Question

is (-2c)^3 simplified -2c^3

Answer 1

no it's not

Answer 2

No.

Generally, \((ab)^{n} = a^{n}\cdot b^{n}\)

In your particular problem, \((-2c)^{3} = (-2)^{3}\cdot c^{3} = -8c^{3}\)

Answer 3

okay thanks i just wasnt sure if you cubed the -2

Answer 4

no, exponentiation happens before subtraction or unary minus. you cube the 2, then stick the minus sign on the front.

Answer 5

if have trouble write it out for example -2c *-2c *-2c

Answer 6

this is a mistake that is made very frequently, even by experienced people who aren't careful!

Answer 7

thanks guys

Answer 8

sorry, I was a bit unclear: I was referring to the -a^b case: you raise a to the b power, then put the minus sign on the front.

Answer 9

how would you simplify (v/w^-2)^-2 then?? :s do you flip it orr

Answer 10

I would work from the inside out: v/w^-2 would become vw^2
then (vw^2)^-2 you can just multiply the exponents giving you v^-2w^-4 or 1/(v^2w^4)

Answer 11

or you could go (v^-1w^-2)^2 = v^-2w^-4 = 1/(v^2w^4)

just pick a method that works reliably for you and stick with it

Answer 12

shortest path to the answer isn't the best if you make more mistakes!

Answer 13

but v^-2w^-4 is not simplied at least not where i go to school

Answer 14

yeah it cant have negative exponents

Answer 15

yup that's what I'm taught

Answer 16

i thought you had to flip the fraction because of the negative exponent outside the parenthesis but what about the negative exponent in the denominator????

Answer 17

Unary minus is OFTEN of higher precedence than exponentiation. It does depend on the working environment. As a very good general rule, check before you write code.

As far as algebra one rules of precedence, I've seen it both ways in different reliable textbooks. Personally, I prefer the unary minus with higher precedence. Obviously, not everyone agrees.

It is a mistake to treat unary minus higher ONLY if it is clearly defined contrary to this position. If it is not clearly defined, I am aware of no authoritative guidance one way or the other.

Answer 18

I dont know then

Answer 19

you can do these problems a few different ways but the answer at the end is the same. u don't have to do a certain step first... u just need to find what works best for u that results in not making silly errors

Answer 20

would i flip the fraction so it becomes (w^2/v)^2?

Answer 21

ur problem isn't (v^-2/w^-2) so if u flipped ur problem inside first it would be (vw^2)^-2

Answer 22

the outside exponent is actually a -3 ugh sorry

Answer 23

when doing these types of problems u might want to rewrite it like (v^1w^-2)^-3

Answer 24

why get rid of the fraction bar?

Answer 25

sorry i just am having troubles with these

Answer 26

sorry my bad (v^1/w^-2)^-3

Answer 27

but to simplify?

Answer 28

me personally i write mine out like V^(1*-3) / w^(-2*-3)

Answer 29

and that's simplified?

Answer 30

no just one step closer ..... then V^-3/w^6 then 1/v^3w^6

Answer 31

lot times problems get larger so doing it that way helps....

Answer 32

ok. why the 1? i thought uou jut flipped the negative exponent to the denominator to get w^6 / v^3

Answer 33

how did u get w^6 on top?

Answer 34

\((v/w^-2)^-2 = \dfrac{1}{\left(v/w^{-2}\right)^{2}} = \dfrac{1}{\left(v\cdot w^{2}\right)^{2}} = \dfrac{1}{v^{2}w^{4}}\)

Answer 35

I thought you always switched the negative exponent to theother side to make it positive

Answer 36

\( (v^{1}/w^{-2})^{-3} = \left(v\cdot w^{2}\right)^{-3} = \dfrac{1}{\left(v\cdot w^{2}\right)^{3}} = \dfrac{1}{v^{3}\cdot w^{6}}\)

Answer 37

if u have 1/w^-2 u can switch that but if u have have v^1/w^-2 u can't switch both only equals v^1w^2

Answer 38

okay >.<

Answer 39

\(\dfrac{v}{w^{-2}} = v\cdot w^{2}\)

Answer 40

when u have v^-1/ w^-2 u switch both w^2/v^1

Answer 41

u don't have to write v^1 i just like to show that because it helps seeing it better for some people

Answer 42

do u think u understand it better now

Answer 43

kind of