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wel
4 years ago
6

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

Mathematics
1 answer:
Ber [7]4 years ago
3 0

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.

_____

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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Help please ill give brainliest answer
mihalych1998 [28]

Answer:

ΔLMN ≅ ΔLQP by (SAA)

Step-by-step explanation:

It is given that line (NM) is congruent to the line (PQ), meaning they have the same measure. This is signified by the small red line on each of these sides.

Moreover, it is also given that angle (MNL) is congruent to angle (QPL), this is shown by the red arc around these angles.

Finally one can figure out that angle (NLM) is congruent to angle (PLQ) by the vertical angles theorem. The verticle angles theorem states that when two lines intersect, the opposite angles are congruent.

Thus the two triangles are congruent by side-angle-angle postulate, abbreviated as (SAA).

6 0
3 years ago
X=-4+y and -2x-5y=15
Scrat [10]

Answer:

x=-5, y=-1

Step-by-step explanation:

Substitute x=-4+y into -2x-5y=15. So we have

-2(-4+y)-5y=15

8-2y-5y=15

-7y=7

y=-1

And x is -4+y so x= -4+(-1)= -5.

7 0
3 years ago
Solve 3m+7/2=−2m+5/2 what is m
GenaCL600 [577]
Solve for m:3 m + 7/2 = 5/2 - 2 m
Put each term in 3 m + 7/2 over the common denominator 2: 3 m + 7/2 = (6 m)/2 + 7/2:(6 m)/2 + 7/2 = 5/2 - 2 m
(6 m)/2 + 7/2 = (6 m + 7)/2:(6 m + 7)/2 = 5/2 - 2 m
Put each term in 5/2 - 2 m over the common denominator 2: 5/2 - 2 m = 5/2 - (4 m)/2:(6 m + 7)/2 = 5/2 - (4 m)/2
5/2 - (4 m)/2 = (5 - 4 m)/2:(6 m + 7)/2 = (5 - 4 m)/2
Multiply both sides by 2:6 m + 7 = 5 - 4 m
Add 4 m to both sides:6 m + 4 m + 7 = (4 m - 4 m) + 5
4 m - 4 m = 0:6 m + 4 m + 7 = 5
6 m + 4 m = 10 m:10 m + 7 = 5
Subtract 7 from both sides:10 m + (7 - 7) = 5 - 7
7 - 7 = 0:10 m = 5 - 7
5 - 7 = -2:10 m = -2
Divide both sides of 10 m = -2 by 10:(10 m)/10 = (-2)/10
10/10 = 1:m = (-2)/10
The gcd of -2 and 10 is 2, so (-2)/10 = (2 (-1))/(2×5) = 2/2×(-1)/5 = (-1)/5:Answer: m = (-1)/5
5 0
4 years ago
Read 2 more answers
the distance between Columbus Ohio and New York City is about 560 miles how many hours would it take the train to travel between
arlik [135]
How fast is the train going??
7 0
4 years ago
Read 2 more answers
A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
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