Math, anyone?

(x^2 + 2 x - 3x)(x^2 +(x^2 +4)

(x^2 +2x - 3)(x^2 +4) = (x^2 +2x - 3) x^2 + (x^2 +2x -3)4

= x^2 x (^2) +2x x (x^2) - 3 x (x^2) + (x ^2) x 4 + (2x) x 4 - 3 x 4

= x^4 + 2x^3 - 3x^2 + 4x^2 +8x -12

= x^4 + 2x^3 + x^2 +8x - 12



Hehe, pretty simple but ah well.

Originally posted by jagdflieger
The Mathematicans would say that 0/0 is an undefined operation. You cannot perfrom such an operation. This is why must of these suprious proofs look like they work, you perform a division by 0 which is undefined. The correct way of defining the operation would be:

lim(0/X) > Y : X => 0 where Y is an abritrary real number.

In short the results gets arbitarily large as X gets smaller

undefined terms are usually left only to fractions,

Such as 9/0 ....

Such is a undefined.

0/9 is however not undefined.....It equals 0

Very cool thread guys. Im gonna show my teacher.

Do you understand the concept of the limit - you will will never get there. The function f(x) = a/x (where a is a constant) is undefined for x = 0 because the function is discontinous for x = 0.

Originally posted by jagdflieger
Do you understand the concept of the limit - you will will never get there. The function f(x) = a/x (where a is a constant) is undefined for x = 0 because the function is discontinous for x = 0.

Thats not undefined, that is simply saying the answer is 0, because z x 0 = o and if g = 0 than z x g = o still therefore,

z x g = g x z

Uh - f(x) = a/x where a is a constant is a division not a multiplication.

f(x) = a*x is a multiplication.

Originally posted by Dreamz

Originally posted by jagdflieger
Do you understand the concept of the limit - you will will never get there. The function f(x) = a/x (where a is a constant) is undefined for x = 0 because the function is discontinous for x = 0.

Thats not undefined, that is simply saying the answer is 0, because z x 0 = o and if g = 0 than z x g = o still therefore,

z x g = g x z

Yes, it is indeterminate. No, your proof does not apply to anything being discussed.

77^3-1434.24314 (133.324) < 654^2 - 143 (543.2499)
+
3432^3 -2341215 (24431.423) + 653442.945 = ~

Yeah BABY!!!!!!

P...
m...

The term 0/0 cannot be defined:

For example:
Sin X = 0 when X = 0 ; however,
Lim (X/[Sin X]) = 1, X=>0 which means that (X/[Sin X]) gets closer and closer to 1 as X gets closer to zero.

However: For F(X) = 2*X and G(X) = 1*X now F(X) = 0 for X = 0 and G(X) = 0 for X =0
But
Lim (F(X)/G(X)) = 2, X=>0 (that as X approaches 0)

Now for:
F(X) = X^2 and G(X) = X^3 we have F(X) = 0 for X = 0 and G(X) = 0 for X = 0;
But
Lim (F(X)/G(X)) becomes arbritarily large X=>0
And
Lim (G(X)/F(X)) = 0 X=>0

Valhall, you said that 0/infinity is indeterminate, but it seems to me that it would be equal to 0.

0/1=0, 0/2=0, 0/3=0 and so on, 0/x as x approaches infiniti would also be equal to 0.

Or is we care to use your definition of infiniti as 1/0 we get:

0/(1/0)=

0*(0/1)=

0


Let me know what you think.

[Edited on 20-10-2003 by greenkoolaid]

Here's a cool Geometry Illusion to play with.


Explanation
I was directed to your site by a friend, and saw your hole puzzle (the one with the two triangles and two other shapes, which comes very close to forming a large triangle. I had never seen this puzzle prior to today, but I love a good thought exercise, and thought I'd try my stab--much different from the other explanations on the site, and I feel it is much more plausible.

First of all, in response to those who believed that 1.2 degrees difference (not even noticeable to the eye) is enough to make enough space for a hole of that magnitude--nice try. Unfortunately, that answer is flawed. The figure is, technically, a quadrilateral, but not enough of one to make the extra space. Therefore, I will call the figure, in this explanation, "the large triangle."

The true answer is actually much simpler, and does not rely on the angles of the triangles, and has little to do with the triangle as a whole (no pun intended), but rather the arrangement of its parts.

First, let's establish what we know about the two triangles--without doing any calculations. The height of the green triangle is 2, and the width of its base is 5. The height of the red triangle is 3, and the width of its base is 8. Just to verify the statement on the site, you can easily count the squares to ensure this.

Okay, now let's take a look at the first large triangle. The green triangle is on top, and its base forms the top of the rectangle (the two shapes inside are irrelevant for right now). The red triangle is to the left, and its right edge forms the left side of the rectangle. This yields a rectangle of length 5 and height 3, and with an area of (5 * 3) = 15.

The second triangle switches the red triangle and the green triangle. This means the bottom of the red triangle forms the top of the rectangle (still ignoring the two shapes inside), and the right side of the green triangle forms the left side of the rectangle. This yields a rectangle of length 8 and height 2, and with an area of (8 * 2) = 16.

Okay, well now this says that there is one extra space, but why is there a hole? This is simply because the two shapes, while forming a rectangle of area 15 in the first triangle, still have a combined area of 15 in the second. Why? The way the shapes were separated allows them to be combined so as to look like a 8 x 2 rectangle, but, since you can't create extra area that didn't exist before, the shapes naturally have a "hole."

The ratios of the smaller triangles are not the same.

The top figure is not a true right triangle. There is a break in the straight line of the hypotenuse where the red and the green figure meet. In the top figure, the opposite side over the adjacent (tan) is:
3/8 = 0.375
However for the entire "triangle", the opposite side over the adjacent (tan) is:
5/13 = 0.3846...
Our eyes see the "hypotenuse" of the top triangle as a straight line, but it is not.

Originally posted by jagdflieger
The top figure is not a true right triangle. There is a break in the straight line of the hypotenuse where the red and the green figure meet. In the top figure, the opposite side over the adjacent (tan) is:
3/8 = 0.375
However for the entire "triangle", the opposite side over the adjacent (tan) is:
5/13 = 0.3846...
Our eyes see the "hypotenuse" of the top triangle as a straight line, but it is not.

I would agree with that. The small dip of the straight line (hypotenuse) therefor also includes a thin slice of "White" space. Then on the lower and true right triangle that thin slice is the "Extra" White square.

Yeah, I was thinking of that too, but I never trust solving something graphically.

It didn't take me too long to figure it out. I showed it to my school's math club and alot of them got it.

Anyways, here's another problem:

Two trains are travelling towards each other on teh same track. One is going at 256 miles per hour while the other one is going at 145 mph. A bee is flying between the two trains at 1 mile an hour. Assuming the bee is equidistant from the two trains and the trains will have to travel four hours to collide, how far can teh be travel before it gets smooshed?

You should be able to solve that if you use a bit of logic.

You are partially correct, neither one is a true right triangle. The hypoteneuse on the top triangle bows in while on the bottom triangle it bows out.

If you add up the areas of the smaller geometric figures on the top drawing you get the triangle having an overall area of 32. i.e. (5*2/2) + (8*3/2) + 7 + 8 = 32

However if it were a true triangle having the dimensions shown then its base*height/2 would be 13*5/2 giving you and area of 32.5.

you will notice that by adding up the shapes in the bottom drawing you get 33. One whole extra square more than the top traingle. Not a half a square more like you should get if the bottom traingle is a true right triangle.

So you see, neither drawing is a true right traingle.

Originally posted by Saucerat
Yeah, I was thinking of that too, but I never trust solving something graphically.

It didn't take me too long to figure it out. I showed it to my school's math club and alot of them got it.

Anyways, here's another problem:

Two trains are travelling towards each other on teh same track. One is going at 256 miles per hour while the other one is going at 145 mph. A bee is flying between the two trains at 1 mile an hour. Assuming the bee is equidistant from the two trains and the trains will have to travel four hours to collide, how far can teh be travel before it gets smooshed?

You should be able to solve that if you use a bit of logic.

What direction is the bee travelling at 1 mile an hour? If he flies up or perpendicular to the trains he doesn't get smooshed. If he is travelling towards one train or the other, it alters the time at which he gets smooshed.

I will assume he travels along the railway track, therefore if he flies toward the slow train he lives a bit longer.

I figured the trains are 1604 miles apart. So the bee is 802 miles from the faster train.

I let x be the distance the bee travels.

the distance the fast train moves is 802 + x

Time = distance/speed

t= x/1 for the bee

t= (802 + x)/256 for the fast train

these eqations equate to x = (802 +x)/256

A little algebra gets x = 802/255

Or the bee travel 3.145 miles

[Edited on 20-10-2003 by greenkoolaid]

[Edited on 21-10-2003 by greenkoolaid]

Train 1 speed = 256 mph
Train 2 speed = 145 mph
Relative closure = 401 mph
Initial distance between trains = 4*401 = 1604 miles
Bee at mid point = 802 miles from start
Travel distance for Train 1 untill collision = 1024 miles
Travel distance for Train 2 untill collision = 580 miles
Assuming bee does not evade trains and flies towards the slower train (since the slower train will never reach his current location by the crash time):
The relative speed of Train 1 and the bee is 255 mph (256 - 1)
By using the relative speed we can simplify and perform the computation using the bee's current location as if the bee were motionless:
So the time for Train 1 to reach the current location of the bee (using the relative speed) would be
802/255 = 3.14509... hours
The bee could fly 3.14509... miles before Train 1 overtakes it and squishes it.

I need to show these off to some of my school's math professors. I could see them losing sleep over stuff like this. LOL