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noname [10]
3 years ago
9

A carpenter leans a 20-foot ladder against a building so that an angle of 60° is formed between the ladder and the ground. How m

any feet away from the
building is the bottom of the ladder?
O 10/3ft
O 20 ft
O 20V2 ft
O 10 ft
Mathematics
2 answers:
iogann1982 [59]3 years ago
8 0

Answer:

The foot of the ladder is

10

feet away from the base of the wall.

Step-by-step explanation:

Ivanshal [37]3 years ago
3 0
10 is the answer hop it is right
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Answer:

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Rectangle PQRS has vertices at P-2,5), Q44,5), R(4, -1), and S(-2,-1). What is the area, in square units, of rectangle PQRS? 21
Dennis_Churaev [7]

Answer:

36 square units.

C

Step-by-step explanation:

Please refer to the provided graph.

So we want to figure out the area of the rectangle. To do so, we simply need to figure out the base and height of the rectangle. We can then multiply them together to acquire the area.

From the graph, we can see that if we find the length of SR and SP, we can use the the base and height, respectively.

Now, we just have to find there lengths. Technically, we could just count. However, I'm going to use the distance formula.

the distance formula is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Find the length of SR:

Point S is (-2,-1) while Point R is (4,-1). Let (-2,-1) be x₁ and y₁ and (4,-1) be x₂ and y₂. Therefore:

d=\sqrt{(4-(-2))^2+(-1-(-1))^2}\\d=\sqrt{(6^2)+(0)^2} \\d=\sqrt{36}=6

So, SR is 6.

Now, find the length of SP:

Point S is (-2,-1) while Point P is (-2,5). Let Let (-2,-1) be x₁ and y₁ and (-2,5) be x₂ and y₂. Therefore:

d=\sqrt{(-2-(-2))^2+(5-(-1))^2}\\d=\sqrt{0^2+6^2}\\ d=\sqrt{36}=6

Therefore, SP is also 6.

Now, the area for a rectangle is:

A=bh

Plug in 6 for the base and 6 for the height:

A=6(6)\\A=36\text{ units}^2

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