Question

5)

1. $\frac{-8^{-2}}{3^{-4}}$
2. $-9x^{0}y^{-3}$
3. $8q^{-2}w^{3}p^{0}$
4. $\frac{10^{0}r^{-11}s}{3^{2}}$

Explanation:

Step1: Simplify problem 5

Any non-zero number to the power of 0 is 1, and $a^{-n}=\frac{1}{a^n}$.
$\frac{6^{-1}}{5^{0}}=\frac{\frac{1}{6}}{1}=\frac{1}{6}$

Step2: Simplify problem 6

Calculate numerator and denominator first.
Numerator: $-8-2=-10$
Denominator: $3-4=-1$
$\frac{-10}{-1}=10$

Step3: Simplify problem 7

Use $x^0=1$ and $y^{-3}=\frac{1}{y^3}$.
$-9x^{0}y^{-3}=-9\times1\times\frac{1}{y^3}=-\frac{9}{y^3}$

Step4: Simplify problem 8

Use $q^{-2}=\frac{1}{q^2}$ and $p^0=1$.
$8q^{-2}w^{3}p^{0}=8\times\frac{1}{q^2}\times w^3\times1=\frac{8w^3}{q^2}$

Step5: Simplify problem 9

Use $10^0=1$, $r^{-11}=\frac{1}{r^{11}}$, and $3^2=9$.
$\frac{10^{0}r^{-11}s}{3^{2}}=\frac{1\times\frac{1}{r^{11}}\times s}{9}=\frac{s}{9r^{11}}$

Answer:

1. $\frac{1}{6}$
2. $10$
3. $-\frac{9}{y^3}$
4. $\frac{8w^3}{q^2}$
5. $\frac{s}{9r^{11}}$