RE: Any Math Wizards Care to Take A Crack At This Problem?

Collarspace Discussion Forums

Home Login Search

RE: Any Math Wizards Care to Take A Crack At This Problem?

View related threads: (in this forum | in all forums)

Logged in as: Guest

All Forums >> [Casual Banter] >> Off the Grid >> RE: Any Math Wizards Care to Take A Crack At This Problem?

Page: << < prev 1 [2] 3 4 next > >>

Login

Message

<< Older Topic Newer Topic >>

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:31:33 AM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

quote:

ORIGINAL: MissAsylum

Photo of the problem

I'll try it by sight :

Each term has X * Y^2, right?
That leaves:

X*(Y^2) * {(3X)^.25 + (3^5*X)^.25 + [(2^4 * 3) * X]^.25}
X*(Y^2) * {(3X)^.25 + 3*(3X)^.25 + 2*(3 X)^.25}
And 1+3+2 = 6,

So: X * (Y^2) * 6* (3X)^.25

Final answer. There are variations, of course. Put the X back into the quarter root and so on.

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to MissAsylum)

Post #: 21

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:34:34 AM

MasterG2kTR

Posts: 6677
Joined: 8/7/2004
From: Wisconsin
Status: offline

Yup...I was just gonna post that answer for you....SERIOUSLY!! I just saw this and worked it out cuz that just happens to be the math chapter I will be testing for tomorrow.

But yes.....6xy^2 4^√3x this is the same answer I came up with.

(in reply to MissAsylum)

Post #: 22

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:35:10 AM

VaguelyCurious

Posts: 5264
Joined: 12/2/2009
From: United Kingdom
Status: offline

I left everything in the quarter root, and got ( 294 (x^5) (y^8) )^(1/4)

_____________________________

Sthetic on FetLife.

(in reply to LanceHughes)

Post #: 23

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:36:26 AM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

quote:

ORIGINAL: MissAsylum

So.... 6xy^2 4^√3x.

I think.

YES! You GO, girl! Except the notation on a keyboard really doesn't allow for fractional roots the way you're trying to show. Either (3x)^.25 OR (3x)^(1/4) - The 2nd is my prefernce.

Notation is EVERYTHING!

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to MissAsylum)

Post #: 24

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:40:16 AM

MissAsylum

Posts: 1863
Joined: 1/9/2009
Status: offline

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: MissAsylum

So.... 6xy^2 4^√3x.

I think.

YES! You GO, girl! Except the notation on a keyboard really doesn't allow for fractional roots the way you're trying to show. Either (3x)^.25 OR (3x)^(1/4) - The 2nd is my prefernce.

Notation is EVERYTHING!

Ugh. Please...don't remind me.

_____________________________

I hate when I'm wearing my apple bottom jeans, but i can't find my boots with the fur.

(in reply to LanceHughes)

Post #: 25

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:41:07 AM

GreedyTop

Posts: 52100
Joined: 5/2/2007
From: Savannah, GA
Status: offline

*sigh* I'm such a bonehead when it comes to math... I wish my Grandfather was around so I could have run this by him...

_____________________________

polysnortatious
Supreme Goddess of Snark
CHARTER MEMBER: Lance's Fag Hags!
Waiting for my madman in a Blue Box.

(in reply to LanceHughes)

Post #: 26

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:41:07 AM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

quote:

ORIGINAL: VaguelyCurious

I left everything in the quarter root, and got ( 294 (x^5) (y^8) )^(1/4)

Well, your integer is WAY off.
6 * (6 ^ .25) = (6 ^4)^.25 * (6^.25) = (6^5)^.25 = 7776 ^ .25

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to VaguelyCurious)

Post #: 27

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:43:29 AM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

quote:

ORIGINAL: MissAsylum

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: MissAsylum

So.... 6xy^2 4^√3x.

I think.

YES! You GO, girl! Except the notation on a keyboard really doesn't allow for fractional roots the way you're trying to show. Either (3x)^.25 OR (3x)^(1/4) - The 2nd is my prefernce.

Notation is EVERYTHING!

Ugh. Please...don't remind me.

And get those parens in there!

6xy^2 could be just about anything!

inc. (6^2)(x^2)(y^2)

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to MissAsylum)

Post #: 28

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:52:53 AM

VaguelyCurious

Posts: 5264
Joined: 12/2/2009
From: United Kingdom
Status: offline

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: VaguelyCurious

I left everything in the quarter root, and got ( 294 (x^5) (y^8) )^(1/4)

Well, your integer is WAY off.
6 * (6 ^ .25) = (6 ^4)^.25 * (6^.25) = (6^5)^.25 = 7776 ^ .25

Where are you getting all the sixes from?

I made the first term (3(x^5)(y^8))^(1/4)
the second term (243(x^5)(y^8))^(1/4)
and the third term (48(x^5)(y^8))^1/4

....and then I'm an idiot, because you can't add them straight up. In my defence, I'm very tired...

Whoops.

Sorry.

_____________________________

Sthetic on FetLife.

(in reply to LanceHughes)

Post #: 29

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:54:19 AM

MissAsylum

Posts: 1863
Joined: 1/9/2009
Status: offline

You have a point. Why can't I type it out the way its supposed to look?

All pretty and confusing-like?

_____________________________

I hate when I'm wearing my apple bottom jeans, but i can't find my boots with the fur.

(in reply to LanceHughes)

Post #: 30

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:55:03 AM

VaguelyCurious

Posts: 5264
Joined: 12/2/2009
From: United Kingdom
Status: offline

quote:

ORIGINAL: MissAsylum

You have a point. Why can't I type it out the way its supposed to look?

All pretty and confusing-like?

If you've got a fairly recent version of microsoft word, it should have an equation builder. That's what you want.

_____________________________

Sthetic on FetLife.

(in reply to MissAsylum)

Post #: 31

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 12:01:33 PM

MissAsylum

Posts: 1863
Joined: 1/9/2009
Status: offline

I don't since my laptop is a dinosaur.

_____________________________

I hate when I'm wearing my apple bottom jeans, but i can't find my boots with the fur.

(in reply to VaguelyCurious)

Post #: 32

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 12:13:15 PM

VaguelyCurious

Posts: 5264
Joined: 12/2/2009
From: United Kingdom
Status: offline

Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

The integer (if we're leaving everything inside that fourth root) is: (3^1/4 + 243^1/4 + 48^1/4)^4

243^1/4 = 3(3^(1/4))

48^1/4 = 2(3^(1/4))

so the part inside the 4th power is:

1(3^(1/4)) + 3(3^(1/4)) + 2(3^(1/4)) = 6(3^(1/4))

And when you ^4 that you get 3(6^4) = 3888

So I don't know why your figure is double mine.

< Message edited by VaguelyCurious -- 10/12/2011 12:21:15 PM >

_____________________________

Sthetic on FetLife.

(in reply to VaguelyCurious)

Post #: 33

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 12:43:10 PM

MasterG2kTR

Posts: 6677
Joined: 8/7/2004
From: Wisconsin
Status: offline

quote:

ORIGINAL: LanceHughes

6xy^2 could be just about anything!

inc. (6^2)(x^2)(y^2)

NO...it can't be that because that would be the same as 36x^2y^2

There are only three expressions in the final answer. (6) (x) & (y^2) the ^2 you see in the solution only applies to the y expression no the 6 & x

quote:

ORIGINAL: VaguelyCurious
And when you ^4 that you get 3(6^4) = 3888

Can't be that either because you have no idea what the value of x or y is. You can only end up with an exponential expression without a definitive solution until you know the actual values of x & y.

ETA: And don't forget you still have the 4th root of 3x as well.

< Message edited by MasterG2kTR -- 10/12/2011 12:44:48 PM >

(in reply to LanceHughes)

Post #: 34

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 12:52:28 PM

shallowdeep

Posts: 343
Joined: 9/1/2006
From: California
Status: offline

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: shallowdeep
6 * x * y^2 * (3 * x)^.25

FAIL!

quote:

ORIGINAL: LanceHughes

X * (Y^2) * 6* (3X)^.25

Final answer.

Ah! I knew I shouldn't have trusted all those teachers when they claimed scalar multiplication was commutative!

Or, if I ask very nicely, do you suppose there might be a chance I could get my grade revised? :)

Of course, I happily admit your pedagogical technique was far better… I was just being impatient.

quote:

ORIGINAL: VaguelyCurious
Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

Lance accidentally started assuming a 6 inside the fourth root, rather than a 3, in the post where he pointed out your mistake. The correct value is 3888. As someone may have mentioned…

(in reply to LanceHughes)

Post #: 35

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 1:10:00 PM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

quote:

ORIGINAL: MasterG2kTR

quote:

ORIGINAL: LanceHughes

6xy^2 could be just about anything!

inc. (6^2)(x^2)(y^2)

NO...it can't be that because that would be the same as 36x^2y^2

There are only three expressions in the final answer. (6) (x) & (y^2) the ^2 you see in the solution only applies to the y expression not to the 6 & x

<snipped>

Well, that's my point. HOW is the reader to KNOW the span of the squaring symbol. Maybe we could compromise on: 6X * Y^2

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to MasterG2kTR)

Post #: 36

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 1:16:36 PM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

quote:

ORIGINAL: shallowdeep

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: shallowdeep
6 * x * y^2 * (3 * x)^.25

FAIL!

quote:

ORIGINAL: LanceHughes

X * (Y^2) * 6* (3X)^.25

Final answer.

Ah! I knew I shouldn't have trusted all those teachers when they claimed scalar multiplication was commutative!

Or, if I ask very nicely, do you suppose there might be a chance I could get my grade revised? :)

Of course, I happily admit your pedagogical technique was far better… I was just being impatient.

HUGE BLUSH! ! ! ! I guess I was seeing (or not seeing) parenthesisessissippi.

quote:

ORIGINAL: VaguelyCurious
Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

Lance accidentally started assuming a 6 inside the fourth root, rather than a 3, in the post where he pointed out your mistake. The correct value is 3888. As someone may have mentioned…

HUGE(r) BLUSH! ! ! !

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to shallowdeep)

Post #: 37

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 1:21:13 PM

MasterG2kTR

Posts: 6677
Joined: 8/7/2004
From: Wisconsin
Status: offline

quote:

ORIGINAL: LanceHughes

Well, that's my point. HOW is the reader to KNOW the span of the squaring symbol. Maybe we could compromise on: 6X * Y^2

If the expression was listed like this (6xy)^2 then the square would apply to all three expressions and your assumption would be correct. Any power indicator only applies to the expression it is directly related to unless shown as above.

(in reply to LanceHughes)

Post #: 38

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 1:31:16 PM

LanceHughes

Posts: 4737
Joined: 2/12/2004
Status: offline

Another! Another! Give us another, Teacher! Pleeeeeeease!

// G'damn, f*ckin' Lance //

_____________________________

"Train 'em the right way - my way." Lance Hughes
"Advice is what we ask for when we already know the answer, but wish we didn't." Erica Jong

10 fluffy points
50 nz points

Member: VAA's posse

(in reply to MasterG2kTR)

Post #: 39

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 1:35:21 PM

MasterG2kTR

Posts: 6677
Joined: 8/7/2004
From: Wisconsin
Status: offline

*looks for ruler*

Hold out those knuckles young man

(you know what's coming next I assume)

Yup....trip to the principle's office

(in reply to LanceHughes)

Post #: 40

Page: << < prev 1 [2] 3 4 next > >>

All Forums >> [Casual Banter] >> Off the Grid >> RE: Any Math Wizards Care to Take A Crack At This Problem?

Page: << < prev 1 [2] 3 4 next > >>

Jump to:

New Messages	No New Messages
Hot Topic w/ New Messages	Hot Topic w/o New Messages
Locked w/ New Messages	Locked w/o New Messages

Post New Thread

Reply to Message

Post New Poll

Submit Vote

Delete My Own Post

Delete My Own Thread

Rate Posts

Collarchat.com © 2025

Terms of Service

0.172