All categories

3 years ago

A cone has a volume of 12 cubic inches. What is the volume of a cylinder that the cone fits exactly inside of?

Mathematics

2 answers:

Viktor [21]3 years ago

7 0

Answer:

The correct option is 3.

Step-by-step explanation:

It is given that the volume of a cone is 12 cubic inches.

The volume of a cone is

$V_1=\frac{1}{3}\pi r^2h$

Put V₁=12 in the value equation.

$12=\frac{1}{3}\pi r^2h$

Multiply both sides by 3.

$36=\pi r^2h$ .... (1)

Where, r is radius of base, h is height.

It is given that the cone fits exactly inside of the cylinder. It means the height and the radius of the cylinder is same.

The volume of a cylinder is

$V_2=\pi r^2h$

Using equation (1) we get

$V_2=36$

The volume of cylinder is 36 cubic inches. Therefore the correct option is 3.

Stells [14]3 years ago

5 0

36in^3
as the volume of a cone is 1/3 that of the volume of a cylinder the smallest cylinder that the cone can fit inside of is one that is exactly 3x its the volume

You might be interested in

Find the point slope equation for the line that passes through the points (2,1) and (-2,5). use the first point in your equation

Delicious77 [7]

Answer:

y - 1 = -(x - 2)

Step-by-step explanation:

y - y1 = m(x - x1)

m = (5-1)/(-2-2) = 4/-4 = -1

Using any one coordinate, construct

y - y1 = m(x - x1)

y - 1 = -1(x - 2)

4 0

2 years ago

Read 2 more answers

HELP ASAP!!

olga2289 [7]

Answer:

62.5%

Step-by-step explanation:

150/240=0.625

you do 0.625x100 to get the percent.

8 0

2 years ago

The constraints of a problem are listed below. What are the vertices of the feasible region?

Mamont248 [21]

Answer:

Option 4 : $(0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)$

Step-by-step explanation:

See the attached figure:

To find the vertices of the feasible region, we need to graph the constraints, then find the area included by them, then calculate the vertices which is the intersection between each two of them.

As shown, the shaded area represents the solution of the constraints

So, the vertices of the feasible region are:

$(0.\frac{3}{2} ) \ , \ (0,2) \ , \ (6,0) \ , \ (\frac{9}{4} ,0)$

8 0

2 years ago

Read 2 more answers

The sixth term is 22 and the common difference is six Ward the question is what is the 15th term

NemiM [27]

The sixth term is 22 and the common difference is is 6.

4 0

3 years ago

Given H(6,7) and I(-7,-6), if point G lies 1/2 of the way along line segment HI, Santiago argues that point G is located at the

irinina [24]

Given the points H(6,7) and I(-7,-6).

If point G lies $\frac{1}{2}$ of the way along line segment HI.

Therefore, we can say that the point G divides the line segment HI in the ratio 1:1.

So, by using the cross section formula we can determine the coordinates of point G.

For the given points say $(x_1, y_1)$ and $(x_2, y_2)$ divided is in the ratio $m_1 : m_2$ , the coordinates are $(\frac{m_1x_2+m_2x_1}{m_1+m_2} , \frac{m_1y_2+m_2y_1}{m_1+m_2} )$

Coordinates G = $(\frac{(1 \times -7)+(1 \times 6)}{2} , \frac{(1 \times -6)+(1 \times 7)}{2})$

= (-0.5 , 0.5)

Hence, the coordinates of G are (-0.5 , 0.5).

So, Santiago argues that point G is located at the origin. The point G is located at (-0.5, 0.5). Therefore, he is not correct.

8 0

3 years ago