All Categories MCQs
Topic Notes: All Categories
General Description
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
3701
What is the distance between the points (a, 0) and (0, b)?
Answer:
√(a² + b²)
Step 1: Apply the distance formula d = √[(x2 - x1)² + (y2 - y1)²]. Step 2: Substitute the given points: d = √[(0 - a)² + (b - 0)²]. Step 3: Simplify the expression to get d = √(a² + b²).
3702
What is the distance between the origin and the point (-8, 6)?
Answer:
10
Step 1: The distance from the origin (0,0) to a point (x,y) is given by d = √(x² + y²). Step 2: Substitute the values: d = √((-8)² + 6²). Step 3: Simplify to d = √[64 + 36] = √100 = 10.
3703
Calculate the distance between the points (-1, -1) and (2, 3).
Answer:
5
Step 1: Use d = √[(x2 - x1)² + (y2 - y1)²]. Step 2: Substitute points: d = √[(2 - (-1))² + (3 - (-1))²]. Step 3: Simplify: d = √[3² + 4²] = √[9 + 16] = √25 = 5.
3704
What is the distance between the points (2, 3) and (2, 7)?
Answer:
4
Step 1: Notice that the x-coordinates are the same, indicating a vertical line segment. Step 2: The distance is simply the absolute difference of the y-coordinates. Step 3: Distance = |7 - 3| = 4.
3705
Find the distance between the points (-5, 0) and (0, 12).
Answer:
13
Step 1: Apply the distance formula d = √[(x2 - x1)² + (y2 - y1)²]. Step 2: Substitute values: d = √[(0 - (-5))² + (12 - 0)²]. Step 3: Calculate d = √[5² + 12²] = √[25 + 144] = √169 = 13.
3706
What is the distance between the points (0, 0) and (3, 4)?
Answer:
5
Step 1: Use the distance formula d = √[(x2 - x1)² + (y2 - y1)²]. Step 2: Substitute the coordinates (0,0) and (3,4) to get d = √[(3 - 0)² + (4 - 0)²]. Step 3: Simplify to d = √[9 + 16] = √25 = 5.
3707
For the quadratic equation x^2 + px + q = 0, if the non-zero roots are p and q, what are the values of p and q?
Answer:
p = 1, q = -2
The sum of the roots is -p, so p + q = -p, leading to q = -2p. The product of the roots is q, so p*q = q. Since the roots are non-zero, q ≠ 0, allowing us to divide by q to get p = 1. Substituting p = 1 into q = -2p gives q = -2. The values are p=1, q=-2.
3708
Which of the following equations has roots -1 and -2?
Answer:
x^2 + 3x + 2 = 0
If the roots are -1 and -2, the sum is -3 and the product is 2. The quadratic equation is x^2 - (sum)x + (product) = 0, which translates to x^2 - (-3)x + 2 = 0, simplifying to x^2 + 3x + 2 = 0.
3709
Solve the equation for x: √x = x - 2
Answer:
4
Squaring both sides gives x = (x - 2)^2. Expanding gives x = x^2 - 4x + 4. Rearranging yields x^2 - 5x + 4 = 0, which factors to (x-4)(x-1) = 0. Potential roots are 4 and 1. Checking the original equation: if x=1, √1 = 1-2, which means 1 = -1 (False). If x=4, √4 = 4-2, which means 2 = 2 (True). The only valid root is 4.
3710
If a quadratic equation has no linear term (b = 0), what can be said about its real roots (if they exist)?
Answer:
They are equal in magnitude and opposite in sign
When b=0, the equation is ax^2 + c = 0, so x^2 = -c/a. If real roots exist, taking the square root gives x = ±√(-c/a). These two values have the same absolute value (magnitude) but opposite signs.