The ball hits the ground when the height h(t) = 0 , so
-16t^2 + 80t = 0
-16t(t - 5) = 0
t- 5 = 0 or -16t^2 = 0 ( here t = 0 which corresponds to when ball is thrown)
t - 5 = 0 gives:-
t = 5 seconds (answer)
Answer:
y = -5 + 6=1 is your answer
Hey!
Hope this helps...
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x + y = 3
x = 3 - y
y = 2x - 15
y = 2(3 - y) - 15
y = 6 - 2y - 15
y = -2y - 9
3y = -9
y = -3
y = 2x - 15
-3 = 2x - 15
-2x = -12
x = 6
So...
The answer is: (6, -3)
I'm pretty sure this is correct
The true question should be like this:
<span>What is the center and radius of the following circle: x^2 + (y _ 6)^2 = 50,
circle quation formula is (x-a)^2 + (y-b)^2 = R², so by identifying each term, we find </span>(x-a)^2=x^2 = (x-0)^2, (y-b)^2= (y _ 6)^2, R² = 50, implies R =5sqrt(2),
it is easy to identify that the center is (a,b)= (0, 6)
the radius is R =5sqrt(2),
