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what is the equation of the line that is parallel to y−5=−13(x+2) and passes through the point (6,−1)?
the equation will be in slope-intercept form.

y=x3+193
y=−x3−1
y=−x3−173
y=−x3+1
y=x3+1

So our equation becomes

for some reason, none of the equations match but you can see this link to show that this answer is correct.

https://www.symbolab.com/solver/parallel-line-calculator/parallel%20y-5%20%3D-13%5Cleft(x%2B2%5Cright)%20%5Cleft(6%2C-1%5Cright)

Step-by-step explanation:

So we have the equation in point-slope form :

y−5=−13(x+2)

So simplifying this to y= mx +b we get:

So now we know that the equation is parallel to the equation we need to find.

Let's call this equation f(x) and the parallel equation as g(x)

So f(x) = -13x -21

this is in the form

y= mx +b

m = - 13

b = - 21

g(x) = ?

Now we know that the equation is parallel this means that the slope is the same. Notice how I broke down the equation above. In that case m means slope. Which means that the parallel function has a slope of -13.

So we can input this in the y = mx + b form as

y = -13x +b

To find b which is the y-intercept, we use the point that we were given (6,-1)

So x = 6 and y = - 1

So we solve for b  using x and y:

So our equation becomes

answer: b.48 for

step-by-step explanation:

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right