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Elodia [21]
4 years ago
14

A water tank has a diameter of 15 ft and is 22 ft high. a. What is the volume of the tank in ft?? b. In m?? c. In cm?

Mathematics
1 answer:
rodikova [14]4 years ago
7 0

Answer:

a) V = 3887.72 ft^{3}

b)V = 104.97 m^{3}

c)V = 104,968,468.538 cm^{3}

Step-by-step explanation:

A tank has the format of a cylinder.

The volume of the cylinder is given by:

V = \pi r^{2}h

In which r is the radius and h is the heigth.

The problem states that the diameter is measured to be 15.00 ft. The radius is half the diameter. So, for this tank

r = \frac{15}{2} = 7.50 ft

The height of the tank is 22 ft, so h = 22.

a) Volume of the tank in ft^{3}:

V = \pi r^{2}h

V = pi*(7.5)^2*22

V = 3887.72 ft^{3}

b) Volume of the tank in m^{3}:

We must convert both the radius and the height to m.

Each feet has 0.30 m, so:

Radius:

1 feet - 0.30m

7.5 feet - r m

r = 7.5*0.30

r = 2.25m

Height

1 feet - 0.30m

22f - h m

h = 22*0.30

r = 6.60m

The volume is:

V = \pi r^{2}h

V = pi*(2.25)^2*6.60

V = 104.97 m^{3}

c) Volume of the tank in cm^{3}:

Each m has 100 cm.

So r = 2.25m = 225cm

h = 6.60m = 660cm

The volume is:

V = \pi r^{2}h

V = pi*(225)^2*660

V = 104,968,468.538 cm^{3}

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Step-by-step explanation:

<u>SOLUTION :-</u>

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Here , x will have two values -

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<u>VERIFICATION :-</u>

When x = 7 ,

4(x-3)^2 - 11 = 4(7 - 3)^2 - 11

                       = 4 \times 4^2 - 11

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When x = -1 ,

4(x-3)^2 - 11 = 4(-1 - 3)^2 - 11

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