Mixed practice: Essential Foundations

A; −12. Evaluate the parentheses first: 5 − 8 = −3. Multiply next: 2(−3) = −6. Finally, −6 + (−6) = −12. A common error is to multiply the initial −6 by 2, but those terms are separated by addition. The expression’s structure, not the order of the printed numbers alone, determines the calculation. Repair: Example 2.01.

9/20. Convert both fractions to twentieths: 7/10 = 14/20 and 1/4 = 5/20. The difference is 14/20 − 5/20 = 9/20. A decimal check gives 0.70 − 0.25 = 0.45 = 9/20. Fractions are subtracted by counting like-sized parts, which is why a common denominator is necessary. Repair: Example 2.04.

D; 48. The statement 0.48 = p% means 0.48 = p/100. Multiply by 100 to obtain p = 48. The question asks for the number before the percent sign, not the original decimal. Choice B would incorrectly turn 0.48 into 0.48% = 0.0048. Repair: Example 2.10.

B; −6x + 13. Distribute the signed factor −2: 5 − 2(3x − 4) = 5 − 6x + 8 = −6x + 13. The product (−2)(−4) is positive 8. At x = 0, the original expression equals 13, which immediately rules out options whose constants are −3 or 1. That check supports the expansion; the distributive argument proves it. Repair: Example 3.05.

6. The output is fixed at 31, so solve 4t + 7 = 31. Subtract 7 to get 4t = 24, then divide by 4 to get t = 6. Verify the input by evaluating f(6) = 31. Computing f (31) would confuse the supplied output with an input. Repair: Example 4.05.

D; 4. The notation g(1) asks for the output paired with input 1. In the column with x = 1, the value of g(x) is 4. Do not read the output under a neighboring input (such as 0, which belongs to input 3), and do not answer with an input value such as 3 or −1. Find input 1 in the x row and read the value directly below it. Use the row labels explicitly. Repair: Example 4.04.

480 meters. A constant speed of 180 meters in 3 minutes is 180/3 = 60 meters per minute. In 8 minutes the distance is 60(8) = 480 meters. The units verify the operation: meters per minute multiplied by minutes gives meters. A proportion 180/3 = d/8 produces the same result. Repair: Example 5.08.

B; 12 balls. Let b be the smaller box’s count. The larger count is 2b − 4, so b + (2b − 4) = 32. Then 3b = 36 and b = 12. The larger count is 2(12) − 4 = 20, and 12 + 20 = 32. Choice D gives the other box’s count, not the target. Repair: Example 5.03.

−24. An exponent is evaluated before a leading negative sign, so −4² = −16. The parenthesized base −2 is cubed: (−2)³ = −8. Add the two terms to obtain −16 − 8 = −24. An even power makes a parenthesized negative base positive; an odd power leaves it negative. Repair: Example 2.02.

C; 30%. The fraction selecting A is 18/60 = 3/10 = 0.3. Convert that proportion to a percent by multiplying by 100: 30%. The denominator is the total number of responses, not the number that chose other options. Choice A mistakes the proportion for the percent number. Repair: Example 2.10.

4/3. Invert the divisor and multiply: (3/5)(20/9) = 60/45 = 4/3. This is greater than 1 because 3/5 is greater than 9/20: more than one divisor-sized group fits in the dividend. Decimal checking gives 0.6/0.45 = 1.333.... Repair: Example 2.07.

A; 2.4 × 10³.

Divide coefficients and subtract exponents:

$\frac{\text{7.2 ×}{\text{10}}^{\text{5}}}{\text{3 ×}{\text{10}}^{\text{2}}}$ = $\frac{\text{7.2}}{\text{3}}$ × 10⁵⁻² = 2.4 × 10³.

The coefficient 2.4 is already in the standard range. Adding the exponents, as in choice B, would use the multiplication rule for a division problem. Repair: Example 2.12.

27. The given equation already supplies the group used in the target. Substitute x −3 = 4 directly: 2(x −3)² −5 = 2(4)² − 5 = 32 − 5 = 27. Solving for x = 7 and then substituting also works, but it adds an unnecessary step. Repair: Example 3.14.

C; 3. A zero of f is an input where the output equals zero. The graph crosses the horizontal axis at (3, 0), so the zero is 3. The point (0, 6) gives the vertical intercept and the value f(0), not a zero. Repair: Example 4.06.

C; x ≥ −4 and x ≠ 2. For a real square root, x + 4 ≥ 0, which gives x ≥ −4. The denominator must also be nonzero: x − 2 ≠ 0, so x ≠ 2. The endpoint x = −4 is allowed because its numerator is zero and its denominator is −6. Both restrictions are necessary. Repair: Example 4.11.

5. Because L is stated to be linear, its rate is constant. From (2, 11) to (6, 23), the rate is (23 − 11)/(6 − 2) = 3. Moving from input 2 back to input 0 lowers the output by 3(2) = 6, so L(0) = 11 − 6 = 5. Equivalently, L(x) = 3x + 5. Repair: Example 4.07.

C; 45. A 20% decrease multiplies by 0.80, producing 45(0.80) = 36. A 25% increase then multiplies the new value by 1.25, producing 36(1.25) = 45. Here 0.80(1.25) = 1, so the changes exactly reverse. They do not cancel merely because one is a decrease and the other an increase; the bases differ. Repair: Example 5.06.

60 pages. The rate is 36/90 = 0.4 page per second. At that constant rate, 150 seconds produces 0.4(150) = 60 pages. You can also scale time by 150/90 = 5/3 and scale pages by the same factor: 36(5/3) = 60. Repair: Example 5.08.

A; 7 items. Let n be the whole-item count. The cost condition 11 + 4n ≤ 40 gives n ≤ 29/4 = 7.25. The greatest nonnegative integer allowed is 7. Seven items cost $39; eight cost $43 and exceed the budget. Rounding 7.25 down here follows the constraint, not a universal rule for all word problems. Repair: Example 5.10.

B.

In option B, the inputs 0, 1, and 2 each have exactly one output. The shared output 4 is allowed. In A, input 1 has outputs 2 and 3; in C, input 0 has outputs 1 and 2; in D, input 2 has outputs 6 and 7. Those repeated-input conflicts violate the function definition. Repair: Example 4.09.

A; −0.7. Convert the fractions for comparison: −5/8 = −0.625 and −2/3 = −0.666.... On the number line, −0.7 lies to the left of both and is therefore least. For negative numbers, the greater absolute value corresponds to the smaller number. Zero is greater than all three negatives. Repair: Example 2.11.

C; x + 6 for x ≠ 6. Factor the numerator as (x − 6)(x + 6) and divide out the nonzero factor x − 6. The original denominator excludes x = 6, and that restriction survives simplification. The value x = −6 is allowed and produces zero, so choice D excludes the wrong input. Repair: Example 3.08.

21. Use the known input-output pair to find the parameter: f(−2) = −2k + 9 = 3. Thus −2k = −6 and k = 3. Now f(4) = 3(4) + 9 = 21. Check the parameter against the original information: 3(−2) + 9 = 3. The target is a function value, not k. Repair: Example 3.13.

11 child tickets. Let c be the number of child tickets, so 20 − c are adult tickets. Revenue gives 7c + 12(20 − c) = 185. Simplifying yields 240 − 5c = 185, so c = 11. Check: 11 child tickets and 9 adult tickets give 77 + 108 = 185. An all-adult baseline gives the same result: the $55 reduction requires eleven $5 substitutions. Repair: Example 5.13.

A; no real solutions. Distribute to get 6x − 2 = 6x + 5. Subtracting 6x leaves −2 = 5, a false statement independent of x. No value can repair that contradiction. Variables disappearing does not automatically mean infinitely many solutions; that requires a true statement such as 5 = 5. Repair: Example 3.11.

B; (−2, 2]. The range consists of outputs. Along the segment, y decreases from the included value 2 to the excluded value −2. Write those outputs from least to greatest: (−2, 2]. The domain would instead be [−3, 1); it answers an input question. Repair: Example 4.12.

120 liters. Let V be the initial volume. After 5/12 is used, 7/12 remains. Using 2/7 of that remainder leaves 5/7 of it. Therefore the final amount is (5/7)(7/12)V = (5/12)V = 50, so V = 120. Check: 50 liters are used first, then 20 of the remaining 70, leaving 50. Repair: Example 5.14.

C; 24,000 square centimeters.

An area conversion uses the squared length factor:

2.4 m² ($\frac{\text{100 cm}}{\text{1 m}}$)² = 2.4(10,000) cm² = 24,000 cm².

A one-meter-by-one-meter square contains 100 × 100 square centimeters, not 100. Repair: Section 5.3.

21. Mean equals total divided by count, so the four measurements total 4(18) = 72. The known values total 16 + 21 + 14 = 51. The missing value is 72 − 51 = 21. Verification: (16 + 21 + 14 + 21)/4 = 18. Do not average the three known values and treat that result as the missing fourth. Repair: the mean relationship in Section 5.3.

B; f is linear. For a linear function, the rate is (10 − 2)/(3 − (−1)) = 8/4 = 2, giving f(x) = 2x + 4 and f(1) = 6. The other conditions are too weak: f(x) = x² + 1 also meets the two given values and has positive outputs at 0 and 2, yet f(1) = 2. Thus those positivity conditions do not determine the requested value. Repair: Example 4.15.