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Reika [66]
2 years ago
13

I need help SHOW YOUR WORK

Mathematics
2 answers:
laila [671]2 years ago
6 0
<h2>Answer:</h2>

\sf\:SA \:  = 4\pi {( {5x}^{4}y {z}^{3})}^{2}.

\sf\:SA \:   = 4\pi25 {x}^{8}{y}^{2} {z}^{6}.

\large\boxed{\sf\:SA \:  = 100\pi{x}^{8}{y}^{2}{z}^{6}.}

GenaCL600 [577]2 years ago
3 0

Answer:

S=100x^8y²z^6 is your answer

Step-by-step explanation:

we have

S=4πr²

when

r=5x⁴yz³

new S

will be

S=4π(5x⁴yz³)²=4×5²×x^(4×2)×y²×z^(3×2)

=100x^8y²z^6

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Answer:

a.(4)

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Which of the following are square roots of —8 + 8i/3? Check all that apply.
8090 [49]

Answer:

Options (2) and (3)

Step-by-step explanation:

Let, \sqrt{-8+8i\sqrt{3}}=(a+bi)

(\sqrt{-8+8i\sqrt{3}})^2=(a+bi)^2

-8 + 8i√3 = a² + b²i² + 2abi

-8 + 8i√3 = a² - b² + 2abi

By comparing both the sides of the equation,

a² - b² = -8 -------(1)

2ab = 8√3

ab = 4√3 ----------(2)

a = \frac{4\sqrt{3}}{b}

By substituting the value of a in equation (1),

(\frac{4\sqrt{3}}{b})^2-b^2=-8

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                     b = ± 2i

b² - 12 = 0 ⇒ b = ±2√3

Since, a = \frac{4\sqrt{3}}{b}

For b = ±2i,

a = \frac{4\sqrt{3}}{\pm2i}

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For b = ±2√3

a = \frac{4\sqrt{3}}{\pm 2\sqrt{3}}

a = ±2

Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)

Options (2) and (3) are the correct options.

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