Math Help

[Trinitis]

Calling all you math wizards!!

I have a two problem assignment for extra credit thats due on Monday. We are allowed to ask anyone we want for help, so..here we go.


I've tried working out the first problem on my own, but keep getting the answer wrong compaired to the back of the book. The teacher wants us to show the work.

#1. Three tired campers stopped for the night. All they had to eat was a bag of apples. During the night, one awake and ate one-third of the apples. Later, the second camper awoke and ate one-third of the apples that remained. Much later, the third camper awake and ate one-third of those apples yet remaining after the other two had eaten. When they got up the next morning, 8 apples were left. How many apples did they begin with?

I tried the following, which came up wrong : X - 1/3x - 1/3(1/3x) - 1/3[1/3(1/3x)] = 8

The answer in the back of the book is 27 apples.


#2. A polynomial in x has degree 3. The coefficient of x^2 is 3 less than the coefficient of x^3. The coefficient of x is three times the coefficient of x^2. The remaining coefficient is 2 more than the coefficent of x^3. The sum of the coefficients is -4. Find the polynomial.

This problem does not have an answer in the back of the book. :(

If anyone can help, PLEASE feel free..and also, if you atempt to explain how your doing it other then just an answer, I'd be greatful (I do like to learn..)

-Adredrin

[Edited on 5-20-2004 by LordAdredrin]

[TheEschaton]

For the first question:

1/3x + 2/9x (which is 1/3rd of the remaining 2/3rds) + 4/27x (1/3rd of the remaining 4/9ths) + 8 = x

x - 9/27x - 6/27x - 4/27x (all made to common denom) = 8

x - 19/27x = 8
x(1-19/27) = 8
x(8/27) = 8

Divide both sides by 8/27, you get

(x(8/27)) * (27/8)) = 8 (27/8)
x = 27

[Xcalibur]

Easy:

#1

27/3
9

remained 18

18/3
6
remained 12

12/3
4
remained 8

X- X/3 =Y

Y - Y/3 =Z

Z - Z/3 = answer

For the other, booh

My answer do look more simple, I know.



[Edited on 20-5-04 by Xcalibur]

[Kitsun]

First Problem:

Well I set it up and worked backwards, did you have to set it up in one formula?

Camper 3:
8/x = 2/3
X = 12

Camper 2:
12/x = 2/3
X = 18

Camper 1:
36/X = 2/3
X = 27

Started with 27 apples. Thats how I'd show the work anyway.

[Kitsun]

ROFLMAO. Goddamnit we were all typing it out at the same damn time.

[Lord Whirlin]

2nd problem, Solution
1x^3+ -2x^2+ -6x+ 3

Logic-ed it out

(mistyped 4 instead of 3, damn laptop)

[Edited on 5-20-2004 by Lord Whirlin]

[TheEschaton]

As for this one:

#2. A polynomial in x has degree 3. The coefficient of x^2 is 3 less than the coefficient of x^3. The coefficient of x is three times the coefficient of x^2. The remaining coefficient is 2 more than the coefficent of x^3. The sum of the coefficients is -4. Find the polynomial.

let y be the coefficient of x^3

y + (y-3) + ((y-3) * 3) + (y+ 2) = -4
y + y - 3 + 3y - 9 + y + 2 = -4
6y - 10 = -4
6y = 6
y = 1, and then you just plug in, and find that the polynomial is:

x^3 - 2x^2 - 6x + 3.

Those fit the rules, and 1 - 2 - 6 + 3 = -4

-TheE-

[Lord Whirlin]

I substituted base value of -1 in for x^2 to begin with, because I knew some values had to be negative for the coefficients to add up to -4. From there, I found the resulting coffecients to add up to 2, so I tried lowering it to -2 to x^2, which worked.

[Latrinsorm]

Eschaton is right, and I need to brush up on algebra (or sleep more).

[Edited on 5-20-2004 by Latrinsorm]

[TheEschaton]

(And Whirlin's answer is wrong, anyways)

:whistle:

-TheE-

[TheEschaton]

And Latrin's is wrong, because 2 less than -1/9 would be -19/9

not -8/9

Edited to add: Not to mention, 2 more than -1/3 for the last coefficient, is 5/3, not 2/3.

-TheE-

[Edited on 5-20-2004 by TheEschaton]

[Latrinsorm]

Originally posted by TheEschaton
And Latrin's is wrong, because 2 less than -1/9 would be -19/9

not -8/9

-TheE- Yeah, yeah, I edited my post already. :bleh:

First one:

x-(x-1\3x)-1\3(X-(x-1\3x))-1\3[X-(x-1\3x)-1\3(x-(x-1\3x))]=8

Second one:

Your gonna have to give an example of the coefficients and polynomials your using because i'm not up on the lingo.

sorry woulda posted early but was on the phone. heh

[Trinitis]

Ok, reading through TheE's stuff on the first one made me understand it.

But anyone feel up to explaining #2 a bit more? That crap was like greek to me..

[TheEschaton]

All your answers are belong to me.

Sirusle, my answers so pwn all of yours. Not only are mine fully correct, they're also lucid (Ahem, X) and easy to follow/comprehend. And I got them in first. Unless you count Whirlin's "I logic'ed them out" ones, which, as we all know, does not fly in the High Court of Mathematics.

-TheE-

[Edaarin]

Okay. For Ax^3+Bx^2+Cx+D

B=A-3
C=3B=3A-9
D=2+A

A+A-3+3A-9+2+A=-4
5A-10=-4
6A=6

A=1
Then just plug in

B=A-3=1-3=-2
C=3A-9=3(1)-9=-6
D=2+A=2+1=3

Then just replace the numbers you get for A, B, C, D!

[TheEschaton]

Look at it this way, Adr:

The "variable" in the second problem is not x, it's the coefficient of x. You know a few things about the various coefficients.

x^3 has a coefficient....name it y.
x^2 has a coefficient which is 3 less than the coefficient of x^3...namely, y-3.
x has a coefficient which is 3 times the coefficient of x^2, namely, ((y-3) * 3).
the remaining coefficient (the constant), is 2 greater than the coefficient of x^3, namely, y+2.

And lastly, what you know is that all these coefficients, if you added them up separately (from the polynomial of x), you would get -4.

Namely,

y + (y-3) + ((y-3) * 3) + (y + 2) = -4.

Then you just solve for y, as I described....you find out y = 1. Then, you just plug one into the above equations, to find the coefficients for the various degress of x:

coefficient of x^3 (y) = 1
coefficient of x^2 (y-3) = -2
coefficient of x ((y-3)*3)= -6
coefficient leftover (y+2) = 3, thus giving you the polynomial (of degree three) x^3 - 2x^2 - 6x + 3.

If you add those coefficients up....you'll get -4, thus proving the answer is correct (the counterproof).

-TheE-

[Edaarin]

Okay, to explain my answer

Let A be the coefficient for x^3.
Let B be the coefficient for x^2.
Let C be the coefficient for x.
Let D be the coefficient for the last number, the constant.

From the problem, we know that

The coefficient B is equal to 3 less than the coefficient A, or A-3
The coefficient C is 3 times the coefficient B, or 3B. Since we want to just have one variable to solve for, plug in A-3 for B to get 3(A-3) = 3A-9
the coefficient D is just 2 more than the coefficient A, or A+2.

Now, since we know the sum of all coefficients is -4, just sum up all the coefficients in terms of A and set it equal to -4.

So...A + (A - 3) + (3A - 9) + (A + 2) = -4
A B C D

which gets you 6A - 10 = -4

Add 10 to both sides and divide by 6 to get the value for A.

Then plug in the result, A = 1, into the equations for coefficients in terms of A to find those values, and then just put that in the original equation

[HarmNone]

Originally posted by LordAdredrin
But anyone feel up to explaining #2 a bit more? That crap was like greek to me..

Nah. It ain't Greek. It's Geek. :P

HarmNone was just glad there were enough apples to go around!

[TheEschaton]

I think HN's just jealous of the amazing computing power of us two Asians.


(and yes, Edaarin, I am an Asian, no matter WHAT you say)

-TheE-

[Trinitis]

ok..but what the hell is a degree of 3?

[TheEschaton]

A degree of three polynomial just means that the highest power you'll see in the polynomial is the third power, IE, x^3

Edited to add: You can have lower powers in a degree 3 polynomial too, such as x^2, x^1, and x^0 (which is a constant).

-TheE-

[Edited on 5-20-2004 by TheEschaton]

[Trinitis]

Now see, if someone would have told me that, I prob would have figured out the problem after a little time workin on it! But I had NO clue on the degree thing.

[TheEschaton]

Heh, the thing with math is to never get caught up in the stupid shit part of the problems....such as the degree of the polynomial. In this problem, the degree doesn't matter one iota. All you have to realize is that the main coefficient is the coefficient of x^3, and that all the other coefficients, are somehow related to this base coefficient.

-TheE-

[CrystalTears]

Do your own damn homework, you cheater!! :P

[TheEschaton]

A) It's extra credit, B) He was told he could ask whoever he wants, and C) I'm still basking in the uberness of my mathematical skillz. Please allow a moment.

-TheE-

[CrystalTears]

Yeah like he's really going to say otherwise. He asks me everyday to do his homework. He's a big lazy ass jerkarama! :bleh:

Note: I was just poking fun at Adredrin. ;)

[Edaarin]

I take back all claims of TheE not being Asian.

TheEschaton: which?
TheEschaton: God, my math is so sexy. I think I'm going to bed.

Go easy on the lubricant bud.

[Edaarin]

Yeah right like we're gonna believe he asked a woman for help in math.

"A girl answered a math question correctly! You all know what that means...a WITCH!"

[CrystalTears]

Holy shit, when did you become so anti-females? :(

[Edaarin]

What would give you that idea...?

[Trinitis]

Hey CT..

::rings his bell::

PWNED! ::cackles::

[CrystalTears]

Oh gee, I don't know. ::stares at your previous post, stares at your signature and smirks::

[Latrinsorm]

Originally posted by CrystalTears
Holy shit, when did you become so anti-females? :( I've noticed it since October 29th, 2003, personally.