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egoroff_w [7]
3 years ago
7

A gas storage tank is in the shape of a right circular cylinder that has a radius of the base of 2ft and a height of 3ft. The fa

rmer wants to paint the tank including both bases but only has 1 gallon of paint. If 1 gallon of paint will cover 162 square​ feet, will the farmer have enough paint to complete the​ job?
Mathematics
1 answer:
bogdanovich [222]3 years ago
3 0

Answer:

Yes, the farmer have enough paint to complete the​ job.

Step-by-step explanation:

It is given that a gas storage tank is in the shape of a right circular cylinder.

The radius of the base is 2 ft and the height of cylinder is 3 ft.

The total surface area of a cylinder is

S=2\pi rh+2\pi r^2

Total surface area of gas storage tank is

S=2\pi (2)(3)+2\pi (2)^2

S=12\pi+8\pi

S=20\pi

S=62.8318530718

S\approx 62.83

The total surface area of gas storage tank is 62.83 square feet.

The farmer has 1 gallon of paint and 1 gallon of paint will cover 162 square​ feet.

Since 62.83<162, therefore 1 gallon of paint is enough to paint the gas storage.

Hence the required statement is Yes, the farmer have enough paint to complete the​ job.

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3 years ago
In a right triangle, the side opposite a 33 degree angle is 17 units. What is the length of the side adjacent to that angle to t
AnnyKZ [126]

Answer:

The length of the side adjacent to that angle  is 26 units

Step-by-step explanation:

We are given

In a right triangle, the side opposite a 33 degree angle is 17 units

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cot(33)=\frac{x}{17}

now, we can solve for x

x=17\times cot(33)

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2 years ago
Given parallelogram MNOP, MA = x + 5, AO = y + 2, PA = 3x and AN = 2y
erma4kov [3.2K]

Answer:

The values of x and y are x = 6 and y = 9

Step-by-step explanation:

MNOP is a parallelogram its diagonal MO and PN intersected at point A

In any parallelogram diagonals:

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  • Meet each other at their mid-point

In parallelogram MNOP

∵ MO and NP are its diagonal

∵ MO intersect NP at point A

- Point A is the mid-point pf them

∴ MO and NP bisect each other

∴ MA = AO

∴ PA = AN

∵ MA = x + 5

∵ AO = y + 2

- Equate them

∴ x + 5 = y + 2 ⇒ (1)

∵ PA = 3x

∵ AN = 2y

- Equate them

∴ 2y = 3x

- Divide both sides by 2

∴ y = 1.5x ⇒ (2)

Now we have a system of equations to solve it

Substitute y in equation (1) by equation (2)

∴ x + 5 = 1.5x + 2

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∴ - 0.5x + 5 = 2

- Subtract 5 from both sides

∴ - 0.5x = -3

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∵ y = 1.5(6)

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