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Luda [366]
3 years ago
6

A craftsman is making a dulcimer with the same dimensions as the one shown. The surface shown requires a special, more durable t

ype of finish. Write a polynomial that represents the area to be finished on the dulcimer shown.

Mathematics
1 answer:
Bogdan [553]3 years ago
8 0

Answer:

\displaystyle A=\frac{3h^2}{2}+h

Step-by-step explanation:

<u>Area of trapezoid</u>

Given a trapezoid shape whose parallel sides measure b1 and b2 and whose height (perpendicular to the base) is h, then the area of the trapezoid is given by

\displaystyle A=\frac{b_1+b_2}{2}\cdot h

The dulcimer has the following dimensions

b_1=2h+1

b_2=h+1

and height h

Thus

\displaystyle A=\frac{2h+1+h+1}{2}\cdot h=\frac{3h+2}{2}\cdot h

Operating

\displaystyle A=\frac{3h^2+2h}{2}=\frac{3h^2}{2}+h

\boxed{A=\frac{3h^2}{2}+h}

Is the required polynomial

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Answer:

\frac{7 \times  {3}^{2} }{3}  \\  \frac{7 \times 3 \times 3}{3}  \\  = 7 \times 3 = 21 \\ thank \: you

7 0
3 years ago
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Pretty Pavers company is installing a driveway. Below is a diagram of the driveway they are
prohojiy [21]

Answer:

The most correct option is;

(B) 958.2 ft.²

Step-by-step explanation:

From the question, the dimension of each square = 3 ft.²

Therefore, the length of the sides of the square = √3 ft.

Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;

The dimension of the diameter of the small semicircle = 10·√3

Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3

Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²

Similarly;

The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3

∴ The dimension of the diameter of the large semicircle = 22·√3

Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3

Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²

Area of rectangle = 11·√3 × 17·√3 = 561

Area, A of large semicircle cutting into the rectangle is found as follows;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2

Where:

\theta = 2\times tan^{-1}( \frac{The \, number \, of  \, vertical  \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle}{The \, number \, of  \, horizontal \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle} )

\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214

Hence;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2

Therefore; t

The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²

Therefor, the most correct option is (B) 958.2 ft.².

4 0
3 years ago
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Answer:

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Hello!

Answer:
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The answer is 0
Because anything times 0 is 0
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2 years ago
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