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Elanso [62]
3 years ago
8

Find the equation of the line parallel to y=-11-5, that passes through the point

Mathematics
2 answers:
Elanso [62]3 years ago
7 0

Answer:

y=41−11x

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=5−11x.

The slope of the parallel line is the same: m=−11.

So, the equation of the parallel line is y=−11x+a.

To find a, we use the fact that the line should pass through the given point: −3=(−11)⋅(4)+a.

Thus, a=41.

Therefore, the equation of the line is y=41−11x.

RSB [31]3 years ago
3 0

Answer:

y=41−11x

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=5−11x.

The slope of the parallel line is the same: m=−11.

So, the equation of the parallel line is y=−11x+a.

To find a, we use the fact that the line should pass through the given point: −3=(−11)⋅(4)+a.

Thus, a=41.

Therefore, the equation of the line is y=41−11x.

Answer: y=41−11x.

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WILL MARK BRAINLIEST
sp2606 [1]

To make this new triangle I am going to pretend that the original triangle was at the points (4,4) (-4,4) and (0,8). So if this was dilated down by 1/2 then the triangles points would be (2,2) (-2,2) and (0,4). Then if it is reflected over the x-axis the new points would be (2,-2) (-2,-2) and (0,-4) and if it was translated left by 2 units then the points would be (0,-2) (-4,-2) and (-2, -4), then if the triangle was translated up by 4 units the points and the new triangle would be (0,2) (-4,2) and (-2,0). And that would be what the new triangle would be.

The triangle went from (4,4) (-4,4) and (0,8) to (0,2) (-4,2) and (-2,0).

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