Formula of the Volume of a hemisphere:
V =
![\frac{2}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20)
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r³
144
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20%20)
=
![\frac{2}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20)
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r³
Multiply by 3 to cancel fraction in the right side
144
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
× 3 = 2
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r³
432
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
= 2
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r³
Divide by 2
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
on either sides to isolate r³
![\frac{432 \pi }{2 \pi }](https://tex.z-dn.net/?f=%20%5Cfrac%7B432%20%5Cpi%20%7D%7B2%20%5Cpi%20%7D%20)
=
![\frac{2 \pi }{2 \pi }](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%20%5Cpi%20%7D%7B2%20%5Cpi%20%7D%20)
r³
2
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
and
![2 \pi](https://tex.z-dn.net/?f=2%20%5Cpi%20)
cancel out
216 = r³
Take cube root to find the radius
![\sqrt[3]{216}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%7D%20)
=
![\sqrt[3]{r^3}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Br%5E3%7D%20)
6 = r
Radius is 6 unitsThe formula of the surface area of a hemisphere is:
S.A = 2
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r² +
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
r²
=
![2\pi](https://tex.z-dn.net/?f=%202%5Cpi%20)
(6)² +
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
(6)²
=2
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
× 36 + 36
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
= 72
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
+ 36
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
= 108
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
units² (in terms of
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
)
≈ 339.12 units²
Surface area = 108
units
<span>No. Its x=4. You couldn't cut the box to much lower without it being absurd.
l
</span>
First, subtract 7 on both sides. Now, you have 4x<1. Divide by 4 on both sides. Finally, you get x<1/4 (a quarter or said as one-fourth). So, x<1/4.
I think we can use the identity sin x/2 = sqrt [(1 - cos x) /2]
cos x - sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1
cos x - sqrt(3/2) sqrt(1 - cos x) = 1
sqrt(3/2)(sqrt(1 - cos x) = cos x - 1 Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1
cos^2 x - 0.5 cos x - 0.5 = 0
cos x = 1 , -0.5
giving x = 0 , 2pi, 2pi/3, 4pi/3 ( for 0 =< x <= 2pi)
because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.
The last 2 results dont fit so the answer is x = 0 , 2pi Answer