Answer:
(4;0)
Step-by-step explanation:
y=0
2x-3×0=8
2x=8
x=4
<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
</span>
Y = -0.01x^2 + 0.5x + 3
when the height y = 8
-0.01x^2 + 0.5x + 3 = 8
-0.01x^2 + 0.5x - 5 = 0
x = 13.82 and x = 36.18 so the ball is at least 8 ft high from 13.82 and 36.18 feet inclusive.- from the player.
If the player is 30 ft from the net then the ball is likely to go over the net.
