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3 years ago

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:

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

$m^{-1}=\dfrac{1}{m}$

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$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

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

$m^{-1}=\dfrac{1}{m}$

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

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

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

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The relation for the distributive property :

a × (b + c)

Expand ;

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According to the distributive property

(2 × 10) + (3 × 10) = 10(2 + 3)

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