If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
Answer:
- 3
Step-by-step explanation:
Answer:
120 degrees
Step-by-step explanation:
Since p is parallel to q, <6 and <8 are congruent because of corresponding angles.
m<8=m<6
120=m<6
Since m<8 is 120, and the two angles are congruent, the m<6 is also 120 degrees.
Answer: 1.11
Step-by-step explanation:
with tangent I guess A+B=C, B goes over A. But like I bet you're wondering how will yk which is a or b? idk either, but I tried my luck and got it right. So yk how C (hypotenuse) is always the biggest number? Well whichever number is bigger (A or B) the bigger number will go over the smaller one and then you round it to the nearest hundredth. I'm not good at rounding, so I used Math//way to get it for me. So it could be like , 12+36=40 (just picking random numbers), 36 will go over 12. And then you round it. I hope thus makes sense, it's 5:40AM rn. I should be sleep because I have a job interview at 3PM today.
