## some.random.cool.kid one year ago New question If anyone has time to spare I have a few questions I need help with

1. some.random.cool.kid

When subtracted, what is the simplified form of (54 − 27i) − (39 + 5i)? Write the answer in standard form, a + bi, where a and b are real numbers.

2. some.random.cool.kid

I know I have to distribute and also I can do like terms

3. Nnesha

distribute 2nd parentheses by negative one

4. some.random.cool.kid

yeah igot tha tpart already im confused as to the rest...

5. Nnesha

alright show ur work

6. some.random.cool.kid

well this is what I have so far 54 − 27i) − (39 + 5i) 27{-1} - 5{-1} thats the part I was stuck on and Dont I have to change the sign when canceling or something like that?

7. some.random.cool.kid

then 54 minus the 39 and now im stuck.

8. some.random.cool.kid

I bleive its 15 addition to something

9. Nnesha

well you didn't distribute the 2nd parentheses

10. Nnesha

why is it 27{-1} = ?

11. some.random.cool.kid

sq to power of 2?

12. some.random.cool.kid

-1 x -1 imaginary number

13. Nnesha

oh well no i = sqrt{-1} i^2 = -1 but there isn't any i^2

14. some.random.cool.kid

so its not power of two?

15. Nnesha

|dw:1442334862171:dw| well we don't have to apply the foil method distribute 2nd parenthesizes by -1 and then combine like terms

16. some.random.cool.kid

uh -5?

17. some.random.cool.kid

i

18. some.random.cool.kid

so 27i + -5i?

19. Nnesha

welll leave the first parentheses alone for first step $\huge\rm -1(39+5i)$ distribute multiply both terms in the parentheses by negative 1

20. some.random.cool.kid

so -39 + -5i?

21. Nnesha

yes right

22. some.random.cool.kid

i thought you needed like terms?

23. Nnesha

yes right now we have like terms |dw:1442335143368:dw| combine like terms

24. some.random.cool.kid

32i btw

25. Nnesha

if we have negative sign outside the parentheses we must hve to distribute first

26. Nnesha

hmm -27i-5i = ?

27. some.random.cool.kid

i guess so

28. some.random.cool.kid

but it has minus

29. Nnesha

yes it supposed to be -32 -# -# = -#

30. some.random.cool.kid

so far i se 15 - 32i? am I steearing in the righ tdirection?

31. Nnesha

yes that's right

32. some.random.cool.kid

- hashtag lol

33. some.random.cool.kid

So am I done now ? because I need help with more if you have time to spare explaining a few with me.

34. Nnesha

yes that's itdone!

35. some.random.cool.kid

would you mind helping me with this one too? The number root of order seven of fifty three cubed can be written as fifty three to the power of start fraction cap a over cap b end fraction end power. What is the value of A?

36. some.random.cool.kid

ooo that came out messy maybe I can attach a file

37. some.random.cool.kid

38. some.random.cool.kid

39. some.random.cool.kid

ok done. :D

40. Nnesha

ah okay so can convert root to an exponent form exponent rule $\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }$ index becomes the denominator of the fraction

41. some.random.cool.kid

sorry my internet went bye bye lol

42. some.random.cool.kid

ok so now what?

43. Nnesha

that's it

44. some.random.cool.kid

im lost?

45. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha ah okay so can convert root to an exponent form exponent rule $\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }$ index becomes the denominator of the fraction $$\color{blue}{\text{End of Quote}}$$ read this

46. some.random.cool.kid

nvm i got it 7/3?

47. Nnesha

no that's not right

48. some.random.cool.kid

hmm?

49. Nnesha

index would be the denominator

50. some.random.cool.kid

ooohh ohhh sorry

51. some.random.cool.kid

get it now

52. Nnesha

alright so A would be what ?

53. some.random.cool.kid

isnt the 3 the den?

54. Nnesha

yes right

55. some.random.cool.kid

because the numer became the power

56. some.random.cool.kid

alright thanks

57. Nnesha

excellent! yw good work!