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Igoryamba
3 years ago
15

A pool measuring 18 meters by 22 meters is surrounded by a path of uniform width. If the area of the pool and the path combined

is 1440 square meters, what is the width of the path?
Mathematics
1 answer:
suter [353]3 years ago
5 0

Answer: the width of the uniform path is 9 meters.

Step-by-step explanation:

Let x represent the width of the uniform path.

A pool measuring 18 meters by 22 meters is surrounded by a path of uniform width. It means that the combined length of the pool and the uniform path is (18 + 2x) meters and the combined width of the pool and the uniform path is (22 + 2x) meters.

If the area of the pool and the path combined is 1440 square meters, it means that

(18 + 2x)(22 + 2x) = 1440

396 + 36x + 44x + 4x² = 1440

4x² + 80x + 396 - 1440 = 0

4x² + 80x - 1044 = 0

Dividing both sides of the equation by 4, it becomes

x² + 20x - 261 = 0

x² + 29x - 9x - 261 = 0

x(x + 29) - 9(x + 29) = 0

x - 9 = 0 or x + 29 = 0

x = 9 or x = 29

Since the width cannot be negative, then x = 9 meters

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Answer:

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7 0
4 years ago
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A. 4x^5+3x+2<br> B. 4x^2+3x+2x^5<br> C. 4x^5+3+2/x^5<br> D. 18x^5+12x+6
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3 years ago
A series of three? separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers. Each
Nadya [2.5K]

Answer:

312.5\pi \text{ km}^3\approx 981.75\text{ km}^3

Step-by-step explanation:

We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.

Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.

Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.

\text{Volume of a semicircular cylinder}=\frac{\pi r^2h}{2}, where,

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Upon substituting our given values in volume formula we will get,

\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

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\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3

Therefore, the volume of earth removed to build the three tunnels is 312.5\pi \text{ km}^3\approx 981.75\text{ km}^3.


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Answer:

25

Step-by-step explanation:

40+76+84 = 200.

200 ÷ 8 = 25

Answer: 25.

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jenyasd209 [6]

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Step-by-step explanation:

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