GIVEN THAT TITAN HAS A RADIUS OF 2575 KM WHICH IS COVERED BY 600 KM ATMOSPHERE. WHAT
PERCENTAGE OF THE MOON'S VOLUME IS ATMOSPHERIC HAZE

  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%.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

the population of other specie of birds will decline or other specie of birds will find some alternate source of food to survive.

explanation:

according to the competitive exclusion principle, which is also known as gause's law, two specie living in the same niche and feeding on the same resource cannot co-exist stably.

as we have well studied the theory of natural selection which states that only those organisms survive which have better tendency to compete for food and adapt to the environmental conditions.

coming towards the question, what wi ll happen when one species outcompetes the other for the small fish , the other specie will not get enough food and will start to die or decline in population. eventually either other specie needs to find another alternative food source for itself or migrate at some other location where there is less competition otherwise it will just not exist anymore at that niche.

hope it !