There is nothing here for anyone to work off, If you have a picture or screenshot that would be very nice!
The answer is X=19/2 y=10
Answer:
130
Step-by-step explanation:
3*10 =30+100=130
Answer:
![\sqrt[5]{13^3} = 13^{\frac{3}{5}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B13%5E3%7D%20%3D%2013%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
Step-by-step explanation:
Answer:
maximum height is 4.058 metres
Time in air = 0.033 second
Step-by-step explanation:
Given that the equation height h
h = -212t^2 + 7t + 4
What is the toy's maximum height?
Let us assume that the equation is a perfect parabola
Time t at Maximum height will be
t = -b/2a
Where b = 7 and a = - 212
t = -7/ - 212 ×2
t = 7/ 424 = 0.0165s
Substitute t in the main equation
h = - 212(7/424)^2 + 7(7/424) + 4
h = - 0.05778 + 0.115567 + 4
h = 4.058 metres
Therefore the maximum height is 4.058 metres
How long is the toy in the air?
The object will go up and return to the ground.
At ground level, h = 0
-212t^2 + 7t + 4 = 0
212t^2 - 7t - 4 = 0
You can factorize the above equation and pick the positive time t since time can't be negative
Or
Since we have assumed that it's a perfect parabola,
Total time in air = (-b/2a) × 2
Time in air = 0.0165 × 2 = 0.033 s
