Prove that < AOD and < BOD are supplementary
1 answer:
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Answer:
Check the explantion
Step-by-step explanation:
H(t) represent the height of the ball after t seconds.
1)
In order to figure the maximum height of the ball we must minupulate this equation into a form where we can easily find the maximum height.
We can also see this parabola will be downward concaved because it has a negative coefficent for the t^2 term.
a) Let's Manipulate!
Vertex form is a form where we can easily tell the maximum and minimum points. To get in into Vertex form we just need to complete the sqaure.
Lets reanrage the terms
h(t)= -16t^2+144t
Undo the distrubitive propety
h(t)= -16(t^2-9t+20.25)+324
Add and subtract 324.
h(t) = -16(t-4.5)^2+324
COmplete the square.
b) Almost done!
We can now see that the stuff with the brackets is always positive/zero because any real number sqaured is a positive/zero. But than we multiply the positive stuff by a negative number (-16) so its now all negative.
The only exception is when its zero.
And off course if we add 324 to zero its always going to be greater than 324 + a negative number. To get the first term to be zero we can have t = 4.5.
c) Proove it!
h(4.5) = -16(4.5-4.5)^2+324
=324
d) Double Check by graphing!
(Its a attach screenshote by the way)
e) Answer:
The ball will reach its maximum height after 4.5 seconds.
Hoped this helped!,
JoeLouis2
