Ch 16 Quizlet Practice Test

Subtracting 5 from both sides of the equation isolates x, yielding x = 5. This simple manipulation demonstrates basic algebraic principles.

Substitute x = 3 into the expression 2x to obtain 2*(3) = 6. This exercise reinforces the concept of variable substitution.

By substituting y = 4 into the expression y + 3, we calculate 4 + 3 = 7. This demonstrates simple addition with variables.

Using the distributive property, multiply 3 by each term inside the parentheses: 3*x + 3*2, resulting in 3x + 6. This reinforces distribution over addition.

According to the order of operations, multiply 3 and 4 first to get 12, then add 2, which results in 14. This question emphasizes the importance of proper sequencing in calculations.

Subtract x from both sides to obtain x - 3 = 4, then add 3 to isolate x, resulting in x = 7. This illustrates basic techniques in solving linear equations.

First, distribute 3 over (x - 2) to get 3x - 6, then add 2 yielding 3x - 4. Equate this to 2x + 4 and solve for x to find that x = 8.

Using the slope formula, (y₂ - y₝) / (x₂ - x₝) = (11 - 3) / (6 - 2) = 8/4, which simplifies to 2. This problem reinforces the calculation of slopes from coordinate pairs.

First, distribute 4 to get 8x + 20, then subtract 3x to combine like terms, resulting in 5x + 20. The problem highlights the use of the distributive property and combining like terms.

Substitute x = 7 into the function f(x) = 2x + 1 to get 2(7) + 1, which equals 15. This question reinforces the process of evaluating a function.

Combine the like terms 2x and 3x to yield 5x, and add the constants -4 and 6 to get 2. The simplified expression is 5x + 2, demonstrating the process of combining like terms.

Substitute a = -2 and b = 3 into the expression: 2(-2) + 4(3) equals -4 + 12, which sums to 8. This question reinforces arithmetic operations with negative numbers.

Multiply both sides of the equation by 2 to obtain 5y = 20, then divide by 5 to solve for y, resulting in y = 4. This stresses the method of clearing fractions in equations.

Distribute 3 over (4x - 5) to get 12x - 15, then add 2x to obtain 14x - 15. This problem demonstrates both the distributive property and combining like terms.

Subtract 3x from both sides to get x + 7 = -5, then subtract 7 to isolate x, resulting in x = -12. This showcases the steps in solving a simple linear equation.

Cross-multiply to eliminate the fractions: 3(x + 3) = 2(2x - 1) leads to 3x + 9 = 4x - 2. Subtracting 3x from both sides and adding 2 gives x = 11. This process emphasizes solving equations with fractions.

Using the distance formula, the distance is √[(5-1)² + (-2-2)²] which simplifies to √(16 + 16) = √32, and in simplest form this is 4√2. This reinforces applying the distance formula in coordinate geometry.

First, compute the slope m using the formula (11 - 7)/(5 - 3) which results in 2. Substitute one of the points into the equation to solve for b, yielding b = 1. This demonstrates solving for parameters in a linear function.

For a quadratic equation ax² + bx + c = 0, the sum of the roots is given by -b/a. Here, -(-5)/1 equals 5. This question reinforces the relationship between the coefficients and the roots of a quadratic.

Let the width be w and the length be 2w. The perimeter is 2(w + 2w) = 6w which equals 36, so w = 6 and the length is 12. The area is then calculated as width * length = 6 * 12 = 72.