Math Help

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 1:38 am

don't think im retarded i spent a half a year in Pre-Calc and a full year in Calc. This stuff is basic algebra and you don't know how much it shames me for being confused but in my defense i always hated math.

So my brothers doing his math HW and he ask me for help heres the problem.

(X^-2)^-2
-------------
X^5

i said the answer was X^4/X^5

his friend said the answer was 1/X^9

i talked to said friend over the phone and he sounded very convincing while i was kinda like >>

Now heres the next problem

(2x^-3)^-2

i said the answer was 2x^6

his friend said the answer was 2/X^6

Im completely confused and lack all confidence in my math abilities at this point but i feel bad for not being able to help especially when i should know this stuff. So who is right here im right don't explain it if im wrong pls tell me why. And plz ppl im letting down my guard here don't lol @ me to much

crazyskwrls · Post by **crazyskwrls** » Fri Oct 26, 2007 1:42 am

((X^-2)^-2 )/X^5 = 1/x = X^4/X^5

(2x^-3)^-2 = 1/4x^6

^

for this one is it (2x^-3)^-2 or ((2x)^-3)^-2

if it is ((2x)^-3)^-2 = 64x^6= (2x)^6

thats my answer i am not exactly sure if it is right

Vandall · Post by **Vandall** » Fri Oct 26, 2007 1:45 am

as I've said before; when in doubt answer is 7.5

Xyzzzy · Post by **Xyzzzy** » Fri Oct 26, 2007 1:46 am

Vandall wrote:as I've said before; when in doubt answer is 7.5

I'm going to pass my geometry test like that tommorow now

thanks

Vandall · Post by **Vandall** » Fri Oct 26, 2007 1:49 am

Xyzzzy wrote:
Vandall wrote:as I've said before; when in doubt answer is 7.5

I'm going to pass my geometry test like that tommorow now

thanks

you can't tell when im joking? :banghead:

Ell · Post by **Ell** » Fri Oct 26, 2007 1:49 am

Never mind for number 2 it's X^6/4

Beanstalk · Post by **Beanstalk** » Fri Oct 26, 2007 1:49 am

If your dealing with negative exponents, then you must take the reciprocal of it, meaning move it to the bottom, and then you should knwo the rest, its basic fractions.

Dystopia · Post by **Dystopia** » Fri Oct 26, 2007 1:58 am

No, when in doubt, x=3, and y=2.

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 2:06 am

Ell wrote:Never mind for number 2 it's X^6/4

are you sure about this?

crazyskwrls · Post by **crazyskwrls** » Fri Oct 26, 2007 2:08 am

i am pretty sure i am right

Dystopia · Post by **Dystopia** » Fri Oct 26, 2007 2:08 am

XemnasXD wrote:
Ell wrote:Never mind for number 2 it's X^6/4

are you sure about this?

Final answer...

Shadowman20818 · Post by **Shadowman20818** » Fri Oct 26, 2007 2:10 am

1/X

(X^6)/4

Ell · Post by **Ell** » Fri Oct 26, 2007 2:13 am

XemnasXD wrote:
Ell wrote:Never mind for number 2 it's X^6/4

are you sure about this?

Yeah, you can test it by setting x as any number and seeing if they equal.

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 2:13 am

say (2X^3)^2

by some of your logic i would square everything int he parenthesis so the answer would be 4X^6

Aiyas · Post by **Aiyas** » Fri Oct 26, 2007 2:15 am

crazyskwrls wrote:i am pretty sure i am right

You are, as far as I can see.

EDIT:
Ell is right on #2

(x^6)/4

Shadowman20818 · Post by **Shadowman20818** » Fri Oct 26, 2007 2:15 am

XemnasXD wrote:say (2X^3)^2

by some of your logic i would square everything int he parenthesis so the answer would be 4X^6

Laws of exponents, yes you would do that.

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 2:17 am

im getting conflicting answers....

i trust Aiyas and Ell but Aiyas trust crazy whose answer is different from Ell's

Ell · Post by **Ell** » Fri Oct 26, 2007 2:17 am

crazyskwrls wrote:for this one is it (2x^-3)^-2 or ((2x)^-3)^-2

if it is ((2x)^-3)^-2 = 64x^6= (2x)^6

thats my answer i am not exactly sure if it is right

Those two equations are different.

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 2:19 am

*explodes* how did i make it through HS....*cries* i took AP calc....i got a 2 thats not good but it means i at least got my name and 1 problem right

Aiyas · Post by **Aiyas** » Fri Oct 26, 2007 2:19 am

Let me clarify:

#1 1/x

#2 (x^6)/4

Ell · Post by **Ell** » Fri Oct 26, 2007 2:20 am

Aiyas wrote:Let me clarify:

#1 1/x

#2 (x^6)/4

Yay

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 2:20 am

Aiyas wrote:Let me clarify:

#1 1/x

#2 (x^6)/4

how do you figure? can you explain in words...

crazyskwrls · Post by **crazyskwrls** » Fri Oct 26, 2007 2:21 am

crazyskwrls wrote:((X^-2)^-2 )/X^5 = 1/x = X^4/X^5

(2x^-3)^-2 = 1/4x^6

^

for this one is it (2x^-3)^-2 or ((2x)^-3)^-2

if it is ((2x)^-3)^-2 = 64x^6= (2x)^6

thats my answer i am not exactly sure if it is right

read the whole thing

Shadowman20818 · Post by **Shadowman20818** » Fri Oct 26, 2007 2:22 am

crazyskwrls wrote:((X^-2)^-2 )/X^5 = 1/x = X^4/X^5

(2x^-3)^-2 = 1/4x^6

^

for this one is it (2x^-3)^-2 or ((2x)^-3)^-2

if it is ((2x)^-3)^-2 = 64x^6= (2x)^6

thats my answer i am not exactly sure if it is right

/sigh, here's where you went wrong

(2x^-3)^-2

1/(2x^-3)^2

you did: 1/(2^2*x^3(2)) rather than 1/(2^2x^-3(2))

crazyskwrls · Post by **crazyskwrls** » Fri Oct 26, 2007 2:23 am

XemnasXD wrote:
Aiyas wrote:Let me clarify:

#1 1/x

#2 (x^6)/4

how do you figure? can you explain in words...

for 1

x^4/X^5 which is what u got = xxxx/xxxxx

the four x cancel out leave 1 x on bottom = 1/x

XemnasXD · Post by **XemnasXD** » Fri Oct 26, 2007 2:27 am

I give up ..... my brother and his math thx for all your help but i can honestly say im more confused now. but i know you guys actually took me serious and i really appreciate that (you don't know how much)

feel free to continue though if it'll bug you but im going back to do my safe chemistry work...where laws and rules are in english and everything makes sense...

crazyskwrls · Post by **crazyskwrls** » Fri Oct 26, 2007 2:28 am

Shadowman20818 wrote:
crazyskwrls wrote:((X^-2)^-2 )/X^5 = 1/x = X^4/X^5

(2x^-3)^-2 = 1/4x^6

^

for this one is it (2x^-3)^-2 or ((2x)^-3)^-2

if it is ((2x)^-3)^-2 = 64x^6= (2x)^6

thats my answer i am not exactly sure if it is right

/sigh, here's where you went wrong

(2x^-3)^-2

1/(2x^-3)^2

you did: 1/(2^2*x^3(2)) rather than 1/(2^2x^-3(2))

i had 2 answer for number 2 u can either consider:
2x as a whole
or
2 and x as separate

Ell · Post by **Ell** » Fri Oct 26, 2007 2:29 am

XemnasXD wrote:
Aiyas wrote:Let me clarify:

#1 1/x

#2 (x^6)/4

how do you figure? can you explain in words...

Anything to a negative power is just the reciprocal. So x^-1 equals 1/x^1, x^-2 equals 1/x^2, x^-3 equals 1/x^3 ect ect.

For problem one, (x^-2)^-2 is just (1/x^2)^-2 which is 1/((1/x^2)^2) and that equals 1/(1/x^4) which equals x^4. Divide that by x^5 and you get 1/x since the x's cancel.

Same thing for problem two. 2x^-3 equals 2/x^3. And (2/x^3)^-2 is 1/((2/x^3)^2) which is 1/(4/x^6) and that equals x^6/4.

Edit: The chemistry class I had was basically half math, so it was like taking one and a half math classes a day :banghead:

Aiyas · Post by **Aiyas** » Fri Oct 26, 2007 2:30 am

#1

(x^-2)^-2 / x^5

Multiply exponents in the numerator. (-2 * -2 = 4)

You should now have:

(x^4)/(x^5)

which is equivalent to:

x^(4-5) = x^-1 = 1/x

#2

(2x^-3)^-2

Rewrite as 1/(2x^-3)^2

Multiply the square to get 1/(4x^-6), or (1/4)*(1/(x^-6))

The second factor can be re-written as (x^-6)^-1, or x^(-6*-1), or x^6

So you have (1/4)*x^6, or x^6/4

Shadowman20818 · Post by **Shadowman20818** » Fri Oct 26, 2007 2:31 am

crazyskwrls wrote:
Shadowman20818 wrote:
crazyskwrls wrote:((X^-2)^-2 )/X^5 = 1/x = X^4/X^5

(2x^-3)^-2 = 1/4x^6

^

for this one is it (2x^-3)^-2 or ((2x)^-3)^-2

if it is ((2x)^-3)^-2 = 64x^6= (2x)^6

thats my answer i am not exactly sure if it is right

/sigh, here's where you went wrong

(2x^-3)^-2

1/(2x^-3)^2

you did: 1/(2^2*x^3(2)) rather than 1/(2^2x^-3(2))

i had 2 answer for number 2 u can either consider:
2x as a whole
or
2 and x as separate

I'm pointing out that the 2 and x^-3 are separate one is done wrong.