A number $x$ factors as $107\cdot109^5$, and $107$ and $109$ are primes. What is the exponent of $109$ in the prime factorization of $x^{11}$?

*55*

Yeah it s very easy Bro!

Step-by-step explanation:

Watch we have something like 109^5=x

then x^11=109^x

11*5=55

⊕⊕⊕

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

answer

you start with $5 and save $10 a week until you have at least $45wer

step-by-step explanation: