Question

Question with exponents!

Question attached below...

Answer 1

\[\huge \frac{ (-xy^4)^{-4} }{ -2x^3 \times x^2y^3 }\]

Answer 2

\[(-2x^3)(x^2y^2) \implies -2 \cdot x^{3+2} \cdot y^2 =~?\]

Answer 3

i got the answer of \[\frac{ 1 }{ 2x^9y^19 }\] but the answer key says that it's negative.. i don't know why

Answer 4

Alright let's check each step.

Answer 5

what would you get for :

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean
\[\large (-2x^3)(x^2y^2) \implies -2 \cdot x^{3+2} \cdot y^2 =~?\]
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 6

\[\large -2x^5y^2\]

Answer 7

Alright, and when you distribute the \(-4\) power into every term in the ( ), you expand it like so: \[\large (-xy^4)^{-4} \implies (-x)^{-4} \cdot (y^4)^{-4} = ~?\]

Answer 8

\[\large -x^{-4} \times y^0\]

Answer 9

which is basically like \[\large -x^{-4}\]

Answer 10

\[\large \times \]

do you mean this symbol? if so, it's multiplication.

Answer 11

be careful, you wouldnt be adding the exponents here, you'd be multiplying them. \[\large (y^4)^{-4} = y^{4( -4)}\qquad (y^4)^{-4} \ne y^{4-4}\]

Answer 12

oh yes

Answer 13

i still don't understand how you got the negative...

Answer 14

yeah or whose ever work that was posted in the 3 parts

Answer 15

pretty fast eh? ^_^ thinking about making it a tool on OS as an extention

Answer 16

defeats the purpose of Openstudy then.

Answer 17

can you please help me understand why it is a negative?

Answer 18

Ok cotinuing from where we left off.

Answer 19

I just realized theres an easier way to simplify the numerator, check this out

Answer 20

Instead of expanding our numerator first, we can basically turn our entire term WITH the negative exponent into a positive one. \[\large (-xy^4)^{-4} \implies \dfrac{1}{(-xy^4)^4}\] you see how the negatives in the numerator would go away?

Answer 21

So now we expand our numerator, and we get: \[\large \frac{1}{x^4y^{16}}\]

Answer 22

Now we put it over our denominator: \[\large \dfrac{\dfrac{1}{x^4y^{16}}}{-2x^5y^3}\qquad \implies \qquad \dfrac{1}{-2x^{4+5}y^{16+3} } \]

Answer 23

Now whether we have the negative in the numerator OR denominator, it doesnt matter, the fraction altogether is STILL negative.

Answer 24

Do you follow, @calculusxy ?

Answer 25

i am still trying to grasp this.

Answer 26

if we have like complex fractions, then we would have to add the exponents n put it over one?

Answer 27

The only reason we did that, was because we treated \((-xy^4)\) as like a variable... let' say all that junk \(=a\). And then... a is raised to a negative power, and how would we turn that power positive? by putting it over 1. \(a^{-1} \iff \dfrac{1}{a}\)

Answer 28

So if we have a FRACTION in the numerator, and we're DIVIDING it by a term or terms in the denominator, we follow this rule: \[\dfrac{\dfrac{a}{b}}{c} \implies \frac{a}{bc}\]

Answer 29

okay...

Answer 30

i have one other question..

Answer 31

thanks for helping me understand that

Answer 32

Which art are you stuck on? WE'll go from there.

Answer 33

part*

Answer 34

so this time i am getting a negative, but it says that it's a positive.

\[\large -\frac{ ba^0 \times -a^3b^3 }{ (ab^0)^{-4} }\]

Answer 35

so i got \[\large -a^7b^4\] but the answer key simply says \[\large a^7b^4\]

Answer 36

@Jhannybean

Answer 37

@jim_thompson5910

Answer 38

@jim_thompson5910 it's the second expression w/ exponents.

Answer 39

oh okay thx !