Answer:
6x^3 - 2x^2 -11x +4
Step-by-step explanation:
(3x-4)(2x^2+2x-1)
Multiply 3x by the second term
3x* (2x^2+2x-1)
6x^3 +6x^2 -3x
Multiply -4 by the second term
-4*(2x^2+2x-1)
-8x^2 -8x+4
Add these together, lining up the terms
6x^3 +6x^2 -3x
-8x^2 -8x+4
----------------------------------
6x^3 - 2x^2 -11x +4
C is the answer because it is
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?
