text stringlengths 22 1.01M |
|---|
# Question Video: Modeling Points, Lines, and Planes Mathematics
Consider the figure. Write another name for the plane π. How many planes are shown in the figure? Which two planes intersect in the line segment π΄π·? Are planes π and π·πΈπΉ parallel or intersecting?
05:28
### Video Transcript
Consider the following figure. Write another name for the plane π. How many planes are shown in the figure? Which two planes intersect in the line segment π΄π·? Are planes π and π·πΈπΉ parallel or intersecting?
Letβs begin by considering this plane π. We can recall that a plane is a two-dimensional surface made up of points that extends infinitely in all directions and there exists exactly one plane through any three noncollinear points. This word, noncollinear, simply means that the three points donβt lie on a straight line. And it is this second sentence in this statement that will help us answer the first part of this question to write another name for the plane π.
Well, because we know that there is one plane through any three noncollinear points, then we can also define a plane as the plane which passes through three given points. In this figure, we can see that the points π΄, π΅, and πΆ lie on plane π. Therefore, another name that we could give to plane π could be plane π΄π΅πΆ. Of course, plane π΅π΄πΆ or plane πΆπ΅π΄ would also be valid answers.
Now, letβs have a look at the second part of this question to identify how many planes there are in this figure. We might think that this is confusing as surely thereβs just one plane, the plane π, or plane π΄π΅πΆ. However, using the fact that a plane is a two-dimensional surface means that we can identify more planes in this figure. We can see, shaded in orange, that there would be another plane at the back of this figure. We could use the points which lie on this plane to identify it as the plane π΄πΆπΉπ·. We can then also identify the plane which passes through the points π΄, π΅, πΈ, and π·. This would be plane π΄π΅πΈπ·. Then we have another plane passing through the points π΅, πΆ, πΉ, πΈ and finally one more plane which passes through the three points at the top of this figure, giving us plane π·πΈπΉ.
Combined with plane π or plane π΄π΅πΆ, that would give us five different planes. As there are no other planes on this figure, we can give the answer for the second part of the question that there are five planes shown in the figure.
We can then look at the third part of this question which asks, which two planes intersect in the line segment π΄π·? Letβs identify that the line segment π΄π· is marked here on the figure in pink. And we will have two planes which intersect here at the line segment π΄π·. Weβll have this plane π΄πΆπΉπ· and this plane π΄π΅πΈπ·. There are no other planes shown in this figure which intersect in the line segment π΄π·. So the answer to this third part is plane π΄πΆπΉπ· and plane π΄π΅πΈπ·.
And finally, letβs take a look at the fourth part of this question which asks, are planes π and π·πΈπΉ parallel or intersecting? The plane π·πΈπΉ will lie at the top of this figure. Letβs recall that there are three possible relationships that can exist between two planes in space. Firstly, they can be parallel. Two parallel planes do not intersect. Secondly, two planes can be intersecting. As we saw in the previous part of this question, when two planes intersect, the intersection is a line. And thirdly, although we are not asked about it in this part of the question, two planes can be coincident. That means that the two planes would share all the same points.
So letβs look at the figure and see if we can identify if planes π and π·πΈπΉ are parallel or intersecting. Now we might look at this diagram and visualize it as a three-dimensional object. We might visualize this as a triangular prism, in which case itβs very likely that the top and the lower base would be parallel. However, we have to be really careful here because actually we donβt have enough information to say that for sure. We donβt know, for example, that the length πΆπΉ is the same as the length π΄π· is the same as the length π΅πΈ. If these three lengths or distances were all the same, then the two planes would be parallel. If the planes are not parallel, then they would be intersecting. But we donβt know for sure. So the statement for the answer needs to reflect this.
We can therefore give the answer for the fourth and final part of this question as βIt cannot be determined without more information.β |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.4: Find Multiples and Factors (Part 2)
[ "article:topic", "authorname:openstax" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
### Identify Prime and Composite Numbers
Some numbers, like 72, have many factors. Other numbers, such as 7, have only two factors: 1 and the number. A number with only two factors is called a prime number. A number with more than two factors is called a composite number. The number 1 is neither prime nor composite. It has only one factor, itself.
Definition: Prime Numbers and Composite Numbers
A prime number is a counting number greater than 1 whose only factors are 1 and itself.
A composite number is a counting number that is not prime.
Figure 2.10 lists the counting numbers from 2 through 20 along with their factors. The highlighted numbers are prime, since each has only two factors.
Figure 2.10 - Factors of the counting numbers from 2 through 20, with prime numbers highlighted
The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. There are many larger prime numbers too. In order to determine whether a number is prime or composite, we need to see if the number has any factors other than 1 and itself. To do this, we can test each of the smaller prime numbers in order to see if it is a factor of the number. If none of the prime numbers are factors, then that number is also prime.
HOW TO: DETERMINE IF A NUMBER IS PRIME.
Step 1. Test each of the primes, in order, to see if it is a factor of the number.
Step 2. Start with 2 and stop when the quotient is smaller than the divisor or when a prime factor is found.
Step 3. If the number has a prime factor, then it is a composite number. If it has no prime factors, then the number is prime.
Example 2.47:
Identify each number as prime or composite: (a) 83 (b) 77
##### Solution
(a) Test each prime, in order, to see if it is a factor of 83, starting with 2, as shown. We will stop when the quotient is smaller than the divisor.
Prime Test Factor of 83?
2 Last digit of 83 is not 0, 2, 4, 6, or 8. No.
3 8 + 3 = 11, and 11 is not divisible by 3. No.
5 The last digit of 83 is not 5 or 0. No.
7 83 ÷ 7 = 11.857…. No.
11 83 ÷ 11 = 7.545… No.
We can stop when we get to 11 because the quotient (7.545…) is less than the divisor. We did not find any prime numbers that are factors of 83, so we know 83 is prime.
(b) Test each prime, in order, to see if it is a factor of 77.
Prime Test Factor of 77?
2 Last digit is not 0, 2, 4, 6, or 8. No.
3 7 + 7 = 14, and 14 is not divisible by 3. No.
5 The last digit is not 5 or 0. No.
7 77 ÷ 11 = 7 Yes.
Since 77 is divisible by 7, we know it is not a prime number. It is composite.
Exercise 2.93:
Identify the number as prime or composite: 91
Exercise 2.94:
Identify the number as prime or composite: 137
Factors
### Practice Makes Perfect
#### Identify Multiples of Numbers
In the following exercises, list all the multiples less than 50 for the given number.
1. 2
2. 3
3. 4
4. 5
5. 6
6. 7
7. 8
8. 9
9. 10
10. 12
#### Use Common Divisibility Tests
In the following exercises, use the divisibility tests to determine whether each number is divisible by 2, 3, 4, 5, 6, and 10.
1. 84
2. 96
3. 75
4. 78
5. 168
6. 264
7. 900
8. 800
9. 896
10. 942
11. 375
12. 750
13. 350
14. 550
15. 1430
16. 1080
17. 22,335
18. 39,075
#### Find All the Factors of a Number
In the following exercises, find all the factors of the given number.
1. 36
2. 42
3. 60
4. 48
5. 144
6. 200
7. 588
8. 576
#### Identify Prime and Composite Numbers
In the following exercises, determine if the given number is prime or composite.
1. 43
2. 67
3. 39
4. 53
5. 71
6. 119
7. 481
8. 221
9. 209
10. 359
11. 667
12. 1771
### Everyday Math
1. Banking Frank’s grandmother gave him $100 at his high school graduation. Instead of spending it, Frank opened a bank account. Every week, he added$15 to the account. The table shows how much money Frank had put in the account by the end of each week. Complete the table by filling in the blanks.
Weeks after graduation Total number of dollars Frank put in the account Simplified Total
0 100 100
1 100 + 15 115
2 100 + 15 • 2 130
3 100 + 15 • 3
4 100 + 15 • []
5 100 + []
6
20
x
1. Banking In March, Gina opened a Christmas club savings account at her bank. She deposited $75 to open the account. Every week, she added$20 to the account. The table shows how much money Gina had put in the account by the end of each week. Complete the table by filling in the blanks.
Weeks after opening the account Total number of dollars Gina put in the account Simplified Total
0 75 75
1 75 + 20 95
2 75 + 20 • 2 115
3 75 + 20 • 3
4 75 + 20 • []
5 75 + []
6
20
x
### Writing Exercises
1. If a number is divisible by 2 and by 3, why is it also divisible by 6?
2. What is the difference between prime numbers and composite numbers?
### Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this? |
# Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross.
## Presentation on theme: "Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross."— Presentation transcript:
Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross
You will soon be assigned problems to test whether you have internalized the material in Lesson 11 of our algebra course. The Keystone Illustrations below are prototypes of the problems you'll be doing. Work out the problems on your own. Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve each problem. Instructions for the Keystone Problems next © 2007 Herbert I. Gross
As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross
The beauty of logical thought is that we can often solve complicated problems by replacing them by a series of equivalent simpler problems. In particular the goal of this set of exercises is to show how using the rules of the game we can replace a rather complex expression such as… Preface next © 2007 Herbert I. Gross 2 {5 [15 – 4 (3 – c)] – 6} by the simpler, but equivalent, expression 40 c + 18 that is much easier to work with. next
We will develop this idea through a step-by-step simplification of the left side of the expression 2 {5 [15 – 4 (3 – c)] – 6}. The two exercises in this set will illustrate our approach. More specifically, the steps that we will use to solve the equation in Question 2 are precisely those which make up the different parts to Question 1. © 2007 Herbert I. Gross
next © 2007 Herbert I. Gross 1. Simplify each of the following expressions… a. - 4(3 + - c) b. 15 + - 4(3 + - c) c. 5[15 + - 4(3 + - c)] d. 5[15 + - 4(3 + - c)] + - 6 e. 2{5[15 + - 4(3 + - c)] + - 6} 2. Use the result of Problem 1e to solve the equation… 2{5[15 - 4(3 - c)] - 6} = 94 next Keystone Problems for Lesson 10
next © 2007 Herbert I. Gross Solution for Problem 1a This exercise asks us to simplify the expression… - 4(3 + - c). By the distributive property… From the arithmetic for signed numbers we know that… If we now replace each term on the right hand side of the first equation by its value in the second equation, we obtain… - 4(3 + - c) = - 4(3) + - 4( - c) - 4(3) = - 12 and - 4( - c) = 4c - 4(3 + - c) = - 12 + 4c next
In problems such as this, we use the word “simplify” rather reluctantly. The reason is that the meaning of simplify is subjective. That is, what seems ”simpler” to one person might not seem simpler to another. So in the spirit of our game of algebra we will use “simplify” as another word for “paraphrase”. In this particular discussion it will mean to paraphrase each expression into a sum consisting of two terms. © 2007 Herbert I. Gross Note 1a
In terms of the above note, someone might prefer to simplify our answer by using the commutative property of addition and rewriting - 12 + 4c in the form 4c + - 12. next © 2007 Herbert I. Gross next And someone else might prefer replacing 4c + - 12 by the “simpler” expression 4c – 12. Note 1a
With respect to the rules of the game, our preference is to write subtraction in the form of addition because our rules are stated in terms of addition. However, once you feel that you’ve internalized the connection between addition and subtraction, you may use expressions such as… next © 2007 Herbert I. Gross Note 1a next 4c + - 12 and 4c –12 interchangeably.
If we were being insistent on playing the game strictly by the rules, we would have to prove such statements as: “The product of two negative numbers is always a positive number”. Such an approach could be tedious and too abstract for our present purposes. Instead, we will assume that the approach we used in Lessons 3, 4, and 5 for developing the arithmetic of signed numbers is sufficiently acceptable to all of our players. next © 2007 Herbert I. Gross Note 1a
next © 2007 Herbert I. Gross Solution for Problem 1b This exercise asks us to simplify the expression… 15 + - 4(3 + - c) In Problem 1a, we showed that… We now replace - 4(3 + c) by - 12 + 4c to obtain… 15 + - 4(3 + - c) - 4(3 + - c) = - 12 + 4c next - 12 + 4c ( )
next © 2007 Herbert I. Gross Solution for Problem 1b and since 15 + - 12 = 3, we may rewrite the above equality as... next 15 + - 4(3 + - c) = 3 + 4c 15 + - 4(3 + - c) = 15 + ( - 12 + 4c) By the associative property of addition we know that… 15 + - 4(3 + - c) = 15 + - 12 + 4c () Thus…
One of the most common mistakes made by beginning students is to begin by adding the 15 and - 4 in the expression next © 2007 Herbert I. Gross to obtain the incorrect expression… next Caution 15 + - 4(3 + - c) 11(3 + - c) next
© 2007 Herbert I. Gross Caution 15 + - 4(3) = 3 For example, if we replace c by 0 in the expression 15 + - 4(3 + - c) we obtain… On the other hand, if we replace c by 0 in 11(3 + - c), we obtain… 11(3) = 33 The fact that the same input gives different outputs shows that the two expressions can’t be equal. next
© 2007 Herbert I. Gross Caution (15 + - 4)(3 + - c) That is, by our agreeing to the PEMDAS convention we do all multiplications before we do any additions. Hence, if we had wanted the 15 and the - 4 to be added first, we would have had to write the expression as…
In words the expression 15 + - 4(3 + - c) tells us… next © 2007 Herbert I. Gross Note 1b *** Start with any number c. *** Replace it by its opposite… - c. *** Add 3… 3 + - c. *** Multiply by - 4… - 4(3 + - c). *** Add 15… 15 + - 4(3 + - c). What we showed in this exercise is that the expression 15 + - 4( 3 + - c) is equivalent to the expression 3 + 4c. next
In words the expression 3 + 4c tells us… © 2007 Herbert I. Gross Note 1b *** Start with c… c. *** Multiply 4… 4c. *** Add 3… 3 + 4c. next
© 2007 Herbert I. Gross Note 1b In this sense, what we mean by “simplified” is that the relatively cumbersome set of instructions… next Start with any number c. Replace it by its opposite. Add 3. Multiply by - 4. Add 15 and then write the answer. …can be replaced by the less cumbersome but equivalent set of instructions… Start with any number c. Multiply it by 4. Add 3 and then write the answer.
next © 2007 Herbert I. Gross Solution for Problem 1c This exercise asks us to simplify the expression… 5[15 + - 4(3 + - c)] From the previous exercise we already know that… So by replacing 15 + - 4(3 + - c) by 3 + 4c, 5[15 + - 4(3 + - c)] becomes… 5[3 + 4c] 15 + - 4(3 + - c) = 3 + 4c next 3 + 4c
next © 2007 Herbert I. Gross By the distributive property, we know that… 5[3 + 4c] = 5(3) + 5(4c) = 15 + 5[4c] next By the associative property of multiplication, we know that… 5[4c] = [5(4)]c = 20c
next © 2007 Herbert I. Gross next If we now replace 5[4c] by 20c, we obtain… 5[3 + 4c] = 15 + 20c next In summary, we have thus far shown that… (1) 5[15 + - 4(3 + - c)] = 5[3 +4c] and (2) 5[3 + 4c] = 15 + 20c 5[15 + - 4(3 + - c)] = 15 + 20c Thus, by transitivity… 5[15 + - 4(3 + - c)] 15 + 20c next
It’s quite possible that in the above equation you looked at 5[4c] and immediately concluded that it was equal to 20c without even thinking consciously about the associative property of multiplication. Hopefully, such observations will become quite common as the course proceeds. For the time being, we’ll try to be as complete as we feel is necessary in demonstrating how certain results follow inescapably from others just by using the accepted rules of the game. © 2007 Herbert I. Gross Note 1c
next © 2007 Herbert I. Gross Solution for Problem 1d This exercise asks us to simplify the expression… 5[15 + - 4(3 + - c)] + - 6 In exercise 1c we showed that… Therefore by substitution 5[15 + - 4(3 + - c)] + - 6 becomes… 5[15 + - 4(3 + - c)] + - 6 (15 + 20c) next 5[15 + - 4(3 + - c)] = 15 + 20c
next © 2007 Herbert I. Gross By the commutative property of addition we know that… So we may replace 15 + 20c by 20c + 15 to obtain… Therefore, 5[15 + - 4(3 + - c)] + - 6 may be replaced by … next 5[15 + - 4(3 + - c)] + - 6 (20c + 15) 5[15 + - 4(3 + - c)] = 15 + 20c 20c + 15 15 + 20c = 20c + 15
next © 2007 Herbert I. Gross Then by the associative property of addition… next 20c + 15 + - 6 In summary… 5[15 + - 4(3 + - c)] + - 6 = (20c + 15) + - 6 and (20c + 15) + - 6 = 20 c + 9 5[15 + - 4(3 + - c)] + - 6 = 20c + 9 next () = 20c + 9 So again by transitivity… 5[15 + - 4(3 + - c)] + - 6 20 c + 9 next
© 2007 Herbert I. Gross Solution for Problem 1e This exercise asks us to simplify the expression… 2{5[15 + - 4(3 + - c)] + - 6} From the previous exercise we know … Hence, we replace 5[15 + - 4(3 + - c)] + - 6 in the first expression by its equivalent value in the above expression, to obtain… 2{5[15 + - 4(3 + - c)] + - 6} = 2{20c + 9} 5[15 + - 4(3 + - c)] + - 6 = 20c + 9 next
© 2007 Herbert I. Gross By the distributive property we know that… 2{20c + 9} = 2{20c} + 2{9} = 40 c + 18 And if we now replace 2{5[15 + - 4(3 + - c)] + - 6} In the previous expression with the value in the above expression we obtain... 2{5[15 + - 4(3 + - c)] + - 6} = 40c + 18 next Solution for Problem 1e
In any game there is usually more than one correct strategy for obtaining a given objective. In the game of mathematics, this may be restated as... next © 2007 Herbert I. Gross Enrichment Note Therefore, do not try to memorize our proofs. Rather, concentrate on how we elected to use a particular strategy to achieve each of the given objectives. next “In a well-defined problem there is only one correct answer, but hardly ever only one correct way to derive the answer”.
next © 2007 Herbert I. Gross Solution for Problem 2 In this exercise we are asked to use the result of Problem 1e to solve the equation… 2 {5[15 – 4(3 – c)] – 6} = 94 We begin our solution by observing that by the “add the opposite” rule, the equation… 2{5[15 + - 4(3 + - c)] + - 6} = 94 next 2 {5[15 – 4(3 – c)] – 6} = 94 …can be rewritten in the equivalent form…
© 2007 Herbert I. Gross Solution for Problem 2 The left side of our equation is precisely the expression we simplified in Exercise 1e. In other words the fact that… …means that we may replace the equation 2{5[15 + - 4(3 + - c)] + - 6} = 94 by the equivalent but simpler equation… next 40c + 18 = 94 2{5[15 + - 4(3 + - c)] + - 6} = 40c + 18
next © 2007 Herbert I. Gross Solution for Problem 2 To solve equation 40c + 18 = 94, we first subtract 18 from both sides to obtain the equivalent equation… Finally, we divide both sides of the above equation by 40 to obtain… next 40c = 76 c = 76 / 40 = 19 / 10 or in decimal form… = 1.9
We could have omitted Exercise 1 in its entirety and just asked you to find the value of c for which... © 2007 Herbert I. Gross Note 2 In that case one of the ways we could have used to answer the question would have been to rewrite the equation in the form… 2 {5[15 – 4(3 – c)] – 6} = 94 2{5[15 + - 4(3 + - c)] + - 6} = 94 next
Notice that the left side of 2{5[15 + - 4(3 + - c)] + - 6} = 94 is the expression that we simplified in Exercise 1e. © 2007 Herbert I. Gross Note 2 Then since grouping symbols are removed starting from the innermost set first, we would have begun the simplifying process by starting with the removal of the parentheses; and this is precisely what we did in Exercises 1a, 1b, 1c, 1d, and 1e. next More visually… 2{ 5 [15 + - 4(3 + - c)] + - 6 } - 4(3 + - c)[15 + - 4(3 + - c)] 5 [15 + - 4(3 + - c)] 5[15 + - 4(3 + - c)] + - 6 next 1a, 2 {5[15 + - 4(3 + - c) + - 6} 1b, 1c, 1d, 1e.
In summary, we used the rules in Lessons 9 and 10 to show that the more complicated expression… next © 2007 Herbert I. Gross Summary …could be replaced by the equivalent, but much simpler expression… next 2{5[15 + - 4(3 + - c)] + - 6} 40c + 18
next © 2007 Herbert I. Gross At least until you become more comfortable with complicated algebraic expressions; it is a good idea to rewrite every subtraction problem in terms of the add-the-opposite rule. That is, wherever we see a minus sign we replace it by a plus sign and change the sign of the number immediately after the minus sign. Feeling Comfortable
next © 2007 Herbert I. Gross *** we first locate the minus signs… *** Then we replace each minus sign by a plus sign… next By way of Review 2 {5[15 – 4(3 – c)] – 6} = 94 2 {5[15 – 4(3 – c)] – 6} = 94 …and we replace each number immediately after the changed minus sign by its opposite… 2 {5[15 + 4(3 + c)] + 6} = 94 - - - next +++
© 2007 Herbert I. Gross *** Notice that the steps we used in solving the equation… …had nothing to do with the fact that the right side was 94. That is, we demonstrated that 2{5[15 – 4(3 – c)] – 6} = 40c + 18 without any reference to the right hand side of the above equation. By way of Review 2 {5[15 – 4(3 – c)] – 6} = 94 next
© 2007 Herbert I. Gross Recall that to evaluate the expression… 2 {5[15 – 4(3 – c)] – 6} …for any given value of c, we start with c and remove the grouping symbols by working our way from the inside to the outside. The above comments may seem more impressive if you look at them in terms of “programs”. next
© 2007 Herbert I. Gross next Program #1 (1) Start with c (2) Subtract it from 3 (3) Multiply the result by 4 (4) Subtract this result from 15 (5) Multiply the result by 5 (6) Subtract 6 (7) Multiply by 2 (8) Write the answer Following this procedure we see that the expression 2 { 5 [ 15 – 4 ( 3 – c ) ] – 6 } is equivalent to Program #1, where… c( 3 – c )4 ( 3 – c )15 – 4 ( 3 – c )5 [ 15 – 4 ( 3 – c ) ]5 [ 15 – 4 ( 3 – c ) ] – 6 2 { 5 [ 15 – 4 ( 3 – c ) ] – 6 } next
© 2007 Herbert I. Gross next Program #2 (1) Start with c (2) Multiply by 40 (3) Add 18 (4) Write the answer On the other hand, the expression 40 c + 18 is equivalent to Program #2, where… c 40+ 18 next
© 2007 Herbert I. Gross If we had not used the rules of the game, how likely is it that we could have guessed that Programs #1 and #2, as different as they look, are equivalent? It is in this sense that the “game” is important. That is, we use our rules to demonstrate that certain things which may not seem obvious are indeed true. Because it contains fewer steps, Program #2 is preferable to Program #1. In other words, the fewer steps we use, the less chance there is that we will make a computational mistake. This is true in other things as well. For example, the fewer “moving parts” a machine has, the easier it is to maintain the machine.
next © 2007 Herbert I. Gross For example, we used Program #2 to prove that if the output of Program #1 was 94 then the input had to be 1.9. To check that our solution was correct, we would replace c by 1.9 in Program #1 and show that we got 94 as the output. Moreover, if we have to evaluate a program for many different values of the input, the shorter program has the potential to save us significant amounts of time. next
© 2007 Herbert I. Gross next Program #1 (1) Start with C (2) Subtract it from 3 (3) Multiply the result by 4 (4) Subtract this result from 15 (5) Multiply the result by 5 (6) Subtract 6 (7) Multiply by 2 (8) Write the answer That is… 1.9 1.1 4.4 10.6 53 47 94
next © 2007 Herbert I. Gross next Program #2 (1) Start with c (2) Multiply by 40 (3) Add 18 (4) Write the answer If now we wanted to find the output in Program #1 when the input was, say, 6.5, we could go through the same tedious process we used above, or we could replace Program #1 by the equivalent Program #2 and see that… 6.5 260 278
next © 2007 Herbert I. Gross The important thing isn't just that Program #2 is simpler than Program #1; but rather that it is equivalent to Program #1. That is, if we replace c by any given value using Program #2, we will find the same answer more simply than we would have found it by using the more cumbersome Program #1. Important Point
next © 2007 Herbert I. Gross Keep in mind that while we simplify expressions by removing the grouping symbols from the inside to the outside; we simplify equations by removing the grouping symbols from the outside to the inside. The point is that when a complicated expression is part of an equation, we may not have to simplify the complicated expression first. For example, another way to solve the equation 2{5[15 + - 4(3 + - c)] + - 6} = 94 without first doing the steps in Exercise 1 is by working on both sides of the equation at the same time. next
© 2007 Herbert I. Gross More specifically, in solving the equation 2{5[15 + - 4(3 + - c)] + - 6} = 94 by the undoing method, we notice that c is inside the parentheses; the parentheses are inside the brackets; and the brackets are inside the braces. So the last step we did to obtain 94 was to multiply by 2. Hence, we divide both sides of the equation by 2 to obtain… 5[15 + - 4(3 + - c)] + - 6 = 472{5[15 + - 4(3 + - c)] + - 6} = 94 next
There is a tendency for beginning students to begin the solution of 2{5[15 + - 4(3 + - c)] + - 6} = 94 by adding 6 to both sides of the equation. However, notice that since - 6 is inside the braces, and the braces are being multiplied by 2, it behaves as 12. More specifically, PEMDAS tells us that we multiply by 2 after we add - 6. © 2007 Herbert I. Gross Caution
To undo subtracting 6, we add 6 to both sides of 5[15 – 4(3 – c)] – 6 = 47 to obtain… next © 2007 Herbert I. Gross To undo multiplying the left side of 5[15 – 4(3 – c)] = 53 by 5, we divide both sides of by 5 to obtain… 5[15 – 4(3 – c)] = 53 15 – 4(3 – c) = 10.6 next
Since it's often easier to deal with undoing addition than with undoing subtraction, we may prefer to replace each subtraction by the add-the-opposite rule and rewrite equation 15 – 4(3 – c) = 10.6 in the form… © 2007 Herbert I. Gross Since the last step we did in the above equation to obtain 10.6 was to add 15, we now subtract 15 from both sides of 15 + - 4(3 + - c) = 10.6 to obtain… 15 + - 4(3 + - c) = 10.6 - 4(3 + - c) = - 4.4 next
Since the last step in obtaining - 4(3 + - c) = - 4.4, was to multiply by - 4, we divide both sides of the equation by - 4 to obtain… © 2007 Herbert I. Gross Next we subtract 3 from both sides of equation 3 + - c = 1.1 to obtain… 3 + - c = 1.1 - c = - 1.9 next Since the opposite of c (that is, - c) is - 1.9, c itself must be equal to 1.9… c = 1.9 next
© 2007 Herbert I. Gross We may solve Question 2 by either of the previous two methods. However, the method we used in Question 1 is the only one we can use if we are asked to simplify an expression rather than to solve an equation. next Important Point
As a final note observe the subtlety involved in the language of subtraction. For example, it makes no difference whether we say “Pick a number and add 7” (in the language of algebra, x + 7) or “Pick a number and add it to 7” (in the language of algebra, 7 + x) because addition obeys the commutative rule. next © 2007 Herbert I. Gross Final Note next
© 2007 Herbert I. Gross However, it makes a big difference whether we say… “Pick a number and subtract 7 (in the language of algebra, x – 7) or “Pick a number and subtract it from 7” (in the language of algebra 7 – x) …because subtraction doesn't obey the commutative property.
next © 2007 Herbert I. Gross This is one reason we prefer to rewrite subtraction in terms of “add the opposite”. That is, while 15 – 7 ≠ 7 – 15; 15 + - 7 = - 7 + 15 In fact, as we’ll discuss in more detail in the Exercise Set, x – 7 and 7 – x are opposites of one another. For example, 15 – 7 = 8 while 7 – 15 = - 8. next |
Comment
Share
Q)
# Using integration, find the area of the region bounded by the parabola $$y^2=4x$$ and the circle $$4x^2+4y^2=9$$
Comment
A)
Toolbox:
• If given two curves represented by y=f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b],the point of intersection of two curves are given by $x=a$ and $x=b$ by taking common values of y from the equation of the two curves.
Step 1:
The curve $x^2=4y$ is the equation of the parabola with vertex at the origin and axis along y-axis and open upwards.
Let $R_1$ be the region lying inside the parabola $x^2=4y\Rightarrow y=\frac{x^2}{4}$
We have $4x^2+4y^2=9$
$x^2+y^2=\large\frac{9}{4}$$\Rightarrow y^2=\large\frac{9}{4}-x^2 y=\sqrt {\large\frac{9}{4}-x^2} Let R_2 be the region lying inside the circle. Hence the area of the required region is bounded by the parabola x^2=4y and the circle x^2+y^2=\large\frac{9}{4}. Step 2: To find the point of intersection let us substitute x^2=4y in the equation of the circle. 4y+y^2=\large\frac{9}{4}$$\Rightarrow 4y^2+16y-9=0$
On factorising we get,
(2y+9)(2y-1)=0
$\Rightarrow x=\large\frac{-9}{2}$
and $y=\large\frac{1}{2}$.
Hence the points of intersection are ($\sqrt 2,\frac{1}{2})(\sqrt 2,\frac{-1}{2}).$
Step 3:
The required area is $A=\int_0^{\Large\frac{1}{\sqrt 2}}(R_2-R_1)dx$
$\Rightarrow\int_0^{\Large\frac{1}{\sqrt 2}}\large\frac{\sqrt {9-4x^2}}{2}-\int_0^\frac{1}{\sqrt 2}\frac{x^2}{4}$$dx. A=\frac{1}{2}\int_0^{\Large\frac{1}{\sqrt 2}}\sqrt {9-4x^2}-\frac{1}{4}\int_0^\sqrt 2x^2dx. On integrating we get, \Rightarrow\large\frac{1}{2}.\frac{1}{2}\begin{bmatrix}\frac{2x}{2}\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}\end{bmatrix}_0^\sqrt 2-\frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^\sqrt 2 On applying the limits we get, \Rightarrow\large\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big). Step 4: But the curves are symmetrical about x-axis the required area is A=2\bigg[\large\frac{\sqrt 2}{12}+\frac{9}{8}$$\sin^{-1}\big(\large\frac{2\sqrt 2}{3}\big)\bigg]$
$\;\;\;=\large\frac{\sqrt 2}{6}+\frac{9}{4}$$\sin^{-1}\big(\frac{2\sqrt 2}{3}\big). Hence the required area is \large\frac{\sqrt 2}{6}+\frac{9}{4}$$\sin^{-1}\big(\large\frac{2\sqrt 2}{3}\big)$. |
# Math is Fun with Triangles
Mathematics, queen of science is a such a subject that has got the number of people hating it more rather than falling in love with it. This subject is the basic foundation for most of the other subjects like mathematical physics, fluid mechanics, probability, electrical engineering and mathematical statistics.
But there are few branches in mathematics which are fun to solve. Again it all depends on the individual. Arithmetic, algebra, geometry, trigonometry are few branches of mathematics.
Triangle:
A triangle is the polygon which has the three edges and the three vertices which is one of the basic shapes in geometry. There are different types of triangles depending upon lengths and internal angles. Equilateral, isosceles and scalene are based on lengths of sides and acute, obtuse and right-angled triangles are based on internal angles.
There are few conditions for any triangle to exist. They are given as:
• Condition on sides: According to this condition sum of the length of any two sides of the triangle must be greater than or equal to the length of the third side. We call this one as triangle-inequality.
• Condition on the angles: According to this condition, each angle of the triangle should be positive and the sum of all the angles must sum up to 180°.
• Trigonometric conditions: There are a few trigonometric formulas that are used to define the existence of any triangle.
Trigonometry:
The trigonometry is defined as one of the branches of the mathematics that is used to study the relationship between lengths and angles of the triangle. The trigonometric functions are used to find the ratios of the sides of the triangle irrespective of the overall size of the triangle.
Using the same trigonometric functions, trigonometric formulas are derived. Following are the important trigonometric functions.
• Sine function(sin) : The ratio of opposite side angle to the hypotenuse.
• Cosine function(cos) : The ratio of the adjacent side angle to the hypotenuse.
• Tangent function(tan) : It is the ratio of opposite side to the adjacent side.
Reciprocals of these functions are also used in trigonometric formulas and they are:
• Cosecant function(cosec): It is the ratio of the hypotenuse to opposite.
• Secant function(sec): The ratio of the hypotenuse to adjacent.
• Cotangent function(cot): It is the ratio of adjacent to opposite.
Trigonometry also finds the applications in scientific fields like probability theory, electrical engineering, and statistics and also in solving Fourier transforms Fourier series.
To know more on the mathematics related concepts, stay tuned with the BYJU’S.
Students can also subscribe to BYJU’S YouTube channel to watch interactive videos of concepts for a better understanding.
https://www.youtube.com/watch?v=jnavB0xgQeU
Please follow and like us:
Subscribe
Notify of |
## How to Solve Quadratic Equations
There are three ways to learn how to solve quadratic equations. Choose your method or scroll down to view all three:
You may also be interested in our quadratic equations solver. You can find it for free at our quadratic formula page, or a larger version here.
An example of a quadratic equation and graph: Once your read our explanation, you will have no problem solving the equation at the right. Let's get started:
A quadratic equation has a term with x2 in it. See the examples below.
Quadratic Equation: x2 + 8x + 15 = 0 Not Quadratic: 5x = 15
The Standard Form of a Quadratic Equation
A quadratic equation is said to be in standard form whenever the exponents go in descending order. In other words, they fit the pattern...
ax2 + bx + c = 0
Now that we have that out of the way, let's solve.
How to Solve Quadratic Equations by Factoring
Once you have factored the polynomial, you will use the zero product property - basically you will set each factor = 0.
To solve by factoring, you will first need to set the quadratic equal to 0. Then, you will factor the equation. For more information, click the links above. After the equation is factored, set each individual factor equal to 0. Then solve the simple equations. Your final answers are x = 6, -7.
Make sure to get your quadratic into standard form (see above), and then plug the values into the quadratic formula.
Details of how to use the quadratic formula.
Solving by Completing the Square
Completing the square is the process of creating a perfect square trinomial.
To do this, you will add an equal amount to both sides of the equation. The amount that you add is determined by the b value. First, convert the equation to the following form: ax2 + bx = c. Next, you will add (b/2)2 to both sides of the equation. Finally, you will factore, take the square root of both sides, and then simplify the equation.
Take a look at the example and step by step instructions below of how to solve quadratic equations by completing the square.
Example: x2 + 6x = -2 Add 9: Factor: Square root: Subtract 3: Final Answer: Find the amount to add
Congratulations, you just learned three different methods of how to solve quadratic equations. |
# Year 9 Maths Chapter 6 - Indices and surds
## Index notation
``````3 x 3 x 3 x 3 x 3 x 3 x 3 = 3⁷
``````
## Scientific notation
science!
``````a x 10ᵐ
where m is an integer
and 1 <= a < 10
``````
## Prime factorisation
Writing a number as a product of its prime factors.
``````108 = 2 x 2 x 3 x 3 x 3
= 2² x 3³
``````
On Casio calculators, you can obtain the prime factorisation of any inputted number by pressing `SHIFT + DMS button`.
On Sharp calculators, well, I don’t know.
## Index laws
### Multiplication and division
``````aᵐ x bⁿ = aᵐ⁺ⁿ
aᵐ ÷ bⁿ = aᵐ⁻ⁿ
``````
### Powers and brackets
``````(aᵐ)ⁿ = aᵐⁿ
(ab)ᵐ = aᵐbᵐ
``````
### Powers and fractions
See Fractional indices for fractional indices.
``````(a/b)ᵐ = aᵐ/bᵐ
``````
### Zero index
``````a⁰ = 1
where a != 0
``````
### Negative indices
``````a⁻ᵐ = 1/aᵐ
``````
### Fractional indices
``````a¹/ᵐ = ᵐ√a
aˣ/ᵐ = ᵐ√aˣ
``````
## Surds
Surds are irrational numbers written with a radical sign. They cannot be expressed as a fraction.
For example, `5√6` is a surd. It means `5 x √6`.
`````` _
5√6
``````
## Surds
You can only add or subtract surds if they are like surds. For example, 5√6, 3√6 and √6 are like surds. If you were to add or subtract these together:
``````5√6 + 3√6 + √6 = 9√6
5√6 - 3√6 - √6 = 5√6
``````
### Multiplication and division
``````a√b x c√d = ab√cd
√a / a \
–– = √| ––– |
√b \ b /
``````
What a beautiful attempt at making a big bracket. I’m not bothered to go to the bracket making website right now so you’re going to have to deal with it.
Copyright © 2017-2020 aidswidjaja and other contributors. CC BY-SA 4.0 Australia unless otherwise stated. |
# Find the domain of a function
I am trying to find the domain of the following equation and I am not sure that I am right or not.
$$f(x) = \sqrt{\dfrac{(2x-4)(5+x)}{(x+3)}}$$
I know that the result in the square root must be positive or 0... and that $x=2$ and $x=-5$ for the numerator.
The square root will fail if I use $x = -3$ .. does it mean that the domain must be $\ge -3$?
Thanks.
You have to deal with two separate failure modes. First, you cannot have a $0$ denominator, so $-3$ cannot be in the domain of the function. Secondly, $$\frac{(2x-4)(5+x)}{x+3}$$ must be non-negative, so you must solve the inequality $$\frac{(2x-4)(5+x)}{x+3}\ge 0\;.$$
The fraction is obviously positive when all three of the expressions $2x-4$, $5+x$, and $x+3$ are positive, but it’s also positive when exactly one of them is positive; in all other cases it is negative or zero. Now $2x-4>0$ when $x>2$, $5+x>0$ when $x>-5$, and $x+3>0$ when $x>-3$, so the relevant break points and intervals are as shown here:
When $x<-5$, none of the expressions is positive, so the fraction is negative; when $-5<x<-3$, only $5+x$ is positive, so the fraction is positive; when $-3<x<2$, $5+x$ and $x+3$ are positive, so the fraction is negative; and when $x>2$, all three are positive, so the fraction is positive. Thus, the fraction is positive on $$(-5,-3)\cup (2,\infty)\;.$$ It is zero at $x=2$ and $x=-5$; neither of these makes the denominator zero, so they also belong in the domain, which is therefore $$[-5,-3)\cup[2,\infty)\;.$$
Notice that $2x-4= 2(x-2)\ge 0$ precisely if $x\ge2$, and in a similar way you need to figure out for which values of $x$ the other factors, $5+x$ and $x+3$, are positive, and for which values of $x$ they are negative. Then you have to think about multiplying and dividing positive and negative numbers, to figure out for which values of $x$ the whole fraction is positive, and for which values of $x$ it is negative. And finally, you need to exclude $-3$ from the domain. |
Ncert class 10 polynomial solutions
# Ncert class 10 polynomial solutions
Are you searching for class 10 chapter – 2 polynomial solutions? you’re at the right place. You can get accurate CBSE NCERT Solutions for all subjects. These solutions will help you a lot in scoring good marks in the exams.
Page No: 28
# Polynomial solutions – Exercises 2.1
1. The graphs of y = p(x) are given in the following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
Page No: 33
## Polynomial solutions – Exercise 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 x – 4
(i) x2 – 2x – 8
= (x – 4) (x + 2)
The value of x2 – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x2
(ii) 4s2 – 4s + 1
= (2s-1)2
The value of 4s2 – 4s + 1 is zero when 2s – 1 = 0, i.e., s = 1/2
Therefore, the zeroes of 4s2 – 4s + 1 are 1/2 and 1/2.
Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2.
(iii) 6x2 – 3 – 7x
= 6x2 – 7x – 3
= (3x + 1) (2x – 3)
The value of 6x2 – 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e., x = -1/3 or x = 3/2
Therefore, the zeroes of 6x2 – 3 – 7x are -1/3 and 3/2.
Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.
(iv) 4u2 + 8u
4u2 + 8u + 0
= 4u(u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = – 2
Therefore, the zeroes of 4u2 + 8u are 0 and – 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.
(v) t2 – 15
t– 0.t – 15
= (t – √15) (t + √15)
The value of t2 – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or = -√15
Therefore, the zeroes of t2 – 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.
(vi) 3x2 – x – 4
= (3x – 4) (x + 1)
The value of 3x2 – x – 4 is zero when 3x – 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.
Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , -1
(ii) √2 , 1/
(iii) 0, √5
(iv) 1,1
(v) -1/4 ,1/
(vi) 4,1
(i) 1/4 , -1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1/4 = –b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2x -4.
(ii) √2 , 1/3
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = √2 = 3√2/3 = –b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.
(iii) 0, √5
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = –b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.
(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 – x +1.
(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = -1/4 = –b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.
(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 – 4x +1.
Page No: 36
## Polynomial solutions – Exercise 2.3
1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Quotient = x-3 and remainder 7x – 9
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Quotient = x2– 3 and remainder 8
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Quotient = –x2 -2 and remainder -5x +10
2. Check whether the first polynomial is a factor of the second polynomial by dividing the
second polynomial by the first polynomial:
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
t2 – 3 exactly divides 2t4 + 3t3 – 2t2 – 9t – 12 leaving no remainder. Hence, it is a factor of 2t4 + 3t3 – 2t2 – 9t – 12.
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
x2 + 3x + 1 exactly divides 3x4 + 5x3 – 7x2 + 2x + 2 leaving no remainder. Hence, it is factor of 3x4 + 5x3 – 7x2 + 2x + 2.
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
x3 – 3x + 1 didn’t divides exactly x5 – 4x3 + x2 + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x5 – 4x3 + x2 + 3x + 1.
3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3)
and – √(5/3).
p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3).
As it has the term (+ 1)2 , therefore, there will be 2 zeroes at x = – 1.
Hence, the zeroes of the given polynomial are √(5/3) and – √(5/3), – 1 and – 1.
4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and
-2x + 4, respectively. Find g(x).
Here in the given question,
Dividend = x3 – 3x2 + x + 2
Quotient = x – 2
Remainder = -2x + 4
Divisor = g(x)
We know that,
Dividend = Quotient × Divisor + Remainder
⇒ x3 – 3x2 + x + 2 = (x – 2) × g(x) + (-2x + 4)⇒ x3 – 3x2 + x + 2 – (-2x + 4) = (x – 2) × g(x)
⇒ x3 – 3x2 + 3x – 2 = (x – 2) × g(x)
g(x) = (x3 – 3x2 + 3x – 2)/(x – 2)
∴ g(x) = (x2 – x + 1)
5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
(i) Let us assume the division of 6x2 + 2x + 2 by 2
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x3+ x by x2,
Here, p(x) = x3 + x
g(x) = x2
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + x = (x2 ) × x + x
x3 + x = x3 + x
Thus, the division algorithm is satisfied.
(iii) Let us assume the division of x3+ 1 by x2.
Here, p(x) = x3 + 1
g(x) = x2
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + 1 = (x2 ) × x + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.
Page No: 36
Read also – Real number solution
## Polynomial solutions – Exercise 2.4 (Optional)
1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5+ 2; 1/2, 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
(i) p(x) = 2x3 + x2 – 5+ 2
Now for zeroes, putting the given value in x.
p(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2)+ 2
= (2×1/8) + 1/4 – 5/2 + 2
= 1/4 + 1/4 – 5/2 + 2
= 1/2 – 5/2 + 2 = 0
p(1) = 2(1)3 + (1)2 – 5(1)+ 2
= (2×1) + 1 – 5 + 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)3 + (-2)2 – 5(-2)+ 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0
Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3 + bx2 + c+ d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 – 2
⇒ 1/2 = 1/2
c/a = αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 – 2 – 1
⇒ -5/2 = -5/2
-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1
Thus, the relationship between zeroes and the coefficients are verified.
(ii) p(x) = x3 – 4x2 + 5x – 2
Now for zeroes, putting the given value in x.
p(2) = 23 – 4(2)2 + 5(2)- 2
= 8 – 16 + 10 – 2
= 0
p(1) = 13 – 4(1)2 + 5(1)- 2
= 1 – 4 + 5 – 2
= 0
p(1) = 13 – 4(1)2 + 5(1)- 2
= 1 – 4 + 5 – 2
= 0
Thus, 2, 1 and 1 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3 + bx2 + c+ d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4
c/a = αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5
-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2
Thus, the relationship between zeroes and the coefficients are verified.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.
Let the polynomial be ax3 + bx+ cx + d and the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a
∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x3 – 2x2 7x + 14
3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a–b, a, a+b, find a and b.
Since, (a – b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b) = -(-3)/1 = 3
⇒ 3a = 3 ⇒ a =1
∴ Sum of the products of is zeroes taken two at a time = a(a – b) + a(a + b) + (a + b) (a – b) =1/1 = 1
a2 – ab + a2 + ab + a2 – b= 1
⇒ 3a2 – b2 =1
Putting the value of a,
⇒ 3(1)2 – b2 = 1
⇒ 3 – b2 = 1
⇒ b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2
4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2±√3, find other zeroes.
2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 – 4x + 4 = 3,
⇒ x2 – 4x + 1= 0
Now, dividing p(x) by x2 – 4x + 1
∴ p(x) = x4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) (x2 – 7x + 5x – 35)
= (x2 – 4x + 1) [x(x – 7) + 5 (x – 7)]
= (x2 – 4x + 1) (x + 5) (x – 7)
∴ (x + 5) and (x – 7) are other factors of p(x).
∴ – 5 and 7 are other zeroes of the given polynomial.
5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
On dividing x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k
∴ Remainder = (2k – 9)x – (8 – k)k + 10
But the remainder is given as x+ a.
On comparing their coefficients,
2k – 9 = 1
⇒ k = 10
⇒ k = 5 and,
-(8-k)k + 10 = a
⇒ a = -(8 – 5)5 + 10 =- 15 + 10 = -5
Hence, k = 5 and a = -5 |
# What Is The Place Value Of 8 In 1278?
## What is the value of 2 in 23?
Let’s look at an example.
What does the 2 in the number 23 represent.
Student: It represents twenty..
## What is the place value of 8?
Learn with the Complete K-5 Math Learning Program 3 is in thousands place and its place value is 3,000, 5 is in hundreds place and its place value is 500, 4 is in tens place and its place value is 40, 8 is in ones place and its place value is 8.
## What is the place value of 6 in 64?
Each digit has a value depending on its place called the place value of the digit. Place value of a digit = (face value of the digit) × (value of the place). Hence, the place value of 6 in 64 = 6 x 10 = 60.
## What is the place value of 7 in 57930?
70007000 is the place value of 7 in 57930……
## What is the place value of 9 in 900?
This 9 is the third from right in the number. There are two digits towards right in the number. We write 9 and put two zeroes to its right. So, the place value of 9 in the number is 900.
## What is the value of 3 in 937?
30 in 937The value of 3 is 30 in 937. Since 937 is a three digit number, the value of 3 forms the ten’s place. Since 3 takes the 10th place and as per the place value of the numbers, the value of 3 will be 3*10=30.
## What is the value of the digit 5 in the number 75?
5Answer Expert Verified Answer is 5 . Explanation; To find out the value of 5 in 75, you should know the place values of digits in numbers.
## What is the difference between the place value of 3 in the number 353?
Answer: The difference between the place values of 3 in the number 353 is 297.
## What is the value of 8 in 80?
The original value of a digit is called its face value. the place value of 8 is 80 and its face value is 8, the place value of 5 is 5 and its face value is 5.
## What is the period in place value?
A place value chart names each place value. When a number is written in standard form, each group of digits separated by a comma is called a period . The number 5,913,603,800 has four periods. Each period is shown by a different color in the place value chart.
## What is the place value of 2 in 123 456?
tenthouandAnswer. the place value of 2 is tenthouand.
## Which number in 39768 has the highest place value?
3The digit 3 has highest place value in the number 39768 Face value is the actual value of the digit.
## What is the value of 3?
632,814 – The value of the digit 3 is 3 ten-thousands, or 30,000. 632,814 – The value of the digit 2 is 2 thousands, or 2,000. 632,814 – The value of the digit 8 is 8 hundreds, or 800. 632,814 – The value of the digit 1 is 1 tens, or 10.
## What is the place value of 4 in 64?
The number 64 has 4 in the ones place and 6 in the tens place.
## What is the place value of 9?
the place value of 9 is 9 × 1 = 9 as 9 is at one’s or unit’s place. the place value of 2 is 2 × 10 = 20 as 2 is at ten’s place. the place value of 1 is 1 × 100 = 100 as 1 is at hundred’s place. the place value of 4 is 4 × 1000 = 4000 as 4 is at thousand’s place.
## What is the value of 8 in 387?
Since the 8 is in the tens place, it represents 8 tens or 80.
## What is the place value of 2 in 93207?
200Place value is the value of the place where the number is present. Say in 93207 the place value of 2 is 200 as number 2 is present in hundreds place.
## What is the place value of 0 in 109?
tensThe place value of 0 in 109 is tens.
## What is the difference between place value and value?
Place value is defined as the digit multiplied wherever it is placed, either by hundreds or thousands. Face value is simply defined as the digit itself within a number. Place value of a digit should be multiplied by the digit value of the position where it is located.
## What is the place value of 9 in the number 291 146?
ten thousandsThe place value of 9 in the number 291 146 is ten thousands place. |
How to solve cos(30+60)
Welcome to my article How to solve cos(30+60). This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve cos(30+60) read and understand it carefully till the end.
Let us know how to solve this question How to solve cos(30+60)
First write the question on the page of the notebook.
How to solve cos(30+60)
This question is based on trigonometry,
in this way, to solve this question,
we will also take the help of trigonometry formula
Let us first write this question in this way,
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})
FORMULA
\displaystyle \cos (A+B)= \displaystyle \cos A\cos B-\sin A\sin B
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=\cos {{30}^{\circ }}\cos {{60}^{\circ }}-\sin {{30}^{\circ }}\sin {{60}^{\circ }}
We know that ,
putting the value,
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=\frac{{\sqrt{3}}}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{{\sqrt{3}}}{2}
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=\frac{{\sqrt{3}\times 1}}{{2\times 2}}-\frac{{1\times \sqrt{3}}}{{2\times 2}}
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=\frac{{\sqrt{3}}}{4}-\frac{{\sqrt{3}}}{4}
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=\cancel{{\frac{{\sqrt{3}}}{4}}}-\cancel{{\frac{{\sqrt{3}}}{4}}}
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=0
or,
\displaystyle \cos ({{30}^{\circ }}+{{60}^{\circ }})=\cos ({{90}^{\circ }})=0 [Answer]
This article How to solve cos(30+60) has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need. |
3 Tutor System
Starting just at 265/hour
Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm.
we have, $$\triangle{ABC}$$,
We have, AB = 3cm, BC = 4cm, CA = 5cm
We have, $${AC}^2 = {AB}^2 + {BC}^2$$
(By Pythagoras theorem)
= $${3}^2 + {4}^2 = 9 + 16 = 25$$
$$\Rightarrow$$ $${AC}^2 = {5}^2$$
$$\Rightarrow$$ AC = 5cm
Hence, $$\triangle{ABC}$$ is a right triangle ...(i)
Now in $$\triangle{ABD}$$, we have,
AC = 5cm, CD = 4cm, DA = 5cm
We know that,
$$s = \frac{a + b + c}{2}$$
$$\therefore$$ $$s = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7cm$$
Now, Area of triangle
= $$\sqrt{7(7 - 5)(7 - 4)(7 - 5)}$$
(Since, Heron's formula [area = $$\sqrt{s(s - a)(s - b)(s - c)}$$])
= $$\sqrt{7 × 2 × 3 × 2}$$
= $$2\sqrt{7 × 3} {cm}^2$$
= $$2\sqrt{21} {cm}^2$$ = 2 × 4.6$${cm}^2$$ = 9.2$${cm}^2$$(approx) ...(ii)
Since, $$\triangle{ABC}$$ is an right angled triangle, ...(from(i))
Area of $$\triangle{ABC}$$ = $$\frac{1}{2}$$ × AB × BC
(Since, Area of triangle = $$\frac{1}{2}$$ × Base × Height)
= $$\frac{1}{2} × 3 × 4 = 6{cm}^2$$ ...(iii)
Area of quadrilateral ABCD = Area of $$\triangle{ABC}$$ + Area of $$\triangle{ACD}$$
Hence from (ii) and (iii),
Area of quadrilateral ABCD = $$9.2{cm}^2 + 6{m}^2 = 15.2{cm}^2$$ |
# The density of air 20 km above Earths surface is 92 g/m^3. The pressure of the atmosphere is 42 mm Hg, and the temperature is -63 °C. What is the average molar mass of the atmosphere at this altitude?
## If the atmosphere at this altitude consists of only ${O}_{2}$ and ${N}_{2}$, what is the mole fraction of each gas?
Dec 8, 2015
Here's what I got.
#### Explanation:
The idea here is that you need to start from the standard form of the ideal gas law equation
$\textcolor{b l u e}{P V = n R T} \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$
and try to manipulate it by incorporating density and molar mass.
As you know, molar mass is defined as mass per unit of mole. This means that you an write the number of moles of substance by using its molar mass and its mass
$\textcolor{b l u e}{{M}_{M} = \frac{m}{n} \implies n = \frac{m}{M} _ M}$
Plug this into equation color(purple)((1)) to get
$P V = \frac{m}{M} _ M \cdot R T$
Rearrange this equation to solve for ${M}_{M}$, which for this problem will represent the average molar mass of air.
${M}_{M} = \frac{m}{V} \cdot \frac{R T}{P} \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$
Now focus on density, which is defined as mass per unit of volume.
$\textcolor{b l u e}{\rho = \frac{m}{V}}$
Plug this into equation $\textcolor{p u r p \le}{\left(2\right)}$ to get
${M}_{M} = \rho \cdot \frac{R T}{P}$
Plug in your values and solve for ${M}_{M}$ - keep in mind that you will need to convert the density of air from grams per cubic meter to grams per liter, the pressure from mmHg to atm, and the temperature from degrees Celsius to Kelvin.
$92 \text{g"/color(red)(cancel(color(black)("m"^3))) * (1color(red)(cancel(color(black)("m"^3))))/(10^3"L") = "0.092 g/L}$
This will get you
M_M = 0.092"g"/color(red)(cancel(color(black)("L"))) * (0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + (-63))color(red)(cancel(color(black)("K"))))/(42/760color(red)(cancel(color(black)("atm"))))
${M}_{M} = \text{28.723 g/mol}$
Now, the average molar mass of air is calculated by taking the weighted average of the molar masses of its constituent gasses.
More specifically, each gas will contribute to the average molar mass proportionally to the mole fraction it has in the mixture.
In your case, air is said to contain oxygen gas, ${\text{O}}_{2}$, and nitrogen gas, ${\text{N}}_{2}$. This means that you can write
${\chi}_{{O}_{2}} + {\chi}_{{N}_{2}} = 1$
The mixture contains two gases, so their mole fractions must add up to give $1$.
${\chi}_{{O}_{2}} \times {M}_{{O}_{2}} + {\chi}_{{N}_{2}} \times {M}_{{N}_{2}} = \text{28.72 g/mol}$
The molar masses of oxygen gas and nitrogen gas are
${M}_{{O}_{2}} = \text{31.999 g/mol" " }$ and $\text{ "M_(N_2) = "28.0134 g/mol}$
Use these two equations to get
${\chi}_{{O}_{2}} = 1 - {\chi}_{{N}_{2}}$
$\left(1 - {\chi}_{{N}_{2}}\right) \cdot 31.999 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g/mol"))) + chi_(N_2) * 28.0134 color(red)(cancel(color(black)("g/mol"))) = 28.723 color(red)(cancel(color(black)("g/mol}}}}$
$31.999 - 3.9856 \cdot {\chi}_{{N}_{2}} = 28.723$
${\chi}_{{N}_{2}} = \frac{3.276}{3.9856} = 0.822$
This means that you have
${\chi}_{{O}_{2}} = 1 - 0.822 = 0.178$
Now, you need to round these answers to two sig figs, the number of sig figs you have for your given values
M_"M air" = color(green)("29 g/mol")
${\chi}_{{O}_{2}} = \textcolor{g r e e n}{0.18}$
${\chi}_{{N}_{2}} = \textcolor{g r e e n}{0.82}$ |
# Solve the equation- 1- m - 7 = 53
1/3m-7=5
One solution was found :
m = 36
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
1/3*m-7-(5)=0
Step by step solution :
Step 1 :
1
Simplify —
3
Equation at the end of step 1 :
1
((— • m) - 7) - 5 = 0
3
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 3 as the denominator :
7 7 • 3
7 = — = —————
1 3
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
2.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
m - (7 • 3) m - 21
——————————— = ——————
3 3
Equation at the end of step 2 :
(m - 21)
———————— - 5 = 0
3
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using 3 as the denominator :
5 5 • 3
5 = — = —————
1 3
Adding fractions that have a common denominator :
3.2 Adding up the two equivalent fractions
(m-21) - (5 • 3) m - 36
———————————————— = ——————
3 3
Equation at the end of step 3 :
m - 36
—————— = 0
3
Step 4 :
When a fraction equals zero :
4.1 When a fraction equals zero ...
Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
m-36
———— • 3 = 0 • 3
3
Now, on the left hand side, the 3 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
m-36 = 0
Solving a Single Variable Equation :
4.2 Solve : m-36 = 0
Add 36 to both sides of the equation :
m = 36
One solution was found :
m = 36
I happen this help
## Related Questions
A person on a moving sidewalk travels 6 feet in 3 seconds. The moving sidewalk has a length of 60 feet. How long will it take to move from one end of the sidewalk to the other?
Speed = distance / time
= 6/3 =2 ft/sec
Time = distance / speed
= 60/2
= 30 seconds
The measure of an inscribed circle of an angle is the measure of the intercepted arc
(the statement does not appear to be true)
Step-by-step explanation:
I don't think the statement is true, but you CAN compute the intercepted arc from the angle.
Note that BFDG is a convex quadrilateral, so its angles sum to 360. Since we know the inscribed circle touches the angle tangentially, angles BFD and BGD are both right angles, with a measure of 90 degrees.
Therefore, adding the angles together, we have:
alpha + 90 + 90 + <FDG = 360
Therefore, <FDG, the inscribed angle, is 180-alpha (ie, supplementary to alpha)
Multiply the polynomials 3(x+7) (show work pls)
3x + 21
Step-by-step explanation:
(3)(x+7)
Now, we distribute the 3 in each term of (x+7)
So, 3*x = 3x and 3*7 = 21.
So our resulting term would be 3x+21.
Tom is the deli manager at a grocery store.He needs to schedule employees to staff the deli department at least 260 person-hours per week.Tom has one part-time employee who works 20 hours per week.Each full-time employee works 40 hours per week.Write an inequality to determine f, the number of full time employees Tom must schedule, so that his employees will work at least 260 person-hours per week. Solve and graph. !PLEASE HELP!
I hope this is right but:
Let n=number of full time employees
260<=20+n*40
260-20<=N*40
240<=n*40
n>=240/40
n>=6 full time employees
which means he would have to hire at least 6 or more fully time employees
Though I'm not sure if the attached file is the graph you are looking for
10/2=5 -9/2=-4.5. To solve the problem take the numerator divided by the denominator. so 10 divided by 2 would be 5.
50 pointsssssss
HELPPPPP NOWWW
The third or fifth answer but mostly third the glaboal depression known as great depression was mostly known for people getting mad because their work checks decreased this caused a lot of people to not work and quit their jobs now that you think of it mabye you can choose a or c please choose both or one these are most recommended........... so i did some research and the answer is #1
the second one is becasue great depression had nothing to do with europe
the third was wrong becasue when the market crash happended peploe did refuse but that not how it affected them
the fourth one during the great depriosn no one was even able to trade except the goverment and its ovidious
the fifth their was no reason for their money to be declined |
Place Value of Whole Numbers
So, what is place value of a whole number?
In a whole number value of digit depends on its place (position) in the number.
The rightmost digit is ones, next to the left tens, next hundreds, then thousands, ten thousands, hundred thousands, millions, ten millions; etc.
Example 1. Determine place value of each digit in 3156.
\underbrace 3_(text(thousands))\quad\underbrace 1_(text(hundreds))\quad\underbrace 5_(text(tens))\quad\underbrace 6_(text(ones)).
Next example involves more digits.
Example 2. Determine place value of each digit in 58365738746.
\underbrace 5_(text(ten billions))\quad\underbrace 8_(text(billions))\quad\underbrace 3_(text(hundred millions))\quad\underbrace 6_(text(ten millions))\quad\underbrace 5_(text(millions))\quad\underbrace 7_(text(hundred thousands))\quad\underbrace 3_(text(ten thousands))\quad\underbrace 8_(text(thousands))\quad\underbrace 7_(text(hundreds))\quad\underbrace 4_(text(tens))\quad\underbrace 6_(text(ones)).
In last example we saw pretty long number 58365738746.
It is better to write them in standard notation, seprating digits into groups of three starting from right, like 58,365,738,746.
Example 3. Write 46736 in standard notation.
Answer: 46,736.
Finally note, that, for example, in number 524 there is no thousands.
But we can easily add it! 524 is same as 0524. This means that thousand is 0.
We can add as many leading zeros (and ignore them) as we want to the whole number.
For example, 524 is same as 0524 or 0000524.
Warning! But we can ignore only leading zeros: 504 IS NOT the same as 54, although 00504 is same as 504.
Now, it is your turn. Take a pen and paper and solve following problems:
Exercise 1. Determine place value of each digit in 56892.
\underbrace 5_(text(ten thousands))\quad\underbrace 6_(text(thousands))\quad\underbrace 8_(text(hundreds))\quad\underbrace 9_(text(tens))\quad\underbrace 2_(text(ones)).
Now, slightly harder exercise.
Exercise 2. Determine place value of each digit in 87587874.
\underbrace 8_(text(ten millions))\quad\underbrace 7_(text(millions))\quad\underbrace 5_(text(hundred thousands))\quad\underbrace 8_(text(ten thousands))\quad\underbrace 7_(text(thousands))\quad\underbrace 8_(text(hundreds))\quad\underbrace 7_(text(tens))\quad\underbrace 4_(text(ones)).
Finally, exercise on standard notation.
Exercise 3. Write 3446736 in standard notation.
Answer: 3,446,736. |
# Simulate many repetitions trials 5 state your
• Notes
• apiccirello
• 14
This preview shows pages 3–5. Sign up to view the full content.
Simulate many repetitions (trials) 5. State your conclusions Example 1: Suppose you left your statistics textbook and calculator in you locker, and you need to simulate a random phenomenon (drawing a heart from a 52-card deck) that has a 25% chance of a desired outcome. You discover two nickels in your pocket that are left over from your lunch money. Describe how you could use the two coins to set up you simulation. Example 2: Suppose that 84% of a university’s students favor abolishing evening exams. You ask 10 students chosen at random. What is the likelihood that all 10 favor abolishing evening exams? Describe how you could use the random digit table to simulate the 10 randomly selected students. Example 3: Use your calculator to repeat example 2 Homework: pg 397 6-1, 4, 5, 8, 15
This preview has intentionally blurred sections. Sign up to view the full version.
Chapter 6: Probability and Simulation: The Study of Randomness Section 6.2: Probability Models Knowledge Objectives: Students will: Explain what is meant by random phenomenon . Explain what it means to say that the idea of probability is empirical . Define probability in terms of relative frequency . Define sample space . Define event . Explain what is meant by a probability model . List the four rules that must be true for any assignment of probabilities. Explain what is meant by equally likely outcomes . Define what it means for two events to be independent . Give the multiplication rule for independent events . Construction Objectives: Students will be able to: Explain how the behavior of a chance event differs in the short- and long-run. Construct a tree diagram . Use the multiplication principle to determine the number of outcomes in a sample space. Explain what is meant by sampling with replacement and sampling without replacement . Explain what is meant by { A B } and { A B }. Explain what is meant by each of the regions in a Venn diagram . Give an example of two events A and B where A B = . Use a Venn diagram to illustrate the intersection of two events A and B . Compute the probability of an event given the probabilities of the outcomes that make up the event. Compute the probability of an event in the special case of equally likely outcomes . Given two events, determine if they are independent. Vocabulary: Empirical – based on observations rather than theorizing Random – individuals outcomes are uncertain Probability – long-term relative frequency Tree Diagram – allows proper enumeration of all outcomes in a sample space Sampling with replacement – samples from a solution set and puts the selected item back in before the next draw Sampling without replacement – samples from a solution set and does not put the selected item back Union – the set of all outcomes in both subsets combined (symbol: ) Empty event – an event with no outcomes in it (symbol: ) Intersect – the set of all in only both subsets (symbol: ) Venn diagram – a rectangle with solution sets displayed within Independent – knowing that one thing event has occurred does not change the probability that the other occurs
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern |
# Math: algebra
## 6+5
Prepare in advance 22 green cubes. Make 2 groups of green cubes. One group should consist of 6 green cubes. Another group should consist of 5 green cubes. The cubes in the groups should be scattered in a random order, not put in a straight line or row. The two groups should be located away from each other to make them distinctive groups but not too far from each other so that the baby can see two groups in one location. Another 11 cubes should be behind you so that the baby does not see them. Smile and say: Look Baby! I see two distinctive groups of cubes! Point to the first and then the second group. Say I wonder how many cubes we have all together? Place your hand over the two groups, make a round in the air above them showing togetherness. Let’s make predictions! I think we have 10! Show ten fingers. Now we can check it out. Place your hand over the first group. Say: This is called summand. Let’s count the cubes in the first group. 1, 2, 3, 4, 5, 6. Say: There are six cubes in this group (show six fingers). Place your hand over the second group. Say; This is called summand too. Now let’s count the cubes in the second group: 1, 2, 3, 4, 5. Say: There are five cubes in this group (show 5 fingers). Say: let’s sum them up (put your 2 hands together as if you were clapping). We will make an equation by putting two signs: “+” and “=”. See how! Make an equation by putting the “+” sign in between the groups. The “+” sign can be made of sticks or cut vegetables. Put the “=” sign after the second group. Say: we are going to count all the cubes before the sign “=”. Start counting by pointing first to the first group, then switch to the second (from left to right): 1,2,3,4,5,6,7,8,9,10,11. There are 11 (show eleven fingers, first 10 and then 1) cubes all together! Take out another 11 cubes you were hiding and put them randomly after the “=” sign. Count them: 1,2,3,4,5,6,7,8,9,10,11! (Show 11 fingers) Point to the group after the sign “=”. Say: This is call the sum!
Say: 6 (show 6 fingers) + 5 (show 5 fingers) equals 11 (show 11 fingers)!
If you've found a typo, mistake, or incorrect information, please let us know! |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Order of Composite Transformations
## Understand the order in which to apply transformations
%
Progress
Progress
%
Order of Composite Transformations
Quadrilateral has coordinates and . Draw the quadrilateral on the Cartesian plane.
The quadrilateral undergoes a dilation centered at the origin of scale factor and then is translated 4 units to the right and 5 units down. Show the resulting image.
### Watch This
First watch this video to learn about the order of composite transformations.
CK-12 Foundation Chapter10OrderofCompositeTransformationsA
Then watch this video to see some examples.
CK-12 Foundation Chapter10OrderofCompositeTransformationsB
### Guidance
In geometry, a transformation is an operation that moves, flips, or changes a shape to create a new shape. A composite transformation is when two or more transformations are performed on a figure (called the preimage) to produce a new figure (called the image).
Imagine if you rotate, then dilate, and then translate a rectangle of vertices , and . You would end up with a diagram similar to that found below:
If you take the same preimage and rotate, translate it, and finally dilate it, you could end up with the following diagram:
Therefore, the order is important when performing a composite transformation. Remember that the composite transformation involves a series of one or more transformations in which each transformation after the first is performed on the image that was transformed. Only the first transformation will be performed on the initial preimage.
#### Example A
Line drawn from (-4, 2) to (3, 2) has undergone a reflection across the -axis. It then undergoes a translation up one unit and over 3 units to the right to produce . Draw a diagram to represent this composite transformation and indicate the vertices for each transformation.
Solution:
#### Example B
For the composite transformation in Example A, suppose the preimage undergoes a translation up one unit and over 3 units to the right and then undergoes a reflection across the -axis. Does the order matter?
Solution:
For this example is not the same as from the previous example (example A). Therefore order does matter.
#### Example C
Triangle is rotated CCW about the origin. The resulting figure is then translated over 3 to the right and down 7. Does order matter?
Order: Rotation then Translation
Order: Translation then Rotation
Solution: The blue triangle represents the final image after the composite transformation. In this example, order does matter as the blue triangles do not end up in the same locations.
#### Concept Problem Revisited
Quadrilateral has coordinates and . Draw the quadrilateral on the Cartesian plane.
The quadrilateral undergoes a dilation centered at the origin of scale factor and then is translated 4 units to the right and 5 units down. Show the resulting image.
### Vocabulary
Image
In a transformation, the final figure is called the image.
Preimage
In a transformation, the original figure is called the preimage.
Transformation
A transformation is an operation that is performed on a shape that moves or changes it in some way. There are four types of transformations: translations, reflections, dilations and rotations.
Dilation
A dilation is a transformation that enlarges or reduces the size of a figure.
Translation
A translation is an example of a transformation that moves each point of a shape the same distance and in the same direction. Translations are also known as slides.
Rotation
A rotation is a transformation that rotates (turns) an image a certain amount about a certain point.
Reflection
A reflection is an example of a transformation that flips each point of a shape over the same line.
Composite Transformation
A composite transformation is when two or more transformations are combined to form a new image from the preimage.
### Guided Practice
1. Line drawn from (-3, 4) to (-3, 8) has undergone a rotation about the origin at CW and then a reflection in the -axis. Draw a diagram with labeled vertices to represent this composite transformation.
2. Line drawn from (-3, 4) to (-3, 8) has undergone a reflection in the -axis and then a rotation about the origin at CW. Draw a diagram with labeled vertices to represent this composite transformation. Is the graph the same as the diagram in #1?
3. The triangle with vertices and has undergone a transformation of up 4 and over to the right 4 and then a reflection in the -axis. Draw and label the composite transformation. Does order matter?
1.
2.
If you compare the graph above to that found in Question 1, you see that the final transformation image has different coordinates than the image in question 2. Therefore order does matter.
3.
Order: Translation then Reflection
Order: Reflection then Transformation
In this problem, order did matter. The final image after the composite transformation changed when the order changed.
### Practice
1. Reflect the above figure across the x-axis and then rotate it CW about the origin.
2. Translate the above figure 2 units to the left and 2 units up and then reflect it across the line
1. Reflect the above figure across the y-axis and then reflect it across the x-axis.
2. What single transformation would have produced the same result as in #3?
1. Reflect the above figure across the line and then across the line .
2. What single transformation would have produced the same result a in #5?
1. Reflect the above figure across the line y=x and then across the x-axis.
2. What single transformation would have produced the same result a in #7?
1. Translate the above figure 2 units to the right and 3 units down and then reflect it across the y-axis.
2. Rotate the above figure CCW about the origin and then translate it over 1 unit to the right and down 1 unit.
1. Reflect the above figure across the line and then translate it 2 units to the left and 3 units down.
2. Translate the above figure 2 units to the left and 3 units down and then reflect it across the line .
1. Translate the above figure 3 units to the right and 4 units down and then rotate it about the origin CW.
2. Rotate the above figure about the origin CW and then translate it 3 units to the right and 4 units down.
3. How did your result to #13 compare to your result to #14?
### Vocabulary Language: English
Dilation
Dilation
To reduce or enlarge a figure according to a scale factor is a dilation. |
### Plum Tree
Label this plum tree graph to make it totally magic!
### Magic W
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### 2-digit Square
A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number?
# Odd Differences
### Why do this problem :
This connection between the set of odd numbers and squares is not often encountered in school and offers a very accessible example of visualisation.
The 'difference of two squares' is a key algebraic transformation and problems like this can lead students into a deeper appreciation of that form through a further visualisation.
### Possible approach :
Start by giving learners time to make sense of the idea that any odd number can be written as the difference of two squares using the dotty grid. Encourage them to investigate with small numbers (for example, $9=5^2 - 4^2$ ) before generalising for larger numbers such as $1155$ and then devising a general rule for anyone to apply.
Remind them of the identity $a^2 - b^2 = (a + b)(a -b)$ and give them time to consider and see the connection with what they have already done. For example: $$9=5^2 - 4^2 = (5+4)(5-4).$$
At this point another visualisation might be useful in order to suggest how further differences might be found.
The problem of writing the number $105$ as the difference of two squares becomes a problem about factor pairs that make a product of $105$.
Draw out from the students how this transformation helps [we now seek factors of $105$ rather than guessing squares and calculating differences].
Moving to the last part, and even pushing beyond that to a general result, may require the group to spend time making sure that they are secure with an algorithm for producing prime factors.
### Key questions :
• Explain how this image shows us the sum of the first n odd numbers - what is that sum ?
• For $105$ (and then for $1155$) how many ways might there be, and why do you think that ?
### Possible extension :
Is it true that no number can be written as the difference of $2$ squares in exactly three ways, and if so why?
### Possible support :
Perhaps try the problem Plus Minus first.
For students not yet ready for this problem, time spent on finding factors will be valuable. When the group are ready, check that they can use an algorithm to find prime factors and invite them to suggest how they can use that to assist factor-finding. |
# What Percent of 44 is 33?
Calculating the percentage of a number can be a difficult task for many people, but luckily, it’s a simple equation that can be solved with a few steps. Knowing what percent of 44 is 33 is a great way to understand the concept of percentages and their uses in the real world. In this article, we’ll discuss how to calculate what percent of 44 is 33 and why it’s important to understand percentages.
## Steps to Calculate What Percent of 44 is 33
It’s easy to calculate what percent of 44 is 33. The first step is to divide 33 by 44. This will give you the decimal equivalent of the percentage. Once you have the decimal equivalent, you can multiply it by 100 to find out what percent of 44 is 33. The answer is 75%.
## Uses of Percentages
Percentages are used in a variety of fields, from finance to engineering. They are a way of expressing a part of a whole in a more understandable way. For example, if you have a budget of \$100 and you spend \$75, you can express that amount as 75%, which makes it easier to understand. In addition, percentages are often used to compare data or to show the relative size of something. For example, if you have two groups of people and one group is twice as large as the other, you can express this as 200%.
## Importance of Understanding Percentages
It’s important to understand percentages, especially when it comes to things like finance, business, and investments. Knowing how to calculate things like what percent of 44 is 33 is a great way to make sure you are making smart decisions. It’s also important to understand how percentages can be used to compare data, as this can help you make better decisions, too. Understanding percentages can also help you make sure you’re not overspending or taking on too much debt.
##
Knowing what percent of 44 is 33 is a great way to understand the concept of percentages and how they can be used in the real world. Percentages are important to understand for a variety of reasons, such as making sure you’re not overspending or taking on too much debt. Calculating the percentage of a number is also a simple equation that can be solved with a few steps. Once you understand the equation, it’s easy to calculate percentages and make better decisions in the future. |
# Solve for x (3x-11)(2x+9)^2x=180
(3x-11)(2x+9)2x=180
Simplify (3x-11)(2x+9)2x.
Rewrite (2x+9)2 as (2x+9)(2x+9).
(3x-11)((2x+9)(2x+9))x=180
Expand (2x+9)(2x+9) using the FOIL Method.
Apply the distributive property.
(3x-11)(2x(2x+9)+9(2x+9))x=180
Apply the distributive property.
(3x-11)(2x(2x)+2x⋅9+9(2x+9))x=180
Apply the distributive property.
(3x-11)(2x(2x)+2x⋅9+9(2x)+9⋅9)x=180
(3x-11)(2x(2x)+2x⋅9+9(2x)+9⋅9)x=180
Simplify and combine like terms.
Simplify each term.
Multiply x by x.
(3x-11)(2⋅2×2+2x⋅9+9(2x)+9⋅9)x=180
Multiply 2 by 2.
(3x-11)(4×2+2x⋅9+9(2x)+9⋅9)x=180
Multiply 9 by 2.
(3x-11)(4×2+18x+9(2x)+9⋅9)x=180
Multiply 2 by 9.
(3x-11)(4×2+18x+18x+9⋅9)x=180
Multiply 9 by 9.
(3x-11)(4×2+18x+18x+81)x=180
(3x-11)(4×2+18x+18x+81)x=180
(3x-11)(4×2+36x+81)x=180
(3x-11)(4×2+36x+81)x=180
Expand (3x-11)(4×2+36x+81) by multiplying each term in the first expression by each term in the second expression.
(3x(4×2)+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Simplify terms.
Simplify each term.
Rewrite using the commutative property of multiplication.
(3⋅4(x⋅x2)+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Multiply x by x2 by adding the exponents.
Multiply x by x2.
Raise x to the power of 1.
(3⋅4(x1x2)+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Use the power rule aman=am+n to combine exponents.
(3⋅4×1+2+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
(3⋅4×1+2+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
(3⋅4×3+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
(3⋅4×3+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Multiply 3 by 4.
(12×3+3x(36x)+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Multiply x by x.
(12×3+3⋅36×2+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Multiply 3 by 36.
(12×3+108×2+3x⋅81-11(4×2)-11(36x)-11⋅81)x=180
Multiply 81 by 3.
(12×3+108×2+243x-11(4×2)-11(36x)-11⋅81)x=180
Multiply 4 by -11.
(12×3+108×2+243x-44×2-11(36x)-11⋅81)x=180
Multiply 36 by -11.
(12×3+108×2+243x-44×2-396x-11⋅81)x=180
Multiply -11 by 81.
(12×3+108×2+243x-44×2-396x-891)x=180
(12×3+108×2+243x-44×2-396x-891)x=180
Simplify terms.
Subtract 44×2 from 108×2.
(12×3+64×2+243x-396x-891)x=180
Subtract 396x from 243x.
(12×3+64×2-153x-891)x=180
Apply the distributive property.
12x3x+64x2x-153x⋅x-891x=180
12x3x+64x2x-153x⋅x-891x=180
12x3x+64x2x-153x⋅x-891x=180
Simplify.
Multiply x3 by x by adding the exponents.
Move x.
12(x⋅x3)+64x2x-153x⋅x-891x=180
Multiply x by x3.
Raise x to the power of 1.
12(x1x3)+64x2x-153x⋅x-891x=180
Use the power rule aman=am+n to combine exponents.
12×1+3+64x2x-153x⋅x-891x=180
12×1+3+64x2x-153x⋅x-891x=180
12×4+64x2x-153x⋅x-891x=180
12×4+64x2x-153x⋅x-891x=180
Multiply x2 by x by adding the exponents.
Move x.
12×4+64(x⋅x2)-153x⋅x-891x=180
Multiply x by x2.
Raise x to the power of 1.
12×4+64(x1x2)-153x⋅x-891x=180
Use the power rule aman=am+n to combine exponents.
12×4+64×1+2-153x⋅x-891x=180
12×4+64×1+2-153x⋅x-891x=180
12×4+64×3-153x⋅x-891x=180
12×4+64×3-153x⋅x-891x=180
Multiply x by x by adding the exponents.
Move x.
12×4+64×3-153(x⋅x)-891x=180
Multiply x by x.
12×4+64×3-153×2-891x=180
12×4+64×3-153×2-891x=180
12×4+64×3-153×2-891x=180
12×4+64×3-153×2-891x=180
Graph each side of the equation. The solution is the x-value of the point of intersection.
x≈-5.08093716,-3.76829935,-0.21025244,3.72615563
Solve for x (3x-11)(2x+9)^2x=180 |
## SectionA.3Absolute Value and Square Root
In this section, we will learn the basics of absolute value and square root. These are actions you can do to a given number, often changing the number into something else.
### SubsectionA.3.1Introduction to Absolute Value
###### DefinitionA.3.1.2.
The absolute value of a number is the distance between that number and $0$ on a number line. For the absolute value of $x\text{,}$ we write $\abs{x}\text{.}$
Let's look at $\abs{2}$ and $\abs{-2}\text{,}$ the absolute value of $2$ and the absolute value of $-2\text{.}$
Since the distance between $2$ and $0$ on the number line is $2$ units, the absolute value of $2$ is $2\text{.}$ We write $\abs{2}=2\text{.}$
Since the distance between $-2$ and $0$ on the number line is also $2$ units, the absolute value of $-2$ is also $2\text{.}$ We write $\abs{-2}=2\text{.}$
###### CheckpointA.3.1.5. Calculating Absolute Value.
Try calculating some absolute values.
###### WarningA.3.1.6. Absolute Value Does Not Exactly “Make Everything Positive”.
Students may see an expression like $\abs{2-5}$ and incorrectly think it is OK to “make everything positive” and write $2+5\text{.}$ This is incorrect since $\abs{2-5}$ works out to be $3\text{,}$ not $7\text{,}$ as we are actually taking the absolute value of $-3$ (the equivalent number inside the absolute value).
### SubsectionA.3.2Square Root Facts
If you have learned your basic multiplication table, you know:
The numbers along the diagonal are special; they are known as perfect squares. And for working with square roots, it will be helpful if you can memorize these first few perfect square numbers.
“Taking a square root” is the opposite action of squaring a number. For example, when you square $3\text{,}$ the result is $9\text{.}$ So when you take the square root of $9\text{,}$ the result is $3\text{.}$ Just knowing that $9$ comes about as $3^2$ lets us realize that $3$ is the square root of $9\text{.}$ This is why memorizing the perfect squares from the multiplication table can be so helpful.
The notation we use for taking a square root is the radical, $\sqrt{\phantom{x}}\text{.}$ For example, “the square root of $9$” is denoted $\sqrt{9}\text{.}$ And now we know enough to be able to write $\sqrt{9}=3\text{.}$
Tossing in a few extra special square roots, it's advisable to memorize the following:
$\sqrt{0}=0$ $\sqrt{1}=1$ $\sqrt{4}=2$ $\sqrt{9}=3$ $\sqrt{16}=4$ $\sqrt{25}=5$ $\sqrt{36}=6$ $\sqrt{49}=7$ $\sqrt{64}=8$ $\sqrt{81}=9$ $\sqrt{100}=10$ $\sqrt{121}=11$ $\sqrt{144}=12$ $\sqrt{169}=13$ $\sqrt{196}=14$ $\sqrt{225}=15$
### SubsectionA.3.3Calculating Square Roots with a Calculator
Most square roots are actually numbers with decimal places that go on forever. Take $\sqrt{5}$ as an example:
\begin{align*} \sqrt{4}\amp=2\amp\sqrt{5}\amp=\mathord{?}\amp\sqrt{9}\amp=3 \end{align*}
Since $5$ is between $4$ and $9\text{,}$ then $\sqrt{5}$ must be somewhere between $2$ and $3\text{.}$ There are no whole numbers between $2$ and $3\text{,}$ so $\sqrt{5}$ must be some number with decimal places. If the decimal places eventually stopped, then squaring it would give you another number with decimal places that stop further out. But squaring it gives you $5$ with no decimal places. So the only possibility is that $\sqrt{5}$ is a decimal between $2$ and $3$ that goes on forever. With a calculator, we can see:
\begin{equation*} \sqrt{5}\approx2.236 \end{equation*}
Actually the decimal will not terminate, and that is why we used the $\approx$ symbol instead of an equals sign. To get $2.236$ we rounded down slightly from the true value of $\sqrt{5}\text{.}$ With a calculator, we can check that $2.236^2=4.999696\text{,}$ a little shy of $5\text{.}$
### SubsectionA.3.4Square Roots of Fractions
We can calculate the square root of some fractions by hand, such as $\sqrt{\frac{1}{4}}\text{.}$ The idea is the same: can you think of a number that you would square to get $\frac{1}{4}\text{?}$ Being familiar with fraction multiplication, we know that $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$ and so $\sqrt{\frac{1}{4}}=\frac{1}{2}\text{.}$
###### CheckpointA.3.4.8. Square Roots of Fractions.
Try calculating some absolute values.
### SubsectionA.3.5Square Root of Negative Numbers
Can we find the square root of a negative number, such as $\sqrt{-25}\text{?}$ That would mean that there is some number out there that multiplies by itself to make $-25\text{.}$ Would $\sqrt{-25}$ be positive or negative? Either way, once you square it (multiply it by itself) the result would be positive. So it couldn't possibly square to $-25\text{.}$ So there is no square root of $-25$ or of any negative number for that matter.
If you are confronted with an expression like $\sqrt{-25}\text{,}$ or any other square root of a negative number, you can state that “there is no real square root” or that the result “does not exist” (as a real number).
### ExercisesA.3.6Exercises
###### 1.
Evaluate the expressions.
1. $1^2$
2. $3^2$
3. $5^2$
4. $7^2$
5. $9^2$
6. $11^2$
###### 2.
Evaluate the expressions.
1. $2^2$
2. $4^2$
3. $6^2$
4. $8^2$
5. $10^2$
6. $12^2$
###### 3.
Evaluate the following.
$\displaystyle{ |{-10}|= }$
###### 4.
Evaluate the following.
$\displaystyle{ |{-7}|= }$
###### 5.
Evaluate the following.
$\displaystyle{ \left\lvert-48.89\right\rvert= }$
###### 6.
Evaluate the following.
$\displaystyle{ \left\lvert-15.76\right\rvert= }$
###### 7.
Evaluate the following.
$\displaystyle{ \left\lvert{-{\frac{3}{59}}}\right\rvert= }$
###### 8.
Evaluate the following.
$\displaystyle{ \left\lvert{-{\frac{19}{11}}}\right\rvert= }$
###### 9.
Evaluate the following.
1. $\displaystyle{ - \lvert 1-9 \rvert = }$
2. $\displaystyle{ \lvert -1-9 \rvert = }$
3. $\displaystyle{ -2 \lvert 9-1 \rvert = }$
###### 10.
Evaluate the following.
1. $\displaystyle{ - \lvert 4-6 \rvert = }$
2. $\displaystyle{ \lvert -4-6 \rvert = }$
3. $\displaystyle{ -2 \lvert 6-4 \rvert = }$
###### 11.
Evaluate the following.
1. $\displaystyle{ |{10}|= }$
2. $\displaystyle{ |{-10}|= }$
3. $\displaystyle{ -|{10}|= }$
4. $\displaystyle{ -|{-10}|= }$
###### 12.
Evaluate the following.
1. $\displaystyle{ |{2}|= }$
2. $\displaystyle{ |{-2}|= }$
3. $\displaystyle{ -|{2}|= }$
4. $\displaystyle{ -|{-2}|= }$
###### 13.
Evaluate the following.
1. $\displaystyle{ \left\lvert 2 \right\rvert = }$
2. $\displaystyle{ \left\lvert -3 \right\rvert = }$
3. $\displaystyle{ \left\lvert 0 \right\rvert = }$
4. $\displaystyle{ \left\lvert {14+\left(-8\right)} \right\rvert = }$
5. $\displaystyle{ \left\lvert {-8-\left(-1\right)} \right\rvert = }$
###### 14.
Evaluate the following.
1. $\displaystyle{ \left\lvert 3 \right\rvert = }$
2. $\displaystyle{ \left\lvert -7 \right\rvert = }$
3. $\displaystyle{ \left\lvert 0 \right\rvert = }$
4. $\displaystyle{ \left\lvert {18+\left(-1\right)} \right\rvert = }$
5. $\displaystyle{ \left\lvert {-8-\left(-3\right)} \right\rvert = }$
###### 15.
Which of the following are square numbers? There may be more than one correct answer.
• $101$
• $80$
• $56$
• $1$
• $16$
• $25$
###### 16.
Which of the following are square numbers? There may be more than one correct answer.
• $36$
• $81$
• $55$
• $64$
• $50$
• $62$
###### 17.
Evaluate the following.
1. $\displaystyle{ \sqrt{64} }$ =
2. $\displaystyle{ \sqrt{16} }$ =
3. $\displaystyle{ \sqrt{1} }$ =
###### 18.
Evaluate the following.
1. $\displaystyle{ \sqrt{81} }$ =
2. $\displaystyle{ \sqrt{144} }$ =
3. $\displaystyle{ \sqrt{25} }$ =
###### 19.
Evaluate the following.
1. $\displaystyle{ \sqrt{{{\frac{100}{49}}}} }$ =
2. $\displaystyle{ \sqrt{{-{\frac{121}{36}}}} }$ =
###### 20.
Evaluate the following.
1. $\displaystyle{ \sqrt{{{\frac{144}{121}}}} }$ =
2. $\displaystyle{ \sqrt{{-{\frac{100}{81}}}} }$ =
###### 21.
Evaluate the following.
Do not use a calculator.
1. $\displaystyle{ \sqrt{4} }$ =
2. $\displaystyle{ \sqrt{0.04} }$ =
3. $\displaystyle{ \sqrt{400} }$ =
###### 22.
Evaluate the following.
Do not use a calculator.
1. $\displaystyle{ \sqrt{9} }$ =
2. $\displaystyle{ \sqrt{0.09} }$ =
3. $\displaystyle{ \sqrt{900} }$ =
###### 23.
Evaluate the following.
Do not use a calculator.
1. $\displaystyle{ \sqrt{16} }$ =
2. $\displaystyle{ \sqrt{1600} }$ =
3. $\displaystyle{ \sqrt{160000} }$ =
###### 24.
Evaluate the following.
Do not use a calculator.
1. $\displaystyle{ \sqrt{25} }$ =
2. $\displaystyle{ \sqrt{2500} }$ =
3. $\displaystyle{ \sqrt{250000} }$ =
###### 25.
Evaluate the following.
Do not use a calculator.
1. $\displaystyle{ \sqrt{36} }$ =
2. $\displaystyle{ \sqrt{0.36} }$ =
3. $\displaystyle{ \sqrt{0.0036} }$ =
###### 26.
Evaluate the following.
Do not use a calculator.
1. $\displaystyle{ \sqrt{64} }$ =
2. $\displaystyle{ \sqrt{0.64} }$ =
3. $\displaystyle{ \sqrt{0.0064} }$ =
###### 27.
Evaluate the following.
Use a calculator to approximate with a decimal.
$\sqrt{61}\approx$
###### 28.
Evaluate the following.
Use a calculator to approximate with a decimal.
$\sqrt{83}\approx$
###### 29.
Evaluate the following.
$\displaystyle{\sqrt{{{\frac{121}{144}}}}={}}$.
###### 30.
Evaluate the following.
$\displaystyle{\sqrt{{{\frac{1}{100}}}}={}}$.
###### 31.
Evaluate the following.
$-\sqrt{9}={}$.
###### 32.
Evaluate the following.
$-\sqrt{16}={}$.
###### 33.
Evaluate the following.
$\sqrt{-25}=$.
###### 34.
Evaluate the following.
$\sqrt{-49}=$.
###### 35.
Evaluate the following.
$\displaystyle{\sqrt{-{{\frac{49}{100}}}}={}}$.
###### 36.
Evaluate the following.
$\displaystyle{\sqrt{-{{\frac{64}{81}}}}={}}$.
###### 37.
Evaluate the following.
$\displaystyle{-\sqrt{{{\frac{81}{100}}}}={}}$.
###### 38.
Evaluate the following.
$\displaystyle{-\sqrt{{{\frac{121}{144}}}}={}}$.
###### 39.
Evaluate the following.
1. $\displaystyle{\sqrt{25}-\sqrt{9}=}$
2. $\displaystyle{\sqrt{25-9}=}$
###### 40.
Evaluate the following.
1. $\displaystyle{\sqrt{25}-\sqrt{9}=}$
2. $\displaystyle{\sqrt{25-9}=}$
###### 41.
Evaluate the following.
$\displaystyle{\frac{3}{\sqrt{{100}}}}$ = .
###### 42.
Evaluate the following.
$\displaystyle{\frac{3}{\sqrt{{49}}}}$ = . |
## Polygon A is similar to polygon B the sides of polygon A are three times larger then the corresponding sides of polygon B the perimeter of p
Question
Polygon A is similar to polygon B the sides of polygon A are three times larger then the corresponding sides of polygon B the perimeter of polygon A is 144 what is the perimeter of polygon B ?
in progress 0
3 years 2021-08-08T13:09:12+00:00 2 Answers 11 views 0
• 48 units
Step-by-step explanation:
Perimeter is the sum of the side lengths
If polygon A has perimeter
• P = a+b+c+d for example
Then polygon B has perimeter of
• P’ = 1/3a+1/3b+1/3+1/3d = 1/3P as each side is 3 times smaller than the corresponding side
Since P = 144
• P’ = 1/3P = 1/3*144 = 48 units
2. We can write a ratio comparing the sides of both polygons.
A:B = 3:1
Since we know that 144 is the perimeter for polygon A, we can substitute it.
144:B = 3:1
~Simplify
144 = 3B
Divide 3 to both sides
48 = B
Therefore, the perimeter of B is 48 units.
Best of Luck! |
# HiSET: Math : Special triangles
## Example Questions
### Example Question #4 : Understand Right Triangles
Two of a triangle's interior angles measure and , respectively. If this triangle's hypotenuse is long, what are the lengths of its other sides?
Possible Answers:
Correct answer:
Explanation:
A triangle that has interior angles of and is necessarily a 30-60-90 triangle—a special right triangle. We can tell that the third angle about which we're not told anything has to be because a triangle's interior angles always sum to , allowing us to solve for the third angle like so:
Since we know this triangle is a 30-60-90 triangle, we can use the special ratios that always hold true for this triangle's sides and angles to figure out the lengths of its other sides. The following ratio holds true for all 30-60-90 triangles, where the side in a fraction with a given angle is the side opposite that angle.
We're told that the hypotenuse of our triangle has a length of . The hypotenuse is the triangle's longest side, so it will be located directly across from its largest angle. In this case, that angle is . So, we need to set equivalent to and solve for .
As you can see, for this particular triangle, . Using this information, we can now calculate the lengths of the other sides of the triangle. The side opposite the angle will be equal to inches; since , this side's length is . The side opposite the angle will be equal to . Substituting in into this expression, we find that this side has a length of .
Thus, the correct answer is .
### Example Question #5 : Understand Right Triangles
Examine the above triangle. Which of the following correctly gives the area of ?
Possible Answers:
None of the other choices gives the correct response.
Correct answer:
Explanation:
Since is a right angle - that is, - and , it follows that
,
making a 30-60-90 triangle.
By the 30-60-90 Triangle Theorem,
,
and
Refer to the diagram below:
The area of a right triangle is equal to half the product of the lengths of its legs, so
,
the correct response.
### Example Question #6 : Understand Right Triangles
Examine the above triangle. Which of the following correctly gives the perimeter of ?
Possible Answers:
Correct answer:
Explanation:
Since is a right angle - that is, - and , it follows that
,
making a 30-60-90 triangle.
By the 30-60-90 Triangle Theorem,
,
and
Refer to the diagram below:
The perimeter - the sum of the sidelengths - is
. |
# Math 25: Solutions to Homework # 5 (6.2 # 8) Show that if p is prime
```Math 25: Solutions to Homework # 5
(6.2 # 8) Show that if p is prime and 2p − 1 is composite, then 2p − 1 is a pseudoprime to
the base 2.
Let m = 2p − 1. Since p is prime, 2p ≡ 2 (mod p), so p | 2p − 2, hence 2p − 2 = kp for
p
some integer k. Then 2m−1 = 22 −2 = 2kp . Now m = 2p − 1 | 2kp − 1 = 2m−1 − 1, so 2m−1 ≡ 1
(mod m). Therefore, m = 2p − 1 is a pseudoprime to the base 2.
(6.2 # 10) Suppose that a and n are relatively prime positive integers. Show that if n is a
pseudoprime to the base a, then n is a pseudoprime to the base a, where a is an inverse of
a modulo n.
Let a be an inverse of a modulo m. Then since an−1 ≡ 1 (mod n),
(a)n−1 ≡ an−1 (a)n−1 ≡ (aa)n−1 ≡ 1n−1 ≡ 1
(mod n),
so n is a pseudoprime to the base a.
(6.2 # 12) Show that 25 is a strong pseudoprime to the base 7.
We can write 25 − 1 = 24 = 23 · 3. Then
72·3 ≡ (72 )3 ≡ (49)3 ≡ (−1)3 ≡ −1
(mod 7),
so 25 passes Miller’s test for the base 7.
(6.3 # 8) Show that if a is an integer such that a is not divisible by 3 or such that a is
divisible by 9, then a7 ≡ a (mod 63).
By the corollary to Fermat’s Little Theorem, a7 ≡ a (mod 7). Suppose that 3 - a. Then
(a, 9) = 1 and φ(9) = 6, so by Euler’s theorem, a6 ≡ aφ(9) ≡ 1 (mod 9), so also a7 ≡ a
(mod 9). If 9 | a then a ≡ 0 (mod 9), so a7 ≡ a ≡ 0 (mod 9). Therefore in either case, we
have a7 ≡ a (mod 7) and a7 ≡ a (mod 9). Since (7, 9) = 1, then a7 ≡ a (mod 63).
(6.3 # 10) Show that aφ(b) + bφ(a) ≡ 1 (mod ab), if a and b are relatively prime positive
integers.
First, ak ≡ 0 (mod a) and bk ≡ 0 (mod b) for any positive integer k. Then by Euler’s
Theorem,
aφ(b) + bφ(a) ≡ bφ(a) ≡ 1 (mod a),
and
aφ(b) + bφ(a) ≡ aφ(b) ≡ 1
φ(b)
Then since (a, b) = 1, a
+b
φ(a)
(mod b).
≡ 1 (mod ab).
(7.1# 8) Show that there is no positive integer n such that φ(n) = 14.
Suppose n is a positive integer with φ(n) = 14. We also know that if n = pa11 · · · pat t then
φ(n) = p1a1 −1 (p1 − 1) · · · pat t −1 (pt − 1). So no prime p > 15 divides n, otherwise φ(n) >
p − 1 > 14. This leaves possible prime factors 2, 3, 5, 7, 11, and 13. But 5, 7, 11 and
13 can all be eliminated since 4, 6, 10, and 12 do not divide 14. But if n = 2a · 3b then
φ(n) = 2a−1 (2 − 1) · 3b−1 (3 − 1) = 2a · 3b−1 , which is not divisible by 7. Therefore there is no
n for which φ(n) = 14.
(7.1 # 32) Show that if m and n are positive integers with m | n, then φ(m) | φ(n).
Suppose that m | n, and write n = pa11 · · · pakk . Then m = pb11 · · · pbkk where 0 ≤ bj ≤ aj for
all 1 ≤ j ≤ k. Then
Qk
aj −1
k
(pj − 1) Y
φ(n)
j=1 pj
paj −bj
= Qk
=
b
−1
j
φ(m)
pj (pj − 1) j=1
j=1
is an integer, so φ(m) | φ(n).
(7.2 # 4) For which positive integers n is the sum of divisors of n odd?
Let n = pa11 · · · pakk . Then σ(n) =
Qk
j=1
paj +1 −1
. In order for σ(n) to be
p−1
2a+1 −1
= 2a − 1 is odd, for any
2−1
a
odd, each term in
this product must be odd. If p = 2, then
positive integer a.
a+1
If p is odd, then p p−1−1 = 1 + p + p2 + · · · + p . Since each power of p is odd, this sum is odd
exactly when a is even. Therefore σ(n) is odd if and only if the power of every odd prime
dividing n is even.
(7.2 # 22) Give a formula for σk (pa ), where p is prime and a is a positive integer.
σk (pa ) = 1k + pk + p2k + · · · + pak =
p(a+1)k − 1
.
pk − 1
(7.3 # 8) Show that any proper divisor of a deficient or perfect number is deficient.
Suppose that a | n and 1 < a < n. We want to prove that if σ(n) ≤ 2n, then σ(a) < 2a.
We will prove the contrapositive, namely, if σ(a) ≥ 2a, then σ(n) > 2n. There must be an
integer k such that ak = n. Then if c | a, then ck | ak = n. Therefore
X
X
σ(n) =
d>
kc = kσ(a) ≥ 2ka = 2n.
d|n
c|a
(7.3 # 20) Find all 3-perfect numbers of the form n = 2k · 3 · p, where p is an odd prime.
First we note that p 6= 3, since then n = 2k · 32 , so 13 = σ(32 ) | σ(n), and hence σ(n) 6= 3n.
So we may assume p 6= 3. If n = 2k · 3 · p is 3-perfect, then σ(n) = 3n = 2k · 32 · p, but also
σ(n) = σ(2k )σ(3)σ(p) = (2k+1 − 1) · 4(p − 1), so we set
2k · 32 · p = (2k+1 − 1) · 4(p − 1).
Since the right and left sides are equal, they must have the same prime power factorization,
which is already given on the left. Then k ≥ 2, so cancelling 4 from both sides, we have
2k−2 · 32 · p = (2k+1 − 1)(p − 1).
Now p and p − 1 are coprime, so p must divide 2k+1 − 1. Similarly, 2k+1 − 1 is odd, so 2k−2
must divide p − 1. Then there are integers m and r such that
2k−2 · 32 · p = (pm)(2k−2 r).
The only remaining factor on the left is 32 , so there are three cases:
(a) m = 9 and r = 1,
(b) m = r = 3, and
(c) m = 1 and r = 9.
In case (a), we have 2k+1 − 1 = 9p and p + 1 = 2k−2 , so that 8(p + 1) = 2k+1 . Substituting
this into the first equation, we have 8(p + 1) − 1 = 9p, and p = 7 is the only solution. Then
8 = 2k−2 , so k = 5. So the only possible n in this case is n = 25 · 3 · 7 = 672. Using a similar
argument for case (b), we get p = 5 and k = 3, so n = 23 · 3 · 5 = 120. Using this method for
case (c) we conclude that p = −1. Since this is not possible, there are no solutions in this
case. Therefore the only numbers of this form that are 3-perfect are 672 and 120.
``` |
Use adaptive quiz-based learning to study this topic faster and more effectively.
# Percentage
## What is a percentage?
A percentage is one-hundredth of a number. It uses the symbol $\%$. A percentage is a fraction where the denominator is $\Tblue{100}$.
$\Tred{50}\% = \frac{\Tred{50}}{\Tblue{100}} = \frac{1}{2} = 0.5,\quad \Tred{40}\% = \frac{\Tred{40}}{\Tblue{100}} = \frac{2}{5} = 0.4.$
To express a number as a percentage, write it in decimal form and multiply it by $\Tblue{100}$.
$0\Torange{.43}2 = \Tblue{100}\times 0\Torange{.43}2\% = \Torange{43.}2\%,\quad \frac{2}{7} = 0\Torange{.28}6 = \Torange{28.}6\%$
Percentages are often used to represent a fraction of a quantity.
Half a cake is the same as $50\%$ of the cake.
$25\%$ of $247$ is $\Torange{25}\%\times 247 = 0\Torange{.25}\times 247 = 61.75$
$26$ is $14.6\%$ of $178$ because $\displaystyle \frac{26}{178} = 0\Torange{.14}6 = \Torange{14.}6\%.$
## Percentage change
Percentages are used to tell how a quantity will increase or decrease.
• Percentage change from an old number $\Tblue{O}$ to a new one $\Tgreen{N}$ is $\Tred{p} = \frac{\Tgreen{N}- \Tblue{O}}{\Tblue{O}} = \frac{\Tgreen{N}}{\Tblue{O}} - 1$
The price of a shirt went from $\Tblue{\11}$ to $\Tgreen{\12}$. It increased by $\frac{\Tgreen{12}- \Tblue{11}}{\Tblue{11}} = \frac{1}{\Tblue{11}} =\Tred{0.09} = \Tred{9\%}.$
• To know a value after the change $p$, multiply by $1+p$ $\Tgreen{N} = \Tblue{O} \times (1+\Tred{p})$
A population of $\Tblue{35}$ million increased by $\Tred{2\%}$. It is now $\Tblue{35}\cdot (1+\Tred{2\%}) = \Tblue{35} \times 1\Tred{.02} = \Tgreen{35.7}$
• To find the value before the change, divide by $1+p$. $\Tblue{O} =\Tgreen{N} \div (1+\Tred{p})$
After a $\Tred{20\%}$ discount, the price of a toy is $\Tgreen{\ 8}$. Before, it was $\frac{\Tgreen{8}}{1\Tred{-20\%}}=\frac{\Tgreen{8}}{1\Tred{-0.2}} = \frac{\Tgreen{8}}{0.8} = \Tblue{\ 10}$
Discounts in sales are often described using percentages. The new price is sometimes not given, so you need to know how to work it out.
## Repeated percentage change
A change in percentage can be repeated several times.
The money I invest in a savings account increases by $4\%$ each year due to interest payment.
For a yearly percentage change of $\Tred{p\%}$, the new value $\Tgreen{N}$ after $\Torange{n}$ years from an old value $\Tblue{O}$ is $\Tgreen{N} = \Tblue{O} \times (1+\Tred{\frac{p}{100}})^\Torange{n}.$
I invest $\Tblue{1000}$ in a bank account with an interest rate of $\Tred{4\%}$. After $\Torange{3}$ years, I will have a balance of $\Tblue{1000} \times (1 + \Tred{4\%})^\Torange{3} = \Tblue{1000} \times 1\Tred{.04}^\Torange{3} = \Tblue{1000}\times 1.125 = \Tgreen{1125}.$
The population in this bacterial colony doubles every generation, an increase of 100%. |
# Laws of Exponents - Multiplying Powers with the Same Base
#### formula
For any non-zero integer a, where m and n are whole numbers, am × an =am + n.
# Multiplying Powers With the Same Base:
For any non-zero integer a, where m and n are whole numbers, am × an =am + n.
(i) Let us calculate 42 × 43.
= 42 × 43
= (4 × 4) × (4 × 4 × 4)
= 4 × 4 × 4 × 4 × 4 = 45 = 42+3...............(am × an = am+n)
Note that the base in 42 and 43 is the same and the sum of the exponents, i.e., 2 and 3 is 5.
(ii) (– 11)4 × (– 11)3
= [(– 11) × (– 11) × (– 11) × (– 11)] × [(– 11) × (– 11) × (– 11)]
= (– 11) × (– 11) × (– 11) × (– 11) × (– 11) × (– 11) × (– 11) = (– 11)7 = (– 11)4+3........(am × an = am+n)
Again, note that the base is the same and the sum of exponents, i.e., 4 and 3, is 7.
(iii) 32 × 34 × 38
= 32 × 34 × 38 = (3)2 + 4 + 8.............(am × an = am+n)
= 314.
#### Example
Solve: (-3)2 × (-3)3.
(-3)2 × (-3)3
= (-3) × (-3) × (-3) × (-3) × (-3)
= (-3)5
Therefore, (-3)2 × (-3)3 = (-3)2 + 3 = (-3)5.
#### Example
Solve: ((-2)/5)^2 xx ((-2)/5)^3
((-2)/5)^2 xx ((-2)/5)^3
= ((-2)/5) xx ((-2)/5) xx ((-2)/5) xx ((-2)/5) xx ((-2)/5)
= ((-2)/5)^5
Therefore, ((-2)/5)^2 xx ((-2)/5)^3 = ((-2)/5)^(2 + 3) = ((-2)/5)^5.
If you would like to contribute notes or other learning material, please submit them using the button below.
### Shaalaa.com
Multiplying Powers with the Same Base [00:05:52]
S
0% |
# Combinations/Probability
Combination problems and Permutation problems are generally solved through some form of calculation, which is usually be done by multiplication. We split the problem into separate parts, calculate the number of instances in each part, and then combine the results of our calculation.
A permutation is an arrangement of a set of objects in a particular order. Permutation problems explore the possible arrangements and number of arrangements of objects selected from the set. They may also involve arrangements where some of the objects are repeated, or where there are multiple sets, each individually with their own permutations, but then combined in a single arrangement. The basic formula for the number of permutations of p things selected from a set of size n is:
\begin{align} \frac {n!}{(n-p)!}\end{align}
Combinations are selections from sets of objects where the order in which the selection is arranged is unimportant. The basic formula for the number of combinations of p things selected from a set of size n is:
\begin{align} \frac {n!}{(n-p)!p!}\end{align}
This is normally written as
\begin{align} n \choose p\end{align}
which is said, “n choose p.”
Probability problems are solved at the basic level by using the simple formula:
\begin{align}probability = \frac {number\hspace{1mm} of\hspace{1mm} favorable\hspace{1mm} outcomes}{number\hspace{1mm} of\hspace{1mm} possible\hspace{1mm} outcomes}\end{align}
Since both the numerator and denominator of this fraction are often determined through some form of “counting,” then probability problems often are solved by solving permutation and combination problems first.
On this page we have 1 Permutations, 3 Combinations, and 2 Probability examples.
Permutations Example
The school orchestra’s string section has 4 chairs designated for violins, 2 for violas, 2 for cellos, and 2 for basses, with the corresponding number of students playing each of these instruments. The conductor wants the students to sit in a different, randomly chosen order each day in practice. How many days can they do this before they repeat a seating order?
A. 32
B. 48
C. 96
D. 128
E. 192
Explanation:
This is a straightforward permutations problem involving 4 different permutations – 1 for each instrument section. Therefore, the solution is the number of permutations of seating order for the violins, times that for violas, times that for cellos, times that for basses. Or
4! 2! 2! 2! = (4 × 3 × 2 × 1) × (2 × 1) × (2 × 1) × (2 × 1) = 24 × 2 × 2 × 2 = 192
So E is the correct answer.
Combinations Example 1)
A state’s license plates consists of 2 letters, followed by 4 numeric digits, with these restrictions: (1) Two letters, O and I, are never used, and (2) 0 is never the first of the 4 digits. How many possible license plates are there for the state?
A. 5,189,184
B. 5,616,000
C. 5,630,625
D. 5,184,000
E. 5,183,424
Explanation:
This is a combination problem whose answer can be calculated by multiplying the correct numbers.
1. We need to know how many different combinations of two letters are possible if the letters O and I are never used. This means there are 24 letters available, and so the number of combinations of letters is 24 x 24 = 576.
2. Next, we need to know how many 4 digit numbers there are if we disallow 0 from being the first digit. What this means is that we want to know how many numbers there are between 1000 and 9999, inclusive. This is determined by subtracting 999 from 9999. The result is 9000.
3. Finally, we multiply the results of 1 and 2: 576 × 9000 = 5,184,000.
So D is the correct answer.
Combinations Example 2)
If a combination lock has 40 numbers on it, and each combination for unlocking the lock uses 3 of the
numbers with repetitions allowed, then how many different combinations are possible?
A. 24,000
B. 44,000
C. 82,000
D. 64,000
E. 36,000
Explanation: The order of the numbers in a combination is important, and each number can be repeated
in a combination 1 or 2 times. So, we have 40 choices for the first number, 40 for the second, and 40 for
the third. The total number of possible combinations then is
40 × 40 × 40 = 64,000
So D is the correct answer.
Combinations Example 3)
If a combination lock has x numbers on it, where x is a multiple of 10, and each combination for unlocking the lock uses 3 of the numbers with repetitions allowed, then what is the smallest possible value for x if it’s desired to have at least 10,000 different combinations?
A. 33
B. 30
C. 20
D. 27
E. 10
Explanation: The number of possible combinations is x\begin{align}^3\end{align}. Since x is a multiple of 10, answers A and D are immediately eliminated. Also, we want the minimum value of x that will yield 10,000 combinations, so we’ll try ascending values for x on the other 3 possible answers:
x = 10: \begin{align}x^3\end{align} = 1000
x = 20: \begin{align}x^3\end{align} = 8000
x = 30: \begin{align}x^3\end{align} = 27,000
Obviously, 30 is the desired number, so B is the correct answer.
Probability Example 1)
When two dice are rolled, the probability that 7 will come up is 1/6, that 4 will come up is 1/12, and that 9 will come up is 1/9. What is the probability that in three rolls of the dice, the first will be 4, the second will not be 9, and the third will not be 7?
A. 5/81
B. 1/8
C. 4/27
D. 5/13
E. 3/16
Explanation: Because each of these rolls is independent of the others, to get the probability that all three events will happen, we need to multiply the following 3 probabilities:
4 Probability = 1/12
not 9 Probability = 1 – 1/9 = 8/9
not 7 Probability = 1 – 1/6 = 5/6
1/12 × 8/9 × 5/6 = 5/81
So A is the correct answer.
Probability Example 2)
In a certain neighborhood the probability that person’s birthday is not in January is .9, and not in July is .85. What’s the probability that a person selected at random from the neighborhood is born in either January or July?
A. 1/9
B. 1/8
C. 1/4
D. 1/6
E. 1/3
Explanation: The probability of an event’s happening is 1 minus the probability of its not happening. Therefore, the probability of a person in the neighborhood being born in January is .1, and in July is .15. Since no one can be born in both months, then the probability of someone being born in either month is additive. So the probability of being born in either January or July is
.1 + .15 = .25 = 1/4
So C is the correct answer. |
## Engage NY Eureka Math Grade 6 Module 5 Lesson 4 Answer Key
### Eureka Math Grade 6 Module 5 Lesson 4 Exercise Answer Key
Opening Exercise:
Draw and label the altitude of each triangle below.
Question a.
Question b.
Question c.
Exploratory Challenge/Exercises 1 – 5:
Question 1.
Use rectangle X and the triangle with the altitude inside (triangle X) to show that the area formula for the triangle is A = $$\frac{1}{2}$$ × base × height.
a. Step One: Find the area of rectangle X.
A = 3 in. × 2.5 in. = 7.5 in2
b. Step Two: What is half the area of rectangle X?
Half of the area of the rectangle is 7.5 in2 ÷ 2= 3.75 in2
c. Step Three: Prove, by decomposing triangle X, that it is the same as half of rectangle X. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle X. What conclusions can you make about the triangle’s area compared to the rectangle’s area?
Because the triangle fits inside half of the rectangle, we know the triangle’s area is half of the rectangle’s area.
Question 2.
Use rectangle Y and the triangle with a side that ¡s the altitude (triangle Y) to show the area formula for the triangle is A = $$\frac{1}{2}$$ × base × height.
a. Step One: Find the area of rectangle Y.
A = 3 in. × 3 in. = 9 in2
b. Step Two: What is half the area of rectangle Y?
Half the area of the rectangle is 9 in2 ÷ 2 = 4.5 in2.
c. Step Three: Prove, by decomposing triangle Y, that it is the same as half of rectangle Y. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Y. What conclusions can you make about the triangle’s area compared to the rectangle’s area?
The right triangle also fits in exactly half of the rectangle, so the triangle’s area is • Students may struggle once again half the size of the rectangle’s area. with this step since they have yet to see an obtuse
Question 3.
Use rectangle Z and the triangle with the altitude outside (triangle Z) to show that the area angle. formula for the triangle is A = $$\frac{1}{2}$$ × base × height.
a. Step One: Find the area of rectangle Z.
A= 3 in. × 2.5 in.=7.5 in2
b. Step Two: What is half the area of rectangle Z?
Half of the area of the rectangle is 7.5 in2 ÷2 = 3.75 in2.
c. Step Three: Prove, by decomposing triangle Z, that it is the same as half of rectangle Z. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Z. What conclusions can you make about the triangle’s area compared to the rectangle’s area?
Similar to the other two triangles, when the altitude is outside the triangle, the area of the triangle is exactly half of the area of the rectangle.
Question 4.
When finding the area of a triangle, does it matter where the altitude is located?
It does not matter where the altitude is located. To find the area of a triangle, the formula is always
A = $$\frac{1}{2}$$ × base × height.
Question 5.
How can you determine which part of the triangle is the base and which is the height?
The base and the height of any triangle form a right angle because the altitude is always perpendicular to the base.
Exercises 6 – 8:
Calculate the area of each triangle. Figures are not drawn to scale.
Question 6.
A = $$\frac{1}{2}$$ × (24 in.) (8 in.) = 96 in2
Question 7.
Question 8.
Draw three triangles (acute, right, and obtuse) that have the same area. Explain how you know they have the same area.
### Eureka Math Grade 6 Module 5 Lesson 4 Problem Set Answer Key
Calculate the area of each figure below. Figures are not drawn to scale.
Question 1.
A = $$\frac{1}{2}$$ (21 in.) (8 in.) = 84 in2
Question 2.
A = $$\frac{1}{2}$$ (72 m) (21 m) = 756 m2
Question 3.
A = $$\frac{1}{2}$$ (75.8 km) (29.2 km) = 1,106.68 km2
Question 4.
A = $$\frac{1}{2}$$ (5 m) (12 m) = 30 m2
A = $$\frac{1}{2}$$ (7 m) (29 m) = 101.5 m2
A = (12 m) (19 m) = 228 m2
A = 30 m2 + 30 m2 + 101.5 m2 + 228 m2
A = 389.5 m2
Question 5.
The Andersons are going on a long sailing trip during the summer. However, one of the sails on their sailboat ripped, and they have to replace it. The sail is pictured below. If the sailboat sails are on sale for $2 per square foot, how much will the new sail cost? Answer: A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (8 ft.) (12 ft.) = 48 ft2 $$\frac{2 \text { dollars }}{\mathrm{ft}^{2}}$$ × 48 ft2 = 96 dollars(or$96)
The cost of the new sail is \$96.
Question 6.
Darnell and Donovan are both trying to calculate the area of an obtuse triangle. Examine their calculations below.
Darnell’s Work Donovan’s Work A = $$\frac{1}{2}$$ × 3 in. × 4 in. A = 6 in2 A = $$\frac{1}{2}$$ × 12 in. × 4 in. A = 24 in2
Which student calculated the area correctly? Explain why the other student is not correct.
Donovan calculated the area correctly. Although Darnell did use the altitude of the triangle, he used the length between the altitude and the base rather than the length of the actual base.
Question 7.
Russell calculated the area of the triangle below. His work is shown.
A = $$\frac{1}{2}$$ bh
A = 150.5 cm2
Although Russell was told his work is correct, he had a hard time explaining why it is correct. Help Russell explain why his calculations are correct.
The formula for the area of triangle is A = bit. Russell followed this formula because 7 cm is the height of the triangle, and 43 cm is the base of the triangle.
Question 8.
The larger triangle below has a base of 10.14 m; the gray triangle has an area of 40.325 m2.
a. Determine the area of the larger triangle if it has a height of 12.2 m.
A = $$\frac{1}{2}$$ (10.14 m) (12.2 m)
= 61.854 m2
b. Let A be the area of the unshaded (white) triangle in square meters. Write and solve an equation to determine the value of A, using the areas of the larger triangle and the gray triangle.
40.325m2 + A = 61.854m2
40.325m2 + A – 40.325 m2 = 61.854 m2 – 40.325 m2
A = 21.529 m2
### Eureka Math Grade 6 Module 5 Lesson 4 Exit Ticket Answer Key
Find the area of each triangle. Figures are not drawn to scale.
Question 1.
A = $$\frac{1}{2}$$ (12.6 cm) (16.8 cm) = 105.84 cm2
A = $$\frac{1}{2}$$ (28 in.) (15 in.) = 210 in2
A = $$\frac{1}{2}$$ (12 ft.) (21 ft.) = 126 ft2 |
Tips for Solving Math Problems
Question
The first thing you want to do is look for
00:00
00:00
Tips for Solving Math Problems
Hi, my name is Bassem Saad. I'm an associate math instructor and a Ph.D. candidate, and I'm here today for about.com to give you some tips for solving math problems.
Find Key Words and Phrases
Let's take a look at an example: Bill ran one mile in 12 minutes, and Cindy ran a mile in 11 minutes. So how much longer did it take Bill to run the mile than Cindy? The first thing you want to do is look for key words and phrases that tell us if we want to add, subtract, multiply, or divide.
So here's a list of key words for addition, subtraction, multiplication, and division:
Addition, some key words are: sum; total; in all; and parameter.
Subtraction: difference; how much more; and exceeds by.
Multiplication: product; area; and times.
And for division, we have quotient and average.
So let's find our key words: “How much longer?” Well that mean subtraction. So the next step is to find out what are we subtracting. How much longer did it take Bill to run than Cindy? So it takes Bill 12 minutes to run the mile, and it took Cindy 11 minutes to run the mile.
So our next step is to write the mathematical expression for the word problem. 12 minus 11, and have one minute remaining.
Use Visual Aids to Help Solve Math Problems
Let's take a look at one more example. John wants to build a fence around his rectangular yard. Find the perimeter if his yard is 30 feet by 40 feet. So let's look for the key term: perimeter. That means we're going to add the length of the sides. So it's helpful to draw a picture to know what we're adding. So a rectangle has a shape like this, with one side being 30 feet and a parallel side also being 30 feet. Another side being 40 feet, and it's parallel side being 40 feet. So the perimeter is just 30 feet, plus 30 feet, plus 40 feet, plus 40 feet, giving us a grand total of 140 feet. So now you have some useful tips for solving math problems.
Thanks for watching, and to learn more visit us on the web at About.com.
#### Discuss
0 comments
characters remaining |
# PROVE THE GIVEN POINTS ARE COLLINEAR USING THE CONCEPT SLOPE
The points which lie on the same line are known as collinear points.
In case we have three point and we need to prove that the given points are collinear, we may follow the steps given below.
## Working Rule
Step 1 :
Choose the points A and B and find the slope.
Step 2 :
Choose the points B and C and find the slope.
Step 3 :
If the slopes found in step 1 and step 2 are equal, then the points A, B and C are collinear.
Formula to find the slope of a line joining the two points (x1, y1) and (x2, y2) :
m = (y2 - y1) / (x- x1)
## Examples
Example 1 :
Using the concept of slope, prove that each of the following set of points are collinear.
(i) (2 , 3), (3 , -1) and (4 , -5)
(ii) (4 , 1), (-2 , -3) and (-5 , -5)
(iii) (4 , 4), (-2 , 6) and (1 , 5)
Solution :
(i) Let the given points be A (2 , 3) B (3 , -1) and C (4 , -5)
If the points A, B and C are collinear then,
Slope of AB = Slope of BC
Slope of AB :m = (y2 - y1)/(x2 - x1)A (2 , 3) B (3 , -1)m = (-1 - 3)/(3 - 2)m = -4/1m = -4 ---(1) Slope of BC :m = (y2 - y1)/(x2 - x1)B (3 , -1) C (4 , -5)m = (-5 + 1)/(4 - 3)m = -4/1m = -4 ---(2)
(1) = (2)
Hence the given points A, B and C are collinear.
(ii) (4 , 1), (-2 , -3) and (-5 , -5)
Solution :
Let the given points be A(4 , 1) B(-2 , -3) and C(-5 , -5)
If the points A, B and C are collinear then,
Slope of AB = Slope of BC
Slope of AB :m = (y2 - y1)/(x2 - x1)A (4, 1) B (-2, -3)m = (-3 - 1)/(-2 - 4)m = -4/(-6)m = 2/3 ---(1) Slope of BC :m = (y2 - y1)/(x2 - x1)B (-2, -3) C (-5, -5)m = (-5 + 3)/(-5 + 2)m = -2/(-3)m = 2/3 ---(2)
(1) = (2)
Hence the given points are collinear.
(iii) (4 , 4), (-2 , 6) and (1 , 5)
Solution :
Let the given points be A(4 , 4) B(-2 , 6) and C(1 , 5)
If the points A, B and C are collinear then,
Slope of AB = Slope of BC
Slope of AB :m = (y2 - y1)/(x2 - x1)A (4, 4) B (-2, 6)m = (6 - 4)/(-2 - 4)m = 2/(-6)m = -1/3 ---(1) Slope of BC :m = (y2 - y1)/(x2 - x1)B (-2, 6) C (1, 5)m = (5 - 6)/(1 + 2)m = -1/3 ---(2)
(1) = (2)
So, the given points are collinear.
Example 2 :
If the points (a, 1), (1, 2) and (0, b+1) are collinear, then prove that (1/a) + (1/b) = 1
Solution :
Let the given points be A(a, 1) B(1, 2) and C(0, b+1). Since the given points are collinear, slope of AB is equal to slope of BC.
Slope of AB :m = (y2 - y1)/(x2 - x1)A (a, 1) B (1, 2)m = (2 - 1)/(1 - a)m = 1/(1 - a) ---(1) Slope of BC :m = (y2 - y1)/(x2 - x1)B (1, 2) C (0, b + 1)m = (b + 1 - 2)/(0 - 1)m = (b - 1)/(-1) ---(2)
1/(1 - a) = -(b - 1)
1 = (1 - b)(1 - a)
1 = 1 - a - b + ab
a + b = ab + 1 - 1
a + b = ab
Dividing by ab, we get
(a/ab) + (b/ab) = (ab/ab)
1/b + 1/a = 1
(1/a) + (1/b) = 1
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Factors of 688: Prime Factorization, Methods, and Examples
Numbers that perfectly divide the given number 688 are known as the 688 factors. The component of the number is its factor.
If the given number is obtained by multiplying two-factor integers, then the factors of the number 688 can be either positive or negative.
### Factors of 688
Here are the factors of number 688.
Factors of 688: 1, 2, 4, 8, 16, 43, 86, 172, 344, and 688
### Negative Factors of 688
The negative factors of 688 are similar to their positive aspects, just with a negative sign.
Negative Factors of 688: -1, -2, -4, -8, -16, -43, -86, -172, -344, and -688
### Prime Factorization of 688
The prime factorization of 688 is the way of expressing its prime factors in the product form.
Prime Factorization: 2 x 2 x 2 x 2 x 43
In this article, we will learn about the factors of 688 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 688?
The factors of 688 are 1, 2, 4, 8, 16, 43, 86, 172, 344, and 688. These numbers are the factors as they do not leave any remainder when divided by 688.
The factors of 688 are classified as prime numbers and composite numbers. The prime factors of the number 688 can be determined using the prime factorization technique.
## How To Find the Factors of 688?
You can find the factors of 688 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 688, create a list containing the numbers that are exactly divisible by688 with zero remainders. One important thing to note is that 1 and 688 are the 688’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 688 are determined as follows:
$\dfrac{688}{1} = 688$
$\dfrac{688}{2} = 344$
$\dfrac{688}{4} = 172$
$\dfrac{688}{8} = 86$
$\dfrac{688}{16} = 43$
$\dfrac{688}{43} = 16$
$\dfrac{688}{86} = 8$
$\dfrac{688}{172} = 4$
$\dfrac{688}{344} = 2$
$\dfrac{688}{688} = 1$
Therefore, 1, 2, 4, 8, 16, 43, 86, 172, 344, and 688are the factors of 688.
### Total Number of Factors of 688
For 688, there are 10 positive factors and 10 negative ones. So in total, there are 20 factors of 688.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 688 is given as:
Factorization of 688 is 1 x2 x 2 x 2 x 2 x 43.
The exponent of 2 is 4, 1 is 1and 43 is 1.
Adding 1 to each and multiplying them together results in 20.
Therefore, the total number of factors of 688 is 20. 10 are positive, and 10 factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 688 by Prime Factorization
The number 688 is a composite/prime number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 688 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 688, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 688 can be expressed as:
688 = 2 x 2 x 2 x 2 x 43
## Factors of 688 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 688, the factor pairs can be found as:
1 x 688 = 688
2 x 344 = 688
4 x 172 = 688
8 x 86 = 688
16 x 43 = 688
The possible factor pairs of 688 are given as (1, 688),(2, 344),(4, 172)(8, 86), and (16, 43 ).
All these numbers in pairs, when multiplied, give 688 as the product.
The negative factor pairs of 688 are given as:
-1 x -688 = 688
-2 x -344 = 688
-4 x -172 = 688
-8 x -86 = 688
-16 x -43 = 688
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -2, -4, -8, -16, -43, -86, -172, -344, and -688are called negative factors of 688.
The list of all the factors of 688, including positive as well as negative numbers, is given below.
Factor list of 688:1, -1,2, -2,4, -4, 8,-8,16, -16,43, -43,86, -86,172, -172,344, -344, 688, and -688
## Factors of 688 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 688 are there?
### Solution
The total number of Factors of 688 is 20.
Factors of 688 are 1, 2, 4, 8, 16, 43, 86, 172, 344, and 688.
### Example 2
Find the factors of688 using prime factorization.
### Solution
The prime factorization of 688 is given as:
688 $\div$ 2 = 344
344 $\div$ 2 = 172
172 $\div$ 2 = 86
86 $\div$ 2 = 43
43 $\div$ 43 = 1
So the prime factorization of 688 can be written as:
2 x 2 x 2 x 2 x 43 = 688 |
Question
# A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km per hour, it takes 2 hours less in the journey. Find the original speed of the train.
Hint: Convert the problem statement into equations based on the speed formula and the conditions given and then solve the equations.
Given, distance to be covered by the train = 300 km
Let us suppose the original speed of the train be $x$ km per hour and $t$ hours be the original time taken by the train to cover 300 km distance with a speed of x km per hour.
As we know that, ${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$
$\Rightarrow x = \dfrac{{300}}{t}{\text{ }} \to {\text{(1)}}$
According to the problem, now the speed of the train is increased from $x$ km per hour to $\left( {x + 5} \right)$ km per hour and the time taken for the journey is reduced from $t$ hours to $\left( {t - 2} \right)$ hours.
For this case, $\left( {x + 5} \right) = \dfrac{{300}}{{\left( {t - 2} \right)}} \Rightarrow x = \dfrac{{300}}{{\left( {t - 2} \right)}} - 5 = \dfrac{{300 - 5\left( {t - 2} \right)}}{{\left( {t - 2} \right)}} = \dfrac{{300 - 5t + 10}}{{\left( {t - 2} \right)}} \\ \Rightarrow x = \dfrac{{310 - 5t}}{{t - 2}}{\text{ }} \to {\text{(2)}} \\$
Now, since the LHS of equations (1) and (2) are equal, their RHS will also be equal.
i.e., $\dfrac{{300}}{t}{\text{ }} = \dfrac{{310 - 5t}}{{t - 2}} \Rightarrow 300\left( {t - 2} \right) = t\left( {310 - 5t} \right) \Rightarrow 300t - 600 = 310t - 5{t^2} \\ \Rightarrow 5{t^2} - 10t - 600 = 0 \\ \Rightarrow {t^2} - 2t - 120 = 0 \\$
Now, by factorising the above quadratic equation we have
$\Rightarrow {t^2} - 2t - 120 = 0 \Rightarrow {t^2} + 10t - 12t - 120 = 0 \Rightarrow t\left( {t + 10} \right) - 12\left( {t + 10} \right) = 0 \\ \Rightarrow \left( {t + 10} \right)\left( {t - 12} \right) = 0 \\$
i.e., Either $t + 10 = 0$ or $t - 12 = 0$
$\Rightarrow t = - 10$ or $t = 12$
Since, time is always positive and never negative.
Therefore, $t = - 10$ is neglected and only possible value is $t = 12$.
Now put $t = 12$ in equation (1), we get
$x = \dfrac{{300}}{{12}} = 25$
Therefore, the original speed of the train is 25 km per hour.
Note- In these types of problems, the problem statement is reduced into important equations which are used further to deduce the unknowns. Here, in this problem the distance covered by train is fixed (i.e., even when the speed of the train and the time taken by the train is changed, the distance covered by the train remains fixed). |
# Understanding of Remainder theorem through divisibility property
First thing you should keep in mind is that in theory of numbers we’ll mostly be talking about numbers from the set of whole or the set of real numbers. One of the most important basics in theory of number is the definition of divisibility and some basic rules.
Definition 1. If there exists such that , we say that a divides b, where a and b are whole numbers and . It is said that b is the multiple of a, and a is the divisor of b. If a divides b we denote it as and if not .
Proposition 1. If and then .
Proof. If and (by the Definition 1.) such that , , .
Since we know that we can affect this equation with absolute value. We get that , and we got where we wanted to be.
We wanted to prove that and we got that is equal to the product of some positive whole number d and which proves our proposition.
Proposition 2. If , then a divides every multiple of b.
Proof. Let’s say that and that c is the multiple of b. According to that, there exists number such that and since we have that . Since and are both whole numbers, so is their product and we got that the multiple of b-c, is the product of some whole number and a. This means that .
Proposition 3. If and , then a \mid (b \pm c) and .
Proof. If and then exist and such that and . Since we want to prove that a divides we’ll simply include b and c into this expression. This means that we’ll have . Since and are whole number, so are their sum and difference and we got that is equal to the product of number with some whole number. This means that .
The same goes for the product of b and c:
and in the same way we got that is equal to the product of number a and some whole number which means that .
This relation – “to be divisible” is relation of partial order. This means that are fulfilled following three conditions:
1. Reflexive property – This means that a certain number is always divisible by itself or:
2. Antisymmetric property – This means that if one number is divisible by another number, and in the same time that other number is divisible by the first number, those two numbers must be equal. and
3. Transitive property – it tells us that if one number divides second number which divides third number, then the first number will also divide third number. and
## The remainder theorem
For real number a and whole number b exist unique whole numbers q and r such that
, where .
Proof. We’ll divide this proof into two parts. First part is in which we will prove existence and second in which we will prove uniqueness.
1. Existence: let’s observe rational number and define number q such that
If we subtract q from these inequalities we’ll get:
Now we can define number r:
This means that
If we multiply the inequality with a we get:
Since we defined r as we now have which proves existence.
2. Uniqueness: here we’ll assume there exist whole numbers and such that
This would mean that .
. If we assume that , then so is .
If .
From (by applying absolute value) we get that
which is in contradiction with the statement that .
This means that if and only if . We say that r is the remainder in division of number b with number a, and q is the quotient of integer .
### Polynomials
The Remainder Theorem for polynomials an application of Euclidean division on polynomials. Let’s explain it closer…
When we divide any polynomial f(x) by a linear polynomial we get another polynomial:
The point of this theorem is realizing what this remainder r really is. Let’s try to replace x with c and see what happens.
When we divide a polynomial by a linear polynomial , where , the remainder r is equal to f(c).
This theorem is very useful, especially if we want to know whether one polynomial is divisible by another, linear polynomial. If the remainder is equal to zero, then he is divisible, if it is any other number, he is not.
Example 5. Is a polynomial divisible by divisible by .
We have a linear polynomial which means that our c is equal to – 2. Now we calculate .
which means that polynomial f(x) is not divisible by (x + 2).
Shares |
Section 3.1
1 / 14
# Section 3.1
## Section 3.1
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Section 3.1 Sets and their operation
2. Definitions A set S is collection of objects. These objects are said to be members or elements of the set, and the shorthand for writing “x is an element of S” is “x S.” The easiest way to describe a set is by simply listing its elements (the “roster method”). For example, the collection of odd one-digit numbers could be written {1, 3, 5, 7, 9}. Note that this is the same as the set {9, 7, 5, 3, 1} since the order elements are listed does not matter in a set.
3. Examples The elements of a set do not have to be numbers as the following examples show: • {Doug, Amy, John, Jessica} • {TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF} • { {A,B}, {A,C}, {B,C} } • { }
4. Common sets of numbers Page 182 • N … set of natural numbers {0, 1, 2, …} • Z … set of integers {…, -2, -1, 0, 1, 2, …} • Q … set of rational numbers • R … set of real numbers
5. Definitions If A and B are sets, then the notation AB (read “A is a subset of B”) means that every element of set A is also an element of set B. Practice. Which is true? • {1, 2, 3, 4} {2, 3, 4} • ZQ • ZN • { } {a, b, c} • {3, 5, 7} {2, 3, 5, 7, 11} • {a, b} { {a, b}, {a, c}, {b, c} } • {a} { {a, b}, {a, c}, {a, b, c} }
6. Set notation Large sets cannot be listed in this way so we need the more compact “set-builder” notation. This comes in two types exemplified by the following: • (Property) {nZ : n is divisible by 4} • (Form) {4k : kZ}
7. Practice with property description List five members of each of the following sets: • {nN : n is an even perfect square } • {xZ : x – 1 is divisible by 3 } • {rQ : r2 < 2 } • {xR : sin(x) = 0 }
8. Practice with form description List five members of each of the following sets: • { 3n2 : nZ } • { 4k + 1 : kN } • { 3 – 2r : rQ and 0 r 5 }
9. Definitions of set operations Let A and B be sets with elements from a specified universal set U. • AB (read “A intersect B”) is the set of elements in both sets Aand B. • A B (read “A union B”) is the set of elements in either set Aor B. • A – B (read “A minus B”) is the set of elements in set Awhich are not in B. • A’ (read “the complement of A”) is the set of elements in the universe Uwhich are not in A.
10. Practice with set operations Let A = {1, 3, 5, 7, 9}, B = {2, 4, 6, 8, 10}, C = {2, 3, 5, 7}, D = {6, 7, 8, 9, 10} be sets with elements from the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Find each of the following: • A C • B D • B – D • B’ • (A B) – C • (A C) B • B’ C’ • (B C)’ • (C D) – A • B D’
11. Venn diagrams
12. Inclusion-Exclusion Principle The notation n(A) means “the number of elements of A.” For example, if A = {2, 3, 6, 8, 9}, then n(A) = 5. Principle of Inclusion/Exclusion for two sets A and B: n(AB) = n(A) + n(B) – n(AB)
13. Inclusion-Exclusion Principle Example. A = { 2, 4, 6, 8, …, 96, 98, 100 } and B = { 5, 10, 15, 20, …, 90, 95, 100} n(AB) = n(A) + n(B) – n(AB) = 50 + 20 – 10 = 60
14. Inclusion-Exclusion Principle Principle of Inclusion/Exclusion for three sets A, B, and C: n(AB C) = n(A) + n(B) + n(C) – n(AB) – n(AC) – n(BC) + n(ABC) |
# Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.3
Last Updated : 09 Mar, 2021
### Question 1. = 1
Solution:
Since denominator of x2/36 is larger than the denominator of y2/16,
the major axis is along the x-axis.
On comparing the given equation with = 1, we get
a2 = 36 and b2 = 16
⇒ a = ±6 and b = ±4
The Foci:
Foci = (c, 0) and (-c, 0) when (a2 > b2)
c = √(a2 – b2) -(when a2 > b2)
c = √(36 – 16)
c = √20 = 2√5
⇒ (2√5, 0) and (-2√5, 0)
Vertices:
Vertices = (a, 0) and (-a, 0) when (a2 > b2)
⇒ (6, 0) and (-6, 0)
Length of major axis:
Length of major axis = 2a (when a2 > b2)
= 2 × 6
⇒ Length of major axis = 12
Length of minor axis:
Length of minor axis = 2b (when a2 > b2)
= 2 × 4
⇒ Length of minor axis = 8
Eccentricity
Eccentricity = c/a (when a2 > b2)
= 2√5/6
= √5/3
Length of the latus rectum:
Length of the latus rectum = 2b2/a (when a2 > b2)
= 2×16/6
= 16/3
### Question 2. = 1
Solution:
Since denominator of y2/25 is larger than the denominator of x2/4,
the major axis is along the y-axis.
Comparing the given equation with = 1, we get
a2 = 4 and b2 = 25
⇒ a = ±2 and b = ±5
The Foci:
Foci = (0, c) and (0, -c) when (a2 < b2)
c = √(b2 – a2) -(when a2 < b2)
c = √(25 – 4)
c = √21
⇒ (0, √21) and (0, -√21)
Vertices:
Vertices = (0, b) and (0, -b) when (a2 < b2)
⇒ (0, 5) and (0, -5)
Length of major axis:
Length of major axis = 2b (when a2 < b2)
= 2 × 5
⇒ Length of major axis = 10
Length of minor axis:
Length of minor axis = 2a (when a2 < b2)
=2 × 2
⇒ Length of minor axis = 4
Eccentricity:
Eccentricity = c/b (when a2 < b2)
= √21/5
Length of the latus rectum:
Length of the latus rectum = 2a2/b (when a2 < b2)
= 2×4/5
= 8/5
### Question 3. = 1
Solution:
Since denominator of x2/16 is larger than the denominator of y2/9,
the major axis is along the x-axis.
Comparing the given equation with = 1, we get
a2 = 16 and b2 = 9
⇒ a = ±4 and b = ±3
The Foci:
Foci = (c, 0) and (-c, 0) when (a2 > b2)
c = √(a2 – b2) -(when a2 > b2)
c = √(16 – 9)
c = √7
⇒ (√7, 0) and (-√7, 0).
Vertices:
Vertices = (a, 0) and (-a, 0) when (a2 > b2)
⇒ (4,0) and (-4,0)
Length of major axis:
Length of major axis = 2a (when a2 > b2)
= 2 × 4
⇒ Length of major axis = 8
Length of minor axis:
Length of minor axis = 2b (when a2 > b2)
= 2 × 3
⇒ Length of minor axis = 6
Eccentricity:
Eccentricity = c/a (when a2>b2)
= √7/4
Length of the latus rectum:
Length of the latus rectum = 2b2/a (when a2 > b2)
= 2 × 9/4
= 9/2
### Question 4. = 1
Solution:
Since denominator of y2/100 is larger than the denominator of x2/25,
the major axis is along the y-axis.
Comparing the given equation with = 1, we get
a2 = 25 and b2 = 100
⇒ a = ±5 and b = ±10
The Foci:
Foci = (0, c) and (0, -c) when (a2 < b2)
c = √(b2 – a2) -(when a2 < b2)
c = √(100 – 25)
c = √75
c = 5√3
⇒ (0, 5√3) and (0, -5√3)
Vertices:
Vertices = (0, b) and (0, -b) when (a2 < b2)
⇒ (0, 10) and (0, -10)
Length of major axis:
Length of major axis = 2b (when a2 < b2)
= 2 × 10
⇒ Length of major axis = 20
Length of minor axis:
Length of minor axis = 2a (when a2 < b2)
= 2 × 5
⇒ Length of minor axis = 10
Eccentricity:
Eccentricity = c/b (when a2 < b2)
= 5√3/10
= √3/2
Length of the latus rectum:
Length of the latus rectum = 2a2/b (when a2 < b2)
= 2 × 25/10
= 5
### Question 5. = 1
Solution:
Since denominator of x2/49 is larger than the denominator of y2/36,
the major axis is along the x-axis.
Comparing the given equation with = 1, we get
a2 = 49 and b2 = 36
⇒ a = ±7 and b = ±6
The Foci:
Foci = (c, 0) and (-c, 0) when (a2 > b2)
c = √(a2 – b2) -(when a2 > b2)
c = √(49 – 36)
c = √13
⇒ (√13, 0) and (-√13, 0).
Vertices:
Vertices = (a, 0) and (-a, 0) when (a2 > b2)
⇒ (7, 0) and (-7, 0)
Length of major axis:
Length of major axis = 2a (when a2 > b2)
= 2 × 7
⇒ Length of major axis = 14
Length of minor axis:
Length of minor axis = 2b (when a2 > b2)
= 2 × 6
⇒ Length of minor axis = 12
Eccentricity:
Eccentricity = c/a (when a2 > b2)
= √13/7
Length of the latus rectum:
Length of the latus rectum = 2b2/a (when a2 > b2)
= 2 × 36/7
= 72/7
### Question 6. = 1
Solution:
Since denominator of y2/400 is larger than the denominator of x2/100,
the major axis is along the y-axis.
Comparing the given equation with = 1, we get
a2 = 100 and b2 = 400
⇒ a = ±10 and b = ±20
The Foci:
Foci = (0, c) and (0, -c) when (a2 < b2)
c = √(b2 – a2) -(when a2 < b2)
c = √(400 – 100)
c = √300
c = 10√3
⇒ (0, 10√3) and (0, -10√3)
Vertices:
Vertices = (0, b) and (0, -b) when (a2 < b2)
⇒ (0, 20) and (0, -20)
Length of major axis:
Length of major axis = 2b (when a2 < b2)
= 2 × 20
⇒ Length of major axis = 40
Length of minor axis:
Length of minor axis = 2a (when a2 < b2)
= 2 × 10
⇒ Length of minor axis = 20
Eccentricity:
Eccentricity = c/b (when a2 < b2)
= 10√3/20
= √3/2
Length of the latus rectum:
Length of the latus rectum = 2a2/b (when a2 < b2)
= 2×100/20
= 10
### Question 7. 36x2 + 4y2 = 144
Solution:
36x2 + 4y2 = 144
Dividing LHS and RHS by144,
= 1 (Obtained Equation)
Since denominator of y2/36 is larger than the denominator of x2/4,
the major axis is along the y-axis.
Comparing the given equation with = 1, we get
a2 = 4 and b2 = 36
⇒ a = ±2 and b = ±6
The Foci:
Foci = (0, c) and (0, -c) when (a2 < b2)
c = √(b2 – a2) -(when a2 < b2)
c = √(36 – 4)
c = √32
c = 4√2
⇒ (0, 4√2) and (0, -4√2)
Vertices:
Vertices = (0, b) and (0, -b) when (a2 < b2)
⇒ (0, 6) and (0, -6)
Length of major axis:
Length of major axis = 2b (when a2<b2)
= 2 × 6
⇒ Length of major axis = 12
Length of minor axis
Length of minor axis = 2a (when a2 < b2)
= 2 × 2
⇒ Length of minor axis = 4
Eccentricity:
Eccentricity = c/b (when a2 < b2)
= 4√2/6
= 2√2/3
Length of the latus rectum:
Length of the latus rectum = 2a2/b (when a2 < b2)
= 2 × 4/6
= 4/3
### Question 8. 16x2 + y2 = 16
Solution:
16x2 + y2 = 16
Dividing LHS and RHS by16,
= 1 (Obtained Equation)
Since denominator of y2/16 is larger than the denominator of x2/1,
the major axis is along the y-axis.
Comparing the given equation with = 1, we get
a2 = 1 and b2 = 16
⇒ a = ±1 and b = ±4
The Foci:
Foci = (0, c) and (0, -c) when (a2 < b2)
c = √(b2 – a2) -(when a2 < b2)
c = √(16 – 1)
c = √15
⇒ (0, √15) and (0, -√15)
Vertices:
Vertices = (0, b) and (0, -b) when (a2 < b2)
⇒ (0, 4) and (0, -4)
Length of major axis:
Length of major axis = 2b (when a2 < b2)
= 2 × 4
⇒ Length of major axis = 8
Length of minor axis:
Length of minor axis = 2a (when a2 < b2)
=2 × 1
⇒ Length of minor axis = 2
Eccentricity:
Eccentricity = c/b (when a2 < b2)
= √15/4
Length of the latus rectum:
Length of the latus rectum = 2a2/b (when a2 < b2)
= 2 × 1/4
= 1/2
### Question 9. 4x2 + 9y2 = 36
Solution:
4x2 + 9y2 = 36
Dividing LHS and RHS by 36,
= 1 (Obtained Equation)
Since denominator of x2/9 is larger than the denominator of y2/4,
the major axis is along the x-axis.
Comparing the given equation with = 1, we get
a2 = 9 and b2 = 4
⇒ a = ±3 and b = ±2
The Foci:
Foci = (c, 0) and (-c, 0) when (a2 > b2)
c = √(a2 – b2) -(when a2 > b2)
c = √(9 – 4)
c = √5
⇒ (√5, 0) and (-√5, 0).
Vertices
Vertices = (a, 0) and (-a, 0) when (a2 > b2)
⇒ (3, 0) and (-3, 0).
Length of major axis
Length of major axis = 2a (when a2 > b2)
= 2 × 3
⇒ Length of major axis = 6
Length of minor axis
Length of minor axis = 2b (when a2 > b2)
= 2 × 2
⇒ Length of minor axis = 4
Eccentricity
Eccentricity = c/a (when a2 > b2)
= √5/3
Length of the latus rectum
Length of the latus rectum = 2b2/a (when a2 > b2)
= 2 × 4/3
= 8/3
### Question 10. Vertices (± 5, 0), foci (± 4, 0).
Solution:
Since the vertices are on x-axis, the equation will be of the form
= 1, where a is the semi-major axis. (where a2 > b2)
Given that a = ±5, c = ±4
As, from the relation
c2 = a2 – b2 (when a2 > b2)
b2 = a2 – c2
b2 = 25 – 16
b2 = 9
So, a2 = 25 and b2 = 9
Hence, the required equation of ellipse,
= 1
### Question 11. Vertices (0, ± 13), foci (0, ± 5).
Solution:
Since the vertices are on y-axis, the equation will be of the form
= 1, where b is the semi-major axis. (where a2 < b2)
Given that b = ±13, c = ±5
As, from the relation
c2 = b2 – a2 (when a2 < b2)
a2 = b2 – c2
a2 = 169 – 25
a2 = 144
So, a2 = 144 and b2 = 169
Hence, the required equation of ellipse,
= 1
### Question 12. Vertices (± 6, 0), foci (± 4, 0).
Solution:
Since the vertices are on x-axis, the equation will be of the form
= 1, where a is the semi-major axis. (where a2 > b2)
Given that a = ±6, c = ±4
As, from the relation
c2 = a2 – b2 (when a2 > b2)
b2 = a2 – c2
b2 = 36 – 16
b2 = 20
So, a2 = 36 and b2 = 20
Hence, the required equation of ellipse,
= 1
### Question 13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2).
Solution:
Since the major axis are on x-axis, and minor axis on the y-axis, the equation will be of the form
= 1, where a is the semi-major axis. (where a2 > b2)
Given that a = ±3, b = ±2
So, a2 = 9 and b2 = 4
Hence, the required equation of ellipse,
= 1
### Question 14. Ends of major axis (0, ±√5), ends of minor axis (± 1, 0).
Solution:
Since the major axis are on y-axis, and minor axis on the x-axis, the equation will be of the form
= 1, where b is the semi-major axis. (where a2 < b2)
Given that a = ±1, b = ±√5
So, a2 = 1 and b2 = 5
Hence, the required equation of ellipse,
= 1
### Question 15. Length of major axis 26, foci (± 5, 0).
Solution:
Since the foci are on x-axis, the equation will be of the form
= 1, where a is the semi-major axis. (where a2 > b2)
Given that c = ±5 and Length of major axis = 26
As, Length of major axis = 2a (when a2 > b2)
2a = 26
a = 13
As, from the relation
c2 = a2 – b2 (when a2 > b2)
b2 = a2 – c2
b2 = 169 – 25
b2 = 144
So, a2 = 169 and b2 = 144
Hence, the required equation of ellipse,
= 1
### Question 16. Length of minor axis 16, foci (0, ± 6).
Solution:
Since the foci are on y-axis, the equation will be of the form
= 1, where b is the semi-major axis. (where a2 < b2)
Given that c = ±6 and Length of minor axis = 16
As, Length of minor axis = 2a (when a2 < b2)
2a = 16
a = 8
As, from the relation
c2 = b2 – a2 (when a2 < b2)
b2 = c2 + a2
b2 = 36 + 64
b2 = 100
So, a2 = 64 and b2 = 100
Hence, the required equation of ellipse,
= 1
### Question 17. Foci (± 3, 0), a = 4.
Solution:
Since the foci are on x-axis, the equation will be of the form
= 1, where a is the semi-major axis. (where a2 > b2)
Given that a = 4 and c = ±3
As, from the relation
c2 = a2 – b2 (when a2 > b2)
b2 = a2 – c2
b2 = 16 – 9
b2 = 7
So, a2 = 16 and b2 = 7
Hence, the required equation of ellipse,
= 1
### Question 18. b = 3, c = 4, centre at the origin; foci on the x axis.
Solution:
Since the foci are on x-axis, the equation will be of the form
= 1, where a is the semi-major axis. (where a2 > b2)
Given that b = 3 and c = 4
As, from the relation
c2 = a2 – b2 (when a2 > b2)
a2 = b2 + c2
a2 = 9 + 16
a2 = 25
So, a2 = 25 and b2 = 9
Hence, the required equation of ellipse,
= 1
### Question 19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
Solution:
The standard equation of ellipse having centre (0, 0) will be of the form
= 1
Since the points (3, 2) and (1, 6) lie on the ellipse, we can have
= 1
= 1 -(1)
= 1
= 1 -(2)
Eq(2) subtracted from (multiplying eq(1) by 9) and, we get
9×() – () = 9 – 1
= 8
80/a2 = 8
a2 = 80/8
a2 = 10
Now, substituting a2 = 10 in eq(1)
= 1
9/10 + 4/b2 = 1
4/b2 = 1 – 9/10
4/b2 = 1/10
b2 = 10 × 4 = 40
So, a2 = 10 and b2 = 40
Hence, the required equation of ellipse,
= 1
### Question 20. Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Solution:
The standard equation of ellipse having centre (0, 0) will be of the form
= 1
Since the points (4,3) and (6,2) lie on the ellipse, we can have
= 1
= 1 -(1)
and, = 1
= 1 -(2)
(multiplying eq(2) by 9) subtracted from (multiplying eq(1) by 4) and, we get
9×( ) – 4×() = 9 – 4
= 5
260/a2 = 5
a2 = 260/5
a2 = 52
Now, substituting a2 = 52 in eq(1)
= 1
9/b2 = 1 – 16/52
9/b2 = 36/52
9/b2 = 36/52
b2 = 9 × 36/52 = 13
So, a2 = 52 and b2 = 13
Hence, the required equation of ellipse,
= 1
Previous
Next |
Maths How To with Anita
# 9 mental math strategies: Tips and Tricks for students
Mental math is an important skill for students of all ages. Applying mental math strategies can help you work faster and more accurately without a calculator.
In this article, we will list 9 strategies that will help you improve your skills. We’ll also include some tips and tricks to make mental math easier and faster!
From my 14+ years of teaching experience, I have observed that there are many students in high school that rely heavily on their calculators instead of using these mental math strategies.
By applying these tips and tricks they can work faster and apply these strategies to other topics such as Algebra and multiplying fractions.
## What are mental math strategies?
Mental math strategies are simply methods or techniques that you can use to do math more quickly and accurately using your brain. These strategies can be used for addition, subtraction, multiplication, division, and more!
Here are my favorite mental math strategies that you can use:
### 1. Bridge to ten
Bridge to ten is when we count on to the next 10 and then add what is left.
Having knowledge of ‘friends of ten‘ which is basically recalling number combinations that add to 10, is essential prior knowledge.
The Bridge to ten strategy is great for simple additions such as 8+6.
You can change this to 8+ 2 + 4.
It is also great for questions such as 47 + 8.
Which can be bridged to ten as follows: 47 + 3 + 5.
Number lines are a helpful tool for this strategy.
### 2. Commutative Property
This mental math strategy can be used with addition and multiplication. It states that you can change the order of the numbers being added or multiplied and get the same answer. For example, 2+3 = 3+2.
Let’s say you’re adding up a list of numbers:
Instead of starting from left to right, try grouping them using ‘friends of ten’.
So you can see the total is 20 + 30 + 4 = 54
This can help you keep track of the numbers more easily and calculate the addition faster using this mental math strategy.
This mental math strategy can be used for addition. To use this strategy, you simply need to add 10 and then subtract 1.
For example, let’s say you’re trying to add 56+9
You can add the numbers like this:
Then, it becomes much easier to calculate the addition mentally because it is very easy to add 10 to a number.
### 4. Near doubles
Near doubles strategy is when you double a number and then adjust.
This is great when adding two consecutive numbers.
For example 7 + 8 is the same as double 7 plus 1, which equals 15.
Or you can adjust by doubling the larger number and subtracting 1.
For example 14+15 is the same as double 15 minus 1 , which equals 29.
### 5. Compensation strategy
The compensation strategy uses rounding up or down to make it easier to calculate an addition mentally.
First you want to round the second number up to the closest ten.
Then you compensate by subracting.
For example 47+19
This works for subtraction too.
For example, 76- 29
### 6. Doubling and halving
I LOVE this strategy! It is genius.
Doubling and halving is a mental math strategy for multiplication.
It works by halving one number (the larger one works best) and doubling the other number.
For example 48 x 5
### 7. Distributive property
The distributive property states that when you are multiplying a number by a certain sum or difference, you can multiply the number by each term in the sum or difference and then add the products together.
For example:
10 x (24 + 16) = 10 x 24 + 10 x 16 = 240 + 160 =400
Algebraically the distributive law looks like this:
This is a great mental math strategy to use for something like 99x 4
### 8. Using landmark numbers
Landmark numbers, such as multiples of ten or a hundred, are familiar with students so they can be used as a mental math strategy when adding.
For example 97 + 68.
97 is so close to 100. So you could add 3 to 97 and then subtract 3 from 68.
### 9. Repeated doubling to multiply by 4 and 8
Students often know how to double or multiply by 2 but their number facts for the 4 and 8 times tables are often not as strong.
Repeated doubling is a mental math strategy to help with this.
To multiply a number by four, double it twice.
For example, 12 x 4
To multiply a number by 8, double it three times.
For example 25 x 8
## Final thoughts and my experience of using mental math strategies in the classroom
There are many more mental math strategies that I use regularly. These are just a few that I teach my high school mathematics students in our numeracy support sessions.
In my experience students need to be reminded of these strategies and given opportunities to practice them regularly.
As you can see, using mental math strategies can help you work faster and more accurately without a calculator. Try out these strategies the next time you’re doing your math homework.
## FAQs
### What are some mental addition strategies?
Some mental addition strategies include bridge to ten, commutative property, using landmark numbers, adding 9, and near doubles. Examples of these mental math strategies are outlined in this article. |
# Permutation And Combination: 7 Complete Quick Facts
## Properties of Permutation and combination
When discussing permutation and combination as we are dealing with selection and arrangement with or without order considerations, depending on the situation there are different types and properties for the permutation and combination, these differences among permutations and combinations we will explain here with justified examples.
## permutations without repetition
This is the normal permutation which arranges n objects taken r at a time i.e nPr
n Pr=n!/(n-r)!
number of orderings of n different objects taken all at a time n Pn =n!
nP0 = n!/n!=1
nPr = n.n-1Pr-1
0!=1
1/(-r)!= 0 or (-r)!=∞
## permutations with repetition
Number of permutations (arrangements) for different items, taken r at a time, where each item can happen once, twice, three times, …….. r-times as many in any arrangement = Number of ways to fill r areas where each item can be filled with any of the n items.
The number of permutations = The number of ways of filling r places = (n)r
The number of orders that can be organized using n objects out of which p are alike (and of one kind) q are alike (and of another kind), r are similar (and of another kind) and the rest are distinct is nPr =n!/(p!q!r!)
Example:
In how many ways can 5 apples be allocated among four boys when every boy can take one or more apples.
Solution: This is the example of permutation with repetition as we know that for such cases we have
The number of permutations = The number of ways of filling r places =nr
Required number of ways are 45 =10, Since each apple can be distributed in 4 ways.
Example: Find the number of words can be organized with the letters of the word MATHEMATICS by regrouping them.
Solution: Here we can observe that there are 2 M’s, 2 A’s and 2T’s this is the example of permutation with repetition
=n!/(p!q!r!)
Required number of ways are =11!/(2!2!2!)=4989600
Example: How many ways in which the number of tails equal to the number of heads if six identical coins are arranged in a row.
Solution: Here we can observe that
No. of tails =3
And since coins are identical this is the example of permutation with repetition =n!/(p!q!r!)
Required number of ways =6!/(3!3!)= 720/(6X6)=20
## Circular permutation:
In circular permutation, most importantly the ordering of the object is respect to the others.
So, in circular permutation, we adjust the position of one object and arrange the other objects in all directions.
Circular permutation is split into two ways:
(i) Circular permutation where clockwise and anti-clockwise settings suggest different permutation, e.g. Arrangements for seating people around the table.
(ii) Circular permutation where clockwise and anti-clockwise settings display same permutation, e.g. arranging certain beads to create a necklace.
## Clockwise and anti-clockwise arrangement
If the anti-clockwise and clockwise order and movement are not different eg, bead arrangement in necklace, flower arrangement in garland etc, then the number of circular permutations of n distinct items is (n-1)!/2
1. The number of circular permutation for n different items, taken r at a time, when the orders for the clockwise and the anti-clockwise are considered to be different by nPr /r
2. The number of circular permutation for n different items, taken r at a time, when clockwise and anti-clockwise orders are not different from nPr /2r
3. The number of circular permutations of n different objects are (n-1)!
4. The number of ways in which n different boys can be seated round a circular table is (n-1)!
5. The number of ways in which n different gems can be set up to form a necklet, is (n-1)!/2
Example:
How many ways can five keys be placed in the ring
Solution:
Since clockwise and anticlockwise are same in case of ring.
If the anti-clockwise and clockwise sequence and movement are not different then the number of circular permutations of n distinct items is
=(n-1)!/2
Required number of ways = (5-1)!/2= 4!/2=12
Example:
What would be the number of arrangements, If eleven members of a committee sit at a round table so that the President and Secretary always sit together.
Solution:
By fundamental property of circular permutation
The count of circular permutations of n different things are (n-1)!
Since two positions are fix so we have
Required number of ways (11-2)!*2=9!*2=725760
Example: What would be number of ways 6 men and 5 women can eat at a round table if no two women can sit together
Solution: By fundamental property of circular permutation.
The count of circular permutations of n different things are (n-1)!
Number of ways in which 6 men can be arranged at a round table = (6 – 1)! =5!
Now women can be arranged in 6! ways and Total Number of ways = 6! × 5!
## Combinations without repetition
This is the usual Combination which is “The number of combinations (selections or groups) that can be formed from n different objects taken r at at a time is nCr =n!/(n-r)!r!
Also nCr =nCr-r
n Pr /r! =n!/(n-r)! =nCr
Example: Find the number of options to fill 12 vacancies if there are 25 candidates and five of them are from the scheduled category, provided that 3 vacancies are reserved for the S.C candidates mean while the remaining are open to everyone.
Solution: Since 3 vacant positions are filled from 5 applicants in 5 C3 ways (i.e 5 CHOOSE 3) and now remaining candidates are 22 and remaining seats are 9 so it would be 22C9 (22 CHOOSE 9) The selection can be made in 5 C3 X 22C9 ={5!/3!(5-3)! }X{22!/9!(22-9)!}
5 C3 X 22C9 = {(3!X4X5)/(3!X2!)}X {22!/(9!X13!)}=4974200
So the selection can be made in 4974200 ways.
Example: There are 10 candidates and three vacancies in the election. in how many ways a voter can cast his or her vote?
Solution: Since there are only 3 vacancies for 10 candidates so this is the problem of 10 CHOOSE 1, 10 CHOOSE 2, and 10 CHOOSE 3 Examples,
A voter can vote in 10C1+10C2+10C3 = {10!/1!(10-1)!}+{10!/2!(10-2)!}+{10!/3!(10-3)!} =10+45+120= 175 ways.
So in 175 ways voter can vote.
Example:There are 9 chairs in a room for 4 people, one of which is a single-seat guest with one specific chair. How many ways can they sit?
Solution: Since 3 chairs can be select in 8C3 and then 3 persons can be arranged in 3! ways.
3 persons are to be seated on 8 chairs 8C3 (i.e 8 CHOOSE 3) arrangement
=8C3 X3! = {8! /3!(8-3)!} X3!
=56X6=336
In 336 ways they can sit.
Example: For five men and 4 women, a group of 6 will be formed. In how many ways this can be done so that the group has more men.
Solution: Here the problem include different combinations like 5 CHOOSE 5, 5 CHOOSE 4, 5 CHOOSE 3 for men and for women it include 4 CHOOSE 1, 4 CHOOSE 2 and 4 CHOOSE 3 as given in the followings
1 woman and 5 men =4C1 X 5C5 ={4!/1!(4-1)!} X{5!/5!(5-5)!}=4
2 women and 4 men =4C2 X 5C4 = {4!/2!(4-2)!} X{5!/4!(5-4)!}=30
3 women and 3 men =4C3 X 5C3 = {4!/3!(4-3)!} X {5!/3!(5-3)!} =40
Hence total ways = 4+30+40=74.
Example: The number of ways 12 boys can travel in three cars so that 4 boys in each car, assuming that three particular boys will not go in the same car.
Solution: First omit three particular boys, remaining 9 boys may be 3 in each car. This can be done in 9 CHOOSE 3 i.e 9C3 ways,
The three particular boys may be placed in three ways one in each car. Therefore total number of ways are =3X9C3.
={9!/3!(9-3)!}X3= 252
so in 252 ways they can be placed.
Example: How many ways did 2 green and 2 black balls come out of a bag containing 7 green and 8 black balls?
Solution: Here bag contains 7 green from that we have to choose 2 so it is 7 CHOOSE 2 problem and 8 black balls from that we have to choose 2 so it is 8 CHOOSE 2 problem.
Hence the Required number = 7C2 X 8C2 = {7!/2!(7-2)!}X{8!/2!(8-2)!}=21X28=588
so in 588 ways we can select 2 green and 2 black from that bag.
Example: Twelve different characters of English words are provided. From these letters, 2 alphabetical names are formed. How many words can be created when at least one letter is repeated.
Solution: here we have to choose 2 letter words from 12 letters so it is 12 CHOOSE 2 problem.
No. of words of 2 letters in which letters have been recurrent any times = 122
But no. of words on having two different letters out of 12 =12C2 = {12!/2!(12-2)!} =66
Required number of words = 122-66=144-66=78.
Example: There are 12 points on the plane where six are collinear, then how many lines can be drawn by joining these points.
Solution: For 12 points in a plane to make line we require 2 points same for six collinear points, so this is 12 CHOOSE 2 and 6 CHOOSE 2 problem.
The number of lines is = 12C26C2 +1={12!/2!(12-2)!}-{6!/2!(6-2)!}+1 =66-15+1=52
So in 52 number of ways it lines can be drawn.
Example: Find the numeral of ways in which a 6-member cabinet can be set up from 8 gentlemen and 4 ladies so that the cabinet consists of at least 3 ladies.
Solution: For forming the committee, we can choose from 3 each men and women and 2men 4 women so the problem includes 8 CHOOSE 3, 4 CHOOSE 3, 8 CHOOSE 2 and 4 CHOOSE 4.
Two types of cabinet can be formed
(i) Having 3 men and 3 ladies
(ii) Having 2 men and 4 ladies
Possible no. of ways = (8C3 X 4C3) + (8C2 X4C4)= {8!/3!(8-3)!}X{4!/3!(4-3)!} +{8!/2!(8-2)!}X{4!/4!(4-4)!} = 56X4+ 28X1 =252
So In 252 ways we can form such cabinet.
These are some examples where we can compare the situation of nPr vs nCr in the case of permutation, the way things are organized is important. However in Combination the order means nothing.
## Conclusion
A brief description of Permutation and combination when repeated and non-repeated with the basic formula and important results is provided in the form of real examples, in this series of articles we will discuss in detail the various outcomes and formulas with relevant examples, if you want to continue reading:
SCHAUM’S OUTLINE OF Theory and Problems of DISCRETE MATHEMATICS
https://en.wikipedia.org/wiki/Permutation
https://en.wikipedia.org/wiki/Combination |
Definite Integral - Definition, Formulas, Properties & Solved Examples - GMS - Learning Simply
Students' favourite free learning app with LIVE online classes, instant doubt resolution, unlimited practice for classes 6-12, personalized study app for Maths, Science, Social Studies, video e-learning, online tutorial, and more. Join Telegram
# Definite Integral - Definition, Formulas, Properties & Solved Examples
Definite integrals are used when the limits are defined to generate a unique value.
Scroll Down and click on Go to Link for destination
# Definite Integral - Calculus
Before going to learn about definite integrals, first, recollect the concept of integral. An integral assigns numbers to functions in mathematics to define displacement, area, volume, and other notions that arise by connecting infinitesimal data. The process of finding integrals is called integration. Definite integrals are used when the limits are defined to generate a unique value. Indefinite integrals are implemented when the boundaries of the integrand are not specified. In case, the lower limit and upper limit of the independent variable of a function are specified, its integration is described using definite integrals. Also, we have several integral formulas to deal with various definite integral problems in maths.
Definite Integral - Definition, Formulas, Properties & Solved Examples
## Definite Integral Definition
The definite integral of a real-valued function f(x) with respect to a real variable x on an interval [a, b] is expressed as
Here,
∫ = Integration symbol
a = Lower limit
b = Upper limit
f(x) = Integrand
dx = Integrating agent
Thus, ∫ab f(x) dx is read as the definite integral of f(x) with respect to dx from a to b.
∫ab f(x) dx is read as the definite integral of f(x) with respect to dx from a to b.
## Definite Integral as Limit of Sum
The definite integral of any function can be expressed either as the limit of a sum or if there exists an antiderivative F for the interval [a, b], then the definite integral of the function is the difference of the values at points a and b. Let us discuss definite integrals as a limit of a sum. Consider a continuous function f in x defined in the closed interval [a, b]. Assuming that f(x) > 0, the following graph depicts f in x.
Definite Integral as Limit of Sum
The integral of f(x) is the area of the region bounded by the curve y = f(x). This area is represented by the region ABCD as shown in the above figure. This entire region lying between [a, b] is divided into n equal subintervals given by [x0, x1], [x1, x2], …… [xr-1, xr], [xn-1, xn].
Let us consider the width of each subinterval as h such that h → 0, x= a, x1 = a + h, x2 = a + 2h,…..,xr = a + rh, xn = b = a + nh
and n = (b – a)/h
Also, n→∞ in the above representation.
Now, from the above figure, we write the areas of particular regions and intervals as:
Area of rectangle PQFR < area of the region PQSRP < area of rectangle PQSE ….(1)
Since. h→ 0, therefore xr– xr-1→ 0. The following sums can be established as;
From the first inequality, considering any arbitrary subinterval [xr-1, xr] where r = 1, 2, 3….n, it can be said that sn< area of the region ABCD <Sn
Since, n→∞, the rectangular strips are very narrow, it can be assumed that the limiting values of sn and Sn are equal, and the common limiting value gives us the area under the curve, i.e.,
From this, it can be said that this area is also the limiting value of an area lying between the rectangles below and above the curve. Therefore,
This is known as the definition of definite integral as the limit of sum.
## Definite Integral Properties
Below is the list of some essential properties of definite integrals. These will help evaluate the definite integrals more efficiently.
• af(x) dx = ∫af(t) d(t)
• af(x) dx = – ∫bf(x) dx
• af(x) dx = 0
• af(x) dx = ∫ac f(x) dx + ∫cb f(x) dx
• af(x) dx = ∫ab f(a + b – x) dx
• 0f(x) dx = f(a – x) dx
Steps for calculating ∫ab f(x) dx
Step 1: Find the indefinite integral ∫f(x) dx. Let this be F(x). There is no need to keep
integration constant C. This is because if we consider F(x) + C instead of F(x), we get
ab f(x) dx = [F(x) + C]ab = [F(b) + C] – [F(a) + C] = F(b) + C – F(a) – C = F(b) – F(a)
Thus, the arbitrary constant will not appear in evaluating the value of the definite integral.
Step 2: Calculate the value of F(b) – F(a) = [F(x)]ab
Hence, the value of ab f(x) dx = F(b) – F(a)
### Definite Integral by Parts
Below are the formulas to find the definite integral of a function by splitting it into parts.
• 02a f (x) dx = ∫0a f (x) dx + ∫0f (2a – x) dx
• 02a f (x) dx = 2 ∫0f (x) dx … if f(2a – x) = f (x).
• 02a f (x) dx = 0 … if f (2a – x) = – f(x)
• -aa f(x) dx = 2 ∫0f(x) dx … if f(- x) = f(x) or it is an even function
• -aa f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function
### Definite Integral Examples
Example 1:
Evaluate the value of ∫23 x2 dx.
Solution:
Let I = ∫2x2 dx
Now, ∫x2 dx = (x3)/3
Now, I = ∫2x2 dx = [(x3)/3]23
= (33)/3 – (23)/3
= (27/3) – (8/3)
= (27 – 8)/3
= 19/3
Therefore, ∫23 x2 dx = 19/3
Example 2:
Calculate: ∫0π/4 sin 2x dx
Solution:
Let I = ∫0 π/4 sin 2x dx
Now, ∫ sin 2x dx = -(½) cos 2x
I = ∫0 π/4 sin 2x dx
= [-(½) cos 2x]0π/4
= -(½) cos 2(π/4) – {-(½) cos 2(0)}
= -(½) cos π/2 + (½) cos 0
= -(½) (0) + (½)
= 1/2
Therefore, ∫0 π/4 sin 2x dx = 1/2
## Frequently Asked Questions on Definite Integral
### What is a definite integral?
The definite integral has a unique value. A definite integral is denoted by ∫ab f(x) dx, where a is called the lower limit of the integral and b is called the upper limit of the integral.
### What is the formula for definite integral?
The formula for calculating the definite integral of a function f(x) in the interval [a, b] is given by,
ab f(x) dx = F(b) – F(a)
### What is a definite integral used for?
We can use definite integrals to find the area under, over, or between curves in calculus. If a function is strictly positive, the area between the curve of the function and the x-axis is equal to the definite integral of the function in the given interval. In the case of a negative function, the area will be -1 times the definite integral.
### Can a definite integral be negative?
Yes, the value of a definite integral can be negative, positive or zero.
### Do definite integrals have C?
No, definite integrals do not have C. As it is not required to add an arbitrary constant, i.e. C in case of definite integrals.
At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio… |
#### Solve the following system of inequalities:
Given: 2x+1/7x-1 > 5
(2x+1)/(7x - 1) – 5 > 0 …….. (on subtracting 5 from both sides)
(2x + 1 – 35x + 5)/(7x – 1) > 0
(6 – 33x)/(7x – 1) >0
Now, either the numerator or the denominator should be greater than 0 or both should be less than 0 for the above fraction to be greater than 0. Thus,
6 – 33x > 0 & 7x – 1 > 0
33x <6 & 7x > 1
X < 2/11 & x > 1/7
i.e., 1/7 < x < 2/11 ……. (i)
Or,
6 – 33x < 0 & 7x – 1 < 0
33x > 6 & 7x < 1
X > 2/11 & x < 1/7
i.e., 2/11 < x < 1/7 ….. viz. impossible
Now,
(x + 7)/(x – 8) > 2 …… (given)
(x + 7)/(x – 8 )– 2 > 0 …… (on subtracting both sides by 2)
(x +7 – 2x + 16)/(x – 8) > 0
(23 – x)/(x – 8) > 0
Now, either the numerator or the denominator should be greater than 0 or both should be less than 0 for the above fraction to be greater than 0. Thus,
23 – x > 0 & x – 8 > 0
X < 23 & x > 8
i.e., 8 < x < 23 …….. (ii)
Or,
23 – x > 0 & 8 > 0
X > 23 & x < 8
i.e., 23 < x < 8 ….. viz. impossible
Therefore, from (i) & (ii) we can say that there is no solution satisfying both inequalities.
Thus, the system has no solution. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# CBSE Class 11 Maths Chapter 13 - Limits and Derivatives Formulas
Last updated date: 10th Sep 2024
Total views: 657.9k
Views today: 18.57k
## Limits and Derivatives Formula for CBSE Class 11 Maths - Free PDF Download
In Mathematics, limits and derivatives are extremely important concepts whose application is not only limited to mathematics but also present in other subjects such as physics. The full definitions of limits and derivatives, along with their properties and formulas, are discussed on this page. This idea is commonly explained in the syllabus of Class 11. Sir Issac Newton laid out the basic laws of differential calculus, based on his principles of rate and change, and integral calculus emerged as the reverse method.
Competitive Exams after 12th Science
## Download Class 11 Maths Chapter 13 - Limits and Derivatives Formulas PDF
### Introduction of Limits and Derivatives
Differentiation and calculus fundamentals serve as the basis for advanced mathematics, modern physics and many other modern science and engineering branches. For CBSE students, Class 11 Limits and Derivatives function as the entry point for calculus.
### Limits of a Function
In Mathematics, a limit is defined as a value approached as the input by a function, and it produces some value. In calculus and mathematical analysis, limits are important and are used to define integrals, derivatives, and continuity.
### Limits Formula
To express a function's limit, we represent it as:
$\lim_{x\to a}f(x)$
Left Hand and Right-Hand Limits
If the function values at the point very close to a, on the left tend to a definite unique number as $x$ tends to $a$, then the unique number so obtained is called the $f(x)$ left-hand limit at $x = a$, we write it as $x = a$.
$f(a-0) = \lim_{x\to a^{-}}f(x) = \lim_{h\to 0}f(a-h)$
Similarly, right hand limit is
$f(a+0) = \lim_{x\to a{+}}f(x) = \lim_{h\to 0}f(a+h)$
### Existence of Limit
$\lim_{x\to a}f(x)$ exists, if
(i) $\lim_{x\to a^{-}}f(x)$ and $\lim_{x\to a^{+}}f(x)$ both exists
(ii) $\lim_{x\to a^{-}}f(x) = \lim_{x\to a^{+}}f(x)$
### Properties of Limits
1. $lim_{x\to a}[p(x) + g(x)] = \lim_{x\to a}p(x) + \lim_{x\to a}g(x)$
2. $lim_{x\to a}[p(x) - g(x)] = \lim_{x\to a}p(x) - \lim_{x\to a}g(x)$
3. For every real number $k$,
$\lim_{x\to a}[kp(x)] = k \lim_{x\to a}p(x)$
4. $\lim_{x\to a}[p(x) q(x)] = \lim_{x\to a}p(x) \times \lim_{x\to a}q(x)$
5. $\lim_{x\to a}\dfrac{p(x)}{q(x)} = \dfrac{\lim_{x\to a}p(x)}{\lim_{x\to a}q(x)}$
Let two functions be $p$ and $q$ and $a$ value be such that
### Derivatives of a Function
A derivative corresponds, in comparison to the other, to the instantaneous rate of change of a quantity. This helps to explore the moment by moment essence of a quantity. A function's derivative is expressed in the formula given below.
### Derivative Formula
Assuming f is a real-valued function, then
$f^\prime (x) = \lim_{x\to a}\dfrac{f(x+h)-f(x)}{h}$ is called the derivative of $f$ at $x$ iff $\lim_{x\to a} \dfrac{f(x+h)-f(x)}{h}$ exists finitely.
Its derivative is said to be $f^\prime(x)$ for function $f$, provided that the above equation exists. Here, search all the derivative formulas relating to trigonometric functions, inverse functions, hyperbolic functions, etc.
### Properties of Derivatives
1. $\dfrac{d}{dx}[p(x) + q(x)] = \dfrac{d}{dx}(p(x)) + \dfrac{d}{dx}(q(x))$
2. $\dfrac{d}{dx}[p(x) - q(x)] = \dfrac{d}{dx}(p(x)) - \dfrac{d}{dx}(q(x))$
3. $\dfrac{d}{dx}[p(x) \times q(x)] = \dfrac{d}{dx}[p(x)] q(x) + p(x) \dfrac{d}{dx}[q(x)]$
4. $\dfrac{d}{dx}\left[\dfrac{p(x)}{q(x)}\right] = \dfrac{\dfrac{d}{dx}[p(x)]q(x) - p(x) \dfrac{d}{dx}[q(x)]}{(q(x))^2}$
### Here Are Some of the Essential Properties of Derivatives:
(i) $\dfrac{d}{dx}(x^n) = nx^{n-1}$
(ii) $\dfrac{d}{dx}(\sin x) = \cos x$
(iii) $\dfrac{d}{dx}(\cos x) = -\sin x$
(iv) $\dfrac{d}{dx}(\tan x) = \sec^2 x$
(v) $\dfrac{d}{dx}(\cot x) = -\text{cosec}^2\,x$
(iv) $\dfrac{d}{dx}(\sec x) = \sec x \tan x$
(v) $\dfrac{d}{dx}(\text{cosec }x) = -\text{cosec }x \cot x$
(vii) $\dfrac{d}{dx}(a^x) = a^x log_ea$
(ix) $\dfrac{d}{dx}(e^x) = e^x$
(x) $\dfrac{d}{dx}(log_ex) = \dfrac{1}{x}$
### Solved Examples On How To Solve Limits
You will come across the following types of limits examples along with step-by-step solutions in the limits question bank chapter provided by Vedantu.
Example: Identify the limit of the following expression?
$\lim_{x \to 5} \dfrac{x^2 - 5}{x^2 + x - 30}$
Solution:
The limit provided is the ratio of two polynomials, $x = 5$. This certainly makes both the numerator as well as the denominator equivalent to zero (0). We are required to factor both the numerator as well as denominator as shown below.
$\lim_{x \to 5} \dfrac{(x - 5)(x + 5)}{(x - 5)(x + 6)}$
Simplify the expression to get:-
$\lim_{x \to 5} \dfrac{x + 5}{x + 6} = \dfrac{10}{11}$
### Introduction to Limits by Factoring
Now, this particular method is quite an interesting way of solving limits. In these types of limits, if you try to substitute, you will obtain an indetermination. For example:
$\lim_{x \to 1} x^2{\dfrac{x^2 - 1}{x - 1}}$
Note that if you simply substitute x by 1 in the algebraic equation, you will have 0/0. So, what do you think can be done? We can take the help of our algebraic skills for the purpose of simplifying the expression. In the example quoted previously, we can factor the numerator:
$\lim_{x \to 1} x^2 \dfrac{x^2 - 1}{x-1} = \lim_{x \to 1} \dfrac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2$
You are going to find these types of numerical problems on limits easily whenever you notice a quotient of two polynomials. You could try your hands on this method given that there is an indetermination.
## FAQs on CBSE Class 11 Maths Chapter 13 - Limits and Derivatives Formulas
1. How do we solve limits?
Get to know the idea behind limits and problem-solving techniques. Following are the various Limits of a Function:
• Evaluate limits using direct substitution
• Evaluate limits using factoring and cancelling
• Evaluate limits by expanding and simplifying
• Evaluate limits by combining fractions
• Evaluate limits by multiplying by the conjugate
2. How are Limits expressed in mathematics?
Mathematically, we write and say “the limit of a function f(x), as x approaches a, is equivalent to L”. If we are able to make the values of the function f(x) arbitrarily close to L by taking x to be sufficiently close to, 'a' (on either side of a) but not equivalent to a.
This is to say that as 'x' becomes closer and closer to the number a (from either side of a), the value of f(x) gets much nearer to the number ‘L’. In computing the limit of f(x) as x approaches, remember that we never take into account x = a.
F(x) is needless to be even defined when x = a. One factor that matters is how f(x) is defined close to 'a'. You will find a collection of Limits solved problems PDF free which will be very helpful for your board exam preparation. |
# LINEAR EQUATION SYSTEM - SOLVING, WORD PROBLEMS MADE EASY
Linear Equations in Two Variables before Linear Equation System
if you have not already done so.
There, we explained the method of solving the equations, with examples.
That knowledge is a prerequisite here.
Here, we apply that knowledge to solve word problems.
Great Deals on School & Homeschool Curriculum Books
### Example 1 of Linear Equation System
Solve the following Word Problem on Linear Equation System
If three times the larger of the two numbers is divided by the smaller,
then the quotient is 4 and remainder is 5. If six times the smaller
is divided by the larger, the quotient is 4 and remainder is 2.
Find the numbers.
Solution to Example 1 of Linear Equation System :
Let x be the larger and y be the smaller numbers.
By data, if three times the larger of the two numbers is divided
by the smaller, then the quotient is 4 and remainder is 5.
i.e. if 3x is divided by y, the quotient is 4 and remainder is 5.
⇒ 3x = 4y + 5 ⇒ 3x - 4y = 5........................................(i)
Also by data, If six times the smaller is divided
by the larger, the quotient is 4 and remainder is 2.
i.e. if 6y is divided by x, the quotient is 4 and remainder is 2.
⇒ 6y = 4x + 2 ⇒ -4x + 6y = 2........................................(ii)
Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.
Now we have to solve these simultaneous equations.
To solve (i) and (ii), Let us make the coefficients of y the same.
(i) x 3 gives 9x - 12y = 15........................................(iii)
(ii) x 2 gives -8x + 12y = 4........................................(iv)
(iii) + (iv) gives 9x - 8x = 15 + 4 ⇒ x = 19.
Using this in (ii), we get
-4(19) + 6y = 2 ⇒ -76 + 6y = 2 ⇒ 6y = 2 + 76 = 78
y = 78⁄6 = 13.
Thus the numbers are 19 and 13. Ans.
Check:
three times the larger ( = 3 x 19 = 57) divided by smaller ( = 13)
gives quotient 4 and remainder 5
( since 4 x 13 = 52 and 57 - 52 = 5) (verified.)
six times the smaller ( = 6 x 13 = 78) divided by larger ( = 19)
gives quotient 4 and remainder 2
( since 4 x 19 = 76 and 78 - 76 = 2) (verified.)
Here is a collection of proven tips,
tools and techniques to turn you into
a super-achiever - even if you've never
thought of yourself as a "gifted" student.
and remember large chunks of information
with the least amount of effort.
If you apply what you read from the above
collection, you can achieve best grades without
giving up your fun, such as TV, surfing the net,
playing video games or going out with friends!
## Speed Study System.
### Example 2 of Linear Equation System
Solve the following Word Problem Linear Equation System
The denominator of a fraction is greater than its numerator by 9.
If 7 is subtracted from both its numerator and denominator,
the fraction becomes 2⁄3. Find the original fraction.
Solution to Example 2 of Linear Equation System :
Let x be the numerator and y be the denominator.
By data, the denominator is greater than its numerator by 9.
y = x + 9 ⇒ -x + y = 9.....................................(i)
Also by data, If 7 is subtracted from both its numerator
and denominator, the fraction becomes 2⁄3.
⇒ (x - 7)⁄(y - 7) = 2⁄3
Crossmultiplying, we get
3(x - 7) = 2(y - 7) ⇒ 3x - 21 = 2y - 14 ⇒ 3x - 2y = -14 + 21
⇒ 3x - 2y = 7.....................................(ii)
Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.
Now we have to solve these simultaneous equations.
To solve (i) and (ii), let us make the coefficients of y the same.
(i) x 2 gives -2x + 2y = 18.....................................(iii)
Adding (ii) and (iii), we get
3x - 2x = 7 + 18 ⇒ x = 25.
Using this in (i), we get
-25 + y = 9 ⇒ y = 9 + 25 = 34.
The required fraction = 25⁄34. Ans.
Check:
Since 34 = 25 + 9,
"the denominator is greater than its numerator by 9" is satisfied.(verified.)
Since (25 - 7)⁄(34 - 7) = 18⁄27 = 2⁄3,
"If 7 is subtracted from both its numerator and denominator,
the fraction becomes 2⁄3" is satisfied.(verified.)
### Example 3 of Linear Equation System
Solve the following Word Problem on Linear Equation System
A and B each have a certain number of marbles. A says to B,
" if you give 30 to me, I will have twice as many as left with you."
B replies "if you give me 10, I will have thrice as many as left with you."
How many marbles does each have?
Solution to Example 3 of Linear Equation System :
Let x be the number of marbles A has.
And Let y be the number of marbles B has.
If B gives 30 to A, then A has x + 30 and B has y - 30.
By data, When this happens, A has twice as many as left with B.
x + 30 = 2(y - 30) = 2y - 2 x 30 = 2y - 60. ⇒ x - 2y = -60 - 30
x - 2y = -90 ..........(i)
If A gives 10 to B, then A has x - 10 and B has y + 10.
By data, When this happens, B has thrice as many as left with A.
y + 10 = 3(x - 10) = 3x - 3 x 10 = 3x - 30 ⇒ y - 3x = -30 -10
⇒ 3x - y = 40 ...........(ii)
Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.
Now we have to solve these simultaneous equations.
To solve (i) and (ii), Let us make y coefficients same.
(ii) x 2 gives 6x - 2y = 80 ...........(iii)
x - 2y = -90 ..........(i)
Subtracting 5x = 80 - (-90) = 80 + 90 = 170
x = 170⁄5 = 34.
Using this in Equation (ii), we get
3(34) - y = 40 ⇒ 102 - y = 40 ⇒ -y = 40 - 102 = -62
y = 62.
Thus A has 34 marbles and B has 62 marbles. Ans.
Check:
If B gives 30 to A from his 62, then A has 34 + 30 = 64
and B has 62 - 30 = 32. Twice 32 is 64. (verified.)
If A gives 10 to B from his 34, then A has 34 - 10 = 24
and B has 62 + 10 = 72. Thrice 24 is 72. (verified.)
Thus the Solution to Example 3 of
Linear Equation System is verified.
### Exercise on Linear Equation System
Solve the following Word Problems on Linear Equation System.
1. Five years hence, a man's age will be three times his son's
age and five years ago, he was seven times as old as his son.
Find their present ages.
2. If the length and breadth of a room are increased by 1 m each,
its area is increased by 21 m2. If the length is increased
by 1 m and breadth decreased by 1m, the area is decreased by
5 m2. Find the area of the room.
3. The students of a class are made to stand in complete rows. If one
student is extra in each row, there would be 2 rows less, and if one
student is less in each row, there would be 3 rows more. Find the
number of students in the class.
## Progressive Learning of Math : Linear Equation System
Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.
This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.
These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.
These also provide teachers and parents with lessons
on how to work with the child on the concepts.
The series is low to reasonably priced and include
1. 40, 10
2. 12, 8
3. 60 |
# 4th Grade NYSE Math FREE Sample Practice Questions
Preparing your student for the 4th Grade NYSE Math test? To succeed on the NYSE Math test, students need to practice as many real NYSE Math questions as possible. There’s nothing like working on NYSE Math sample questions to measure your student’s exam readiness and put him/her more at ease when taking the 4th Grade NYSE Math test. The sample math questions you’ll find here are brief samples designed to give students the insights they need to be as prepared as possible for their 4th Grade NYSE Math test.
Check out our sample 4th Grade NYSE Math practice questions to find out what areas your student needs to practice more before taking the 4th Grade NYSE Math test!
Start preparing your student for the 2022 NYSE Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
## 10 Sample 4th Grade NYSE Math Practice Questions
1- What is the value of A in the equation $$64 ÷ A = 8$$
A. 2
B. 4
C. 6
D. 8
2- Jason’s favorite sports team has won 0.62 of its games this season. How can Jason express this decimal as a fraction?
A. $$\frac{6}{2}$$
B.$$\frac{62}{10}$$
C. $$\frac{62}{100}$$
D. $$\frac{6.2}{10}$$
3- Use the models below to answer the question.
$$\img{https://appmanager.effortlessmath.com/public/images/questions/taar422d.JPG}$$
Which statement about the models is true?
A. Each shows the same fraction because they are the same size.
B. Each shows a different fraction because they are different shapes.
C. Each shows the same fraction because they both have 3 sections shaded.
D. Each shows a different fraction because they both have 3 shaded sections but a different number of total sections.
4- Sophia flew 2,448 miles from Los Angeles to New York City. What is the number of miles Sophia flew rounded to the nearest thousand?
A. 2,000
B. 2,400
C. 2,500
D. 3,000
5- Write $$\frac{124}{1000}$$ as a decimal number.
A. 1.24
B. 0.124
C. 12.4
D. 0.0124
6- Erik made 12 cups of juice. He drinks 3 cups of juice each day. How many days will Erik take to drink all of the juice he made?
A. 2 days
B. 4 days
C. 8 days
D. 9 days
7- Jason has prepared $$\frac{4}{10}$$ of his assignment. Which decimal represents the part of the assignment Jason has prepared?
A. 4.10
B. 4.01
C. 0.4
D. 0.04
8- Emma described a number using these clues.
• 3 digits of the number are 4, 7, and 9
• The value of the digit 4 is $$(4 × 10)$$
• The value of the digit 7 is $$(7 × 1000)$$
• The value of the digit 9 is $$(9 × 10000)$$
Which number could fit Emma’s description?
A. 9,724.04
B. 90,734.40
C. 97,040.04
D. 98,740.70
9- There are 18 boxes and each box contains 26 pencils. How many pencils are in the box in total?
A. 108
B. 208
C. 468
D. Not here
10- Emily and Ava were working on a group project last week. They completed $$\frac{7}{10}$$ of their project on Tuesday and the rest on Wednesday.
Ava completed $$\frac{3}{10}$$ of their project on Tuesday. What fraction of the group project did Emily complete on Tuesday?
A. $$\frac{2}{10}$$
B. $$\frac{3}{10}$$
C. $$\frac{4}{10}$$
D. $$\frac{5}{10}$$
## Answers:
1- D
$$A = 64 ÷ 8$$
$$A=8$$
2- C
0.62 is equal to $$\frac{62}{100}$$
3- D
the model for the first fraction is divided into 6 equal parts. We shade 3_6 to show the same amount as 1_2. The model for the second fraction is divided into 8 equal parts. We shade 3_ 8 which shows these two models are different fractions.
4- A
When rounding to the nearest thousand, you will need to look at the last three digits. If the last three digits are 449 or less round to the next number that is smaller than the number given and end with three zeros.
On the other hand, If the last three digits are 500 or more, round to the next number bigger than the given number and end with three zeros.
5- B
$$\frac{124}{1000}$$ is equal to 0.124
6- B
$$12 ÷ 3 = 4$$
7- C
$$\frac{4}{10}=0.4$$
8- C
only choice C fits Emma’s description
9- C
$$18 × 26 = 468$$
10- C
$$\frac{7}{10}-\frac{3}{10}=\frac{4}{10}$$
Looking for the best resource to help you succeed on the NYSE Math test?
## Related to This Article
### What people say about "4th Grade NYSE Math FREE Sample Practice Questions - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.
X
51% OFF
Limited time only!
Save Over 51%
SAVE $15 It was$29.99 now it is \$14.99 |
# algebraic expression 265 results
algebraic expression - A way to write a mathematical expression from a phrase
-65 times the difference between a number and 79 is equal to the number plus 98
-65 times the difference between a number and 79 is equal to the number plus 98 The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. The first expression, [I]the difference between a number and 79[/I] means we subtract 79 from our arbitrary variable of x: x - 79 Next, -65 times the difference between a number and 79 means we multiply our result above by -65: -65(x - 79) The phrase [I]the number[/I] refers to the arbitrary variable x earlier. The number plus 98 means we add 98 to x: x + 98 Now, let's bring it all together. The phrase [I]is equal to[/I] means an equation. So we set -65(x - 79) equal to x + [B]98: -65(x - 79) = x + 98[/B] <-- This is our algebraic expression If the problem asks you to take it a step further and solve for x, then you [URL='https://www.mathcelebrity.com/1unk.php?num=-65%28x-79%29%3Dx%2B98&pl=Solve']type this equation into our search engine[/URL], and you get: x = [B]76.31818[/B]
1/2 of x and 10 is 30. Find the x.
1/2 of x and 10 is 30. Find the x. x and 10 means we add: x + 10 1/2 of this: 1/2(x + 10) The phrase is means equal to, so we set 1/2(x + 10) equal to 30 for our algebraic expression [B]1/2(x + 10) = 30[/B]
1/4 of the difference of 6 and a number is 200
1/4 of the difference of 6 and a number is 200 Take this [B]algebraic expression[/B] in 4 parts: [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]The difference of 6 and a number means we subtract x from 6: 6 - x [*]1/4 of the difference means we divide 6 - x by 4: (6 - x)/4 [*]Finally, the phrase [I]is[/I] means an equation, so we set (6 - x)/4 equal to 200 [/LIST] [B](6 - x)/4 = 200[/B]
10 divided by the sum of 4 and u
10 divided by the sum of 4 and u Take this algebraic expression in parts: The sum of 4 and u means we add 4 to u: 4 + u Next, we divide 10 by this sum: [B]10/(4 + u)[/B]
10 times a number is 420
10 times a number is 420 A number denotes an arbitrary variable, let's call it x. 10 times a number: 10x The phrase is means equal to, so we set 10x equal to 420 [B]10x = 420 <-- This is our algebraic expression [/B] If you want to solve for x, use our [URL='http://www.mathcelebrity.com/1unk.php?num=10x%3D420&pl=Solve']equation calculator[/URL] We get x = 42
108 times a, reduced by 147 is k subtracted from 56
108 times a, reduced by 147 is k subtracted from 56 Take this algebraic expression in pieces: Step 1: 108 times a: 108a Step 2: Reduced by means subtract, so we subtract 47 from 108a: 108a - 47 Step 3: ksubtracted from 56: 56 - k Step 4: The phrase [I]is[/I] means equal to, so we set 108a - 47 equal to 56 - k [B]108a - 47 = 56 - k [MEDIA=youtube]KrY6uzKeeB0[/MEDIA][/B]
12 divided into groups of s
12 divided into groups of s We build our algebraic expression as follows: [B]12/s[/B]
12 plus 6 times a number is 9 times the number
12 plus 6 times a number is 9 times the number The phrase [I]a number [/I]means an arbitrary variable. Let's call it x. 6 times a number is written as: 6x 12 plus 6 times the number means we add 6x to 12: 12 + 6x 9 times a number is written as: 9x The phrase [I]is[/I] means an equation, so we set 12 + 6x equal to 9x [B]12 + 6x = 9x <-- This is our algebraic expression[/B] [B][/B] If the problem asks you to solve for x, then you [URL='https://www.mathcelebrity.com/1unk.php?num=12%2B6x%3D9x&pl=Solve']type this expression into our search engine[/URL] and you get: x = [B]4[/B]
132 is 393 multiplied by y
132 is 393 multiplied by y 393 multiplied by y 393y The word [I]is[/I] means equal to, so we set 393y equal to 132 as our algebraic expression [B]393y = 132 [/B] If you need to solve for y, use our [URL='http://www.mathcelebrity.com/1unk.php?num=393y%3D132&pl=Solve']equation calculator[/URL]
15 added to the quotient of 8 and a number is 7.
15 added to the quotient of 8 and a number is 7. Take this algebraic expression in pieces: [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. [*]The quotient of 8 and a number: 8/x [*]15 added to this quotient: 8/x + 15 [*]The word [I]is[/I] means an equation, so we set 8/x + 15 equal to 7 [/LIST] [B]8/x + 15 = 7[/B]
16 decreased by 3 times the sum of 3 and a number
16 decreased by 3 times the sum of 3 and a number Take this algebraic expression in parts: [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. [*]The sum of 3 and a number: 3 + x [*]3 times the sum: 3(3 + x) [*]16 decreased by... means we subtract 3(3 + x) from 16 [/LIST] [B]3(3 + x) from 16[/B]
18 seconds faster than Tina’s time
18 seconds faster than Tina’s time Let Tina's time be t. Speaking in terms of time, faster means less. So we have an algebraic expression of: [B]t - 18[/B]
19 decreased by the absolute value of c
19 decreased by the absolute value of c Take this algebraic expression in parts: [LIST] [*]Absolute value of c: |c| [*]19 decreased by the absolute value of c is found by subtracting |c| from 19 [/LIST] [B]19 - |c|[/B]
2 times a number equals that number plus 5
2 times a number equals that number plus 5 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 2 times a number means we multiply 2 by x: 2x That number plus 5 means we add 5 to the number x x + 5 The phrase [I]equals[/I] means we set both expressions equal to each other [B]2x = x + 5[/B] <-- This is our algebraic expression If you want to take this further and solve this equation for x, [URL='https://www.mathcelebrity.com/1unk.php?num=2x%3Dx%2B5&pl=Solve']type this expression in the search engine[/URL] and we get: [B]x = 5[/B]
2 times the sum of 7 times a number and 4
2 times the sum of 7 times a number and 4 This is an algebraic expression. Let's take it in 4 parts: [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]7 times a number means we multiply x by 7: 7x [*]The sum of 7 times a number and 4 means we add 4 to 7x: 7x + 4 [*]Finally, we multiply the sum in #3 by 2 [/LIST] Build our final algebraic expression: [B]2(7x + 4)[/B]
20 yards longer than p
20 yards longer than p We want to build an algebraic expression. Longer means we add 20 to p: [B]p + 20[/B]
217 times u, reduced by 180 is the same as q
217 times u, reduced by 180 is the same as q. Take this algebraic expression pieces: Step 1: 217 times u We multiply the variable u by 217 217u Step 2: reduced by 180 Subtract 180 from 217u 217u - 180 The phrase [I]is the same as[/I] means an equation, so we set 217u - 180 equal to q [B]217u - 180 = q[/B]
223 subtracted from the quantity 350 times a is equal to b
223 subtracted from the quantity 350 times a is equal to b Take this algebraic expression in parts: [LIST] [*]the quantity 350 times a: 350a [*]223 subtracted from the quantity: 350a - 223 [*]The phrase [I]is equal to[/I] means an equation, so we set 350a - 223 equal to b [/LIST] [B]350a - 223 = b[/B]
231 is 248 subtracted from the quantity h times 128
231 is 248 subtracted from the quantity h times 128 Let's take this algebraic expression in parts: [LIST=1] [*]h times 128: 128h [*]24 subtracted from this: 128h - 248 [*]The word [I]is[/I] means an equation, so we set 128h - 248 equal to 231 [/LIST] [B]128h - 248 = 231[/B] <-- This is our algebraic expression If the problem asks you to solve for h, then you [URL='https://www.mathcelebrity.com/1unk.php?num=128h-248%3D231&pl=Solve']type in this equation into our search engine[/URL] and get: h = [B]3.742[/B]
249 equals 191 times c, decreased by 199
249 equals 191 times c, decreased by 199 [U]Take this in pieces:[/U] 191 times c: 191c The phrase [I]decreased by[/I] means we subtract 199 from 191c: 191c - 199 We set this expression equal to 249: [B]191c - 199 = 249[/B] <-- This is our algebraic expression If you want to solve for c, type this equation into the search engine and we get: [B]c = 2.346[/B]
2x decreased by 15 is equal to -27
2x decreased by 15 is equal to -27 The phrase [I]decreased by[/I] 15 means we subtract 15 from 2x: 2x - 15 The phrase [I]is equal to[/I] means an equation, so we set 2x - 15 equal to -27 [B]2x - 15 = -27 [/B] <-- This is our algebraic expression To solve this, [URL='https://www.mathcelebrity.com/1unk.php?num=2x-15%3D-27&pl=Solve']type 2x - 15 = -27 into the search engine[/URL].
2x plus 4 increased by 15 is 57
2x plus 4 increased by 15 is 57 Take this algebraic expression in parts: [LIST] [*]2x plus 4: 2x + 4 [*][I]Increased by[/I] means we add 15 to 2x + 4: 2x + 4 + 15 = 2x + 19 [*]The word [I]is[/I] means an equation, so we set 2x + 19 equal to 57: [/LIST] Our final algebraic expression is: [B]2x + 19 = 57 [/B] To solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B19%3D57&pl=Solve']type this equation into our search engine [/URL]and we get x = [B]19[/B]
3 is subtracted from the square of x
3 is subtracted from the square of x Let's take this algebraic expression in two parts: Part 1: The square of x means we raise x to the power of 2: x^2 Part 2: 3 is subtracted means we subtract 3 from x^2 [B]x^2 - 3[/B]
3 more than the product of 7 and a number x is less than 26
The product of 7 and a number x is written as 7x. 3 more than that product is written as 7x + 3. Finally, that entire expression is less than 26, so we have: 7x + 3 < 26 as our algebraic expression. [MEDIA=youtube]ESuHovml5WQ[/MEDIA]
3 times a number increased by 1 is between -8 and 13
3 times a number increased by 1 is between -8 and 13. Let's take this algebraic expression in [U]4 parts[/U]: Part 1 - The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Part 2 - 3 times this number means we multiply x by 3: 3x Part 3 - Increased by 1 means we add 1 to 3x: 3x + 1 The phrase [I]between[/I] means we have an inequality: [B]-8 <= 3x + 1 <=13[/B]
3 times larger than the sum of 4 and 9
The sum of 4 and 9: 4 + 9 3 times larger than this sum [B]3(4 + 9) <-- This is our algebraic expression [/B] Evaluating this amount: 3(13) [B]39[/B]
3 times the square of a number x minus 12
3 times the square of a number x minus 12. Build the algebraic expression piece by piece: [LIST] [*]Square of a number x: x^2 [*]3 times this: 3x^2 [*]Minus 12: [B]3x^2 - 12[/B] [/LIST]
3 times the sum of x and 9y
3 times the sum of x and 9y The sum of x and 9y means we add 9y to x: x + 9y Now we take this sum, and multiply by 3 to get our final algebraic expression: 3(x + 9y)
3 times the width plus 2 times the length
3 times the width plus 2 times the length Let w be the width Let l be the length We have an algebraic expression of: [B]3w + 2l[/B]
3 times x minus y is 5 times the sum of y and 2 times x
3 times x minus y is 5 times the sum of y and 2 times x Take this algebraic expression in pieces: 3 times x: 3x Minus y means we subtract y from 3x 3x - y The sum of y and 2 times x mean we add y to 2 times x y + 2x 5 times the sum of y and 2 times x: 5(y + 2x) The word [I]is[/I] means an equation, so we set 3x - y equal to 5(y + 2x) [B]3x - y = 5(y + 2x)[/B]
30% larger then 1/3 of twice q
30% larger then 1/3 of twice q Take this algebraic expression in 3 parts: [LIST=1] [*]Twice q means multiply q by 2: 2q [*]1/3 of twice q means we multiply 2q in Step 1 by 1/3: 2q/3 [*]30% larger means we multiply 2q/3 in step 2 by 1.3, since 30% = 0.3: 1.3(2q/3) [/LIST] [B]1.3(2q/3)[/B]
300 reduced by 5 times my age is 60
300 reduced by 5 times my age is 60 Let my age be a. We have: 5 times my age = 5a 300 reduced by 5 times my age means we subtract 5a from 300: 300 - 5a The word [I]is[/I] means an equation, so we set 300 - 5a equal to 60 to get our final algebraic expression: [B]300 - 5a = 60 [/B] If you have to solve for a, you [URL='https://www.mathcelebrity.com/1unk.php?num=300-5a%3D60&pl=Solve']type this equation into our search engine[/URL] and you get: a = [B]48[/B]
309 is the same as 93 subtracted from the quantity f times 123
309 is the same as 93 subtracted from the quantity f times 123. The quantity f times 123: 123f Subtract 93: 123f - 93 The phrase [I]is the same as[/I] means an equation, so we set 123f - 93 equal to 309 [B]123f - 93 = 309[/B] <-- This is our algebraic expression If you wish to solve for f, [URL='https://www.mathcelebrity.com/1unk.php?num=123f-93%3D309&pl=Solve']type this equation into the search engine[/URL], and we get f = 3.2683.
324 times z, reduced by 12 is z
324 times z, reduced by 12 is z. Take this algebraic expression in pieces: 324 [I]times[/I] z means we multiply 324 by the variable z. 324z [I]Reduced by[/I] 12 means we subtract 12 from 324z 324z - 12 The word [I]is[/I] means we have an equation, so we set 324z - 12 equal to z [B]324z - 12 = z [/B] <-- This is our algebraic expression
339 equals 303 times w, minus 293
339 equals 303 times w, minus 293 Take this algebraic expression in pieces: 303 times w: 303w Minus 293: 303w - 293 The phrase [I]equals[/I] means we have an equation. We set 303w - 293 = 339 [B]303w - 293 = 339[/B] <-- This is our algebraic expression To solve for w, [URL='https://www.mathcelebrity.com/1unk.php?num=303w-293%3D339&pl=Solve']we type this equation into our search engine[/URL] to get: [B]w = 2.086[/B]
35 added to n is greater than or equal to the sum of k and 21
35 added to n is greater than or equal to the sum of k and 21 Take this algebraic expression in 3 parts: [LIST=1] [*]35 added to n means we have a sum: n + 35 [*]The sum of k and 21 means we add 21 to k: k +21 [*]The phrase [I]greater than or equal to[/I] means an inequality using this sign (>=), so we write this as follows: [/LIST] [B]n + 35 >= k + 21[/B]
3f,subtract g from the result, then divide what you have by h
3f,subtract g from the result, then divide what you have by h Take this algebraic expression in pieces: 3f subtract g means we subtract the variable g from the expression 3f: 3f - g Divide what we have by h, means we take the result above, 3f - g, and divide it by h: [B](3f - g)/h[/B]
3x less than 2 times the sum of 2x and 1 is equal to the sum of 2 and 5
3x less than 2 times the sum of 2x and 1 is equal to the sum of 2 and 5 This is an algebraic expression. Let's take this algebraic expression in 5 parts: [LIST=1] [*]The sum of 2x and 1 means we add 1 to 2x: 2x + 1 [*]2 times the sum of 2x and 1: 2(2x + 1) [*]3x less than the sum of 2x and 1 means we subtract 3x from 2(2x + 1): 2(2x + 1) - 3x [*]The sum of 2 and 5 means we add 5 to 2: 2 + 5 [*]Finally, the phrase [I]equal[/I] means an equation, so we set #3 equal to #4 [/LIST] Our algebraic expression is: [B]2(2x + 1) - 3x = 2 + 5[/B] Now, some problems may ask you to simplify. In this case, we multiply through and group like terms: 4x + 2 - 3x = 7 [B]x + 2 = 7 <-- This is our simplified algebraic expression [/B] Now, what if the problem asks you to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=x%2B2%3D7&pl=Solve']you type this into our search engine[/URL] and get: x =[B] 5 [MEDIA=youtube]3hzyc2NPCGI[/MEDIA][/B]
4 times a number added to 8 times a number equals 36
4 times a number added to 8 times a number equals 36 Let [I]a number[/I] be an arbitrary variable, let us call it x. 4 times a number: 4x 8 times a number: 8x We add these together: 4x + 8x = 12x We set 12x equal to 36 to get our final algebraic expression of: [B]12x = 36 [/B] If the problem asks you to solve for x, you [URL='https://www.mathcelebrity.com/1unk.php?num=12x%3D36&pl=Solve']type this algebraic expression into our search engine[/URL] and get: x = [B]3[/B]
4 times a number is the same as the number increased by 78
4 times a number is the same as the number increased by 78. Let's take this algebraic expression in parts: [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]4 times a number is written as 4x [*]The number increased by 78 means we add 78 to x: x + 78 [*]The phrase [I]the same as[/I] mean an equation, so we set #2 equal to #3 [/LIST] [B]4x = x + 78[/B] <-- This is our algebraic expression If the problem asks you to take it a step further, then [URL='https://www.mathcelebrity.com/1unk.php?num=4x%3Dx%2B78&pl=Solve']we type this equation into our search engine [/URL]and get: x = 26
4 times b increased by 9 minus twice y
4 times b increased by 9 minus twice y Take this algebraic expression in parts: Step 1: 4 times b means we multiply the variable b by 4: 4b Step 2: Increased by 9 means we add 9 to 4b: 4b + 9 Step 3: Twice y means we multiply the variable y by 2: 2y Step 4: The phrase [I]minus[/I] means we subtract 2y from 4b + 9 [B]4b + 9 - 2y[/B]
4 times the number of cows plus 2 times the number of ducks
4 times the number of cows plus 2 times the number of ducks Let c be the number of cows. Let d be the number of ducks. We've got an algebraic expression below: [B]4c + 2d[/B]
4subtractedfrom6timesanumberis32
4 subtracted from 6 times a number is 32. Take this algebraic expression in pieces. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 6 times this number means we multiply by x by 6 6x 4 subtracted from this expression means we subtract 4 6x - 4 The phrase [I]is[/I] means an equation, so we set 6x - 4 equal to 32 [B]6x - 4 = 32 [/B] If you need to solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=6x-4%3D32&pl=Solve']type it in the search engine here[/URL].
5 added to x is 11
x + 5 = 11 for the algebraic expression. Plug that into the [URL='http://www.mathcelebrity.com/1unk.php?num=x%2B5%3D11&pl=Solve']search engine[/URL], and solve for x. x = 6.
5 added to x means we use the plus sign for a sum. x + 5 "is" means equals, so we set that equal to 11. x + 5 = 11 <-- This is our algebraic expression.
5 diminished by twice the sum of a and b
5 diminished by twice the sum of a and b Take this algebraic expression in parts: [LIST] [*]The sum of a and b: a + b [*]Twice the sum means we multiply a + b by 2: 2(a + b) [*]5 diminished by twice the sum means we subtract 2(a + b) from 5 [/LIST] [B]5 - 2(a + b)[/B]
5 more than the reciprocal of a number
5 more than the reciprocal of a number Take this algebraic expression in pieces: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The reciprocal of this number means we divide 1 over x: 1/x 5 more means we add 5 to 1/x [B]1/x + 5[/B]
5 more than twice the cube of a number
5 more than twice the cube of a number. Take this algebraic expression in pieces. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The cube of a number means we raise it to a power of 3 x^3 Twice the cube of a number means we multiply x^3 by 2 2x^3 5 more than twice the cube of a number means we multiply 2x^3 by 5 5(2x^3) Simplifying, we get: 10x^3
5 subtracted from 3 times a number is 44
5 subtracted from 3 times a number is 44. The problem asks for an algebraic expression. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 3 times this number is 3x. 5 subtracted from this is written as 3x - 5. The phrase [I]is[/I] means an equation, so we set 3x - 5 equal to 44 [B]3x - 5 = 44[/B]
5 times a number increased by 4 is divided by 6 times the same number
5 times a number increased by 4 is divided by 6 times the same number Take this algebraic expression in parts. Part 1: 5 times a number increased by 4 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x [*]5 times the number means multiply x by 5: 5x [*][I]Increased by 4[/I] means we add 4 to 5x: 5x + 4 [/LIST] Part 2: 6 times the same number [LIST] [*]From above, [I]a number[/I] is x: x [*]6 times the number means we multiply x by 6: 6x [/LIST] The phrase [I]is divided by[/I] means we have a quotient, where 5x + 4 is the numerator, and 6x is the denominator. [B](5x + 4)/6x[/B]
5 times g reduced by the square of h
5 times g reduced by the square of h Take this algebraic expression in pieces: [LIST=1] [*]5 times g means we multiply g by 5: 5g [*]The square of h means we raise h to the 2nd power: h^2 [*]5 times g reduced by the square of h means we subtract h^2 from 5g: [/LIST] [B]5g - h^2[/B]
50 is more than the product of 4 and w
50 is more than the product of 4 and w Take this algebraic expression in pieces: The product of 4 and w mean we multiply the variable w by 4: 4w The phrase [I]is more than[/I] means an inequality using the (>) sign, where 50 is greater than 4w: [B]50 > 4w[/B]
54 is the sum of 15 and Vidyas score
54 is the sum of 15 and Vidyas score. Let Vida's score be s. The sum of 15 and s: s + 15 When they say "is", they mean equal to, so we set s + 15 equal to 54. Our algebraic expression is below: [B]s + 15 = 54 [/B] To solve this equation for s, use our [URL='http://www.mathcelebrity.com/1unk.php?num=s%2B15%3D54&pl=Solve']equation calculator[/URL]
54 is the sum of 24 and Julies score. Use the variable J to represent Julies score.
54 is the sum of 24 and Julies score. Use the variable J to represent Julies score. Sum of 24 and Julie's score: 24 + J The phrase [I]is[/I] means an equation, so we set 24 + J equal to 54 to get an algebraic expression: [B]24 + J = 54[/B]
59 is the sum of 16 and Donnie's saving. Use the variable d to represent Donnie's saving.
59 is the sum of 16 and Donnie's saving. Use the variable d to represent Donnie's saving. The phrase [I]the sum of[/I] means we add Donnie's savings of d to 16: d + 16 The phrase [I]is[/I] means an equation, so we set d + 16 equal to 59 d + 16 = 59 <-- [B]This is our algebraic expression[/B] Now, if the problem asks you to solve for d, then you[URL='https://www.mathcelebrity.com/1unk.php?num=d%2B16%3D59&pl=Solve'] type the algebraic expression into our search engine to get[/URL]: d = [B]43[/B]
6 subtracted from the product of 5 and a number is 68
6 subtracted from the product of 5 and a number is 68 Take this algebraic expression in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The product of 5 and this number is: 5x We subtract 6 from 5x: 5x - 6 The phrase [I]is[/I] means an equation, so we set 5x - 6 equal to 68 [B]5x - 6 = 68[/B]
6 times a number multiplied by 3 all divided by 4
6 times a number multiplied by 3 all divided by 4 Take this algebraic expression in parts: [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]6 times a number: 6x [*]Multiplied by 3: 3(6x) = 18x [*]All divided by 4: 18x/4 [/LIST] We can simplify this: We type 18/4 into our search engine, simplify, and we get 9/2. So our answer is: [B]9x/2[/B]
6 times a number, x, is at least 22.
6 times a number, x, is at least 22. 6 times a number x: 6x The phrase [I]is at least[/I] means greater than or equal to. So we have an inequality: [B]6x >= 22[/B] <-- This is our algebraic expression [URL='https://www.mathcelebrity.com/interval-notation-calculator.php?num=6x%3E%3D22&pl=Show+Interval+Notation']To solve this for x, paste this into the search engine[/URL] and we get: [B]x >= 3.666667[/B]
6 times the reciprocal of a number equals 2 times the reciprocal of 7. What is the number
6 times the reciprocal of a number equals 2 times the reciprocal of 7. What is the number We've got two algebraic expressions here. Let's take it in parts: Term 1: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The reciprocal is 1/x Multiply this by 6: 6/x Term 2: Reciprocal of 7: 1/7 2 times this: 2/7 We set these terms equal to each other: 6/x = 2/7 [URL='https://www.mathcelebrity.com/prop.php?num1=6&num2=2&den1=x&den2=7&propsign=%3D&pl=Calculate+missing+proportion+value']Type this proportion into the search engine[/URL], and we get: [B]x = 21[/B]
6 times the reciprocal of a number equals 3 times the reciprocal of 7 .
6 times the reciprocal of a number equals 3 times the reciprocal of 7 . This is an algebraic expression. Let's take it in parts: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The reciprocal of a number x means we divide 1 over x: 1/x 6 times the reciprocal means we multiply 6 by 1/x: 6/x The reciprocal of 7 means we divide 1/7 1/7 3 times the reciprocal means we multiply 1/7 by 3: 3/7 Now, the phrase [I]equals[/I] mean an equation, so we set 6/x = 3/7 [B]6/x = 3/7[/B] <-- This is our algebraic expression If the problem asks you to solve for x, then [URL='https://www.mathcelebrity.com/prop.php?num1=6&num2=3&den1=x&den2=7&propsign=%3D&pl=Calculate+missing+proportion+value']we type this proportion in our search engine[/URL] and get: x = 14
6 times the sum of a number and 3 is equal to 42. What is this number?
6 times the sum of a number and 3 is equal to 42. What is this number? The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The sum of a number and 3 means we add 3 to x: x + 3 6 times the sum: 6(x + 3) The word [I]is[/I] means an equation, so we set 6(x + 3) equal to 42 to get our [I]algebraic expression[/I] of: [B]6(x + 3) = 42[/B] [B][/B] If the problem asks you to solve for x, then [URL='https://www.mathcelebrity.com/1unk.php?num=6%28x%2B3%29%3D42&pl=Solve']you type this equation into our search engine[/URL] and you get: x = [B]4[/B]
6 times the sum of a number and 5 is 16
6 times the sum of a number and 5 is 16 A number represents an arbitrary variable, let's call it x x The sum of x and 5 x + 5 6 times the sum of x and 5 6(x + 5) Is means equal to, so set 6(x + 5) equal to 16 [B]6(x + 5) = 16 <-- This is our algebraic expression Solve for x[/B] Multiply through: 6x + 30 = 16 Subtract 30 from each side: 6x - 30 + 30 = 16 - 30 6x = -14 Divide each side by 6 6x/6 = -14/6 Simplify this fraction by dividing top and bottom by 2: x = [B]-7/3 [MEDIA=youtube]oEx5dsYK7DY[/MEDIA][/B]
60 percent of a number minus 17 is -65
60 percent of a number minus 17 is -65 Using our [URL='https://www.mathcelebrity.com/perc.php?num=+5&den=+8&num1=+16&pct1=+80&pct2=+90&den1=+80&pct=60&pcheck=4&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']percent to decimal calculator[/URL], we see that 60% is 0.6, so we have: 0.6 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. So 60% of a number is: 0.6x Minus 17: 0.6x - 17 The word [I]is[/I] means an equation, so we set 0.6x - 17 equal to -65 to get our algebraic expression of: [B]0.6x - 17 = -65[/B] [B][/B] If you want to solve for x in this equation, you [URL='https://www.mathcelebrity.com/1unk.php?num=0.6x-17%3D-65&pl=Solve']type it in our search engine and you get[/URL]: [B]x = -80[/B]
64 is 4 times the difference between Sarah's age, a, and 44. Assume Sarah is older than 44.
64 is 4 times the difference between Sarah's age, a, and 44. Assume Sarah is older than 44. The phrase [I]difference between[/I] means we subtract 44 from a: a - 44 The phrase [I]64 is[/I] means an equation, so we set a - 44 equal to 64 [B]a - 44 = 64 <-- This is our algebraic expression [/B] If you want to solve for a, then we [URL='https://www.mathcelebrity.com/1unk.php?num=a-44%3D64&pl=Solve']type this expression into our search engine[/URL] and get: [B]a = 108[/B]
64 is 4 times the difference between Sarah’s age a, and 44.Assume Sarah is older than 44
64 is 4 times the difference between Sarah’s age a, and 44.Assume Sarah is older than 44 Difference between Sarah's age (a) and 44 (Assuming Sarah is older than 44): a - 44 4 times the difference: 4(a - 44) The word [I]is[/I] means equal to, so we set 4(a - 44) equal to 64 to get our algebraic expression: [B]4(a - 44) = 64[/B] If the problem asks you to solve for a, we [URL='https://www.mathcelebrity.com/1unk.php?num=4%28a-44%29%3D64&pl=Solve']type this equation into our search engine[/URL] and we get: a = [B]60[/B]
7 multiplied by the quantity 7 take away 6
7 multiplied by the quantity 7 take away 6 Take this algebraic expression in pieces: [LIST] [*]7 take away 6: 7 - 6 [*]7 multiplied by the quantity: [B]7(7 - 6)[/B] [/LIST] This is our algebraic expression. If you need to evaluate this expression, we [URL='https://www.mathcelebrity.com/order-of-operations-calculator.php?num=7%287-6%29&pl=Perform+Order+of+Operations']type it in the search engine[/URL] and we get; [B]7[/B]
7 times a number increased by 4 times the number
7 times a number increased by 4 times the number Let [I]a number[/I] and [I]the number[/I] be an arbitrary variable. Let's call it x. We have an algebraic expression. Let's take it in pieces: [LIST] [*]7 times a number: 7x [*]4 times the number: 4x [*]The phrase [I]increased by[/I] means we add 4x to 7x: [*]7x + 4x [*]Simplifying, we get: (7 + 4)x [*][B]11x[/B] [/LIST]
7 times a number is the same as 12 more than 3 times a number
7 times a number is the same as 12 more than 3 times a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [B][U]Algebraic Expression 1:[/U][/B] 7 times a number means we multiply 7 by x: 7x [B][U]Algebraic Expression 2:[/U][/B] 3 times a number means we multiply 3 by x: 3x 12 more than 3 times a number means we add 12 to 3x: 3x + 12 The phrase [I]is the same as[/I] means an equation, so we set 7x equal to 3x + 12 [B]7x = 3x + 12[/B] <-- Algebraic Expression
7 times a positive number n is decreased by 3, it is less than 25
7 times a positive number n is decreased by 3, it is less than 25 7 times a positive number n: 7n Decreased by 3: 7n - 3 The phrase [I]it is less than [/I]means an inequality. So we relate 7n - 3 less than 25 using the < sign to get our algebraic expression of: [B]7n - 3 < 25[/B]
7 times the cube of the sum of x and 8
7 times the cube of the sum of x and 8 Take this algebraic expression in 3 parts: [LIST=1] [*]The sum of x and 8 means we add 8 to x: x + 8 [*]The cube of this sum means we raise the sum to the 3rd power: (x + 8)^3 [*]7 times this cubed sum means we multiply (x + 8)^3 by 7: [/LIST] [B]7(x + 8)^3[/B]
7 times the number of lions plus 4 times the number of tigers
7 times the number of lions plus 4 times the number of tigers Let the number of lions be l Let the number of tigers be t We have an algebraic expression of: [B]7l + 4t[/B]
72 pounds and increases by 3.9 pounds per month
72 pounds and increases by 3.9 pounds per month Let m be the number of months. We write the algebraic expression below: [B]3.9m + 72[/B]
75% of x is 25 dollars and 99 cents
75% of x is 25 dollars and 99 cents [URL='https://www.mathcelebrity.com/perc.php?num=+5&den=+8&num1=+16&pct1=+80&pct2=+90&den1=+80&pct=75&pcheck=4&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']Since 75%[/URL] is 0.75 as a decimal, we rewrite this as an algebraic expression: 0.75x = 25.99 If we want to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.75x%3D25.99&pl=Solve']type this equation into our search engine[/URL] and we get: x = [B]34.65[/B]
76 subtracted from p is equal to the total of g and 227
76 subtracted from p is equal to the total of g and 227 We've got two algebraic expressions. Take them in pieces: Part 1: 76 subtracted from p We subtract 76 from the variable p p - 76 Part 2: The total of g and 227 The total means a sum, so we add 227 to g g + 227 Now the last piece, the phrase [I]is equal to[/I] means an equation. So we set both algebraic expressions equal to each other: [B]p - 76 = g + 227[/B]
8 is subtracted from the square of x
8 is subtracted from the square of x Take this algebraic expression in parts: [LIST] [*]The square of x means we raise x to the power of 2: x^2 [*]8 subtracted from the square of x is found by subtracting 8 from x^2 [/LIST] [B]x^2 - 8[/B]
8 more than the product of x and 2 equals 4
8 more than the product of x and 2 equals 4 The product of x and 2: 2x 8 more than this, means we add 8: 2x + 8 Set this equal to 4: [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B8%3D4&pl=Solve']2x + 8 = 4[/URL] <-- Algebraic expression to solve for x, type this into the search engine and we get [B]x = -2[/B].
8 more than twice a number is less than 6 more than the number
8 more than twice a number is less than 6 more than the number. This is an algebraic expression, let's take it in pieces... The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 8 more than twice a number: Twice a number means multiply x by 2: 2x Then add 8: 2x + 8 6 more than the number, means we add 6 to x x + 6 The phrase [I]is less than[/I] means an inequality, where we set 2x + 8 less than x + 6 [B]2x + 8 < x + 6[/B]
8 taken away from y
8 taken away from y This is an algebraic expression. The phrase [I]taken away[/I] means we subtract 8 from y: [B]y - 8[/B]
8 times the sum of 5 times a number and 9
8 times the sum of 5 times a number and 9 Take this algebraic expression in parts: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 5 times a number means: 5x The sum of this and 9 means we add 9 to 5x: 5x + 9 Now we multiply 8 times this sum: [B]8(5x + 9)[/B]
9 divided by the sum of x and 4 is equal to 6 divided by x minus 4
9 divided by the sum of x and 4 is equal to 6 divided by x minus 4. Build our two algebraic expressions first: 9 divided by the sum of x and 4 9/(x + 4) 6 divided by x minus 4 6/(x - 4) The phrase [I]is equal to[/I] means and equation, so we set the algebraic expressions equal to each other: [B]9/(x + 4) = 6/(x - 4) <-- This is our algebraic expression[/B] [B][/B] If the problem asks you to solve for x, we cross multiply: 9(x - 4) = 6(x + 4) To solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=9%28x-4%29%3D6%28x%2B4%29&pl=Solve']type this equation into our search engine[/URL] and we get: x = [B]20[/B]
9 is one-third of a number x
9 is one-third of a number x A number x can be written as x x one-third of a number x means we multiply x by 1/3: x/3 The phrase [I]is[/I] means an equation, so we set 9 equal to x/3 to get our final algebraic expression of: [B]x/3 = 9[/B] If the problem asks you to solve for x, you [URL='https://www.mathcelebrity.com/prop.php?num1=x&num2=9&den1=3&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']type this algebraic expression into our search engine[/URL] and you get: [B]x = 27[/B]
9 less than 5 times a number is 3 more than 2x
9 less than 5 times a number is 3 more than 2x The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x 5 times a number means we multiply x by 5: 5x 9 less than 5x means we subtract 9 from 5x: 5x - 9 3 more than 2x means we add 3 to 2x: 2x + 3 The word [I]is[/I] means an equation, so we set 5x - 9 equal to 2x + 3: [B]5x - 9 = 2x + 3 <-- This is our algebraic expression[/B] [B][/B] If you want to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=5x-9%3D2x%2B3&pl=Solve']type this equation into the search engine[/URL], and we get: x = [B]4[/B]
9 subtracted from the product of 3 and a number is greater than or equal to 16
9 subtracted from the product of 3 and a number is greater than or equal to 16 [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]The product of 3 and a number means we multiply 3 times x: 3x [*]9 subtracted from the product: 3x - 9 [*]The phrase is greater than or equal to means an inequality. So we set up an inequality with >= for the greater than or equal to sign in relation to 3x - 9 and 16 [/LIST] Our algebraic expression (inequality) becomes: [B]3x - 19 >= 16[/B]
9 times a number is that number minus 3
9 times a number is that number minus 3 Let [I]a number[/I] be an arbitrary variable, let's call it x. We're given: 9 times a number is 9x The number minus 3 is x - 3 The word [I]is[/I] means an equation, so we set 9x equal to x - 3 to get our [I]algebraic expression[/I]: [B]9x = x - 3[/B] To solve for x, we type this equation into our search engine and we get: x = [B]-0.375 or -3/8[/B]
A man's age (a) 10 years ago is 43
A man's age (a) 10 years ago is 43 [U]10 years ago means we subtract 10 from a:[/U] a - 10 [U]The word [I]is[/I] means an equation. So we set a - 10 equal to 43 to get our algebraic expression[/U] [B]a - 10 = 43[/B] If the problem asks you to solve for a, [URL='https://www.mathcelebrity.com/1unk.php?num=a-10%3D43&pl=Solve']we type this equation into our search engine[/URL] and we get: a = 53
a mans age (a) ten years ago
a mans age (a) ten years ago The problem asks for an algebraic expression for age. The phrase [I]ago[/I] means before now, so they were younger. And younger means we [B]subtract[/B] from our current age: [B]a - 10[/B]
A number y increased by itself
A number y increased by itself increased by itself means we add the variable y to itself to get our final algebraic expression of: [B]y + y [/B] [I]If[/I] the problem asks you to simplify, we group like terms and get: [B]2y[/B]
a printer charges a \$30 setup fee plus \$2.00 per ticket. Write an algebraic expression for the cost
a printer charges a \$30 setup fee plus \$2.00 per ticket. Write an algebraic expression for the cost of t tickets. What is the cost of 225 tickets? Algebraic Expression: Cost per ticket * t + set up fee [B]2t + 30[/B] How much for t = 225? 2(225) + 30 450 + 30 [B]480[/B]
A quarter of a number is greater than or equal to 38
A quarter of a number is greater than or equal to 38. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. A quarter of a number means 1/4, so we have: x/4 The phrase [I]is greater than or equal to[/I] means an inequality, so we use the >= sign in relation to 38: [B]x/4 >= 38 <-- This is our algebraic expression [/B] If you want to solve this inequality, [URL='https://www.mathcelebrity.com/prop.php?num1=x&num2=38&propsign=%3E%3D&den1=4&den2=1&pl=Calculate+missing+proportion+value']we type it in the search engine[/URL] to get: x >= [B]152[/B]
A restaurant is going to raise all their prices by 7%. If the current price of an item is p dollars
A restaurant is going to raise all their prices by 7%. If the current price of an item is p dollars, write an expression for the price after the increase. A 7% increase on price means we multiply the current price of p by 1.07. So our algebraic expression is: [B]1.07p[/B]
A secret number is added to 6. The total is multiplied by 5 to get 50. What is the secret number?
A secret number is added to 6. The total is multiplied by 5 to get 50. What is the secret number? Take this algebraic expression in pieces: [LIST] [*]Let the secret number be n. [*]Added to means we add 6 to n: n + 6 [*]The total is multiplied by 5: 5(n + 6) [*]The phrase [I]to get[/I] means equal to, so we set 5(n + 6) equal to 50 [/LIST] 5(n + 6) = 50 To solve this equation for n, we type it in our search engine and we get: n = [B]4[/B]
A state park charges a \$6.00 entry fee plus \$7.50 per night of camping. Write an algebraic expressio
A state park charges a \$6.00 entry fee plus \$7.50 per night of camping. Write an algebraic expression for the cost in dollars of entering the park and camping for n nights. The cost in dollars C is found below: [B]C = 7.50n + 6[/B]
A state park charges a \$6.00 entry fee plus \$7.50 per night of camping. Write an algebraic expressio
A state park charges a \$6.00 entry fee plus \$7.50 per night of camping. Write an algebraic expression for the cost in dollars of entering the park and camping for n nights. We write this as: cost per night of camping * n nights + entry fee [B]7.50n + 6[/B]
absolute value of x is less than or equal to 4
absolute value of x is less than or equal to 4 Absolute value of x: |x| Set up an inequality where this is less than or equal to 4: [B]|x| <= 4 [/B] <-- This is our algebraic expression To solve this, we have the following compound inequality: -4 < x < 4
add 7 and 2, raise the result to the 6th power, then add what you have to s
add 7 and 2, raise the result to the 6th power, then add what you have to s Add 7 and 2: 7 + 2 Simplify this, we get:9 Raise the result to the 6th power: 9^6 [URL='https://www.mathcelebrity.com/powersq.php?sqconst=+6&num=9%5E6&pl=Calculate']Simplifying this using our exponent calculator[/URL], we get: 531,441 Now, we add what we have (our result) to s to get our final algebraic expression: [B]s + 531,441[/B]
add c and b, multiply the result by a, then double what you have
add c and b, multiply the result by a, then double what you have Take this algebraic expression in pieces: [LIST] [*]add c and b: c + b [*]Multiply the result by a: a(c + b) [*]Double what you have means take the last step result, and multiply it by 2: [/LIST] [B]2a(c + b)[/B]
Add q and t, subtract s from the result, then multiply by r
Add q and t, subtract s from the result, then multiply by r Take this algebraic expression in parts: [LIST] [*]Add q and t: q + t [*]Subtract s from the result: q + t - s [*]Multiply by r means we multiply the entire expression by r: [/LIST] [B]r(q + t - s)[/B]
add r and s, add the result to q, then subtract what you have from p
add r and s, add the result to q, then subtract what you have from p Take this algebraic expression in 3 parts: [LIST=1] [*]Add r and s: r + s [*]Add the result to q: r + s + q [*]Subtract what we have from p: [/LIST] [B]p - (r + s + q)[/B]
add r to 3, triple the result, then divide s by what you have
add r to 3, triple the result, then divide s by what you have Take this algebraic expression in parts: [LIST=1] [*]Add r to 3: 3 + r [*]Triple the result means multiply the result above by 3: 3(3 + r) [*]Then divide s by what you have. [B]s/3(3 + r)[/B] [/LIST]
add s and t, multiply the result by u, then add r to what you have
add s and t, multiply the result by u, then add r to what you have. Take this algebraic expression in 3 parts: [LIST=1] [*]Add s and t: s + t [*]Multiply the result by u means me multiply (s + t) times u: u(s + t) [*]Then add r to what you have. [I]what you have means the result in #2.[/I] [/LIST] [B]u(s + t) + r[/B]
add s to v, multiply the result by u, then multiply t by what you have
add s to v, multiply the result by u, then multiply t by what you have Take this algebraic expression in parts: [LIST] [*]Add s to v: v + s [*]Multiply the result by u: u(v + s) [*]Then multiply t by what you have: [/LIST] [B]tu(v + s)[/B]
add u and t divide s by the result then triple what you have
add u and t divide s by the result then triple what you have Take this algebraic expression in parts: [LIST] [*]Add u and t: u + t [*]Divide s by the result: s/(u + t) [*]Triple what you have means we you multiply s/(u + t) by 3 [/LIST] [B]3s/(u + t)[/B]
add w to t, add u to the result, then divide what you have by v
add w to t, add u to the result, then divide what you have by v Take this algebraic expression in parts: [LIST] [*]Add w to t: t + w [*]Add u to the result: t + w + u [*]Divide what you have by v: [/LIST] ([B]t + w + u)/v[/B]
add w to u, triple the result, then add v to what you have
add w to u, triple the result, then add v to what you have Take this algebraic expression in parts: [LIST] [*]add w to u: w + u [*]triple the result means we multiply w + u by 3: 3(w + u) [*]Then add v to what you have [/LIST] [B]3(w + u) + v[/B]
algebraic expression for the sum of x and double the value of y
algebraic expression for the sum of x and double the value of y Double the value of y means we multiply y by 2: 2y The sum of x and 2y means we add 2y to x: [B]x + 2y[/B]
Algebraic Expressions
Free Algebraic Expressions Calculator - This calculator builds algebraic expressions based on word representations of numbers using the four operators and the words that represent them(increased,product,decreased,divided,times) Also known as Mathematical phrases
algexpress: letthefirstnumberequalx.thesecondnumberis3morethantwicethefirstnumber.expressthesecondnu
Let the first number equal x. The second number is 3 more than twice the first number. Express the second number in terms of the first number x. [LIST] [*]Let the second number be y. [*]Twice means multiply by 2 [*]3 more than means we add 3 [/LIST] So we have the following algebraic expression: [B]y = 2x + 3[/B]
Alyssa has \$952 and is spending \$27 each week (w) for math tutoring write an algebraic expression to
Alyssa has \$952 and is spending \$27 each week (w) for math tutoring write an algebraic expression to model the situation Alyssa's balance is found by using this expression: [B]952 - 27w[/B]
Amount you spend if you buy a shirt for \$20 and jeans for j dollars
Amount you spend if you buy a shirt for \$20 and jeans for j dollars We want an algebraic expression for our total spend. We add the \$20 for a shirt plus j for the jeans: [B]20 + j[/B]
b to the fifth power decreased by 7
b to the fifth power decreased by 7 Take this algebraic expression in steps: [LIST] [*]b to the fifth power: b^5 [*]Decreased by 7 means we subtract 7 from b^5: [B]b^5 - 7[/B] [/LIST]
Charmaine’s fish tank has 16 liters of water in it. she plans to add 6 liters per minute until the t
Charmaine’s fish tank has 16 liters of water in it. she plans to add 6 liters per minute until the tank has at least 58 liters. What are the possible numbers of minutes Charmaine could add water? This is an algebraic inequality. The phrase [I]at least[/I] means greater than or equal to. So we have: 6m + 16 >= 58 <-- This is our algebraic expression/inequality. To solve this, [URL='https://www.mathcelebrity.com/1unk.php?num=6m%2B16%3E%3D58&pl=Solve']we type this into our search engine [/URL]and we get: [B]m >= 7[/B]
Cindy is c years old. Cindy is 5 years younger than half Jennifer's age ( j )
Cindy is c years old. Cindy is 5 years younger than half Jennifer's age ( j ) Build an algebraic expression: [B]c = j/2 - 5[/B] <-- Half means we divide by 2 and [I]younger[/I] means we subtract
Difference between 23 and y is 12
Difference between 23 and y 23 - y Is, means equal to, so we set 23 - y equal to 12 [B]23 - y = 12 [/B] If you need to solve this algebraic expression, use our [URL='http://www.mathcelebrity.com/1unk.php?num=23-y%3D12&pl=Solve']equation calculator[/URL]: [B]y = 11[/B]
divide 8 by t, raise the result to the 7th power
divide 8 by t, raise the result to the 7th power. We take this algebraic expression in two parts: 1. Divide 8 by t 8/t 2. Raise the result to the 7th power. (This means we use an exponent of 7) [B](8/t)^7[/B]
Divide a by b, double the result, then multiply c by what you have
Divide a by b, double the result, then multiply c by what you have Take this algebraic expression in parts: [LIST] [*]Divide a by b: a/b [*]Double the result means multiply by 2: 2a/b [*]Then multiply c by what you have: [/LIST] [B]2ac/b[/B]
divide a by c, triple the result, then subtract what you have from b
divide a by c, triple the result, then subtract what you have from b Let's take this algebraic expression in parts: [LIST=1] [*]Divide a by c: a/c [*]Triple the result. This means we multiply a/c by 3: 3a/c [*]Then subtract what you have (the result) from b: b - 3a/c [/LIST] [B]b - 3a/c[/B]
divide b by a, subtract the result from c, then add what you have to d
divide b by a, subtract the result from c, then add what you have to d Take this algebraic expression in 3 parts: [U]1) Divide b by a:[/U] b/a [U]2) Subtract the result from c:[/U] c - b/a [U]3) Then add what you have to d:[/U] [B]c - b/a + d[/B]
divide the difference of q and s by the sum of p and r
divide the difference of q and s by the sum of p and r Take this algebraic expression in pieces: [LIST] [*]The difference of q and s: q - s [*]The sum of p and r: p + r [*]The word [I]divide[/I] means we divide q - s by p + r [/LIST] [B](q - s)/(p + r)[/B]
Divide the sum x and y by the difference of subtracting a from b
Divide the sum x and y by the difference of subtracting a from b The sum x and y is written as: x + y The difference of subtracting a from b is written as: b - a We divide and get the algebraic expression: [B](x + y)/(b - a)[/B]
Divide v by the sum of 4 and w
Divide v by the sum of 4 and w The sum of 4 and w means we add w to 4: 4 + w Next, we divide v by this sum to get our final algebraic expression: [B]v/(4 + w)[/B]
Divya has 70 rocks. She donates half of the rocks to a science center. Then she collects 3 rocks on
Divya has 70 rocks. She donates half of the rocks to a science center. Then she collects 3 rocks on each of her nature hikes. Write an expression to represent the number of rocks Divya has after she collects rocks on n nature hikes. For each hike, we have: [LIST=1] [*]Start with 70 rocks [*]She donates half which is 35, which means she's left with 35 [/LIST] Since each nature hike gives her 3 more rocks, and she goes on n nature hikes, we have the following algebraic expression: [B]3n + 35[/B]
Each piece of candy costs 25 cents. The cost of x pieces of candy is \$2.00. Use variable x to transl
Each piece of candy costs 25 cents. The cost of x pieces of candy is \$2.00. Use variable x to translate the above statements into algebraic equation. Our algebraic expression is: [B]0.25x = 2 [/B] To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.25x%3D2&pl=Solve']type it in our search engine[/URL] and we get: x = [B]8[/B]
Expand Master and Build Polynomial Equations
Free Expand Master and Build Polynomial Equations Calculator - This calculator is the ultimate expansion tool to multiply polynomials. It expands algebraic expressions listed below using all 26 variables (a-z) as well as negative powers to handle polynomial multiplication. Includes multiple variable expressions as well as outside multipliers.
Also produces a polynomial equation from a given set of roots (polynomial zeros). * Binomial Expansions c(a + b)x
* Polynomial Expansions c(d + e + f)x
* FOIL Expansions (a + b)(c + d)
* Multiple Parentheses Multiplications c(a + b)(d + e)(f + g)(h + i)
Express the fact that x differs from 3 by less than 2/7 as an inequality involving absolute value. S
Express the fact that x differs from 3 by less than 2/7 as an inequality involving absolute value. Solve for x. Let's build this algebraic expression in pieces: The phrase [I]differs from[/I] means a difference. x - 3 By less than 2/7 means we use the < sign compared to 2/7 x - 3 < 2/7 Finally, the problem says we involve absolute value. So we write this as: [B]|x - 3| < 2/7[/B]
Five times Kim's age plus 13 equals 58. How old is Kim?
Five times Kim's age plus 13 equals 58. How old is Kim? Let Kim's age be a. We have: Five times Kim's age: 5a Plus 13 means we add 13 5a + 13 Equals 58 means we set the expression 5a + 13 equal to 58 5a + 13 = 58 <-- This is our algebraic expression To solve this equation for a, [URL='https://www.mathcelebrity.com/1unk.php?num=5a%2B13%3D58&pl=Solve']we type it into our search engine[/URL] and get: a = [B]9[/B]
Four times the quantity six plus two
six plus two: 6 + 2 Four times the quantity six plus two [B]4(6 + 2) [/B]<-- This is our algebraic expression If we need to evaluate this, we have: 4(8) [B]32[/B]
The phrase a number means an arbitrary variable, let's call it x. Three times a number: 3x And 18 means we add 18 3x + 18 The word is means equal to, so we set 3x + 18 equal to -39 3x + 18 = -39 This is your algebraic expression. If you want to solve for x, plug it into the [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B18%3D-39&pl=Solve']search engine[/URL] and you get x = -19
How old am I if 400 reduced by 3 times my age is 124?
How old am I if 400 reduced by 3 times my age is 124? Let my age be a. We're given an algebraic expression: [LIST] [*]3 times my age means we multiply a by 3: 3a [*]400 reduced by 3 times my age means we subtract 3a from 400: [*]400 - 3a [*]The word [I]is[/I] mean an equation, so we set 400 - 3a equal to 124 [/LIST] 400 - 3a = 124 To solve for a, [URL='https://www.mathcelebrity.com/1unk.php?num=400-3a%3D124&pl=Solve']we type this equation into our search engine[/URL] and we get: a = [B]92[/B]
How old am I of 400 reduced by 2 times my age is 224
How old am I of 400 reduced by 2 times my age is 224 [LIST=1] [*]Let my age be a. [*]2 times my age: 2a [*]400 reduced by 2 times my age: 400 - 2a [*]The phrase [I]is [/I]means an equation. So we set 400 - 2a equal to 224 for our algebraic expression [/LIST] [B]400 - 2a = 224 [/B] If the problem asks you to solve for a, we [URL='https://www.mathcelebrity.com/1unk.php?num=400-2a%3D224&pl=Solve']type this equation into our search engine[/URL] and we get: a = [B]88[/B]
If 11 times a number is added to twice the number, the result is 104
If 11 times a number is added to twice the number, the result is 104 Let [I]the number[/I] be an arbitrary variable we call x. 11 times a number: 11x Twice the number (means we multiply x by 2): 2x The phrase [I]is added to[/I] means we add 2x to 11x: 11x + 2x Simplify by grouping like terms: (11 + 2)x = 13x The phrase [I]the result is[/I] means an equation, so we set 13x equal to 104: 13x = 104 <-- This is our algebraic expression To solve this equation for x, [URL='https://www.mathcelebrity.com/1unk.php?num=13x%3D104&pl=Solve']we type it in our search engine[/URL] and we get: x = [B]8[/B]
If 4 times a number is added to 9, the result is 49
If 4 times a number is added to 9, the result is 49. [I]A number[/I] means an arbitrary variable, let's call it x. 4 [I]times a number[/I] means we multiply x by 4 4x [I]Added to[/I] 9 means we add 9 to 4x 4x + 9 [I]The result is[/I] means we have an equation, so we set 4x + 9 equal to 49 [B]4x + 9 = 49[/B] <-- This is our algebraic expression To solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=4x%2B9%3D49&pl=Solve']we type it in the search engine[/URL] and get x = 10
If 72 is added to a number it will be 4 times as large as it was originally
If 72 is added to a number it will be 4 times as large as it was originally The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. x 72 added to a number: x + 72 4 times as large as it was originally means we take the original number x and multiply it by 4: 4x Now, the phrase [I]it will be[/I] means an equation, so we set x + 72 equal to 4x to get our final algebraic expression: [B]x + 72 = 4x[/B] [B][/B] If the problem asks you to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x%2B72%3D4x&pl=Solve']type this equation into our search engine[/URL] and we get: x = [B]24[/B]
if a number is tripled the result is 60
if a number is tripled the result is 60 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Triple the number means we multiply by 3: 3x The phrase [I]the result is[/I] means an equation, so we set 3x equal to 60: [B]3x = 60 <-- This is our algebraic expression [/B] If you want to solve this equation, then [URL='https://www.mathcelebrity.com/1unk.php?num=3x%3D60&pl=Solve']you type in 3x = 60 into the search engine[/URL] and get: x = 20
If from twice a number you subtract four, the difference is twenty
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number means we multiply x by 2: 2x Subtract four: 2x - 4 The word [I]is [/I]means equal to. We set 2x - 4 equal to 20 for our algebraic expression: [B]2x - 4 = 20 [/B] If the problem asks you to solve for x: we [URL='https://www.mathcelebrity.com/1unk.php?num=2x-4%3D20&pl=Solve']plug this equation into our calculator [/URL]and get x = [B]12[/B]
If i triple the number then subtract 7 the answer is 2. What is the number
If i triple the number then subtract 7 the answer is 2. What is the number Let the number be x. Triple the number: 3x Subtract 7 3x - 7 The answer is 2 means we set: [B]3x - 7 = 2[/B] This is our algebraic expression. To solve this, [URL='https://www.mathcelebrity.com/1unk.php?num=3x-7%3D2&pl=Solve']we type this problem into the search engine[/URL] and get [B]x = 3[/B].
If sales tax is currently 8.2%, write an algebraic expression representing the amount of sales tax y
If sales tax is currently 8.2%, write an algebraic expression representing the amount of sales tax you would have to pay for an item that costs D dollars. 8.2% is 0.082 as a decimal. So we have: Sales Tax Paid = [B]0.082D[/B]
If thrice a number is increased by 11,the result is 35. What is the number
If thrice a number is increased by 11,the result is 35. What is the number? [LIST] [*]The phrase [I]a number [/I]means an arbitrary variable. Let's call it x. [*]Thrice means multiply by 3, so we have 3x [*]Increased by 11 means we add 11, so we have 3x + 11 [*]The [I]result is[/I] means an equation, so we set 3x + 11 equal to 35 [/LIST] 3x + 11 = 35 <-- This is our algebraic expression The problem ask us to solve the algebraic expression. [URL='https://www.mathcelebrity.com/1unk.php?num=3x%2B11%3D35&pl=Solve']Typing this problem into our search engine[/URL], we get [B]x = 8[/B].
If twice a number is divided by 7, the result is -28
If twice a number is divided by 7, the result is -28. The phrase [I]a number[/I] means an arbitrary variable, let's call it "x". Twice x means we multiply x by 2: 2x Divide this by 7: 2x/7 We set this equal to -28, and we have our algebraic expression: [B]2x/7 = -28 [/B]
if you add 7 to 2x, the result is 17
if you add 7 to 2x, the result is 17 Add 7 to 2x: 2x + 7 The phrase [I]the result is[/I] means an equation, so we set 2x + 7 = 17 [B]2x + 7 = 17 [/B] <-- This is our algebraic expression Now, if you want to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B7%3D17&pl=Solve']type in 2x + 7 = 17 into the search engine[/URL], and we get [B]x = 5[/B].
is 6x a monomial?
[B]Yes[/B]. It's an algebraic expression consisting of one term. The constant is 6, and the variable is x.
Last year, Greg biked 524 miles. This year, he biked m miles. Using m , write an expression for the
Last year, Greg biked 524 miles. This year, he biked m miles. Using m , write an expression for the total number of miles he biked. We add both years to get our algebraic expression of miles biked: [B]m + 524[/B]
Lauren's savings increased by 12 and is now 31
Lauren's savings increased by 12 and is now 31 [LIST] [*]Let Lauren's savings be s. [*]The phrase increased by means we add. [*]The phrase [I]is now[/I] means an equation. [*]We have an algebraic expression of: [/LIST] [B]s + 12 = 31 [/B] To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=s%2B12%3D31&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]19[/B]
M is halved, then 7 is added
M is halved, then 7 is added Take this algebraic expression in parts: [LIST] [*]M is halved. This means we divide M by 2: M/2 [*]Then 7 is added. We add 7 to M/2 [/LIST] [B]M/2 + 7[/B]
Multiply 0 by 3 and add 4
Multiply 0 by 3 and add 4 multiply 0 by 3: 0 * 3 Then add 4: [B]0 * 3 + 4 <--- [/B][I]This is our algebraic expression.[/I] If we want to evaluate this expression, we [URL='https://www.mathcelebrity.com/order-of-operations-calculator.php?num=0%2A3%2B4&pl=Perform+Order+of+Operations']type it in our search engine[/URL] and we get: [B]4[/B]
multiply 3 by the difference of u and t
multiply 3 by the difference of u and t Take this algebraic expression in parts: The difference of u and t means we subtract t from u u - t Multiply this difference by 3: [B]3(u - t)[/B]
multiply k by 5.8, and then subtract 3.09 from the product
multiply k by 5.8, and then subtract 3.09 from the product Take this algebraic expression in pieces: [U]Multiply k by 5.8:[/U] 5.8k [U]Then subtract 3.09 from the product[/U] [B]5.8k - 3.09[/B]
multiply m by 5, double the result, then multiply 10 by what you have
multiply m by 5, double the result, then multiply 10 by what you have Take this algebraic expression in parts: [LIST] [*]Multiply m by 5: 5m [*]double the result means multiply 5m by 2: 2(5m) = 10m [*]Multiply 10 by what you have means multiply 10 by the result of 10m above: [/LIST] 10(10m) = [B]100m[/B]
multiply r by t, add the result to u, then multiply what you have by s
multiply r by t, add the result to u, then multiply what you have by s Take this algebraic expression in parts: [LIST=1] [*]Multiply r by t: rt [*]Add the result to u means we add rt to u: u + r [*]Multiply what you have by s. This means we take the result in #2, u + r, and multiply it by s: [/LIST] [B]s(u + r)[/B]
Multiply the difference of 3 and q by p
Multiply the difference of 3 and q by p. Take this algebraic expression in pieces: [B][U]Step 1: The difference of 3 and q[/U][/B] The word [I]difference[/I] means we subtract the variable q from 3 3 - q [B][U]Step 2: Multiply the expression 3 - q by p:[/U] p(3 - q)[/B]
multiply u by s, multiply the result by v, then multiply t
multiply u by s, multiply the result by v, then multiply t Take this algebraic expression in parts: [LIST] [*]Multiply u by s: us [*]Multiply the result by v: usv [*]Then multiply by t: [B]usvt[/B] [/LIST]
Multiplying a number by 6 is equal to the number increased by 9
Multiplying a number by 6 is equal to the number increased by 9. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Multiply it by 6 --> 6x We set this equal to the same number increased by 9. Increased by means we add: [B]6x = x + 9 <-- This is our algebraic expression [/B] To solve this equation, we [URL='https://www.mathcelebrity.com/1unk.php?num=6x%3Dx%2B9&pl=Solve']type it into the search engine [/URL]and get x = 1.8.
n increased by the difference between 10 times n and 9
n increased by the difference between 10 times n and 9 Take this algebraic expression in pieces: [LIST] [*]10 times n: 10n [*]The difference between 10 times n and 9: 10n - 9 [*]n increased by the difference...: [B]n + (10n - 9)[/B] [/LIST]
n subtract m, multiply by c, then add w
n subtract m, multiply by c, then add w Take this algebraic expression in pieces: [LIST] [*]n subtract m: n - m [*]multiply by c: c(n - m) [*]Then add w: [B]c(n - m) + w[/B] [/LIST]
Nine less than twelve
less than means we subtract 9 from 12: [B]12 - 9[/B] <-- This is our algebraic expression IF we want to evaluate this, it's: 12- 9 = [B]3[/B]
nine times x is twice the sum of x and five
nine times x is twice the sum of x and five Take this algebraic expression in 4 pieces: [U]Step 1: nine time x:[/U] 9x [U]Step 2: The sum of x and five means we add 5 to x:[/U] x + 5 [U]Step 3: The word [I]twice[/I] means we multiply the sum x + 5 by 2:[/U] 2(x + 5) [U]Step 4: The word [I]is[/I] means equal to, so we set 9x equal to 2(x + 5) to get our final algebraic expression of:[/U] [B]9x = 2(x + 5)[/B]
One fifth of the square of a number
One fifth of the square of a number We have an algebraic expression. Let's break this into parts. [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]The square of a number means we raise it to the power of 2. So we have x^2 [*]One-fifth means we have a fraction, where we divide our x^2 in Step 2 by 5. So we get our final answer below: [/LIST] [B]x^2/5[/B]
One-fourth the sum of m and p
One-fourth the sum of m and p Take this algebraic expression in parts: [LIST] [*]The sum of m and p means we add p to m: m + p [*]1/4 of the sum mean we divide m + p by 4 [/LIST] [B](m + p)/4[/B]
One-half a number is fifty
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. One-half a number means we divide x by2: x/2 The word [I]is[/I] means equal to. We set x/2 equal to 50 for our algebraic expression [B]x/2 = 50 [/B] If the problem asks us to solve for x, we cross multiply: x = 2 * 50 x = [B]100[/B]
p more than the square of q
p more than the square of q Take this algebraic expression in parts: Step 1: Square of q means raise q to the 2nd power: q^2 Step 2: The phrase [I]more[/I] means we add p to q^2 [B]q^2 + p[/B]
Peter was thinking of a number. Peter doubles it and adds 0.8 to get an answer of 31. Form an equati
Peter was thinking of a number. Peter doubles it and adds 0.8 to get an answer of 31. Form an equation with x from the information. Take this algebraic expression in parts, starting with the unknown number x: [LIST] [*]x [*][I]Double it [/I]means we multiply x by 2: 2x [*]Add 0.8: 2x + 0.8 [*]The phrase [I]to get an answer of[/I] means an equation. So we set 2x + 0.8 equal to 31 [/LIST] Build our final algebraic expression: [B]2x + 0.8 = 31[/B] [B][/B] If you have to solve for x, then we [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B0.8%3D31&pl=Solve']type this equation into our search engine[/URL] and we get: x = 15.1
q increased by the difference between 18 times q and 5
q increased by the difference between 18 times q and 5 Take this algebraic expression in parts. 18 times q: 18q The difference between 18 times q and 5 means we subtract 5 from 18q: 18q - 5 q increased by the difference between 18 times q and 5 means we add 18q - 5 to q: q + (18q - 5) [B]q + 18q - 5[/B] IF we want to simplify, we group like terms: [B]19q - 5[/B]
q is equal to 207 subtracted from the quantity 4 times q
q is equal to 207 subtracted from the quantity 4 times q 4 time q 4q 207 subtracted from 4 times q: 4q - 207 Set this equal to q: [B]4q - 207 = q [/B]<-- This is our algebraic expression To solve for q, [URL='https://www.mathcelebrity.com/1unk.php?num=4q-207%3Dq&pl=Solve']type this equation into the search engine[/URL]. We get: [B]q = 69[/B]
Raise 9 to the 3rd power, subtract d from the result, then divide what you have by c
Raise 9 to the 3rd power, subtract d from the result, then divide what you have by c. This is an algebraic expression, let's take in parts (or chunks). Raise 9 to the 3rd power. This means we take 9, and raise it to an exponent of 3 9^3 Subtract d from the result, means we subtract d from 9^3 9^3 - d Now we divide 9^3 - d by c [B](9^3 - d) / c[/B]
Raise c to the 7th power, divide the result by 4, then triple what you have
Raise c to the 7th power, divide the result by 4, then triple what you have. Take this algebraic expression in pieces. Raise c to the 7th power: c^7 Divide the result by 4, means we divide c^7 by 4 c^7 / 4 Triple what you have means multiply c^7 / 4 by 3 [B]3(c^7 / 4)[/B]
raise f to the 3rd power, then find the quotient of the result and g
raise f to the 3rd power, then find the quotient of the result and g Take this algebraic expression in two parts: [LIST=1] [*]Raise f to the 3rd power means we take f, and write it with an exponent of 3: f^3 [*]Find the quotient of the result and g. We take f^3, and divide it by g [/LIST] [B]f^3/g[/B]
Raise f to the 8th power, divide the result by 5, then multiply 10
Raise f to the 8th power, divide the result by 5, then multiply 10 f to the 8th power means we raise f to the power of 8 using an exponent: f^8 Divide f^8 by 5 (f^8)/5 Now multiply this by 10: 10(f^8)/5 We can simplify this algebraic expression by dividing 10/5 to get 2 on top: 2[B](f^8)[/B]
raise q to the 5th power add the result to p then divide what you have by r
raise q to the 5th power add the result to p then divide what you have by r Take this algebraic expression in parts: [LIST] [*]Raise q to the 5th power: q^5 [*]Add the result to p: p + q^5 [*]Divide what you have by r. This means we take our result above and divide it by r: [/LIST] [B](p + q^5)/r[/B]
raise z to the 2nd power, multiply 8 by the result then subtract what you have from 4
raise z to the 2nd power, multiply 8 by the result then subtract what you have from 4 Take this algebraic expression in pieces: [LIST] [*]Raise z to the 2nd power: z^2 [*]Multiply by 8: 8z^2 [*]Subtract what you have from 4: [/LIST] [B]4 - 8z^2[/B]
ratio of the squares of t and u
ratio of the squares of t and u Ratio is also known as quotient in algebraic expression problems. The square of t means we raise t to the power of 2: t^2 The square of u means we raise u to the power of 2: u^2 ratio of the squares of t and u means we divide t^2 by u^2: [B]t^2/u^2[/B]
Sales tax is currently 9.1%. Write an algebraic expression to represent the total amount paid for an
Sales tax is currently 9.1%. Write an algebraic expression to represent the total amount paid for an item that costs d dollars after tax is added to the purchase. We need to increase the price by 9.1%. Our expression is: [B]1.091d[/B]
Simplest Exponent Form
Free Simplest Exponent Form Calculator - This expresses repeating algebraic expressions such as 3*a*a*a*b*b into simplest exponent form.
Six less two
less means subtract: [B]6 - 2[/B] <-- This is our algebraic expression If we need to evaluate we have: 6 - 2 = [B]4[/B]
Sixteen subtracted from five times a number equals the number plus four
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. 5 times a number 5x Sixteen subtracted from five times a number 5x - 16 the number plus 4: x + 4 Equals means we set 5x - 16 equals to x + 4 for our algebraic expression: [B]5x - 16 = x + 4[/B] [B][/B] If you have to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=5x-16%3Dx%2B4&pl=Solve']type this expression into our math solver[/URL] and we get: x = [B]5[/B]
Squaring a number equals 5 times that number
Squaring a number equals 5 times that number. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Squaring this number: x^2 5 times this number means we multiply by 5: 5x The phrase [I]equals[/I] means we set both expressions equal to each other: [B]x^2 = 5x [/B] <-- This is our algebraic expression If you want to solve for x, then we subtract 5x from each side: x^2 - 5x = 5x - 5x Cancel the 5x on the right side, leaving us with 0: x^2 - 5x = 0 Factor out x: x(x - 5) So we get x = 0 or [B]x = 5[/B]
Start with x , subtract 6, then times by 3. We start with x: x Subtract 6: x - 6 The phrase [I]times by[/I] means we multiply (x - 6) by 3 [B]3(x - 6) [/B] <-- This is our algebraic expression If the problem asks you to multiply through, then you'd have: 3x - 18
Subtract 4 from the sum of 2x and 5y
Subtract 4 from the sum of 2x and 5y. The sum of 2x and 5y means we add both terms: 2x + 5y Subtract 4 from this sum to get our algebraic expression: [B](2x + 5y) - 4[/B]
subtract 5 from the sum of 3x and 8y
subtract 5 from the sum of 3x and 8y Take this algebraic expression in parts: [U]The sum of 3x and 8y means we add 8y to 3x:[/U] 3x + 8y [U]Subtract 5 from this sum above:[/U] [B]3x + 8y - 5[/B]
subtract w from u, triple the result, then multiply v by what you have
subtract w from u, triple the result, then multiply v by what you have Take this algebraic expression in 3 parts: [U]1) subtract w from u:[/U] u - w [U]2) Triple the result means we multiply u - w by 3:[/U] 3(u - w) [U]3) Multiply v by what you have. [I]What you have[/I] means the result from step 2:[/U] [B]3v(u - w)[/B]
subtract w from v, add the result to u, then triple what you have
subtract w from v, add the result to u, then triple what you have Take this algebraic expression in parts: [LIST=1] [*]Subtract w from v: v - w [*]Add the result to u (the result is #1): u + v - w [*]Triple what you have. This means multiply the result in #2 by 3: [/LIST] [B]3(u + v - w)[/B]
sum of 5 times h and twice g is equal to 23
sum of 5 times h and twice g is equal to 23 Take this [U]algebraic expressions[/U] problem in pieces. Step 1: 5 times h: 5h Step 2: Twice g means we multiply g by 2: 2g Step 3: sum of 5 times h and twice g means we add 2g to 5h 5h + 2g Step 4: The phrase [I]is equal to[/I] means an equation, so we set 5h + 2g equal to 23: [B]5h + 2g = 23[/B]
sum of a number and 7 is subtracted from 15 the result is 6.
Sum of a number and 7 is subtracted from 15 the result is 6. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We take this expression in pieces. Sum of a number and 7 x + 7 Subtracted from 15 15 - (x + 7) The result is means an equation, so we set this expression above equal to 6 [B]15 - (x + 7) = 6 <-- This is our algebraic expression[/B] If the problem asks you to solve for x, we Group like terms 15 - x - 7 = 6 8 - x = 6 [URL='https://www.mathcelebrity.com/1unk.php?num=8-x%3D6&pl=Solve']Type 8 - x = 6 into the search engine[/URL], and we get [B]x = 2[/B]
Sum of a number and it's reciprocal is 6. What is the number?
Sum of a number and it's reciprocal is 6. What is the number? Let the number be n. The reciprocal is 1/n. The word [I]is[/I] means an equation, so we set n + 1/n equal to 6 n + 1/n = 6 Multiply each side by n to remove the fraction: n^2 + 1 = 6n Subtract 6n from each side: [B]n^2 - 6n + 1 = 0 [/B]<-- This is our algebraic expression If the problem asks you to solve for n, then you [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-6n%2B1%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this quadratic equation into our search engine[/URL].
The area of a desert in Africa is 12 times the area of a desert in Asia. If the area of a desert in
The area of a desert in Africa is 12 times the area of a desert in Asia. If the area of a desert in Asia is Y square miles, express the area of a desert in Africa as an algebraic expression in Y. [B]Africa Area = 12Y[/B]
The cube of the difference of 5 times the square of y and 7 divided by the square of 2 times y
The cube of the difference of 5 times the square of y and 7 divided by the square of 2 times y Take this in algebraic expression in parts: [U]Term 1[/U] [LIST] [*]The square of y means we raise y to the 2nd power: y^2 [*]5 times the square of y: 5y^2 [/LIST] [U]Term 2[/U] [LIST] [*]2 times y: 2y [*]The square of 2 times y: (2y)^2 = 4y^2 [*]7 divide by the square of 2 times y: 7/4y^2 [/LIST] [U]The difference of these terms is written as Term 1 minus Term 2:[/U] [LIST] [*]5y^2/4y^2 [/LIST] [U]The cube of the difference means we raise the difference to the power of 3:[/U] [B](5y^2/4y^2)^3[/B]
the cube of the difference of 5 times x and 4
the cube of the difference of 5 times x and 4 Take this algebraic expression in pieces: 5 times x: 5x The difference of 5x and 4 means we subtract 4 from 5x: 5x - 4 We want to cube this difference, which means we raise the difference to the power of 3. [B](5x - 4)^3[/B]
The difference between a and b is 10
The difference between a and b is 10. The problem asks for an algebraic expression. Let's take each piece one by one: [I]Difference between[/I] means we subtract: a - b The phrase [I]is [/I]means an equation, so we set a - b equal to 10 [B]a - b = 10[/B]
The difference in Julies height and 9 is 48 letting j be Julie's height
The difference in Julies height and 9 is 48 letting j be Julie's height Step 1: If Julie's height is represented with the variable j, then we subtract 9 from j since the phrase [I]difference[/I] means we subtract: j - 9 Step 2: The word [I]is[/I] means an equation, so we set j - 9 equal to 48 for our final algebraic expression: [B]j - 9 = 48[/B]
The difference of a number times 3 and 6 is equal to 7 . Use the variable w for the unknown n
The difference of a number times 3 and 6 is equal to 7 . Use the variable w for the unknown number. The phrase a number uses the variable w. 3 times w is written as 3w The difference of 3w and 6 is written as 3w - 6 Set this equal to 7 [B]3w - 6 = 7 [/B] This is our algebraic expression. To solve this equation for w, we [URL='http://www.mathcelebrity.com/1unk.php?num=3w-6%3D7&pl=Solve']type the algebraic expression into our search engine[/URL].
The difference of five and five y is the same as eight and two y
The difference of five and five y 5 - 5y eight and two y 8 + 2y The phrase [I]is the same as[/I] means equal to. Set 5 - 5y equal to 8 + 2y for our final algebraic expression [B]5 - 5y = 8 + 2y[/B] [B][/B] If the problem asks you to solve for y: Add 5y to each side: 5 = 8 + 7y Subtract 8 from each side: 7y = -3 Divide each side by 7: [B]y= -3/7[/B]
The difference of twice a number and 6 is at most 28
The difference of twice a number and 6 is at most 28 This is an algebraic expression. Let's take it in parts: [LIST=1] [*]The phrase [I]a number[/I], means an arbitrary variable, let's call it x [*]Twice this number means we multiply x by 2: 2x [*][I]The difference of[/I] means subtract, so we subtract 6 to 2x: 2x - 6 [*][I]Is at most [/I]means less than or equal to, so we create an inequality where 2x - 6 is less than or equal to 28, using the <= sign [/LIST] [B]2x - 6 <= 28 [/B] If you wish to solve this inequality, [URL='https://www.mathcelebrity.com/1unk.php?num=2x-6%3C%3D28&pl=Solve']click this link[/URL].
the difference of twice a number and 8 is at most -30
the difference of twice a number and 8 is at most -30. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Twice this number means we multiply by 2, so we have 2x. We take the difference of 2x and 8, meaning we subtract 8: 2x - 8 Finally, the phrase [I]at most[/I] means an inequality, also known as less than or equal to: [B]2x - 8 <= 30 <-- This is our algebraic expression [/B] To solve this, we [URL='https://www.mathcelebrity.com/1unk.php?num=2x-8%3C%3D30&pl=Solve']type it into the search engine[/URL] and get x <= 19.
the difference of x and y added to twice the sum of a and b
the difference of x and y added to twice the sum of a and b Take this algebraic expression in parts: [LIST] [*]The difference of x and y: x - y [*]The sum of a and b: a + b [*]Twice the sum of a and b means we multiply a + b by 2: 2(a + b) [*]The phrase [I]added to[/I] means we add: [/LIST] [B]x - y + 2(a + b)[/B]
The larger of 2 numbers is 1 more than 3 times the smaller number
The larger of 2 numbers is 1 more than 3 times the smaller number. Let the larger number be l. Let the smaller number be s. The algebraic expression is: 3 times the smaller number is written as: 3s 1 more than that means we add 1 3s + 1 Our final algebraic expression uses the word [I]is[/I] meaning an equation. So we set l equal to 3s + 1 [B]l = 3s + 1[/B]
The length of a train car is 50.6 feet. This is 5.8 feet less than 6 times the width. What is the wi
The length of a train car is 50.6 feet. This is 5.8 feet less than 6 times the width. What is the width? 5.8 feet less than 6 times the width is an algebraic expression: 6w - 5.8 We set this equal to the length of 50.6 6w - 5.8 = 50.6 Add 5.8 to each side: 6w - 5.8 + 5.8 = 50.6 + 5.8 Cancel the 5.8 on the left side: 6w = 56.4 Divide each side by 6: 6w/6 = 56.4/6 [URL='http://www.mathcelebrity.com/1unk.php?num=6w-5.8%3D50.6&pl=Solve']Typing this problem into the search engine[/URL], we get [B]w = 9.4[/B]. [MEDIA=youtube]gfM-d_Ej728[/MEDIA]
the product of 2 less than a number and 7 is 13
the product of 2 less than a number and 7 is 13 Take this algebraic expression in [U]4 parts[/U]: Part 1 - The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Part 2 - 2 less than a number means we subtract 2 from x x - 2 Part 3 - The phrase [I]product[/I] means we multiply x - 2 by 7 7(x - 2) Part 4 - The phrase [I]is[/I] means an equation, so we set 7(x - 2) equal to 13 [B]7(x - 2) = 13[/B]
the product of 8 and 15 more than a number
the product of 8 and 15 more than a number. Take this algebraic expression in pieces. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 15 more than x means we add 15 to x: x + 15 The product of 8 and 15 more than a number means we multiply 8 by x + 15 [B]8(x + 15)[/B]
the product of a number and 15 is not less than 15
the product of a number and 15 is not less than 15 The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. x the product of a number and 15 means we multiply x by 15 15x The phrase [I]not less than[/I] means greater than or equal to. We set 15x greater than prequel to 15 [B]15x >= 15 <-- This is our algebraic expression [/B] [U]If the problem asks you to solve for x:[/U] Divide each side by 15: 15x/15 >= 15/15 [B]x >= 1[/B]
the product of k and 70, minus 15
the product of k and 70, minus 15 Take this algebraic expression in pieces: The product of k and 70 means we multiply 70 times k 70k The word [I]minus[/I] means we subtract 15 from 70k [B]70k - 15[/B]
The product of two consecutive integers is greater than 100
The product of two consecutive integers is greater than 100 Take an integer x. Next consecutive integer is x + 1 The product of those integers is: x(x + 1) This product is greater than 100 which gives us the algebraic expression of: x(x + 1) > 100 IF we want to solve for x: x^2 + x > 100 Subtract 100 from each side: x^2 + x - 100 > 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-100%3E0&pl=All&hintnum=+0']Solve this quadratic:[/URL]
The product of x and 7 is not greater than 21
The product of x and 7 is not greater than 21 The product of x and 7: 7x Is not greater than means less than or equal to, so we have our algebraic expression: 7x <= 21 If you want to solve this inequality and interval notation, use our [URL='http://www.mathcelebrity.com/interval-notation-calculator.php?num=7x%3C%3D21&pl=Show+Interval+Notation']calculator[/URL]. [MEDIA=youtube]pyYaiKSI0ZQ[/MEDIA]
the quotient of 3 and u is equal to 52 divided by u
the quotient of 3 and u is equal to 52 divided by u Take this algebraic expression in 3 parts: [LIST=1] [*]The quotient of 3 and u means we divide 3 by u: 3/u [*]52 divided by u means we divide 52 by u: 52/u [*]The phrase [I]is equal to[/I] means an equation, so we set (1) equal to (2) [/LIST] [B]3/u = 52/u[/B]
the quotient of 4 more than a number and 7 is 10
the quotient of 4 more than a number and 7 is 10 Take this algebraic expression in pieces: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 4 more than a number means we add 4 to x: x + 4 The quotient of 4 more than a number and 7 means we divide x + 4 by 7 (x + 4)/7 The word [I]is[/I] means an equation, so we set (x + 4)/7 equal to 10 to get our algebraic expression of: [B](x + 4)/7 = 10 [/B] If the problem asks you to solve for x, then we cross multiply: x + 4 = 10 * 7 x + 4 = 70 Subtract 4 from each side: x + 4 - 4 = 70 - 4 x = [B]66 [MEDIA=youtube]j-GZLPVKbTM[/MEDIA][/B]
the quotient of d and 182 is the same as w minus 137
The quotient of d and 182 is the same as w minus 137 Take this algebraic expression in 3 parts: The quotient of d and 182 d/182 w minus 137 w - 137 The phrase [I]is the same as[/I] means we set d/182 equal to w - 137 [B]d/182 = w - 137[/B]
the ratio of 50 and a number added to the quotient of a number and 10
the ratio of 50 and a number added to the quotient of a number and 10 Take this algebraic expression in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The ratio of 50 and x means we divide by 50 by x 50/x The quotient of a number and 10 means we have a fraction: x/10 The phrase [I]added to[/I] means we add 50/x to x/10 [B]50/x + x/10[/B]
the ratio of a number x and 4 added to 2
the ratio of a number x and 4 added to 2 Take this algebraic expression in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The ratio of this number and 4 means we have a fraction: x/4 The phrase [I]added to[/I] means we add 2 to x/4 [B]x/4 + 2[/B]
the reciprocal of the product a and b
the reciprocal of the product a and b Take this algebraic expression in pieces: The product a and b means we multiply a times b ab The [I]reciprocal[/I] means we take 1 over ab [B]1/ab[/B]
The square of a number added to its reciprocal
The square of a number added to its reciprocal The phrase [I]a number [/I]means an arbitrary variable, let's call it x. the square of x mean we raise x to the power of 2. It's written as: x^2 The reciprocal of x is 1/x We add these together to get our final algebraic expression: [B]x^2 + 1/x [MEDIA=youtube]ZHut58-AoDU[/MEDIA][/B]
The square of the sum of twice a number x and y
The square of the sum of twice a number x and y Take this in algebraic expression in 3 parts: [LIST=1] [*]Twice a number x means we multiply x by 2: 2x [*]The sum of twice a number x and y means we add y to 2x above: 2x + y [*]The square of the sum means we raise the sum (2x + y) to the second power below: [/LIST] [B](2x + y)^2[/B]
The sum of 2 and w is less than or equal to 27.
The sum of 2 and w is less than or equal to 27. Take this algebraic expression in parts: [LIST] [*]The sum of 2 and w: 2 + w [*]The phrase [I]less than or equal to[/I] means an inequality, using the <= sign. [/LIST] [B]2 + w <= 27[/B]
the sum of 23 and victor age is 59
the sum of 23 and victor age is 59 Let's Victor's age be a. The sum of 23 and Victor's age (a) mean we add a to 23: 23 + a The word [I]is[/I] means an equation, so we set 23 + a equal to 59: [B]23 + a = 59[/B] <-- This is our algebraic expression Now if the problem asks you to take it a step further and solve this for a, [URL='https://www.mathcelebrity.com/1unk.php?num=23%2Ba%3D59&pl=Solve']we type this equation into our search engine[/URL] and we get: [B]a = 36[/B]
The sum of 4 and x is multiplied by 5. The result is then taken away from 16
The sum of 4 and x is multiplied by 5. The result is then taken away from 16. Take this algebraic expression in 3 parts: [U]Part 1: The sum of 4 and x:[/U] 4 + x [U]Part 2: Multiplied by 5:[/U] 5(4 + x) [U]Part 3: The result is then taken away from 16:[/U] [B]16 - 5(4 + x)[/B]
the sum of 5 and y is less than or equal to -21
the sum of 5 and y is less than or equal to -21 Take this algebraic expression in parts: The sum of 5 and y means we add y to 5 5 + y The phrase [I]less than or equal to[/I] -21 means an inequality. We use the <= sign to relate 5 + y to -21 [B]5 + y <= -21[/B]
the sum of 6 and 7, plus 5 times a number, is -12
the sum of 6 and 7, plus 5 times a number, is -12 The sum of 6 and 7 means we add the two numbers: 6 + 7 This evaluates to 13 Next, we take 5 times a number. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. So we multiply x by 5: 5x The first two words say [I]the sum[/I], so we add 13 and 5x 13 + 5x The word [I]is[/I] means an equation, so we set 13 + 5x equal to -12 [B]13 + 5x = -12[/B] <-- This is our algebraic expression If the problem asks you to take it a step further and solve for x, then you [URL='https://www.mathcelebrity.com/1unk.php?num=13%2B5x%3D-12&pl=Solve']type this algebraic expression into our search engine[/URL] and you get: [B]x = -5[/B]
The sum of 9 and victors age is 55
The sum of 9 and victors age is 55 Let v be Victor's age. We have the algebraic expression: [B]v + 9 = 55 [/B] If you want to solve or v, use our [URL='http://www.mathcelebrity.com/1unk.php?num=v%2B9%3D55&pl=Solve']equation calculator[/URL].
The sum of a and b divided by their product
The sum of a and b divided by their product The sum of a and b means we add b to a: a + b The product of a and b means we multiply a by b: ab To get our final algebraic expression, we divide the sum (a + b) by the product ab: [B](a + b)/ab[/B]
The sum of a number and 5 divided by 8
The sum of a number and 5 divided by 8. Let's take this algebraic expression in parts. Part 1: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Part 2: The sum of a number and 5 means we add 5 to the number x x + 5 Part 3: Next, we divide this expression by 8 [B](x + 5)/8[/B]
the sum of a number and itself is 8
A number means an arbitrary variable, let's call it x. The sum of a number and itself means adding the number to itself x + x Simplified, we have 2x The word is means equal to, so we have an algebraic expression of: [B]2x= 8 [/B] IF you need to solve this equation, divide each side by 2 [B]x = 4[/B]
The sum Of a number squared and 14
The sum Of a number squared and 14. A number means an arbitrary variable, let's call it x. Squared means we raise x to the 2nd power: x^2 The sum means we add x^2 to 14 to get our algebraic expression below: [B]x^2 + 14[/B]
The sum of d and v, all multiplied by 8
The sum of d and v, all multiplied by 8 This is an algebraic expression. The sum of d and v: d + v Multiply this sum by 8: [B]8(d + v)[/B]
the sum of doubling a number and 100 which totals to 160
the sum of doubling a number and 100 which totals to 160 Take this algebraic expression in pieces: [LIST=1] [*]Let the number be n. [*]Double it, means we multiply n by 2: 2n [*]The sum of this and 100 means we add 100 to 2n: 2n + 100 [*]The phrase [I]which totals[/I] means we set 2n + 100 equal to 160 [/LIST] [B]2n + 100 = 160[/B] <-- This is our algebraic expression If the question asks you to solve for n, then we [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B100%3D160&pl=Solve']type this equation into our search engine[/URL] and we get: [B]n = 30[/B]
The sum of five and twice a number is 17
The sum of five and twice a number is 17 [U]The phrase a number means an arbitrary variable, let's call it x[/U] x [U]Twice a number means we multiply x by 2:[/U] 2x [U]The sum of five and twice a number means we add 5 to 2x:[/U] 2x + 5 [U]The phrase [I]is[/I] means an equation, so we set 2x + 5 equal to 17 to get our algebraic expression[/U] [B]2x + 5 = 17[/B] [B][/B] As a bonus, if the problem asks you to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B5%3D17&pl=Solve']type in this algebraic expression into our math engine[/URL] and we get: x = 6
the sum of the cube of a number and 12
the sum of the cube of a number and 12 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The cube of a number means we raise x to the power of 3: x^3 Finally, we take the sum of x^3 and 12. Meaning, we add 12 to x^3. This is our final algebraic expression. [B]x^3 + 12[/B]
the sum of the squares of a and b
the sum of the squares of a and b Square of a means we raise a to the 2nd power: a^2 Square of b means we raise b to the 2nd power: b^2 The sum of squares means we add these terms together to get our algebraic expression: [B]a^2 + b^2[/B]
The sum of the sum of x and z and the difference of y and z
The sum of the sum of x and z and the difference of y and z Take this algebraic expression in 3 parts: Step 1: The sum of x and z is written as: x + z Step 2: The difference of y and z is written as: y - z Step 3: the sum of the sum and difference is written as: x + z + (y - z) x + z + y - z Cancelling the z terms, we get: [B]x + y [MEDIA=youtube]bmoZXImYCrg[/MEDIA][/B]
The sum of two consecutive integers plus 18 is 123
The sum of two consecutive integers plus 18 is 123. Let our first integer be n and our next integer be n + 1. We have: n + (n + 1) + 18 = 123 Group like terms to get our algebraic expression: 2n + 19 = 123 If we want to solve the algebraic expression using our [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B19%3D123&pl=Solve']equation solver[/URL], we get n = 52. This means the next integer is 52 + 1 = 53
The sum of two y and the quantity of three plus y plus twice the quantity two y minus two equals fif
The sum of two y and the quantity of three plus y plus twice the quantity two y minus two equals fifteen The sum of two y and the quantity of three plus y 2y + (3 + y) twice the quantity two y minus two 2(2y - 2) The sum of two y and the quantity of three plus y plus twice the quantity two y minus two 2y + (3 + y) + 2(2y - 2) Equals 15 to get our algebraic expression [B]2y + (3 + y) + 2(2y - 2) = 15[/B] [B][/B] If the problem asks you to solve for yL 2y + 3 + y + 4y - 4 = 15 Group like terms: 7y - 1 = 15 Add 1 each side: 7y = 16 Divide each side by 7: y = [B]16/7[/B]
The sum of two y and three is the same as the difference of three y and one
The sum of two y and three 2y + 3 the difference of three y and one 3y - 1 is the same as means equal to. Set 2y + 3 equal to 3y - 1 for our final algebraic expression: [B]2y + 3 = 3y - 1[/B] [B][/B] If the problem asks you to solve for y, subtract 2y from each side: 3 = y - 1 Add 1 to each side: y = [B]4[/B]
The sum of x and one half of x
The sum of x and one half of x To write this algebraic expression correctly, we have (1 + 1/2)x To get common denominators, we write 1 as 2/2. So we have: (2/2 + 1/2)x [B]3/2x[/B]
The sum of y and z decreased by the difference of m and n
The sum of y and z decreased by the difference of m and n. Take this algebraic expression in 3 parts: [LIST=1] [*]The sum of y and z means we add z to y: y + z [*]The difference of m and n means we subtract n from m: m - n [*]The phrase [I]decreased by[/I] means we subtract the quantity (m - n) from the sum (y + z) [/LIST] [B](y + z) - (m - n)[/B]
the total of a and 352 equals a divided by 195
the total of a and 352 equals a divided by 195 Take this algebraic expression in 3 parts: [LIST=1] [*]The total of a and 352 means we add 352 to a: a + 352 [*]a divided by 195: a/195 [*]The phrase [I]equals[/I] means we set (1) equal to (2) to get our final algebraic expression: [/LIST] [B]a + 352 = a/195[/B]
There are 12 eggs in a dozen. Write an algebraic expression for the number of eggs in d dozen.
There are 12 eggs in a dozen. Write an algebraic expression for the number of eggs in d dozen. [B]12d[/B]
There are thrice as many girls (g) as there are boys (b)
There are thrice as many girls (g) as there are boys (b) Thrice means we multiply by 3, so we have the following algebraic expression: [B]g = 3b[/B]
Think of a number. Double the number. Subtract 6 from the result and divide the answer by 2. The quo
Think of a number. Double the number. Subtract 6 from the result and divide the answer by 2. The quotient will be 20. What is the number Let's call our number n. Double the number means we multiply n by 2: 2n Subtract 6 from the result means we subtract 6 from 2n: 2n - 6 Divide the answer by 2: (2n - 6)/2 We can simplify this as n - 3 The quotient will be 20. This means the simplified term above is set equal to 20: [B]n - 3 = 20 [/B] <-- This is our algebraic expression If you want to take it a step further, and solve for n in the algebraic expression above, we [URL='https://www.mathcelebrity.com/1unk.php?num=n-3%3D20&pl=Solve']type this expression into our calculator[/URL], and get: n = 23
Thirty is half of the sum of 4 and a number
Thirty is half of the sum of 4 and a number. The phrase [I]a number[/I] represents an arbitrary variable, let's call it x. The sum of 4 and a number: 4 + x Half of this sum means we divide by 2: (4 + x)/2 Set this equal to 30: [B](4 + x)/2 = 30[/B] <-- This is our algebraic expression
Three x is five less than twice x
Twice x means we multiply x by 2: 2x five less than twice x 2x - 5 Three x 3x The word [I]is[/I] means equal to. Set 2x - 5 equal to 3x for our algebraic expression: [B]2x - 5 = 3x [/B] If the problem asks you to solve for x, subtract 2x from each side [B]x = -5[/B]
Tickets to the amusement park cost \$12 for adults and \$8 for kids. Write on algebraic expression to
Tickets to the amusement park cost \$12 for adults and \$8 for kids. Write on algebraic expression to show the cost of a adult and k kids Since cost = price * quantity, we have: [B]12a + 8k[/B]
Translate this phrase into an algebraic expression. 57 decreased by twice Mais savings Use the varia
Translate this phrase into an algebraic expression. 57 decreased by twice Mais savings Use the variable m to represent Mais savings. Twice means multiply by 2 2m 57 decreased by means subtract 2m from 57 [B]57 - 2m[/B]
Translate this phrase into an algebraic expression. 58 decreased by twice Gails age. Use the variabl
Translate this phrase into an algebraic expression. 58 decreased by twice Gails age. Use the variable g to represent Gails age. Twice Gail's age: 2g 58 decreased by twice Gail's age [B]58 - 2g[/B]
Translate this sentence into an equation. 43 is the sum of 17 and Gregs age. Use the variable g to
Translate this sentence into an equation. 43 is the sum of 17 and Gregs age. Use the variable g to represent Gregs age. The sum of 17 and Greg's age: g + 17 The word [I]is[/I] means equal to, so we set g + 17 equal to 43 [B]g + 17 = 43[/B] <-- This is our algebraic expression If you want to solve this equation for g, use our [URL='http://www.mathcelebrity.com/1unk.php?num=g%2B17%3D43&pl=Solve']equation calculator[/URL]. [B]g = 26[/B]
triple c divide the result by a
triple c divide the result by a Take this algebraic expression in pieces. Triple c means we multiply the variable c by 3 3c Divide the result by a, means we take 3c, and divide by a [B]3c/a[/B]
triple the sum of 36 and 6 then add 4
triple the sum of 36 and 6 then add 4 Take this algebraic expression in parts: The sum of 36 and 6: 36 + 6 Triple the sum means we multiply the sum by 3: 3(36 + 6) Then add 4: [B]3(36 + 6) + 4[/B] If the problem asks you to simplify the algebraic expression, we have: 3(42) + 4 126 + 4 [B]130[/B]
Twenty-five is nine more than four times a number
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Four times a number: 4x nine more than four times a numbrer 4x + 9 The phrase [I]is[/I] means equal to. We set 4x + 9 equal to 25 as our algebraic expression: [B]4x + 9 = 25 [/B] If the problem asks you to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=4x%2B9%3D25&pl=Solve']type it in our math solver[/URL] and get: x = [B]4[/B]
Twenty-five is the same as ten added to twice a number
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number means we multiply x by 2: 2x ten added to twice a number 2x + 10 The phrase [I]is the same as [/I]means equal to. Set 25 equal to 2x + 10 to get our algebraic expression [B]25 = 2x + 10 [/B] If the problem asks you to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=25%3D2x%2B10&pl=Solve']type it in our math solver [/URL]and get x = [B]7.5[/B]
Twice a number decreased by eight is zero
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number: 2x decreased by eight 2x - 8 [I]is [/I]means equal to. Set 2x - 8 equal to zero for our algebraic expression: [B]2x - 8 = 0 [/B] If the problem asks you to solve for x, add 8 to each side: 2x = 8 Divide each side by 2: x= [B]4[/B]
twice the difference of a number and 3 is equal to 3 times the sum of a number and 2
twice the difference of a number and 3 is equal to 3 times the sum of a number and 2. We've got 2 algebraic expressions here. Let's take them in parts. Left side algebraic expression: twice the difference of a number and 3 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]The word [I]difference[/I] means we subtract 3 from the variable x [*]x - 3 [*]Twice this difference means we multiply (x - 3) by 2 [*]2(x - 3) [/LIST] Right side algebraic expression: 3 times the sum of a number and 2 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]The word [I]sum[/I] means we add 2 to the variable x [*]x + 2 [*]3 times the sum means we multiply (x + 2) by 3 [*]3(x + 2) [/LIST] Now, we have both algebraic expressions, the problem says [I]is equal to[/I] This means we have an equation, where we set the left side algebraic expression equal to the right side algebraic expression using the equal sign (=) to get our answer [B]2(x - 3) = 3(x + 2)[/B]
twice the difference of a number and 55 is equal to 3 times the sum of a number and 8
twice the difference of a number and 55 is equal to 3 times the sum of a number and 8 Take this algebraic expression in pieces. Left side: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The difference of this number and 55 means we subtract 55 from x x - 55 Twice the difference means we multiply x - 55 by 2 2(x - 55) Right side: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The sum of a number and 8 means we add 8 to x x + 8 3 times the sum means we multiply x + 8 by 3 3(x + 8) Now that we have the left and right side of the expressions, we see the phrase [I]is equal to[/I]. This means an equation, so we set the left side equal to the right side: [B]2(x - 55) = 3(x + 8)[/B]
Twice the quantity of seven plus x is the same as the difference of x and seven
seven plus x 7 + x Twice the quantity of seven plus x 2(7 + x) Difference of x and seven x - 7 The phrase [I]is the same as[/I] means equal to. This is our algebraic expression: [B]2(7 + x) = x - 7 [/B] If the problem asks you to solve for x, distribute 2 on the left side: 14 + 2x = x - 7 Subtract x from the right side 14 + x = -7 Subtract 14 from each side [B]x = -21[/B]
twice the square of the product of x and y
twice the square of the product of x and y Take this algebraic expression in pieces: [LIST] [*]The product of x and y means we multiply x and y: xy [*]The square of the product means we raise xy to the power of 2: (xy)^2 = x^2y^2 [*]Twice the square means we multiply the square by 2: [B]2x^2y^2[/B] [/LIST]
Verbal Phrase
Free Verbal Phrase Calculator - Given an algebraic expression, this translates back to a verbal phrase
vw^2+y=x for w
vw^2+y=x for w This is an algebraic expression. Subtract y from each side: vw^2 + y - y = x - y The y's cancel on the left side, so we're left with: vw^2 = x - y Divide each side by v w^2 = (x - y)/v Take the square root of each side: w = [B]Sqrt((x - y)/v)[/B]
What is the sum of a number x and y raised to the power of two in algebraic expression
What is the sum of a number x and y raised to the power of two in an algebraic expression? The sum of a number x and y: x + y Raise this to the power of 2 (x + y)^2
When 20 is subtracted from 3 times a certain number, the result is 43
A certain number means an arbitrary variable, let's call it x x 3 times x 3x 20 is subtracted from 3 time x 3x - 20 The result is means equal to, so we set 3x - 20 equal to 43 for our algebraic expression [B]3x - 20 = 43 [/B] If you need to solve this, use our [URL='http://www.mathcelebrity.com/1unk.php?num=3x-20%3D43&pl=Solve']equation calculator[/URL]: [B]x = 21[/B]
When 54 is subtracted from the square of a number, the result is 3 times the number.
When 54 is subtracted from the square of a number, the result is 3 times the number. This is an algebraic expression. Let's take it in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it "x". x Square the number, means raise it to the 2nd power: x^2 Subtract 54: x^2 - 54 The phrase [I]the result[/I] means an equation, so we set x^2 - 54 equal to 3 [B]x^2 - 54 = 3[/B]
When 9 is subtracted from 5 times a number ,the result is 31
When 9 is subtracted from 5 times a number ,the result is 31 A number means an arbitrary variable, let's call it x. 5 times this number is written as 5x. 9 subtracted from this is written as 5x - 9 [I]The result[/I] means we have an equation, so we set [B]5x - 9 = 31[/B]. This is our algebraic expression. Now if we want to solve for x, we [URL='http://www.mathcelebrity.com/1unk.php?num=5x-9%3D31&pl=Solve']plug this equation into the search engine [/URL]and get [B]x = 8[/B].
When a number is doubled, the result is 36
Excited to announce these types of algebraic expressions can be [URL='http://www.mathcelebrity.com/algexpress.php?num=whenanumberisdoubled,theresultis36&pl=Write+Expression']typed directly in our search engine[/URL].
When twice a number is reduced by 15 you get 95 what is the number
When twice a number is reduced by 15 you get 95 what is the number? The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x [I]Twice[/I] x means we multiply x by 2 2x [I]Reduced by[/I] 15 means we subtract 15 2x - 15 [I]You get[/I] means equal to, as in an equation. Set 2x - 15 = 95 2x - 15 = 95 <-- This is our algebraic expression. [URL='https://www.mathcelebrity.com/1unk.php?num=2x-15%3D95&pl=Solve']Type 2x - 15 = 95 into the search engine[/URL] and we get [B]x = 55[/B].
write an algebraic expression for 197 times y
write an algebraic expression for 197 times y [B]197y [/B] This can also be found by typing 197 times y into our search engine
Write an algebraic expression for 8 multiplied by the result of u reduced by 11.
Write an algebraic expression for 8 multiplied by the result of u reduced by 11. u [I]reduced by[/I] 11 Reduced by means subtract 11 from u. So we have: u - 11 We multiply this expression by 8 to get our algebraic expression of: [B]8(u - 11)[/B]
x add y, multiply by z then subtract d
x add y, multiply by z then subtract d Take this algebraic expression in pieces: [LIST] [*]x add y: x + y [*]multiply by z: z(x + y) [*]Subtract d: [B]z(x + y) - d[/B] [/LIST]
X is at least as large as 4
X is at least as large as 4. This is an algebraic expression, where the phrase [I]at least as large as[/I] means greater than or equal to: [B]x >=4[/B]
y minus 10 is equal to the product of y and 8
y minus 10 is equal to the product of y and 8. Take this algebraic expression in 3 parts: Part 1: y minus 10 Subtract 10 from the variable y y - 10 Part 2: The product of y and 8 We multiply 8 by the variable y 8y Part 3: The phrase [I]is equal to[/I] means an equation, so we set y - 10 equal to 8y [B]y - 10 = 8y[/B]
You started this year with \$491 saved and you continue to save an additional \$11 per month. Write an
You started this year with \$491 saved and you continue to save an additional \$11 per month. Write an algebraic expression to represent the total amount of money saved after m months. Set up a savings function for m months [B]S(m) = 491 + 11m[/B]
Your job pays you \$7 per hour. What is the algebraic expression if you worked h hours?
Your job pays you \$7 per hour. What is the algebraic expression if you worked h hours? If your pay is rate times hours, we have: [B]7h[/B]
z , subtract 5 then times by 3
z , subtract 5 then times by 3 Take this algebraic expression two parts: [LIST] [*]z subtract 5: z - 5 [*][I]Then times by 3[/I] means we multiply the expression z - 5 by 3 [/LIST] [B]3(z - 5)[/B]
z fewer than the difference of 5 and y
z fewer than the difference of 5 and y Take this algebraic expression in parts: The difference of 5 and y means we subtract y from 5 5 - y z fewer than this difference means we subtract z from 5 - y [B]5 - y - z[/B] |
# Speed
Speed is calculated using the equation:
$\text{speed}=\frac{\text{distance}}{\text{time}}$
An equation triangle is a tool that can be used to help rearrange equations
This is an example of a compound measure because distance and time can’t be measured in the same unit. Speed can be measured in metres per second, kilometres per hour and several other forms too.
Different measures have different sizes. 10 km/h is not the same as 10 m/s.
Question
Calculate the average speed of a mongoose if it travels 4 km in 0.25 hours.
$\text{speed}=\frac{\text{distance}}{\text{time}}$
$\text{speed}=\frac{\text{4~km}}{\text{0.25~hours}}$
$\text{speed = 16 km/h}$
Question
Calculate the average speed of a cheetah if it travels 300 m in 10 s.
$\text{speed}=\frac{\text{distance}}{\text{time}}$
$\text{speed}=\frac{\text{300~m}}{\text{10~s}}$
$\text{speed = 30 m/s}$
The equation for speed can also be rearranged into either of the following forms:
$\text{distance}=\text{speed}\times{\text{time}}$
or
$\text{time}=\frac{\text{distance}}{\text{speed}}$
This allows you to calculate either distance or time.
Distance and time are not compound measures as they are both measured in a single unit.
Question
A car travels from Cardiff to London with an average speed of 50 . If the journey takes three hours, estimate the distance from Cardiff to London.
$\text{distance}=\text{speed}\times{\text{time}}$
$\text{distance}=\text{50~mph}\times{\text{3~hours}}$
$\text{distance}=\text{150~miles}$ |
## Derivative Patterns
Read this section to learn about patterns of derivatives. Work through practice problems 1-8.
In Section 1.2 we saw that the "holey" function $\mathrm{h}(\mathrm{x})= \begin{cases}2 & \text { if } \mathrm{x} \text { is a rational number } \\ 1 & \text { if } \mathrm{x} \text { is an irrational number }\end{cases}$
is discontinuous at every value of $x$, so at every $x$ $h(x)$ is not differentiable. We can create graphs of continuous functions that are not differentiable at several places just by putting corners at those places, but how many corners can a continuous function have? How badly can a continuous function fail to be differentiable?
In the mid–1800s, the German mathematician Karl Weierstrass surprised and even shocked the mathematical world by creating a function which was continuous everywhere but differentiable nowhere - a function whose graph was everywhere connected and everywhere bent! He used techniques we have not investigated yet, but we can start to see how such a function could be built.
Start with a function $f_1$ (Fig. 4) which zigzags between the values $+1/2$ and $–1/2$ and has a "corner" at each integer. This starting function $f_1$ is continuous everywhere and is differentiable everywhere except at the integers. Next create a list of functions $f_{2}, f_{3}, f_{4}, \ldots$, each of which is a lot shorter but with many more "corners" than the previous ones. For example, we might make $\mathrm{f}_{2}$ zigzag between the values $+1 / 4$ and $-1 / 4$ and have "corners" at $\pm 1 / 2, \pm 3 / 2$, $\pm 5 / 2$, etc., and $\mathrm{f}_{3}$ zigzag between $+1 / 9$ and $-1 / 9$ and have "corners" at $\pm 1 / 3$, $\pm 2 / 3, \pm 4 / 3$, etc. If we add $\mathrm{f}_{1}$ and $\mathrm{f}_{2}$, we get a continuous function (since the sum of two continuous functions is continuous) which will have corners at $0, \pm 1 / 2, \pm 1$, $\pm 3 / 2, \ldots$ If we then add $f_{3}$ to the previous sum, we get a new continuous function with even more corners. If we continue adding the functions in our list "indefinitely", the final result will be a continuous function which is differentiable nowhere.
We haven't developed enough mathematics here to precisely describe what it means to add an infinite number of functions together or to verify that the resulting function is nowhere differentiable, but we will. You can at least start to imagine what a strange, totally "bent" function it must be.
Until Weierstrass created his "everywhere continuous, nowhere differentiable" function, most mathematicians thought a continuous function could only be "bad" in a few places, and Weierstrass' function was (and is) considered "pathological", a great example of how bad something can be. The mathematician Hermite expressed a reaction shared by many when they first encounter Weierstrass' function:
"I turn away with fright and horror from this lamentable evil of functions which do not have derivatives".
##### IMPORTANT RESULTS
Power Rule For Functions: $\quad D\left(f^{n}(x)\right)=n \cdot f^{n-1}(x) \cdot D(f(x))$
Derivatives of the Trigonometric Functions:
\begin{align*} \begin{array}{lll} \mathbf{D}(\sin (\mathrm{x}))=\cos (\mathrm{x}) & \mathrm{D}(\tan (\mathrm{x}))=\sec ^{2}(\mathrm{x}) & \mathrm{D}(\sec (\mathrm{x}))=\sec (\mathrm{x}) \tan (\mathrm{x}) \\ \mathbf{D}(\cos (\mathrm{x}))=-\sin (\mathrm{x}) & \mathrm{D}(\cot (\mathrm{x}))=-\csc ^{2}(\mathrm{x}) & \mathrm{D}(\csc (\mathrm{x}))=-\csc (\mathrm{x}) \cot (\mathrm{x}) \end{array} \end{align*}
Derivatives of the Exponential Function: $\quad D\left(e^{x}\right)=e^{x}$ |
# Grade 3: Meaning of multiplication
Multiplication is one of Four fundamental operations in Arithmetic Math. The fundamental operations of Arithmetic are Addition, Subtraction, Multiplication, and Division. Addition and Subtraction are the first steps in understanding the operation of Arithmetic. Grade 1 and 2 are basically focused on these concepts because to understand multiplication and division, having a thorough knowledge of addition and subtraction is crucial.
## What is Multiplication?
In simpler way, multiplication is adding a number to itself multiple times.
The second number in multiplication sum is the number to be added to itself and the first number shows “how many times” it is to be added. The answer to multiplication is called “Product”.
Multiplication can be explained in number of ways.
• Array: Arranging objects, pictures, or numbers in rows and columns is called an array. for example: when it is said, 2 x 4, it means there are 2 rows and there are 4 objects in each row. it can be shown like:
• Making Groups: Multiplication concept can also be explained by making equal groups of objects. when we say 2 x 4, it essentially means, there are 2 groups, and there are 4 objects in each group.
• Repeated addition: Multiplication is nothing but adding the same number over and over again. While understanding the concept , keeping in mind that multiplication is nothing new but adding the same number repeatedly helps to great extend. For example: 2 x 4 is nothing but 4 + 4.
we are adding number 4 twice here because first part of our sum is showing number 2.
Number line / Skip counting: multiplication can be done quickly if we know how to skip count or can be done with the help of a number line. The number line can be used in the initial stage because as the number gets bigger, it is a little time-consuming to draw the number line. But to understand and to get comfortable with the concept of multiplication, Number line helps to visualize the action of multiplication. 2 x 4 can be solved with the help of number line like; |
Sorry, you do not have a permission to add a post.
Please briefly explain why you feel this question should be reported.
# What is the additive inverse of 4 by 7?
What is the additive inverse of 4 by 7? The additive inverse of 4/7 is -4/7. Hence, the additive inverse of 4/7 is -4/7.
## What is the opposite of the additive inverse of 4?
Additive inverse is the number from zero on a number line but in the opposite direction. It is used to describe the opposites like the reflection of the number in the mirror. For example 3, the opposite of it in the number line is -3 and so for the 4, the additive inverse will be -4.
## What is the additive inverse of 7 by 5?
The additive inverse of 7/5 is -7/5.
## What is the additive inverse of 1 3?
Explanation: The opposite (also known as the additive inverse) is the number we have to add to get an answer equal to the additive identity, 0 . Since 13+(−13)=(−13)+13=0 , the opposite of 13 is −13 .
## What is the additive inverse of 3 by 5?
The additive inverse of -3/5 is 3/5.
## What is the additive inverse of 3?
Additive Inverse. Two numbers are additive inverses if they add to give a sum of zero. 3 and -3 are additive inverses since 3 + (-3) = 0. -3 is the additive inverse of 3.
## What is the multiplicative inverse of 7?
Dividing by a number is equivalent to multiplying by the reciprocal of the number. Thus, 7 ÷7=7 × 17 =1. Here, 17 is called the multiplicative inverse of 7.
## How do you find the additive inverse?
For a number (and more generally in any ring), the additive inverse can be calculated using multiplication by −1; that is, −n = −1 × n. Examples of rings of numbers are integers, rational numbers, real numbers, and complex numbers.
## What is the additive inverse of 3 7?
Additive inverse of 3/7 is -3/7
## What is the multiplicative inverse of 1?
The multiplicative inverse of 1 is 1.
## Does 0 have a multiplicative inverse?
The short answer is that 0 has no multiplicative inverse, and any attempt to define a real number as the multiplicative inverse of 0 would result in the contradiction 0 = 1. Some people find these points to be confusing. These notes may be useful for anyone with questions about dividing by 0.
## What is the inverse of 3?
3 * 1/3 = 1.
Thus the multiplicative inverse of 3 is 1/3.
## What is the additive inverse of the multiplicative inverse of 3 1 3?
The additive inverse is 1/3. multiplicative inverse is -3. = -1.
## What is the additive inverse of * 2 points?
The additive inverse of 2 is -2.
## What is the additive inverse of 3 upon 2?
So, keeping these concepts in mind, we can say that additive inverse of 3/2 is the negative of 3/2 = -3/2. We can also say that multiplicative inverse of 3/2 is 2/3.
## What will be the additive inverse of 3 by 5?
The additive inverse of -3/5 is 3/5.
## What is the formula of multiplicative inverse?
The multiplicative inverse of a fraction a/b is b/a. For the multiplicative inverse of a real number, divide 1 by the number. For example, the reciprocal of 5 is one fifth (1/5 or 0.2), and the reciprocal of 0.25 is 1 divided by 0.25, or 4.
## What is the multiplicative inverse of 5 7?
The multiplicative inverse of 5/7 is 7/5 = 1 2/5.
## What is the multiplicative inverse of 1 2?
Answer: The multiplicative inverse or reciprocal of 1/2 is 2.
## Which of the following is the additive inverse of 1 2?
SOLUTION: What is the additive and multiplicative inverse of 1/2. is 2.
## Do all matrices have an additive inverse?
« Every vector must have an additive inverse, the sum of these two vectors being the zero vector. » …
## What is the additive inverse of 2?
the number in the set of real numbers that when added to a given number will yield zero: The additive inverse of 2 is −2.
## How do you calculate additive inverse?
Think about what number needs to be added to the given number in order for the sum to equal 0. To find the additive inverse of a the given number, take the given number and change the sign. The resulting number is the additive inverse.
## What is the product of additive inverse and multiplicative inverse of 3 7?
So, Additive inverse of -3/7 is 3/7 . Multiplicative inverse of -3/7 is -7/3 . |
# Equation of a Circle
We will learn how to find the equation of a circle whose centre and radius are given.
Case I: If the centre and radius of a circle be given, we can determine its equation:
To find the equation of the circle whose centre is at the origin O and radius r units:
Let M (x, y) be any point on the circumference of the required circle.
Therefore, the locus of the moving point M = OM = radius of the circle = r
OM$$^{2}$$ = r$$^{2}$$
x$$^{2}$$ + y$$^{2}$$ = r$$^{2}$$, which is the required equation of the circle.
Case II: To find the equation of the circle whose centre is at C (h, k) and radius r units:
Let M (x, y) be any point on the circumference of the requited circle. Therefore, the locus of the moving point M = CM = radius of the circle = r
CM$$^{2}$$ = r$$^{2}$$
(x - h)$$^{2}$$ + (y - k)$$^{2}$$ = r$$^{2}$$, which is the required equation of the circle.
Note:
(i) The above equation is known as the central from of the equation of a circle.
(ii) Referred to O as pole and OX as initial line of polar co-ordinate system, if the polar co-ordinates of M be (r, θ) then we shall have,
r = OM = radius of the circle = a and ∠MOX = θ.
Then, from the above figure we get,
x = ON = a cos θ and y = MN = a sin θ
Here, x = a cos θ and y = a sin θ represent the parametric equations of the circle x$$^{2}$$ + y$$^{2}$$ = r$$^{2}$$.
Solved examples to find the equation of a circle:
1. Find the equation of a circle whose centre is (4, 7) and radius 5.
Solution:
The equation of the required circle is
(x - 4)$$^{2}$$ + (y - 7)$$^{2}$$ = 5$$^{2}$$
x$$^{2}$$ - 16x + 16 + y$$^{2}$$ - 14y + 49 = 25
x$$^{2}$$ + y$$^{2}$$ - 16x - 14y + 40 = 0
2. Find the equation of a circle whose radius is 13 and the centre is at the origin.
Solution:
The equation of the required circle is
x$$^{2}$$ + y$$^{2}$$ = 13$$^{2}$$
x$$^{2}$$ + y$$^{2}$$ = 169
The Circle
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Share this page: What’s this?
## Recent Articles
1. ### Tangrams Math | Traditional Chinese Geometrical Puzzle | Triangles
Apr 17, 24 01:53 PM
Tangram is a traditional Chinese geometrical puzzle with 7 pieces (1 parallelogram, 1 square and 5 triangles) that can be arranged to match any particular design. In the given figure, it consists of o…
Read More
2. ### Time Duration |How to Calculate the Time Duration (in Hours & Minutes)
Apr 17, 24 01:32 PM
We will learn how to calculate the time duration in minutes and in hours. Time Duration (in minutes) Ron and Clara play badminton every evening. Yesterday, their game started at 5 : 15 p.m.
Read More
3. ### Worksheet on Third Grade Geometrical Shapes | Questions on Geometry
Apr 16, 24 02:00 AM
Practice the math worksheet on third grade geometrical shapes. The questions will help the students to get prepared for the third grade geometry test. 1. Name the types of surfaces that you know. 2. W…
Read More
4. ### 4th Grade Mental Math on Factors and Multiples |Worksheet with Answers
Apr 16, 24 01:15 AM
In 4th grade mental math on factors and multiples students can practice different questions on prime numbers, properties of prime numbers, factors, properties of factors, even numbers, odd numbers, pr…
Read More
5. ### Worksheet on Factors and Multiples | Find the Missing Factors | Answer
Apr 15, 24 11:30 PM
Practice the questions given in the worksheet on factors and multiples. 1. Find out the even numbers. 27, 36, 48, 125, 360, 453, 518, 423, 54, 58, 917, 186, 423, 928, 358 2. Find out the odd numbers.
Read More |
# Find the value
Question:
Find the
(i) lengths of major axes,
(ii) coordinates of the vertices,
(iii) coordinates of the foci,
(iv) eccentricity, and
(v) length of the latus rectum of each of the following ellipses.
$9 x^{2}+16 y^{2}=144$
Solution:
Given:
$9 x^{2}+16 y^{2}=144$
Divide by 144 to both the sides, we get
$\frac{9}{144} x^{2}+\frac{16}{144} y^{2}=1$
$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ …(i)
Since, $16>9$
So, above equation is of the form,
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ …(ii)
Comparing eq. (i) and (ii), we get
$a^{2}=16$ and $b^{2}=9$
$\Rightarrow a=\sqrt{16}$ and $b=\sqrt{9}$
$\Rightarrow \mathrm{a}=4$ and $\mathrm{b}=3$
(i) To find: Length of major axes
Clearly, $a>b$, therefore the major axes of the ellipse is along $x$ axes.
$\therefore$ Length of major axes $=2 \mathrm{a}$
$=2 \times 4$
$=8$ units
(ii) To find: Coordinates of the Vertices
Clearly, $a>b$
$\therefore$ Coordinate of vertices $=(a, 0)$ and $(-a, 0)$
$=(4,0)$ and $(-4,0)$
(iii) To find: Coordinates of the foci
We know that,
Coordinates of foci $=(\pm c, 0)$ where $c^{2}=a^{2}-b^{2}$
So, firstly we find the value of $c$
$c^{2}=a^{2}-b^{2}$
$=16-9$
$c^{2}=7$
$c=\sqrt{7} \ldots(l)$
$\therefore$ Coordinates of foci $=(\pm \sqrt{7}, 0)$
(iv) To find: Eccentricity
We know that,
Eccentricity $=\frac{\mathrm{c}}{\mathrm{a}}$
$\Rightarrow \mathrm{e}=\frac{\sqrt{7}}{4}[$ from $(l)]$
(v) To find: Length of the Latus Rectum
We know that,
Length of Latus Rectum $=\frac{2 b^{2}}{a}$
$=\frac{2 \times(3)^{2}}{4}$
$=\frac{9}{2}$ |
009a-identity of awareness and representational equivalence
Let's consider the numeral 5 as an object of awareness.
AW: 5 = 5. Awareness of 5 is 5.
Our awareness of Roman numerals should follow the same pattern.
AW:V = V
As an example, this is the question that needs to be answered: How does our AW:V get "equated" to our AW:5?
We can describe 5 as a representation of some number that we call "five"
5; A NUMBER --> "5" represents a number.
How is our awareness of 5 different than our awareness of 5 as A NUMBER or 5 as A SYMBOL?
AW:5 = 5
AW:(5; A NUMBER) = (5; A NUMBER)
AW:(5; A SYMBOL) = (5; A SYMBOL)
Clearly (5; A SYMBOL) is not the same as (5; A NUMBER). Or in english, 5 as a symbol is not the same as 5 as a number. But both 5 as a symbol and 5 as a number are both 5.
How do we show this?
AW:5 = 5
5 = 5; "A NUMBER"
(5; "A NUMBER") = AW:(5; "A NUMBER")
5; "A NUMBER" = "A NUMBER"
AW:"A NUMBER" = "A NUMBER"
Through this series of steps we modeled how the awareness of "5" is associated to "5" as "a number". And how "5" itself is equivalent to "5" as "a number". And that "number" is the same as awareness of that "number". The " are for clarity, but are they necessary?
For the association of 5 as a symbol, we can show this shift in awareness with a single string of identities, associations, and equivalence.
reading left to right or right to left:
AW:5 = 5 = (5;A SYMBOL) = AW:(5; A SYMBOL) = (5; A SYMBOL) = A SYMBOL = AW:A SYMBOL
We have three objects of awareness. 5, 5 as a symbol, and a symbol. And through association and equivalence we can see a path or connection between the three.
We know that objects are the same as objects of awareness. And we also know from experience that representations occur. So how do different objects relate to each other? They relate to each other by being representationally equivalent.
To ask again: How does our AW:V get "equated" to our AW:5?
AW:V = V = (V;NUMBER FIVE) = AW:(V;NUMBER FIVE) = (V;NUMBER FIVE) = NUMBER FIVE = AW:NUMBER FIVE = NUMBER FIVE = (A NUMBER FIVE;5) = AW:(NUMBER FIVE;5) = (NUMBER FIVE;5) = 5 = AW:5
note that NUMBER FIVE refers to the mathematical concept of the number or quantity we call five.
in english:
awareness of roman numeral five is the same as roman numeral five which is equivalent to roman numeral five as the number five. Roman numeral five as the number five is the same as the awareness of roman numeral five as the number five. the awareness of roman numeral five as the number five is the same as roman numeral five as the number five which is equivalent to the number five. The number five is identical to the awareness of the number five. The awareness of the number five is identical to the number five which is equivalent to the number five as the english numeral five. the number five as the english numeral five is identical to the awareness of the number five as the english numeral five. The awareness of the number five as the english numeral five is the same as the number five as the english numeral five which is equivalent to the numeral five. the numeral five is the same as the awareness of the numeral five.
Once we learn this association from V to number five to 5 we can make a new association. V;5. Once a string of associations are made in awareness, we often make direct associations to the things at the end of those strings. thus:
AW:V = V = (V;5) = AW(V;5) = (V;5) = 5 = AW:5
we see at the center of this string of equality and equivalence the awareness of the association of V to 5. And for most people, once they have learned Roman numerals, they no longer make the association of "V" to the number five and then to the english symbol for 5 but associate "V" directly to "5". It's not that the association of "V" to the number 5 disappears, but it is no longer needed for "translation". A new association is made because of the associative or representational equivalence.
What is critical here is that all the associations or representations must exist in awareness. If the associations between objects do not exist in awareness, the association does not exist*. What is being connected through representational equivalence are different objects of awareness. It is because they are objects of awareness that we know they are objects and that they have associations.
*note: associations of objects must be associations of objects of awareness. there are not objects that exist apart from awareness, and there are not representations (which are objects) that exists apart from awareness. In practice, this must mean that representational objects must be local to the subject or shared among subjects.
In our ordinary way of speaking or writing we could leave out the steps of awareness and model the associations and equivalence like so:
V = V;NUMBER FIVE = NUMBER FIVE = NUMBER FIVE; 5 = 5 showing the long connection or
V = V;5 = 5 showing the short connection to the ends of the longer associative string. Once a longer string is made, the shorter one can be made creating 6 objects: V, 5, NUMBER 5, V;NUMBER FIVE, NUMBER FIVE;5, V;5
But V;5, and V;NUMBER FIVE, and NUMBER FIVE; 5 all must be objects of awareness to connect the concept V, 5 to the mathematical idea of five . We often speak about associations or representations as being external to awareness, but representations must be objects of awareness.
Representational objects may not be considered objects of direct experience.
We may not think of representational objects as being objects of direct awareness, or experienced directly, because we experience some objects only in representational ways. Names of objects often demonstrate the problem of direct experience or representational experience. And we can flit between the sense of a name being a thing itself, and being a stand-in for something else.
-----
We can add representation to the awareness because representation is part of experience itself we can show how objects relate to various representations of objects. This function has two "sides" to it. The awareness side (AW:X) and the non-awareness side (X).
The simple form of representation is some object X as some other object Y. Say a flower is object X. and Red Flower is object Y. so X = X;Y A flower is the equivalent to a flower in the form of a red flower. the semi-colon can be expressed as "in the form of" or simply "as".
I'm using the symbol for equivalence, but this may be a conceit. On the awareness side of the equation, it's not the same. But on the non-awareness side of the equation it is. We are aware of the difference between a flower versus a flower as a red flower.
If the representation were somehow separate from our awareness of it, there may not be a difference between a flower and a flower as a red flower. The representational shift seems to occur on the awareness side of the function. But the there is an identity on the non-awareness side between a thing and the thing in a form.
We can ordinarily say this (read ";" with "as" or "as a" or "in the form of")
flower = flower;(red flower) = (red flower)
In the ordinary way, we say a flower is a red flower.
But on the awareness side we should say this: AW:(flower) = AW:(flower;(red flower)) = AW:(red flower)
We must distinguish the object and the object as a representation on the awareness side because awareness of a flower, awareness of a red flower, and awareness of a flower as a red flower are three different instances of awareness. And so on the awareness side, the different awarenesses are not identical, but they are equivalent.
Now lets put these three parts together.
I am aware of a flower:
AW:flower
which is the same as the flower:
AW:flower = flower
A flower is the same as a red flower:
flower = flower;(red flower)
Which is the same as awareness of a flower in the form of a red flower:
flower;(red flower) = AW:(flower;(red flower))
And an awareness of a flower in the form a red flower is equivalent to an awareness of a red flower;
AW:(flower;(red flower)) = AW:(red flower)
This can be shorthanded to:
AW:flower = flower = flower;(red flower) = (red flower) = AW:(red flower).
I use the = because I'm skipping the AW:(flower;(red flower) portion. The equivalence indicates the awareness is implied.
It can be shorthanded even more using variables below:
AW:X = X;Y = AW:Y
Alternatively, if it makes you feel better you could say: AW:(flower) = AW:(flower;(red flower)) = AW:(red flower)
This all 'awareness side' relationship is true. But showing how representation is taking place on the non-awareness side of the equation will be useful later in explaining how to code for changes in experience. Because, the awareness has to shift as the objects of experience shift.
The awareness function is an identity. That produce changes in awareness that match the changes of objects. The changes flow both ways. It's important that changes "in the world" correspond to "changes in awareness". This mystical notion that "changes in awareness" correspond to changes "in the world" is important, because it will let us model and explain experience that would be impossible to model otherwise. The awareness function is a two-way street between the awareness side (AW:X) and the non-awareness side (X) of the function.
previous next |
hopeforchildren foundation inc. is happy to coordinate all of our related services with our Partners.
b
tthemathpage@nyc.rr.comhemathpage@nyc.rr.com
SOME RULES
OF
ALGEBRA
The rule of symmetry
The commutative rules
Two rules for equations
ALGEBRA, we can say, is a body of formal rules. They are rules that show how something written one form may be rewritten in another form. For what is a calculation if not transforming one set of symbols into another? In arithmetic we may replace the symbols '2 + 2' with this symbol '4.' In algebra, we may replace 'a + (−b)' with 'a − '
Here are some of the basic rules of algebra:
1· a = a.
(1 times any number does not change it. Therefore 1 is called the identity of multiplication.)
(−1)a = −a.
−(−a) = a. (Lesson 2)
a + (−b) = a − b. (Lesson 3)
a − (−b) = a + b. (Lesson 3)
Associated with these -- and with any rule -- is the rule of symmetry:
If a = b, then b = a.
For one thing, this means that a rule of algebra goes both ways.
Since we may write
p + (−q) = p − q
-- that is, in a calculation we may replace p + (−q) with p − q -- then, symmetrically:
p − q = p + (−q).
We may replace p − q with p + (−q).
The rule of symmetry also means that in any equation, we may exchange the sides.
If
15 = 2x + 7,
then we are allowed to write
2x + 7 = 15.
And so the rules of algebra tell us what we are allowed to write. They tell us what is legal.
A statement of equality between two algebraic expressions that is true for only certain values of the variables involved is called an equation. The values are called the solutions of the equation.
The following are examples of some basic types of equations.
3x + 5 = -2 A Linear equation in one variable, x
x -3y = 10 A Linear equation in two variables,x and y
2
20y + 6y - 17 = 0 A quadratic equation in one variable, y
Rules of Exponents:
a
In the algebraic expression x where x is raised to the power a, x is called the base and a is called the exponent.
For some equations involving bases and exponents, the following property is very useful:
y 6
For Example: If 2 = 64 since 64 = 2 , we have
y 6
2 = 2 and can conclude that y = 6
Some basic rules of exponents:
- a 1
x = ------ a 1
a and 2 = ----
x -a
2
a b a + b
(x ) (x ) = (x )
a
x a - b 1
__ = x = ___
b b - a
x x
The commutative rules
The order of terms does not matter. We express this in algebra by writing
a + b = b + a
That is called the commutative rule of addition. It will apply to any number of terms.
a + b − c + d = b + d + a − c = −c + a + d + b
The order does not matter.
Example 1. Apply the commutative rule to p − q.
Solution. The commutative rule for addition is stated for the operation + . Here, though, we have the operation − . But we can write
p − q = p + (−q).
Therefore,
p − q = −q + p.
Here is the commutative rule of multiplication:
ab = ba
The order of factors does not matter.
abcd = dbac = cdba.
The rule applies to any number of factors.
What is more, we may associate factors in any way:
Solving Linear Equations:
A linear equation is an equation involving one or more variables in which each term in the equation is either a constant term or a variable multiplied by a coefficient. None of the variables are multiplied together or raised to a power greater than 1.
For example,2x + 1 = 7x and 10x - 9y - z = 3
Linear Equations in One Variable:
To solve a linear equation in one variable, simplify each side of the equation by combining like terms. Then use the rules for producing simpler equivalent equations.
11 x - 4 - 8 x = 2 (x - 4 ) - 2 x
3x - 4 = 2x + 8 - 2 x (like terms combined)
3x - 4 = 8 (Simplified)
3x - 4 + 4 = 8 + 4 (4 added to both sides)
3x = 12
3x = 12
___ ___ (Both sides divided by 3)
3 3
Linear Equations in Two Variables:
There are two basic methods for solving systems of linear equations, by substitution or by elimination. In the substitution method, one equation is manipulated to express one variable in terms of the other. Then the expression is substituted in the other equation. For example, to solve the system of equations.
4 x + 3 y = 13
x + 2y = 2
we can express x in the second equation in terms of y, x = 2 - 2y then substitute 2 - 2y for x
in the first equation to find the value of y.
4(2 - 2y) + 3y = 13
8 - 8y + 3y = 13
- 8y + 3y = 5 ( 8 subtracted from both sides)
- 5y = 5 (like terms combined)
y = - 1 (both sides divided by - 5)
And - 1 can be substituted for y in either equation to find the value of x .Using second equation
x + 2 y = 2
x + 2 (- 1) = 2
x - 2 = 2 now add 2 to both sides to get value of x
In the elimination method, the object is to make the coefficients of one variable the same in both equations so that one variable can be eliminated either by adding the equations together or by subtracting one from the other.
Using the same example,multiply both sides of the equation 4 yields 4(x = 2y) or 4x + 8y = 8
now we have 2 equations with the same coefficient of x
4x + 3y = 13
4x + 8y = 8
subtract second equation from the first thus, x = 4 , y = - 1
Home About Us Services DONATIONS Products Science Page
Math Page Technology Jobs Math page 1 Reading & Writing PHYSICS
CHEMISTRY BIOLOGY SANDY HOOK ELEMENTARY VIDEOS Math Page 2
A quadratic equation in the variable x is an equation that can be written in the form
2
ax + bx + c = 0
where a, b, and c are real numbers and a =/= 0 When such an equation has solutions, they can be found using the
2
x == _- b ± _sqroot b_ -4ac
_____________________
2a
Where the notation ± is shorthand for indicating two solutions one that uses the plus sign and the other that uses the minus sign. ±
2
Example: In the quadratic equation 2x – x – 6 = 0
We have a = 2, b = - 1 and c = -6
X = -(-1) ± Sq.Root (-1) – 4 (2) (-6)
--------------------------------------------------
2 (2)
= 1 ± Sq Root 49
-----------------------------
4
= 1 ± 7
--------
4
Hence, the two solutions are : x = 1 + 7
---------- = 2
4
and x = 1 - 7 3
--------- = --------
4 2
Factorization is simplifying a quadratic equation
into 2 brackets.
2
Y = X + 4X + 3 .........(1)
Y = (X + 1) ( X + 3) .......(2)
Solving a quadratic equation involves finding
2 values of X that make equation = 0
In the example:
2
X + 4X + 3 = 0
(X + 1) ( X + 3) = 0
Since X + 1 = 0; X = - 1 and X + 3 0, = -3
This is true as the 2 equations are a factor of 0
their products.Solved by finding the 2 roots.
An algebraic expression has one or more variables and can be written as a single term or as a sum of terms. Here are some examples of algebraic expressions.
3 2 2
2x ; y - 1/4 ; W Z - 5 Z - Z + 6 ; 8/n - p
In the examples above, 2x is a single term,
y - 1/4 has two terms,
3 2 2
W Z - 5 Z - Z + 6 has four terms, and
8/n - p has one term.
3 2 2
In the expression W Z - 5 Z - Z + 6
2 2
5 Z and - Z are called like terms because they have the same variables (Z), and the corresponding variables have the same exponents (2).
A term that has no variable is called a constant term.
what's the constant term in the above expression ?
A number that is multiplied by variables is called the coefficient of a term. 2
For example, in the expression 2 X + 7 X - 5
2
2 is the coefficient of the term 2 X and 7 is the coefficient of the term 7 X and - 5 is a constant term.
The same rules that govern operations with numbers apply to operations with algebraic expressions.
One additional rule, which helps in simplifying algebraic expressions, is that like terms can be combined by simply adding their coefficients, as the following examples show.
3 2 2 3 2
2 X + 5X = 7X and W Z + 5 Z - Z + 6 = W Z + 4 Z + 6
3xy + 2x - xy -3x = 2xy - x
Factorization:
A number or variable that is a factor of each term in an algebraic
expression can be factored out as in the following examples.
2
4x + 12 = 4(x + 3) and 15y - 9y = 3y(5y - 3)
To multiply two algebraic expressions,each term of the first
expression is multiplied by each term of the second expression
and the results are added.As in the following examples.
(x + 2)(3x - 7) = x(3x) + x(- 7) + 2(3x) + 2(- 7)
2 2
= 3x - 7x + 6x - 14 which = 3x - x - 14
LESSON 2.
HOW TO FACTORIZE AND SOLVE QUADRATIC EQUATIONS.
Consider factorizing only quadratic equation where the co-efficient of the
2 2
X term , a = 1 and X + bX + C = 0
Question:Factorize and solve the quadratic equation below;
2
X + 5X + 4 = 0
Factorization: (X + X ) (X + X ) = 0
1 2
Step 1.Find the values of a,b,&c where a =1;b = 5; c = 4
step 2.The c term is the product of the roots c = x x
1 2
List the pair of factors that multiply to make the c term: c = 4 (1 x 4 ) (2 x 2 )
Step 3.b term is the sum of the roots x x = 5 therefore b = 5 (1 +4) = 5
1 2
Step 4.Insert the values in the pair of brackets x = 1; x = 4 (x+10 (x + 4) = 0
1 2
Solution:equate the 1st brackets = 0 using substitution method to find x
1
x = 1 = 0 therefore,x = -1 Equate the 2nd bracket x + 4 = 0 ;x = -4
1 1 2 2 2
the solution of the equation x + 5x + 4 = 0 is x = -1; or x = -4
HOME SCHOOL ONLINE
Attend a home school online at the comfort of your home. |
Maths
Week Beginning: 10th June 2018
This week we are learning how to:
1. Multiply using the formal written method. Previously, we have used the expanded method and now we are moving onto the compact method. Here is a video to show you how we teach it! https://www.youtube.com/watch?v=eUUrV5onhyo
2. Divide using the formal written method. Previously, we have used the bus stop method with numbers that can divided by the number exactly. This week we are going to be finding remainders. Here is a video explaining it https://www.youtube.com/watch?v=FApcjdAhnrY Watch for the first minute only as the rest is taught further up the school.
3. Make quarter, three quarter and half turns.
Week Commencing: 7th May 2018
This week we are learning how to add and subtract fractions. Again I have put the powerpoint below as this is what the homework is based on. Just remember to emphasise that it is the numerator we add or subtract NOT the denominator. We are also looking again at 3D shapes and their properties as they struggled with this last time.
Week Commencing: 30th April 2018
This week we are learning about equivalent fractions. We have learned that sometimes fractions have the same amount shaded, for example 3/6 can also be 1/2. We also learned how to multiply and divide the numerator and denominator to find other equivalent fractions, for example with 4/12 we know that 4 goes in each one so it could be 1/3. Below is the powerpoint we used to explain equivalent fractions.
Week Commencing 23rd April 2018
This week we are looking at fractions. We are focussing on the numerator and denominator and what they both mean. The bottom number is the denominator and tells you how many parts there are and the numerator tells you how many parts you are focussing on. An example of this would be: ¼ of 24 = 6. 24 is split into 4 parts and there are 6 in one part. 2/4 of 24 would be 12 as 24 is split into 4 parts and there are 12 in 2 parts. The homework is based on this. It is double sided so please do both.
Week Commencing: 16th April 2018
This week we are learning about time and the children really do find this very hard. They will need a lot of practise at home to consolidate this. These are the three skills we will be learning:-
1. How to tell the time using Roman numerals.
2. How to tell the time to the exact minute, for example 22 minutes to 3 or 12 minutes past 4.
3. How to convert the 12 hour clock to the 24 hour clock, for example 3:10pm would be 15:10 in the 24 hour clock and 4.20am would be 04:20 in the 24 hour clock. They do find this tricky, especially the am times!
Below are some powerpoints that might help. Please practise skills 2 and 3 at home. In terms of their homework, they can write the time in words (i.e. 5 minutes to 3) or digitally (2:55), it is up to them and what they feel most comfortable with.
Week Commencing: 12th March 2018
This week we are learning about 3D shapes and will be looking at the names and properties of them by making them out of card and labelling them. This is what the homework is based on. If they are finding this difficult, try using 3D shapes from around the house so that they can count the vertices (corners), edges and faces. Finally, we looked at turns and how two right angle turns is 1/2, 3 right angle turns is ¾ and 1 right angle turn is ¼.
Week Commencing: 5th March 2018
This week we are focussing on division and learning about the formal written method which is also known as the 'bus stop' method. Please see the route through calculation on the front page of the Year 3 page to show you how we do it. Please be aware, however, that we are not learning about remainders until the summer term. This is a challenging concept for them. The homework is all about seeing the relationships between multiplication and division and realising that they are interlinked. If they are struggling, remind them that with multiplication, the big number goes at the end and with division, it goes at the start.
Week Commencing: 26th February 2018
This week we are completing work on multiplication using the formal written method. We will teach them how to do this using the expanded method. This is what their homework is about and there are instructions for how to do this on the front sheet of paper. There is also a good video you can watch: https://www.youtube.com/watch?v=MFSf52TFBwo but only watch the first minute as we do not do 3 digits by 1 digit yet! Hopefully with this video and the intructions on the sheet, they will get this! If you do it this weekend, it will hopefully be fresh in their heads! The children will probably find this hard but unfortunately this is a skill they have to know by the end of year 3.
Week Commencing: 12th February 2018
This week we are completing the end of half term testing. We are also learning about mass. We will be weighing items and then answering questions using the data based on addition and subtraction.
Week Commencing: 5th February 2018
This week we are continuing our work on capacity and are going to subtract capacities. We are then going to move onto length and will measure each other's hands, legs, arms, head etc and then compare the lengths working on addition and subtraction.
Week Commencing: 29th January 2018
This week, the children have been learning about adding and subtracting 1, 10, 100 to a number and finding the missing number. They always find this very difficult as they struggle with basic calculations, for example, 27+4. They often make careless mistakes when crossing the boundaries i.e. knowing that the answer will be in the 30s. They can also struggle with place value and knowing where the numbers fit in. For homework I have given a missing number activity and there are instructions to help at the top. Please focus on what is happening to the digits when going down, up, left and right!
We also started work on capacity focusing on reading scales and adding capacities together.
Week Commencing: 22nd January 2018
This week, we are learning how to add and subtract using the formal written method using the dienes and recording it on paper. This is what the homework is based on and you can find posters to help you below. If you follow this step by step, the children will be able to do it.
Week Commencing: 15th January 2018
This week we are learning about the following:-
1. Counting in 50s and 100s from 0
2. Estimating where numbers go on a numberline
3. Calculating 10/100 more and 10/100 less.
The homework is based on estimating where a number goes on a numberline.
Week Commencing: 8th January 2018
In Maths, we have continued our work on shape. We have identified parallel and perpendicular lines in shapes. If you are not sure what these are, see the powerpoint below – this is what the homework is based on. We have also looked at making turns using right angles and have recognised 1 right angle turn to be a quarter turn. To end the week and our shape topic, we have learned about perimeter.
Week Commencing: 1st January 2018
In Maths, we have been looking at shape and angles. I have taught them that an angle occurs where two lines meet. We have looked at the different angles also – there is a powerpoint underneath to help you with them and they loved this song:https://www.youtube.com/watch?v=NVuMULQjb3o.
We looked at what a right angle is and angles that are greater and less than a right angle. The children had great difficulty with this.
They used right angle checker to see if the angle was bigger than it or smaller than it. The homework is based on identifying a right angle. They can use a checker to help them (a post it note or a corner of a piece of paper will do) If anyone wants an extra challenge, they can label the acute and obtuse angles as well!
Week Commencing: 4th December 2017
This week we are finishing our work on fractions for this term and ordering them from smallest to largest.
We are also looking at 2D shape today and are drawing and labelling them according to their names and properties.
Week Commencing: 27th December 2017
We are continuing our work on fractions this week. We are looking at finding fractions of a shape and of an object. See below for the powerpoints we will be using to help them with this understanding. The homework is based on this and should be quite self explanatory!
Week Commencing: 20th November 2017
This week we are continuing our work on time and the children always find this very difficult. We have looked at the difference betweenam and pm and read the time to the exact minute. This is an excellent video to demonstrate what I mean: https://www.youtube.com/watch?v=f1AavpvRLvo. If you get any time at home, ask them periodically what the time is on an analogue clock as they need a lot of practise on this. We have also looked at tenths and have counted up and down in tenths, looked at what they mean and finally have identified what 2/10, 3/10 etc are. See the powerpoint below for more information. This will help you with their homework!
Week Commencing: 13th November 2017
This week we are converting time from minutes into seconds. For example if you were to convert 2 minutes and 15 seconds into seconds you would need to do the following:-
1. Convert the minutes by multiplying the number by 60. In this case 60 x 2 - 120 seconds
2. Add 120 seconds and 15 seconds together.
3. The answer would be 135 seconds.
The children also did some reasoning on this where they had to compare times. For example, in a race Chloe took 68 seconds and David took 1 minute and 3 seconds. Who was the quickest? In this case, they would have to convert both into seconds so that David took 63 seconds. Then they would have to say that David was the quickest because 63 seconds is quicker than 68. The children often get stuck with this as they assume the highest number wins!
We also read the time on an analogue clock to the hour, half past the hour, quarter to the hour and quarter past the hour.
Week Commencing: 6th November 2017
This week we will be starting our work on telling the time. These are the activities we have completed:
1. We have looked at time facts i.e. how many seconds in a minute, how many minutes in an hour, how many hours in a day etc.
2. We have looked at what activities can be done in a second, a minute and an hour to give an understanding of the value of each unit of time.
3. We have looked at how much time has passed between two times and used the numberline to help us. For example, when looking at how much time has passed between 1.45 and 3.15, this is how you would work it out
First, see if you can move any hours. So in this example, you can jump at hour and you get to 2.45. Emphasise that you can’t jump another hour as that would be 3.45 and that would be too much.
Secondly, try and get to the next hour. In this case it is easy as you know that there are 15 minutes to the next hour (45+15=60)
Thirdly, what’s left – in this case it is 15 minutes which is easy to see from the minutes part.
Finally, count up the jumps. In this case 1 hr plus 15 + 15. So, it would be 1hr and 30 minutes. Emphasise it would not be 130.
Week Commencing: 30th October 2017
This week we are dividing using arrays. This is very similar to multiplication but this time looking at how many are in each group. So in this case there are 18 altogether, they are divided into 3 groups and there are 6 in each group.
We have also looked at how to work them out using a numberline.
Week Commencing: 16th October 2017
This week we are learning how to multiply two numbers and we have done this in 2 ways. The first way is using arrays – see powerpoint below to show you how to do this. This will be what the homework is based on so hopefully this will help you!
The second way is through partitioning – these are the steps we use:-
36 x 4 = 144
Step 1. 30 x 4 = 120
Step 2. 6 x 4 = 24
Step 3. 120 + 24 = 144
Week Commencing: 9th October 2017
In maths we have been learning about capacity and money.
With capacity we have learned how to read different scales in order to be able to find out how much liquid is in a container. We have also put containers in order from the smallest amount of liquid to the largest and answered reasoning questions. For example, if I added 350ml to container a, would that change which had the largest amount of water? Prove it.
Money is an excellent skill that they can practise at home. Give them a random set of coins and see if they can count how much they have got. Ask them reasoning questions, for example, would you rather have 108p or £1.18? Why? This is what we have been doing this week.
Week Commencing: 2nd October 2017
This week we have been looking at data handling and have discussed the best scales to use for pictograms and bar charts. For example, it is not always best to use 1s as your scale, especially if there are a lot of people to collect data from. The children have done well at this and have been able to interpret the data as well.
Week Commencing: 25th September 2017
This week we have used our knowledge of place value to put numbers on a numberline. For example, if the start of the numberline is 600 and the end of the numberline is 700, where might 660 be? We have talked about how it is easier to find the middle number, in this case 650 and then work backwards or forwards from there. Sounds easy but it is quite challenging for the children. This is what the homework is based on.
We have also started the process of teaching the children how to add and subtract using the formal written method. Below are two posters explaining how we teach it. Please note that we have only done this using the dienes equipment so we are not expecting them to be able to record it like this yet!
Week Ending: 23rd September 2017
This week we have been adding ones, tens and hundreds to a 3 digit number. They have found this quite difficult. We have used the dienes equipment to try and demonstrate what happens to each number, especially when crossing the boundaries. For example, when working out 357 + 4, we not only have to change the units but also the tens to. We will be continuing work on this during our mental starters as it is a very important skill.
As well as this we have been counting in 50s and recognising how it is similar to counting in 5s. For example, 0, 50, 100, 150, 200 is similar to 0, 5, 10, 15, 20. We have also looked at numbers and compared them, putting them in order from smallest to largest using our knowledge of place value.
Week Ending: 15th September 2017
In Maths, we have been looking a lot at place value, for example, in 458 we have focussed on what each number represents, in this case 400, 50 and 8. We have used this knowledge to add and take away 10 and 100 from a number, write numbers in words and numerals and we have partitioned numbers. They are getting there but can find it difficult to reason using these skills. We have also practised counting in 4s and 8s.
See below for a number generator where you can put in the starting number and the steps you want to jump and it generates numbers. You can hide or show the numbers so it is nice to get the children to work out which one comes next! In year 3, one of the targets is to count forwards and backwards in 4s, 8s, 50s and 100s.
The homework this week should be self explanatory!
Peterbrook Primary School, High Street, Solihull Lodge, Shirley, Solihull, West Midlands, B90 1HR
Tel: +44(0)121 4302545 office@peterbrook.solihull.sch.uk
Top |
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4.
## Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.4
Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
In ∆ABC, ∠ABC = 90°
∠A + ∠C = 90°
∠ABC > ∠BAC and ∠ABC < ∠BCA
∴ D is the largest side of ∠ABC.
∴ AC is opposite side of larger angle.
∴ AC hypoternuse is largest side of ∆ABC.
Question 2.
In sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
Data: AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively.
∠PBC < ∠QCB.
To Prove: AC > AB
Proof: ∠PBC < ∠QCB
Now, ∠PBC + ∠ABC = 180°
∠ABC = 180 – ∠PBC ………. (i)
Similarly, ∠QCB + ∠ACB = 180°
∠ACB = 180 – ∠QCB …………. (ii)
But, ∠PBC < ∠QCB (Data)
∴ ∠ABC > ∠ACB Comparing (i) and (ii), AC > AB
(∵ Angle opposite to larger side is larger)
Question 3.
In fig ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Solution:
Data : ∠B < ∠A and ∠C < ∠D.
To Prove: AD < BC
Proof: ∠B < ∠A
∴ OA < OB …………. (i)
Similarly, ∠C < ∠D
∴ OD < OC …………. (ii)
Adding (i) and (ii), we have
OA + OD < OB + OC
∴ AD < BC.
Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.
Solution:
Data : AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.
To Prove :
(i) ∠A > ∠C
(ii) ∠B > ∠D
Construction: Join AC and BD
Proof: In ∆ABC, AB < BC (data)
∴ ∠3 < ∠4 ………. (i)
∴ ∠2 <∠1 …………. (ii)
Adding (i) and (ii),
∠3 + ∠2 < ∠4 + ∠1
∠C < ∠A
∠A > ∠C
In ∆ABD, AB < AD
∠5 < ∠8
In ∆BCD, BC < CD
∠6 < ∠7
∴ ∠5 + ∠6 < ∠8 + ∠7
∠D < ∠B
OR ∠B > ∠D.
Question 5.
In Fig. PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.
Solution:
Data: PR > PQ and PS bisect ∠QPR.
To Prove: ∠PSR > ∠PSQ
Proof: In ∆PQR, PR > PQ
∴ ∠PQR > ∠PRQ ………. (i)
PS bisects ∠QPR.
∴ ∠QPS = ∠RPS ………… (ii)
By adding (i) and (ii),
∠PQR + ∠QPS > ∠PRQ + ∠RPS ………. (iii)
But, in ∆PQS, ∠PSR is an exterior angle.
∴ ∠PSR = ∠PQR + ∠QPS ………… (iv)
Similarly, in ∆PRS, ∠PSQ is the exterior angle.
∴ ∠PSQ = ∠PRS + ∠RPS ……… (v)
In (iii), subtracting (iv) and (v),
∠PSR > ∠PSQ.
Question 6.
Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.
Solution:
P is a point on straight line l.
PR is the segment for l drawn such that PQ ⊥ l.
Now, in ∆PQR,
∠PQR = 90° (Construction)
∴ ∠QPR + ∠QRP = 90°
∴ ∠QRP < ∠PQR
∴ PQ < PR
Any line segment is drawn from P to l they form a small angle.
∴ PQ ⊥ l is smaller.
We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4, drop a comment below and we will get back to you at the earliest. |
## College Algebra (10th Edition)
(a) $C(x)=90x+1800$, where $C(x)$ represents the cost (in dollars) of making $x$ bicycles in a day. (b) See the image given. (c) Cost of manufacturing $14$ bicycles a day: $C(14)=\$3060$. (d)$22$bicycles can be manufactured for$\$3780$.
Step-1: It is given that a linear cost function is of the form $$C(x) = mx+b$$ According to the question, the manufacturer has a fixed daily cost of $\$1800$, and it costs$\$90$ to manufacture a $single$ bicycle. Suppose the manufacturer made $4$ bicycles in a day. His cost of manufacturing will be $(90\times 4)+1800=\$2160$. Thus, replacing the bicycle quantity$4$with a more general number,$x$, for the number of bicycles, we obtain the required cost function: $$C(x)=90x+1800$$ Step-2: Let us take values from$x=0$to$x=4$bicycles, to obtain the desired data points for the graph. For$C(x=0)=\$1800$ For $C(x=1)=\$1890$For$C(x=2)=\$1980$ For $C(x=3)=\$2070$For$C(x=4)=\$2160$ The obtained graph is given below. Step-3: $C(x=14)=(90 \times 14) + 1800 = \$3060$Step-4: Suppose$x$bicycles can be manufactured for$\$3780$ a day. Thus, $$3780=90x+1800$$ $$\implies 90x=3780-1800$$ $$\implies x=\frac{1980}{90}=22$$ |
# Math Snap
## lete the problems by writing and solving an algebraic equation. Copy Center charges $\ 0.08$ a page for machine-fed copies and $\ 0.20$ per page for hand-fed copies. If Logan's bill for 80 copies of his movie script is $\ 13.00$, how many copies of each type were made?
#### STEP 4
We now have a system of two equations, which we can solve simultaneously to find the values of $x$ and $y$.$\begin{cases} x + y =80 \\0.08x +0.20y =13.00 \end{cases}$
#### STEP 5
To make the system easier to solve, we can multiply the second equation by100 to get rid of the decimals.$\begin{cases} x + y =80 \\8x +20y =1300 \end{cases}$
#### STEP 6
To eliminate one of the variables, we can multiply the first equation by8 and subtract it from the second equation. $8x +20y -8x -8y =1300 -640$
#### STEP 7
implify the equation to find the value of $y$. $12y =660$$y =660 /12$
#### STEP 8
Calculate the value of $y$. $y =660 /12 =55$
#### STEP 9
Now that we have the value of $y$, we can substitute it into the first equation to find the value of $x$. $x +55 =80$
#### STEP 10
olve the equation to find the value of $x$. $x =80 -55$
##### SOLUTION
Calculate the value of $x$. $x =80 -55 =25$So, Logan made25 machine-fed copies and55 hand-fed copies. |
### Complementary Angles
#### Lesson Objective
In this lesson, we will learn about complementary angles.
In this lesson, we will learn:
1. What are complementary angles
2. See some examples on identifying them
The study tips below will give you a short summary on this.
The math video below will explain in depth about this. Furthermore, it will show some examples so that you can understand this lesson better.
### Study Tips
#### Tip #1
When we add two angles and get 90°, we can say that these angles are complementary angles. See an example below:
The math video below will explain more on this.
### Math Video
#### Math Video Transcript
00:00:03.220
In this lesson, we will learn about comp. angles.
00:00:08.110
Let's take a look at these two angles.
00:00:12.130
Notice that, when we add these two angles, we get 90 degrees.
00:00:18.200
Seeing this, we can say that these two angles are comp. angles.
00:00:24.210
Now, when we rotate this arm, these two angles must add to 90 degrees, to remain as comp. angles.
00:00:39.070
Adding 15 degrees with 75 degrees, gives 90 degrees.
00:00:51.020
Adding 67 degrees with 23 degrees, gives 90 degrees.
00:00:58.030
Now, let's see some examples.
00:01:02.140
Are these comp. angles?
00:01:06.030
To find out, we just need to add 12 degrees with 78 degrees. This gives 90 degrees.
00:01:15.220
Hence, these are angles that are complementary.
00:01:21.110
Next example, these two angles are comp. angles. Find x.
00:01:28.240
We know that, when we add these 2 angles, we must get 90 degrees.
00:01:36.240
Hence, we can find 'x', by subtracting 90 degrees, with 73 degrees.
00:01:44.210
Therefore, ‘x’ will be equals to 17 degrees.
00:01:50.190
That is all for this lesson.
### Practice Questions & More
#### Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on complementary angles or pick your choice of question below.
#### Site-Search and Q&A Library
Please feel free to visit the Q&A Library. You can read the Q&As listed in any of the available categories such as Algebra, Graphs, Exponents and more. Also, you can submit math question, share or give comments there. |
It is currently 24 Jun 2017, 21:09
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# M15#18
Author Message
Intern
Joined: 19 Jun 2008
Posts: 20
### Show Tags
19 Nov 2008, 19:24
7
This post was
BOOKMARKED
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .
I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
SVP
Joined: 29 Aug 2007
Posts: 2473
### Show Tags
19 Nov 2008, 23:44
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
(C) 2008 GMAT Club - m15#18
* $$\frac{2}{21}$$
* $$\frac{3}{25}$$
* $$\frac{1}{6}$$
* $$\frac{9}{28}$$
* $$\frac{11}{24}$$
We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .
I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
= (9c1/9c1) (6c1/8c1) (3c1/7c1)
= 1 (6/8) (3/7)
= 9/28
_________________
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
SVP
Joined: 17 Jun 2008
Posts: 1547
### Show Tags
20 Nov 2008, 00:26
1
KUDOS
(3C1 * 3C1 * 3C1) / 9C3
Intern
Joined: 19 Jun 2008
Posts: 20
### Show Tags
20 Nov 2008, 11:08
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3
Why are the combinations in the numerator multiplied with each other?
CIO
Joined: 02 Oct 2007
Posts: 1218
### Show Tags
17 Dec 2008, 08:27
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3
Why are the combinations in the numerator multiplied with each other?
This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.
Hope this helps.
_________________
Welcome to GMAT Club!
Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?
Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.
GMAT Club Premium Membership - big benefits and savings
Senior Manager
Joined: 01 Feb 2010
Posts: 264
### Show Tags
02 Mar 2010, 07:17
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .
I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
p = (3c1*3c1*3c1)/9c3 = 9/28 hence D.
Intern
Joined: 06 Jan 2010
Posts: 17
Schools: Wake Forest Evening
WE 1: ~12 years total
### Show Tags
02 Mar 2010, 07:58
Sorry, how is (3c1*3c1*3c1)/9c3 = 9/28. I keep getting 9/84? 3c1 cubed is 9. Why is 9c3 28? I am sure it has to do with order or something or am I just missing something?
Edit: Crap because 3 cubed is not 9. 27/84 == 9/28. I really hope I don't make stupid mistakes like this on the 17th when I take the test. (It is taking some will power not to edit my question above.)
Intern
Joined: 11 Jan 2010
Posts: 38
### Show Tags
02 Mar 2010, 08:25
2
KUDOS
Out of 3 blue marbles, 3 red marbles, and 3 yellow marbles, probability of selecting one marble of each color is as follows:
B R Y
(3/9) * (3/8) * (3/7)
B, R, and Y can be arranged in factorial (3) = 3 * 2 * 1 ways.
Thus, resulting probability is:
(3/9) * (3/8) * (3/7) * (3 * 2 * 1) = 9/28
Intern
Joined: 30 Oct 2009
Posts: 4
### Show Tags
02 Mar 2010, 09:18
3/9 * 3/8 * 3/7 * 3! = 9/28
Intern
Joined: 23 Dec 2009
Posts: 47
Schools: HBS 2+2
WE 1: Consulting
WE 2: Investment Management
### Show Tags
02 Mar 2010, 22:41
I'm still a bit confused. I got the denominator correct (9C3), which enumerates all the possible outcomes. But I only see 6 (3!) desired outcomes...not 27:
BRY
BYR
YRB
YBR
RBY
RYB
What am I missing?
EDIT: I think this is only taking into account only one of three of each color...i.e. B1 Y1 R1, etc.
Let me know if I am right!
_________________
My GMAT quest...
...over!
Senior Manager
Joined: 13 Dec 2009
Posts: 262
### Show Tags
13 Mar 2010, 09:15
(3C1 * 3C1 * 3C1)/9C3
_________________
My debrief: done-and-dusted-730-q49-v40
Manager
Status: A continuous journey of self-improvement is essential for every person -Socrates
Joined: 02 Jan 2011
Posts: 71
### Show Tags
07 Mar 2011, 21:36
dzyubam wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3
Why are the combinations in the numerator multiplied with each other?
This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.
Hope this helps.
Please explain why we have written 9C3 in the denominator.
CIO
Joined: 02 Oct 2007
Posts: 1218
### Show Tags
08 Mar 2011, 01:30
Denominator contains the number of total possible outcomes. In this case, the number of possible outcomes equals the number of ways 3 marbles can be drawn from the total of 9 marbles available (9C3).
_________________
Welcome to GMAT Club!
Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?
Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.
GMAT Club Premium Membership - big benefits and savings
Intern
Joined: 11 Jan 2010
Posts: 6
### Show Tags
08 Mar 2011, 23:44
4
KUDOS
The faster way to calculate the probability is to follow events manually:
1. The first marble will be of any one color. Probability (P1) is 100% or 1;
2. The second marble should be of other two colors (any of 6 out of 8 remaining marbles). Probability (P2) is 6/8 or 3/4;
3. The third marble should be of the last color (any of 3 out of 7 remaining marbles). Probability (P3) is 3/7;
P=P1*P2*P3=1*3/4*3/7=9/28
Intern
Joined: 15 Apr 2011
Posts: 49
### Show Tags
10 Mar 2012, 01:35
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .
I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
Got D!!!
First Draw - (9 marbles/ 9 marbles) = 1
Second Draw - Since one color is selected & we dont want that same one again... (6 marbles/ 8 remaining marbles) = 3/4
Third Draw - Since two colors are selected & we dont want the same ones again... (3 marbles/ 7 remaining marbles) = 3/7
Multiply all three to get the probability...
1 * (3/4) * (3/7) = 9/28
Intern
Joined: 10 Aug 2012
Posts: 19
Location: India
Concentration: General Management, Technology
GPA: 3.96
### Show Tags
11 Mar 2013, 02:41
1
KUDOS
For first ball...we can pick any one out of nine..(lets say first is blue)
For second ball of different color...we have choose one ball out of 8 balls
and color of ball should not blue..it may be of red or yellow color...
so probability is 6(3 red and 3 yellow)/8...(lets say we pick a red one here)
For third ball of different color...we have choose one ball out of 7 balls
and color of ball should yellow..
so probability is 3 (yellow)/7
So probability that a marble of each color is among the extracted
$$= \frac{6}{8}*\frac{3}{7}$$
$$= \frac{9}{28}$$
Math Expert
Joined: 02 Sep 2009
Posts: 39662
### Show Tags
11 Mar 2013, 02:49
1
KUDOS
Expert's post
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .
I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
A. $$\frac{2}{21}$$
B. $$\frac{3}{25}$$
C. $$\frac{1}{6}$$
D. $$\frac{9}{28}$$
E. $$\frac{11}{24}$$
$$P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$. Numerator represents ways to select 1 blue marble out of 3, 1 red marble out of 3 and 1 yellow marble out of 3. Denominator represents total ways to select 3 marbles out of 9.
_________________
Intern
Joined: 04 Sep 2012
Posts: 15
Location: India
WE: Marketing (Consumer Products)
### Show Tags
12 Mar 2013, 23:36
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3
Why are the combinations in the numerator multiplied with each other?
So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.
Instead if this was a case of 'OR', we would have added the events.
Hope that helped.
Can someone tell me what is the difficulty level of this question?
Math Expert
Joined: 02 Sep 2009
Posts: 39662
### Show Tags
13 Mar 2013, 02:35
1
KUDOS
Expert's post
ranjitaarao wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3
Why are the combinations in the numerator multiplied with each other?
So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.
Instead if this was a case of 'OR', we would have added the events.
Hope that helped.
Can someone tell me what is the difficulty level of this question?
The difficulty level is ~650.
_________________
Current Student
Joined: 28 Feb 2013
Posts: 15
Location: United States
Concentration: Statistics, Strategy
GMAT Date: 03-12-2014
GRE 1: 324 Q154 V170
WE: Information Technology (Other)
### Show Tags
24 Feb 2014, 11:03
nevermind... i realize i was forgetting the parenthetical part of the selection. in the denominator... its follish lapses like these that affect my scores and keep me in the 600s
Re: M15#18 [#permalink] 24 Feb 2014, 11:03
Go to page 1 2 Next [ 23 posts ]
Display posts from previous: Sort by
# M15#18
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. |
## Logarithms
### Exponential Functions
A function in the form of y = ax, where a >= 1 is called an exponential function.
E.g.
y = 2x; y = 3x
The following image shows the nature of two exponential functions:
Exponential functions show the following features, regardless of the number:
• They go through (0,1) on a grid.
• For a small increase in x, there is a steep increase in y - hence, the change becomes exponential.
• As x approaches negative infinity, y approaches 0 - creating an asymptote.
• The values of the functions remain positive, regardless of the value of x.
### The Exponential Function - ex
This is a special exponential function, which is widely used in mathematics. It's special, because the gradient of the curve at any point is the same as the value of the function at the same point. It's still an exponential function, because it shows all the features that we discussed before.
The following image shows this feature:
### The Exponential Function - ex - and the Natural Logarithm Function - ln(x)
The following image shows the exponential function and the natural logarithm function on the same grid. You can see they are closely related and worth studying together.
The following features can be seen in the two graphs:
• They swap around x and y values - e0 = 1; ln(1) = 0.
• They are symmetrical around y = x line.
• ln(x) is not defined for x = 0 or negative x values.
• The values of the function ln(x) can be positive, negative or even 0.
### Logarithms
Consider the following indices and how they produce the corresponding logarithm values:
82 = 64 => we say that log864 = 2; - this is read as log 64 to the base 8 is 2.
43 = 64 => we say that log464 = 3; - this is read as log 64 to the base 4 is 3.
Note the positions of the bases and indices in each case.
You may have noticed that in both cases the function is log(64), yet the values are different, because of the different bases. It shows the significance of the base, when it comes to dealing with logarithms. In short, log without a base is like a laptop without a keypad! The base is really important in logarithms.
In theory, a logarithm function can take any base value; the most used bases, however, are 10 and e.
log10x = log(x) or lg(x); logex = ln(x);
Let's solve some problems involving logarithms now; the following approach is highly recommended to master this topic:
E.g.1
Find the value of log327
Let log327 = x
3x = 27
Now write down 27 in index form with the same base.
3x = 33
So, x = 3.
log327 = 3.
E.g.2
Find the value of log232
Let log232 = x
2x = 32
Now write down 32 in index form with the same base.
2x = 25
So, x = 5.
log232 = 5.
E.g.3
Find the value of log10100
Let log10100 = x
10x = 100
Now write down 100 in index form with the same base.
10x = 102
So, x = 2.
log10100 = 2.
E.g.4
Find the value of log22
Let log22 = x
2x = 2
Now write down 2 in index form with the same base.
2x = 21
So, x = 1.
log22 = 1.
E.g.5
Find the value of log31
Let log31 = x
3x = 1
Now write down 1 in index form with the same base.
3x = 30
So, x = 0.
log31 = 0.
E.g.6
Find the value of log21/2
Let log21/2 = x
2x = 1/2
Now write down 1/2 in index form with the same base.
2x = 2-1
So, x = -1.
log21/2 = -1.
E.g.7
Find the value of log48
Let log48 = x
4x = 8
Now write down both 4 and 8 in a common base, such as 2, in index form.
(22)x = 23
22x = 23
So, 2x = 3.
x = 3/2
log48 = 3/2.
E.g.8
Find the value of log366
Let log366 = x
36x = 6
Now write down both in index form with the base 6.
36x = 61
(62)x = 61
So, 62x = 61
2x = 1 => x = 1/2
log366 = 1/2.
E.g.9
Find the value of log0.254
Let log0.254 = x
0.25x = 4
Now write down both in index form with the base 4.
(1/4)x = 41
(4-1)x = 41
So, 4-x = 41
-x = 1 => x = -1
log0.254 = -1.
E.g.10
Find the value of log168
Let log168 = x
16x = 8
Now write down both in index form with the base 2.
(24)x = 23
So, 24x = 23
4x = 3 => x = 3/4
log168 = 3/4.
E.g.11
Solve logx125 = 3
x3 = 125
Now write down 125 with the base 5.
x3 = 53
x = 5
E.g.12
Solve logx9 = 0.5
x1/2 = 9
Now write down 9 with the base 81.
x1/2 = 811/2
x = 81
Laws of Indices
Indices and logarithms go hand in hand. So, it's really important to understand the rules of indices, in order to deal with logarithms. The following examples may be helpful, when it comes to revising the basic rules of indices:
1. ax . ay = ax + y
E.g.
23 . 24 = 23 + 4 = 27
2. ax / ay = ax - y
E.g.
27 / 24 = 27 - 4 = 23
3. (ax)y = axy
E.g.
(23)2 = 23X2 = 26
### Laws of Logarithms
1. loga(xy) = logax + logay
Proof:
Let loga(x) = k and loga(y) = l
So, ak = x and al = y
By multiplying together, xy = ak X al = ak + l
loga(xy) = k + l = loga(x) + loga(y).
E.g.
Find loga(4X3).
loga(4X3) = loga4 + loga3.
2. loga(x/y) = logax - logay
Proof:
Let loga(x) = k and loga(y) = l
So, ak = x and al = y
By dividing, x/y = ak :- al = ak - l
loga(x/y) = k - l = loga(x) - loga(y).
E.g.
Find loga(5/2).
loga(5/2) = loga5 - loga2.
3. loga(x)y = ylogax
Proof:
Let loga(x) = k
ak = x
(ak)y = xy
aky = xy
loga(xy) = ky
loga(xy) = yloga(x).
E.g.
Find loga(4)3.
loga(4)3 = 3loga4.
Proof:
loga(1/x) = loga(1) - loga(x)
= 0 - loga(x)
- loga(x).
E.g.
Express loga(1/4) in an alternate form.
loga(1/4) = -loga4.
Proof:
Let logb(x) = k
bk = x
Taking log to the base a on both sides,
loga(bk) = loga(x)
k loga(b) = loga(x)
k = loga(x)/loga(b)
logb(x) = loga(x)/loga(b).
E.g.
Find log4(8) as a logarithm of the base 2.
log48 = log28/log24
Let x = log28 => 2x = 8 => x = 3
Let y = log24 => 2y = 4 => y = 2
log48 = 3/2
Proof:
logx(y) = logy(y)/logy(x)
= 1/logy(x).
E.g.
Express log4(8) in an alternate form.
log4(8) = 1/log84.
Logarithms Problem Solving
E.g.1
Write log38 + log37 as a single logarithm.
log38 + log37 = log38X7
log356.
E.g.2
Write log35 + log34 - log32 as a single logarithm.
log35 + log34 - log32 = log3(5X4/2)
log310.
E.g.3
Simplify log381 - log39
log381 - log39 = log3(81/9)
= log39
Let x = log39
3x = 9 = 32
x = 2
log381 - log39 = 2.
E.g.4
Simplify 2 log39 - log327
2 log39 - log327 = log392 - log327
= log3(81/27)
= log33
Let x = log33
3x = 3 = 31
x = 1
2 log39 - log327 = 1.
E.g.5
Simplify 3 log35 - 2 log35
3 log35 - 2 log35 = log353 - log352
= log3(125/25)
= log35.
E.g.6
Simplify 3 log66 - 2 log66
3 log66 - 2 log66 = log663 - log662
= log6(216/36)
= log66
Let x = log66
6x = 6 = 61
x = 1
3 log66 - 2 log66 = 1.
E.g.7
Simplify 3 log44 + 1/2 log44 - 2 log44
3 log44 + 1/2 log44 - 2 log44 = log443 + log441/2 - log442
= log4(64X2/16)
= log48
Let x = log48
4x = 8
(22)x = 23
2x = 3 => x = 3/2
3 log44 + 1/2 log44 - 2 log44 = 3/2.
E.g.8
Simplify 4 log1010 - (2 log105 + 2 log102)
4 log1010 - (2 log105 + 2 log102 = log10104 - (log1052 + log1022)
= log10(10000/25X4)
= log10100
Let x = log10100
10x = 100
10x = 102
x = 2
4 log1010 - (2 log105 + 2 log102) = 2.
E.g.9
Write in terms of logxa, logxb and logxc, the simplified term of logx(a2b / √c).
logx(a2b / √c) = logxa2 + logxb - logx√ c
= 2 logxa + logxb - 1/2 logxc
E.g.10
Write in terms of logxa and logxb, the simplified term of logx(a2b2).
logx(a2b2) = logxa2 + logxb2
= 2 logxa + 2 logxb
= 2(logxa + logxb).
### Solving equations involving logarithms
E.g.1
Solve 10x = 102.
Before finding the value of x, we can estimate it; since 102 = 100, x must be slightly bigger than 2.
10x = 102 => log10102 = x
From calculator, log10102 = 2.009
So, x = 2.009(3 d.p.).
E.g.2
Solve 102x - 1 = 400.
102x - 1 = 400 => log10400 = 2x - 1
From calculator, log10400 = 2.6
So, 2x - 1 = 2.6
2x = 3.6
x = 1.8
E.g.3
Solve 5x = 130.
Since 53 = 125, x must be slightly bigger than 3. Let's solve this by logarithms of different bases.
Method 1 - using log to the base 10
log105x = log10130
x log105 = 10130
x = 10130/105
= 3.02(2dp)
Method 2 - using log to the base e
loge5x = loge130
x loge5 = loge130
x ln(5) = ln(130)
x = ln(130)/ln(5)
= 3.02(2 d.p.)
The answer is almost the same regardless of the log base.
E.g.4
Solve 5(2x + 1) = 3(x - 1) .
log(52x + 1) = 3(x - 1)
(2x + 1)log 5 = (x - 1)log 3
(2x + 1)/(x - 1) = log 3 / log 5
(2x + 1)/(x - 1) = log 3 / log 5 = 0.6826
2x + 1 = 0.6826x - 0.6826
1.3174x = -1.6826
x = -1.28
E.g.5
Solve 32x + 4(3x) - 12 = 0.
Let y = 3x => 32x = 3x . 3x = y2
So, the equation becomes, y2 + 4y - 12 = 0
(y + 6)(y - 2) = 0
y = -6 or y = 2
Since, y, 3x - an exponential function - cannot be negative,
y = 2 => 3x = 2
log 3x = log 2
x log 3 = log 2
x = log 2/ log 3
x = 0.63(2 d.p.)
E.g.6
log3x + 8/log3x = 6
Let y = log3x
y + 8/y = 6
y2 + 8 = 6y
y2 - 6y + 8 = 0
(y - 4)(y - 2)=0
y = 4 or y = 2
log3x = 4 or log3x = 2
x = 34 or x = 32
x = 81 or x = 9
E.g.6
log5(3 - 2x) = log25(5x2 - 13x + 3)
= log5(5x2 - 13x + 3) / log525
Since log525 = 2,
log5(5x2 - 13x + 3) / log525 = log5(5x2 - 13x + 3) / 2
2 log5(3 - 2x) = log5(5x2 - 13x + 3)
log5(3 - 2x)2 = log5(5x2 - 13x + 3)
log5(3 - 2x)2 - log5(5x2 - 13x + 3) = 0
log5[(3 - 2x)2 / (5x2 - 13x + 3)] = 0
Since log(1) = 0 => [(3 - 2x)2 / (5x2 - 13x + 3)] = 1
9 - 12x + 4x2 = 5x2 - 13x + 3
x2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 or x = -2
Since x = 3 is not valid for log(3 - 2x),
x = -2
### Sketching Straight Lines from Curves
Thanks to logarithms, a complex mathematical relationship can be sketched as straight line, after a transformation in the format. In practice, it is tremendously useful and easy to analyse the relationship.
Suppose, the relationship involved is y = xn, where n is a constant.
So, log y = log xn
log y = n log x
This is in the form of y = mx, which is a straight line, as follows:
In the same way, we can turn the following curves into straight lines easily by converting the variables into logarithms.
y = x2 => log y = 2 log x
y = x3 => log y = 3 log x
y = x4 => log y = 4 log x
The gradient of each line is the index number of each function of the corresponding curve. The straight lines are as follows:
In addition, even a complex relationship can be turned into a corresponding logarithm function.
Suppose, y = pxn
log y = log pxn
log y = log p + log xn
log y = log p + n log x
log y = n log x + log p
y = mx + c
The line is as follows:
The gradient, m = n; y-intercept = log p
Do you want this tutorial with more worked examples? It's available at Amazon:
### Resources at Fingertips
This is a vast collection of tutorials, covering the syllabuses of GCSE, iGCSE, A-level and even at undergraduate level. They are organized according to these specific levels.
The major categories are for core mathematics, statistics, mechanics and trigonometry. Under each category, the tutorials are grouped according to the academic level.
This is also an opportunity to pay tribute to the intellectual giants like Newton, Pythagoras and Leibniz, who came up with lots of concepts in maths that we take for granted today - by using them to serve mankind.
Email:
### Stand Out - from the crowd
"There's no such thing as a free lunch."
The best things in nature are free with no strings attached - fresh air, breathtakingly warm sunshine, scene of meadow on the horizon...
Vivax Solutions, while mimicking nature, offers a huge set of tutorials along with interactive tools for free.
Please use them and excel in the sphere of science education.
Everything is free; not even registration is required.
" |
# Fig.19 shows a crane lifting a car of mass 850 kg at constant velocity through a height of 12m in a time of 40s. The crane has a working efficiency of 60%.
i) Calculate the tension in the lifting cable (in N)
ii) Calculate the total input power required by the crane to lift the car (in W)
iii) Suggest and explain many ways the crane can be modified to improve its efficiency
## I will give you the meaning of the abbreviations, but you will have to write them into the equations.
F = Force in newtons.
2. M = Mass in kilograms.
3. g = Acceleration due to gravity in m/s^2.
4. d = Distance in meters.
5. J = Joules.
6. s = Time in seconds.
7. Pin = Power input.
8. Eff. = Efficiency.
## I. F = M*g = 850 * 9.8 = 8330 N. = Tension in the cable.
II. Po = F*d/t = 8330 * 12/40 = 2499 J./s = 2499 Watts. = Power out.
Pin = Po/Eff. = 2499/0.60 = 4165 Watts.
= Input power.
## i) To calculate the tension in the lifting cable, we can use the equation: Tension = Weight + Friction.
First, let's calculate the weight of the car:
Weight = mass * gravity
Weight = 850 kg * 9.8 m/s^2 = 8330 N
Since the car is moving at a constant velocity, the net force on it is zero, which means the force due to friction is equal to the tension in the cable.
Friction = Tension
Therefore:
Tension = Weight + Friction = 2 * Weight
Tension = 2 * 8330 N = 16660 N
ii) The total input power required by the crane can be calculated using the equation: Power = Work Done / Time
The work done is equal to the force applied multiplied by the distance it acts over:
Work Done = Force * Distance
In this case, the force is equal to the tension in the cable, and the distance is the height through which the car is lifted:
Work Done = Tension * Distance = Tension * Height
Work Done = 16660 N * 12 m = 199,920 J (Joules)
Now we can calculate the total input power:
Power = Work Done / Time
Power = 199,920 J / 40 s = 4998 W (Watts)
iii) To improve the efficiency of the crane, we can suggest the following modifications:
1. Reduce friction: By using lubricants or improving the lubrication system, the amount of friction between the moving parts of the crane can be reduced. This would decrease the force due to friction and therefore increase the efficiency.
2. Upgrade the motor: By using a more powerful motor, the crane can lift the car more efficiently, reducing the required input power.
3. Optimize the design: The crane can be modified to have a more aerodynamic shape or reduce unnecessary weight. These modifications would reduce air resistance and make the crane more efficient.
4. Improve the pulley system: By using higher-quality pulleys with lower friction, the efficiency of the crane can be improved.
5. Use lighter materials: Constructing the crane using lighter materials can reduce the weight that needs to be lifted, leading to increased efficiency.
Remember, improving the efficiency of a crane involves a combination of reducing friction, optimizing design, upgrading components, and using more efficient materials and systems.
## i) To calculate the tension in the lifting cable, we can use the equation:
Force = mass * acceleration
Since the car is being lifted at constant velocity, the acceleration is zero. Therefore, the force needed to lift the car is equal to its weight.
Weight = mass * gravity
where gravity is the acceleration due to gravity (approximately 9.8 m/s^2).
Weight = 850 kg * 9.8 m/s^2 = 8330 N
However, since the crane has a working efficiency of 60%, we need to account for the fact that only 60% of the input energy is converted into useful work. So, we divide the calculated weight by the efficiency:
Tension in lifting cable = (Weight / efficiency) = (8330 N / 0.6) = 13883.33 N (to 2 decimal places)
ii) The total input power required by the crane can be calculated using the formula:
Power = work done / time
The work done by the crane is equal to the force applied (tension in the lifting cable) multiplied by the distance covered (height).
Work done = force * distance
Work done = (Tension in lifting cable) * (distance)
Work done = 13883.33 N * 12 m = 166599.96 J (to 2 decimal places)
Now, we can calculate the total input power using the formula:
Power = work done / time
Power = 166599.96 J / 40 s = 4164.999 W (to 3 decimal places)
iii) To improve the efficiency of the crane, several modifications can be made:
1. Reduce friction: Minimizing friction in the crane's moving parts by proper lubrication and maintenance can reduce energy losses.
2. Increase mechanical advantage: Increasing the mechanical advantage of the crane's pulley system or using a more efficient lifting mechanism can reduce the force required, leading to lower energy consumption.
3. Streamline design: Reducing air resistance by designing the crane with aerodynamics in mind can help improve efficiency.
4. Improve motor efficiency: Upgrading to a more efficient motor can reduce energy losses and improve the overall efficiency of the crane.
5. Optimize cable length: Adjusting the length of the lifting cable to minimize unnecessary weight and friction can improve the crane's efficiency.
6. Use lightweight materials: Constructing the crane using lightweight, strong materials can reduce its overall weight and the force required to lift loads.
By implementing these modifications, the crane can become more energy-efficient and require less power to perform the same tasks. |
Mathura Name Aman Chaudhary Class 8 th Section
• Slides: 11
Mathura. Name: Aman Chaudhary Class: 8 th Section: B
Vedic Ganit: - Introduction: Vedic Ganit is the collection of easy mathematic sutra by which we can solve mathematical question/problems easily. We also find solution orally. It help me in examination time to calculate multiplication, division etc. in less time.
Vedic Ganit help me in my mathematics syllabus in following Topic: ü Multiplication ü Division ü Square ü Cube & Cube root
Multiplication: - In multiplication Two Sutra are use. The Sutra are: - 1. Nikhilam Navatas’caramam 2. Urdhva-tiryagbhyam Das’atah Example by Nikhilam Navatas’caramam Das’atah: - v 91 multiply by 91. By general method. 91*91 91 819* 8281 v By Vedic method- Nikhilam Navatas’caramam Das’atah 91 -9 82/81 In this method, Firstly we are count the complement of 91 -9; second, add them ; third, subtract the sum from nearest base-100 and multiply the complement. So, answer is 8281. 56 multiply by 98. By general method. 56*98 448 504* 5488 By Vedic method-Nikhilam Navatas’caramam Das’atah 56 -44 98 - 2 54/88 In this method, Firstly, we are count the complement of 56 -44 & 98 -2; second, add them ; third , subtract the sum from nearest base-100 and multiply the complement. So, answer is 5488.
Example by Urdhva-tiryagbhyam: v 12 multiply by 11. By General method. By Vedic Ganit method- Urdhva-tiryagbhyam. 12*11 12 12* 132 12 *11 1/1+2/2 =132 v 23 multiply by 21. By General method. 23*21 23 46* 483 By Vedic Ganit method- Urdhva-tiryagbhyam. 23 *21 4/2+6/3 =483
Division: In division, we are use Dhwajanka method. For example : v 98374 divided by 87. By general method. 87)98374(1130. 8 87 113 87 261 640 By Vedic Ganit method-Dhwajanka. 87 9837: 4 1130: 8 In this method , we are imagine Dhwaj of first digit of divider. After we divide 9 by 8 i. e. 1 and 1 reminder. After we subtract 7*1 from 18 and divide answer by 8 i. e. (18 -7)/8=1 and 3 reminder. After we subtract 7*1 from 33 and divide answer by 8 i. e. (33 -7)/8=3 and 2 reminder. After we subtract 7*3 from 27 and divide answer by 8 i. e. (27 - 21)/8=0 and 6 reminder. After we subtract 7*0 from 64 and divide by 8 i. e. (64 -0)/8=8. So, answer is 1130. 8
Square: We are use Yavdunam Sutra to calculate square. For Example: v. Square of 997. =994/009 =994009 In this method, first, we are subtract number from nearest base-1000 i. e. 1000 -997=3; second, calculate square of 3 in three digit : we get first part i. e. 009 & to find last part, subtract 3 from number i. e. 9973=994. So, number =994009.
v. Square of 113. =126/169 =12769 There number is larger than nearest base. So, first, we are subtract nearest base-100 from number i. e. 113 -100=13; second, calculate square of 13 in two number : we get first part i. e. 69 and 1 carry & to find another part, add 13 to number i. e. 13+113=126 and add carry. So, number =12769.
Cube: We are use Yavdunam Sutra to calculate cube. For Example: v Cube of 104. =112/48/64 =1124864 In this method, first, we find nearest base-100, there number is larger than nearest base. So, second, subtract nearest base from number i. e. 104 -100=4 & we find mass. To find first part, we write cube of mass i. e. 43=64; to find next part, we multiply mass*3 i. e. 4*4*3=48 & to find last part, add number to 2*mass i. e. 104+2*4=112. So, number is 1124864.
v. Cube of 996. = 988/048/064 =988047936 In this method, first, we find nearest base-1000, there number is smaller than nearest base. So, second, subtract number from nearest base i. e. 1000 -996=4 & we find mass. To find first part, we write cube of mass i. e. 43=064; to find next part, we multiply mass*3 i. e. 4*4*3=048 & to find last part, subtract 2*mass from number i. e. 996 -2*4=998. There number is smaller than nearest base. So, we subtract first part from nearest base i. e. 1000 -064=936 & subtract 1 from second part i. e. 048 -1=047. Note: This type question we subtract carry of first part from second part.
Cube root: For example: v Cube root of 32768. Step 1 From groups of three starting from the rightmost digit of 32768. 32 768. In this case one group i. e. , 768 has three digits whereas 32 has only two digits. Step 2 Take 768. One’s place of cube of 8 is 2. So, we take the one’s place of the required cube root as 2. Step 3 Take the other group, i. e. , 32. Cube of 3 is 27 and cube of 4 is 64. 32 lies between 27 and 64. The smaller number among 3 and 4 is 3. Take 3 as ten’s place of the cube root of 32768. Thus, cube root of 32768 is 32. |
reasoning (43) Maths (40) Home (8) integration (6)
## Ten most important problems of Reasoning for competitive exam Part1
Ten most important problems of Reasoning for competitive exam which includes missing numbers and missing terms and their solutions in reasoning analogy . These questions are very very important for upcoming competitive exams like SSC CGL ,SSC CHSL and RRB NTPC Etc
# Ten most important problems of Reasoning for competitive exams
Solution
6 is related to 29 in the same way 24 will be related to ? , It means we have to apply same mathematical operations to 24 to get ?.
So if we multiply 6 with 5 and then subtract 1 from result obtained in previous step like this ( 6 × 5 ) - 1 which will be equal to 29.
Same operation we have to apply to 24.
( 24 × 5 ) - 1 = 120 - 1 = 119
So Correct option is ( c ) 119
## Problem # 2
5 : 100 :: 7 : ?
(a) 135 (b) 91 (c) 196 (d) 49
Solution
Because in 1st case if we take Square of 5 then multiply it with 4 we shall have 100.
5² × 4 = 25 × 4 = 100
Same procedure will be applied in 3rd number by taking square of 7 then multiply it with 4
7² × 4 = 49 × 4 = 196
Correct option is ( c ) 196
Problem # 3
6 : 18 :: 4 : ?
(a)4 (b)6 (c)8 (d)10
Solution
Divide 6² by 2
6² ÷ 2 = 36 ÷ 2 = 18
Similarly divide 4² with 2
i.e. 4² ÷ 2 = 16 ÷2 = 8
Divide the square of 1st number by 2 to get 2nd number. Similarly take square of third number and then divide it by 2 to to get the number equal to? Since square of 4 is 16 then divided by 2 to get it.
6² ÷ 2 = 36 ÷ 2 = 18
4² ÷ 2 = 16 ÷ 2 = 8
Correct option is ( c ) 8
Problem # 4
18 : 30 :: 36 : ?
(a)64 (b) 62 (c)54 (d) 66
Solution
(18 × 2 ) - 6 = 36 - 6 = 30
(36 × 2 ) - 6 = 72 - 6 = 66
Multiply 1st number (18) with 2 and then subtract 6 from it
18 × 2 = 36 - 6 = 30
Similarly Multiply 3rd number ( 36 ) with 2 and then subtract 6 from it ,
36 × 2 = 72 - 6 = 66
correct option is ( d) 66
Problem # 5
12 : 20 :: 30 : ?
(a)48 (b)42 (c)15 (d)35
Solution
Method 1
Split 12 = 3 × 4
Split 20 = 4 × 5,
Split 30 = 5 × 6,
Study these factors (3 , 4 ) , (4 ,5 ) ,(5 ,6 ) so next pair will be ( 6 , 7 ) , It means ? Will be replaced by the number 6 × 7 = 42
Method 2
12 will be written as square of 3 plus 3 , 20 will be written as square of 4 plus 4 ,30 will be written as square of 5 plus 5, so next number will be written as square of 6 plus 6 which is equal to 42 , Therefore required option will be 42.
Or
Make continuous factors 3 and 4 , 4 and 5 ,5 and 6 and 6 and 7 of 12, 20, 30 and 42 respectively
12 = 3 × 4 ,
20 = 4 × 5 ,
30 = 5 × 6 so
42 = 6 × 7
Add same number to its square to get next number
Or multiply next number to the number
9 = 3² + 3 ,
20 = 4² +4 ,
30 = 5² + 5
42 = 6² + 6
Correct option is ( b ) 42
Problem # 6
12 : 54 :: 15 : ?
(a)64 (b)69 (c)56 (d)67
Solution
{(1st number ) × 5} - 6 = 2nd number
(12 × 5) - 6 = 54
Multiply 3rd number with 5 then subtract 6 from it
{(3rd number ) × 5} - 6 = 4th number
(15 × 5) - 6 = 69
Multiply first number ( 12 ) with 5 then subtract 6 from it to get 2nd number ( 54 ) , Similarly Multiply third number ( 15 ) with 5 then subtract 6 from it.
( 1 2 × 5 ) - 6 = 60 - 6 = 54
( 1 5 × 5 ) - 6 = 75 - 6 = 69
Correct option is ( b) 69
Problem # 7
6 : 5 :: 8 : ?
(a) 6 (b)10 (c) 2 (d)4
Solution
Add 4 to 1st number and divide it with 2 to get 2nd number( 6 + 4 )/2 = 5
Similarly in 2nd case Add 4 to 3rd number then divide the resultant with 2 to get 4th number
( 8 + 4 )/2 = 6
Correct option is ( a ) 6
Problem # 8
29 : 319 :: 23 : ?
(a)115 (b)252 (c)151 (d)46
Solution
Add both the digits of first number i.e. 2 and 9 then multiply it with first number to get second number. similarly in the third number add both the digits I.e 2 and 3 and multiply it with 3rd number to get fourth number.
29 × ( 2 + 9 ) = 29 × 11 = 319
23 × ( 2 + 3 ) = 23 × 5 = 115
So 253 is the right option (a) 115
Problem # 9
6 : 64 :: 11 : ?
(a) 127 (b) 124 (c) 144 (d) 169
Solution
Add 2 to 1st given number and take its Square to get 2nd number. Similarly add 2 to 3rd number and take it square to get fourth number.
(6 + 2 )² = 8² = 64
(11 + 2)² = 13² = 169
Correct option is ( d ) 169
Problem # 10
36 : 50 :: 64 : ?
(a)70 (b) 82 (c)78 (d) 72
Solution
Take Square Root of 1st number and add 1 to it then add 1 to its Square to get 2nd number.
Similarly take square root of 64 add 1 to its square root and take it square and add 1 to get 4th number.
36 = 6² ---> (6 + 1)² + 1 = 50 ,
64 = 8² ------> ( 8 + 1)² + 1 = 82
√36 = 6 add 1 to it = 7 , square this number = 49 + 1 = 50
√64 = 8 add 1 to it = 9 , square this number = 81 + 1 = 82
So Correct option is ( b ) 82
### Do you like Missing number in box, Reasoning problem
To buy cell phones and books for competitive Exams click here
Share:
#### Post a Comment
Your valuable suggestions are always acceptable to us for betterment of this website |
## Monday, August 18, 2014
### Pascal's triangle: fitting polynomials
Taking another look at Pascal's triangle is always worth it. One will always notice something new. There's a neat limit in Pascal's triangle. And here's a neat example of a Pascal's triangle pattern:
The first diagonal is constant.
The second diagonal is linear. If one indexes it as f (0)=1, then the function is f (x)= x+1.
The third diagonal is quadratic. It presents a good challenge to Algebra 2 students to find the quadratic that fits. (Thanks to Tara Slesar for the idea.) Again, if f (0)=1, then the quadratic is $f\left(x\right)=\frac{1}{2}{x}^{2}+1.5x+1$.
The pattern continues. The next diagonal is given by a cubic. This represents a good problem for calculus students. They usually start by stating that $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+1$, since f (0)=1. They also know that the third derivative of f (x) will be a constant 1. Since ${f}^{\left(3\right)}\left(x\right)=6a=1$, which means $a=\frac{1}{6}$. The students are usually tempted to try using the same method to find b and c, but it fails because we don't know how to index them. In other words, ${f}^{\left(2\right)}\left(x\right)=6ax+2b$, and we know a, but we can't say for what value of x the second derivative is 1, or 2, or 3. Of course, they can solve it using known points in f (x) such as (1, 4) and (2, 10) and just making a system of equations. Another method I mentioned in this post.
Anyway, it gets even neater. The linear diagonal is f (x)=x+1. The quadratic diagonal is $f\left(x\right)=\frac{1}{2}\left(x+1\right)\left(x+2\right)$. The cubic diagonal is $f\left(x\right)=\frac{1}{6}\left(x+1\right)\left(x+2\right)\left(x+3\right)$. This pattern continues. Why? Well, let's stop and make the connection to combinatorics. Here is Pascal's triangle in terms of combinations.
The quadratic diagonal cuts through 2C2, 3C2, 4C2, etc. Given that a combination nCr can be calculated by taking $\frac{n!}{r!\left(n-r\right)!}$, the quadratic pattern, starting at x = 0, will go $\frac{2!}{2!\left(2-2\right)!}$, $\frac{3!}{2!\left(3-2\right)!}$, $\frac{4!}{2!\left(4-2\right)!}$, ... $\frac{\left(x+2\right)!}{2!\left(x+2-2\right)!}$. The general term simplifies down to $\frac{\left(x+2\right)\left(x+1\right)}{2·1}$, which is what I found before.
One other neat pattern: because of the way the diagonals cut across Pascal's triangle, every polynomial function is guaranteed at one point to exactly double. |
# Find the distance between the points :
Question:
Find the distance between the points :
(i) $A(5,1,2)$ and $B(4,6,-1)$
(ii) $\mathbf{P}(1,-1,3)$ and $\mathbf{Q}(2,3,-5)$
(iii) $R(1,-3,4)$ and $S(4,-2,-3)$
(iv) $C(9,-12,-8)$ and the origin
Solution:
Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by
$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$
(i) $A(5,1,2)$ and $B(4,6,-1)$
Here, $\left(x_{1}, y_{1}, z_{1}\right)=(5,1,2)$
$\left(x_{2}, y_{2}, z_{2}\right)=(4,6,-1)$
Therefore,
$D=\sqrt{(4-5)^{2}+(6-1)^{2}+(-1-2)^{2}}$
$=\sqrt{(-1)^{2}+(5)^{2}+(-3)^{2}}$
$=\sqrt{1+25+9}$
$=\sqrt{35}$
Distance between points A and B is
$\sqrt{35}$
(ii) $P(1,-1,3)$ and $Q(2,3,-5)$
Here, $\left(x_{1}, y_{1}, z_{1}\right)=(1,-1,3)$
$\left(x_{2}, y_{2}, z_{2}\right)=(2,3,-5)$
Therefore,
$D=\sqrt{(2-1)^{2}+(3-(-1))^{2}+(-5-3)^{2}}$
$=\sqrt{(1)^{2}+(4)^{2}+(-8)^{2}}$
$=\sqrt{1+16+64}$
$=\sqrt{81}=9$
Distance between points P and Q are 9 units
(iii) $R(1,-3,4)$ and $S(4,-2,-3)$
Here, $\left(x_{1}, y_{1}, z_{1}\right)=(1,-3,4)$
$\left(x_{2}, y_{2}, z_{2}\right)=(4,-2,-3)$
Therefore,
$D=\sqrt{(4-1)^{2}+(-2-(-3))^{2}+(-3-4)^{2}}$
$=\sqrt{(3)^{2}+(1)^{2}+(-7)^{2}}$
$=\sqrt{9+1+49}$
$=\sqrt{59}$
Distance between points $R$ and $S$ is $\sqrt{59}$ units.
(iv) $C(9,-12,-8)$ and the origin
Coordinates of origin are $(0,0,0)$
Here, $\left(x_{1}, y_{1}, z_{1}\right)=(9,-12,-8)$
$\left(x_{2}, y_{2}, z_{2}\right)=(0,0,0)$
Therefore,
$D=\sqrt{(0-9)^{2}+(0-(-12))^{2}+(0-(-8))^{2}}$
$=\sqrt{(-9)^{2}+(12)^{2}+(8)^{2}}$
$=\sqrt{81+144+64}$
$=\sqrt{289}=17$
Distance between points $C$ and origin is 17 units. |
Set theoretic operations using Venn diagrams
## Objective
To represent set theoretic operations using Venn diagrams
## Set & Set Operations
A set is a collection of well defined objects or groups of objects.
These objects are often called elements or members of a set. For example, a group of players in a cricket team is a set.
The symbols used while representing the operations of sets
• Union of sets symbol: ∪
• Intersection of sets symbol: ∩
• Complement of set: A’ or Ac
• Subset of set: ⊂
### Operations for sets
• Union of sets : A∪B
• Intersection of Sets : A∩B
• Complement of Sets : A’
• Difference of Sets : A-B
## Venn Diagrams
Most of the relationships between sets can be represented by means of diagrams which are known as Venn diagrams.
These diagrams consist of rectangles and closed curves usually circles. The universal set is represented by rectangle and its subsets by circles.
A Venn diagram is a diagram that helps us visualize the logical relationship between sets and their elements and helps us solve examples based on these sets.
## Examples of different operations of sets using Venn Diagrams
A ∪ B : The union of two sets A and B is the set C which consists of all those elements which are either in A or in B (including those which are in both).
Symbolically we write A ∪ B = {x : x ∈A or x ∈B}
The union of two sets can be represented by a Venn diagram as shown in following figure.
The shaded portion in figure represents A ∪ B.
Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and all the elements of B, the common elements being taken only once.
The symbol ‘∪’ is used to denote the union.
Symbolically, we write A ∪ B and usually read as ‘A union B’.
Example : Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A ∪ B.
Solution : We have A ∪ B = {2, 4, 6, 8, 10, 12}
A ∩ B The intersection of two sets A and B is the set of all those elements which belong to both A and B.
Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B}.
The shaded portion in Figure indicates the intersection of A and B.
Example: Consider the sets A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A ∩ B.
Solution: We see that 6, 8 are the only elements which are common to both A and B.
A ∩ B = { 6, 8 }
#### (A ∪ B)' A union B Complement: consists of those elements of the universal set U which are not in A U B.
The required formula can be written in any of the following forms: A Union B Complement is equal to the intersection of the complements of the two sets A and B.
Example: Determine the elements of A union B complement if U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, and B = {1, 3, 5}
Solution: We have A = {2, 4, 6}, and B = {1, 3, 5}, then A U B is given by, A U B = {1, 2, 3, 4, 5, 6}, then A union B complement is given by, (A ∪ B)' = U - (A ∪ B) = {1, 2, 3, 4, 5, 6, 7} - {1, 2, 3, 4, 5, 6} = {7}
Answer: (A ∪ B)' = {7}
#### (A ∩ B)' A Intersection B Complement consists of all elements of the universe except the elements in A Intersection B. A Intersection B Complement is equal to the union of the complements of the sets A and B.
Mathematically, it is written as (A ∩ B)' = A' U B'.
Example: Consider U = {1, 2, 3, 4, 5, 6, 7, 8, 9} and A' = {2, 3, 5} and B' = {1, 2, 3, 4, 5}. Find the elements in A Intersection B Complement.
Solution: We know that (A ∩ B)' = A' U B' Now, A' U B' = {2, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} ⇒ (A ∩ B)' = {1, 2, 3, 4, 5}
Answer:The elements in A Intersection B Complement are {1, 2, 3, 4, 5}.
#### A′ : If A is a subset of the universal set U, then its complement A′ is also a subset of U.
Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A.
Symbolically, we write A′ to denote the complement of A with respect to U.
Thus, A′ = {x : x ∈ U and x ∉ A}.
Obviously A′ = U – A. We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A.
Let U be the universal set which consists of all prime numbers and A be the subset of U which consists of all those prime numbers that are not divisors of 42.
Thus, A = {x : x ∈ U and x is not a divisor of 42}. We see that 2 ∈ U but 2 ∉ A, because 2 is divisor of 42.
Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are the only elements of U which do not belong to A.
The set of these three prime numbers, i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by A′. So we have A′ = {2, 3, 7}.
Example: Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.
Solution: We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Answer : A′ = {2, 4, 6, 8,10} |
Chapter 6 Uniform Circular Motion and Gravitation
# 6.3 Centripetal Force
### Summary
• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration— a = ac. Thus, the magnitude of centripetal force Fc is
$\boldsymbol{F_{\textbf{c}}=ma_{\textbf{c}}}.$
By using the expressions for centripetal acceleration ac from $\boldsymbol{a_{\textbf{c}}=\frac{v^2}{r};\:a_{\textbf{c}}=r\omega^2},$ we get two expressions for the centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:
$\boldsymbol{F_{\textbf{c}}=\:m}$$\boldsymbol{\frac{v^2}{r}}$$\boldsymbol{;\:F_{\textbf{c}}=mr\omega^2}.$
You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.
Note that if you solve the first expression for r, you get
$\boldsymbol{r\:=}$$\boldsymbol{\frac{mv^2}{F_{\textbf{c}}}}.$
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
### Example 1: What Coefficient of Friction Do Care Tires Need on a Flat Curve?
(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 2).
Strategy and Solution for (a)
We know that $\boldsymbol{{F}_{\textbf{c}}=\frac{mv^2}{r}}.$Thus,
$\boldsymbol{{F}_{\textbf{c}}\:=}$$\boldsymbol{\frac{mv^2}{r}}$$\boldsymbol{=}$$\boldsymbol{\frac{(900\textbf{ kg})(25.0\textbf{ m/s})^2}{(500\textbf{ m})}}$$\boldsymbol{=\:1125\textbf{ N}}.$
Strategy for (b)
Figure 2 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is μsN, where μs is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that N = mg. Thus the centripetal force in this situation is
$\boldsymbol{F_{\textbf{c}}=f=\mu_{\textbf{s}}\textbf{N}=\mu_{\textbf{s}}mg}.$
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for Fc from the equation
$\begin{array}{c} \boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}} \\ \boldsymbol{F_{\textbf{c}}=mr\omega^2} \end{array}$$\rbrace$,
$\boldsymbol{m}$$\boldsymbol{\frac{v^2}{r}}$$\boldsymbol{=}$$\boldsymbol{\mu_{\textbf{s}}mg.}$
We solve this for μs, noting that mass cancels, and obtain
$\boldsymbol{\mu_{\textbf{s}}\:=}$$\boldsymbol{\frac{v^2}{rg}}.$
Solution for (b)
Substituting the knowns,
$\boldsymbol{\mu_{\textbf{s}}\:=}$$\boldsymbol{\frac{(25.0\textbf{ m/s})^2}{(500\textbf{ m})(9.80\textbf{ m/s}^2)}}$$\boldsymbol{=\:0.13}.$
(Because coefficients of friction are approximate, the answer is given to only two digits.)
Discussion
We could also solve part (a) using the first expression in $\begin{array}{l} \boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}} \\ \boldsymbol{F_{\textbf{c}}=mr\omega^2} \end{array}$$\rbrace$, because m, v, and r are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than μsN. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.
Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 3. The greater the angle θ, the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle θ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for θ for an ideally banked curve and consider an example related to it.
For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.
Figure 3 shows a free body diagram for a car on a frictionless banked curve. If the angle θ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w and the normal force of the road N. (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude mv2/r. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,
$\boldsymbol{N\textbf{sin} \;\theta\:=}$$\boldsymbol{\frac{mv^2}{r}}.$
Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is N cos θ, and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,
$\boldsymbol{N\textbf{cos} \;\theta=mg}.$
Now we can combine the last two equations to eliminate N and get an expression for θ, as desired. Solving the second equation for N = mg/(cos θ), and substituting this into the first yields
$\boldsymbol{mg}$$\boldsymbol{\frac{\textbf{sin} \;\theta}{\textbf{cos} \;\theta}}$$\boldsymbol{=}$$\boldsymbol{\frac{mv^2}{r}}$
$\boldsymbol{mg \;\textbf{tan}(\theta)\:=}$$\boldsymbol{\frac{mv^2}{r}}$
$\boldsymbol{\textbf{tan}\theta\:=}$$\boldsymbol{\frac{v^2}{rg}.}$
Taking the inverse tangent gives
$\boldsymbol{\theta=\textbf{tan}^{-1}}$$\boldsymbol{(\frac{v^2}{rg})}$$\textbf{(ideally banked curve, no friction).}$
This expression can be understood by considering how θ depends on v and r. A large θ will be obtained for a large v and a small r. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that θ does not depend on the mass of the vehicle.
### Example 2: What Is the Ideal Speed to Take a Steeply Banked Tight Curve?
Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.
Strategy
We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.
Solution
Starting with
$\boldsymbol{\textbf{tan}\theta\:=}$$\boldsymbol{\frac{v^2}{rg}}$
we get
$\boldsymbol{v=(rg\:\textbf{tan}\:\theta)^{1/2}}.$
Noting that tan 65.0º = 2.14, we obtain
$\begin{array}{lcl} \boldsymbol{v} & = & \boldsymbol{[(100\textbf{ m})(9.80\textbf{ m/s}^2)(2.14)]^{1/2}} \\ \boldsymbol{} & = & \boldsymbol{45.8}\textbf{ m/s.} \end{array}$
Discussion
This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.
Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.
### TAKE-HOME EXPERIMENT
Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.
### PHET EXPLORATIONS: GRAVITY AND ORBITS
Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!
# Section Summary
• Centripetal force Fc is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude
$\boldsymbol{F_{\textbf{c}}=ma_{\textbf{c}}},$
which can also be expressed as
$\boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}}$
or
$\boldsymbol{F_{\textbf{c}}=mr\omega^2}$
# Conceptual Questions
### Conceptual Questions
1: If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.
2: Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?
3: If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain.
4: Race car drivers routinely cut corners as shown in Figure 5. Explain how this allows the curve to be taken at the greatest speed.
5: A number of amusement parks have rides that make vertical loops like the one shown in Figure 6. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:
(a) The car goes over the top at faster than this speed?
(b)The car goes over the top at slower than this speed?
7: What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6 under the following circumstances:
(a) The car goes over the top at such a speed that the gravitational force is the only force acting?
(b) The car goes over the top faster than this speed?
(c) The car goes over the top slower than this speed?
8: As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.
9: Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 7 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.
10: Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat?
11: Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.
### Problems & Exercises
1: (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?
(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?
(c) Compare each force with her weight.
2: Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.
3: What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?
4: What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?
5: (a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?
(b) Calculate the centripetal acceleration.
(c) Does this acceleration seem large to you?
6: Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 9. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).
(a) Show that θ (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, θ = tan-1 v2/rg.
(b) Calculate θ for a 12.0 m/s turn of radius 30.0 m (as in a race).
7: A large centrifuge, like the one shown in Figure 10(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.
(a) At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation?
(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 10(b). At what angle θ below the horizontal will the cage hang when the centripetal acceleration is 10 g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle θ should be.)
8: Integrated Concepts
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?
9: Modern roller coasters have vertical loops like the one shown in Figure 11. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?
10: Unreasonable Results
(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s.
(b) What is unreasonable about the result?
(c) Which premises are unreasonable or inconsistent?
## Glossary
centripetal force
any net force causing uniform circular motion
ideal banking
the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction
ideal speed
the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road
ideal angle
the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed
banked curve
the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve
### Solutions
Problems & Exercises
1: (a) $$\boldsymbol{483\textbf{ N}}$$ (b) $$\boldsymbol{17.4\textbf{ N}}$$ (c) 2.24 times her weight, 0.0807 times her weight
3: $\boldsymbol{4.14^0}$
5: (a) $$\boldsymbol{24.6\textbf{ m}}$$ (b) $\boldsymbol{36.6\textbf{ m/s}^2}$ (c) $\boldsymbol{a_{\textbf{c}}=3.73\textbf{ g}.}$ This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns.
7: (a) $$\boldsymbol{2.56\textbf{ rad/s}}$$ (b) $\boldsymbol{5.71^0}$
8: (a) $$\boldsymbol{16.2\textbf{ m/s}}$$ (b) $$\boldsymbol{0.234}$$
10: (a) $$\boldsymbol{1.84}$$ (b) A coefficient of friction this much greater than 1 is unreasonable . (c) The assumed speed is too great for the tight curve. |
# 4/29/2015Section 8.31 Section 8.3 Compound Interest Objectives 1.Use the compound interest formulas. 2.Calculate present value. 3.Understand and compute.
## Presentation on theme: "4/29/2015Section 8.31 Section 8.3 Compound Interest Objectives 1.Use the compound interest formulas. 2.Calculate present value. 3.Understand and compute."— Presentation transcript:
4/29/2015Section 8.31 Section 8.3 Compound Interest Objectives 1.Use the compound interest formulas. 2.Calculate present value. 3.Understand and compute effective annual yield.
4/29/2015Section 8.32 Compound Interest Compound interest is interest computed on the original principal as well as on any accumulated interest. To calculate the compound interest paid once a year we use A = P(1 + r) t, where A is called the account’s future value, the principal P is called its present value, r is the rate, and t is the years.
4/29/2015Section 8.33 Example: You deposit \$2000 in a savings account at Hometown bank, which has a rate of 6%. a.Find the amount, A, of money in the account after 3 years subject to compound interest. b.Find the interest. Solution: a.Principal P is \$2000, r is 6% or 0.06, and t is 3. Substituting this into the compound interest formula, we get A = P(1 + r) t = 2000(1 + 0.06) 3 = 2000(1.06) 3 ≈ 2382.03 Compound Interest
4/29/2015Section 8.34 Rounded to the nearest cent, the amount in the savings account after 3 years is \$2382.03. b.The amount in the account after 3 years is \$2382.03. So, we take the difference of this amount and the principal to obtain the interest amount. \$2382.03 – \$2000 = \$382.03 Thus, the interest you make after 3 years is \$382.03. Compound Interest Example Continued
4/29/2015Section 8.35 Compound Interest To calculate the compound interest paid more than once a year we use where A is called the account’s future value, the principal P is called its present value, r is the rate, n is the number of times the interest is compounded per year, and t is the years.
4/29/2015Section 8.36 Example: You deposit \$7500 in a savings account that has a rate of 6%. The interest is compounded monthly. a.How much money will you have after five years? b.Find the interest after five years. Solution: a.Principal P is \$7500, r is 6% or 0.06, t is 5, and n is 12 since interest is being compounded monthly. Substituting this into the compound interest formula, we get Compound Interest
4/29/2015Section 8.37 Rounded to the nearest cent, you will have \$10,116.38 after five years. b.The amount in the account after 5 years is \$10,116.38. So, we take the difference of this amount and the principal to obtain the interest amount. \$ 10,116.38 – \$7500 = \$2616.38 Thus, the interest you make after 5 years is \$ 2616.38. Compound Interest Example Continued
4/29/2015Section 8.38 Compound Interest Continuous Compounding Some banks use continuous compounding, where the compounding periods increase infinitely. After t years, the balance, A, in an account with principal P and annual interest rate r (in decimal form) is given by the following formulas: 1.For n compounding per year: 2.For continuous compounding: A = P e rt.
4/29/2015Section 8.39 Example: You decide to invest \$8000 for 6 years and you have a choice between two accounts. The first pays 7% per year, compounded monthly. The second pays 6.85% per year, compounded continuously. Which is the better investment? Solution: The better investment is the one with the greater balance at the end of 6 years. 7% account: The balance in this account after 6 years is \$12,160.84. Compound Interest Continuous Compounding
4/29/2015Section 8.310 6.85% account: A = P e rt = 8000 e 0.0685(6) ≈ 12,066.60 The balance in this account after 6 years is \$12,066.60. Thus, the better investment is the 7% monthly compounding option. Compound Interest Continuous Compounding Example Continued
4/29/2015Section 8.311 Planning for the Future with Compound Interest Calculating Present Value If A dollars are to be accumulated in t years in an account that pays rate r compounded n times per year, then the present value P that needs to be invested now is given by
4/29/2015Section 8.312 Example: How much money should be deposited in an account today that earns 8% compounded monthly so that it will accumulate to \$20,000 in five years? Solution: We use the present value formula, where A is \$20,000, r is 8% or 0.08, n is 12, and t is 5 years. Approximately \$13,424.21 should be invested today in order to accumulate to \$20,000 in five years. Planning for the Future with Compound Interest
4/29/2015Section 8.313 Effective Annual Yield The effective annual yield, or the effective rate, is the simple interest rate that produces the same amount of money in an account at the end of one year as when the account is subjected to compound interest at a stated rate. Example: You deposit \$4000 in an account that pays 8% interest compounded monthly. a.Find the future value after one year. b.Use the future value formula for simple interest to determine the effective annual yield.
4/29/2015Section 8.314 Solution: a.We use the compound interest formula to find the account’s future value after one year. Rounded to the nearest cent, the future value after one year is \$4332.00. Effective Annual Yield Example Continued
Section 8.315 b.The effective annual yield is the simple interest rate. So, we use the future value formula for simple interest to determine rate r. Thus, the effective annual yield is 8.3%. This means that an account that earns 8% interest compounded monthly has an equivalent interest rate of 8.3% compounded annually. Effective Annual Yield Example Continued
EFFECTIVE YIELD Effective yield formula Where r is the rate and n is the number compounded times per year. 4/29/2015Section 8.316
EXAMPLE A passbook savings account has a rate of 5%. The interest rate is compounded daily. Find the account’s effective annual yield. (Assume 360 days in a rear) Solution Assignment – 13 Quiz - 7 Section 8.317
Download ppt "4/29/2015Section 8.31 Section 8.3 Compound Interest Objectives 1.Use the compound interest formulas. 2.Calculate present value. 3.Understand and compute."
Similar presentations |
Special right triangle
(Redirected from Special right triangles)
Position of some special triangles in an Euler diagram of types of triangles, using the definition that isosceles triangles have at least two equal sides, i.e. equilateral triangles are isosceles.
A special right triangle is a right triangle with some regular feature that makes calculations on the triangle easier, or for which simple formulas exist. For example, a right triangle may have angles that form simple relationships, such as 45°–45°–90°. This is called an "angle-based" right triangle. A "side-based" right triangle is one in which the lengths of the sides form ratios of whole numbers, such as 3 : 4 : 5, or of other special numbers such as the golden ratio. Knowing the relationships of the angles or ratios of sides of these special right triangles allows one to quickly calculate various lengths in geometric problems without resorting to more advanced methods.
Angle-based
Special angle-based triangles inscribed in a unit circle are handy for visualizing and remembering trigonometric functions of multiples of 30 and 45 degrees.
"Angle-based" special right triangles are specified by the relationships of the angles of which the triangle is composed. The angles of these triangles are such that the larger (right) angle, which is 90 degrees or π/2 radians, is equal to the sum of the other two angles.
The side lengths are generally deduced from the basis of the unit circle or other geometric methods. This approach may be used to rapidly reproduce the values of trigonometric functions for the angles 30°, 45°, and 60°.
Special triangles are used to aid in calculating common trigonometric functions, as below:
degrees radians gons turns sin cos tan cotan
0 0g 0 0/2 = 0 4/2 = 1 0 undefined
30° π/6 33+1/3g 1/12 1/2 = 1/2 3/2 1/3 3
45° π/4 50g 1/8 2/2 = 1/2 2/2 = 1/2 1 1
60° π/3 66+2/3g 1/6 3/2 1/2 = 1/2 3 1/3
90° π/2 100g 1/4 4/2 = 1 0/2 = 0 undefined 0
45°–45°–90°
30°–60°–90°
The 45°–45°–90° triangle, the 30°–60°–90° triangle, and the equilateral/equiangular (60°–60°–60°) triangle are the three Möbius triangles in the plane, meaning that they tessellate the plane via reflections in their sides; see Triangle group.
45°–45°–90° triangle
The side lengths of a 45°–45°–90° triangle
In plane geometry, constructing the diagonal of a square results in a triangle whose three angles are in the ratio 1 : 1 : 2, adding up to 180° or π radians. Hence, the angles respectively measure 45° (π/4), 45° (π/4), and 90° (π/2). The sides in this triangle are in the ratio 1 : 1 : 2, which follows immediately from the Pythagorean theorem.
Of all right triangles, the 45°–45°–90° degree triangle has the smallest ratio of the hypotenuse to the sum of the legs, namely 2/2.[1]:p.282,p.358 and the greatest ratio of the altitude from the hypotenuse to the sum of the legs, namely 2/4.[1]:p.282
Triangles with these angles are the only possible right triangles that are also isosceles triangles in Euclidean geometry. However, in spherical geometry and hyperbolic geometry, there are infinitely many different shapes of right isosceles triangles.
30°–60°–90° triangle
Set square
The side lengths of a 30°–60°–90° triangle
This is a triangle whose three angles are in the ratio 1 : 2 : 3 and respectively measure 30° (π/6), 60° (π/3), and 90° (π/2). The sides are in the ratio 1 : 3 : 2.
The proof of this fact is clear using trigonometry. The geometric proof is:
Draw an equilateral triangle ABC with side length 2 and with point D as the midpoint of segment BC. Draw an altitude line from A to D. Then ABD is a 30°–60°–90° triangle with hypotenuse of length 2, and base BD of length 1.
The fact that the remaining leg AD has length 3 follows immediately from the Pythagorean theorem.
The 30°–60°–90° triangle is the only right triangle whose angles are in an arithmetic progression. The proof of this fact is simple and follows on from the fact that if α, α + δ, α + 2δ are the angles in the progression then the sum of the angles 3α + 3δ = 180°. After dividing by 3, the angle α + δ must be 60°. The right angle is 90°, leaving the remaining angle to be 30°.
Side-based
Right triangles whose sides are of integer lengths, with the sides collectively known as Pythagorean triples, possess angles that cannot all be rational numbers of degrees.[2] (This follows from Niven's theorem.) They are most useful in that they may be easily remembered and any multiple of the sides produces the same relationship. Using Euclid's formula for generating Pythagorean triples, the sides must be in the ratio
m2n2 : 2mn : m2 + n2
where m and n are any positive integers such that m > n.
Common Pythagorean triples
There are several Pythagorean triples which are well-known, including those with sides in the ratios:
3: 4 :5
5: 12 :13
8: 15 :17
7: 24 :25
9: 40 :41
The 3 : 4 : 5 triangles are the only right triangles with edges in arithmetic progression. Triangles based on Pythagorean triples are Heronian, meaning they have integer area as well as integer sides.
The possible use of the 3 : 4 : 5 triangle in Ancient Egypt, with the supposed use of a knotted rope to lay out such a triangle, and the question whether Pythagoras' theorem was known at that time, have been much debated.[3] It was first conjectured by the historian Moritz Cantor in 1882.[3] It is known that right angles were laid out accurately in Ancient Egypt; that their surveyors did use ropes for measurement;[3] that Plutarch recorded in Isis and Osiris (around 100 AD) that the Egyptians admired the 3 : 4 : 5 triangle;[3] and that the Berlin Papyrus 6619 from the Middle Kingdom of Egypt (before 1700 BC) stated that "the area of a square of 100 is equal to that of two smaller squares. The side of one is ½ + ¼ the side of the other."[4] The historian of mathematics Roger L. Cooke observes that "It is hard to imagine anyone being interested in such conditions without knowing the Pythagorean theorem."[3] Against this, Cooke notes that no Egyptian text before 300 BC actually mentions the use of the theorem to find the length of a triangle's sides, and that there are simpler ways to construct a right angle. Cooke concludes that Cantor's conjecture remains uncertain: he guesses that the Ancient Egyptians probably did know the Pythagorean theorem, but that "there is no evidence that they used it to construct right angles".[3]
The following are all the Pythagorean triple ratios expressed in lowest form (beyond the five smallest ones in lowest form in the list above) with both non-hypotenuse sides less than 256:
11: 60 :61
12: 35 :37
13: 84 :85
15: 112 :113
16: 63 :65
17: 144 :145
19: 180 :181
20: 21 :29
20: 99 :101
21: 220 :221
24: 143 :145
28: 45 :53
28: 195 :197
32: 255 :257
33: 56 :65
36: 77 :85
39: 80 :89
44: 117 :125
48: 55 :73
51: 140 :149
52: 165 :173
57: 176 :185
60: 91 :109
60: 221 :229
65: 72 :97
84: 187 :205
85: 132 :157
88: 105 :137
95: 168 :193
96: 247 :265
104: 153 :185
105: 208 :233
115: 252 :277
119: 120 :169
120: 209 :241
133: 156 :205
140: 171 :221
160: 231 :281
161: 240 :289
204: 253 :325
207: 224 :305
Almost-isosceles Pythagorean triples
Isosceles right-angled triangles cannot have sides with integer values, because the ratio of the hypotenuse to either other side is 2, but 2 cannot be expressed as a ratio of two integers. However, infinitely many almost-isosceles right triangles do exist. These are right-angled triangles with integral sides for which the lengths of the non-hypotenuse edges differ by one.[5][6] Such almost-isosceles right-angled triangles can be obtained recursively,
a0 = 1, b0 = 2
an = 2bn−1 + an−1
bn = 2an + bn−1
an is length of hypotenuse, n = 1, 2, 3, .... Equivalently,
${\displaystyle ({\tfrac {x-1}{2}})^{2}+({\tfrac {x+1}{2}})^{2}=y^{2}}$
where {x, y} are the solutions to the Pell equation x2 − 2y2 = −1, with the hypotenuse y being the odd terms of the Pell numbers 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378... (sequence A000129 in the OEIS).. The smallest Pythagorean triples resulting are:[7]
3 : 4 : 5
20 : 21 : 29
119 : 120 : 169
696 : 697 : 985
4,059 : 4,060 : 5,741
23,660 : 23,661 : 33,461
137,903 : 137,904 : 195,025
803,760 : 803,761 : 1,136,689
4,684,659 : 4,684,660 : 6,625,109
Alternatively, the same triangles can be derived from the square triangular numbers.[8]
Arithmetic and geometric progressions
A Kepler triangle is a right triangle formed by three squares with areas in geometric progression according to the golden ratio.
The Kepler triangle is a right triangle whose sides are in a geometric progression. If the sides are formed from the geometric progression a, ar, ar2 then its common ratio r is given by r = φ where φ is the golden ratio. Its sides are therefore in the ratio 1 : φ : φ. Thus, the shape of the Kepler triangle is uniquely determined (up to a scale factor) by the requirement that its sides be in a geometric progression.
The 3–4–5 triangle is the unique right triangle (up to scaling) whose sides are in an arithmetic progression.[9]
Sides of regular polygons
The sides of a pentagon, hexagon, and decagon, inscribed in congruent circles, form a right triangle
Let a = 2 sin π/10 = −1 + 5/2 = 1/φ be the side length of a regular decagon inscribed in the unit circle, where φ is the golden ratio. Let b = 2 sin π/6 = 1 be the side length of a regular hexagon in the unit circle, and let c = 2 sin π/5 = ${\displaystyle {\sqrt {\tfrac {5-{\sqrt {5}}}{2}}}}$ be the side length of a regular pentagon in the unit circle. Then a2 + b2 = c2, so these three lengths form the sides of a right triangle.[10] The same triangle forms half of a golden rectangle. It may also be found within a regular icosahedron of side length c: the shortest line segment from any vertex V to the plane of its five neighbors has length a, and the endpoints of this line segment together with any of the neighbors of V form the vertices of a right triangle with sides a, b, and c.[11]
References
1. ^ a b Posamentier, Alfred S., and Lehman, Ingmar. The Secrets of Triangles. Prometheus Books, 2012.
2. ^ Weisstein, Eric W. "Rational Triangle". MathWorld.
3. Cooke, Roger L. (2011). The History of Mathematics: A Brief Course (2nd ed.). John Wiley & Sons. pp. 237–238. ISBN 978-1-118-03024-0.
4. ^ Gillings, Richard J. (1982). Mathematics in the Time of the Pharaohs. Dover. p. 161.
5. ^ Forget, T. W.; Larkin, T. A. (1968), "Pythagorean triads of the form x, x + 1, z described by recurrence sequences" (PDF), Fibonacci Quarterly, 6 (3): 94–104.
6. ^ Chen, C. C.; Peng, T. A. (1995), "Almost-isosceles right-angled triangles" (PDF), The Australasian Journal of Combinatorics, 11: 263–267, MR 1327342.
7. ^ (sequence A001652 in the OEIS)
8. ^ Nyblom, M. A. (1998), "A note on the set of almost-isosceles right-angled triangles" (PDF), The Fibonacci Quarterly, 36 (4): 319–322, MR 1640364.
9. ^ Beauregard, Raymond A.; Suryanarayan, E. R. (1997), "Arithmetic triangles", Mathematics Magazine, 70 (2): 105–115, doi:10.2307/2691431, MR 1448883.
10. ^
11. ^ |
# II. Fractions
O'k, the site has a focus on Algebra 1, and we're talking about fractions. Didn't we already cover this in 5th grade? Well - yes, and no. It turns out a lot of people still have problems with fractions. So, I thought it might be helpful to bring you up to speed. I'm calling this chapter -1 because, it's stuff you should have had before Algebra 1.
## Where to start.
There's a couple of concepts you need to agree with before we get started. The first one is that if you cut a pie into 5 equal slices, and you still have all 5 of those slices, then you still have the whole pie. Mathematically that would be 5/5 = 1. In fact, we can expand that to say that anything divided by itself is equal to one.
Are you with me? O'k, lets try the second concept. If I have 1 of something, then I have that thing. Mathematically that would mean that multiplying any number times 1 doesn't change that number. 1*5 = 5.
If we can agree on these two facts - we're ready to start.
## What are fractions?
Fractions are also called ratios and both of these are just different ways to write division problems. 5 divided by 7 could be written as 5/7 or as 5:7. The 5 (on the top of the line) is called the numerator and the 7 (on the bottom of the line) is called the denominator.
## Multiplication
Multiplication is really easy. You multiply the numerators together for the new numerator, and the denominators together for the new denominator. So, (5/7)(2/3) would be (5*2)/(7*3) = 10/21.
Note: When parenthesis touch - as in (5/7)(2/3) that means we're multiplying 5/7 times 2/3. Also, I'm using * as a multiplication symbol because we use x for other things in Algebra.
## Division
There's kind of a trick to division. If one fraction is divided by another, you flip the second and multiply it times the first. (5/7) /(2/3) would be (5/7)(3/2) or 15/14.
Now, if you like - that's all you need to know about division. But, if you want to know why that works - keep reading.
What we're actually doing is multiplying the inverse of the denominator times the numerator and denominator. So, in our example we have (5/7)/(2/3) and we're going to multiply that times (3/2)/(3/2). Hopefully you noticed that (3/2)/(3/2) is a number divided by itself. So, it equals 1, and 1 times anything is just that thing. So we can multiply by this and not change the value.
So, if we're multiplying these two together, we multiply the numerator times the numerator and get (5/7)(3/2). And we multiply the denominators and get (2/3)(3/2) which is (2*3)/(3*2) which is 6/6 which is 1. If we divide a number by 1, we just get the original number. So, the denominator drops to 1. The numerator (5/7)(3/2) becomes our original problem with an answer of 15/14.
Subtraction is just addition with negative numbers. Addition with fractions is kind of a pain.
You need to have the same denominators for each of the ratios. So, if I'm adding 1/3 to 1/4, I have to change the denominators to something that is evenly divisible by both 3 and 4. In this case, I'd go with 12. Now, remember the rule where we can multiply anything by 1 and not change it's value? And the rule that anything divided by itself is 1. So, we're going to multiply (1/3) by (4/4) and get 4/12. Can we agree that 4/12 is the same as 1/3? O'k, then we multiply 1/4 times 3/3 we get 3/12 which is the same as 1/4.
Now we can add 4/12 + 3/12 = 7/12.
III. Algebra
VI Equations
VII. Exponents
IX Fractions
XIII. Linear Equations
XVII. Real Numbers
XIV. Math Terms
XVIII. Percentages
XV. Matrices
XIX. PEMDAS Order of Operations
XXII. Slope |
# Difference between linear and quadratic
## Difference Between Linear Equation and Quadratic Equation
Apart from the adding complexity of solving a quadratic equation compared to a linear one, the two equations produce different types of graphs. A linear function produces a straight line while a quadratic function produces a parabola. Graphing a linear function is straightforward.
and do you need steady hands to be a neurosurgeon
Use linear equations and quadratic equations to find a formula that will allow you to find any term in a sequence or the nth term. If you have a sequence with a constant first common difference, it can be defined using a linear equation. The following formula can be used to find any nth term in the sequence. When you cannot find a constant first common difference, look at the second differences. If that second difference is constant, you can define the sequence using a quadratic equation. The following formula can be used to find any term in the sequence. Create Account Sign In.
Students are often tripped up by the difference between quadratic and linear graphs. However, the shapes and equations of linear and quadratic graphs are very easy to recognize with practice. The graph shapes are dictated by the equations that create them. Following some simple guidelines will help you to recognize the differences between these equations and their graph shapes. Linear graphs are always shaped like straight lines, which can have either positive or negative slopes. If "m" is positive, then the line slopes upward from left to right. If "m" is negative, then the line slopes downward from left to right.
A linear equation in two variables doesn't involve any power higher than one for either variable. A quadratic equation, on the other hand, involves one of the variables raised to the second power. Apart from the adding complexity of solving a quadratic equation compared to a linear one, the two equations produce different types of graphs. Linear functions are one-to-one while quadratic functions are not. A linear function produces a straight line while a quadratic function produces a parabola. Graphing a linear function is straightforward while graphing a quadratic function is a more complicated, multi-step process. A linear equation produces a straight line when you graph it.
A quadratic equation is defined as an equation in which one or more of the terms is squared but raised to no higher power. In a linear- quadratic system where only one variable in the quadratic is squared, the graphs will be a parabola and a straight line. When graphing a parabola and a straight line on the same set of axes, three situations are possible. When prompted for the "First curve? Hit Enter. When prompted for the "Second curve? Ignore the prompt for "Guess?
Receive free math worksheets via email:. Article Summary: "Math textbooks are loaded with a lot of terms. But unlike other subjects, math terms are not always easy to understand based on their name alone. Many students learn dozens of mathematical skills without actually learning the use of the skill or even the name of the skill. Math textbooks are loaded with a lot of terms. American literature is pretty self explanatory.
## How Do You Determine if a Graph Represents a Linear, Exponential, or Quadratic Function?
Linear Equation vs Quadratic Equation. In mathematics, algebraic equations are equations which are formed using polynomials.
## 3.1 - Linear & Quadratic nth Terms
.
.
who won dillashaw vs garbrandt
.
.
. |
# Find the path of least time given these conditions? Starting point = (2,6) End point = (8,-3) Velocity above the line y=2 is 1 unit/second Velocity below the line y=2 is 7 units/second
Feb 11, 2017
There could be simpler way of doing this but, this is what I followed.
#### Explanation:
As the intermediate point must lie at the interface $y = 2$
Hence the general point must be $\left(x , 2\right)$
Path of Displacement is $\left(a\right) \left(2 , 6\right) \to \left(x , 2\right) \mathmr{and} \left(b\right) \left(x , 2\right) \to \left(8 , - 3\right)$
(a) Displacement$= \sqrt{{\left(x - 2\right)}^{2} + {\left(2 - 6\right)}^{2}}$
Time taken $= \text{Displacement"/"Velocity} = \frac{\sqrt{{\left(x - 2\right)}^{2} + 16}}{1} s$
(b) Displacement$= \sqrt{{\left(8 - x\right)}^{2} + {\left(- 3 - 2\right)}^{2}}$
Time taken $= \frac{\sqrt{{\left(8 - x\right)}^{2} + 25}}{7} s$
Total Time $t = \sqrt{{\left(x - 2\right)}^{2} + 16} + \frac{\sqrt{{\left(8 - x\right)}^{2} + 25}}{7}$
$\implies t = \sqrt{\left({x}^{2} - 4 x + 4 + 16\right)} + \frac{\sqrt{\left(64 - 16 x + {x}^{2}\right) + 25}}{7}$
Differentiating with respect to $x$ and setting it equal to $0$
$\frac{d}{\mathrm{dx}} \left({\left({x}^{2} - 4 x + 20\right)}^{\frac{1}{2}} + \frac{1}{7} {\left({x}^{2} - 16 x + 89\right)}^{\frac{1}{2}}\right) = 0$
$\implies \frac{1}{2} {\left({x}^{2} - 4 x + 20\right)}^{- \frac{1}{2}} \left(2 x - 4\right) +$
$\frac{1}{7} \times \frac{1}{2} {\left({x}^{2} - 16 x + 89\right)}^{- \frac{1}{2}} \left(2 x - 16\right) = 0$
$\implies {\left({x}^{2} - 4 x + 20\right)}^{- \frac{1}{2}} \left(x - 2\right) + \frac{1}{7} {\left({x}^{2} - 16 x + 89\right)}^{- \frac{1}{2}} \left(x - 8\right) = 0$
Plotting this equation with the help of inbuilt graphics utility, gives us
$x = 2.43$, rounded to two decimal places, as there is only one point of inflection.
Hence path with least time is through the point $\left(2.43 , 2\right)$ |
mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community
Community expand_more
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
{{ searchError }}
search
{{ courseTrack.displayTitle }}
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
# Writing and Using Explicit Rules for Arithmetic Sequences
Arithmetic sequences have a common difference, which is the same as them having a constant rate of change. Thus, even though arithmetic sequences have domains that are discrete, they are similar to linear functions. Just as linear function can be described and analyzed using function rules, so can arithmetic sequences.
## Explicit Rule
An explicit rule of a sequence is a function rule where the input is a term's position, and the output is the value of the term. This is opposed to recursive rules, where a term's value is found using previous terms. Explicit rules are often written as where is the function rule. An example would be the sequence which can be described by the explicit rule
where is the position of a term in the sequence. It says that the first term, has the value the second term, has the value and so on.
## Explicit Rule of Arithmetic Sequences
All arithmetic sequences have some common difference, Using this common difference, and the value of the first term, it's possible to find an explicit rule that describes the sequence. By thinking of the terms in a sequence using and a pattern emerges.
Using and
When increases by the coefficient of increases by as well. Due to this, and that the coefficient is when is the coefficient is always less than Expressing this in a general form gives the explicit rule.
Thus, knowing and is enough to write the explicit rule of an arithmetic sequence. Note that an arithmetic sequence is a linear function where the domain is the positive integers. The difference is then the slope, and is a point on the graph. Substituting this into the point-slope form is an alternative way of finding the rule:
This equality can be rearranged into the explicit rule previously stated.
fullscreen
Exercise
The first five terms of an arithmetic sequence are Find an explicit rule describing the arithmetic sequence. Then, use the rule to find the twelfth term of the sequence.
Show Solution
Solution
To find the rule, we first need to find the common difference, of the sequence. We can do this by subtracting one term from the next: We also know that the first term of the sequence is Substituting these pieces of information into the general form of the explicit rule gives the desired rule.
To find the twelfth term of the sequence, we can now substitute into the rule.
The twelfth term is
fullscreen
Exercise
Pelle is an avid collector of pellets. During his spare time, he likes to arrange his pellets in different patterns. Today, he's chosen to place them in the shape of a triangle. The top row consists of one pellet, the second of three pellets, the third of five pellets, and so on. Write a rule where is the amount of pellets in row Then, use the rule to find which row has pellets.
Show Solution
Solution
To begin, we can make sense of the given information. For every row, the amount of pellets increase by Thus, we know that It is also given that the first row consists of pellet, Using this information, we can find the rule.
Thus, the explicit rule is Next, we can find the row that contains pellets. In other words, we are looking for the that gives This is done by substituting into the rule and solving the resulting equation for
Row has pellets.
fullscreen
Exercise
For an arithmetic sequence, Write an explicit rule of the sequence and give its first six terms.
Show Solution
Solution
To begin, we must determine the common difference, Since we do not know the values of two consecutive terms, we cannot directly find However, the terms and are positions apart, so they must differ by
This gives the equation which we can solve for
Now that we know the common difference, we have to find the first term of the sequence, to write the explicit rule. Using and we can find Knowing one term, a subsequent term can be found by adding Similarly, a previous term is found by subtracting Repeating this, we find We now know both and so we can find the explicit rule.
Thus, the explicit rule is We have already found the terms and Finding the last two can be done either by using the explicit rule, or by adding to and then to For simplicity's sake, let's add to to find Then, is found using Thus, the first six terms of the sequence are |
# Math Expressions Grade 3 Student Activity Book Unit 2 Lesson 10 Answer Key Write First Step Questions for Two-Step Problems
This handy Math Expressions Grade 3 Student Activity Book Answer Key Unit 2 Lesson 10 Write First Step Questions for Two-Step Problems provides detailed solutions for the textbook questions.
## Math Expressions Grade 3 Student Activity Book Unit 2 Lesson 10 Write First Step Questions for Two-Step Problems Answer Key
Use Order of Operations
This exercise involves subtraction and multiplication:
10 – 3 × 2
Question 1.
What do you get if you subtract first and then multiply? ______
14
Explanation:
Given, 10 – 3 × 2
If we subtract first, we get 10 – 3 = 7
When we multiply, we get 7 x 2 = 14
Question 2.
What do you get if you multiply first and then subtract? ______
4
Explanation:
Given, 10 – 3 × 2
If we multiply first, we get 3 x 2 = 6
When we subtract, we get 10 – 6 = 4 = 14
To make sure everyone has the same answer to problems like this one, people have decided that multiplication and division will be done before addition and subtraction. The answer you found in question 2 is correct.
If you want to tell people to add or subtract first, you must use parentheses. Parentheses mean “Do this first.” For example, if you want people to subtract first in the exercise above, write it like this:
(10 – 3) × 2
Question 3.
5 + 4 × 2 = ____
13
Explanation:
multiplication and division will be done before addition and subtraction.
5 + 4 x 2
= 5 + 8
= 13
Question 4.
(9 – 3) × 6 = ____
36
Explanation:
The order of operations is a rule that tells the correct sequence of steps for evaluating a math expression.
We can remember the order using PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
(9 – 3) × 6
= 6 x 6
= 36
Question 5.
8 ÷ 2 + 2 = ____
6
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
8 ÷ 2 + 2
= 4 + 2
= 6
Question 6.
6 × (8 – 1) = _____
42
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
6 × (8 – 1)
= 6 x 7
= 42
Rewrite each statement, using symbols and numbers instead of words.
Question 7.
Add 4 and 3, and multiply the total by 8. ________
(4 + 3) x 8
Explanation:
(4 + 3)
and multiply the total by 8.
(4 + 3) x 8
Question 8.
Multiply 3 by 8, and add 4 to the total. _______
(3 x 8) + 4
Explanation:
Multiply 3 by 8,
3 x 8
and add 4 to the total.
(3 x 8) + 4
What’s the Error?
Today I found the answer to 6 + 3 × 2. Here is how I found the answer.
Is my answer correct? If not. please correct my work and tell me what I did wrong.
Puzzled Penguin
Question 9.
Write an answer to the Puzzled Penguin.
No, its wrong
6 + 3 × 2 = 12
Explanation:
3 x 2 = 6
6 + 3 × 2
= 6 + 6
= 12
Question 10.
4 + 3 × 5 = ____
19
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, 4 + 3 × 5
= 4 + 15
= 19
Question 11.
10 ÷ 2 + 3 = _____
8
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, 10 ÷ 2 + 3
= 5 + 3
= 8
Question 12.
12 – 9 ÷ 3 = ____
9
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, 12 – 9 ÷ 3
= 12 – 3
= 9
Question 13.
3 × 5 – 2 = ____
13
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, 3 × 5 – 2
=15 – 2
=13
Question 14.
(4 + 3) × 5 = ____
35
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, (4 + 3) × 5
= 7 × 5
= 35
Question 15.
10 ÷ (2 + 3) = ____
2
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, 10 ÷ (2 + 3)
= 10 ÷ 5
= 2
Question 16.
(12 – 9) ÷ 3 = ____
1
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, (12 – 9) ÷ 3
= 3 ÷ 3
= 1
Question 17.
3 × (5 – 2) = ____
9
Explanation:
By using the order PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
Given, 3 × (5 – 2)
= 3 × 3
= 9
Write First Step Questions
Write the first step question and answer. Then solve the problem.
Question 18.
A roller coaster has 7 cars. Each car has 4 seats. If there were 3 empty seats, how many people were on the roller coaster?
25 people.
Explanation:
A roller coaster has 7 cars.
Each car has 4 seats
7 x 4 = 28 seats
There were 3 empty seats.
Number of people were on the roller coaster are,
28 – 3 = 25
Question 19.
Each week, Marta earns $10 babysitting. She always spends$3 and saves the rest. How much does she save in 8 weeks?
$56 Explanation: Marta earns$10 baby sitting,
she spends $3 and saves the rest. (10 – 3) x 8 Total amount she save in 8 weeks,$7 – 8 = \$56
Question 20.
Abu bought 6 packs of stickers. Each pack had 8 stickers. Then Abu’s friend gave him 10 more stickers. How many stickers does Abu have now?
58 stickers.
Explanation:
Abu bought 6 packs of stickers.
6 x 8
Abu’s friend gave him 10 more stickers.
(6 x 8) + 10
Number of stickers Abu have now,
= 48 + 10
= 58
Question 21.
Zoe made some snacks. She put 4 apple slices and 2 melon slices on each plate. She prepared 5 plates. How many slices of fruit did Zoe use in all?
30 slices.
Explanation:
Zoe put 4 apple slices and 2 melon slices on each plate.
4 + 2
She prepared 5 plates
(4 + 2 ) x 5
= 6 x 5
= 30
Question 22.
Kyle ordered 8 pizzas for his party. Each pizza was cut into 8 slices. 48 of the slices were plain cheese, and the rest had mushrooms. How many slices of pizza had mushrooms?
16 slices of pizza had mushrooms.
Explanation:
Kyle ordered 8 pizzas,
each pizza was cut into 8 slices.
8 x 8 = 64
48 of the slices were plain cheese,
(8 x 8) – 48
Number of mushrooms slice,
= 64 – 48
= 16
Write the first step question and answer. Then solve the problem.
Question 23.
Nadia counted 77 birds on the pond. 53 were ducks, and the rest were geese. Then the geese flew away in 4 equal flocks. How many geese were in each flock?
6 geese in each flock.
Explanation:
Nadia counted 77 birds on the pond.
53 were ducks, and the rest were geese.
Then the geese flew away in 4 equal flocks.
Number of geese in each flock,
(77 – 53) ÷ 4
= 24 ÷ 4
= 6
Question 24.
Kagami baked 86 blueberry muffins. Her sisters ate 5 of them. Kagami divided the remaining muffins equally among 9 plates. How many muffins did she put on each plate?
9 muffins.
Explanation:
Kagami baked 86 blueberry muffins.
Her sisters ate 5 of them.
Kagami divided the remaining muffins equally among 9 plates.
Number of muffins she put on each plate,
(86 – 5) ÷ 9
= 81 ÷ 9
= 9
Question 25.
Lucia had 42 plums. Jorge had 12 more plums than Lucia. Jorge divided his plums equally among 6 people. How many plums did each person get?
9 plums.
Explanation:
Jorge had 12 more plums than Lucia.
Jorge divided his plums equally among 6 people.
Number of plums did each person get,
(42 + 12) ÷ 6
= 54 ÷ 6
= 9
Question 26.
On his way to school, Kevin counted 5 mountain bikes and 3 road bikes. How many wheels were on the bikes altogether?
16 wheels.
Explanation:
Kevin counted 5 mountain bikes and 3 road bikes.
Number of wheels were on the bikes altogether,
(5 + 3) x 2
= 8 x 2
= 16
Question 27.
Juana has 21 shirts. Leslie had 7 less shirts than Juana, but then she bought 4 more. How many shirts
does Leslie have now? |
# Aptitude Problems on Numbers Test Paper 2
6) Two different natural numbers are such that their product is less than their sum. One of the numbers must be
1. 1
2. 2
3. 3
4. None of these.
Explanation:
As per the question:
Let one natural number = x
Since, 1*x < 1 + x, so the other number would be 1.
7) Two-third of one-fifth of one-fourth of a number is 10. What is 30% of that number?
1. 60
2. 100
3. 270
4. 90
Explanation:
x = 10 × 3 × 5 × 2 = 300
Now,
30% of x = × 300=90
8) The sum of two consecutive odd numbers in a set of three consecutive odd numbers is 5 more than the third number. What is the second of these numbers?
1. 5
2. 7
3. 9
4. 11
Explanation:
Let the numbers be x, x+2, x+4,
∴ (x + x+2) - (x+4) = 5
2x + 2 - x - 4 = 5
X - 2 = 5
X = 5+2 = 7
∴ Second number: 7+2= 9
9) The ratio between a two-digit number and the sum of the digits of that number is 4:1. If the digit in the unit place is 3 more than the digit in the ten's place, what is that number?
1. 24
2. 63
3. 36
4. None of these.
Explanation:
Let ten's digit be x. Then, unit's digit = (x+3).
Sum of the digits = x +(x+3) = 2x+3
Number = 10x+(x+3) = 11x+3
Then,
11x + 3 = 8x + 12
3x = 9
X = 3
∴ Number = (11x+3)
= 33 + 3 = 36
10) The difference between a two-digit number and the number obtained by interchanging the digits is 27. What is the difference between the digits of the number?
1. 3
2. 6
3. 9
4. 5
Explanation:
Let the ten's digit be = x
Unit's digit = y
So, the number is = 10x +y
The number obtained by interchanging the digits = 10y + x
According to the question:
(10x + y) - (10y + x) = 27
10x + y - 10y - x = 27
9x - 9y = 27
(x - y) = 3
∴The difference between the digits of the numbers, i.e., (x - y) = 3.
Aptitude Problems on Numbers Test Paper 1
Aptitude Problems on Numbers Test Paper 3
Aptitude Problems on Numbers Test Paper 4
Problems on Numbers Concepts |
## How Much are 6 Times 12?
Do you want to know how to multiply 6×12? Right here, you can learn how much six times 12 is. For this, you must first understand the terms of six × twelve.
The parts of this multiplication are the first factor 6 and the second factor twelve. The first factor, 6, is known as the multiplicand term, and the second factor, twelve, is called the multiplier. The multiplication sign is x and is read as “by.”
The result of that mathematical operation, six x twelve, is called the product of six and twelve. Therefore, six × twelve is pronounced as six multiplied by twelve or six times twelve.
## How Much are 6 Times 12?
Now you know how to multiply 6×12 and how much is six times twelve: Construct the sum of six (adding six-twelve times) or the sum of twelve (adding twelvesix times).
If you have been checking what six times twelve or six x twelve is, you have also found the page. Also, if you have entered 6×12 or 6×12 in your browser preferences, this is the right page to answer your question.
Here we mainly unit the letter x (x) to indicate the multiplication of two by three (6×12). But the proliferation of six times twelve is often characterized as a cross (×): six × twelve.
When those characters are absent, you can use the semi-colon (·): 6·12—the asterisk (*) could also be used: six*twelve.
This brings us to the end of our article on multiplying six by 12, which we can summarize with the following lines:
The multiplication of six×twelve is an abbreviated addition. The six and twelve are called the multiplicand and the multiplier, respectively.
The product of six x twelve is equal to 72. Likewise 6×12 = 12×6.
## What is Twelve Percent of Six?
Using this percentage calculator, it is possible to make a percentage count in three ways. So, we think you found us looking for answers to questions like this: 1) What is 12% of 6? 2) 0.72 is how many per cent of 6? Or maybe: What is twelve per cent of six?
## How to Calculate Percent – Step by Step
Set the answers to the questions posed above:
### 1) What is 12% of 6?
To calculate this percentage, we suggest using this formula:
% / 100 = Part / Total
Substituting the supplied values:
twelve / 100 = Part / six.
Performing Cross Multiplication:
twelve x six = 100 x Part, or
72 = 100 x Part
Now it’s just divide by 100 and get the answer:
Part = 72 / 100 = 0.72
### 2) Twelve is How many Percent of Six?
• Reuse the same formula:
• % / 100 = Part / Total
• Substituting the values:
• % / 100 = twelve / six
• Performing Cross Multiplication:
• % x six = 12 x 100
• Divide by 6 and here is the answer:
• % = twelve x 100 / six = 200%
See more examples of percentage calculations below.
### Per cent Examples
• 20 per cent of 14.99
• 60 per cent of 520,000
• 30 per cent of 220,000
• 4 per cent of 7,100
• 15 per cent of 402.9
• 20 per cent of 26.7
• 5 per cent of 33
• 25 per cent of 985
• 35 per cent of 24
• 45 per cent of 18.97
• 1 per cent of 17.98
• 70 per cent of 3.79
• Thirty-five per cent of 124.9.
To solve the problem, we start by labelling 12/4 of 6/8 like this:
N1/D1 of N2/D2
Now we use this formula fraction of a fraction to solve the problem:
(N1 x N2) / (D1 x D2)
When we enter 12/4 of 6/8 in the formula, we get:
(twelve x 6) / (4 x 8) = 72/32
Therefore, the answer to 12/4 of 6/8 in its simplest form is:
12/4 of 6/8 = 2 1/4
The answer above is 12/4 of 6/8 in its fractional form. We have converted the answer into the decimal form below:
12/4 of 6/8 = 2.25
If you need to solve a fraction of a fraction problem like 12/4 of 6/8 in the future, remember that we get the answer by multiplying the numerators together and the denominators together.
## Fraction of a Fraction
Please enter another fraction of a fraction new problem so we can solve it.
### How much is it
How much is 12/4 of 6/9?
Here is the following fraction of a fraction problem on our list that we have solved.
## Conclusion
How much is 12/4 of 6/8? What is the fractional answer to 12/4 of 6/8? What is the decimal answer to 12/4 of 6/8? Here we explain what 12/4 of 6/8 means and how to calculate the answer.
12/4 of 6/8 is the same as asking: How much do we get if we take 12/4 of 6/8? Also, 12/4 is 300 per cent. Therefore, we solve for 300 per cent of 6/8 when we solve for 12/4 of 6/8. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Infinite and Non-Existent Limits
## Functions where output values continue to get larger or smaller without limit or where limits do not exist.
0%
Progress
Practice Infinite and Non-Existent Limits
Progress
0%
Infinite and Non-Existent Limits
You have heard time and again that it is "against the rules" to divide by the number 0. Even the most basic calculator will return some form of "ERROR" if you try to divide even the smallest of numbers by 0.
Do you really understand why it is "against the rules"? What is really wrong with dividing by nothing?
Embedded Video:
### Guidance
Infinite limits
Functions can exhibit a number of different behaviors as the input value gets very large or very small.
As x\begin{align*}x\end{align*} approaches \begin{align*} \infty\end{align*}, some functions output values closer and closer to a single number, some approach zero, and some continue to get larger and larger or smaller and smaller without limit.
In this lesson, we will explore functions of the last type, functions with infinite limits, and the different types of asymptotes they may have.
#### Example A
Evaluate the function h(x)=x22x1\begin{align*}h(x)=\frac{x^2}{2x-1}\end{align*}.
Solution
To evaluate this function, consider the behavior of the function as larger and larger values are Inserted for x. As x approaches \begin{align*} \infty\end{align*}, the function values also approach \begin{align*}\infty\end{align*}. Therefore the limit of the function as x approaches \begin{align*} \infty\end{align*} is: limxx22x1=\begin{align*}\lim_{x \to \infty} \frac{x^2} {2x - 1} = \infty\end{align*}. Similarly, as x approaches \begin{align*}-\infty\end{align*}, f(x) approaches \begin{align*}-\infty\end{align*}. Therefore we have limxx22x1=.\begin{align*}\lim_{x \to -\infty} \frac{x^2} {2x - 1} = -\infty.\end{align*}
We can also understand this limit if we analyze the equation for h(x). As x gets larger and larger, the value of the expression 2x - 1 gets closer and closer to the value of the expression 2x. That is, for sufficiently large values of x, 2x - 1 \begin{align*}\approx\end{align*} 2x. Therefore the values of h(x) approach x22x=x2\begin{align*}\frac{x^2}{2x}=\frac{x}{2}\end{align*}. As x gets larger and larger, so does x2\begin{align*}\frac{x}{2}\end{align*}. For large values of x, the function h(x) gets closer and closer to x2\begin{align*}\frac{x}{2}\end{align*}. Therefore the limit is infinity.
#### Example B
Approximate the function f(x) = x2 + 2x - 3.
Solution
This function has an infinite limit as x approaches infinity. However, as x gets larger and larger, f(x) \begin{align*}\approx\end{align*} x2, since the x2 value grows much more quickly than the 2x value, particularly apparent at very large +/- values of x. If this is not immediately apparent, evaluate the function for x = 1,000,000, and you will quickly get the idea!
Therefore we can use the function y = x2 to describe the end behavior of f(x).
#### Example C
Describe the end behavior of each graph. That is, determine if the function has a limit L, if the limit is infinite, or if the limit does not exist.
a.) y=x2\begin{align*}\,\! y=x^2 \end{align*}
b.) y=2(1)x\begin{align*}\,\! y=2(-1)^x\end{align*}
c.) y=11|x|\begin{align*}y=1 - \frac{1} {|x|}\end{align*}
Solution:
a.) y=x2\begin{align*}\,\! y=x^2 \end{align*}
As x approaches +\begin{align*}+\infty\end{align*}, x2 also approaches +\begin{align*}+\infty\end{align*}. As x approaches \begin{align*}-\infty\end{align*}, x2 approaches +\begin{align*}+\infty\end{align*}. Therefore limxx2=limxx2=\begin{align*}\lim_{x \to \infty} x^2=\lim_{x \to -\infty} x^2=\infty\end{align*}.
b.) y=2(1)x\begin{align*}\,\! y=2(-1)^x\end{align*}
This function is difficult to understand without producing a graph. The table shows that the function only takes on two values: 2, and -2, and is undefined at non-integer values of x. As x approaches +\begin{align*}+\infty\end{align*} or \begin{align*}-\infty\end{align*}, the function values alternate between 2 and -2. Therefore the limit does not exist.
c.) y=11|x|\begin{align*}y=1 - \frac{1} {|x|}\end{align*}
If you look at the table of this function, which has negative and positive values of x, you can see that as x approaches +\begin{align*}+\infty\end{align*} or \begin{align*}-\infty\end{align*}, the function values approach 1.
Therefore limx(11|x|)=limx(11|x|)=1\begin{align*}\lim_{x \to \infty} \left (1 - \frac{1} {|x|}\right)=\lim_{x \to -\infty} \left (1 - \frac{1} {|x|}\right)=1\end{align*}.
We can also determine this limit analytically. For large values of x, |x| is also large, and so 1|x|\begin{align*}\frac{1}{|x|}\end{align*} is small (since dividing 1 by a large number results in a very small number). Therefore, for large values of x, 11|x|10=1\begin{align*}1 - \frac{1} {|x|} \approx 1 - 0 = 1 \end{align*}. We can make the same argument for x approaching \begin{align*}-\infty\end{align*}.
Concept question wrap-up
Dividing by zero is "against the rules" because there is no definition for the answer you would get.
Consider what happens as you take a given value and divide it by smaller and smaller numbers:
2 / 10 = 1/5
2 / 1 = 2
2 / .1 = 20
2 / .001 = 2,000
2 / .000000001 = 2,000,000,000
As we divide by progressively smaller numbers, the quotient gets larger and larger. Also, we can see that in each case, the problem could be reversed by multiplying the product by the dividend to get the divisor, for instance: 2 / .1 = 20 ==> 20 * .1 = 2.
Unfortunately, this doesn't work if you actually divide by 0, even if you assume that dividing by zero resulted in infinity! No matter how big the number you multiply by zero, even infinity, you will never be able to get back to 2.
x/0=undefined\begin{align*}\therefore x/0 = undefined\end{align*}
### Vocabulary
An infinite function is one whose output approaches infinity or negative infinity as very large or very small values are calculated for the input variable (usually "\begin{align*}x\end{align*}").
An asymptote is a line representing a value toward which the value of a function may approach, but never actually reach.
### Guided Practice
Questions
1) Evaluate \begin{align*}\lim_{x \to \infty} \frac{x^3}{x-7}\end{align*}
2) Evaluate \begin{align*}\lim_{x \to \infty} \frac{x^2 +2x}{x - 3}\end{align*}
3) Evaluate \begin{align*}\lim_{x \to \infty} x^3\end{align*}
4) Describe the end behavior of \begin{align*}y = 2x^2 + 3x - 7\end{align*}
5) Evaluate \begin{align*}-2x^3 - 5x^2 + 8x\end{align*}
Solutions
1) The degree of the numerator is greater than the degree of the denominator, so the function will grow without bound. Since the denominator is x - 7, the function cannot include x = 7, because the function cannot be defined where the denominator = 0. Logically, as x gets huge, the -7 matters less and less, and we end up with just x2.
2) Similar to the last problem, the numerator is of greater degree than the denominator, so the function does not approach a limit. The denominator is x - 3, so the graph cannot include 3.
3) As \begin{align*}x \to \infty\end{align*}, or in other words, "as x gets huge", the value of x3 grows even faster, either positive or negative, so there is no limit.
4) As x grows huge, x2 grows much faster than the rest of the expression, therefore, we can approximate the end behavior of \begin{align*}2x^2 + 3x - 7\end{align*} with \begin{align*}y = x^2\end{align*}
5) \begin{align*}-2x^3 - 5x^2 + 8x\end{align*} is a 3rd degree equation, so it will turn twice, since it is not a rational function, there are no concerns about numerator or denominator. The function will have no limits, and will grow without bound in both the positive and negative directions. If you use a graphing calculator to graph the function, you will see that \begin{align*}y = x^3\end{align*} can be used to approximate it.
### Practice
Problems 1-3: The limit of \begin{align*} f(x)\end{align*} as \begin{align*}x \to t\end{align*} cannot exist if which conditions are true? List three conditions.
1. Give an example of a limit that does not exist
Problems 5-7: Assuming that \begin{align*}f(x)\end{align*} is a rational function:
1. What is \begin{align*}\lim_{x \to \infty} f(x)\end{align*} when the degree of the numerator is less than the degree of the denominator?
2. What is \begin{align*}\lim_{x \to \infty} f(x)\end{align*} when the degrees of the numerator and the denominator are equal?
3. What is \begin{align*}\lim_{x \to \infty} f(x)\end{align*} when the degree of the numerator is greater than the degree of the denominator?
1. In general, if r is a positive real number, what is \begin{align*}\lim_{x \to \infty}\frac{1}{x^r}\end{align*}?
2. In general, if r is a positive real number, what is \begin{align*}\lim_{x \to \infty} x^r\end{align*}?
Problems 10-13: Let a and b be real numbers and let t be a positive integer. Complete each of the following properties of limits.
1. \begin{align*}\lim_{x \to a} x^t =\end{align*}
2. If f is a polynomial, \begin{align*}\lim_{x \to a} f(x) =\end{align*}
3. \begin{align*}\lim_{x \to a} k \cdot f(x) =\end{align*}
4. \begin{align*}\lim_{x \to \infty} t^x =\end{align*}
Problems 14-16: Evaluate.
1. \begin{align*}\lim_{x \to -5}\frac{(5 + x)^2 - 25}{x}\end{align*}
2. \begin{align*}\lim_{x \to 3}\frac{x^3 - 6x + 2}{x^2 + 2x - 3}\end{align*}
3. \begin{align*}\lim_{x \to 0}\frac{(3 + 3y)^{-1} - 3y^{-1}}{x}\end{align*}
### Vocabulary Language: English
$\infty$
$\infty$
The symbol "$\infty$" means "infinity", and is an abstract concept describing a value greater than any countable number.
Asymptotes
Asymptotes
An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).
infinite limit
infinite limit
A function has an infinite limit if it's output approaches infinity or negative infinity as very large or very small values are calculated for the input variable (usually "$x$").
infinity
infinity
Infinity is an unbounded quantity that is greater than any countable number. The symbol for infinity is $\infty$.
limit
limit
A limit is the value that the output of a function approaches as the input of the function approaches a given value. |
# NCERT Solutions for Class 10 Math Chapter 12 - Areas Related to Circles
Aasoka is an online learning platform that provides top-quality educational resources including NCERT Solutions for Class 10th. The study material is helpful in getting a better grasp of the concepts included in the chapter. The entire chapter is covered accurately and students can easily rely on the NCERT Solutions to clear their exams successfully. These solutions will clear all doubts of students of Class 10th related to any topic of the chapter “Areas Related to Circles”. Grab the latest solutions now and kick start your learning journey.
Students will learn about the area of a circle and its perimeter from the chapter “Areas Related to Circles”. The chapter helps in understanding the segment of a circle, the area of a sector, and the areas of the figures that contain a part of a circle or a circle.
##### Question 1:
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Radius of first circle (r1) = 19 cm
Radius of second circle (r2) = 9 cm
Let radius of third circle be R cm.
According to question,
Circumference of first circle
+ Circumference of second circle
= Circumference of third circle
2$\mathrm{\pi }$r1 + 2$\mathrm{\pi }$r2 = 2$\mathrm{\pi }$R
2$\mathrm{\pi }$ [r1 + r2] = 2$\mathrm{\pi }$R
19 + 9 = R
∴ R = 28
∴ Radius of third circle (R) = 28 cm
##### Question 2:
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of two circles.
Radius of first circle (r1) = 8 cm
Radius of second circle (r2) = 6 cm
Let radius of third circle be R cm.
According to question,
Area of third circle = Area of first circle
+ Area of second circle
##### Question 3:
Fig., depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
∴ Area of Red region = 1039.5 cm2
[Area of Blue scoring region] = [Combined area of Red, Blue and Gold region] − [Combined area of Gold and Red region]
##### Question 4:
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
##### Question 5:
Tick the correct answer in the following and justify your choice: If the perimeter and area of a circle are numerically equal, then the radius of the circle is
1. 2 units
2. $\mathrm{\pi }$ units
3. 4 units
4. 7 units
(i) 2 units
##### Question 6:
Find the area of a sector of a circle with radius 6 cm, if angle of the sector is 60$°$.
##### Question 7:
Find the area of a quadrant of a circle whose circumference is 22 cm.
##### Question 8:
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
##### Question 9:
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use $\mathrm{\pi }$ = 3.14)
##### Question 10:
In a circle of radius 21 cm, an arc subtends an angle of 60$°$ at the centre. Find : (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.
Radius of circle (R) = 21 cm
##### Question 11:
Radius of circle = (R) = 15 cm
Central angle ($\mathrm{\theta }$) = 60$°$
In ΔOAB, central angle $\mathrm{\theta }$ = 60$°$
OA = OB = 15 cm
$\angle$A = $\angle$B
Now, $\angle$A + $\angle$B + $\angle$O = 180$°$
2$\angle$A + 60$°$ = 180$°$
$\angle$A = 60$°$
$\angle$A = $\angle$B = 60$°$
∴ ΔOAB is equilateral triangle.
##### Question 12:
Radius of circle (R) = 12 cm
Central angle ($\mathrm{\theta }$) = 120$°$
= 150.72 – 36 × 1.73
= (150.72 – 62.28) cm2 = 88.44 cm2
∴ Area of the segment = 88.44 cm2.
##### Question 13:
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see fig.). Find : the area of that part of the field in which the horse can graze.
Side of square = 15 m
Length of Peg = Radius of rope (R) = 5 m
Central angle ($\mathrm{\theta }$) = 90$°$
[Each angle of square]
##### Question 14:
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see fig.). Find :
the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\mathrm{\pi }$ = 3.14)
Side of square = 15 m
Length of Peg = Radius of rope (R) = 5 m
Central angle ($\mathrm{\theta }$) = 90$°$
[Each angle of square]
##### Question 15:
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see fig.). Find :
the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\mathrm{\pi }$ = 3.14)
##### Question 16:
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in fig. Find : the total length of the silver wire required.
##### Question 17:
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in fig. Find :
the area of each sector of the brooch.
##### Question 18:
An umbrella has 8 ribs which are equally spaced (see fig.). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
##### Question 19:
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115$°$. Find the total area cleaned at each sweep of the blades.
Length of blade (R) = 25 cm
Sector angle ($\mathrm{\theta }$) = 115$°$
Wiper moves in form of sector.
##### Question 20:
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80$°$ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $\mathrm{\pi }$ = 3.14)
##### Question 22:
Area of a sector of angle p (in degrees) of a circle with radius R is
(iv)
##### Question 23:
Find the area of the shaded region in fig., if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
##### Question 24:
Find the area of the shaded region in fig., if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and $\angle$AOC = 40$°$.
Radius of smaller circle (r) = 7 cm Radius of bigger circle (R) = 14 cm Central angle $\angle$AOC ($\mathrm{\theta }$) = 40$°$
##### Question 25:
Find the area of the shaded region in fig., if ABCD is a square of side 14 cm and APD and BPC are semicircles.
##### Question 26:
Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
##### Question 27:
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in fig. Find the area of the remaining portion of the square.
##### Question 28:
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in fig. Find the area of the design.
Radius of table cover (R) = 32 cm
OA = OB = OC = 32 cm
##### Question 29:
In fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
##### Question 30:
Fig. depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find: the distance around the track along its inner edge.
##### Question 31:
Fig. depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
Area of track
= Area of rectangle ABEF
+ Area of region BEMGCRB
+ Area of rectangle CGHD
+ area of region (II).
##### Question 32:
In fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
##### Question 33:
The area of an equilateral triangle ABC is
17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half of the length of the side of the triangle (see fig.). Find the area of the shaded region.
##### Question 34:
On a square handkerchief, nine circular designs each of radius 7 cm are made (see fig.). Find the area of the remaining portion of the handkerchief.
##### Question 35:
In fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the quadrant OACB,
Sector angle ($\mathrm{\theta }$) = 90º
OD = 2 cm.
##### Question 36:
In fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.
Sector angle ($\mathrm{\theta }$) = 90º
OD = 2 cm.
##### Question 37:
In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use $\mathrm{\pi }$ = 3.14)
##### Question 38:
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see fig.) If ∠AOB = 30º, find the area of the shaded region.
##### Question 39:
In fig., ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
##### Question 40:
Calculate the area of the designed region in fig., common between the two quadrants of circles of radius 8 cm each.
= Area of sector ABPD – Area of $∆$ABD |
Review question
# Can we sketch the curve $y=(x-a)e^x/(x-b)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R9898
## Solution
The curve $y=\left(\dfrac{x-a}{x-b}\right)e^x,$ where $a$ and $b$ are constants, has two stationary points. Show that $a-b<0 \quad \text{ or } \quad a-b>4.$
Let’s find the stationary points of $y=\left(\frac{x-a}{x-b}\right)e^x.$
We plan to differentiate using the product rule here, so it helps us to find the derivative of $\dfrac{x-a}{x-b}$ first.
If $z = \dfrac{x-a}{x-b}$, then by the quotient rule, $z' = \dfrac{a-b}{(x-b)^2}$. Returning to our original problem, we have $y'=0$ when $\frac{dy}{dx}=\left(\frac{x-a}{x-b}\right)e^x+\left[\dfrac{a-b}{(x-b)^2}\right]e^x=0$ $\implies \frac{e^x(x^2-(a+b)x+(a+ab-b))}{(x-b)^2}=0.$
We know that $e^x > 0$ for any $x$, so we need $x^2-(a+b)x+(a+ab-b)=0.$
This has two solutions as long as the discriminant is positive, which happens if and only if $(a+b)^2 -4(a+ab-b) > 0.$
Multiplying out and rearranging here gives $(a-b)(a-b-4)>0,$ which holds when $a-b<0 \quad \text{ or } \quad a-b>4.$
1. Show that, in the case $a=0$ and $b=\dfrac{1}{2}$, there is one stationary point on either side of the curve’s vertical asymptote, and sketch the curve.
The vertical asymptote to the curve is at $x=\dfrac{1}{2}$. Now let’s substitute in the values of $a$ and $b$ to find the stationary points: $x^2-(a+b)x+ab-b+a=x^2-\frac{1}{2}x-\frac{1}{2}=0.$ Solving this, we find that \begin{align*} &x^2-\frac{1}{2}x-\frac{1}{2}=0 \\ &\implies 2x^2-x-1=0\\ &\implies (2x+1)(x-1)=0\\ &\implies x = 1, -\dfrac{1}{2}, &\end{align*}
which lie on either side of the asymptote $x=\dfrac{1}{2}$.
We can now sketch the graph. We know two stationary points, $\left(-\dfrac{1}{2},\dfrac{e^{-1/2}}{2}\right)$ and $(1,2e)$, and by inspection we can see that it also passes through $(0,0)$.
We also have that $y\to\infty$ as $x\to+\infty$, and $y\to 0$ as $x\to -\infty$.
Also, as $x\to \dfrac{1}{2}$ from the right, $y\to+\infty$ and as $x \to \dfrac{1}{2}$ from the left $y \to -\infty$.
We may now give an approximate sketch of the graph.
Note that the curve is close to $y = e^x$ when $\vert x \vert$ is large.
1. Sketch the curve in the case $a=\dfrac{9}{2}$ and $b=0$.
We again find the stationary points of the curve, this time solving \begin{align*} &x^2-\frac{9}{2}x+\frac{9}{2}=0 \\ &\implies 2x^2-9x+9=0\\ &\implies (2x-3)(x-3)=0\\ &\implies x = 3, \dfrac{3}{2}. \end{align*}
So our two stationary points are $\left(\frac{3}{2},-2e^{3/2}\right)$ and $\left(3,-2e^3\right)$.
The vertical asymptote is at $x=0$. As $x\to 0$ from the right, we have that $y\to-\infty$, since in this case $x-\tfrac{9}{2} < 0$ and the denominator $x$ is positive, so $y$ is negative. As $x\to 0$ from the left, we have $y\to +\infty$ as $x-\tfrac{9}{2} < 0$ again, but this time the denominator $x$ is negative.
The graph crosses the $x$-axis once, at $x=\dfrac{9}{2}$, and as $x \to +\infty$, $y \to +\infty$ and as $x \to -\infty$, $y\to 0$. |
# What is the integral of (x^2 +2x-1)/(x^2+9)?
Mar 1, 2016
$= x + \ln \left({x}^{2} + 9\right) - \frac{10}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) + C$
#### Explanation:
First of all split the fraction up like so:
$\frac{{x}^{2} + 2 x - 1}{{x}^{2} + 9} = {x}^{2} / \left({x}^{2} + 9\right) + \frac{2 x}{{x}^{2} + 9} - \frac{1}{{x}^{2} + 9}$
We can now integrate each fraction one by one, i.e:
$\int \frac{{x}^{2} + 2 x - 1}{{x}^{2} + 9} \mathrm{dx} = \int {x}^{2} / \left({x}^{2} + 9\right) + \frac{2 x}{{x}^{2} + 9} - \frac{1}{{x}^{2} + 9} \mathrm{dx}$
We will have to do a bit of rearranging of the first fraction (add and subtract 9):
$\int {x}^{2} / \left({x}^{2} + 9\right) + \frac{2 x}{{x}^{2} + 9} - \frac{1}{{x}^{2} + 9} \mathrm{dx} =$
$= \int \frac{{x}^{2} + 9 - 9}{{x}^{2} + 9} + \frac{2 x}{{x}^{2} + 9} - \frac{1}{{x}^{2} + 9} \mathrm{dx}$
This can now be rearranged like so:
$= \int \frac{{x}^{2} + 9}{{x}^{2} + 9} + \frac{2 x}{{x}^{2} + 9} - \frac{1}{{x}^{2} + 9} - \frac{9}{{x}^{2} + 9} \mathrm{dx}$
$= \int 1 + \frac{2 x}{{x}^{2} + 9} - \frac{10}{{x}^{2} + 9} \mathrm{dx}$
-The first term obviously just integrates to $x$.
-For the second term we should apply: $\int \frac{f ' \left(x\right)}{f} \left(x\right) = \ln \left(f \left(x\right)\right) + C$
$\int \frac{2 x}{{x}^{2} + 9} \mathrm{dx} = \ln \left({x}^{2} + 9\right) + C$
-And for the third term:
$\int \frac{10}{{x}^{2} + 9} \mathrm{dx}$
Use the substitution $x = 3 \tan \left(u\right)$
$\to \mathrm{dx} = 3 {\sec}^{2} \left(u\right) \mathrm{du}$
We also need the trig identity: ${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$ Putting the substitution in:
$10 \int \frac{3 {\sec}^{2} \left(u\right)}{9 {\tan}^{2} \left(u\right) + 9} \mathrm{du} = \frac{10}{3} \int {\sec}^{2} \frac{u}{\sec} ^ 2 \left(u\right) \mathrm{du}$
$= \frac{10}{3} \int \mathrm{du} = \frac{10}{3} u + C$
Reverse the substitution and we get:
$\frac{10}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) + C$
So returning to our original integral if apply what have found we get that:
$\int \frac{{x}^{2} + 2 x - 1}{{x}^{2} + 9} \mathrm{dx} = \int 1 + \frac{2 x}{{x}^{2} + 9} - \frac{10}{{x}^{2} + 9} \mathrm{dx}$
$= x + \ln \left({x}^{2} + 9\right) - \frac{10}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) + C$ |
## Envision Math 4th Grade Textbook Answer Key Topic 2.2 Rounding Whole Numbers
Rounding Whole Numbers
How can you round numbers?
Round 293,655,404 to the nearest thousand and to the nearest hundred thousand. You can use place value to round numbers.
Guided Practice*
Do you know HOW?
In 1 through 6, round each number to the place of the underlined digit.
Question 1.
128,955
Answer:
Question 2.
85,639
Answer:
Question 3.
9,924
Answer:
Question 4.
1,194,542
Answer:
Question 5.
160,656
Answer:
Question 6.
149,590
Answer:
Do You Understand ?
Question 7.
Writing to Explain Explain how to round a number when 7 is the digit to the right of the rounding place.
Answer:
Question 8.
In 2000 the population of the United States was 281,421,906. Round 281,421,906 to the nearest hundred thousand.
Answer:
independent Practice
Leveled Practice In 9 through 28, round each number to the place of the underlined digit. You may use a number line to help you.
Question 9.
493,295
Answer:
Question 10.
39,230
Answer:
Question 11.
77,292
Answer:
Question 12.
54,846
Answer:
Question 13.
4,028
Answer:
Question 14.
6,668,365
Answer:
Question 15.
453,280
Answer:
Question 16.
17,909
Answer:
Question 17.
1,406
Answer:
Question 18.
55,560
Answer;
Question 19.
21,679
Answer;
Question 20.
3,417,547
Answer:
Question 21.
117,821
Answer:
Question 22.
75,254
Answer:
Question 23.
9,049
Answer:
Question 24.
1,666,821
Answer:
Question 25.
2,420
Answer:
Question 26.
9,000,985
Answer:
Question 27.
9,511
Answer:
Question 28.
73,065
Answer:
Round 293,655,404 to the nearest thousand.
If the digit to the right of the rounding place is 5 or more, add 1 to the rounding digit. If it is less than 5, leave the rounding digit alone.
293,655,000
Since 4 < 5, leave the rounding digit as is. Change the digits to the right of the rounding place to zeros.
So, 293,655,404 rounds to 293,655,000.
Round 293,655,404 to the nearest hundred thousand.
The digit to the right of the rounding place is 5.
Since the digit is 5, round by adding 1 to the digit in the hundred thousands place.
So, 293,655,404 rounds to 293,700,000.
Problem Solving
Question 29.
For each zoo in the chart, round the attendance to the nearest hundred thousand.
Answer:
Question 30.
Reasoning Which zoo had the greatest number of visitors?
Answer:
Question 31.
Number Sense Write four numbers that round to 700 when rounded to the nearest hundred.
Answer:
Question 32.
Reasoning Write a number that when rounded to the nearest thousand and hundred will have a result that is the same.
Answer:
Question 33.
Jonas read that about 1,760,000 people will graduate from high school in California in the next four years. Jonas thinks this number
is rounded to the nearest ten thousand. What would the number be if it was rounded to the nearest hundred thousand?
Answer:
Question 34.
Liz had attended class every day since she started school as a kindergartner. She said she had been in school for about 1,000 days. What could the actual number of school days be if she rounded to the nearest ten?
Answer:
Question 35.
When rounded to the nearest ten thousand, which number would be rounded to 120,000?
A. 123,900
B. 126,480
C. 128,770
D. 130,000
Answer:
Question 36.
A fruit market sold 3,849 apples, 3,498 oranges, and 3,894 pears in one day. Write these numbers in order from greatest to least.
Answer: |
In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!
# Ex.14.4 Q3 Statistics Solution - NCERT Maths Class 9
Go back to 'Ex.14.4'
## Question
The following observations have been arranged in ascending order. If the median of the data is $$63,$$ find the value of $$x.$$
$$29, 32, 48, 50, x, x + 2, 72, 78, 84, 95$$
Video Solution
Statistics
Ex exercise-14-4 | Question 3
## Text Solution
What is known?
Ungrouped data and median of data is $$63.$$
What is unknown?
Value of $$x.$$
Reasoning:
The median is that value of the given number of observations, which divides it into exactly two parts. So, when the data is arranged in ascending (or descending) order the median of ungrouped data can be calculated based on no. of observation are even or odd.
Steps:
It can be observed that the total number of observations in the given data is $$10$$ (even number).
Therefore, the median of this data will be the mean of \begin{align} \frac{{10}}{2}\end{align} i.e., $$5$$th and \begin{align} \frac{{10}}{2}\end{align} $$+ 1$$ i.e., $$6$$th observation.
Therefore, median of the data
\begin{align} &= \frac{ \begin{Bmatrix} 5^{\rm th} \text{observation } + \\ 6^{\rm th} \text{ observation} \end{Bmatrix} }{2}\\& \Rightarrow \,\,63 = \,\frac{{x + x + 2}}{2}\\&\Rightarrow \,\,63 = \,\frac{{2x + 2}}{2}\,\\&\Rightarrow \,\,63\, = \,x + 1\\& \Rightarrow \,\,x\, = \,62\end{align}
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school |
Upcoming SlideShare
×
# Notes 3-5
892 views
Published on
The Graph Scale-Change Theorem
Published in: Technology
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total views
892
On SlideShare
0
From Embeds
0
Number of Embeds
257
Actions
Shares
0
7
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Notes 3-5
1. 1. Section 3-5 The Graph Scale-Change Theorem
2. 2. Warm-up 1 1. Multiplying by three gives the same answer as dividing by 3 4 2. Multiplying by .75 gives the same answer as dividing by 3 3. Make a general statement that summarizes numbers 1 and 2 above. 1 Multiplying by n (n ≠ 0) gives the same answer as dividing by n
3. 3. Scale Change: A transformation that stretches or shrinks a graph both vertically and horizontally Horizontal Scale Factor: The value that changes the horizontal values of a graph Vertical Scale Factor: The value that changes the vertical values of a graph Size Change: When the horizontal and vertical scale factors are the same
4. 4. Graph Scale-Change Theorem When dealing with a graph, the following transformations will give the same graphs: x y 1. Replacing x with and y with in the equation a b 2. Applying the scale change (x, y) → (ax, by) to the graph, where a is the horizontal scale factor and b is the vertical scale factor
5. 5. Example 1 Compare the graphs of y = |x| and y = |6x|
6. 6. So, what can we say? The graph of the second is the image of the graph of the first under a horizontal scale change of magnitude 1 . 6 This will shrink the graph horizontally. The angle at the vertex becomes acute, but the vertex does not change. The y-coordinates will now be 6 times as large for the corresponding x-coordinates.
7. 7. Example 2 y = |6x| Sketch the graph of 4
8. 8. Negative Scale Factor: When using a negative scale factor, the values will be reflected over an axis
9. 9. Example 3 a. Sketch the image of y = x3 under S(x, y) = (-2x, y)
10. 10. Example 3 b. Give an equation for the image Since the x-coordinate is being multiplied by -2 in the scale change, we take the opposite for the equation. Thus, we need to divide x by -2. x y = ( -2 )3 1 y = - x3 8
11. 11. Example 4 The line 41x - 29y = 700 contains the points (39, 31) and (10, -10). Use this information to obtain two points on the line with equation 20.5x - 87y = 700. x has changed by: 1/2 y has changed by: 3 S(x, y) = (2x, y/3) (2×39, 31÷3) = (78, 31/3) (2×10, -10÷3) = (20, -10/3)
12. 12. Homework p. 191 # 1 - 16, skip 13 |
Polynomial Factoring with Synthetic Divisionand Solving
A common and effective approach to solving polynomials that we have focused on a lot thus far is using synthetic division.
This page will look at further examples of factoring with synthetic division and solving with polynomials of degree 3 and higher.
Factoring with Synthetic Division and Solving,Examples
(1.1)
Show that (x + 1) is a factor of f(x) = 3x^3 \space {\text{–}} \space 7x^2 \space {\text{–}} \space 18x \space {\text{–}} \space 8,
then factorise fully and solve for f(x) = 0.
Solution
x + 1 \space = \space 0 \space => \space x = {\text{-}}1
A remainder of 0 shows that (x + 1) is a factor,
and we can now factorise fully and solve.
(3x^2 \space {\text{–}} \space 10x \space {\text{–}} \space 8)(x + 1)
The larger quadratic turns out to factorise nicely into two neat binomials.
(3x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space \space \space \space )(x + 1) => (3x + 2)(x \space {\text{–}} \space 4)(x + 1)
Setting the factors equal to 0 and solving will give us the solutions.
3x + 2 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}\frac{2}{3} \space \space \space , \space \space \space x \space {\text{–}} \space 4 = 0 \space \space {\scriptsize{=>}} \space \space x = 4
x + 1 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}1
The solutions of 3x^3 \space {\text{–}} \space 7x^2 \space {\text{–}} \space 18x \space {\text{–}} \space 8 are x = {\text{-}}1 \space , \space {\text{-}}\frac{2}{3} \space , \space 4.
(1.2)
If (x \space {\text{–}} \space 2) is a factor of f(x) = x^3 + 3x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 12.
Use synthetic division to find the other two factors, and subsequently all roots.
Solution
We’re given the information that (x {\text{–}} \space 2) is a factor, so can procced straight away with synthetic division without having to make any guesses at values to try.
x \space {\text{–}} \space 2 \space = \space 0 \space => \space x = 2
A remainder of 0 is obtained as expected,
and we can now factorise fully and find the roots/solutions.
(x^2 + 5x + 6)(x \space {\text{–}} \space 2)
The larger quadratic turns out to factorise nicely into two neat binomials.
(x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space 2) => (x + 3)(x + 2)(x \space {\text{–}} \space 2)
Setting the factors equal to 0 and solving will give us the solutions.
x + 3 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}3 \space \space \space , \space \space \space x \space {\text{–}} \space 2 = 0 \space \space {\scriptsize{=>}} \space \space x = 2
x + 2 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}2
The solutions of f(x) = x^3 + 3x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 12 are x = {\text{-}}3 \space , \space {\text{-}}2 \space , \space 2.
(1.3)
Find the roots/zeros of h(x) = x^3 \space {\text{–}} \space 3x^2 \space {\text{–}} \space 10x + 24.
Solution
We can apply the rational roots test, but we can also just proceed with a few basic initial guesses to find a first factor and root.
h(1) = 1^3 \space {\text{–}} \space 3(1)^2 \space {\text{–}} \space 10(1) + 24 \space = \space 1 \space {\text{–}} \space 3 \space {\text{–}} \space 10 + 24 \space = \space 13
h({\text{-}}1) = ({\text{-}}1)^3 \space {\text{–}} \space 3({\text{-}}1)^2 \space {\text{–}} \space 10({\text{-}}1) + 24 \space = \space {\text{-}}1 \space {\text{–}} \space 3 + 10 + 24 \space = \space 13
h(2) = 2^3 \space {\text{–}} \space 3(2)^2 \space {\text{–}} \space 10(2) + 24 \space = \space 8 \space {\text{–}} \space 12 \space {\text{–}} \space 20 + 24 \space = \space 0
2 is a root, so with that information we can proceed with finding the other roots.
A remainder of 0 is obtained as expected,
and we can now factorise fully and find the roots/solutions.
(x^2 \space {\text{–}} \space x \space {\text{–}} \space 12)(x \space {\text{–}} \space 2)
(x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space 2) => (x + 3)(x \space {\text{–}} \space 4)(x \space {\text{–}} \space 2)
x + 3 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}3 \space \space \space , \space \space \space x \space {\text{–}} \space 4 = 0 \space \space {\scriptsize{=>}} \space \space x = 4
The solutions of h(x) = x^3 \space {\text{–}} \space 3x^2 \space {\text{–}} \space 10x + 24 are x = {\text{-}}3 \space , \space 2 \space , \space 4.
(1.4)
Show that (x + 2) is a factor of f(x) = x^4 + 8x^3 + 17x^2 \space {\text{–}} \space 2x \space {\text{–}} \space 24.
Hence factorise fully and solve.
Solution
f(x) here is a polynomial of degree 4, so there are 4 factors and solutions.
Firstly we deal with showing that (x + 2) is a factor.
f({\text{-}}2) = ({\text{-}}2)^4 + 8({\text{-}}2)^3 + 17({\text{-}}2)^2 \space {\text{–}} \space 2({\text{-}}2) \space {\text{–}} \space 24
= 16 \space {\text{–}} \space 64 + 68 + 4 \space {\text{–}} \space 24 \space = \space 0
(x + 2) is a factor, with -2 being a zero.
With that information we can proceed with finding the other roots.
A remainder of 0 is obtained as expected,
we can now look to factorise further.
(x + 2)(x^3 + 6x^2 + 5 \space {\text{–}} \space 12)
We can proceed with synthetic division for x^3 + 6x^2 + 5 \space {\text{–}} \space 12, but we don’t know any exact zeros/roots to start with.
The rational roots theorem can help us find values to try.
Though sometimes we can try some basic values first, as we may get lucky and find a zero/root.
A good value to try initially is usually 0 or 1.
Factoring further.
Success on this occasion, we have quickly found another zero/root.
It doesn’t always work out like this, but it’s nice when it does.
Now we can factor fully and solve.
(x^2 + 7x + 12)(x \space {\text{–}} \space 1)(x + 2)
(x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space \space \space \space )(x \space {\text{–}} \space 1)(x + 2) => (x + 3)(x + 4)(x \space {\text{–}} \space 1)(x + 2)
x + 3 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}3 \space \space \space , \space \space \space x + 4 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}4
x \space {\text{–}} \space 1 = 0 \space \space {\scriptsize{=>}} \space \space x = 1 \space \space \space , \space \space \space x + 2 = 0 \space \space {\scriptsize{=>}} \space \space x = {\text{-}}2
The solutions of f(x) = x^4 + 8x^3 + 17x^2 \space {\text{–}} \space 2x \space {\text{–}} \space 24 are x = {\text{-}}4 \space , \space {\text{-}}3 \space , \space {\text{-}}2 \space , \space 1.
1. Home
2. ›
3. Algebra 2
4. › Synthetic Division Factoring |
Upcoming SlideShare
×
# Lesson 15: Inverse Functions And Logarithms
15,000 views
Published on
Inverse functions in general, and the inverses of very important exponential functions
Published in: Education, Technology
1 Like
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
Views
Total views
15,000
On SlideShare
0
From Embeds
0
Number of Embeds
119
Actions
Shares
0
124
0
Likes
1
Embeds 0
No embeds
No notes for slide
### Lesson 15: Inverse Functions And Logarithms
1. 1. Section 3.2 Inverse Functions and Logarithms V63.0121, Calculus I March 4/9/10, 2009 . . Image credit: Roger Smith . . . . . .
2. 2. Outline Inverse Functions Derivatives of Inverse Functions Logarithmic Functions . . . . . .
3. 3. What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. . . . . . .
4. 4. What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. So f−1 (f(x)) = x, f(f−1 (x)) = x . . . . . .
5. 5. What functions are invertible? In order for f−1 to be a function, there must be only one a in D corresponding to each b in E. Such a function is called one-to-one The graph of such a function passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is continuous, then f−1 is continuous. . . . . . .
6. 6. Graphing an inverse function The graph of f−1 interchanges the x and y f . coordinate of every point on the graph of f . . . . . . .
7. 7. Graphing an inverse function The graph of f−1 interchanges the x and y f . coordinate of every point on the graph of f .−1 f The result is that to get the graph of f−1 , we . need only reflect the graph of f in the diagonal line y = x. . . . . . .
8. 8. How to find the inverse function 1. Write y = f(x) 2. Solve for x in terms of y 3. To express f−1 as a function of x, interchange x and y . . . . . .
9. 9. How to find the inverse function 1. Write y = f(x) 2. Solve for x in terms of y 3. To express f−1 as a function of x, interchange x and y Example Find the inverse function of f(x) = x3 + 1. . . . . . .
10. 10. How to find the inverse function 1. Write y = f(x) 2. Solve for x in terms of y 3. To express f−1 as a function of x, interchange x and y Example Find the inverse function of f(x) = x3 + 1. Answer √ y = x3 + 1 =⇒ x = y − 1, so 3 √ f−1 (x) = 3 x−1 . . . . . .
11. 11. Outline Inverse Functions Derivatives of Inverse Functions Logarithmic Functions . . . . . .
12. 12. derivative of square root √ dy Recall that if y = x, we can find by implicit differentiation: dx √ x =⇒ y2 = x y= dy =⇒ 2y =1 dx dy 1 1 =√ =⇒ = dx 2y 2x d2 y , and y is the inverse of the squaring function. Notice 2y = dy . . . . . .
13. 13. Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) . . . . . .
14. 14. Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) “Proof”. If y = f−1 (x), then f(y) = x, So by implicit differentiation dy dy 1 1 f′ (y) = 1 =⇒ =′ = ′ −1 dx dx f (y) f (f (x)) . . . . . .
15. 15. Outline Inverse Functions Derivatives of Inverse Functions Logarithmic Functions . . . . . .
16. 16. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . . . . . . .
17. 17. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . Facts (i) loga (x · x′ ) = loga x + loga x′ . . . . . .
18. 18. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . Facts (i) loga (x · x′ ) = loga x + loga x′ (x) (ii) loga ′ = loga x − loga x′ x . . . . . .
19. 19. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . Facts (i) loga (x · x′ ) = loga x + loga x′ (x) (ii) loga ′ = loga x − loga x′ x (iii) loga (xr ) = r loga x . . . . . .
20. 20. Logarithms convert products to sums Suppose y = loga x and y′ = loga x′ ′ Then x = ay and x′ = ay ′ ′ So xx′ = ay ay = ay+y Therefore loga (xx′ ) = y + y′ = loga x + loga x′ . . . . . .
21. 21. Example Write as a single logarithm: 2 ln 4 − ln 3. . . . . . .
22. 22. Example Write as a single logarithm: 2 ln 4 − ln 3. Solution 42 2 ln 4 − ln 3 = ln 42 − ln 3 = ln 3 ln 42 not ! ln 3 . . . . . .
23. 23. Example Write as a single logarithm: 2 ln 4 − ln 3. Solution 42 2 ln 4 − ln 3 = ln 42 − ln 3 = ln 3 ln 42 not ! ln 3 Example 3 Write as a single logarithm: ln + 4 ln 2 4 . . . . . .
24. 24. Example Write as a single logarithm: 2 ln 4 − ln 3. Solution 42 2 ln 4 − ln 3 = ln 42 − ln 3 = ln 3 ln 42 not ! ln 3 Example 3 Write as a single logarithm: ln + 4 ln 2 4 Answer ln 12 . . . . . .
25. 25. “. . lawn” . Image credit: Selva . . . . . . .
26. 26. Graphs of logarithmic functions y . . = 2x y y . = log2 x . . 0, 1) ( ..1, 0) . x . ( . . . . . .
27. 27. Graphs of logarithmic functions y . . = 3x= 2x y. y y . = log2 x y . = log3 x . . 0, 1) ( ..1, 0) . x . ( . . . . . .
28. 28. Graphs of logarithmic functions y . . = .10x 3x= 2x y y=. y y . = log2 x y . = log3 x . . 0, 1) ( y . = log10 x ..1, 0) . x . ( . . . . . .
29. 29. Graphs of logarithmic functions y . . = .10=3xx 2x yxy y y. = .e = y . = log2 x y . = ln x y . = log3 x . . 0, 1) ( y . = log10 x ..1, 0) . x . ( . . . . . .
30. 30. Change of base formula for exponentials Fact If a > 0 and a ̸= 1, then ln x loga x = ln a . . . . . .
31. 31. Change of base formula for exponentials Fact If a > 0 and a ̸= 1, then ln x loga x = ln a Proof. If y = loga x, then x = ay So ln x = ln(ay ) = y ln a Therefore ln x y = loga x = ln a . . . . . . |
# Math U See Multiplication Chart
Studying multiplication right after counting, addition, as well as subtraction is ideal. Children understand arithmetic via a normal progression. This progression of learning arithmetic is often the pursuing: counting, addition, subtraction, multiplication, and lastly section. This assertion brings about the concern why understand arithmetic in this particular series? More importantly, why discover multiplication after counting, addition, and subtraction but before section?
## The next facts respond to these questions:
1. Youngsters understand counting first by associating graphic items using their hands and fingers. A concrete instance: The amount of apples are there within the basket? A lot more abstract instance is the way old have you been?
2. From counting amounts, the following reasonable stage is addition followed by subtraction. Addition and subtraction tables can be quite helpful teaching aids for children because they are visual instruments creating the move from counting simpler.
3. Which will be acquired following, multiplication or section? Multiplication is shorthand for addition. At this stage, young children have a organization knowledge of addition. Therefore, multiplication may be the following reasonable method of arithmetic to find out.
## Evaluate basic principles of multiplication. Also, assess the basics utilizing a multiplication table.
Allow us to review a multiplication case in point. Utilizing a Multiplication Table, grow 4 times three and have a solution 12: 4 x 3 = 12. The intersection of row about three and line four of the Multiplication Table is twelve; twelve may be the answer. For the kids beginning to discover multiplication, this is certainly straightforward. They could use addition to eliminate the situation therefore affirming that multiplication is shorthand for addition. Example: 4 x 3 = 4 4 4 = 12. It is really an excellent overview of the Multiplication Table. The additional advantage, the Multiplication Table is graphic and mirrors back to discovering addition.
## Where will we begin studying multiplication making use of the Multiplication Table?
1. First, get familiar with the table.
2. Start out with multiplying by 1. Start at row primary. Proceed to line primary. The intersection of row one and column the initial one is the solution: one.
3. Repeat these steps for multiplying by 1. Grow row a single by posts a single via a dozen. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Replicate these actions for multiplying by two. Flourish row two by columns a single through 5 various. The responses are 2, 4, 6, 8, and 10 respectively.
5. We will leap ahead. Repeat these steps for multiplying by several. Increase row 5 by posts one by means of twelve. The answers are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now allow us to raise the amount of trouble. Recurring these techniques for multiplying by three. Flourish row three by columns a single via twelve. The replies are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. If you are comfortable with multiplication thus far, use a test. Resolve the next multiplication troubles in your head and then compare your answers to the Multiplication Table: grow six as well as 2, flourish 9 and three, grow 1 and eleven, increase a number of and 4, and increase six as well as 2. The problem solutions are 12, 27, 11, 16, and 14 respectively.
In the event you received a number of from 5 various troubles appropriate, create your individual multiplication assessments. Compute the answers in your head, and view them while using Multiplication Table. |
# What is alternate angle explain with diagram?
## What is alternate angle explain with diagram?
Definition of alternate angle : one of a pair of angles with different vertices and on opposite sides of a transversal at its intersection with two other lines: a : one of a pair of angles inside the two intersected lines. — called also alternate interior angle.
## Is there such thing as alternate corresponding angles?
Corresponding angles are congruent. Angles that are on the opposite sides of the transversal are called alternate angles e.g. 1 + 8. All angles that are either exterior angles, interior angles, alternate angles or corresponding angles are all congruent.
Are alternate angles the same?
Alternate angles are equal. (c and f are also alternate). Alternate angles form a ‘Z’ shape and are sometimes called ‘Z angles’. Any two angles that add up to 180 degrees are known as supplementary angles.
READ ALSO: Is brown sugar and coconut oil a good exfoliator?
What is the difference between alternate angles and corresponding angles?
One of corresponding angles is always interior (in between parallel lines) and another – exterior (outside of the area in between parallel lines). The alternate angles are either both interior or both exterior.
### What does corresponding angles mean in math?
Definition of corresponding angles : any pair of angles each of which is on the same side of one of two lines cut by a transversal and on the same side of the transversal.
### Why are corresponding angles equal?
The corresponding angle postulate states that the corresponding angles are congruent if the transversal intersects two parallel lines. In other words, if a transversal intersects two parallel lines, the corresponding angles will be always equal.
What are the alternate angles of a transversal?
Alternate angles are shaped by the two parallel lines crossed by a transversal. Consider the given figure, EF and GH are the two parallel lines. RS is the transversal line that cuts EF at L and GH at M. If two parallel lines are cut by a transversal, then the alternate angles are equal. Therefore, ∠3 = ∠ 5 and ∠4 = ∠6.
READ ALSO: How do you become a somatic psychotherapist?
How do you find the alternate angles of two parallel lines?
If two parallel lines are cut by a transversal, then the alternate angles are equal. Therefore, ∠3 = ∠ 5 and ∠4 = ∠6. ∠2 = ∠8 and ∠1 = ∠7. Types of Alternate Angles. Based on the position of the angles, the alternate angles are classified into two types, namely
## What is the shape of the alternate angles?
Alternate angles are shaped by the two parallel lines crossed by a transversal. EF and GH are the two parallel lines. If two parallel lines are cut by a transversal, then the alternate angles are equal.
## What are the alternate interior angles of a line?
Alternate Interior Angles – Alternate interior angles are the pair of angles on the inner side of the two parallel lines but on the opposite side of the transversal. From the above-given figure, ∠3, ∠4, ∠5, ∠6 are the alternate interior angles |
Geometry is an Ancient Greek word where "geo" means earth and "metron" means measurement. Geometry is a branch of mathematics that deals with the study of shapes, sizes, area, position of figures and properties of points, angles, lines.
Our geometry calculator poses many calculators below that are used to find the midpoint of the line segment, equation of line, slope of triangle, distance between the two points etc.
Some of the Calculators are given below:
1. Distance Formula Calculator
2. Slope Calculator
3. Midpoint Calculator
4. Equation of a line Calculator
5. Vertex Calculator
Basic Geometry Terms
These are some basic terms to be known for the calculators under this geometry calculator.
Points:
We use points to specify locations in geometry. These points are denoted by letter or number,they are zero-dimensional. Point is thought as "dot" on a plane paper, have no length, width and height. P, Q, and R are the points in the below diagram.
Lines:
In geometry, lines are the straight lines that can be drawn on a paper. Line can be extended infinitely in both directions, is one dimensional having length but no width and height. Lines are determined by two points.
Considering the below diagram, we denote the line as $\overleftrightarrow{PQ}$ that passes through the points P and Q.
Line Segments:
Line Segment is also a straight line that can be drawn on a plane paper but with the defined poinTs. These defined points are known as end points.
The line segment, in the below diagram, is denoted as $\overline{PQ}$ the passes through two end points P and Q.
Midpoints:
A point in the middle of the line segment which divides the line segment into two equal segments is known as Midpoints. For example, if M is the midpoint of the line segment $\overline{PQ}$, then $\overline{PM}$ = $\overline{MQ}$.
Intersection:
When lines, line segment or rays meet or cross at a certain point, that point is known as an intersection point.
I is the intersection point in the below diagram.
Angle Calculator
Analytic Geometry Angle Geometry Chord Geometry Circle Geometry Cone Geometry Congruence Geometry
Geometry Area Calculator Geometry Distance Formula Calculator Midpoint Calculator Geometry Calculator Calculate Compound Interest Calculator Calculate Simple Interest Calculator
vertex calculator hyperbola calculator asymptotes calculator asymptote calculator tangent line calculator |
# What Do The Stars Say About Jonah Hill? (11/08/2019)
How will Jonah Hill do on 11/08/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is not at all guaranteed – do not take this too seriously. I will first find the destiny number for Jonah Hill, and then something similar to the life path number, which we will calculate for today (11/08/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology people.
PATH NUMBER FOR 11/08/2019: We will consider the month (11), the day (08) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 08 we do 0 + 8 = 8. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 8 + 12 = 22. This still isn’t a single-digit number, so we will add its digits together again: 2 + 2 = 4. Now we have a single-digit number: 4 is the path number for 11/08/2019.
DESTINY NUMBER FOR Jonah Hill: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Jonah Hill we have the letters J (1), o (6), n (5), a (1), h (8), H (8), i (9), l (3) and l (3). Adding all of that up (yes, this can get tiring) gives 44. This still isn’t a single-digit number, so we will add its digits together again: 4 + 4 = 8. Now we have a single-digit number: 8 is the destiny number for Jonah Hill.
CONCLUSION: The difference between the path number for today (4) and destiny number for Jonah Hill (8) is 4. That is higher than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t let this get you down! As mentioned earlier, this is of questionable accuracy. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
# Prove that
Question:
Prove that $1+1 .{ }^{1} P_{1}+2 .{ }^{2} P_{2}+3 .{ }^{3} P_{3}+\ldots . n .{ }^{n} P_{n}={ }^{n+1} P_{n+1}$
Solution:
To Prove: $1+1 .{ }^{1} \mathrm{P}_{1}+2 .{ }^{2} \mathrm{P}_{2}+3 .{ }^{3} \mathrm{P}_{3}+\ldots . . \mathrm{n} .{ }^{n} \mathrm{P}_{\mathrm{n}}={ }^{\mathrm{n}+1} \mathrm{P}_{\mathrm{n}+1}$
Formula Used:
Total number of ways in which $\mathrm{n}$ objects can be arranged in $\mathrm{r}$ places (Such that no object is replaced) is given by,
${ }_{n}{ }_{P_{r}}=\frac{n !}{(n-r) !}$
$1+1 .{ }^{1} P_{1}+2 .{ }^{2} P_{2}+3 .{ }^{3} P_{3}+\ldots . . n .{ }^{n} P_{n}={ }^{n+1} P_{n+1}$
$1+(2 !-1 !)+(3 !-2 !)+(4 !-3 !)+\ldots \ldots((n+1) !-n !)=(n+1) !$
$1+((n+1) !-1 !)=(n+1) !$
$(n+1) !=(n+1) !$
Hence proved. |
# Limit definition of derivative calculator with steps
Limit definition of derivative calculator with steps can help students to understand the material and improve their grades. We can solving math problem.
## The Best Limit definition of derivative calculator with steps
This Limit definition of derivative calculator with steps supplies step-by-step instructions for solving all math troubles. Next, use algebraic methods to isolate the variable on one side of the equation. Finally, substitute in values from the other side of the equation to solve for the variable. With practice, solving equations will become second nature. And with a little creativity, you might even find that equations can be fun. After all, there's nothing quite as satisfying as finding the perfect solution to a challenging problem.
This can be especially helpful when working with complex problems or when trying to learn a new concept. By seeing the step-by-step process that was used to solve the problem, students can better understand the material and develop their own problem-solving skills. In addition, a math solver with work can often be used to check answers that have been arrived at using other methods. This can help to ensure that the solution is correct and also help identify any mistakes that were made along the way. Whether you are a student who is struggling with math or a teacher who is looking for a way to check answers, a math solver with work can be an invaluable tool.
While they may seem daunting at first, there are a number of ways to solve quadratic equations. One popular method is known as factoring. This involves breaking down the equation into smaller factors that can be more easily solved. For example, if we have the equation ax^2 + bx + c = 0, we can factor it as (ax + c)(bx + c) = 0. This enables us to solve for x by setting each factor equal to zero and then solving for x. While factoring is a popular method for solving quadratic equations, it is not always the most straightforward approach. In some cases, it may be easier to use the quadratic formula, which is a formula specifically designed to solve quadratic equations. The quadratic formula can be used to solve any quadratic equation, regardless of how complex it may be. However, it is important to note that the quadratic formula only provides one solution for x. In some cases, there may be multiple solutions, so it is important to check all possible values of x before settling on a final answer. Regardless of which method you use, solving a quadratic equation can be an satisfying way to apply your math skills to real-world problems.
When it comes to solving math problems, there is no one-size-fits-all solution. The best approach depends on the nature of the problem, as well as the skills and knowledge of the person solving it. However, there are a few general tips that can help make solving math problems easier. First, it is important to take the time to understand the problem. What is being asked for? What information is given? Once you have a clear understanding of the problem, you can begin to consider different approaches. Sometimes, visual aids such as charts or diagrams can be helpful in solving math problems. Other times, it may be helpful to break the problem down into smaller steps. And sometimes, simply taking a step back and looking at the problem from a different perspective can make all the difference. There is no single right way to solve math problems. However, by taking the time to understand the problem and trying different approaches, it is usually possible to find a solution that works.
For example, the equation 2 + 2 = 4 states that two plus two equals four. To solve an equation means to find the value of the unknown variable that makes the equation true. For example, in the equation 2x + 3 = 7, the unknown variable is x. To solve this equation, we would need to figure out what value of x would make the equation true. In this case, it would be x = 2, since 2(2) + 3 = 7. Solving equations is a vital skill in mathematics, and one that can be used in everyday life. For example, when baking a cake, we might need to figure out how many eggs to use based on the number of people we are serving. Or we might need to calculate how much money we need to save up for a new car. In both cases, solving equations can help us to get the answers we need.
## Math solver you can trust
The app asked me for a rating and I initially said no, but then it gave me messages saying "no problem, we won't ask again" and this made me happy. So, I had to go ahead a give a rating
Waneeta Harris
Extremely helpful! This app allows me to format an equation in a really nice way, it has different features to make problems less confusing rather than typing them in to a conventional calculator. plus, the photo is really nice for quick easy problems that would just be a nuisance to type in. Only thing I'd change is maybe add an advanced mode for higher levels of math (Algebra 2, Trig., Pre-Calc, etc.)
Fayth Cook
Solving maths questions Scatter plot solver Algebraic formula solver Math problem solver app How to solve an absolute value equation |
# Assigning Numbers to Points in the Plane
Here is Problem 6 from the 2001 USAMO:
Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number.
Solution
Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number.
### Solution
The solution is by Michael Hamburg, St. Joseph's HS, South Band, Indiana, for which he was awarded The Clay Olympiad Scholar Award from the Clay Mathematics Institute. The judges observed that only 9 out of 260 participants solved the problem correctly. They praised Michael's ingenuity and elegance noting that the simple diagram practically amounts to a proof without words.
Let A and B be two arbitrary points. Form a regular hexagon ABCDEF and let G be the intersection of CD and EF. Note that points A, F, E are the reflections of points B, C, D in the line through G parallel to both AE and BD. It follows that the triangles ACE and BDF share the incircle and hence the incenter. The same holds for triangles CEG and DFG. Denoting the number assigned to a point by the corresponding lower case letter, we have two equations:
a + c + e = b + d + f and
d + f + g = c + e + g.
Adding the two and simplifying yields a = b; and we are done.
If we replace in the formulation of the problem the "inscribed" with "circumscribed" the statement holds and here is a
### Solution
We first show that, for any circle, the numbers assigned to its points are all the same.
So pick circle C(O), with center O. Choose three points P, Q, X on C(O).
p + q + x = 3o.
Fix, C(O), P, and Q, and let X change over C(O). Clearly, x = 3o - (p + q) does not depend on a specific position of X on C(O). Fixing X and Q, and then also X and P, we see that p and q take up the same common value, which proves that the number assignment is constant on any circle.
Any two points lie on a multitude of circles; but just one will suffice. Since on any circle the number assignment is constant, the numbers assigned to the given two points are equal.
### Remark
It may be observed that the fact that the assigned numbers were real played no role in the solution. Any set of numbers that is closed under the operation of taking the average of any three of them would do as well. In particular, the numbers in the problem could be rational, imaginary, infinitesimal, hyperreal or even surreal or quaternion. Naturally, vectors and matrices could be used as well. To use whole numbers we'd need to replace the "average" with the "sum". If 0 is excluded, the "product" could be used with the same degree of success.
### References
1. C. Alsina, R. B. Nelsen, Charming Proofs, MAA, 2010, pp. 47-48 |
# Question Video: Using Trigonometric Ratios to Solve Problems Involving Circles Mathematics • 10th Grade
Line segment π΄π΅ is a diameter of a circle with a radius of 62.5 cm. Point πΆ is on the circumference of the circle where line segment π΄πΆ β₯ line segment πΆπ΅ and π΄πΆ = 75 cm. Find the exact values of cos π΄ and sin π΅.
03:06
### Video Transcript
Line segment π΄π΅ is a diameter of a circle with a radius of 62 and a half centimeters. Point πΆ is on the circumference of the circle where line segment π΄πΆ is perpendicular to line segment πΆπ΅ and π΄πΆ equals 75 centimeters. Find the exact values of cos π΄ and sin π΅.
The first thing we can do is label this image with the information we know. The image already has points π΄, π΅, and πΆ. It also already has the perpendicular intersection of π΄πΆ and πΆπ΅. That would be this right angle. Weβre given that the radius is 62 and a half. We could use that to find the diameter. The diameter equals two times the radius. Two times 62 and a half equals 125. And that means, line segment π΄π΅ measures 125 centimeters. π΄πΆ measures 75 centimeters. We want to know the cos of π΄ and the sin of π΅.
We sometimes use the memory device SOH CAH TOA to help us remember the sine, cosine, and tangent relationships. The sin of π is its opposite side length over the hypotenuse. And the cos of π equals the adjacent side length over the hypotenuse.
Letβs first consider the cos of angle π΄. The adjacent side length to angle π΄ measures 75 centimeters. 75 should be the numerator. The hypotenuse of a right triangle is its longest side. And itβs always opposite the right angle. In this case, the hypotenuse is 125. So, 125 is the denominator. We can reduce this fraction. Both 75 and 125 are divisible by 25. So, we divide the numerator and the denominator by 25. 75 divided by 25 equals three. And 125 divided by 25 equals five. The cos of π΄ is three-fifths.
Now weβll consider sin of angle π΅. The opposite side length of angle π΅ is 75 centimeters. 75 is the numerator. And because angle π΅ is part of the same triangle as angle π΄, they have the same hypotenuse, which measures 125. 75 over 125 can, again, be reduced to three-fifths. The cos of π΄ and the sin of π΅ are equal to each other. Theyβre both three over five. |
# 2.10 Inverses of Exponential Functions
## 2.10 Inverses of Exponential Functions
We’ll now continue our conversation about logarithmic expressions, which mathematically translate into… logarithmic functions! Before we dive deeper into the nooks and crannies of such functions, let’s first connect the dots between them and a familiar face in this part of town: exponential functions. 📝
A logarithmic function can be defined as $f(x) = alog_b(x)$, where b is the base of the logarithm and a is the coefficient of the function. The base b must be greater than 0 and not equal to 1, and the coefficient a must not be 0. 🥴
The inverse relationship between the input and output values in exponential and logarithmic functions can be observed by comparing the general forms of these functions. The general form of an exponential function is $f(x) = ab^x$, where a is the coefficient and b is the base. As the input value x increases, the output value f(x) increases proportionately in equal-length intervals.
On the other hand, the general form of a logarithmic function is $f(x) = alog_b(x)$, where a is the coefficient and b is the base. As the output value f(x) increases, the input value x increases proportionately in equal-length intervals.
Coincidence? Definitely not! 😼
## 📖 A Story of Inverses
Exponential growth is characterized by output values changing multiplicatively as input values change additively, while logarithmic growth is characterized by output values changing additively as input values change multiplicatively. This is because in exponential functions, the input value is the exponent, and in logarithmic functions, the input value is the argument of the logarithm.
Graph displaying three different functions: y=a^x, y=x, and y=log_a(x).
Source: GitHub
### Reflections and the Identity Function h(x)
The graph of the logarithmic function $f(x) = log_b(x)$, where b > 0 and b ≠ 1, is a reflection of the graph of the exponential function $g(x) = b^x$, where b > 0 and b ≠ 1, over the graph of the identity function $h(x) = x$. This is due to the inverse relationship between logarithmic and exponential functions. 🪞
The exponential function $g(x) = b^x$ has a curve that increases rapidly as the x-value increases. The base b > 1, the greater the increase, while the base b < 1 the smaller the increase. The graph of the function g(x) has a vertical asymptote at x = 0 and has no horizontal asymptote.
On the other hand, the logarithmic function $f(x) = log_b(x)$ has a curve that increases slowly as the x-value increases. The base b > 1, the greater the slope of the curve, while the base b < 1 the smaller the slope of the curve. The graph of the function f(x) has a horizontal asymptote at y = 0 and has no vertical asymptote.
The identity function h(x) = x is a straight line with slope 1 and y-intercept 0. The graph of the function h(x) has no asymptotes. 🙅🏽
When the graphs of g(x) and f(x) are reflected over the graph of h(x), they become the same graph, but with the x and y coordinates switched. The resulting graph will have the shape of f(x) on one side of the identity line and the shape of g(x) on the other side of the identity line. This is why the graph of f(x) is a reflection of the graph of g(x) over the graph of h(x). 🔷
### Applying the Relationship: Ordered Pairs
If (s, t) is an ordered pair of the exponential function $g(x) = b^x$, where b > 0 and b ≠ 1, then (t, s) is an ordered pair of the logarithmic function $f(x) = log_b(x)$, where b > 0 and b ≠ 1. This is because exponential and logarithmic functions are inverses of each other.
The exponential function $g(x) = b^x$ takes the input value x and raises the base b to the power of x, resulting in the output value t. If we have an ordered pair (s, t) where s is the input value and t is the output value, we can say that $t = b^s$.
On the other hand, the logarithmic function $f(x) = log_b(x)$ takes the input value x and finds the exponent to which the base b must be raised to equal x, resulting in the output value s. If we have an ordered pair (t, s) where t is the input value and s is the output value, we can say that $s = log_b(t)$. 🗣️
Since exponential functions and logarithmic functions are inverses of each other, we can say that if (s, t) is an ordered pair of the exponential function $g(x) = b^x$, then (t, s) is an ordered pair of the logarithmic function $f(x) = log_b(x)$. 🤩 |
# Infinite and Non-Existent Limits
You have heard time and again that it is "against the rules" to divide by the number 0. Even the most basic calculator will return some form of "ERROR" if you try to divide even the smallest of numbers by 0.
Do you really understand why it is "against the rules"? What is really wrong with dividing by nothing?
# Infinite and Non-Existent Limits
Functions can exhibit a number of different behaviors as the input value gets very large or very small.
As x approaches ∞, some functions output values get closer and closer to a single number, some approach zero, and some continue to get larger and larger or smaller and smaller without limit.
In this lesson, we will explore functions of the last type, functions with infinite limits, and the different types of asymptotes they may have.
# Examples
Example 1
Solution
Earlier, you were given a question about dividing by zero.
Dividing by zero is "against the rules" because there is no definition for the answer you would get.
Consider what happens as you take a given value and divide it by smaller and smaller numbers:
2 / 10 = 1/5
2 / 1 = 2
2 / .1 = 20
2 / .001 = 2,000
2 / .000000001 = 2,000,000,000
As we divide by progressively smaller numbers, the quotient gets larger and larger. Also, we can see that in each case, the problem could be reversed by multiplying the product by the dividend to get the divisor, for instance: 2 / .1 = 20 ==> 20 * .1 = 2.
Unfortunately, this doesn't work if you actually divide by 0, even if you assume that dividing by zero resulted in infinity! No matter how big the number you multiply by zero, even infinity, you will never be able to get back to 2.
$$\ \therefore \frac{x}{0}=\text { undefined }$$
Example 2
Evaluate the function $$\ h(x)=\frac{x^{2}}{2 x-1}$$
Solution
To evaluate this function, consider the behavior of the function as larger and larger values are inserted for x. As x approaches ∞, the function values also approach ∞. Therefore the limit of the function as x approaches ∞ is: $$\ \lim _{x \rightarrow \infty} \frac{x^{2}}{2 x-1}=\infty$$. Similarly, as x approaches −∞, f(x) approaches −∞. Therefore we have $$\ \lim _{x \rightarrow-\infty} \frac{x^{2}}{2 x-1}=-\infty$$.
We can also understand this limit if we analyze the equation for h(x). As x gets larger and larger, the value of the expression 2x - 1 gets closer and closer to the value of the expression 2x. That is, for sufficiently large values of x, 2x - 1 ≈ 2x. Therefore the values of h(x) approach $$\ \frac{x^{2}}{2 x}=\frac{x}{2}$$.
As x gets larger and larger, so does $$\ x\over 2$$. For large values of x, the function h(x) gets closer and closer to $$\ x\over 2$$. Therefore the limit is infinity.
Example 3
Approximate the function f(x) = x2 + 2x - 3.
Solution
This function has an infinite limit as x approaches infinity. However, as x gets larger and larger, f(x) ≈ x2, since the x2 value grows much more quickly than the 2x value, particularly apparent at very large +/- values of x. If this is not immediately apparent, evaluate the function for x = 1,000,000, and you will quickly get the idea!
Therefore we can use the function y = x2 to describe the end behavior of f(x).
Example 4
Describe the end behavior of each graph. That is, determine if the function has a limit L, if the limit is infinite, or if the limit does not exist.
1. y=x2
2. y=2(−1)x
3. $$\ y=1-\frac{1}{|x|}$$
Solution
1. As x approaches +∞, x2 also approaches +∞. As x approaches −∞, x2 approaches +∞. Therefore $$\ \lim _{x \rightarrow \infty} x^{2}=\lim _{x \rightarrow-\infty} x^{2}=\infty$$.
2. This function is difficult to understand without producing a graph. The table shows that the function only takes on two values: 2, and -2, and is undefined at non-integer values of x. As x approaches +∞ or −∞, the function values alternate between 2 and -2. Therefore the limit does not exist.
3. If you look at the table of this function, which has negative and positive values of x, you can see that as x approaches +∞ or −∞, the function values approach 1.
Therefore $$\ \lim _{x \rightarrow \infty}\left(1-\frac{1}{|x|}\right)=\lim _{x \rightarrow-\infty}\left(1-\frac{1}{|x|}\right)=1$$.
We can also determine this limit analytically. For large values of x, |x| is also large, and so $$\ \frac{1}{|x|}$$ is small (since dividing 1 by a large number results in a very small number). Therefore, for large values of $$\ x, 1-\frac{1}{|x|} \approx 1-0=1$$. We can make the same argument for x approaching −∞.
Example 5
Evaluate $$\ \lim _{x \rightarrow \infty} \frac{x^{2}+2 x}{x-3}$$.
Solution
The numerator is of greater degree than the denominator, so the function does not approach a limit. The denominator is x - 3, so the graph cannot include 3.
Example 6
Evaluate −2x3−5x2+8x.
Solution
−2x3−5x2+8x is a 3rd degree equation, so it will turn twice, since it is not a rational function, there are no concerns about numerator or denominator. The function will have no limits, and will grow without bound in both the positive and negative directions. If you use a graphing calculator to graph the function, you will see that y=x3 can be used to approximate it.
# Review
For questions 1-3: The limit of f(x) as x→t cannot exist if which conditions are true? List three conditions.
1. One condition is...
2. A second condition is...
3. A third condition is...
1. Give an example of a limit that does not exist
Questions 5-7: Assuming that f(x) is a rational function:
1. What is $$\ \lim _{x \rightarrow \infty} f(x)$$ when the degree of the numerator is less than the degree of the denominator?
2. What is $$\ \lim _{x \rightarrow \infty} f(x)$$ when the degrees of the numerator and the denominator are equal?
3. What is $$\ \lim _{x \rightarrow \infty} f(x)$$ when the degree of the numerator is greater than the degree of the denominator?
4. In general, if r is a positive real number, what is $$\ \lim _{x \rightarrow \infty} \frac{1}{x^{r}}$$?
5. In general, if r is a positive real number, what is $$\ \lim _{x \rightarrow \infty} x^{r}$$?
Questions 10-13: Let a and b be real numbers and let t be a positive integer. Complete each of the following properties of limits.
10. $$\ \lim _{x \rightarrow a} x^{t}=$$
11. If f is a polynomial, $$\ \lim _{x \rightarrow a} f(x)=$$
12. $$\ \lim _{x \rightarrow a} k \cdot f(x)=$$
13. $$\ \lim _{x \rightarrow \infty} t^{x}=$$
Questions 14-16: Evaluate.
14. $$\ \lim _{x \rightarrow-5} \frac{(5+x)^{2}-25}{x}$$
15. $$\ \lim _{x \rightarrow 3} \frac{x^{3}-6 x+2}{x^{2}+2 x-3}$$
16. $$\ \lim _{x \rightarrow 0} \frac{(3+3 y)^{-1}-3 y^{-1}}{x}$$
To see the Review answers, open this PDF file and look for section 1.9.
# Vocabulary
Term Definition
The symbol "∞" means "infinity", and is an abstract concept describing a value greater than any countable number.
Asymptotes An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).
infinite limit A function has an infinite limit if it's output approaches infinity or negative infinity as very large or very small values are calculated for the input variable (usually "x").
infinity Infinity is an unbounded quantity that is greater than any countable number. The symbol for infinity is ∞.
limit A limit is the value that the output of a function approaches as the input of the function approaches a given value. |
# What is the equation of the normal line of f(x)=e^x/x at x=1 ?
Jan 24, 2016
$x = 1$
#### Explanation:
First, find the point the normal line will intercept.
$f \left(1\right) = {e}^{1} / 1 = e$
The normal line will pass through the point $\left(1 , e\right)$.
Now, to find the slope of the normal line, we must first find the slope of the tangent line. Thinking ahead, once we find the slope of the tangent line at $x = 1$, we can take the opposite reciprocal of that to find the slope of the normal line, since the normal line is perpendicular to the tangent line and perpendicular lines have opposite reciprocal slopes.
First, find the derivative of the function, will require the quotient rule.
$f ' \left(x\right) = \frac{x \frac{d}{\mathrm{dx}} \left({e}^{x}\right) - {e}^{x} \frac{d}{\mathrm{dx}} \left(x\right)}{x} ^ 2$
$f ' \left(x\right) = \frac{x {e}^{x} - {e}^{x}}{x} ^ 2$
$f ' \left(x\right) = \frac{\left(x - 1\right) {e}^{x}}{x} ^ 2$
The slope of the tangent line is
$f ' \left(1\right) = \frac{\left(1 - 1\right) {e}^{1}}{1} ^ 2 = 0$
This is a rather odd case. This is a horizontal line, since it has a slope of $0$. Thus, the normal line, which is perpendicular, will have an undefined slope.
The vertical line that passes through the point $\left(1 , e\right)$ is $x = 1$.
Graphed are the function and its normal line: |
Home / The science / Mathematics / How to multiply and divide fractions
How to multiply and divide fractions
/
130 Views
</a>
A fraction in mathematical sciences is a number that consists of a single or several parts of a unit, which in turn are called fractions.
The number of shares to which the unit was divided is the denominator of fractions, the number of fractions taken is the numerator of the fraction.
You will need
• - knowledge of the multiplication table or calculator-
• - paper-
• - a pen.
Instructions
1
To multiply a fraction and a natural number,Multiply the numerator of the fraction you multiply by this number, and leave the denominator of the fraction unchanged. The numerator of a fraction is the dividend, or that part of it that is above the horizontal or slash, which means the sign of division when the fraction is written. The denominator of a fraction is a divisor, or that part of it that is under a horizontal or slash.
2
To multiply a mixed fraction and a naturalNumber, multiply the whole fraction of this fraction, as well as its numerator, by this number, and leave the denominator of the multiply fraction unchanged. A fraction is mixed if it is written in the form of a proper fraction and an integer, and in this case it is understood as the sum of a fraction and an integer.
3
To multiply one fraction by another, firstMultiply the numerator of the first fraction by the numerator of the second fraction, then multiply the denominator of the first fraction by the denominator of the second fraction. Write the first found product as the numerator of the fraction, which is the desired result of the work, and write the second found product as the denominator of the same fraction.
4
To multiply mixed fractions,Note each mixed fraction as an irregular fraction, and then apply the multiplication rule. Correct is the fraction in which the numerator is modulo less than the denominator modulo. If the fraction does not fit this definition, then it is incorrect. In order to move from the mixed fraction to the wrong one, multiply the whole fraction of this fraction by the denominator, and then add the numerator to the resulting product. Write the final amount as the numerator of the received irregular fraction, and leave the denominator unchanged.
5
To divide one fraction into another, multiply the first fraction by the fraction that is the reciprocal of the second. To obtain the reverse fraction, swap the numerator and denominator.
How to multiply and divide fractions Was last modified: June 21st, 2017 By
It is main inner container footer text |
# My Favorite One-Liners: Part 17
In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.
Sometimes it’s pretty easy for students to push through a proof from beginning to end. For example, in my experience, math majors have little trouble with each step of the proof of the following theorem.
Theorem. If $z, w \in \mathbb{C}$, then $\overline{z+w} = \overline{z} + \overline{w}$.
Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then
$\overline{z + w} = \overline{(a + bi) + (c + di)}$
$= \overline{(a+c) + (b+d) i}$
$= (a+c) - (b+d) i$
$= (a - bi) + (c - di)$
$= \overline{z} + \overline{w}$
$\square$
For other theorems, it’s not so easy for students to start with the left-hand side and end with the right-hand side. For example:
Theorem. If $z, w \in \mathbb{C}$, then $\overline{z \cdot w} = \overline{z} \cdot \overline{w}$.
Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then
$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$
$= \overline{ac + adi + bci + bdi^2}$
$= \overline{ac - bd + (ad + bc)i}$
$= ac - bd - (ad + bc)i$
$= ac - bd - adi - bci$.
A sharp math major can then provide the next few steps of the proof from here; however, it’s not uncommon for a student new to proofs to get stuck at this point. Inevitably, somebody asks if we can do the same thing to the right-hand side to get the same thing. I’ll say, “Sure, let’s try it”:
$\overline{z} \cdot \overline{w} = \overline{(a + bi)} \cdot \overline{(c + di)}$
$= (a-bi)(c-di)$
$= ac -adi - bci + bdi^2$
$= ac - bd - adi - bci$.
$\square$
I call working with both the left and right sides to end up at the same spot the Diamond Rio approach to proofs: “I’ll start walking your way; you start walking mine; we meet in the middle ‘neath that old Georgia pine.” Not surprisingly, labeling this with a catchy country song helps the idea stick in my students’ heads.
Though not the most elegant presentation, this is logically correct because the steps for the right-hand side can be reversed and appended to the steps for the left-hand side:
Proof (more elegant). Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then
$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$
$= \overline{ac + adi + bci + bdi^2}$
$= \overline{ac - bd + (ad + bc)i}$
$= ac - bd - (ad + bc)i$
$= ac - bd - adi - bci$
$= ac -adi - bci + bdi^2$
$= (a-bi)(c-di)$
$= \overline{(a + bi)} \cdot \overline{(c + di)}$
$\overline{z} \cdot \overline{w}$.
$\square$
For further reading, here’s my series on complex numbers. |
# Probability question (5000 tires, 200 defective, 4800 good)
Today Nicole from Lenmore, CA, asked this question:
There are 5000 tires made including 200 that are defective if 4 tires are randomly selected, what is the probability that they are good?
Asked to use the method of redundancy but I'm having a hard time understanding how to write the equation and solve the problem.
The Wheeling Tire Company produced a batch of 5000 tires that includes exactly 200 that are defective.
a) If 4 tires are randomly selected for installation on a car, what is the probability that they are all good?
b) If 100 tires are randomly selected for shipment to an outlet, what is the probability that they are all good? Should the outlet plan to deal with defective tires returned by their customers?
Hi Nicole, thanks for asking. Here's how I would do this problem.
T= total number of tires
D = total number of defective tires
G = total number of good tires
2. Write an equation that represents the relationship between these variables
T = D + G
(Total number of tires) = (number of defective tires) + (number of good tires)
From this equation we can conclude that the probability of selecting a good tire is going to be:
P = G/T
3. Find the probability of selecting one good tire from the total number of tires
For this problem it helps to remember you are working with four tires, so let's label them individually as t1, t2, y3, and t4. In order to calculate the probability that ALL tires are good, you have to first calculate the individual probabilities then multiply them together.
Remember to think practically here. Every time we select a tire for use, we must subtract one from the total
P(t1, good) = 4800/5000
P(t2, good) = 4799/4999
P(t3, good) = 4798/4998
P(t4, good) = 4797/4997
Let's go ahead and keep them in fraction form to avoid miscalculating a decimal.
4. Multiply the individual probabilities to find the combined probability.
P(t1234, good) = (4800/5000)x(4799/4999)x(4798/4998)x(4797/4997) = 0.849
Make sure to go back and check that all your numbers are correct.
I hope this helps! Honestly part B) is above my level, but I found the answer here:
http://fredmath.net/Statistics/lecture2/sec2.2/sec-2.2_sol.pdf
It looks like for part B you have to use the 5% guideline, which I'm hoping your teacher can explain better than I can.
Cheers,
John W.
My questions to other tutors:
1) How would you go about solving this problem?
2) Do you agree with the methods I used?
3) Any suggestions for improvement? I am fairly new to tutoring, seeking to refine my skills in any way that I can. I grow every day through constructive feedback
\$40p/h
John W.
Private tutor - proficient in Math, Language Arts, and Music
50+ hours
if (isMyPost) { } |
# Pythagoras Theorem
In Glogpedia
by pjsk2202
Last updated 5 years ago
Discipline:
Math
Subject:
Geometry
8
Pythagoras Theorem
The Rule: c² = a² + b²where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides in a right angled triangle.
Scroll down for the answer.c² = a² + b²c² = 3² + 4²c² = 9 + 16c² = 25c = √25c = 5
The life of Pythagoras• 570 BC – 495 BC• Also known as Pythagoras of Samos.• A Greek philosopher and mathematician.• Most famous for Pythagorean theorem.• Founder of Pythagoreanism, a religious movement.• Pythagoreans believed that it was sinful to eat beans.• Pythagoras spent 22 years in Egypt, where he mastered mathematics.• In Egypt, he was taken to Babylon as a prisoner by the Persians.• After 12 years he was set free and returned to his birthplace, Samos, where he began teaching his philosophy.• People in Samos were not welcoming of his self established ideas.• He then moved to Croton in Greece, now in South Italy, where he founded his religion.
When do you use the Pythagoras Theorem?You can use the Pythagoras Theorem when you are presented with a problem involving a right-angled triangle, of which you know 2 of the side lengths and you want to find the 3rd side.You can also use pythagoras to find the length of the diagonal of a square or rectangle (if the length and width are known) and to prove whether a triangle has a right angle or not (if all sides are known).There is many different uses for pythagoras outside of maths, such as finding the length of a ladder leaned against a wall (assuming you know how high the wall is and the distance from the wall to the base of the ladder) and finding the length of the diagonal of a ramp (given the height and length).
The hypotenuse is the side opposite the right angle.
If a right angled triangle with the hypotenuse unknown, and the other two sides are 3m and 4m, what is the length of the unknown side?
A right-angled triangle with all sides as positive integers is called a Pythagorean triple. |
# How do you find the arc length of the curve y=x^3 over the interval [0,2]?
Mar 9, 2018
A first-order approximation to the arc length gives $8 + {\tan}^{- 1} \left(2 \sqrt{3}\right)$ units.
#### Explanation:
$y = {x}^{3}$
$y ' = 3 {x}^{2}$
Arc length is given by:
$L = {\int}_{0}^{2} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$
Apply the substitution $\sqrt{3} x = u$:
$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \sqrt{1 + {u}^{4}} \mathrm{du}$
Complete the square in the square root:
$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \sqrt{{\left({u}^{2} + 1\right)}^{2} - 2 {u}^{2}} \mathrm{du}$
Factor out the larger piece:
$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left({u}^{2} + 1\right) \sqrt{1 - \frac{2 {u}^{2}}{{u}^{2} + 1} ^ 2} \mathrm{du}$
For $x \in \left[0 , 2 \sqrt{3}\right]$, $\frac{2 {u}^{2}}{{u}^{2} + 1} ^ 2 < 1$. Take the series expansion of the square root:
$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left({u}^{2} + 1\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{2 {u}^{2}}{{u}^{2} + 1} ^ 2\right)}^{n}\right\} \mathrm{du}$
Isolate the $n = 0$ and $n = 1$ cases:
$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left\{\left({u}^{2} + 1\right) - {u}^{2} / \left({u}^{2} + 1\right)\right\} \mathrm{du} + \frac{1}{\sqrt{3}} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 2\right)}^{n} {\int}_{0}^{2 \sqrt{3}} {u}^{2 n} / {\left({u}^{2} + 1\right)}^{2 n - 1} \mathrm{du}$
Hence
$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left({u}^{2} + \frac{1}{{u}^{2} + 1}\right) \mathrm{du} + \frac{1}{\sqrt{3}} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 2\right)}^{n} {\int}_{0}^{2 \sqrt{3}} {u}^{2 n} / {\left({u}^{2} + 1\right)}^{2 n - 1} \mathrm{du}$
Ignoring the higher-order terms, a first-order approximation gives:
$L \approx \frac{1}{\sqrt{3}} {\left[\frac{1}{3} {u}^{3} + {\tan}^{- 1} u\right]}_{0}^{2 \sqrt{3}}$
Insert the limits of integration:
$L \approx 8 + {\tan}^{- 1} \left(2 \sqrt{3}\right)$ |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.