tan(θ) = 12/5; θ = tan^(-1)(12/5) which is approximately 67.38 degrees, closest to 60 degrees.

K + L = (2i + 2j) + (-i + 3j) = (2 - 1)i + (2 + 3)j = 1i + 5j = i + 5j.

A · B = x*1 + 2*y + 3*4 = 0. Thus, x + 2y + 12 = 0. Solving gives x + y = -6.

A · B = (3)(2) + (4)(-1) = 6 - 4 = 2.

A · B = (3)(2) + (4)(-1) = 6 - 4 = 2.

If A and B are perpendicular, then by definition A · B = |A||B|cos(90°) = 0.

The arrangement does not provide enough information to determine who is opposite X.

Since x is between 3 and 7, x - 1 will be between 2 and 6, thus x - 1 < 5 is true.

Since Y = Z and X > Y, it follows that X > Z.

Since Y = Z, and X < Y, it follows that X < Z.

To find x, subtract 2 from both sides of the equation: x = 5 - 2, which gives x = 3.

From the first equation, x = 10 - 2y. Substituting into the second gives 2(10 - 2y) - y = 3, solving gives y = 4, x = 2.

Rearranging gives 3 + 1 > 2x - x, thus 4 > x, so x < 4.

Rearranging gives 4 < x, thus the minimum integer value is 5.

To find x, subtract 3 from both sides: x = 7 - 3, which gives x = 4.

x > 6, so the minimum integer value of x is 7.

Rearranging gives 4 + 1 < 2x - x, thus 5 < x, or x > 5.

Rearranging gives 4 + 3 < x, thus x > 7.

Subtracting 4 from both sides gives x = 6.

We know that x^2 + y^2 = (x + y)^2 - 2xy. Substituting the values, we get (10)^2 - 2(21) = 100 - 42 = 58.

x² + y² = (x + y)² - 2xy = 10² - 2*21 = 100 - 42 = 58.

From x - 2 < 3, we get x < 5. From x + 1 > 0, we get x > -1. Thus, 0 < x < 5.

|-3| = 3, so |x| + x = 3 - 3 = 0.

The absolute value |x| of -5 is 5.

1/x = 1/(1/2) = 2.

Substituting x and y gives us 2^(2+3) = 2^5 = 32.

Calculating, we find 2^3 + 3^2 = 8 + 9 = 17.

Substituting x = 2: 3(2^2) + 4(2) - 5 = 3(4) + 8 - 5 = 12 + 8 - 5 = 15.

Substituting x = 2, we get 3(2²) + 4 = 3(4) + 4 = 12 + 4 = 16.

We have x = 8 and y = 4. Thus, x/y = 8/4 = 2.