The integral of e^(2x) is (1/2)e^(2x) + C, where C is the constant of integration.

The integral of x^2 is (1/3)x^3 + C, where C is the constant of integration.

The inverse of D is (1/det(D)) * adj(D) = (1/(4*6 - 7*2)) * [[6, -7], [-2, 4]] = [[3/2, -7/4], [-1/2, 2/4]].

The inverse of F is given by (1/det(F)) * adj(F). Here, det(F) = 4*6 - 7*2 = 10, and adj(F) = [[6, -7], [-2, 4]]. Thus, F^(-1) = (1/10) * [[6, -7], [-2, 4]] = [[3/5, -7/10], [-1/5, 2/5]].

Using the fact that sin(x) approaches x as x approaches 0, we have lim (x -> 0) (x^3)/(sin(x)) = 0.

As x approaches infinity, the leading terms dominate. Thus, lim (x -> ∞) (3x^2)/(5x^2) = 3/5.

f'(x) = -2x + 6; setting to 0 gives x = 3; f(3) = -3^2 + 6(3) - 8 = 1.

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, so local minima is (2, 1).

The maximum occurs at x = -b/(2a) = 10/(2*2) = 2.5. f(2.5) = -2(2.5^2) + 10(2.5) - 12 = 6.

The maximum occurs at x = -b/(2a) = -12/(-6) = 2. f(2) = -3(2^2) + 12(2) - 5 = 7.

The vertex is at x = -4/(2*(-1)) = 2. The maximum value is f(2) = -2^2 + 4*2 + 5 = 7.

First, arrange the numbers: 10, 11, 12, 13, 14, 15. Since there are 6 numbers (even), the median is the average of the 3rd and 4th numbers: (12 + 13) / 2 = 12.5.

First, arrange the numbers: 10, 11, 12, 13, 14, 15. Since there are 6 numbers (even), the median is the average of the 3rd and 4th numbers: (12 + 13) / 2 = 12.5.

First, arrange the numbers: 10, 11, 12, 13, 14, 15. The median is the average of the 3rd and 4th numbers: (12 + 13) / 2 = 12.5.

Arrange the numbers: 10, 20, 30, 40, 50, 60, 70, 80. The median is the average of the 4th and 5th values: (40 + 50) / 2 = 45.

Arrange the numbers: 2, 3, 6, 8, 9. The median is the middle number, which is 6.

Arrange the numbers: 10, 12, 15, 18, 20, 25. Since there are 6 numbers, the median is the average of the 3rd and 4th values: (15 + 18) / 2 = 16.5.

The minimum occurs at x = -b/(2a) = 16/(2*4) = 2. f(2) = 4(2^2) - 16(2) + 15 = 1.

The vertex is at x = 8/(2*4) = 1. The minimum value is f(1) = 4(1)^2 - 8(1) + 3 = -1.

The minimum occurs at x = -b/(2a) = -6/(2*1) = -3. f(-3) = (-3)^2 + 6(-3) + 10 = 1.

The minimum occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

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.