**Question 1:**

### What is the magnitude of the vector A = (3, 4) in two-dimensional space?

Explanation: The correct answer is C) 5. The magnitude of a vector A = (x, y) in two-dimensional space is given by the formula √(x^2 + y^2). In this case, the magnitude of vector A = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.

**Question 2:**

### What are the components of a vector B with magnitude 6 and an angle of 30 degrees with the positive x-axis?

Explanation: The correct answer is D) (3√3, 3). The x-component of the vector B can be calculated as Bx = B * cos(θ), where θ is the angle with the positive x-axis. In this case, Bx = 6 * cos(30°) = 6 * √3/2 = 3√3. The y-component of the vector B can be calculated as By = B * sin(θ), where θ is the angle with the positive x-axis. In this case, By = 6 * sin(30°) = 6 * 1/2 = 3. Therefore, the components of vector B are (3√3, 3).

**Question 3:**

### A vector C has components Cx = -5 and Cy = 12. What is the magnitude of vector C?

Explanation: The correct answer is C) 13. The magnitude of a vector C with components Cx and Cy is given by the formula √(Cx^2 + Cy^2). In this case, the magnitude of vector C = √((-5)^2 + 12^2) = √(25 + 144) = √169 = 13.

**Question 4:**

### What are the x and y components of a vector D with magnitude 8 and an angle of 45 degrees with the positive x-axis?

Explanation: The correct answer is C) (4√2, 4√2). The x-component of the vector D can be calculated as Dx = D * cos(θ), where θ is the angle with the positive x-axis. In this case, Dx = 8 * cos(45°) = 8 * 1/√2 = 4√2. The y-component of the vector D can be calculated as Dy = D * sin(θ), where θ is the angle with the positive x-axis. In this case, Dy = 8 * sin(45°) = 8 * 1/√2 = 4√2. Therefore, the components of vector D are (4√2, 4√2).

**Question 5:**

### A vector E has components Ex = 5 and Ey = -3. What is the angle that vector E makes with the positive x-axis?

Explanation: The correct answer is A) 30 degrees. The angle that vector E makes with the positive x-axis can be calculated as the inverse tangent of Ey/Ex. In this case, the angle = arctan((-3)/5) = -30°. Since the angle is negative, we add 180° to obtain the positive angle, resulting in 180° - 30° = 150°. However, we also know that the angle with the positive x-axis is always measured in the counterclockwise direction. Therefore, the angle that vector E makes with the positive x-axis is 360° - 150° = 210°. However, this angle is equivalent to 210° - 180° = 30°. Hence, the answer is 30 degrees.

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