3 years ago

HELP PLSSS THIS IS HARD SOMEONE

Mathematics

1 answer:

Rashid [163]3 years ago

5 0

Answer:

Scale factor = 3

Step-by-step explanation:

A = (4,8) and A’ = (12,24)

4x = 12

X = 12/4

X = 3

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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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3 years ago

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