What is the slope of a line perpendicular to the line whose equation is 12x-9y=135

Mathematics

1 answer:

trasher [3.6K]3 years ago

7 0

Answer:

$\frac{-9}{12}$

Step-by-step explanation:

First, we need to isolate y.

1. Subtract 12x from both sides.

-9y=-12x+135

2. Divide by -9 in order to isolate y.

$y=\frac{12}{9} x-15$

3. Understand that perpendicular equations' slopes are opposite and reciprocal to the original equation. The slope of the original equation is $\frac{12}{9}$ , so the reciprocal and opposite (flip and change the sign) slope would be $\frac{-9}{12}$ .

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2x^{3}+30xx^{2} +72x=0 solve using zero product property

Goshia [24]

Answer:

Values of x are: x=0 or x=-3 or x=-12

Step-by-step explanation:

The equation given to solve using zero product property is $2x^3+30x^2+72x=0$

Zero property rule states that if ab=0 then a=0 or b=0

Taking x common from the equation:

$2x^3+30x^2+72x=0\\x(2x^2+30x+72)=0\\$

Applying zero product rule

$x(2x^2+30x+72)=0\\x=0 \ or \ 2x^2+30x+72=0$

Now, solving $2x^2+30x+72=0$

Using quadratic formula to find value of x

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Putting values

$$x=\frac{-30\pm\sqrt{(30)^2-4(2)(72)}}{2(2)}$\\$x=\frac{-30\pm\sqrt{324}}{4}$\\$x=\frac{-30\pm18}{4}$\\$x=\frac{-30+18}{4} \ or \ x=\frac{-30-18}{4}$ \\x=-3 \ or \ x=-12$