Think of a squared plus b squared equals c squared.  :-)
121903
Bob, I used to enjoy your show before you started doing all those really expensive houses.  Then, of course, you left.\n\nOh, well.\n\nDon't you know, Bob, that isosceles triangles don't have "right sides?"  An isosceles triangle has two "legs" of equal length.  If it is a right isosceles triangle, the side opposite the right angle is called the hypotenuse.\n\nSay hi to Norm for me!
31994
A*B=(3i+5j)*(2i+4j)=2k
513126
v = b - u/k\nv - b = u/k\nk(v-b) = u
507885
Suppose C() replaces circle (so circle x = C(x) = √x)\nSuppose T() replaces triangle (so triangle x = T(x) = x + 5)\nSuppose S() replaces square (so square x = S(x) = 7x - 2)\n\nSo S o C o T(x) = S o C (x + 5) = S( √(x + 5)) = 7 √(x + 5) - 2\n\nThus  S o C o T(4) = 7 √(4 + 5) - 2 = 19
510925
it really doesnt matter...\nyou could do a x b then the answer times c\nb x c and then the answer times a\nor c x a and then the answer times c\n\nif you have a calculator just enter it as a x b x c\n\nno matter how you multiply them you will get the same answer
283573
Figure out what other number times itself equals your number.  That's its square root.
480145
for example\na/b\n---- = ad/bc\nc/d\n \n\na/b + c/d = (ad +cb)/ad\n\na/b -c/d =( ad- bc) /ad\n\n(a/b)(c/d)= ac/bd\n\nx
288911
In (a+b)(a-b)=b(a-b)\n\nyou divided by (a-b)\n\nHowever a-b = 0 so you can't divide by it
333425
find whole (integer) number a,b, and c such that a^3+b^3=c^3.\nDon't remember the answer anymore,but it should be easy - all numbers were less that 10.
170230
1 = b\n2 = d\n3 = d\n4 = c\n5 = c\n\n4/5
349671
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
69467
2x-1(xsquared-x+10)squared
353768
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).
483905
Gary, that's a good one, but one mistake...\nyou said that a=b, thus you cannot divide by (a-b) b/c that equals zero, and you cant divide by zero.
257605
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
439617
I would change cos 2 theta to 2 cos squared theta -1.\n\nYou then have a quadratic in cos theta.
177827
ya it is a^2 +b ^2 +2abcos(c)