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IrinaK [193]
3 years ago
12

HELP Please don’t answer if you don’t know it.

Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
6 0

Answer: d= 5

Step-by-step explanation:

I did this test on edgunity if you need a step by step I can leave it in the comments below. Good luck

inysia [295]3 years ago
4 0

Answer:

d=5

Step-by-step explanation:

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Step-by-step explanation:

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How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
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\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

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Read 2 more answers
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