Using the determinant formula, we find it equals 10.

The determinant evaluates to 0.

The determinant is \( 2*4 - 3*1 = 8 - 3 = 5 \).

The determinant is given by the formula \( ad - bc \).

This is the identity matrix, and its determinant is 1.

The determinant evaluates to 0 as the third row can be expressed as a linear combination of the first two.

det = (1*5) - (2*3) = 5 - 6 = -1.

For the parabola y^2 = 4px, here 4p = -8, so p = -2. The directrix is given by x = -p, which is x = 2.

Distance = √[(4-1)² + (6-2)²] = √[9 + 16] = √25 = 5.

Distance = √[(7-3)² + (1-4)²] = √[4 + 9] = √13 ≈ 3.6, closest option is 4.

Distance = √[(5-2)² + (7-3)²] = √[3² + 4²] = √[9 + 16] = √25 = 5.

Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2) = |3(1) + 4(2) - 12| / sqrt(3^2 + 4^2) = |3 + 8 - 12| / 5 = 1.

Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2) = |2*3 + 3*4 - 6| / sqrt(2^2 + 3^2) = |6 + 12 - 6| / sqrt(13) = 12 / sqrt(13).

The characteristic polynomial is det(A - λI) = (2-λ)(2-λ) - 1 = λ^2 - 4λ + 3 = 0, giving eigenvalues 1 and 3.

Equation of circle: (x-h)² + (y-k)² = r² => (x-2)² + (y+3)² = 5².

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

The slope m = (4-2)/(3-1) = 1. Using point-slope form: y - 2 = 1(x - 1) gives y = x + 1.

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.

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.

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.