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?
X= 3r+5/2 is the answer I believe
Answer:
see explanation
Step-by-step explanation:
To find the y- intercept let x = 0 in the function
g(0) = 0 - 0 - 84 = - 84 ← y- intercept
To find the x- intercepts let y = 0, that is
x² - 5x - 84 = 0
To factor the quadratic
Consider the factors of the constant term (- 84) which sum to give the coefficient of the x- term
The factors are - 12 and + 7, since
- 12 × 7 = - 84 and - 12 + 7 = - 5, thus
(x - 12)(x + 7) = 0
Equate each factor to zero and solve for x
x - 12 = 0 ⇒ x = 12
x + 7 = 0 ⇒ x = - 7
x- intercepts are x = - 7 and x = 12
Answer:
No
Step-by-step explanation:
