Question

simplify. express your answer using positive exponents.
$\frac{7c^{8}d^{0}}{c^{8}d^{-1}}$

Explanation:

Step1: Use the quotient - rule of exponents

For the same base $a$, $\frac{a^m}{a^n}=a^{m - n}$. Also, recall that any non - zero number to the power of 0 is 1, i.e., $a^0 = 1$ ($a
eq0$). Here, $d^0 = 1$. The expression becomes $\frac{7c^{8}\times1}{c^{8}d^{-1}}=\frac{7c^{8}}{c^{8}d^{-1}}$.

Step2: Simplify the $c$ terms

Using the quotient - rule $\frac{c^{8}}{c^{8}}=c^{8 - 8}=c^{0}=1$. So the expression is now $\frac{7}{d^{-1}}$.

Step3: Rewrite with positive exponent

Using the rule $a^{-n}=\frac{1}{a^{n}}$, then $\frac{7}{d^{-1}} = 7d^{1}=7d$.

Answer:

$7d$