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NemiM [27]
3 years ago
15

Which expression is equivalent to 3 sqrt32x8y10?

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
6 0

Answer: 2x^{2} y^{3} (\sqrt[3]{4x^{2} y} )

Step-by-step explanation:

\sqrt[3]{32x^{8} y^{10} } =\sqrt[3]{2^{3} \cdot 2^{2}  \cdot x^{2} \cdot (x^{2} )^{3} \cdot y \cdot (y^{3})^{3}  } =2x^{2} y^{3} (\sqrt[3]{4x^{2} y} )

brilliants [131]3 years ago
6 0

Answer:

Option C :

               \sqrt[3]{32x^8y^10} = 2x^2 y^3 \sqrt[3]{4x^2 y}

Step-by-step explanation:

\sqrt[3]{32x^8y^{10}}}  = ( 32 x^8 y^{10})^{\frac{1}{3}}

              = ( 2^5 \times x^8  \times y^{10})^{\frac{1}{3}}\\\\= ( 2^{5\times\frac{1}{3}} \times x^{8 \times \frac{1}{3}} \times y^{10 \times \frac{1}{3}})\\\\=(2^{\frac{3}{3} + \frac{2}{3}}} \times x^{\frac{6}{3} + \frac{2}{3}} \times y^{\frac{9}{3} + \frac{1}{3}})\\\\= 2 \times 2^{\frac{2}{3}} x^2 \times x^{\frac{2}{3}} \times y^3 \times y^\frac{1}{3}\\\\=2x^2 y^3  \times 2^{\frac{2}{3} \times }x^{\frac{2}{3}} \times y^{\frac{1}{3}}\\\\=2x^2 y^3 (2^2x^2 y)^{\frac{1}{3}}\\\\= 2x^2y^3 \ \sqrt[3]{ \ 4 x^2 \ y }

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VARVARA [1.3K]

Hey there! :)

Answer:

g^{-1} (g(10))=10

Step-by-step explanation:

Begin by calculating g(10):

g(x) = 3x - 6

Substitute in 10 for x:

g(10) = 3(10) - 6

g(10) = 30-6

g(10) = 24.

Plug '24' into 'x' into g^{-1} (x)

g^{-1} (24)=\frac{(24) + 6}{3}

Simplify:

g^{-1} (24)=\frac{30}{3}

g^{-1} (24)=10

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3 years ago
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Step-by-step explanation:

Replace "z" with "2":

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3 years ago
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Answer:

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Step-by-step explanation:

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