First find the time that the ball is level with the top of the building on its descent. You can do this by solving 280 = -16^2 + 48t + 280 for t. This gives t = 3 seconds .
Then when the ball reaches the ground the time t is obtained by solving 0 = -16t^2 + 48t + 280 This gives t = 5.94 seconds.
Answer in interval notation is (3, 5.94].
The equation that you are looking for is y = -2x + 3
Answer:
I cannot answer all of this, so I will only answer what I can:
Q1: 13
Q3: 22.5
IQR: 13.5
IQR(1.5): 20.25
Q3 + IQR(1.5) = 43
Q1 - IQR(1.5) = -7.25
I don't know if my calculations are all right but I hope this helps! :)
Answer: 112°
Step-by-step explanation:
a straight line only has 180° at all times so it is a matter of subtraction. 180-68 is 112° so from there you can find #2 as 68°, #3 as 90° and #4 as 90°, and so on.
