Answer:
if you are lookng for C it is 17
Step-by-step explanation:
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?
The parallelogram will translate along the arrow and reach the head of the arrow.
The translation is a category of motion in which one body moves along towards a particular direction and reaches a particular point.
In translation, the rotational motion is not present. Also, the translational motion is a one-dimensional motion.
In the given figure, the parallelogram is present at the tail of the arrow.
The arrow depicts the direction of the translational motion.
So, the parallelogram will reach the end of the arrow as shown in the picture present in the attachment.
For more explanation about translation or rotation, refer to the following link:
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615.44 = (3.14)(radius)^2
<span>radius = square-root of (615.44/3.14) </span>
<span>diameter = radius x 2 </span>
<span>diameter = 28 inches
3.14
</span>