Question

Which expression is equivalent to 2^a+1/10a-5/10a/4a^2-1?

A 2a/(2a-1)^2
B 50a/(2a+1)^2
C (2a - 1)^2/2a
D (2a + 1)^2/50a

Which expression is equivalent to the expression below?
(x/x+4)/x
A x/x+4 * x/1
B x/x+4 * 1/x
C x+4/x * 1/x
D x+4/x * x/1

What is the quotient? n+3/2n-6/n+3/3n-9
A 2/3
B 3/2
C (n+3)^2/6(n-3)^2
D 6(n-3)^2/(n+3)^2

What is the product? a-3/15a * 5/a-3
A 1/3
B 1/3a
C 3a
D 3

What is the quotient? n+3/2n-6/n+3/3n-9
A 2/3
B 3/2
C (n+3)^2/6(n-3)^2
D 6(n-3)^2/(n+3)^2

What is the product? a-3/15a * 5/a-3
A 1/3
B 1/3a
C 3a
D 3

What is the product? 2a-7/a * 3a^2/2a^2-11a+14
A 3/a-2
B 3a/a-2
C 3a/a+2
D 3/a+2

What is the quotient? a-3/7 / 3-a/21
A -(a-3)^2/147
B (a-3)^2/147
C 3
D -3

Answer 1

1. D

2.B

3.B

4.B

5. B

6.D

Explanation:

Answer 2

1. D

2.B

3.B

4.B

5. B

6.D

Explanation:

QUESTION 1

We want to simplify
`((2a+1)/(10a-5))/((10a)/(4a^2-1))`.

Let us change the middle bar to a normal division sign to obtain,

`(2a+1)/(10a-5)/(10a)/(4a^2-1)`.

We now multiply, the first fraction by the reciprocal of the second fraction to get,

`(2a+1)/(10a-5)* (4a^2-1)/(10a)`.

We now factor to obtain,

`(2a+1)/(5(2a-1))* ((2a-1)(2a+1))/(10a)`.

We cancel out common factors to get,

`(2a+1)/(5(1))* ((1)(2a+1))/(10a)`.

We now multiply out to get,

`((2a+1)^2)/(50a)`.

Ans:D

QUESTION 2

The given expression is
`((x)/(x+4) )/(x)`

We change the middle bar to a normal division sign to get,

`(x)/(x+4)/ x`

We now multiply by the reciprocal of the second fraction to obtain,

`(x)/(x+4)* (1)/(x)`

Ans:B

QUESTION 3

We want to simplify the quotient,

`((n+3)/(2n-6))/((n+3)/(3n-9))`.

We change the middle to a normal division sign to obtain,

`(n+3)/(2n-6)/(n+3)/(3n-9)`.

We now multiply the first fraction by the reciprocal of the second fraction to get,

`(n+3)/(2n-6)* (3n-9)/(n+3)`.

We factor to get,

`(n+3)/(2(n-3))* (3(n-3))/(n+3)`.

We cancel out common factors to obtain,

`(1)/(2(1))* (3(1))/(1)`.

We simplify to get,

`(3)/(2)`.

Ans:B

QUESTION 4

We want to find the product
`(a-3)/(15a) *(5)/(a-3)`.

This multiplication is very simple to do. We just have to cancel out common factors to get,

`(1)/(3a) *(1)/(1)`.

We now multiply out to get,

`(1)/(3a)`

Ans:B

QUESTION 5.

We want to find the product
`(2a-7)/(a) * (3a^2)/(2a^2-11a+14)`.

We factor the denominator of the second fraction to obtain,

`(2a-7)/(a) * (3a^2)/((2a-7)(a-2))`.

We now cancel out common factors to obtain,

`(1)/(1) * (3a)/((1)(a-2))`.

We multiply out to get,

`(3a)/(a-2)`.

Ans:B

QUESTION 6

We want to find the quotient
`((a-3)/(7))/((3-a)/(21) )`.

We need to change the middle bar to a normal division sign to get,

`(a-3)/(7)/ (3-a)/(21)`

We multiply by the reciprocal of the second function to get,

`(a-3)/(7) * (21)/(3-a)`

We factor the negative 1 from the denominator of the second fraction to get,

`(a-3)/(7) * (21)/(-1(a-3))`

We now cancel common factors to get,

`(1)/(1) * (3)/(-1(1))`

This simplifies to,

`-3`.

Ans: D

Two questions repeated