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what is an equation of the line that is perpendicular to 3x+y=−5 and passes through the point (3, −7) ?
enter your equation in the box.

y = 1/3x -8

Step-by-step explanation:

3x+y=−5

First we need to get the equation in slope intercept form to find the slope

Subtract 3x from each side

3x-3x+y=-3x−5

y = -3x-5

The slope is -3

We want a line perpendicular.  Perpendicualr lines have slopes that are the negative reciprocal.  Take the negative reciprocal

m new = - (1/-3) = 1/3

The slope of the new line is 1/3

We have the slope (1/3) and a point (3,-7)

We can use the point slope form for the equation of a line

y-y1 = m(x-x1)

y--7 = 1/3 (x-3)

y+7 = 1/3 (x-3)

or if we want the slope intercept form

Distribute

y+7 = 1/3x -1

Subtract 7 from each side

y+7-7 = 1/3x -1-7

y = 1/3x -8

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

d is the answer

step-by-step explanation: