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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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# 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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# 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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# 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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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:
_
/
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Minimum point:
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Point of inflexion:
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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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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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# 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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# Find the square root of: $12\sqrt 5 + 2\sqrt {55}$ A. $\left( {\sqrt {11} + 1} \right)\sqrt[4]{5}$ B. $\sqrt[4]{5}\left( {1 + \sqrt 5 } \right)$ C. $\sqrt[4]{5}\left( {\sqrt {11} + \sqrt 5 } \right)$ D. $\sqrt 5 \left( {\sqrt {11} + 1} \right)$
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Hint: In this question, we will use factorization, and expansion of algebraic identities. For this problem, we will use the algebraic identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$ .
Complete step by step solution:
Now, in this question, we have to find the square root of $12\sqrt 5 + 2\sqrt {55}$.
So it will become: $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, which on simplification will become:
$\sqrt {12\sqrt 5 + 2\sqrt {55} } \\ = \sqrt {12\sqrt 5 + 2\sqrt {5 \times 11} } \\ = \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\$
Now, to solve $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, we will take $\sqrt 5$ common within the under root and get:
$= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\$
Now we will change the term $(12 + 2\sqrt {11} )$ inside the under root sign to express it in terms of ${(a + b)^2}$ .
Now we can write $(12 + 2\sqrt {11} )$ as:
$(1 + 11 + 2\sqrt {11} )$ which can we reframed as:
$({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$, comparing it with the RHS of the expansion of the algebraic identity ${(a + b)^2}$ which is given as:
${a^2} + 2ab + {b^2}$
We will get $({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$= ${a^2} + 2ab + {b^2}$
So that,
${a^2} = {1^2}, \\ 2ab = 2.1.\sqrt {11} \\ {b^2} = {(\sqrt {11} )^2} \\$
Such that we get :
$a = 1, \\ 2ab = 2.1.\sqrt {11} \\ b = \sqrt {11} \\$
Now, since
${a^2} + 2ab + {b^2} = {(a + b)^2}$
Then putting the values obtained above:
$({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} ) = {(1 + \sqrt {11} )^2}$
Therefore $\sqrt {12\sqrt 5 + 2\sqrt {55} }$ will now become:
$= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 ({1^2} + {{(\sqrt {11} )}^2} + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 {{(1 + \sqrt {11} )}^2}} \\ = \sqrt {\sqrt 5 } (1 + \sqrt {11} ) \\ = \sqrt[4]{5}(1 + \sqrt {11} ) \\$
So, finally we can say that :
Square root of $12\sqrt 5 + 2\sqrt {55}$
$= \sqrt[4]{5}(\sqrt {11} + 1)$
Hence, the correct answer is option A.
Note: We cannot afford to forget the square root operation throughout the solution of this problem. For such problems, which require us to find the square root of another square root, we need to identify the algebraic expansion accurately so that we can get the correct corresponding algebraic identity to simplify and evaluate the square root.
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maths > commercial-arithmetics
Proportions
what you'll learn...
overview
Two ratios are said to be in proportion, if the ratios are equivalent. For example 2:4$2 : 4$ and 3:6$3 : 6$ are equivalent. Such equivalent ratios are formally represented as a proportion. The representation is 2:4::3:6$2 : 4 : : 3 : 6$.
illustrative example
A ratio of two quantities helps: to understand and to use the comparative measure of quantities.
Let us consider making dough for chappati or pizza. The recipe gives that for 400$400$ gram of flour, 150$150$ml water is used.
One person has only 200$200$ gram of flour. In that case, the person can reduce the water to $75$ml.
For $400$ gram flour $150$ml water is used. This is in $400 : 150$ ratio.
For $200$ gram flour $75$ml water is used. This is in $200 : 75$ ratio.
These two ratios can be simplified to $8 : 3$.
To denote that two ratios are identical, they are said to be in same proportion. That is $400 : 150$ is in the same proportion as $200 : 75$. This is given as $400 : 150 : : 200 : 75$.
proportion
The word "proportion" means: comparative measurement of quantities".
Pro-portion was from root word meaning "person's portion or share".
Proportion : Two ratios are said to be in proportion if the corresponding terms of ratio are identical in the simplified form.
Consider the example $2 : 3 : : 4 : 6$ proportion.
• The numbers in the proportion are called first term, second term, third term, and fourth term in the order.
• The first and fourth terms are called the extremes of the proportion.
• The second and third terms are called the means of the proportion.
The word "extreme" means: farthest from the center.
The root word is from "exter" meaning outer.
The word "mean" means: average. The word is derived from a root word meaning "middle".
illustrative example
• Fruit basket $A$ has $4$ apples and $16$ oranges. Ratio of apples to oranges is $4 : 16$ which is $1 : 4$.
• Fruit basket $B$ has $20$ apples and $80$ oranges. Ratio of apples to oranges is $20 : 80$ which is $1 : 4$.
The proportion of apples to oranges in the two baskets is $4 : 16 : : 20 : 80$.
The proportion of apples to oranges in basket $A$ and basket $B$ is $4 : 16 : : 20 : 80$.
The word "Proportion" is also used to specify the simplified ratio, as in the following.
The proportion of apples to oranges in basket $A$ and basket $B$ is $1 : 4$.
Students are reminded to note the context in which the word "proportion" is used.
The proportion of count in basket A to count in basket B for apples and oranges is $4 : 20 : : 16 : 80$.
This proportion is given as
apples of basket $A$ to basket $B$ is in the same proportion as oranges of basket $A$ to basket $B$
• Fruit basket $A$ has $4$ apples and $16$ oranges.
• Fruit basket $B$ has $20$ apples and $80$ oranges.
The proportion of apples to oranges in basket $A$ and basket $B$ is $4 : 16 : : 20 : 80$.
The proportion of apples and oranges in basket $A$ to basket $B$ is $4 : 20 : : 16 : 80$.
Students are reminded to note the context in which proportion is defined.
proportion to fractions
We learned that "Ratio can be equivalently represented as a fraction.".
A basket has $3$ apples and $4$ oranges.
• The ratio of the number of apples to number of oranges is $3 : 4$.
• Number of apples are $\frac{3}{4}$ of the number of oranges.
The number of apples to number of oranges in basket A and B is in proportion $3 : 4 : : 6 : 8$.The following are all true.
the number of apples is $\frac{3}{4}$ of the number of oranges in basket A
the number of apples is $\frac{6}{8}$ of the number of oranges in basket B
the number of apples is $\frac{3}{4}$ of the number of oranges -- in both basket A and basket B
Note that in a proportion, the two fractions are equivalent fractions.
$\frac{6}{8}$, when simplified, is $\frac{3}{4}$.
Given the proportion $3 : 4 : : 6 : 8$,
$\frac{3}{4} = \frac{6}{8}$, that is, the two ratios given as fractions are always equal.
formula
Given a proportion, $a : b : : c : d$, it is understood that $\frac{a}{b} = \frac{c}{d}$. In that case,
$a \times d = b \times c$.
Given a proportion, $a : b : : c : d$, it is understood that $\frac{a}{b} = \frac{c}{d}$.
Given that
$\frac{a}{b} = \frac{c}{d}$
Note that $a , b , c , d$ are numbers. As per the properties of numbers, if two numbers are equal, then the numbers multiplied by another number are equal. (eg: if $4 = 2 \times 2$, then multiplying by $5$ we get $4 \times 5 = 2 \times 2 \times 5$.)
$\frac{a}{b}$ and $\frac{c}{d}$ are two numbers that are equal. On multiplying these numbers by $b d$, we get the two numbers $\frac{a}{b} \times b d$ and $\frac{c}{d} \times b d$. Simplifying these two numbers we get, $a d$ and $b c$. As per the property, these two numbers are equal.
$a d = b c$.
That is product of extremes and product of means are equal.
examples
Given the proportion $3 : 4 : : x : : 12$ find the value of $x$.
The answer is "$9$". Products of extremes equals product of means.
$3 \times 12 = 4 \times x$
$x = \frac{36}{4}$
$x = 9$
summary
Proportion : Two ratios are said to be in proportion if the corresponding terms of ratio are identical in the simplified form.
Mean-Extreme Property of Proportions : product of extremes = product of means
If $a : b : : c : d$ is a proportion, then $a d = b c$
Outline
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# Geometric Probability
## Use geometric properties to evaluate probability
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Basic Geometric Probabilities
A rectangular dartboard that measures 12 inches by 24 inches has a 2-inch by 2-inch red square painted at its center. What is the probability that a dart that hits the dartboard will land in the red square?
### Geometric Probabilities
Sometimes we need to use our knowledge of geometry to determine the likelihood of an event occurring. We may use areas, volumes, angles, polygons or circles.
Let's solve the following problems.
1. A game of pin-the-tale-on-the-donkey has a rectangular poster that is 2ft by 2ft. The area in which the tale should be pinned is shown as a circle with radius 1 inch. Assuming that the pinning of the tale is completely random and that it will be pinned on the poster (or the player gets another try), find the probability of pinning the tale in the circle?
This probability can be found by dividing the area of the circle target by the area of the poster. We must have the same units of measure for each area so we will convert the feet to inches.
\begin{align*}\frac{1^2 \pi}{24^2} \thickapprox 0.005454 \ or \ \text{about} \ 0.5 \% \ \text{chance}.\end{align*}
1. In a game of chance, a pebble is dropped onto the board shown below. If the radius of each of blue circle is 1 cm, find the probability that the pebble will land in a blue circle.
The area of the square is \begin{align*}16 \ cm^2\end{align*}. The area of each of the 16 circles is \begin{align*}1^2 \pi=\pi\end{align*}. The probability of the pebble landing in a circle is the sum of the areas of the circles divided by the area of the square.
\begin{align*}P(\text{blue circle}) = \frac{16 \pi}{64} \thickapprox 0.785\end{align*}
1. What is the probability that a randomly thrown dart will land in a red area on the dart board shown? What is the probability that exactly two of three shots will land in the red? The radius of the inner circle is 1 unit and the radius of each annulus is 1 unit as well.
First we need to determine the probability of landing in the red. There are four rings of width 1 and the radius of the center circle is also 1 so the total radius is 5 units. The area of the whole target is thus \begin{align*}25 \pi\end{align*} square units. Now, we need to find the areas of the two red rings and the red circular center. The center circle area is \begin{align*}\pi\end{align*} square units. The outside ring area can be found by subtracting the area inside from the entire circle’s area. The inside circle will have a radius of 4 units, the area of the outer ring is \begin{align*}25 \pi -16 \pi=9 \pi\end{align*} square units. This smaller red ring’s area can be found similarly. The circle with this red ring on the outside has a radius of 3 and the circle inside has a radius of 2 so, \begin{align*}9 \pi -4 \pi=5 \pi\end{align*} square units. Finally, we can add them together to get the total red area and divide by the area of the entire target. \begin{align*}\frac{9 \pi+5 \pi+ \pi}{25 \pi}=\frac{15 \pi}{25 \pi}=\frac{3}{5}\end{align*}. So the probability of hitting the red area is \begin{align*}\frac{3}{5}\end{align*} or 60%.
For the second part of the problem we will use a binomial probability. There are 3 trials, 2 successes and the probability of a success is 0.6: \begin{align*}\dbinom{3}{2}(0.6)^2(0.4)=0.432\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the probability that a dart that hits the dartboard will land in the red square.
This probability can be found by dividing the area of the red square by the area of the dartboard. The area of the dartboard is \begin{align*}12\times24=288\end{align*}. The area of the red square is \begin{align*}2\times2=4\end{align*}. Therefore, the probability of the dart landing in the red square is
\begin{align*}\frac{4}{288}\\ \frac{1}{72}\\ \approx 0.0139\end{align*}
Therefore, there is about a 1.39% chance the dart will hit the red square.
#### Example 2
Consider the picture below. If a “circle” is randomly chosen, what is the probability that it will be:
1. red
\begin{align*}\frac{29}{225}\end{align*}
1. yellow
\begin{align*}\frac{69}{225}\end{align*}
1. blue or green
\begin{align*}\frac{84}{225}\end{align*}
1. not orange
\begin{align*}\frac{182}{225}\end{align*}
#### Example 3
If a dart is randomly thrown at the target below, find the probability of the dart hitting in each of the regions and show that the sum of these probabilities is 1. The diameter of the center circle is 4 cm and the diameter of the outer circle is 10 cm. What is the probability that in 5 shots, at least two will land in the 4 region?
\begin{align*}P(1)&=\frac{2^2 \pi}{5^2 \pi}=\frac{4}{25} \\ P(2)&=\frac{120}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{3} \times \frac{21 \pi}{25 \pi}=\frac{7}{25} \\ P(3)&=\frac{90}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{4} \times \frac{21 \pi}{25 \pi}=\frac{21}{100} \\ P(4)&=\frac{150}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{5}{12} \times \frac{21 \pi}{25 \pi}=\frac{35}{100}\end{align*}
\begin{align*}& P(1) + P(2) + P(3) + P(4) = \\ &=\frac{4}{25}+\frac{7}{25}+\frac{21}{100}+\frac{35}{100} \\ &=\frac{16}{100}+\frac{28}{100}+\frac{21}{100}+\frac{35}{100} \\ &=\frac{100}{100}=1\end{align*}
The probability of landing in region 4 at least twice in five shots is equivalent to \begin{align*}1 - \left [ P(0) + P(1) \right ]\end{align*}.
Use binomial probability to determine these probabilities:
\begin{align*}&1-\left [ \dbinom{5}{0} \left(\frac{35}{100}\right)^0 \left(\frac{65}{100}\right)^5+ \dbinom{5}{1} \left(\frac{35}{100}\right)^1 \left(\frac{65}{100}\right)4\right ] \\ &=1-(0.116029+0.062477) \\ & \thickapprox 0.821\end{align*}
### Review
Use the diagram below to find the probability that a randomly dropped object would land in a triangle of a particular color.
1. yellow
2. green
3. plum
4. not yellow
5. not yellow or light blue
The dart board to the right has a red center circle (bull’s eye) with area \begin{align*}\pi \ cm^2\end{align*}. Each ring surrounding this bull’s eye has a width of 2 cm. Use this information to answer the following questions.
1. Given a random throw of a dart, what is the probability that it will land in a white ring?
2. What is the probability of a bull’s eye?
3. What is the probability that in 10 throws, exactly 6 land in the black regions?
4. What is the probability that in 10 throws, at least one will land in the bull’s eye?
5. How many darts must be thrown to have a 95% chance of making a bull’s eye?
To see the Review answers, open this PDF file and look for section 12.11.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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# A Fraction as a Percent
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The authors
Chris S.
## Basics on the topicA Fraction as a Percent
After this lesson you will be able to convert fractions to percents and back again, using visual models and double number lines.
The lesson begins with a review of the definition of percent, which leads to a method for writing any percent as a fraction over 100. It continues by applying this method to solve real world problems, with the help of visual models and your knowledge of equivalent fractions.
Learn how to convert fractions to percents and back again, by helping Olivia battle wildfires!
This video includes key concepts, notation, and vocabulary such as: percent (a quantity out of 100); equivalent fractions (two fractions which represent the same proportional relationship); and double number line (a visual representation of a ratio, in which the units are represented by two simultaneous number lines).
Before watching this video, you should already be familiar with the relationship between percents and part-to-whole ratios, and rates where the whole is 100.
After watching this video, you will be prepared to solve problems by finding the percent of a quantity.
Common Core Standard(s) in focus: 6.RP.A.3.c A video intended for math students in the 6th grade Recommended for students who are 11-12 years old
### TranscriptA Fraction as a Percent
Olivia has spent the last months training herself physically and mentally to be ready to battle wildfires. She’s passed every test so far and now she’s reached the last challenge. The firefighters have set a controlled burn in a field at their training center and she must extinguish 85% of the flames, by herself, before sundown. In order to track her progress and accomplish the mission, Olivia will need to convert fractions to percents, and back again. Time is ticking but before grabbing a firehose, Olivia needs to perform some calculations. The whole field is divided into 20 squares. If she needs to extinguish 85% of the field, how many squares IS that? Remember, a percent is a quantity out of 100. That means 85% is the same as the fraction 85 over 100. We can use equivalent fractions to rewrite 85 over 100 to what we need. For this problem, we want to find the equivalent fraction that has 20 in the denominator. 85 over 100 equals WHAT over 20? In order to change the denominator from 100 to 20, we need to divide by 5. But since we want an equivalent fraction, we also have to divide the numerator by 5. Dividing 85 by 5 gives us 17 which tells us 85 percent is equivalent to the fraction seventeen twentieths. Olivia needs to extinguish 17 out of the 20 squares. After a few hours of hard work, she takes a short break giving us time to calculate her progress so far. Olivia has extinguished the fire in 8 squares out of 20. What percentage has she extinguished so far? Let's take a look at the grid. Here are the eight squares where the fire has already been extinguished. On the right side we can write the number of squares in each row: 4, 8, 12, 16 and 20. Hmm, this reminds us of a double number line, so let's put the strip right here and turn it on it's side. The whole is always equal to 100 percent, so we can write that 20 squares is equal to 100 percent. We need to divide the 100 percent into 5 parts. Since 100 divided by 5 is 20, each part is 20 percent. That gives us 20 percent, 40 percent, 60 percent, and 80 percent. Looking at the double number line, that shows us that 8 squares represents 40 percent. That's 40 percent of the fire already extinguished. Olivia just remembered that she can earn a special commendation if she extinguishes at least 90 percent of the field before noon and now that seems like a real possibility. It looks like Olivia has extinguished all the squares but one! Has she earned the special commendation? What is the fraction of the fire that is still burning? One twentieth. What percent is that? Again, a percent is a quantity out of 100. That means we have to find an equivalent fraction with 100 in the denominator. What can we multiply 20 by to get 100? 100 divided by 20 is 5, so we need to multiply the denominator by 5. But since we want to find an equivalent fraction, we also have to multiply the numerator by 5. That gives us 5 over 100, which is equal to 5%. If only 5% is still burning, what percent did Olivia extinguish? 100%, which represents one whole, minus 5% gives us 95%. She did it! While Olivia takes a well-earned rest, let's review. Because percents are a quantity out of 100, we can convert fractions to percents, and vice versa. To convert a percent to a fraction, first write the percent as a fraction with 100 in the denominator. Then multiply or divide the numerator and denominator by the same number to find the equivalent fraction with the denominator you want. To convert a fraction to a percent, we first find what number we need to multiply or divide the denominator by in order to get 100. Then we multiply or divided both the numerator and denominator by that value to get a fraction out of 100. The numerator shows us the percent. This allows us to compare numbers even when they are given in different forms. Congratulations Olivia! With careful conversions of fractions to percents and back, you'll always be able to track your firefighting progress. But what’s this? Hmm, I’m not sure fractions and percents will help here. Olivia has to rescue 100% of the cat!
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## Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions)
### Locus and Its Constructions Exercise 16A – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC.
Prove— Q is equidistant from A and B.
Solution:
Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90 )
PQ = PQ (Common)
By Side-Angle-Side criterian of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.
Question 2.
Given— CP is bisector of angle C of ∆ ABC.
Prove— P is equidistant from AC and BC.
Solution:
Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.
Prove—
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:
Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than $$\frac{1}{2} \mathrm{BC}$$ draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Hence, D is equidistant from B and C.
Question 5.
In each of the given figures: PA = PH and QA = QB.
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
Construction: Join PQ which meets AB in D.
Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.
Question 6.
Construct a right angled triangle PQR, in which ∠ Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
∆PQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Hence, T is equidistant from PQ and QR.
Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Hence P is equidistant from AC and BC.
Solution:
Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Solution:
Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that –
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:
Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M
Question 10.
Use ruler and compasses only for this question:
(i) construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Solution:
Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60o. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
⇒ BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
PB=3.6 cm
Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E
Prove that:
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:
Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC …. (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD …. (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.
Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:
Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.
Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B.
Solution:
Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.
Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.
Solution:
Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.
Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Solution:
Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.
Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Solution:
Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.
Question 19.
On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.
Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.
### Locus and Its Constructions Exercise 16B – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Describe the locus of a point at a distance of 3 cm from a fixed point.
Solution:
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle.
Question 2.
Describe the locus of a point at a distance of 2 cm from a fixed line.
Solution:
The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.
Question 3.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:
The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.
Question 4.
Describe the locus of the moving end of the minute hand of a clock.
Solution:
The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.
Question 5.
Describe the locus of a stone dropped from the top of a tower.
Solution:
The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.
Question 6.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:
The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.
Question 7.
Describe the locus of the door handle as the door opens.
Solution:
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.
Question 8.
Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.
Question 9.
Describe the locus of the centers of all circles passing through two fixed points.
Solution:
The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.
Question 10.
Describe the locus of vertices of all isosceles triangles having a common base.
Solution:
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.
Question 11.
Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.
Question 12.
Describe the locus of a point P so that:
AB2 = AP2 + BP2, where A and B are two fixed points.
Solution:
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2 + BP2.
Question 13.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
i) AB and BC
ii) B and D.
Solution:
i)
The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.
ii)
The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.
Question 14.
The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
Solution:
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.
Question 15.
Describe:
i) The locus of points at distances less than 3 cm from a given point.
ii) The locus of points at distances greater than 4 cm from a given point.
iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
iv) The locus of points at distances greater than or equal to 35 mm from a given point.
v)The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.
vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.
Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.
Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Solution:
Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively.
Hence, a, b, c and d are the required four points.
Question PQ.
By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Solution:
Steps of construction:
i) Draw a linen.
ii) Take a point Lonn and draw a perpendicular to n.
iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
iv) At M, draw a line m making an angle of 90°.
v) Produce LM and mark a point P such that PM = 2 cm.
vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.
A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.
Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
i) Draw an angle of 60° with AB = BC = 8 cm
ii) Draw the angle bisector BX of ∠ABC
iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
iv) Join MP, NP
BMPN is a rhombus since MP = BM = NB = NP = 4 cm
Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:
Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.
Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠ A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:
Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.
Question PQ.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
P moves along AB, and Qmoves in such a way that PQ is always equal to OP.
But Pis the mid-point of OQ
Now in ∆OQQ’
P’and P” are the mid-points of OQ’ and OQ”
Therefore, AB||Q’Q”
Therefore, Locus of Q is a line CD which is parallel to AB.
Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:
Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.
Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:
Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.
Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:
Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.
Question 25.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.
Question 26.
State the locus of a point in a rhombus ABCD, which is equidistant
(ii) from the vertices A and C.
Solution:
Question 27.
Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Steps of Construction:
i) Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm
Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Solution:
Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.
Question PQ.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Solution:
Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at Mand meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof:
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore. Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.
Question 29.
Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Solution:
Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A.
Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Solution:
i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
ii) On measuring AB = 5.7 cm
Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
i. Steps of construction:
1. Draw BC = 6.5 cm using a ruler.
2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
3. With Q as center and same radius, cut the previous arc at P.
4. Join BP and extend it.
5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
6. Join AC to obtain ΔABC.
ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.
iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.
iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)
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Question Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point | Nagwa Question Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point | Nagwa
# Question Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point Mathematics • Second Year of Secondary School
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Given that lim_(π₯ β β2) (π(π₯))/(3π₯Β²) = β3, determine lim_(π₯ β β2) π(π₯).
02:04
### Video Transcript
Given that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to negative three, determine the limit as π₯ tends to negative two of π of π₯.
In this question, weβ²ve been given the limit as π₯ tends to negative two of π of π₯ over three π₯ squared. We can break this limit down using the properties of limits. We have the property for limits of quotients of functions, which tells us that the limit as π₯ tends to π of π of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as π₯ tends to π of π of π₯. For our limit, weβre taking the limit as π₯ tends to negative two. Therefore, π is equal to negative two. And we have a quotient of functions. In the numerator, we have π of π₯. And in the denominator, we have three π₯ squared. Applying this rule for limits of quotients of functions, we obtain that our limit is equal to the limit as π₯ tends to negative two of π of π₯ over the limit as π₯ tends to negative two of three π₯ squared.
Letβ²s now consider the limit in the denominator of the fraction. Thatβ²s the limit as π₯ tends to negative two of three π₯ squared. We can apply direct substitution to this limit, giving us that the limit as π₯ tends to negative two of three π₯ squared is equal to three times negative two squared. Negative two squared is equal to four. We then simplify to obtain that this limit is equal to 12. We can substitute this value of 12 back in to the denominator of our fraction, giving us that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to the limit as π₯ tends to negative two of π of π₯ all over 12.
However, weβ²ve been given in the question that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to negative three. And since this is on the left-hand side of our equation, we can set our equation equal to negative three. So we now have that the limit as π₯ tends to negative two of π of π₯ over 12 is equal to negative three. We simply multiply both sides of the equation by 12. Here, we reach our solution which is that the limit as π₯ tends to negative two of π of π₯ is equal to negative 36.
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Modular Arithmetic/What is a Modulus?
Modular Arithmetic What is a Modulus? Modular Arithmetic →
In modular arithmetic, 38 can equal 14 — what??
You might be wondering how so! (Or you might already know how so, but we will assume that you don't.) Well, modular arithmetic works as follows:
Think about military time which ranges from 0000 to 2359. For the sake of this lesson, we will only consider the hour portion, so let us consider how hourly time works from 0 to 23. After a 24 hour period, the time restarts at 0 and builds again to 23. So, using this reasoning, the 27th hour would be equivalent to the third hour in military time, as the remainder when 27 is divided by 24 is 3 (make sure you understand why this is true as it is vital to understanding everything that follows in this book). So to get any hour into military time we just find the number's remainder when divided by 24. What we really are doing is finding that number modulo 24.
To write the earlier example mathematically, we would write:
${\displaystyle 27\equiv 3{\pmod {24}}}$
which reads as "27 is congruent to 3 modulo 24". Let us try one more example before setting you loose: what is 34 congruent to modulo 4, which can also be written as "34 (mod 4)" which is read as "what is 34 congruent to modulo 4?". We find the remainder when 34 is divided by 4; the remainder is 2. So, ${\displaystyle 34\equiv 2{\pmod {4}}}$.
Now, try some on your own:
Exercises: Definition of Modular Arithmetic Replace the '?' by the correct value. ${\displaystyle 25\equiv {\text{?}}{\pmod {7}}}$ ${\displaystyle 10^{10}\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {729}}}$
Challenge: Definition of Modular Arithmetic These two are a lot harder. Simplify: ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {11}}}$ You do not need to work out what (936 + 3) is to solve these, but they will require some calculation.
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# Properties of a Parallelogram
In Geometry, a parallelogram is a type of quadrilateral. It is a two-dimensional figure with four sides. The most important properties of a parallelogram are that the opposite sides are parallel and congruent and the opposite angles are also equal. In this article, let us discuss all the properties of a parallelogram with a complete explanation and many solved examples.
Table of Contents:
## Important Properties of a Parallelogram
A parallelogram is a closed four-sided two-dimensional figure in which the opposite sides are parallel and equal in length. Also, the opposite angles are also equal. Learning the properties of a parallelogram is useful in finding the angles and sides of a parallelogram. The four most important properties of a parallelogram are:
• The opposite sides of a parallelogram are equal in measurement and they are parallel to each other.
• The opposite angles of a parallelogram are equal.
• The sum of interior angles of a parallelogram is equal to 360°.
• The consecutive angles of a parallelogram should be supplementary (180°).
## Theorems on Properties of a Parallelogram
The 7 important theorems on properties of a parallelogram are given below:
### Theorem 1:
A diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Proof:
Assume that ABCD is a parallelogram and AC is a diagonal of a parallelogram.
The diagonal AC divides the parallelogram into two congruent triangles, such as ∆ABC and ∆CDA. Now, we need to prove that the triangles ∆ABC and ∆CDA are congruent.
From the triangles, ∆ABC and ∆CDA, AD || BC, and AC is a transversal.
Hence, ∠ BCA = ∠ DAC (By using the property of pair of alternate angles)
Also, we can say that AB || DC and line AC is transversal.
Thus, ∠ BAC = ∠ DCA (Using the pair of alternate angles)
Also, AC = CA (Common side)
By using the Angle side Angle rule (ASA rule), we can conclude that ∆ABC is congruent to ∆CDA.
(i.e) ∆ABC ≅ ∆CDA.
Thus, the diagonal AC divides a parallelogram ABCD into two congruent triangles ABC and CDA.
Hence, proved.
### Theorem 2:
The opposite sides of a parallelogram are equal.
Proof:
From theorem 1, it is proved that the diagonals of a parallelogram divide it into two congruent triangles. When you measure the opposite sides of a parallelogram, it is observed that the opposite sides are equal.
Hence, we conclude that the sides AB = DC and AD = BC.
### Theorem 3:
If each pair of opposite sides of a quadrilateral is equal, then the quadrilateral is a parallelogram.
Proof:
Assume that the sides AB and CD of the quadrilateral ABCD are equal and also AD = BC
Now, draw the diagonal AC.
Clearly, we can say that ∆ ABC ≅ ∆ CDA (From theorem 1)
Therefore, ∠ BAC = ∠ DCA and ∠ BCA = ∠ DAC.
From this result, we can say that the quadrilateral ABCD is a parallelogram because each pair of opposite sides is equal in measurement.
Thus, conversely, we can say that if each pair of opposite sides of a quadrilateral is equal, then the quadrilateral is a parallelogram.
Hence, proved.
### Theorem 4:
The opposite angles are equal in a parallelogram.
Proof:
Using Theorem 3, we can conclude that the pairs of opposite angles are equal.
(i.e) ∠A = ∠C and ∠B = ∠D
Thus, each pair of opposite angles is equal, a quadrilateral is a parallelogram.
### Theorem 5:
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof:
We can say that Theorem 5 is the converse of Theorem 4.
### Theorem 6:
The diagonals of a parallelogram bisect each other.
Proof:
Consider a parallelogram ABCD and draw both the diagonals and it intersects at the point O.
Now, measure the lengths, such as OA, OB, OC and OD.
You will observe that OA = OC and OB = OD
So, we can say that “O” is the midpoint of both the diagonals.
Hence, proved.
### Theorem 7:
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Proof:
This theorem is the converse of Theorem 6.
Now, consider a parallelogram ABCD which is given below:
From the figure, we can observe that OA = OC and OB = OD.
Hence, we can say ∆ AOB ≅ ∆ COD.
∠ ABO = ∠ CDO
Thus, we can conclude that AB || CD and BC || AD.
Hence, ABCD is a parallelogram.
Also, read:
### Properties of a Parallelogram Example
Question:
Prove that the bisectors of angles of a parallelogram form a rectangle.
Solution:
Assume that ABCD is a parallelogram. Let P, Q, R, S be the point of intersection of bisectors of
∠A and ∠B, ∠B and ∠C, ∠C and ∠D, and ∠D and ∠A, respectively.
From the triangle, ASD, we can observe that DS bisects ∠D and AS bisects ∠A.
Therefore,
∠ADS + ∠DAS = (½)∠D + (½)∠A
∠ADS + ∠DAS = ½ (∠A + ∠D)
∠ADS + ∠DAS = ½ (180°) [Since ∠A and ∠D are the interior angles on the same side of the transversal]
Therefore, ∠ADS + ∠DAS = 90°.
Using the angle sum property of a triangle, we can write:
∠ DAS + ∠ ADS + ∠ DSA = 180°
Now, substitute ∠ADS + ∠DAS = 90° in the above equation, we get
90° + ∠DSA = 180°
Therefore, ∠DSA = 180° – 90°
∠DSA = 90°.
Since, ∠PSR is being vertically opposite to ∠DSA,
We can say ∠PSR = 90°
Likewise, we can be shown that ∠ APB = 90° or ∠ SPQ = 90°
Similarly, ∠ PQR = 90° and ∠ SRQ = 90°.
Since all the angles are at right angles, we can say that PQRS is a quadrilateral.
Now, we need to conclude that the quadrilateral is a rectangle.
We have proved that ∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°.
As, both the pairs of opposite angles are equal to 90°, we can conclude that PQRS is a rectangle.
Hence, proved.
### Practice Problems
1. A quadrilateral ABCD is a parallelogram where AP and CQ are perpendiculars from vertices A and C on diagonal BD as shown in the figure.
2. Prove that (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ
3. If the diagonals of a parallelogram are equal, prove that it is a rectangle.
Stay tuned with BYJU’S – The Learning App and learn all the class-wise concepts easily by exploring more videos.
## Frequently Asked Questions on Properties of a Parallelogram
### What type of polygon is a parallelogram?
A parallelogram is a quadrilateral.
### What are the properties of the parallelogram?
The properties of the parallelogram are:
The opposite sides of a parallelogram are parallel and congruent
The consecutive angles of a parallelogram are supplementary
The opposite angles are equal
A diagonal bisect the parallelogram into two congruent triangles
Diagonals bisect each other
### What are the two special types of a parallelogram?
The two special types of a parallelogram are square and rectangle.
### What is the order of rotational symmetry of a parallelogram?
The order of rotational symmetry of a parallelogram is 2.
### Does a parallelogram have reflectional symmetry?
No, a parallelogram does not have reflectional symmetry.
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By accessing our 180 Days of Math for Fifth Grade Answers Key Day 174 regularly, students can get better problem-solving skills.
Directions: Solve each problem.
Question 1.
Calculate the difference between 348 and 96.
Subtraction is one of the four basic arithmetic operations in mathematics. We can observe the applications of subtraction in our everyday life in different situations. For example, when we purchase fruits and vegetables for a certain amount of money say Rs. 200 and we have given an Rs. 500 note to the vendor. Now, the vendor returns the excess amount by performing subtraction such as 500 – 200 = 300. Then, the vendor will return Rs. 300.
Now we need to calculate the above-given question:
we need to subtract 348 from 96
348 = Minuend; 96 = Subtrahend; 252 = Difference
Question 2.
In mathematics, multiplication is a method of finding the product of two or more numbers. It is one of the basic arithmetic operations, that we use in everyday life. The major application we can see in multiplication tables.
In arithmetic, the multiplication of two numbers represents the repeated addition of one number with respect to another. These numbers can be whole numbers, natural numbers, integers, fractions, etc. If m is multiplied by n, then it means either m is added to itself ‘n’ number of times or vice versa.
The formula for multiplication:
The multiplication formula is given by:
Multiplier × Multiplicand = Product
– The multiplicand is the total number of objects in each group
– A multiplier is the number of equal groups
– Product is the result of multiplication of multiplier and multiplicand
Question 3.
Is 129 evenly divisible by 9?
No, it is not divisible by 9.
The division is breaking a number into an equal number of parts. The division is an arithmetic operation used in Maths. It splits a given number of items into different groups.
There are a number of signs that people may use to indicate division. The most common one is ÷, but the backslash / and a horizontal line (-) is also used in the form of Fraction, where a Numerator is written on the top and a Denominator on the bottom.
The division formula is:
Dividend ÷ Divisor = Quotient (or) Dividend/Divisor=quotient
14 is the quotient; 3 is the remainder.
Question 4.
What digit is in the thousands place in the number 95,387?
Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on.
The above-given number is 95,387
There are five digits in the number 95,387.
7 is in the unit’s place.
8 is in the tens place
3 is in the hundreds place
5 is in the thousands place
9 is in the ten thousand place
Question 5.
Write $$\frac{5}{2}$$ as a mixed number.
Step 1: Find the whole number
Calculate how many times the denominator goes into the numerator. To do that, divide 5 by 2 and keep only what is to the left of the decimal point:
5 / 2 = 2.5000 = 2
Step 2: Find a new numerator
Multiply the answer from Step 1 by the denominator and deduct that from the original numerator.
5-(2×2)
=5-4
=1
Step 3: Get a solution
Keep the original denominator and use the answers from Step 1 and Step 2 to get the answer. 5/2 as a mixed number is:
2 1/2
Question 6.
5 × 5 – 3 × 5 = ____________
The above equation can be solved as:
5 × 5 = 25
3 × 5 = 15
Now subtract the numbers 25 and 15
=25 – 15
=10
Therefore, 5 × 5 – 3 × 5 = 10.
Question 7.
165 – b = 87
b = ___________
We need to find out the value of b
The above-given equation is: 165 – b = 87
If we ‘b’ to right-hand side then it becomes ‘+b’
Now the equation will be:
165=87+b
Now get 87 to the left-hand side then it will subtract. Here the equation will be:
165-87=b
78=b
Therefore, the value of b is 78
Now verify the answer. Put the number 78 in the place of b in the above-given equation.
165 – b = 87
165-78=87
87=87
Question 8.
2$$\frac{1}{2}$$ hours = _________ minutes
2$$\frac{1}{2}$$ hours can be written as: 2 1/2 hours
1 hour is equal to 60 minutes
for 2 hours:
2*60=120 minutes.
1/2 (half an hour)=1/2*60
=30 minutes.
Now add 120 and 30 then we get:
120+30=150
Therefore, 2 1/2 hours = 150 minutes.
Question 9.
Which 3-dimensional figure has only square faces?
Three-dimensional shapes are those figures that have three dimensions, such as height, width, and length, like any object in the real world. It is also known as 3D.
Polyhedral or polyhedrons are straight-sided solid shapes. They are based on polygons, two- dimensional plane shapes with straight lines. Polyhedrons are defined as having straight edges, corners called vertices, and flat sides called faces. The polyhedrons are often defined by the number of edges, faces and vertices, as well as their faces are all the same size and same shape. Like polygons, polyhedrons can be regular and irregular based on regular and irregular polygons. Polyhedrons can be convex or concave. The cube is the most familiar and basic polyhedron. It has 6 square faces, 12 edges and eight vertices.
Question 10.
What is the mean of these numbers?
528, 455, 537
Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean=sum of observations/number of observations
The above-given numbers are 528, 455, 537
528+455+537=1520
The number of observations is 3
Now divide by 3 (total number of observations)
Mean=1520/3
Mean=506.66667
Therefore, the mean is 506.6
Question 11.
If the probability that someone knows how to swim is $$\frac{5}{6}$$, how many people in a group of 100 will likely not know how to swim?
P(S)=5/6=0.83
The probability of knowing how to swim is 0.83 that is nothing but 83%
The number of people is there in a group is 100
We know the probability of swimming. Now we need to find out the number will likely not know how to swim. Assume it as X
Now subtract the total group percentage and the probability we know how to swim
X=100-83
X=17
therefore, 17 people do not know how to swim.
Question 12.
Quadruple 46, then divide by 2.
First, we need to know what is quadruple.
to become four times as big, or to multiply a number or amount by four.
This means we need to multiply 46 with 4
46 * 4 = 184
And asked we need to divide by 2.
Therefore, quotient is 92 and remainder is 0
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## Which applies the power of a power rule property to simplify the expression?
The power of a power rule says, when a number with an exponent is raised to another exponent, we can simplify the exponent by keeping the base and multiplying the exponents. The general form of the rule is (am)n=am·n. For example, to find the power of the power of the expression, (x2)7=x2·7=x14.
## What is the rule for power of product?
The power of a product rule tells us that we can simplify a power of a power by multiplying the exponents and keeping the same base.
## How do you rewrite an expression as a power of a product?
To find a power of a product, find the power of each factor and then multiply. In general, (ab)m=am⋅bm. am⋅bm=(ab)m. In other words, you can keep the exponent the same and multiply the bases.
## What is the difference between the product rule and the power rule?
The power rule to follow when finding the derivative of a variable base, raised to a fixed power. … How the product rule allows us to find the derivative of a function that is defined as the product of another two (or more) functions.
## What is the product of powers rule for exponents?
Lesson Summary
When you are multiplying like terms with exponents, use the product of powers rule as a shortcut to finding the answer. It states that when you are multiplying two terms that have the same base, just add their exponents to find your answer.
## What is the power rule in exponents?
The Power Rule for Exponents: (am)n = am*n. To raise a number with an exponent to a power, multiply the exponent times the power. Negative Exponent Rule: xn = 1/xn. Invert the base to change a negative exponent into a positive.
## Which rule is used to simplify 34 2?
The power rule says that if we have an exponent raised to another exponent, you can just multiply the exponents together. For example, suppose we wanted to simplify (34)2. Our rule tells us: (34)2=34⋅2=38 And to prove this we can write out the multiplication.
## How do you simplify?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## Why can’t you use the product of powers rule to simplify this expression explain 34 28?
Why can’t you use the product of powers rule to simplify this expression? Explain. They are not the same number. to do that it would have to be 3 and a 3 or 2 and a 2 can not combine unlike terms.
## Which rules of exponents will be used to evaluate the expression?
Divide the coefficients and subtract the exponents of matching variables. All of these rules of exponents—the Product Rule, the Power Rule, and the Quotient Rule—are helpful when evaluating expressions with common bases.
## Which is the value of this expression when a =- 2 and B =- 3?
IF your expression “a-2 b-3” is meant to be “(a-2)(b-3)” then Bill is correct, and the answer is -5 when a is 3 and b is -2. However, if you mean”a – 2b -3″ then, substituting the values you gave for a=3 and b=-2, that would be “3 – 2(-2) -3” which is the same as “3 +4 -3” and the expression evaluates to 4.
## How does the use of exponents simplify the way we write expressions?
Any non-zero number or variable raised to a power of 0 is equal to 1. When dividing two terms with the same base, subtract the exponent in the denominator from the exponent in the numerator: Power of a Power: To raise a power to a power, multiply the exponents.
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# Unit Fraction
In mathematics, when we divide a complete part into some equal parts then each part shows the fraction of whole part. Like pizza, pizza is a circular in shape, if we divide it in 8 equal parts then each part is the fraction of whole part.
Fractions are written in the form of ratio of two numbers. The upper number is called as numerator and the number below is called as denominator.
For example: a/b is the fraction, in which a is numerator and b is the denominator.
## Unit fractions:
Unit fractions are the fractions in which numerator is always one and denominator is any whole number except zero.
For example:
a/b is the unit fraction only if a= 1 and b≠0.
1/5, 1/8, 1/12 all are the unit fractions, because here numerator is one.
Here, numerator is 1 and denominator is 8≠0.
If a complete circle is divided into 4 equal parts then each part will be the fraction of whole part. And each part takes the value as ¼ and it is the unit fraction.
• Here, complete part / total equal parts = 1/4
• Here, each part contributes ¼ th part of the whole part and it is in the unit fraction form.
• And total part of circle = ¼ + ¼ + ¼ + ¼ = 1
• Similarly, if we divide a complete circle into 8 equal parts then each part will contribute the 1/8th part of the complete part and it is in the form of proper fraction as numerator is less than denominator.
• Here, each fraction takes value 1/8 because, complete part/ number of equal parts = 1/8
• And hence, 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 = 1
The following figures shows the different fraction on the basis of the portion selected out of complete part.
Here, in figure there is a complete circle hence it takes Value 1
• In fig. 2 we divided the complete circle into 2 equal parts hence each part takes the value ½.
Since, complete part/ number of equal parts = ½, and it is the proper fraction.
• In fig. 3 the whole circle is divided into 3 equal parts and hence each part takes value 1/3 and it is the proper fraction.
• In fig. 4 the whole circle is divided into 4 equal parts hence each part takes the value ¼ and it is the proper fraction.
• In fig. 5, the whole circle is divided into 5 equal parts and each part takes the value 1/5, and it is the proper fraction.
• Also in fig. 6, the whole circle is divided into 6 equal parts hence, each part takes the value 1/6 and it is the proper fraction.
Note:
As the numerator is less than denominator in proper fraction, the decimal value of proper fraction is always less than 1.
Updated: July 15, 2021 — 12:06 am
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# Solving equations symbolically
When Solving equations symbolically, there are often multiple ways to approach it. We can solve math word problems.
## Solve equations symbolically
We will also give you a few tips on how to choose the right app for Solving equations symbolically. The disparities between minority groups and the majority is a major problem in the United States. Exact statistics on how many minorities are unemployed and how many people of lower income are living in poverty are hard to track, but it’s clear that there is still much to be done. One way that the inequality gap can be closed is by encouraging more minorities to go into STEM fields. This will not only help them to earn more money, but it will also give them more recognition in the workplace and make it easier for them to get raises and promotions. Another way that inequality can be closed is by improving access to education. If more minorities have access to quality education, they will be less likely to end up stuck in low-paying jobs or trapped in poverty.
The intercept is the value that represents the y value of each data point when plotted on a graph. Sometimes it is useful to know the value of x at which y = 0. This is called the x-intercept and it can be used to estimate where y will be when x = 0. There are two main ways to determine the intercept: 1) The easiest way is to use a line of best fit. The line shows that when x increases, y increases by the same amount. Therefore, if you know x, you can calculate y based on that value and then plot the resulting line on your graph (see figure 1 below). If there is more than one data point, you can select the one that has the highest y value and plot that point on your graph (see figure 2 below). When you do this for all data points, you get an approximation of where the line of best fit crosses zero. This is called the x-intercept and it is equal to x minus y/2 (see figure 3). 2) Another way to find x-intercept involves using the equation y = mx + b. The left side is equation 1 and the right side is equation 2. When solving for b, remember that b depends on both m and x, so make sure to factor in your other values as well (for example, if you have both
If you want to calculate an individual’s natural log, then you need to measure their height and multiply it by three. The basic idea behind natural log is that trees grow in all directions, so if you take the total diameter of a tree and divide it by its height, you will get 1, 2 or 3. The more branches there are on a tree and the longer they are, the higher the log will be. The thicker a tree trunk is, the more logs it has. The larger a tree grows in diameter, the more logs it has, but only up to a certain point as it would have to have more branches and trunks to offset the increased surface area of each branch. There are two main ways to get around this problem: 1) Take out one branch in order to get less branches and increase your natural log. A common example of this is grafting where one sapling is grafted onto another sapling that has fewer branches. 2) Grow multiple trunks from one original trunk so that each new trunk has equal or
Linear equations are very common in every grade. They are used to show the relationship between two numbers or values. There are a few different ways to solve linear equations by graphing. You can graph the equation on a coordinate grid, plot points on a coordinate grid, or plot points on an axes grid. When graphing, always follow the order of operations. To graph an equation, start with an ordered pair (x, y). Then put points in between the coordinates that indicate how you want your equation to look. For example, if x = 2 and y = -8, then your graphed equation would look like this: (2,-8). Starting from the left and working from one point to the next will help you visualize how you want your graph to look.
It actually is a great tool for learning how to do problems, the steps are clear and there even is information about every single step, really quick and works just as promised. Great app.
Marlee Gonzales
The best math app I have seen so far, definitely recommend it to others. The photo feature is more than amazing and the step-by-step detailed explanation is quite on point. Gave it a try never regretted.
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# Power of a product rule
##### Intros
###### Lessons
1. What are exponent rules?
##### Examples
###### Lessons
1. Simplify the following:
1. $(-4xy)^4$
##### Practice
###### Topic Notes
We use the power of a product rule when there are more than one variables being multiplied together and raised to a power. The power of a product rule tells us that we can simplify a power of a power by multiplying the exponents and keeping the same base.
## Introduction to the Power of a Product Rule
The power of a product rule is a fundamental concept in exponent laws, essential for mastering algebra and higher mathematics. Our introduction video serves as a crucial starting point, offering a clear and concise explanation of this rule and its applications. By watching this video, students gain a solid foundation in understanding how exponents work when multiplying terms with the same base. Rather than simply memorizing formulas, the video emphasizes the importance of grasping the underlying principles of exponents. This approach enables learners to apply the power of a product rule confidently across various mathematical scenarios. Understanding this rule is key to unlocking more complex exponent laws and algebraic expressions. By focusing on the logic behind the rule, students develop a deeper, more intuitive understanding of exponents, setting the stage for advanced mathematical concepts and problem-solving skills.
## Understanding the Basics of Exponents
Exponents are a fundamental concept in mathematics that represent repeated multiplication. This powerful notation allows us to express large numbers concisely and perform complex calculations efficiently. At its core, an exponent indicates how many times a number, called the base, is multiplied by itself.
Let's start with a simple example to illustrate the concept of exponents. Consider the expression 2³. This means we multiply 2 by itself three times: 2 × 2 × 2 = 8. In this case, 2 is the base, and 3 is the exponent. We read this as "2 to the power of 3" or "2 cubed."
Here are a few more examples of positive integer exponents:
• 3² = 3 × 3 = 9
• 5 = 5 × 5 × 5 × 5 = 625
• 10³ = 10 × 10 × 10 = 1,000
As we can see, exponents provide a shorthand way to express repeated multiplication. This notation becomes especially useful when dealing with larger numbers or variables.
Speaking of variables, exponents work the same way with algebraic expressions with exponents. For instance:
• x² = x × x
• y = y × y × y × y × y
Understanding exponents as repeated multiplication naturally leads us to one of the most important rules in exponent arithmetic: the product of powers rule. This rule states that when multiplying expressions with the same base, we keep the base and add the exponents.
Mathematically, we express the product of powers rule as:
x^a × x^b = x^(a+b)
This rule makes sense when we think about exponents as repeated multiplication. Let's break it down with an example:
2³ × 2 = (2 × 2 × 2) × (2 × 2 × 2 × 2) = 2
We can see that we're simply combining all the 2s being multiplied, resulting in 2 to the power of 3 + 4 = 7.
The product rule for exponents works with any base, including variables:
• x² × x³ = x
• y × y² = y
• 5³ × 5² = 5
This rule is incredibly useful in simplifying algebraic expressions and solving complex mathematical problems. It allows us to quickly combine like terms and reduce expressions to their simplest form.
As we delve deeper into the world of exponents, we'll encounter more rules and properties that build upon this fundamental understanding. The product of powers rule is just the beginning, but it serves as a crucial foundation for mastering exponent operations.
In conclusion, exponents represent repeated multiplication, providing a concise way to express large numbers and repeated operations. The product of powers rule naturally extends from this concept, allowing us to simplify expressions by adding exponents when multiplying terms with the same base. This understanding of exponents and their properties is essential for advancing in algebra, calculus, and many other areas of mathematics and science.
## The Power of a Product Rule Explained
The power of a product rule, also known as the product of powers law, is a fundamental concept in algebra that simplifies the process of raising a product to a power. This rule states that when multiplying two or more factors raised to the same power, we can simply multiply the bases and keep the exponent the same. Mathematically, it can be expressed as (ab)^n = a^n * b^n, where a and b are the bases and n is the exponent.
Let's explore this rule with numerical examples first. Consider (2 * 3)^4. Without the rule, we would need to multiply 2 and 3, then raise the result to the fourth power: 6^4 = 1296. Using the power of a product rule, we can simplify this process: (2 * 3)^4 = 2^4 * 3^4 = 16 * 81 = 1296. This method is often more efficient, especially with larger numbers or variables.
The rule also applies to algebraic expressions simplification. For instance, (xy)^3 = x^3 * y^3. This simplification is particularly useful when dealing with complex algebraic expressions simplification. Consider (2ab)^5. Using the rule, we can expand this to 2^5 * a^5 * b^5 = 32a^5b^5, which is much easier to work with in further calculations.
To understand why this rule works, let's expand an expression like (ab)^3:
(ab)^3 = (ab)(ab)(ab) = a*a*a * b*b*b = a^3 * b^3
This expansion demonstrates that each base is indeed multiplied by itself as many times as the exponent indicates, validating the rule.
The versatility of the power of a product rule becomes evident when we apply it to different bases and exponents. For example:
(3x^2y)^4 = 3^4 * (x^2)^4 * y^4 = 81x^8y^4
Here, we see the rule applied to a numerical coefficient (3), a variable with its own exponent (x^2), and another variable (y).
The rule also works with fractional and negative exponents in algebra. For instance:
(2ab)^(1/2) = 2^(1/2) * a^(1/2) * b^(1/2) = 2 * a * b
(xy)^(-3) = x^(-3) * y^(-3) = 1/(x^3 * y^3)
These examples showcase how the rule simplifies expressions with various types of exponents, making it a powerful tool in algebra.
It's important to note that the power of a product rule is closely related to other exponent rules interaction, such as the power of a power rule ((a^m)^n = a^(mn)) and the power of a quotient rule ((a/b)^n = a^n / b^n). Understanding how these rules interact can greatly enhance one's ability to manipulate and simplify complex exponent rules interaction.
In practical applications, the power of a product rule is invaluable in fields such as physics, engineering, and computer science, where complex calculations involving powers are common. For instance, in calculating compound interest or analyzing exponential growth, this rule can significantly streamline computations.
To further illustrate the rule's applicability, consider a problem in physics where we need to calculate the force of gravity between two objects. The formula F = G(m1m2)/r^2 involves a product in the numerator. If we needed to cube this entire expression, the power of a product rule would allow us to distribute the exponent: (G(m1m2)/r^2)^3 = G^3 * m1^3 * m2^3 / r^6, greatly simplifying the calculation process.
In conclusion, the power of a product rule
## Applications and Examples of the Power of a Product Rule
The power of a product rule is a fundamental concept in algebra that simplifies expressions involving exponents. This rule states that when raising a product to a power, we can raise each factor to that power and then multiply the results. Let's explore various examples and applications of this rule, ranging from simple to complex scenarios.
Simple Examples:
1. (xy)² = x²y²
This basic example shows how the rule applies to a product of two variables.
2. (3a)³ = 3³a³ = 27a³
Here, we see how the rule works with a coefficient and a variable.
3. (2xy) = 2xy = 16xy
This example demonstrates the rule applied to a product with a coefficient and two variables.
Complex Examples:
4. (4abc)³ = 4³a³b³c³ = 64a³b³c³
The rule extends easily to products with multiple variables.
5. (x²y³z) = (x²)(y³)z = xy¹²z
This example shows how the rule works with variables already raised to powers.
6. ((a²b)(cd³))² = (a²b)²(cd³)² = ab²c²d
Here, we apply the rule to a more complex expression with grouped terms.
Negative Exponents:
7. (xy)² = x²y²
The rule applies similarly to negative exponents in algebra, simplifying reciprocal expressions.
8. (2ab¹)³ = 2³a³(b¹)³ = a³b³
This example combines negative exponents in algebra with the power of a product rule.
Applications in Different Scenarios:
9. Area of a rectangle: If the length and width of a rectangle are doubled, the new area is (2l)(2w) = 4lw, which is four times the original area.
10. Volume of a cube with tripled side length: If the side length of a cube is tripled, the new volume is (3s)³ = 27s³, which is 27 times the original volume.
11. Compound interest growth factor: In finance, (1 + r) represents the growth factor for compound interest, where r is the interest rate and n is the number of compounding periods.
12. Scientific notation multiplication: (5.2 × 10)(3.1 × 10²) = (5.2 × 3.1)(10 × 10²) = 16.12 × 10²
13. Simplifying algebraic fractions with exponents: (x²y³)/(xy²) = x²y³/(xy) = x²y
14. Physics equations: In kinetic energy (KE = ½mv²), doubling the velocity results in (½m(2v)²) = ½m(4v²) = 2mv², quadrupling the energy.
15. Probability calculations: If the probability of an event occurring twice independently is (0.3)², the probability of it not occurring twice is (1 - 0.3)² = 0.7² = 0.49.
These examples demonstrate the versatility and power of the product rule
## Common Mistakes and How to Avoid Them
When applying the power of a product rule, students often encounter several common mistakes that can lead to incorrect solutions. Understanding these errors and learning how to avoid them is crucial for mastering this important mathematical concept.
One of the most frequent mistakes is misapplying the rule by distributing the exponent to each factor individually. For example, students might incorrectly write (xy)² as x²y². This error stems from confusing the power of a product rule with the product rule for exponents. To avoid this, always remember that (xy)² means (xy)(xy), not x²y².
Another common error is forgetting to include all factors when raising a product to a power. For instance, (xyz)³ should be expanded to (xyz)(xyz)(xyz), not just x³y³. Students often overlook one or more factors, leading to incomplete or incorrect answers. To prevent this, carefully count the number of factors and ensure each is included in the expansion.
Students also frequently struggle with negative exponents in product expressions. For example, (xy)² is sometimes mistakenly written as x²y². The correct application would be 1/(xy)², which can be further simplified to 1/(x²y²). To avoid this error, remember that negative exponents indicate reciprocals, and the entire product should be treated as a single unit.
Confusion often arises when dealing with fractional exponents in product expressions. For instance, (xy)½ is not equal to x½y½. The correct approach is to treat the entire product as a single term and apply the fractional exponent to it, resulting in the square root of xy. To prevent this mistake, always consider the product as a whole when applying fractional exponents.
Another pitfall is incorrectly applying the rule to sums or differences. The power of a product rule does not apply to expressions like (x+y)². Students sometimes erroneously write this as x²+y². To avoid this, recognize that the rule only applies to products, not sums or differences. For expressions involving sums or differences, use the binomial theorem or FOIL method instead.
To master the power of a product rule, it's essential to understand its underlying principle rather than blindly applying a formula. Visualize the repeated multiplication of the entire product, and practice expanding various expressions step-by-step. For example, expand (ab)³ as (ab)(ab)(ab) = a³b³, reinforcing the correct application of the rule.
Regular practice with diverse examples, including those with multiple factors, negative exponents, and fractional powers, will help solidify understanding and reduce errors. Additionally, always double-check your work by expanding the expression manually to verify the result. By focusing on comprehension and careful application, students can significantly improve their accuracy when using the power of a product rule in mathematical problem-solving.
## Related Exponent Rules and Their Connections
When exploring exponent rules connections, it's essential to understand that they are interconnected and build upon one another. While the power of a product rule is fundamental, other rules like the quotient rule exponent and the power of a power rule are equally important in mastering exponents. By grasping these rules and their relationships, students can tackle complex problems with confidence.
The quotient rule exponent is closely related to the power of a product rule. It states that when dividing expressions same base, we subtract the exponents. Mathematically, this is expressed as (x^a) / (x^b) = x^(a-b). This rule complements the power of a product rule, which involves addition of exponents when multiplying expressions with the same base.
The power of a power rule, on the other hand, deals with exponents raised to another power. It states that (x^a)^b = x^(ab). This rule is particularly useful when simplifying nested exponents and can be seen as an extension of the power of a product rule. Understanding the power of a power rule helps in breaking down complex exponential expressions into simpler forms.
These exponent rules connections are interconnected in several ways. For instance, the quotient rule exponent can be derived from the power of a product rule by considering division as multiplication by the reciprocal. Similarly, the power of a power rule can be understood as repeated application of the power of a product rule. By recognizing these connections, students can develop a more intuitive understanding of exponents and their properties.
Applying multiple exponent rules to solve complex problems is a crucial skill. For example, consider the expression ((x^3)^2 * (x^4)) / (x^5). To simplify this, we can start by applying the power of a power rule to (x^3)^2, giving us x^6. Then, we can use the power of a product rule to combine x^6 and x^4, resulting in x^10. Finally, we apply the quotient rule exponent to divide x^10 by x^5, yielding x^5 as the simplified result.
Another example that combines multiple rules is (y^-2 * y^5)^3 / y^4. Here, we first use the power of a product rule inside the parentheses to get y^3. Then, we apply the power of a power rule to (y^3)^3, resulting in y^9. Lastly, we use the quotient rule exponent to divide y^9 by y^4, giving us y^5 as the final answer.
Understanding these interconnections helps in developing problem-solving strategies. When faced with a complex exponential expression, students can break it down into smaller parts and apply the appropriate rules step by step. This approach not only simplifies the problem-solving process but also reinforces the relationships between different exponent rules.
Moreover, recognizing the patterns in exponent rules can lead to a deeper understanding of mathematical concepts. For instance, the quotient rule exponent can be extended to negative exponents explanation, explaining why x^-n is equivalent to 1/(x^n). Similarly, the power of a power rule helps in understanding the concept of roots, as (x^(1/n))^n = x.
In conclusion, the quotient rule exponent, power of a power rule, and other exponent rules connections are closely interconnected. By understanding these relationships and practicing with diverse problems, students can enhance their mathematical skills and tackle complex exponential expressions with ease. The key is to recognize the patterns, apply the rules systematically, and always be mindful of how different rules interact with each other in various mathematical contexts.
## Practical Applications in Algebra and Beyond
The power of a product rule in algebra is a fundamental concept with wide-ranging practical applications across various mathematical fields and real-world scenarios. This rule states that when raising a product to a power, we can raise each factor to that power and then multiply the results. In algebra, this rule is crucial for solving equations and simplifying complex expressions efficiently.
In solving polynomial equations, the power of a product rule allows mathematicians to break down complicated terms into more manageable components. For instance, when dealing with equations involving variables raised to powers, this rule enables us to distribute the exponent across multiple factors, making it easier to isolate variables and find solutions. This technique is particularly useful in polynomial equations and exponential functions, which are common in scientific and engineering calculations.
The rule's application extends beyond basic algebra into more advanced mathematical concepts. In calculus, it plays a vital role in differentiating and integrating complex functions. When working with derivatives, the power rule combines with the product rule to simplify the process of finding rates of change for intricate expressions. In integral calculus, this rule aids in breaking down complex integrands, making integration more approachable.
Real-world applications of the power of a product rule are abundant. In physics, it's used to calculate work done by varying forces or to determine the kinetic energy calculations of objects in motion. Engineers apply this rule when designing structures, calculating stress and strain on materials, or optimizing energy systems. In finance, the rule is crucial for compound interest modeling, helping investors and economists model growth over time.
In computer science and cryptography, the power of a product rule is fundamental to many algorithms, particularly in public-key encryption systems. These systems rely on the difficulty of factoring large numbers, a process that involves extensive use of exponents and products. The rule also finds applications in probability theory, where it's used to calculate the likelihood of multiple independent events occurring simultaneously.
Environmental scientists use this rule when modeling population growth or decay, considering factors like birth rates, death rates, and environmental carrying capacities. In chemistry, it's applied in reaction kinetics in chemistry to understand how the concentration of reactants affects reaction rates. The versatility of this algebraic principle demonstrates its importance across diverse fields, making it a cornerstone of mathematical problem-solving in both theoretical and applied contexts.
## Conclusion
In summary, the product rule is a powerful tool in exponent laws, allowing us to simplify expressions by adding exponents when multiplying terms with the same base. Understanding the principle behind this rule is crucial, as it enables you to apply it confidently across various mathematical scenarios. Rather than memorizing formulas, focus on grasping the underlying concept. We encourage you to practice applying the product rule regularly, as this will reinforce your understanding and improve your problem-solving skills. For those seeking to deepen their knowledge, explore further resources on exponent laws and related topics. Remember, the introduction video serves as an excellent foundation for mastering exponent laws, including the product rule. By building on this knowledge, you'll develop a strong mathematical toolkit that will serve you well in future studies and real-world applications. Keep practicing, stay curious, and don't hesitate to revisit the video for a refresher on these fundamental concepts.
### Example:
Simplify the following: $(-4xy)^4$
#### Step 1: Understand the Power of a Product Rule
The Power of a Product Rule states that when you have a product raised to a power, you can distribute the exponent to each factor in the product. In mathematical terms, this means: $(ab)^n = a^n \cdot b^n$ In this example, we have $(-4xy)^4$. According to the Power of a Product Rule, we need to distribute the exponent 4 to each factor inside the parentheses.
#### Step 2: Separate the Terms
First, let's separate the terms inside the parentheses. We have three factors: -4, x, and y. So, we can rewrite the expression as: $(-4xy)^4 = (-4)^4 \cdot (x)^4 \cdot (y)^4$ This step helps us to handle each factor individually.
#### Step 3: Simplify the Negative Base
Next, we need to simplify $(-4)^4$. It's important to note that raising a negative number to an even power results in a positive number. This is because multiplying an even number of negative factors results in a positive product. Therefore: $(-4)^4 = 4^4$ Now, we only need to calculate $4^4$.
#### Step 4: Calculate the Power of 4
Now, let's calculate $4^4$. This means multiplying 4 by itself four times: $4^4 = 4 \cdot 4 \cdot 4 \cdot 4 = 256$ So, $(-4)^4$ simplifies to 256.
### FAQs
Here are some frequently asked questions about the power of a product rule:
#### What is the power of a product rule?
The power of a product rule states that when raising a product to a power, you can raise each factor to that power and then multiply the results. Mathematically, it's expressed as (ab)^n = a^n * b^n, where a and b are the factors and n is the exponent.
#### What is an example of the power of a product rule?
A simple example is (2x)^3 = 2^3 * x^3 = 8x^3. Here, we raise both 2 and x to the power of 3 separately and then multiply the results.
#### How do you apply the power of a product rule to simplify expressions?
To simplify expressions using this rule, identify the product within parentheses and the power it's raised to. Then, apply the power to each factor individually. For example, (3ab)^4 simplifies to 3^4 * a^4 * b^4 = 81a^4b^4.
#### Does the power of a product rule work with negative exponents?
Yes, the rule works with negative exponents. For instance, (xy)^-2 = x^-2 * y^-2 = 1/(x^2 * y^2). Remember that a negative exponent means the reciprocal of the positive exponent.
#### How is the power of a product rule different from the product rule for exponents?
The power of a product rule deals with raising a product to a power, while the product rule for exponents involves multiplying terms with the same base and adding their exponents. For example, the power of a product rule is (ab)^n = a^n * b^n, while the product rule for exponents is x^a * x^b = x^(a+b).
### Prerequisite Topics for Understanding the Power of a Product Rule
Mastering the power of a product rule in mathematics requires a solid foundation in several key areas. One of the most crucial prerequisites is combining the exponent rules. Understanding how exponents work and how to manipulate them is essential when dealing with products raised to powers.
Equally important is the ability to simplify rational expressions and understand their restrictions. This skill helps in breaking down complex expressions that often arise when applying the power of a product rule. Additionally, familiarity with the negative exponent rule is crucial, as it allows for the proper handling of expressions with negative powers.
When working with the power of a product rule, you'll often encounter situations that require solving polynomial equations. This prerequisite topic provides the tools needed to manipulate and solve equations that result from applying the rule. Moreover, understanding scientific notation is beneficial, especially when dealing with very large or small numbers in product expressions.
The power of a product rule has practical applications in various fields. For instance, in finance, it's used in compound interest calculations. Understanding this connection can provide real-world context and motivation for mastering the rule. Similarly, in chemistry, the rule is applied in reaction kinetics, demonstrating its importance beyond pure mathematics.
By building a strong foundation in these prerequisite topics, students can approach the power of a product rule with confidence. Each of these areas contributes to a deeper understanding of how products and exponents interact, making it easier to grasp and apply the rule in various contexts. Whether you're solving complex algebraic problems or applying the concept in scientific or financial scenarios, a solid grasp of these prerequisites will significantly enhance your ability to work with the power of a product rule effectively.
Remember, mathematics is a cumulative subject. Each new concept builds upon previous knowledge. By taking the time to review and master these prerequisite topics, you're not just preparing for understanding the power of a product rule, but also laying the groundwork for more advanced mathematical concepts that you'll encounter in your future studies.
$({a^mb^n} {)^p} = {a^{mp}b^{np}}$
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# Inverse 3×3 matrix online
Calcula la matriz inversa 3x3 online step by step with our calculator that will allow you to find the inverse of a matrix instantly.
Sólo tienes que rellenar correctamente todos los valores que componen la matriz 3x3 y pulsar sobre el botón de calcular para obtener el resultado paso a paso cómo se ha calculado su matriz inversa a partir de la fórmula que veremos más adelante. Descubre cómo se calcula la inversa de una matriz 3x3.
## Fórmula para calcular la matriz inversa 3x3 y nxn
Para calcular la inversa de una matriz 3x3, 4x4 o del tipo nxn, sólo tienes que aplicar la fórmula que tienes encima de estas líneas.
In the formula to calculate the inverse matrix we can see that we have to calculate:
• The determinant of the starting matrix nxn
• Transpose the attached matrix
Cuando tengamos ambos datos, hacemos la división correspondiente y obtendremos la matriz inversa 3x3 o la del tipo nxn que busquemos.
## Ejemplos de matriz inversa 3x3
Let's take a practical example cómo calculamos la matriz inversa 3x3 that we have just put.
### 1 - Calcular determinante de matriz 3x3
As we have said in the previous point, the first thing we must do in order to sacar la matriz inversa 3x3 is calculate the determinant of the matrix.
When we have the result of the determinant, two things can happen:
• Si el valor del determinante 3x3 is equal to zero, then there is no inverse matrix.
• If the determinant has a non-zero heat, we can calculate its inverse.
### 2 - Calcular matriz adjunta
Now let's calculate the attached matrix and to do so, we must eliminate the row and the column in which each element of the matrix is located. With the remaining elements we will form a determinante 2x2 to be solved. This must be done with each and every one of the elements that form the matrix.
In addition, you must remember that the attached matrix has a specific order of symbols associated with each of the positions of the matrix as shown in the following figure:
Based on our example, it remains that the step by step to calculate the attached matrix is as follows:
Once we group the results obtained by calculating the adjoint determinant of each element of the matrix, we are left with the following matriz adjunta 3x3 is as follows:
### 3 - Transponemos la matriz adjunta
This step is very fast, because in order to make the transpose of the adjoining matrixIn the table below, we only have to exchange rows for columns. This leaves us with the following result:
### 4 - Aplicamos la fórmula para sacar la matriz inversa 3x3
The fourth and final step is to apply the fórmula para calcular la matriz inversa 3x3 we saw above.
We copy it, compile the results obtained in the previous operations and solve.
## Cómo sacar la matriz inversa 3x3 en Excel
If you want to make your own calculadora de matriz inversa 3x3 con ExcelYou can get it in a very simple way if you follow these steps:
1. Abre una nueva hoja de cálculo y escribe en ella la matriz 3x3 para la cual quieres calculate its inverse.
2. Cuando la tengas escriba, busca un rango 3x3 de celdas vacías y selecciónalo. Cuando lo tengas, escribe la siguiente función de Excel que te permitirá calcular la matriz inversa 3x3 (recuerda que entre los paréntesis irá el rango de celdas 3x3 en los que has escrito tu matriz):
=MINVERSE()
1. Now press simultaneously the CTRL + Shift keys on your keyboard and without releasing, press the ENTER key to accept the formula. If you have done it right, Excel calculará la matriz inversa 3x3 automatically.
If you have any doubts about the procedure to follow, we recommend you to watch the video and you will find out. If you still have any doubts, leave us a comment and we will try to help you as soon as possible.
### 1 thought on “Matriz inversa 3×3 online”
1. WILFRIDO RAMIREZ
The video on how to calculate the inverse matrix in excel is missing or not available.
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# Finding the Median Value
The median is the middle value of a list of data values. To find the median of a list of data values:
1. Place the numbers in increasing or decreasing order by value.
2. Identify the number that appears in the middle position in the sorted list.
## Example 1
Find the median of the list
8, 2, 15, 3 and 6
Solution: Start by sorting these scores into increasing order. The sorted list is:
2, 3, 6, 8, 15.
As there are five values in the list, the middle value is the third one in the list. So, the median is 6.
## Example 2
Find the median of the list
15, 7, 8, 11, 23, 25, 17, 12, 15, 16, 19, 24, 21, 22, 31, 36, 47
Solution: Start by sorting these scores into increasing order. The sorted list is:
7, 8, 11, 12, 15, 15, 16, 17, 19, 21, 22, 23, 24, 25, 31, 36, 47
Now identify the middle value. There are 17 numbers in this list, so the median is the 9th value, or 19.
The repetition of values in the list (e.g. 15) does not affect the way we calculate the median. We simply include the repeated values in our sorted list.
## What About Lists With Even Numbers of Elements?
If a list has an even number of elements, we can still find the median, even though we can't identify the middle element exactly.
When a list has an even number of elements, it has two middle numbers. The median is defined to be the average of these values (add the two numbers together and divide by 2).
### Example 3
Find the median of the following list of values:
15, 7, 8, 11, 23, 25, 17, 12, 15, 16, 19, 24, 21, 22, 31, 36, 47, 54
This list has an even number (18) of entries, so it has a pair of middle values. Let's write the list in increasing order:
7, 8, 11, 12, 15, 15, 16, 17, 19, 21, 22, 23, 24, 25, 31, 36, 47, 54
The two middle entries are the 9th and 10th entries (19 and 21). So, to find the median of this list, we add them together and divide by 2 to give
$\dfrac{19 + 21}{2} = \dfrac{40}{2} = 20$.
The median of our list is $20$.
## Finding the Median of Longer Lists
Often, the most challenging part of finding the median is identifying the middle value in the list. However, there's an easy trick to get you out of trouble. Simply count the number of elements in your list, add one, and then divide the result by 2.
For example, if a list has $253$ elements, the middle element will be the $\dfrac{253 + 1}{2} = \dfrac{254}{2} = 127$th element in the list after it has been sorted into increasing order.
But, what if the list has an even number of elements? That's OK. Let's suppose we have a list with $366$ elements. When we apply our little trick, we end up with the middle element being element number $\dfrac{366 + 1}{2} = \dfrac{367}{2} = 183.5$, which doesn't exist. However, 183.5 is halfway between 183 and 184, and we know that we have to find the average of the corresponding list elements to find our median. So, to find the median of the numbers in this list, we sort the list into increasing or decreasing order, identify the 183rd and 184th elements, add them together and divide the result by 2.
## When is the Median Useful?
Mrs Fitzsimmons asks Sam's class to evaluate her teaching of Shakespeare. She receives an average score of 5.36 out of 10 from her 11 students, which makes her pretty happy until she decides to look at the data more closely. Then she sees that the 11 scores were 1,3,3,3,3,3,5,9,9,10,10. The median score of this list is a not very impressive 3 out of 10. This tells her that at least half of her students were not very happy with her Shakespeare teaching. The average hides this fact because the unexpected 9s and 10s (possibly from students who were trying to get a higher mark on their exam, or maybe they really liked Shakespeare) skewed the data, making the average higher than it should have been. In this case, the median is a much better choice for data analysis.
If you've ever looked at the property pages of a newspaper, you might have noticed that median house prices are more commonly reported than average house prices. This is because there are always a few house prices that are completely over the top, and which could over-inflate the apparent value of property if the mean was used as the reported measure.
The median is a better choice than the mean when you are looking for a more representative value for data sets that contain extreme values.
### Description
This chapter series is on Data and is suitable for Year 10 or higher students, topics include
• Accuracy and Precision
• Calculating Means From Frequency Tables
• Correlation
• Cumulative Tables and Graphs
• Discrete and Continuous Data
• Finding the Mean
• Finding the Median
• FindingtheMode
• Formulas for Standard Deviation
• Grouped Frequency Distribution
• Normal Distribution
• Outliers
• Quartiles
• Quincunx
• Quincunx Explained
• Range (Statistics)
• Skewed Data
• Standard Deviation and Variance
• Standard Normal Table
• Univariate and Bivariate Data
• What is Data
### Audience
Year 10 or higher students, some chapters suitable for students in Year 8 or higher
### Learning Objectives
Learn about topics related to "Data"
Author: Subject Coach
You must be logged in as Student to ask a Question.
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# 3.06 Unit rate
Lesson
### What is unit rate?
A rate is a measure of how quickly one measurement changes with respect to another. Some commonly used rates in our everyday lives are speed, which measures distance per time, and the price of food, which is often measured in dollars per pound.
Notice how both of these examples combines two different units into a single compound unit. We can write these compound units using a slash ( / ) between the different units, so "meters per second" becomes "m/s" and "dollars per pound" becomes "$/lb". This compound unit represents the division of one measurement by another to get a rate. When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates Let's have a look at an example. #### Exploration Consider an Olympic sprinter who runs$100$100 meters in$10$10 seconds. How fast does he run? We can find how far the sprinter runs in$1$1 second by dividing the$100$100 meters evenly between the$10$10 seconds. This calculation tells us that the sprinter runs$10$10 meters in one second. We can write this as a unit rate for the sprinter's speed in meters per second using the compound unit m/s to give us: Sprinter's speed$=$=$10$10 m/s Now let's try a more direct method to finding the sprinter's speed. Since the sprinter runs$100$100 meters in$10$10 seconds we can say that he runs at a rate of$100$100 meters per$10$10 seconds. Writing this as a fraction gives us: Sprinter's speed$=$=$100$100m/$10$10s$=$=$\frac{100}{10}$10010 m/s$=$=$10$10 m/s After some simplifying we find that the speed of the sprinter matches that from the previous method. Notice that we were able to separate the numbers and the units into separate fractions, this is the core concept we use for turning fractions of measurements into rates. Did you know? Not all compound units are written using a slash and instead use the letter "p" to represent "per". For example, "beats per minute" uses the compound unit bpm and "frames per second" uses fps. ### Finding unit rates Rates have two components, the numeric value and the compound unit. The compound unit tells us which units are being measured and the numeric value tells us how quickly the numerator unit changes with respect to the denominator unit. When constructing a rate we usually start with just a fraction of measurements. For example, let's find the speed of a car that travels$180$180 miles in$3$3 hours. We start by setting up the fraction as distance per time, written: Speed$=$=$180$180mi/$3$3hr Then we can separate the fraction into its numeric value and its compound unit. This gives us: Numeric value Compound unit$=$=$\frac{180}{3}$1803$=$=$60$60$=$= mi/hr We can then combine them again to get the unit rate which is: Speed$=$=$60$60 mi/hr Whenever we can, simplify the fraction to get the unit rate. This is much nicer to work with as we can now say that the car travels$60$60 miles per$1$1 hour, rather than$180$180 miles per$3$3 hours. ### Applying unit rate Now that we know how to make unit rates, it's time to use them. Rates are very similar to ratios in that we can use them to calculate how much one measurement changes based on the change in another. #### Worked examples ##### question 1 Returning to our sprinter, we found that they could run at a speed of$10$10 m/s. Assuming that they can maintain this speed, how far will the sprinter run in$15$15 seconds? Think: One way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by$15$15. This will give us: Do: Speed$=$=$10$10 m/s$\times$×$\frac{15}{15}$1515$=$=$150$150m/$15$15s By turning the rate back into a fraction we can see that the sprinter will run$150$150 meters in$15$15 seconds. Reflect: Another way to solve this problem is to apply the rate directly to the question. We can do this by multiplying the time by the rate. This gives us: Distance$=$=$15$15$\times$×$10$10m/s$=$=$\left(15\times10\right)$(15×10) m$\times$×s/s$=$=$150$150m Notice that the units for seconds from the time canceled with the units for second in the compound unit leaving only meters as the unit for distance. ##### question 2 We can also ask the similar question, how long will it take for the sprinter to run$220$220 meters? Think: Again, one way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by$22$22. This gives us: Do: Speed$=$=$10$10 m/s$\times$×$\frac{22}{22}$2222$=$=$220$220m/$22$22s By turning the rate back into a fraction we can see that the sprinter will take$22$22 seconds to run$220$220 meters. Reflect: Another way to solve this problem is to apply the rate directly to the question. We can do this by dividing the distance by the rate. This gives us: Time$=$=$220$220$\div$÷$10$10 m/s$=$=$\frac{220}{10}$22010 s$\times$×m/m =$22$22s This time we divided by the rate so that the compound unit would be flipped and the meters units would cancel out to leave only seconds as the unit for time. Careful! A rate of$10$10 meters per second ($10$10 m/s) is not the same as a rate of$10$10 seconds per meter ($10$10 s/m). In fact,$10$10 m/s$=$=$\frac{1}{10}$110 s/m. When we flip the compound unit we also need to take the reciprocal of the numeric value. ### Converting units When applying rates it's important to make sure that we are applying the right one. #### Worked example ##### question 3 Consider the car from before that traveled at a speed of$60$60 mi/hr. How many miles will the car travel in$7$7 minutes? Think: Before we use one of the methods we learned for applying rates we should first notice that the units in the question don't quite match up with our unit rate. Specifically, the question is asking for minutes as the units for time instead of hours. We can fix this by converting hours into minutes for our unit rate. Using the fact that$1$1 hour$=$=$60$60 minutes we can convert our speed from mi/hr to mi/min like so: Do: Speed$=$=$60$60mi/hr$=$=$60$60mi/$60$60min$=$=$\frac{60}{60}$6060 mi/min$=$=$1$1 mi/min Now that we have a speed with the appropriate units we can apply the unit rate to the question to find how far the car will travel in$7$7 minutes: Distance$=$=$7$7min$\times$×$1$1 mi/min$=$=$\left(7\times1\right)$(7×1) mi$\times$×min/min$=$=$7$7mi Now that we have some experience with this type of question you can try one yourself. #### Practice question ##### question 4 A car travels$320$320 km in$4$4 hours. 1. Complete the table of values. Time taken (hours) Distance traveled (kilometers)$4$4$2$2$1$1$320$320$\editable{}\editable{}$2. What is the speed of the car in kilometers per hour? ##### question 5 Adam really likes apples and eats them at a rate of$4$4 per day. 1. How many apples does Adam eat in one week? 2. If Adam buys$44$44 apples how many days will this last him? ##### question 6 A cyclist travels$70$70 meters per$10$10 seconds. 1. How many groups of$10$10 seconds are in$1\$1 minute?
2. What is the speed of the cyclist in meters per minute?
### Outcomes
#### 6.12b
Determine the unit rate of a proportional relationship and use it to find a missing value in a ratio table
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Calculus Volume 2
# 5.1Sequences
Calculus Volume 25.1 Sequences
## Learning Objectives
• 5.1.1 Find the formula for the general term of a sequence.
• 5.1.2 Calculate the limit of a sequence if it exists.
• 5.1.3 Determine the convergence or divergence of a given sequence.
In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.
## Terminology of Sequences
To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form
$a1,a2,a3,…,an,….a1,a2,a3,…,an,….$
Each of the numbers in the sequence is called a term. The symbol $nn$ is called the index variable for the sequence. We use the notation
${an}n=1∞,or simply{an},{an}n=1∞,or simply{an},$
to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number $anan$ exists for each positive integer $n,n,$ we can also define a sequence as a function whose domain is the set of positive integers.
Let’s consider the infinite, ordered list
$2,4,8,16,32,….2,4,8,16,32,….$
This is a sequence in which the first, second, and third terms are given by $a1=2,a1=2,$ $a2=4,a2=4,$ and $a3=8.a3=8.$ You can probably see that the terms in this sequence have the following pattern:
$a1=21,a2=22,a3=23,a4=24,anda5=25.a1=21,a2=22,a3=23,a4=24,anda5=25.$
Assuming this pattern continues, we can write the $nthnth$ term in the sequence by the explicit formula $an=2n.an=2n.$ Using this notation, we can write this sequence as
${2n}n=1∞or{2n}.{2n}n=1∞or{2n}.$
Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the $nthnth$ term $anan$ in terms of the previous term $an−1.an−1.$ In particular, we can define this sequence as the sequence ${an}{an}$ where $a1=2a1=2$ and for all $n≥2,n≥2,$ each term $anan$ is defined by the recurrence relation$an=2an−1.an=2an−1.$
## Definition
An infinite sequence${an}{an}$ is an ordered list of numbers of the form
$a1,a2,…,an,….a1,a2,…,an,….$
The subscript $nn$ is called the index variable of the sequence. Each number $anan$ is a term of the sequence. Sometimes sequences are defined by explicit formulas, in which case $an=f(n)an=f(n)$ for some function $f(n)f(n)$ defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.
Note that the index does not have to start at $n=1n=1$ but could start with other integers. For example, a sequence given by the explicit formula $an=f(n)an=f(n)$ could start at $n=0,n=0,$ in which case the sequence would be
$a0,a1,a2,….a0,a1,a2,….$
Similarly, for a sequence defined by a recurrence relation, the term $a0a0$ may be given explicitly, and the terms $anan$ for $n≥1n≥1$ may be defined in terms of $an−1.an−1.$ Since a sequence ${an}{an}$ has exactly one value for each positive integer $n,n,$ it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence ${an}{an}$ consists of all points $(n,an)(n,an)$ for all positive integers $n.n.$ Figure 5.2 shows the graph of ${2n}.{2n}.$
Figure 5.2 The plotted points are a graph of the sequence ${2n}.{2n}.$
Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence
$3,7,11,15,19,….3,7,11,15,19,….$
You can see that the difference between every consecutive pair of terms is $4.4.$ Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation
${a1=3an=an−1+4forn≥2.{a1=3an=an−1+4forn≥2.$
Note that
$a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.$
Thus the sequence can also be described using the explicit formula
$an=3+4(n−1)=4n−1.an=3+4(n−1)=4n−1.$
In general, an arithmetic sequence is any sequence of the form $an=cn+b.an=cn+b.$
In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence
$2,−23,29,−227,281,….2,−23,29,−227,281,….$
We see that the ratio of any term to the preceding term is $−13.−13.$ Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as
$a1=2an=−13·an−1forn≥2.a1=2an=−13·an−1forn≥2.$
Alternatively, since
$a2=−13·2a3=(−13)(−13)(2)=(−13)2·2a4=(−13)(−13)(−13)(2)=(−13)3·2,a2=−13·2a3=(−13)(−13)(2)=(−13)2·2a4=(−13)(−13)(−13)(2)=(−13)3·2,$
we see that the sequence can be described by using the explicit formula
$an=2(−13)n−1.an=2(−13)n−1.$
The sequence ${2n}{2n}$ that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2.2.$ In general, a geometric sequence is any sequence of the form $an=crn.an=crn.$
## Example 5.1
### Finding Explicit Formulas
For each of the following sequences, find an explicit formula for the $nthnth$ term of the sequence.
1. $−12,23,−34,45,−56,…−12,23,−34,45,−56,…$
2. $34,97,2710,8113,24316,…34,97,2710,8113,24316,…$
## Checkpoint5.1
Find an explicit formula for the $nthnth$ term of the sequence ${15,−17,19,−111,…}.{15,−17,19,−111,…}.$
## Example 5.2
### Defined by Recurrence Relations
For each of the following recursively defined sequences, find an explicit formula for the sequence.
1. $a1=2,a1=2,$ $an=−3an−1an=−3an−1$ for $n≥2n≥2$
2. $a1=12,a1=12,$ $an=an−1+(12)nan=an−1+(12)n$ for $n≥2n≥2$
## Checkpoint5.2
Find an explicit formula for the sequence defined recursively such that $a1=−4a1=−4$ and $an=an−1+6.an=an−1+6.$
## Limit of a Sequence
A fundamental question that arises regarding infinite sequences is the behavior of the terms as $nn$ gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as $n→∞.n→∞.$ For example, consider the following four sequences and their different behaviors as $n→∞n→∞$ (see Figure 5.3):
1. ${1+3n}={4,7,10,13,…}.{1+3n}={4,7,10,13,…}.$ The terms $1+3n1+3n$ become arbitrarily large as $n→∞.n→∞.$ In this case, we say that $1+3n→∞1+3n→∞$ as $n→∞.n→∞.$
2. ${1−(12)n}={12,34,78,1516,…}.{1−(12)n}={12,34,78,1516,…}.$ The terms $1−(12)n→11−(12)n→1$ as $n→∞.n→∞.$
3. ${(−1)n}={−1,1,−1,1,…}.{(−1)n}={−1,1,−1,1,…}.$ The terms alternate but do not approach one single value as $n→∞.n→∞.$
4. ${(−1)nn}={−1,12,−13,14,…}.{(−1)nn}={−1,12,−13,14,…}.$ The terms alternate for this sequence as well, but $(−1)nn→0(−1)nn→0$ as $n→∞.n→∞.$
Figure 5.3 (a) The terms in the sequence become arbitrarily large as $n→∞.n→∞.$ (b) The terms in the sequence approach $11$ as $n→∞.n→∞.$ (c) The terms in the sequence alternate between $11$ and $−1−1$ as $n→∞.n→∞.$ (d) The terms in the sequence alternate between positive and negative values but approach $00$ as $n→∞.n→∞.$
From these examples, we see several possibilities for the behavior of the terms of a sequence as $n→∞.n→∞.$ In two of the sequences, the terms approach a finite number as $n→∞.n→∞.$ In the other two sequences, the terms do not. If the terms of a sequence approach a finite number $LL$ as $n→∞,n→∞,$ we say that the sequence is a convergent sequence and the real number $LL$ is the limit of the sequence. We can give an informal definition here.
## Definition
Given a sequence ${an},{an},$ if the terms $anan$ become arbitrarily close to a finite number $LL$ as $nn$ becomes sufficiently large, we say ${an}{an}$ is a convergent sequence and $LL$ is the limit of the sequence. In this case, we write
$limn→∞an=L.limn→∞an=L.$
If a sequence ${an}{an}$ is not convergent, we say it is a divergent sequence.
From Figure 5.3, we see that the terms in the sequence ${1−(12)n}{1−(12)n}$ are becoming arbitrarily close to $11$ as $nn$ becomes very large. We conclude that ${1−(12)n}{1−(12)n}$ is a convergent sequence and its limit is $1.1.$ In contrast, from Figure 5.3, we see that the terms in the sequence $1+3n1+3n$ are not approaching a finite number as $nn$ becomes larger. We say that ${1+3n}{1+3n}$ is a divergent sequence.
In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.
## Definition
A sequence ${an}{an}$ converges to a real number $LL$ if for all $ε>0,ε>0,$ there exists an integer $NN$ such that $|an−L|<ε|an−L|<ε$ if $n≥N.n≥N.$ The number $LL$ is the limit of the sequence and we write
$limn→∞an=Loran→L.limn→∞an=Loran→L.$
In this case, we say the sequence ${an}{an}$ is a convergent sequence. If a sequence does not converge, it is a divergent sequence, and we say the limit does not exist.
We remark that the convergence or divergence of a sequence ${an}{an}$ depends only on what happens to the terms $anan$ as $n→∞.n→∞.$ Therefore, if a finite number of terms $b1,b2,…,bNb1,b2,…,bN$ are placed before $a1a1$ to create a new sequence
$b1,b2,…,bN,a1,a2,…,b1,b2,…,bN,a1,a2,…,$
this new sequence will converge if ${an}{an}$ converges and diverge if ${an}{an}$ diverges. Further, if the sequence ${an}{an}$ converges to $L,L,$ this new sequence will also converge to $L.L.$
Figure 5.4 As $nn$ increases, the terms $anan$ become closer to $L.L.$ For values of $n≥N,n≥N,$ the distance between each point $(n,an)(n,an)$ and the line $y=Ly=L$ is less than $ε.ε.$
As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences ${1+3n}{1+3n}$ and ${(−1)n}{(−1)n}$ shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The sequence ${(−1)n}{(−1)n}$ diverges because the terms alternate between $11$ and $−1,−1,$ but do not approach one value as $n→∞.n→∞.$ On the other hand, the sequence ${1+3n}{1+3n}$ diverges because the terms $1+3n→∞1+3n→∞$ as $n→∞.n→∞.$ We say the sequence ${1+3n}{1+3n}$ diverges to infinity and write $limn→∞(1+3n)=∞.limn→∞(1+3n)=∞.$ It is important to recognize that this notation does not imply the limit of the sequence ${1+3n}{1+3n}$ exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence ${−5n+2}{−5n+2}$ diverges to negative infinity because $−5n+2→−∞−5n+2→−∞$ as $n→∞.n→∞.$ We write this as $limn→∞(−5n+2)=→−∞.limn→∞(−5n+2)=→−∞.$
Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence ${an}{an}$ and a related function $ff$ defined on all positive real numbers such that $f(n)=anf(n)=an$ for all integers $n≥1.n≥1.$ Since the domain of the sequence is a subset of the domain of $f,f,$ if $limx→∞f(x)limx→∞f(x)$ exists, then the sequence converges and has the same limit. For example, consider the sequence ${1n}{1n}$ and the related function $f(x)=1x.f(x)=1x.$ Since the function $ff$ defined on all real numbers $x>0x>0$ satisfies $f(x)=1x→0f(x)=1x→0$ as $x→∞,x→∞,$ the sequence ${1n}{1n}$ must satisfy $1n→01n→0$ as $n→∞.n→∞.$
## Theorem5.1
### Limit of a Sequence Defined by a Function
Consider a sequence ${an}{an}$ such that $an=f(n)an=f(n)$ for all $n≥1.n≥1.$ If there exists a real number $LL$ such that
$limx→∞f(x)=L,limx→∞f(x)=L,$
then ${an}{an}$ converges and
$limn→∞an=L.limn→∞an=L.$
We can use this theorem to evaluate $limn→∞rnlimn→∞rn$ for $0≤r≤1.0≤r≤1.$ For example, consider the sequence ${(1/2)n}{(1/2)n}$ and the related exponential function $f(x)=(1/2)x.f(x)=(1/2)x.$ Since $limx→∞(1/2)x=0,limx→∞(1/2)x=0,$ we conclude that the sequence ${(1/2)n}{(1/2)n}$ converges and its limit is $0.0.$ Similarly, for any real number $rr$ such that $0≤r<1,0≤r<1,$ $limx→∞rx=0,limx→∞rx=0,$ and therefore the sequence ${rn}{rn}$ converges. On the other hand, if $r=1,r=1,$ then $limx→∞rx=1,limx→∞rx=1,$ and therefore the limit of the sequence ${1n}{1n}$ is $1.1.$ If $r>1,r>1,$ $limx→∞rx=∞,limx→∞rx=∞,$ and therefore we cannot apply this theorem. However, in this case, just as the function $rxrx$ grows without bound as $n→∞,n→∞,$ the terms $rnrn$ in the sequence become arbitrarily large as $n→∞,n→∞,$ and we conclude that the sequence ${rn}{rn}$ diverges to infinity if $r>1.r>1.$
We summarize these results regarding the geometric sequence ${rn}:{rn}:$
$rn→0if01.rn→0if01.$
Later in this section we consider the case when $r<0.r<0.$
We now consider slightly more complicated sequences. For example, consider the sequence ${(2/3)n+(1/4)n}.{(2/3)n+(1/4)n}.$ The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences ${(2/3)n}{(2/3)n}$ and ${(1/4)n}.{(1/4)n}.$ As we describe in the following algebraic limit laws, since ${(2/3)n}{(2/3)n}$ and ${1/4)n}{1/4)n}$ both converge to $0,0,$ the sequence ${(2/3)n+(1/4)n}{(2/3)n+(1/4)n}$ converges to $0+0=0.0+0=0.$ Just as we were able to evaluate a limit involving an algebraic combination of functions $ff$ and $gg$ by looking at the limits of $ff$ and $gg$ (see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of $anan$ and $bnbn$ by evaluating the limits of ${an}{an}$ and ${bn}.{bn}.$
## Theorem5.2
### Algebraic Limit Laws
Given sequences ${an}{an}$ and ${bn}{bn}$ and any real number $c,c,$ if there exist constants $AA$ and $BB$ such that $limn→∞an=Alimn→∞an=A$ and $limn→∞bn=B,limn→∞bn=B,$ then
1. $limn→∞c=climn→∞c=c$
2. $limn→∞can=climn→∞an=cAlimn→∞can=climn→∞an=cA$
3. $limn→∞(an±bn)=limn→∞an±limn→∞bn=A±Blimn→∞(an±bn)=limn→∞an±limn→∞bn=A±B$
4. $limn→∞(an·bn)=(limn→∞an)·(limn→∞bn)=A·Blimn→∞(an·bn)=(limn→∞an)·(limn→∞bn)=A·B$
5. $limn→∞(anbn)=limn→∞anlimn→∞bn=AB,limn→∞(anbn)=limn→∞anlimn→∞bn=AB,$ provided $B≠0B≠0$ and each $bn≠0.bn≠0.$
### Proof
We prove part iii.
Let $ϵ>0.ϵ>0.$ Since $limn→∞an=A,limn→∞an=A,$ there exists a constant positive integer $N1N1$ such that $|an-A|<ε2|an-A|<ε2$ for all $n≥N1.n≥N1.$ Since $limn→∞bn=B,limn→∞bn=B,$ there exists a constant $N2N2$ such that $|bn−B|<ε/2|bn−B|<ε/2$ for all $n≥N2.n≥N2.$ Let $NN$ be the larger of $N1N1$ and $N2.N2.$ Therefore, for all $n≥N,n≥N,$
$|(an+bn)−(A+B)|≤|an−A|+|bn−B|<ε2+ε2=ε.|(an+bn)−(A+B)|≤|an−A|+|bn−B|<ε2+ε2=ε.$
The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence ${1n2}.{1n2}.$ As shown earlier, $limn→∞1/n=0.limn→∞1/n=0.$ Similarly, for any positive integer $k,k,$ we can conclude that
$limn→∞1nk=0.limn→∞1nk=0.$
In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.
## Example 5.3
### Determining Convergence and Finding Limits
For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.
1. ${5−3n2}{5−3n2}$
2. ${3n4−7n2+56−4n4}{3n4−7n2+56−4n4}$
3. ${2nn2}{2nn2}$
4. ${(1+4n)n}{(1+4n)n}$
## Checkpoint5.3
Consider the sequence ${(5n2+1)/en}.{(5n2+1)/en}.$ Determine whether or not the sequence converges. If it converges, find its limit.
Recall that if $ff$ is a continuous function at a value $L,L,$ then $f(x)→f(L)f(x)→f(L)$ as $x→L.x→L.$ This idea applies to sequences as well. Suppose a sequence $an→L,an→L,$ and a function $ff$ is continuous at $L.L.$ Then $f(an)→f(L).f(an)→f(L).$ This property often enables us to find limits for complicated sequences. For example, consider the sequence $5−3n2.5−3n2.$ From Example 5.3a. we know the sequence $5−3n2→5.5−3n2→5.$ Since $xx$ is a continuous function at $x=5,x=5,$
$limn→∞5−3n2=limn→∞(5−3n2)=5.limn→∞5−3n2=limn→∞(5−3n2)=5.$
## Theorem5.3
### Continuous Functions Defined on Convergent Sequences
Consider a sequence ${an}{an}$ and suppose there exists a real number $LL$ such that the sequence ${an}{an}$ converges to $L.L.$ Suppose $ff$ is a continuous function at $L.L.$ Then there exists an integer $NN$ such that $ff$ is defined at all values $anan$ for $n≥N,n≥N,$ and the sequence ${f(an)}{f(an)}$ converges to $f(L)f(L)$ (Figure 5.5).
### Proof
Let $ϵ>0.ϵ>0.$ Since $ff$ is continuous at $L,L,$ there exists $δ>0δ>0$ such that $|f(x)−f(L)|<ε|f(x)−f(L)|<ε$ if $|x−L|<δ.|x−L|<δ.$ Since the sequence ${an}{an}$ converges to $L,L,$ there exists $NN$ such that $|an−L|<δ|an−L|<δ$ for all $n≥N.n≥N.$ Therefore, for all $n≥N,n≥N,$ $|an−L|<δ,|an−L|<δ,$ which implies $|f(an)−f(L)|<ε.|f(an)−f(L)|<ε.$ We conclude that the sequence ${f(an)}{f(an)}$ converges to $f(L).f(L).$
Figure 5.5 Because $ff$ is a continuous function as the inputs $a1,a2,a3,…a1,a2,a3,…$ approach $L,L,$ the outputs $f(a1),f(a2),f(a3),…f(a1),f(a2),f(a3),…$ approach $f(L).f(L).$
## Example 5.4
### Limits Involving Continuous Functions Defined on Convergent Sequences
Determine whether the sequence ${cos(3/n2)}{cos(3/n2)}$ converges. If it converges, find its limit.
## Checkpoint5.4
Determine if the sequence ${2n+13n+5}{2n+13n+5}$ converges. If it converges, find its limit.
Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.
## Theorem5.4
### Squeeze Theorem for Sequences
Consider sequences ${an},{an},$ ${bn},{bn},$ and ${cn}.{cn}.$ Suppose there exists an integer $NN$ such that
$an≤bn≤cnfor alln≥N.an≤bn≤cnfor alln≥N.$
If there exists a real number $LL$ such that
$limn→∞an=L=limn→∞cn,limn→∞an=L=limn→∞cn,$
then ${bn}{bn}$ converges and $limn→∞bn=Llimn→∞bn=L$ (Figure 5.6).
### Proof
Let $ε>0.ε>0.$ Since the sequence ${an}{an}$ converges to $L,L,$ there exists an integer $N1N1$ such that $|an−L|<ε|an−L|<ε$ for all $n≥N1.n≥N1.$ Similarly, since ${cn}{cn}$ converges to $L,L,$ there exists an integer $N2N2$ such that $|cn−L|<ε|cn−L|<ε$ for all $n≥N2.n≥N2.$ By assumption, there exists an integer $NN$ such that $an≤bn≤cnan≤bn≤cn$ for all $n≥N.n≥N.$ Let $MM$ be the largest of $N1,N2,N1,N2,$ and $N.N.$ We must show that $|bn−L|<ε|bn−L|<ε$ for all $n≥M.n≥M.$ For all $n≥M,n≥M,$
$−ε<−|an−L|≤an−L≤bn−L≤cn−L≤|cn−L|<ε.−ε<−|an−L|≤an−L≤bn−L≤cn−L≤|cn−L|<ε.$
Therefore, $−ε and we conclude that $|bn−L|<ε|bn−L|<ε$ for all $n≥M,n≥M,$ and we conclude that the sequence ${bn}{bn}$ converges to $L.L.$
Figure 5.6 Each term $bnbn$ satisfies $an≤bn≤cnan≤bn≤cn$ and the sequences ${an}{an}$ and ${cn}{cn}$ converge to the same limit, so the sequence ${bn}{bn}$ must converge to the same limit as well.
## Example 5.5
### Using the Squeeze Theorem
Use the Squeeze Theorem to find the limit of each of the following sequences.
1. ${cosnn2}{cosnn2}$
2. ${(−12)n}{(−12)n}$
## Checkpoint5.5
Find $limn→∞2n−sinnn.limn→∞2n−sinnn.$
Using the idea from Example 5.5b. we conclude that $rn→0rn→0$ for any real number $rr$ such that $−1 If $r<−1,r<−1,$ the sequence ${rn}{rn}$ diverges because the terms oscillate and become arbitrarily large in magnitude. If $r=−1,r=−1,$ the sequence ${rn}={(−1)n}{rn}={(−1)n}$ diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.
$rn→0if|r|<1rn→0if|r|<1$
(5.1)
$rn→1ifr=1rn→1ifr=1$
(5.2)
$rn→∞ifr>1rn→∞ifr>1$
(5.3)
${rn}diverges ifr≤−1{rn}diverges ifr≤−1$
(5.4)
## Bounded Sequences
We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.
## Definition
A sequence ${an}{an}$ is bounded above if there exists a real number $MM$ such that
$an≤Man≤M$
for all positive integers $n.n.$
A sequence ${an}{an}$ is bounded below if there exists a real number $MM$ such that
$M≤anM≤an$
for all positive integers $n.n.$
A sequence ${an}{an}$ is a bounded sequence if it is bounded above and bounded below.
If a sequence is not bounded, it is an unbounded sequence.
For example, the sequence ${1/n}{1/n}$ is bounded above because $1/n≤11/n≤1$ for all positive integers $n.n.$ It is also bounded below because $1/n≥01/n≥0$ for all positive integers n. Therefore, ${1/n}{1/n}$ is a bounded sequence. On the other hand, consider the sequence ${2n}.{2n}.$ Because $2n≥22n≥2$ for all $n≥1,n≥1,$ the sequence is bounded below. However, the sequence is not bounded above. Therefore, ${2n}{2n}$ is an unbounded sequence.
We now discuss the relationship between boundedness and convergence. Suppose a sequence ${an}{an}$ is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms $anan$ that are arbitrarily large in magnitude as $nn$ gets larger. As a result, the sequence ${an}{an}$ cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.
## Theorem5.5
### Convergent Sequences Are Bounded
If a sequence ${an}{an}$ converges, then it is bounded.
Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence ${(−1)n}{(−1)n}$ is bounded, but the sequence diverges because the sequence oscillates between $11$ and $−1−1$ and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.
Consider a bounded sequence ${an}.{an}.$ Suppose the sequence ${an}{an}$ is increasing. That is, $a1≤a2≤a3….a1≤a2≤a3….$ Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that ${an}{an}$ converges. For example, consider the sequence
${12,23,34,45,…}.{12,23,34,45,…}.$
Since this sequence is increasing and bounded above, it converges. Next, consider the sequence
${2,0,3,0,4,0,1,−12,−13,−14,…}.{2,0,3,0,4,0,1,−12,−13,−14,…}.$
Even though the sequence is not increasing for all values of $n,n,$ we see that $−1/2<−1/3<−1/4<⋯.−1/2<−1/3<−1/4<⋯.$ Therefore, starting with the eighth term, $a8=−1/2,a8=−1/2,$ the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.
## Definition
A sequence ${an}{an}$ is increasing for all $n≥n0n≥n0$ if
$an≤an+1for alln≥n0.an≤an+1for alln≥n0.$
A sequence ${an}{an}$ is decreasing for all $n≥n0n≥n0$ if
$an≥an+1for alln≥n0.an≥an+1for alln≥n0.$
A sequence ${an}{an}$ is a monotone sequence for all $n≥n0n≥n0$ if it is increasing for all $n≥n0n≥n0$ or decreasing for all $n≥n0.n≥n0.$
We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.
## Theorem5.6
### Monotone Convergence Theorem
If ${an}{an}$ is a bounded sequence and there exists a positive integer $n0n0$ such that ${an}{an}$ is monotone for all $n≥n0,n≥n0,$ then ${an}{an}$ converges.
The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 5.7).
Figure 5.7 Since the sequence ${an}{an}$ is increasing and bounded above, it must converge.
In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.
## Example 5.6
### Using the Monotone Convergence Theorem
For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.
1. ${4nn!}{4nn!}$
2. ${an}{an}$ defined recursively such that
$a1=2andan+1=an2+12anfor alln≥2.a1=2andan+1=an2+12anfor alln≥2.$
## Checkpoint5.6
Consider the sequence ${an}{an}$ defined recursively such that $a1=1,a1=1,$ $an=an−1/2.an=an−1/2.$ Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.
## Student Project
### Fibonacci Numbers
The Fibonacci numbers are defined recursively by the sequence ${Fn}{Fn}$ where $F0=0,F0=0,$ $F1=1F1=1$ and for $n≥2,n≥2,$
$Fn=Fn−1+Fn−2.Fn=Fn−1+Fn−2.$
Here we look at properties of the Fibonacci numbers.
1. Write out the first twenty Fibonacci numbers.
2. Find a closed formula for the Fibonacci sequence by using the following steps.
1. Consider the recursively defined sequence ${xn}{xn}$ where $xo=cxo=c$ and $xn+1=axn.xn+1=axn.$ Show that this sequence can be described by the closed formula $xn=canxn=can$ for all $n≥0.n≥0.$
2. Using the result from part a. as motivation, look for a solution of the equation
$Fn=Fn−1+Fn−2Fn=Fn−1+Fn−2$
of the form $Fn=cλn.Fn=cλn.$ Determine what two values for $λλ$ will allow $FnFn$ to satisfy this equation.
3. Consider the two solutions from part b.: $λ1λ1$ and $λ2.λ2.$ Let $Fn=c1λ1n+c2λ2n.Fn=c1λ1n+c2λ2n.$ Use the initial conditions $F0F0$ and $F1F1$ to determine the values for the constants $c1c1$ and $c2c2$ and write the closed formula $Fn.Fn.$
3. Use the answer in 2 c. to show that
$limn→∞Fn+1Fn=1+52.limn→∞Fn+1Fn=1+52.$
The number $ϕ=(1+5)/2ϕ=(1+5)/2$ is known as the golden ratio (Figure 5.8 and Figure 5.9).
Figure 5.8 The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)
Figure 5.9 The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon appears to have these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)
## Section 5.1 Exercises
Find the first six terms of each of the following sequences, starting with $n=1.n=1.$
1.
$an=1+(−1)nan=1+(−1)n$ for $n≥1n≥1$
2.
$an=n2−1an=n2−1$ for $n≥1n≥1$
3.
$a1=1a1=1$ and $an=an−1+nan=an−1+n$ for $n≥2n≥2$
4.
$a1=1,a1=1,$ $a2=1a2=1$ and $an+2=an+an+1an+2=an+an+1$ for $n≥1n≥1$
5.
Find an explicit formula for $anan$ where $a1=1a1=1$ and $an=an−1+nan=an−1+n$ for $n≥2.n≥2.$
6.
Find a formula $anan$ for the $nthnth$ term of the arithmetic sequence whose first term is $a1=1a1=1$ such that $an+1−an=17an+1−an=17$ for $n≥1.n≥1.$
7.
Find a formula $anan$ for the $nthnth$ term of the arithmetic sequence whose first term is $a1=−3a1=−3$ such that $an+1−an=4an+1−an=4$ for $n≥1.n≥1.$
8.
Find a formula $anan$ for the $nthnth$ term of the geometric sequence whose first term is $a1=1a1=1$ such that $an+1an=10an+1an=10$ for $n≥1.n≥1.$
9.
Find a formula $anan$ for the $nthnth$ term of the geometric sequence whose first term is $a1=3a1=3$ such that $an+1an=1/10an+1an=1/10$ for $n≥1.n≥1.$
10.
Find an explicit formula for the $nthnth$ term of the sequence whose first several terms are ${0,3,8,15,24,35,48,63,80,99,…}.{0,3,8,15,24,35,48,63,80,99,…}.$ (Hint: First add one to each term.)
11.
Find an explicit formula for the $nthnth$ term of the sequence satisfying $a1=0a1=0$ and $an=2an−1+1an=2an−1+1$ for $n≥2.n≥2.$
Find a formula for the general term $anan$ of each of the following sequences.
12.
${1,0,−1,0,1,0,−1,0,…}{1,0,−1,0,1,0,−1,0,…}$ (Hint: Find where $sinxsinx$ takes these values)
13.
${ 1 , − 1 / 3 , 1 / 5 , − 1 / 7 ,… } { 1 , − 1 / 3 , 1 / 5 , − 1 / 7 ,… }$
Find a function $f(n)f(n)$ that identifies the $nthnth$ term $anan$ of the following recursively defined sequences, as $an=f(n).an=f(n).$
14.
$a1=1a1=1$ and $an+1=−anan+1=−an$ for $n≥1n≥1$
15.
$a1=2a1=2$ and $an+1=2anan+1=2an$ for $n≥1n≥1$
16.
$a1=1a1=1$ and $an+1=(n+1)anan+1=(n+1)an$ for $n≥1n≥1$
17.
$a1=2a1=2$ and $an+1=(n+1)an/2an+1=(n+1)an/2$ for $n≥1n≥1$
18.
$a1=1a1=1$ and $an+1=an/2nan+1=an/2n$ for $n≥1n≥1$
Plot the first $NN$ terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.
19.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ and for $n≥2,n≥2,$ $an=12(an−1+an−2);an=12(an−1+an−2);$ $N=30N=30$
20.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ $a3=3a3=3$ and for $n≥4,n≥4,$ $an=13(an−1+an−2+an−3),an=13(an−1+an−2+an−3),$ $N=30N=30$
21.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ and for $n≥3,n≥3,$ $an=an−1an−2;an=an−1an−2;$ $N=30N=30$
22.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ $a3=3,a3=3,$ and for $n≥4,n≥4,$ $an=an−1an−2an−3;an=an−1an−2an−3;$ $N=30N=30$
Suppose that $limn→∞an=1,limn→∞an=1,$ $limn→∞bn=−1,limn→∞bn=−1,$ and $0<−bn for all $n.n.$ Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.
23.
$lim n → ∞ ( 3 a n − 4 b n ) lim n → ∞ ( 3 a n − 4 b n )$
24.
$lim n → ∞ ( 1 2 b n − 1 2 a n ) lim n → ∞ ( 1 2 b n − 1 2 a n )$
25.
$lim n → ∞ a n + b n a n − b n lim n → ∞ a n + b n a n − b n$
26.
$lim n → ∞ a n − b n a n + b n lim n → ∞ a n − b n a n + b n$
Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.
27.
$n 2 2 n n 2 2 n$
28.
$( n − 1 ) 2 ( n + 1 ) 2 ( n − 1 ) 2 ( n + 1 ) 2$
29.
$n n + 1 n n + 1$
30.
$n1/nn1/n$ (Hint: $n1/n=e1nlnn)n1/n=e1nlnn)$
For each of the following sequences, whose $nthnth$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.
31.
$n/2n,n/2n,$ $n≥2n≥2$
32.
$ln ( 1 + 1 n ) ln ( 1 + 1 n )$
33.
$sin n sin n$
34.
$cos ( n 2 ) cos ( n 2 )$
35.
$n1/n,n1/n,$ $n≥3n≥3$
36.
$n−1/n,n−1/n,$ $n≥3n≥3$
37.
$tan n tan n$
38.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
$a1=2,a1=2,$ $a2=22,a2=22,$ $a3=222a3=222$ etc.
39.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
$a1=3,a1=3,$ $an=2an−1,an=2an−1,$ $n=2,3,….n=2,3,….$
Use the Squeeze Theorem to find the limit of each of the following sequences.
40.
$n sin ( 1 / n ) n sin ( 1 / n )$
41.
$cos ( 1 / n ) − 1 1 / n cos ( 1 / n ) − 1 1 / n$
42.
$a n = n ! n n a n = n ! n n$
43.
$a n = sin n sin ( 1 / n ) a n = sin n sin ( 1 / n )$
For the following sequences, plot the first $2525$ terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.
44.
[T] $an=sinnan=sinn$
45.
[T] $an=cosnan=cosn$
Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.
46.
$a n = tan −1 ( n 2 ) a n = tan −1 ( n 2 )$
47.
$a n = ( 2 n ) 1 / n − n 1 / n a n = ( 2 n ) 1 / n − n 1 / n$
48.
$a n = ln ( n 2 ) ln ( 2 n ) a n = ln ( n 2 ) ln ( 2 n )$
49.
$a n = ( 1 − 2 n ) n a n = ( 1 − 2 n ) n$
50.
$a n = ln ( n + 2 n 2 − 3 ) a n = ln ( n + 2 n 2 − 3 )$
51.
$a n = 2 n + 3 n 4 n a n = 2 n + 3 n 4 n$
52.
$a n = ( 1000 ) n n ! a n = ( 1000 ) n n !$
53.
$a n = ( n ! ) 2 ( 2 n ) ! a n = ( n ! ) 2 ( 2 n ) !$
Newton’s method seeks to approximate a solution $f(x)=0f(x)=0$ that starts with an initial approximation $x0x0$ and successively defines a sequence $xn+1=xn−f(xn)f′(xn).xn+1=xn−f(xn)f′(xn).$ For the given choice of $ff$ and $x0,x0,$ write out the formula for $xn+1.xn+1.$ If the sequence appears to converge, give an exact formula for the solution $x,x,$ then identify the limit $xx$ accurate to four decimal places and the smallest $nn$ such that $xnxn$ agrees with $xx$ up to four decimal places.
54.
[T] $f(x)=x2−2,f(x)=x2−2,$ $x0=1x0=1$
55.
[T] $f(x)=(x−1)2−2,f(x)=(x−1)2−2,$ $x0=2x0=2$
56.
[T] $f(x)=ex−2,f(x)=ex−2,$ $x0=1x0=1$
57.
[T] $f(x)=lnx−1,f(x)=lnx−1,$ $x0=2x0=2$
58.
[T] Suppose you start with one liter of vinegar and repeatedly remove $0.1L,0.1L,$ replace with water, mix, and repeat.
1. Find a formula for the concentration after $nn$ steps.
2. After how many steps does the mixture contain less than $10%10%$ vinegar?
59.
[T] A lake initially contains $20002000$ fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by $6%6%$ each month. However, factoring in all causes, $150150$ fish are lost each month.
1. Explain why the fish population after $nn$ months is modeled by $Pn=1.06Pn−1−150Pn=1.06Pn−1−150$ with $P0=2000.P0=2000.$
2. How many fish will be in the pond after one year?
60.
[T] A bank account earns $5%5%$ interest compounded monthly. Suppose that $10001000$ is initially deposited into the account, but that $1010$ is withdrawn each month.
1. Show that the amount in the account after $nn$ months is $An=(1+.05/12)An−1−10;An=(1+.05/12)An−1−10;$ $A0=1000.A0=1000.$
2. How much money will be in the account after $11$ year?
3. Is the amount increasing or decreasing?
4. Suppose that instead of $10,10,$ a fixed amount $dd$ dollars is withdrawn each month. Find a value of $dd$ such that the amount in the account after each month remains $1000.1000.$
5. What happens if $dd$ is greater than this amount?
61.
[T] A student takes out a college loan of $10,00010,000$ at an annual percentage rate of $6%,6%,$ compounded monthly.
1. If the student makes payments of $100100$ per month, how much does the student owe after $1212$ months?
2. After how many months will the loan be paid off?
62.
[T] Consider a series combining geometric growth and arithmetic decrease. Let $a1=1.a1=1.$ Fix $a>1a>1$ and $0 Set $an+1=a.an−b.an+1=a.an−b.$ Find a formula for $an+1an+1$ in terms of $an,an,$ $a,a,$ and $bb$ and a relationship between $aa$ and $bb$ such that $anan$ converges.
63.
[T] The binary representation $x=0.b1b2b3...x=0.b1b2b3...$ of a number $xx$ between $00$ and $11$ can be defined as follows. Let $b1=0b1=0$ if $x<1/2x<1/2$ and $b1=1b1=1$ if $1/2≤x<1.1/2≤x<1.$ Let $x1=2x−b1.x1=2x−b1.$ Let $b2=0b2=0$ if $x1<1/2x1<1/2$ and $b2=1b2=1$ if $1/2≤x<1.1/2≤x<1.$ Let $x2=2x1−b2x2=2x1−b2$ and in general, $xn=2xn−1−bnxn=2xn−1−bn$ and $bn−1=0bn−1=0$ if $xn<1/2xn<1/2$ and $bn−1=1bn−1=1$ if $1/2≤xn<1.1/2≤xn<1.$ Find the binary expansion of $1/3.1/3.$
64.
[T] To find an approximation for $π,π,$ set $a0=2+1,a0=2+1,$ $a1=2+a0,a1=2+a0,$ and, in general, $an+1=2+an.an+1=2+an.$ Finally, set $pn=3.2n+12−an.pn=3.2n+12−an.$ Find the first ten terms of $pnpn$ and compare the values to $π.π.$
For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of $NN$ integers $a1,a2,…,aNa1,a2,…,aN$ by fixing two special integers $KK$ and $MM$ and letting $an+1an+1$ be the remainder after dividing $K.anK.an$ into $M,M,$ then creates a bit sequence of zeros and ones whose $nthnth$ term $bnbn$ is equal to one if $anan$ is odd and equal to zero if $anan$ is even. If the bits $bnbn$ are pseudorandom, then the behavior of their average $(b1+b2+⋯+bN)/N(b1+b2+⋯+bN)/N$ should be similar to behavior of averages of truly randomly generated bits.
65.
[T] Starting with $K=16,807K=16,807$ and $M=2,147,483,647,M=2,147,483,647,$ using ten different starting values of $a1,a1,$ compute sequences of bits $bnbn$ up to $n=1000,n=1000,$ and compare their averages to ten such sequences generated by a random bit generator.
66.
[T] Find the first $10001000$ digits of $ππ$ using either a computer program or Internet resource. Create a bit sequence $bnbn$ by letting $bn=1bn=1$ if the $nthnth$ digit of $ππ$ is odd and $bn=0bn=0$ if the $nthnth$ digit of $ππ$ is even. Compute the average value of $bnbn$ and the average value of $dn=|bn+1−bn|,dn=|bn+1−bn|,$ $n=1,...,999.n=1,...,999.$ Does the sequence $bnbn$ appear random? Do the differences between successive elements of $bnbn$ appear random?
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## Geometric Sequences and Series
Introduction
We have discussed about arithmetic sequences, its characteristics and its connection to linear functions. In this post, we will discuss another type of sequence.
The sequence of numbers 2, 6, 18, 54, 162, … is an example of an geometric sequence. The first term 2 is multiplied by 3 to get the second term, the second term is multiplied by 3 to get the third term, the third term is multiplied by 3 to get the fourth term, and so on. The same number that we multiplied to each term is called the common ratio. Expressing the sequence above in terms of the first term and the common ratio, we have 2, 2(3), 2(32), 2(33), …. Hence, a geometric sequence, also known as a geometric progression, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.
The Sierpinski triangle below is an example of a geometric representation of a geometric sequence. The number of blue triangles, the number of white triangles, their areas, and their side lengths form different geometric sequences. It is left to the reader, as an exercise, to find the rules of these geometric sequences.
Figure 1 - The Seriepinski Triangles.
To generalize, if a1 is its first term and the common ratio is r, then the general form of a geometric sequence is a1, a1r, a1r2, a1r3,…, and the nth term of the sequence is a1rn-1.
A geometric series, on the other hand, is the sum of the terms of a geometric sequence. Given a geometric sequence with terms a1r, a1r2, a1r3,…, the sum Sn of the geometric sequence with n terms is the geometric series a1 + a1r + a1r2, a1r3 + … + arn-1. Multiplying Sn by -r and adding it to Sn, we have
Hence, the sum of a geometric series with n terms, and $r \neq 1 = \displaystyle\frac{a_1(1-r^n)}{1-r}$.
Sum of Infinite Geometric Series and a Little Bit of Calculus
Note: This portion is for those who have already taken elementary calculus.
The infinite geometric series $\{a_n\}$ is the the symbol $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots$. From above, the sum of a finite geometric series with $n$ terms is $\displaystyle \sum_{k=1}^n \frac{a_1(1-r^n)}{1-r}$. Hence, to get the sum of the infinite geometric series, we need to get the sum of $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r}$. However, $\displaystyle \sum_{k=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n\to \infty} \frac{a_1(1-r^n)}{1-r}$.
Also, that if $|r| < 1$, $r^n$ approaches $0$ (try $(\frac{2}{3})^n$ or any other proper fraction and increase the value of $n$), thus, $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r} = \frac{a_1}{1-r}$. Therefore, sum of the infinite series $\displaystyle a_1 + a_2r + a_2r^2 + \cdots = \frac{a_1}{1-r}$.
One very common infinite series is $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$, or the sum of the areas of the partitions of the square with side length 1 unit shown below. Using the formula above,
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{a_1}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$.
Figure 2 - A representation of an infinite geometric series.
This is evident in the diagram because the sum of all the partitions is equal to the area of a square. We say that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n}$ converges to 1.
## Is 0.999… really equal to 1?
Introduction
Yes it is. 0.999… is equal to 1.
Before we begin our discussion, let me make a remark that the symbol “…” in the decimal 0.999… means that the there are infinitely many 9’s, or putting it in plain language, the decimal number has no end.
For non-math persons, you will probably disagree with the equality, but there are many elementary proofs that could show it, some of which, I have shown below. A proof is a series of valid, logical and relevant arguments (see Introduction to Mathematical Proofs for details), that shows the truth or falsity of a statement.
Proof 1
$\frac{1}{3} = 0.333 \cdots$
$\frac{2}{3} = 0.666 \cdots$
$\frac{1}{3} + \frac{2}{3} = 0.333 \cdots + 0.666 \cdots$
$\frac{3}{3} =0.999 \cdots$
But $\frac{3}{3} = 1$, therefore $1 =0.999 \cdots$
Proof 2
$\frac{1}{9} = 0.111 \cdots$
Multiplying both sides by 9 we have
$1 = 0.999 \cdots$
Proof 3
Let $x = 0.999 \cdots$
$10x = 9.999 \cdots$
$10x - x = 9.999 \cdots - 0.9999 \cdots$
$9x = 9$
$x = 1$
Hence, $0.999 \cdots = 1$
Still in doubt?
Many will probably be reluctant in accepting the equality $1 = 0.999 \cdots$ because the representation is a bit counterintuitive. The said equality requires the notion of the real number system, a good grasp of the concept of limits, and knowledge on infinitesimals or calculus in general. If, for instance,you have already taken sequences (in calculus), you may think of the $0.999 \cdots$ as a sequence of real numbers $(0.9, 0.99, 0.999,\cdots)$. Note that the sequence gets closer and closer to 1, and therefore, its limit is 1.
Infinite Geometric Sequence
My final attempt to convince you that $0.999 \cdots$ is indeed equal to $1$ is by the infinite geometric sequence. For the sake of brevity, in the remaining part of this article, we will simply use the term “infinite sequence” to refer to an infinite geometric sequence. We will use the concept of the sum of an infinite sequence, which is known as an infinite series, to show that $0.999 \cdots = 1$.
One example of an infinite series is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$. If you add its infinite number of terms, the answer is equal to 1. Again, this is counterintuitive.
How can addition of numbers with infinite number of terms have an exact (or a finite) answer?
There is a formula to get the sum of an infinite geometric sequence, but before we discuss the formula, let me give the geometric interpretation of the sum above. The sum $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ can be represented geometrically using a 1 unit by 1 unit square as shown below. If we divide the square into two, then we will have two rectangles, each of which has area $\frac{1}{2}$ square units. Dividing the other half into two, then we have three rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{4}$ square units. Dividing the one of the smaller rectangle into two, then we have four rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{8}$. Again, dividing one of the smallest rectangle into two, we have five rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$, and $\frac{1}{16}$ Since this process can go on forever, the sum of all the areas of all the rectangles will equal to 1, which is the area of the original square.
Now that we have seen that an infinite series can have a finite sum, we will now show that $0.999 \cdots$ can be expressed as a finite sum by expressing it as an infinite series. The number $0.999 \cdots$ can be expressed as an infinite series $0.9 + 0.09 + 0.009 + \cdots$. Converting it in fractional form, we have $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$.
We have learned that the sum of the infinite series with first term $\displaystyle a_1$ and ratio $r$ is described by $\displaystyle\frac{a_1}{1-r}$. Applying the formula to our series above, we have
$\displaystyle\frac{\frac{9}{10}}{1-\frac{1}{10}} = 1$
Therefore, the sum our infinite series is 1.
Implication
This implication of the equality $0.999 \cdots =1$ means that any rational number that is a non-repeating decimal can be expressed as a repeating decimal. Since $0.999 \cdots =1$, it follows that $0.0999 \cdots =0.1, 0.00999 \cdots=0.01$ and so on. Hence, any decimal number maybe expressed as number + 0.00…01. For example, the decimal $4.7$, can be expressed as $4.6 + 0.1 = 4.6 + 0.0999 \cdots = 4.6999 \cdots$. The number $0.874$ can also be expressed as $0.873 + 0.001 = 0.873 + 0.000999 \cdots = 0.873999 \cdots$
Conclusion
Any of the four proofs above is actually sufficient to show that $0.999 \cdots = 1$. Although this concept is quite hard to accept, we should remember that in mathematics, as long as the steps of operations or reasoning performed are valid and logical, the conclusion will be unquestionably valid.
There are many counterintuitive concepts in mathematics and the equality $0.999 \cdots = 1$ is only one of the many. In my post, Counting the Uncountable: A Glimpse at the Infinite, we have also encountered one: that the number of integers (negative, 0, positive) is equal to the number of counting numbers (positive integers) and we have shown it by one-to-one pairing. We have also shown that the number of counting numbers is the same as the number of rational numbers. Thus, we have shown that a subset can have the same element as the “supposed” bigger set. I guess that is what makes mathematics unique; intuitively, some concepts do not make sense, but by valid and logical reasoning, they perfectly do.
Notes:
1. You can find discussions about 0.999… = 1 here and here.
2. There is another good post about it here and here.
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# Resolve the following into partial fractions. 2x−1(1−x−x2)(x+2)
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## To resolve the expression 2x−1(1−x−x2)(x+2) into partial fractions, we will follow these steps:Step 1: Set up the partial fraction decompositionWe express 2x−1(1−x−x2)(x+2) as a sum of partial fractions. Since 1−x−x2 is a quadratic polynomial, we can write:2x−1(1−x−x2)(x+2)=Ax+2+Bx+C1−x−x2where A, B, and C are constants to be determined.Step 2: Combine the fractionsTo combine the fractions on the right side, we find a common denominator:A(1−x−x2)+(Bx+C)(x+2)(1−x−x2)(x+2)=2x−1(1−x−x2)(x+2)Step 3: Eliminate the denominatorsSince the denominators are the same, we can equate the numerators:2x−1=A(1−x−x2)+(Bx+C)(x+2)Step 4: Expand the right sideNow we expand the right-hand side:A(1−x−x2)=A−Ax−Ax2(Bx+C)(x+2)=Bx2+2Bx+Cx+2C=(B)x2+(2B+C)x+2CCombining these gives:2x−1=(B−A)x2+(2B+C−A)x+(A+2C)Step 5: Compare coefficientsNow we compare coefficients from both sides:1. Coefficient of x2: B−A=0 (1)2. Coefficient of x: 2B+C−A=2 (2)3. Constant term: A+2C=−1 (3)Step 6: Solve the system of equationsFrom equation (1), we have:B=ASubstituting B=A into equation (2):2A+C−A=2⟹A+C=2(4)Substituting B=A into equation (3):A+2C=−1(5)Now we solve equations (4) and (5):From (4): C=2−ASubstituting into (5):A+2(2−A)=−1A+4−2A=−1−A+4=−1−A=−5⟹A=5Now substituting A=5 back into (4):5+C=2⟹C=2−5=−3And since B=A:B=5Step 7: Write the final partial fractionsNow we can substitute A, B, and C back into the partial fraction decomposition:2x−1(1−x−x2)(x+2)=5x+2+5x−31−x−x2Final AnswerThus, the partial fraction decomposition is:2x−1(1−x−x2)(x+2)=5x+2+5x−31−x−x2
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# Factors of 911: Prime Factorization, Methods, and Examples
The numbers that can be divided by 911 are called the factors of 911. It is an odd integer because it is 911.
It only has two elements because it is a prime number. Learn more about the causes of 911 now.
### Factors of 911
Here are the factors of the number 911.
Factors of 911: 1 and 911
### Negative Factors of 911
The negative factors of 911 are similar to its positive aspects, just with a negative sign.
Negative Factors of 911: -1, and – 911
### Prime Factorization of 911
The prime factorization of 911 is the way of expressing its prime factors in the product form.
Prime Factorization: 1 x 911
In this article, we will learn about the factors of 911 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 911?
The factors of 911 are 1 and 911. These numbers are the factors as they do not leave any remainder when divided by 911.
The factors of 911 are classified as prime numbers and composite numbers. The prime factors of the number 911 can be determined using the prime factorization technique.
## How To Find the Factors of 911?
You can find the factors of 911 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 911, create a list containing the numbers that are exactly divisible by 911 with zero remainders. One important thing to note is that 1 and 911 are the 911’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 911 are determined as follows:
$\dfrac{ 911}{1} = 911$
$\dfrac{ 911}{ 911} = 1$
Therefore, 1 and 911 are the factors of 911.
### Total Number of Factors of 911
For 911, there are 2 positive factors and 2 negative ones. So in total, there are 4 factors of 911.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 911 is given as:
The factorization of 911 is 1 x 911.
The exponent of 1, and 911 is 1.
Adding 1 to each and multiplying them together results in 4.
Therefore, the total number of factors of 911 is 4. 2 are positive, and 2 factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 911 by Prime Factorization
The number 911 is a prime number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 911 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 911, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 911 can be expressed as:
911 = 1 x 911
## Factors of 911 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 911, the factor pairs can be found as:
1 x 911 = 911
The possible factor pairs of 911 are given as (1, 911).
All these numbers in pairs, when multiplied, give 911 as the product.
The negative factor pairs of 911 are given as:
-1 x -911 = 911
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1 and -911 are called negative factors of 911.
The list of all the factors of 911, including positive as well as negative numbers, is given below.
Factor list of 911: 1,-1,911 and – 911
## Factors of 911 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 911 are there?
### Solution
The total number of Factors of 911 is 4.
Factors of 911 are 1 and 911.
### Example 2
Find the factors of 911 using prime factorization.
### Solution
The prime factorization of 911 is given as:
911 $\div$ 1 = 911
911 $\div$ 911 = 1
So the prime factorization of 911 can be written as:
1 x 911 = 911
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Ex
# Real Numbers
#### Examples of Euclids Division Lemma or Euclids Division Algorithm
Example: Using Euclids division algorithm, find the H.C.F. (Highest Common Factor)
(i) 135 and 225 (ii) 196 and 38220 and (iii) 867 and 255
Solution:
(i) Starting with the larger number i.e., 225, we get:
225 = 135 × 1 + 90
Now taking divisor 135 and remainder 90, we get 135 = 90 × 1 + 45
Further taking divisor 90 and remainder 45, we get 90 = 45 × 2 + 0
∴ Required H.C.F. (Highest Common Factor) is equal to 45 (Answer)
(ii) Starting with larger number 38220, we get:
38220 = 196 × 195 + 0
Since, the remainder is 0
↠ Required H.C.F. (Highest Common Factor) is equal to 196 (Answer.)
(iii) Given number are 867 and 255
↠ 867 = 255 × 3 + 102 (Step-1)
255 = 102 × 2 + 51 (Step-2)
102 = 51 × 2 + 0 (Step-3)
↠ Highest Common Factor (H.C.F.) = 51 (Ans.)
Example: Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some integer.
Solution. According to Euclids division lemma, if a and b are two positive integers such that a is greater than b; then these two integers can be expressed as a = bq + r; where 0 ≤ r < b
Now consider
b = 2; then a = bq + r will reduce to a = 2q + r; where 0 ≤ r < 2,
i.e., r = 0 or r = 1
If r = 0, a = 2q + r ↠ a = 2q
i.e., a is even
and, if r = 1, a = 2q + r ↠ a = 2q + 1
as if the integer is not even; it will be odd.
Since, a is taken to be any positive integer so it is applicable to the every positive integer that when it can be expressed as a = 2q.
∴ a is even and when it can expressed as a = 2q + 1; a is odd. Hence the required result.
Example: Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution: Let a is b be two positive integers in which a is greater than b. According to Euclids division algorithm; a and b can be expressed as a = bq + r, where q is quotient and r is remainder and 0 ≤ r < b.
Taking b = 4, we get: a = 4q + r, where 0 ≤ r < 4 i.e., r = 0, 1, 2 or 3
r = 0 ↠ a = 4q, which is divisible by 2 and so is even.
r = 1 ↠ a = 4q + 1, which is not divisible by 2 and so is odd.
r = 2 ↠ q = 4q + 2, which is divisible by 2 and so is even.
and r = 3 ↠ q = 4q + 3, which is not divisible by 2 and so is odd.
∴ Any positive odd integer is of the form 4q + 1 or 4q + 3; where q is an integer.
Hence the required result.
Example: Show that one and only one out of n; n + 2 or n + 4 is divisible by 3, where n is any positive integer.
Solution: Consider any two positive integers a and b such that a is greater than b, then according to Euclids division algorithm:
a = bq + r; where q and r are positive integers and 0 ≤ r < b
Let a = n and b = 3, then
a = bq + r ↠ n = 3q + r; where 0 £ r < 3.
r = 0 ↠ n = 3q + 0 = 3q
r = 1 ↠ n = 3q + 1 and r = 2 ↠ n = 3q + 2
If n = 3q; n is divisible by 3
If n = 3q + 1; then n + 2 = 3q + 1 + 2
= 3q + 3; which is divisible by 3
↠ n + 2 is divisible by 3
If n = 3q + 2; then n + 4 = 3q + 2 + 4
= 3q + 6; which is divisible by 3
↠ n + 4 is divisible by 3
Hence, if n is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3.
Hence the required result.
×
#### NTSE Mathematics (Class X)
• Real Numbers
• Polynomials
• Linear Equation in Two Variables
• Trigonometry
• Similar Triangles
• Statistics
• Arithmetic Progressions
• Application of Trigonometry
• Circle
• Co-ordinate Geometry
• Area related to Circle
• Surface Area & Volume
• Constructions
• Probability
#### NTSE Mathematics (Class IX)
• Real Numbers
• Polynomials
• Linear Equation in Two Variables
• Trigonometry
• Similar Triangles
• Statistics
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Lesson
The additive inverse of a number is a number that has the same distance from $0$0 on the number line, but is on the opposite side of $0$0. That sounds a bit confusing but, if you remember when we learned about absolute value, you'll know that there is a positive value and a negative value that are equal distances from zero.
Another way to think about an additive inverse is what value do I add to the number so that my answer is zero.
The picture below shows an example of this using the term $3$3 and its additive inverse, $-3$3.
### Calculating a term's additive inverse
Any term's additive inverse can be calculated by multiplying the term by $-1$1.
For example, the additive inverse of $8$8 is $-8$8 ($8\times\left(-1\right)=-8$8×(1)=8),
the additive inverse of $-12$12 is $12$12 ($-12\times\left(-1\right)=12$12×(1)=12)
and the additive inverse of $a$a is $-a$a ($a\times\left(-1\right)=-a$a×(1)=a).
Remember!
A number and its additive inverse should sum to zero. e.g. $7+\left(-7\right)=0$7+(7)=0.
### Using additive inverses to solve equations
It's helpful to imagine the adding or subtracting as moving up or down the number line.
Moving in a positive direction (i.e. if we're adding a positive number) means moving to the right along a number line.
Conversely, moving in a negative direction (i.e. subtracting a positive number) means moving to the left along a number line.
If we're solving these kinds of questions mentally, using the jump strategy for example, using additive inverses can help.
Remember!
There are important rules to following when adding or subtracting negative terms:
• Adding a negative number is the same as subtracting its inverse, so we can solve it as a subtraction problem, e.g. $4+\left(-5\right)=4-5$4+(5)=45$=$=$-1$1.
• Subtracting a negative number is equivalent to adding its inverse, so we can solve it as an addition problem, e.g. $2-\left(-10\right)=2+10$2(10)=2+10$=$=$12$12.
#### Worked examples
##### Question 1
Evaluate: $2-3$23
Think: This is a subtraction problem so we are moving to the left down the number line.
Do: $2-2=0$22=0. Then we still have $1$1 left to take away. So, $2-3=-1$23=1.
##### Question 2
Evaluate: $-2-8$28
Think: Like question 1, this is a subtraction problem so we are moving to the left down the number line.
Do: $-2-8$28 will be the same distance away from $0$0 as $2+8$2+8. $2+8=10$2+8=10 so $-2-8=-10$28=10.
##### Question 3
Evaluate: $3-\left(-8\right)$3(8)
Think: Two negative signs together become a positive.
Do:
$3-\left(-8\right)$3−(−8) $=$= $3+8$3+8 $=$= $11$11
##### Question 4
What is the additive inverse of $26$26?
Think: the additive inverse of $26$26 is the number that is the same distance from $0$0 on the number line as $26$26, but is on the opposite side of $0$0.
Do: The additive inverse of $26$26 is $-26$26.
##### Question 5
In the last financial year, Delicious Donuts had an overall loss of $\$88000$$88000. a) What integer is used to represent how much the company made? Think: The key word here is loss. What type of number would represent losing money? Do: We need to use a negative number to represent the loss of money, so the integer -8800088000 represents how much the company made. b) What is the additive inverse of this result? Think: -8800088000 lies to the left of 00 on the number line, What number would be the same distance from zero on the number line, but to the right instead? Do: The additive inverse of -8800088000 is 8800088000. c) -88000+88000=\editable{}88000+88000= Think: What do additive inverses always add up to? Do: Additive inverses always add to 00 so -88000+88000=088000+88000=0 d) What does the additive inverse represent here? • A) the amount, in dollars, the company made the year before • B) the amount, in dollars, the company needs to make to turn a profit of \88000$$88000
• C) the amount, in dollars, the company is expected to make this financial year
• D) the amount, in dollars, the company needs to make to break even
Think: The additive inverse is $88000$88000. Which scenario could a positive $88000$88000 represent?
Do: Since we know that adding $-88000+88000=0$88000+88000=0 what would $0$0 represent financially? $0$0 would represent breaking even because the company has not lost money or earned money. So the correct answer here is D) the amount, in dollars, the company needs to make to break even.
#### Practice questions
##### Question 6
What is the additive inverse of $-23$23?
##### Question 7
Fill in the blank to make the statement true.
1. $8-13=8+\editable{}$813=8+
##### Question 8
You have two numbered cards in your hand. The sum of the numbers is $0$0. If the number on one of the cards is $6$6, what is the number on the other card?
##### Question 9
Nanga Parbat is a mountain that is $8126$8126 meters above sea level.
1. What integer is used to represent the elevation?
2. What is the additive inverse of this elevation?
3. $8126+\left(-8126\right)$8126+(8126) $=$= $\editable{}$
4. What does the additive inverse represent here?
The amount we would have to travel to get back to sea level.
A
The number of meters above sea level.
B
The number of meters the mountain is below Mt Everest.
C
The width of the mountain.
D
The amount we would have to travel to get back to sea level.
A
The number of meters above sea level.
B
The number of meters the mountain is below Mt Everest.
C
The width of the mountain.
D
##### Question 10
On the number line attached, the numbers $x$x and $y$y are the same distance from $0$0. What is the sum $x+y$x+y?
### Outcomes
#### 7.NS.1.c
Understand subtraction of rational numbers as adding the additive inverse, p - q = p (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
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# Identify The Properties Of Addition And Multiplication Worksheet
## 6 problems
Identifying the properties of addition and multiplication involves recognizing and understanding fundamental rules or characteristics that govern these operations in mathematics. These properties include commutative, associative, distributive, identity, and inverse properties, which describe how numbers can be combined or manipulated under addition and multiplication. Understanding these properties is crucial for solving equations, simplifying expressions, and mastering algebraic concepts.
Algebra 1
Expressions
## How Will This Worksheet on "Identify the Properties of Addition and Multiplication" Benefit Your Students' Learning?
• Students grasp fundamental rules governing addition and multiplication.
• Practice improves their ability to manipulate numbers effectively.
• Helps simplify complex expressions and equations confidently.
• Builds a foundation for algebraic concepts and problem-solving.
• Applies math to everyday scenarios, enhancing practical skills.
• Encourages analytical skills in applying properties to solve problems.
## How to Identify the Properties of Addition and Multiplication?
• Commutative Property
Addition: You can add numbers in any order.
Multiplication: You can multiply numbers in any order.
• Associative Prop...
Show all
## Solved Example
Q. Which property of multiplication is shown?$\newline$$k \cdot (m + n) = k \cdot m + k \cdot n$
Solution:
1. Identify Property of Multiplication: Identify the property of multiplication demonstrated by the equation $k \cdot (m + n) = k \cdot m + k \cdot n$.$\newline$In this equation, a single number $k$ is being multiplied by the sum of two numbers $m$ and $n$. The result is the same as when $k$ is multiplied by each of $m$ and $n$ separately, and then those products are added together. This is a classic example of the distributive property of multiplication over addition.
2. Match with Correct Choice: Match the property identified in Step $1$ with the correct choice from the given options.$\newline$The distributive property allows us to multiply a number by a sum by distributing the multiplication over each addend and then adding the results. This matches choice $(B)$ distributive from the given options.
### What teachers are saying about BytelearnWhat teachers are saying
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# 11.1 Systems of linear equations: two variables (Page 5/20)
Page 5 / 20
Solve the system of equations by addition.
$\begin{array}{c}2x+3y=8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x+5y=10\end{array}$
$\left(10,-4\right)$
## Identifying inconsistent systems of equations containing two variables
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different $\text{\hspace{0.17em}}y$ -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as $\text{\hspace{0.17em}}12=0.$
## Solving an inconsistent system of equations
Solve the following system of equations.
We can approach this problem in two ways. Because one equation is already solved for $\text{\hspace{0.17em}}x,$ the most obvious step is to use substitution.
$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+2y=13\hfill \\ \text{\hspace{0.17em}}\left(9-2y\right)+2y=13\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9+0y=13\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9=13\hfill \end{array}$
Clearly, this statement is a contradiction because $\text{\hspace{0.17em}}9\ne 13.\text{\hspace{0.17em}}$ Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
We then convert the second equation expressed to slope-intercept form.
Comparing the equations, we see that they have the same slope but different y -intercepts. Therefore, the lines are parallel and do not intersect.
$\begin{array}{l}\begin{array}{l}\\ y=-\frac{1}{2}x+\frac{9}{2}\end{array}\hfill \\ y=-\frac{1}{2}x+\frac{13}{2}\hfill \end{array}$
Solve the following system of equations in two variables.
$\begin{array}{l}2y-2x=2\\ 2y-2x=6\end{array}$
No solution. It is an inconsistent system.
## Expressing the solution of a system of dependent equations containing two variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as $\text{\hspace{0.17em}}0=0.$
## Finding a solution to a dependent system of linear equations
Find a solution to the system of equations using the addition method .
$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+3y=2\\ 3x+9y=6\end{array}$
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If we multiply both sides of the first equation by $\text{\hspace{0.17em}}-3,$ then we will be able to eliminate the $\text{\hspace{0.17em}}x$ -variable.
$\begin{array}{l}\underset{______________}{\begin{array}{ll}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3x-9y\hfill & =-6\hfill \\ +\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x+9y\hfill & =6\hfill \end{array}}\hfill \\ \begin{array}{ll}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\hfill & =0\hfill \end{array}\hfill \end{array}$
We can see that there will be an infinite number of solutions that satisfy both equations.
Solve the following system of equations in two variables.
The system is dependent so there are infinite solutions of the form $\text{\hspace{0.17em}}\left(x,2x+5\right).$
## Using systems of equations to investigate profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation $\text{\hspace{0.17em}}R=xp,$ where $\text{\hspace{0.17em}}x=$ quantity and $\text{\hspace{0.17em}}p=$ price. The revenue function is shown in orange in [link] .
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
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# Multiplication Chart For 4: Everything You Need To Know
## Introduction
In this year 2023, we are here to provide you with all the information you need about the multiplication chart for 4. Whether you are a student, a teacher, or just someone looking to brush up on their math skills, this article will cover everything you need to know about this specific multiplication chart.
### What is a Multiplication Chart?
A multiplication chart is a table that displays the multiplication facts for a specific number. It is a useful tool for students to memorize and understand the multiplication facts quickly and easily.
### Why is the Multiplication Chart for 4 Important?
The multiplication chart for 4 is important because it helps students learn and understand the multiplication facts for the number 4. By using this chart, students can easily see the patterns and relationships between the numbers, making it easier for them to solve multiplication problems.
## How to Use the Multiplication Chart for 4
Using the multiplication chart for 4 is simple. Start by finding the number 4 in the leftmost column of the chart. Then, find the number you want to multiply by 4 in the top row of the chart. Finally, locate the cell where the row and column intersect to find the product of the multiplication problem.
### Tips for Memorizing the Multiplication Facts for 4
Memorizing the multiplication facts for 4 can be made easier with a few simple tips:
1. Practice regularly: Set aside a few minutes each day to practice the multiplication facts for 4. Repetition is key to memorization.
2. Use visual aids: Use the multiplication chart for 4 as a visual aid to help you see the patterns and relationships between the numbers.
3. Create mnemonics: Create fun and memorable mnemonics or rhymes to help you remember the multiplication facts for 4.
## Common Questions about the Multiplication Chart for 4
### 1. What is 4 times 4?
4 times 4 equals 16. You can find this answer by locating the cell where the row of 4 and the column of 4 intersect on the multiplication chart for 4.
### 2. What is 4 times 6?
4 times 6 equals 24. To find this answer, locate the cell where the row of 4 and the column of 6 intersect on the multiplication chart for 4.
### 3. How can I use the multiplication chart for 4 to solve multiplication problems?
The multiplication chart for 4 can be used as a reference tool to quickly find the product of any multiplication problem involving the number 4. Simply locate the row and column that correspond to the two numbers you want to multiply, and find the cell where they intersect to find the product.
### 4. Can the multiplication chart for 4 be used for other numbers?
No, the multiplication chart for 4 is specifically designed for the multiplication facts involving the number 4. For other numbers, you would need a different multiplication chart.
## Conclusion
The multiplication chart for 4 is a valuable tool for learning and understanding the multiplication facts for the number 4. By using this chart and following the tips provided, you can easily memorize the multiplication facts for 4 and solve multiplication problems with ease. So, start practicing and make math fun with the multiplication chart for 4!
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Succeed with maths – Part 2
Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.
Free course
# 1.1 Pascal’s triangle
Numbering the tiles in Activity 1 gave a number pattern that described a geometric pattern. This next example is a number pattern that appeared in China and Persia more than 700 years ago, but is still used by students in mathematical and statistical problems today – it even appears in chemistry. It is known as ‘Pascal’s triangle’ after the French mathematician Blaise Pascal who studied the properties of this triangle. The first part of Pascal’s triangle is shown in Figure 2:
Figure _unit4.1.2 Figure 2 Pascal's triangle
You can continue the triangle indefinitely by following the pattern. So how is it created?
If you look at the triangle you can see that each row of numbers starts and ends with the number 1. Now look at the numbers in the second to last line (1 4 6 4 1) and the last line of the triangle. Look carefully at them both - can you see a way that you can make the numbers in the last line from the line above? If you add each pair of numbers in the second to last line, this gives you numbers in the last line.
Starting from the left-hand side the pairs of numbers are 1 and 4, 4 and 6, 6 and 4, and 4 and 1. So:
This process, plus adding a 1 at end of the line, generates the next row of the triangle. Have a go at creating the next line for yourself in this next activity.
## Activity _unit4.1.2 Activity 2 Pascal’s triangle – next line
Timing: Allow approximately 5 minutes
Create the next line of Pascal’s triangle by adding the pairs of numbers in the last line. Remember to add the 1s to the end of the rows when you are done.
Starting from the left-hand side the pairs of numbers are 1 and 5, 5 and 10, 10 and 10, 10 and 5, and lastly 5 and 1.
So the next line in the triangle should look like the bottom one in Figure 3:
Figure _unit4.1.3 Figure 3 How to generate row 7 of Pascal’s triangle from row 6
You can watch a larger version of Pascal’s triangle being built in this video:
Interactive feature not available in single page view (see it in standard view).
Now you’ve seen how to build Pascal’s triangle by adding pairs of numbers it’s time to see if there are more patterns hiding in this number triangle in the next section.
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# 3.3 Quadratic equations: applications
Page 1 / 4
This module is from Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr. Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2, and applied problems from the areas of manufacturing, population, physics, geometry, mathematics (numbers and volumes), and astronomy, which are solved using the five-step method.Objectives of this module: become more proficient at using the five-step method for solving applied problems.
## Overview
• The Five-Step Method
• Examples
## The five-step method
We are now in a position to study some applications of quadratic equations. Quadratic equations can arise from a variety of physical (applied) and mathematical (logical) problems.
We will, again, apply the five-step method for solving word problems.
## Five-step method of solving word problems
• Step 1: Let $x$ (or some other letter) represent the unknown quantity.
• Step 2: Translate the verbal expression to mathematical symbols and form an equation.
• Step 3: Solve this equation.
• Step 4: Check the solution by substituting the result into the equation found in step 2.
• Step 5: Write a conclusion.
Remember, step 1 is very important.
ALWAYS START BY INTRODUCING A VARIABLE.
Once the quadratic equation is developed (step 2), try to solve it by factoring. If factoring doesn’t work, use the quadratic formula. A calculator may help to make some of the calculations a little less tedious.
## Sample set a
A producer of personal computer mouse covers determines that the number $N$ of covers sold is related to the price $x$ of a cover by $N=35x-{x}^{2}.$ At what price should the producer price a mouse cover in order to sell 216 of them?
$\begin{array}{l}\begin{array}{ccc}\text{Step\hspace{0.17em}}1:& & \text{Let}\text{\hspace{0.17em}}x=\text{the}\text{\hspace{0.17em}}\text{price}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{mouse}\text{\hspace{0.17em}}\text{cover}\text{.}\end{array}\hfill \\ \begin{array}{lll}\text{Step\hspace{0.17em}}2:\hfill & \hfill & \text{Since}\text{\hspace{0.17em}}N\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{216,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{equation}\text{\hspace{0.17em}}\text{is}\hfill \\ \hfill & \hfill & 216=35x-{x}^{2}\hfill \end{array}\hfill \\ \begin{array}{lllllll}\text{Step\hspace{0.17em}}3:\hfill & \hfill & 216\hfill & =\hfill & 35x-{x}^{2}\hfill & \hfill & \text{Rewrite}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{standard}\text{\hspace{0.17em}}\text{form}\text{.}\hfill \\ \hfill & \hfill & {x}^{2}-35x+216\hfill & =\hfill & 0\hfill & \hfill & \text{Try}\text{\hspace{0.17em}}\text{factoring}\text{.}\hfill \\ \hfill & \hfill & \left(x-8\right)\left(x-27\right)\hfill & =\hfill & 0\hfill & \hfill & \hfill \\ \hfill & \hfill & x-8=0\hfill & or\hfill & \hfill & x-27=0\hfill & \hfill \\ \hfill & \hfill & x=8\hfill & or\hfill & \hfill & x=27\hfill & \hfill \end{array}\hfill \\ \begin{array}{ccc}& & \text{Check}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{potential}\text{\hspace{0.17em}}\text{solutions}\text{.}\end{array}\hfill \\ \begin{array}{llllllllll}\text{Step\hspace{0.17em}}4:\hfill & \hfill & \text{If}\text{\hspace{0.17em}}x=8,\hfill & \hfill & \hfill & \hfill & \text{If}\text{\hspace{0.17em}}x=27,\hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 35\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8-{8}^{2}& =\hfill & 216\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill 35\text{\hspace{0.17em}}·\text{\hspace{0.17em}}27-{27}^{2}& =\hfill & 216\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 280-64& =\hfill & 216\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill 945-729& =\hfill & 216\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 216& =\hfill & 216\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill & \hfill 216& =\hfill & 216\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}\hfill \\ \begin{array}{ccc}& & \text{These}\text{\hspace{0.17em}}\text{solutions}\text{\hspace{0.17em}}\text{check}\text{.}\end{array}\hfill \\ \hfill \begin{array}{ccc}\text{Step}\text{\hspace{0.17em}}5:& & \text{The}\text{\hspace{0.17em}}\text{computer}\text{\hspace{0.17em}}\text{mouse}\text{\hspace{0.17em}}\text{covers}\text{\hspace{0.17em}}\text{can}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{priced}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{either}\text{\hspace{0.17em}}\text{8}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{27}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{order}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{sell}\text{\hspace{0.17em}}\text{216}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{them}\text{.}\text{\hspace{0.17em}}\end{array}\end{array}$
## Practice set a
A manufacturer of cloth personal computer dust covers notices that the number $N$ of covers sold is related to the price of covers by $N=30x-{x}^{2}.$ At what price should the manufacturer price the covers in order to sell 216 of them?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5: In order to sell 216 covers, the manufacturer should price them at either or .
12 or 18
It is estimated that $t$ years from now the population of a particular city will be
$P={t}^{2}-24t+96,000.$
How many years from now will the population be 95,865?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
In 9 and 15 years, the population of the city will be 95,865.
#### Questions & Answers
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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# AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4
AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Questions and Answers.
## AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.4
Question 1.
Find the value of ’x’ so that l || m
Solution:
Given l|| m. Then 3x – 10° = 2x + 15°
[Vertically opposite angles and corresponding angles are equal.]
⇒ 3x – 10 = 2x + 15
⇒ 3x – 2x = 15 + 10
∴ x = 25°
Question 2.
Eight times of a number reduced by 10 is equal to the sum of six times the number and 4. Find the number.
Solution:
Let the number be ‘x’ say.
8 times of a number = 8 × x = 8x
¡f10 is reduced from 8x then 8x – O
6 times of a number = 6 × x = 6x
If 4 is added to 6x then 6x + 4
According to the sum,
8x – 10 = 6x + 4
⇒ 8x – 6x = 4 + 10
⇒ 2x = 14
⇒ x = 7
∴ The required number = 7
Question 3.
A number consists of two digits whose sum is 9. If 27 is subtracted from the number its digits are reversed. Find the number.
Solution:
Let a digit of two digit number be x.
The sum of two digits = 9
∴ Another digit = 9 – x
The number = 10 (9 – x) + x
= 90 – 10x + x
= 90 – 9x
If 27 is subtraçted from the number its digits are reversed.
∴ (90 – 9x) – 27 = 10x + (9 – x)
63 – 9x = 9x + 9
9x + 9x = 63 – 9
18x = 54
∴ x = $$\frac { 54 }{ 18 }$$ = 3
∴ Units digit = 3
Tens digit = 9 – 3 = 6
∴ The number = 63
Question 4.
A number is divided into two parts such that one part is 10 more than the other. If the two parts are in the ratio 5:3, find the number and the two parts.
Solution:
If a number is divided into two parts in he ratio of 5 : 3, let the parts be 5x, 3x say.
According to the sum,
5x = 3x + 10
⇒ 5x – 3x = 10
⇒ 2x = 10
∴ x = $$\frac { 10 }{ 2 }$$
∴ x = 5
∴ The required number be
x + 3x = 8x
= 8 × 5 = 40
And the parts of number are
5 = 5 × 5 = 25
3 = 3 × 5 = 15
Question 5.
When I triple a certain number and add 2, I get the same answer as I do when I subtract the number from 50. Find the number.
Solution:
Let the number be x’ say.
3 times of a number = 3 × x = 3x
If 2 is added to 3x then 3x + 2
If ‘xis subtracted from 50 then it becomes 50 – x.
According to the sum,
3x + 2 = 50 – x
3x + x = 50 – 2
4x = 48 .
x = 12
∴ The required number 12
Question 6.
Mary is twice older than her sister. In 5 years time, she will be 2 years older than her sister. Find how old are they both now.
Solution:
Let the age of Marys sister = x say.
Mary’s age = 2 × x = 2x
After 5 years her sister’s age
= (x + 5) years
After 5 years Mary’s age
= (2x + 5) years
According to the sum,
2x + 5 = (x + 5) + 2
= 2x – x = 5 + 2 – 5
∴ The age of Mary’s sister = x = 2 years
Mary’s age = 2x = 2 x 2 = 4 years
Question 7.
In 5 years time, Reshma will be three times old as she was 9 years ago. How old is she now?
Solution:
Reshma’s present age = ‘x’ years say.
After 5 years Reshmats age
= (x + 5) years
Before 9 years Reshma’s age
=(x – 9) years
According to the sum
= x+ 5 = 3(x – 9) = 3x – 27
x – 3x = -27-5
-2x = -32
x = $$\frac{-32}{-2}$$ = 16
∴ x = 16
∴ Reshma’s present age = 16 years.
Question 8.
A town’s population increased by 1200 people, and then this new population decreased 11%. The town now had 32 less people than it did before the 1200 increase. Find the original population.
Solution:
Let th population of a town after the increase of 1200 is x say.
11% of present population
The present population of town
= 11,200 – 1200 = 10,000
Question 9.
A man on his way to dinner shortly after 6.00 p.m. observes that the hands of his watch form an angle of 110°. Returning before 7.00 p.m. he notices that again the hands of his watch form an angle of 1100. Find the number of minutes that he has been away.
Solution:
Let the number be ‘x ray.
$$\frac { 1 }{ 3 }$$ rd of a number = $$\frac { 1 }{ 3 }$$ x x = $$\frac { x }{ 3 }$$
$$\frac { 1 }{ 5 }$$ th of a number = $$\frac { 1 }{ 5 }$$ x x = $$\frac { x }{ 5 }$$
According to the sum
∴ x = 30
∴ The required number is 30.
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This is “Solving Equations and Inequalities”, chapter 6 from the book Advanced Algebra (v. 1.0). For details on it (including licensing), click here.
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# Chapter 6 Solving Equations and Inequalities
## 6.1 Extracting Square Roots and Completing the Square
### Learning Objectives
1. Solve certain quadratic equations by extracting square roots.
2. Solve any quadratic equation by completing the square.
## Extracting Square Roots
Recall that a quadratic equation is in standard formAny quadratic equation in the form $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0.$ if it is equal to 0:
$ax2+bx+c=0$ where a, b, and c are real numbers and $a≠0.$ A solution to such an equation is a root of the quadratic function defined by $f(x)=ax2+bx+c.$ Quadratic equations can have two real solutions, one real solution, or no real solution—in which case there will be two complex solutions. If the quadratic expression factors, then we can solve the equation by factoring. For example, we can solve $4x2−9=0$ by factoring as follows:
$4x2−9=0(2x+3)(2x−3)=0 2x+3=0or2x−3=02x=−3 2x=3x=−32x=32$
The two solutions are $±32.$ Here we use ± to write the two solutions in a more compact form. The goal in this section is to develop an alternative method that can be used to easily solve equations where $b=0$, giving the form
$ax2+c=0$
The equation $4x2−9=0$ is in this form and can be solved by first isolating $x2.$
$4x2−9=04x2=9x2=94$
If we take the square root of both sides of this equation, we obtain the following:
$x2=94|x|=32$
Here we see that $x=±32$ are solutions to the resulting equation. In general, this describes the square root propertyFor any real number k, if $x2=k$, then $x=±k.$; for any real number k,
$if x2=k, then x=±k$
Applying the square root property as a means of solving a quadratic equation is called extracting the rootApplying the square root property as a means of solving a quadratic equation.. This method allows us to solve equations that do not factor.
### Example 1
Solve: $9x2−8=0.$
Solution:
Notice that the quadratic expression on the left does not factor. However, it is in the form $ax2+c=0$ and so we can solve it by extracting the roots. Begin by isolating $x2.$
$9x2−8=09x2=8x2=89$
Next, apply the square root property. Remember to include the ± and simplify.
$x=±89=±223$
For completeness, check that these two real solutions solve the original quadratic equation.
$Check x=−223$
$Check x=223$
$9x2−8=09(−223)2−8=09(4⋅29)−8=08−8=00=0 ✓$
$9x2−8=09(223)2−8=09(4⋅29)−8=08−8=00=0 ✓$
Answer: Two real solutions, $±223$
Sometimes quadratic equations have no real solution. In this case, the solutions will be complex numbers.
### Example 2
Solve: $x2+25=0.$
Solution:
Begin by isolating $x2$ and then apply the square root property.
$x2+25=0x2=−25x=±−25$
After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation; the solutions are complex. We can write these solutions in terms of the imaginary unit $i=−1.$
$x=±−25=±−1⋅25=±i⋅5=±5i$
$Check x=−5i$
$Check x=5i$
$x2+25=0(−5i)2+25=025i2+25=025(−1)+25=0−25+25=00=0 ✓$
$x2+25=0(5i)2+25=025i2+25=025(−1)+25=0−25+25=00=0 ✓$
Answer: Two complex solutions, $±5i.$
Try this! Solve: $2x2−3=0.$
Answer: The solutions are $±62.$
Consider solving the following equation:
$(x+5)2=9$
To solve this equation by factoring, first square $x+5$ and then put the equation in standard form, equal to zero, by subtracting 9 from both sides.
$(x+5)2=9x2+10x+25=9x2+10x+16=0$
Factor and then apply the zero-product property.
$x2+10x+16=0(x+8)(x+2)=0x+8=0orx+2=0x=−8x=−2$
The two solutions are −8 and −2. When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots.
### Example 3
Solve by extracting roots: $(x+5)2=9.$
Solution:
The term with the square factor is isolated so we begin by applying the square root property.
$(x+5)2=9Apply the square root property.x+5=±9Simplify.x+5=±3x=−5±3$
At this point, separate the “plus or minus” into two equations and solve each individually.
$x=−5+3orx=−5−3x=−2x=−8$
Answer: The solutions are −2 and −8.
In addition to fewer steps, this method allows us to solve equations that do not factor.
### Example 4
Solve: $2(x−2)2−5=0.$
Solution:
Begin by isolating the term with the square factor.
$2(x−2)2−5=02(x−2)2=5(x−2)2=52$
Next, extract the roots, solve for x, and then simplify.
$x−2=±52Rationalize the denominator.x=2±52⋅22x=2±102x=4±102$
Answer: The solutions are $4−102$ and $4+102.$
Try this! Solve: $2(3x−1)2+9=0.$
Answer: The solutions are $13±22i.$
## Completing the Square
In this section, we will devise a method for rewriting any quadratic equation of the form
$ax2+bx+c=0$ as an equation of the form
$(x−p)2=q$
This process is called completing the squareThe process of rewriting a quadratic equation to be in the form $(x−p)2=q.$. As we have seen, quadratic equations in this form can be easily solved by extracting roots. We begin by examining perfect square trinomials:
$(x+3)2=x2+ 6x+9 ↓↑(62)2=(3)2=9$
The last term, 9, is the square of one-half of the coefficient of x. In general, this is true for any perfect square trinomial of the form $x2+bx+c.$
$(x+b2)2=x2+2⋅b2x+(b2)2=x2+bx+(b2)2$
In other words, any trinomial of the form $x2+bx+c$ will be a perfect square trinomial if
$c=(b2)2$
Note: It is important to point out that the leading coefficient must be equal to 1 for this to be true.
### Example 5
Complete the square: $x2−6x+ ? =(x+ ? )2.$
Solution:
In this example, the coefficient b of the middle term is −6. Find the value that completes the square as follows:
$(b2)2=(−62)2=(−3)2=9$
The value that completes the square is 9.
$x2−6x + 9=(x−3)(x−3)=(x−3)2$
Answer: $x2−6x+9=(x−3)2$
### Example 6
Complete the square: $x2+x+ ? =(x+ ? )2.$
Solution:
Here b = 1. Find the value that will complete the square as follows:
$(b2)2=(12)2=14$
The value $14$ completes the square:
$x2+x +14=(x+12)(x+12)=(x+12)2$
Answer: $x2+x+14=(x+12)2$
We can use this technique to solve quadratic equations. The idea is to take any quadratic equation in standard form and complete the square so that we can solve it by extracting roots. The following are general steps for solving a quadratic equation with leading coefficient 1 in standard form by completing the square.
### Example 7
Solve by completing the square: $x2−8x−2=0.$
Solution:
It is important to notice that the leading coefficient is 1.
Step 1: Add or subtract the constant term to obtain an equation of the form $x2+bx=c.$ Here we add 2 to both sides of the equation.
$x2−8x−2=0x2−8x=2$
Step 2: Use $(b2)2$ to determine the value that completes the square. In this case, b = −8:
$(b2)2=(−82)2=(−4)2=16$
Step 3: Add $(b2)2$ to both sides of the equation and complete the square.
$x2−8x=2x2−8x + 16=2 + 16(x−4) (x−4)=18(x−4)2=18$
Step 4: Solve by extracting roots.
$(x−4)2=18x−4=±18x=4±9⋅2x=4±32$
Answer: The solutions are $4−32$ and $4+32.$ The check is left to the reader.
### Example 8
Solve by completing the square: $x2+2x−48=0.$
Solution:
Begin by adding 48 to both sides.
$x2+2x−48=0x2+2x=48$
Next, find the value that completes the square using b = 2.
$(b2)2=(22)2=(1)2=1$
To complete the square, add 1 to both sides, complete the square, and then solve by extracting the roots.
$x2+2x=48Complete the square.x2+2x + 1=48 + 1(x+1)(x+1)=49(x+1)2=49Extract the roots.x+1=±49x+1=±7x=−1±7$
At this point, separate the “plus or minus” into two equations and solve each individually.
$x=−1−7orx=−1+7 x=−8x=6$
Answer: The solutions are −8 and 6.
Note: In the previous example the solutions are integers. If this is the case, then the original equation will factor.
$x2+2x−48=0(x−6)(x+8)=0$
If an equation factors, we can solve it by factoring. However, not all quadratic equations will factor. Furthermore, equations often have complex solutions.
### Example 9
Solve by completing the square: $x2−10x+26=0.$
Solution:
Begin by subtracting 26 from both sides of the equation.
$x2−10x+26=0x2−10x=−26$
Here b = −10, and we determine the value that completes the square as follows:
$(b2)2=(−102)2=(−5)2=25$
To complete the square, add 25 to both sides of the equation.
$x2−10x=−26x2−10x + 25=−26 + 25x2−10x + 25=−1$
Factor and then solve by extracting roots.
$x2−10x+25=−1(x−5)(x−5)=−1(x−5)2=−1x−5=±−1x−5=±ix=5±i$
Answer: The solutions are $5±i.$
Try this! Solve by completing the square: $x2−2x−17=0.$
Answer: The solutions are $x=1±32.$
The coefficient of x is not always divisible by 2.
### Example 10
Solve by completing the square: $x2+3x+4=0.$
Solution:
Begin by subtracting 4 from both sides.
$x2+3x+4=0x2+3x=−4$
Use b = 3 to find the value that completes the square:
$(b2)2=(32)2=94$
To complete the square, add $94$ to both sides of the equation.
$x2+3x=−4x2+3x + 94=−4 + 94(x+32)(x+32)=−164+94(x+32)2=−74$
Solve by extracting roots.
$(x+32)2=−74x+32=±−1⋅74x+32=±i72x=−32±72i$
Answer: The solutions are $−32±72i.$
So far, all of the examples have had a leading coefficient of 1. The formula $(b2)2$ determines the value that completes the square only if the leading coefficient is 1. If this is not the case, then simply divide both sides by the leading coefficient before beginning the steps outlined for completing the square.
### Example 11
Solve by completing the square: $2x2+5x−1=0.$
Solution:
Notice that the leading coefficient is 2. Therefore, divide both sides by 2 before beginning the steps required to solve by completing the square.
$2x2+5x−12=022x22+5x2−12=0x2+52x−12=0$
Add $12$ to both sides of the equation.
$x2+52x−12=0x2+52x=12$
Here $b=52$, and we can find the value that completes the square as follows:
$(b2)2=(5/22)2=(52⋅12)2=(54)2=2516$
To complete the square, add $2516$ to both sides of the equation.
$x2+52x=12x2+52x + 2516=12 + 2516(x+54)(x+54)=816+2516(x+54)2=3316$
Next, solve by extracting roots.
$(x+54)2=3316x+54=±3316x+54=±334x=−54±334x=−5±334$
Answer: The solutions are $−5±334.$
Try this! Solve by completing the square: $3x2−2x+1=0.$
Answer: The solutions are $x=13±23i.$
### Key Takeaways
• Solve equations of the form $ax2+c=0$ by extracting the roots.
• Extracting roots involves isolating the square and then applying the square root property. Remember to include “±” when taking the square root of both sides.
• After applying the square root property, solve each of the resulting equations. Be sure to simplify all radical expressions and rationalize the denominator if necessary.
• Solve any quadratic equation by completing the square.
• You can apply the square root property to solve an equation if you can first convert the equation to the form $(x−p)2=q.$
• To complete the square, first make sure the equation is in the form $x2+bx=c.$ The leading coefficient must be 1. Then add the value $(b2)2$ to both sides and factor.
• The process for completing the square always works, but it may lead to some tedious calculations with fractions. This is the case when the middle term, b, is not divisible by 2.
### Part A: Extracting Square Roots
Solve by factoring and then solve by extracting roots. Check answers.
1. $x2−16=0$
2. $x2−36=0$
3. $9y2−1=0$
4. $4y2−25=0$
5. $(x−2)2−1=0$
6. $(x+1)2−4=0$
7. $4(y−2)2−9=0$
8. $9(y+1)2−4=0$
9. $(u−5)2−25=0$
10. $(u+2)2−4=0$
Solve by extracting the roots.
1. $x2=81$
2. $x2=1$
3. $y2=19$
4. $y2=116$
5. $x2=12$
6. $x2=18$
7. $16x2=9$
8. $4x2=25$
9. $2t2=1$
10. $3t2=2$
11. $x2−40=0$
12. $x2−24=0$
13. $x2+1=0$
14. $x2+100=0$
15. $5x2−1=0$
16. $6x2−5=0$
17. $8x2+1=0$
18. $12x2+5=0$
19. $y2+4=0$
20. $y2+1=0$
21. $x2−49=0$
22. $x2−925=0$
23. $x2−8=0$
24. $t2−18=0$
25. $x2+8=0$
26. $x2+125=0$
27. $5y2−2=0$
28. $3x2−1=0$
29. $(x+7)2−4=0$
30. $(x+9)2−36=0$
31. $(x−5)2−20=0$
32. $(x+1)2−28=0$
33. $(3t+2)2+6=0$
34. $(3t−5)2+10=0$
35. $4(3x+1)2−27=0$
36. $9(2x−3)2−8=0$
37. $2(3x−1)2+3=0$
38. $5(2x−1)2+2=0$
39. $3(y−23)2−32=0$
40. $2(3y−13)2−52=0$
41. $−3(t−1)2+12=0$
42. $−2(t+1)2+8=0$
43. Solve for x: $px2−q=0$, $p,q>0$
44. Solve for x: $(x−p)2−q=0$, $p,q>0$
45. The diagonal of a square measures 3 centimeters. Find the length of each side.
46. The length of a rectangle is twice its width. If the diagonal of the rectangle measures 10 meters, then find the dimensions of the rectangle.
47. If a circle has an area of $50π$ square centimeters, then find its radius.
48. If a square has an area of 27 square centimeters, then find the length of each side.
49. The height in feet of an object dropped from an 18-foot stepladder is given by $h(t)=−16t2+18$, where t represents the time in seconds after the object is dropped. How long does it take the object to hit the ground? (Hint: The height is 0 when the object hits the ground. Round to the nearest hundredth of a second.)
50. The height in feet of an object dropped from a 50-foot platform is given by $h(t)=−16t2+50$, where t represents the time in seconds after the object is dropped. How long does it take the object to hit the ground? (Round to the nearest hundredth of a second.)
51. How high does a 22-foot ladder reach if its base is 6 feet from the building on which it leans? Round to the nearest tenth of a foot.
52. The height of a triangle is $12$ the length of its base. If the area of the triangle is 72 square meters, find the exact length of the triangle’s base.
### Part B: Completing the Square
Complete the square.
1. $x2−2x+ ? =(x− ? )2$
2. $x2−4x+ ? =(x− ? )2$
3. $x2+10x+ ? =(x+ ? )2$
4. $x2+12x+ ? =(x+ ? )2$
5. $x2+7x+ ? =(x+ ? )2$
6. $x2+5x+ ? =(x+ ? )2$
7. $x2−x+ ? =(x− ? )2$
8. $x2−12x+ ? =(x− ? )2$
9. $x2+23x+ ? =(x+ ? )2$
10. $x2+45x+ ? =(x+ ? )2$
Solve by factoring and then solve by completing the square. Check answers.
1. $x2+2x−8=0$
2. $x2−8x+15=0$
3. $y2+2y−24=0$
4. $y2−12y+11=0$
5. $t2+3t−28=0$
6. $t2−7t+10=0$
7. $2x2+3x−2=0$
8. $3x2−x−2=0$
9. $2y2−y−1=0$
10. $2y2+7y−4=0$
Solve by completing the square.
1. $x2+6x−1=0$
2. $x2+8x+10=0$
3. $x2−2x−7=0$
4. $x2−6x−3=0$
5. $y2−2y+4=0$
6. $y2−4y+9=0$
7. $t2+10t−75=0$
8. $t2+12t−108=0$
9. $u2−23u−13=0$
10. $u2−45u−15=0$
11. $x2+x−1=0$
12. $x2+x−3=0$
13. $y2+3y−2=0$
14. $y2+5y−3=0$
15. $x2+3x+5=0$
16. $x2+x+1=0$
17. $x2−7x+112=0$
18. $x2−9x+32=0$
19. $t2−12t−1=0$
20. $t2−13t−2=0$
21. $4x2−8x−1=0$
22. $2x2−4x−3=0$
23. $3x2+6x+1=0$
24. $5x2+10x+2=0$
25. $3x2+2x−3=0$
26. $5x2+2x−5=0$
27. $4x2−12x−15=0$
28. $2x2+4x−43=0$
29. $2x2−4x+10=0$
30. $6x2−24x+42=0$
31. $2x2−x−2=0$
32. $2x2+3x−1=0$
33. $3u2+2u−2=0$
34. $3u2−u−1=0$
35. $x2−4x−1=15$
36. $x2−12x+8=−10$
37. $x(x+1)−11(x−2)=0$
38. $(x+1)(x+7)−4(3x+2)=0$
39. $y2=(2y+3)(y−1)−2(y−1)$
40. $(2y+5)(y−5)−y(y−8)=−24$
41. $(t+2)2=3(3t+1)$
42. $(3t+2)(t−4)−(t−8)=1−10t$
Solve by completing the square and round the solutions to the nearest hundredth.
1. $(2x−1)2=2x$
2. $(3x−2)2=5−15x$
3. $(2x+1)(3x+1)=9x+4$
4. $(3x+1)(4x−1)=17x−4$
5. $9x(x−1)−2(2x−1)=−4x$
6. $(6x+1)2−6(6x+1)=0$
### Part C: Discussion Board
1. Create an equation of your own that can be solved by extracting the roots. Share it, along with the solution, on the discussion board.
2. Explain why the technique of extracting roots greatly expands our ability to solve quadratic equations.
3. Explain why the technique for completing the square described in this section requires that the leading coefficient be equal to 1.
4. Derive a formula for the diagonal of a square in terms of its sides.
1. −4, 4
2. $−13,13$
3. 1, 3
4. $12,72$
5. 0, 10
6. ±9
7. $±13$
8. $±23$
9. $±34$
10. $±22$
11. $±210$
12. $±i$
13. $±55$
14. $±24i$
15. $±2i$
16. $±23$
17. $±22$
18. $±2i2$
19. $±105$
20. −9, −5
21. $5±25$
22. $−23±63i$
23. $−2±336$
24. $13±66i$
25. $4±326$
26. −1, 3
27. $x=±pqp$
28. $322$ centimeters
29. $52$ centimeters
30. 1.06 seconds
31. 21.2 feet
1. $x2−2x+1=(x− 1)2$
2. $x2+10x+25=(x+5)2$
3. $x2+7x+494=(x+ 72)2$
4. $x2−x+14=(x−12)2$
5. $x2+23x+19=(x+ 13)2$
6. −4, 2
7. −6, 4
8. −7, 4
9. $−2,12$
10. $−12,1$
11. $−3±10$
12. $1±22$
13. $1±i3$
14. −15, 5
15. $−13,1$
16. $−1±52$
17. $−3±172$
18. $−32±112i$
19. $7±332$
20. $1±174$
21. $2±52$
22. $−3±63$
23. $−1±103$
24. $3±262$
25. $1±2i$
26. $1±174$
27. $−1±73$
28. $2±25$
29. $5±3$
30. $1±52$
31. $5±212$
32. 0.19, 1.31
33. −0.45, 1.12
34. 0.33, 0.67
### Learning Objectives
2. Use the determinant to determine the number and type of solutions to a quadratic equation.
In this section, we will develop a formula that gives the solutions to any quadratic equation in standard form. To do this, we begin with a general quadratic equation in standard form and solve for x by completing the square. Here a, b, and c are real numbers and $a≠0$:
$ax2+bx+c=0Standard form of a quadratic equation.ax2+bx+ca=0a Divide both sides by a.x2+bax+ca=0Subtract ca from both sides.x2+bax=−ca$
Determine the constant that completes the square: take the coefficient of x, divide it by 2, and then square it.
$(b/a2)2=(b2a)2=b24a2$
Add this to both sides of the equation to complete the square and then factor.
$x2+bax+b24a2=−ca+b24a2(x+b2a)(x+b2a)=−ca+b24a2(x+b2a)2=−4ac4a2+b24a2(x+b2a)2=b2−4ac4a2$
Solve by extracting roots.
$(x+b2a)2=b2−4ac4a2x+b2a=±b2−4ac4a2x+b2a=±b2−4ac2ax=−b2a±b2−4ac2ax=−b±b2−4ac2a$
This derivation gives us a formula that solves any quadratic equation in standard form. Given $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0$, the solutions can be calculated using the quadratic formulaThe formula $x=−b±b2−4ac2a$, which gives the solutions to any quadratic equation in the standard form $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0.$:
$x=−b±b2−4ac2a$
### Example 1
Solve using the quadratic formula: $2x2−7x−15=0.$
Solution:
Begin by identifying the coefficients of each term: a, b, and c.
$a=2 b=−7 c=−15$
Substitute these values into the quadratic formula and then simplify.
$x=−b±b2−4ac2a=−(−7)±(−7)2−4(2)(−15)2(2)=7±49+1204=7±1694=7±134$
Separate the “plus or minus” into two equations and simplify further.
$x=7−134orx=7+134x=−64x=204x=−32x=5$
Answer: The solutions are $−32$ and 5.
The previous example can be solved by factoring as follows:
$2x2−7x−15=0(2x+3)(x−5)=02x+3=0orx−5=02x=−3x=5x=−32$
Of course, if the quadratic expression factors, then it is a best practice to solve the equation by factoring. However, not all quadratic polynomials factor so easily. The quadratic formula provides us with a means to solve all quadratic equations.
### Example 2
Solve using the quadratic formula: $3x2+6x−2=0.$
Solution:
Begin by identifying a, b, and c.
$a=3 b=6 c=−2$
Substitute these values into the quadratic formula.
$x=−b±b2−4ac2a=−(6)±(6)2−4(3)(−2)2(3)=−6±36+246=−6±606$
At this point we see that $60=4×15$ and thus the fraction can be simplified further.
$=−6±606=−6±4×156=−6±2156=2(−3±15)63=−3±153$
It is important to point out that there are two solutions here:
$x=−3−153 or x=−3+153$
We may use ± to write the two solutions in a more compact form.
Answer: The solutions are $−3±153.$
Sometimes terms are missing. When this is the case, use 0 as the coefficient.
### Example 3
Solve using the quadratic formula: $x2−45=0.$
Solution:
This equation is equivalent to
$1x2+0x−45=0$
And we can use the following coefficients:
$a=1 b=0 c=−45$
Substitute these values into the quadratic formula.
$x=−b±b2−4ac2a=−(0)±(0)2−4(1)(−45)2(1)=0±0+1802=±1802=±36×52=±652=±35$
Since the coefficient of x was 0, we could have solved this equation by extracting the roots. As an exercise, solve it using this method and verify that the results are the same.
Answer: The solutions are $±35.$
Often solutions to quadratic equations are not real.
### Example 4
Solve using the quadratic formula: $x2−4x+29=0.$
Solution:
Begin by identifying a, b, and c. Here
$a=1 b=−4 c=29$
Substitute these values into the quadratic formula and then simplify.
$x=−b±b2−4ac2a=−(−4)±(−4)2−4(1)(29)2(1)=4±16−1162=4±−1002Negative radicand=4±10i2Two complex solutions=42±10i2=2±5i$
Check these solutions by substituting them into the original equation.
$Check x=2−5i$
$Check x=2+5i$
$x2−4x+29=0(2−5i)2−4(2−5i)+29=04−20i+25i2−8+20i+29=025i2+25=025(−1)+25=0−25+25=0 ✓$
$x2−4x+29=0(2+5i)2−4(2+5i)+29=04+20i+25i2−8−20i+29=025i2+25=025(−1)+25=0−25+25=0 ✓$
Answer: The solutions are $2±5i.$
The equation may not be given in standard form. The general steps for using the quadratic formula are outlined in the following example.
### Example 5
Solve: $(5x+1)(x−1)=x(x+1).$
Solution:
Step 1: Write the quadratic equation in standard form, with zero on one side of the equal sign.
$(5x+1)(x−1)=x(x+1)5x2−5x+x−1=x2+x5x2−4x−1=x2+x4x2−5x−1=0$
Step 2: Identify a, b, and c for use in the quadratic formula. Here
$a=4 b=−5 c=−1$
Step 3: Substitute the appropriate values into the quadratic formula and then simplify.
$x=−b±b2−4ac2a=−(−5)±(−5)2−4(4)(−1)2(4)=5±25+168=5±418$
Answer: The solution is $5±418.$
Try this! Solve: $(x+3)(x−5)=−19$
Answer: $1±i3$
## The Discriminant
If given a quadratic equation in standard form, $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0$, then the solutions can be calculated using the quadratic formula:
$x=−b±b2−4ac2a$
As we have seen, the solutions can be rational, irrational, or complex. We can determine the number and type of solutions by studying the discriminantThe expression inside the radical of the quadratic formula, $b2−4ac.$, the expression inside the radical, $b2−4ac.$ If the value of this expression is negative, then the equation has two complex solutions. If the discriminant is positive, then the equation has two real solutions. And if the discriminant is 0, then the equation has one real solution, a double root.
### Example 6
Determine the type and number of solutions: $2x2+x+3=0.$
Solution:
We begin by identifying a, b, and c. Here
$a=2 b=1 c=3$
Substitute these values into the discriminant and simplify.
$b2−4ac=(1)2−4(2)(3)=1−24=−23$
Since the discriminant is negative, we conclude that there are no real solutions. They are complex.
If we use the quadratic formula in the previous example, we find that a negative radicand introduces the imaginary unit and we are left with two complex solutions.
$x=−b±b2−4ac2a=−(1)±−232(2)=−1±i234=−14±234i Two complex solutions$
Note: Irrational and complex solutions of quadratic equations always appear in conjugate pairs.
### Example 7
Determine the type and number of solutions: $6x2−5x−1=0.$
Solution:
In this example,
$a=6 b=−5 c=−1$
Substitute these values into the discriminant and simplify.
$b2−4ac=(−5)2−4(6)(−1)=25+24=49$
Since the discriminant is positive, we conclude that the equation has two real solutions. Furthermore, since the discriminant is a perfect square, we obtain two rational solutions.
Because the discriminant is a perfect square, we could solve the previous quadratic equation by factoring or by using the quadratic formula.
Solve by factoring: Solve using the quadratic formula:
$6x2−5x−1=0(6x+1)(x−1)=06x+1=0orx−1=06x=−1x=1x=−16$ $x=−b±b2−4ac2a=−(−5)±492(6)=5±712x=5−712orx=5+712x=−212x=1212x=−16x=1$
Given the special condition where the discriminant is 0, we obtain only one solution, a double root.
### Example 8
Determine the type and number of solutions: $25x2−20x+4=0.$
Solution:
Here $a=25$, $b=−20$, and $c=4$, and we have
$b2−4ac=(−20)2−4(25)(4)=400−400=0$
Since the discriminant is 0, we conclude that the equation has only one real solution, a double root.
Since 0 is a perfect square, we can solve the equation above by factoring.
$25x2−20x+4=0(5x−2)(5x−2)=05x−2=0or5x−2=05x=25x=2x=25x=25$
Here $25$ is a solution that occurs twice; it is a double root.
### Example 9
Determine the type and number of solutions: $x2−2x−4=0.$
Solution:
Here $a=1$, $b=−2$, and $c=−4$, and we have
$b2−4ac=(−2)2−4(1)(−4)=4+16=20$
Since the discriminant is positive, we can conclude that the equation has two real solutions. Furthermore, since 20 is not a perfect square, both solutions are irrational.
If we use the quadratic formula in the previous example, we find that a positive radicand in the quadratic formula leads to two real solutions.
$x=−b±b2−4ac2a=−(−2)±202(1)Positive discriminant=2±4×52=2±252=2(1±5)21=1±5 Two irrational solutions$
The two real solutions are $1−5$ and $1+5.$ Note that these solutions are irrational; we can approximate the values on a calculator.
$1−5≈−1.24 and 1+5≈3.24$
In summary, if given any quadratic equation in standard form, $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0$, then we have the following:
$Positive discriminant:b2−4ac>0Two real solutionsZero discriminant:b2−4ac=0One real solutionNegative discriminant:b2−4ac<0Two complex solutions$
Furthermore, if the discriminant is nonnegative and a perfect square, then the solutions to the equation are rational; otherwise they are irrational. As we will see, knowing the number and type of solutions ahead of time helps us determine which method is best for solving a quadratic equation.
Try this! Determine the number and type of solutions: $2x2=x−2.$
### Key Takeaways
• We can use the quadratic formula to solve any quadratic equation in standard form.
• To solve any quadratic equation, we first rewrite it in standard form $ax2+bx+c=0$, substitute the appropriate coefficients into the quadratic formula, $x=−b±b2−4ac2a$, and then simplify.
• We can determine the number and type of solutions to any quadratic equation in standard form using the discriminant, $b2−4ac.$ If the value of this expression is negative, then the equation has two complex solutions. If the discriminant is positive, then the equation has two real solutions. And if the discriminant is 0, then the equation has one real solution, a double root.
• We can further classify real solutions into rational or irrational numbers. If the discriminant is a perfect square, the roots are rational and the equation will factor. If the discriminant is not a perfect square, the roots are irrational.
### Part A: The Quadratic Formula
Identify the coefficients, a, b and c, used in the quadratic formula. Do not solve.
1. $x2−x+3=0$
2. $5x2−2x−8=0$
3. $4x2−9=0$
4. $x2+3x=0$
5. $−x2+2x−7=0$
6. $−2x2−5x+2=0$
7. $px2−qx−1=0$
8. $p2x2−x+2q=0$
9. $(x−5)2=49$
10. $(2x+1)2=2x−1$
Solve by factoring and then solve using the quadratic formula. Check answers.
1. $x2−6x−16=0$
2. $x2−3x−18=0$
3. $2x2+7x−4=0$
4. $3x2+5x−2=0$
5. $4y2−9=0$
6. $9y2−25=0$
7. $5t2−6t=0$
8. $t2+6t=0$
9. $−x2+9x−20=0$
10. $−2x2−3x+5=0$
11. $16y2−24y+9=0$
12. $4y2−20y+25=0$
Solve by extracting the roots and then solve using the quadratic formula. Check answers.
1. $x2−18=0$
2. $x2−12=0$
3. $x2+12=0$
4. $x2+20=0$
5. $3x2+2=0$
6. $5x2+3=0$
7. $(x+2)2+9=0$
8. $(x−4)2+1=0$
9. $(2x+1)2−2=0$
10. $(3x+1)2−5=0$
1. $x2−5x+1=0$
2. $x2−7x+2=0$
3. $x2+8x+5=0$
4. $x2−4x+2=0$
5. $y2−2y+10=0$
6. $y2−4y+13=0$
7. $2x2−10x−1=0$
8. $2x2−4x−3=0$
9. $3x2−x+2=0$
10. $4x2−3x+1=0$
11. $5u2−2u+1=0$
12. $8u2−20u+13=0$
13. $−y2+16y−62=0$
14. $−y2+14y−46=0$
15. $−2t2+4t+3=0$
16. $−4t2+8t+1=0$
17. $12y2+5y+32=0$
18. $3y2+12y−13=0$
19. $2x2−12x+14=0$
20. $3x2−23x+13=0$
21. $1.2x2−0.5x−3.2=0$
22. $0.4x2+2.3x+1.1=0$
23. $2.5x2−x+3.6=0$
24. $−0.8x2+2.2x−6.1=0$
25. $−2y2=3(y−1)$
26. $3y2=5(2y−1)$
27. $(t+1)2=2t+7$
28. $(2t−1)2=73−4t$
29. $(x+5)(x−1)=2x+1$
30. $(x+7)(x−2)=3(x+1)$
31. $2x(x−1)=−1$
32. $x(2x+5)=3x−5$
33. $3t(t−2)+4=0$
34. $5t(t−1)=t−4$
35. $(2x+3)2=16x+4$
36. $(2y+5)2−12(y+1)=0$
Assume p and q are nonzero integers and use the quadratic formula to solve for x.
1. $px2+x+1=0$
2. $x2+px+1=0$
3. $x2+x−p=0$
4. $x2+px+q=0$
5. $p2x2+2px+1=0$
6. $x2−2qx+q2=0$
Solve using algebra.
1. The height in feet reached by a baseball tossed upward at a speed of 48 feet per second from the ground is given by $h(t)=−16t2+48t$, where t represents time in seconds after the ball is tossed. At what time does the baseball reach 24 feet? (Round to the nearest tenth of a second.)
2. The height in feet of a projectile launched upward at a speed of 32 feet per second from a height of 64 feet is given by $h(t)=−16t2+32t+64.$ At what time after launch does the projectile hit the ground? (Round to the nearest tenth of a second.)
3. The profit in dollars of running an assembly line that produces custom uniforms each day is given by $P(t)=−40t2+960t−4,000$ where t represents the number of hours the line is in operation. Determine the number of hours the assembly line should run in order to make a profit of \$1,760 per day.
4. A manufacturing company has determined that the daily revenue R in thousands of dollars is given by $R(n)=12n−0.6n2$ where n represents the number of pallets of product sold. Determine the number of pallets that must be sold in order to maintain revenues at 60 thousand dollars per day.
5. The area of a rectangle is 10 square inches. If the length is 3 inches more than twice the width, then find the dimensions of the rectangle. (Round to the nearest hundredth of an inch.)
6. The area of a triangle is 2 square meters. If the base is 2 meters less than the height, then find the base and the height. (Round to the nearest hundredth of a meter.)
7. To safely use a ladder, the base should be placed about $14$ of the ladder’s length away from the wall. If a 32-foot ladder is used safely, then how high against a building does the top of the ladder reach? (Round to the nearest tenth of a foot.)
8. The length of a rectangle is twice its width. If the diagonal of the rectangle measures 10 centimeters, then find the dimensions of the rectangle. (Round to the nearest tenth of a centimeter.)
9. Assuming dry road conditions and average reaction times, the safe stopping distance in feet of a certain car is given by $d(x)=120x2+x$ where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 50 feet. (Round to the nearest mile per hour.)
10. The width of a rectangular solid is 2.2 centimeters less than its length and the depth measures 10 centimeters.
Determine the length and width if the total volume of the solid is 268.8 cubic centimeters.
11. An executive traveled 25 miles in a car and then another 30 miles on a helicopter. If the helicopter was 10 miles per hour less than twice as fast as the car and the total trip took 1 hour, then what was the average speed of the car? (Round to the nearest mile per hour.)
12. Joe can paint a typical room in 1.5 hours less time than James. If Joe and James can paint 2 rooms working together in an 8-hour shift, then how long does it take James to paint a single room? (Round to the nearest tenth of an hour.)
### Part B: The Discriminant
Calculate the discriminant and use it to determine the number and type of solutions. Do not solve.
1. $x2−x+1=0$
2. $x2+2x+3=0$
3. $x2−2x−3=0$
4. $x2−5x−5=0$
5. $3x2−1x−2=0$
6. $3x2−1x+2=0$
7. $9y2+2=0$
8. $9y2−2=0$
9. $2x2+3x=0$
10. $4x2−5x=0$
11. $12x2−2x+52=0$
12. $12x2−x−12=0$
13. $−x2−3x+4=0$
14. $−x2−5x+3=0$
15. $25t2+30t+9=0$
16. $9t2−12t+4=0$
Find a nonzero integer p so that the following equations have one real solution. (Hint: If the discriminant is zero, then there will be one real solution.)
1. $px2−4x−1=0$
2. $x2−8x+p=0$
3. $x2+px+25=0$
4. $x2−2x+p2=0$
### Part C: Discussion Board
1. When talking about a quadratic equation in standard form $ax2+bx+c=0$, why is it necessary to state that $a≠0$? What would happen if a is equal to zero?
2. Research and discuss the history of the quadratic formula and solutions to quadratic equations.
3. Solve $mx2+nx+p=0$ for x by completing the square.
1. $a=1$; $b=−1$; $c=3$
2. $a=4$; $b=0$; $c=−9$
3. $a=−1$; $b=2$; $c=−7$
4. $a=p$; $b=−q$; $c=−1$
5. $a=1$; $b=−10$; $c=−24$
6. −2, 8
7. $−4,12$
8. $±32$
9. $0,65$
10. 4, 5
11. $34$
12. $±32$
13. $±2i3$
14. $±i63$
15. $−2±3i$
16. $−1±22$
17. $5±212$
18. $−4±11$
19. $1±3i$
20. $5±332$
21. $16±236i$
22. $15±25i$
23. $8±2$
24. $2±102$
25. $−5±22$
26. $18±78i$
27. $x≈−1.4$ or $x≈1.9$
28. $x≈0.2±1.2i$
29. $−3±334$
30. $±6$
31. $−1±7$
32. $12±12i$
33. $1±33i$
34. $12±i$
35. $x=−1±1−4p2p$
36. $x=−1±1+4p2$
37. $x=−1p$
38. 0.6 seconds and 2.4 seconds
39. 12 hours
40. Length: 6.22 inches; width: 1.61 inches
41. 31.0 feet
42. 23 miles per hour
43. 42 miles per hour
1. −3; two complex solutions
2. 16; two rational solutions
3. 25; two rational solutions
4. −72; two complex solutions
5. 9; two rational solutions
6. −1; two complex solutions
7. 25; two rational solutions
8. 0; one rational solution
9. $p=−4$
10. $p=±10$
## 6.3 Solving Equations Quadratic in Form
### Learning Objectives
1. Develop a general strategy for solving quadratic equations.
2. Solve equations that are quadratic in form.
## General Guidelines for Solving Quadratic Equations
Use the coefficients of a quadratic equation to help decide which method is most appropriate for solving it. While the quadratic formula always works, it is sometimes not the most efficient method. If given any quadratic equation in standard form,
$ax2+bx+c=0$ where c = 0, then it is best to factor out the GCF and solve by factoring.
### Example 1
Solve: $12x2−3x=0.$
Solution:
In this case, c = 0 and we can solve by factoring out the GCF 3x.
$12x2−3x=03x(4x−1)=0$
Then apply the zero-product property and set each factor equal to zero.
$3x=0or4x−1=0x=04x=1x=14$
Answer: The solutions are 0 and $14.$
If b = 0, then we can solve by extracting the roots.
### Example 2
Solve: $5x2+8=0.$
Solution:
In this case, b = 0 and we can solve by extracting the roots. Begin by isolating the square.
$5x2+8=05x2=−8x2=−85$
Next, apply the square root property. Remember to include the ±.
$x=±−85Rationalize the denominator.=±−4⋅25⋅55Simplify.=±2i2⋅55=±2i105$
Answer: The solutions are $±2i105.$
When given a quadratic equation in standard form where a, b, and c are all nonzero, determine the value for the discriminant using the formula $b2−4ac.$
1. If the discriminant is a perfect square, then solve by factoring.
2. If the discriminant is not a perfect square, then solve using the quadratic formula.
Recall that if the discriminant is not a perfect square and positive, the quadratic equation will have two irrational solutions. And if the discriminant is negative, the quadratic equation will have two complex conjugate solutions.
### Example 3
Solve: $(3x+5)(3x+7)=6x+10.$
Solution:
Begin by rewriting the quadratic equation in standard form.
$(3x+5)(3x+7)=6x+109x2+21x+15x+35=6x+109x2+36x+35=6x+109x2+30x+25=0$
Substitute a = 9, b = 30, and c = 25 into the discriminant formula.
$b2−4ac=(30)2−4(9)(25)=900−900=0$
Since the discriminant is 0, solve by factoring and expect one real solution, a double root.
$9x2+30x+25=0(3x+5)(3x+5)=03x+5=0or3x+5=03x=−53x=−5x=−53x=−53$
Answer: The solution is $−53.$
It is good to know that the quadratic formula will work to find the solutions to all of the examples in this section. However, it is not always the best solution. If the equation can be solved by factoring or by extracting the roots, you should use that method.
## Solving Equations Quadratic in Form
In this section we outline an algebraic technique that is used extensively in mathematics to transform equations into familiar forms. We begin by defining quadratic formAn equation of the form $au2+bu+c=0$ where a, b and c are real numbers and u represents an algebraic expression.,
$au2+bu+c=0$
Here u represents an algebraic expression. Some examples follow:
$(t+2t)2+8(t+2t)+7=0⇒u = t+2tu2+8u+7=0x2/3−3x1/3−10=0⇒u = x1/3u2−3u−10=03y−2+7y−1−6=0⇒u = y−13u2+7u−6=0$
If we can express an equation in quadratic form, then we can use any of the techniques used to solve quadratic equations. For example, consider the following fourth-degree polynomial equation,
$x4−4x2−32=0$
If we let $u=x2$ then $u2=(x2)2=x4$ and we can write
$x4−4x2−32=0⇒(x2)2−4(x2)−32=0↓↓u2−4u−32=0$
This substitution transforms the equation into a familiar quadratic equation in terms of u which, in this case, can be solved by factoring.
$u2−4u−32=0(u−8)(u+4)=0 u=8 or u=−4$
Since $u=x2$ we can back substitute and then solve for x.
$u=8oru=−4↓ ↓x2=8x2=−4x=±8x=±−4x=±22x=±2i$
Therefore, the equation $x4−4x2−32=0$ has four solutions ${±22,±2i}$, two real and two complex. This technique, often called a u-substitutionA technique in algebra using substitution to transform equations into familiar forms., can also be used to solve some non-polynomial equations.
### Example 4
Solve: $x−2x−8=0.$
Solution:
This is a radical equation that can be written in quadratic form. If we let $u=x$ then $u2=(x)2=x$ and we can write
$x−2x−8=0 ↓ ↓ u2− 2u − 8=0$
Solve for u.
$u2−2u−8=0(u−4)(u+2)=0u=4 or u=−2$
Back substitute $u=x$ and solve for x.
$x=4orx=−2(x)2=(4)2(x)2=(−2)2x=16x=4$
Recall that squaring both sides of an equation introduces the possibility of extraneous solutions. Therefore we must check our potential solutions.
$Check x=16Check x=4x−2x−8=016−216−8=016−2⋅4−8=016−8−8=00=0✓x−2x−8=04−24−8=04−2⋅2−8=04−4−8=0−8=0✗$
Because $x=4$ is extraneous, there is only one solution, $x=16.$
### Example 5
Solve: $x2/3−3x1/3−10=0.$
Solution:
If we let $u=x1/3$, then $u2=(x1/3)2=x2/3$ and we can write
$x2/3−3x1/3−10=0 ↓ ↓ u2 − 3u − 10=0$
Solve for u.
$u2−3u−10=0(u−5)(u+2)=0u=5 or u=−2$
Back substitute $u=x1/3$ and solve for x.
$x1/3=5orx1/3=−2(x1/3)3=(5)3(x1/3)3=(−2)3x=125x=−8$
Check.
$Check x=125Check x=−8x2/3−3x1/3−10=0(125)2/3−3(125)1/3−10=0(53)2/3−3(53)1/3−10=052−3⋅5−10=025−15−10=00=0✓x2/3−3x1/3−10=0(−8)2/3−3(−8)1/3−10=0[(−2)3]2/3−3[(−2)3]1/3−10=0(−2)2−3⋅(−2)−10=04+6−10=00=0✓$
Answer: The solutions are −8, 125.
### Example 6
Solve: $3y−2+7y−1−6=0.$
Solution:
If we let $u=y−1$, then $u2=(y−1)2=y−2$ and we can write
$3y−2+7y−1−6=0 ↓ ↓3u2 + 7u − 6=0$
Solve for u.
$3u2+7u−6=0(3u−2)(u+3)=0u=23 or u=−3$
Back substitute $u=y−1$ and solve for y.
$y−1=23ory−1=−31y=231y=−3y=32y=−13$
The original equation is actually a rational equation where $y≠0.$ In this case, the solutions are not restrictions; they solve the original equation.
Answer: The solutions are $−13, 32.$
### Example 7
Solve: $(t+2t)2+8(t+2t)+7=0.$
Solution:
If we let $u=t+2t$, then $u2=(t+2t)2$ and we can write
$(t+2t)2+8(t+2t)+7=0 ↓↓u2+8u+7=0$
Solve for u.
$u2+8u+7=0(u+1)(u+7)=0u=−1 or u=−7$
Back substitute $u=t+2t$ and solve for t.
$t+2t=−1ort+2t=−7t+2=−tt+2=−7t2t=−28t=−2t=−1t=−14$
Answer: The solutions are −1, $−14.$ The check is left to the reader.
Try this! Solve: $12x−2−16x−1+5=0$
Answer: The solutions are $65$, 2.
So far all of the examples were of equations that factor. As we know, not all quadratic equations factor. If this is the case, we use the quadratic formula.
### Example 8
Solve: $x4−10x2+23=0.$ Approximate to the nearest hundredth.
Solution:
If we let $u=x2$, then $u2=(x2)2=x4$ and we can write
$x4−10x2+23=0↓ ↓u2−10u+23=0$
This equation does not factor; therefore, use the quadratic formula to find the solutions for u. Here $a=1$, $b=−10$, and $c=23.$
$u=−b±b2−4ac2a=−(−10)±(−10)2−4(1)(23)2(1)=10±82=10±222=5±2$
Therefore, $u=5±2.$ Now back substitute $u=x2$ and solve for x.
$u=5−2oru=5+2↓↓x2=5−2x2=5+2x=±5−2x=±5+2$
Round the four solutions as follows.
$x=−5−2≈−1.89x=−5+2≈−2.53x=5−2≈1.89x=5+2≈2.53$
Answer: The solutions are approximately ±1.89, ±2.53.
If multiple roots and complex roots are counted, then the fundamental theorem of algebraIf multiple roots and complex roots are counted, then every polynomial with one variable will have as many roots as its degree. implies that every polynomial with one variable will have as many roots as its degree. For example, we expect $f(x)=x3−8$ to have three roots. In other words, the equation
$x3−8=0$ should have three solutions. To find them one might first think of trying to extract the cube roots just as we did with square roots,
$x3−8=0x3=8x=83x=2$
As you can see, this leads to one solution, the real cube root. There should be two others; let’s try to find them.
### Example 9
Find the set of all roots: $f(x)=x3−8.$
Solution:
Notice that the expression $x3−8$ is a difference of cubes and recall that $a3−b3=(a−b)(a2+ab+b2).$ Here $a=x$ and $b=2$ and we can write
$x3−8=0(x−2)(x2+2x+4)=0$
Next apply the zero-product property and set each factor equal to zero. After setting the factors equal to zero we can then solve the resulting equation using the appropriate methods.
$x−2=0orx2+2x+4=0x=2x=−b±b2−4ac2a=−(2)±(2)2−4(1)(4)2(1)=−2±−122=−2±2i32=−1±i3$
Using this method we were able to obtain the set of all three roots ${2,−1±i3}$, one real and two complex.
Answer: ${2,−1±i3}$
Sometimes the roots of a function will occur multiple times. For example, $g(x)=(x−2)3$ has degree three where the roots can be found as follows:
$(x−2)3=0(x−2)(x−2)(x−2)=0x−2=0orx−2=0orx−2=0x=2x=2x=2$
Even though g is of degree 3 there is only one real root {2}; it occurs 3 times.
### Key Takeaways
• The quadratic formula can solve any quadratic equation. However, it is sometimes not the most efficient method.
• If a quadratic equation can be solved by factoring or by extracting square roots you should use that method.
• We can sometimes transform equations into equations that are quadratic in form by making an appropriate u-substitution. After solving the equivalent equation, back substitute and solve for the original variable.
• Counting multiple and complex roots, the fundamental theorem of algebra guarantees as many roots as the degree of a polynomial equation with one variable.
### Part A: Solving Quadratic Equations
Solve.
1. $x2−9x=0$
2. $x2+10x=0$
3. $15x2+6x=0$
4. $36x2−18x=0$
5. $x2−90=0$
6. $x2+48=0$
7. $2x2+1=0$
8. $7x2−1=0$
9. $6x2−11x+4=0$
10. $9x2+12x−5=0$
11. $x2+x+6=0$
12. $x2+2x+8=0$
13. $4t2+28t+49=0$
14. $25t2−20t+4=0$
15. $u2−4u−1=0$
16. $u2−2u−11=0$
17. $2(x+2)2=11+4x−2x2$
18. $(2x+1)(x−3)+2x2=3(x−1)$
19. $(3x+2)2=6(2x+1)$
20. $(2x−3)2+5x2=4(2−3x)$
21. $4(3x−1)2−5=0$
22. $9(2x+3)2−2=0$
### Part B: Solving Equations Quadratic in Form
Find all solutions.
1. $x4+x2−72=0$
2. $x4−17x2−18=0$
3. $x4−13x2+36=0$
4. $4x4−17x2+4=0$
5. $x+2x−3=0$
6. $x−x−2=0$
7. $x−5x+6=0$
8. $x−6x+5=0$
9. $x2/3+5x1/3+6=0$
10. $x2/3−2x1/3−35=0$
11. $4x2/3−4x1/3+1=0$
12. $3x2/3−2x1/3−1=0$
13. $5x−2+9x−1−2=0$
14. $3x−2+8x−1−3=0$
15. $8x−2+14x−1−15=0$
16. $9x−2−24x−1+16=0$
17. $(x−3x)2−2(x−3x)−24=0$
18. $(2x+1x)2+9(2x+1x)−36=0$
19. $2(xx+1)2−5(xx+1)−3=0$
20. $3(x3x−1)2+13(x3x−1)−10=0$
21. $4y−2−9=0$
22. $16y−2+4y−1=0$
23. $30y2/3−15y1/3=0$
24. $y2/3−9=0$
25. $81y4−1=0$
26. $5(1x+2)2−3(1x+2)−2=0$
27. $12(x2x−3)2−11(x2x−3)+2=0$
28. $10x−2−19x−1−2=0$
29. $x1/2−3x1/4+2=0$
30. $x+5x−50=0$
31. $8x2/3+7x1/3−1=0$
32. $x4/3−13x2/3+36=0$
33. $y4−14y2+46=0$
34. $x4/3−2x2/3+1=0$
35. $2y−2−y−1−1=0$
36. $2x−2/3−3x−1/3−2=0$
37. $4x−1−17x−1/2+4=0$
38. $3x−1−8x−1/2+4=0$
39. $2x1/3−3x1/6+1=0$
40. $x1/3−x1/6−2=0$
1. $x4−6x2+7=0$
2. $x4−6x2+6=0$
3. $x4−8x2+14=0$
4. $x4−12x2+31=0$
5. $4x4−16x2+13=0$
6. $9x4−30x2+1=0$
Find the set of all roots.
1. $f(x)=x3−1$
2. $g(x)=x3+1$
3. $f(x)=x3−27$
4. $g(x)=x4−16$
5. $h(x)=x4−1$
6. $h(x)=x6−1$
7. $f(x)=(2x−1)3$
8. $g(x)=x2(x−4)2$
9. $f(x)=x3−q3$, $q>0$
10. $f(x)=x3+q3$, $q>0$
Find all solutions.
1. $x6+7x3−8=0$
2. $x6−7x3−8=0$
3. $x6+28x3+27=0$
4. $x6+16x3+64=0$
5. $| x2+2x−5 |=1$
6. $| x2−2x−3 |=3$
7. $| 2x2−5 |=4$
8. $| 3x2−9x |=6$
Find a quadratic function with integer coefficients and the given set of roots. (Hint: If $r1$ and $r2$ are roots, then $(x−r1)(x−r2)=0.$)
1. ${±3i}$
2. ${±i5}$
3. ${±3}$
4. ${±26}$
5. ${1±3}$
6. ${2±32}$
7. ${1±6i}$
8. ${2±3i}$
### Part C: Discussion Board
1. On a note card, write out your strategy for solving a quadratic equation. Share your strategy on the discussion board.
2. Make up your own equation that is quadratic in form. Share it and the solution on the discussion board.
1. 0, 9
2. $−25$, 0
3. $±310$
4. $±22i$
5. $12,43$
6. $−12±232i$
7. $−72$
8. $2±5$
9. $−32,12$
10. $±23$
11. $2±56$
1. $±22$, $±3i$
2. ±2, ±3
3. 1
4. 4, 9
5. −27, −8
6. $18$
7. $−12$, 5
8. $−25,43$
9. $±35$
10. $−32,−13$
11. $±23$
12. 0, $18$
13. $±13,±i3$
14. $−32$, 6
15. 1, 16
16. −1, $1512$
17. $±7−3$, $±7+3$
18. −2, 1
19. $116$, 16
20. $164$, 1
21. ±1.26, ±2.10
22. ±1.61, ±2.33
23. ±1.06, ±1.69
24. ${1,−12±32i}$
25. ${3,−32±332i}$
26. ${±1,±i}$
27. ${ 12 }$
28. ${q,−q2±q32i}$
29. −2, 1, $1±i3$, $−12±32i$
30. −3, −1, $32±332i$, $12±32i$
31. $−1±7$, $−1±5$
32. $±22$, $±322$
33. $f(x)=x2+9$
34. $f(x)=x2−3$
35. $f(x)=x2−2x−2$
36. $f(x)=x2−2x+37$
## 6.4 Quadratic Functions and Their Graphs
### Learning Objectives
1. Graph a parabola.
2. Find the intercepts and vertex of a parabola.
3. Find the maximum and minimum y-value.
4. Find the vertex of a parabola by completing the square.
## The Graph of a Quadratic Function
A quadratic function is a polynomial function of degree 2 which can be written in the general form,
$f(x)=ax2+bx+c$
Here a, b and c represent real numbers where $a≠0.$ The squaring function $f(x)=x2$ is a quadratic function whose graph follows.
This general curved shape is called a parabolaThe U-shaped graph of any quadratic function defined by $f(x)=ax2+bx+c$, where a, b, and c are real numbers and $a≠0.$ and is shared by the graphs of all quadratic functions. Note that the graph is indeed a function as it passes the vertical line test. Furthermore, the domain of this function consists of the set of all real numbers $(−∞,∞)$ and the range consists of the set of nonnegative numbers $[0,∞).$
When graphing parabolas, we want to include certain special points in the graph. The y-intercept is the point where the graph intersects the y-axis. The x-intercepts are the points where the graph intersects the x-axis. The vertexThe point that defines the minimum or maximum of a parabola. is the point that defines the minimum or maximum of the graph. Lastly, the line of symmetryThe vertical line through the vertex, $x=−b2a$, about which the parabola is symmetric. (also called the axis of symmetryA term used when referencing the line of symmetry.) is the vertical line through the vertex, about which the parabola is symmetric.
For any parabola, we will find the vertex and y-intercept. In addition, if the x-intercepts exist, then we will want to determine those as well. Guessing at the x-values of these special points is not practical; therefore, we will develop techniques that will facilitate finding them. Many of these techniques will be used extensively as we progress in our study of algebra.
Given a quadratic function $f(x)=ax2+bx+c$, find the y-intercept by evaluating the function where $x=0.$ In general, $f(0)=a(0)2+b(0)+c=c$, and we have
$y-intercept (0,c)$
Next, recall that the x-intercepts, if they exist, can be found by setting $f(x)=0.$ Doing this, we have $a2+bx+c=0$, which has general solutions given by the quadratic formula, $x=−b±b2−4ac2a.$ Therefore, the x-intercepts have this general form:
$x-intercepts(−b−b2−4ac2a,0) and (−b+b2−4ac2a,0)$
Using the fact that a parabola is symmetric, we can determine the vertical line of symmetry using the x-intercepts. To do this, we find the x-value midway between the x-intercepts by taking an average as follows:
$x=(−b−b2−4ac2a+−b+b2−4ac2a) ÷2=(−b−b2−4ac−b+b2−4ac2a)÷(21)=−2b2a⋅12=−b2a$
Therefore, the line of symmetry is the vertical line $x=−b2a.$ We can use the line of symmetry to find the the vertex.
$Line of symmetryVertexx=−b2a(−b2a,f(−b2a))$
Generally three points determine a parabola. However, in this section we will find five points so that we can get a better approximation of the general shape. The steps for graphing a parabola are outlined in the following example.
### Example 1
Graph: $f(x)=−x2−2x+3.$
Solution:
Step 1: Determine the y-intercept. To do this, set $x=0$ and find $f(0).$
$f(x)=−x2−2x+3f(0)=−(0)2−2(0)+3=3$
The y-intercept is $(0,3).$
Step 2: Determine the x-intercepts if any. To do this, set $f(x)=0$ and solve for x.
$f(x)=−x2−2x+3Set f(x)=0.0=−x2−2x+3Multiply both sides by −1.0=x2+2x−3Factor.0=(x+3)(x−1)Set each factor equal to zero.x+3=0orx−1=0x=−3x=1$
Here where $f(x)=0$, we obtain two solutions. Hence, there are two x-intercepts, $(−3,0)$ and $(1,0).$
Step 3: Determine the vertex. One way to do this is to first use $x=−b2a$ to find the x-value of the vertex and then substitute this value in the function to find the corresponding y-value. In this example, $a=−1$ and $b=−2.$
$x=−b2a=−(−2)2(−1)=2−2=−1$
Substitute −1 into the original function to find the corresponding y-value.
$f(x)=−x2−2x+3f(−1)=−(−1)2−2(−1)+3=−1+2+3=4$
The vertex is $(−1,4).$
Step 4: Determine extra points so that we have at least five points to plot. Ensure a good sampling on either side of the line of symmetry. In this example, one other point will suffice. Choose $x=−2$ and find the corresponding y-value.
$x yPoint−2 3f(−2)=−(−2)2−2(−2)+3=−4+4+3=3(−2,3)$
Our fifth point is $(−2,3).$
Step 5: Plot the points and sketch the graph. To recap, the points that we have found are
$y-intercept:(0,3)x-intercepts:(−3,0) and (1,0)Vertex:(−1,4)Extra point:(−2,3)$
The parabola opens downward. In general, use the leading coefficient to determine if the parabola opens upward or downward. If the leading coefficient is negative, as in the previous example, then the parabola opens downward. If the leading coefficient is positive, then the parabola opens upward.
All quadratic functions of the form $f(x)=ax2+bx+c$ have parabolic graphs with y-intercept $(0,c).$ However, not all parabolas have x-intercepts.
### Example 2
Graph: $f(x)=2x2+4x+5.$
Solution:
Because the leading coefficient 2 is positive, we note that the parabola opens upward. Here c = 5 and the y-intercept is (0, 5). To find the x-intercepts, set $f(x)=0.$
$f(x)=2x2+4x+50=2x2+4x+5$
In this case, a = 2, b = 4, and c = 5. Use the discriminant to determine the number and type of solutions.
$b2−4ac=(4)2−4(2)(5)=16−40=−24$
Since the discriminant is negative, we conclude that there are no real solutions. Because there are no real solutions, there are no x-intercepts. Next, we determine the x-value of the vertex.
$x=−b2a=−(4)2(2)=−44=−1$
Given that the x-value of the vertex is −1, substitute −1 into the original equation to find the corresponding y-value.
$f(x)=2x2+4x+5f(−1)=2(−1)2+4(−1)+5=2−4+5=3$
The vertex is (−1, 3). So far, we have only two points. To determine three more, choose some x-values on either side of the line of symmetry, x = −1. Here we choose x-values −3, −2, and 1.
$x yPoints−3 11f(−3)=2(−3)2+4(−3)+5=18−12+5=11(−3,11)−2 5f(−2)=2(−2)2+4(−2)+5=8−8+5=5(−2,5)1 11f(1)=2(1)2+4(1)+5=2+4+5=11(1,11)$
To summarize, we have
$y-intercept:(0,5)x-intercepts:NoneVertex:(−1,3)Extra points:(−3,11),(−2,5),(1,11)$
Plot the points and sketch the graph.
### Example 3
Graph: $f(x)=x2−2x−1.$
Solution:
Since a = 1, the parabola opens upward. Furthermore, c = −1, so the y-intercept is $(0,−1).$ To find the x-intercepts, set $f(x)=0.$
$f(x)=x2−2x−10=x2−2x−1$
In this case, solve using the quadratic formula with a = 1, b = −2, and c = −1.
$x=−b±b2−4ac2a=−(−2)±(−2)2−4(1)(−1)2(1)=2±82=2±222=2(1±2)2=1±2$
Here we obtain two real solutions for x, and thus there are two x-intercepts:
$(1−2,0)and(1+2,0)Exact values(−0.41,0)(2.41,0)Approximate values$
Approximating the x-intercepts using a calculator will help us plot the points. However, we will present the exact x-intercepts on the graph. Next, find the vertex.
$x=−b2a=−(−2)2(1)=22=1$
Given that the x-value of the vertex is 1, substitute into the original equation to find the corresponding y-value.
$y=x2−2x−1=(1)2−2(1)−1=1−2−1=−2$
The vertex is (1, −2). We need one more point.
$x yPoint2 −1f(2)=(2)2−2(2)−1=4−4−1=−1(2,−1)$
To summarize, we have
$y-intercept:(0,−1)x-intercepts:(1−2,0) and (1+2,0)Vertex:(1,−2)Extra point:(2,−1)$
Plot the points and sketch the graph.
Try this! Graph: $g(x)=−4x2+12x−9.$
## Finding the Maximum or Minimum
It is often useful to find the maximum and/or minimum values of functions that model real-life applications. To find these important values given a quadratic function, we use the vertex. If the leading coefficient a is positive, then the parabola opens upward and there will be a minimum y-value. If the leading coefficient a is negative, then the parabola opens downward and there will be a maximum y-value.
### Example 4
Determine the maximum or minimum: $y=−4x2+24x−35.$
Solution:
Since a = −4, we know that the parabola opens downward and there will be a maximum y-value. To find it, first find the x-value of the vertex.
$x=−b2ax-value of the vertex.=−242(−4)Substitute a = -4 and b = 24.=−24−8Simplify.=3$
The x-value of the vertex is 3. Substitute this value into the original equation to find the corresponding y-value.
$y=−4x2+24x−35Substitute x = 3.=−4(3)2+24(3)−35Simplify.=−36+72−35=1$
The vertex is (3, 1). Therefore, the maximum y-value is 1, which occurs where x = 3, as illustrated below:
Note: The graph is not required to answer this question.
### Example 5
Determine the maximum or minimum: $y=4x2−32x+62.$
Solution:
Since a = 4, the parabola opens upward and there is a minimum y-value. Begin by finding the x-value of the vertex.
$x=−b2a=−−322(4)Substitute a = 4 and b = -32.=−−328Simplify.=4$
Substitute x = 4 into the original equation to find the corresponding y-value.
$y=4x2−32x+62=4(4)2−32(4)+62=64−128+62=−2$
The vertex is (4, −2). Therefore, the minimum y-value of −2 occurs where x = 4, as illustrated below:
### Example 6
The height in feet of a projectile is given by the function $h(t)=−16t2+72t$, where t represents the time in seconds after launch. What is the maximum height reached by the projectile?
Solution:
Here $a=−16$, and the parabola opens downward. Therefore, the y-value of the vertex determines the maximum height. Begin by finding the time at which the vertex occurs.
$t=−b2a=−722(−16)=7232=94$
The maximum height will occur in $94$ seconds (or $214$ seconds). Substitute this time into the function to determine the maximum height attained.
$h(94)=−16(94)2+72(94)=−16(8116)+72(94)=−81+162=81$
Answer: The maximum height of the projectile is 81 feet.
## Finding the Vertex by Completing the Square
In this section, we demonstrate an alternate approach for finding the vertex. Any quadratic function $f(x)=ax2+bx+c$ can be rewritten in vertex formA quadratic function written in the form $f(x)=a(x−h)2+k.$,
$f(x)=a(x−h)2+k$
In this form, the vertex is $(h,k).$ To see that this is the case, consider graphing $f(x)=(x−2)2+3$ using the transformations.
$y=x2Basic squaring functiony=(x−2)2Horizontal shift right 2 unitsy=(x−2)2+3Vertical shift up 3 units$
Use these translations to sketch the graph,
Here we can see that the vertex is (2, 3).
$f(x)=a(x−h)2+k↓↓f(x)=(x−2)2+3$
When the equation is in this form, we can read the vertex directly from it.
### Example 7
Determine the vertex: $f(x)=2(x+3)2−2.$
Solution:
Rewrite the equation as follows before determining h and k.
$f(x)=a(x−h)2+k↓↓f(x)=2[x−(−3)]2+(−2)$
Here h = −3 and k = −2.
Answer: The vertex is (−3, −2).
Often the equation is not given in vertex form. To obtain this form, complete the square.
### Example 8
Rewrite in vertex form and determine the vertex: $f(x)=x2+4x+9.$
Solution:
Begin by making room for the constant term that completes the square.
$f(x)=x2+4x+9=x2+4x+___+9−___$
The idea is to add and subtract the value that completes the square, $(b2)2$, and then factor. In this case, add and subtract $(42)2=(2)2=4.$
$f(x)=x2+4x+9Add and subtract 4.=x2+4x +4+9 −4Factor.=(x2+4x+4)+5.=(x+3)(x+2)+5=(x+2)2+5$
Adding and subtracting the same value within an expression does not change it. Doing so is equivalent to adding 0. Once the equation is in this form, we can easily determine the vertex.
$f(x)=a(x−h)2+k↓↓f(x)=(x−(−2))2+5$
Here h = −2 and k = 5.
Answer: The vertex is (−2, 5).
If there is a leading coefficient other than 1, then we must first factor out the leading coefficient from the first two terms of the trinomial.
### Example 9
Rewrite in vertex form and determine the vertex: $f(x)=2x2−4x+8.$
Solution:
Since a = 2, factor this out of the first two terms in order to complete the square. Leave room inside the parentheses to add and subtract the value that completes the square.
$f(x)=2x2−4x+8=2(x2−2x )+8$
Now use −2 to determine the value that completes the square. In this case, $(−22)2=(−1)2=1.$ Add and subtract 1 and factor as follows:
$f(x)=2x2−4x+8=2(x2−2x+__−__)+8Add and subtract 1.=2(x2−2x +1−1)+8 Factor.=2[ (x−1)(x−1)−1 ]+8=2[ (x−1)2−1 ]+8Distribute the 2.=2(x−1)2−2+8=2(x−1)2+6$
In this form, we can easily determine the vertex.
$f(x)=a(x−h)2+k↓↓f(x)=2(x−1)2+6$
Here h = 1 and k = 6.
Answer: The vertex is (1, 6).
Try this! Rewrite in vertex form and determine the vertex: $f(x)=−2x2−12x+3.$
Answer: $f(x)=−2(x+3)2+21$; vertex: $(−3,21)$
### Key Takeaways
• The graph of any quadratic function $f(x)=ax2+bx+c$, where a, b, and c are real numbers and $a≠0$, is called a parabola.
• When graphing a parabola always find the vertex and the y-intercept. If the x-intercepts exist, find those as well. Also, be sure to find ordered pair solutions on either side of the line of symmetry, $x=−b2a.$
• Use the leading coefficient, a, to determine if a parabola opens upward or downward. If a is positive, then it opens upward. If a is negative, then it opens downward.
• The vertex of any parabola has an x-value equal to $−b2a.$ After finding the x-value of the vertex, substitute it into the original equation to find the corresponding y-value. This y-value is a maximum if the parabola opens downward, and it is a minimum if the parabola opens upward.
• The domain of a parabola opening upward or downward consists of all real numbers. The range is bounded by the y-value of the vertex.
• An alternate approach to finding the vertex is to rewrite the quadratic function in the form $f(x)=a(x−h)2+k.$ When in this form, the vertex is $(h,k)$ and can be read directly from the equation. To obtain this form, take $f(x)=ax2+bx+c$ and complete the square.
### Part A: The Graph of Quadratic Functions
Does the parabola open upward or downward? Explain.
1. $y=x2−9x+20$
2. $y=x2−12x+32$
3. $y=−2x2+5x+12$
4. $y=−6x2+13x−6$
5. $y=64−x2$
6. $y=−3x+9x2$
Determine the x- and y-intercepts.
1. $y=x2+4x−12$
2. $y=x2−13x+12$
3. $y=2x2+5x−3$
4. $y=3x2−4x−4$
5. $y=−5x2−3x+2$
6. $y=−6x2+11x−4$
7. $y=4x2−27$
8. $y=9x2−50$
9. $y=x2−x+1$
10. $y=x2−6x+4$
Find the vertex and the line of symmetry.
1. $y=−x2+10x−34$
2. $y=−x2−6x+1$
3. $y=−4x2+12x−7$
4. $y=−9x2+6x+2$
5. $y=4x2−1$
6. $y=x2−16$
Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.
1. $f(x)=x2−2x−8$
2. $f(x)=x2−4x−5$
3. $f(x)=−x2+4x+12$
4. $f(x)=−x2−2x+15$
5. $f(x)=x2−10x$
6. $f(x)=x2+8x$
7. $f(x)=x2−9$
8. $f(x)=x2−25$
9. $f(x)=1−x2$
10. $f(x)=4−x2$
11. $f(x)=x2−2x+1$
12. $f(x)=x2+4x+4$
13. $f(x)=−4x2+12x−9$
14. $f(x)=−4x2−4x+3$
15. $f(x)=x2−2$
16. $f(x)=x2−3$
17. $f(x)=−4x2+4x−3$
18. $f(x)=4x2+4x+3$
19. $f(x)=x2−2x−2$
20. $f(x)=x2−6x+6$
21. $f(x)=−2x2+6x−3$
22. $f(x)=−4x2+4x+1$
23. $f(x)=x2+3x+4$
24. $f(x)=−x2+3x−4$
25. $f(x)=−2x2+3$
26. $f(x)=−2x2−1$
27. $f(x)=2x2+4x−3$
28. $f(x)=3x2+2x−2$
### Part B: Finding the Maximum or Minimum
Determine the maximum or minimum y-value.
1. $y=−x2−6x+1$
2. $y=−x2−4x+8$
3. $y=25x2−10x+5$
4. $y=16x2−24x+7$
5. $y=−x2$
6. $y=1−9x2$
7. $y=20x−10x2$
8. $y=12x+4x2$
9. $y=3x2−4x−2$
10. $y=6x2−8x+5$
11. $y=x2−5x+1$
12. $y=1−x−x2$
Given the following quadratic functions, determine the domain and range.
1. $f(x)=3x2+30x+50$
2. $f(x)=5x2−10x+1$
3. $g(x)=−2x2+4x+1$
4. $g(x)=−7x2−14x−9$
5. $f(x)=x2+x−1$
6. $f(x)=−x2+3x−2$
7. The height in feet reached by a baseball tossed upward at a speed of 48 feet per second from the ground is given by the function $h(t)=−16t2+48t$, where t represents the time in seconds after the ball is thrown. What is the baseball’s maximum height and how long does it take to attain that height?
8. The height in feet of a projectile launched straight up from a mound is given by the function $h(t)=−16t2+96t+4$, where t represents seconds after launch. What is the maximum height?
9. The profit in dollars generated by producing and selling x custom lamps is given by the function $P(x)=−10x2+800x−12,000.$ What is the maximum profit?
10. The profit in dollars generated from producing and selling a particular item is modeled by the formula $P(x)=100x−0.0025x2$, where x represents the number of units produced and sold. What number of units must be produced and sold to maximize revenue?
11. The average number of hits to a radio station Web site is modeled by the formula $f(x)=450t2−3,600t+8,000$, where t represents the number of hours since 8:00 a.m. At what hour of the day is the number of hits to the Web site at a minimum?
12. The value in dollars of a new car is modeled by the formula $V(t)=125t2−3,000t+22,000$, where t represents the number of years since it was purchased. Determine the minimum value of the car.
13. The daily production cost in dollars of a textile manufacturing company producing custom uniforms is modeled by the formula $C(x)=0.02x2−20x+10,000$, where x represents the number of uniforms produced.
1. How many uniforms should be produced to minimize the daily production costs?
2. What is the minimum daily production cost?
14. The area in square feet of a certain rectangular pen is given by the formula $A=14w−w2$, where w represents the width in feet. Determine the width that produces the maximum area.
### Part C: Finding the Vertex by Completing the Square
Determine the vertex.
1. $y=−(x−5)2+3$
2. $y=−2(x−1)2+7$
3. $y=5(x+1)2+6$
4. $y=3(x+4)2+10$
5. $y=−5(x+8)2−1$
6. $y=(x+2)2−5$
Rewrite in vertex form $y=a(x−h)2+k$ and determine the vertex.
1. $y=x2−14x+24$
2. $y=x2−12x+40$
3. $y=x2+4x−12$
4. $y=x2+6x−1$
5. $y=2x2−12x−3$
6. $y=3x2−6x+5$
7. $y=−x2+16x+17$
8. $y=−x2+10x$
Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.
1. $f(x)=x2−1$
2. $f(x)=x2+1$
3. $f(x)=(x−1)2$
4. $f(x)=(x+1)2$
5. $f(x)=(x−4)2−9$
6. $f(x)=(x−1)2−4$
7. $f(x)=−2(x+1)2+8$
8. $f(x)=−3(x+2)2+12$
9. $f(x)=−5(x−1)2$
10. $f(x)=−(x+2)2$
11. $f(x)=−4(x−1)2−2$
12. $f(x)=9(x+1)2+2$
13. $f(x)=(x+5)2−15$
14. $f(x)=2(x−5)2−3$
15. $f(x)=−2(x−4)2+22$
16. $f(x)=2(x+3)2−13$
### Part D: Discussion Board
1. Write down your plan for graphing a parabola on an exam. What will you be looking for and how will you present your answer? Share your plan on the discussion board.
2. Why is any parabola that opens upward or downward a function? Explain to a classmate how to determine the domain and range.
3. Research and discuss ways of finding a quadratic function that has a graph passing through any three given points. Share a list of steps as well as an example of how to do this.
1. Upward
2. Downward
3. Downward
4. x-intercepts: (−6, 0), (2, 0);y-intercept: (0, −12)
5. x-intercepts: (−3, 0), $(12, 0)$; y-intercept: (0, −3)
6. x-intercepts: (−1, 0), $(25, 0)$; y-intercept: (0, 2)
7. x-intercepts: $(−332,0)$, $(332,0)$; y-intercept: (0, −27)
8. x-intercepts: none; y-intercept: (0, 1)
9. Vertex: (5, −9); line of symmetry: $x=5$
10. Vertex: $(32, 2)$; line of symmetry: $x=32$
11. Vertex: (0, −1); line of symmetry: $x=0$
1. Maximum: y = 10
2. Minimum: y = 4
3. Maximum: y = 0
4. Maximum: y = 10
5. Minimum: $y=−103$
6. Minimum: $y=−214$
7. Domain: $(−∞,∞)$; range: $[−25,∞)$
8. Domain: $(−∞,∞)$; range: $(−∞,3]$
9. Domain: $(−∞,∞)$; range: $[−54,∞)$
10. The maximum height of 36 feet occurs after 1.5 seconds.
11. \$4,000
12. 12:00 p.m.
1. 500 uniforms
2. \$5,000
1. (5, 3)
2. (−1, 6)
3. (−8, −1)
4. $y=(x−7)2−25$; vertex: (7, −25)
5. $y=(x+2)2−16$; vertex: (−2, −16)
6. $y=2(x−3)2−21$; vertex: (3, −21)
7. $y=−(x−8)2+81$; vertex: (8, 81)
### Learning Objectives
1. Check solutions to quadratic inequalities with one variable.
2. Understand the geometric relationship between solutions to quadratic inequalities and their graphs.
A quadratic inequalityA mathematical statement that relates a quadratic expression as either less than or greater than another. is a mathematical statement that relates a quadratic expression as either less than or greater than another. Some examples of quadratic inequalities solved in this section follow.
$x2−2x−11≤0$ $2x2−7x+3>0$ $9−x2>0$
A solution to a quadratic inequality is a real number that will produce a true statement when substituted for the variable.
### Example 1
Are −3, −2, and −1 solutions to $x2−x−6≤0$?
Solution:
Substitute the given value in for x and simplify.
$x2−x−6≤0x2−x−6≤0x2−x−6≤0(−3)2−(−3)−6≤0(−2)2−(−2)−6≤0(−1)2−(−1)−6≤09+3−6≤04+2−6≤01+1−6≤06≤0✗0≤0✓−4≤0✓$
Answer: −2 and −1 are solutions and −3 is not.
Quadratic inequalities can have infinitely many solutions, one solution, or no solution. If there are infinitely many solutions, graph the solution set on a number line and/or express the solution using interval notation. Graphing the function defined by $f(x)=x2−x−6$ found in the previous example we have
The result of evaluating for any x-value will be negative, zero, or positive.
$f(−3)=6Positivef(x)>0f(−2)=0Zerof(x)=0f(−1)=−4Negativef(x)<0$
The values in the domain of a function that separate regions that produce positive or negative results are called critical numbersThe values in the domain of a function that separate regions that produce positive or negative results.. In the case of a quadratic function, the critical numbers are the roots, sometimes called the zeros. For example, $f(x)=x2−x−6=(x+2)(x−3)$ has roots −2 and 3. These values bound the regions where the function is positive (above the x-axis) or negative (below the x-axis).
Therefore $x2−x−6≤0$ has solutions where $−2≤x≤3$, using interval notation $[−2,3].$ Furthermore, $x2−x−6≥0$ has solutions where $x≤−2$ or $x≥3$, using interval notation $(−∞,−2]∪[−3,∞).$
### Example 2
Given the graph of $f$ determine the solutions to $f(x)>0$:
Solution:
From the graph we can see that the roots are −4 and 2. The graph of the function lies above the x-axis ($f(x)>0$) in between these roots.
Because of the strict inequality, the solution set is shaded with an open dot on each of the boundaries. This indicates that these critical numbers are not actually included in the solution set. This solution set can be expressed two ways,
${x|−4
In this textbook, we will continue to present answers in interval notation.
Answer: $(−4,2)$
Try this! Given the graph of $f$ determine the solutions to $f(x)<0$:
Answer: $(−∞,−4)∪(2,∞)$
Next we outline a technique used to solve quadratic inequalities without graphing the parabola. To do this we make use of a sign chartA model of a function using a number line and signs (+ or −) to indicate regions in the domain where the function is positive or negative. which models a function using a number line that represents the x-axis and signs (+ or −) to indicate where the function is positive or negative. For example,
The plus signs indicate that the function is positive on the region. The negative signs indicate that the function is negative on the region. The boundaries are the critical numbers, −2 and 3 in this case. Sign charts are useful when a detailed picture of the graph is not needed and are used extensively in higher level mathematics. The steps for solving a quadratic inequality with one variable are outlined in the following example.
### Example 3
Solve: $−x2+6x+7≥0.$
Solution:
It is important to note that this quadratic inequality is in standard form, with zero on one side of the inequality.
Step 1: Determine the critical numbers. For a quadratic inequality in standard form, the critical numbers are the roots. Therefore, set the function equal to zero and solve.
$−x2+6x+7=0−(x2−6x−7)=0−(x+1)(x−7)=0x+1=0orx−7=0x=−1x=7$
The critical numbers are −1 and 7.
Step 2: Create a sign chart. Since the critical numbers bound the regions where the function is positive or negative, we need only test a single value in each region. In this case the critical numbers partition the number line into three regions and we choose test values $x=−3$, $x=0$, and $x=10.$
Test values may vary. In fact, we need only determine the sign (+ or −) of the result when evaluating $f(x)=−x2+6x+7=−(x+1)(x−7).$ Here we evaluate using the factored form.
$f(−3)=−(−3+1)(−3−7)=−(−2)(−10)=−Negativef(0)=−(0+1)(0−7)=−(1)(−7)=+Positivef(10)=−(10+1)(10−7)=−(11)(3)=−Negative$
Since the result of evaluating for −3 was negative, we place negative signs above the first region. The result of evaluating for 0 was positive, so we place positive signs above the middle region. Finally, the result of evaluating for 10 was negative, so we place negative signs above the last region, and the sign chart is complete.
Step 3: Use the sign chart to answer the question. In this case, we are asked to determine where $f(x)≥0$, or where the function is positive or zero. From the sign chart we see this occurs when x-values are inclusively between −1 and 7.
Using interval notation, the shaded region is expressed as $[−1,7].$ The graph is not required; however, for the sake of completeness it is provided below.
Indeed the function is greater than or equal to zero, above or on the x-axis, for x-values in the specified interval.
Answer: $[−1,7]$
### Example 4
Solve: $2x2−7x+3>0.$
Solution:
Begin by finding the critical numbers, in this case, the roots of $f(x)=2x2−7x+3.$
$2x2−7x+3=0(2x−1)(x−3)=02x−1=0orx−3=02x=1x=3x=12$
The critical numbers are $12$ and 3. Because of the strict inequality > we will use open dots.
Next choose a test value in each region and determine the sign after evaluating $f(x)=2x2−7x+3=(2x−1)(x−3).$ Here we choose test values −1, 2, and 5.
$f(−1)=[2(−1)−1](−1−3)=(−)(−)=+f(2)=[2(2)−1](2−3)=(+)(−)=−f(5)=[2(5)−1](5−3)=(+)(+)=+$
And we can complete the sign chart.
The question asks us to find the x-values that produce positive results (greater than zero). Therefore, shade in the regions with a + over them. This is the solution set.
Answer: $(−∞,12)∪(3,∞)$
Sometimes the quadratic function does not factor. In this case we can make use of the quadratic formula.
### Example 5
Solve: $x2−2x−11≤0.$
Solution:
Find the critical numbers.
$x2−2x−11=0$
Identify a, b, and c for use in the quadratic formula. Here $a=1$, $b=−2$, and $c=−11.$ Substitute the appropriate values into the quadratic formula and then simplify.
$x=−b±b2−4ac2a=−(−2)±(−2)2−4(1)(−11)2(1)=2±482=2±432=1±23$
Therefore the critical numbers are $1−23≈−2.5$ and $1+23≈4.5.$ Use a closed dot on the number to indicate that these values will be included in the solution set.
Here we will use test values −5, 0, and 7.
$f(−5)=(−5)2−2(−5)−11=25+10−11=+f(0)=(0)2−2(0)−11=0+0−11=−f(7)=(7)2−2(7)−11=49−14−11=+$
After completing the sign chart shade in the values where the function is negative as indicated by the question ($f(x)≤0$).
Answer: $[1−23,1+23]$
Try this! Solve: $9−x2>0.$
Answer: $(−3,3)$
It may be the case that there are no critical numbers.
### Example 6
Solve: $x2−2x+3>0.$
Solution:
To find the critical numbers solve,
$x2−2x+3=0$
Substitute $a=1$, $b=−2$, and $c=3$ into the quadratic formula and then simplify.
$x=−b±b2−4ac2a=−(−2)±(−2)2−4(1)(3)2(1)=2±−82=2±2i22=1+i2$
Because the solutions are not real, we conclude there are no real roots; hence there are no critical numbers. When this is the case, the graph has no x-intercepts and is completely above or below the x-axis. We can test any value to create a sign chart. Here we choose $x=0.$
$f(0)=(0)2−2(0)+3=+$
Because the test value produced a positive result the sign chart looks as follows:
We are looking for the values where $f(x)>0$; the sign chart implies that any real number for x will satisfy this condition.
Answer: $(−∞,∞)$
The function in the previous example is graphed below.
We can see that it has no x-intercepts and is always above the x-axis (positive). If the question was to solve $x2−2x+3<0$, then the answer would have been no solution. The function is never negative.
Try this! Solve: $9x2−12x+4≤0.$
Answer: One solution, $23.$
### Example 7
Find the domain: $f(x)=x2−4.$
Solution:
Recall that the argument of a square root function must be nonnegative. Therefore, the domain consists of all real numbers for x such that $x2−4$ is greater than or equal to zero.
$x2−4≥0$
It should be clear that $x2−4=0$ has two solutions $x=±2$; these are the critical values. Choose test values in each interval and evaluate $f(x)=x2−4.$
$f(−3)=(−3)2−4=9−4=+f(0)=(0)2−4=0−4=−f(3)=(3)2−4=9−4=+$
Shade in the x-values that produce positive results.
Answer: Domain: $( −∞,−2 ]∪[ 2,∞ )$
### Key Takeaways
• Quadratic inequalities can have infinitely many solutions, one solution or no solution.
• We can solve quadratic inequalities graphically by first rewriting the inequality in standard form, with zero on one side. Graph the quadratic function and determine where it is above or below the x-axis. If the inequality involves “less than,” then determine the x-values where the function is below the x-axis. If the inequality involves “greater than,” then determine the x-values where the function is above the x-axis.
• We can streamline the process of solving quadratic inequalities by making use of a sign chart. A sign chart gives us a visual reference that indicates where the function is above the x-axis using positive signs or below the x-axis using negative signs. Shade in the appropriate x-values depending on the original inequality.
• To make a sign chart, use the function and test values in each region bounded by the roots. We are only concerned if the function is positive or negative and thus a complete calculation is not necessary.
### Part A: Solutions to Quadratic Inequalities
Determine whether or not the given value is a solution.
1. $x2−x+1<0$; $x=−1$
2. $x2+x−1>0$; $x=−2$
3. $4x2−12x+9≤0$; $x=32$
4. $5x2−8x−4<0$; $x=−25$
5. $3x2−x−2≥0$; $x=0$
6. $4x2−x+3≤0$; $x=−1$
7. $2−4x−x2<0$; $x=12$
8. $5−2x−x2>0$; $x=0$
9. $−x2−x−9<0$; $x=−3$
10. $−x2+x−6≥0$; $x=6$
Given the graph of f determine the solution set.
1. $f(x)≤0$;
2. $f(x)≥0$;
3. $f(x)≥0$;
4. $f(x)≤0$;
5. $f(x)>0$;
6. $f(x)<0$;
7. $f(x)>0$;
8. $f(x)<0$;
9. $f(x)≥0$;
10. $f(x)<0$;
Use the transformations to graph the following and then determine the solution set.
1. $x2−1>0$
2. $x2+2>0$
3. $(x−1)2>0$
4. $(x+2)2≤0$
5. $(x+2)2−1≤0$
6. $(x+3)2−4>0$
7. $−x2+4≥0$
8. $−(x+2)2>0$
9. $−(x+3)2+1<0$
10. $−(x−4)2+9>0$
### Part B: Solving Quadratic Inequalities
Use a sign chart to solve and graph the solution set. Present answers using interval notation.
1. $x2−x−12>0$
2. $x2−10x+16>0$
3. $x2+2x−24<0$
4. $x2+15x+54<0$
5. $x2−23x−24≤0$
6. $x2−12x+20≤0$
7. $2x2−11x−6≥0$
8. $3x2+17x−6≥0$
9. $8x2−18x−5<0$
10. $10x2+17x+6>0$
11. $9x2+30x+25≤0$
12. $16x2−40x+25≤0$
13. $4x2−4x+1>0$
14. $9x2+12x+4>0$
15. $−x2−x+30≥0$
16. $−x2−6x+27≤0$
17. $x2−64<0$
18. $x2−81≥0$
19. $4x2−9≥0$
20. $16x2−25<0$
21. $25−4x2≥0$
22. $1−49x2<0$
23. $x2−8>0$
24. $x2−75≤0$
25. $2x2+1>0$
26. $4x2+3<0$
27. $x−x2>0$
28. $3x−x2≤0$
29. $x2−x+1<0$
30. $x2+x−1>0$
31. $4x2−12x+9≤0$
32. $5x2−8x−4<0$
33. $3x2−x−2≥0$
34. $4x2−x+3≤0$
35. $2−4x−x2<0$
36. $5−2x−x2>0$
37. $−x2−x−9<0$
38. $−x2+x−6≥0$
39. $−2x2+4x−1≥0$
40. $−3x2−x+1≤0$
Find the domain of the function.
1. $f(x)=x2−25$
2. $f(x)=x2+3x$
3. $g(x)=3x2−x−2$
4. $g(x)=12x2−9x−3$
5. $h(x)=16−x2$
6. $h(x)=3−2x−x2$
7. $f(x)=x2+10$
8. $f(x)=9+x2$
9. A robotics manufacturing company has determined that its weekly profit in thousands of dollars is modeled by $P(n)=−n2+30n−200$ where n represents the number of units it produces and sells. How many units must the company produce and sell to maintain profitability. (Hint: Profitability occurs when profit is greater than zero.)
10. The height in feet of a projectile shot straight into the air is given by $h(t)=−16t2+400t$ where t represents the time in seconds after it is fired. In what time intervals is the projectile under 1,000 feet? Round to the nearest tenth of a second.
### Part C: Discussion Board
1. Does the sign chart for any given quadratic function always alternate? Explain and illustrate your answer with some examples.
2. Research and discuss other methods for solving a quadratic inequality.
3. Explain the difference between a quadratic equation and a quadratic inequality. How can we identify and solve each? What is the geometric interpretation of each?
1. No
2. Yes
3. No
4. Yes
5. Yes
6. $[−4,2]$
7. $[−1,3]$
8. $(−∞,∞)$
9. $(−∞,4)∪(8,∞)$
10. ${−10}$
11. $(−∞,−1)∪(1,∞)$
12. $(−∞,1)∪(1,∞)$
13. $[−3,−1]$
14. $[−2,2]$
15. $(−∞,−4)∪(−2,∞)$
1. $(−∞,−3)∪(4,∞)$
2. $(−6,4)$
3. $[−1,24]$
4. $(−∞,−12]∪[6,∞)$
5. $(−14,52)$
6. $−53$
7. $(−∞,12)∪(12,∞)$
8. $[−6,5]$
9. $(−8,8)$
10. $(−∞,−32]∪[32,∞)$
11. $[−52,52]$
12. $(−∞,−22)∪(22,∞)$
13. $(−∞,∞)$
14. $(0,1)$
15. Ø
16. $32$
17. $(−∞,−23]∪[1,∞)$
18. $(−∞,−2−6)∪(−2+6,∞)$
19. $(−∞,∞)$
20. $[2−22,2+22]$
21. $(−∞,−5]∪[5,∞)$
22. $(−∞,−23]∪[1,∞)$
23. $[−4,4]$
24. $(−∞,∞)$
25. The company must produce and sell more than 10 units and fewer than 20 units each week.
## 6.6 Solving Polynomial and Rational Inequalities
### Learning Objectives
1. Solve polynomial inequalities.
2. Solve rational inequalities.
## Solving Polynomial Inequalities
A polynomial inequalityA mathematical statement that relates a polynomial expression as either less than or greater than another. is a mathematical statement that relates a polynomial expression as either less than or greater than another. We can use sign charts to solve polynomial inequalities with one variable.
### Example 1
Solve: $x(x+3)2(x−4)<0.$
Solution:
Begin by finding the critical numbers. For a polynomial inequality in standard form, with zero on one side, the critical numbers are the roots. Because $f(x)=x(x+3)2(x−4)$ is given in its factored form the roots are apparent. Here the roots are: 0, −3, and 4. Because of the strict inequality, plot them using open dots on a number line.
In this case, the critical numbers partition the number line into four regions. Test values in each region to determine if f is positive or negative. Here we choose test values −5, −1, 2, and 6. Remember that we are only concerned with the sign (+ or −) of the result.
$f(−5)=(−5)(−5+3)2(−5−4)=(−)(−)2(−)=+Positivef(−1)=(−1)(−1+3)2(−1−4)=(−)(+)2(−)=+Positivef(2)=(2)(2+3)2(2−4)=(+)(+)2(−)=−Negativef(6)=(6)(6+3)2(6−4)=(+)(+)2(+)=+Positive$
After testing values we can complete a sign chart.
The question asks us to find the values where $f(x)<0$, or where the function is negative. From the sign chart we can see that the function is negative for x-values in between 0 and 4.
We can express this solution set in two ways:
${x|0
In this textbook we will continue to present solution sets using interval notation.
Answer: $(0,4)$
Graphing polynomials such as the one in the previous example is beyond the scope of this textbook. However, the graph of this function is provided below. Compare the graph to its corresponding sign chart.
Certainly it may not be the case that the polynomial is factored nor that it has zero on one side of the inequality. To model a function using a sign chart, all of the terms should be on one side and zero on the other. The general steps for solving a polynomial inequality are listed in the following example.
### Example 2
Solve: $2x4>3x3+9x2.$
Solution:
Step 1: Obtain zero on one side of the inequality. In this case, subtract to obtain a polynomial on the left side in standard from.
$2x4>3x3+9x22x4−3x3−9x2>0$
Step 2: Find the critical numbers. Here we can find the zeros by factoring.
$2x4−3x3−9x2=0x2(2x2−3x−9)=0x2(2x+3)(x−3)=0$
There are three solutions, hence, three critical numbers $−32$, 0, and 3.The strict inequality indicates that we should use open dots.
Step 3: Create a sign chart. In this case use $f(x)=x2(2x+3)(x−3)$ and test values −2, −1, 1, and 4 to determine the sign of the function in each interval.
$f(−2)=(−2)2[2(−2)+3](−2−3)=(−)2(−)(−)=+f(−1)=(−1)2[2(−1)+3](−1−3)=(−)2(+)(−)=−f(1)=(1)2[2(1)+3](1−3)=(+)2(+)(−)=−f(4)=(4)2[2(4)+3](4−3)=(+)2(+)(+)=+$
With this information we can complete the sign chart.
Step 4: Use the sign chart to answer the question. Here the solution consists of all values for which $f(x)>0.$ Shade in the values that produce positive results and then express this set in interval notation.
Answer: $(−∞,−32)∪(3,∞)$
### Example 3
Solve: $x3+x2≤4(x+1).$
Solution:
Begin by rewriting the inequality in standard form, with zero on one side.
$x3+x2≤4(x+1)x3+x2≤4x+4x3+x2−4x−4≤0$
Next find the critical numbers of $f(x)=x3+x2−4x−4$:
$x3+x2−4x−4=0Factor by grouping.x2(x+1)−4(x+1)=0(x+1)(x2−4)=0(x+1)(x+2)(x−2)=0$
The critical numbers are −2, −1, and 2. Because of the inclusive inequality ($≤$) we will plot them using closed dots.
Use test values −3, $−32$, 0, and 3 to create a sign chart.
$f(−3)=(−3+1)(−3+2)(−3−2)=(−)(−)(−)=−f(−32)=(−32+1)(−32+2)(−32−2)=(−)(+)(−)=+f(0)=(0+1)(0+2)(0−2)=(+)(+)(−)=−f(3)=(3+1)(3+2)(3−2)=(+)(+)(+)=+$
And we have
Use the sign chart to shade in the values that have negative results ($f(x)≤0$).
Answer: $(−∞,−2]∪[−1,2]$
Try this! Solve: $−3x4+12x3−9x2>0.$
Answer: $(1,3)$
## Solving Rational Inequalities
A rational inequalityA mathematical statement that relates a rational expression as either less than or greater than another. is a mathematical statement that relates a rational expression as either less than or greater than another. Because rational functions have restrictions to the domain we must take care when solving rational inequalities. In addition to the zeros, we will include the restrictions to the domain of the function in the set of critical numbers.
### Example 4
Solve: $(x−4)(x+2)(x−1)≥0.$
Solution:
The zeros of a rational function occur when the numerator is zero and the values that produce zero in the denominator are the restrictions. In this case,
$Roots (Numerator)Restriction (Denominator)x−4=0orx+2=0x=4x=−2x−1=0x=1$
Therefore the critical numbers are −2, 1, and 4. Because of the inclusive inequality ($≥$) use a closed dot for the roots {−2, 4} and always use an open dot for restrictions {1}. Restrictions are never included in the solution set.
Use test values $x=−4,0,2,6.$
$f(−4)=(−4−4)(−4+2)(−4−1)=(−)(−)(−)=−f(0)=(0−4)(0+2)(0−1)=(−)(+)(−)=+f(2)=(2−4)(2+2)(2−1)=(−)(+)(+)=−f(6)=(6−4)(6+2)(6−1)=(+)(+)(+)=+$
And then complete the sign chart.
The question asks us to find the values for which $f(x)≥0$, in other words, positive or zero. Shade in the appropriate regions and present the solution set in interval notation.
Answer: $[−2,1)∪[4,∞)$
Graphing such rational functions like the one in the previous example is beyond the scope of this textbook. However, the graph of this function is provided below. Compare the graph to its corresponding sign chart.
Notice that the restriction $x=1$ corresponds to a vertical asymptote which bounds regions where the function changes from positive to negative. While not included in the solution set, the restriction is a critical number. Before creating a sign chart we must ensure the inequality has a zero on one side. The general steps for solving a rational inequality are outlined in the following example.
### Example 5
Solve: $7x+3<2.$
Solution:
Step 1: Begin by obtaining zero on the right side.
$7x+3<27x+3−2<0$
Step 2: Determine the critical numbers. The critical numbers are the zeros and restrictions. Begin by simplifying to a single algebraic fraction.
$7x+3−21<07−2(x+3)x+3<07−2x−6x+3<0−2x+1x+3<0$
Next find the critical numbers. Set the numerator and denominator equal to zero and solve.
$RootRestriction−2x+1=0−2x=−1x=12x+3=0x=−3$
In this case, the strict inequality indicates that we should use an open dot for the root.
Step 3: Create a sign chart. Choose test values −4, 0, and 1.
$f(−4)=−2(−4)+1−4+3=+−=−f(0)=−2(0)+10+3=++=+f(1)=−2(1)+11+3=−+=−$
And we have
Step 4: Use the sign chart to answer the question. In this example we are looking for the values for which the function is negative, $f(x)<0.$ Shade the appropriate values and then present your answer using interval notation.
Answer: $(−∞,−3)∪(12,∞)$
### Example 6
Solve: $1x2−4≤12−x$
Solution:
Begin by obtaining zero on the right side.
$1x2−4≤12−x1x2−4−12−x≤0$
Next simplify the left side to a single algebraic fraction.
$1x2−4−12−x≤01(x+2)(x−2)−1−(x−2)≤01(x+2)(x−2)+1(x+2)(x−2)(x+2)≤01+x+2(x+2)(x−2)≤0x+3(x+2)(x−2)≤0$
The critical numbers are −3, −2, and 2. Note that ±2 are restrictions and thus we will use open dots when plotting them on a number line. Because of the inclusive inequality we will use a closed dot at the root −3.
Choose test values −4, $−212=−52$, 0, and 3.
$f(−4)=−4+3(−4+2)(−4−2)=(−)(−)(−)=−f(−52)=−52+3(−52+2)(−52−2)=(+)(−)(−)=+f(0)=0+3(0+2)(0−2)=(+)(+)(−)=−f(3)=3+3(3+2)(3−2)=(+)(+)(+)=+$
Construct a sign chart.
Answer the question; in this case, find x where $f(x)≤0.$
Answer: $(−∞,−3]∪(−2,2)$
Try this! Solve: $2x22x2+7x−4≥xx+4.$
Answer: $(−4,0]∪(12,∞)$
### Key Takeaways
• When a polynomial inequality is in standard form, with zero on one side, the roots of the polynomial are the critical numbers. Create a sign chart that models the function and then use it to answer the question.
• When a rational inequality is written as a single algebraic fraction, with zero on one side, the roots as well as the restrictions are the critical numbers. The values that produce zero in the numerator are the roots, and the values that produce zero in the denominator are the restrictions. Always use open dots for restrictions, regardless of the given inequality, because restrictions are not part of the domain. Create a sign chart that models the function and then use it to answer the question.
### Part A: Solving Polynomial Inequalities
Solve. Present answers using interval notation.
1. $x(x+1)(x−3)>0$
2. $x(x−1)(x+4)<0$
3. $(x+2)(x−5)2<0$
4. $(x−4)(x+1)2≥0$
5. $(2x−1)(x+3)(x+2)≤0$
6. $(3x+2)(x−4)(x−5)≥0$
7. $x(x+2)(x−5)2<0$
8. $x(2x−5)(x−1)2>0$
9. $x(4x+3)(x−1)2≥0$
10. $(x−1)(x+1)(x−4)2<0$
11. $(x+5)(x−10)(x−5)2≥0$
12. $(3x−1)(x−2)(x+2)2≤0$
13. $−4x(4x+9)(x−8)2>0$
14. $−x(x−10)(x+7)2>0$
Solve.
1. $x3+2x2−24x≥0$
2. $x3−3x2−18x≤0$
3. $4x3−22x2−12x<0$
4. $9x3+30x2−24x>0$
5. $12x4+44x3>80x2$
6. $6x4+12x3<48x2$
7. $x(x2+25)<10x2$
8. $x3>12x(x−3)$
9. $x4−5x2+4≤0$
10. $x4−13x2+36≥0$
11. $x4>3x2+4$
12. $4x4<3−11x2$
13. $9x3−3x2−81x+27≤0$
14. $2x3+x2−50x−25≥0$
15. $x3−3x2+9x−27>0$
16. $3x3+5x2+12x+20<0$
### Part B: Solving Rational Inequalities
Solve.
1. $xx−3>0$
2. $x−5x>0$
3. $(x−3)(x+1)x<0$
4. $(x+5)(x+4)(x−2)<0$
5. $(2x+1)(x+5)(x−3)(x−5)≤0$
6. $(3x−1)(x+6)(x−1)(x+9)≥0$
7. $(x−8)(x+8)−2x(x−2)≥0$
8. $(2x+7)(x+4)x(x+5)≤0$
9. $x2(2x+3)(2x−3)≤0$
10. $(x−4)2−x(x+1)>0$
11. $−5x(x−2)2(x+5)(x−6)≥0$
12. $(3x−4)(x+5)x(x−4)2≥0$
13. $1(x−5)4>0$
14. $1(x−5)4<0$
Solve.
1. $x2−11x−12x+4<0$
2. $x2−10x+24x−2>0$
3. $x2+x−302x+1≥0$
4. $2x2+x−3x−3≤0$
5. $3x2−4x+1x2−9≤0$
6. $x2−162x2−3x−2≥0$
7. $x2−12x+20x2−10x+25>0$
8. $x2+15x+36x2−8x+16<0$
9. $8x2−2x−12x2−3x−14≤0$
10. $4x2−4x−15x2+4x−5≥0$
11. $1x+5+5x−1>0$
12. $5x+4−1x−4<0$
13. $1x+7>1$
14. $1x−1<−5$
15. $x≥30x−1$
16. $x≤1−2xx−2$
17. $1x−1≤2x$
18. $3x+1>−1x$
19. $4x−3≤1x+3$
20. $2x−9x+49x−8<0$
21. $x2(x+2)−1x+2≤12x(x+2)$
22. $12x+1−92x−1>2$
23. $3xx2−4−2x−2<<$
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# Generalise Number Patterns
## Generalising Linear Number Patterns
We can use algebra to create a general rule for a number sequence. For linear sequences, this is very easy. A linear sequence is a number sequence that goes up by the same amount every time. So for example 1, 4, 7, 10 (+3). The advantage of finding a rule for a number sequence is that we can determine any number in that sequence. This is how we do it. Look at the first few numbers of the sequence and see by how much it goes up or down each time. Then use this value and put it in front of n. In the example, this is 3. “n” stands for the term in the sequence (eg: 2 is the first number in the sequence and 5 the second in the example). Now calculate any number in the sequence using xn (3n in this example). There might be a difference between the actual number and the number that you get using your formula. Modify the formula accordingly (in the example, we had to take away 1). Now you can find any number in the sequence. The 100th number in this sequence would be 299.
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Example 1 In this number sequence, the numbers increase by two each time. Consequently, we must write down 2n. However, we need to add 2 to make the formula work.
Example 2 In this number sequence, the numbers increase by four each time. Consequently, we must write down 4n. However, we need to subtract 3 to make the formula work.
## Generalising Quadratic Number Patterns
Generalising quadratic number patterns can be a bit more challenging as generalising linear ones. Quadratic number patterns do not go up or down by the same amount every time. An example of a quadratic number sequence is: 3, 6, 12 & 18. However, it is still possible to generalise these types of sequences by finding the second difference as shown in the example. First find the first difference as we did with the linear sequences and then solve the second difference (the difference of the differences). Here is a rule for the second differences: If the 2nd difference is 2, your formula starts with n² If the 2nd difference is 4, your formula starts with 2n² If the 2nd difference is 6, your formula starts with 3n² In our example, the second difference is two, so our formula starts with n². Then we need to modify the formula to make it work. In this case, we had to add 3. These formulas can be more complicated to modify (eg: n² + 3n -1 or n² +2n). You need to use your maths skills and common sense to change the formula so that works. These number patterns can get very complicated. You can even have n³ depending on the sequence. However, by knowing how to generalise linear and quadratic number sequences you are at a pretty decent level.
## Presentation
Example 1 In this number sequence, the second difference is +4. Consequently, the first part of the formula is 2n². We had to add 2n in this example in order for the formula to work.
Example 2 In this number sequence, the second difference is +6. Consequently, the first part of the formula is 3n². We had to subtract 1 in this example in order for the formula to work.
## More Algebra
You should try to do some linear equations now as this is another basic algebra skill that you should be able to master. If you want to have a look at all our topics, make sure to visit our library. Please share this page if you like it or found it helpful!
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Chapter 11.0: Learning Outcomes Students will be introduced to number patterns! Students will learn how to generalise linear and quadratic number patterns!
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Unit 3: Subtraction
# Topic D: Subtraction with Borrowing
When you subtract, the digit that you are taking away may be larger than the top digit in that same column. You must borrow from the column on the left. First, let’s look at two examples using the place value shapes.
Example A: 243 − 128 =
Step 1: 3 ones − 8 ones cannot be done
Borrow one ten and rename it as ten ones. Add the ten ones to the three ones.
Now you can subtract: 13 ones − 8 ones = 5 ones
Step 2: Subtract the tens. 3 tens − 2 tens = 1 ten
Step 3: Subtract the hundreds. 2 hundreds − 1 hundred = 1 hundred
Here is the question using numerals.
Example B: 350 − 124 =
Step 1: 0 ones − 4 ones cannot be done
Borrow one ten and rename it as ten ones.
10 ones − 4 ones = 6 ones
Step 2: 4 tens − 2 tens = 2 tens
Step 3: 3 hundreds − 1 hundred = 2 hundreds
This is how the question looks using numerals.
Exercise One
You may need to borrow 1 ten and rename it as 10 ones to do these subtractions. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}&53\\-&16\\ \hline &37 \end{array}$
2. $\begin{array}[t]{rr}&82\\-&45\\ \hline&37 \end{array}$
3. $\begin{array}[t]{rr}&37\\-&9\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}&28\\-&4\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}&63\\-&7\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}&54\\-&5\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}&25\\-&7\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}&84\\-&6\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}&45\\-&15\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}&40\\-&38\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}&45\\-&20\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}&70\\-&21\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}&645\\-&26\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}&258\\-&14\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}&786\\-&47\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}&895\\-&29\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}&747\\-&109\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}&642\\-&420\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}&438\\-&215\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}&953\\-&838\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}&532\\-&314\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}&795\\-&238\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}&956\\-&348\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}&574\\-&218\\ \hline \\ \end{array}$
1. 37
2. 37
3. 28
4. 24
5. 56
6. 49
7. 18
8. 78
9. 30
10. 2
11. 25
12. 49
13. 619
14. 244
15. 739
16. 866
17. 638
18. 222
19. 223
20. 115
21. 218
22. 557
23. 608
24. 356
To check your subtraction, add the answer (the difference) to the number you took away. If your subtracting was correct, the result of the adding will equal the number you started with in the subtraction question.
Example C: 726 − 317
Difference: $\begin{array}[t]{r}726\\-317\\ \hline 409\end{array}$
To check, add 409 to 317
$\begin{array}[t]{r} 409\\+317\\ \hline 726\end{array}$
Exercise Two
You may need to borrow 1 ten and rename it as 10 ones to do these subtractions. Use the method for checking your answer beside each question. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}&42\\-&5\\ \hline &37 \end{array}$ check: $\begin{array}[t]{rr}&37\\+&5\\ \hline &42 \end{array}$
2. $\begin{array}[t]{rr}&83\\-&6\\ \hline \\ \end{array}$ check:
3. $\begin{array}[t]{rr}&91\\-&7\\ \hline \\ \end{array}$ check:
4. $\begin{array}[t]{rr}&70\\-&4\\ \hline \\ \end{array}$ check:
5. $\begin{array}[t]{rr}&64\\-&37\\ \hline \\ \end{array}$ check:
6. $\begin{array}[t]{rr}&32\\-&16\\ \hline \\ \end{array}$ check:
7. $\begin{array}[t]{rr}&65\\-&16\\ \hline \\ \end{array}$ check:
8. $\begin{array}[t]{rr}&98\\-&39\\ \hline \\ \end{array}$ check:
9. $\begin{array}[t]{rr}&775\\-&49\\ \hline \\ \end{array}$ check:
10. $\begin{array}[t]{rr}&974\\-&26\\ \hline \\ \end{array}$ check:
11. $\begin{array}[t]{rr}&483\\-&75\\ \hline \\ \end{array}$ check:
12. $\begin{array}[t]{rr}&896\\-&57\\ \hline \\ \end{array}$ check:
13. $\begin{array}[t]{rr}&785\\-&627\\ \hline \\ \end{array}$ check:
14. $\begin{array}[t]{rr}&961\\-&543\\ \hline \\ \end{array}$ check:
15. $\begin{array}[t]{rr}&941\\-&319\\ \hline \\ \end{array}$ check:
16. $\begin{array}[t]{rr}&850\\-&434\\ \hline \\ \end{array}$ check:
1. 37
2. 77
3. 84
4. 66
5. 27
6. 16
7. 49
8. 59
9. 726
10. 948
11. 408
12. 839
13. 158
14. 418
15. 622
16. 416
Use this same method of borrowing when you subtract the hundreds, thousands, ten thousands, and so on. Look at the place value shapes as you work through these examples.
Example D: 225 − 162
Step 1: 5 ones − 2 ones = 3 ones
Step 2: 2 tens − 6 tens (can’t be done)
Borrow one hundred and rename it as 10 tens which you add onto the 2 tens.
12 tens − 6 tens = 6 tens
Step 3: 1 hundred – 1 hundred = 0 hundreds
Note: The 0 in the hundreds is not needed in the answer (063) because it is the first digit and does not have to hold the place.
Example E: 331 − 145
Step 1: 1 one − 5 ones (can’t be done)
Borrow 1 ten and rename it as 10 ones which you add onto the 1 one.
11 ones − 5 ones = 6 ones
Step 2: 2 tens − 4 tens (can’t be done)
Borrow one hundred and rename it as 10 tens which you add onto the 2 tens.
Step 3: 2 hundreds − 1 hundred = 1 hundred
Exercise Three
Subtract the following. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}& 286\\-&138\\ \hline& 148 \end{array}$
2. $\begin{array}[t]{rr}& 481\\-&225 \\ \hline& 256 \end{array}$
3. $\begin{array}[t]{rr}& 390\\-&135\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 390\\-&135\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}& 734\\-&582\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}& 281\\-&175\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}& 925\\-&68\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}& 260\\-&154\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}& 379\\-&235\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}& 532\\-&290\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}& 82\\-&79\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}& 262\\-&39\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}& 427\\-&183\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}& 452\\-&173\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}& 692\\-&473\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}& 634\\-&273\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}& 465\\-&374\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}& 785\\-&147\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}& 937\\-&258\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}& 946\\-&463\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}& 734\\-&208\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}& 563\\-&154\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}& 782\\-&254\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}& 621\\-&442\\ \hline \\ \end{array}$
1. 148
2. 256
3. 255
4. 152
5. 152
6. 106
7. 857
8. 106
9. 144
10. 242
11. 3
12. 223
13. 244
14. 279
15. 219
16. 361
17. 91
18. 638
19. 679
20. 483
21. 526
22. 409
23. 528
24. 179
Exercise Four
Subtract the following. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}&776\\-&382\\ \hline \\ \end{array}$
2. $\begin{array}[t]{rr}&426\\-&327\\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}&957\\-&234\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 845\\-&416\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}&967\\-&173\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}&406\\-&257\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}&857\\-&143\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}&757\\-&129\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}&567\\-&182\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}&952\\-&278\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}&863\\-&389\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}&689\\-&434\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}&754\\-&526\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}&572\\-&493\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}&714\\-&588\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}&795\\-&497\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}&390\\-&256\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}&745\\-&649\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}&639\\-&484\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}&811\\-&173\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}&678\\-&290\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}&740\\-&272\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}&983\\-&876\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}&839\\-&653\\ \hline \\ \end{array}$
1. 394
2. 109
3. 723
4. 429
5. 794
6. 149
7. 714
8. 628
9. 385
10. 674
11. 474
12. 255
13. 228
14. 79
15. 126
16. 298
17. 134
18. 96
19. 155
20. 638
21. 388
22. 468
23. 107
24. 186
Now work through this example, where you must also rename one thousand as ten hundreds to do the subtraction.
Example F: 3 245 − 1 678
Step 1: Subtract the ones.
Step 2: Subtract the tens.
Step 3: Subtract the hundreds.
Step 4: Subtract the thousands and check.
Exercise Five
Find the differences. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}&4\, 295\\-& 724\\ \hline \\ \end{array}$
2. $\begin{array}[t]{rr}&8\, 281 \\-& 470\\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}&5\, 564 \\-& 644\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}&6\, 382\\-& 882\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}&8\, 513 \\-& 829\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}&3\, 527 \\-& 758\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}&3\, 154\\-& 205\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}&2\, 640 \\-& 834\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}&7\, 355 \\-& 4\, 038\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}&5\, 189 \\-& 2\, 348\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}&4\, 289 \\-& 2\, 534\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}&6\, 753\\-& 1\, 942\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}&8\, 684\\-& 2\, 916\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}&7\, 459 \\-& 3\, 927\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}&8\, 360\\-& 6\, 376\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}&9\, 418 \\-& 4\, 739\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}&75\, 762\\-& 9\, 351\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}&72\, 641\\-& 8\, 736\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}&16\, 793\\-& 7\, 325\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}&12\, 533\\-& 9\, 362\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}&72\, 209\\-& 9\, 786\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}&34\, 092 \\-& 4\, 538\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}&42\, 126 \\-& 24\, 762\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}&52\, 750 \\-& 14\, 789\\ \hline \\ \end{array}$
1. 3571
2. 7811
3. 4920
4. 5500
5. 7684
6. 2769
7. 2949
8. 1806
9. 3317
10. 2841
11. 1755
12. 4811
13. 5768
14. 3532
15. 1984
16. 4679
17. 66411
18. 63905
19. 9468
20. 3171
21. 62423
22. 29554
23. 17364
24. 37961
Exercise Six
Find the differences. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}& 4\,262 \\-& 2\, 738 \\ \hline& 1\, 524 \end{array}$
2. $\begin{array}[t]{rr}& 3 \,236 \\-& 1\, 594\\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}& 4\, 697 \\-& 3\, 268\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 8 \,321 \\-& 4\, 543\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}& 2\, 831 \\-& 289\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}& 5\, 623 \\-& 3\, 352\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}& 8 \,428 \\-& 6\, 309\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}& 9 \,629\\-& 7\, 258\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}& 5 \,230\\-& 2 \,456\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}& 3 \,682 \\-& 963\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}& 29\, 285 \\-& 18\, 357\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}& 43\, 325 \\-& 3\, 187\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}& 81\, 328 \\-& 22\, 595\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}& 58\, 234 \\-& 23\, 678\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}& 28\, 243\\-& 9 \,578\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}& 3\, 245 \\-& 1\, 678\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}& 6\, 254 \\-& 1 \,733\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}& 5\, 214 \\-& 1 \,783\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}& 23\, 244 \\-& 15\, 534\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}& 16\, 121 \\-& 12\, 768\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}& 53\, 507 \\-& 14 \,421\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}& 31\, 582 \\-& 14 \,413\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}& 71\, 629 \\-& 12 \,350\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}& 44\, 610 \\-& 13 \,071\\ \hline \\ \end{array}$
1. 1524
2. 1642
3. 1429
4. 3778
5. 2542
6. 2271
7. 2119
8. 2371
9. 2774
10. 2719
11. 10928
12. 40138
13. 58733
14. 34556
15. 18665
16. 1567
17. 4521
18. 3431
19. 7710
20. 3353
21. 39086
22. 17169
23. 59279
24. 31539
# Zeroes in Subtracting
You will have subtraction questions with a zero in the place that you want to borrow from. You have to do a double borrowing. Look carefully at the example.
Example G: 2 405 − 368 =
Step 1: 5 ones – 8 ones (can’t be done)
Borrow one ten – whoops – no tens!
Borrow one hundred and rename it as 10 tens…
Now, borrow a ten. 15 ones − 8 ones = 7 ones
Step 2: 9 tens − 6 tens = 3 tens
Step 3: 3 hundreds − 3 hundreds = 0 hundreds
Step 4: 2 thousands − no thousands = 2 thousands
Exercise Seven
Find the differences. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}& 102\\-&23\\ \hline \\ \end{array}$
2. $\begin{array}[t]{rr}& 508\\-&39\\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}& 804\\-&37\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 607\\-&48\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}& 406\\-&178\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}& 302\\-&218\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}& 203\\-&157\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}& 601\\-&296\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}& 2\, 075\\-&436\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}& 3\, 076\\-&594\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}& 4\, 037\\-&289\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}& 6\, 032\\-&764\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}& 4\, 057 \\-& 2\, 049\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}& 6\, 035 \\-& 2\, 634\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}& 9\, 025 \\-& 4\, 603\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}& 5\, 075 \\-& 2\, 364\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}& 50\, 398\\-& 4\, 247\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}& 40\, 683 \\-& 3 \,162\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}& 50\, 216 \\-& 5 \,183\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}& 60\, 831 \\-& 7\, 081\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}& 40\, 465 \\-& 21\, 528\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}& 30 \,429 \\-& 14\, 953\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}& 70\, 543 \\-& 37\, 835\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}& 80\, 106 \\-& 47\, 297\\ \hline \\ \end{array}$
1. 79
2. 469
3. 767
4. 559
5. 228
6. 84
7. 46
8. 305
9. 1639
10. 2482
11. 3748
12. 5268
13. 2008
14. 3401
15. 4422
16. 2711
17. 46151
18. 37521
19. 45033
20. 53750
21. 18937
22. 15476
23. 32708
24. 32809
Exercise Eight
Find the differences. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}& 400\\-&197\\ \hline \\ \end{array}$
2. $\begin{array}[t]{rr}& 307\\-&138\\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}& 800\\-&475\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 608\\-&439\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}& 307\\-&168\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}& 200\\-&99\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}& 400\\-&43\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}& 208\\-&126\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}& 3\, 000\\-& 2\, 678\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}& 7\, 205 \\-& 2\, 306\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}& 2\, 048 \\-& 281\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}& 6\, 005\\-& 2 \,368\\ \hline \\ \end{array}$
13. $\begin{array}[t]{rr}& 5\, 000 \\-& 3 \,468\\ \hline \\ \end{array}$
14. $\begin{array}[t]{rr}& 4 \,006 \\-& 2 \,179\\ \hline \\ \end{array}$
15. $\begin{array}[t]{rr}& 3 \,007 \\-& 1 \,930\\ \hline \\ \end{array}$
16. $\begin{array}[t]{rr}& 2 \,007 \\-& 237\\ \hline \\ \end{array}$
17. $\begin{array}[t]{rr}& 43\, 004 \\-& 2 \,873\\ \hline \\ \end{array}$
18. $\begin{array}[t]{rr}& 20\, 038 \\-& 9 \,156\\ \hline \\ \end{array}$
19. $\begin{array}[t]{rr}& 60 \,125 \\-& 8 \,421\\ \hline \\ \end{array}$
20. $\begin{array}[t]{rr}& 40 \,063 \\-& 2 \,734\\ \hline \\ \end{array}$
21. $\begin{array}[t]{rr}& 70 \,059 \\-& 38 \,423\\ \hline \\ \end{array}$
22. $\begin{array}[t]{rr}& 80 \,062 \\-& 35 \,087\\ \hline \\ \end{array}$
23. $\begin{array}[t]{rr}& 90 \,035 \\-& 68 \,746\\ \hline \\ \end{array}$
24. $\begin{array}[t]{rr}& 60 \,063 \\-& 55 \,895\\ \hline \\ \end{array}$
1. 203
2. 169
3. 325
4. 169
5. 139
6. 101
7. 357
8. 82
9. 322
10. 4899
11. 1767
12. 3637
13. 1532
14. 1827
15. 1077
16. 1770
17. 40131
18. 10882
19. 51704
20. 37329
21. 31636
22. 44975
23. 21289
24. 4168
If a subtraction question has the numbers side by side, rewrite the question in columns. Put the ones under the ones, the tens under the tens, the hundreds under the hundreds, etc.
Example H: 5 625 − 2 468
Exercise Nine
Rewrite each question in columns and find the difference. Check your work using the answer key at the end of the exercise.
1. $\begin{array}[t]{rr}& 5\, 042 \\-& 3\, 185 \\ \hline \\ \end{array}$
2. $\begin{array}[t]{rr}& 8 \,042 \\-&6 \,368 \\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}& 2 \,630 \\-&95 \\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 1 \,201 \\-&159 \\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}& 34 \,582\\-& 6\, 121 \\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}& 44 \,610\\-& 4\, 527 \\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}& 54 \,507 \\-& 13\, 421 \\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}& 7 \,050 \\-& 2 \,144 \\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}& 71 \,629 \\-& 12 \,350 \\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}& 64 \,182 \\-& 28 \,934 \\ \hline \\ \end{array}$
1. 1857
2. 1674
3. 2535
4. 1042
5. 28461
6. 40083
7. 41086
8. 4906
9. 59279
10. 35248
# Topic D: Self-Test
Mark /15 Aim 11/15
1. $\begin{array}[t]{rr}& 71\\-&32\\ \hline \\ \end{array}$
2. $\begin{array}[t]{rr}& 704\\-&325\\ \hline \\ \end{array}$
3. $\begin{array}[t]{rr}& 400\\-&208\\ \hline \\ \end{array}$
4. $\begin{array}[t]{rr}& 8\,923\\-& 3\, 061\\ \hline \\ \end{array}$
5. $\begin{array}[t]{rr}& 5\, 211\\-& 4 \,390\\ \hline \\ \end{array}$
6. $\begin{array}[t]{rr}& 8 \,204 \\-& 3 \,461\\ \hline \\ \end{array}$
7. $\begin{array}[t]{rr}& 9 \,074 \\-& 5 \,482\\ \hline \\ \end{array}$
8. $\begin{array}[t]{rr}& 8 \,092 \\-& 6 \,578\\ \hline \\ \end{array}$
9. $\begin{array}[t]{rr}& 49\, 053\\-& 8\, 954\\ \hline \\ \end{array}$
10. $\begin{array}[t]{rr}& 86 \,502 \\-& 6 \,590\\ \hline \\ \end{array}$
11. $\begin{array}[t]{rr}& 47 \,293 \\-& 26\, 349\\ \hline \\ \end{array}$
12. $\begin{array}[t]{rr}& 73 \,050 \\-& 27\, 455\\ \hline \\ \end{array}$
2. Subtract. (3 marks)
1. 5302 − 3981 =
2. 7043 − 95 =
3. 6000 − 989 =
1. 39
2. 379
3. 192
4. 5 862
5. 821
6. 4743
7. 3562
8. 1514
9. 40099
10. 79912
11. 20944
12. 45595
1. 1321
2. 6948
3. 5011
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# 1.4 Right triangle trigonometry (Page 2/12)
Page 2 / 12
## Relating angles and their functions
When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in [link] . The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.
We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.
Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.
1. If needed, draw the right triangle and label the angle provided.
2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
3. Find the required function:
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side
## Evaluating trigonometric functions of angles not in standard position
Using the triangle shown in [link] , evaluate $\mathrm{sin}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{cos}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{tan}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{sec}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{csc}\text{\hspace{0.17em}}\alpha ,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\alpha .$
Using the triangle shown in [link] , evaluate and
## Finding trigonometric functions of special angles using side lengths
We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of $\text{\hspace{0.17em}}30°,$ $60°,$ and $\text{\hspace{0.17em}}45°,$ however, remember that when dealing with right triangles, we are limited to angles between
Suppose we have a $\text{\hspace{0.17em}}30°,60°,90°\text{\hspace{0.17em}}$ triangle, which can also be described as a triangle. The sides have lengths in the relation $\text{\hspace{0.17em}}s,\sqrt{3}s,2s.\text{\hspace{0.17em}}$ The sides of a $\text{\hspace{0.17em}}45°,45°,90°$ triangle, which can also be described as a $\text{\hspace{0.17em}}\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\text{\hspace{0.17em}}$ triangle, have lengths in the relation $\text{\hspace{0.17em}}s,s,\sqrt{2}s.\text{\hspace{0.17em}}$ These relations are shown in [link] .
We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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Adding and subtracting fractions may seem tricky at first, but if we follow a few simple steps, we will have the hang of it in no time. Nevertheless, we recommend starting with 'simplifying a fraction’ if this lesson has not been read yet ;)
After this course, we will be able, among other things, to:
1. Better understand fractions and how to manipulate them.
2. Add or subtract simple fractions.
3. Transform a fraction to an equivalent one.
4. Add or subtract fractions with different denominators.
Let’s have a quick reminder of what a fraction is and how it can be represented.
Reminder
The number above, at the numerator, represents the number of shares we take.
The number below, at the denominator, represents the number of total shares.
If we multiply or divide the numerator and the denominator of a fraction by the same number: we get an equivalent fraction (equal).
We never divide by 0!
The minus sign can be moved all over the fraction: $$\frac{-a}{b}=-\frac{a}{b}=\frac{a}{-b}$$ $$\frac{-a}{-b}=\frac{a}{b}$$
The game
When fractions have the same denominators we just add or subtract the numerators and place the result over the common denominator. Then, we simplify the final fraction if needed (cf. simplifying a fraction) to keep the result as neat as possible.
Let’s see first some simple examples where denominators are the same:
Things start getting interesting when denominators are different: it happens for instance when one friend bring a pizza cut in six while we have one cut in 8. The difficulty is only about getting the same denominator.
If adding or subtracting is our aim, the bottom numbers must be the same!
When there is different denominators, there are a few different ways to change the frations to equivalent ones in order to have a common denominator. We will learn here the easiest way, then a quick trick, and finally, the traditional way.
Easiest method
For this method we only have to multiply the numerator and the denominator of each fraction by the denominator of the other; we call it cross-multiplying the fractions. Be careful though: this method may be inefficient if we are dealing with big numbers. Here is some examples:
Let's visualize our first example using pineapple pizzas to fully master the concept :)
Quick trick
This trick is the quickest way to add fractions. However, it only works in special cases: when one denominator is a multiple of the other. In such case, we only have to multiply the fraction having the smallest denominator by the multiple (good to know the multiplication tables) that reaches the other one:
Traditional method using the LCM (or LCD)
We should use this method only when we can’t use either of the other methods as it can be quite long. Unless... we get the way just by looking at the denominators thanks to some practice.
The least common multiple (or LCM) is the smallest number divisible by the two denominators. Suppose we want to add the fractions 3/6 and 12/15, we find the LCM by listing their multiples and taking the first (smallest) one they have in common:
$$6 * 1 = 6$$ $$6 * 2 = 12$$ $$6 * 3 = 18$$ $$6 * 4 = 24$$ $$6 * \color{blue}{5} = \color{green}{\textbf{30}}$$ $$15 * 1 = 15$$ $$15 * \color{blue}{2} = \color{green}{\textbf{30}}$$
So 30 is the LCM of 6 and 15!. We also call this number the Least Common Denominator (LCD), which is the smallest common denominator for both fractions.
Now we just have to multiply each fraction with the blue number to get the same denominator: $$\frac{3}{6} + \frac{12}{15} = \frac{3 * \pmb{5}}{6 * \pmb{5}} + \frac{12 * \pmb{2}}{15 * \pmb{2}} = \frac{15 + 24}{30} = \frac{39}{30} \quad = \quad \frac{13}{10}$$
Congratulations!
Always find the LCM with Euclide
To always find the LCM we first need to find the GCD using Euclid's algorithm. Do not panic, it's not complicated! We just need to know how to divide a number.
The procedure is as follows:
- Perform the Euclidean division of the greatest number (noted a) of the fraction on the smallest number (noted b) and keep the rest (noted r).
- As long as the rest is different from 0, we reiterate the division replacing a by b and b by r.
Using the GCD we can calculate the LCM using the following: $$LCM = (a * b) / GCD(a,b)$$ In our case, with 6 and 15 as denominators: $$LCM = (6 * 15) / GCD(6,15) = 90 / 3 = 30$$ Sweet... !
In brief
To add fractions there are three steps:
1. Make sure the bottom numbers (the denominators) are the same.
2. Add the top numbers (the numerators) and put that answer over the denominator.
3. Simplify the fraction if needed.
We can now add or subtract any fractions!
Here is some tools if you want to get further with mastering the vast world of fractions. Feel free to use Globo and its explanations to solve your problems.
Tools - Cheatsheets
$$\frac{a}{a}=1$$ $$\require{cancel} \frac{a \cdot b}{a \cdot c} = \frac{\cancel{a} \cdot b}{\cancel{a} \cdot c} = \frac{b}{c}$$ $$\frac{a}{b} \cdot c = \frac{a \cdot c}{b} = \frac{c}{b} \cdot a$$ $$\frac{-a}{b}=-\frac{a}{b}$$ $$\frac{1}{\frac{b}{c}}=\frac{c}{b}$$ $$\frac{a}{1}=a$$ $$\frac{-a}{-b}=\frac{a}{b}$$ $$\frac{a}{-b}=-\frac{a}{b}$$ $$\frac{a}{\frac{b}{c}}=\frac{a\cdot c}{b}$$ $$\frac{\frac{b}{c}}{a}=\frac{b}{c\cdot a}$$ $$\frac{0}{a}=0\:,\:a\ne 0$$
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# 1’s, 2’s’, 9’s, and 10’s Complements
In this tutorial, you will learn how to find the complement in any base system. We will see complements in decimal, binary, octal, and hexadecimal number systems. Also, you will learn how computers perform signed subtraction using complements along with addition.
Contents:
## R and R-1’s Complement
In Digital Electronics, we have the concept of complements to make certain operations like subtraction easier to perform. Complement in general means adding a feature to something. Any number system with radix R has mainly two forms of complements: – R’s Complement and (R-1)’s Complement. This is how they are found: –
• R’s Complement: For any number with Radix R and N digits, its R’s Complement is given by the formula: (RN – number)
• (R-1)’s Complement: – For any number with Radix R and N-1 digits, its R-1’s Complement is given by the formula: (RN – 1 – number)
In finding both the kind of complements, the subtraction is done as per the particular bases. Also, the interrelation between R and R-1’s Complement is
(R-1)’s Complement + 1 = R’s Complement
The addition is simpler to perform than subtraction. The advantage of using complements is that we can convert subtraction arithmetically into addition.
## Complements in Decimal System
In the Decimal system, we have the base as 10. Thus we have 10 and (10-1)=9’s Complements. For any digit of Decimal, its complements are given as: –
• 10’s Complement: – 10’s Complement of any digit is 10 – that digit. For a n-bit decimal number, its 10’s complement will be (10n – number).
• 9’s Complement: – 9’s Complement of any digit is (10 –1 – that digit). For a n-bit decimal number, its 10’s complement will be (10n – 1 – number).
Here is a table that shows the unique Complements.
10’s Complement 9’s Complement
0 10-0=10 0 9-0=9
1 10-1=9 1 9-1=8
2 10-2=8 2 9-2=7
3 10-3=7 3 9-3=6
4 10-4=6 4 9-4=5
5 10-5=5
As shown in the table, the 10’s complement and 9’s complement of a digit are given till 5 and 4 respectively as after that the complements repeat. For example, 4 is the 10’s complement of 6, and 6 is the 10’s complement of 4.
## Complements in Binary
In the binary number system, we have 2 as the base. Thus Binary system has 2’s complement and 1’s complement. The complements are given as: –
• 1’s Complement: – For any n-bit binary number, its 1’s complement is found by inverting all of its digits. 1 becomes 0 and 0 becomes 1.
• 2’s Complement: – For any n-bit binary number, its 2’s complement is found by adding 1 to its 1’s complement
Here is a table that shows 1’s complement of binary digits.
1’s Complement
0 1
1 0
For binary numbers, finding out the 1’s complement and 2’s complement of an n-bit number is more significant as binary has just 2 digits.
## Complements in Octal
In the octal number system, we have 8 as the base. Thus, Octal has 8’s complement and 7’s complement. For Octal digits, the complement is given as; –
• 8’s Complement: – For any digit, its 8’s Complement is given by (8 – that digit).
• 7’s Complement: – For any digit, its 7’s Complement is given as (8 – 1 – that digit)
Here is a table that shows the unique Complements in Octal System.
8’s Complement 7’s Complement
0 8-0=8 0 7-0=7
1 8-1=7 1 7-1=6
2 8-2=6 2 7-2=5
3 8-3=5 3 7-3=4
4 10-4=6 4 9-4=5
5 8-4=4
As shown in the table, the 8’s complement and 7’s complement of a digit are given till 4 and 3 respectively as after that the complements repeat. For example, 2 is the 8’s complement of 6, and 6 is the 8’s complement of 2.
In the hexadecimal number system, we have 10 as the base. Thus, Hexadecimal has 16’s complement and 15’s complement. For Hexadecimal digits, the complement is given as; –
• 16’s Complement: – For any digit, its 16’s Complement is given by (16 – that digit).
• 15s Complement: – For any digit, its 15’s Complement is given as (16 – 1 – that digit). 15 in hexadecimal is F.
Here is a table that shows the unique Complements in Octal System.
Digit 16’s Complement 15’s Complement
0 16 – 0 = 16 15 – 0 = 15 (F)
1 16 – 1 = 15 (F) 15 – 1 = 14 (E)
2 16 – 2 = 14 (E) 15 – 2 = 13 (D)
3 16 – 3 = 13 (D) 15 – 3 = 12 (C)
4 16 – 4 = 12 (C) 15 – 4 = 11 (B)
5 16 – 5 = 11 (B) 15 – 5 = 10 (A)
6 16 – 6 = 10 (A) 15 – 6 = 9
7 16 – 7 = 9 15 – 7 = 8
8 16 – 8 = 8 15 – 8 = 7
As shown in the table, the 16’s complement and 15’s complement of a digit are given till 8 and 7 respectively as after that the complements repeat. For example, 4 is the 16’s complement of 12(C) and C is the 16’s complement of 4.
## Subtraction using 10’s and 9’s Complement
For subtraction of any two given signed decimal numbers, we use 9’s complement and 10’s complement to find out the value of subtraction. The steps to do so are given as: –
• Using 10’s Complement
• For two numbers A and B, if we have to subtract B from A, we find out the 10’s complement of B by subtracting it from 10N, where N is the number of digits in B.
10’s complement of B = 10N – B
• We then add the 10’s complement of B to the number A and remove any carry which occurs and the remaining digits give the difference.
For example, (75)10 – (43)10, Here 75 is A and 43 is B
1st step: 10’s Complement of B is (102 – 43 = 57)
2nd step: A + 10’s complement of B = 75 + 57 = 132
3rd step: Remove the extra carry bit.
So, we get 32 as the answer.
• Using 9’s Complement
• For two numbers A and B, if we have to subtract B from A, we find out the 9’s complement of B by subtracting it from (10N-1), where N is the number of digits in B.
9’s complement of B = 10N – 1 – B
• We then add the (9’s complement of B + 1) to the number A.
• If any carry occurs, we add the carry bit to the remaining generated number, and the result that occurs is our required difference. This will give us the 10’s complement only but finding 9’s complement of a number is easier.
In the same example, (75)10 – (43)10, Here 75 is A and 43 is B
1st step: 9’s Complement of B is (102 – 1 – 43 = 99 – 43 = 56)
2nd step: A + 9’s complement of B + 1 = 75 + 56 = 131.
3rd step: Add the extra carry bit to the remaining result. 31 + 1 = 32
So, we get 32 as the answer.
## Subtraction Using 1’s Complement
To subtract two n-bit signed binary numbers, we can use 1’s complement method of subtraction. If we have our numbers in decimal, we can convert them into signed binary form. Here are the steps to follow to subtract two numbers A and B.
• First, the number that we have to subtract(B), we take out its 1’s Complement.
• Next, we add the 1’s complement of the number to the first number(A).
• If any carry bit is generated, we add that carry bit to the remaining digits of the result and then we get our final difference. Also, this will give us a positive difference.
• If no carry bit is generated, complement the result of step2. That will give us the result, Also, the number will be a negative difference.
For example, (0111)2 – (0110)2 , here we have A as 0111 [7 in decimal] and B as 0110 [6 in decimal]
1st step: The 1’s complement of B is 1001.
2nd step: We add 1’s complement of B to A 0111 + 1001 = 10000
3rd step: As a carry bit is generated, we add this carry bit to the remaining result
0000 + 1 =0001
Thus we have our answer as (0001) and the result obtained is +1 as we had a carry.
But if we had to subtract (0110)2 – (0111)2 , here we have A as 0110 and B as 0111.
1st step: The 1’s complement of B is 1000.
2nd step: We add 1’s complement of B to A 0110 + 1000 = 111
3rd step: As no carry bit is generated, we complement this obtained result to understand which number it is.
1110 ‘s complement —> 0001
Thus we have our answer as (1110) and we understand that the result obtained is -1 as we don’t have carry.
The disadvantage of using 1’s complement is that we need to add the carry bit.
## Subtraction Using 2’s Complement
To subtract two n-bit signed binary numbers, we can also use 2’s complement method of subtraction. Here are the steps to follow to subtract two numbers A and B.
• First, we take out the 2’s complement of the number(B) that we have to subtract.
• Next, we add the 2’s complement of that number to the first number(A).
• If any carry bit is generated, we consider the difference to be positive and we ignore the carry bit from our result.
• If no carry bit is generated, we consider the difference as our result and it to be negative. We then find again the 2’s complement of the obtained result to get the equivalent signed number.
For example, (01101)2 – (01000)2, here we have A as 01101 and B as 01000
1st step: The 2’s complement of B is its 1’s complement + 1= 10111 + 1 = 11000.
2nd step: We add 2’s complement of B to A 01101 + 11000 = 100101
3rd step: As a carry bit is generated, we ignore this carry bit in our result.
Thus we have our answer as (00101) and the result obtained is positive as we had a carry.
But if we had to subtract, (01000)2 – (01101)2, here we have A as 01000 and B as 01101
1st step: The 2’s complement of B is its 1’s complement + 1= 10010 + 1 = 10011.
2nd step: We add 2’s complement of B to A 01000 + 10011 = 11011
3rd step: As no carry bit is generated, we find out the 2’s complement of this number to understand which number it is in signed form.
11011’s 2’s complement = 00100 + 1 = 00101
Thus, we have our answer as (11011) which is -5 in signed form.
## Key Points to Remember
Here are the key points to remember in “Complements in Number Systems”.
• In any number system, we can have R’s complement and R-1’s complement, where R represents the radix of the system.
• We use complements exclusively to perform signed subtraction using addition.
• The R’s complement of an n-bit number can be found out by (RN– the number) and the R-1’s complement of an n-bit number can be found out by (RN -1 – the number).
• It is easier to find R-1’s complement of an n-bit number as we directly need to subtract each digit from Rn-1, which never needs a borrow.
• In the decimal number system, we have 10’s and 9’s complement and in the binary system, we have 2’s and 1’s complements.
• In the octal number system, we have 8’s and 7’s complements and in the hexadecimal number system, we have 16’s and 15’s complement.
If you find any mistake above, kindly email to [email protected]
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# Dividing algebraic fractions
In this post we will learn to divide two algebraic fractions using solved examples.
To understand this chapter you should have basic knowledge about factorization of polynomials.
## How to divide algebraic fractions ?
Two fractions can be easily divided by converting division sign into multiplication by taking reciprocal of divisor and then multiplying the fractions.
To divide two fractions, follow the below step;
(a) Convert division into multiplication by taking reciprocal of divisor.
(b) Now multiply numerator and denominator separately
(c) If possible, simplify the fraction by cancelling the common terms.
Generally, the above division can be expressed as;
\mathtt{\Longrightarrow \ \frac{Numerator\ 1}{Denominator\ 1} \ \div \frac{Numerator\ 2}{Denominator\ 2}}\\\ \\ \mathtt{\Longrightarrow \ \frac{Numerator\ 1}{Denominator\ 1} \times \frac{Denominator\ 2}{Numerator\ 2}}\\\ \\ \mathtt{\Longrightarrow \ \frac{Numerator\ 1\ \times Denominator\ 2}{Denominator\ 1\times Numerator\ 2}}
I hope you understand the above concept. Let us solve some examples for further clarity.
## Dividing algebraic fractions – Solved examples
Example 01
Divide the below algebraic fraction
\mathtt{\Longrightarrow \ \frac{x}{y} \ \div \frac{7}{3x}}
Solution
Take reciprocal of divisor to convert division into multiplication.
\mathtt{\Longrightarrow \ \frac{x}{y} \ \div \frac{7}{3x}}\\\ \\ \mathtt{\Longrightarrow \ \frac{x}{y} \times \frac{3x}{7}}
Now multiply the numerator and denominator separately.
\mathtt{\Longrightarrow \ \frac{x\times 3x}{y\times 7}}\\\ \\ \mathtt{\Longrightarrow \ \frac{3x^{2}}{7y}}
Example 02
Divide the algebraic fractions.
\mathtt{\Longrightarrow \ \frac{3xy}{2z} \ \div \frac{5x}{7z}}
Solution
Take reciprocal of divisor to convert division into multiplication.
\mathtt{\Longrightarrow \ \frac{3xy}{2z} \ \div \frac{5x}{7z}}\\\ \\ \mathtt{\Longrightarrow \ \frac{3xy}{2z} \times \frac{7z}{5x}}
Cancelling the common terms from numerator and denominator.
\mathtt{\Longrightarrow \ \frac{3\ \cancel{x} \ y}{2\ \cancel{z}} \times \frac{7\ \cancel{z}}{5\ \cancel{x}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{3y\times 7}{2\times 5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{21y}{10}}
Hence, the above expression is the solution.
Example 03
Divide the below algebraic fractions.
\mathtt{\Longrightarrow \ \frac{5\left( x^{2} -y^{2}\right)}{11x^{2} y} \ \div \frac{3( x+y)}{2y}}
Solution
Take reciprocal of divisor to convert division into multiplication.
\mathtt{\Longrightarrow \frac{5\left( x^{2} -y^{2}\right)}{11x^{2} y} \times \frac{2y}{3\ ( x+y)}}
Referring formula;
\mathtt{\left( a^{2} -b^{2}\right) =( a-b)( a+b)}
Using the formula in above expression, we get;
\mathtt{\Longrightarrow \ \frac{5( x-y)( x+y)}{11x^{2} y} \ \times \frac{2y}{3( x+y)}}
Cancelling out common term from numerator and denominator.
\mathtt{\Longrightarrow \ \frac{5( x-y) \ \cancel{( x+y)}}{11x^{2} \ \cancel{y}} \ \times \frac{2\ \cancel{y}}{3\ \cancel{( x+y)}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{10\ ( x-y)}{33\ x^{2}}}
Hence, the above expression is the solution.
Example 04
Divide the below algebraic fractions.
\mathtt{\Longrightarrow \ \frac{x^{2} -6x+9}{( x+5)} \ \div \frac{x^{2} -9}{x^{2} +10x+25}}
Solution
Take reciprocal of divisor to convert division into multiplication.
\mathtt{\Longrightarrow \ \frac{x^{2} -6x+9}{( x+5)} \ \times \frac{x^{2} +10x+25}{x^{2} -9}}
Now try to factorize the given polynomial into simple terms.
\mathtt{\Longrightarrow \frac{x^{2} -2.x.3+3^{2}}{( x+5)} \times \frac{x^{2} +2.5.x+5^{2}}{x^{2} -3^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{( x-3)^{2}}{( x+5)} \times \frac{( x+5)^{2}}{( x-3)( x+3)}}
Cancel the common terms from numerator and denominator.
\mathtt{\Longrightarrow \ \frac{( x-3)^{\cancel{2}}}{\cancel{( x+5)}} \times \frac{( x+5)^{\cancel{2}}}{\cancel{( x-3)}( x+3)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{( x-3)( x+5)}{( x+3)}}
Hence, the above expression is the solution.
Example 05
Divide the below algebraic fractions.
\mathtt{\Longrightarrow \ \frac{x^{2} +11x}{6xy} \div \frac{\left( x^{2} -121\right)}{3xy}}
Solution
Take reciprocal of divisor to convert division into multiplication.
\mathtt{\Longrightarrow \frac{x^{2} +11x}{6xy} \times \frac{3xy}{\left( x^{2} -121\right)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{x( x+11)}{6xy} \times \frac{3xy}{( x-11)( x+11)}}
Cancel out the common terms.
\mathtt{\Longrightarrow \ \frac{x\ \cancel{( x+11)}}{\mathbf{2} \ \cancel{6xy}} \times \frac{\cancel{3xy}}{( x-11)\cancel{( x+11)}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{x}{2( x-11)}}
Hence, the above expression is the solution.
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# Difference between revisions of "2017 AMC 8 Problems/Problem 22"
## Problem 22
In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle? $[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("A", (0,0), W); label("C", (12,0), E); label("B", (12,5), NE); label("12", (6, 0), S); label("5", (12, 2.5), E);[/asy]$
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}$
## Solution
We can reflect triangle $ABC$ on line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$
We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The area of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$
## Solution
We immediately see that $AB=13$, and we label the center of the semicircle $O$. Drawing radius $OD$ with length $x$ such that $OD$ is tangent to $AB$, we immediately see that $ODB\cong OCB$ because of HL congruence, so $BD=5$ and $DA=8$. By similar triangles $ODA$ and $BCA$, we see that $\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{D}}$.
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## NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.2
Chapter 13 Limits and Derivatives Exercise 13.2 NCERT Solutions for Class 11 Maths will help you in solving your doubts as these Class 11 Maths NCERT Solutions are prepared by Studyrankers experienced subject matter experts. These NCERT Solutions are very important for the purpose of examinations as it will help you knowing the basic concepts of the chapter.
1. Find the derivative of x 2 – 2 at x = 10
Let y = x 2 – 2
dy/dx = 2x dy/dx at x = 10 is equal to 20
2. Find the derivative of x 2 – 2 at x = 10
Find the derivative of 99x at x = 100 (By first principle)
Let f (x) = 99x. Derivative of f (x) at x = 100 is
Now, f (x) = 99x
f (100 + h)= 99(100 + h)
f (100)= 99 × 100
∴ f (100 + h) – f (100) = 99 (100 + h) – 99 × 100
= 99 [100 + h – 100] = 99 × h
3. Find the derivative of x at x = 1
Derivative of f (x) = x at x = 1
4. Find the derivative of the following functions from first principle:
(i) x 3 – 27 (ii) (x – 1)(x – 2)
(iii) 1/x2 (iv) (x + 1)/(x – 1)
5. For the function
f(x) = x100/100 + x99/99 + ……..+ x2/2 + x + 1
prove that f’(1) = 100 f’(0)
We know that d/dx (xn) = nxn-1
∴ For f(x) = x100/100 + x99/99 + ……+ x2/2 + x+ 1
f'(x) = 100x99/100 + 99. x98/99 + …..+ 2x/2 + 1
= x99 + x98 + ………….+ x + 1
Now, f’(x) = 1 + 1+ …….. to 100 term = 100
f'(0) = 1
∴ f'(1) = 100 × 1 = 100f’(0)
hence, f’(1) = 100f’(0)
6. Find the derivative of x n + axn – 1 + a 2n – 2 + ....+ a n – 1x + a n for some fixed real number a?
Let f (x) = xn + axn – 1 + a 2 x n – 2+ ........ + a n – 1x + a n
Now, d/dx xn = nxn-1 , d/dx xn-1 = (n – 1)xn-2.
etc and d/dx [ag(x)] = ag’(x), d/dx an = 0
f’(x) = nxn-1 + (n – 1)axn – 2 + (n – 2)a2xn-3 + ………..+ an-1
7. For some constants a and b, find the derivative of:
(i) (x – a)(x – b)
(ii) For some constant a and b find the derivation of (ax2 + b) 2 .
(iii) (x – a)/( x – b)
(i) Let f (x) = (x – a)(x – b)
Using product rule, we have
df(x)/dx = (x – a) d(x – b)/dx + (x-b) d(x-a)/dx
= (x – a)[d(x)/dx - d(b)/dx] + (x – b)[d(x)/dx - d(a)/dx]
= (x – a)[1 – 0] + (x – b)[1 – 0]
(ii) f(x) = (ax2 + b)2 = a2x+ 2abx2 + b2
now, d/dx x4 = 4x3 and
d/dx x2 = 2x, d/dx b2 = 0
f’(x) = a2. 4x3 + 2ab.2x + 0
= 4a2x3 + 4abx
(iii) Let f(x) = (x- a)/(x – b)
Using quotient rule, we have
= ((x – b)(d(x-a))/dx – (x – a) (d(x – b))/dx)/(x – b)2
8. Find the derivative of (xn – an)/(x – a) for some constant ‘a’.
We know d/dx (u/v) = (u’v – uv’)/v2
d/dx ((xn – an)/(x – a))
= ([d/dx (xn – an)](x – a)-(xn – an)d/dx (x – a))/(x – a)2
= (nxn-1(x – a)- (xn – an). 1)/(x – a)2
= (nxn – n.xn – 1 a – xn + an)/(x – a)2
= ((n – 1 )xn – naxn-1 + an)/(x – a)2
9. Find the derivative of
(i) 2x – 3/4
(ii) (5x3 + 3x – 1)(x – 1)
(iii) x-3(5 + 3x)
(iv) x5(3 – 6x-9)
(v) x-4(3 – 4x-5)
(vi) 2/(x + 1) - x2/(3x – 1)
(i) Let f(x) = 2x – 3/4
f'(x) = 2 (d/dx) x = 1, (d/dx)(3/4) = 0
(ii) let f(x) = (5x3 + 3x – 1)(x – 1)
(uv)’ = u’v + uv’
f'[(5x3 + 3x – 1)(x – 1)]
= [d/dx (5x3 + 3x – 1)](x – 1) + (5x3 + 3x – 1) d/dx (x – 1)
= (15x2 + 3)(x – 1) + (15x3 + 3x – 1).1
= x(15x2 + 3) – (15x2 + 3) + (5x3 + 3x – 1))
= 20x3 – 15x2 + 6x - 4
(iii) let f(x) = x-3 (5 + 3x) = 5x-3 + 3x-2
f’(x) = 5(-3)x-4 + 3(-2)x-3
= (-15/x4) – (6/x3) = (-(6x + 15))/x4 = (-3(2x + 5))/x4
(iv) Let f(x) = x5(3 – 6x-9) = 3x5 – 6x-4
f'(x) = (d(3x5 – 6x-4))/dx
= (d(3x5)/dx) – (d(6x-4)/dx)
= 3.(dx5/dx) – 6.(dx-4/dx)
= (3)(5)x4 – (6)(-4)x-5
= 15x4 + 24x-5
(v) let f(x) = x-4(3 – 4x-5)
f'(x) = x-4 . (d(3 – 4x-5)/dx) + (3 – 4x-5)(dx-4/dx)
= x-4[(d(3)/dx) – (d(4x-5)/dx)] + (3 – 4x-5)(-4)x-5
= x-4[0 – (4)(-5)x-6] + (3 – 4x-5)(-4)x-5
= x-4[20x-6]-x-5(12 – 16x-5)
= 20x-10 – 12x-5 + 16x-10 = 36x10 – 12x-5
10. Find the derivative of cos x from first principle.
f (x) = cos x By first Principle,
11. Find the derivative of the following functions:
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x + 7
(vii) 2tanx – 7 sec x
(i) Let f(x) = sinx cosx
f’(x) = u’v + uv’
= (d/dx sin x) cosx + sinx d/dx (cos x)
= cos x. cos x + sin x(-sin x)
cos2x – sin2x = cos2x
(ii)
(iii) Let f(x) = 5secx + 4cos x
f’(x) = (d(5sec x + 4cos x))/dx
= (d(5sec x)/dx) + (d(4cosx)/dx)
= 5.(d(sec x)/dx) + 4.(d(cos x)/dx)
= 5sec x tan x – 4 sin x
(iv) Let f(x) = cosec x
f'(x) = (d(cosec x))/dx = (d. 1/sinx)/dx
= (sinx (d(1)/dx) – (1.d(sin x)/dx))/(sin x)2
= (0 – cos x)/(sin 2 x) = (-cosx/sin x). (1/sin x)
= - cosec x cot x
(v) Now,
d/dx (3 cot x) = 3. d/dx (cot x) = -3 cosec2 x
Also, 5.d/dx (cosec x) = - 5cosec x cot x
d/dx (3cot x + 5 cosec x)
= -3 cosec2x – 5cosec x cot x
= -cosec x (3cosec x + 5 cot x)
(vi) Let f(x) = 5 sin x – 6 cos x + 7
f'(x) = (d(5 sin x – 6 cos x + 7))/dx
= (d(5 sin x))/dx - (d(6 cos x))/dx + (d(7))/dx
= 5cos x – 6(- sin x) + 0
= 5 cos x + 6 sin x
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# Breaking down Equations one step - Yr9AMM
```Breaking down Equations – 0ne step
When there is an unknown letter, we would want to solve it or basically get a
value for this letter. Another word used to call this letter is a pronumeral.
Hence, you may see such statements in textbooks which says, ‘solve the
unknown’ or ‘solve for the pronumeral’.
The letter or pronumeral would indicate that we would like to find the value of
something that we do not know.
An example for this maybe, 2 dvds cost \$21.00, so how much does one dvd
cost?
This can be represented as 2y= 21.50, y would represent the cost of a dvd, so 1
dvd would cost 21.50 divided 2, which would be \$10.50.
In solving equations, remember, the opposite operations
Opposite operation
+
x
÷
minus, subtract, take away
times, multiply,product
divide, over, quotient
-, minus
+, plus
÷, divide
x, times
In solving equations, what we are doing when using opposite operations is
simply to balance the equation so that we do not change the equation, what we
do to one side of the equal sign, we need to do the same to the other.
One step Equation
2y = 10, meaning 2 x y
Opposite operation
÷ or written as __
2 x y = 10
2
2
y=5
t + 7 = 12
t + 7 -7 = 12 -7
t= 5
opposite of +7 is -7
-
r =4
opposite of divide 3 is X3, so
3
3 X r = 4 X 3 , so r = 12
3
Solve for the pronumeral:
Opposite operation:
3t = 15
t = __________________
2g = 4
g = ___________________
r=6
2
r = ___________________
j – 5 = 11
j = ___________________
y = 10
3
y = ___________________
g+5=9
g = ___________________
```
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### Simplifying Square Roots (Review)
Let's review the steps involved in simplifying square roots:
1. Factor the number inside the square root sign.
2. If a factor appears twice, cross out both and write the factor one time to the left of the square root sign. If the factor appears three times, cross out two of the factors and write the factor outside the sign, and leave the third factor inside the sign.
Note: If a factor appears 4, 6, 8, etc. times, this counts as 2, 3, and 4 pairs, respectively.
3. Multiply the numbers outside the sign. Multiply the numbers left inside the sign.
4. Check: The outside number squared times the inside number should equal the original number inside the square root.
To simplify the square root of a fraction, simplify the numerator and simplify the denominator.
Example 1: Simplify
1. =
2. = 2×2
3. 2×2 = 4
4. Check: 42(3) = 48
Thus, = 4.
Example 2: Simplify .
First, reduce the fraction to lowest terms: =
=
Numerator:
1. =
2. = 2
3. 2 = 2
4. Check: 22(3) = 12
Denominator:
1. =
2. = 5
3. 5 = 5
4. Check: 52(7) = 175
Thus, = .
### Rationalizing the Denominator
In addition to simplifying the numerator and the denominator in a fraction, it is mathematical convention to rationalize the denominator--that is, to write the fraction as an equivalent expression with no roots in the denominator.
To rationalize a denominator, multiply the fraction by a "clever" form of 1--that is, by a fraction whose numerator and denominator are both equal to the square root in the denominator. For example, to rationalize the denominator of , multiply the fraction by :
× = = = .
Thus, = .
Often, the fraction can be reduced:
Rationalize the denominator of :
× = = = = 3.
Thus, = 3.
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# RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions
The RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions is based on Algebraic Identities. Students will get to learn about these expressions in detail. In this article, we will explain about using Algebraic Identities in the problems. Our experts prepare the questions in the PDF we attached below with easy solutions.
As we know, algebra is the part of Maths, to represent the quantities and number the several symbols and letters will be used in the form of formulas and equations. It is compulsory for the students to learn all the formulas and should have the knowledge when to use the particular formula in the questions.
Learn about RD Sharma Class 9 Chapter 4- Algebraic Identities
## Download RD Sharma Chapter 9 4 Class 9 Maths Exercise 4.1 Solutions PDF
Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.1
## Important Definition RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions
RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions is based on the following formulas-
Look down for some examples based on the formulas of algebraic identities mentioned above-
Ques)- (3p – 1/p)²
Solution- (3p – 1/p)²
= Algebraic Identity- (a-b)² = a² – 2ab + b²
= (3p – 1/p)² = (3p)² – (2 x 3p x 1/p) + (1/p)²
= 9p² – 6 + 1/p²
Ques)- (3x + y) (3x – y)
Solution- (3x + y) (3x – y)
= Algebric Identity- (a + b) (a – b) = a² – b²
= (3x + y) (3x – y) = 3x² – y²
= 3x² – y²
Ques)- 175 x 175 +2 x 175 x 25 + 25 x 25
Solution- 175 x 175 +2 x 175 x 25 + 25 x 25
= Algebraic Identity- a² + 2ab + b² = (a+b)²
= (175)² + 2 (175) (25) + (25)²
= (175 + 25)²
= (200)²
= 40000
### Benefits of RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions
Go to the following points to know the Benefits of the RD Sharma Chapter 4 Class 9 Maths Exercise 4.1 Solutions-
1. The solutions we have prepared in the above PDF is based on the NCERT Syllabus with detailed information.
2. With the help of RD Sharma, students will get to learn tips and tricks to solve the questions by applying algebraic identities.
3. The answers to the questions are presented organized stepwise in an easy way.
Ques 1- How many algebraic identities are there?
Ans)- There are majorly four algebraic identities. The algebraic identities are used by using basic formulas mentioned below-
Ques 2- What is algebraic identity?
Ans)- An algebraic identity is an identity that exists for any values of its variables.
Ques 3- Are all identity equations?
Ans)- An equation satisfied for all variable values determining the expressions is termed as an identity. So, the identity is a unique case of an equation.
In other words, all identities can be equations, but not all equations will be identities.
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## What is Arithmetic Sequence
### Arithmetic Sequences Definition
A list of numbers with a clear pattern is known as an arithmetic sequence. We can tell if a series of numbers is arithmetic by taking any number and subtracting it by the number before it. In an arithmetic sequence, the difference between two numbers stays the same.
The letter $$d$$ stands for the common difference, the constant difference between the numbers in an arithmetic sequence. We have an increasing arithmetic sequence if $$d$$ is positive. We have a decreasing arithmetic sequence if $$d$$ is negative.
### Arithmetic Sequences Formula
We can use the following formula to find different terms for the arithmetic sequence:
$$a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)$$, where:
• $$a_1$$: The first term
• $$a_n$$: The $$n$$th term
• $$n$$: The term poistion
• $$d$$: Common difference
### How to figure out the terms in an arithmetic sequence?
To find any term in the arithmetic sequence, we need to know a common difference, the position of the term we want to find, and a term in the sequence.
### Example
Consider the arithmetic sequence: $$3, \ 7, \ 11, \ 15, \ 19, \ ...$$, Find $$a_{28} \ = \ ?$$
Solution
Before using the arithmetic sequence formula, we need to know the first term, the common difference, and the position of the term we want to find:
• The first term: $$a_1 \ = \ 3$$
• common difference: $$d \ = \ 4$$
• Term's position: $$n \ = \ 28$$
Now, we put these numbers into the formula and find the answer:
$$a_n \ = \ a_1 \ + \ d \ (n \ –\ 1) \ ⇒ \ a_{28} \ = \ 3 \ + \ 4(28 \ - \ 1) \ ⇒ \ a_{28} \ = \ 111$$
### Summing an Arithmetic Series
To summing the terms in an arithmetic sequence, use this formula:
$$\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} (2a \ + \ (n \ - \ 1)d)$$
### Example
Find the sum of the first $$12$$ terms for the sequence: $$1, \ 6, \ 11, \ 16, \ 21, \ ...$$
Solution
• The first term: $$a_1 \ = \ 1$$
• common difference: $$d \ = \ 5$$
• $$n \ = \ 12$$
So:
$$\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} (2a \ + \ (n \ - \ 1)d) \ ⇒$$
$$\sum_{k \ = \ 0}^{11} \ (1 \ + \ 5k) \ = \ \frac{12}{2} (2 \ + \ (11)5) \ = \ 6 \times (57) \ = \ 342$$
### Exercises for Arithmetic Sequences
1) Find the explicit formula: $$5, \ 14, \ 23, \ 32, \ 41, \ ...$$
2) Find the explicit formula: $$-7, \ -2, \ 3, \ 8, \ 13, \ ...$$
3) Find the explicit formula and $$a_{12}$$: $$-22, \ -16, \ -10, \ -4, \ 2, \ ...$$
4) Find the explicit formula and $$a_{21}$$: $$a_{30} \ = \ 72, \ d \ = \ -5$$
5) Find the first $$10$$ terms: $$a_{15} \ = \ 50, \ d \ = \ 4.5$$
6) Find the first $$7$$ terms: $$a_{41} \ = \ 178, \ d \ = \ 4$$
7) Find $$a_{30}$$: $$45, \ , \ 41.8, \ 38.6, \ 35.4, \ ...$$
8) Find $$a_{27}$$: $$a_{41} \ = \ 167, \ d \ = \ 6.4$$
9) Find the explicit formula and the sum of the first five terms: $$a_1 \ = \ 10, \ d \ = \ 3$$
10) Find the explicit formula and the sum of the first $$16$$ terms: $$-55, \ -48, \ -41, \ -34, \ -27, \ ...$$
1) Find the explicit formula: $$5, \ 14, \ 23, \ 32, \ 41, \ ...$$
$$\color{red}{a_1 \ = \ 5, \ d \ = \ 14 \ - \ 5 \ = \ 9, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ 5 \ + \ 9 \ (n \ – \ 1)}$$
2) Find the explicit formula: $$-7, \ -2, \ 3, \ 8, \ 13, \ ...$$
$$\color{red}{a_1 \ = \ -7, \ d \ = \ -2 \ - \ (-7) \ = \ 5, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ -7 \ + \ 5 \ (n \ – \ 1)}$$
3) Find the explicit formula and $$a_{12}$$: $$-22, \ -16, \ -10, \ -4, \ 2, \ ...$$
$$\color{red}{a_1 \ = \ -22, \ d \ = \ -16 \ - \ (-22) \ = \ 6, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ -22 \ + \ 6 \ (n \ – \ 1) \ ⇒ \ a_{12} \ = \ -22 \ + \ 6 \ (12 \ – \ 1) \ = \ 44}$$
4) Find the explicit formula and $$a_{21}$$: $$a_{30} \ = \ 72, \ d \ = \ -5$$
$$\color{red}{a_{30} \ = \ 72, \ d \ = \ -5, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{30} \ = \ a_1 \ + \ (-5) \ (30 \ – \ 1) \ ⇒ \ a_1 \ = \ a_{30} \ + \ 5(29) \ = \ 72 \ + \ 145 \ = \ 217}$$
$$\color{red}{⇒ \ a_n \ = \ 217 \ - \ 5 \ (n \ – \ 1) \ ⇒ \ a_{21} \ = \ 217 \ - \ 5(20) \ = \ 117}$$
5) Find the first $$10$$ terms: $$a_{15} \ = \ 50, \ d \ = \ 4.5$$
$$\color{red}{a_{15} \ = \ 50, \ d \ = \ 4.5, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{15} \ = \ a_1 \ + \ 4.5 \ (15 \ – \ 1) \ ⇒ \ a_1 \ = \ 50 \ - \ 4.5(14) \ = \ -13}$$
$$\color{red}{⇒ -13, \ -8.5, \ -4, \ 0.5, \ 5, \ 9.5, \ 14, \ 18.5, \ 23, \ 27.5, \ ...}$$
6) Find the first $$7$$ terms: $$a_{41} \ = \ 178, \ d \ = \ 4$$
$$\color{red}{a_{41} \ = \ 178, \ d \ = \ 4, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{41} \ = \ a_1 \ + \ 4 \ (41 \ – \ 1) \ ⇒ \ a_1 \ = \ 178 \ - \ 4(40) \ = \ 18}$$
$$\color{red}{⇒ 18, \ 22, \ 26, \ 30, \ 34, \ 38, \ 42, \ ...}$$
7) Find $$a_{30}$$: $$45, \ , \ 41.8, \ 38.6, \ 35.4, \ ...$$
$$\color{red}{a_1 \ = \ 45, \ d \ = \ 41.8 \ - \ 45 \ = \ -3.2, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ 45 \ - \ 3.2 \ (n \ – \ 1) \ ⇒ \ a_{30} \ = \ 45 \ - \ 3.2 \ (30 \ – \ 1) \ = \ -47.8}$$
8) Find $$a_{27}$$: $$a_{41} \ = \ 167, \ d \ = \ 6.4$$
$$\color{red}{a_{41} \ = \ 167, \ d \ = \ 6.4, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_{41} \ = \ a_1 \ + \ 6.4 \ (41 \ – \ 1) \ ⇒ \ a_1 \ = \ 167 \ - \ 6.4(40) \ = \ -98}$$
$$\color{red}{⇒ \ a_n \ = \ -98 \ + \ 6.4 \ (n \ – \ 1) \ ⇒ \ a_{27} \ = \ -98 \ + \ 6.4(26) \ = \ 68.4}$$
9) Find the explicit formula and the sum of the first five terms: $$a_1 \ = \ 10, \ d \ = \ 3$$
$$\color{red}{a_1 \ = \ 10, \ d \ = \ 3, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ 10 \ + \ 3 \ (n \ – \ 1)}$$
$$\color{red}{\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} \ (2a \ + \ (n \ - \ 1)d)}$$
$$\color{red}{⇒ \ \sum_{k \ = \ 0}^{4} \ (10 \ + \ 3k) \ = \ \frac{5}{2} \ (20 \ + \ (4)3) \ = \ \frac{5}{2} \times (32) \ = \ 80}$$
10) Find the explicit formula and the sum of the first $$16$$ terms: $$-55, \ -48, \ -41, \ -34, \ -27, \ ...$$
$$\color{red}{a_1 \ = \ -55, \ d \ = \ -48 \ - \ (-55) \ = \ 7, \ a_n \ = \ a_1 \ + \ d \ (n \ – \ 1)}$$
$$\color{red}{⇒ \ a_n \ = \ -55 \ + \ 7 \ (n \ – \ 1)}$$
$$\color{red}{\sum_{k \ = \ 0}^{n \ - \ 1} \ (a \ + \ kd) \ = \ \frac{n}{2} \ (2a \ + \ (n \ - \ 1)d)}$$
$$\color{red}{⇒ \ \sum_{k \ = \ 0}^{15} \ (-55 \ + \ 7k) \ = \ \frac{16}{2} \ (-110 \ + \ (15)7) \ = \ 8 \times (-5) \ = \ -40}$$
## Arithmetic Sequences Practice Quiz
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How to Factor Algebraic Equations
In mathematics, factoring is the act of finding the numbers or expressions that multiply together to make a given number or equation. Factoring is a useful skill to learn for the purpose of solving basic algebra problems; the ability to competently factor becomes almost essential when dealing with quadratic equations and other forms of polynomials. Factoring can be used to simplify algebraic expressions to make solving simpler. Factoring can even give you the ability to eliminate certain possible answers much more quickly than you would be able to by solving manually.[1]
Method 1
Method 1 of 3:
Factoring Numbers and Basic Algebraic Expressions
1. 1
Understand the definition of factoring when applied to single numbers. Factoring is conceptually simple, but, in practice, can prove to be challenging when applied to complex equations. Because of this, it's easiest to approach the concept of factoring by starting with simple numbers, then move on to simple equations before finally proceeding to more advanced applications. A given number's factors are the numbers that multiply to give that number. For example, the factors of 12 are 1, 12, 2, 6, 3 and 4, because 1 × 12, 2 × 6, and 3 × 4 all equal 12.[2]
• Another way to think of this is that a given number's factors are the numbers by which it is evenly divisible.
• Can you find all the factors of the number 60? We use the number 60 for a wide variety of purposes (minutes in an hour, seconds in a minute, etc.) because it's evenly divisible by a fairly wide range of numbers.
• The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
2. 2
Understand that variable expressions can also be factored. Just as lone numbers can be factored, so too can variables with numeric coefficients be factored. To do this, simply find the factors of the variable's coefficient. Knowing how to factor variables is useful for simplifying algebraic equations that the variables are a part of.[3]
• For example, the variable 12x can be written as a product of the factors of 12 and x. We can write 12x as 3(4x), 2(6x), etc., using whichever factors of 12 are best for our purposes.
• We can even go as far as to factor 12x multiple times. In other words, we don't have to stop with 3(4x) or 2(6x) - we can factor 4x and 6x to give 3(2(2x) and 2(3(2x), respectively. Obviously, these two expressions are equal.
3. 3
Apply the distributive property of multiplication to factor algebraic equations. Using your knowledge of how to factor both lone numbers and variables with coefficients, you can simplify simple algebraic equations by finding factors that the numbers and variables in an algebraic equation have in common. Usually, to make the equation as simple as possible, we try to search for the greatest common factor. This simplification process is possible because of the distributive property of multiplication, which states that for any numbers a, b, and c, a(b + c) = ab + ac.[4]
• Let's try an example problem. To factor the algebraic equation 12 x + 6, first, let's try to find the greatest common factor of 12x and 6. 6 is the biggest number that divides evenly into both 12x and 6, so we can simplify the equation to 6(2x + 1).
• This process also applies to equations with negatives and fractions. x/2 + 4, for instance, can be simplified to 1/2(x + 8), and -7x + -21 can be factored to -7(x + 3).
Method 2
Method 2 of 3:
1. 1
Ensure the equation is in quadratic form (ax2 + bx + c = 0). Quadratic equations are of the form ax2 + bx + c = 0, where a, b, and c are numeric constants and a does not equal 0 (note that a can equal 1 or -1). If you have an equation containing one variable (x) that has one or more terms of x to the second power, you can usually shift the terms in the equation around using basic algebraic operations to get 0 on one side of equals sign and ax2, etc. on the other side.[5]
• For example, let's consider the algebraic equation. 5x2 + 7x - 9 = 4x2 + x - 18 can be simplified to x2 + 6x + 9 = 0, which is in the quadratic form.
• Equations with greater powers of x, like x3, x4, etc. can't be quadratic equations. They are cubic equations, quartic equations, and so on, unless the equation can be simplified to eliminate these terms of x above the power of 2.
2. 2
In quadratic equations where a = 1, factor to (x+d )(x+e), where d × e = c and d + e = b. If your quadratic equation it is in the form x2 + bx + c = 0 (in other words, if the coefficient of the x2 term = 1), it's possible (but not guaranteed) that a relatively simple shortcut can be used to factor the equation. Find two numbers that both multiply to make c and add to make b. Once you find these two numbers d and e, place them in the following expression: (x+d)(x+e). These two terms, when multiplied together, produce your quadratic equation - in other words, they are your quadratic equation's factors.
• For example, let's consider the quadratic equation x2 + 5x + 6 = 0. 3 and 2 multiply together to make 6 and also add up to make 5, so we can simplify this equation to (x + 3)(x + 2).
• Slight variations on this basic shortcut exist for slight variations in the equation itself:
• If the quadratic equation is in the form x2-bx+c, your answer is in this form: (x - _)(x - _).
• If it is in the form x2+bx+c, your answer looks like this: (x + _)(x + _).
• If it is in the form x2-bx-c, you answer is in the form (x + _)(x - _).
• Note: the numbers in the blanks can be fractions or decimals. For example, the equation x2 + (21/2)x + 5 = 0 factors to (x + 10)(x + 1/2).
3. 3
If possible, factor by inspection. Believe it or not, for uncomplicated quadratic equations, one of the accepted means of factoring is simply to examine the problem, then just consider possible answers until you find the right one. This is also known as factoring by inspection. If the equation is in the form ax2+bx+c and a>1, your factored answer will be in the form (dx +/- _)(ex +/- _), where d and e are nonzero numerical constants that multiply to make a. Either d or e (or both) can be the number 1, though this is not necessarily so. If both are 1, you've essentially used the shortcut described above.[6]
• Let's consider an example problem. 3x2 - 8x + 4 at first seems intimidating. However, once we realize that 3 only has two factors (3 and 1), it becomes easier, because we know that our answer must be in the form (3x +/- _)(x +/- _). In this case, adding a -2 to both blank spaces gives the correct answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x add to -8x. -2 × -2 = 4, so we can see that the factored terms in parentheses multiply to become the original equation.
4. 4
Solve by completing the square. In some cases, quadratic equations can be quickly and easily factored by using a special algebraic identity. Any quadratic equation of the form x2 + 2xh + h2 = (x + h)2. So, if, in your equation, your b value is twice the square root of your c value, your equation can be factored to (x + (sqrt(c)))2.[7]
• For example, the equation x2 + 6x + 9 fits this form. 32 is 9 and 3 × 2 is 6. So, we know that the factored form of this equation is (x + 3)(x + 3), or (x + 3)2.
5. 5
Use factors to solve quadratic equations. Regardless of how you factor your quadratic expression, once it is factored, you can find possible answers for the value of x by setting each factor equal to zero and solving. Since you're looking for values of x that cause your equation to equal zero, a value of x that makes either of your factors equal zero is a possible answer for your quadratic equation.[8]
• Let's return to the equation x2 + 5x + 6 = 0. This equation factored to (x + 3)(x + 2) = 0. If either of the factors equals 0, the entire equation equals 0, so our possible answers for x are the numbers that make (x + 3) and (x + 2) equal 0. These numbers are -3 and -2, respectively.
6. 6
Check your answers - some of them may be extraneous! When you've found your possible answers for x, plug them back in to your original equation to see if they are valid. Sometimes, the answers you find don't cause the original equation to equal zero when plugged back in. We call these answers extraneous and disregard them.[9]
• Let's plug -2 and -3 into x2 + 5x + 6 = 0. First, -2:
• (-2)2 + 5(-2) + 6 = 0
• 4 + -10 + 6 = 0
• 0 = 0. This is correct, so -2 is a valid answer.
• Now, let's try -3:
• (-3)2 + 5(-3) + 6 = 0
• 9 + -15 + 6 = 0
• 0 = 0. This is also correct, so -3 is also a valid answer.
Method 3
Method 3 of 3:
Factoring Other Forms of Equations
1. 1
If the equation is in the form a2-b2, factor it to (a+b)(a-b). Equations with two variables factor differently than basic quadratics. For any equation a2-b2 where a and b do not equal 0, the equation factors to (a+b)(a-b).
• For example, the equation 9x2 - 4y2 = (3x + 2y)(3x - 2y).
2. 2
If the equation is in the form a2+2ab+b2, factor it to (a+b)2. Note that, If the trinomial is in the form a2-2ab+b2, the factored form is slightly different: (a-b)2.
• The equation 4x2 + 8xy + 4y2 can be re-written as 4x2 + (2 × 2 × 2)xy + 4y2. We can now see that it's in the correct form, so we can say with confidence that our equation factors to (2x + 2y)2
3. 3
If the equation is in the form a3-b3, factor it to (a-b)(a2+ab+b2). Finally, it bears mentioning that cubics and even higher-order equations can be factored, though the factoring process quickly becomes prohibitively complicated.
• For instance, 8x3 - 27y3 factors to (2x - 3y)(4x2 + ((2x)(3y)) + 9y2)
Community Q&A
Search
• Question
How do I factor simple addition?
Find a common factor for the two numbers. For example, for 6 + 8, 6 and 8 share a factor of two. You can then rewrite it as 2 (3 + 4).
• Question
How would I factor -24x+4x^2?
Donagan
Both terms have 4x as a factor. Therefore, -24x + 4x² = 4x(-6 + x) = 4x(x - 6).
• Question
Can you demonstrate an easier problem? I have problems like 42r - 18 to factor.
Find a common factor of 42r and 18, e.g. 6. This number will go on the outside of the bracket so 6(...). Then divide the original number(s) you had by 6. We end up with 7r-3. This will go on the inside of the bracket to make the final answer: 6(7r-3). You can check your answer by expanding the brackets again: if the answer matches what you started with, then the answer is correct!
200 characters left
Tips
• If you have a trinomial in the form x2+bx+ (b/2)2, the factored form is (x+(b/2))2. (You may have this situation while completing the square.)
⧼thumbs_response⧽
• a2-b2 is factorable, a2+b2 isn't factorable.
⧼thumbs_response⧽
• Remember how to factor constants- it might help.
⧼thumbs_response⧽
Things You'll Need
• Paper
• Pencil
• Math Book (if necessary)
Co-authored by:
This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. This article has been viewed 642,266 times.
Co-authors: 44
Updated: September 30, 2022
Views: 642,266
Categories: Algebra
Article SummaryX
To factor a basic algebraic equation, start by looking for the largest factor that all the numbers in the equation have in common. For instance, if your equation is 6x + 2 = 0, the largest common factor that can be divided evenly into both terms on the left side of the equation is 2. Divide each term by the largest common factor, then rewrite the expression in the form a(b + c), where a is the largest common factor. After factoring, our example equation would become 2(3x + 1) = 0. If the equation contains variables that are common factors in multiple terms, you can factor those out as well. For instance, in the equation 4x² – 2x = 0, each term contains the common factors 2 and x. To factor the left side of the equation, divide each term by those factors to get 2x(2x – 1) = 0. If you want to learn how to factor quadratic equations, keep reading the article!
Thanks to all authors for creating a page that has been read 642,266 times.
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# Perimeter and Area of an Equilateral Triangle – Formulas and Examples
The perimeter of an equilateral triangle is the length of the triangle’s outline. On the other hand, the area is a measure of the space occupied by the triangle. We can find the perimeter of an equilateral triangle by adding the lengths of its three sides, and we can find its area by multiplying the length of its base by its height and dividing by 2.
In this article, we will learn all about the perimeter and area of an equilateral triangle. We will explore its formulas and apply them to solve some practice problems.
##### GEOMETRY
Relevant for
Learning about the perimeter and area of an equilateral triangle.
See examples
##### GEOMETRY
Relevant for
Learning about the perimeter and area of an equilateral triangle.
See examples
## How to find the perimeter of an equilateral triangle?
To calculate the perimeter of an equilateral triangle, we have to add the lengths of its three sides. Recalling that an equilateral triangle has all its sides of the same length, we only have to multiply the length of one of the sides by 3:
where, a is the length of one side of the triangle.
This means that we only need to know the length of one of the sides of the equilateral triangle to calculate its perimeter.
## How to find the area of an equilateral triangle?
To calculate the area of any triangle, we can multiply its base by its height and divide by 2. In the case of equilateral triangles, we can use the following formula to calculate their area:
where, a is the length of one of the sides of the equilateral triangle.
### Proof of the formula for the area of an equilateral triangle
To prove the formula for the area of an equilateral triangle, we are going to use the following diagram, where we draw a bisector perpendicular to the base with height h:
We recall that the area of any triangle can be calculated with the following formula:
$latex \text{Area}= \frac{1}{2} \times \text{base} \times \text{height}$
Here, the base is equal to “a” and the height is equal to “h“.
Using the Pythagorean theorem to calculate the height, we have:
$latex {{a}^2}={{h}^2}+{{( \frac{a}{2})}^2}$
⇒ $latex {{h}^2}={{a}^2}- \frac{{{a}^2}}{4}$
⇒ $latex {{h}^2}=\frac{3{{a}^2}}{4}$
⇒ $latex h=\frac{\sqrt{3}~a}{2}$
Now that we have an expression for h, we can use it in the formula for the area of a triangle:
$latex \text{Area}= \frac{1}{2} \times \text{base} \times \text{height}$
$latex A=\frac{1}{2}\times a \times \frac{\sqrt{3}~a}{2}$
⇒ $latex A=\frac{\sqrt{3}~{{a}^2}}{4}$
## Perimeter and area of an equilateral triangle – Examples with answers
The following examples are solved using the perimeter and area formulas of an equilateral triangle. Try to solve the problems yourself before looking at the answer.
### EXAMPLE 1
Find the perimeter of an equilateral triangle that has sides with a length of 5 inches.
We use the formula for the perimeter with the value $latex a=5$. Therefore, we have:
$latex p=3a$
$latex p=3(5)$
$latex p=15$
The perimeter of the equilateral triangle is equal to 15 inches.
### EXAMPLE 2
What is the area of an equilateral triangle that has sides with a length of 10 feet?
We use the formula for the area with the length a=10:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex A= \frac{ \sqrt{3}}{4}({{10}^2})$
$latex A= \frac{ \sqrt{3}}{4}(100)$
$latex A=43.3$
The area of the equilateral triangle is equal to 43.3 ft².
### EXAMPLE 3
Find the perimeter of an equilateral triangle that has sides with a length of 8 yards.
Using the value $latex a=8$ in the formula for the perimeter, we have:
$latex p=3a$
$latex p=3(8)$
$latex p=24$
The perimeter of the equilateral triangle is equal to 24 yd.
### EXAMPLE 4
Find the area of an equilateral triangle that has sides with a length of 14 inches.
Applying the formula for the area with the given length, we have:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex A= \frac{ \sqrt{3}}{4}({{14}^2})$
$latex A= \frac{ \sqrt{3}}{4}(196)$
$latex A=84.87$
The area of the equilateral triangle is equal to 84.87 in².
### EXAMPLE 5
What is the perimeter of an equilateral triangle that has sides with a length of 15 inches?
Applying the formula for the perimeter with the value $latex a=15$:
$latex p=3a$
$latex p=3(15)$
$latex p=45$
The perimeter of the triangle is equal to 45 inches.
### EXAMPLE 6
What is the area of an equilateral triangle that has sides with a length of 15 feet?
We use the length $latex a=15$ in the formula for the area:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex A= \frac{ \sqrt{3}}{4}({{15}^2})$
$latex A= \frac{ \sqrt{3}}{4}(225)$
$latex A=97.43$
The area of the equilateral triangle is equal to 97.43 ft².
### EXAMPLE 7
Find the length of the sides of an equilateral triangle that has a perimeter of 39 ft.
In this example, we know the perimeter of the triangle, and we have to find the length of one side. Therefore, we use the perimeter formula and solve for a:
$latex p=3a$
$latex 39=3a$
$latex a=13$
The sides of the triangle have a length of 13 ft.
### EXAMPLE 8
Find the length of the sides of an equilateral triangle that has an area of 35.07 ft².
In this case, we know the area, and we need to find the length of the sides. Therefore, we use the formula for the area and solve for a:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 35.07= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 35.07=0.433{{a}^2}$
$latex {{a}^2}=81$
$latex a=9$
The sides of the triangle have a length of 9 ft.
### EXAMPLE 9
Find the length of the sides of an equilateral triangle that has a perimeter of 102 in.
We are going to use the formula for the perimeter with the value $latex p=102$ and solve for a:
$latex p=3a$
$latex 102=3a$
$latex a=34$
The length of one side of the triangle is 34 in.
### EXAMPLE 10
Find the length of the sides of an equilateral triangle that has an area of 73.18 ft².
We use the formula for the area with the given value and solve for a:
$latex A= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 73.18= \frac{ \sqrt{3}}{4}{{a}^2}$
$latex 73.18=0.433{{a}^2}$
$latex {{a}^2}=169$
$latex a=13$
The length of the sides of the triangle is 13 ft.
## Perimeter and area of an equilateral triangle – Practice problems
Solve the following problems by applying everything you have learned about the perimeter and area of equilateral triangles. Click “Check” to make sure that you got the correct answer.
Choose an answer
Choose an answer
Choose an answer
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Choose an answer
## See also
Interested in learning more about perimeters and areas of geometric figures? Take a look at these pages:
LEARN MORE
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# Recursion. Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms.
## Presentation on theme: "Recursion. Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms."— Presentation transcript:
Recursion
Understand how the Fibonacci series is generated Recursive Algorithms Write simple recursive algorithms Analyze simple recursive algorithms Understand the drawbacks of recursion Name other recursive algorithms and data structures John Edgar2
What happens if you put a pair of rabbits in a field? More rabbits! Assume that rabbits take one month to reach maturity and that Each pair of rabbits produces another pair of rabbits one month after mating. John Edgar3
How many pairs of rabbits are there after 5 months? Month 1: start – 1 Month 2: the rabbits are now mature and can mate – 1 Month 3: – the first pair give birth to two babies – 2 Month 4: the first pair give birth to 2 babies, the pair born in month 3 are now mature – 3 Month 5: the 3 pairs from month 4, and 2 new pairs – 5 John Edgar4
AAfter 5 months there are 5 pairs of rabbits ii.e. the number of pairs at 4 months (3) plus the number of pairs at 3 months (2) WWhy? WWhile there are 3 pairs of bunnies in month 4 only 2 of them are able to mate tthe ones alive in month 3 TThis series of numbers is called the Fibonacci series 5 month: pairs: 1 1 2 2 3 3 4 4 5 5 6 6 1 1 1 1 2 2 3 3 5 5 8 8
TThe n th number in the Fibonacci series, fib(n), is: 00 if n = 0, and 1 if n = 1 ffib(n – 1) + fib(n – 2) for any n > 1 ee.g. what is fib(23) EEasy if we only knew fib(22) and fib(21) TThe answer is fib(22) + fib(21) WWhat happens if we actually write a function to calculate Fibonacci numbers like this? John Edgar6
Let's write a function just like the formula fib(n) = 0 if n = 0, 1 if n = 1, otherwise fib(n) = fib(n – 1) + fib(n – 2) John Edgar7 int fib(int n){ if(n == 0 || n == 1){ return n; }else{ return fib(n-1) + fib(n-2); } int fib(int n){ if(n == 0 || n == 1){ return n; }else{ return fib(n-1) + fib(n-2); } The function calls itself C++
The Fibonacci function is recursive A recursive function calls itself Each call to a recursive method results in a separate call to the method, with its own input Recursive functions are just like other functions The invocation is pushed onto the call stack And removed from the call stack when the end of a method or a return statement is reached Execution returns to the previous method call John Edgar8
int fib(int n){ if(n == 0 || n == 1) return n; else return fib(n-1) + fib(n-2); } John Edgar9 fib(5) fib(4) fib(3) fib(2) fib(1) fib(0) fib(2) fib(1) fib(0) fib(1) fib(2) fib(1) fib(0) fib(1) 1111100011 21 3 2 5
When a function is called it is pushed onto the call stack This applies to each invocation of that function When a recursive call is made execution switches to that method call The call stack records the line number of the previous method where the call was made from Once a method call execution finishes, returns to the previous invocation John Edgar10
January 2010Greg Mori11
Recursive functions do not use loops to repeat instructions But use recursive calls, in if statements Recursive functions consist of two or more cases, there must be at least one Base case, and one Recursive case John Edgar12
The base case is a smaller problem with a simpler solution This problem’s solution must not be recursive ▪ Otherwise the function may never terminate There can be more than one base case John Edgar13
The recursive case is the same problem with smaller input The recursive case must include a recursive function call There can be more than one recursive case John Edgar14
Define the problem in terms of a smaller problem of the same type The recursive part e.g. return fib(n-1) + fib(n-2); And the base case where the solution can be easily calculated This solution should not be recursive e.g. if (n == 0 || n == 1) return n; John Edgar15
How can the problem be defined in terms of smaller problems of the same type? By how much does each recursive call reduce the problem size? By 1, by half, …? What are the base cases that can be solved without recursion? Will a base case be reached as the problem size is reduced? John Edgar16
January 2010Greg Mori17
Linear Search Binary Search Assume sorted array John Edgar18
John Edgar19 C++ int linSearch(int *arr, int n, int x){ for (int i=0; i < n; i++){ if(x == arr[i]){ return i; } } //for return -1; //target not found } int linSearch(int *arr, int n, int x){ for (int i=0; i < n; i++){ if(x == arr[i]){ return i; } } //for return -1; //target not found } The algorithm searches the array one element at a time using a for loop
Base cases Target is found at first position in array The end of the array is reached Recursive case Target not found at first position ▪ Search again, discarding the first element of the array John Edgar20
John Edgar21 int linSearch(int *arr, int n, int x){ return recLinSearch(arr,n,0,x); } int recLinSearch(int *arr, int n, int i, int x){ if (i >= n){ return -1; } else if (x == arr[i]){ return i; } else return recLinSearch(arr, n, i + 1, x); } int linSearch(int *arr, int n, int x){ return recLinSearch(arr,n,0,x); } int recLinSearch(int *arr, int n, int i, int x){ if (i >= n){ return -1; } else if (x == arr[i]){ return i; } else return recLinSearch(arr, n, i + 1, x); } C++
Of course, if it’s a sorted array we wouldn’t do linear search John Edgar22
Each sub-problem searches a subarray Differs only in the upper and lower array indices that define the subarray Each sub-problem is smaller than the last one In the case of binary search, half the size There are two base cases When the target item is found and When the problem space consists of one item ▪ Make sure that this last item is checked John Edgar23
John Edgar24 C++ int binSearch(int *arr, int lower, int upper, int x){ int mid = (lower + upper) / 2; if (lower > upper){ return - 1; //base case } else if(arr[mid] == x){ return mid; //second base case } else if(arr[mid] < x){ return binSearch(arr, mid + 1, upper, x); } else { //arr[mid] > target return binSearch(arr, lower, mid - 1, x); } int binSearch(int *arr, int lower, int upper, int x){ int mid = (lower + upper) / 2; if (lower > upper){ return - 1; //base case } else if(arr[mid] == x){ return mid; //second base case } else if(arr[mid] < x){ return binSearch(arr, mid + 1, upper, x); } else { //arr[mid] > target return binSearch(arr, lower, mid - 1, x); }
January 2010Greg Mori25
Merge Sort Quicksort John Edgar26
January 2010Greg Mori27
What’s the easiest list to sort? A list of 1 number John Edgar28
Let’s say I have 2 sorted lists of numbers How can I merge them into 1 sorted list? John Edgar29 1 1 3 3 5 5 12 22 23 42 99 output 1 1 12 22 23 3 3 5 5 42 99 List 1List 2
If I have a list of n numbers, how should I sort them? I know two things How to sort a list of 1 number How to merge 2 sorted lists of numbers into 1 sorted list Smells like recursion John Edgar30
John Edgar31 mergeSort (array) if (array is length 1) // base case, one element return the array else arr1 = mergeSort(first half of array) arr2 = mergeSort(second half of array) return merge(arr1,arr2) mergeSort (array) if (array is length 1) // base case, one element return the array else arr1 = mergeSort(first half of array) arr2 = mergeSort(second half of array) return merge(arr1,arr2)
What is the time complexity of a merge? John Edgar32 1 1 3 3 5 5 12 22 23 42 99 output 1 1 12 22 23 3 3 5 5 42 99 List 1List 2
How many recursive steps are there? How large are the merges at each recursive step? Merge takes O(n) time for n elements John Edgar33
John Edgar34 2341338107191145 Sort entire array 2341338107191145 Sort halves 2341338107191145 Sort quarters 2341338107191145 Sort eighths 2341338107191145 Sorted quarters 2333418107111945 Sorted halves 0711192333414581 Sorted entire array
John Edgar35 2341338107191145 Sort entire array 2341338107191145 Sort halves 2341338107191145 Sort quarters 2341338107191145 Sort eighths How many recursive steps are there? How large are the merges at each recursive step? Merge takes O(n) time for n elements
John Edgar36 2341338107191145 Sort entire array 2341338107191145 Sort halves 2341338107191145 Sort quarters 2341338107191145 Sort eighths How many recursive steps are there? O(log n) steps: split array in half each time How large are the merges at each recursive step? In total, merge n elements each step Time complexity is O(n log n)
Mergesort Best case: O(n(log 2 n)) Average case: O(n(log 2 n)) Worst case: O(n(log 2 n)) John Edgar37
Quicksort is a more efficient sorting algorithm than either selection or insertion sort It sorts an array by repeatedly partitioning it We will go over the basic idea of quicksort and an example of it See text / on-line resources for details John Edgar39
Partitioning is the process of dividing an array into sections (partitions), based on some criteria "Big" and "small" values Negative and positive numbers Names that begin with a-m, names that begin with n-z Darker and lighter pixels Quicksort uses repeated partitioning to sort an array John Edgar40
John Edgar41 Partition this array into small and big values using a partitioning algorithm 31 12 07 23 93 02 11 18
John Edgar42 Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot 3112072393021118 smalls < 18 bigs > 18 pivot 18
John Edgar43 31120723930211 18 arr[low ] is greater than the pivot and should be on the right, we need to swap it with something We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high Partition this array into small and big values using a partitioning algorithm
John Edgar44 31120723930211 18 arr[low ] (31) is greater than the pivot and should be on the right, we need to swap it with something arr[high] (11) is less than the pivot so swap with arr[low ] Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high
John Edgar45 31120723930211183111 Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high
John Edgar46 120723930218 repeat this process until: 31230211 Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high 1207
John Edgar47 12 07 93 18 repeat this process until: 31 23 02 11 high and low are the same Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high
John Edgar48 repeat this process until: high and low are the same We'd like the pivot value to be in the centre of the array, so we will swap it with the first item greater than it Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high 12 07 93 18 31 23 02 11 93 18
John Edgar49 smallsbigs pivot Partition this array into small and big values using a partitioning algorithm We will partition the array around the last value (18), we'll call this value the pivot Use two indices, one at each end of the array, call them low and high 12 07 93 18 31 23 02 11
John Edgar50 Use the same algorithm to partition this array into small and big values 00 08 07 01 06 02 05 09 bigs! pivot 00 08 07 01 06 02 05 09 smalls
John Edgar51 Or this one: 09 08 07 06 05 04 02 01 bigs pivot 01 08 07 06 05 04 02 09 smalls
The quicksort algorithm works by repeatedly partitioning an array Each time a subarray is partitioned there is A sequence of small values, A sequence of big values, and A pivot value which is in the correct position Partition the small values, and the big values Repeat the process until each subarray being partitioned consists of just one element John Edgar52
How long does quicksort take to run? Let's consider the best and the worst case These differ because the partitioning algorithm may not always do a good job Let's look at the best case first Each time a subarray is partitioned the pivot is the exact midpoint of the slice (or as close as it can get) ▪ So it is divided in half What is the running time? John Edgar53
John Edgar54 08 01 02 07 03 06 04 05 bigs pivot 04 01 02 03 05 06 08 07 smalls First partition
John Edgar55 big1 pivot1 02 01 04 05 06 08 sm1 04 01 02 03 05 06 08 07 Second partition 07 03 pivot1pivot2 big2sm2
John Edgar56 pivot1 02 03 04 05 06 07 08 Third partition 02 01 03 04 05 06 07 08 pivot1done 01
Each subarray is divided exactly in half in each set of partitions Each time a series of subarrays are partitioned around n comparisons are made The process ends once all the subarrays left to be partitioned are of size 1 How many times does n have to be divided in half before the result is 1? log 2 (n) times Quicksort performs around n * log 2 (n) operations in the best case John Edgar57
John Edgar58 09 08 07 06 05 04 02 01 bigs pivot 01 08 07 06 05 04 02 09 smalls First partition
John Edgar59 bigs pivot 01 08 07 06 05 04 02 09 smalls 01 08 07 06 05 04 02 09 Second partition
John Edgar60 bigs pivot 01 02 07 06 05 04 08 09 Third partition 01 08 07 06 05 04 02 09
John Edgar61 pivot 01 02 07 06 05 04 08 09 smalls Fourth partition 01 02 07 06 05 04 08 09
John Edgar62 bigs pivot 01 02 04 06 05 07 08 09 Fifth partition 01 02 07 06 05 04 08 09
John Edgar63 pivot 01 02 04 06 05 07 08 09 smalls Sixth partition 01 02 04 06 05 07 08 09
John Edgar64 pivot 01 02 04 05 06 07 08 09 Seventh(!) partition 01 02 04 06 05 07 08 09
Every partition step results in just one partition on one side of the pivot The array has to be partitioned n times, not log 2 (n) times So in the worst case quicksort performs around n 2 operations The worst case usually occurs when the array is nearly sorted (in either direction) John Edgar65
With a large array we would have to be very, very unlucky to get the worst case Unless there was some reason for the array to already be partially sorted ▪ In which case first randomize the position of the array elements! The average case is much more like the best case than the worst case John Edgar66
January 2010Greg Mori67
Recursive algorithms have more overhead than similar iterative algorithms Because of the repeated method calls This may cause a stack overflow when the call stack gets full It is often useful to derive a solution using recursion and implement it iteratively Sometimes this can be quite challenging! John Edgar68
Some recursive algorithms are inherently inefficient e.g. the recursive Fibonacci algorithm which repeats the same calculation again and again Look at the number of times fib(2) is called Such algorithms should be implemented iteratively Even if the solution was determined using recursion John Edgar69
It is useful to trace through the sequence of recursive calls This can be done using a recursion tree Recursion trees can be used to determine the running time of algorithms Annotate the tree to indicate how much work is performed at each level of the tree And then determine how many levels of the tree there are John Edgar70
January 2010Greg Mori71
Recursion is similar to induction Recursion solves a problem by Specifying a solution for the base case and Using a recursive case to derive solutions of any size from solutions to smaller problems Induction proves a property by Proving it is true for a base case and Proving that it is true for some number, n, if it is true for all numbers less than n John Edgar72
Prove, using induction that the algorithm returns the values fact(0) = 0! =1 fact(n) = n! = n * (n – 1) * … * 1 if n > 0 John Edgar73 int fact (int x){ if (x == 0){ return 1; } else return n * fact(n – 1); } int fact (int x){ if (x == 0){ return 1; } else return n * fact(n – 1); } C++
Basis: Show that the property is true for n = 0, i.e. that fact(0) returns 1 This is true by definition as fact(0) is the base case of the algorithm and returns 1 Establish that the property is true for an arbitrary k implies that it is also true for k + 1 Inductive hypothesis: Assume that the property is true for n = k, that is assume that fact(k) = k * ( k – 1) * ( k – 2) * … * 2 * 1 John Edgar74
IInductive conclusion: Show that the property is true for n = k + 1, i.e., that fact(k + 1) returns ((k + 1) * k * (k – 1) * (k – 2) * … * 2 * 1 BBy definition of the function: fact(k + 1) returns ((k + 1) * fact(k) – the recursive case AAnd by the inductive hypothesis: fact(k) returns kk * (k – 1) * (k – 2) * … * 2 * 1 TTherefore fact(k + 1) must return ((k + 1) * k * (k – 1) * (k – 2) * … * 2 * 1 WWhich completes the inductive proof John Edgar75
Recursive sum Towers of Hanoi – see text Eight Queens problem – see text Sorting Mergesort Quicksort John Edgar76
Linked Lists are recursive data structures They are defined in terms of themselves There are recursive solutions to many list methods List traversal can be performed recursively Recursion allows elegant solutions of problems that are hard to implement iteratively ▪ Such as printing a list backwards John Edgar77
January 2010Greg Mori78
Recursion as a problem-solving tool Identify base case where solution is simple Formulate other cases in terms of smaller case(s) Recursion is not always a good implementation strategy Solve the same problem many times Function call overhead Recursion and induction Induction proves properties in a form similar to how recursion solves problems John Edgar79
Carrano Ch. 2, 5 John Edgar80
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## Poissonizing the Multinomial
### Poissonizing the Multinomial¶
Bernoulli trials come out in one of two ways. But many trials come out in multiple different ways, all of which we might want to track. A die can land six different ways. A jury member can have one of several different identities. In general, an individual might belong to one of several classes.
The multinomial distribution is an extension of the binomial to the case where there are more than two possible outcomes of each trial. Let's look at it first in an example, and then we will define it in general.
A box contains 2 blue tickets, 5 green tickets, and 3 red tickets. Fifteen draws are made at random with replacement. To find the chance that there are 4 blue, 9 green, and 2 red tickets drawn, we could start by writing all possible sequences of 4 B's, 9 G's, and 2 R's.
Each sequence would have chance $0.2^4 0.5^9 0.3^2$, so all we need to complete the probability calculation is the number of sequences we could write.
• There are $\binom{15}{4}$ ways of choosing places to write the B's.
• For each of these ways, there are $\binom{11}{9}$ ways of choosing 9 of the remaining 11 places to write the G's.
• The remaining 2 places get filled with R's.
So
\begin{align*} P(\text{4 blue, 9 green, 2 red}) &= \binom{15}{4} \cdot \binom{11}{9} 0.2^4 0.5^9 0.3^2 \\ \\ &= \frac{15!}{4!11!} \cdot \frac{11!}{9!2!} 0.2^4 0.5^9 0.3^2 \\ \\ &= \frac{15!}{4!9!2!} 0.2^4 0.5^9 0.3^2 \end{align*}
Notice how this simply extends the binomial probability formula by including a third category in exactly the same way.
Analogously or by induction, you can extend the formula to any finite number of categories or classes.
### Multinomial Distribution¶
Fix a positive integer $n$. Suppose we are running $n$ i.i.d. trials where each trial can result in one of $k$ classes. For each $i = 1, 2, \ldots, k$, let the chance of getting Class $i$ on a single trial be $p_i$, so that $\sum_{i=1}^k p_i = 1$.
For each $i = 1, 2, \ldots , k$, let $N_i$ be the number of trials that result in Class $i$, so that $N_1 + N_2 + \ldots + N_k = n$.
Then the joint distribution of $N_1, N_2, \ldots , N_k$ is given by
$$P(N_1 = n_1, N_2 = n_2, \ldots , N_k = n_k) = \frac{n!}{n_1!n_2! \ldots n_k!}p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k}$$
where $n_i \ge 0$ for $1 \le i \le k$ and $\sum_{i=1}^k n_i = n$.
When there just two classes then $k = 2$ and the formula reduces to the familiar binomial formula, written as the joint distribution of the number of successes and the number of failures:
$$P(N_1 = n_1, N_2 = n_2) = \frac{n!}{n_1!n_2!} p_1^{n_1}p_2^{n_2} ~~ \text{where } p_1+p_2=1 \text{ and } n_1+n_2=n$$
Notice that the marginal distribution of each $N_i$ is binomial $(n, p_i)$. You don't have to sum the joint distributions to work this out. $N_i$ is the number of Class $i$ individuals in the sample; each sampled individual is in Class $i$ with probability $p_i$; and there are $n$ independent draws. That's the binomial setting.
### Poissonization¶
If you replace the fixed number $n$ of trials by a Poisson $(\mu)$ random number of trials, then the multinomial gets Poissonized as follows:
• For each $i = 1, 2, \ldots , k$, the distribution of $N_i$ is Poisson $(\mu p_i)$.
• The counts $N_1, N_2, \ldots , N_k$ in the $k$ different categories are mutually independent.
We won't go through the proof, which is a straightforward extension of the proof in the case $k=2$ given in an earlier section. Rather, we will look at why the result matters.
When the number of trials is fixed, $N_1, N_2, \ldots , N_k$ are all dependent on each other in complicated ways. But when you let the sample size be a Poisson random variable, then the independence of the counts $N_1, N_2, \ldots , N_k$ lets you quickly calculate the chance of any particular configuration of classes in the sample.
For example, if in your population the distribution of classes is as follows:
• Class 1: 20%
• Class 2: 30%
• Class 3: 50%
and you draw $N$ independent times where $N$ has the Poisson $(20)$ distribution, then the chance that you will get at least 3 individuals in each class is about 71.27%.
(1 - stats.poisson.cdf(2, 4))*(1-stats.poisson.cdf(2, 6))*(1-stats.poisson.cdf(2, 10))
0.71270362753222383
The number of factors in the answer is equal to the number of classes, unlike the inclusion-exclusion formula in which the amount of work increases much more with each additional class, as you have seen in exercises.
This helps data scientists tackle questions like, "How many times must I sample so that my chance of seeing at least one individual of each class exceeds a given threshold?" The answer depends on the distribution of classes in the population, of course, but allowing the sample size be a Poisson random variable can make calculations much more tractable. For applications, see for example the Abstract and References of this paper.
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# Solving problems involving surface area, volume and capacity
## Worked Example 17.15: Calculating the height of a prism
The volume of the prism is $$4\ 032\text{ cm}^3$$. The area of the shaded region is $$224 \text{ cm}^2$$. Calculate the value of the length, $$x$$.
### Use the general formula for volume to find height of prism.
$\text{Volume of prism} = \text{area of base} \times h$
We know the volume of the prism, and we are given the area of the base of the prism, so we can solve for the height, $$x$$:
\begin{align} 4\ 032 &= 224 \times x \\ \frac{4\ 032}{224} &= x \\ 18 &= x \end{align}
The height of the prism $$x = 18 \text{ cm}$$.
## Worked Example 17.16: Calculating the surface area of a rectangular prism
Dintle wants to paint a wooden crate. If she uses $$\text{0,25} \text{ ml}$$ of paint per square centimetre, calculate how much paint she will need to paint the crate. Assume that Dintle will paint every outside surface of the crate. Express your answer in litres.
### Find the surface area of the crate.
\begin{align} \text{Surface area of crate} &= 2(\text{area of sides}) + 2(\text{area of ends}) + 2(\text{area of top/bottom}) \\ &= 2(85 \times 32) + 2(48 \times 32) + 2(85 \times 48) \\ &= 2(2\ 720) + 2(1\ 536) + 2(4\ 080) \\ &= 5\ 440 + 3\ 072 + 8\ 160 \\ &= 16\ 672 \end{align}
The surface area of the crate is $$16\ 672 \text{ cm}^2$$.
### Calculate the volume of paint needed.
Dintle will use $$\text{0,25} \text{ ml}$$ of paint per square centimetre.
So, we can calculate:
\begin{align} \text{Volume of paint} &= \text{0,25} \times \text{16 672} \\ &= 4\ 168 \end{align}
Volume of paint needed is $$\text{4 168} \text{ ml}$$.
\begin{align} \text{4 168} \text{ ml} &= \frac{\text{4 168}}{\text{1 000}} \\ &= \text{4,2} \text{ litres} \end{align}
Dintle will need $$\text{4,2} \text{ litres}$$ of paint.
## Worked Example 17.17: Calculating the volume of a complex solid
Calculate the volume of the solid. All measurements are in centimetres.
### Find the volume of the cuboid.
\begin{align} \text{Volume of cuboid} &= l \times b \times h \\ &= 8 \times 15 \times 11 \\ &= 1\ 320 \end{align}
### Calculate the volume of the triangular prism.
\begin{align} \text{Volume of triangular prism} &= \frac{1}{2}(b \times h) \\ &= \frac{1}{2}(8 \times 6) \\ &= 24 \end{align}
\begin{align} \text{Volume of solid} &= \text{Volume of cuboid} + \text{Volume of the triangular prism} \\ &= 1\ 320 + 24 \\ &= 1\ 344 \text{ cm}^3 \end{align}
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A Basic Linear Algebra Principle
2. Implication of size on existence/uniqueness
By combining the numerical implications of the existence and the uniqueness on the size, we have
Ax = b has a unique solution for any b
⇒ number of rows of A = number of columns of A (A is a square matrix)
⇔ number of equations = number of variables
Conversely, assume A is an n by n matrix. Then we have (if you find the general argument too difficult, try 3 by 3 matrix first)
Ax = b has solutions for any b
⇒ All rows are pivot
⇒ Number of pivot rows is n (because there are n rows)
⇒ Number of pivot columns is n (by this equality)
⇒ All columns are pivot (because there are n columns)
⇒ The solution of a consistent system Ax = b is unique
In other words, if the number of equations is equal to the number of variables, then always existence implies uniqueness. By similar argument, we can also prove that uniqueness implies always existence.
In summary, we have the following basic principle of linear algebra.
For a square matrix A, the following are equivalent
Always Existence + Uniqueness
Ax = b has a unique solution for any b
Always Existence
Ax = b has solutions for any b
Uniqueness
The solution of a consistent system Ax = b is unique
Our discussion also tells us when the above happens from computational viewpoint.
For a square matrix A, the following are equivalent
• Ax = b has a unique solution for any b
• All rows of A are pivot
• All columns of A are pivot
• A can be row operated to become I
For the claim that A can be row operated to become I, please check out more details in this exercise.
Example The system
x1 - x2 + 2x3 = 1 3x1 + x2 - 2x3 = 3 2x1 - x2 + 2x3 = 2
has x1 = 1, x2 = x3 = 0 as an obvious solution. It is also easy to see that x1 = 1, x2 = 2, x3 = 1 is another solution. Therefore the system has many solutions. Since the system
x1 - x2 + 2x3 = b1 3x1 + x2 - 2x3 = b2 2x1 - x2 + 2x3 = b3
has the same coefficient matrix, by the basic principle, it does not always have solutions.
The significance of the basic principle is the following: We may consider (always) existence and the uniqueness as two complementary aspects of systems of linear equations. In general, there is no relation between the two aspects. However, in case the size is right (square coefficient matrix, or number of variables = number of equations), the two aspects are equivalent.
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# Cool Math Equations That Mean Something latest 2023
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Page Contents
## Elimination Method To Solve System Of Linear Equations
The method of elimination is most often used by students to solve a system of linear equations. Moreover, this method is easy to understand and involves the addition and subtraction of polynomials. Students should be able to add and subtract polynomials involving two or three variables.
In the method of elimination, the coefficients of the same variable are identical, then the two equations are subtracted to eliminate this variable. The resulting equation involves only one variable and can be easily simplified. For example; Consider that there are two equations in the system of linear equations with variables “x” and “y” as shown below:
2x – 5a = 11
3x + 2a = 7
To solve the above equation by the method of elimination, we need to make the coefficients of one of the variables (either “x” or “y”) the same by multiplying the equation with some numbers, and these numbers can be obtained by finding the least common multiple of the coefficients. Consider that we want the coefficients of “x” to be the same in both equations. For this we need to find the least common multiple of “2” and “3” which is “6”.
To get “6” as the coefficient of the “x” variable in the equations, we need to multiply the first equation by “3” and the second equation by “2”, as shown below:
(2x – 5a = 11) * 3
(3x + 2y = 7) * 2
The new set of equations after multiplication is obtained as shown below:
6x – 15y = 33
6x + 4a = 14
Now we have the same coefficient of variable “x” in both equations. Once a variable has obtained the same coefficient, subtract one equation from the other. We will subtract the second equation from the first as shown below:
(6x – 15y = 33) – (6x + 4y = 14)
In the next step, combine similar terms:
6x – 6x – 15y – 4y = 33 – 14
– 19a = 19
y = – 1
So far we have solved the equations for one variable. To find the value of the other variable “x”, we will substitute the value of “y” into one of the equations given in the question.
Substitute the value of “y = – 1” in the equation 2x – 5y = 11 to find the value of “x” as shown in the next step:
2x – 5 (- 1) = 11
2x + 5 = 11
2x = 11 – 5
2x = 6
x = 3
Therefore, we have solved the two equations to find the value of the variables and our solution is x = 3 and y = – 1. You can take the same approach to solve the system of linear equations by eliminating one of the variables.
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## Cool Math Don T Look Back latest 2023
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## Find the roots of basic equations containing roots
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Last week Sherri bought 324 square yards of sod to grass a square play area for the children of her day care. Now she has to fence the area to keep the children safe but does not know how many yards of fencing she needs to buy. All Sherri can figure out is that all the sides are the same length because the grassy area is a square.
How can she determine how many yards of fencing to buy?
In this concept, you will learn to solve equations involving radicals.
When you solve an equation you are trying to find the value for the variable that will make the equality statement true. The steps applied to solving an equation are inverse operations. To solve an equation involving radicals, inverse operations are used to solve for the variable.
A radical involving the square root of a number can be evaluated by determining the square root of the number under the radical sign. If the radicand is a perfect square then its square will be the number which multiplied by itself twice will give the value of the radicand. For example the square root of 81 can be denoted by . What number times itself twice gives 81?
Taking the square root of a number is the inverse operation of squaring and vice versa.
Let’s look at an equation involving radicals.
The variable ‘’ is squared and its value is 121. To solve this equation the value of needs to be determined. The inverse operation of squaring is taking the square root. Remember, whatever operation is applied to one side of the equation must also be applied to the other side.
First, take the square root of both sides of the equation.
Next, verify the answer by substituting the value for ‘’ into the original equation.
Then, perform any indicated operations.
The value of 11 made the equality statement true – both sides of the equation are the same.
Let’s look at one more.
Solve the following equation involving radicals:
Notice the left side of the equation is a radical. The inverse operation of taking the square root is squaring.
Remember, the square of the square root of anything is the anything. In other words
First, square both sides of equation.
Next, perform any indicated operations and simplify the equation.
Then, subtract 2 from both sides of the equation to solve the equation for ‘’.
Next, verify the answer by substituting the value for ‘’ into the original equation.
Then, perform any indicated operations and simplify the equation.
The value of 34 made the equation true.
### Examples
#### Example 1
Earlier, you were given a problem about Sherri and the fence for the grassy square. She needs to figure out the perimeter of the square.
First, write an equation for the area of the grassy square.
Such that is the area and is the length of any side of the square.
Next, fill in the value for the area.
Next, solve the equation for ‘’ by taking the square root of both sides of the radical equation.
The length of each side of the square is 18 yds.
Next, write an equation for finding the perimeter of the square. The perimeter is the distance all around the grassy area.
Such that ‘’ is the perimeter and ‘’ is the length of one side of the square.
Next, fill in the value for ‘’ and perform the indicated operation.
Sherri needs to buy 72 yards of fencing.
#### Example 2
Solve the following equation involving radicals to the nearest tenth.
First, the opposite operation of squaring is taking the square root. Take the square root of both sides of the equation.
Notice that the number 26 is not a perfect square. Use the TI calculator to find the square root of 26.
Next, on the calculator press 2nd enter.
Then, round the value shown on the screen to one place after the decimal.
#### Example 3
Solve the following radical equation for the variable ‘’.
First, take the square root of both sides of the equation.
Next, simplify both sides of the equation.
#### Example 4
Solve the following radical equation for the variable ‘’.
First, apply the inverse operation of taking the square root to both sides of the equation.
Next, perform any indicated operations and simplify the equation.
Then, add eight to both sides of the equation to solve for the variable ‘’.
Next, verify the answer by substituting the value for ‘’ into the original equation.
Then, perform any indicated operations and simplify the equation.
#### Example 5
Solve the following equation for the variable.
First, isolate the variable by subtracting 4 from both sides of the equation.
Next, simplify both sides of the equation.
Next, perform the inverse operation of cubing – taking the cube root, on both sides of the equation.
Then, perform the indicated operations.
### Review
Solve each equation involving radical expressions.
1.
2.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Base When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Cubed The cube of a number is the number multiplied by itself three times. For example, "two-cubed" = $2^3 = 2 \times 2 \times 2 = 8$.
Exponent Exponents are used to describe the number of times that a term is multiplied by itself.
Extraneous Solution An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.
Fractional Power A fractional power is an exponent in fraction form. A fractional exponent of $\frac{1}{2}$ is the same as the square root of a number. A fractional exponent of $\frac{1}{3}$ is the same as the cube root of a number.
Perfect Square A perfect square is a number whose square root is an integer.
Quadratic Equation A quadratic equation is an equation that can be written in the form $=ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.
Quadratic Formula The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Squared Squared is the word used to refer to the exponent 2. For example, $5^2$ could be read as "5 squared". When a number is squared, the number is multiplied by itself.
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Lorem ipsum dolor sit amet, conse ctetur adip elit, pellentesque turpis.
## Lesson Tutor : Algebra Lesson 12 : Multiplying Positives and Negatives
/ Lesson Tutor : Algebra Lesson 12 : Multiplying Positives and Negatives
Algebra Lesson 12: Multiplying Positives and Negatives
by Elaine Ernst Schneider
Before starting: Review/ complete Algebra Lesson 11
As we learned last lesson: Additive Inverses are opposites. Two numbers are opposites if their sum equals zero. For example, -8 and 8 are additive inverses because their sums total zero. This makes them opposites. You can do the same thing with variables. For example, -(-x) = x.
When this principle is used in multiplication, these rules of thumb emerge:
1. A negative times a negative gives a positive answer.
2. A negative times a positive is negative.
3. A positive times a positive is a positive result.
Examples
(-6)(8) = -48
(-3)(-12) = 36
(12)(11) = 132
Now, what happens when there are three numbers to multiply? Simply work in sequence, following the rules you have learned.
Example: (-5) (-4) (-20)
(20) (-20) [multiply –5 times –4 to get positive 20]
-400 [multiply positive 20 from last step by –20 to get –400]
And what about exponents? Just write them out and follow the rules.
Example: -4 = (-4) (-4) (-4) (-4) (-4)
(16) (-4) (-4) (-4)
-64 (-4) (-4)
256 (-4)
-1024
Exercises:
1. (10) (-8)
2. (-5) (-26)
3. -5³
4. (-2) (-3)²
5. (-1.5) (4)
6. (-5) (22) (-2)
7. (-3) (-5) (-4)
8. (-1) (-3)
9. (-1)
10. (-2)² (-3)
1. –80
2. 130
3. –125
4. –18
5. –6
6. 220
7. –60
8. –81
9. –1 *
10. 324
*Note: Did you learn something about exponents? If the exponent is even, the answer is positive. If the exponent is odd, the answer is negative.
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# How kids can quickly learn multiplication tables?
Multiplication is one of four basic rules of mathematics (addition, subtraction, multiplication, and division). A multiplication table is used to get the product of two numbers. If we learn the basic concept of multiplication, it should be very interesting to solve any multiplication.
In this post, we learn about the definition, notation, rules, and how to find the multiplication table.
## What is a Multiplication table?
In mathematics, a multiplication table is a mathematical table used to define a multiplication operation for an algebraic operation. It is also known as a times table. In simple algebra, multiplication is the process of calculating the results when a number is taken times. The result of the multiplication is called product and each number is known as the factors of the product.
Multiplication sign also known as times sign or dimension sign is the symbol x used in mathematics to denote the multiplication operation and its resulting product.
We can also use the multiplication chart. A multiplication chart is a table that shows the product of two numbers.
A 12×12 multiplication table is given below.
We can use the multiplication chart up to infinite numbers but here we made a multiplication chart for few tables in order to understand the main concept of the multiplication chart.
## Rules of multiplication table
Some basic rules are very important in the multiplication table.
1. Any number multiplied by zero is zero e.g., 2×0 = 0, 5×0 = 0, 100×0 = 0.
2. Any number multiplied by one stays the same or give the result itself e.g., 2×1 = 2, 5×1 = 5, 100×1 = 100.
3. When a number is multiplied by two, we are doubling that number e.g., 2×2 = 4, 5×2 = 10, 100×2 = 200.
4. When a number is multiplied by ten, we simply write zero at the end of that number e.g., 2×10 = 20, 5×10 = 50.
5. When a number is multiplied by a hundred, we simply write two zeros at the end of that number, and in the case of a thousand we simply write three zeros at the end of that number.
## Tips for students
Many educators believe it is necessary to memorize the table up to 9×9. You should learn l, 2, 5, and 10 times tables first. These tables are pretty easy to learn for the students as these tables are simple as compared to others. After this, you should learn 3, 4, 6, 7, 8, 9 times tables. If you want to learn and practice the multiplication tables, use an online multiplication table whichwill help you learn and practice the times tables easily.
## How to learn times tables?
If you want to work out the times table for 2, start with 2 and then add 2 in each step. The result obtained in every step is a multiple of 2 and is known as multiplication fact.
• 2 x 1 = 2
• 2 x 2 = 4
• 2 x 3 = 6
• 2 x 4 = 8
• 2 x 5 = 10
• 2 x 6 = 12
• 2 x 7 = 14
• 2 x 8 = 16
• 2 x 9 = 18
• 2 x 10 = 20
If you want to work out the times table for 5, start with 5 and then add 5 in each step. The result obtained in every step is a multiple of 5.
If you want to work out the times table for 10, start with10and then add 10 in each step. The result obtained in every step is a multiple of 10.
## Benefits of learning Multiplication table
Learning basic multiplication tables will make it easier to learn more challenging ones e.g. if kids know their 2 times table, they will be able to work out their 4 times table by doubling the results.
• 2 x 1 = 2 4 x 1 = 4
• 2 x 2 = 4 4 x 1 = 8
• 2 x 3 = 6 4 x 1 = 12
• 2 x 4 = 8 4 x 1 = 16
• 2 x 5 = 10 4 x 1 = 20
• 2 x 6 = 12 4 x 1 = 24
• 2 x 7 = 14 4 x 1 = 28
• 2 x 8 = 16 4 x 1 = 32
• 2 x 9 = 18 4 x 1 = 36
• 2 x 10 = 20 4 x 1 = 40
When kids successfully recall their time tables they will grow in confidence, which will not only tackle more challenging math problems but will help to keep them motivated and engaged across other objects too.
Recalling times tables improves memory skills, which is a transferrable skill that will help kids throughout school and into adult life.
Kids will find it easier to solve math problems and to do mental arithmetic if they have already memorized their times tables.
Multiplication is used throughout adulthood, whether it is working out price reductions, doubling recipes, or splitting bills.
## Summary
In this post, we have learned the definition of multiplication tables, rules, and some basic concepts. Once you grab the knowledge of this topic you will be master in it.
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Home Polynomial Functions Introduction Investigating Polynomial Functions Polynomial Division Remainder & Factor Theorems Polynomial Equations Polynomial Inequalities Rational Functions Review&Test
UNIT 4 : POLYNOMIAL AND RATIONAL FUNCTIONS
LESSON 1: POLYNOMIAL FUNCTIONS INTRODUCTION
Example 1:
f(x) = x3 4x2 + x + 6 is a cubic polynomial function because the largest exponent of the variable is 3. Make a table of values or use
a graphing calculator to draw its graph [see below].
Note: The function has 3 zeros (x intercepts), x = -1, 2, 3 and two
turning points A and B. Turning point A is a local maximum point as
the function changes from increasing to decreasing at A. Turning point B
is a local minimum point as the function changes from decreasing to increasing
at B. The function has a local maximum value of y = 6.1 at point A and a
local minimum value of y = -0.9 at point B.
Example 2:
Use your graphing calculator to graph and study the cubic functions below.
a) f(x) = x3 x2 + 5x 3 b) f(x) = (x 3)3 + 1
c) f(x) = x3 + 3x2 + 3x + 2
Note: From the examples above we can make the following observations.
• A cubic function can have 1, 2 or 3 zeros (x-intercepts)
• It can have 0 or 2 turning points.
• The coefficient of x3 is called the leading coefficient (k). In (a) the leading coefficient is negative(k = -1) and the function rises to the left and falls to the right.
• The leading coefficient(k) determines the end behaviours. If k > 0, the function falls to the left and rises to the right (ex. 1, 2 b,c)
If k < 0, the function rises to the left and falls to the right (ex. 2a). This characteristic is true for all polynomial functions of odd degree (1, 3, 5,
).
Example 3:
Use your graphing calculator to graph and study the quartic functions below.
a) f(x) = x4 + x3 5x2 3x b) y = x4 + x3 2x2 3x
c) y = -0.5x4 x3 + 2x2 5 d) y = x4 + x3 2x2 3x + 3
Note: From the examples above we can make the following observations.
• A quaric function can have 0,1, 2, 3 or 4 zeros (x-intercepts)
• It can have 1 or 3 turning points.
• The coefficient of x4 is called the leading coefficient (k). In (c) the leading coefficient is negative(k = -0.5) and the function falls to the left and right.
In (a,b,d) the leading coefficient is positive and the function rises to the left and right
• The leading coefficient(k) determines the end behaviours. If k > 0, the function rises to the left and right (ex. 3 a,b,d)
If k < 0, the function falls to the left and right (ex. 3c). This characteristic is true for all polynomial functions of even degree (2, 4, 6,
).
Polynomial Functions in Factored Form f(x) = k(x a)(x b)(x c) . . . etc.
# Linear Functions in factored Form:f(x) = k(x a)
# Cubic Functions in factored Form:f(x) = k(x a)(x b)(x c)
x y -2 6 0 -6
# Quartic Functions in factored Form:f(x) = k(x a)(x b)(x c)(x d)
x y -1 16 2 16
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# RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions is concerned with Heron’s formula. The students will study “how to find the area of a triangle using heron’s formula?” with the help of several solved problems. The experts have used the step-by-step problem-solving methods of the examples mentioned in this article.
Students can freely download RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions PDF and grind their skills. The PDF consists of varieties of questions based on Heron’s Formula. The questions were explained stepwise, which is easily understandable to the students. By practicing through PDF, learners will get to know the several types of questions that can be formed. Practicing with these problems help students to score well in the exam.
## Download RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions PDF
Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1
## Important Definitions RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
We can calculate any triangle if we know the length of all three (3) sides of that triangle by using Heron’s Formula, which has been known for approximately 2000 years. It is known as “Heron’s Formula” after Hero of Alexandria.
### Heron’s Formula
Area of a triangle= √s (s-a) (s-b) (s-c)
Semi Perimeter= s=(a+b+c)2, where, a, b, and c are the sides of the triangle.
### Examples of Heron’s Formula RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
Ques- In a triangle PQR, PQ = 15cm, QR = 13cm, and PR = 14cm. Find the area of a triangle PQR and hence its altitude on PR.
Solution-
Let the sides of the given triangle be PQ = p, QR = q, PR = r, respectively.
Here,
= p = 15 cm
= q = 13 cm
= r = 14 cm
From Heron’s Formula;
Area of a triangle= √s (s-p) (s-q) (s-r)
Semi Perimeter= s=(p+ q+ r)2
Where, p, q, and r are sides of a triangle.
= s= (15 + 13 + 14)/ 2
= s= 21
= Area= √21 (21-15) (21-13) (21-14)
= √21 (6 x 7 x 8)
= √7056
= 84
= Area = 84 cm2
= Let, QT is a perpendicular on PR
Now, area of triangle = ½ x Base x Height
= ½ × QT × PR = 84
= QT = 12cm
= The altitude is 12 cm (Hence Proved).
## Frequently Asked Questions (FAQs) of RD Sharma Chapter 12 Class 9 Maths Exercise 12.1 Solutions
Ques- What is S in a Triangle of Hero’s Formula?
Ans- The other is Heron’s formula, which gives the area in terms of the three sides of a triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c), where, ‘s’ is the semi perimeter of a triangle. So, s = (a + b + c)/2.
Ques- What is the semi perimeter of a triangle?
Ans- A semi perimeter of a triangle is equivalent to the perimeter of its medial triangle.
Ques- Who gave Heron’s formula?
Ans- Hero of Alexandria, a great mathematician who derived the formula to calculate the area of the triangle using the length of all three (3) sides.
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# Integers
```Section 7.3 – Ratio and Proportion
Definition. A ratio is an ordered pair of numbers, denoted as either a : b or
a
and read as the ratio of a to b.
b
Generally, ratios relate values in one of three types: part-to-part, part-to-whole, or whole-to-part.
For example, suppose a + b = c, we may write the following ratios as:
a
b
part-to-part
a : b or b : a In the other notation,
or .
b
a
a
b
part-to-whole
a : c or b : c In the other notation,
or .
c
c
c
c
whole-to-part
c : a or c : b In the other notation,
or .
a
b
Examples.
1.
In a school with 420 students and 35 teachers:
(a) What is the ratio of students to teachers?
(b) What is the ratio of teachers to students?
(c) Which type of a ratio did you use for each of the above?
2.
A school had 180 boys and 240 girls attending.
(a) What is the ratio of boys to girls?
(b) What is the ratio of girls to boys?
(c) What is the ratio of students to boys?
(d) What is the ratio of girls to students?
(e) Which type of ratio was used for each of the above?
3.
A recipe calls for 2 parts sugar, 5 parts flour, and 3 parts milk.
(a) What is the ratio of sugar to flour to milk?
(b) What is the ratio of sugar to the whole recipe?
Note. The ratio a : b is equivalent to the ratio an : bn,. We can use this relationship to solve some types of problems that
involve ratios.
Example. The ratio of boys to girls in a class is 3:4. The class has 35 students.
How many boys and girls are in the class?
Solution: The ratio 3:4 is equivalent to the ratio 3n : 4n where 3n represents the number of boys and 4n
represents the number of girls.
3n + 4n = 35
7n = 35
n=5
The number of boys is 3n = 3 ∙ 5 = 15.
The number of girls is 4n = 4 ∙ 5 = 20.
-1-
The class had 15 boys and 20 girls.
Definition. A proportion is a statement that gives the equality of two ratios, denoted as either a : b :: c : d or
a c
.
b d
The a and d are called the extremes and the b and c are called the means.
We use two methods for solving problems involving proportions. We will call the first method Equivalent Ratios. The
procedure is the same as finding equivalent fractions. The second method is called Cross-Product Algorithm (note the
comment on this method on p. 480 in the book—the discussion gives several reasons why the method is not good for
teaching proportional reasoning).
Equivalent Ratios
The method is the similar to finding equivalent fractions. Write the ratios in the fractional form and then change the
fractions with a common denominator.
Examples.
1. The dog to cat ratio for Mathville is two to three. If there are 18 cats, how many dogs are there?
Solution:
We use the ratio
dogs
and let N represent the number of dogs.
cats
2 N
3 18
26 N
3 6 18
12 N
18 18
N = 12, since the numerators must be equal.
There are twelve dogs in Mathville.
2.
Kim was assessed property taxes of \$1420 on a house valued at \$62,000. Approximately, what would
the assessment be on a \$140,000 house?
Solution:
We use the ratio of taxes to value and let N represent taxes on a \$140,000 house.
1420
N
62,000 140,000
1420 140
62 N
Note that 62,000 ∙ 140 = 62 ∙ 1,000 ∙ 140 = 62 ∙ 140,000.
62,000 140 62 140,000
62N = 1420 ∙ 140, since the numerators are equal when the denominators are equal.
N
1420 140 198800
3206.45
62
62
The \$140,000 house would have assessed property taxes of \$3206.45.
3.
Pat used three gallons of gas to drive seventy miles. If Pat used eight gallons of gas, approximately
how many miles could Pat drive?
Solution:
We use the ratio of gallons to miles. (Note this is opposite of the usual ratio of miles per gallon.)
3
8
70 N
3 N
70 8
70 N 70 N
-2-
3N = 560
2
N = 186 .
3
Pat would drive approximately 187 miles.
How could we simplify this process? What is method that requires fewer steps?
Examine problem 3 above: does the solution give a method that would require fewer steps?
Why does this method work?
Cross-Product Algorithm
The second method for solving problems is often called Cross-Multiplication. The procedure is essentially the same as
the shortcut used to show fractions are equivalent or to compare two fractions where we did not write the common
denominator. The justification follows from the equivalent ratios method.
Cross-Multiplication. a : b :: c : d if and only if ad = bc.
As your grandparents or great-grandparents would have said,
"the product of the means is equal to the product of the extremes."
a c
In the other notation, if and only if ad = bc.
b d
Examples.
4.
If a six-foot man standing near a flag pole casts a shadow four and one-half feet on a sunny day, how tall is a
flag pole that casts a shadow of eighteen and one-third feet?
5.
A store has two sizes of peanut butter, an economy size of 41.2 ounces for \$3.60 and a regular size of
25.5 ounces for \$2.30. Which size offers the better deal?
6.
ASU has 5200 students and 390 faculty members; whereas, BSU has 17,800 students and 1480 faculty
members. Assuming the institutions have similar missions, which college should be able to offer the
students the most individual attention?
Property of Proportions
3 9
7 21
3 7
9 21
,
,
and
.
7 21
3 9
9 21
3 7
Note that all four represent the same proportional relationships between the values 3, 7, 9, and 21.
Consider the four proportions:
Property of Proportions.
a c
b d
a b
c d
if and only if
if and only if
if and only if
b d
a c
c d
a b
where a, b, c, and d are nonzero.
This property can often simplify the steps in solving a proportion.
320 960
.
731
N
960 3
N
3
, we may rewrite the above as
.
Since
320 1
731 1
Hence, N = 3 ∙ 731 = 2193.
By using the Property of Proportions, a person could do this problem mentally.
-3-
Example. Solve
Problems and Exercises
1.
The distribution of final grades in a mathematics class showed 4 A's, 6 B's, 12 C's, 8 D's, and 2 F's. What is the ratio
of: A's to the class, A's to F's, C's to A's, the class to C's, and B's to D's?
2.
The pitch of a roof is the ratio of the rise to the half-span. What is the pitch of a roof with a rise of six feet and a span of
eighteen feet?
3.
What is the ratio of: a nickel to a dime? a dozen to a gross? 20 minutes to 45 minutes? 1 foot 6 inches to 4 yards?
4.
What is the ratio of 325 miles to 5 hours? Interpret this ratio.
5.
Kim saved \$25 in 9 weeks. At that rate, how long will it take Kim to save \$175?
6.
A picture two and one-half inches wide and three and one-half inches high is to be enlarged so that the height will be
seven inches. How wide will it be?
7.
Solve each proportion.
3 n
(a)
8 56
(b)
14 n
18 27
(c)
10 5
d 16
(d)
9 3
c 53
(e)
16 96
7
x
(f)
4 15
13 y
-4-
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# Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x.
## Presentation on theme: "Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x."— Presentation transcript:
Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x + 6)(x – 1) What do you notice? Are they the same graph? Discuss what you can conclude from the graphs.
Roots What is the value of 2 2 ? What is x? However: x = In this case 2 is considered the principal root or the nonnegative root, when there is more than one real root. Finding the square root of a number and squaring a number are inverse operations. WHY? Should be 2, right?
Roots What is the value of The values found are known as the nth roots and are also principal roots. The following is the format for a radical expression. index radical sign radicand
Roots of Summary of the real nth roots. Real nth Roots of b,, or – nb > 0b < 0b = 0 even one positive root one negative root no real roots one real root, 0 odd one positive root no negative roots no positive roots one negative root
Practice with Roots Simplify the following. 1. 2. 3. 4. 5. has to be absolute value to identify principal root Estimated between 5 and 6 because 5 2 = 25 and 6 2 = 36.
Radical Expressions Radical “like expressions” have the same index and same radicand. Product and Quotient Properties:
Simplifying Radical Expressions 1.The index, n, has to be as small as possible 2.radicand NO factors, nth roots 3.radicand NO fractions 4.NO radical expressions in denominator For example:
More examples: Simplifying Radical Expressions
More examples: Simplifying Radical Expressions
Radical Expressions Conjugates ± same terms Multiply the following:
Radical Expressions Deduction about conjugates: Product of conjugates is always a rational number. For example:
Radical Expressions Simplify the following
Radical Expressions In-Class work Rationalize the denominator:
Download ppt "Checking Factoring The checking of factoring can be done with the calculator. Graph the following expressions: 1.x 2 + 5x – 6 2.(x – 3)(x – 2) 3.(x."
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# 3.5 Addition of velocities (Page 4/12)
Page 4 / 12
## Calculating relative velocity: an airline passenger drops a coin
An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?
Strategy
Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.
Solution for (a)
Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:
${{v}_{y}}^{2}={{v}_{0y}}^{2}-2g\left(y-{y}_{0}\right)\text{.}$
Substituting known values into the equation, we get
${{v}_{y}}^{2}={0}^{2}-2\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(-1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{m}-0 m\right)=\text{29}\text{.}4\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}{\text{/s}}^{2}$
yielding
${v}_{y}=-5\text{.}\text{42 m/s.}$
We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.
Solution for (b)
Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is ${v}_{y}=-5.42\phantom{\rule{0.25em}{0ex}}\text{m/s}$ , the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and . The x - and y -components of velocity can be combined to find the magnitude of the final velocity:
$v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\text{.}$
Thus,
$v=\sqrt{\left(\text{260 m/s}{\right)}^{2}+\left(-5\text{.}\text{42 m/s}{\right)}^{2}}$
yielding
$v=\text{260}\text{.}\text{06 m/s.}$
The direction is given by:
$\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)={\text{tan}}^{-1}\left(-5\text{.}\text{42}/\text{260}\right)$
so that
$\theta ={\text{tan}}^{-1}\left(-0\text{.}\text{0208}\right)=-1\text{.}\text{19º}\text{.}$
Discussion
In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not ; rather, it is . The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See [link] .) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
how does a model differ from a theory
what is vector quantity
Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
Besmellah
Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?
what's electromagnetic induction
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
yes it was an assignment question "^"represent raise to power pls
Gabriel
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
method of polarization
Ajayi
What is atomic number?
The number of protons in the nucleus of an atom
Deborah
type of thermodynamics
oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm
why the satellite does not drop to the earth explain
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Yinka
what is matter
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Mohamed
because space is a vacuum and anything outside the earth 🌎 can not come back without an act of force applied to it to leave the vacuum and fall down to the earth with a maximum force length of 30kcm per second
Clara
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# Polynomial Exponents Lessons
The previous lesson explained how to simplify exponents of a single term inside parentheses, like the problem below.
(x3y4)5
This lesson covers how to simplify exponents on parentheses that contain a polynomial (more than one term), like the problem below.
(x3 + y4)2
Because the two terms inside parentheses are not being multiplied or divided, the exponent outside the parentheses can not just be "distributed in". Instead, a 1 must be multiplied by the entire polynomial the number of times indicated by the exponent. In this problem the exponent is 2, so it is multiplied two times:
1(x3 + y4)(x3 + y4)
Use the FOIL Method to simplify the multiplication above, then combine like terms.
x6 + x3y4 + x3y4 + y8
x6 + 2x3y4 + y8
## Multiplication and Division
Examine the problem below.
(x3y4)5
Recall that multiplication is implied when there is no sign between a variable or set of parentheses and a number, another variable, or another set of parentheses. Therefore in this problem, the x3 and y4 are being multiplied.
In the next problem the x2 and x are being multiplied. The difference is that a * is present which explicitly indicates multiplication. We will solve this problem, then return to the first problem on the page.
(x2 * x)3
Because there is no addition or subtraction inside the parentheses, the exponent can be just "distributed" in and simplified:
(x2*3 * x3)
x6 * x3
x9
Notice that this gives the same result as if we had simplified the inside of the parentheses first, as we have done below.
(x2 * x)3
(x3)3
x3*3
x9
So why are there two different methods of solving this problem? The first method, where the exponent was distributed in can be applied to the first problem on this page, whereas the second method cannot.
We will now apply the "distribute in" method to the first problem presented on this page.
(x3y4)5
(x3*5y4*5)
x15y20
This method will also work when the terms are being divided, like the problem below:
(x2 / x)3
Again, the exponent is just "distributed" in:
(x2*3 / x3)
(x6 / x3)
x3
## Fractions
Fractions are really just a division problem which is shown in a special form. Since we can just "distribute" in the exponents for an ordinary division problem, we can do the same for a fraction. Look over the example below:
We can just distribute in the 3, as in the other problems.
As you can see, once the 3 was distributed, the parentheses could be removed. Then the 23 was simplified.
## Exponents of Polynomials (Parentheses) Resources
Practice Problems / WorksheetPractice all of the methods you learned in this lesson. Next Lesson: Order of OperationsLearn how to use the Order of Operations to simplify expressions containing more than one operation.
## Tutoring
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## Evaluate and estimate numerical square and cube roots
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Suppose that a shoemaker has determined that the optimal weight in ounces of a pair of running shoes is \begin{align*}\sqrt[4]{20000}\end{align*}. How many ounces would this be? Is there a way that you could rewrite this expression to make it easier to grasp? In this Concept, you'll learn how to simplify radical expressions like this one so that you can write them in multiple ways.
### Guidance
Radicals are the roots of values. In fact, the word radical comes from the Latin word “radix,” meaning “root.” You are most comfortable with the square root symbol \begin{align*}\sqrt{x}\end{align*}; however, there are many more radical symbols.
A radical is a mathematical expression involving a root by means of a radical sign.
Some roots do not have real values; in this case, they are called undefined.
Even roots of negative numbers are undefined.
\begin{align*}\sqrt[n]{x}\end{align*} is undefined when \begin{align*}n\end{align*} is an even whole number and \begin{align*}x<0\end{align*}.
#### Example A
• \begin{align*}\sqrt[3]{64}\end{align*}
• \begin{align*}\sqrt[4]{-81}\end{align*}
Solution:
\begin{align*}\sqrt[3]{64} = 4\end{align*} because \begin{align*}4^3=64\end{align*}
\begin{align*}\sqrt[4]{-81}\end{align*} is undefined because \begin{align*}n\end{align*} is an even whole number and \begin{align*}-81<0\end{align*}.
In a previous Concept, you learned how to evaluate rational exponents:
This can be written in radical notation using the following property.
Rational Exponent Property: For integer values of \begin{align*}x\end{align*} and whole values of \begin{align*}y\end{align*}:
#### Example B
Rewrite \begin{align*}x^{\frac{5}{6}}\end{align*} using radical notation.
Solution:
This is correctly read as the sixth root of \begin{align*}x\end{align*} to the fifth power. Writing in radical notation, \begin{align*}x^{\frac{5}{6}}=\sqrt[6]{x^5}\end{align*}, where \begin{align*}x^5>0\end{align*}.
You can also simplify other radicals, like cube roots and fourth roots.
#### Example C
Simplify \begin{align*}\sqrt[3]{135}\end{align*}.
Solution:
Begin by finding the prime factorization of 135. This is easily done by using a factor tree.
-->
### Guided Practice
Evaluate \begin{align*}\sqrt[4]{4^2}\end{align*}.
Solution: This is read, “The fourth root of four to the second power.”
The fourth root of 16 is 2; therefore,
### Explore More
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Radical Expressions with Higher Roots (8:46)
1. For which values of \begin{align*}n\end{align*} is \begin{align*}\sqrt[n]{-16}\end{align*} undefined?
1. \begin{align*}\sqrt{169}\end{align*}
2. \begin{align*}\sqrt[4]{81}\end{align*}
3. \begin{align*}\sqrt[3]{-125}\end{align*}
4. \begin{align*}\sqrt[5]{1024}\end{align*}
Write each expression as a rational exponent.
1. \begin{align*}\sqrt[3]{14}\end{align*}
2. \begin{align*}\sqrt[4]{zw}\end{align*}
3. \begin{align*}\sqrt{a}\end{align*}
4. \begin{align*}\sqrt[9]{y^3}\end{align*}
Write the following expressions in simplest radical form.
1. \begin{align*}\sqrt{24}\end{align*}
2. \begin{align*}\sqrt{300}\end{align*}
3. \begin{align*}\sqrt[5]{96}\end{align*}
4. \begin{align*}\sqrt{\frac{240}{567}}\end{align*}
5. \begin{align*}\sqrt[3]{500}\end{align*}
6. \begin{align*}\sqrt[6]{64x^8}\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 11.2.
### Vocabulary Language: English Spanish
A mathematical expression involving a root by means of a radical sign. The word radical comes from the Latin word radix, meaning root.
Rational Exponent Property
Rational Exponent Property
For integer values of $x$ and whole values of $y$: $a^{\frac{x}{y}}= \sqrt[y]{a^x}$
A radical expression is an expression with numbers, operations and radicals in it.
Rationalize the denominator
Rationalize the denominator
To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
Variable Expression
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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# Solving Quadratic Equations – Methods and Examples
Quadratic equations have the form ax²+bx+c=0. These equations can be solved using various methods depending on the type of quadratic equation we have. We can use methods for incomplete equations, solve equations by factoring, by completing the square, or with the quadratic formula.
Here, we will learn how to solve quadratic equations using different methods. We will use several examples to improve the learning process of the methods.
##### ALGEBRA
Relevant for
Learning to solve quadratic equations with various methods.
See methods
##### ALGEBRA
Relevant for
Learning to solve quadratic equations with various methods.
See methods
An incomplete quadratic equation is an equation that does not have a term from the form $latex ax^2+bx+c=0$, as long as the x² term is always present. When this is the case, we have two types of incomplete quadratic equations depending on the missing term.
### Solving quadratic equations that do not have the term bx
To solve quadratic equations of the form $latex ax^2+c=0$ that do not have the bx term, we need to isolate x² and take the square root of both sides of the equation.
For example, suppose we want to solve the equation $latex x^2-9=0$. First, we have to write it as follows:
$latex x^2=9$
Now that we have x² on the left-hand side, we can take the square root of both sides of the equation:
$latex x=\sqrt{9}$
$latex x=\pm 3$
Note: We must consider both the positive solution and the negative solution, since $latex (-3)^2=9$.
### Solving quadratic equations that do not have the term c
To solve equations of the form $latex ax^2+bx=0$ that do not have the constant term c, we have to factor the x on the left-hand side of the equation. Then, we form two equations with the factors and solve them.
For example, suppose we want to solve the equation $latex x^2-5x=0$. First, we factor it as follows:
$latex x(x-5)=0$
Since we have two factors, we can form an equation with each factor and solve:
$latex x=0~~$ or $latex ~~x-5=0$
$latex x=0~~$ or $latex ~~ x=5$
Note: In this type of equation, one of the solutions will always be $latex x=0$.
## Solving quadratic equations by factoring
The factorization method consists of finding the factors of the quadratic equation so that we have the roots of the equation exposed. By forming an equation with each factor, we can find the roots.
We can solve quadratic equations by factoring by following these steps:
Step 1: Simplify the equation if possible and write it in the form $latex ax^2+bx+c=0$.
Step 2: Find the factors of the equation using any method and write it in the form $latex (x+p)(x+q)=0$.
Step 3: Take each factor and set it equal to zero to form an equation. For example, $latex x+p=0$.
Step 4: Solve the equation for each factor.
There are several methods we can use to factor quadratic equations. The general idea is to find two factors of the form $latex (x+p)(x+q)=0$, which result in the form $latex x^2+bx+c=0$ when multiplied.
For example, the equation $latex x^2+2x-3=0$ can be factored into the form $latex (x+3)(x-2)=0$, since multiplying the factors gives us the original equation.
You can learn or review how to factor quadratic equations by visiting our article: Factoring Quadratic Equations.
This method allows us to find both roots of the equation relatively easily. However, it is not always possible to factor a quadratic equation.
## Solving quadratic equations by completing the square
Completing the square is a factoring technique that allows us to write an equation from the form $latex ax^2+bx+c=0$ to the form $latex (x-h)^2+k=0$. Thus, we can solve quadratic equations that cannot be easily factored.
To solve quadratic equations using the method of completing the square, we can follow the steps below:
Step 1: Simplify and write the equation in the form $latex ax^2+bx+c=0$.
Step 2: When a is different from 1, the entire equation must be divided by a to obtain an equation with a value of a equal to 1:
$latex x^2+bx+c=0$
Step 3: Divide the coefficient b by 2 to obtain:
$$\left(\frac{b}{2}\right)$$
Step 4: Square the expression from step 2:
$$\left(\frac{b}{2}\right)^2$$
Step 5: Add and subtract the expression obtained in step 4 to the equation obtained in step 2:
$$x^2+bx+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c=0$$
Step 6: Factor the equation using the identity $latex x^2+2xy+y^2=(x+y)^2$:
$$\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c=0$$
Step 7: Simplify to obtain an equation of the following form:
$latex (x-h)^2+k=0$
Step 8: Rearrange the equation as follows:
$latex (x-h)^2=-k$
Step 9: Take the square root of both sides of the equation:
$latex x-h=\sqrt{-k}$
Step 10: Solve for x:
$latex x=h\pm \sqrt{-k}$
When it is not possible to solve quadratic equations with any other method, we can use the quadratic formula, since this method allows us to find both solutions of any quadratic equation.
To use the general quadratic formula, we have to write the equation in the form $latex a{{x}^2}+bx+c=0$. This will allow us to identify the values of the coefficients a, b and c easily. We then use those values in the quadratic formula:
Note: We must not forget the ± sign, since in this way we will obtain both solutions to the quadratic equation when it is the case.
The expression inside the square root of the quadratic formula ($latex b^2-4ac$) is the discriminant of the quadratic equation. The discriminant determines the type of root that the quadratic equation will have.
Therefore, depending on the value of the discriminant, we have the following:
• When $latex b^2-4ac>0$, the equation has two real roots.
• When $latex b^2-4ac<0$, the equation has no real roots.
• When $latex b^2-4ac=0$, the equation has a repeated root.
When the value inside the square root of the formula is positive, we will have two real roots. When that value is negative, we won’t have real roots (but we will have imaginary or complex roots). When that value is equal to zero, we will have a single root.
The following examples are solved using all the methods for solving quadratic equations studied above. Try to solve the problems yourself before looking at the solution.
### EXAMPLE 1
What are the solutions of the equation $latex x^2-4=0$?
This equation is an incomplete quadratic equation that does not have the bx term. Therefore, we can find the solutions by isolating the quadratic term and taking the square root of both sides of the equation:
$latex x^2-4=0$
$latex x^2=4$
$latex x=\pm\sqrt{4}$
$latex x=\pm 2$
The solutions of the equation are $latex x=2$ and $latex x=-2$.
### EXAMPLE 2
Solve the equation $latex x^2-7x=0$.
This equation is an incomplete quadratic equation that does not have the constant term c. We can solve it by factoring the x and forming an equation with each factor:
$latex x^2-7x=0$
$latex x(x-7)=0$
$latex x=0 ~~$ or $latex ~~x-7=0$
$latex x=0 ~~$ or $latex ~~x=7$
The solutions of the equation are $latex x=0$ and $latex x=-7$.
### EXAMPLE 3
Solve the equation $latex x^2+2x-8=0$ using the factoring method.
Factoring the left-hand side of the equation, we have:
$latex x^2+2x-8=0$
$latex (x+4)(x-2)=0$
Now, we form an equation with each factor and solve:
$latex x+4=0~~$ or $latex ~~x-2=0$
$latex x=-4~~$ or $latex ~~x=2$
The solutions of the equation are $latex x=-4$ and $latex x=2$.
### EXAMPLE 4
Solve the equation $latex 2x^2-13x-24=0$ using the factoring method.
We can factor the left-hand side of the equation as follows:
$latex 2x^2-13x-24=0$
$latex (2x+3)(x-8)=0$
Now, we form an equation with each factor and solve:
$latex 2x+3=0~~$ or $latex ~~x-8=0$
$latex x=-\frac{3}{2}~~$ or $latex ~~x=8$
The solutions of the equation are $latex x=-\frac{3}{2}$ and $latex x=8$.
### EXAMPLE 5
Use the method of completing the square to solve the equation $latex x^2+4x-6=0$.
In this equation, the coefficient b is equal to 4. Therefore, we have:
$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$
$$=2^2$$
Now, we have to add and subtract that value to the quadratic equation:
$$x^2+4x-6=x^2+4x+2^2-2^2-6$$
Completing the square and simplifying, we have:
$latex = (x+2)^2-4-6$
$latex = (x+2)^2-10$
Rearranging, we form the equation:
$latex (x+2)^2=10$
Taking the square root of both sides, we have:
⇒ $latex x+2=\sqrt{10}$
Solving, we have:
⇒ $latex x=-2\pm \sqrt{10}$
### EXAMPLE 6
Solve the equation $latex 2x^2+8x-10=0$ by completing the square.
We can divide the entire equation by 2 to make the coefficient of the quadratic term equal to 1:
⇒ $latex x^2+4x-5=0$
Now, we can see that the coefficient b is equal to 4. Therefore, we have:
$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$
$$=2^2$$
If we add and subtract that expression to the quadratic equation, we have:
$$x^2+4x-5=x^2+4x+2^2-2^2-5$$
Completing the square and simplifying, we have:
$latex = (x+2)^2-4-5$
$latex = (x+2)^2-9$
Now, we rearrange the equation as follows:
⇒ $latex (x+2)^2=9$
And we take the square root of both sides:
⇒ $latex x+2=3~~$ or $latex ~~x+2=-3$
Solving, we have:
⇒ $latex x=1~~$ or $latex ~~x=-5$
### EXAMPLE 7
Use the general formula to solve the equation $latex 2x^2+3x-4=0$. Express the solutions to two decimal places.
This equation has the coefficients $latex a=2$, $latex b=3$, and $latex c=-4$. Therefore, using the quadratic formula with those values, we have:
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-(3)\pm \sqrt{( 3)^2-4(2)(-4)}}{2(2)}$$
$$=\frac{-3\pm \sqrt{9+32}}{4}$$
$$=\frac{-3\pm \sqrt{41}}{2}$$
$$=-2.35 \text{ or }0.85$$
The solutions of the equation are $latex x=-2.35$ and $latex x=0.85$.
## Solving quadratic equations – Practice problems
Solve the following problems using any of the methods we studied above.
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# Vector Equations – Explanation and Examples
In vector geometry, one of the most important concepts in solving real-world problems is using vector equations. The vector equation is defined as:
“The vector equation is an equation of vectors which is when solved, gives the result in the form of a vector.”
In this topic, we shall briefly discuss the following mentioned concepts:
• What is a vector equation?
• How to solve a vector equation?
• What is a vector equation of a straight line?
• What is a vector equation of a circle?
• Examples
• Problems
## What Is A Vector Equation?
A vector equation is an equation involving n numbers of vectors. More formally, it can be defined as an equation involving a linear combination of vectors with possibly unknown coefficients, and upon solving, it gives a vector in return.
Generally, a vector equation is defined as “Any function that takes anyone or more variable and in return gives a vector.”
Any vector equation involving vectors with n number of coordinates is similar to the linear equation system with n number of coordinates involving numbers. For example,
Consider a vector equation,
r <4,5,6> + t<3,4,1> = <8,5,9>
It can also be written as
<4r,5r,6r> + <3t,4t,1t> =<8,5,9>
Or
<4r+3t, 5r+4t, 6r+1t> = <8,5,9>
For two vectors to be equal, all the coordinates must be equal, so it can also be written as a system of linear equations. Such a representation is as follows:
4r+3t = 8
5r+4t = 5
6r+1t = 9
So, the vector equation can be solved by converting it into a system of linear equations. Hence, it simplifies and becomes easier to solve.
In our daily life, vectors play a vital role. Most of the physical quantities used are vector quantities. Vectors have many true applications, including the situations designated by force and velocity. For example, if a car is moving on a road, various forces will be acting on it. Some forces act in the forward direction and some in the backward direction to balance the system. So, all these forces are vector quantities. We use vector equations to find out various physical quantities in 2-D or 3-D, such as velocity, acceleration, momentum, etc.
Vector equations give us a diverse and more geometric way of viewing and solving the linear system of equations.
Overall, we can conclude that the vector equation is:
x1.t1+x2.t2+···+xk.tk = b
where t 1,t 2,…,t k,b are vectors in Rn and x 1,x 2,…,xk are unknown scalars, has the same solution set as the linear system with an augmented matrix of the given equation.
Therefore, the vector equation is given as,
r = r0 +kv
Let’s understand this concept with the help of examples.
Example 1
A car moves with a constant velocity on a straight road initially at time t=2 the position vector of the car is (1,3,5) then after some time at t=4, the car’s position vector is described as (5,6,8). Write down the vector equation of the position of the object. Also, express it in the form of parametric equations.
Solution
Since the vector equation of a straight-line is given as
r = r0 +tv
Since,
r0 = <1,3,5>
r = <5,6,8>
<5,6,8> = <1,3,5> + 4v
<5,6,8> – <1,3,5> = 4v
<4,3,3> = 4v
v = <1,3/4,3/4>
Now, finding vector equation of object’s position
r = r0 +tv
r = <1,3,5> + t<1,3/4,3/4>
where vector r is <x,y,z>
<x,y,z> = <1,3,5> + <1t,3/4t,3/4t>
Expressing in the form of the parametric equation:
As two vectors are only equivalent if their coordinates are equal. So, due to equality, we can write as,
x = 1+t
y = 3+3/4t
z = 5+3/4t
The vector equation of lines identifies the position vector of line with reference to the origin and direction vector and we can find out the dimensions of vectors corresponding to any length. This works for the straight lines and curves.
Note: The position vector is used to describe the position of the vector. It is a straight line having one end fixed and the other attached to the moving vector to specify its position.
Let’s understand this concept with the help of examples.
Example 2
Write down the following equations as vector equations
1. x=-2y+7
2. 3x=-8y+6
3. x=-3/5-8
Solution
Let’s consider equation 1 first:
x = -2y+7
Since the equation given above is an equation of a straight-line:
y = mx+c
Firstly, we will select two points on the given line.
Let’s simplify the equation,
x = -2y+7
let y = 0
x = 7
So, the first point is s (7,0) or OS (7,0)
Now let find out the second point that is halfway through the first point then,
Let x = 14
14 = -2y + 7
-2y = 7
y = -3.5
So, the second point T (14, -3.5) or OT (14, -3.5)
Then,
OS OT = (7,0) – (14, -3.5)
OS OT = (-7, 3.5)
So, the vector equation form of the above equation is,
R = <7,0> + k<-7,3.5>
R = <7-7k, 3.5k>
Now, let’s solve equation 2:
3x = -8y+6
Since the equation given above is an equation of a straight-line
y = mx+c
Firstly, we will select two points on the given line.
Let’s simplify the equation,
3x = -8y+6
let y = 0
x = 2
So, the first point is s (2,0) or OS (2,0)
Now let find out the second point that is halfway through the first point then,
Let x = 4
12 = -2y+7
-2y = 12-7
y = -5/2
So, the second point T (4, -5/2) or OT (4, -5/2)
Then,
OS OT = (2,0) – (4, -5/2)
OS OT = (-2, 5/2)
So, the vector equation form of the above equation is,
R = <2,0> + k<-2,5/2>
R = <2-2k, 5/2k>
Now, let’s do equation 3:
x = -3/5-8
Since the equation given above is an equation of a straight-line
y = mx+c
Firstly, we will select two points on the given line.
Let’s simplify the equation,
x = -3/5y+8
let y = 0
x = 8
So, the first point is s (8,0) or OS (8,0)
Now let find out the second point that is halfway through the first point then,
Let x=16
16 = -3/5y+8
-3/5y = 16-8
y = -13.33
So, the second point T (16, -13.33) or OT (16, -13.33)
Then,
OS OT = (8,0) – (16, -13.33)
OS OT = (-8, 13.33)
So, the vector equation form of the above equation is,
R = <8,0> + k<-8,13.33>
R = <8-8k, 13.33k>
## Vector Equation Of A Straight Line
We all are familiar with the equation of the line that is y=mx+c, generally called a slope-intercept form where m is the slope of the line and x and y are the point coordinates or intercepts defined on the x and y axes. However, this form of the equation is not enough to completely explain the line’s geometrical features. That’s why we use a vector equation to describe the position and direction of the line completely.
To find the points on the line, we will use the method of vector addition. We need to find out the position vector and the direction vector. For the position vector, we will add the position vector of the known point on the line to the vector v that lies on the line, as shown in the figure below.
So, the position vector r for any point is given as r = op + v
Then, the vector equation is given as
R = op + kv
Where k is a scalar quantity that belongs from RN, op is the position vector with respect to the origin O, and v is the direction vector. Basically, k tells you how many times you will go the distance from p to q in the specified direction. It can be ½ if half of the distance would be covered and so on.
If two points on the line are known, we can find out the line’s vector equation. Similarly, if we know the position vectors of two points op and oq on a line, we can also determine the vector equation of the line by using the vector subtraction method.
Where,
v = opoq
Therefore, the equation of vector is given as,
R = op +kv
Let’s solve some examples to comprehend this concept.
Example 3
Write down the vector equation of a line through points P (2,4,3) and Q (5, -2,6).
Solution
Let the position vector of the given points P and Q with respect to the origin is given as OP and OQ, respectively.
OP = (2,4,3) – (0,0,0)
OP = (2,4,3)
OQ = (5, -2,6) – (0,0,0)
OQ = (5, -2 ,6)
Since we know that the vector equation of a line is defined as,
R = OP + kv
Where v = OQOP
v = (5, -2,6) – (2,4,3)
v = (3, -6, 3)
So, the vector equation of the straight line is given as,
R = <2,4,3> + k<3, -6,3>
Example 4
Determine the vector equation of the line where k=0.75. If the points given on the line are defined as A (1,7) and B (8,6).
Solution:
k is the scale that can vary from -∞ to +∞. In this case, k is given as 0.75, which is the distance covered on AB in the given direction.
Let the position vector of the given points A and B with respect to the origin are OA and OB, respectively.
OA = (1,7) – (0,0)
OA = (1,7)
OB = (8,6) – (0,0)
OB = (8,6)
Since we know that the vector equation of a line is defined as,
R = OA +kv
Where v = OBOA
v = (8,6) – (1,7)
v = (7, -1)
So, the vector equation of the straight line is given as,
Where k=0.75
R = <1,7> + 0.75<7, -1>
Example 5
Write down the vector equation of a line through points P (-8,5) and Q (9,3).
Solution
Let the position vector of the given points P and Q with respect to the origin is given as OP and OQ, respectively.
OP = (-8,5) – (0,0)
OP = (-8,5)
OQ = (9,3) – (0,0)
OQ = (9,3)
Since we know that the vector equation of a line is defined as,
R = OP + kv
Where v = OQOP
v = (9,3) – (-8,5)
v = (17, -2)
So, the vector equation of the straight line is given as,
R = <-8,5> + k<17, -2>
## Vector Equation Of A Circle
Earlier, we have discussed the vector equation of a straight line. Now we will discuss the vector equation of a circle having radius r and with some center c, which we generally say that the circle is centered at c (0,0), but it may be located at any other point in the plane.
The vector equation of a circle is given as
r (t) = <x(t), y(t)>
where x(t) = r.cos(t) and y(t) = r.sin(t), r is the radius of the circle and t is the defined as the angle.
Let us consider a circle with center c and radius r, as shown in the figure below.
.
The position vector of the radius and center c is given as r and c, respectively. Then the radius of the circle is represented by vector CR, where CR is given as r c.
Since the radius is given as r so magnitude if CR can be written as
|CR| = r^2
Or
(r c). (r c) = r^2
Or
| r c| = r
This can also be called a vector equation of a circle.
Example 5
Write down the vector equation and the cartesian equation of a circle with center c at (5,7) and radius 5m.
Solution
Vector equation of a circle:
| r c| = r
| r – <5,7>| = 5
(r – <5,7>)^2 = 25
Cartesian equation of a circle:
(x-h)^2 +(y-k)^2 = r2
(x-5)^2 + (y-7)^2 = 25
Example 6
Determine if the point (2,5) lies on the circle with the vector equation of a circle given as |r -<-6,2>| = 3.
Solution
We must find out whether the given point lies inside the circle or not provided the circle’s vector equation.
Since putting the value of the point in the given vector equation
= |<2,5>-<-6,2>|
= |<2+6,5-2>|
= |<8,3>|
= √ ((8)^2+(3)^2)
= (64+9)
= (73) ≠ 3
Hence, the point does not lies inside the circle.
### Practice Problems
1. Write down the following equations as vector equations : x=3y+5 x=-9/5y+3 x+9y=4
2. Determine the equation for the line defined by points A (3,4,5) and B (8,6,7). Find the position vector for a point, half-way between the two points.
3. Write a vector equation of the line parallel to vector Q and passing through point o with the given position vector P.
Q = <-2,6> P = <3, -1>
Q = <1,8> P = <9, -3>
1. Write down the vector equation of a line through points P (-8/3,5) and Q (5,10).
2. A car moves with a constant velocity on a straight road initially at time t=2 the position vector of the car is (1/2,8) then after some time at t=4, the car’s position vector is described as (5,10). Write down the vector equation of the position of the object. Also, express it in the form of parametric equations.
3. Write down the vector equation and the cartesian equation of a circle with center c at (8,0) and radius 7m.
4. Determine if the point (3,-5) lies on the circle with the vector equation of a circle given as |r -<-3,4>| = 4.
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# Section 4.2: Least-Squares Regression
## Objectives
By the end of this lesson, you will be able to...
1. find the least-squares regression (LSR) line
2. use the LSR line to make predictions
3. interpret the slope and y-intercept of the LSR line
For a quick overview of this section, watch this short video summary:
Because we'll be talking about the linear relationship between two variables, we need to first do a quick review of lines.
## The Slope and Y-intercept
If there's one thing we all remember about lines, it's the slope-intercept form of a line:
The slope-intercept form of a line is
y = mx + b
where m is the slope of the line and b is the y-intercept.
Knowing the form isn't enough, though. We also need to know what each part means. Let's start with the slope. Most of us remember the slope as "rise over run", but that only helps us graph lines. What we really need to know is what the slope represents in terms of the original two variables. Let's look at an example to see if we can get the idea.
Example 1
The equation T = 6x + 53 roughly approximates the tuition per credit at ECC since 2001. In this case, x represents the number of years since 2001 and T represents the tuition amount for that year.
The graph below illustrates the relationship.
In this example, we can see that both the 6 and the 53 have very specific meanings:
The 6 is the increase per year. In other words, for every additional year, the tuition increases \$6.
The 53 represents the initial tuition, or the tuition per credit hour in 2001.
As we progress into the relationship between two variables, it's important to keep in mind these meanings behind the slope and y-intercept.
## Finding the Equation for a Line
Another very important skill is finding the equation for a line. In particular, it's important for us to know how to find the equation when we're given two points.
A very useful equation to know is the point-slope form for a line.
The point-slope form of a line is
y - y1 = m(x - x1)
where m is the slope of the line and (x1, y1) is a point on the line.
Let's practice using this form to find an equation for the line.
Example 2
In Example 1 from section 4.1, we talked about the relationship between student heart rates (in beats per minute) before and after a brisk walk.
before after before after before after 86 98 58 128 60 70 62 70 64 74 80 92 52 56 74 106 66 70 90 110 76 84 80 92 66 76 56 96 78 116 80 96 72 82 74 114 78 86 72 78 90 116 74 84 68 90 76 94
Let's highlight a pair of points on that plot and use those two points to find an equation for a line that might fit the scatter diagram.
Using the points (52, 56) and (90, 116), we get a slope of
m = 116-56 = 60 ≈ 1.58 90-52 38
So an equation for the line would be:
y - y1 = m(x - x1)
y - 56 = 1.58(x - 52)
y - 56 = 1.58x - 82.16
y = 1.58x - 26.16
It's interesting to note the meanings behind the slope and y-intercept for this example. A slope of 1.58 means that for every additional beat per minute before the brisk walk, the heart rate after the walk was 1.58 faster.
The y-intercept, on the other hand, doesn't apply in this case. A y-intercept of -26.16 means that if you have 0 beats per minute before the walk, you'll have -26.16 beats per minute after the walk. ?!?!
This brings up a very important point - models have limitations. In this case, we say that the y-intercept is outside the scope of the model.
Now that we know how to find an equation that sort of fits the data, we need a strategy to find the best line. Let's work our way up to it.
## Residuals
Unless the data line up perfectly, any line we use to model the relationship will have an error. We call this error the residual.
The residual is the difference between the observed and predicted values for y:
residual = observed y - predicted y
residual =
Notice here that we used the symbol (read "y-hat") for the predicted. This is standard notation in statistics, using the "hat" symbol over a variable to note that it is a predicted value.
Example 3
Let's again use the data from Example 1 from section 4.1. In Example 2 from earlier this section, we found the model:
= 1.58x - 30.16
Let's use this model to predict the "after" heart rate for a particular students, the one whose "before" heart rate was 86 beats per minute.
The predicted heart rate, using the model above, is:
= 1.58(86) - 26.16 = 109.72
Using that predicted heart rate, the residual is then:
residual = = 98 - 109.72 = -11.72
Here's that residual if we zoom in on that particular student:
Notice here that the residual is negative, since the predicted value was more than the actual observed "after" heart rate.
## The Least-Squares Regression (LSR) line
So how do we determine which line is "best"? The most popular technique is to make the sum of the squares of the residuals as small as possible. (We use the squares for much the same reason we did when we defined the variance in Section 3.2.) The method is called the method of least squares, for obvious reasons!
#### The Equation for the Least-Squares Regression line
The equation of the least-squares is given by
where
is the slope of the least-squares regression line
and
is the y-intercept of the least squares regression line
Let's try an example.
Example 4
Let's again use the data from Example 1 in Section 4.1, but instead of just using two points to get a line, we'll use the method of least squares to find the Least-Squares Regression line.
before after before after before after 86 98 58 128 60 70 62 70 64 74 80 92 52 56 74 106 66 70 90 110 76 84 80 92 66 76 56 96 78 116 80 96 72 82 74 114 78 86 72 78 90 116 74 84 68 90 76 94
Using computer software, we find the following values:
≈72.16667
sx ≈10.21366
= 90.75
sy
≈17.78922
r ≈
0.48649
Note: We don't want to round these values here, since they'll be used in the calculation for the correlation coefficient - only round at the very last step.
Using the formulas for the LSR line, we have
= 0.8473x + 29.60
(A good general guideline is to use 4 digits for the slope and y-intercept, though there is no strict rule.)
One thought that may come to mind here is that this doesn't really seem to fit the data as well as the one we did by picking two points! Actually, it does do a much better job fitting ALL of the data as well as possible - the previous line we did ourselves did not address most of the points that were above the main cluster. In the next section, we'll talk more about how outliers like the (58, 128) point far above the rest can affect a model like this one.
## Technology
Here's a quick overview of how to find the Least-Squares Regression line in StatCrunch.
Select Stat > Regression > Simple Linear Select the predictor variable for X & the response variable for Y Select Calculate The fourth line shows the equation of the regression line. Note that it will not have x and y shown, but rather the names that you've given for x and y. For example: Avg. Final Grade = 88.73273 - 2.8272727 Num. Absences
You can also go to the video page for links to see videos in either Quicktime or iPod format.
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# Rounding Numbers – Definition, Place-value Chart & Examples
## What is rounding off numbers?
Rounding off numbers is a mathematical technique of adjusting the number’s digits to make the number easier to use during calculations. Numbers are rounded off to a particular degree of accuracy to make calculations simpler and the results easier to understand.
Before rounding off any number, you should know the place of all digits of a number. Below is the place value chart.
## How to Round off the Whole Numbers?
It is fundamentally important to understand the term “rounding digit” when rounding off numbers.
For example, when rounding off numbers 100 to tens, the rounding digit is the second number from the right. Similarly, the rounding digit is in the third place when rounding to the nearest hundred, which is 1. Therefore, the first step when rounding a number is identifying the rounding digit and looking at the next digit to the right side.
• If the digit to the right side of the rounding digit is 0, 1, 2, 3, or 4, then the rounding digit doesn’t change. All digits to the right of the rounding digit become zero.
• If the digit to the right side of the rounding digit is 5, 6, 7, 8, or 9, the rounding digit increases by one digit. All the digits to the right are dropped to zero.
### Practice questions
1. Round off the following numbers to the nearest tens.
29
95
43
75
2. Round off these numbers to the nearest tens.
164
1,989
765
9,999,995
3. Round the following list of numbers to the nearest hundreds.
439
2,950
109,974
562
4. Round off the numbers below to the nearest thousands.
5,280
1,899,999
77,777
1,234,567
### Solutions
1. Rounding off to the nearest ten:
The digit to the right of the rounding digit in 29 is 9. Therefore, one digit is added to the rounding digit, 2, and the other digit is dropped to zero.
29 → 30
The digit to the right of the rounding digit in 43 is 3. The number does not affect the rounding digit, 4 and 3 are dropped to zero.
43 → 40
The digit to the right of the rounding digit in 75 is 5. One digit is added to the rounding digit, and 5 is dropped to zero.
75 → 80.
The digit to the right of the rounding digit in 95 is 5. One digit is added to 9, and the rest dropped to zero.
95 → 100.
2. Round to the nearest ten:
• 164 has the rounding digit as 6 and 4 as the right-hand side digit.
• 164 will become 160
• 765 will become 770.
• 1,989 → 1,990.
• 9,999,995 has 5 as the digit to the right of the rounding digit.
• 9,999,995 → 10,000,000
3. Rounding off to the nearest hundreds:
Identify the digit to the right (tens digit) of the rounding digit.
The tens digit in 439 is 3, so there is no effect on the hundreds digit:
439 →400
Number 6 is the tens digit or digit to the right of the rounding digit:
562 → 600.
5 is the tens digit in 2,950, so,
2,950 → 3,000.
109,974 has the tens digit as 7.
109,974 then becomes 110,000.
4. To round off to the nearest thousands, the hundreds digit is considered.
The hundred digit in 5,280 is 2, so it does not affect the rounding digit.
5,280, therefore, becomes 5,000.
77,777 becomes 78,000.
1,234, 567 becomes 1,235,000.
1,899,999 becomes 1,900,000.
## How to Round Off Decimal Numbers?
Decimal numbers are rounded off to estimate an answer quickly and easily. Decimal numbers can be rounded off to the nearest integer or whole number, tenths, hundredths, thousandths, etc.
### Rounding off to the nearest integer
The following rules are followed when rounding off a decimal to the nearest whole number:
• The number to be rounded off is identified.
• The digit in one’s place is marked.
• The first digit to the right of the decimal, or in the tenths place, is checked.
• If the tenths digit is less than or equal to 4, then the number in the one’s place is rounded off to a whole number.
• Similarly, if the tenths digit is greater than or equal to 5, add 1 digit to the number in the one’s place.
• Drop all digits after the decimal point to result in the desired whole number.
### Rounding to the nearest tenths
A similar procedure is applied when rounding a number to the nearest tenths.
• The number to be rounded off is first of all identified.
• The digit in the tenths place is marked.
• The digit in the hundredths place is checked.
• If the hundredths digit is less than or equal to 4, then the number in the tenths digit remains the same.
• Similarly, if the hundredths digit is greater than or equal to 5, add 1 digit to the number in the tenths place.
• Drop all digits to the right of the tenths column to result in the desired number.
### Rounding off to the nearest hundredths
The first step is to identify the required number. The digit in the hundredths place is marked. Check whether the digit in the thousandths place is 0, 1, 2, 3, 4 or 5, 6, 7, 8, 9. Round off the number to the nearest hundredths and drop the other numbers to the right of the hundredths place.
### Practice Questions
1. When rounded off to the nearest whole number, $975.6539$ is equal to which number?
2. When rounded off to the nearest whole number, $1240.68$ is equal to which number?
3. When rounded off to the nearest tenths, $845.324$ is equal to which decimal?
4. The length of a road is $27.56$ km. What is the length to the nearest km?
5. The weight of a boy is $37.35$ kg. Calculate the weight to the nearest kg.
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# Aim What are the arithmetic series and geometric
• Slides: 9
Aim: What are the arithmetic series and geometric series? Do Now: Find the sum of each of the following sequences: a) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 b) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 +. . . + 98 + 99 + 100 HW: p. 265 # 12, 14, 20 p. 272 # 6, 8, 10, 16 p. 278 # 8, 10
The sum of an arithmetic sequence is called arithmetic series Although we can find the arithmetic series one after the other, there is a formula to find the series faster.
Find the sum of the first 150 terms of the arithmetic sequence 5, 16, 27, 38, 49, . . . First we need to determine what the last term of the 150 terms (or the 150 th term) is. a 1 = 5; d = 16 – 5 = 11. a 150 = a 1 + d(150 – 1), a 150 = 5 + 11(149) = 1644
Write the sum of the first 15 terms of the arithmetic series 1 + 4 + 7 + · · · in sigma notation and then find the sum First of all, we need to find the recursive formula a 1 = 1 and d = 3 To find the sum, we need to find a 15
The sum of an geometric sequence is called geometric series The formula to find the finite (limited number of terms) geometric sequence is 3, 15, 75, 375, 1875, 9375, 46875, 234375, 1171875 is a geometric sequence, find the sum of sequence.
Write as a series and then find the sum
Infinite series : The last term of the series is the infinity An infinite arithmetic series has no limit An infinite geometric series has no limit when An infinite geometric series has a finite limit when the limit can be found by the formula
Find the sum of the following infinite geometric sequence: 4, 4(0. 6)2, 4(0. 6)3, . . . , 4(0. 6)n - 1 , . . . a 1 = 4 and r = 0. 6 Find
Find the sum of the first 10 terms of the geometric series Find the sum of five terms of the geometric series whose first term is 2 and fifth term is 162 S 5 = 242
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Read- Identify Arithmetic Patterns Video for Kids | 3th, 4th & 5th Grade
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# Read About Identifying Number Patterns (In Arithmetic and the Multiplication Table)
WHAT IS IDENTIFYING NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE)
Having explored patterns with shapes, you will now discover that patterns can be found in numbers as well! Specifically, you will explore outcomes when even and odd numbers are added, subtracted, and multiplied.
To better understand identifying number patterns…
WHAT IS IDENTIFYING NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE). Having explored patterns with shapes, you will now discover that patterns can be found in numbers as well! Specifically, you will explore outcomes when even and odd numbers are added, subtracted, and multiplied. To better understand identifying number patterns…
## LET’S BREAK IT DOWN!
### Odd and Even Socks
Adesina, April, and Marcos are folding laundry. Each person has several socks, and they want to know how they can tell if it’s an even or odd number. If we can put all the socks into pairs, then there is an even number. April has 6 socks, and they can be put into 3 pairs, so 6 is even. Marcos has 8 socks, and they can all be paired up as well, so 8 is even. Combined, they have 14 pairs of socks, and 14 socks can all be paired up as well, so that number is even. They have just discovered a pattern: Even + even = even, since there are no left overs. Next, they fold socks that are a different color. This time, Marcos has 5 socks and April has 9 socks. 5 is an odd number because one of the socks is not paired. 9 is also an odd number because one of the socks is unpaired. But, if Marcos and Adesina put their socks together, now their unpaired socks make a pair together! Therefore, odd + odd = even. Next, they work on a new pile of socks. This time, Marcos has 6 socks and April has 5 socks. 6 is even, since Marcos' socks can all be paired. 5 is odd, since April has one sock left unpaired. If they combine their socks, one sock is still unpaired, and the answer is still odd. Therefore, odd + even = odd! Now you try: Without calculating, can you tell if 7 + 12 is odd or even?
Odd and Even Socks Adesina, April, and Marcos are folding laundry. Each person has several socks, and they want to know how they can tell if it’s an even or odd number. If we can put all the socks into pairs, then there is an even number. April has 6 socks, and they can be put into 3 pairs, so 6 is even. Marcos has 8 socks, and they can all be paired up as well, so 8 is even. Combined, they have 14 pairs of socks, and 14 socks can all be paired up as well, so that number is even. They have just discovered a pattern: Even + even = even, since there are no left overs. Next, they fold socks that are a different color. This time, Marcos has 5 socks and April has 9 socks. 5 is an odd number because one of the socks is not paired. 9 is also an odd number because one of the socks is unpaired. But, if Marcos and Adesina put their socks together, now their unpaired socks make a pair together! Therefore, odd + odd = even. Next, they work on a new pile of socks. This time, Marcos has 6 socks and April has 5 socks. 6 is even, since Marcos' socks can all be paired. 5 is odd, since April has one sock left unpaired. If they combine their socks, one sock is still unpaired, and the answer is still odd. Therefore, odd + even = odd! Now you try: Without calculating, can you tell if 7 + 12 is odd or even?
### Googly Eyes
Adesina starts by putting 12 googly eyes into pairs. 12 can be put into 6 pairs, so 12 is even. If we subtract 4, which is an even number, then we have 8 googly eyes left, which is also even. Subtracting an even number from an even number doesn't leave any googly eyes unpaired. Even – even = even. If we try subtracting an odd number from an odd number, like 11 – 5 = 6, we end up with an even number! The unpaired googly eye in 11 is taken away by the unpaired eye in 5. Odd – odd = even. If we subtract an odd number from an even number, like 14 – 3 = 11, we get an odd number. Likewise, if we subtract an even number from an odd number, like 13 – 4 = 9, we also get an odd number. Then, even – odd = odd, and odd – even = odd. Now you try: Without calculating, can you tell if 21 – 9 is even or odd?
Googly Eyes Adesina starts by putting 12 googly eyes into pairs. 12 can be put into 6 pairs, so 12 is even. If we subtract 4, which is an even number, then we have 8 googly eyes left, which is also even. Subtracting an even number from an even number doesn't leave any googly eyes unpaired. Even – even = even. If we try subtracting an odd number from an odd number, like 11 – 5 = 6, we end up with an even number! The unpaired googly eye in 11 is taken away by the unpaired eye in 5. Odd – odd = even. If we subtract an odd number from an even number, like 14 – 3 = 11, we get an odd number. Likewise, if we subtract an even number from an odd number, like 13 – 4 = 9, we also get an odd number. Then, even – odd = odd, and odd – even = odd. Now you try: Without calculating, can you tell if 21 – 9 is even or odd?
### Multiplication Table
Adesina, Marcos, and April are looking at a multiplication table. They notice some interesting number patterns: The row showing multiples of 2 contains all even numbers. All multiples of 2 are even! Looking at the row with multiples of 5, they notice that you add 5 each time you move along the row. All multiples of 5 end in 5 or 0. This is how you know you cannot divide 23 by 5: it doesn’t end in 5 or 0. Looking at the row with multiples of 10, we can see that all multiples of 10 end in 0. Discovering these patterns can help us check our work. Now you try: Is 73 divisible by 5? Is 90 divisible by 10? Is 43 divisible by 2?
Multiplication Table Adesina, Marcos, and April are looking at a multiplication table. They notice some interesting number patterns: The row showing multiples of 2 contains all even numbers. All multiples of 2 are even! Looking at the row with multiples of 5, they notice that you add 5 each time you move along the row. All multiples of 5 end in 5 or 0. This is how you know you cannot divide 23 by 5: it doesn’t end in 5 or 0. Looking at the row with multiples of 10, we can see that all multiples of 10 end in 0. Discovering these patterns can help us check our work. Now you try: Is 73 divisible by 5? Is 90 divisible by 10? Is 43 divisible by 2?
### Multiplying Evens and Odds
Let’s see what happens when we multiply using even and odd numbers. To find the product of two numbers, we can look up the two numbers on the multiplication table to see what they multiply to. 2 and 4 are even, and when we multiply them, we get 8, which is also an even number. Then, even × even = even. 1 and 3 are odd, and if we multiply them, we get 3, an odd number. Same with 5 × 7 = 35. Then, odd × odd = odd. 2 is even and 3 is odd, and if we multiply them, we get 6, an even number. Same with 6 × 9 = 54. Then, even × odd = even. Also, odd × even = even. Now you try: Without calculating, can you tell if 11 × 16 is odd or even?
Multiplying Evens and Odds Let’s see what happens when we multiply using even and odd numbers. To find the product of two numbers, we can look up the two numbers on the multiplication table to see what they multiply to. 2 and 4 are even, and when we multiply them, we get 8, which is also an even number. Then, even × even = even. 1 and 3 are odd, and if we multiply them, we get 3, an odd number. Same with 5 × 7 = 35. Then, odd × odd = odd. 2 is even and 3 is odd, and if we multiply them, we get 6, an even number. Same with 6 × 9 = 54. Then, even × odd = even. Also, odd × even = even. Now you try: Without calculating, can you tell if 11 × 16 is odd or even?
## IDENTIFY NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE) VOCABULARY
Pattern
Something that follows a rule.
Combined
Pair
Two of the same thing.
Sequence
When one thing follows another according to a rule.
Equation
Mathematical expressions that are equal to the same number.
Multiplication table
A tool that helps us quickly see the product of two numbers.
Can be put into two equal groups, or into groups of two.
Cannot be put into two equal groups, or into groups of two.
## IDENTIFY NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE) DISCUSSION QUESTIONS
### If I have 4 socks and you have 8 socks, can we pair them all up if we put them together?
Yes! 4 can be put into pairs, and 8 can be put into pairs, so if we put them together, we are just adding pairs.
### If I have 5 socks and you have 11 socks, can we pair them all up if we put them together?
Yes. All my socks can be put into pairs except one. All your socks can be put into pairs except one. Together, we have several pairs of socks and two unpaired socks. Those unpaired socks become a pair together!
### If you start with 12 socks and take away 3 socks, you have 9 socks left. Which rule can you learn from this story?
I started with an even number of socks that can be put into pairs. Then I took away an odd number of socks, which means that I take away some pairs and also break up a pair to take one sock. That means that I have an unpaired sock left over, so the amount of leftover socks is an odd number. The rule is: even – odd = odd.
### What happens if you multiply an even number by an odd number? Explain.
I get an even number! I can think of it as an odd number of groups of even numbers (which can be paired).
### What happens if you multiply an odd number by an odd number? Explain.
I get an odd number. For example, if I have 3 groups that each contain 5 socks, the socks in each group make 2 pairs with 1 sock unpaired. When I combine the groups, there are 3 unpaired socks, which still leaves 1 unpaired.
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# A-level Mathematics/OCR/C3/Numerical Methods
< A-level Mathematics | OCR | C3
## Finding Roots of an Equation
In this module we will explore approximating where an equation has a root. Below we have two graphs.
### Finding A Root Using a Graph
In the graph we have two functions. If we want to approximate where the roots are we have to look at where the function cross the x-axis. The lilac function crosses the x-axis somewhere between 2.05 and 2.25. The green function has roots around 1 and 3.5. If we want to know where the green function is equivalent to the lilac function we need to look at the graph. When the two graphs cross they are equivalent. This number will be the root. In this case it is around 1.75. We can also say that on this domain, the functions will only cross once.
### Finding A Root By Searching for A Sign Change
When a function has a root the value of the function will change from a positive value to a negative value or vice versa. If below is the table of values for the lilac function we can say that the root occurs between 2.05 and 2.10:
x f(x) 2 -1 2.05 -.4849 2.1 .061 2.15 .63838 2.2 1.248 2.25 1.08906
## Iterative Formulae
An iterative formula is a formula that is composed of itself. That is the output of the function is the next input of the function. ${\displaystyle x_{n+1}=F(n)}$. These functions are a sequence of approximations that usually converge to the value of a function. You can tell if a function converges by the fact that the outputs become closer and closer to each other, if this does not happen then the function diverges and there is no value. The output of an iterative formula is written as:
${\displaystyle x_{2}\,}$
${\displaystyle x_{3}\,}$
${\displaystyle x_{4}\,}$
The first x is provided for you. The output of the iterative formula is written as the Greek letter alpha: α. α can be found to the required degree of accuracy when ${\displaystyle x_{n}=x_{n+1}}$ to the required number of decimal places. Given an iterative function you can find the root of an equation. Also you find the function from a given iterative function by setting x = iterative function, then solving so that you have a zero on one side. The aforementioned process can be done in reverse.
### Example
Given the sequence that is defined by the iterative formula ${\displaystyle \left(2x+5\right)^{\frac {1}{3}}}$ with ${\displaystyle x_{1}=2}$ converges to ${\displaystyle \alpha }$
1. Find ${\displaystyle \alpha }$ correct to 4 decimal places.
2. Find equation that has ${\displaystyle \alpha }$ as a root.
3. Does the equation have any more roots?
With an iterative formula we have:
1. We plug in ${\displaystyle x_{1}}$ to get ${\displaystyle x_{2}}$ and so on
1. ${\displaystyle x_{2}=\left(2\times (2)+5\right)^{\frac {1}{3}}=2.0801}$
2. ${\displaystyle x_{3}=\left(2\times (2.0801)+5\right)^{\frac {1}{3}}=2.0924}$
3. ${\displaystyle x_{4}=\left(2\times (2.0924)+5\right)^{\frac {1}{3}}=2.0942}$
4. ${\displaystyle x_{5}=\left(2\times (2.0942)+5\right)^{\frac {1}{3}}=2.0945}$
5. ${\displaystyle x_{6}=\left(2\times (2.0945)+5\right)^{\frac {1}{3}}=2.0945}$
6. ${\displaystyle \alpha =2.0945\,}$
2. ${\displaystyle x=\left(2x+5\right)^{\frac {1}{3}}}$
1. ${\displaystyle x^{3}=2x+5\,}$
2. ${\displaystyle x^{3}-2x-5=0\,}$
3. To determine if the function has any roots just graph the graph of the highest power of x and then the rest. The number of times they cross is the number of roots you have.
1. If we plot ${\displaystyle y=x^{3}}$ and ${\displaystyle y=2x+5}$ we can see that they only cross once.
2. The function only has one root.
## Simpson's Rule for Area
To find the area beneath a curve we have already learned the trapezium rule. The trapezium rule is not very accurate it takes a very large number of trapezoids to get an very accurate area. The Simpson rule is much more accurate to find the area underneath a curve. Simpson's Rule states:
${\displaystyle \int _{a}^{b}ydx\approx {\frac {1}{3}}h\left\{\left(y_{0}+y_{n}\right)+4\left(y_{1}+y_{3}+\ldots +y_{n-1}\right)+2\left(y_{2}+y_{4}+\ldots +y_{n-2}\right)\right\}}$
where${\displaystyle h={\frac {b-a}{n}}}$ and n is even
#### Example
Use Simpson's Rule to evaluate ${\displaystyle \int _{1}^{5}x^{2}+2x\,dx}$ with h = 1.
${\displaystyle \int _{1}^{5}x^{2}+2x\,dx\approx {\frac {1}{3}}\left[\left(1^{2}+2\times 1\right)+4\left(2^{2}+2\times 2\right)+2\left(3^{2}+2\times 3\right)+4\left(4^{2}+2\times 4\right)+\left(5^{2}+2\times 5\right)\right]}$
${\displaystyle \int _{1}^{5}x^{2}+2x\,dx=65{\frac {1}{3}}}$
The area under the curve ${\displaystyle x^{2}+2x\,}$is equal to ${\displaystyle 65{\frac {1}{3}}}$.
This is the true value. If we compare it to the value of 66 that we trapezium rule and to the value of 65 that we got from the midpoint rule we can see that Simpson's rule is the most accurate.
This is part of the C3 (Core Mathematics 3) module of the A-level Mathematics text.
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# 7.11: Factorization of Special Cubics
Difficulty Level: Advanced Created by: CK-12
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Factor the following cubic polynomial: 375x3+648\begin{align*}375x^3+648\end{align*}.
### Guidance
While many cubics cannot easily be factored, there are two special cases that can be factored quickly. These special cases are the sum of perfect cubes and the difference of perfect cubes.
• Factoring the sum of two cubes follows this pattern: x3+y3=(x+y)(x2xy+y2)\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}
• Factoring the difference of two cubes follows this pattern: x3y3=(xy)(x2+xy+y2)\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}
#### Example A
Factor: x3+27\begin{align*}x^3+27\end{align*}.
Solution: This is the sum of two cubes and uses the factoring pattern: x3+y3=(x+y)(x2xy+y2)\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
x3+33=(x+3)(x23x+9)\begin{align*}x^3+3^3=(x+3)(x^2-3x+9)\end{align*}.
#### Example B
Factor: x3343\begin{align*}x^3-343\end{align*}.
Solution: This is the difference of two cubes and uses the factoring pattern: x3y3=(xy)(x2+xy+y2)\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
x373=(x7)(x2+7x+49)\begin{align*}x^3-7^3=(x-7)(x^2+7x+49)\end{align*}.
#### Example C
Factor: 64x31\begin{align*}64x^3-1\end{align*}.
Solution: This is the difference of two cubes and uses the factoring pattern: x3y3=(xy)(x2+xy+y2)\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
(4x)313=(4x1)(16x2+4x+1)\begin{align*}(4x)^3-1^3=(4x-1)(16x^2+4x+1)\end{align*}.
#### Concept Problem Revisited
Factor the following cubic polynomial: 375x3+648\begin{align*}375x^3+648\end{align*}.
First you need to recognize that there is a common factor of 3\begin{align*}3\end{align*}. 375x3+648=3(125x3+216)\begin{align*}375x^3+648=3(125x^3+216)\end{align*}
Notice that the result is the sum of two cubes. Therefore, the factoring pattern is x3+y3=(x+y)(x2xy+y2)\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
375x3+648=3(5x+6)(25x230x+36)\begin{align*}375x^3 +648 = 3(5x+6)(25x^2-30x+36)\end{align*}
### Vocabulary
Difference of Two Cubes
The difference of two cubes is a special polynomial in the form of x3y3\begin{align*}x^3-y^3\end{align*}. This type of polynomial can be quickly factored using the pattern: (x3y3)=(xy)(x2+xy+y2)\begin{align*}(x^3-y^3)=(x-y)(x^2+xy+y^2)\end{align*}
Sum of Two Cubes
The sum of two cubes is a special polynomial in the form of x3+y3\begin{align*}x^3+y^3\end{align*}. This type of polynomial can be quickly factored using the pattern: (x3+y3)=(x+y)(x2xy+y2)\begin{align*}(x^3+y^3)=(x+y)(x^2-xy+y^2)\end{align*}
### Guided Practice
Factor each of the following cubics.
1. x3+512\begin{align*}x^3+512\end{align*}
2. 8x3+125\begin{align*}8x^3+125\end{align*}
3. x3216\begin{align*}x^3-216\end{align*}
1. x3+83=(x+8)(x28x+64)\begin{align*}x^3+8^3=(x+8)(x^2-8x+64)\end{align*}.
2. (2x)3+53=(2x+5)(4x210x+25)\begin{align*}(2x)^3+5^3=(2x+5)(4x^2-10x+25)\end{align*}.
3. x363=(x6)(x2+6x+36)\begin{align*}x^3-6^3=(x-6)(x^2+6x+36)\end{align*}.
### Practice
Factor each of the following cubics.
1. x3+h3\begin{align*}x^3+h^3\end{align*}
2. a3+125\begin{align*}a^3+125\end{align*}
3. 8x3+64\begin{align*}8x^3+64\end{align*}
4. x3+1728\begin{align*}x^3+1728\end{align*}
5. 2x3+6750\begin{align*}2x^3+6750\end{align*}
6. h364\begin{align*}h^3-64\end{align*}
7. s3216\begin{align*}s^3-216\end{align*}
8. p3512\begin{align*}p^3-512\end{align*}
9. 4e332\begin{align*}4e^3-32\end{align*}
10. 2w3250\begin{align*}2w^3-250\end{align*}
11. x3+8\begin{align*}x^3+8\end{align*}
12. y31\begin{align*}y^3-1\end{align*}
13. 125e38\begin{align*}125e^3-8\end{align*}
14. 64a3+2197\begin{align*}64a^3+2197\end{align*}
15. 54z3+3456\begin{align*}54z^3+3456\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Cubed
The cube of a number is the number multiplied by itself three times. For example, "two-cubed" = $2^3 = 2 \times 2 \times 2 = 8$.
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What Is 10/55 as a Decimal + Solution With Free Steps
The fraction 10/55 as a decimal is equal to 0.181.
A fractional form is the way of representation of a decimal operation. This is expressed as a fractional number o/p, where o is the numerator (upper value) and p is the denominator (lower value).
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 10/55.
Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 10
Divisor = 55
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 10 $\div$ 55
This is when we go through the Long Division solution to our problem. Given is the long division process in Figure 1:
Figure 1
10/55 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 10 and 55, we can see how 10 is Smaller than 55, and to solve this division, we require that 10 be Bigger than 55.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 10, which after getting multiplied by 10 becomes 100.
We take this 100 and divide it by 55; this can be done as follows:
100 $\div$ 55 $\approx$ 1
Where:
55 x 1 = 55
This will lead to the generation of a Remainder equal to 100 – 55 = 45. Now this means we have to repeat the process by Converting the 45 into 450 and solving for that:
450 $\div$ 55 $\approx$ 8
Where:
55 x 8 = 440
This, therefore, produces another Remainder which is equal to 450 – 440 = 10. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 100.
100 $\div$ 55 $\approx$ 1
Where:
55 x 1 = 55
Finally, we have a Quotient generated after combining the three pieces of it as 0.181, with a Remainder equal to 45.
Images/mathematical drawings are created with GeoGebra.
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# Box & Whisker Plot: Lesson for Kids
I have a Bachelor's degree in Elementary Education and Spanish. I have taught for 5 years in bilingual classrooms of various elementary grade levels.
What do you do when you have a lot of numbers that are spread out, and you want to group them into sets and graph them? In this lesson, you will learn how to create and read a box and whisker plot which is perfect to use with that type of data.
## Uses of a Box and Whisker Plot
Cookies, cupcakes, fruit pies! Oh my! Your school just had its annual bake sale, and each class had to make and sell their sweets. Your teacher wants to see how well the 16 students in her class did and suggests using a box and whisker plot, but what exactly is that?
A box and whisker plot is a special type of graph that is used to show groups of number data and how they are spread. It shows the median, which is the middle value of the numbers in your data, the lowest number, the highest number and the quartiles, which divides the data into four equal groups. Quartiles are like quarters. There are 4 quarters in a dollar, each of which is worth 25 cents, so there are 4 quartiles in a box and whisker plot, each showing 25% of the number data. Let's take a look at our bake sale numbers and put this information to use.
## Creating a Box and Whisker Plot
Take a look at the image which shows the numbers of goodies each classmate sold. Now it's time to create your own box and whisker plot! Your teacher tells you to:
1. Order the numbers from least to greatest:
• 0, 3, 7, 9, 10, 15, 22, 27, 28, 30, 32, 35, 36, 40, 50, 55
2. Find the median and record it on the line. If there is an odd amount of number data, this will be the middle number. If there is an even amount of number data, take the average of the two numbers in the middle. This will create halves.
• We have 16 numbers, so our two middle numbers are 27 and 28. Add them together and divide by two to find the average = 27.5. This is our median.
3. Plot the lowest number at the beginning and the highest number at the end of the number line.
• 0 is our lower extreme and 55 is our upper extreme
4. Find the median for each half of the data - this is dividing the data into quartiles. To do this, you must repeat the process for finding the median, but do it for each half. Record both numbers on the line.
• Our first half of the data are the first eight numbers. The two middle numbers are 9 and 10. So, the average is 9.5. Our second half of data are the last eight numbers. The middle numbers are 35 and 36, so the average is 35.5.
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## Numerical Summaries of Data
### Learning Outcomes
• Calculate the mean, median, and mode of a set of data
• Calculate the range of a data set, and recognize it’s limitations in fully describing the behavior of a data set
• Calculate the standard deviation for a data set, and determine it’s units
• Identify the difference between population variance and sample variance
• Identify the quartiles for a data set, and the calculations used to define them
• Identify the parts of a five number summary for a set of data, and create a box plot using it
It is often desirable to use a few numbers to summarize a data set. One important aspect of a set of data is where its center is located. In this lesson, measures of central tendency are discussed first. A second aspect of a distribution is how spread out it is. In other words, how much the data in the distribution vary from one another. The second section of this lesson describes measures of variability.
## Measures of Central Tendency
### Mean, Median, and Mode
Let’s begin by trying to find the most “typical” value of a data set.
Note that we just used the word “typical” although in many cases you might think of using the word “average.” We need to be careful with the word “average” as it means different things to different people in different contexts. One of the most common uses of the word “average” is what mathematicians and statisticians call the arithmetic mean, or just plain old mean for short. “Arithmetic mean” sounds rather fancy, but you have likely calculated a mean many times without realizing it; the mean is what most people think of when they use the word “average.”
### Mean
The mean of a set of data is the sum of the data values divided by the number of values.
### examples
Marci’s exam scores for her last math class were 79, 86, 82, and 94. What would the mean of these values would be?
The number of touchdown (TD) passes thrown by each of the 31 teams in the National Football League in the 2000 season are shown below.
37 33 33 32 29 28 28 23 22 22 22 21 21 21 20
20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6
What is the mean number of TD passes?
Both examples are described further in the following video.
The price of a jar of peanut butter at 5 stores was $3.29,$3.59, $3.79,$3.75, and $3.99. Find the mean price. ### examples The one hundred families in a particular neighborhood are asked their annual household income, to the nearest$5 thousand dollars. The results are summarized in a frequency table below.
Income (thousands of dollars) Frequency 15 6 20 8 25 11 30 17 35 19 40 20 45 12 50 7
What is the mean average income in this neighborhood?
Extending off the last example, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand).
What is the new mean of this neighborhood’s income?
Both situations are explained further in this video.
While 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a “typical” value. Imagine the data values on a see-saw or balance scale. The mean is the value that keeps the data in balance, like in the picture below. If we graph our household data, the$5 million data value is so far out to the right that the mean has to adjust up to keep things in balance.
For this reason, when working with data that have outliers – values far outside the primary grouping – it is common to use a different measure of center, the median.
### Median
The median of a set of data is the value in the middle when the data is in order.
• To find the median, begin by listing the data in order from smallest to largest, or largest to smallest.
• If the number of data values, N, is odd, then the median is the middle data value. This value can be found by rounding N/2 up to the next whole number.
• If the number of data values is even, there is no one middle value, so we find the mean of the two middle values (values N/2 and N/2 + 1)
### example
Returning to the football touchdown data, we would start by listing the data in order. Luckily, it was already in decreasing order, so we can work with it without needing to reorder it first.
37 33 33 32 29 28 28 23 22 22 22 21 21 21 20
20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6
What is the median TD value?
Find the median of these quiz scores: 5 10 8 6 4 8 2 5 7 7
The price of a jar of peanut butter at 5 stores was $3.29,$3.59, $3.79,$3.75, and $3.99. Find the median price. ### Example Let us return now to our original household income data Income (thousands of dollars) Frequency 15 6 20 8 25 11 30 17 35 19 40 20 45 12 50 7 What is the mean of this neighborhood’s household income? If we add in the new neighbor with a$5 million household income, then there will be 101 data values, and the 51st value will be the median. As we discovered in the last example, the 51st value is 35 thousand. Notice that the new neighbor did not affect the median in this case. The median is not swayed as much by outliers as the mean is. View more about the median of this neighborhood’s household incomes here. In addition to the mean and the median, there is one other common measurement of the “typical” value of a data set: the mode. ### Mode The mode is the element of the data set that occurs most frequently. The mode is fairly useless with data like weights or heights where there are a large number of possible values. The mode is most commonly used for categorical data, for which median and mean cannot be computed. ### Example In our vehicle color survey earlier in this section, we collected the data Color Frequency Blue 3 Green 5 Red 4 White 3 Black 2 Grey 3 Which color is the mode? Mode in this example is explained by the video here. It is possible for a data set to have more than one mode if several categories have the same frequency, or no modes if each every category occurs only once. ### Try It Reviewers were asked to rate a product on a scale of 1 to 5. Find 1. The mean rating 2. The median rating 3. The mode rating Rating Frequency 1 4 2 8 3 7 4 3 5 1 ## Measures of Variation ### Range and Standard Deviation Consider these three sets of quiz scores: Section A: 5 5 5 5 5 5 5 5 5 5 Section B: 0 0 0 0 0 10 10 10 10 10 Section C: 4 4 4 5 5 5 5 6 6 6 All three of these sets of data have a mean of 5 and median of 5, yet the sets of scores are clearly quite different. In section A, everyone had the same score; in section B half the class got no points and the other half got a perfect score, assuming this was a 10-point quiz. Section C was not as consistent as section A, but not as widely varied as section B. In addition to the mean and median, which are measures of the “typical” or “middle” value, we also need a measure of how “spread out” or varied each data set is. There are several ways to measure this “spread” of the data. The first is the simplest and is called the range. ### Range The range is the difference between the maximum value and the minimum value of the data set. ### example Using the quiz scores from above, For section A, the range is 0 since both maximum and minimum are 5 and 5 – 5 = 0 For section B, the range is 10 since 10 – 0 = 10 For section C, the range is 2 since 6 – 4 = 2 In the last example, the range seems to be revealing how spread out the data is. However, suppose we add a fourth section, Section D, with scores 0 5 5 5 5 5 5 5 5 10. This section also has a mean and median of 5. The range is 10, yet this data set is quite different than Section B. To better illuminate the differences, we’ll have to turn to more sophisticated measures of variation. The range of this example is explained in the following video. ### Standard deviation The standard deviation is a measure of variation based on measuring how far each data value deviates, or is different, from the mean. A few important characteristics: • Standard deviation is always positive. Standard deviation will be zero if all the data values are equal, and will get larger as the data spreads out. • Standard deviation has the same units as the original data. • Standard deviation, like the mean, can be highly influenced by outliers. Using the data from section D, we could compute for each data value the difference between the data value and the mean: data value deviation: data value – mean 0 0-5 = -5 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 10 10-5 = 5 We would like to get an idea of the “average” deviation from the mean, but if we find the average of the values in the second column the negative and positive values cancel each other out (this will always happen), so to prevent this we square every value in the second column: data value deviation: data value – mean deviation squared 0 0-5 = -5 (-5)2 = 25 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 10 10-5 = 5 (5)2 = 25 We then add the squared deviations up to get 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 25 = 50. Ordinarily we would then divide by the number of scores, n, (in this case, 10) to find the mean of the deviations. But we only do this if the data set represents a population; if the data set represents a sample (as it almost always does), we instead divide by n – 1 (in this case, 10 – 1 = 9).[1] So in our example, we would have 50/10 = 5 if section D represents a population and 50/9 = about 5.56 if section D represents a sample. These values (5 and 5.56) are called, respectively, the population variance and the sample variance for section D. Variance can be a useful statistical concept, but note that the units of variance in this instance would be points-squared since we squared all of the deviations. What are points-squared? Good question. We would rather deal with the units we started with (points in this case), so to convert back we take the square root and get: \begin{align}&\text{populationstandarddeviation}=\sqrt{\frac{50}{10}}=\sqrt{5}\approx2.2\\&\text{or}\\&\text{samplestandarddeviation}=\sqrt{\frac{50}{9}}\approx2.4\\\end{align} If we are unsure whether the data set is a sample or a population, we will usually assume it is a sample, and we will round answers to one more decimal place than the original data, as we have done above. ### To compute standard deviation 1. Find the deviation of each data from the mean. In other words, subtract the mean from the data value. 2. Square each deviation. 3. Add the squared deviations. 4. Divide by n, the number of data values, if the data represents a whole population; divide by n – 1 if the data is from a sample. 5. Compute the square root of the result. ### example Computing the standard deviation for Section B above, we first calculate that the mean is 5. Using a table can help keep track of your computations for the standard deviation: data value deviation: data value – mean deviation squared 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 Assuming this data represents a population, we will add the squared deviations, divide by 10, the number of data values, and compute the square root: $\sqrt{\frac{25+25+25+25+25+25+25+25+25+25}{10}}=\sqrt{\frac{250}{10}}=5$ Notice that the standard deviation of this data set is much larger than that of section D since the data in this set is more spread out. For comparison, the standard deviations of all four sections are: Section A: 5 5 5 5 5 5 5 5 5 5 Standard deviation: 0 Section B: 0 0 0 0 0 10 10 10 10 10 Standard deviation: 5 Section C: 4 4 4 5 5 5 5 6 6 6 Standard deviation: 0.8 Section D: 0 5 5 5 5 5 5 5 5 10 Standard deviation: 2.2 See the following video for more about calculating the standard deviation in this example. ### Try It The price of a jar of peanut butter at 5 stores was3.29, $3.59,$3.79, $3.75, and$3.99. Find the standard deviation of the prices.
Where standard deviation is a measure of variation based on the mean, quartiles are based on the median.
### Quartiles
Quartiles are values that divide the data in quarters.
The first quartile (Q1) is the value so that 25% of the data values are below it; the third quartile (Q3) is the value so that 75% of the data values are below it. You may have guessed that the second quartile is the same as the median, since the median is the value so that 50% of the data values are below it.
This divides the data into quarters; 25% of the data is between the minimum and Q1, 25% is between Q1 and the median, 25% is between the median and Q3, and 25% is between Q3 and the maximum value.
While quartiles are not a 1-number summary of variation like standard deviation, the quartiles are used with the median, minimum, and maximum values to form a 5 number summary of the data.
### Five number summary
The five number summary takes this form:
Minimum, Q1, Median, Q3, Maximum
To find the first quartile, we need to find the data value so that 25% of the data is below it. If n is the number of data values, we compute a locator by finding 25% of n. If this locator is a decimal value, we round up, and find the data value in that position. If the locator is a whole number, we find the mean of the data value in that position and the next data value. This is identical to the process we used to find the median, except we use 25% of the data values rather than half the data values as the locator.
### To find the first quartile, Q1
1. Begin by ordering the data from smallest to largest
2. Compute the locator: L = 0.25n
3. If L is a decimal value:
• Round up to L+
• Use the data value in the L+th position
4. If L is a whole number:
• Find the mean of the data values in the Lth and L+1th positions.
### To find the third quartile, Q3
Use the same procedure as for Q1, but with locator: L = 0.75n
Examples should help make this clearer.
### examples
Suppose we have measured 9 females, and their heights (in inches) sorted from smallest to largest are:
59 60 62 64 66 67 69 70 72
What are the first and third quartiles?
Suppose we had measured 8 females, and their heights (in inches) sorted from smallest to largest are:
59 60 62 64 66 67 69 70
What are the first and third quartiles? What is the 5 number summary?
The 5-number summary combines the first and third quartile with the minimum, median, and maximum values.
What are the 5-number summaries for each of the previous 2 examples?
More about each set of women’s heights is in the following videos.
Returning to our quiz score data: in each case, the first quartile locator is 0.25(10) = 2.5, so the first quartile will be the 3rd data value, and the third quartile will be the 8th data value. Creating the five-number summaries:
Section and data 5-number summary Section A: 5 5 5 5 5 5 5 5 5 5 5, 5, 5, 5, 5 Section B: 0 0 0 0 0 10 10 10 10 10 0, 0, 5, 10, 10 Section C: 4 4 4 5 5 5 5 6 6 6 4, 4, 5, 6, 6 Section D: 0 5 5 5 5 5 5 5 5 10 0, 5, 5, 5, 10
Of course, with a relatively small data set, finding a five-number summary is a bit silly, since the summary contains almost as many values as the original data.
A video walkthrough of this example is available below.
### Try It
The total cost of textbooks for the term was collected from 36 students. Find the 5 number summary of this data.
$140$160 $160$165 $180$220 $235$240 $250$260 $280$285
$285$285 $290$300 $300$305 $310$310 $315$315 $320$320
$330$340 $345$350 $355$360 $360$380 $395$420 $460$460
### Example
Returning to the household income data from earlier in the section, create the five-number summary.
Income (thousands of dollars) Frequency 15 6 20 8 25 11 30 17 35 19 40 20 45 12 50 7
This example is demonstrated in this video.
Note that the 5 number summary divides the data into four intervals, each of which will contain about 25% of the data. In the previous example, that means about 25% of households have income between $40 thousand and$50 thousand.
For visualizing data, there is a graphical representation of a 5-number summary called a box plot, or box and whisker graph.
### Box plot
A box plot is a graphical representation of a five-number summary.
To create a box plot, a number line is first drawn. A box is drawn from the first quartile to the third quartile, and a line is drawn through the box at the median. “Whiskers” are extended out to the minimum and maximum values.
### examples
The box plot below is based on the 9 female height data with 5 number summary:
59, 62, 66, 69, 72.
The box plot below is based on the household income data with 5 number summary:
15, 27.5, 35, 40, 50
Box plot creation is described further here.
### Try It
Create a box plot based on the textbook price data from the last Try It.
Box plots are particularly useful for comparing data from two populations.
### examples
The box plot of service times for two fast-food restaurants is shown below.
While store 2 had a slightly shorter median service time (2.1 minutes vs. 2.3 minutes), store 2 is less consistent, with a wider spread of the data.
At store 1, 75% of customers were served within 2.9 minutes, while at store 2, 75% of customers were served within 5.7 minutes.
Which store should you go to in a hurry?
The box plot below is based on the birth weights of infants with severe idiopathic respiratory distress syndrome (SIRDS)[2]. The box plot is separated to show the birth weights of infants who survived and those that did not.
Comparing the two groups, the box plot reveals that the birth weights of the infants that died appear to be, overall, smaller than the weights of infants that survived. In fact, we can see that the median birth weight of infants that survived is the same as the third quartile of the infants that died.
Similarly, we can see that the first quartile of the survivors is larger than the median weight of those that died, meaning that over 75% of the survivors had a birth weight larger than the median birth weight of those that died.
Looking at the maximum value for those that died and the third quartile of the survivors, we can see that over 25% of the survivors had birth weights higher than the heaviest infant that died.
The box plot gives us a quick, albeit informal, way to determine that birth weight is quite likely linked to survival of infants with SIRDS.
The following video analyzes the examples above.
1. The reason we do this is highly technical, but we can see how it might be useful by considering the case of a small sample from a population that contains an outlier, which would increase the average deviation: the outlier very likely won't be included in the sample, so the mean deviation of the sample would underestimate the mean deviation of the population; thus we divide by a slightly smaller number to get a slightly bigger average deviation.
2. van Vliet, P.K. and Gupta, J.M. (1973) Sodium bicarbonate in idiopathic respiratory distress syndrome. Arch. Disease in Childhood, 48, 249–255. As quoted on http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398296§ion=1.1.3
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https://socratic.org/questions/circle-a-has-a-radius-of-3-and-a-center-at-1-2-circle-b-has-a-radius-of-5-and-a--2
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# Circle A has a radius of 3 and a center at (1 ,2 ). Circle B has a radius of 5 and a center at (3 ,7 ). If circle B is translated by <2 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Jun 3, 2018
$\text{no overlap } \approx 2.82$
#### Explanation:
$\text{what we have to do here is compare the distance (d)}$
$\text{between the centres to the sum of the radii}$
• " if sum of radii">d" then circles overlap"
• " if sum of radii"< d" then no overlap"
$\text{before calculating d we require to find the new centre}$
$\text{of B under the given translation}$
$\text{under the translation } < 2 , 4 >$
$\left(3 , 7\right) \to \left(3 + 2 , 7 + 4\right) \to \left(5 , 11\right) \leftarrow \textcolor{red}{\text{new centre of B}}$
$\text{to calculate d use the "color(blue)"distance formula}$
•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
$\text{let "(x_1,y_1)=(1,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 11\right)$
$d = \sqrt{{\left(5 - 1\right)}^{2} + {\left(11 - 2\right)}^{2}} = \sqrt{16 + 81} = \sqrt{117} \approx 10.82$
$\text{sum of radii } = 3 + 5 = 8$
$\text{since sum of radii"< d" then no overlap}$
$\text{minimum distance "=d-" sum of radii}$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} = 10.82 - 8 = 2.82$
graph{((x-1)^2+(y-2)^2-9)((x-5)^2+(y-11)^2-25)=0 [-20, 20, -10, 10]}
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https://www.brighthubeducation.com/elementary-school-activities/3123-teach-equivalent-fractions-using-simple-paper-and-crayon-activity/
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# Teach Equivalent Fractions Using This Simple Paper and Crayon Activity
Page content
## Needed Materials
rectangular pieces of paper
crayons
## Description of the Activity
• Provide each student with a piece of rectangular paper. Fold the paper in half. After you have folded the paper in half, instruct students to do the same. Explain that a fraction is a part of a whole. You have divided a whole piece of paper into two equal parts.
• Instruct students to color one of the two equal parts. Ask a student to write 1/2 on the board to show that one out of the two equal parts is now shaded.
• Introduce the vocabulary words numerator and denominator. The numerator is the number of parts shaded and the denominator is the total number of equal parts. (For those students who have difficulty remembering which is the numerator and which is the denominator, try this memory association technique - In a fraction, one number is UP above the line and one is DOWN below the line. Numerator has a “u” in it and so does up; denominator begins with “d” and so does down.)
• Demonstrate and have students fold additional pieces of paper (one for each fractional amount) to represent 1/4, 3/4, 1/3, 2/3, and 1/8. Each time, a student should write the fraction on the board and identify its numerator and denominator. If you prefer, project a rectangle onto the board and divide the rectangle into the same fractions as those in the paper-folding demonstration.
• Equivalent Fractions: Ask students to fold a rectangular sheet of paper in half and color one of the two equal parts. Ask what fraction of the paper is colored. (1/2) Now have them refold the same paper and then fold it in half once again. Unfold. How many equal parts now? (4) What fraction is shaded? (2/4 or 1/2) Since the amount of shading has not changed, this means that 1/2=2/4. Tell students that 1/2 and 2/4 are two names for the same amount. Therefore, they are equivalent. Now have students refold the papers and then fold in half a third time. Unfold. What new fraction have they found that is equivalent to 1/2 and 2/4? (4/8) These three fractions (1/2, 2/4, 4/8) name the same amount.
## Assessment
• Students can demonstrate fractions and equivalent fractions using paper strips.
• Students can write the fractions for the amounts demonstrated using the paper strips.
## Integration
Fractions can be integrated with music by connecting whole, half, quarter, eighth, and sixteenth notes with the appropriate fractions.
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11.4 - Fractional equations
Before reading this section you may want to review the following topics:
A fractional equation is one that contains fraction terms. In section 4.2 we saw how to solve a linear equation that contains fractions. The steps for solving any fractional equation are exactly the same:
• Look at the denominators of all the fraction terms and find their lowest common multiple (LCM) (this is also called the lowest common denominator (LCD) of the fractions).
• Multiply both sides of the equation by the LCM.
• Distribute the LCM over both sides of the equation.
• The equation no longer contains fraction terms and you can continue solving it by using the basic procedures for solving equations.
• Check the solution. This is especially important with fractional equations. There are two possible problems:
• If the denominator of any fraction term contains x, then the LCM will also contain x, and multiplying both sides of the equation by the LCM will increase the degree of x in the equation. This often leads to extraneous solutions.
• When substituting the solutions back into the original equation to check them, any solution that causes any fraction term to have a denominator of zero must be dropped because division by zero is forbidden in mathematics.
Example 1: Solve this fractional equation for x:
Solution: The fraction terms have denominators of 3, 2 and 6. The LCM of these numbers is 6. Multiply both sides of the equation by 6. (Don’t forget to put brackets around both sides of the equation.)
Distribute on both sides of the equation:
x − 3 = 6 x + 7.
The fractions are now cleared so this is no longer a fractional equation. Finish solving the equation by collecting linear terms on the left-hand-side and constant terms on the right-hand-side. This gives:
−2 x = 10.
Divide both sides by −2. This gives the solution:
x = −5.
Check it by substituting it back into the original equation. This gives −23 / 6 = −23 / 6, so the solution checks out.
Example 2: Solve this fractional equation for x:
Solution: The fraction terms have denominators of x 2 + x − 2, x + 2, and x − 1. It might appear that the LCM is just the product of all three, but because x 2 + x − 2 can be factored as (x + 2)(x − 1), the LCM is actually just (x + 2)(x − 1). Multiply both sides of the equation by it. (Don’t forget to put brackets around both sides of the equation.)
Distribute on both sides of the equation:
9 = 3 (x − 1) + 7 (x + 2).
The fractions are now cleared so this is no longer a fractional equation; it is a linear equation. Solve it using the usual techniques. Distribute once more on the right-hand-side:
9 = 10 x + 11.
Collect constant terms on the left-hand-side:
−2 = 10 x.
Divide both sides by 10. This gives the solution:
x = −1/5.
Check it by substituting it back into the original equation. This gives −25 / 6 = −25 / 6, so the solution checks out.
Example 3: The purpose of this example is to illustrate a solution that must be rejected because it causes a division by zero. The equation is identical to the one in the previous example except that it differs in the sign of one term. Solve this fractional equation for x:
Solution: Compare each step here with the corresponding step in the example above. Multiply both sides of the equation by the LCM, which again is (x + 2)(x − 1):
Distribute on both sides of the equation:
9 = −3 (x − 1) + 7 (x + 2).
Distribute once more on the right-hand-side:
9 = 4 x + 17.
This time the solution is x = −2. If we try to substitute it back into the original equation we get divisions by zero in two of the fractions. Therefore we must reject this solution and state that the equation has no solution.
Algebra Coach Exercises
If you found this page in a web search you won’t see the
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## The Ratio Word Problems Tutorial Series
This is a series of tutorials regarding ratio word problems. Ratio is defined as the relationship between two numbers where the second number is how many times the first number is contained. In this series of problems, we will learn about the different types of ratio word problems.
How to Solve Word Problems Involving Ratio Part 1 details the intuitive meaning of ratio. It uses arithmetic calculations in order to explain its meaning. After the explanation, the algebraic solution to the problem is also discussed.
How to Solve Word Problems Involving Ratio Part 2 is a continuation of the first part. In this part, the ratio of three quantities are described. Algebraic methods is used as a solution to solve the problem.
How to Solve Word Problems Involving Ratio Part 3 in this post, the ratio of two quantities are given. Then, both quantities are increased resulting to another ratio.
How to Solve Word Problems Involving Ratio Part 4 involves the difference of two numbers whose ratio is given.
If you have more math word problems involving ratio that are different from the ones mention above, feel free to comment below and let us see if we can solve them.
## How to Solve Word Problems Involving Ratio Part 4
This is the fourth and the last part of the solving problems involving ratio series. In this post, we are going to solve another ratio word problem.
Problem
The ratio of two numbers 1:3. Their difference is 36. What is the larger number?
Solution and Explanation
Let x be the smaller number and 3x be the larger number.
3x – x = 36
2x = 36
x = 18
So, the smaller number is 18 and the larger number is 3(18) = 54.
Check:
The ratio of 18:54 is 1:3? Yes, 3 times 18 equals 54.
Is their difference 36? Yes, 54 – 18 = 36.
Therefore, we are correct.
## How to Solve Word Problems Involving Ratio Part 3
In the previous two posts, we have learned how to solve word problems involving ratio with two and three quantities. In posts, we are going to learn how to solve a slightly different problem where both numbers are increased.
Problem
The ratio of two numbers is 3:5 and their sum is 48. What must be added to both numbers so that the ratio becomes 3:4?
Solution and Explanation
First, let us solve the first sentence. We need to find the two numbers whose ratio is 3:5 and whose sum is 48.
Now, let x be the number of sets of 3 and 5.
3x + 5x = 48
8x = 48
x = 6
Now, this means that the numbers are 3(6) = 18 and 5(6) = 30.
Now if the same number is added to both numbers, then the ratio becomes 3:4.
Recall that in the previous posts, we have discussed that ratio can also be represented by fraction. So, we can represent 18:30 as $\frac{18}{30}$. Now, if we add the same number to both numbers (the numerator and the denominator), we get $\frac{3}{4}$. If we let that number y, then
$\dfrac{18 + y}{30 + y} = \dfrac{3}{4}$.
Cross multiplying, we have
$4(18 + y) = 3(30 + y)$.
By the distributive property,
$72 + 4y = 90 + 3y$
$4y - 3y = 90 - 72$
$y = 18$.
So, we add 18 to both the numerator and denominator of $\frac{18}{30}$. That is,
$\dfrac{18 + 18}{30 + 18} = \dfrac{36}{48}$.
Now, to check, is $\dfrac{36}{48} = \frac{3}{4}$? Yes, it is. Divide both the numerator and the denominator by 12 to reduce the fraction to lowest terms.
## How to Solve Word Problems Involving Ratio Part 2
This is the second part of a series of post on Solving Ratio Problems. In the first part, we have learned how to solve intuitively and algebraically problems involving ratio of two quantities. In this post, we are going to learn how to solve a ratio problem involving 3 quantities.
Problem 2
The ratio of the red, green, and blue balls in a box is 2:3:1. If there are 36 balls in the box, how many green balls are there?
Solution and Explanation
From the previous, post we have already learned the algebraic solutions of problems like the one shown above. So, we can have the following:
Let $x$ be the number of grous of balls per color.
$2x + 3x + x = 36$
$6x = 36$
$x = 6$
So, there are 6 groups. Now, since we are looking for the number of green balls, we multiply x by 3.
So, there are 6 groups (3 green balls per group) = 18 green balls.
Check:
From above, $x = 6(1)$ is the number of blue balls. The expression 2x represent the number of red balls, so we have 2x = 2(6) = 12 balls. Therefore, we have 12 red balls, 18 green balls, and 6 blue balls.
We can check by adding them: 12 + 18 + 6 = 36.
This satisfies the condition above that there are 36 balls in all. Therefore, we are correct.
## How to Solve Word Problems Involving Ratio Part 1
In a dance school, 18 girls and 8 boys are enrolled. We can say that the ratio of girls to boys is 18:8 (read as 18 is to 8). Ratio can also be expressed as fraction so we can say that the ratio is 18/8. Since we can reduce fractions to lowest terms, we can also say that the ratio is 9/4 or 9:4. So, ratio can be a relationship between two quantities. It can also be ratio between two numbers like 4:3 which is the ratio of the width and height of a television screen.
Problem 1
The ratio of boys and girls in a dance club is 4:5. The total number of students is 63. How many girls and boys are there in the club?
Solution and Explanation
The ratio of boys is 4:5 means that for every 4 boys, there are 5 girls. That means that if there are 2 groups of 4 boys, there are also 2 groups of 5 girls. So by calculating them and adding, we have
4 + 5 = 9
4(2) +5(2) =18
4(3) +5(3) =27
4(4) +5(4) = 36
4(5) +5(5) = 45
4(6) +5(6) =54
4(7) +5(7) =63
As we can see, we are looking for the number of groups of 4 and, and the answer is 7 groups of each. So there are 4(7) = 28 boys and 5(7) = 35 girls.
As you can observe, the number of groups of 4 is the same as the number of groups of 5. Therefore, the question above is equivalent to finding the number of groups (of 4 and 5), whose total number of persons add up to 63.
Algebraically, if we let x be the number of groups of 4, then it is also the number of groups of 5. So, we can make the following equation.
4 x number of groups + 5 x number of groups of 5 = 63
Or
4x + 5x = 63.
Simplifying, we have
9x = 63
x = 7.
So there are 4(7) = 28 boys and 5(7) = 35 girls. As we can see, we confirmed the answer above using algebraic methods.
## How to Solve Investment Word Problems in Algebra
Investment word problems in Algebra is one of the types of problems that usually come out in the Civil Service Exam. In solving investment word problems, you should know the basic terms used. Some of these terms are principal (P) or the money invested, the rate (R) or the percent of interest, the interest (I) or the return of investment (profit), and the time or how long the money is invested. Investment is the product of the principal, the rate, and the time, and therefore, we have the formula
I = PRT.
This tutorial series discusses the different types of problems in investment and discussed the method and strategies used in solving them.
How to Solve Investment Problems Part 1 discusses the common terminology used in investment problems. It also discusses an investment problem where the principal is invested at two different interest rates.
How to Solve Investment Problems Part 2 is a discussion of another investment problem just like in part 1. In the problem, the principal is invested at two different interest rates and the interest in one investment is larger than the other.
How to Solve Investment Problems Part 3 is very similar to part 2, only that the smaller interest amount is described.
How to Solve Investment Problems Part 4 discusses an investment problem with a given interest in one investment and an unknown amount of investment at another rate to satisfy a percentage of interest for the entire investment.
## How to Solve Investment Problems Part 4
This is the fourth part of the Solving Investment Problems Series. In this part, we discuss a problem which is very similar to the third part. We discuss an investment at two different interest rates.
Problem
Mr. Garett invested a part of $20 000 at a bank at 4% yearly interest. How much does he have to invest at another bank at a 8% yearly interest so that the total interest of the money is 7%. Solution and Explanation Let x be the money invested at 8% (1) We know that the interest of 20,000 invested at 4% yearly interest is 20,000(0.04) (2) We also know that the interest of the money invested at 8% is (0.08)(x) (3) The interest of total amount of money invested is 7%. So, (20,000 + x)(0.07) Now, the interest in (1) added to the interest in (2) is equal to the interest in (3). Therefore, 20,000(0.04) + (0.08)(x) = (20,000 + x)(0.07) Simplifying, we have 800 + 0.08x = 1400 + 0.07x To eliminate the decimal point, we multiply both sides by 100. That is 80000 + 8x = 140000 + 7x 8x – 7x = 140000 – 80000 x = 60000 This means that he has to invest$60,000 at 8% interest in order for the total to be 7% of the entire investment.
Check:
$20,000 x 0.04 =$800
$60,000 x 0.08 = 4800 Adding the two interest, we have$5600. We check if this is really 7% of the total investment.
Our total investment is $80,000. Now,$80,000 x 0.07 = \$5600.
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# Scrambled Egg Math
## Curriculum Goal
#### Primary: Number Sense
• Use objects, diagrams, and equations to represent, describe, and solve situations involving addition and subtraction of whole numbers that add up to no more than 50.
## Context
• Educator working with a group of four children.
## Materials
• One egg cartons per student (size: 12 eggs)
• Two small game tokens per student (e.g., beans, pennies)
• One die per student
• Lined papers or math notebooks
## Lesson
• Open the egg cartons and write the numbers 1 through to 9 on the bottom of each rounded space. Choose three numbers to be written down twice to fill all 12 spaces.
• Hand out an egg carton, two game tokens, and a die to each child. Tell them we will be playing a game to practice addition and subtraction.
• Explain to students that they will put the game tokens inside the carton, close the lid and “scramble the eggs” by shaking the carton. Once they finish shaking the carton, tell them to set it on the table.
• Instruct children to then roll their die to determine whether they will be adding or subtracting in this round: an even number means that they will add, an odd number means that they will subtract. Students who are just beginning to solve equations can play the game using one operation.
• Ask students to open the carton lid and see which spaces the tokens fell into. Have them create a math equation using the two numbers and the chosen operation and then solve it.
• E.g., if the child rolled a four (even = addition) and their tokens landed in the seven and three spaces, they can make the equation “7 + 3 = 10” or “3 + 7 = 10”.
• Instruct students to write their equation down on their sheet of paper or notebook.
• Repeat the game for 10 rounds and then discuss with students how they approached the task. Some questions to extend student thinking:
• How did you solve your equation?
• Do you find that certain numbers make it easier to find the answer?
• Can you tell me something about the two numbers that the tokens landed on?
## Look Fors
• What strategies does the child use to solve the equations? Do they count up, count on, or find the answer through retrieval? Do children recognize certain math strategies? For example, do they recognize the different ways to make the number 10 or know the doubles fact to help them solve equations faster?
• Griffin (2003) states that as children develop stronger number sense and computation abilities, the strategy they employ will also become more sophisticated (instead of counting up, children may use counting on or retrieval).
• Do children demonstrate the ability to manipulate a mental number line?
• This activity requires children to solve equations without using any concrete objects. According to Griffin (2005), a major learning stage occurs when children are able to solve addition and subtraction questions in their minds by manipulating their “mental counting line structure” forwards and backwards.
• Do children use appropriate math language? Are they using words such as sum, difference, bigger than, smaller than, etc. when asked questions?
• During the activity, talk with students and ask them to verbalize their thoughts or talk about the numbers to listen to their vocabulary.
## Extension
• As children become proficient, introduce three-step codes and encourage children to create pathways with more steps
## References
Griffin, S. (2005). Fostering the development of whole number sense: Teaching mathematics in the primary grades. In M.S. Donovan & J. D. Bransford (Eds.), How students learn: History, mathematics and science in the classroom (pp. 257308). Washington, DC: The National Academies Press.
Griffin, S. (2003). Laying the foundations for computational fluency in early childhood. Teaching Children Mathematics, 9(6), 306309.
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## Pre-Algebra Tutorial
#### Intro
A typical worded math problem is the ‘two working together’ problem, which is described here in detail. This is a classic test problem encountered in algebra and both quick and in-depth explanations are provided. Other examples are included.
#### Sample Problem
Jim can paint a room in 3 hours and Tom can paint the same room in 4 hours. How long with it take to paint that room if they work together?
#### Solution
For this “working together” problem, savvy students will recognize it and think, “Oh, the product over the sum solution”, and write down (or punch in) (3×4)/7, or about 1.7 hours (how many digits you include may depend on your teacher’s demands). This is correct but, what is behind this technique?
The first thing to realize is that this is a “Rate, Time and Distance” type of problem. I always remember that Rate X Time = Distance, or “RTD”. More generally, Rate X Time = Task Complete. Rate is always in ‘units of progress’ over time, such as miles per hour or other things done per time. Example: If you travel at 60 miles per hour for 2 hours, you have gone 60 x 2 = 120 miles. If you didn’t know how long it would take to travel 120 miles at a rate of 60 miles per hour, you would solve 60 X T = 120, or T = 120/60 = 2 hours. Any one of the three terms can be solved for if you know the others.
For the above word problem, first think of Jim’s rate to paint a room as 1 room in 3 hours and denote Rj = 1/3 or 1 room in 3 hours. If you care to divide 1 by 3, you see that he can paint one third of a room in one hour. Tom’s rate, Rt, is 1/4 and he can paint a quarter of a room in one hour.
OK, we figure that the sum of these two rates multiplied by the time working together is equal to 1 (room complete). So, using the RTD format, (Rj + Rt) * T = 1.
[Note: I sometimes use spaces in place of parentheses as it makes typed equations easier to read.]
Now, cross dividing to solve for T, T = 1/(Rj + Rt) = 1/(1/3 + 1/4), or simplifying to a common denominator, T = 1/(4/12 + 3/12) = 1/(7/12); multiplying numerator and denominator by 12/7 we get T = 12/7 / (12/7 * 7/12)), = 12/7! The answer is about 1.7 hours. 12/7 is simply the product over the sum. In the actual effort of painting the room, Jim paints 1.7 * 1/3 = .57– or about 57% of the room and Tom paints 1.7 * 1/4 = .43– or about 43% of the room.
What if Jim’s little brother helps them? It takes him 10 hours to paint a room! OK, T = 1/(1/3 + 1/4 + 1/10) = about 1.46 hours. That helps a bit, but little brother drips a lot of paint. The ‘product over the sum’ technique doesn’t work here (why?) so be careful – it only works for two! For more than two, you have to do it the long way.
Here’s a quick calculator sequence to solve the inverse of such fractional sums for the three workers:
3,1/x,+4,1/x,+10,1/x,=,1/x. The result is 1.46…
Try this other example: You drive half way around a race track at 60 mph and then the remaining half at 50 mph. What is your average speed? Is it 55 mph? No!
Solution: You can’t use the product over the sum technique here, as that only works to solve for time (the T part of RTD), given the other terms. This problem is to solve for the combined rate (R). So, what do we know? Let’s say D is 1 (track length) and the rates are as listed. But we can’t just average the rates and say it’s 55 mph because the times for each half of the track are different. Well, for the first half, T = D/R = .5/60 = .008333— hours and for the second half, .5/50 = .01 hours, for a total of .018333— hours. Finally, solving, R * .018333— = 1 (track), or R = 1/.018333— = 54.5454– mph. This was the first tricky algebra problem to given us in the 8th grade and everyone was mystified that the answer was not 55 mph and unsure how to get the right answer!
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# 4.08 Polynomials
Lesson
## Introduction
An expression of the form Ax^{n}, where A is any number and n is any non-negative integer, is called a monomial. When we take the sum of multiple monomials, we get a polynomial.
## Parts of polynomials
In the monomial Ax^{n}:
• A is the coefficient
• x is the variable
• n is the index
A polynomial is a sum of any number of monomials (and consequently, each term of a polynomial is a monomial). The highest index is called the degree of the polynomial. For example, x^{3}+4x+3 is a polynomial of degree three. The coefficient of the term with the highest index is called the leading coefficient. The coefficient of the term with index 0 is called the constant.
We often name polynomials using function notation. For example P(x) is a polynomial where x is the variable. If we write a constant instead of x, that means that we substitute that constant for the variable. For example, if P(x)=x^{3}+4x+3 then P(3)=3^{3}+4\times 3+3=42.
Polynomials of particular degrees are given specific names. Some of these we have seen before.
### Examples
#### Example 1
For the polynomial P(x)=\dfrac{x^7}{5}+\dfrac{x^6}{6}+5.
a
What's the degree of the polynomial?
Worked Solution
Create a strategy
The degree of a polynomial is the highest power of x.
Apply the idea
The highest power of x is 7, so the degree of the polynomial is 7.
b
What's the leading coefficient of the polynomial?
Worked Solution
Create a strategy
The term containing the highest power of x is called the leading term, and its coefficient is the leading coefficient.
Apply the idea
The leading term of the polynomial is \dfrac{x^7}{5} which can be written as \dfrac{1}{5}x^7. So the leading coefficient of the polynomial is \dfrac{1}{5}.
c
What's the constant term of the polynomial?
Worked Solution
Create a strategy
The constant term is the term that is independent of x.
Apply the idea
The term that is independent of x is 5, so the constant term of the polynomial is 5.
#### Example 2
Consider P(x)=4x^5+3x^{6}-8.
a
Find P(0).
Worked Solution
Create a strategy
Substitute x=0 into the polynomial.
Apply the idea
b
Find P(-4).
Worked Solution
Create a strategy
Substitute x=-4 into the equation.
Apply the idea
Idea summary
A monomial is an expression of the form:
\displaystyle Ax^{n}
\bm{A}
is the coefficient
\bm{x}
is the variable
\bm{n}
is the index
A polynomial is a sum of any number of monomials. The highest index is called the degree of the polynomial. The coefficient of the term with the highest index is called the leading coefficient. The coefficient of the term with index 0 is called the constant.
## Operations on polynomials
We apply operations to polynomials in the same way as we apply operations to numbers. For addition and subtraction we add or subtract all of the terms in both polynomials and we simplify by collecting like terms. For multiplication we multiply each term in one polynomial by each term in the other polynomial similar to how we expand binomial products. Division is a more complicated case that we will look at in the next lesson .
A polynomial is a sum of any number of monomials. In a polynomial:
• The highest index is the degree
• The coefficient of the term with the highest index is the leading coefficient
• The coefficient of the term with index 0 is the constant
We apply operations to polynomials in the same way that we apply operations to numbers.
### Examples
#### Example 3
If P(x)=-5x^{2}-6x-6 and Q(x)=-7x+7, form a simplified expression for P(x)-Q(x).
Worked Solution
Create a strategy
Substitute the expressions for P(x) and Q(x), and then subtract the like terms.
Apply the idea
#### Example 4
Simplify \left(3x^{3}-9x^{2}-8x-7\right)+\left(-7x^{3}-9x\right).
Worked Solution
Create a strategy
Apply the idea
Idea summary
A polynomial is a sum of any number of monomials. In a polynomial:
• The highest index is the degree
• The coefficient of the term with the highest index is the leading coefficient
• The coefficient of the term with index 0 is the constant
### Outcomes
#### VCMNA357 (10a)
Investigate the concept of a polynomial and apply the factor and remainder theorems to solve problems.
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# McRuffy Math Blog
## Marvelous Multiplication #2
Marvelous Multiplication Trick #2: A Multiplication Challenge Using Easy Squares.
If you haven’t already, check out our previous three blogs:
Quick Math: Find the Squares of Numbers That End With 5
Quick Math: Squaring Numbers 51 to 59
In the previous post (Marvelous Multiplication #1) we generated and easily solved problems based on what we termed “easy squares” of two-digit numbers. This included the techniques learned in the two Quick Math blogs and squares of multiples of ten (10, 20, 30, 40…). We generated multipliers by adding and subtracting 1 to and from the easy squares.
In this post, we’ll expand the number of multiplication problems that can be generated and suggest a math challenge game.
The Trick:
Generate 2 two-digit multipliers around an easy square by adding and subtracting the same number to and from the easy square. In the last post, we subtracted 1, but the trick works with adding and subtracting any number. Let’s use the easy square of 30 for an example and add and subtract 2 for the multipliers. We’ll call the number 2 the “distance” between either multiplier and the easy square.
30 -2 = 28 and 30 + 2 = 32 Our multipliers will be 28 and 32.
To quickly find the product of 28 x 32 find the product of the easy square (302 = 900) and subtract the square of the distance. (22 = 4)
28 x 32 = 900 – 4 = 896
Let’s try it with a “distance” of 3:
30 – 3 = 27 and 30 + 3 = 33 Our multipliers will be 27 and 33.
27 x 33 = 900 – 9 = 891
Teaching Sheet 1 provides a form children can use to practice stepping through the process. Chose an easy square and write the digits in the boxes. Choose a “distance” number. Add and subtract the “distance” to generate multipliers for the problem.
Write the product of the easy square on the bottom row and the distance square. Subtract the distance squared to find the answer.
Challenge Game:
Given a two-digit number, generate a problem around an Easy Square and use the math tricks to solve the problem. The ultimate goal would be doing all the steps using mental math, but teaching sheets 2 and 3 present the steps involved in solving the problems to practice thinking through the steps.
For example, try the number 43.
You can make an easy square based problem around the Easy Square of 45. The distance between 43 and 45 is 2. (45 – 43 = 2) The second multiplier will be 47 (45 + 2 = 47). The problem generated will be 43 x 47.
The Easy Square of 45 is 2025. The Distance of 2 squared is 4.
2025 – 4 = 2021 so 43 x 47 = 2021
You could also make an easy square based problem around the Easy Square of 40. The distance between 43 and 40 is 3. (43 – 40 = 3) The second multiplier will be 37 (40 - 3 = 37). The problem generated will be 43 x 37.
Easy Square is 402 = 1600 and the Distance squared is 32 = 9
43 x 37 = 1600 – 9 = 1591
You could also make an easy square based problem around the Easy Square of 53. The distance between 43 and 53 is 10. The second multiplier will be 63. The problem generated will be 43 x 63.
The Easy Square is 532 = 2809 and the Distance squared is 102 = 100
43 x 63 = 2809 – 100 = 2709
Sheet 2 shows all the steps. Start with a random two-digit number and write it on the top line of the “Generate a Problem” section. Choose a nearby easy square. Write the number that is to be added or subtracted to equal the easy square. Next, write the easy square. Generate the second multiplier by doing the opposite operation to the Easy Square. Write the second multiplier in the boxes.
On the bottom row write the products of the squares of the Easy Square and Distance. Subtract for the answer to the generated multiplication problem.
Sheet 3 visually simplifies the process. The first section is used to write the multiplication problem starting with any given two-digit number. The ES boxes are for the Easy Square. The D boxes are for the Distance. Most of the time it will be a single digit distance, but it can be any number of digits. The sheet is set up for a distance with a maximum of two-digits. The far right boxes are for the squares of the Easy Square and Distance.
Encourage children to skip any boxes they can do mentally. For example, if they can write the product of the Easy Square directly in the far right boxes, skip the center box. If the distance and distance square are smaller numbers and easy to remember those boxes can be skipped.
Why it Works:
Let’s build an algebra problem. We’ll represent the Easy Square with the letter a and the “distance” with the letter b. We generated the multipliers by adding and subtracting the “distance” to and from an Easy Square.
(a – b)(a + b) = a2 + ab – ab – b2 = a2 – b2
This is true for any “distance” even when it makes numbers negative and multipliers have more than two digits. For example, with the number 30 we can make the “distance” 200.
The multipliers are 230 and -170
Easy Square is 302 = 900 and The distance squared is 2002 = 40,000
230 x (-170) = 900 - 40,000 = -39,100
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## McRuffy Press offers complete math curriculum by grade level, math activity books, manipulatives, games and more! Find more McRuffy Math products here.
Improve multiplication facts skills with our Multiplication Fast Fact books:
Multiplication Facts (Book 1) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
Multiplication Practice (Book 2) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
## Marvelous Multiplication #1
Marvelous Multiplication Trick #1: Using Math Squaring Tricks to Multiply Other Numbers.
If you haven’t already, check out our previous two blogs:
Quick Math: Find the Squares of Numbers That End With 5
Quick Math: Squaring Numbers 51 to 59
We can use these techniques to do some Quick Math multiplying two-digit numbers that aren't squares. The tricks in this post offer more of an entertainment value than everyday practical value. Helping children feel empowered to do math is a great thing. These tricks do that! Once after teaching these techniques to some children, they replied, “Wow! That’s what algebra does.”
Like the previous two tricks, the minimum a child has to understand is basic multiplication facts. The Why it Works section is great to help students who have some understanding of algebra see how math mysteries can be resolved.
The Trick:
We’ll start with the example of 64 x 66. After learning the trick you’ll immediately see that the product is 4224. First, let’s explore lists of problems we can use for the trick.
We can apply the trick to all these problems:
14 x 16, 24 x 26, 34 x 36, 44 x 46, 54 x 56, 64 x 66, 74 x 76, 84 x 86, 94 x 96
Do you see the pattern?
It also applies to these problems:
50 x 52, 51 x 53, 52 x 54, 53 x 55, 54 x 56, 55 x 57, 56 x 58, 57 x 59, 58 x 60, 59 x 61
And these problems:
19 x 21, 29 x 31, 39 x 41, 49 x 51, 59 x 61, 69 x 71, 79 x 81, 89 x 91
The lists of problems do not need to be memorized because there is a simple method to generate the problems:
Start with a number that the child has learned to easily square. We’ll call a number to be squared an Easy Square.
The problems were generated by adding and subtracting one from the numbers that were squared. For example, start with the Easy Square of 65.
Generate the problem by subtracting and adding one: (65 – 1)(65 + 1) = 64 x 66.
The solution is simply the product of the Easy Square minus 1. Applying the Quick Math trick from the first blog we can quickly find the solution. 4225 – 1 = 4224.
The middle row of problems was generated from the Easy Squares demonstrated in the second blog.
For example, using the trick we can quickly know that 532 = 2809. From that we generate the problem 52 x 54 = 2809 – 1 = 2808
We can call the squares of two-digit numbers ending with zero Easy Squares, too. That was how we generated the last row of problems with 9’s and 1’s in the units place.
For example: 802 = 6400, so 79 x 81 = 6400 – 1 = 6399.
Practice the trick, and we’ll explore how to greatly expand the list of problems that can be generated in our next blog.
The Teaching Sheet provides a form children can use to practice stepping through the process. Chose an easy square and write the digits in the boxes. Skip to the other side of the page and generate a -1, +1 problem. Use an easy square trick to find the square of the Easy Square. Write the answer in the Th, H, T, O boxes. Subtract 1 to find the product of the generated problem.
Why it Works:
Let’s build an algebra problem. We’ll represent the Easy Square with the letter a. We generated the multipliers by adding and subtracting one to an Easy Square.
(a – 1)(a + 1) = a2 + a – a – 12 = a2 – 1
The algebra problem shows we can substitute any of our Easy Square numbers. Apply the easy square trick and subtract one.
The formula is not limited to two-digit numbers. For example:
2002 = 40,000 So 199 x 201 = 39,999
Nor is it limited to Easy Squares. It’s just more difficult to apply. For example generating a problem from 392:
38 x 40 = 392 -1 = 1521 – 1 = 1520. It’s actually easier to multiply 38 x 40 than to first square 39 and then subtract one, so the trick just isn’t helpful. Nevertheless, there is a trick involving this problem, but that’s for another blog.
## Thanks for Visting Free Gift:
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Subscribe to this blog's RSS feed using http://feeds.feedburner.com/McruffyPress-McruffyMathBlog
## McRuffy Press offers complete math curriculum by grade level, math activity books, manipulatives, games and more! Find more McRuffy Math products here.
Improve multiplication facts skills with our Multiplication Fast Fact books:
Multiplication Facts (Book 1) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
Multiplication Practice (Book 2) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
## Quick Math: Find the Squares of Numbers That End With 5
Want to teach your child some amazing math shortcuts? Follow our new blog series. The first blog features a math trick to quickly square numbers that end with 5.
It’s simple enough for children who have knowledge of basic multiplication facts.
For younger children simply teach the trick. For students learning algebra, explore in the Why it Works section for the more mathematically curious. In future blogs, we’ll do some interesting things with this trick to multiply numbers that aren’t squares.
The Trick:
You can quickly find the square of any two digit number ending with 5, such as 15, 25, 35, 45, etc. The trick actually works for any number that ends with five, but the mental math becomes increasingly difficult with numbers greater than 95.
Simply look at the digit in the tens place and multiply by the next digit. The product will be the digits for the thousands and hundreds place (Note the thousands place will be zero for 152 and 252). Add 25 to that product for the product of the number squared.
For example: 352
Multiply the number in the tens place, 3 by the next number, 4.
3 x 4 = 12 the product is the thousands and hundreds place digits. The 12 represents 1200. Then add 25:
352 = 1225
Print Teaching Sheets for the trick. On the first two sheets ask students to put in a random digit in the first box to choose a number to square. Students use that digit and that digit + 1 to make a multiplication problem for the thousands and hundreds place values. If the product is a single digit, the thousands place is left blank (152 and 252). The third page is a practice sheet for the numbers 15 to 95.
Why it works:
Think of any square of a two digit number as the square of a sum of the tens and ones places.
Let’s make the tens place X and the units (ones) place Y.
So, for the number 35, x = 30 and y = 5
The square of 35 is the square of our x and y values. Let’s work with the variables first.
(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
When there are 5 ones (y = 5) then 2xy is 10x .
So the problem simplifies to x2 + 10x + y2 Furthermore, y2 will be 52 = 25
So, x2 + 10x + 25
We can group using the associative property and factor: (x2 +10x) + 25 = x(x +10) + 25
X+10 is the current tens place plus ten more, which is the next ten. In our example, it will be 30 and ten more, 40. So the product of X times the next ten in our example is 1200.
Add 25 to 1200 in our example of 352 = 1200 + 25 = 1225.
Try it with all the squares that end with 5 from 15 to 95.
152 = 225
252 = 625
352 = 1225
452 = 2025
552 = 3025
652 = 4225
752 = 5625
852 = 7225
952 = 9025
You may continue forever, but the mental math becomes more cumbersome. For 1052 The product of 10 x 11 generates the ten thousands, thousands, and hundreds places. Tens and Units will still be 25
1052 = 11025
1152 = 13225
## Thanks for Visting Free Gift:
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Subscribe to this blog's RSS feed using http://feeds.feedburner.com/McruffyPress-McruffyMathBlog
## McRuffy Press offers complete math curriculum by grade level, math activity books, manipulatives, games and more! Find more McRuffy Math products here.
Improve multiplication facts skills with our Multiplication Fast Fact books:
Multiplication Facts (Book 1) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
Multiplication Practice (Book 2) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
## Quick Math: Squaring Numbers 51 to 59
In our previous post, we found a quick shortcut to square two-digit numbers with 5 in the ones place. There is an even simpler shortcut to square two-digit numbers with the digit 5 in the tens place.
For younger students, you can simply teach the trick. For older students explore the Why It Works section.
The Trick:
To get the thousands and hundreds place value add the digit in the ones place (units) to 25.
To get the tens and ones place, square the number in the ones place. The tens place will be zero for the numbers 51, 52, and 53.
For example, 542 Add the ones digit to 25 to get the thousands and hundreds places. 25 + 4 = 29, which represents 2900.
Square the digit in the ones place to get the digits for the tens and units. 4 x 4 = 16
542 = 2900 + 16 = 2916
Print the Teaching Sheets for the trick. On the first sheet ask students to put in a random digit in the box to complete the two-digit number to square. Follow the steps. Remind students that if the product of squaring the ones digit is a single digit answer, the tens place is zero.
The second page is a practice sheet for the numbers 51 to 59.
Why It Works
Like the previous post, let’s make the tens place X and the units (ones) place Y and solve for variables first. The algebra is the same equation as last time. In fact, this algebraic equation is the same for the square of any two-digit number.
(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
In our example of 542, x = 50 and y = 4
The 25 that we start with for the thousands and hundreds place is simply x2 .
Since all the numbers from 50 to 59 have the digit 5 in the tens place, it will be 25 for all the numbers. Remember this is actually 50 times 50 which equals 2500.
The step for adding the units digit to 25 comes from the 2xy. We can use the associative property of multiplication to first find the product of 2x and then multiply by y. For all the numbers in the 50's, x=50 Two times that will equal 100. 2(50) = 100.
So the 2xy part of the algebra equation will always yield a product that puts the units digit into the hundreds place.
For our example of 54 x = 50, y = 4, so 2xy = (two times 50) times y. Which is 100 times y.
100 times 4 is 400.
Add that to the x2 which will be 2500. 2500 + 400 = 2900
Next, add the digits for the tens and units. This is the ypart of the algebra equation. Remember that we designated y as the units place. In our example of 54, the units (ones) place is 4, so we square that number and add it to 2900 for the square are 54.
y= 4 x 4 = 16
54= 2900 + 16 = 2916.
Try it with all the squares that begin with 5 from 50 to 59.
502 = 2500
512 = 2601
522 = 2704
532 = 2809
542 = 2916
552 = 3025
562 = 3136
572 = 3249
582 = 3364
592 = 3481
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# All Factors of 16: Prime Factors, Pair Factors
The factors of 16 are the numbers that completely divide 16 without leaving a remainder. Thus we can say that the factors of 16 are the divisors of 16. In this section, we will learn about the factors of 16 and the prime factors of 16.
Table of Contents
## Highlights of Factors of 16
• 16=2×2×2×2 is the prime factorization of 16.
• The factors of 16 are 1, 2, 4, 8, and 16.
• 2 is the only prime factor of 16.
• The negative factors of 16 are -1, -2, -4, -8, and -16.
## What are the factors of 16?
If 16=a×b, then both a and b are the factors of 16. Thus to find the factors of 16, we need to write the number 16 multiplicatively in all possible ways. Note that we have:
As there are no other ways we can express 16 multiplicatively, we will stop now, and we have obtained all the factors of 16. So the factors of 16 are 1, 2, 4, 8, and 16.
It is known that if m is a factor of 16, then -m is also a factor of 16. So the negative factors of 16 are -1, -2, -4, -8, and -16.
## Pair Factors of 16
At first, we will find the positive pair factors of 16.
Thus the positive pair factors of 16 are (1, 16), (2, 8), and (4, 4). We know that if (a, b) is a positive pair factor of 16, then (-a, -b) is a negative pair factor of 16. Thus, all the negative pair factors of 16 are given below. They are:
(-1, -16), (-2, -8) and (-4, -4).
For more details of factors, visit the page: Basic concepts of factors.
Also Read: Square root of 16
## Number of factors of 16
From above we see that the factors of 16 are 1, 2, 4, 8, and 16.
Therefore the total number of factors of 16 is five.
## Prime Factors of 16
We have found above that the factors of 16 are 1, 2, 4, 8, and 16. Among those factors, we observe that only 2 is a prime number as it does not have any proper divisors.
∴ the only prime factor of 16 is 2.
Question: What are the factors of 16?
Video Solution:
Also Read:
## How to find factors of 16?
By division method, we will determine the factors of 16. In this method, we will find the numbers that can divide 16 completely without a remainder.
Note that no numbers other than 1, 2, 4, 8, and 16 can divide 16. So the numbers 1, 2, 4, 8, and 16 are the complete list of factors of 16.
BACK to HOME PAGE
## FAQs on Factors of 16
Q1: What are the all factors of 16?
Answer: All factors of 16 are 1, 2, 4, 8, and 16.
Q2: Find the prime factors of 16.
Answer: 2 is the only prime factor of 16.
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# Solving Linear Inequalities
An inequality is a statement indicating that two expressions are not equal to one another in a particular way (e.g. one expression is larger or smaller than the other). In the case of a linear inequality, it can be simplified into the form
ax ≥ b, ax b, ax ≤ b, or ax b
where a and b are real numbers and a ≠ 0.
## Solving a Linear Inequality
To solve a linear inequality, you have to isolate for the variable by doing the following steps:
• Expand (if applicable)
• Group like terms (if applicable)
• Rearrange so that all terms with the variable in them are on one side of the inequality while all the terms without the variable in them (i.e. just number terms) are on the other side. That is, rearrange into the form ax > b or ax b, etc. (to do this, you're just adding/subtracting terms from both sides)
• Divide by the coefficient of the variable to solve for the variable (i.e., once you've got ax > b or ax b, etc., divide both sides by a to solve for x). IMPORTANT: if you divide by a negative number, the inequality switches direction (i.e., > becomes < and so on)
Example: Solve the linear inequality 5x - 1 3x + 2
Solution:
5x - 1 3x + 2
5x - 3x 2+ 1
2x 3
x 3/2
Example: Solve the inequality -6x + 1 < 3x + 4
Solution:
-6x +1 < 3x + 4
-6x - 3x < 4 - 1
-9x < 3
x > -1/3
Note that we could have avoided the situation having to divide by a negative number (and hence changing the sign of the inequality) by collecting the terms with the variable on the right of the inequality and the terms with only numbers on the left. Let's do the question this way as well so that you can see the difference:
-6x +1 < 3x + 4
1 - 4 < 3x + 6x
-3 < 9x
-3/9 x
-1/3 x
Of course, even though it might not look like it at first glance, -1/3 < x and x -1/3 are saying the exact same thing. What we're saying in both cases is: x is greater than - 1/3
Linear Example 1:
Linear Example 2:
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# Solving Systems of Equations by Substitution
Related Topics:
Worksheets to practice solving systems of equations
More Algebra Lessons
These algebra lessons introduce the technique of solving systems of equations by substitution.
In some word problems, we may need to translate the sentences into more than one equation. If we have two unknown variables then we would need at least two equations to solve the variable. In general, if we have n unknown variables then we would need at least n equations to solve the variable.
The following example show the steps to solve a system of equations using the substitution method. Scroll down the page for more examples and solutions.
In the Substitution Method, we isolate one of the variables in one of the equations and substitute the results in the other equation. We usually try to choose the equation where the coefficient of a variable is 1 and isolate that variable. This is to avoid dealing with fractions whenever possible. If none of the variables has a coefficient of 1 then you may want to consider the Addition Method or Elimination Method.
Steps to solving Systems of Equations by Substitution:
1. Isolate a variable in one of the equations. (Either y = or x =).
2. Substitute the isolated variable in the other equation.
3. This will result in an equation with one variable. Solve the equation.
4. Substitute the solution from step 3 into another equation to solve for the other variable.
5. Recommended: Check the solution.
Example:
3x + 2y = 2 (equation 1)
y + 8 = 3x (equation 2)
Solution:
Step 1: Try to choose the equation where the coefficient of a variable is 1.
Choose equation 2 and isolate variable y
y = 3x - 8 (equation 3)
Step 2: From equation 3, we know that y is the same as 3x - 8
We can then substitute the variable y in equation 1 with 3x - 8
3x + 2(3x - 8) = 2
3x + 6x - 16 = 2
9x - 16 = 2
Step 5: Isolate variable x
9x = 18
Step 6: Substitute x = 2 into equation 3 to get the value for y
y = 3(2) - 8
y = 6 - 8 = -2
3(2) + 2(-2) = 6 - 4 = 2
Answer: x = 2 and y = -2
Solving systems of equations using Substitution Method through a series of mathematical steps to teach students algebra
Example:
2x + 5y = 6
9y + 2x = 22 How to solve systems of equations by substitution?
Example:
y = 2x + 5
3x - 2y = -9 Steps to solve a linear system of equations using the substitution method
Example:
x + 3y = 12
2x + y = 6 Example of a system of equations that is solved using the substitution method.
Example:
2x + 3y = 13
-2x + y = -9
Solving Linear Systems of Equations Using Substitution
Include an explanation of the graphs - one solution, no solution, infinite solutions
Examples: 2x + 4y = 4
y = x - 2
x + 3y = 6
2x + 6y = -12
2x - 3y = 6
4x - 6y = 12 Example of how to solve a system of linear equation using the substitution method.
x + 2y = -20
y = 2x
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# How do you use synthetic division to show that x=sqrt2 is a zero of x^3+2x^2-2x-4=0?
Jan 24, 2017
By showing that the division gives no remainder, i.e. $r \left(x\right) = 0$.
The basic setup of synthetic division is:
ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")
where the coefficients correspond to the function ${x}^{3} + 2 {x}^{2} - 2 x - 4$. Since we are dividing by a linear factor, $x - \sqrt{2}$, we should get a quadratic back.
(Note that the root you use in your divisor would be the root you acquire from $x - r = 0$.)
The general steps are:
1. Bring down the first coefficient.
2. Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
4. Repeat 2-3 until you reach the last column and have evaluated the final addition.
You should then get:
ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "sqrt2" "(2^"3/2" + 2)" "2^("1/2" + "3/2")" "" ")
$\text{ "" "" "" "1" "(2+sqrt2)" "2^"3/2"" "" "" } 0$
$= \textcolor{b l u e}{{x}^{2} + \left(2 + \sqrt{2}\right) x + {2}^{\text{3/2}}}$
$\frac{{x}^{3} + 2 {x}^{2} - 2 x - 4}{x - \sqrt{2}} = \stackrel{p \left(x\right) \text{/"q(x))overbrace(x^2 + (2 + sqrt2)x + 2^"3/2}}{+} \stackrel{r \left(x\right)}{\overbrace{\frac{0}{x - \sqrt{2}}}}$
And you can check to see that it properly expands to give the original cubic. Furthermore, we now see that $r \left(x\right) = 0$, so it is evident that $x = \sqrt{2}$ is a root of ${x}^{3} + 2 {x}^{2} - 2 x - 4$.
CHALLENGE: Can you show that $- \sqrt{2}$ is also a root?
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# Mean value theorem – Conditions, Formula, and Examples
The mean value theorem is an important application of derivatives. In fact, it is one of the most important and helpful tools in Calculus, so we need to understand the theorem and learn how we can apply it to different problems.
The mean value theorem helps us understand the relationship shared between a secant and tangent line that passes through a curve.
This theorem also influences the theorems that we have for evaluating first and second derivatives. Knowing about its origin, formula, and application can help us get a head start we need to excel in other Calculus topics.
In fact, the mean value theorem is sometimes coined as “one of the fundamental theorems in differential calculus.” We’ll learn the reason behind this in this article.
## What is the mean value theorem?
According to the mean value theorem, if the function, $f(x)$, is continuous for a closed interval, $[a, b]$, there is at least one point at $x = c$, where the tangent line passing through $f(x)$ will be parallel with the secant line that passes through the points, $(a, f(a))$ and $(b, f(b))$.
This graph is a good depiction of what we want to show in the mean value theorem. This means that the slopes of this pair of tangent and secant lines are equal, and we can use this to derive the equation we use to find the value of $f’(c)$ – the slope of the tangent line.
### Mean value theorem equation
Recall that the slope a line can be calculated using the formula, $m = \dfrac{y_2 – y_1}{x_2 – x _1}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the two given coordinate pairs. We can apply this to find the slope of the secant line that passes through the points, $(a, f(a))$ and $(b, f(b))$. Equate the result to the tangent line slope, and we’ll have the equation representing the mean value theorem.
\begin{aligned}m_{\text{tangent}} &= f'(c)\\m_{\text{secant}} &= \dfrac{f(b) – f(a)}{b -a}\\\\m_{\text{tangent}} &= m_{\text{secant}}\\f'(c) &= \dfrac{f(b) – f(a)}{b -a} \end{aligned}
This form comes in handy, especially when we want to find the average value of the rate of change of function given an interval we can work on. A special form of the mean value theorem is the Rolles theorem, and this occurs when the value of $f(x)$ is the same when $x = a$ and $x = b$.
### Understanding the Rolles theorem
The Rolles theorem is a special application of the mean value theorem. Recall that this theorem states that when $f(x)$ is continuous and differentiable within $[a, b]$, the rate of change between $f(a)$ and $f(b)$ is zero when $f(a) = f(b)$.
We can confirm this by using the mean value theorem:
• Find $f’(c)$ by taking the difference between $f(a)$ and $f(b)$.
• Divide the result by $b – a$.
Since $f(a) = f(b)$, we have zero at the numerator as shown below.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a}\\&=\dfrac{0}{b -a}\\&= 0 \end{aligned}
This shows that through the mean value theorem, we can confirm that $f’(c)$ or the change rate is equal to $0$ when the function’s end values are equal.
## How to use the mean value theorem?
We’ve shown you how we can use the mean value theorem to prove the Rolles theorem. There are other theorems in differential calculus where we’ll need to apply this important theorem, so we need to learn how we can use this theorem properly.
• Make sure to check if the function is continuous within the interval.
• Check if the function is differentiable within the interval as well.
• Apply the formula to attain our desired result.
For the second bullet, there are two ways for us to check if the function is differentiable. We’ll show you the two ways, and your preferred method will depend on whether you’ve mastered your derivative rules.
Using Limits Using Derivatives Check if the limit shown below exists.\begin{aligned}\lim_{h\rightarrow 0} \dfrac{f(x +h) – f(x)}{h}\end{aligned}If this limit exists, then the function is said to be differentiable. When given $f(x)$, use the different derivative rules to find the expression for $f’(x)$.
As you can see, the second route (using the different derivative rules) is much easier and will help the next step most of the time. For now, let us list down some important derivative rules we might need in this section. Of course, making sure that you know this by heart at this point makes a lot of difference.
Constant Rule \begin{aligned}\dfrac{d}{dx} c = 0\end{aligned} Power Rule \begin{aligned}\dfrac{d}{dx} x^n = nx^{n -1}\end{aligned} Constant Multiple Rule \begin{aligned}\dfrac{d}{dx} c\cdot f(x) = c \cdot f’(x)\end{aligned} Sum/Difference Rules \begin{aligned}\dfrac{d}{dx} f(x) \pm g(x) = f’(x) \pm g’(x)\end{aligned} Product Rule \begin{aligned}\dfrac{d}{dx} [f(x) \cdot g(x)] = f’(x) \cdot g(x) + g’(x) \cdot f(x)\end{aligned} Quotient Rule \begin{aligned}\dfrac{d}{dx} \left[\dfrac{f(x)}{g(x)}\right] =\dfrac{g(x)f’(x) – f(x) g’(x)}{[g(x)]^2}\end{aligned} Chain Rule \begin{aligned}\dfrac{d}{dx} f(g(x))= f’(g(x)) g’(x)\end{aligned}
These are just some of the rules that might come in handy in our examples but don’t worry, we’ll recall other important derivative rules when the need arises in our discussion.
Once we’ve confirmed that the function is continuous and differentiable, we can then apply the equation representing the mean value theorem.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a} \end{aligned}
We can then use this to apply the mean value theorem in whatever problem we’re given and derive the output we’re being asked.
### Using the mean value theorem to prove a corollary
We can try using the mean value theorem to prove important corollaries. Why don’t we prove the corollary about increasing and decreasing functions?
Let’s say we have a continuous and differentiable function, $f(x)$, within $[a, b]$. The corollary states that:
i) When $f’(x) >0$, the function is said to be increasing within the given interval.
ii)When $f’(x) <0$, the function is said to be decreasing within the given interval.
We can use the mean value theorem to prove this by contradiction. We’ll show how i) can be proven, and you can work on your own to show ii).
To prove the first point by contradiction, we can let $a$ and $b$ be part of the interval so that $a < b$ but $f(a) \geq f(b)$. This means that we assume that $f(x)$ is not increasing throughout the interval. It’s given that $f(x)$ is continuous and differentiable so that we can proceed with the mean value theorem’s equation.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a} \end{aligned}
We’ve estbalished that $f(a) \geq f(b)$, so we’re expecting $f(b) – f(a)$ to be less than or equation to $0$. In addition, we’ve assumed that $a <b$, so $b -a$ must be greater than zero. Let’s observe the nature of $f’(c)$ given our assumptions.
Sign of $\boldsymbol{f(b) – f(a)}$ 0 or Negative Sign of $\boldsymbol{b – a}$ Always Positive Sign of $\boldsymbol{\dfrac{f(b) – f(a)}{b – a}}$ 0 or Negative
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a} \\&\leq 0\end{aligned}
But, we want the derivative to be greater than zero, so for this this statement is a contradiction. For $f(x) >0$, we’ll need $f(b) \geq f(a)$, so $f(x)$ will have to be an increasing function.
This is just one of the many corollaries and theorems that rely on the mean value theorem. This shows how important it is for us to master this theorem and learn the common types of problems we might encounter and require to use the mean value theorem.
Example 1
If $c$ is within the interval, $[2, 4]$, find the value of $c$ so that $f’(c)$ represents the slope within the endpoints of $y= \dfrac{1}{2}x^2$.
Solution
The function, $y = \dfrac{1}{2}x^2$ is continuous throughout the the interval, $[2, 4]$. We want to find the value of $f’(c)$ to be within $[2, 4]$. Before we can find the value of $c$, we’ll have to confirm that the function is differentiable. In general, quadratic expressions are differentiable, and we can confirm $y = \dfrac{1}{2}x^2$’s nature by either:
• Evaluating limit of the expression,$\lim_{h\rightarrow 0} \dfrac{f(x +h) – f(x)}{h}$, where $f(x + h) = \dfrac{1}{2}(x + h)^2$ and $f(x) = \dfrac{1}{2}x^2$.
• Taking the derivative of $y = \dfrac{1}{2}x^2$ using the power and constant rule.
Just for this example, we can you both approaches – but for the rest, we’ll stick with finding the derivatives and you’ll see why.
ing Limits \begin{aligned}\lim_{h\rightarrow 0} \dfrac{f(x +h) – f(x)}{h} &= \lim_{h\rightarrow 0}\dfrac{\dfrac{1}{2}(x + h)^2 – \dfrac{1}{2}(x)^2}{h}\\&= \lim_{h\rightarrow 0}\dfrac{\dfrac{1}{2}(x^2 + 2xh + h) – \dfrac{1}{2}x^2}{h}\\&=\lim_{h\rightarrow 0} \dfrac{\cancel{\dfrac{1}{2}x^2} + xh + \dfrac{1}{2}h- \cancel{\dfrac{1}{2}x^2}}{h}\\&= \lim_{h\rightarrow 0}\dfrac{\cancel{h}\left(x+\dfrac{1}{2}h\right )}{\cancel{h}}\\&= x + \dfrac{1}{2}h\\&= x + \dfrac{1}{2}(0)\\&= x\end{aligned} Using Derivatives \begin{aligned}\dfrac{d}{dx} \dfrac{1}{2}x^2 &= \dfrac{1}{2} \dfrac{d}{dx} x^2\phantom{xxx}\color{green}\text{Constant Rule: } \dfrac{d}{dx} c\cdot f(x) = c\cdot f'(x)\\&=\dfrac{1}{2}(2)x^{2 -1}\phantom{xxx}\color{green}\text{Power Rule: } \dfrac{d}{dx} x^n = nx^{n -1}\\&= 1x^{2 -1}\\&= x\end{aligned}
See how using limits to confirm that a given function is differentiable is very tedious? This is why it pays to use the derivative rules you’ve learned in the past to speed up the process. From the two, we have $f’(x) = x$, so this means that $f’(c)$ is equal to $c$.
Using the mean value theorem’s equation, $f'(c) = \dfrac{f(b) – f(a)}{b -a}$, we can replace $f’(c)$ with $c$. We can use the intervals, $[2, 4]$, for $a$ and $b$, respectively.
\begin{aligned}\boldsymbol{f(x)}\end{aligned} \begin{aligned}\boldsymbol{a = 2}\end{aligned} \begin{aligned}f(a) &= \dfrac{1}{2}(2)^2\\&=\dfrac{1}{2} \cdot 4\\&= 2\end{aligned} \begin{aligned}\boldsymbol{b = 4}\end{aligned} \begin{aligned}f(b) &= \dfrac{1}{2}(4)^2\\&=\dfrac{1}{2} \cdot16\\&= 8\end{aligned}
Let’s use these values into the equation we’re given to find the value of $c$.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a}\\c &= \dfrac{8 – 2}{4 – 2}\\c &= \dfrac{6}{2}\\c&=3\end{aligned}
This means that for $c$ to be within the interval and still satisfy the mean value theorem, $c$ must be equal to $3$.
Example 2
Let’s say that we have a differentiable function, $f(x)$, so that $f’(x) \leq 4$ for all values of $x$. What is largest value that is possible for $f(18)$ if $f(12) = 30$?
Solution
We are given the following values from the problem:
• $a = 12$, $f(12) = 30$
• $b = 18$, $f(18) = ?$
From the mean value theorem, we know that there exists $c$ within the interval of $x \in [12, 18]$, so we can use the mean value theorem to find the value of $f(12)$.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a}\\a&= 12\\b&=18\\\\f'(c) &= \dfrac{f(18)- f(12)}{18 – 12}\\&= \dfrac{f(18) – 30}{6}\end{aligned}
Let’s isolate $f(18)$ on the left-hand side of the equation, as shown below.
\begin{aligned}f'(c) \cdot {\color{green} 6}&= \dfrac{f(18)- 30}{6}\cdot {\color{green} 6}\\6f'(c) &= f(18) -30\\f(18) &=6f'(c) + 30 \end{aligned}
Although the value of $c$ and $f’(c)$ are not given, we know the fact that $f’(x) \leq 4$ for all values of $x$, so we’re sure that $f(c) \leq 4$. Let’s use this to rewrite the relationship of $f(18)$ and $f(c)$ in terms of an inequality.
\begin{aligned}f(18) &= 6f'(c) + 30\\ f'(c) &\leq 4\\\\f(18) &\leq 6(4) + 30\\f(18) &\leq 24 + 30\\f(18) &\leq 54\end{aligned}
This means that the maximum possible value for $f(18)$ is $54$.
Example 3
Verify that the function, $f(x) = \dfrac{x}{x – 4}$, satisfies the conditions of the mean value theorem on the interval $[-2, 2]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
Solution
When given a rational function, it’s important to see if the interval is within its larger domain to confirm continuity. For $f(x) = \dfrac{x}{x – 4}$, we can see that as long as $x \neq 4$, all values of $x$ are valid. This means that $f(x)$ is continuous within the interval, $[-2, 2]$.
Now, we can find the derivative of $f(x) = \dfrac{x}{x – 4}$ and confirm its differentiability. We can begin by using the quotient rule, $\dfrac{d}{dx} \left[\dfrac{f(x)}{g(x)}\right] =\dfrac{g(x)f'(x) -f(x) g'(x)}{[g(x)]^2}$.
\begin{aligned}\dfrac{d}{dx} \dfrac{x}{x – 4} &= \dfrac{(x -4)\dfrac{d}{dx}x – x \dfrac{d}{dx} (x -4)}{(x -4)^2}\\&=\dfrac{(x – 4)(1) – x(1 – 0)}{(x – 4)^2}\\&= \dfrac{x – 4 – x}{(x – 4)^2}\\&= -\dfrac{4}{(x – 4)^2}\end{aligned}
We can see that it’s possible for us to find the derivative of $f(x)$ and with this, we have $f’(c) = -\dfrac{4}{(c – 4)^2}$.
We’ve shown that $f(x)$ meets the mean value theorem’s conditions, so we can use the concluding equation to find $c$.
• We can let $a = -2$ and $b = 2$, so we can substitute these values into $f(x)$ to find $f(-2)$ and $f(2)$.
• On the left-hand side of the equation, we can then replace $f’(c)$ with $-\dfrac{4}{(c – 4)^2}$.
\begin{aligned}f(-2)&=\dfrac{-2}{-2 – 4}\\&= \dfrac{1}{3}\\f(2) &= \dfrac{2}{2 – 4}\\&=-1\\\\f'(c) &= \dfrac{f(b) – f(a)}{b – a}\\-\dfrac{4}{(c -4)^2} &= \dfrac{-1 – \dfrac{1}{3}}{2 – (-2)}\\-\dfrac{4}{(c – 4)^2} &= \dfrac{-5 – (-1)}{5 – 2}\\-\dfrac{4}{(c – 4)^2}&= -\dfrac{1}{3}\end{aligned}
Cross-multiply then isolate $(c – 4)^2$ on one side of the equation. Solve for $c$ and only take note of the root of $c$ that will be within the interval of $[-2, 2]$.
\begin{aligned}-\dfrac{4}{(c -4)^2}&= -\dfrac{1}{3}\\12 &= (c – 4)^2\\c -4 &= \pm \sqrt{12}\\c&= 4 + 2\sqrt{3}\\&\approx 7.46\\c&= 4 – 2\sqrt{3}\\ &\approx 0.54\end{aligned}
This means that $c$ is equal to $4- 2\sqrt{3}$ or approximately, $0.54$.
Example 4
Verify that the function, $f(x) = \dfrac{x^4}{2 – \sqrt{x}}$, satisfies the conditions of the mean value theorem on the interval $[1, 4]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
Solution
First, let’s check if the function is continuous before we apply the mean value theorem. We can equate the denominator, $2 – \sqrt{x}$, to check for any restrictions on the function. Once we have the restricted values for $x$, check if these values are within the interval, $[1, 4]$.
\begin{aligned}2 – \sqrt{x} &\neq 0\\\sqrt{x} &\neq 2\\x&\neq 4\end{aligned}
This means that as long as the interval does not contain $x = 4$ within it, the function can be considered continuous for the said interval. Since $4 \in [1, 4]$, $f(x)$ is not continuous within the interval.
Hence, we can’t apply the mean value theorem for this problem since it is not continuous within the interval.
This problem is a good example of why we must always ensure the function meets the conditions for us to apply the mean value theorem.
Example 5
Verify that the function, $f(x) = 2x + \sqrt{x – 4}$, satisfies the conditions of the mean value theorem on the interval $[4, 8]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
Solution
Since we have a radical expression for $f(x)$, we have to ensure that the restricted values are outside the interval, $[4, 8]$. To check for restrictions, we have to ensure that $x – 4$ is greater than or equal to $0$.
\begin{aligned}x – 4 \geq 0 \\x \geq 4\end{aligned}
Since $x \geq 4$ and $x\in [4, 8]$, we can see that the function remains continuous within our given interval. This means that we can then find the derivative of the equation.
To do so, recall that $\sqrt{b} = b^{\frac{1}{2}}$, so we can rewrite $f(x)$ using this property to differentiate it faster.
• Apply the sum rule to differentiate the two terms, first making the process easier.
• Use the power rule to find the derivative of $2x$ and $\sqrt{x – 4} = (x – 4)^{\frac{1}{2}}$.
\begin{aligned} \dfrac{d}{dx} 2x + \sqrt{x – 4} &= \dfrac{d}{dx} 2x + \dfrac{d}{dx} \sqrt{x – 4}\\&= 2\dfrac{d}{dx} x + \dfrac{d}{dx} (x- 4)^{\frac{1}{2}}\\ &=2(1) + \dfrac{1}{2}(x – 4)^{1/2 – 1}\\&= 2 + \dfrac{1}{2} (x-4)^{-\frac{1}{2} }\\&= 2 + \dfrac{1}{2\sqrt{x – 4}}\end{aligned}
From this, we can say that the function is continuous and differentiable, so we can confirm that the conditions for the mean value theorem have been met.
Now that we have the expression for $f’(x)$, what we can do is find $f(4)$ and $f(8)$ then apply the mean value theorem’s equation.
\begin{aligned}f(4)&=2(4) + \sqrt{4 – 4}\\&=8 \\f(8)&=2(8) + \sqrt{8 – 4}\\&= 18 \\\\f'(c) &= \dfrac{f(b) – f(a)}{b – a}\\2 + \dfrac{1}{2\sqrt{c -4}} &= \dfrac{18 – 8}{8 – 4}\\2 + \dfrac{1}{2\sqrt{c -4}} &= \dfrac{10}{4}\\\dfrac{1}{2\sqrt{c -4}}&= \dfrac{1}{2}\end{aligned} Simplify the equation further and square $\sqrt{c – 4}$ once isolated on either side of the equation.
\begin{aligned}\dfrac{1}{\sqrt{c -4}}&= 1\\\sqrt{c – 4} &= 1\\(\sqrt{c – 4} )^2 &= (1)^2\\c – 4 &= \pm 1\\c &= 3, 5\end{aligned}
Choose the value of $c$ that is within the given interval, $[4, 8]$, so we have $c = 5$.
Example 6
Use the mean value theorem to show that $|\sin a – \sin b| \leq |a – b|$ for the interval, $0 \leq a < b \leq 2\pi$.
Solution
The sine function, $f(x) = \sin x$, is known to be continuous within the interval, $[0, 2\pi]$, and we’ve learned in the past that the derivative of $\sin x$ is equal to $\cos x$.
\begin{aligned}f’(x) = \cos x\end{aligned}
This shows that the function $f(x) = \sin x$ meets the conditions for the mean value theorem, so we can use the conclusion to establish the relationship shown below.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b – a}\\f'(c) &= \dfrac{\sin b – \sin a}{b – a}\\\cos c &= \dfrac{\sin b – \sin a}{b -a}\end{aligned}
Recall that the values of $\cos c$ will be within the range of $[-1, 1]$, so $|\cos c| \leq 1$. Since $\cos c$ is equal to $\dfrac{\sin b – \sin a}{b – a}$ using the mean value theorem, we have the following inequality:
\begin{aligned}\cos c &= \dfrac{\sin b – \sin a}{b -a}\\|\cos c| &= \left|\dfrac{\sin b – \sin a}{b -a}\right|&\leq 1\end{aligned}
Let’s focus on the inequality, $\left|\dfrac{\sin b – \sin a}{b -a}\right| \leq 1$.
\begin{aligned}\left|\dfrac{\sin b – \sin a}{b -a}\right|&\leq 1\\\dfrac{|\sin b – \sin a}{|b -a|} &\leq 1\\|\sin b – \sin a| &\leq |b – a|\end{aligned}
This confirms the statement we want to prove:$|\sin b – \sin a| \leq |b – a|$. Hence, we’ve shown how it’s possible to prove mathematical statements using the mean value theorem.
### Practice Questions
1. If $c$ is within the interval, $[4, 8]$, find the value of $c$ so that $f’(c)$ represents the slope within the endpoints of $y= -2x^3$.
2. Let’s say that we have a differentiable function, $f(x)$, so that $f’(x) \leq 12$ for all values of $x$. What is largest value that is possible for $f(20)$ if $f(15) = 60$?
3. Verify that the function, $f(x) = \dfrac{2x}{x – 3}$, satisfies the conditions of the mean value theorem on the interval $[-1, 1]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
4. Verify that the function, $f(x) = 3x + \sqrt{x – 9}$, satisfies the conditions of the mean value theorem on the interval $[9, 15]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
5. Verify that the function, $f(x) = \dfrac{x^3}{3 – 3\sqrt{x}}$, satisfies the conditions of the mean value theorem on the interval $[1, 5]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
1. $\dfrac{4\sqrt{21}}{3} \approx 6.11$
2. $120$
3. $3 – 2\sqrt{2} \approx 0.17$
4. $\dfrac{21}{2} = 10.5$
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Trigonometric Differentiation
Learn about trigonometric differentiation, its formula along with different examples. Also find ways to calculate using trigonometric differentiation.
Alan Walker-
Published on 2023-05-26
Introduction to Trigonometric Differentiation
There are six basic trigonometric functions in geometry. The rate of change of these functions is known as trigonometric differentiation. This method follows the derivative rules and formulas to calculate the rate of change. Let’s understand how to apply trigonometric differentiation step-by-step and learn the difference between log and trig differentiation.
Understanding the Trigonometric Differentiation
A function that relates an angle of a right-angled triangle to its two sides is known as a trigonometric function. There are six basic trigonometric functions: cosine, tangent, cosecant, secant, and cotangent. In calculus, the derivative of trigonometric functions can be calculated using derivative rules.
By definition the trigonometric differentiation is defined as:
“The process of finding the derivative of a trigonometric function is called trigonometric differentiation.”
Trigonometric Differentiation Formulas
The derivative of a trigonometric function can be calculated by finding the rate of change of the sine and cosine functions. It is because knowing these two derivatives leads to the derivative of all other trig functions. The list of derivative formulas for trigonometric functions is as follows:
• $\frac{d}/dx(\sin x) = \cos x$
• $\frac{d}{dx}(\cos x) =\sin x$
• $\frac{d}{dx}(\sec x) = \sec x\tan x$
• $\frac{d}{dx}(\csc x) = -\csc x\cot x$
• $\frac{d}{dx}(\tan x) = \sec^2x$
• $\frac{d}{dx}(\cot x) = \csc²x$
The trigonometric differentiation formula can be modified to use it with other derivative rules. Let’s discuss the trig differentiation with the product rule, quotient, and power rules.
Trigonometric Differentiation and Product Rule
If a trig function is a product of two trig functions, the product rule with the trigonometric differentiation is used to calculate rate of change. For example, the derivative of secx tanx can be calculated as;
$\frac{d}{dx}(\sec x\tan x) = \sec x\frac{d}{dx}[\tan x] + \tan x\frac{d}{dx}[\sec x]$
Which is equal to,
$\frac{d}{dx}(\sec x\tan x) = \sec x(\sec^2x) + \tan x[\sec x\tan x]=\sec^3x+\tan^2xsec x$
Since $\sec^2x + \tan^2x = 1$ then,
$\frac{d}{dx}(\sec x\tan x)=\sec x(\sec^2x+\tan^2x)=\sec x$
Trigonometric Differentiation and Power Rule
Since the trigonometric differentiation is used to calculate derivatives of a trigonometric function. It can be used along with the power rule if the function contains a trig function with power n. The relation between power rule and trig differentiation for a function $f(x) = \cos^3x$ is expressed as;
$f’(x)=\frac{d}{dx}[\cos^3x]$
By using power rule formula,
$f’(x,y)=3\cos^{3-1}\frac{d}{dx}(\cos x)= -3\cos^2x\sin x$
Where, the cos^3x derivative with respect to x is $–3\cos^2x\sin x$.
Trigonometric Differentiation and Quotient Rule
If a trigonometric function is divided by another function, the trigonometric differentiation along with the quotient rule to find derivative. For a quotient of a function f(x) = tan x = sin x/cos x, the relation between trig derivative and quotient rule is,
$\frac{df}{dx}=\frac{\cos x\frac{d}{dx}[\sin x] – \sin x\frac{d}{dx}[\cos x]}{\cos^2x}$
And,
$\frac{df}{dx}=\frac{\cos^2x + \sin^2x}{\cos^2x} = \frac{1}{\cos^2x}=\sec^2x$
Hence the derivative of tan x is $\sec^2x$.
How do you do Trigonometric differentiation step by step?
The implementation of trigonometric derivatives is divided into a few steps. These steps assist us in calculating the derivative of a function having a trigonometric identity. These steps are:
1. Write the expression of the function.
2. Identify the trig function.
3. Differentiate the function with respect to the variable involved.
4. Use the trigonometric differentiation formula to calculate derivatives. For example, the derivative of sec x is tanx secx.
5. Simplify if needed.
Applying logarithmic differentiation formula by using calculator
The derivative of a log function can also be calculated using the derivative calculator. The online tool follows the log differentiation formula to find the derivative. You can find it online by searching for a derivative calculator. For example, to calculate the derivative of ln x, the following steps are used by using this calculator.
1. Write the expression of the function in the input box, such as ln x.
2. Choose the variable to calculate the rate of change, which will be x in this example.
3. Review the input so there will be no syntax error in the function.
4. Now at the last step, click on the calculate button. By using this step, the derivative calculator will provide the derivative of ln x quickly and accurately, which will be 1/x
Comparison between Trigonometric and Logarithmic differentiation
The comparison between the logarithmic and trigonometric differentiation can be easily analysed using the following difference table.
Logarithmic Differentiation Trigonometric Differentiation The logarithmic differentiation is used to calculate the derivative of a logarithmic function. The trig differentiation is used to calculate derivative of a trigonometric function. The derivative of a logarithmic function ln x is defined as;$f’(x) = \frac{1}{x}\frac{d}{dx}(x)$ There are different formulas to calculate derivative of trigonometric functions. The logarithmic differentiation can be used along with different derivative formulas. The derivative of all trigonometric functions can be calculated by using product rule, quotient rule and power rule.
Conclusion
The trigonometric functions are the functions that relate an angle with the right-angle triangle. Calculating the derivative of such functions is known as trig differentiation. In conclusion, the trigonometric differentiation of all trig functions can be calculated by using the derivatives of sine and cosine.
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### Infant Maths Evening
```Infant Maths Evening
Helping Your Child with Their Maths at Home
Maths in the Infants
Progression in number work
Methods we use for the operations
Other maths in the infants
Counting using Objects
•The first step in children’s number work is counting up to 10 and
beyond.
• Children then need to understand how to relate the numbers to
objects.
• They need to come up with a system so that they do not miss
objects.
• We encourage children to put the objects in a line and start from
one side.
• We also encourage them to touch the objects as they count them.
Relating amounts to number
Children then need to be able to recognise the numbers that
they are using to count. Relating the numbers to a numeral is
quite a big jump for some children. The more familiar they are
with the numerals, the quicker they will learn them.
=5
=6
Recognising and Writing numbers
This is how we write numbers in our school. The earlier children
practise writing numbers the right way round the less likely they are
to get into the habit of writing them incorrectly. In early number
formation 2 and 5 are easily confused.
2 3
6 7 8
5
0
Ordering numbers
2 5 3
2 3
5
Key Words:
• More than
• Less / fewer than
Ordering numbers
9 3 2 7 2
2 3 7 9 2
Key Words:
• More than
• Less / fewer than
Place Value
• A child having a deep understanding of place value is integral to
their progression in maths.
•Once they are familiar with numbers over 10 we work on
identifying the ‘tens digit’ and the ‘units digit’ in each number.
•It is important that the children know the value of each digit.
• In this example 13 is made up of ‘1 ten’ and ‘3 units’
•Place Value cards are one resource we use to support this
concept.
0
Place Value
In school we also use tens rods and unit cubes to help
children understand that 10 units is the same as one set of 10.
= 10
=1
= 36
You could support this idea at home when they are counting numbers
greater than 10, by grouping objects together in tens as they count up.
Place Value
To further support this idea we have 100 squares which are
the size of 10 tens rods.
= 124
Place Value
The children need to be
able to locate given
numbers in a hundred
square by identifying the
tens digit of that number
first then finding the
corresponding row. They
should also know that
the higher the tens digit,
the lower the row is
located in the hundred
square.
Key Words:
• tens /units digit
• teens number
Number Facts
A ‘number bond’ is two numbers which are added together to
make another number.
Children need to work towards a quick recall of number bonds
for 5 e.g. 1 + 4, 2 + 3......
They will also need to know the number bonds for 10 off by
heart e.g. 0 + 10, 1 + 9, 2 + 8.....
As their understanding of place value improves they will start to
be able to recall number bonds for larger numbers using the
above number bonds to help them.
We do work on this in class; however once your child
understands what a number bond is, quick recall comes from
frequent practice.
Another vital mental maths skill is doubling numbers up to 5 /10
/ 20. This is first taught using hands and then pictures.
After this, the children will then learn the inverse of doubling:
halving.
We often get asked what objects children should use to
help them add up at home...... ANYTHING!!!
objects, combine them and see how much there is
‘altogether’.
For subtraction, encourage children to count out the larger
group then ‘take away’ the smaller number and see ‘how
many are left’.
We use lots of different words for addition and subtraction,
and we do not introduce the + and – symbols until children
are very confident with the operations.
Using a number line to add
•Children can start to use a number line for addition
and subtraction when they start to have a better
understanding of abstract number.
•It is important that they relate addition to ‘counting
on’ and subtraction to ‘counting back’ on the number
lines.
•They must understand that, with addition, the total
amount will be the largest and, when taking away,
the result will be smaller than the initial amount.
1
2
3 4 5
6 7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
12 + 8 =
Using a number line to subtract
When subtracting, children will need to understand
back.
Some children prefer to ‘find the difference’ to solve
subtraction number sentences – where they start
with the lower number.
1
2
3 4 5
6 7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
20 – 8 =
A hundred square
•When dealing with
larger numbers children
progress from using a
number line to a
hundred square.
•The methods of
are the same as on a
number line.
•Children soon learn
that, to add 10, they can
simply ‘jump down’ 1
place.
subtraction, we use
Multiplication and Division
We do not use the symbols for multiplication or division
until children are confident with the concept of ‘lots of’ as
repeated addition and division as ‘sharing’.
2 + 2 + 2 + 2 =8
Key Words:
• Lots of ...
•Sets of …
•Groups of …
• Shared between…
Word problems
Once the children are confident with using the methods
of each operation we use word problems so they can
apply their skills to ‘real life’ situations.
The problem:
Bob had 24 sweets. He ate 6
• When the children are familiar
with more than one operation
an important part of word
problems is deciding what
operation to use.
many sweets does Bob
What do I need to do?
Write the number sentence and solve it:
Data handling
Tally chart
Pictogram
Bar graph
Venn diagram
Carroll diagram
Other Maths in The Infants
Patterns
Sorting
Other Maths in The Infants
2D Shape
3D Shape
Key Words:
• Faces
• Edges
• Vertices
Key Words:
• Corners
• Sides
•Straight
•Curved
Other Maths in The Infants
Measuring
Key Words:
• Estimate
• Length – long, tall, wide
thick thin......not ‘big’
• Mass – weigh, light, heavy
• Capacity – full, empty
Other Maths in The Infants
Time
• Begin by sequencing events.
• Distinguish between times of day,
e.g. morning, afternoon, night.
• Learn days, then months, in order.
• Analogue clock to tell the time.
• Events that happen at o’clock times.
• Hour hand points to an o’clock, or
tells us where we are in relation to an
o’clock.
• Minute hand tells us if it is o’clock
now, or how many minutes past an
o’clock or coming up to an o’clock.
• Once confident, move onto 12 hour
Money
Need to recognise coins and know the value of
each.
When counting small amounts, tap the coin the
correct amount of times.
Making totals, first with 1ps, then using other
coins.
Finding change.
The more opportunity
money, the easier they
will find maths related
to it.
Language in Maths
We have included a vocabulary list in your
packs to show the words that get used in
maths lessons in the infants.
We encourage children to verbalise their
understanding and explain how they have
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# What Is 71/100 as a Decimal + Solution With Free Steps
The fraction 71/100 as a decimal is equal to 0.71.
Long division is a mathematical operation that enables you to solve challenging and intricate division-related problems. The Long division approach simplifies difficult division by breaking enormous numbers into manageable pieces.
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 71/100.
## 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 = 71
Divisor = 100
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 = 71 $\div$ 100
This is when we go through the Long Division solution to our problem. Figure 1 shows how Long Division is done:
Figure 1
## 71/100 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 71 and 100, we can see how 71 is Smaller than 100, and to solve this division, we require that 71 be Bigger than 100.
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 71, which after getting multiplied by 10 becomes 710.
We take this 710 and divide it by 100; this can be done as follows:
710 $\div$ 100 $\approx$ 7
Where:
100 x 7 = 700
This will lead to the generation of a Remainder equal to 710 – 700 = 10. Now this means we have to repeat the process by Converting the 10 into 100 and solving for that:
100 $\div$ 100 $=$ 1
Where:
100 x 1 = 100
This, therefore, produces another Remainder which is equal to 100 – 100 = 0.
Finally, we have a Quotient generated after combining the three pieces of it as 0.71=z, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
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Section 4: Inverse Functions
# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)
• Notes
• davidvictor
• 12
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Section 8.4 Inverse Functions 803 Version: Fall 2007 8.4 Inverse Functions As we saw in the last section, in order to solve application problems involving expo- nential functions, we will need to be able to solve exponential equations such as 1500 = 1000 e 0 . 06 t or 300 = 2 x . However, we currently don’t have any mathematical tools at our disposal to solve for a variable that appears as an exponent, as in these equations. In this section, we will develop the concept of an inverse function, which will in turn be used to define the tool that we need, the logarithm, in Section 8.5. One-to-One Functions Definition 1. A given function f is said to be one-to-one if for each value y in the range of f , there is just one value x in the domain of f such that y = f ( x ) . In other words, f is one-to-one if each output y of f corresponds to precisely one input x . It’s easiest to understand this definition by looking at mapping diagrams and graphs of some example functions. l⚏ Example 2. Consider the two functions h and k defined according to the mapping diagrams in Figure 1 . In Figure 1 (a), there are two values in the domain that are both mapped onto 3 in the range. Hence, the function h is not one-to-one. On the other hand, in Figure 1 (b), for each output in the range of k , there is only one input in the domain that gets mapped onto it. Therefore, k is a one-to-one function. 1 2 3 h 1 2 3 4 k (a) (b) Figure 1. Mapping diagrams help to determine if a function is one-to-one. l⚏ Example 3. The graph of a function is shown in Figure 2 (a). For this function f , the y -value 4 is the output corresponding to two input values, x = 1 and x = 3 (see the corresponding mapping diagram in Figure 2 (b)). Therefore, f is not one-to-one. Graphically, this is apparent by drawing horizontal segments from the point (0 , 4) on the y -axis over to the corresponding points on the graph, and then drawing vertical segments to the x -axis. These segments meet the x -axis at 1 and 3 . Copyrighted material. See: 1
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804 Chapter 8 Exponential and Logarithmic Functions Version: Fall 2007 x y f 4 3 1 1 3 4 f (a) (b) Figure 2. A function which is not one-to-one. l⚏ Example 4. In Figure 3 , each y -value in the range of f corresponds to just one input value x . Therefore, this function is one-to-one. Graphically, this can be seen by mentally drawing a horizontal segment from each point on the y -axis over to the corresponding point on the graph, and then drawing a vertical segment to the x -axis. Several examples are shown in Figure 3 . It’s apparent that this procedure will always result in just one corresponding point on the x -axis, because each y -value only corresponds to one point on the graph. In fact, it’s easiest to just note that since each horizontal line only intersects the graph once, then there can be only one corresponding input to each output.
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• '
• NoProfessor
• Inverse function
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# How do you simplify ${{\left( 2+5\iota \right)}^{2}}$?
Last updated date: 19th Jul 2024
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Hint: In this problem, we have to find the square of a complex number. This can also be done by multiplying the complex number by itself. After multiplication, we shall group all the terms and further simplify them. Then we will use the basic properties of complex numbers such as the different values of iota when it is raised to various powers.
In order to simplify the given expression, we must have prior knowledge of complex numbers. A complex number is of the form, $x+\iota y$. It comprises two parts. One is the real number part which lies on the x-axis of the cartesian plane and the other is a complex number part which lies on the y-axis of the cartesian plane.
The expression given can also be written as:
${{\left( 2+5\iota \right)}^{2}}=\left( 2+5\iota \right)\left( 2+5\iota \right)$
Thus, we shall multiply the complex function by itself.
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+10\iota +10\iota +25{{\iota }^{2}}$
Adding the like terms of iota, we get
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota +25{{\iota }^{2}}$
There are predefined values assigned to iota when it is raised to certain powers. They are as follows.
${{\iota }^{1}}=\iota$
${{\iota }^{2}}=-1$
${{\iota }^{3}}=-\iota$
${{\iota }^{4}}=1$
From the above results, we see that ${{\iota }^{2}}=-1$. Substituting this value in our equation, we get
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota +25\left( -1 \right)$
\begin{align} & \Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota -25 \\ & \Rightarrow {{\left( 2+5\iota \right)}^{2}}=-21+20\iota \\ \end{align}
Therefore, the given expression ${{\left( 2+5\iota \right)}^{2}}$ is simplified to $-21+20\iota$.
Note:
The generalized rule for iota raised to any power is that iota raised to the power of 4 or multiples of 4 is equal to 1. Otherwise, the value of iota in terms of multiples 4 is given as, ${{\iota }^{4k+1}}=\iota$, ${{\iota }^{4k+2}}=-1$ and ${{\iota }^{4k+3}}=-\iota$ where $k$ is a constant. Using this, we can easily find the value of iota even when its power is very large natural numbers.
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# FIND QUOTIENT AND REMAINDER USING SYNTHETIC DIVISION
## About "Find Quotient and Remainder Using Synthetic Division"
Find Quotient and Remainder Using Synthetic Division :
Here we are going to see some practice questions to understand the concept of finding quotient and remainder using synthetic division.
## Find Quotient and Remainder Using Synthetic Division - Practice questions
Question 1 :
Find the quotient and remainder for the following using synthetic division:
(i) (x3 + x2 - 7x - 3) ÷ (x - 3)
Solution :
Quotient = x2 + 4x - 3
Remainder = -12
(ii) (x3 + 2x2 - x - 4) ÷ (x + 2)
Solution :
Quotient = x2 + 0x - 1
Remainder = -2
(iii) (3x3 - 2x2 + 7x - 5) ÷ (x + 3)
Solution :
Quotient = 3x2 - 11x + 40
Remainder = -125
(iv) (8x4 - 2x2 + 6x + 5) ÷ (4x + 1)
Solution :
Quotient = 8x2 - x + 40
Remainder = -125
(iv) (8x4 - 2x2 + 6x + 5) ÷ (4x + 1)
Solution :
Quotient : 8x3 - 2x2 -(3x/2) + (51/8)
Remainder : 109/32
Question 2 :
If the quotient obtained on dividing (8x4 - 2x2 + 6x - 7) by (2x + 1) is (4x3 + px2 - qx + 3) then find p, q and also the remainder.
Solution :
8x3 - 4x2 + 0x + 6
Dividing the quotient by 2, we get
4x3 - 2x2 + 0x + 3
The value of p and q are -2 and 0 respectively and remainder is -10.
Question 3 :
If the quotient obtained on dividing 3x3 + 11x2 + 34x + 106 by x - 3 is 3x2 + ax + b, then find a, b and also the remainder.
Solution :
Quotient = 3x2 + 20x + 94
Given quotient = 3x2 + ax + b
a = 20 and b = 94.
After having gone through the stuff given above, we hope that the students would have understood, "Find Quotient and Remainder Using Synthetic Division"
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https://www.jobilize.com/trigonometry/course/2-7-linear-inequalities-and-absolute-value-inequalities-by-openstax?qcr=www.quizover.com
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# 2.7 Linear inequalities and absolute value inequalities
Page 1 / 11
In this section you will:
• Use interval notation.
• Use properties of inequalities.
• Solve inequalities in one variable algebraically.
• Solve absolute value inequalities.
It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.
## Using interval notation
Indicating the solution to an inequality such as $\text{\hspace{0.17em}}x\ge 4\text{\hspace{0.17em}}$ can be achieved in several ways.
We can use a number line as shown in [link] . The blue ray begins at $\text{\hspace{0.17em}}x=4\text{\hspace{0.17em}}$ and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.
We can use set-builder notation : $\text{\hspace{0.17em}}\left\{x|x\ge 4\right\},$ which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.
The third method is interval notation , in which solution sets are indicated with parentheses or brackets. The solutions to $\text{\hspace{0.17em}}x\ge 4\text{\hspace{0.17em}}$ are represented as $\text{\hspace{0.17em}}\left[4,\infty \right).\text{\hspace{0.17em}}$ This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.
The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval , or a set of numbers in which a solution falls, are $\text{\hspace{0.17em}}\left[-2,6\right),$ or all numbers between $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}6,$ including $\text{\hspace{0.17em}}-2,$ but not including $\text{\hspace{0.17em}}6;$ $\left(-1,0\right),$ all real numbers between, but not including $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0;$ and $\text{\hspace{0.17em}}\left(-\infty ,1\right],$ all real numbers less than and including $\text{\hspace{0.17em}}1.\text{\hspace{0.17em}}$ [link] outlines the possibilities.
Set Indicated Set-Builder Notation Interval Notation
All real numbers between a and b , but not including a or b $\left\{x|a $\left(a,b\right)$
All real numbers greater than a , but not including a $\left\{x|x>a\right\}$ $\left(a,\infty \right)$
All real numbers less than b , but not including b $\left\{x|x $\left(-\infty ,b\right)$
All real numbers greater than a , including a $\left\{x|x\ge a\right\}$ $\left[a,\infty \right)$
All real numbers less than b , including b $\left\{x|x\le b\right\}$ $\left(-\infty ,b\right]$
All real numbers between a and b , including a $\left\{x|a\le x $\left[a,b\right)$
All real numbers between a and b , including b $\left\{x|a $\left(a,b\right]$
All real numbers between a and b , including a and b $\left\{x|a\le x\le b\right\}$ $\left[a,b\right]$
All real numbers less than a or greater than b $\left\{x|xb\right\}$ $\left(-\infty ,a\right)\cup \left(b,\infty \right)$
All real numbers $\left(-\infty ,\infty \right)$
## Using interval notation to express all real numbers greater than or equal to a
Use interval notation to indicate all real numbers greater than or equal to $\text{\hspace{0.17em}}-2.$
Use a bracket on the left of $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and parentheses after infinity: $\text{\hspace{0.17em}}\left[-2,\infty \right).$ The bracket indicates that $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ is included in the set with all real numbers greater than $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ to infinity.
Use interval notation to indicate all real numbers between and including $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}5.$
$\left[-3,5\right]$
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Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
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tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
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If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
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1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
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https://www.cuemath.com/ncert-solutions/q-7-exercise-13-6-surface-areas-and-volumes-class-9-maths/
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# Ex.13.6 Q7 Surface Areas and Volumes Solution - NCERT Maths Class 9
## Question
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is $$7 \; \rm mm$$ and the diameter of the graphite is $$1\; \rm mm$$. If the length of the pencil is $$14 \; \rm cm$$, find the volume of the wood and that of the graphite.
Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-6 | Question 7
## Text Solution
Reasoning:
Volume of cylinder $$\pi {r^2}h$$
What is known?
Diameter of the pencil, diameter of graphite and length of the pencil.
What is unknown?
Volume of the wood and that if graphite.
Steps:
For cylinder graphite.
Diameter $$(2r) = 1 \; \rm mm$$
Radius \begin{align}(r) = \frac{1}{2}\,\, \rm mm \end{align}
Length of the pencil $$(h) = 14 \; \rm cm = 140 \;\rm mm$$
\begin{align}\text{Capacity}&= \text{Volume} \\ &=\,\pi {r^2}h\\ &= \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times 140\\ &= 110\,\, \rm mm^3 \end{align}
\begin{align} \rm{In} \,\,c{m^3} &= \frac{{110}}{{10 \times 10 \times 10}} = 0.11\,\, \rm cm^3 \end{align}
For cylinder of wood:
To find the volume of the wood:
Total volume of the pencil $$-$$ Volume of graphite
\begin{align}\pi {R^2}h - \pi {r^2}h \pi h({R^2} - {r^2})\end{align}
Diameter of pencil $$(2R) = 7 \; \rm mm$$
Radius $$(r) = \frac{7}{2} \rm mm$$
Length of the pencil $$(h) = 14 \; \rm cm = 140 \; \rm mm$$
Volume of wood \begin{align}=\,\pi h({R^2} - {r^2}) \end{align}
\begin{align} &= \frac{{22}}{7} \times 140 \times [{(\frac{7}{2})^2} - {(\frac{1}{2})^2}]\\ &= \frac{{22}}{7} \times 140 \times [\frac{{49}}{4} - \frac{1}{2}]\\ &= 5280\,\, \rm mm^3\\ &= \frac{{5280}}{{10 \times 10 \times 10}}\, \rm cm^3 \\ &= 5.28\,\, \rm cm^3 \end{align}
Volume of the wood $$= 5.28\,\, \rm cm^3$$
Volume of graphite $$= 0.11\,\,\, \rm cm^3$$
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# ORDER OF OPERATIONS WORKSHEETS
## About "Order of operations worksheets"
Order of operations worksheets are much useful to the kids who would like to practice problems on the binary operations like add, subtract, multiply, divide, squaring,etc.
When we have two or more operations in the same expression, we may have question about which one has to be done first, which one has to be done next.
But order of operations or bodmas rule or pemdas rule tells us in which order we have to do the operations one by one.
What is BODMAS rule ?
The rule or order that we use to simplify expressions in math is called "BODMAS" rule.
Very simply way to remember BODMAS rule!
B -----> Brackets first (Parentheses)
O -----> Of (orders :Powers and radicals)
D -----> Division
M -----> Multiplication
S -----> Subtraction
Important notes :
1. In a particular simplification, if you have both multiplication and division, do the operations one by one in the order from left to right.
2. Division does not always come before multiplication. We have to do one by one in the order from left to right.
3. In a particular simplification, if you have both addition and subtraction, do the operations one by one in the order from left to right.
Examples :
12 ÷ 3 x 5 = 4 x 5 = 20
13 - 5 + 9 = 8 + 9 = 17
In the above simplification, we have both division and multiplication. From left to right, we have division first and multiplication next. So we do division first and multiplication next.
To have better understanding on "Order of operations", let us look at order of operations worksheets problems.
## Order of operations worksheets problems
1. Evaluate : 6 + 7 x 8
2. Evaluate : 10² - 16 ÷ 8
3. Evaluate : (25 + 11) x 2
4. Evaluate : 3 + 6 x (5+4) ÷ 3 -7
5. Evaluate : 36 - 2(20+12÷4x3-2x2) + 10
6. Evaluate : 6+[(16-4)÷(2²+2)]-2
7. Evaluate : (96÷12)+14x(12+8)÷2
8. Evaluate : (93+15) ÷ (3x4) - 24 + 8
9. Evaluate : 55 ÷ 11 + (18 - 6) x 9
10. Evaluate : (7 + 18) x 3 ÷(2+13) - 28
Here, they are
## Step by step solution
Problem 1 :
Evaluate : 6 + 7 x 8
Expression6 + 7 x 8 Evaluation= 6 + 7 x 8= 6 + 56 = 62 OperationMultiplicationAdditionResult
Problem 2 :
Evaluate : 10² - 16 ÷ 8
Expression10² - 16 ÷ 8 Evaluation= 10² - 16 ÷ 8= 100 - 16 ÷ 8= 100 - 2= 98 OperationPowerDivisionSubtractionResult
Problem 3 :
Evaluate : (25 + 11) x 2
Expression(25 + 11) x 2 Evaluation= (25 + 11) x 2= 36 x 2= 72 OperationParenthesisMultiplicationResult
Problem 4 :
Evaluate : 3 + 6 x (5+4) ÷ 3 -7
Expression3 + 6 x (5+4) ÷ 3 -7 Evaluation= 3 + 6 x (5+4) ÷ 3 -7= 3 + 6 x 9 ÷ 3 -7= 3 + 54 ÷ 3 -7= 3 + 18 -7= 21 - 7= 14 OperationParenthesisMultiplicationDivisionAdditionSubtractionResult
Problem 5 :
Evaluate : 36 - 2(20+12÷4x3-2x2) + 10
Problem 6 :
Evaluate : 6+[(16-4)÷(2²+2)]-2
Expression 6+[(16-4)÷(2²+2)]-2 Evaluation= 6+[(16-4)÷(2²+2)]-2= 6+[12÷(2²+2)]-2= 6+[12÷(4+2)]-2= 6+[12÷6]-2= 6+2 - 2= 8 - 2=6 OperationParenthesisPowerParenthesisParenthesisAdditionSubtractionResult
Problem 7 :
Evaluate : (96÷12)+14x(12+8)÷2
Expression (96÷12)+14x(12+8) ÷ 2 Evaluation=(96÷12)+14x(12+8) ÷ 2= 8 + 14x20 ÷ 2= 8 + 280 ÷ 2= 8 + 140 = 148 OperationParenthesesMultiplicationDivisionAdditionResult
Problem 8 :
Evaluate : (93+15) ÷ (3x4) - 24 + 8
Expression (93+15)÷(3x4)-24+8 Evaluation= (93+15)÷(3x4)-24+8 = 108 ÷ 12 - 24 + 8 = 9 - 24 + 8= -15 + 8= -7 OperationParenthesisDivisionSubtractionSubtractionResult
Problem 9 :
Evaluate : 55 ÷ 11 + (18 - 6) x 9
Expression 55÷11+(18-6)x9 Evaluation= 55÷11+(18-6)x9 = 55÷11 + 12x9= 5 + 12x9= 5 + 108= 113 OperationParenthesisDivisionMultiplicationAdditionResult
Problem 10 :
Evaluate : (7 + 18) x 3 ÷(2+13) - 28
Expression(7+18)x3÷(2+13)- 28 Evaluation= (7+18)x3÷(2+13)-28= 25x3 ÷ 15 - 28= 75 ÷ 15 - 28= 5 - 28= -23 OperationParenthesesMultiplicationDivisionSubtractionResult
We hope that the students would have understood the stuff given on "Order of operations worksheets".
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