Prove by induction that for every n > = 1 we have 19^n - 13 is a multiple of 6.

Explanation:

For n=1, 19^n -13 = 19 -13 = 6, a multiple of 6.

For n = n+1, ...

  19^(n+1) -13 = 19(19^n -13) +18·13

  the factor (19^n -13) is divisible by 6 (for n=1), and ...

  the factor 18·13 = 6·39 is divisible by 6.

Hence 19^n -13 is divisible by 6 for all n ≥ 1.

Another way to look at this is to do modulo 6 arithmetic on the expression:

  19^n -13 mod 6 = (19 mod 6)^n - (13 mod 6) = 1^n - 1 = 0

If the expression mod 6 is 0, then the expression is divisible by 6.

the image of (6, 9) under a dilation is (4, 6). the scale factor is

1. 8

2. 9

3. 20

4. 3

5. 17

6. 2.2

7. 11.25

8. 1

9. 14

10. 2

step-by-step explanation:

thats the first half