About 7,460 punds if you add everything together
We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3)
whole area=A1+A2=25/2+75/2=100/2=
50
We can write a proportion to resemble the problem;
AE/ED = AB/BC
AE = 9
ED = 6
AB = x
BC = 10
Substitute with the given values.
9/6 = x/10
9/6 * 10 = x/10 * 10
90/6 = x
15 = x
Therefore, the answer is 15.
Best of Luck!
Step-by-step explanation:
I count four 6s, four 4s and three b.
so, we have
6⁴×4⁴×b³ or (using the same symbols as above)
6⁴.4⁴.b³