Sumi lives 2 miles from school. valerie lives 10,542 feet from school. who lives closer to school ?

2 answers:

8 0

Valerie lives closer because when you convert the 2 miles to feet from where Sumi is it’s 10,560 feet so the answer is Valerie.

nasty-shy [4]3 years ago

5 0

Sumi because 10452 feets are 1.997 miles

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Answer:

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

Step-by-step explanation:

Given:

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

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

$q^0=1$

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

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

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

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

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

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

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Answer:

A. It would become a different atom.

C. Its charge would increase by 1

Step-by-step explanation:

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The given equation is presented as follows;

$\frac{3}{(x - 4) \cdot (x - 7) } + \frac{6}{(x - 7) \cdot (x - 13) } + +\frac{15}{(x - 13) \cdot (x - 28) } - \frac{1}{(x - 28) } = - \frac{1}{20}$

Factoring the common denominator, we have;

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