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-   -   Snake eats four golf balls (http://www.shreveport.com/forums/showthread.php?t=3350)

piemaker720 01-03-2008 09:09 AM

Snake eats four golf balls
 
1 Attachment(s)
http://www.smh.com.au/news/unusual-t...949879965.html


Quote:

A carpet python has been operated on by Gold Coast vets to remove four golf balls from its intestine.

"People have been putting golf balls under their brooding hens pretending that they are eggs to make the chickens happy," senior veterinarian at Currumbin Wildlife Sanctuary Dr Michael Pyne said.

"The snake has got in thinking that he has got a free meal and swallowed up these four golf balls."

The snake was found by residents at a house at Nobby's Creek near Murwillumbah in northern New South Wales.

"During the surgery we could see the name of the golf balls through the intestine because they were so stretched from these large golf balls being in there."
The old golf balls work everytime. We use to do that years ago, or we would put plastic easter eggs in the nest so when a chicken snake ate them it would kill the snake.

joepole 01-04-2008 12:04 AM

That snake actually ate three golf balls:

http://www.futilitycloset.com/2007/0...at-4-equals-3/

Al Swearengen 01-04-2008 12:17 AM

My condolences to Joe and his fam. I hope he fully recovers.

piemaker720 01-04-2008 09:39 AM

Quote:

Originally Posted by Al Swearengen (Post 27227)
My condolences to Joe and his fam. I hope he fully recovers.

I don't know Al it could be heredity. [yicks] I wonder if he used these kind of formulas when they were making a family.:D

AnimeSpirit 01-04-2008 10:24 AM

Quote:

Originally Posted by joepole (Post 27226)
That snake actually ate three golf balls:
http://www.futilitycloset.com/2007/0...at-4-equals-3/

I know you meant this as a joke, but let me clarify it anyway.

This equation was not properly calculated. In step 3, the author chose to shift the numbers around and factor them by reversing the distributive property. Combining like terms would've been the proper means of solving the equation. If he had properly combined like terms, he would've ended up right back at "a+b=c". Therefore, step 3 was pretty much pulled out of nowhere, invalidating the rest of the equation.

BUT, let's say he didn't screw up in step 3. Given the answer that the author received, the answer to this equation would be expressed as "4≠3," making the statement true and the equation solved.

I hate college algebra, but I've seen quite a bit of it this past year.

Pocahontas 01-04-2008 11:28 AM

Why didn't they let the grotesque creature and his poor eating habits die??:confused:I hate all snakes especially really large ones!:p

joepole 01-04-2008 11:48 AM

Actually, step 3 is fine, the distributive property was used correctly.

The problem is that between 3 and 4 you divide both sides by (a + b - c) which, according to the initial equation, is 0. You can't divide by 0.

AnimeSpirit 01-04-2008 01:54 PM

Quote:

Originally Posted by joepole (Post 27245)
Actually, step 3 is fine, the distributive property was used correctly.

The problem is that between 3 and 4 you divide both sides by (a + b - c) which, according to the initial equation, is 0. You can't divide by 0.

Yes, the distributive property was used correctly, but it shouldn't have been used at all. Combining like terms is a fundamental of algebra and the author avoided it entirely in order to get the outcome that he did.

Go to http://www.algebrahelp.com/calculators/equation/ and run the equation from step 2 and you'll see what I mean.

Improper calculation = invalid outcome!

j.nc 01-04-2008 03:52 PM

yeah, I like this one though. As JP says...
if a+b=c then (a+b-c) = 0

So in the last step you are actually saying x divided by zero equals y divided by zero:
3/(a+b-c)=4/(a+b-c)
or
3/0=4/0

heh, tricky

AnimeSpirit 01-04-2008 05:16 PM

Quote:

Originally Posted by j.nc (Post 27251)
yeah, I like this one though. As JP says...
if a+b=c then (a+b-c) = 0

So in the last step you are actually saying x divided by zero equals y divided by zero:
3/(a+b-c)=4/(a+b-c)
or
3/0=4/0

heh, tricky

Yes, that's another couple of rules of algebra broken.

(a+b-c) would equal zero. That, I can agree with, but if you are to add a 3 to one side of the equation, you must add a 3 to the other in order to keep the equation balanced, getting (3/0=3/0). This, in fact, would be a mathematical impossibility because, as Joe said, you can't divide by zero. However, discussing division is a side issue anyway, because no division took place in this equation nor should it have.

So let's see where the author started. He started with this, (a+b=c), which equals zero when you rearrange it as (a+b-c=0). Now, if you wanted to distribute 3 and 4 across (a+b-c), you can NOT put 3 on one side and 4 on the other as the author did. This imbalances the equation so no wonder he got such an impossible outcome.

When you add an element to an equation, you have to add equally to BOTH sides of the equation in order to keep it balanced, getting this (3*4(a+b-c)=3*4(0)). If (a+b-c) equals zero, then while following proper order of operations, this would work down to (0=0), a perfectly acceptable outcome.


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