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Allisa [31]
3 years ago
8

He floor of a moving van is 3 feet off of the ground. The ramp into the van makes a 12° angle with the ground. About how long is

the ramp?
Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Given :

Hight of the van from the ground, h = 3 feet.

The ramp into the van makes a 12° angle with the ground.

To Find :

How long is the ramp.

Solution :

Let, length of ramp is l .

So, height of ramp in terms of length is given by :

h = l \times sin\ 12^o\\\\l = \dfrac{h}{sin \ 12^o}\\\\l = \dfrac{3}{sin \ 12^o}\\\\l = 14.429 \ feet

Hence, this is the required solution.

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An octagonal swimming pool has a base area of 22 square feet The pool is 3 feet deep How many cubic feet of water can the pool h
LenKa [72]

The pool can hold 65.84 ft³ of water

<u>Explanation:</u>

Given:

Shape of pool = octagonal

Base area of the pool = 22 ft²

Depth of the pool = 3 feet

Volume, V = ?

We know:

Area of octagon = 2 ( 1 + √2) a²

22 ft² = 2 ( 1 + √2 ) a²

\frac{11}{1+\sqrt{2} } = a^2

a² = \frac{11}{2.42}

a² = 4.55

a = 2.132 ft

Side length of the octagon is 2.132 ft

We know:

Volume of octagon = 2(1+\sqrt{2} ) X (a)^2 X h

V = 2(1+\sqrt{2})X (2.132)^2 X 3\\ \\V = 2 ( 2.414) X 4.5454 X 3\\\\V = 65.84 ft^3

Therefore, the pool can hold 65.84 ft³ of water

8 0
3 years ago
Please please help me. i’ll literally send you money. i need to pass please help
mihalych1998 [28]
<h3>Answer: Approximately 6.4 units</h3>

==================================================

Explanation:

The origin is the point (0,0)

Use the distance formula to find the distance from (0, 0) to (4, -5)

Let

(x_1,y_1) = (0,0)\\\\(x_2,y_2) = (4,-5)\\\\

be our two points. Plug those values into the distance formula below and use a calculator to compute

d = \sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0-(-5)\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0+5\right)^2}\\\\d = \sqrt{\left(-4\right)^2+\left(5\right)^2}\\\\d = \sqrt{16+25}\\\\d = \sqrt{41} \ \text{ exact distance}\\\\d \approx 6.40312 \ \text{ approximate distance}\\\\d \approx 6.4\\\\

The distance between the two points (0,0) and (4,-5) is approximately 6.4 units.

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4 years ago
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