Question

Plz help I'll give brainy!!!

Answer 1

Answer:

see below

Step-by-step explanation:

we are given a exponential function

$\displaystyle y = \left(1/5\right) ^{x}$

we want to figure out the missing values of y for x i.e (-2,-1, and 2) to do we can consider substituting so,

when x is -2 y should be

$\displaystyle y = \left(1/5\right) ^{ - 2}$

rewrite ⅕ in exponent notation:

$\displaystyle y = \left((5) ^{ - 1} \right) ^{ - 2}$

by law of exponent we obtain:

$\displaystyle y = (5) ^{ 2}$

simplify square:

$\displaystyle y = \boxed{ 25}$

when x is -1:

$\displaystyle y = \left(1/5\right) ^{ - 1}$

rewrite ⅕ in exponent notation:

$\displaystyle y = \left((5) ^{ - 1} \right) ^{ - 1}$

use law of exponent which yields:

$\displaystyle y = \boxed5$

when x is 3:

$\displaystyle y = \left(1/5\right) ^{ 3}$

by law of exponent we obtain:

$\displaystyle y = 1/ {5}^{3}$

simplify square which yields:

$\displaystyle y = \frac{1}{ \boxed{125}}$