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If [tex]10^{2y} = 25[/tex], then [tex]10^{-y} = ?[/tex]

A. [tex]-\frac{1}{5}[/tex]
B. [tex]\frac{1}{5}[/tex]
C. [tex]\frac{1}{25}[/tex]
D. [tex]-\frac{1}{25}[/tex]

Answer :

Answer:

b. 1/5

Step-by-step explanation:

[tex] {10}^{2y} = 25[/tex]

[tex] { ({10}^{y}) }^{2} = {5}^{2} [/tex]

SQUARE ROOT ON BOTH SIDES

[tex] {10}^{y} = 5[/tex]

RECIPROCAL ON BOTH SIDES

[tex] \frac{1}{ {10}^{y} } = \frac{1}{5} [/tex]

[tex] {10}^{ - y} = \frac{1}{5} [/tex]

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Rewritten by : Batagu

Answer:

y=.04 Does not change just because power of does, and sum does. Answer C

Step-by-step explanation:

10^2y=25

100y=25

100 x .04=25

10^-2y=-25