Select all the expressions that are equivalent to 5^5 * 8^5. Multi-select Choices: (A)40^5 (B)1/40^5 (C)1/40^-5 (D)40^25

Simplify using property of exponents: We need to find expressions equivalent to $5^5 \times 8^5$. We can start by simplifying the expression using the property of exponents that states $a^m \times b^m = (a \times b)^m$. So, $5^5 \times 8^5 = (5 \times 8)^5 = 40^5$.Compare with choices: Now let's compare the given choices with our simplified expression $40^5$. (A) $40^5$ is exactly what we found, so this is equivalent. (B) $1/40^5$ is the reciprocal of $40^5$, so it is not equivalent. (C) $1/40^{-5}$ is the reciprocal of $40^{-5}$, which is $40^5$, so this is equivalent. (D) $40^{25}$ is not equivalent because it represents $40$ raised to the 25th power, not the 5th power.

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Select all the expressions that are equivalent to $5^5 \times 8^5$ . $\newline$ Multi-select Choices: $\newline$ (A) $40^5$ $\newline$ (B) $\frac{1}{40^5}$ $\newline$ (C) $\frac{1}{40^{-5}}$ $\newline$ (D) $40^{25}$

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Math Problems

Algebra 2

Find trigonometric ratios using reference angles

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Q. Select all the expressions that are equivalent to

5^5 \times 8^5

\newline

Multi-select Choices:

\newline

(A)

40^5

\newline

(B)

\frac{1}{40^5}

\newline

(C)

\frac{1}{40^{-5}}

\newline

(D)

40^{25}

Simplify using property of exponents: We need to find expressions equivalent to $5^5 \times 8^5$ . We can start by simplifying the expression using the property of exponents that states $a^m \times b^m = (a \times b)^m$ . $\newline$ So, $5^5 \times 8^5 = (5 \times 8)^5 = 40^5$ .
Compare with choices: Now let's compare the given choices with our simplified expression $40^5$ . $\newline$ (A) $40^5$ is exactly what we found, so this is equivalent. $\newline$ (B) $1/40^5$ is the reciprocal of $40^5$ , so it is not equivalent. $\newline$ (C) $1/40^{-5}$ is the reciprocal of $40^{-5}$ , which is $40^5$ , so this is equivalent. $\newline$ (D) $40^{25}$ is not equivalent because it represents $40$ raised to the $25$ th power, not the $5$ th power.