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%.

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°

i believe that the answer is

b. y-1=-2(x-4)

hope this !