−5=−3And since B=A:B=5Step 7: Write the final partial fractionsNow we can substitute A, B, and C back into the partial fraction decomposition:2x−1(1−x−x2)(x+2)=5x+2+5x−31−x−x2Final AnswerThus, the partial fraction decomposition is:2x−1(1−x−x2)(x+2)=5x+2+5x−31−x−x2
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# 10th CBSE math solution for exercise 2.2 part 1
This page 10th CBSE math solution for exercise 2.2 part 1is going to provide you solution for every problems that you find in the exercise no 2.2
## 10th CBSE math solution for exercise 2.2 part 1
(iii) 6 x² – 3 – 7 x
Solution:
Find let us find the zeroes of the quadratic polynomial.
6 x² – 3 – 7 x = 0
6 x² - 7 x – 3 = 0
(3 x + 1) ( 2 x - 3) = 0
3 x + 1 = 0 2 x – 3 = 0
3 x = -1 2 x = 3
x = -1/3 x = 3/2
So the values of α = -1/3 and β = 3/2
Now we are going to verify the relationship between these zeroes and coefficients
6 x² - 7 x – 3 = 0
ax² + b x + c = 0
a = 6 b = -7 c = -3
Sum of zeroes α + β = -b/a
(-1/3) + (3/2) = -(-7)/6
(- 2 + 9)/6 = 7/6
7/6 = 7/6
Product of zeroes α β = c/a
(-1/3)(3/2) = -3/6
-1/2 = -1/2
(iv) 4 u² + 8 u
Solution:
Find let us find the zeroes of the quadratic polynomial.
4 u² + 8 u = 0
4 u (u + 2) = 0
4 u = 0 u + 2 = 0
u = 0 u = -2
So the values of α = 0 and β = -2
Now we are going to verify the relationship between these zeroes and coefficients
4 u² + 8 u = 0
ax² + b x + c = 0
a = 4 b = 8 c = 0
Sum of zeroes α + β = -b/a
0 + (-2) = -8/4
- 2 = -2
Product of zeroes α β = c/a
(0)(-2) = 0/4
0 = 0
(v) t² - 15
Solution:
Find let us find the zeroes of the quadratic polynomial.
t^2 - 15 = 0
t² = 15
t = √15
t = ± √15
t = √15 t = - √15
So the values of α = √15 and β = -√15
Now we are going to verify the relationship between these zeroes and coefficients
t² - 15 = 0
ax² + b x + c = 0
a = 1 b = 0 c = -15
Sum of zeroes α + β = -b/a
√15 + (-√15) = -0/1
0 = 0
Product of zeroes α β = c/a
(√15)( -√15) = -15/1
-15 = -15
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# Inter 2nd Year Maths 2A De Moivre’s Theorem Formulas
Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 2 De Moivre’s Theorem to solve questions creatively.
## Intermediate 2nd Year Maths 2A De Moivre’s Theorem Formulas
Statement:
→ If ‘n’ is an integer, then (cos θ + i sin θ)n = cos nθ + i sin nθ
If n’ is a rational number, then one of the values of
(cos θ + i sin θ)n is cos nθ + i sin nθ
nth roots of unity:
→ nth roots of unity are {1, ω, ω2 …….. ωn – 1}.
Where ω = $$\left[\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right]$$ k = 0, 1, 2 ……. (n – 1).
If ω is a nth root of unity, then
• ωn = 1
• 1 + ω + ω2 + ………… + ωn – 1 = 0
Cube roots of unity:
→ 1, ω, ω2 are cube roots of unity when
• ω3 = 1
• 1 + ω + ω2 = 0
• ω = $$\frac{-1+i \sqrt{3}}{2}$$, ω2 = $$\frac{-1-i \sqrt{3}}{2}$$
• Fourth roots of unity roots are 1, – 1, i, – i
→ If Z0 = r0 cis θ0 ≠ 0, then the nth roots of Z0 are αk = (r0)1/n cis$$\left(\frac{2 k \pi+\theta_{0}}{n}\right)$$ where k = 0, 1, 2, ……… (n – 1)
→ If n is any integer, (cos θ + i sin θ)n = cos nθ + i sin nθ
→ If n is any fraction, one of the values of (cosθ + i sinθ)n is cos nθ + i sin nθ.
→ (sinθ + i cosθ)n = cos($$\frac{n \pi}{2}$$ – nθ) + i sin($$\frac{n \pi}{2}$$ – nθ)
→ If x = cosθ + i sinθ, then x + $$\frac{1}{x}$$ = 2 cosθ, x – $$\frac{1}{x}$$ = 2i sinθ
→ xn + $$\frac{1}{x^{n}}$$ = 2cos nθ, xn – $$\frac{1}{x^{n}}$$ = 2i sin nθ
→ The nth roots of a complex number form a G.P. with common ratio cis$$\frac{2 \pi}{n}$$ which is denoted by ω.
→ The points representing nth roots of a complex number in the Argand diagram are concyclic.
→ The points representing nth roots of a complex number in the Argand diagram form a regular polygon of n sides.
→ The points representing the cube roots of a complex number in the Argand diagram form an equilateral triangle.
→ The points representing the fourth roots of complex number in the Argand diagram form a square.
→ The nth roots of unity are 1, w, w2,………. , wn-1 where w = cis$$\frac{2 \pi}{n}$$
→ The sum of the nth roots of unity is zero (or) the sum of the nth roots of any complex number is zero.
→ The cube roots of unity are 1, ω, ω2 where ω = cis$$\frac{2 \pi}{3}$$, ω2 = cis$$\frac{4 \pi}{3}$$ or
ω = $$\frac{-1+i \sqrt{3}}{2}$$
ω2 = $$\frac{-1-i \sqrt{3}}{2}$$
1 + ω + ω2 = 0
ω3 = 1
→ The product of the nth roots of unity is (-1)n-1 .
→ The product of the nth roots of a complex number Z is Z(-1)n-1 .
→ ω, ω2 are the roots of the equation x2 + x + 1 = 0
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# Introduction
Often, you will be asked to solve problems involving geometric relationships or other shapes. For real-world problems, those geometric relationships mostly involve measurable attributes, such as length, area, or volume.
Sometimes, those problems will involve the perimeter or circumference, or the area of a 2-dimensional figure.
For example, what is the distance around the track that is shown? Or, what is the area of the portion of the field that is covered with grass?
You may also see problems that involve the volume or surface area of a 3-dimensional figure. For example, what is the area of the roof of the building that is shown?
Another common type of geometric problem involves using proportional reasoning.
For example, an artist created a painting that needs to be reduced proportionally for the flyer advertising an art gallery opening. If the dimensions of the painting are reduced by a factor of 40%, what will be the dimensions of the image on the flyer?
In this resource, you will investigate ways to apply a problem-solving model to determine the solutions for geometric problems like these.
A basic problem solving model contains the following four steps:
Step 1: Read, understand, and interpret the problem. What information is presented? What is the problem asking me to find? What information may be extra information that I do not need? Step 2: Make a plan. draw a picture look for a pattern systematic guessing and checking acting it out making a table working a simpler problem working backwards Step 3: Implement your plan. What formulas do I need? What information can I interpret from the diagram, table, or other given information? Solve the problem. Step 4: Evaluate your answer. Does my answer make sense? Did I answer the question that was asked? Are my units correct?
# Solving Problems Using Perimeter and Circumference
You may recall that the perimeter of an object is the distance around the edge of the object. If the object contains circles, then you may need to think about the circumference of a circle, which is the perimeter of the circle.
Problem
A tire on a passenger car has a diameter of 18 inches. When the tire has rotated 5 times, how far will the car have traveled?
Step 1: Read, understand, and interpret the problem.
• What information is presented?
• What is the problem asking me to find?
• What information may be extra information that I do not need?
Step 2: Make a plan.
• Draw a picture.
• Use a formula: Which formula do I need to use?
(Hint: Look at your Mathematics Reference Materials)
• What formulas do I need?
• What information can I interpret from the diagram, table, or other given information?
• Solve the problem.
• Does my answer make sense?
• Are my units correct?
Practice
A cylindrical barrel with a diameter of 20 inches is used to hold fuel for a barbecue cook off. The chef rolls the barrel so that it completes 7 rotations. How many feet did the chef roll the barrel?
# Solving Problems Using Area and Surface Area
You may also encounter real-world geometric problems that ask you to find the area of 2-dimensional figures or the surface area of 3-dimensional figures. The key to solving these problems is to look for ways to break the region into smaller figures of which you know how to find the area.
Problem
Mr. Elder wants to cover a wall in his kitchen with wallpaper. The wall is shown in the figure below.
If wallpaper costs $1.75 per square foot, how much will Mr. Elder spend on wallpaper to completely cover this wall, excluding sales tax? To solve this problem, let's use the 4-step problem solving model. Step 1: Read, understand, and interpret the problem. • What information is presented? • What is the problem asking me to find? • What information may be extra information that I do not need? Step 2: Make a plan. • Draw a picture. Step 3: Implement your plan. • What formulas do I need? • What information can I interpret from the diagram, table, or other given information? • Solve the problem. Step 4: Evaluate your answer. • Does my answer make sense? • Did I answer the question that was asked? • Are my units correct? Practice Mrs. Nguyen wants to apply fertilizer to her front lawn. A bag of fertilizer that covers 1,000 square feet costs$18. How many bags of fertilizer will Mrs. Nguyen need to purchase?
Surface Area Problem
After a storm, the Serafina family needs to have their roof replaced. Their house is in the shape of a pentagonal prism with the dimensions shown in the diagram.
A roofing company gave Mr. Serafina an estimate based on a cost of $2.75 per square foot to replace the roof. How much will it cost the Serafina family to have their roof replaced? To solve this problem, let's use the 4-step problem solving model. Step 1: Read, understand, and interpret the problem. • What information is presented? • What is the problem asking me to find? • What information may be extra information that I do not need? Step 2: Make a plan. • Draw a picture. Step 3: Implement your plan. • What formulas do I need? • What information can I interpret from the diagram, table, or other given information? • Solve the problem. Step 4: Evaluate your answer. • Does my answer make sense? • Did I answer the question that was asked? • Are my units correct? Practice To match their new roof, Mrs. Serafina decided to have both pentagonal sides of their house covered in aluminum siding. Their house is in the shape of a pentagonal prism with the dimensions shown in the diagram. A contractor gave Mrs. Serafina an estimate based on a cost of$3.10 per square foot to complete the aluminum siding. How much will it cost the Serafina family to have the aluminum siding installed?
# Solving Problems Using Proportionality
Proportional relationships are another important part of geometric problem solving.
A woodblock painting has dimensions of 60 centimeters by 79.5 centimeters. In order to fit on a flyer advertising the opening of a new art show, the image must be reduced by a scale factor of 1/25.
What will be the final dimensions of the image on the flyer?
Measuring Problem
For summer vacation, Jennifer and her family drove from their home in Inlandton to Beachville. Their car can drive 20 miles on one gallon of gasoline. Use the ruler to measure the distance that they drove to the nearest 1/4 inch, and then calculate the number of gallons of gasoline their car will use at this rate to drive from Inlandton to Beachville.
Step 1: Read, understand, and interpret the problem.
• What information is presented?
• What is the problem asking me to find?
• What information may be extra information that I do not need?
Step 2: Make a plan.
• Draw a picture.
• What formulas do I need?
• What information can I interpret from the diagram, table, or other given information?
• Solve the problem.
• Does my answer make sense?
• Are my units correct?
Practice #1
A blueprint for a rectangular tool shed has dimensions shown in the diagram below.
Todd is using this blueprint to build a tool shed, and he wants to surround the base of the tool shed with landscaping timbers as a border. How many feet of landscaping timbers will Todd need?
Practice #2
A scale model of a locomotive is shown. Use the ruler to measure the dimensions of the model to the nearest 1/4 inch, and then calculate the actual dimensions of the locomotive.
Scale: 1 inch = 5 feet
# Summary
Solving geometric problems, such as those found in art and architecture, is an important skill. As with any mathematical problem, you can use the 4-step problem solving model to help you think through the important parts of the problem and be sure that you don't miss key information.
There are a lot of different applications of geometry to real-world problem solving. Some of the more common applications include the following:
Perimeter or Circumference Applications
What is the perimeter of the base of the cup, if the cup is in the shape of an octagonal prism?
In West Texas, farmers use circular sprinklers that pivot around a center point to irrigate their crops. From the air, you can see distinct circles from the resulting vegetation. If there are four sprinklers anchored in a field and each sprinkler extends 100 feet from the pivot point, how much area does the farmer irrigate?
Surface Area Applications
The JP Morgan Chase Bank Tower in downtown Houston, Texas, is one of the tallest buildings west of the Mississippi River. It is in the shape of a pentagonal prism. If 40% of each face is covered with glass windows, what is the amount of surface area covered with glass?
Proportional Reasoning
The dimensions of Vincent van Gogh's Starry Night are 29 inches by $36\frac{1}{4}$ inches. If a print reduces these dimensions by a scale factor of 30%, what will be the dimensions of the print?
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# Factors of 911: Prime Factorization, Methods, and Examples
The numbers that can be divided by 911 are called the factors of 911. It is an odd integer because it is 911.
It only has two elements because it is a prime number. Learn more about the causes of 911 now.
### Factors of 911
Here are the factors of the number 911.
Factors of 911: 1 and 911
### Negative Factors of 911
The negative factors of 911 are similar to its positive aspects, just with a negative sign.
Negative Factors of 911: -1, and – 911
### Prime Factorization of 911
The prime factorization of 911 is the way of expressing its prime factors in the product form.
Prime Factorization: 1 x 911
In this article, we will learn about the factors of 911 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 911?
The factors of 911 are 1 and 911. These numbers are the factors as they do not leave any remainder when divided by 911.
The factors of 911 are classified as prime numbers and composite numbers. The prime factors of the number 911 can be determined using the prime factorization technique.
## How To Find the Factors of 911?
You can find the factors of 911 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 911, create a list containing the numbers that are exactly divisible by 911 with zero remainders. One important thing to note is that 1 and 911 are the 911’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 911 are determined as follows:
$\dfrac{ 911}{1} = 911$
$\dfrac{ 911}{ 911} = 1$
Therefore, 1 and 911 are the factors of 911.
### Total Number of Factors of 911
For 911, there are 2 positive factors and 2 negative ones. So in total, there are 4 factors of 911.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 911 is given as:
The factorization of 911 is 1 x 911.
The exponent of 1, and 911 is 1.
Adding 1 to each and multiplying them together results in 4.
Therefore, the total number of factors of 911 is 4. 2 are positive, and 2 factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 911 by Prime Factorization
The number 911 is a prime number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 911 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 911, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 911 can be expressed as:
911 = 1 x 911
## Factors of 911 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 911, the factor pairs can be found as:
1 x 911 = 911
The possible factor pairs of 911 are given as (1, 911).
All these numbers in pairs, when multiplied, give 911 as the product.
The negative factor pairs of 911 are given as:
-1 x -911 = 911
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1 and -911 are called negative factors of 911.
The list of all the factors of 911, including positive as well as negative numbers, is given below.
Factor list of 911: 1,-1,911 and – 911
## Factors of 911 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 911 are there?
### Solution
The total number of Factors of 911 is 4.
Factors of 911 are 1 and 911.
### Example 2
Find the factors of 911 using prime factorization.
### Solution
The prime factorization of 911 is given as:
911 $\div$ 1 = 911
911 $\div$ 911 = 1
So the prime factorization of 911 can be written as:
1 x 911 = 911
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