Question

plz help

Answer 1

hi

Answer 2

whats the question

Answer 3

\[(3x)^-1 \over 4\]

Answer 4

write each expression so that it contains only positive exponents

Answer 5

I dunno this 1 im so sorry

Answer 6

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

Answer 7

how did you do that

Answer 8

\(b^{-1}=\frac{1}{b}\)

Answer 9

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

Answer 10

how did you get b

Answer 11

i picked \(b\) as a variable
you can use anyone you prefer

Answer 12

oh ok

Answer 13

satellite are you there:(

Answer 14

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

Answer 15

idk i just dont know how to do it

Answer 16

what happened to the -1

Answer 17

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

Answer 18

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

Answer 19

yes

Answer 20

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

Answer 21

no

Answer 22

i thought x-1=1/x1

Answer 23

Same thing. \[x^1=x\] just like \[3^{1}=3\]

Answer 24

ok

Answer 25

but isnt negative 1

Answer 26

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.

Answer 27

oh ok sorry

Answer 28

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

Answer 29

yes

Answer 30

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

Answer 31

oh ok

Answer 32

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

Answer 33

yeah i know

Answer 34

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

Answer 35

uh \[1\over x\]

Answer 36

i think

Answer 37

am i right

Answer 38

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

Answer 39

?

Answer 40

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 \[x\]

Answer 41

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

Answer 42

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

Answer 43

so I fliped x^21 over to the numerator and get
x−2

then I take away the negative sign which gets me
x2

Answer 44

-2

Answer 45

x^2 is right. Next one:\[x ^{-3}\]

Answer 46

\[1\over x^2\]

Answer 47

Are you sure?

Answer 48

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

Answer 49

Might be because the number was so small.

Answer 50

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

Answer 51

\[1 \over x^3\]

Answer 52

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

Answer 53

idk 4/x^2

Answer 54

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

Answer 55

so it would be 4x^2

Answer 56

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

Answer 57

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

Answer 58

im still thinking

Answer 59

3x/4

Answer 60

i got it wrong

Answer 61

Then change it :P

Answer 62

change what

Answer 63

remember what we did with \[x^{-2}\]

Answer 64

1/x^2

Answer 65

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

Answer 66

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

Answer 67

but 3 is a numerator too it can turn to 1

Answer 68

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

Answer 69

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

Answer 70

\[1\over (3x)^1\]

Answer 71

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

Answer 72

But you agree with it, right?

Answer 73

yes

Answer 74

i should've looked at it as a whole

Answer 75

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

Answer 76

yes

Answer 77

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

Answer 78

and that simplifies to\[\frac{ 1 }{ 12x }\]

Answer 79

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