Respuesta :
Simplifying
F(x) = -0.05(x2 + -26x + -120)
Multiply F * x
xF = -0.05(x2 + -26x + -120)
Reorder the terms:
xF = -0.05(-120 + -26x + x2)
xF = (-120 * -0.05 + -26x * -0.05 + x2 * -0.05)
xF = (6 + 1.3x + -0.05x2)
Solving
xF = 6 + 1.3x + -0.05x2
Reorder the terms:
-6 + -1.3x + xF + 0.05x2 = 6 + 1.3x + -0.05x2 + -6 + -1.3x + 0.05x2
-6 + -1.3x + xF + 0.05x2 = 6 + -6 + 1.3x + -1.3x + -0.05x2 + 0.05x2
Combine like terms:
6 + -6 = 0
-6 + -1.3x + xF + 0.05x2 = 0 + 1.3x + -1.3x + -0.05x2 + 0.05x2
-6 + -1.3x + xF + 0.05x2 = 1.3x + -1.3x + -0.05x2 + 0.05x2
1.3x + -1.3x = 0.0
-6 + -1.3x + xF + 0.05x2 = 0.0 + -0.05x2 + 0.05x2
-6 + -1.3x + xF + 0.05x2 = -0.05x2 + 0.05x2
-0.05x2 + 0.05x2 = 0.00
-6 + -1.3x + xF + 0.05x2 = 0.00
Answer:
1.The distance between the cannon and the ground is represented by f(x).
2. When f(x) = 0, there will be two 'x' values. The point where the graph crosses the positive x-axis represents the point where the cannon hits the earth.
3.It would, absolutely. We may position the net at the spot where the cannon will hit the ground if we know where the cannon will hit the earth.
4f(x) = 0
-0.05 (x² - 26x -120) = 0
x² - 26x - 120 = 0
Explanation: