For each linear equation write the slope of a line perpendicular to the given line. 1) y=-2/7x + 5 2) 3x + 5y = –15

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The slope of a perpendicular line is always the negative reciprocal of the original.

QUESTION 1

For the first one, we have a slope of -2/7. To find the reciprocal you flip the numerator and denominator, giving us -7/2. Then, we want the negative reciprocal, so since two negatives make a positive our perpendicular slope is 7/2.

QUESTION 2

Let's first rearrange our equation to be in slope intercept form.

3x+5y=-15

Subtract 3x from both sides.

5y=-3x-15

Divide both sides by 5.

y=-3/5x-3

First we find the reciprocal of -3/5, which is -5/3. This is the normal reciprocal, but we want the negative one, and since two negatives makes a positive, our slope is 5/3.

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answer: your gay

step-by-step explanation:

you dont know math

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15