Answer:
Based on the given information, we know the following:
Solid A is formed by joining one hemisphere to the frustum of a cone.
Solid B is formed by joining the other hemisphere to the small cone that was removed from the large cone.
The volume of solid A is 6 times the volume of solid B.
We can use the formulas for the volume of a hemisphere and a frustum of a cone to represent the volumes of solids A and B.
The volume of a hemisphere is (2/3)πr^3, where r is the radius of the hemisphere.
The volume of a frustum of a cone is (1/3)πh(R^2+r^2+Rr), where h is the height of the frustum, R is the radius of the larger base and r is the radius of the smaller base.
The volume of a cone is (1/3)πr^2h
Let's call the height of the frustum h, the radius of the large base R and the radius of the small base r
We can set up the equation: (2/3)πr^3 = 6((1/3)πr^2h + (1/3)πR^2h + (1/3)πRrh)
After simplifying, we get:
2h = 12r + 12Rh + 6Rr
We know k = (R/r)^(1/3)>(7)^1/3, we can substitute it into the equation:
2h = 12r + 12rk + 6rk^2
Solving for h, we get:
h = 6rk + 6rk^2
So h is expressed in terms of k and r.