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=-\frac{196}{2(-16)}](https://tex.z-dn.net/?f=h%3D-%5Cfrac%7B196%7D%7B2%28-16%29%7D)
∴ 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
Answer:
The answer is "25".
Step-by-step explanation:
Please find the graph in the attachment.
![Longest \ side= 12](https://tex.z-dn.net/?f=Longest%20%5C%20side%3D%2012)
Calculating the triangle two other sides:
![=\sqrt{3^2+6^2}\\\\=(9+36)\\\\=45](https://tex.z-dn.net/?f=%3D%5Csqrt%7B3%5E2%2B6%5E2%7D%5C%5C%5C%5C%3D%289%2B36%29%5C%5C%5C%5C%3D45)
let
![\to \sqrt{45}= 6.7.\\\\\to 6.7\times 2=13.4\\\\\to 13.4+12=25.4 \approx 25](https://tex.z-dn.net/?f=%5Cto%20%5Csqrt%7B45%7D%3D%206.7.%5C%5C%5C%5C%5Cto%206.7%5Ctimes%202%3D13.4%5C%5C%5C%5C%5Cto%2013.4%2B12%3D25.4%20%5Capprox%2025)
Answer:
Vertical
Step-by-step explanation:
Intersecting lines form Vertical angles.
Answer:
82/125
Step-by-step explanation:
add all of them for denoinator the numeratior is the p you need to find out