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goldenfox [79]
3 years ago
11

A mailbox 4 ft high casts a 5 ft shadow. How tall is a street light that casts a 26 ft shadow at the same time of the day? (to t

he nearest tenth) A) 10.4 ft B) 18.6 ft C) 20.8 ft D) 32.5 ft
Mathematics
1 answer:
zysi [14]3 years ago
8 0
Proportions!
4/5 = x/26
Cross multiply
26*4=5x
104=5x
x=20.8 ft
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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

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

=> \cos 55 = \frac{x}{6}

=> \frac{x}{6} = 0.57357......

=> x = 6 \times 0.57357.... = 3.44145.... ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

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

=> \frac{8}{x} = 0.29237......

=> x = \frac{8}{0.29237.....} = 27.36242..... ≈ 27.3

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

5 0
2 years ago
A piece of cloth is folded into a square with a side length measuring x inches. When it is unfolded, it has an area of x2 + 27x
Dmitriy789 [7]
Our task here is to find possible factors of <span>x^2 + 27x + 162.
We need to check out possible factors of 162 first.  Examples are:
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7 0
2 years ago
the area of a gymnasium floor is 264 square yard. the floor is 11 yards wide . how long is the floor?
Alekssandra [29.7K]
Area = Length x Width (A = L x W)

A = 264, W = 11, L = ?

Rewrite the equation to find L. In order to reverse multiplication, you must divide.

L = A / W

L = 264 / 11 = 24 yards
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What are the inputs of the following function?
igor_vitrenko [27]

Answer:

  A. {-4, -3, 7, 8}

Step-by-step explanation:

The ordered pairs representing a function are always written ...

  (input, output)

<h3>Inputs</h3>

The set of inputs for the given function is the list of first-numbers of the ordered pairs. Those numbers are -3, -4, 8, 7. When we express them as a set, we like to have the elements of the set in increasing order:

  inputs = {-4, -3, 7, 8} . . . . . . matches the first choice

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grin007 [14]

Answer:

x=2.5

Step-by-step explanation:

First, you do 5 divided by 2, because you are trying to find a missing factor. After that, your answer, which is 2.5, is what x equals.

Hope that helps!

7 0
3 years ago
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