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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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Answer:  Radius = 10 cm and Arc length = 5 cm

Step-by-step explanation:

The area of a sector with radius r and central angle \theta (In radian) is given by :-

A=\dfrac{1}{2}r^2\theta

Given : A sector of a circle has area 25 cm^2 and central angle  0.5 radians.

Let r be the radius , then we have

25=\dfrac{1}{2}r^2(0.5)\\\\\Rightarrow\ r^2=\dfrac{2\times25}{0.5}\\\\\Rightarrow\ r^2=\dfrac{50}{0.5}=100\\\\\Rightarrow\ r=\sqrt{100}=10\ cm

Thus, radius = 10 cm

The length of arc is given by :-

l=r\theta=10\times0.5=5\ cm

Hence, the length of the arc = 5 cm

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