The slope of the perpendicular line is -1/5. Using point-slope form: y - 3 = -1/5(x - 2) gives y = -1/5x + 13/5.

The slope of the given line is 5. The slope of the perpendicular line is -1/5. Using y = mx + c, we get y = -1/5x.

Using the slope-intercept form: y = mx + b, where b = 0, we have y = -2x.

Using point-slope form: y - 2 = 3(x - 1) => y = 3x - 1.

Using point-slope form: y - 3 = -1(x - 2) => y = -x + 5.

Using the quadratic formula for the slopes gives m1 = -2 and m2 = -1/3.

Factoring the equation gives (x - 2y)(x + 2y) = 0, which represents the lines x = 2y and x = -2y.

Using the vertex form and substituting the point (2, -4), we find that the equation is y = -2x^2.

The distance from the focus to the directrix is 6, so p = -3. The equation is x^2 = 4py, which gives x^2 = -12y.

The vertex is at (0, 0) and p = 2. The equation is y^2 = 4px, which gives y^2 = 8x.

The vertex form of a parabola is given by (x - h)^2 = 4p(y - k). Here, h = 2, k = 3, and p = 1 (distance from vertex to focus). Thus, the equation is (x - 2)^2 = 4(1)(y - 3) or y = (1/4)(x - 2)^2 + 3.

f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 3). The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.

The equation y = mx + c represents straight lines where m is the slope and c is the y-intercept.

The equation x^2 = 12y can be rewritten as (y - 0) = (1/3)(x - 0)^2, indicating the focus is at (0, 3).

The standard form of a parabola is y^2 = 4px. Here, 4p = 12, so p = 3. The focus is at (p, 0) = (3, 0).

This is a separable equation. Integrating gives ln|y| = 2x + C, hence y = Ce^(2x).

This is a separable equation. Integrating gives ln|y| = x + C, hence y = Ce^x.

The characteristic equation is r^2 - 5r + 6 = 0, giving roots 2 and 3. Thus, y = C1 e^(2x) + C2 e^(3x).

The general solution is x = (2n+1)π/4, where n is any integer.

Factoring gives sin(x)(1 + 2cos(x)) = 0, leading to x = nπ or cos(x) = -1/2.

The general solution is x = (2n+1)π/3, where n is an integer.

The general solution is x = (2n+1)π/3, where n is an integer.

The general solutions are x = 7π/6 + 2nπ and x = 11π/6 + 2nπ.

Using the identity sin(a) = sin(b) gives x = nπ or x = (2n+1)π/3.

The general solution is x = nπ + (-1)^n π/4, where n is any integer.

This is a first-order linear differential equation. The integrating factor is e^(-3x).

The characteristic equation is r^2 - 5r + 6 = 0, giving roots 2 and 3. Thus, y = C1 e^(2x) + C2 e^(3x).

The integral ∫(2x + 3)dx = x^2 + 3x + C.

The integral ∫(2x^3 - 4x + 1)dx = (1/2)x^4 - 2x^2 + x + C.

The integral of 1/x is ln|x| + C.