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larisa [96]
3 years ago
12

Need to find volume, step by step explanation pls ​

Mathematics
1 answer:
worty [1.4K]3 years ago
7 0

Given:

A figure of combination of hemisphere, cylinder and cone.

Radius of hemisphere, cylinder and cone = 6 units.

Height of cylinder = 12 units

Slant height of cone = 10 units.

To find:

The volume of the given figure.

Solution:

Volume of hemisphere is:

V_1=\dfrac{2}{3}\pi r^3

Where, r is the radius of the hemisphere.

V_1=\dfrac{2}{3}(3.14)(6)^3

V_1=\dfrac{6.28}{3}(216)

V_1=452.16

Volume of cylinder is:

V_2=\pi r^2h

Where, r is the radius of the cylinder and h is the height of the cylinder.

V_2=(3.14)(6)^2(12)

V_2=(3.14)(36)(12)

V_2=1356.48

We know that,

l^2=r^2+h^2                               [Pythagoras theorem]

Where, l is length, r is the radius and h is the height of the cone.

(10)^2=(6)^2+h^2

100-36=h^2

\sqrt{64}=h

8=h

Volume of cone is:

V_3=\dfrac{1}{3}\pi r^2h

Where, r is the radius of the cone and h is the height of the cone.

V_3=\dfrac{1}{3}(3.14)(6)^2(8)

V_3=\dfrac{25.12}{3}(36)

V_3=301.44

Now, the volume of the combined figure is:

V=V_1+V_2+V_3

V=452.16+1356.48+301.44

V=2110.08

Therefore, the volume of the given figure is 2110.08 cubic units.

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Answer:

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Answer:

(1) The slope of the line segment AB is 1.\bar 3

(2) The length of the line segment AB is 10

(3) The coordinates of the midpoint of AB is (5, 7)

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Step-by-step explanation:

The coordinates of the line segment AB are;

A(2, 3) and B(8, 11)

(1) The slope of a line segment is given by the following equation;

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

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(x₁, y₁) and (x₂, y₂) are two points on the line segment

Therefore;

The slope, m, of the line segment AB is given as follows;

A(2, 3) = (x₁, y₁) and B(8, 11) = (x₂, y₂)

Slope, \, m_{AB} =\dfrac{11-3}{8-2} = \dfrac{8}{6}  = 1 \frac{1}{3} = 1.\bar3

The slope of the line segment AB = 1.\bar 3

(2) The length, l, of the line segment AB is given by the following equation;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Therefore, we have;

l_{AB} = \sqrt{\left (11-3  \right )^{2}+\left (8-2  \right )^{2}} = \sqrt{64 +36} = 10

The length of the line segment AB is 10

(3) The coordinates of the midpoint of AB is given as follows;

Midpoint, M = \left (\dfrac{x_1 + x_2}{2} , \ \dfrac{y_1 + y_2}{2} \right )

Therefore;

Midpoint, M_{AB} = \left (\dfrac{2 + 8}{2} , \ \dfrac{3 + 11}{2} \right ) = (5, \ 7)

The coordinates of the midpoint of AB is (5, 7)

(4) The relationship between the slope, m₁, of a line AB perpendicular to another line DE with slope m₂, is given as follows;

m_1 = -\dfrac{1}{m_2}

Therefore, the slope, m₁, of the line perpendicular to the line AB, that has a slope m₂ = 4/3 = 1.\bar 3 is given as follows;

m_1 = -\left (\dfrac{1}{\frac{4}{3} } \right ) = -\dfrac{3}{4}  = -0.75

The slope, m₁, of the line perpendicular to the line AB is m₁ = -0.75.

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