PLEASE HELP ME!! Write an equation of the line that passes through (4,−1) and is parallel to the line y=3x+7.
An equation of the parallel line is _

y = 3x-13

Step-by-step explanation:

For parallel lines, their slopes/gradients of the line are equal.

Given the parallel line, y = 3x+7 implies the slope say m1 = 3.

Hence, the slope of the new line say m2 should be 3 too.

Now, substitute the point (4,-1) into the general formula y-y1 = m2 (X-X1)

This implies, y - (-1) = 3 (X-4) = 3X-12

                     y + 1 = 3X - 12.   Subtract 1 from both sides gives y = 3x - 13

Therefore, the equation of the new parallel line is y = 3x - 13

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3