The sum of the roots is -1 + (-2) = -3, and the product is (-1)(-2) = 2. Thus, k = 2.

If the roots are both 2, then k = 2^2 - 4*2 = 4 - 8 = -4. Thus, k = 4.

Mean = (20 + 30 + 40 + 50) / 4 = 35. Variance = [(20-35)² + (30-35)² + (40-35)² + (50-35)²] / 4 = 125. SD = √125 = 11.18 (approx 10).

Using Vieta's formulas, the sum of the roots is 7 and the product is k = 10.

Using Vieta's formulas, the sum of the roots is -b/a = -12/3 = -4.

To find the vertex, use the formula x = -b/(2a). Here, a = 2, b = -4, so x = 1. Plugging x = 1 into the equation gives y = -1.

The distance between the centers of two externally tangent circles is the sum of their radii: 3 + 4 = 7 cm.

If two lines are perpendicular, the sum of their angles is 90 degrees. Therefore, if one line makes an angle of 30 degrees with the horizontal, the other line must make an angle of 90 - 30 = 60 degrees with the horizontal.

Alternate interior angles are equal when two parallel lines are cut by a transversal. Therefore, if one angle is 65 degrees, the other alternate interior angle is also 65 degrees.

The absolute value of -7 is 7.

The slopes are m1 = 2 and m2 = -1/2. The angle θ between the lines is given by tan(θ) = |(m1 - m2) / (1 + m1*m2)|, which results in 90 degrees.

Area = πr² = 3.14 * (4)² = 3.14 * 16 = 50.24 cm²

The area of a triangle is given by A = 1/2 * base * height. Here, A = 1/2 * 8 * 5 = 20 cm².

The area under the curve is given by ∫(from 1 to 4) (1/x) dx = [ln(x)] from 1 to 4 = ln(4) - ln(1) = ln(4).

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.

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

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.

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.

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

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

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 * ln(x) + x^2.

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

The determinant of a 2x2 matrix is calculated as ad - bc.

Using the distance formula: d = √[(3 - 0)² + (4 - 0)²] = √[9 + 16] = √25 = 5.