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GrogVix [38]
3 years ago
10

In the previous problem, how does the angle of depression from the top of the taller building relate to the angle of elevation f

rom the top of the shorter building?

Mathematics
1 answer:
tia_tia [17]3 years ago
3 0
The angles of elevation and depression are formed by the line of sight and the horizontal line. When the line of vision is above the horizontal line, the angle is of elevation, and if the line of sight is below the horizontal, the angle is of depression.

 The angle of depresion from the top of the taller building and the angle of elevation from the top of the shorter building are alternate interior angles. Then, if the angle of depression of the taller building is 15° the angle of elevation of the shorter building is 15° too. To understand this, you should see the diagram attached. 

 In the diagram you can notice that both angles, of elevation and depression, have the same value.

 Then, the answer is: The angle of depresion from the top of the taller building and the angle of elevation from the top of the shorter building are alternate interior angles.

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How many bucketloads would it take to bucket out the world’s oceans? Write your answer in scientific notation.
kodGreya [7K]

Answer: 7\times10^{19} buckets

Step-by-step explanation:

Given:  A cubic kilometer=10^{15}cubic centimeters

The volume of  world’s oceans=1.4\times10^{9} cubic kilometers of water.

⇒ The volume of  world’s oceans=1.4\times10^{9}\times10^{15} cubic centimeters of water.

Volume of a bucket = 20,000 cubic centimeters of water.

The number of bucket-loads would it take to bucket out the world’s oceans

n=\frac{\text{volume of ocean}}{\text{volume of bucket}}=\frac{1.4\times10^{9}\times10^{15}}{20000}

\Rightarrow\ n=\frac{1.4\times10^{9+15}}{0.2\times10^5}......[a^n\times a^m=a^{m+n}]\\\Rightarrow\ n=7\times10^{24-5}.....[\frac{a^m}{a^n}=a^{m-n}]\\\Rightyarrow\ n=7\times10^{19}

hence,  7\times10^{19} bucketloads would it take to bucket out the world’s oceans.

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3 years ago
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Step-by-step explanation:

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