Write the equation of the line perpendicular to 5y - 5x = -20 that passes through the point (-3, 8)

y = 5x - 8 is in slope intercept format:

y = mx + b where m is slope, b is y-intercept.

Slope is 5

Perpendicular lnes have slopes that are

negative reciprocals... mp = -1/m

You need equation of line through (2, 2)

with slope -1/5

y = (-1/5)x + b where (2,2) is on the line.

2 = (-1/5)(2) + b

2 = -2/5 + b

b = 2 2/5 or 2.4 or 12/5

Your equation is: y = (-1/5)x + 12/5

This can be reworked to standard form or left

as is depending on your needs.

5y = -x + 12

x + 5y = 12

x+y-5=0

Step-by-step explanation:

let eqn Perpendicular to eqn 5y-5x=-20 be -5x-5y+k=0(1)

eqn (1) passes through (-3,8)

-5×-3-5×8+k=0

15-40+k=0

K=25

since required eqn perpendicular to 5y-5x=-20 is -5x-5y+25=0

x+y-5=0ans.

3x + 18 = 147°

step-by-step explanation:

3x + 18 = 147°, because the value of the central angle is equal to the value of the arc formed by it.

i hope i you.

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step-by-step explanation: