The temperature was -12 degrees
V= 3.14 r2 h
1. 3.14 x 14 x 14 x 18 = 11083.54
2. 3.14 x 13 x 13 x 17 = 9025.8
3. 3.14 x 15 x 15 x 14 = 9896.02
4 .3.14 x 12 x 12 x 20 = 9047.79
so the order would be the second then the forth then the third then the first
Answer:
m = 54
Step-by-step explanation:
The secant- secant angle 20° is half the difference of the intercepted arcs, that is
(160 - 2m - 12) = 20° ( multiply both sides by 2 to clear the fraction )
- 2m + 148 = 40° ( subtract 148 from both sides )
- 2m = - 108 ( divide both sides by - 2 )
m = 54
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
