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2 years ago

(R^3)^-3 what is theAnswer

Mathematics

1 answer:

Nastasia [14]2 years ago

3 0

$(r^{3})^{-3}$
$\frac{1}{(r^{3})^{3}}$

$\frac{1}{(r^{3})(r^{3})(r^{3})}$

$\frac{1}{r^{9}}$

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The resulting equation after simplyfing the given equation is 3x^2 + 20x + 12 = 0 and whoose solutions are x = -2/3 or x = -6.

According to the given question.

We have an equation

$\frac{x^{2}-x-6 }{x^{2} } = \frac{x-6}{2x} +\frac{2x+12}{x}$

So, to find the resulting equation of the above equation we need to simplify.

First we will take LCD

$\frac{x^{2} -x - 6 }{x^{2} } = \frac{x -6+2(2x + 12)}{2x}$

$\implies \frac{x^{2}-x-6 }{x^{2} } =\frac{x-6+4x+24}{2x}$

$\implies \frac{x^{2}-x-6 }{x^{2} } = \frac{5x +18}{2x}$

Multiply both the sides by x.

$\frac{x^{2}-x-6 }{x} = \frac{5x+18}{2}$

Again multiply both the sides by x

$2x^{2} -2x-12 = 5x^{2} +18x$

$\implies 5x^{2} -2x^{2} +18x +2x +12 = 0$

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Factorize the above equation

⇒3x(x+6)+2(x+6) = 0

⇒(3x + 2)(x+6) = 0

⇒ x = -2/3 or x = -6

Hence, the resulting equation after simplyfing the given equation is 3x^2 + 20x + 12 = 0 and whoose solutions are x = -2/3 or x = -6.

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