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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

User JHurrah
by
7.7k points

2 Answers

0 votes

Answer:

1. D

2.B

3.B

4.B

5. B

6.D

Explanation:

6 votes

Answer:

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











User Darrell Plank
by
8.0k points

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