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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.
Format
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