f(x) is a polynomial function, which is differentiable everywhere, including at x = 1.

The left limit is 0 and the right limit is undefined. Thus, f(x) is not continuous at x = 0.

At x = 1, f(1) = 3, but lim x->1- f(x) = 1 and lim x->1+ f(x) = 2. Thus, it is not continuous.

Both sides equal 2 at x = 1, hence it is continuous.

The left-hand derivative is -1 and the right-hand derivative is 1. Since they are not equal, f(x) is not differentiable at x = 1.

The slopes are m1 = 2 and m2 = -1/2. The angle θ = tan⁻¹(|(m1 - m2) / (1 + m1*m2)|) = tan⁻¹(5/3), which is approximately 90 degrees.

The slopes are m1 = 2 and m2 = -1/2. The angle θ = tan^(-1) |(m1 - m2)/(1 + m1*m2)| = tan^(-1)(5/4) which is approximately 60 degrees.

The area is given by the integral from 0 to 1 of (x - x^3) dx. This evaluates to [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.

The area enclosed is found by integrating from -2 to 2: ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 16/3.

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

The area under the curve y = 1/x from x = 1 to x = 2 is given by ∫(from 1 to 2) (1/x) dx = [ln(x)] from 1 to 2 = ln(2) - ln(1) = ln(2).

The area under the curve y = e^x from 0 to 1 is given by ∫(from 0 to 1) e^x dx = [e^x] from 0 to 1 = e - 1.

The coefficient of x^2 is C(4,2) * (3)^2 * (-4)^2 = 6 * 9 * 16 = 864.

The coefficient of x^2 is C(6,2) * (3)^2 * (-4)^4 = 15 * 9 * 256 = 34560.

The coefficient of x^2 is C(6,2)(-2)^4 = 15 * 16 = 240.

The coefficient of x^4 is given by 6C4 * (2)^4 * (-3)^2 = 15 * 16 * 9 = 2160.

The coefficient of x^5 in (3x - 4)^7 is C(7, 5) * (3)^5 * (-4)^2 = 21 * 243 * 16 = 68016.

The lines are perpendicular if 2h = a + b, which leads to h^2 = -ab.

The lines are parallel if the discriminant of the quadratic equation is zero, which leads to the condition h^2 = ab.

For the lines to be coincident, the discriminant must be zero, i.e., b^2 - 4ac = 0.

The lines are perpendicular if the condition a + b = 0 holds true.

The limit as x approaches 0 does not exist, hence f(x) is not continuous at x = 0.

The left limit is 0, the right limit is 2, and f(1) = 3. Thus, it is discontinuous.

The left limit as x approaches 1 is 1, the right limit is 2, and f(1) = 2. Since the left and right limits do not match, f(x) is not continuous at x = 1.

The left limit as x approaches 1 is 1, and the right limit is also 1. Thus, f(1) = 1, making it continuous.

At x = 1, f(1) = 2(1) - 1 = 1 and lim x→1- f(x) = 1, lim x→1+ f(x) = 1. Thus, f(x) is continuous at x = 1.

The function f(x) = |x| is continuous at x = 0 since both the left-hand limit and right-hand limit equal f(0) = 0.

Centroid G = ((0+4+0)/3, (0+0+3)/3, (0+0+0)/3) = (4/3, 1, 0).

Centroid = ((0+6+0)/3, (0+0+8)/3, (0+0+0)/3) = (2, 2.67, 0).

Centroid = ((0+0+3)/3, (0+4+0)/3, (0+0+0)/3) = (1, 1.33, 0).