To find where on the hill the canonball lands
So 0.15x = 2 + 0.12x - 0.002x^2
Taking the LHS expression to the right and rearranging we have -0.002x^2 + 0.12x -
0.15x + 2 = 0.
So we have -0.002x^2 - 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x^2 + 0.03x -2 = 0.
This is a quadratic eqn with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75
If you would like to solve 2x + 5y = - 13 and 3x - 4y = -8, you can do this using the following steps:
<span>2x + 5y = -13 /*4
3x - 4y = -8 /*5
</span>_________________
8x + 20y = -52
15x - 20y = -40
_________________
8x + 15x + 20y - 20y = -52 - 40
23x = -92 /23
x = -92 / 23
x = -4
<span>2x + 5y = -13
</span>2 * (-4) + 5y = -13
-8 + 5y = -13
5y = -13 + 8
5y = -5
y = -1
(x, y) = (-4, -1)
The correct result would be D.) <span>(-4, -1).</span>
The key lies in translating "1/10 percent." If the unknown number is n, then 0.001n = 10. Mult. both sides by 1000: n = 10000 (answer)
The different group of 2 players that the coach can pick is 45 groups.
<h3>Selection of groups</h3>
The selection of groups of 2 player can be done using the method of combination.
<h3>Different groups of 2 players</h3>
The different group of 2 players that the coach can pick from the 10 players is calculated as follows;
n = 10C2
Thus, the different group of 2 players that the coach can pick is 45 groups.
Learn more about combinations here: brainly.com/question/25821700
