Answer:
<em>The time it takes the ball to hit the ground is 3.05secs</em>
step - by - step explanation:
<em>In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05 </em>
Answer:
3) x-axis: 1 unit
y-axis: 10 units
4) quadrant II
coordinates: (-2, 4)
7) Quadrants I and IV; time is always positive , but temperature can be positive and negative
8) W = (-0.75, -1)
Step-by-step explanation:
3) x-axis: 1 unit
y-axis: 10 units
4) quadrant II
coordinates: (-2, 4)
7) Quadrants I and IV; time is always positive , but temperature can be positive and negative
8) W = (-0.75, -1)
Answer:
Vertex of the parabola is (2,3)
Step-by-step explanation:
The parabola is a negative, it opens down and the maximum place it can reach is 3 (y).
D., Because 29 x 15 is 435 but if you throw the .07 in the mix you'll have 436.05 which breaks even.