Find BC (triangle) sides = 20mi, 22mi inside = 95 degrees

The volume of a right circular cone is 300 cubic inches. What is the volume, in cubic inches, of a right cylinder th

AlexFokin [52]

Answer:

The volume of the cylinder is 900 cubic inches

Step-by-step explanation:

Given

Let V1 represent the volume of a cone and V2 represents the volume of a cylinder

$V_1 = 300in^3$

Required

Determine the volume of a cylinder with same dimension

The volume of a cone is calculated as:

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

The volume of a cylinder is calculated as:

$V_2 = \pi r^2h$

Since, the cone and cylinder have the same dimensions, we can substitute $\pi r^2h$ for $V_2$ in $V_1 = \frac{1}{3}\pi r^2h$

So, we have:

$V_1 = \frac{1}{3} * V_2$

Multiply both sides by 3

$3 * V_1 = \frac{1}{3} * V_2 * 3$

$3 * V_1 = V_2$

$V_2 = 3 * V_1$

Substitute $300in^3$ for $V_1$

$V_2 = 3 * 300in^3$

$V_2 = 900in^3$

Hence, the volume of the cylinder is 900 cubic inches

6 0

3 years ago

Mr.rodriguez bought 420 pencils for the school.if there are 10 pencils In each box how many boxes are there'

Sophie [7]

$420 \div 10 = 42$

3 0

4 years ago

Read 2 more answers

How do you write 53% as a decimal?

Cerrena [4.2K]

Percents are parts of 100, so we divide 53 by 100 to find the decimal form of 53%.

53 / 100 = 0.53

8 0

3 years ago

Read 2 more answers

Jack’s shipping company has to transport boxes of industrial nuts and industrial bolts to its clients. Each box of industrial nu

Ipatiy [6.2K]

Answer:

The correct option is B.

Step-by-step explanation:

Let the boxes of nuts be x and the boxes of bolts by y.

The weight of nut box is 100 kg and the weight of bolt box is 200 kg.The maximum weight the company’s cargo container can hold is 24,000 kilograms.

$100x+200y\leq 24000$

$x+2y\leq 240$ .... (1)

The company needs to transport a minimum of 160 boxes per cargo container.

$x+y\geq 160$ .... (2)

Plot the above inequalities on the coordinate plane.

The related equations are

$x+2y=240$ .... (3)

$x+y=160$ .... (4)

The x-intercept of equation 3 is (240,0) and the y-intercept is (0,120).

The x-intercept of equation 4 is (160,0) and the y-intercept is (0,160).

Check the inequality by (0,0).

$0+2(0)\leq 240$

$0\leq 240$

This statement is true, therefore the shaded region of inequality (1) contains (0,0).

$x+y\geq 160$

$0+0\geq 160$ \

$0\geq 160$

This statement is false, therefore the shaded region of inequality (2) does not contains (0,0).

Therefore graph (2) represents the solution and option B is correct.

4 0

3 years ago

Which graph represent the solution set of this inequality

VashaNatasha [74]

Hi!

First, let me explain that open circles mean that the answer starts after or before that number. So if the answer is q < 4, the answer would be C, and not B.

If the answer was q ≤ 4, the answer would be B, and not C.

I hope that makes sense. Now let's solve the inequality.

Remember that whatever we do to the inequality, we have to do it to both sides.

We need to isolate q on one side.
11q + 5 ≤ 49
Start by subtracting 5 from both sides.
11q + 5 - 5 ≤ 49 - 5
11q ≤ 44
Divide by 11 on both sides.
11q/11 ≤ 44/11
q ≤ 4

The answer is B.

Hope this helps! :)

3 0

3 years ago