Math Is Fun Forum

I think you mean putting the lines into the form y = mx + c, i would suggest against this as it will not deal with vertical lines for a start.
There is an easier way working with lines parametrically as rays.

You 'can' solve for s and t in this manner, but it is quite messy.
It easier to think about it another way:

If you take the vector from one of the starting points 'a' or 'c', to the intersection point, that vector will be parallel to the corresponding starting point, so if the intersection point is f, then (f-a) is parallel to v, and (f-c) is parallel to q
to know if two vectors are parallel, we take the dot product and if they are paralell the dot product will be equal to (plus/minus) the length of the two vectors multiplied together. However, we can think of this another way again, and by rotating (f-a) and (f-c) by 90 degrees, we can instead check that the perpendicular vector to (f-a) is perpendicular to v, and the perpendicular vector to (f-c) is perpendicular to q, this is easier to do since to check if two vectors are perpendicular we are only checking if their dot product is 0.

All that remains to test whether the two line segements intersect is to check that both s and t are betweeen 0 and 1, and if both of them are, then you can find the point of intersection by plugging either s or t into their respective ray equations for the lines

The equations have been set out so that the least amount of computation is needed, so to calculate s,t i would have:

// a {x,y}, b {x,y}, c {x,y}, d {x,y}
\\ v {}, q {}
\\ amc {}, denom, s, t

v.x = b.x-a.x;
v.y = b.y-a.y;
q.x = d.x-c.x;
q.y = d.y-c.y;

denom = v.y*q.x - v.x*q.y;
if(denom ~=~ 0) { /*no intersection, or the lines are colinear*/ return; }
denom = 1/denom;

amc.x = c.x-a.x;
amc.y = c.y-a.y;

t = (amc.y*q.x - amc.x*q.y)*denom;
if( t < 0 || t > 1 ) { /*no intersection*/ return; }

s = (amc.y*v.x - amc.x*v.y)*denom;
if( s < 0 || s > 1 ) { /*no intersection*/ return; }

/*intersection*/ return {a.x + t*v.x, a.y + t*v.y }

How can you call this logical? the elipses are a notation used to denote that there are an infinite number of the preceeding digit in this case, you can not put anything after this: You say there is an 8 after the end of the infinite number of 7's? That makes no sense, because placing an 8 at any point along the length of 7's would be to give the number of 7's a finite value, and the number of 7's would no longer be infinite.

No, the comparison you should be making if you want to make such a silly comparison; is that

999999999999...9999.0 = 10^infinity = infinity, aka an infinite number of 9's preceeding the decimal point is equal to infinity; which is by definition infinity anyways and therefore true, since that length of 9's is infinitely long, it is infinite big, and therefore equal to infinity.

Nothing in the wikipedia article agrees with you; beyond that 0.999... = 1, which is something we (bar george) agree with anyways.

do you mean:

or

: eitherway just remember that you can multiply and divide both sides like normal so:

if you mean the first one:

or second one:

to put them in order, make them all into single fractions with the same denominator, so in your example:

from which you can now easily put them in order

Or, since you didn't specify precisely what we could us.

Let's use the natural numbers under mod 12
9+10+11 = 6

Jane's numbers DO satisfy that constraint.

a)
asymptotes of

b) solving it:

which as before gives:

Are you thinking of:

which can be rearranged to give:

or how to prove that those are true?

To prove it in the second form (Since that's the one you showed in your post)

Jane, 827 is not given as one of the possible answers so whilst your solution may work (I havn't tested it) It can't be the correct solution.

The result will indeed be complex, both inverse sine, and inverse cosine can be expanded to all real numbers: