The area under the curve is given by ∫(from 0 to 2) (2x^2 + 3) dx = [(2/3)x^3 + 3x] from 0 to 2 = (16/3 + 6) = 10.

The axis of symmetry for a parabola in vertex form y = a(x - h)^2 + k is x = h. Here, h = 2.

The characteristic polynomial is given by det(G - λI) = det([[2-λ, 1], [1, 2-λ]]) = (2-λ)(2-λ) - 1 = λ^2 - 3λ + 3.

The circumference C of a circle is given by C = π * diameter. Here, C = π * 14 cm ≈ 3.14 * 14 ≈ 44 cm.

Circumference = 2 * π * radius = 2 * π * 7 = 14π cm

The coefficient of x^0 in (x - 1)^5 is given by 5C5 * (-1)^5 = -1.

Using the binomial theorem, the coefficient of x^2 in (2x + 5)^4 is given by 4C2 * (2)^2 * (5)^2 = 6 * 4 * 25 = 600.

The coefficient of x^2 in (2x - 5)^5 is given by 5C2 * (2)^2 * (-5)^3 = 10 * 4 * (-125) = -5000.

The coefficient of x^3 in (3x + 2)^5 is given by 5C3 * (3)^3 * (2)^2 = 10 * 27 * 4 = 1080.

The conjugate of z is given by z* = 7 + 4i.

The function f(x) = sqrt(x) is continuous at x = 0 as it is defined and the limit exists.

Set f'(x) = 2x - 4 = 0; solving gives x = 2.

Find f'(x) = 2x - 4. Set f'(x) = 0, giving 2x - 4 = 0, hence x = 2.

A × B = |i  j  k| |1  2  3| |4  5  6| = -3i + 6j - 3k.

A × B = |i  j  k| |1  2  0| |3  4  0| = (0 - 0)i - (0 - 0)j + (4 - 6)k = -2k.

E × F = |i  j  k| |1  2  0| |3  4  0| = (0 - 0)i - (0 - 0)j + (1*4 - 2*3)k = -2k.

Using the power rule, f'(x) = 12x^3 - 10x.

Using the power rule, f'(x) = 12x^3 - 15x^2 + 2.

Using the power rule, f'(x) = -4/x^2.

Using the power rule, f'(x) = 10x - 4.

The derivative f'(x) = 15x^2 - 2.

Using the power rule, f'(x) = 25x^4 - 3.

Using the product rule, f'(x) = e^x * ln(x) + e^x * (1/x) = e^x * (ln(x) + 1/x).

Using the product rule, f'(x) = e^x * x^2 + e^x * 2x = e^x * (x^2 + 2x).

The derivative f'(x) = d/dx(ln(x)) = 1/x.

Using the product rule, f'(x) = 2x * sin(x) + x^2 * cos(x).

Using the product rule, f'(x) = 3x^2 * e^x + x^3 * e^x.

Using the product rule, f'(x) = 3x^2 * ln(x) + x^2.

Using the power rule, f'(x) = 4x^3 - 12x.

The derivative f'(x) = d/dx(3x^2 + 5x - 7) = 6x + 5.