Subtracting 3 from both sides of the equation yields 2x = 4 and then dividing by 2 gives x = 2. This is the straightforward application of basic algebraic principles.

The slope is calculated as the change in y over the change in x: (6 - 2) / (3 - 1) = 4/2 = 2. This basic concept is essential for understanding linear functions.

First, distribute 3 to get 3x + 12, then subtract 2x to combine like terms which results in x + 12. This simplification exercise reinforces distribution and combining like terms.

The sine of 30° is a well-known trigonometric value equal to 1/2. This fundamental result is often memorized in introductory trigonometry courses.

Adding the fractions 1/4 and 1/4 gives 2/4, which simplifies to 1/2. This question reinforces basic operations with fractions.

The discriminant is calculated using the formula b² - 4ac. For this equation, a = 1, b = -4, and c = 3, so the discriminant equals 16 - 12 = 4.

Factoring the quadratic gives (x - 2)(x - 3) = 0, so the solutions are x = 2 and x = 3. This tests the ability to factor and solve simple quadratic equations.

An exponential function features a constant base raised to a variable exponent, as seen in f(x) = 2^x. The other options represent a linear function, a quadratic function, and a logarithmic function.

The logarithmic equation log₂(x) = 3 can be rewritten as x = 2³, which means x = 8. This conversion between logarithmic and exponential forms is a key concept in algebra.

The numerator factors as (x - 3)(x + 3), so the (x + 3) cancels with the denominator, leaving x - 3. This demonstrates the technique of factoring and simplifying rational expressions.

The vertex of a parabola defined by a quadratic in standard form can be found using the formula x = -b/(2a). For y = x² - 6x + 5, x = 3 and substituting back gives y = -4, so the vertex is (3, -4).

Substituting x = 5 into f(x) = 2x - 1 yields 2(5) - 1 = 10 - 1 = 9. This question checks the ability to evaluate a linear function.

Adding the two equations eliminates y: (x+y) + (x-y) = 7 + 3 leads to 2x = 10 so x = 5, and substituting back gives y = 2. This problem tests solving linear systems using elimination.

Since (x - 4)² is always nonnegative, its minimum value is 0, which occurs when x = 4. This question demonstrates understanding of the properties of squared functions.

The equation demonstrates that how numbers are grouped during addition does not affect the sum, which is known as the associative property. Recognizing this property is fundamental to algebraic manipulation.

To find the inverse function, write y = 3x - 7, then swap x and y to get x = 3y - 7, and finally solve for y. The correct inverse is f❻¹(x) = (x + 7)/3.

Recognize that 16 can be written as 2^4. Equating the exponents gives x + 1 = 4, therefore x = 3. This question reinforces the application of exponents in solving equations.

The vertex of an absolute value function in the form |x - h| + k is located at (h, k). Here, h = 2 and k = 3, so the vertex is (2, 3).

The slope between any two points is the change in y divided by the change in x. Calculating for any pair (for example, (4, 10) and (1, 4)) gives (10 - 4) / (4 - 1) = 6/3 = 2.

The probability of not seeing a shooting star in one observation is 1 - 0.2 = 0.8. Since the observations are independent, multiply 0.8 by itself to get 0.64.