Someone please HELP!?

Mathematics

1 answer:

jeka942 years ago

7 0

A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side.

For example

$x^{-2}=\frac{1}{x^2}$

or

$( \frac{2}{5} )^{-4}=( \frac{5}{2} )^4$

***************************************
$(8r^{-5})^{-3}$

$(8* \frac{1}{r^5} )^{-3} \\\\( \frac{8}{r^5} )^{-3}
 \\ \\ ( \frac{r^5}{8} )^3
\\\\ \mathrm{Apply\:exponent\:rule}:\quad *\left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} \ \ \ \ \ \ \ \ \ \ \ \ *(a^b)^c=a^{bc}
 \\\\( \frac{r^{15}}{8^3} )
\\\\( \frac{r^{15}}{512} )$

The answer is "D"

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If the numbers 4, 5 and 6 are each used exactly once to replace the letters in the expression A ( B − C ), what is the least pos

guapka [62]

Answer:

The least possible result is <em>-10</em>.

Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

First of all,

Let A = 4, B = 5 and C = 6

$A(B-C) = 4 \times (5-6) = 4 \times -1 = -4$

Let A = 4, B = 6 and C = 5

$A(B-C) = 4 \times (6-5) = 4 \times 1 = 4$

Let A = 5, B =4 and C = 6

$A(B-C) = 5 \times (4-6) = 5 \times -2 = -10$

Let A = 5, B = 6 and C = 4

$A(B-C) = 4 \times (6-4) = 4 \times 2 = 8$

Let A = 6, B = 4 and C = 5

$A(B-C) = 6 \times (4-5) = 6 \times -1 = -6$

Let A = 6, B = 5 and C = 4

$A(B-C) = 6 \times (6-5) = 6 \times 1 = 6$

Summarizing the above values in the form of a table:

$\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\ 4 & 6 & 5 & 4\\ 5 & 4 & 6 & -10\\ 5 & 6 & 4 & 10\\ 6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}$