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timurjin [86]
3 years ago
6

Jerry has a miniature model of a boat. He knows that the boat is 3 3/4 inches wide and 5 1/2 inches long. What is the actual len

gth of the boat if the actual width is 15 feet?
Mathematics
1 answer:
FromTheMoon [43]3 years ago
6 0

Width of the miniature model = 3\frac{3}{4} inches

Length of the miniature model = 5\frac{1}{2} inches

The actual width of the boat = 15 feet

Let us find the actual length of the boat.

Write it in a proportion form.

width of model : length of model : : width of actual : length of actual

$\Rightarrow 3\frac{3}{4}:5\frac{1}{2}::15:x

$\Rightarrow\frac{15}{4}:\frac{11}{2}::15:x

$\Rightarrow\frac{\frac{15}{4}}{\frac{11}{2}} =\frac{15}{x}

Do cross multiplication.

$\Rightarrow{\frac{15}{4}x =15\times\frac{11}{2}

$\Rightarrow{\frac{15}{4}x =\frac{165}{2}

$\Rightarrow x =\frac{165}{2}\times{\frac{4}{15}

⇒ x = 22

Hence the actual length of the boat is 22 feet.

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Explanation:

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(37/x)*x=(100/8)*x       - we multiply both sides of the equation by x

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Answer:

288 ft³

Step-by-step explanation:

Given that

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And, Hollies required 8 of these layers

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3 years ago
Complete the assignment on a separate sheet of paper<br><br> Please attach pictures of your work.
Irina18 [472]

Answer:

<u>TO FIND :-</u>

  • Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

  • \sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}
  • \cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}
  • \tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

=> \sin 16 = \frac{7}{x}

=> \frac{7}{x} = 0.27563......

=> x = \frac{7}{0.27563....} = 25.39568..... ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

=> \sin 29 = \frac{6}{x}

=> \frac{6}{x} = 0.48480......

=> x = \frac{6}{0.48480....} = 12.37599..... ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

=> \sin 30 = \frac{x}{11}

=> \frac{x}{11} = 0.5

=> x = 0.5 \times 11 = 5.5

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

=> \cos 43 = \frac{x}{12}

=> \frac{x}{12} = 0.73135......

=> x = 12 \times 0.73135.... = 8.77624.... ≈ 8.8

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Length of side adjacent to θ = x

Hypotenuse = 6

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Length of side adjacent to θ = 8

Hypotenuse = x

=> \cos 73 = \frac{8}{x}

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7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

=> \tan 69 = \frac{12}{x}

=> \frac{12}{x} = 2.60508......

=> x = \frac{12}{2.60508....}  = 4.60636.... ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

=> \tan 20 = \frac{11}{x}

=> \frac{11}{x} = 0.36397......

=> x = \frac{11}{0.36397....}  =30.22225.... ≈ 30.2

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