The eccentricity of a parabola is always equal to 1.

Substituting x = 1 gives 2(1)^2 + 3(1) + k = 0 => 2 + 3 + k = 0 => k = -5.

Using Vieta's formulas, k = 2*1*(-2) = -4.

Using Vieta's formulas, r1 + r2 = -b/a = -3/2.

In the equation y^2 = 4px, p = 3, hence the value of 'p' is 3.

Rewriting the equation in standard form gives (x^2/16) + (y^2/9) = 1, so semi-major axis a = 4 and semi-minor axis b = 3.

Using Vieta's formulas, p = -(2 + 3) = -5.

The term is given by C(n, 3) * a^2 * x^3. Setting C(n, 3) * a^2 = 15 gives n = 6.

The equation y = k/x represents a family of hyperbolas.

a_n = a + (n-1)d = 5 + (15-1) * 3 = 5 + 42 = 47.

Range = Maximum - Minimum = 14 - 3 = 11

Evaluating f(0) = e^0 + 0^2 = 1 + 0 = 1.

The area of a triangle is given by (1/2) * base * height = (1/2) * 10 * 12 = 60 cm².

The distance between the foci is given by 2c, where c = √(a^2 - b^2). Here, c = √(5^2 - 3^2) = √16 = 4, so the distance is 2c = 8.

Set y = 0 in the equation: 3x = 12 => x = 4. The point is (4, 0).

Set y = 0 in the equation: 3x = 12 => x = 4.

Rearranging gives y = -3/4x + 3. The slope is -3/4.

Rearranging gives y = -3/4x + 3. The y-intercept is 3.

The slope of the line is given by -A/B = -3/-4 = 3/4. Parallel lines have the same slope.

Rearranging gives y = (3/4)x + 3, so the slope is -3/4.

Rearranging gives y = -5/12 x + 5, so the slope is -5/12.

Setting x = 0 in the equation gives 2y = 10, hence y = 5.

Reflecting about the x-axis changes the sign of y-coefficient: 5x + 2y + 10 = 0.

Slope of 2x + 3y = 6 is -2/3, so m = 3/2 (negative reciprocal).

The lines intersect if the discriminant D = b^2 - 4ac > 0.

The sum of the slopes of the lines can be found using the relation -b/a, which gives -3.

Using the formula for the angle between two lines, we can derive the coefficient of xy that satisfies the angle condition.

The product of the slopes of the lines can be found from the equation. Here, the product of the slopes is given by -c/a, where c is the coefficient of xy and a is the coefficient of x^2.

For the lines to be perpendicular, the condition 4 - 4(3)(2) = 0 must hold, leading to k = 0.

The product of the slopes of the lines represented by ax^2 + bxy + cy^2 = 0 is given by c/a. Here, c = 2 and a = 3, so the product is 2/3.