2 years ago

Look for a pattern in the first three equations

Mathematics

2 answers:

lbvjy [14]2 years ago

5 0

Answer:

Where?

Step-by-step explanation:

tia_tia [17]2 years ago

4 0

Answer:

<em>Is there a graph orrrrr?</em>

Step-by-step explanation:

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Which expression is equivalent

gtnhenbr [62]

Answer:

Option B is correct.

$\frac{81m^2n^5}{8}$ is equivalent to $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$

Step-by-step explanation:

Given expression: $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$

Using exponents power:

$(ab)^n = a^nb^n$

$(a^n)^m = a^{nm}$

$a^m \cdot a^n = a^{m+n}$

Given: $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$

Apply exponent power :

⇒ $\frac{3^4 (m^{-1})^4(n^2)^4}{2^3(m^{-2})^3 n^3}$

⇒ $\frac{81 m^{-4}n^8}{8m^{-6}n^3} = \frac{81 m^{-4} \cdot m^6 n^8 \cdot n^{-3}}{8}$

⇒ $\frac{81 m^{-4+6} n^{8-3}}{8} = \frac{81 m^2 n^5}{8} = \frac{81m^2 n^5}{8}$

Therefore, the expression which is equivalent to $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$ is, $\frac{81m^2 n^5}{8}$

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3 years ago

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