The mode is 4, as it appears 3 times, which is more than any other number.

The number 10 appears 3 times and 15 appears 3 times. Since both have the highest frequency, the mode is 10 and 15 (bimodal). However, if only one mode is required, we can take the first one, which is 10.

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

The mode is 30, as it appears 3 times, more than any other number.

The mode is 15, which appears 3 times in the data set.

The number 15 appears most frequently (three times), making it the mode.

The mode is 4, as it appears four times, which is more than any other number.

The mode is 10, as it appears 3 times, which is more than any other number.

The number 10 appears 3 times and 15 appears 3 times. Both are modes, but since we need to choose one, we can say the mode is 10.

The mode is 10, which appears 3 times, more than any other number.

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

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

Solving the equations simultaneously, we find the intersection point is (3, 4).

A · B = (2)(1) + (3)(2) + (1)(3) = 2 + 6 + 3 = 11

A · B = (6)(2) + (8)(3) = 12 + 24 = 36.

A · B = (7)(3) + (-2)(4) + (1)(-5) = 21 - 8 - 5 = 8.

A · B = (7)(3) + (1)(4) + (2)(5) = 21 + 4 + 10 = 35.

First derivative f'(x) = 16x^3 - 6x^2 + 1. Second derivative f''(x) = 48x^2 - 12x.

First derivative f'(x) = 3x^2 - 6x; second derivative f''(x) = 6x - 6.

First derivative f'(x) = 3x^2 - 6x; Second derivative f''(x) = 6x - 6.

Magnitude |A| = √(6^2 + (-8)^2) = √(36 + 64) = 10. Unit vector = (6/10)i + (-8/10)j = (3/5)i - (4/5)j.

Magnitude |D| = √((-3)^2 + 4^2) = √(9 + 16) = √25 = 5. Unit vector = D/|D| = (-3/5)i + (4/5)j = -0.6i + 0.8j.

(1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i - 1 = 2i.

Using the binomial theorem, (1 + 2)^6 = 3^6 = 729.

Using the binomial theorem, (a + b)^4 = C(4, 0)a^4b^0 + C(4, 1)a^3b^1 + C(4, 2)a^2b^2 + C(4, 3)a^1b^3 + C(4, 4)a^0b^4. Substituting a = 2 and b = 3 gives 16 + 4*6 + 6*9 + 4*27 + 81 = 81.

The discriminant must be less than zero: k^2 - 4*1*16 < 0 leads to k < -8.

The coefficient of x^2 in (x + k)^4 is C(4, 2) * k^2 = 6. Thus, 6k^2 = 6, giving k^2 = 1, so k = 1 or -1.

The term containing x^4 is C(6,4) * (2)^2 * x^4 = 15 * 4 * x^4 = 60x^4. Setting 60 = 240 gives k = 4.

C(7, 4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7*6*5)/(3*2*1) = 35.

Using the binomial theorem, the coefficient of x^4 in (a + b)^n is given by nCk * a^(n-k) * b^k. Here, n=6, a=x, b=-2, and k=2. Thus, the coefficient is 6C2 * (1)^4 * (-2)^2 = 15 * 4 = 60.