If you want to know the answer you need to give us the question.
If the line is perpendicular, that means the slope has to be opposite, so it will have a slope of -1/2
Fill in the point that it has to pass through on y and x
1=-1/2(0) + b
1=0+b
1=b
Then go back to the standard form y=mx + b and fill in the slope and y intercept
Y= -1/2x + 1
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.
PQ peependicular to PS
Thus angle QPS = 90⁰

7x - 9 + 4x + 22 = 90⁰
11x = 77
x = 7
Thus angle QPR = 7(7) - 9 = 40⁰
Answer: .05
Step-by-step explanation: 3.60 divided by 12: 0.3
5.00 divided by 20: 0.25
0.3-0.25=0.05