Anewly discovered planet orbits a distant star with the same mass as the sun at an average distance of 101 million kilometers. its orbital eccentricity is 0.4. find the planet's orbital period.

Kepler's third law described the relation between semi-major axis (or average distance to the star) and the orbital period (how long it takes to complete one lap) as follows: 
                                           a^3 / p^2 = constant
In the case of our Solar system the constant is 1 
This means that, for this problem:
a^3 / p^2 = 1
p^2 = a^3
p = a^(3/2)
The semi major axis is given as 101 million km. We need to convert this into AU where 1 AU is approximately 150 million Km
101 million Km = (101x1) / 150 = 0.67 AU

Now, we substitute in the equation to get the orbital period as follows:
p = (0.67)^(3/2) = 0.548 earth years

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post