No, the cone and the cylinder can't have congruent heights and bases.
<h3>
is it possible that the two cones have congruent bases and congruent heights?</h3>
The volume of a cylinder of radius R and height H is:
V = pi*R^2*H
And for a cone of radius R and height H is:
V = pi*R^2*H/3
So, for the same dimensions R and H, the cone has 1/3 of the volume of the cylinder.
Here, the cylinder has a volume of 120cm³ and the cone a volume of 360cm³, so the cone has 3 times the volume of the cylinder.
This means that the measures must be different, so the cone and the cylinder can't have congruent heights and bases.
If you want to learn more about volumes:
brainly.com/question/1972490
#SPJ1
Its 2 since its in the 10's place
39/8. When you change this to an improper fraction, that is what you get.
Volume of a cone: 1/3
r²h
r is the radius (diameter ÷ 2) and h is the height.
Find the radius first using the diameter: 2
÷ 2 = 1.25
Now plug in the numbers into the formula.
1/3
1.25²10 = <span>16.4
</span>
The volume of the cone is 16.4 cubic feet.
Answer:
y = 4(x + 11)² - 484
Step-by-step explanation:
y = 4x² + 88x
factor the expression
y = 4(x² + 22x)
complete the square
y + ? = 4(x² + 22x + ?)
y + ? = 4(x² + 22x + 121)
add 4 • 121 to the left side
y + 4 • 121 = 4(x² + 22x + 121)
multiply
y + 484 = 4(x² + 22x + 121)
y + 484 = 4(x + 11)²
subtract both sides by 484
y = 4(x + 11)² - 484
