## anonymous one year ago math

1. pooja195

2. anonymous

|dw:1440720978922:dw|

3. jim_thompson5910

are you familiar with the rule $\Large x^{-k} = \frac{1}{x^k}$ ??

4. anonymous

no

5. jim_thompson5910

the rule is that if you have a negative exponent, you take the reciprocal of the base to make the exponent positive in this case,$\Large 3^{-3} = \frac{1}{3^3}$

6. jim_thompson5910

other examples $\Large 2^{-7} = \frac{1}{2^7}$ $\Large 9^{-4} = \frac{1}{9^4}$

7. jim_thompson5910

|dw:1440721304965:dw|

8. jim_thompson5910

now you need to simplify the following |dw:1440721342332:dw|

9. anonymous

|dw:1440721303495:dw|

10. jim_thompson5910

is that a p? or a q? or a 9? (see circled) |dw:1440721502919:dw|

11. anonymous

that was a p sorry got connfused |dw:1440721443290:dw|

12. jim_thompson5910

ok now simplify that and you'll be done

13. jim_thompson5910

think of p as p/1

14. jim_thompson5910

|dw:1440721767667:dw|

15. jim_thompson5910

when dividing 2 fractions, flip the second fraction and multiply

16. anonymous

what would i get?

17. anonymous

like p=27?

18. jim_thompson5910

|dw:1440721912391:dw|

19. anonymous

2 and 27p

20. jim_thompson5910

|dw:1440721937428:dw|

21. anonymous

yea but i still get 1/27p

22. jim_thompson5910

$\Large \frac{1}{27p}$ is the answer

23. anonymous

oh ok thank you

24. anonymous

if i have a diff question later ill post it

25. jim_thompson5910

no problem

26. anonymous

i need help

27. anonymous

|dw:1440724201473:dw|

28. anonymous

needs positive exponents,simplefy

29. jim_thompson5910

use the same rule to get |dw:1440724750331:dw|

30. jim_thompson5910

so you need to simplify $\LARGE \frac{1}{(-8v)^2}*w^3$

31. anonymous

so no multiplying?

32. jim_thompson5910

you do multiply think of the w^3 as w^3 over 1

33. jim_thompson5910

$\LARGE \frac{1}{(-8v)^2}*w^3 = \frac{1}{(-8v)^2}*\frac{w^3}{1}$

34. jim_thompson5910

then multiply straight across

35. jim_thompson5910

also, simplify (-8v)^2

36. anonymous

|dw:1440725573970:dw|

37. jim_thompson5910

it should be $\Large \frac{w^3}{64v^2}$

38. jim_thompson5910

you square the -8 to get 64 you square the v to get v^2

39. anonymous

ok

40. anonymous

41. jim_thompson5910

Use the rule $\LARGE \frac{x^a}{x^b} = x^{a-b}$

42. jim_thompson5910

|dw:1440726015179:dw|

43. jim_thompson5910

so for example $\Large \frac{B^{12}}{B^{8}} = B^{12-8} = B^4$ this is just an example and not the answer

44. anonymous

ok

45. anonymous

i think thats all thank you

46. jim_thompson5910

47. anonymous

can i ask you 2 more questions?

48. anonymous

|dw:1440729886308:dw|

49. anonymous

@jim_thompson5910

50. jim_thompson5910

think of x as x^1 y as y^1 then use that last rule I posted above

51. jim_thompson5910

|dw:1440730407539:dw|

52. jim_thompson5910

do the same for y and z

53. jim_thompson5910

and of course, 32 over -8 = -4