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2 years ago

Can someone answer 55 ÷ (9 - 4) ⋅ 8 for me please,

Mathematics

1 answer:

allochka39001 [22]2 years ago

7 0

Answer:

Step-by-step explanation:

PE M/D A/S

55/ (9-4) *8

9-4=5

55/5*8

11*8

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Answer:

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

Step-by-step explanation:

Another complex expression, let's simplify it step by step...

We'll start by re-writing 256 as 4^4

$\sqrt[3]{256 x^{10} y^{7} } = \sqrt[3]{4^{4} x^{10} y^{7} }$

Then we'll extract the 4 from the cubic root. We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.

$\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }$

Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.

$4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }$

For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.

$4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}$

The answer is then:

$4x^{3} y^{2} \sqrt[3]{4 x y} = 4x^{3} y^{2} (\sqrt[3]{4 x y})$

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