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AlexFokin [52]
3 years ago
8

PLEASE NEED HELP ON THE QUESTION (20 POINTS) THANK YOU :D

Mathematics
2 answers:
vlada-n [284]3 years ago
6 0
What the person below said very smart
murzikaleks [220]3 years ago
3 0

Let's start by knowing what information we are given in the problem and what we can infer.

  1. We can see that \angle OAD is an central angle, meaning that m\angle OAD = m \stackrel \frown{AC}. Essentially, the measure of the central angle (\angle OAD) is the same as the measure of the arc (\stackrel \frown{AC}) of which it relates to. Since we figured out that \angle OAD = 62^{\circ}, we can also say m \stackrel \frown{AC} = 62^{\circ}.
  2. \angle ABC is an inscribed angle, meaning that m\angle ABC = \frac{1}{2} \,m\stackrel \frown{AC}. Essentially, the measure of \angle ABC is equal to one-half of the measure of \stackrel \frown{AC}.

Using this information, we can say that m \angle ABC = \frac{1}{2} (62^{\circ}) = 31^{\circ}. Thus, \boxed{m \angle ABC = 31^{\circ}}

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GIVEN THAT TITAN HAS A RADIUS OF 2575 KM WHICH IS COVERED BY 600 KM ATMOSPHERE. WHAT
Ber [7]

Answer:

  87.4%

Step-by-step explanation:

The radius to the top of the atmosphere as a fraction of the moon's radius is ...

  (2575 +600)/2575 ≈ 1.23301

The ratio of the moon's volume with atmosphere to the moon's volume without is the cube of this, or 1.87456

Then the ratio of the atmosphere's volume to the moon's volume is ...

  (1.87456 -1)/1 = 0.87456

Atmospheric haze is about 87.4% of the moon's volume.

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We have assumed that the desired ratio is to the solid moon's volume, not the volume of moon and atmosphere together. The latter ratio would be 0.875 to 1.875 or about 46.7%.

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