You didn’t add a screenshot. I’m sorry for answering but someone else will if I don’t.
The answer is d because 4+8=12 and they are 4 units away from eachother:)
The digital time for quarter past three is 3:15, it is 15 minutes past 3 o'clock.
<span>The maxima of a differential equation can be obtained by
getting the 1st derivate dx/dy and equating it to 0.</span>
<span>Given the equation h = - 2 t^2 + 12 t , taking the 1st derivative
result in:</span>
dh = - 4 t dt + 12 dt
<span>dh / dt = 0 = - 4 t + 12 calculating
for t:</span>
t = -12 / - 4
t = 3
s
Therefore the maximum height obtained is calculated by
plugging in the value of t in the given equation.
h = -2 (3)^2 + 12 (3)
h =
18 m
This problem can also be solved graphically by plotting t
(x-axis) against h (y-axis). Then assigning values to t and calculate for h and
plot it in the graph to see the point in which the peak is obtained. Therefore
the answer to this is:
<span>The ball reaches a maximum height of 18
meters. The maximum of h(t) can be found both graphically or algebraically, and
lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball
to reach maximum height, and the y-coordinate, 18, is the max height in meters.</span>
