Answer:
<em>Proof below</em>
Step-by-step explanation:
Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that
a=3n+1
We need to prove 
is a type 1 integer
Expanding
If is a type 1 integer, then we should be able to find an integer m such as
Equating
solving for m
Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven
Answer:
![\left[\begin{array}{cccc}1&3&4&5\&3&1&5&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%264%265%5C%263%261%265%266%5Cend%7Barray%7D%5Cright%5D)
The only ones are on the "diagonal", where i = j. Else the position is simply the sum.
Answer:
30
Step-by-step explanation:
Follow the correct order of operations.
There are only multiplications and divisions, so do them in the order they appear from left to right.
625 ÷ 62.5 × 30 ÷ 10 =
= 10 × 30 ÷ 10
= 300 ÷ 10
= 30
Answer:
I cant see the whole question...
