512=2w^2 is the best answer
H(t)=-16t²+160
0=-16t²+160
16t²=160
t²=10
t=√10 or 3.16 seconds
☺☺☺☺
Q11
Suppose the two numbers are x and y so xy = 3000
the HCF = 10
we know that product of two numbers = product of their HCF and LCM
XY = HCF * LCM
3000 = 10 * LCM
3000/10 = LCM
300 = LCM
Answer 300
Q12 : Let us say second number is x so
product of x and 160 = product of HCF and LCM of ( x and 160 )
160 x = 32 * 1760
x= 32 * 1760 / 160
x= 352
Answer : 352
Answer:
Step-by-step explanation:
For this case we are interested on the region shaded on the figure attached.
And we can find the volume with the method of rings.
The area on this case is given by:
![A(x) = \pi [f(x)]^2 = \pi r^2 = \pi [3x]^2 = 9\pi x^2](https://tex.z-dn.net/?f=%20A%28x%29%20%3D%20%5Cpi%20%5Bf%28x%29%5D%5E2%20%3D%20%5Cpi%20r%5E2%20%3D%20%5Cpi%20%5B3x%5D%5E2%20%3D%209%5Cpi%20x%5E2)
And the volume is given by the following formula:

For our case our limits are x=0 and x=2 so we have this:

And if we solve the integral we got this:
![V= \pi [\frac{x^3}{3}]\Big|_0^{2}](https://tex.z-dn.net/?f=%20V%3D%20%5Cpi%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5D%5CBig%7C_0%5E%7B2%7D)
And after evaluate we got this:
![V=\pi [(\frac{8}{3} )-(\frac{0}{3} )]](https://tex.z-dn.net/?f=%20V%3D%5Cpi%20%5B%28%5Cfrac%7B8%7D%7B3%7D%20%29-%28%5Cfrac%7B0%7D%7B3%7D%20%29%5D)