The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet.
We know there will be a constant downward acceleration of 32.15 ft/s^2.
Use the following equation:
v_final^2 = v_initial^2 + 2ah
v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2
v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.
2x+3y-10=0 (1)
+
4x-3y-2=0 (2)
____________
6x -12=0 (if you just want the resulting equation. It is 6x-12=0)
x=2
take x=2 and put it into equation (2)
4(2) -3y -2=0
-3y= 2-8
y= 2
(x=2,y=2)
Answer is B, the graph increases by about 0.28 per year
