Math Is Fun Forum

rickyoswaldiow, (s)he's asking:

Not just how.  I guess one reason why is because it's hard to divide fractions, but it's easy to multiply them.  I don't think that one was ever said.

Boy, if only I could.  But Texas Instruments protects stuff like that.

But it isn't krisper.  Factoring a quartic equation is only an exercise if you are bored or insane.  No math teacher woudl ever have you do this, nor would you learn anything from doing it.

The chain rule strikes again!

Using the chain rule, the derivative of 1 - p is -1.

Square both sides, mutiply by the denominator, then subtract.  You end up with:

x^4 - x^3 - 10x^2 - 5x + 7 = 0

Which factors into:

(x^2 + x - 1)(x^2 - 2x - 7) = 0

And you use the quadratic from there.  It comes out to (-√5 + 1)/2, (√5- 1)/2, -2√2 + 1, and 2√2 - 1.

Boy, I thought this thread was finally dead...

I think there is, if you know enough about Hamiltonian Circuits.  I don't.  But I wouldn't quite call it guessing, just making observations.

Whoops, I wrote the solution, not the problem.  What I meant was:

(1-p)-¹

Red:

cos(xy) = 0
cos(θ) = 0 when θ = pi/2 or 3pi/2.  So xy = pi/2, y = pi/(2x) or xy = 3pi/2, y = 3pi/(2x).

Green:

sin(xy) = 0
sin(θ) = 0 when θ = 0 or pi.  So xy = 0, x = 0 or y = 0, or xy = pi so y = pi/x.

Blue:

tan(xy) = 0
tan has periodic 0's at n*Pi.  This means that xy = n*Pi, so y = nPi / x, where n is any integer. 
I believe your graph of this is wrong, it should just be curves of y = pi/x, 2pi/x, 3pi/x, etc.

And remember that each one of these is periodic, which means it repeats.

What software did you use to make them?

I don't think that's right.

Let's start with one digit.  You can go from 0 to 9, so you have 10 choices.

Now two digits.  Each are 0-9, with which you can represent all numbers from 00 to 99, 100 choices.

Same for three digits.  1000 choices.

And so on.  The answer is 10^x, where x is the number of digits.  10^6 in this case.

Huh?  If you think you can do the problems, try them first before asking for help.  If you want us to check your answers, then post all your work.  It makes it easier for us to help you.

Bad algebra.  It's not (4x)^5, it's 4 (x)^5.

A good way to rewrite it is:

4^1 * x^5

So 3 / (4^1 * x^5) = 3*4^-1 * x^-5.  Of course, this is just (3/4) x^-5.

Edit: Oh, and my point in my first post was that you should just take the fraction out before you integrate, then multiply.  I forgot to say that though.

Ok, here's me thinking outside the box.

Take a protractor, measure the angle of the triangle, use this and tan to find A.

Problem solved.

May I also suggest an image which may help you: