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Vsevolod [243]
3 years ago
9

if you can answer the rest i will give you brainless. i already anwer some if them so can you answer the rest. plss​ no link pls

s!!!​

Mathematics
1 answer:
otez555 [7]3 years ago
8 0
6= 1040 and 7= 380 and 8= A
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12.Divide. (x^2+2x+1/x^2-8x+16)/(x+1/x^2-16
lina2011 [118]
For this case we have the following expression:
 (x ^ 2 + 2x + 1 / x ^ 2-8x + 16) / (x + 1 / x ^ 2-16)
 Rewriting we have:
 (((x + 1) (x + 1)) / ((x-4) (x-4))) / (x + 1 / ((x + 4) (x-4)))
 Then, we cancel similar terms:
 ((x + 1) / (x-4)) / (1 / (x + 4))
 Rewriting:
 ((x + 1) (x + 4)) / (x-4)
 Answer:
 
((x + 1) (x + 4)) / (x-4)
3 0
3 years ago
Find 8 /10 + 7 /100
Sergio [31]

Answer:

87          

 ——— = 0.87000

 100          

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The location of the mid point between (-15,4) (2,-10)
frozen [14]

Answer:

The answer is

<h2>( -  \frac{13}{2} , \:  - 3)</h2>

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

<h3>M  = ( \frac{x1 + x2}{2} , \:  \frac{y1 + y2}{2} )</h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-15,4) (2,-10)

The midpoint is

<h3>M = ( \frac{ - 15 + 2}{2}  , \:  \frac{4 - 10}{2}  ) \\  = ( -  \frac{13}{2} , \:  -  \frac{6}{2} )</h3>

We have the final answer as

<h3>( -  \frac{13}{2} , \:  - 3)</h3>

Hope this helps you

5 0
3 years ago
Which expression is equivalent to x+x+x+5+5 no matter what value is substituted in for x
MA_775_DIABLO [31]

Answer:

3x+10

Step-by-step explanation:

x+x+x+5+5=3x+5+5=3x+10

4 0
3 years ago
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Explain how to determine if a proportional relationship exist between two variables in a table
Rufina [12.5K]

Answer:

  a proportional relationship exists if the ratio of values is a constant

Step-by-step explanation:

The equation describing a proportional relationship is ...

   y = kx

If you are given a table of values of x and y, they will be proportional if ...

  y/x = k

for all (x, y) pairs.

3 0
3 years ago
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