Subtracting 3 from both sides of the equation gives 2x = 4, and dividing by 2 yields x = 2. This demonstrates a basic method for solving linear equations.

Adding 4 and 5 gives 9, and subtracting 3 yields 6. This problem reinforces basic arithmetic operations.

For two lines to be perpendicular, the slope of one must be the negative reciprocal of the other. Therefore, the negative reciprocal of 2 is -1/2.

A square is defined by having 4 equal sides and 4 right (90-degree) angles. These properties distinguish it clearly from other quadrilaterals.

Performing the arithmetic sequentially, 9 minus 4 equals 5, and adding 2 gives 7. This question checks basic numerical operations.

Subtracting x from both sides results in 2x - 5 = 3, and then adding 5 gives 2x = 8. Dividing by 2 yields x = 4.

The quadratic factors by finding two numbers that multiply to 6 and add to -5, which are -2 and -3. This leads to the factorization (x - 2)(x - 3).

Using the slope formula (y2 - y1)/(x2 - x1), we calculate (11 - 3)/(6 - 2) = 8/4, which simplifies to 2. This reinforces the concept of slope.

Adding the equations eliminates y, resulting in 2x = 12 and x = 6. Substituting back into one of the original equations yields y = 4.

There are a total of 10 marbles, with 2 being blue. Thus, the probability of drawing a blue marble is 2/10, which simplifies to 1/5.

Calculating the exponents, 2^3 equals 8 and 2^2 equals 4. Their sum is 8 + 4 = 12.

Dividing both the numerator and denominator by their greatest common factor, 6, reduces the fraction to 3/4.

Adding 10 to both sides gives 5y = 25, and dividing by 5 results in y = 5. This question reinforces solving simple linear equations.

Twenty percent of a number is computed as 0.2 times that number. Multiplying 50 by 0.2 gives 10.

Substituting x = 5 into f(x) gives f(5) = 3(5) - 4, which simplifies to 15 - 4 = 11. This tests function evaluation.

Recognizing 4x^2 and 9 as perfect squares reveals a difference of squares. Factoring it accordingly results in (2x - 3)(2x + 3).

Factoring the quadratic into (x - 5)(x + 1) = 0 yields the solutions x = 5 and x = -1. This is a standard method for solving quadratics.

Combining the terms gives (1/2 + 1/3)x = (5/6)x, so (5/6)x = 5. Multiplying both sides by 6/5 results in x = 6.

First, evaluate g(3) = 3 + 2 = 5. Then, using f(x) = x^2, compute f(5) = 25. This demonstrates function composition.

By solving the second equation for y (y = 4x - 5) and substituting into the first, the resulting equation gives x = 27/14. Substituting back yields y = 19/7.