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Vera_Pavlovna [14]
3 years ago
14

8x^3y^4-22x^5y^6 gcf

Mathematics
1 answer:
schepotkina [342]3 years ago
5 0
Don't trust those link my cuz
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Delia measured her bathtub to be 2 meters long. What is an equivalent measurement?
allochka39001 [22]


An equivalent measurement to 2 meters is 200 cm. Since 1 meter is equivalent to 100 cm, multiply both of them by two. So, your answer is - 2 meters is equivalent to 200 cm.

Hope it helps :)

4 0
3 years ago
Type the single exponent that would appear on the 7: *
liq [111]

You add the exponent. You never multiply it.

Hope this helps!

<em>Say, mind doing me a favor and clicking the brainliest button for me? It would help me tons.</em>

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<h2><em>~~~PicklePoppers~~~</em></h2>
6 0
2 years ago
How many is 7= 12 + 5h<br> With procedure
Arlecino [84]
H= -1 here's why:

7 = 12 + 5h 

First you subtract 12 from both sides then the equation would look like:

-5 = 5h

Then you would divide the equation by 5 from both sides it would then look like:

-1 = h 
8 0
2 years ago
Can someone help me out
LiRa [457]

3/196

Step-by-step explanation:

multiply everything

5 0
3 years ago
Read the instructions
nordsb [41]

Answer:

Step-by-step explanation:

\frac{5x - 2}{4}+\frac{1}{2}=\frac{3y+2}{2}

Multiply the equation by 4

4*\frac{5x - 2}{4}+4*\frac{1}{2}=4*\frac{3y+2}{2}\\

5x - 2 + 2 = 2*(3y + 2)

5x +0 = 2*3y + 2*2

5x = 6y + 4

5x - 6y = 4 --------------------(I)

\frac{7y+3}{3}=\frac{x}{2}+\frac{7}{3}\\

Multiply the equation by 6

6*\frac{7y+3}{3}=6*\frac{x}{2}+6*\frac{7}{3}\\

2*(7y + 3) = 3x + 2*7

14y + 6 = 3x + 14

14y = 3x + 14 - 6

14y = 3x + 8

-3x + 14y = 8 ------------------------(II)

Multiply equation (I) by 3 and equation (II) by 5 and then add

(I)*3              15x - 18y = 12

(II)*5           <u>-15x  + 70y = 40</u>     {Now add}

                          52y = 52

                              y = 52/52

                            y = 1

Substitute y =1 in equation (I)

5x - 6*1 =  4

5x - 6 = 4

      5x = 4 +6

      5x = 10

         x = 10/5

x = 2

7 0
3 years ago
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