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pickupchik [31]
2 years ago
10

A circular swimming pool has a radius of 28 ft. There is a path all the way around the pool that is 4 ft wide. A fence is going

to be built around the outside edge of the pool path.
A circle has a radius of 28 feet. A larger circle goes around the smaller circle that is 4 feet wide.

About how many feet of fencing are needed to go around the pool path? Use 3.14 for Pi.
88 ft
100 ft
176 ft
201 ft
Mathematics
1 answer:
Alexxandr [17]2 years ago
7 0

The number of feet of fencing that is needed to go around the pool path is D. 201 ft.

<h3>How to calculate the number of feet?</h3>

We have a pool with a radius of 28 feet. The radius of the outer edge of the pool path would be equal to the radius of the pool plus the width of the path.

= (28 + 4) feet

= 32 feet

The circumference of the outer edge of the pool path would be equal to the perimeter.

p = 2 * pi * r

p = 2 * 3.14 * 32

p = 200.96 feet

p = 201 feet

Learn more about distance on:

brainly.com/question/15973912

#SPJ1

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A rectangle box with a square base and no top needs to be made using 300ft^2 of material. Find the dimensions of the box with th
Yuri [45]

The dimensions of the box are 10 ft and 5 ft

The maximum volume is 500 ft³

Step-by-step explanation:

A rectangle box with

  • A square base and no top
  • It needs to be made using 300 ft² of material
  • It has greatest volume

Surface area of a box without top (SA) = perimeter of base × height + area of the base

Volume of a box (V) = base area × height

Assume that the length of the side of the square base is x and the height of the box is y

∵ It needs to be made using 300 ft² of material

∴ The surface area of the box is 300 ft²

∵ Its base is a square of side length x ft

∴ Perimeter of the base = 4 × x = 4 x

∴ Area of the base = x²

∵ The height of the box = y ft

∵ SA = perimeter of base × height + area of the base

∵ SA = (4x)(y) + x²

∴ SA = 4xy + x²

∵ SA of the box = 300 ft²

- Equate the two expressions of SA

∴ 4xy + x² = 300

Now let us find y in terms of x

- Subtract x² from both sides

∴ 4xy = 300 - x²

- Divide each term by 4x to find y

∴ y=\frac{75}{x}-\frac{1}{4}x

∵ V = area of the base × height

∴ V = x² × y = x²y

- Substitute y by the equation of it above

∴ V=x^{2}(\frac{75}{x}-\frac{1}{4}x)

∴ V=75x-\frac{1}{4}x^{3}

∵ The volume of the box is greatest

- That means differentiate V and equate it by 0

∵ \frac{dV}{dx}=75-\frac{3}{4}x^{2}

∵ \frac{dV}{dx}=0 ⇒ greatest volume

∴ 75-\frac{3}{4}x^{2}=0

- Subtract 75 from both sides

∴ -\frac{3}{4}x^{2}=-75

- Divide both sides by -\frac{3}{4}

∴ x² = 100

- Take √ for both sides

∴ x = 10

Substitute the value of x in the equation of y

∵ y=\frac{75}{10}-\frac{1}{4}(10)

∴ y = 5

The dimensions of the box are 10 ft and 5 ft

∵ V=75x-\frac{1}{4}x^{3}

∵ x = 10

∴ V=75(10)-\frac{1}{4}(10)^{3}

∴ V=750-\frac{1}{4}(1000)

∴ V = 750 - 250

∴ V = 500 ft³

The maximum volume is 500 ft³

Learn more:

You can learn more about the volume in brainly.com/question/6443737

#LearnwithBrainly

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