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2 years ago

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The volume of a cylinder is

1 answer:

Dima020 [189]2 years ago

7 0

The volume of a cylinder is equal to $\frac{1}{3}\pi r^{2} h$ , where r is the radius of the cylinder and h is its height.

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

324 = $\frac{1}{3}$ x 3.14 x $r^{2}$ x 1

309.55 = $r^{2}$

r = $\sqrt{309.55}$

r = approx. 17.59 inches

I hope this answer helped you! If you have any questions, feel free to ask away.

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The slope is 2/1

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2 years ago

Which value of m will create a system of parallel lines with no solution? y=mx-6 8x-4y=12 A coordinate grid with one line labele

nydimaria [60]

Answer:

A system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2

Step-by-step explanation:

The equation of the given line is 8·x - 4·y = 12

Which gives;

8·x- 12= 4·y

y = 2·x - 3

Given that the line passes through the points (0, -3) and (1, -1), we have;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

When (x₁, y₁) = (0. -3) and (x₂, y₂) = (1, -1), we have;

$Slope, \, m =\dfrac{(-1)-(-3)}{1-(0)} = 2$

y - (-3) = 2×(x - 0)

y = 2·x - 3 which is the equation of the given line

For the lines 8·x - 4·y = 12, which is the sane as y = 2·x - 3 and the line y = m·x - 6 to have no solution, the slope of the two lines should be equal that is m = 2

Given that the line passes through the point (1.5, 0), we have;

y - 0 = 2×(x - 1.5)

y = 2·x - 3...................(1)

For the equation, y = m·x - 6, when m = 2, we have;

y = 2·x - 6..................(2)

Solving equations (1) and (2) gives;

2·x - 3 = 2·x - 6, which gives;

2·x - 2·x= - 3 - 6

0 = 9

Therefore, a system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2.

4 0

3 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

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2 years ago