Math Homework Help

[Kuiper7986]

Doing some Algebra II math here, I need help.

True or False:

1.) If n distinct objects are arranged in a circle, then there are (n -1)! permutations of the objects around the circle.

2.) Fractal is a geometric figure that has self-similiarity, is created using a recursive process, and is infinite in structure.


Solve:

3.) What\'s the antilogorithm for the logarithm 0.3142, rounded to four decimal places.

4.) If Log 6500 = 3.8129 find each number:
   
        a.) Mantissa of log 0.065
   
        b.) Antilog 3.8129

[Abemore]

... false

[ampathy]

this much i know: fractals are not necessarily created with a recursive process (eg. mandelbrot)

[Xalthar]

Gah! 8o I\'m having enough trouble with my own math homework, (which is quite like that of yours) so I haven\'t the time(read: resources) to help...

[TheGeneral]

Here is what I bothered to plug into my calc. Don\'t hold me to my answers, cause by the looks of it, you seem to be going into depth about this stuff, and I briefly covered it.

Solve:
#1) 10^(0.314)
#2)I have no clue what you are asking for, but I think it\'s the base, so....around 10....X^3.8129 = 6500
#3)Matissa....? don\'t tell me you did not smoke something here...
#4)6499.8

[Kuiper7986]

Mathematically,

Mantissa is the logarithm of a number between 1 and 10.

[Xandria]

Originally posted by ampathy
this much i know: fractals are not necessarily created with a recursive process (eg. mandelbrot)

I hate to disappoint you, but all fractals are created using a recursive process, and all are infinitely complex.  Not sure where you learned that they weren\'t  ?(

Ex: the Mandelbrot set is developed using the following equation:

z -> z^2 + c

So pick a complex number, square it, and add to that your original number.  Then take this new number, square it, and add the original number (the same original number you used the first time), etc, etc.  Hense, a recursive process.

And if you\'d like to know how you actually get the image that is associated with the Mandelbrot set, just take a complex number graph and put every single point into the above equation.  If, after say 2000 iterations, you\'re still no farther away from the origin by 2 (in either the real or the imaginary axis) then that point is in the set, and you plot a black point.  If it gets to be greater than 2 in less than 2000 iterations, then you can either plot a white point, or a colored point based on how many iterations have passed (makes a cool graphical pattern, kind of like this:



Note that this was a program that I wrote using C++/MFC in my spare time, nothing fancy    )