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Alex777 [14]
2 years ago
12

Mr. Chen is painting one wall of a room in his attic. He needs to detemine the area of the wall so he will know how much paint h

e will buy. The drawing below shows the dimensions of the wall.
Part A: Given the drawing, what is the area of the wall?



The area of the wall is

square feet.


Part B: If one quart of paint will cover 22 square feet, how many whole quarts will Mr. Chen need to buy?



He will need to buy
whole quarts.

Mathematics
1 answer:
MAVERICK [17]2 years ago
8 0

Area of the figure

= (area of rectangle) + (area of triangle)

= (8 × 10) + (½ × 4 × 8)

= 80 + 16

= 96

Area = 96ft²

Quarts of paint needed

= 96/22

= 48/11

= 4.36

Mr. Chen will have to buy 5 quarts of paint.

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Find the volume of the rectangular pyramid below.
VMariaS [17]

Answer:

315 yd cubed

Step-by-step explanation:

To find the volume of the pyramid, we first need to consult the equation below. Thebase is for the rectangular portion, then after multiplied by the height it is divided by 3 to get the pyramid.

\frac{1}{3} (B * H )

To get the base, we multiply the length by the width.

21 * 9 = 189

189 * 5 = 945

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2 years ago
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A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can cost
Advocard [28]

Answer:

Radius = 1.12 inches and Height = 4.06 inches

Step-by-step explanation:

A soup can is in the shape of a right circular cylinder.

Let the radius of the can is 'r' and height of the can is 'h'.

It has been given that the can is made up of two materials.

Material used for side of the can costs $0.015 and material used for the lids costs $0.027.

Surface area of the can is represented by

S = 2πr² + 2πrh ( surface area of the lids + surface are of the curved surface)

Now the function that represents the cost to construct the can will be

C = 2πr²(0.027) + 2πrh(0.015)

C = 0.054πr² + 0.03πrh ---------(1)

Volume of the can = Volume of a cylinder = πr²h

16 = πr²h

h=\frac{16}{\pi r^{2}} -------(2)

Now we place the value of h in the equation (1) from equation (2)

C=0.054\pi r^{2}+0.03\pi r(\frac{16}{\pi r^{2}})

C=0.054\pi r^{2}+0.03(\frac{16}{r})

C=0.054\pi r^{2}+(\frac{0.48}{r})

Now we will take the derivative of the cost C with respect to r to get the value of r to get the value to construct the can.

C'=0.108\pi r-(\frac{0.48}{r^{2} })

Now for C' = 0

0.108\pi r-(\frac{0.48}{r^{2} })=0

0.108\pi r=(\frac{0.48}{r^{2} })

r^{3}=\frac{0.48}{0.108\pi }

r³ = 1.415

r = 1.12 inch

and h = \frac{16}{\pi (1.12)^{2}}

h = 4.06 inches

Let's check the whether the cost is minimum or maximum.

We take the second derivative of the function.

C"=0.108+\frac{0.48}{r^{3}} which is positive which represents that for r = 1.12 inch cost to construct the can will be minimum.

Therefore, to minimize the cost of the can dimensions of the can should be

Radius = 1.12 inches and Height = 4.06 inches

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