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

Collarspace Discussion Forums

Home Login Search

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

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

Login

Message

<< Older Topic Newer Topic >>

Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 7:18:45 AM

MissAsylum

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

Word-for-word problem

Combine the expressions. (Assume the variables are nonnegative numbers)

x^4√3xy^8 + y^4√243x^5y^4 + y^2 4^√48x^5

i'm stumpped 3 ways til sunday. Can anybody help?

_____________________________

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

Post #: 1

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 7:37:58 AM

MadAxeman

Posts: 4171
Joined: 8/28/2008
From: UK
Status: offline

Isn't that the Binomial Expansion Theorem?

If you google Pascal's Triangle there should be a comprehensive deconstruction

_____________________________

Hitman for the Subby Mafia

(in reply to MissAsylum)

Post #: 2

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 9:51:33 AM

SoulAlloy

Posts: 2106
Joined: 8/23/2009
From: Preston, UK
Status: offline

Where do the "power" and "square route" symbols end?

For example the middle section y^4√243x^5y^4 - is it y to the power of 4√243x times x to the power of 5y to the power of 4?

Or is it y to the power of 4 times the square route of 243x to the power of 5 times y to the power of 5?

There's other combinations from reading it too, a bit hard to combine without knowing the beginning/ends

_____________________________

"Better to be a diamond with a flaw than a pebble without" - Confucius

"It'll be alright in the end - if it isn't alright, it's not the end." - unknown

Kinky crossdressing Whovian

Host of the Preston (UK) Munch, 2nd Wednesday each month

(in reply to MadAxeman)

Post #: 3

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:02:04 AM

MissAsylum

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

quote:

ORIGINAL: SoulAlloy

Where do the "power" and "square route" symbols end?

For example the middle section y^4√243x^5y^4 - is it y to the power of 4√243x times x to the power of 5y to the power of 4?

Or is it y to the power of 4 times the square route of 243x to the power of 5 times y to the power of 5?

There's other combinations from reading it too, a bit hard to combine without knowing the beginning/ends

Its the second possibility that you mentioned.

_____________________________

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

(in reply to SoulAlloy)

Post #: 4

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:19:49 AM

LanceHughes

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

quote:

ORIGINAL: MissAsylum

Word-for-word problem

Combine the expressions. (Assume the variables are nonnegative numbers)

x^4√3xy^8 + y^4√243x^5y^4 + y^2 4^√48x^5

i'm stumpped 3 ways til sunday. Can anybody help?

PARENTHESIS, please! So far, (I'm guessing) we have:

x^4√3xy^8 + (y^4)(√(243x)^5)(y^4) + y^2 4^√48x^5

So, the FIRST thing that jumps out of the middle term is (y^4)(y^4) = y^8, so the first term and second have at least one common factor. BUT!

What is the subject at hand? Factoring trinomials? The subject material WILL give a HUGE clue.

We have:

x^4√3xy^8 + (y^8)(√(243x)^5) + y^2 4^√48x^5

Same idea with
(x^4)x in first term = x^5 in both 2nd and third terms. BUT! Really, really need to know where pieces begin end for middle term. I'm hoping the radical sign does NOT cover x^5 in second term. If it doesn't we can write (x^5)x(bunch of stuff) and then see if that y term might break-down. I suspect (x^5 ) x {y^2 x [√48+y^6 x (√(243) ] } or.... something like that. One term got lost while I was trying to get []'s and {}'s lined up.... and still need parens. Maybe re-write with numerical roots at back of each term... that is, if they are just numerical roots.

< Message edited by LanceHughes -- 10/12/2011 10:42:49 AM >

_____________________________

"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 #: 5

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

VaguelyCurious

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

Well the first two terms simplify down to (x^5)(y^8)(10√3) pretty easily, but I have no idea what the third term is supposed to mean. Write it out using actual brackets? Because I suspect it's also supposed to be in terms of (X^5)(y^8) and you've just typed it out wrong.

_____________________________

Sthetic on FetLife.

(in reply to MissAsylum)

Post #: 6

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:40:39 AM

MissAsylum

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

error.

< Message edited by MissAsylum -- 10/12/2011 10:41:38 AM >

_____________________________

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

(in reply to MissAsylum)

Post #: 7

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:41:54 AM

mnottertail

Posts: 60698
Joined: 11/3/2004
Status: offline

you need to post reply, load the picture and embed in post.

_____________________________

Have they not divided the prey; to every man a damsel or two? Judges 5:30

(in reply to MissAsylum)

Post #: 8

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:42:23 AM

MissAsylum

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

Photo of the problem

Attachment (1)

_____________________________

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

(in reply to MissAsylum)

Post #: 9

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 10:53:37 AM

LanceHughes

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

YIKES! Those are fourth roots! Square roots of square roots!
First term: X * {3 * X * (Y^8)}^ (1/4)
First term: X * {[3 * X]^ (1/4)} * { (Y^8)^ (1/4)} because any root of a multiplicative expression is the same root of each part multiplied
First term: X * {[3 * X]^ (1/4)} * { Y^2 } because the quarter root of an eight power is a square root.
First term: X(Y^2) * {[3X]^ (1/4)}

Second term: Y * {243 * X^5 * Y^4} ^ (1/4) That 4th root of three makes me wonder about 243. 243 / 3 = 81 * 3, so we write:
Second term: Y * {3^5 * X^5 * Y^4} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] * [3 * X]} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] ^ (1/4)} * {[3 * X] ^ (1/4)}
Second term: Y * {3 * X * Y} * {[3 * X] ^ (1/4)}
Second term: 3 X( Y^2) * {[3X] ^ (1/4)}

Third term: Left for the student

< Message edited by LanceHughes -- 10/12/2011 11:11:24 AM >

_____________________________

"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 #: 10

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:06:06 AM

VaguelyCurious

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

quote:

ORIGINAL: MissAsylum

Photo of the problem

Ah. Then ignore what I just said about the first two terms, ok? Brb.

ETA: also, that bears absolutely no relation to what you actually typed out. You can see that, right?

< Message edited by VaguelyCurious -- 10/12/2011 11:10:08 AM >

_____________________________

Sthetic on FetLife.

(in reply to MissAsylum)

Post #: 11

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:08:58 AM

shallowdeep

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

6 * x * y^2 * (3 * x)^.25

or, equivalently,

(3888 * x^5 * y^8)^.25

(in reply to MissAsylum)

Post #: 12

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:10:59 AM

VaguelyCurious

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

I love you, man. Really I do.

_____________________________

Sthetic on FetLife.

(in reply to shallowdeep)

Post #: 13

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:13:40 AM

LanceHughes

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

quote:

ORIGINAL: shallowdeep

6 * x * y^2 * (3 * x)^.25

or, equivalently,

(3888 * x^5 * y^8)^.25

FAIL!

ETA: There are no factors of 6 across the three original terms. Not even factors of 2.

< Message edited by LanceHughes -- 10/12/2011 11:17:44 AM >

_____________________________

"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 #: 14

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:16:50 AM

VaguelyCurious

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

Although I disagree with you. I got 294^0.25 as the integer. Where'd you get such a big number?

_____________________________

Sthetic on FetLife.

(in reply to shallowdeep)

Post #: 15

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:18:11 AM

MissAsylum

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

quote:

ORIGINAL: VaguelyCurious

quote:

ORIGINAL: MissAsylum

Photo of the problem

Ah. Then ignore what I just said about the first two terms, ok? Brb.

ETA: also, that bears absolutely no relation to what you actually typed out. You can see that, right?

I tried. Isn't that the most important thing? Lol

_____________________________

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 #: 16

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:19:37 AM

VaguelyCurious

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

Not in maths, honey. Do you need the answer explaining, or do you just need to know what it is?

_____________________________

Sthetic on FetLife.

(in reply to MissAsylum)

Post #: 17

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:20:05 AM

MissAsylum

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

quote:

ORIGINAL: LanceHughes

YIKES! Those are fourth roots! Square roots of square roots!
First term: X * {3 * X * (Y^8)}^ (1/4)
First term: X * {[3 * X]^ (1/4)} * { (Y^8)^ (1/4)} because any root of a multiplicative expression is the same root of each part multiplied
First term: X * {[3 * X]^ (1/4)} * { Y^2 } because the quarter root of an eight power is a square root.
First term: X(Y^2) * {[3X]^ (1/4)}

Second term: Y * {243 * X^5 * Y^4} ^ (1/4) That 4th root of three makes me wonder about 243. 243 / 3 = 81 * 3, so we write:
Second term: Y * {3^5 * X^5 * Y^4} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] * [3 * X]} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] ^ (1/4)} * {[3 * X] ^ (1/4)}
Second term: Y * {3 * X * Y} * {[3 * X] ^ (1/4)}
Second term: 3 X( Y^2) * {[3X] ^ (1/4)}

Third term: Left for the student

Thank you! Now I can actually work out the problem!

_____________________________

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 #: 18

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:22:05 AM

MissAsylum

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

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: shallowdeep

6 * x * y^2 * (3 * x)^.25

or, equivalently,

(3888 * x^5 * y^8)^.25

FAIL!

ETA: There are no factors of 6 across the three original terms. Not even factors of 2.

I have the mathematical ability of a 3 year old, and even I knew that wasn't right.

_____________________________

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 #: 19

RE: Any Math Wizards Care to Take A Crack At This Problem? - 10/12/2011 11:23:59 AM

MissAsylum

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

So.... 6xy^2 4^√3x.

I think.

_____________________________

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

(in reply to MissAsylum)

Post #: 20

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

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

Page: [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

0.109