1) 10/6 = H/ 4 or 5/3 = H/2 or H = 10/3 = 3.3 m
<span>2) sin P = 3/5 (given) = sin (90-Q) = cos Q = 3/5</span>
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?
Answer:
Step-by-step explanation:
The slope-intercept form of an equation of a line:
m - slope
b - y-intercept → (0, b)
The formula of a slope:
We have two points (8, -4) and (0, 2) → b = 2.
Calculate the slope:
Therefore the equation of a line t in the slope-intercept form is:
