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Answers by Roman C.

Triple both sides and then regroup based on pairs which look like a product rule.    3(x2 - 4xy - 2y2)dx + 3(y2 - 4xy - 2x2) dy = 0   3y2 dy - (6y2 dx + 12xy dy) - (12xy dx + 6x2 dy) + 3x2 dx   x3 - 6x2y - 6xy2 + y3 = C &n...

f(x) = 4 - 3x - x2   [f(x) - f(a)] / (x - a)   = [(4 - 3x - x2) - (4 - 3a - a2)] / (x - a)   = (-3x - x2 + 3a + a2) / (x - a)   = [-3(x - a) - (x + a)(x - a)] / (x - a)   = -3 - x - a    

(z2 - 3z + 2)(z2 + 6z + 8) / (z2 + 3z - 4)(z2 - 4)   = (z-1)(z-2)(z+2)(z+4) / (z+3)(z-1)(z+2)(z-2)   = (z+4) / (z+3)  

You can only give bounds for this number.   Let's assume, the n hexagons must form a polyhex (like a polyomino but with regular hexagons).   Maximum:   The first hexagon contributes 6 vertices and each new hexagon after that contributes at most 4 vertices.   MAX...

calculus help (answer)

1.   The total area of the squares is 13.   The area of the disk is 4π < 4 × 3.25 = 13.   Thus clearly, some squares must overlap   2. This is actually a Number Theory problem.   If we choose 12 different two digit numbers...

If F = ∇U then we find that:   U = xey + f(y,z)   U = xey + yez + g(x,z)   U = yez + h(x,y)   We can take:   f(y,z) = yez   g(x,z) = 0   h(x,y) = xey   This means that F(x,y,z)...

Let's neaten things up:   f(x) = 2x3 - 6x2 + 3x + 1   f'(x) = 6x2 - 12x + 3   We are iterating:   xn+1 = xn - (2x3 - 6x2 + 3x + 1)/(6x2 - 12x + 3)   Note that convergence is quadratic. That is, there is...

First let's get the local maximums over the real number line.   The the critical points of y = ax3 + bx2 + cx + d are at   x = [-b ±√(b2 - 3ac)] / (3a).   Note that it looks almost identical to the quadratic formula. There is a reason for this...