Using the order of operations, you first multiply 3 by 4 to get 12 and then add 2, resulting in 14. This reinforces the importance of following PEMDAS.

Combine like terms by subtracting the coefficients (4 - 2) to obtain 2x. This demonstrates a basic technique in simplifying algebraic expressions.

The perimeter of a rectangle is calculated as 2 times the sum of its length and width, so 2 × (5 + 3) equals 16. This applies the basic perimeter formula.

Subtracting 7 from both sides of the equation yields x = 5. This problem illustrates the process of isolating the variable in a simple linear equation.

Subtracting fractions with the same denominator involves subtracting the numerators: (3 - 1)/4 results in 2/4, which simplifies to 1/2. This question reinforces basic fraction subtraction.

Expanding the equation gives 3x - 6 + 4, which simplifies to 3x - 2. Adding 2 to both sides yields 3x = 21, so x = 7. This problem tests use of the distributive property and combining like terms.

Distribute 5 into the parentheses to get 10x + 15 and then subtract 4x to combine like terms, resulting in 6x + 15. This demonstrates proper use of the distributive property.

The slope is calculated by dividing the difference in y-values by the difference in x-values: (11 - 3) / (6 - 2) = 8/4 = 2. This problem applies the slope formula to find the rate of change.

To solve (3/5)x = 18, multiply both sides by the reciprocal 5/3, yielding x = 30. This reinforces solving equations that involve fractions.

Calculate 2³ (which is 8) and 3² (which is 9), then multiply them to get 72. This question emphasizes working with exponents and the proper order of operations.

25% of 200 is found by multiplying 0.25 by 200, which equals 50. This problem tests the ability to convert percentages to decimals and perform multiplication.

Adding 7 to both sides gives 4y = 16, and dividing both sides by 4 results in y = 4. This reinforces solving basic linear equations.

The correct expression, a(b + c) = ab + ac, illustrates the distributive property, which allows multiplication to be distributed over addition. This is a fundamental algebraic tool.

Compute the squares separately: 4² is 16 and 3² is 9. Subtracting 9 from 16 gives 7, demonstrating how to evaluate expressions with exponents.

The average is calculated by summing the numbers (4 + 8 + 12 = 24) and then dividing by the number of values (3), resulting in 8. This reinforces the concept of mean.

Find a common denominator to combine the terms: (4x + 3x)/12 equals 7, so 7x/12 = 7. Multiplying both sides by 12 and then dividing by 7 yields x = 12. This problem emphasizes solving equations involving fractions.

Let the width be w and the length be 2w; then the perimeter is 2(w + 2w) = 6w, so 6w = 36 leads to w = 6. The area is calculated as width × length = 6 × 12, resulting in 72. This problem integrates ratio relationships with perimeter and area calculations.

First, evaluate g(3) by substituting x = 3 to get 3 + 4 = 7. Then, substitute 7 into f(x) to obtain 3(7) - 2 = 19. This tests the concept of function composition.

Distribute -2 to obtain -2x + 6 ≥ 4. Subtracting 6 from both sides gives -2x ≥ -2, and dividing by -2 (and reversing the inequality sign) results in x ≤ 1. This problem highlights the rule of reversing the inequality when dividing by a negative number.

Since 30 students represent the 1/6 who did not attend, multiplying 30 by 6 yields a total of 180 students. This problem applies proportional reasoning to find the total from a part.