The pattern is multiplying by 2. 20 * 2 = 40.

'are' should be 'is' because 'team' is a collective noun treated as singular.

The mode is 10, as it appears most frequently (three times) in the data set.

The pattern is n(n+1) where n = 1, 2, 3, 4. Next is 5(5+1) = 30.

Water is a liquid, while the others are metals.

All numbers except 9 are prime numbers.

Car is an inanimate object; the others are animals.

Sunday is a day of the week; the others are months.

Integrating gives y = x^2 + C. Using the initial condition y(0) = 1, we find C = 1.

The general solution is y = Ce^(2x). Using the initial condition y(0) = 1 gives C = 1, so y = e^(2x).

The general solution is y = Ce^(4x). Using the initial condition y(0) = 2, we find C = 2, thus y = 2e^(4x).

The general solution is y = Ce^(4x). Using the initial condition y(0) = 2, we find C = 2, thus y = 2e^(4x).

The general solution is y = e^x + C. Using the initial condition y(0) = 1, we find C = 1.

f''(x) = 6x - 12. Setting f''(x) = 0 gives x = 2. f(2) = 3.

f''(x) = 6x - 12. Setting f''(x) = 0 gives x = 2. The point of inflection is (2, f(2)) = (2, 3).

f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. The point of inflection is at (2, f(2)) = (2, 2).

Solving the equations simultaneously, we find the intersection point is (1, 1).

Setting 2x + 1 = -x + 4 gives 3x = 3, thus x = 1. Substituting x back gives y = 3, so the point is (1, 3).

Set x + 1 = -x + 5. Solving gives x = 2, y = 3. Thus, the point is (2, 3).

Setting x + 2 = -x + 4 gives 2x = 2, so x = 1. Substituting x back gives y = 3. Thus, the point is (1, 3).

To find horizontal tangents, set the derivative y' = 3x^2 - 6x = 0. This gives x = 0 and x = 2. The point (1, 2) has a horizontal tangent.

To find horizontal tangents, set dy/dx = 0. dy/dx = 3x^2 - 6x = 0 gives x = 0 and x = 2. At x = 1, y = 2.

Projection of A onto B = (A · B) / |B|^2 * B; A · B = 2*1 + 3*1 = 5; |B|^2 = 1^2 + 1^2 = 2; Projection = (5/2)(1, 1) = (2.5, 2.5).

Projection of A onto B = (A · B) / |B|^2 * B. A · B = 3*1 + 4*2 = 11, |B|^2 = 1^2 + 2^2 = 5. Thus, projection = (11/5) * (1, 2) = (11/5, 22/5).

Range = Maximum - Minimum = 30 - 10 = 20.

Range = Maximum - Minimum = 30 - 12 = 18.

Range = max - min = 30 - 5 = 25.

Range = Maximum - Minimum = 25 - 8 = 17.

The real part of the complex number 4 + 5i is 4.

The real part of z = 2 + 3i is 2.