First step is Add 3 to both sides
hope it helps
The coordinates of point K(7, 0).
Given, JK has midpoint M(7, 2).
And, the coordinates of point J(7, 4).
We have to find the coordinates of point K.
As M is the midpoint, therefore by using the midpoint formula.
(xₙ + yₙ) = (x₁ + x₂/2 , y₁ + y₂/2)
Using x-coordinates,
xₙ = x₁ + x₂/2
7 = x₁ + 7/2
14 = x₁ + 7
or x₁ = 7
Nos using y-coordinates,
yₙ = y₁ + y₂/2
2 = y₁ + 4/2
4 = y₁ + 4
y₁ = 0
Therefore, the coordinates of point K(7, 0).
The coordinates of J(7, 4); K(7, 0); and M(7, 2).
To learn more about midpoints, visit: brainly.com/question/4637646
#SPJ9
<span>the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>