All categories

9 days ago

A cone has a volume of about 28 cubic inches. Which are possible dimensions for the cone?

1 answer:

8 0

The volume, V of a cone is calculated using the formula V = (1/3)*π*(radius)^2 * height. This indicates that there are two variables involved: radius and height, resulting in multiple radius and height combinations producing the same volume. Therefore, the question lacks certain details. In my research, I came across a list of options for answers: a) radius 6 inches, height 3 inches b) diameter 6 inches, height 3 inches c) diameter 4 inches, height 6 inches d) diameter 6 inches, height 6 inches. Now, you can evaluate these options to determine which configuration results in a volume of roughly 28 cubic inches. a) with a radius of 6 in and height of 3 in yields V = (1/3)*3.14*(6in)^2 * 3in = 113.04 in^3, which is not feasible. b) with a diameter of 6 in and a height of 3 in, implies a radius of 3 in, which results in V = (1/3)*3.14*(3in)^2 * 3in = 28.26 in^3, making this the suitable choice. The other alternatives do not come close to matching the 28 in^3 volume target. The correct response is: diameter 6 inches, height 3 inches.

You might be interested in

Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)

Leona [9271]

Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

$\Rightarrow x^2+y^2 = 16900$

$\Rightarrow y^2 = 16900-x^2$

$\Rightarrow y = \sqrt{16900-x^2}$

Since the point lies on the circle, its coordinates must be,

$(x,\sqrt{16900-x^2})$

The distance "d" from the point to (30,40) can be calculated as,

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}$

$=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}$

$=\sqrt{9400-60x-80\sqrt{16900-x^2}}$

Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when $9400-60x-80\sqrt{16900-x^2}$ is at its lowest value.

Let’s set up the equation,

$\Rightarrow f(x)=9400-60x-80\sqrt{16900-x^2}$

$\Rightarrow f'(x)=-60+80\dfrac{x}{\sqrt{16900-x^2}}$

$\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}$

We find the critical points,

$\Rightarrow f'(x)=0$

$\Rightarrow-60+80\dfrac{x}{\sqrt{16900-x^2}}=0$

$\Rightarrow 80\dfrac{x}{\sqrt{16900-x^2}}=60$

$\Rightarrow 80x=60\sqrt{16900-x^2}$

$\Rightarrow 80^2x^2=60^2(16900-x^2)$

$\Rightarrow 6400x^2=3600(16900-x^2)$

$\Rightarrow \dfrac{16}{9}x^2=16900-x^2$

$\Rightarrow \dfrac{25}{9}x^2=16900$

$\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78$

$\Rightarrow x=78$

Then,

$\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2$

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

5 0

1 month ago

Three classes of school children are selling tickets to the school play. the number of tickets sold by these classes, and the nu

lawyer [9248]

Aqe4hjwrs4jyjreajrsjwtj

6 0

1 month ago

Read 2 more answers

A frog is sitting exactly in the middle of a board that is five feet long. Every ten seconds he jumps one foot to the left or on

PIT_PIT [9132]

Answer:

Given that the frog jumps every 10 seconds

(using digits from a random number table)

It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Step-by-step explanation:

A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.

Every 10 seconds, the frog jumps left or right.

If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.

If it jumps as LLRLL, it will jump off the board after fifty seconds.

Given that the frog jumps every 10 seconds

(using digits from a random number table)

It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

4 0

1 month ago

Malcolm is driving 1,373 miles from Wichita to Charleston for a family Reunion. He drives 468 miles the first day and 434 miles

zzz [9093]

Malcolm's remaining distance is roughly
1370 - 470 - 430 = 470

This coincides with selection...
D) 470 miles

8 0

1 day ago

A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each r

Leona [9271]

Response:

The answer to the inquiry is 8 hours.

Step-by-step breakdown:

Information

Ladder length = 200 cm

Distance between rungs = 20 cm

Tide rise rate = 10 cm/h

Fifth rung =?

Procedure

1.- Determine the total height the tide must reach

Height = 20 cm x 4

= 80 cm as the first rung is touching the water.

2.- Calculate the time needed

Rate = distance / time

-Solve for time

Time = distance / rate

-Substitute values

Time = 80 cm / 10 cm/h

-Final outcome

Time = 8 hours.

3 0

16 days ago