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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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