We're not supposed to help with exams. This isn't an exam, is it?
A(0)=202, A(37)=half that.


Last choice.
Answer:
If it is at -8, -8 then p is at 0,0
Step-by-step explanation:
Sorry, there is not much information given to answer.
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Answer:
[2,10,18]
y = 4x + 2
0,2
1,6
2,10
3,14
4,18
Step-by-step explanation:
Answer:
The rocket will reach its maximum height after 6.13 seconds
Step-by-step explanation:
To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height
∵ y is the height of the rocket after launch, x seconds
∵ y = -16x² + 196x + 126
- Differentiate y with respect to x
∴ y' = -16(2)x + 196
∴ y' = -32x + 196
- Equate y' by 0
∴ 0 = -32x + 196
- Add 32x to both sides
∴ 32x = 196
- Divide both sides by 32
∴ x = 6.125 seconds
- Round it to the nearest hundredth
∴ x = 6.13 seconds
∴ The rocket will reach its maximum height after 6.13 seconds
There is another solution you can find the vertex point (h , k) of the graph of the quadratic equation y = ax² + bx + c, where h =
and k is the value of y at x = h and k is the maximum/minimum value
∵ a = -16 , b = 196
∴ 
∴ h = 6.125
∵ h is the value of x at the maximum height
∴ x = 6.125 seconds
- Round it to the nearest hundredth
∴ x = 6.13 seconds