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amid [387]
3 years ago
10

Which of the following is the product of the rational expression 1/x+2 × x/x-2

Mathematics
1 answer:
pav-90 [236]3 years ago
6 0

Answer:

the answer of your question

\frac{1}{x + 2}  \times  \frac{x}{x - 2}  =  \frac{x}{x ^{2}  - 4}

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Peter is trying to buy fencing for the perimeter of his garden. His garden is in the shape of a rectangle with a length of 2(x+6
RideAnS [48]

Given dimensions of a rectangular garden are:

length = 2(x+6) feet

width = 3.5x feet

using the distributive property on length we get, length = 2x+12

Perimeter of a rectangle is 2(length+width)

So, P = 2(2x+12+3.5x)

Solving it we get,

P = 2(5.5x +12)   ..............adding like terms

P = (2\times5.5)+(2\times12) .............multiplying and adding

P = 11x+24

So, perimeter is 11x+24 feet. This value of fencing will be needed.

4 0
3 years ago
A water tank that holds 30 gallons of water when full is leaking 3/4 gallon of water every hour. If no one notices and stops the
viva [34]
It will take 40 hours for the water tank to become empty.

Explanation:
Divide 30 by 0.75 (equivalent to 3/4). 
7 0
3 years ago
Select all of the quotients that have the same value as 5.04 divided by 7
defon

Answer: 2 and 4

Step-by-step explanation: the quick way is by dividing all the answer options

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4%7D%20" id="TexFormula1" title=" \sqrt{4} " alt=" \sqrt{4} " align="absmiddle" c
Elan Coil [88]

Answer:

<h2>2</h2>

Step-by-step explanation:

\sqrt{4}  = 2 \\ 2 \times 2 = 4

3 0
3 years ago
Read 2 more answers
Find the perimeter of shape #8
Wittaler [7]
The squiggly lines and the 2 straight lines through to sides of the pentagon mean that they are congruent (equal in length).

So, to find the perimeter of a regular polygon, you add all of its sides together.


So, the perimeter= 4x-2+x-2+x-2+x+2+x+2

To simplify it, add all of the x's together.

4x+x+x+x+x=8x

Perimeter=8x-2-2-2+2+2

-2-2=-4-2=-6+2=-4+2=-2

Perimeter=8x-2

Reach out to me if you have any questions. I'd be glad to help! :)
4 0
3 years ago
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