Question #1

quote:

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Originally posted by atomosXIII:
Same as the astronomy thread. Come in, solve the question/problem left by the last person to enter, then leave a problem yourself. Please relate the question to mathematics, it doesn't necessarily have to be an equation, but can be a question. I will ask the first question, then somebody answers it and asks a question of their own.

Question #1

1+1=

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This is a repeat thread by atmos.

Ans=2

Source: http://www.math.princeton.edu/~arbooker/nthprime.html

so we can't always say 1+1=2

It's true!

2.4 rounds to 2, 4.8 rounds to 5...

2.4 + 2.4 = 4.8
2 + 2 = 5!

Hurkyl

quote:

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Originally posted by MSI:
(1+1) doesn't -usually- equal 2...
when we say 1gm+1kgm the awnser will be:
* 1001 gm or 1.001 kgm
* 2 m : m = 500.5 gm

so we can't always say 1+1=2

IF YOU BRING 100 MAN TO UNDERSTAND 1 WOMAN THEY CAN NOT DO !!
the dead man wises
forums HERE

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Now to the next question (if i may ..)
Let
y_n=y_n-1*5
y₀=1
rewrite the previous equation as a function f(n), where f(n) is not a reccursion function (it is really easy )

EDIT:
I forgot to express how much i liked Hurkyl's idea, i think i will use it like a million time in my life (if Hurkyl gives me the copyright permission ), really impressive !

quote:

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Originally posted by STAii:
y_n=y_n-1*5
y₀=1
rewrite the previous equation as a function f(n), where f(n) is not a reccursion function

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f(n) = 5^n

Question (still easy - are we going to get progressively harder?):

Find the set of quadratics to fit the points

okz, you can write a quadratic f(x) on the form of x² + b*x + c
so
f(x) = x²+bx+c
f(-1)=(-1)²-b+c=-11.75
c-b=-10.75 ------(1)
f(0)=0²+0*b+c=-9.6
c=-9.6
put it back in (1)
-9.6-b=-10.75
-b=-10.75+9.6
b=10.75-9.6=1.15

so the function you are asking for is
f(x)=x²+1.15*x-9.6

Okz, now it is my turn to ask ...

You have a clock in front of you (not a digital one, a normal one), it is 12:00, how many seconds (to the nearest second) you will need to wait to see the two lines of the clock meeting again ?
(I don't know the exact word, but what i meant with lines is actually the the things that point to the hours and minutes).

If no one understands my question please forget about it.

After 1 hour (3600 sec.) the minute hand points to 12 again, and the hour hand points to 1.

Now, the hour hand moves at 2pi/43200 radians/sec
The minute hand moves at 2pi/3600 rad/sec
And the hour hand is 2pi/12 radians ahead

So, let t be the number of seconds after 1 o’clock that the minute hand catches the hour hand.

t*2pi/3600 = t*2pi/43200 + 2pi/12
simplifies to:
t/3600 = t/43200 +1/12
or:
12t = t + 3600
t = 327.2727...

So, 3600 + 327 = 3927 seconds after 12 o’clock the hands coincide.

Next question:

You have N bags each containing an unknown number of 10 ounce kiwis. Buuuuut...in one of them each kiwi weighs only 9 ounces. You have a DIGITAL scale; but unfortunately, it cannot weigh more than 10 kiwis at one time: put any more than 10 on it and it just won’t work at all. You are allowed to use the scale 5 times, which permits you to identify the bag of illegal 9 ounce kiwis. What is the maximum N?

first weigh number rounded down to 22
second weigh number rounded down to 12 -> assign to batches of 4
third weigh 1 of bag 1A, 2 of 1B, 3 of 1C, 4 of 1D
fourth weigh 1 of bag 2A, 2 of 2B, 3 of 2C, 4 of 2D
fifth weigh 1 of bag 3A, 2 of 3B, 3 of 3C, 4 of 3D mass missing shows illegal bag.

Am I right?

Prove 3 is irrational.

In one weighing, you can tell which of 5 bags the light lemon is. You put 1 from 1, 2 from 2, 3 from 3, 4 from 4. If it comes out even you know the 5th has light lemons. :)

After doing some initial thinking, I don't think (but can't prove yet) that you can gain any more information from a weighing other than identifying whether or not a particular groups of bags has the light lemon in it, in a manner similar to above...

So, for the second weighing, we must narrow down the choices to being one of 5 bags.

I think the greedy algorithm is optimal; if we need to narrow down the choices to a particular group of 5, we should put 1 lemon from 5 bags on the scale, then 2 lemons from 2 bags on the scale allowing us to work with 7 lemons. We can't work with any more than 7, because that would require doing 1 lemon from 6 bags, which could be rendered unidentifiable in the last weighing. There are other ways to select the 7, though.

So for 2 weighings, we can distinguish betwee, 12 bags of lemons; it's either among the 5 single lemon bags, the 2 double lemon bags, or the 5 we didn't weigh.

The above comments on the greedy algorithm apply to 3+ weighing scenarios too. Since in 2 weighings we can distingish bewteen 12 lemons, the greedy algorithm says put 1 lemon from 10 bags on the scale until you're down to your last 2 weighings.

So if you're given M weighings, you can identify the light bag of lemons amongst 12 + 10 * (M - 2) = 10 M - 8 bags for M >= 3.

So for 5 weighings, you can pick the light bag out of 42 bags. Weigh 10 at a time in the first 3 weighings. For the 4th weighing you've narrowed it down to no more than 12 bags, and you can use the algorithm mentioned above for 2 weighings (5 singles & 2 doubles for the second to last weighing, then 1&2&3&4 to identify which if the last 5 has the light lemon)

Hurkyl

Alternatively, you can do fourth weighing of 1,1,1,1,2,2,2. That
identifies the group of 3, 4 or 5 bags containing the
light ones.
Last weighing 1,2,3 kiwis if group of 3, 1,2,3,4 if
group of 4 or 5. Last weight will be 96,97,98,99, or
100 yada yada yada ...

quote:

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Originally posted by FZ:
Prove 3 is irrational.

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First you asume that (3) is rational, and therefor can be expressed on the form of a/b (where both a and b are integers)
You also assume that there is no common factor between a and b
(3)=a/b ---(0)
3=a²/b² ----(1)
a²=3*b² -----(2)
(there is a rule that says if the square of an integer is from the multiplies of a prime, then the number itself will be one of the multiplies of the prime).
Applying this to (2) (and sicne b² is an integer (Since b itself is an integer))
a=3*c (where c is just another integer)
now put this conclusion back in (2)
9*c²=3*b²
b²=3*c²
(applying the previous rules again)
b=3*d (where d is another integer)
now put the values of b and a back into (0)
(3)=3*c/(3*d)
Therefor there is a common factor between a and b, which is against the original assumption.
Therefore you reached a contrast between assumptions and conclusions.
Therefore the original assumption (That (3) is rational) is wrong.
Therefore (3) is irrational

(This proof is so flakey), actually i am not even sure if it is right , therefore i will not put a question.

quote:

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Originally posted by STAii:
I hate prooving this one ... but anyway ...

First you asume that (3) is rational, and therefor can be expressed on the form of a/b (where both a and b are integers)
You also assume that there is no common factor between a and b
(3)=a/b ---(0)
3=a²/b² ----(1)
a²=3*b² -----(2)
(there is a rule that says if the square of an integer is from the multiplies of a prime, then the number itself will be one of the multiplies of the prime).
Applying this to (2) (and sicne b² is an integer (Since b itself is an integer))
a=3*c (where c is just another integer)
now put this conclusion back in (2)
9*c²=3*b²
b²=3*c²
(applying the previous rules again)
b=3*d (where d is another integer)
now put the values of b and a back into (0)
(3)=3*c/(3*d)
Therefor there is a common factor between a and b, which is against the original assumption.
Therefore you reached a contrast between assumptions and conclusions.
Therefore the original assumption (That (3) is rational) is wrong.
Therefore (3) is irrational

(This proof is so flakey), actually i am not even sure if it is right , therefore i will not put a question.

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So it is my turn to actually put a question ...
find the interval that solves
|x+3|+|x-3|=6

quote:

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Originally posted by STAii:
Thanks for the tip (i will work on it, thanks a lot, i really need tips like those ones, cause i rarely find anyone to tell them to me).

So it is my turn to actually put a question ...
find the interval that solves
|x+3|+|x-3|=6

You've gotta be japaneese !

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Just to clarify things:
How are you defining |x|?
Is |x| integer compenent of x?
i.e. |-2.3| = -2,

|x|= x,x>0
       -x.x<0
       0,x=0

BTW, if no one answers this soon, i will have to answer it myself to keep the thread running (although this is really easy !)

{x: -3 x 3 }

My question is write down the next three terms in the following series:

2, 13, 28, 1, 126, -19, 344, -47, 730, -83.......

how could we mathmaticly solve this :
|x+3|+|x-3|=6

i tryed but no use

I believe a valid way of solving it would be to say

(|x + 3 + x - 3|)² = 6²

(2x)² = 36

to give x = 3 or -3

Or plot y = |x+3| and y = |x-3| graphically and see where they cross the x-axis. You can do similar for an inequality.