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3 years ago

Help can't figure this problem out

Mathematics

1 answer:

Darya [45]3 years ago

8 0

(x^1/2 y^-2/3)^-6
= x^(1/2*-6) y^[(-2/3) *(-6)]
= x^-3 y ^4
= y^4 / x^3

answer is D. y^4 / x^3

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To solve this question, we use the factor theorem, and using it, the polynomial function is:

$f(x) = x^3 - 2x^2 - 3x + 6$

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The factor theorem means that if k is a root of f(x), f(k) = 0.

Thus, applying the factor theorem for this question, we have to choose the function for which: $f(2) = 0, f(\sqrt{3}) = 0, f(-\sqrt{3}) = 0$

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Function 1:

$f(x) = x^3 - 2x^2 - 3x + 6$

Testing the values:

$f(2) = 2^3 - 2(2)^2 - 3(2) + 6 = 8 - 8 - 6 + 6 = 0$

$f(\sqrt{3}) = \sqrt{3^3} - 2(\sqrt{3})^2 - 3\sqrt{3} + 6 = \sqrt{3^2\times3} - 6 - 3\sqrt{3} + 6 = 3\sqrt{3} - 3\sqrt{3} = 0$

$f(-\sqrt{3}) = -\sqrt{3^3} - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 = -\sqrt{3^2\times3} - 6 + 3\sqrt{3} + 6 = -3\sqrt{3} + 3\sqrt{3} = 0$

Thus, since all three conditions are satisfied, $f(x) = x^3 - 2x^2 - 3x + 6$ is the polynomial function.

A similar question is given at brainly.com/question/11378552

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