12th Grade Math Quiz - Medium Difficulty

To find the derivative of the function f(x) = 3x^2 + 5x - 4, we apply the power rule. The derivative of 3x^2 is 6x (since the exponent 2 is multiplied by the coefficient 3, and then the exponent is reduced by 1). The derivative of 5x is simply 5, as the derivative of x is 1. The constant term -4 has a derivative of 0. Combining these results gives us the derivative f'(x) = 6x + 5.

The integral of a function represents the area under the curve of that function. For f(x) = 2x, we apply the power rule of integration, which states that the integral of x^n is (x^(n+1))/(n+1) + C. Here, n is 1, so we increase the exponent by 1, resulting in x^2, and then divide by the new exponent (2). Therefore, the integral of 2x is x^2 + C, where C is the constant of integration, representing any constant value that could be added to the function.

To solve the equation 2x + 3 = 11, first subtract 3 from both sides to isolate the term with x: 2x = 11 - 3, which simplifies to 2x = 8. Next, divide both sides by 2 to solve for x: x = 8 / 2. This results in x = 4. Thus, the solution to the equation is 4.

Sin(30°) equals 0.5 because, in a right triangle where one angle is 30 degrees, the ratio of the length of the side opposite the angle (1) to the hypotenuse (2) is 1/2. This fundamental property of trigonometric functions can also be derived from the unit circle, where the y-coordinate of the point corresponding to 30 degrees is 0.5. Thus, sin(30°) is consistently defined as 0.5 in trigonometry.

The quadratic formula is used to find the solutions of a quadratic equation in the form ax² + bx + c = 0. It derives from the process of completing the square and provides the values of x where the equation equals zero. The term under the square root, b² - 4ac, is called the discriminant and indicates the nature of the roots. The formula allows for both real and complex solutions depending on the value of the discriminant, making it a fundamental tool in algebra for solving quadratic equations.

As x approaches 0, the ratio of sin x to x approaches 1. This can be understood through the squeeze theorem, where both sin x and x converge to the same value, leading to their ratio stabilizing at 1. Additionally, using Taylor series expansion for sin x around 0 shows that sin x is approximately equal to x for small values of x, reinforcing that the limit of (sin x)/x is indeed 1 as x approaches 0.

To find f'(1), we first need to calculate the derivative of the function f(x) = x^3 - 3x + 2. The derivative, f'(x), is found using standard differentiation rules: f'(x) = 3x^2 - 3. Next, we evaluate the derivative at x = 1: f'(1) = 3(1)^2 - 3 = 3 - 3 = 0. Thus, the value of f'(1) is 0, indicating that there is a horizontal tangent to the curve at this point.

To find the area of a circle, the formula used is A = πr², where A is the area and r is the radius. In this case, the radius is 5. Plugging this value into the formula, we calculate the area as A = π(5)² = π(25) = 25π. Thus, the area of the circle with a radius of 5 is 25π.