Math Is Fun Forum

I wouldn't have thought so, because one could quite easily define a language where every name for a number, has 1 more letter than the number value, and so you could never reach a finite loop.

Well not sure about theorem to describe it, but is it not enough to drawn an equilaterial triangle and dividing one side into 6 using the ruler, and drawn the diagonal making a 10 and 50 degree angle?

make a point, draw a circle, from an edge of the circle, draw another circle with same radius so that it passes through centre of the first circle

at one of the points where the two circles meet, draw lines to the two centres, draw line between centres and that gives you an equilateral triangle, using ruler divide one side into 6 equal segments, and then as above.

Your solution is 'still' very convoluted, with so much unnecessary calculations, There is absolutely no need to test the circle against each 4 edges of the rectangle (Not to mention that your code snippet there will not work for vertical lines, nor is it complete, as it is only intersecting against an infinite line, and not the bounded line segment which a rectangle edge is) And even testing against the 4 edges is not enough if you go down this road, because you then also need to handle the case of the circle being completely within the rectangle.

My solution get's around all of this, to give a very fast, very stable solution, which then also opens the door to calculate the minimum displacement needed to seperate the rectangle and the circle so that they no longer intersect, and also the normal of the collision which would be used to resolve the collision and change the velocities of the circle etc appropriately etc.

Let me guess:

you are attempting to render the circle via the rearrangement:

y = b + sqrt(r^2 - (x-a)^2))

?

in which case, remember that there are two roots, so you need to render both

y = b + sqrt(r^2 - (x-a)^2))

AND

y = b - sqrt(r^2 - (x-a)^2))

There are ofcourse, simpler ways of rendering a circle which will give nicer results, using the polar equation for example with a displacement:

centre of circle C, radius r, by iterating over the full circle for θ render curves through the points

P = C + (r.cosθ i + r.sinθ j)

do you mean:
(x^2-36) / (x+6)

or:
x^2 - 36/x + 6

or:
x^2 - 36/(x+6)

or something else?

without brackets, or any form of formatting it is impossible to know.

if you have a sequence

then the term for n = 1 will be equal to a + b + c

..

the first sequence of differences is:

and the second sequence of differences is:

..

if we start with f_1 and have f_2 and f_3, then n = 1 and we have the following:

...

so for example, the following sequence of 3 numbers, starting from n = 1

2,3,10

the sequence of differences is

1,7

which gives a second difference of 6

so we have:

spin and charge are inherit properties of basic particles, which cannot be changed, only observed. quantum entanglement describes what you are talking about though, if two particles for example are entangled, and you observe one to have a positive spin, then the other particle MUST have a negative spin.

But like i said, information cannot be transferred, because you cannot control what the spin on your particle is when it is observed, all you know is that the other particle must have the opposite.

There are many occurences of things travelling faster than the speed of light; for example the path of a beam of light emitted from a distant fast rotating pulsar.

The difference, is that in none of these occurences of things travelling past c, is any information able to be transmitted.

submit

the two equations are exactly the same in the end; both C and K are both constants, so the product CK is also a constant.

thus e^k(t-c) is essentialy exactly the same as e^(tk-c), only that c has a different value because it is not multiplied by k

You could say that your teachers solution is simplified.

--

In the same way, you might end up with an integration where when you manipulate it you end with a constant like:

I(x) = f(x) + 4.5C

which ofcourse, you can equally just write as

I(x) + f(x) + C'

the C is written with the apostrophe here just to explicitly show that it is not the same constant as the first equation, but both are correct

moving at 10km/h, after 30 minutes Peter will have travelled 10*0.5 = 5km  (since 30minutes = 0.5hours)

at time t (hours), the sister will have moved 12t, whilst peter moves 10t, but peter is also 5km ahead of his sister

so you have the equation

12t = 5 + 10t
2t = 5
t = 2.5hrs