How do you evaluate $\frac{{(2 p^{3} q^{4})}^{4}}{{(- 3 q^{5})}^{2}} \div \frac{{(4 p^{2} q)}^{2}}{9} = \frac{p^{a + b}}{q^{a - b}}$ $\frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}$ ?

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How do you evaluate $\frac{{(2 p^{3} q^{4})}^{4}}{{(- 3 q^{5})}^{2}} \div \frac{{(4 p^{2} q)}^{2}}{9} = \frac{p^{a + b}}{q^{a - b}}$ $\frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}$ ?

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Answer 1 · 2017-12-15T14:20:07

$b = 6$ $b=6$
$a = 2$ $a=2$

$\frac{p^{a + b}}{q^{a - b}} = \frac{p^{2 + 6}}{q^{2 - 6}} = \frac{p^{8}}{q^{- 4}}$ $(p^(a+b))/(q^(a-b))=(p^(2+6))/(q^(2-6))=(p^(8))/(q^(-4))$

Explanation:

$\frac{{(2 p^{3} q^{4})}^{4}}{{(- 3 q^{5})}^{2}} \div \frac{{(4 p^{2} q)}^{2}}{9} = \frac{p^{a + b}}{q^{a - b}}$ $((2p^3q^4)^4)/((-3q^5)^2)-:((4p^2q)^2)/9=(p^(a+b))/(q^(a-b))$

$\frac{2^{4} \cdot p^{(3 \cdot 4)} \cdot q^{(4 \cdot 4)}}{{(- 3)}^{2} \cdot q^{(5 \cdot 2)}} \cdot \frac{9}{4^{2} \cdot p^{(2 \cdot 2)} \cdot q^{2}}$ $(2^4*p^((3*4))*q^((4*4)) )/((-3)^2*q^((5*2)))*9/(4^2*p^((2*2))*q^2)$

$(cancel16*p^(12)*q^(16) )/(cancel9*q^(10))*cancel9/(cancel16*p^(4)*q^2)$

$(p^(12)*q^(16) )/(q^(10))*1/(p^(4)*q^2)$

$(p^(12)*q^(16) )/(q^(10)*p^(4)*q^2)$

$(p^(12)*q^(16) )/(q^((10+2))*p^(4))$

$(p^((12-4)))/(q^((12-16)))=(p^(8))/(q^(-4))$

$a+b=8$
$a-b=-4quad=>quada=b-4$

$b-4+b=8quad=>quad2b=12quad=>quadb=6$
$a=b-4quad=>quada=6-4quad=>quada=2$

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How do you evaluate $\frac{{(2 p^{3} q^{4})}^{4}}{{(- 3 q^{5})}^{2}} \div \frac{{(4 p^{2} q)}^{2}}{9} = \frac{p^{a + b}}{q^{a - b}}$ $\frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}$ ?

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Explanation:

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How do you evaluate \frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}(2p3q4)4(−3q5)2÷(4p2q)29=pa+bqa−b\frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}?

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How do you evaluate $\frac{{(2 p^{3} q^{4})}^{4}}{{(- 3 q^{5})}^{2}} \div \frac{{(4 p^{2} q)}^{2}}{9} = \frac{p^{a + b}}{q^{a - b}}$ $\frac{(2p^{3}q^{4})^{4}}{(-3q^{5})^{2}}\div \frac{(4p^{2}q)^{2}}{9}=\frac{p^{a+b}}{q^{a-b}}$ ?