Recent content by larstuff

The function for the ellipse is: f(x)= 4/pi * sqrt ( 1 + ((x - 0,5)/0,5)^2 ) The TEX on these pages is difficult to get ... But again, though this is the correct function for the semi-ellipse between x=0 and x=1, with area exactly 1, you will get a hard time computing the arc length...

There seems to be a problem with the cut ellipsis too, I'm afraid. The equation for calculating arc length is given by the integral formula: So when trying to calculate this integral, we get similar problems to those of the rectangle. All ellipsis functions derived from simple...

Yes, it does. There are 3x4 of each kind then in the three decks, that's to say 12 Fours, 12 Fives and so on. This is not a simple task, as far as I can see. Try to rule out the "oposite", that's to say, find the probability of getting only 1 of each kind + 1 or 2 of each kind + 2 of each kind...

That's not what I am getting here. I would have you start with the 1x1 matrix, find the det for that one, continue with the 2x2 matrix. Just use the simplest way of calculating the determinant (diagonal --> right --> down = + and diagonal --> left -->down = -).

This problem can not be solved because the function is not strictly converging towards infinity. You can calculate definite integrals, f.ex. [2,100 000], [2, 1 000 000] and so on, but there is no way this integral reaches a certain limit as x goes towards infinity. I agree with Random Variable...

Try downloading the program Geogebra (Web Start) - it's free math software, then let it draw the graph of this "weird" thing. You'll probably see what the answer is..

I'm not quite sure what the question is here. What does 3-of-a-kind mean in this situation? Does it mean 3 of the ace of Spades, 3 spades, 3 aces... In other words, precicely what is the outcome we seek to get? And - anyway, do not forget all possible combinations in your calculation.

Remember that Df for f(x) is [0,1]. So the sides are there, as given by condition 2), the latter part... :wink: But don't let's argue. I will give you right in the part of sides on a rectangle. Usually there are 4 of them..

Guesswork is a good thing in mathematics, but guesswork often need some kind of reasoning. Drawing a graph by hand / on computer is not guesswork in this case. You should come up with a more sophisticated way of showing what your guesswork builds on. One way to go by, is to look at the last...

No. Perhaps not, but what about the function f(x) = 8 ? Is that not a rectangle function on at clearly defined area, like this? So a rectangle, in this respect ,might be considered a function, howewer for this exercise, not the correct one, as f(x) needs to be 0 both at x=0 and x=1...

Le Cavalieris principle is about functions in the plane (2d) that are "hightened" into space (3d) by revolving them about the x-axis. A cut half circle would do here.

Triangle or trapezoid is simplest. I'm not sure if a rectangle would satisfy the conditions.

Try solving for the det. for matrixes from smaller to bigger. This will give an interesting result. Start with a 1x1 matrix, then a 2x2 matrix, and so on. A pattern will then emerge. The tricky part is to show why this pattern will continue. It all has to do with the diagonals - how many of...

3/(1+4x)^0.5 = 3 * (1+4x)^(-0,5) Now you can use the common powers law with the exponent (-0,5). Adding 1 to this gives (0,5).

The determinant must eighter be 0 og -1. If the number of rows are even, then the determinant will be zero. If the number of rows are odd, the determinant will be -1.