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Answers by Roman C.

x8 - 28x5 + 27x2 = 0   x2(x6 - 28x3 + 27) = 0   x2(x3 - 1)(x3 - 27) = 0   x2(x - 1)(x2 + x + 1)(x - 3)(x2 + 3x + 9) = 0   So setting each factor to 0 and solving gives all roots, real and complex.   Real: x...

We can use the adjugate matrix relation: A adj(A) = In det(A)   Then we have:   kA adj(kA) = In det(kA)   kA adj(kA) = kn In det(A)   adj(kA) = kn-1 A-1 det(A)   det(adj(kA)) = det(kn-1 A-1 det(A))   det(adj(kA)) = kn(n-1)...

To find it by the definition of derivatives, do this:   dy/dx = limh→0 [((x+h)2 + 5(x+h) - 2) - (x2 + 5x - 2)]/h   = limh→0 [(x2 +(2h+5)x + 5h + h2 - 2) - (x2 + 5x - 2)]/h   = limh→0 (2hx + 5h + h2)/h   =...

All you have to do is determine what the equation will be in each quadrant when the absolute values are removed.   To do this, note that |A| = A for all A ≥ 0, and |A| = -A for all A < 0.   You should get the following equations for the quadrants.   Q1:...

A) x = sin y + cos x   1 = (cos y) dy/dx - sin x   (cos y) dy/dx = 1 + sin x   dy/dx = (1 + sin x) / cos y     B) xy - y3 = sin x   y + x dy/dx - 3y2 dy/dx = cos x   (x - 3y2) dy/dx = -y +...

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...