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Whitepunk [10]
3 years ago
15

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

Mathematics
1 answer:
jonny [76]3 years ago
4 0

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

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