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Igoryamba
3 years ago
10

1. A circular swimming pool has a diameter of 60 feet, and is surrounded by a concrete sidewalk that is 5 feet wide. What is the

area of the sidewalk? Round to the nearest square foot.
Correct area of the pool and surrounding sidewalk: 4 points
Correct area of the pool water surface: 3 points
Correct area of the side walk: 1 point
Sentences explaining the process used:
Mathematics
1 answer:
Ivenika [448]3 years ago
6 0
We first calculate the area of the entire pool and the side walk.

Given that the pool has a diameter of 60 feet and that the side walk surrounds the pool with a width of 5 feet. This means that the diameter of the entire pool and the side walk is 60 + 5 + 5 = 70 feet and the radius is 70 / 2 = 35 feet

Thus the area of the entire pool and the side walk is obtained as follows:

Area=\pi r^2 \\  \\ =\pi(35)^2=1,225\pi\ square\ feet

Given that the pool has a diameter of 60 feet, this means that the radius of the pool is 30 feet.

Thus the area of the pool is given by:

Area_{pool}=\pi r^2 \\  \\ =\pi(30)^2=900\pi\ square\ feet

Therefore, the area of the sidewark is the area of the entire pool and side walk minus the area of the pool, this is given by:

Area_{side walk}=1,225\pi-900\pi \\ \\ =325\pi=1,021 \ square\ feet.
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Please can someone help me?
Feliz [49]

Answer:

weight of a water=0.5

spherical ball is filled With water=0.95.

so dear for

22\7*0.5*0.95=1.49

7 0
3 years ago
In a shipment of 22 smartphones, 2 are defective. How many ways can a quality control inspector randomly test 4 smartphones, of
Eddi Din [679]

Answer:

Therefore the required ways are =190

Step-by-step explanation:

Combination: Combination is the number of selection of items from a collection of items where the order of selection does not matter.

Total number of smartphones =22

Defective phone = 2

Non defective phone = (22-2) =20

The control inspector randomly test 4 smartphones of which 2 are defective.

Non defective phone = 2

The ways to select 2 non defective phone is

^{20}C_2=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!}=\frac{19\times 20}{2} =190

The ways to select 2 defective phone is= ^2C_2=1

Therefore the required ways are = (190×1) =190

5 0
4 years ago
Quadrilateral ABCD has the following vertices.
Marizza181 [45]
A(-5,5)
B(4,5)
C(2,0)
D(-5,-2)

AB,BC,CD,DA

AB = [4-(-5)),5-5]=[9,0]
Lenght  AB= \sqrt{9^2+0^2}= \sqrt{81}  =9

BC = [2-4,0-5]=[-2,-5]
Lenght  BC=\sqrt{(-2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}

CD = [-5-2,-2-0]=[-7,-2]
Lenght  CD=\sqrt{(-7)^2+(-2)^2}=\sqrt{49+4}=\sqrt{53}

DA =[-5-(-5),-2-5]=[0,-7]
Lenght DA=\sqrt{0^2+(-7)^2}=\sqrt{49}=7

sorted from longest to shortest:
AB, CD,DA,BC
\sqrt{81}, \sqrt{53}, \sqrt{49}, \sqrt{29}

3 0
3 years ago
A can in the shape of a cylinder has a diameter of 10 cm and a height of 6 cm. how much wrapping paper is needed to cover the cy
Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

To determine the amount of wrapping paper needed to cover the can, we would determine the total surface area of the cylindrical can. The formula for determining the total surface area of a cylinder is expressed as

Total surface area = 2πr² + 2πrh

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Diameter = 10 cm

Radius = diameter/2 = 10/2

Radius = 5 cm

Height = 6 cm

Therefore,

Total surface area = (2 × 3.14 × 5²) + (2 × 3.14 × 5 × 6)

= 157 + 188.4 = 345.4 cm²

345 cm² of wrapping paper is needed to cover the cylinder.

2) Volume of cylinder = πr²h

Volume of cone = 1/3πr²h

3 × Volume of cone = πr²h

The volume of the cylinder is 3 times greater than the volume of the cone. Therefore, the volume of the cone is 1/3 times lesser than the volume if the cylinder.

5 0
3 years ago
If k is a constant, what is the value of k such that the polynomial k^2x^3 - 6kz+9 is
Tresset [83]

x-1 divides p(x)=k^2x^3-6kx+9 if the remainder is p(1)=0. (This is the remainder theorem)

Then we must have

p(1)=k^2-6k+9=(k-3)^2=0\implies\boxed{k=3}

3 0
4 years ago
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