Please help me this is my question. 14 > u + 10
1 answer:
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<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>3</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
Answer:
- hits the ground at x = -0.732, and x = 2.732
- only the positive solution is reasonable
Step-by-step explanation:
The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.
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<h3>graphing</h3>The attachment shows a graphing calculator solution to the equation
-3x^2 + 6x + 6 = 0
The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.
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<h3>completing the square</h3>We can also solve the equation algebraically. One of the simplest methods is "completing the square."
-3x^2 +6x +6 = 0
x^2 -2x = 2 . . . . . . . . divide by -3 and add 2
x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square
(x -1)^2 = 3 . . . . . . . . written as a square
x -1 = ±√3 . . . . . . . take the square root
x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground
The numerical values of these solutions are approximately ...
x ≈ {-0.732, 2.732}
The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.
