## anonymous 3 years ago plz help

1. anonymous

hi

2. anonymous

whats the question

3. anonymous

$(3x)^-1 \over 4$

4. anonymous

write each expression so that it contains only positive exponents

5. anonymous

I dunno this 1 im so sorry

6. anonymous

$\frac{(3x)^{-1}}{ 4}=\frac{1}{4(3x)}$

7. anonymous

how did you do that

8. anonymous

$$b^{-1}=\frac{1}{b}$$

9. anonymous

in general, $b^{-n}=\frac{1}{b^n}$

10. anonymous

how did you get b

11. anonymous

i picked $$b$$ as a variable you can use anyone you prefer

12. anonymous

oh ok

13. anonymous

satellite are you there:(

14. anonymous

I don't know what else there is to say. What don't you understand?

15. anonymous

idk i just dont know how to do it

16. anonymous

what happened to the -1

17. anonymous

This is what your problem looks like, right? $\frac{ (3x)^{-1} }{ 4 }$

18. anonymous

This is what your problem looks like, right? $\frac{ (3x)^{-1} }{ 4 }$

19. anonymous

yes

20. anonymous

Did you know that $x^{-1}=\frac{ 1 }{ x }$?

21. anonymous

no

22. anonymous

i thought x-1=1/x1

23. anonymous

Same thing. $x^1=x$ just like $3^{1}=3$

24. anonymous

ok

25. anonymous

but isnt negative 1

26. anonymous

One thing first. Whenever you want to say that x has an exponential value of -1, you should type x^-1 or x to the power of -1 rather than x-1 because x-1 is just subtraction.

27. anonymous

oh ok sorry

28. anonymous

Is this what you're saying?$x^{-1}=\frac{ 1 }{ x^{1} }$

29. anonymous

yes

30. anonymous

Then you are right. But, as I said, $x^{1}=x$ So you can just change the $x^{1}$ to $x$

31. anonymous

oh ok

32. anonymous

This means that $x^{−1}=\frac{ 1 }{ x }$

33. anonymous

yeah i know

34. anonymous

Rewrite this so that there are only positive exponents:$\frac{ 1 }{ x^{-1} }$

35. anonymous

uh $1\over x$

36. anonymous

i think

37. anonymous

am i right

38. anonymous

No. General rule of thumb. If you see a negative number as an exponent, flip the term(with the exponent) over the fraction bar and make the exponent positive

39. anonymous

?

40. anonymous

so I flip x^-1 over to the numerator and get $x^{-1}$ then I take away the negative sign which gets me $x^{1}$ or $x$

41. anonymous

so every time there an equation like this $1 \over x ^-1$ you have to flip

42. anonymous

Yes. But ONLY if the exponent is negative. Next problem:$\frac{ 1 }{ x^{-2} }$

43. anonymous

so I fliped x^21 over to the numerator and get x−2 then I take away the negative sign which gets me x2

44. anonymous

-2

45. anonymous

x^2 is right. Next one:$x ^{-3}$

46. anonymous

$1\over x^2$

47. anonymous

Are you sure?

48. anonymous

I think it's a silly mistake, but it was x^3, not x^2

49. anonymous

Might be because the number was so small.

50. anonymous

oh yeah it was a little as i looked closer it was a 3

51. anonymous

$1 \over x^3$

52. anonymous

Next one(and the important one to help you with your problem):$\frac{ 4 }{ x^{-2} }$

53. anonymous

idk 4/x^2

54. anonymous

Why don't you bring x^-2 to the numerator and take away the negative sign?

55. anonymous

so it would be 4x^2

56. anonymous

Exactly. Now can you do this?$\frac{ (3x)^{-1} }{ 4 }$

57. anonymous

If you really don't understand, no is a good answer too.

58. anonymous

im still thinking

59. anonymous

3x/4

60. anonymous

i got it wrong

61. anonymous

Then change it :P

62. anonymous

change what

63. anonymous

remember what we did with $x^{-2}$

64. anonymous

1/x^2

65. anonymous

We flipped it into the denominator and took away the negative sign to get $\frac{ 1 }{ x^{2} }$

66. anonymous

Can you apply this to your problem or should I just explain?

67. anonymous

but 3 is a numerator too it can turn to 1

68. anonymous

yeah you should explain i get everything you taught me i just can put it all together

69. anonymous

Do you agree that $(3x)^{-1}=\frac{ 1 }{ (3x)^{1} }$or $\frac{ 1 }{ (3x) }$

70. anonymous

$1\over (3x)^1$

71. anonymous

Errr I mean that it equals both but the lower one without the exponent of 1 is simplified

72. anonymous

But you agree with it, right?

73. anonymous

yes

74. anonymous

i should've looked at it as a whole

75. anonymous

Now do you agree that $(3x)^{−1}\times \frac{ 1 }{ 4 }=\frac{ (3x)^{−1} }{ 4 }$

76. anonymous

yes

77. anonymous

so if you substitute the (3x)^-1 with $\frac{ 1 }{ 3x}$ in the expression, you get $\frac{ 1 }{ 4(3x) }$

78. anonymous

and that simplifies to$\frac{ 1 }{ 12x }$

79. anonymous

But if you think about it, all I actually did was flip the (3x)^-1 over to the denominator and take away the negative sign|dw:1363833108473:dw|

80. anonymous

is it the same thing as multiplying 1/3x and 1/4

81. anonymous

Yes, but the point is that I just flipped it over and that it's incredibly simple.

82. anonymous

Everyone else does it like:|dw:1363833370136:dw|

83. anonymous

you know what your right its really easy well thank you very much i truly get thank you very much you should be a teacher and again thank you very much 100x:)

84. anonymous

lolwut. Just understand it. and REMEMBER IT.

85. anonymous

yes sir

86. anonymous

lol

87. anonymous

hahahadont make me take out my shoe