A pass code contains 6 digits. The first 3 digits of the code are all even (2,4,6, or 8) and the last 3 are all odd (1,3,5,7, or 9). If digits can be used more than once, how many possible codes are there?

8000

Step-by-step explanation:

The first 3 digits each have 4 options.

The last 3 digits each have 5 options.

The number of possible codes is:

4 × 4 × 4 × 5 × 5 × 5

= 4³ × 5³

= 8000

10x

step-by-step explanation:

the greatest common factor 30 and 20 is 10

5 is a common factor of 30 and 20 but 10 is also a factor and 10 is greater than 5 so 5 will definitely not be apart of our answer

now the most amount of x's you factor our is just 1 factor of 1

so 10x the gcf of the terms in the polynomial