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Algebra II : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #11 : Other Factorials

What is $\frac{x!}{(x+1)!}$ ?

Possible Answers:

x^2+x

x+1

$\frac{1}{x+1}$

Correct answer:

$\frac{1}{x+1}$

Explanation:

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach . Since we are dealing with variables, let's analyze each the numerator and the denominator. x+1 is definitely greater than in this situation because factorials are always positive numbers. If we took the difference between x+1 and we would get . This means that if we were to expand both the numerator and denominator, we cancel everything out except the extra term in the denominator which is x+1 .

$\frac{x!}{(x+1)!)}=\frac{x\cdot 1}{(x+1)(x)(1)}=\frac{1}{x+1}$

So final answer is $\frac{1}{x+1}$ .

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Example Question #231 : Mathematical Relationships And Basic Graphs

Simplify.

$\frac{(x+2)!}{(x-2)!}$

Possible Answers:

(x+1)x(x-1)

(x+2)(x+1)

(x+2)(x+1)x(x-1)

x(x-1)

(x+2)(x+1)x

Correct answer:

(x+2)(x+1)x(x-1)

Explanation:

A factorial means you are multiplying the number, with one less than previous number and you keep multiplying until you reach . Since we are dealing with variables, let's analyze each the numerator and the denominator. x+2 is definitely greater than x-2 in this situation because factorials are always positive numbers. If we took the difference between x+2 and x-2 we would get . This means that if we were to expand both the numerator and denominator, we cancel everything out except the extra terms in the numerator which are x+2, x+1, x, x-1 . So final answer is (x+2)(x+1)x(x-1) .

If you need convincing, let x=6 . So we have $\frac{(6+2)!}{(6-2)!}=\frac{8!}{4!}=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1}$ .

You can cancel out the $4\cdot 3\cdot 2\cdot1$ from top and bottom to get $8\cdot 7\cdot 6\cdot 5$ . We need to express them into expressions.

Since was , to get , you need to add to or x+2.

To get , you add to or x+6.

x=6 To get you subtract from or x-1 .

You still get the same answer of (x+2)(x+1)x(x-1) .

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Example Question #81 : Factorials

Simplify: $\frac{2! + 0!}{3!-0!}$

Possible Answers:

$\frac{2}{3}$

$\frac{3}{5}$

$\frac{3}{4}$

$\frac{1}{5}$

$\frac{1}{2}$

Correct answer:

$\frac{3}{5}$

Explanation:

Write out the terms of each factorial in expanded form. The only exception is the zero factorial, which is equal to one.

$\frac{2! + 0!}{3!-0!} =\frac{2\times 1+1}{3\times 2\times 1-1}$

Simplify the terms. The values in the numerator and denominator cannot cancel out!

$\frac{2\times 1+1}{3\times 2\times 1-1} = \frac{2+1}{6-1} = \frac{3}{5}$

The correct answer is: $\frac{3}{5}$

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Example Question #84 : Factorials

What is the value of ?

Possible Answers:

3268800

5040

40320

69120

Correct answer:

40320

Explanation:

Step 1: We must define factorial. Factorial is defined as , where n!=n(n-1)(n-2)(n-3)...1

Step 2: Evaluate

8!=8(8-1)(8-2)(8-3)(8-4)(8-5)(8-6)(8-7)
$8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=40320$

$40320\equiv 8!$

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Example Question #85 : Factorials

Compute: $\frac{(3!)!+3(3!)}{3!}$

Possible Answers:

123

456

$\textup{The answer is not given.}$

246

345

Correct answer:

123

Explanation:

Simplify the factorials in the numerator and denominator.

$\frac{(3!)!+3(3!)}{3!}=\frac{(3 \times 2 \times 1)!+3(3 \times 2 \times 1)}{3 \times 2 \times 1}$

Simplify the terms on the top and bottom.

$\frac{(3 \times 2 \times 1)!+3(3 \times 2 \times 1)}{3 \times 2 \times 1} =\frac{(6)!+3(6)}{6}$

$\frac{(6)!+3(6)}{6} = \frac{(6\times 5\times4\times3\times2\times1)+3(6)}{6}$

$\frac{720+18}{6} = \frac{738}{6} = 123$

The answer is: 123

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Example Question #231 : Mathematical Relationships And Basic Graphs

Multiply: $[(2!)! \cdot (1-1)!]^2$

Possible Answers:

Correct answer:

Explanation:

Simplify all the terms in the parentheses first.

$(2!) = 2\times 1 =2$

(1-1)! = 0! = 1

This indicates that:

$(2!)! \cdot (1-1)! = 2\cdot 1 =2$

$[(2!)! \cdot (1-1)!]^2 = [2]^2 = 4$

The answer is:

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Example Question #92 : Factorials

Evaluate:

$\sum_{k=1}^{4} \frac{1}{k!}$

Possible Answers:

$2 \frac{1}{12}$

$1\frac{17}{24}$

$1\frac{3}{4}$

None of the other choices gives the correct response.

$1\frac{1}{33}$

Correct answer:

$1\frac{17}{24}$

Explanation:

$\sum_{k=1}^{4} \frac{1}{k!}$ is equal to the sum of the expressions formed by substituting 1, 2, 3 and 4, in turn, for in the expression $\frac{1}{k!}$ , as follows:

- or factorial - is defined to be the product of the integers from 1 to . Therefore, each term can be calculated by multiplying the integers from 1 to , then taking the reciprocal of the result.

k = 1 :

$\frac{1}{k!} = \frac{1}{1!} = \frac{1}{1} = 1$

k = 2 :

$\frac{1}{k!} = \frac{1}{2!} = \frac{1}{1 \cdot 2} = \frac{1}{2}$

k = 3 :

$\frac{1}{k!} = \frac{1}{3!} = \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}$

k = 4 :

$\frac{1}{k!} = \frac{1}{4!} = \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} = \frac{1}{24}$

Add the terms:

$\sum_{k=1}^{4} \frac{1}{k!} = 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} = 1\frac{17}{24}$

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Example Question #2891 : Algebra Ii

Try without a calculator:

True or false: $\sqrt{9!}= 3!$

Possible Answers:

True

False

Correct answer:

False

Explanation:

- or factorial - is defined to be the product of the integers from 1 to . Therefore,

$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 362,880$

$3! = 1\times 2 \times 3 = 6$

Since $6^{2} = 36$ ,

$(3!)^{2} = 36 \ne 362,880 = 9!$

and

$\sqrt{9!}= 3!$ is a false statement.

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Example Question #21 : Other Factorials

Try without a calculator.

10,000 < N! < 100,000

Give the value of that makes this three-part inequality true.

Possible Answers:

N = 9

N = 10

The inequality has no solution.

N = 8

N = 7

Correct answer:

N = 8

Explanation:

Given a nonnegative integer , - or factorial - is defined to be the product of the integers from 1 to . After some exploration, multiplying 1 by 2, the result by 3, that result by 4, and so forth, we find that

$7! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5,040$

Multiplying by 8, we find that

$8! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8= 40,320$

Multiplying by 9, we find that

$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9= 362,880$

Therefore,

7! < 10,000 < 8! < 100,000 < 9!

Since only integers may have factorial values, it follows that the only value of that makes the inequality true is N = 8 .

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Example Question #95 : Factorials

Try without a calculator:

is equal to which expression?

Possible Answers:

$7! \cdot 7$

7! + 7

7! + 8

$7! \cdot 8$

None of these

Correct answer:

$7! \cdot 8$

Explanation:

- or factorial - is defined to be the product of the integers from 1 to . Therefore,

$7 ! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$

and

$8 ! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$

$= (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7) \cdot 8$

$= 7! \cdot 8$ ,

the correct response.

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Other Factorials

Example Question #231 : Mathematical Relationships And Basic Graphs

Example Question #81 : Factorials

Example Question #84 : Factorials

Example Question #85 : Factorials

Example Question #231 : Mathematical Relationships And Basic Graphs

Example Question #92 : Factorials

Example Question #2891 : Algebra Ii

Example Question #21 : Other Factorials

Example Question #95 : Factorials

All Algebra II Resources

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