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Sophia factored
81y^(6) as

(9y^(3))(9y^(2))". "
Ahmed factored
81y^(6) as
(3y^(6))(27 y).
Which of them factored
81y^(6) correctly?
Choose 1 answer:
(A) Only Sophia
(B) Only Ahmed
(c) Both Sophia and Ahmed
(D) Neither Sophia nor Ahmed

Sophia factored $81 y^{6}$ as $\newline$ $\left(9 y^{3}\right)\left(9 y^{2}\right) \text {. }$ $\newline$ Ahmed factored $81 y^{6}$ as $\left(3 y^{6}\right)(27 y)$ . $\newline$ Which of them factored $81 y^{6}$ correctly? $\newline$ Choose $1$ answer: $\newline$ (A) Only Sophia $\newline$ (B) Only Ahmed $\newline$ (C) Both Sophia and Ahmed $\newline$ (D) Neither Sophia nor Ahmed

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Quantities that combine to zero: word problems

Full solution

Q. Sophia factored

81 y^{6}

as

\newline

\left(9 y^{3}\right)\left(9 y^{2}\right) \text {. }

\newline

Ahmed factored

81 y^{6}

as

\left(3 y^{6}\right)(27 y)

.

\newline

Which of them factored

81 y^{6}

correctly?

\newline

Choose

1

answer:

\newline

(A) Only Sophia

\newline

(B) Only Ahmed

\newline

(C) Both Sophia and Ahmed

\newline

(D) Neither Sophia nor Ahmed

Sophia's factorization: Sophia's factorization of $81y^{6}$ is $(9y^{3})(9y^{2})$ . Let's check if this is correct by multiplying the factors together. $\newline$ $(9y^{3})(9y^{2}) = 81y^{3+2} = 81y^{5}$
Checking Sophia's factorization: Ahmed's factorization of $81y^{6}$ is $(3y^{6})(27y)$ . Let's check if this is correct by multiplying the factors together. $\newline$ $(3y^{6})(27y) = 81y^{6+1} = 81y^{7}$

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