Question

PLEEASSEEEE Answer ASAP. Due in 5 Minutes. 25 points!!!!!!!!!

Answer 1

Answer:

Here we will use the relationships:

$a^n*a^m = a^{n + m}$

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

$\frac{a^n}{a^m} = a^{n - m}$

And a number:

$a^n$

is between 0 and 1 if a is positive and larger than 1, and n is negative.

if a is positive and 0 < a < 1, then we need to have n positive such that:

0 < a^n < 1

A) $5^3*5^{-4} = 5^{3 + (-4)} = 5^{-1}$

This is between zero and 1,

B) $\frac{3^5}{3^{-6}} = 3^{5 - (-6)} = 3^{11}$

This is greater than 1, because the exponent is positive.

C) $(\frac{1}{4})^3*( \frac{1}{4})^2 = (\frac{1}{4})^{2 + 3} = (\frac{1}{4})^5$

Because a is smaller than 1, and the exponent is positive, then the expression is between 0 and 1.

D) $\frac{(-7)^5}{(-7)^7} = (-7)^{5 - 7} = -7^{-2}$

The exponent is negative (and pair) then the expression is between 0 and 1.

Remember that when the exponent is pair, we always have that:

(-N)^m = (N)^m

So (-7)^-2 = 7^-2