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#126 Re: Help Me ! » How to solve these equations » 2016-12-11 20:55:54

182-165=17    ........................(1)
If we multiply both sides by 10 (to get R.H.S.>165) 10(182-165)=170=1*165+5 which gives us
10*182-11*165=5 ..................(2)
If we can make R.H.S of the required equation as linear combination of (1) and (2) we get the solution.
(1) Since 89=2*17+11*5
2*(182-165)+11(10*182-11*165)=89 which leads to
182*112-165*123=89. which givesi=112  and j=123.
However this is not unique combination. i=112+165n and j=123+182n will also foot the bill.
I am aware that this may not work out in  cases where linear combination of 17 and 5 may not be possible.But still I am sure some more relations like (2) could be formed which will help linear combination.

mr.wong,
It is o.k. I found the answer for any number on R.H.S.
1=3*17-10*5=3*{182-165}-10*{10*182-11*165}=182*(-97)+165*107  i=-97 and j=-107
e.g.(5) could be written
182*{-97*929}-165*{-107*929}=929
i=-97*929  and j=-107*929

#127 Re: Help Me ! » train accelerating » 2016-12-09 13:34:19

Hi abhishek1996,
I get a=4/3.

#128 Re: Help Me ! » train accelerating » 2016-12-09 06:22:56

Average speed =v/2. Does it help?

#129 Re: Help Me ! » Knights and knaves » 2016-12-08 21:11:34

What I would suggest is ask David's question to one person.You would come to know whether he is knight or knave. then you ask him to list who are the knights.Two questions are enough.

#133 Re: Help Me ! » Knights and knaves » 2016-12-07 21:32:51

Can you ask the same person more than once?

#134 Re: Help Me ! » Knights and knaves » 2016-12-07 21:20:19

David wrote:

The minimum number of yes-no questions that he must ask the inhabitants is 9. (5-yes 4-no/ 5-no 4-yes) At 9, there's enough information to find out the 5 naves.

To be precise,
The minimumof minimum number of yes-no questions that he must ask the inhabitants is 5.
The maximumof minimum number of yes-no questions that he must ask the inhabitants is 9.

#135 Re: Help Me ! » Algebra Problems » 2016-12-05 18:37:48

Hi dazzle1230,
According to your earlier comment, you substituted x=y+1 ,expanded and got 3 simultaneous equations in a,b,c. It is o.k. when you are not writing exam or if you are allowed to use programmable calculators but it is awfully simple if you can solve them one by one.

#136 Re: Help Me ! » Minimum surface area » 2016-12-04 19:24:42

sisyphus wrote:

Hi. My snail speed math progression has carried me to a bunch of word exercises in the precalculus book that i'm trying to go through for most of this year, and i am pitifully terrible at word problems:

The box for the new Sasquatch-themed cereal, ‘Crypt-Os’, is to have a volume of 140 cubic inches. For aesthetic reasons, the height of the box needs to be 1.62 times the width of the base of the box. Find the dimensions of the box which will minimize the surface area of the box. What is the minimum surface area? Round your answers to two decimal places.

So my attempt at solving this is
1) find height function (i don't know another way of finding this without making width and length equal):
        h = 1.62x
        x^2 * 1.62x = 140
        h = 140/1.62x^2
2) find surface area function and plug in the height function:
        2x^2 + 4xh
        2x^2 + 4x(140/1.62x^2)
        2x^2 + 560x/1.62x^2
3) I feel like i should find the slant of this function, then find the vertex of this slant, which will give me x, then i should plug that number in surface area function, and get the minimal surface area. Of course none of this is working out for me, so could anyone lend me a helping hand here?

#137 Re: Help Me ! » Algebra Problems » 2016-12-04 19:03:32

thickhead wrote:
dazzle1230 wrote:
thickhead wrote:

Can you illustrate the method in detail?

First I knew that f(x) had to be in form ax^5+bx^4+cx^3 in order to be divisible by x^3.  I let y=x-1,
so x=y+1.  So then f(x)-1 is only divisible by (x-1)^3 if f(y+1)-1 is divisible by (x-1)^3.  After expanding, I solved a system of equations to find a, b, and c.

1|	a	b	1-a-b	0	0	-1
		a	a+b	1	1	1
	………………	………………	………………	………………	………………	………………
	a	a+b	1	1	1	0

1|	a	`-3-a	1	1	1
		a	-3	-2	-1
	………………	………………	………………	………………	………………
	a	-3	-2	-1	0 

#138 Re: Help Me ! » Algebra Problems » 2016-12-04 16:41:30

dazzle1230 wrote:
thickhead wrote:
dazzle1230 wrote:

and for part one of the second problem, I said that g(x) has to be divisible by f(x):
g(x)=a(x+5)(x+3)(x-2)(x-4)(x-8)
f(x)=b(x+3)(x-4)(x-8)
Once we divide f(x) from g(x) and simplify the quadratic, we get that (a/b)x^2+3x10...is that correct?
If so, how do we proceed part 2 of the problem?

I do not know the technicalities but is it not necessary that a has to be divisible by b also?
In my opinion the coefficient of highest degree in f(x) must be divisible by the coefficient of highest degree in g(x) and f(x) should contain all the zeroes of g(x) ,even the repeated ones.

So are you saying that g(x) doesnt have to be divisible by f(x)?  Should we take multiplicity of roots into account?  Also, for part 2, I'm thinking that the degree of f(x) has to be 3 and the degree of g(x) has to be 5, using the fundamental theorem of algebra, so there won't be multiplicity of roots.

O.K. It is g(x)/f(x) My comments were for f(x)/g(x). g(x) has to be divisible by f(x). All the zeroes of f(x) should be there in g(x) In addition a/b that you have taken is to be integer.That is my opinion.

#140 Re: Help Me ! » Algebra Problems » 2016-12-03 18:38:42

dazzle1230 wrote:

Yes, I got 6x^5-15x^4+10x^3
Thanks!

Can you illustrate the method in detail?

#141 Re: Help Me ! » Algebra Problems » 2016-12-03 18:34:15

dazzle1230 wrote:

and for part one of the second problem, I said that g(x) has to be divisible by f(x):
g(x)=a(x+5)(x+3)(x-2)(x-4)(x-8)
f(x)=b(x+3)(x-4)(x-8)
Once we divide f(x) from g(x) and simplify the quadratic, we get that (a/b)x^2+3x10...is that correct?
If so, how do we proceed part 2 of the problem?

I do not know the technicalities but is it not necessary that a has to be divisible by b also?
In my opinion the coefficient of highest degree in f(x) must be divisible by the coefficient of highest degree in g(x) and f(x) should contain all the zeroes of g(x) ,even the repeated ones.

#142 Re: Help Me ! » Algebra Problems » 2016-12-02 19:39:43

Hi dazzle1230,
How did you solve prob (1) of  #5?

#143 Re: Help Me ! » Algebra Problems » 2016-11-30 22:07:37

bob bundy wrote:

Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8.

Number of factors and number of zeros are not the same thing.

Bob

I think they are 2 zeroes superposed one over the other.

#144 Re: Help Me ! » Algebra Problems » 2016-11-30 20:21:36

bob bundy wrote:

hi dazzle1230

But, for example, f could have a factor (x+3)^2 and still satisfy the requirements.

Bob

Then f will have 4 factors.(-3 twice)

#146 Re: Help Me ! » a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c » 2016-11-21 15:41:20

Trial and error no dobt,but systematically. e.g. if you assume a<b<c  since there is nothing to choose among them, c has very limited value.To start with c<=6. once c=6 fails you come to c=5.

#147 Re: Help Me ! » a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c » 2016-11-20 23:53:51

But then there is the condition that they are only natural numbers.That limits the scope of the solution and gives only one solution.

#148 Re: Coder's Corner » Is Combinations and Permutations Calculator working with "<" and ">" » 2016-11-20 17:21:01

I am also withholding it as it could be a homework problem and the ready solution could kill the urge to think.

#149 Re: Coder's Corner » Is Combinations and Permutations Calculator working with "<" and ">" » 2016-11-20 16:31:55

bobbym,
I don't think it needs programming but only some thinking/awareness of some classical method.

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