The coordinates of the midpoint of line GH are M(−132,−6) and the coordinates of one endpoint are G(−4, 1). The coordinates of the other endpoint are

Since, the coordinates of the midpoint of line GH are M(\frac{-13}{2}, -6)(

2

−13

,−6) .

The coordinates of endpoint G are (-4,1)

We have to determine the coordinates of endpoint H.

The midpoint of the line segment joining the points (x_1, y_1)(x

1

,y

1

) and (x_2, y_2)(x

2

,y

2

) is given by the formula (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})(

2

x

1

+x

2

,

2

y

1

+y

2

) .

Here, The endpoint G is (-4,1) So, x_1 = -4 , y_1=1x

1

=−4,y

1

=1

Let the endpoint H be (x_2,y_2)(x

2

,y

2

)

The midpoint coordinate M is (\frac{-13}{2}, -6)(

2

−13

,−6) .

So, \frac{-13}{2} = \frac{-4+x_2}{2}

2

−13

=

2

−4+x

2

{-13} = {-4+x_2}−13=−4+x

2

{-13}+4 = {x_2}−13+4=x

2

{x_2}=9x

2

=9

Now, -6 = \frac{1+y_2}{2}−6=

2

1+y

2

-12 = {1+y_2}−12=1+y

2

y_2= -13y

2

=−13

So, the other endpoint H is (-9,-13).

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204