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Ivahew [28]
4 years ago
12

Please help with this problem. I keep getting -25

Mathematics
1 answer:
katrin [286]4 years ago
4 0
\bf -\sqrt[5]{-(400^2)}\implies -\sqrt[5]{-[(20\cdot 20)^2]}\implies -\sqrt[5]{-[(20^2)^2]}
\\\\\\
-\sqrt[5]{-(20^{2\cdot 2})}\implies -\sqrt[5]{-(20)^4}\implies -\sqrt[5]{-(2^2\cdot 5)^4}
\\\\\\
-\sqrt[5]{-2^{2\cdot 4}\cdot 5^4}\implies -\sqrt[5]{-2^8\cdot 5^4}\implies -\sqrt[5]{-2^{5+3}\cdot 5^4}
\\\\\\
-\sqrt[5]{-2^5\cdot 2^3\cdot 5^4}\implies 2\sqrt[5]{2^3\cdot 5^4}\implies 2\sqrt[5]{5000}
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Based on the work shown on the right, check all of the possible solutions of the equation.
Mazyrski [523]

Incomplete question. However, here is a similar question attached.

Solve 3x^2 + 17x - 6 = 0.

Based on the work shown to the left, which of these values are possible solutions of the equation? Check all of the boxes that apply

A X=-6

B X=6

C X=-1/3

D X=1/3

E X=0

Answer:

A and D

Step-by-step explanation:

Note that such question requires using completing the square method of solving equations. By using the values X= -6 and X= 1/3 we arrive at a solution.

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3 years ago
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What are u asking, the difference?

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3 years ago
Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.
____ [38]

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z lies in the second quadrant, so

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So we have

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2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}

2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}

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3 years ago
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The answer is (D) or 1728 In³
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