Module 7-10 Test Practice Quiz

By subtracting 3 from both sides, you get 2x = 4, and dividing by 2 yields x = 2. This is a straightforward application of basic algebraic manipulation.

The slope-intercept form y = mx + b reveals the slope, m, as the coefficient of x, which in this case is -2.

Squaring a number means multiplying it by itself. Since 5 x 5 equals 25, that is the correct answer.

The distributive property multiplies 2 by both x and 3, resulting in 2x + 6. This is the proper application of distribution.

Combining like terms, 3x minus 2x simplifies directly to x. This demonstrates a basic understanding of algebraic simplification.

Distribute 3 to get 3x - 12 = 12. Adding 12 to both sides gives 3x = 24, and dividing by 3 results in x = 8.

Subtracting 2x from both sides gives 5 = x - 7. Adding 7 to both sides results in x = 12, which is the correct solution.

The factors of 6 that add up to 5 are 2 and 3, so the quadratic factors as (x + 2)(x + 3).

The vertex x-coordinate of a quadratic in the form ax² + bx + c is given by -b/(2a). Here, -(-4)/(2*1) equals 2.

First add 5 to both sides to obtain 2x < 12, then divide by 2 to get x < 6. This inequality solution is straightforward.

Evaluate the expression inside the parentheses first: 3 - 2 equals 1. Multiplying by 4 gives 4, and adding 5 results in 9.

When multiplying powers with the same base, you add the exponents: 3 + 2 equals 5, so the product simplifies to x❵.

Substitute x = 4 into the function: 2(4) + 3 equals 11. This is a direct computation of the function's output.

Substituting x = 3 into the equation yields y = 2(3) - 1, which simplifies to 5. Thus, the point (3, 5) satisfies the equation.

The distance formula gives √[(4-1)² + (6-2)²] = √(9 + 16) = √25 = 5. This calculation confirms the distance between the two points.

Use substitution by expressing x from the second equation, x = y + 1, and plug into the first equation. Solving yields y = 2 and consequently x = 3.

The quadratic can be factored as (x + 1)² = 0. Thus, the only solution is x = -1.

After squaring both sides and rearranging, the quadratic equation x² - 4x - 2 = 0 is obtained. Checking potential solutions shows that x = 2 + √6 is the only valid answer.

By Vieta's formulas, the sum of the roots for ax² + bx + c = 0 is -b/a. Here, -(-12)/3 equals 4.

Although direct substitution results in a 0/0 indeterminate form, factoring x² - 9 as (x - 3)(x + 3) and then canceling yields f(x) = x + 3 for x ≠ 3. By continuity, f(3) is 6.