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Nitella [24]
3 years ago
12

Each story of a building is 14.3 feet tall. If the height of the building is 185.9 feet, how many floors does the building have?

A. 9 B. 10 C. 17 D. 13
Mathematics
2 answers:
Leya [2.2K]3 years ago
5 0

Answer:

your answer is D.

Step-by-step explanation:

if you times each number (9, 10, 17, 13) by 14.3 you would find your number like so:

A. 14.3 · 9 = 128.7

B. 14.3 · 10 = 143

C. 14.3 · 17 = 243

D. 14.3 · 13 = 185.9

which gives you your answer and shows explanation of how you got your answer.

zvonat [6]3 years ago
3 0

Answer:

D 13

Step-by-step explanation:

To find out how many floors the building has, we take the height of the building and divide by the size of each floor

floors = 185.9 ft / 14.3ft per floor

           = 13 floors

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To solve this problem, you have to write a proportion using the ratios formed by the similar triangles.

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x/(x+2) = 10/15

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10x + 20 = 15x
20 = 5x
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The value of x is 4.

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(9/ 4)/ 18 = 9/ (4 /18) A.True B.False
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Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A
alekssr [168]

Answer:

\theta_{CAB}=128.316

\theta_{ABC}=25.842

\theta_{BCA}=25.842

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

AB= \sqrt{(-3-1)^2+(0-3)^2}=5

BC= \sqrt{(1-1)^2+(3-(-3))^2}=9

CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}

\theta_{CAB}=\arccos{-\dfrac{0.62}}

\theta_{CAB}=128.316

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

180

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