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BaLLatris [955]
3 years ago
6

Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Su

perscript 5 )squared (8 n Superscript 4 Baseline q )Superscript negative 2 Over (negative 3 nq Superscript 9 )Superscript negative 1 Baseline (4 n cubed q Superscript 9 )cubed EndFraction StartFraction
Mathematics
1 answer:
Alja [10]3 years ago
4 0

Answer:

\frac{ - 3}{ 256  {q}^{10} {n}^{8}  }

Step by step explanation:

\frac{ {(4 {n}^{4} {q}^{5})}^{2}  {(8 {n}^{4} q)}^{-2} }{  {(- 3 {nq}^{9})}^{ - 1}   {(4 {n}^{3} {q}^{9})  }^{3} }

first we will change the terms with negative superscrips to the other side of the fraction

\frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} }

then we will distribute the superscripts

\frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2}  {q}^{2} }

\frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8}  {q}^{2} }

as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents

{4}^{2 - 3}  {q}^{10  + 9 - 2 - 27}  {n}^{8 + 1 - 8 - 9}  {8}^{ - 2}  { (- 3)}^{1}

{4}^{ - 1}  {q}^{ - 10}  {n}^{ - 8}  {8}^{ - 2}  { (- 3)}^{1}

then we will change again the terms with negative superscrips to the other side of the fraction

\frac{ - 3}{ 4 \times  {8}^{2}  {q}^{10} {n}^{8}  }

\frac{ - 3}{ 256  {q}^{10} {n}^{8}  }

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