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].
Let the required equation be y = mx + c; where y = 1, m = f'(0), x = 0
f(a) = sec(a)
f'(a) = sec(a)tan(a)
f'(0) = sec(0)tan(0) = 0
y = mx + c
1 = 0(0) + c
c = 1
Therefore, required equation is
y = 1.
Answer:
y=x÷20x19
Step-by-step explanation:
x÷20=5% then 5%x19=95% basically 5% off
Answer:
take a screen shot and then post it
Step-by-step explanation:
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32x^2y
8xy^6
the gcf is 8xy
