Answer:
The last graph best represents the story
Step-by-step explanation:
Mary left for school from rest. So the initial speed is zero.
So we start from the origin and draw a straight line with a positive slope to running, that is as time increases , distance covered also increases.
When she is resting, her speed is zero so we draw a horizontal line to indicate this second phase of her journey.
After resting she still continued her journey to school.
This part also has a positive slope.
The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?
Let x be the distance (in feet) along the road that the car has traveled and h be the distance (in feet) between the car
and the observer.
(a) Before the car passes the observer, we have dh/dt < 0; after it passes, we have dh/dt > 0. So at the instant it passes the observer we have
dh/dt = 0, given that dh/dt varies continuously since the car travels at a constant velocity.
