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iren2701 [21]
3 years ago
10

Absolute value to show the distance between -70 and 15

Mathematics
1 answer:
castortr0y [4]3 years ago
7 0

To find the absolute value, change all into positives, and add them

|-70| + 15 = answer

|-70| = 70

70 + 15 = 85

The distance in the number line between -70 and 15 is 85

85 is your answer

hope this helps

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
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Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

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2.

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3.

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(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

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2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

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The expression for the greatest common factor of 20 and 30 using the distributive property is 10(2 + 3)

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The relation for the distributive property :

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Learn more :brainly.com/question/15263211

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