Question

Please help with my questions, it would really help me out! (The numbers on the top are exponents" Please help!!!

Answer 1

Problem 1

Answer: $\frac{q^4}{p^2}$

This is the fraction of $q^4$ over top $p^2$

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Explanation:

The $r^0$ term becomes 1 since raising any nonzero value to the 0 exponent is 1.

So $r^0 = 1$ as long as r is nonzero.

The negative exponent over the variable q will mean we apply the reciprocal to make that exponent positive. Meaning that $\frac{1}{q^{-4}} = \frac{q^4}{1} = q^4$

The $p^2$ term stays as it is.

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Problem 2

Answer:  4

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Explanation:

$2^{-4}*2^{6} = 2^{-4+6} = 2^2 = \boldsymbol{4}$

The exponent rule used here is $a^b*a^c = a^{b+c}$. You add the exponents together when multiplying exponential expressions of the same base.

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Problem 3

Answer: $\frac{t^{11}}{27m^2}$

This is one single fraction with $t^{11}$ over top $27m^2$

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Explanation:

The rule we use is  $a^b \div a^c = a^{b-c}$  which is similar to the last rule, but now we're dividing the terms and subtracting the exponents. The bases must be the same.

With that rule in mind we can say:

$\frac{3^2}{3^5} = 3^{2-5} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27}$

$\frac{m^5}{m^7} = m^{5-7} = m^{-2} = \frac{1}{m^2}$

$\frac{t^6}{t^{-5}} = t^{6-(-5)} = t^{6+5} = t^{11}$

which when put together forms this

$\frac{t^{11}}{27m^2}$

In other words,

$\frac{3^2m^5t^6}{3^5m^7t^{-5}} = \frac{t^{11}}{27m^2} \ \text{ where } \ m \ne 0, \ t \ne 0$