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Liono4ka [1.6K]
4 years ago
8

Need help pls someone

Mathematics
1 answer:
dimaraw [331]4 years ago
7 0

\sqrt[3]{250a^{11}}=\sqrt[3]{125\cdot2a^{2+9}}=\sqrt[3]{125}\cdot\sqrt[3]{2a^2\cdot a^9}=5\sqrt[3]{2a^2}\cdot\sqrt[3]{a^9}\\\\=5\sqrt[3]{2a^2}\cdot\sqrt[3]{a^{3\cdot3}}=5\sqrt[3]{2a^2}\cdot\sqrt[3]{(a^3)^3}=5\sqrt[3]{2^2}\cdot a^3\\\\=\boxed{5a^3\sqrt[3]{2a^2}}

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Step-by-step explanation:

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4 0
3 years ago
OQ is a diagonal of quadrilateral NOPQ. If NO=PQ, NQO=7y-37, and POQ=2y+93, find POQ such that NOPQ is a parallelogram. A.26° B.
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Step-by-step explanation:

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