How do you simplify $\frac{2 p^{3} q^{2}}{8 p^{4} q} \div \frac{4 p q^{2}}{16 p^{4}}$ $(2p^3q^2)/(8p^4q) div (4pq^2)/(16p^4)$ ?

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How do you simplify $\frac{2 p^{3} q^{2}}{8 p^{4} q} \div \frac{4 p q^{2}}{16 p^{4}}$ $(2p^3q^2)/(8p^4q) div (4pq^2)/(16p^4)$ ?

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Answer 1 · 2015-07-16T22:49:24

$\frac{\frac{2 p^{3} q^{2}}{8 p^{4} q}}{\frac{4 p q^{2}}{16 p^{4}}} = \frac{p^{2}}{q}$ $((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=p^2/q$

Explanation:

We start off with what's given:

$\frac{\frac{2 p^{3} q^{2}}{8 p^{4} q}}{\frac{4 p q^{2}}{16 p^{4}}}$ $((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))$

Recall: $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{\frac{a}{b}}{\frac{c}{d}} \cdot \frac{\frac{d}{c}}{\frac{d}{c}} = \frac{a}{b} \cdot \frac{d}{c}$ $(a/b)/(c/d)=(a/b)/(c/d)*(d/c)/(d/c)=a/b*d/c$

So,

$\frac{\frac{2 p^{3} q^{2}}{8 p^{4} q}}{\frac{4 p q^{2}}{16 p^{4}}} = (\frac{2 p^{3} q^{2}}{8 p^{4} q}) \cdot (\frac{16 p^{4}}{4 p q^{2}})$ $((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=((2p^3q^2)/(8p^4q))*((16p^4)/(4pq^2))$

Recall: $\frac{a^{5}}{a^{3}} = a^{5} \cdot a^{- 3} = a^{5 - 3} = a^{2}$ $a^5/a^3=a^5*a^-3=a^(5-3)=a^2$

So,

$(\frac{2 p^{3} q^{2}}{8 p^{4} q}) (\frac{16 p^{4}}{4 p q^{2}})$ $((2p^3q^2)/(8p^4q))((16p^4)/(4pq^2))$ $=$ $=$ $(\frac{p^{3} q^{2}}{4 p^{4} q}) (\frac{4 p^{4}}{p q^{2}})$ $((p^3q^2)/(4p^4q))((4p^4)/(pq^2))$

The $(\frac{p^{3} q^{2}}{4 p^{4} q})$ $((p^3q^2)/(4p^4q))$ part becomes:

$(\frac{p^{3} q^{2}}{4 p^{4} q}) = (p^{3} q^{2}) (4^{- 1} p^{- 4} q^{- 1}) = 4^{- 1} p^{3 - 4} q^{2 - 1} = 4^{- 1} p^{- 1} q^{1}$ $((p^3q^2)/(4p^4q))=(p^3q^2)(4^(-1)p^(-4)q^(-1))=4^(-1)p^(3-4)q^(2-1)=4^(-1)p^(-1)q^1$

$= \frac{q}{4 p}$ $=q/(4p)$

The $(\frac{4 p^{4}}{p q^{2}})$ $((4p^4)/(pq^2))$ part becomes:

$(\frac{4 p^{4}}{p q^{2}}) = (4 p^{4}) (p^{- 1} q^{- 2}) = 4 p^{4 - 1} q^{- 2} = 4 p^{3} q^{- 2}$ $((4p^4)/(pq^2))=(4p^4)(p^(-1)q^(-2))=4p^(4-1)q^(-2)=4p^3q^(-2)$

$= \frac{4 p^{3}}{q^{2}}$ $=(4p^3)/q^2$

So, our original setup was:

$\frac{\frac{2 p^{3} q^{2}}{8 p^{4} q}}{\frac{4 p q^{2}}{16 p^{4}}} = (\frac{2 p^{3} q^{2}}{8 p^{4} q}) \cdot (\frac{16 p^{4}}{4 p q^{2}}) = (\frac{p^{3} q^{2}}{4 p^{4} q}) (\frac{4 p^{4}}{p q^{2}})$ $((2p^3q^2)/(8p^4q))/((4pq^2)/(16p^4))=((2p^3q^2)/(8p^4q))*((16p^4)/(4pq^2))=((p^3q^2)/(4p^4q))((4p^4)/(pq^2))$

$= (\frac{q}{4 p}) (\frac{4 p^{3}}{q^{2}}) = \frac{4 p^{3} q}{4 p q^{2}} = \frac{p^{2}}{q}$ $=(q/(4p))((4p^3)/q^2)=(4p^3q)/(4pq^2)=p^2/q$

Answer 2 · 2015-07-17T00:09:32

The answer is $\frac{p^{2}}{q}$ $p^2/q$ .

Explanation:

$\frac{2 p^{3} q^{2}}{8 p^{4} q}$ $(2p^3q^2)/(8p^4q)$ $\div$ $-:$ $\frac{4 p q^{2}}{16 p^{4}}$ $(4pq^2)/(16p^4)$

Reduce the numerical part of each fraction.

$(cancel2^1p^3q^2)/(cancel8^4p^4q)$ $-:$ $(cancel4^1pq^2)/(cancel16^4p^4)$ =

$(p^3q^2)/(4p^4q)$ $-:$ $(pq^2)/(4p^4)$ =

Simplify each fraction.

Use the exponent rule $x^m/x^n=x^(m-n)$ for each fraction.

$(p^3q^2)/(4p^4q)=(p^(3-4)q^(2-1))/(4)=(p^(-1)q)/4$

$(pq^2)/(4p^4)=(p^(1-4)q^2)/(4)=(p^-3q^2)/4$

Use exponent rule $x^(-m)=1/(x^m)$ for each fraction.

$(q)/(4p)-:(q^2)/(4p^3)$

In order to divide fractions, invert the second fraction to get its reciprocal, and then multiply.

$(q)/(4p)xx(4p^3)/(q^2)$ =

$(4p^3q)/(4pq^2)$ =

Cancel the $4$ .

$(cancel4^1p^3q)/(cancel4^1pq^2)$

Apply the exponent rules $x^m/x^n=x^(m-n)$ and $x^(-m)=1/(x^m)$ .

$(p^(3-1)q^(1-2))=p^2q^(-1)=(p^2)/q$

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How do you simplify $\frac{2 p^{3} q^{2}}{8 p^{4} q} \div \frac{4 p q^{2}}{16 p^{4}}$ $(2p^3q^2)/(8p^4q) div (4pq^2)/(16p^4)$ ?

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