Solve for x under the assumption that x > 0. x-36/x <-9

Mathematics

1 answer:

ycow [4]3 years ago

4 0

$\dfrac{x-36}{x}$
as $x>0$
$x-36$
$x+9x$
$10x$
$x$
$\boxed{x$

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First step is to simplify the inside of the radical. There are x's on both top and bottom of the fraction. So we subtract their exponents. x³ - x¹ = x². Because the x³ was in the denominator and it was bigger, the x² goes on the bottom.

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$\sqrt{\frac{126y^{5}}{32x^{2}} }$

Now, I'd suggest factoring each term to correspond with the square root.

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