Answer:
Answer:
164.32 earth year
Step-by-step explanation:
distance of Neptune, Rn = 4.5 x 10^9 km
distance of earth, Re = 1.5 x 10^8 km
time period of earth, Te = 1 year
let the time period of Neptune is Tn.
According to the Kepler's third law
T² ∝ R³


Tn = 164.32 earth years
Thus, the neptune year is equal to 164.32 earth year.
Step-by-step explanation:
Using the properties of arcs and inscribed angles, B is 110/2=55.
All the numbers in this range can be written as

with

and

. Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)
so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.
For each number that occupies an entire diagonal in the table, it's easy to see that that number

shows up

times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.
So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.
Answer:
This is the sample answer for edmentum.
Answer:
6) B distributive property 7) B distributive property 8) C associative property
Step-by-step explanation: