What is an equation of the line that is perpendicular to 3x+y=−5 and passes through the point (3, −7) ?

y = x - 8

Step-by-step explanation:

the equation of a line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 3x + y = - 5 into this form

subtract 3x from both sides

y = - 3x - 5 ← in slope-intercept form

with slope m = - 3

given a line with slope m then the slope of a lin e perpendicular to it is

= - , hence

= - =

y = x + c ← is the partial equation

to find c substitute (3, - 7 ) into the partial equation

- 7 = 1 + c ⇒ c = - 7 - 1 = - 8

y = x - 8 ← equation of perpendicular line

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.