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barxatty [35]
3 years ago
13

If the height of Mount Everest is about 8.8×103 meters, and the height of the Empire State Building is about 3.8×102 meters, whi

ch of these statements is true?
Mathematics
2 answers:
Luda [366]3 years ago
6 0
Mount Everest is about 23<span> times as tall as the Empire State Building. Hope that helps!</span>
Colt1911 [192]3 years ago
6 0

Answer:

option C is correct, i.e. Mount Everest is about 23 times as tall as the Empire State Building .

Step-by-step explanation:

Given options are:-

A. There is no way to compare these heights .

B. Mount Everest is about 40 times as tall as the Empire State Building .

C. Mount Everest is about 23 times as tall as the Empire State Building .

D. The Empire State Building is about 23 times as tall as Mount Everest.

Given the height of Mount Everest is about 8.8×10^3 meters, and the height of the Empire State Building is about 3.8×10^2 meters.

Finding ratios of given heights:-

\frac{Mount\;\;Everest}{Empire\;\;State\;\;Building} =\frac{8.8*10^3m}{3.8*10^2m} \\\\\frac{Mount\;\;Everest}{Empire\;\;State\;\;Building} =23.15789474 \approx 23 \\\\Mount\;\;Everest=23*(Empire\;\;State\;\;Building)

Hence, option C is correct, i.e. Mount Everest is about 23 times as tall as the Empire State Building .

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OleMash [197]

The angle between u = (-2,-5) and v = (5,2) is 134 degrees approximately.

<u>Solution:</u>

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(-2 \vec{\imath}-5 \vec{\jmath}) \cdot(5 \vec{\imath}+2 \vec{\jmath})=\sqrt{(-2)^{2}+(-5)^{2}} \times \sqrt{5^{2}+2^{2}} \times \cos \theta

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\begin{array}{l}{-10-10=\sqrt{29} \times \sqrt{29} \times \cos \theta} \\\\ {-20=29 \times \cos \theta} \\\\ {\cos \theta=\frac{-20}{29}} \\\\ {\theta=\cos ^{-1} \frac{-20}{29}} \\\\ {\theta=133.60}\end{array}

Hence, the angle between given two vectors is 134 degrees approximately.

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Step-by-step explanation:

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