Math Problems!........share It Here!

[borgyX]

i remember my algebra 2...once lang ako natulog sa class buong college..sa algebra 2 pa...huli ni prof...tinawag tuloy ako sa board to answer a problem na wala ako idea

[keymaker]

Karamihan ng knowledge ko about math nakalimutan na.

[ninj@]

There are three kinds of mathematicians: those who can count and those who can't.

[b_junzo]

There are three kinds of mathematicians: those who can count and those who can't

ninj@ kala ko three kinds bat dalawa lang ata?

[Guest wackeen]

There are three kinds of mathematicians: those who can count and those who can't

ninj@ kala ko three kinds bat dalawa lang ata?

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and of course, you're kidding when you ask that right?

 

tatlo nga: kami ni ninja dalawa -- hulaan mo kung sino yung isa pa?

 

peace, brother!

[dem0liti0nman]

Solid Geometry take 2!

 

Engineering Mechanics take 2!

[dem0liti0nman]

There are three kinds of mathematicians: those who can count and those who can't

ninj@ kala ko three kinds bat dalawa lang ata?

<{POST_SNAPBACK}>

 

 

and of course, you're kidding when you ask that right?

 

tatlo nga: kami ni ninja dalawa -- hulaan mo kung sino yung isa pa?

 

peace, brother!

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b_junzo, wag kang mag-alala hindi kaw ung pangatlo peace man!

[Moonwalker]

Beatrice spent all of her money in five stores. In each store, she spent 1 peso more than half of what she had when she came in. How much money did Beatrice have when she entered the first store?

[redshift]

Beatrice spent all of her money in five stores.  In each store, she spent 1 peso more than half of what she had when she came in.  How much money did Beatrice have when she entered the first store?

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P62.00?

[floppydrive]

62 din kwenta ko para wala na siyang pera paglabas ng huling tindahan.

[Moonwalker]

tumpak po 62 pesos

 

may solution po ba kayo?

[mick8]

can anyone give me a simpler explanations about Euler's identity? the explanation that can be found over the net are very complex. thanks.

[Guest wackeen]

can anyone give me a simpler explanations about Euler's identity? the explanation that can be found over the net are very complex. thanks.

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There's an extension of the real number system called the complex numbers. The numbers in this larger system can be expressed in a number of ways:

 

1) a+bI

2) r cis t

 

In handling the more powerful second type of notation, there are 'rules' to follow. But note that this is not just shorthand, but has an actual meaning.

 

The Euler identity is the foundation for such a notation to be possible.

[mick8]

There's an extension of the real number system called the complex numbers. The numbers in this larger system can be expressed in a number of ways:

 

1) a+bI

2) r cis t

 

In handling the more powerful second type of notation, there are 'rules' to follow. But note that this is not just shorthand, but has an actual meaning.

 

The Euler identity is the foundation for such a notation to be possible.

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bro, what do you mean by, "an extension of the real number system"? & can you give an example if possible, thanks.

[PageDown]

246

 

I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ):

 

The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression):

 

    b        b*r      b*r^2    b*r^3

b*r^3    b*r^2      b*r        b

 

(where each digit <=9)

 

With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number).

 

So now we have either of these two conditions (since the leftmost digit is a 1):

 

b=1

b*r^3=1

 

In the first case, if b=1 then r=1 or 2  (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false.

 

If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well.

 

So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number.

 

Ok, so now we know the sum is a 3-digit number, hence in the following form:

 

    b      b*r      b*r^2

b*r^2    b*r      b

 

We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus:

 

9 >=    b    >= 6

9 >= b*r^2 >= 6

 

Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r:

 

b*r^2    b (r=1)    b (r=2)    b (r=3)

-------    --------    ---------    ---------

  6            6              X              X

  7            7              X              X

  8            8              2              X

  9            9              X              1

 

('X' marks non-valid combinations)

 

Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596:

 

666 (b=6,r=1)

777 (b=7,r=1)

888 (b=8,r=1)

999 (b=9,r=1)

842 (b=2,r=2)

931 (b=1,r=3)

 

Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress.

 

Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer

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Nope! you have to use a 4 x 4 matrix to get to it (easier solution).