Aimathic – Absolute value applications

Absolute value applications

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Difficulty: 1 – 5

5510307

Let $A$ and $B$ be the values marked on the number line. a) What are $A$ and $B$? b) Find $|A|$ and $|B|$. c) Why are the two absolute values equal even though $A\ne B$?

Hints

- Read the coordinate of each marked point from the ticks. - Think of absolute value as distance from zero. - Compare the two distances without using left or right direction as part of the distance.

Solution

1. The number line shows $A=-3$ and $B=3$. 2. Absolute value gives distance from zero, so $|A|=|-3|=3$ and $|B|=|3|=3$. 3. The points lie on opposite sides of zero, but each is $3$ units from zero. Therefore, their absolute values are equal.

Answer

a) $A=-3$, $B=3$ b) $|A|=3$, $|B|=3$ c) Both values are $3$ units from zero, so they have the same absolute value.

5545697

A point is located at $-7.5$ on a number line. How far is the point from $0$? Write an absolute-value expression and evaluate it.

Hints

- Represent distance from zero using absolute value. - A distance is nonnegative even when the coordinate is negative.

Solution

1. Distance from zero is $|-7.5|$. 2. $|-7.5|=7.5$.

Answer

$|-7.5|=7.5$. The point is $7.5$ units from $0$.

5122117

Each statement about rational numbers is false. Give a counterexample for each statement. a) The absolute value of a sum is always equal to the sum of the absolute values. b) Adding a rational number to its absolute value always gives $0$. c) If $a < b$, then $|a| < |b|$.

Hints

- For a), try one positive number and one negative number. - For b), test a positive number. - For c), choose a negative number far from zero and a small positive number. - One valid counterexample is enough to disprove a universal statement.

Solution

1. For a), let $a = 2$ and $b = -5$. Then $|2 + (-5)| = 3$, but $|2| + |-5| = 7$. Therefore, the statement is false. 2. For b), let $a = 4$. Then $4 + |4| = 8 \ne 0$, so the statement is false. 3. For c), let $a = -10$ and $b = 2$. Although $-10 < 2$, $|-10| = 10 > 2 = |2|$. Therefore, the statement is false.

Answer

a) For example, $|2 + (-5)| = 3$, but $|2| + |-5| = 7$. b) For example, $4 + |4| = 8$, not $0$. c) For example, $-10 < 2$, but $|-10| > |2|$.

5122837

For each pair, write and evaluate an absolute-difference expression for the distance between the numbers. Then decide which pair is farther apart. Pair A: $-12$ and $8$ Pair B: $-5.5$ and $-20$

Hints

- Use the same absolute-difference structure for both pairs. - The signed difference may depend on subtraction order, but its absolute value does not. - Compare the two nonnegative distances after evaluating them.

Solution

1. Pair A: $|8-(-12)|=|20|=20$. 2. Pair B: $|-5.5-(-20)|=|14.5|=14.5$. 3. Since $20>14.5$, Pair A is farther apart.

Answer

Pair A: $|8-(-12)|=20$ Pair B: $|-5.5-(-20)|=14.5$ Pair A is farther apart.

5182817

How much less is $-245$ than $-112$? Write an absolute-difference expression for the distance between the two values, evaluate it, and use that distance to answer the question.

Hints

- Represent the separation of the two signed numbers with an absolute difference. - The distance is nonnegative even though both coordinates are negative. - Use the distance to state how much less one value is than the other.

Solution

1. The distance is $|-245-(-112)|=|-133|=133$. 2. Therefore, $-245$ is $133$ less than $-112$.

Answer

$|-245-(-112)|=133$. Therefore, $-245$ is $133$ less than $-112$.

5216437

Evaluate $|30-75|-20+15$.

Hints

- Evaluate the expression inside the absolute-value bars first. - Absolute value makes $-45$ into its distance from zero. - Then complete addition and subtraction from left to right.

Solution

1. Evaluate inside the absolute-value bars: $30-75=-45$. 2. Take the absolute value: $|-45|=45$. 3. Complete the remaining operations from left to right: $45-20+15=25+15=40$.

Answer

$40$

5545707

A laboratory freezer has a target temperature of $-18\,^\circ\text{C}$. Its current reading is $-15.6\,^\circ\text{C}$. How far is the current reading from the target temperature? Write and evaluate an absolute-difference expression.

Hints

- Keep both signed temperatures inside one absolute-difference expression. - Simplify the signed difference before taking its absolute value. - Report the result as a nonnegative temperature difference.

Solution

1. The distance from target is $|-15.6-(-18)|$. 2. The signed difference is $2.4$, so the distance is $2.4\,^\circ\text{C}$.

Answer

$|-15.6-(-18)|=2.4$. The reading is $2.4\,^\circ\text{C}$ from the target.

5545717

A machine is set to cut a part to a target length of $12.50\,\text{mm}$. A part is acceptable when its length is within $0.30\,\text{mm}$ of the target. Three measured lengths are $12.20\,\text{mm}$, $12.45\,\text{mm}$, and $12.85\,\text{mm}$. Which measurements are acceptable? Show an absolute-difference check for each one, and identify any measurement exactly on the tolerance boundary.

Hints

- Write one absolute difference for each measurement and the target. - Compare each nonnegative distance with $0.30\,\text{mm}$. - “Within” includes equality with the tolerance boundary.

Solution

1. $|12.20-12.50|=0.30\,\text{mm}$, so $12.20\,\text{mm}$ is acceptable and exactly on the boundary. 2. $|12.45-12.50|=0.05\,\text{mm}$, so $12.45\,\text{mm}$ is acceptable. 3. $|12.85-12.50|=0.35\,\text{mm}$, so $12.85\,\text{mm}$ is not acceptable.

Answer

$12.20\,\text{mm}$: $|12.20-12.50|=0.30\,\text{mm}$ — acceptable, exactly on the boundary $12.45\,\text{mm}$: $|12.45-12.50|=0.05\,\text{mm}$ — acceptable $12.85\,\text{mm}$: $|12.85-12.50|=0.35\,\text{mm}$ — not acceptable

5103777

Use positions on a number line. a) Find the number exactly halfway between $-2.4$ and $5.6$. b) Find the distance between $-\frac{3}{4}$ and $\frac{5}{4}$. c) Point $P$ is at $-2$. Point $Q$ is on the positive side of the number line and is twice as far from $0$ as $P$. What number does $Q$ represent?

Hints

- The midpoint is the average of two endpoints. - Distance is always nonnegative. - Use absolute value to find a point’s distance from $0$. - Pay attention to which side of $0$ point $Q$ is on.

Solution

1. For a), find the midpoint: $\frac{-2.4+5.6}{2}=\frac{3.2}{2}=1.6$. 2. For b), subtract the lesser number from the greater number: $\frac{5}{4}-\left(-\frac{3}{4}\right)=\frac{8}{4}=2$. 3. For c), $P$ is $2$ units from $0$. Twice that distance is $2\cdot2=4$. Since $Q$ is positive, it represents $4$.

Answer

a) $1.6$ b) $2$ c) $4$

5104837

Which fraction is closest to $0.41$ on a number line: $\frac{3}{8}$, $\frac{2}{5}$, or $\frac{4}{9}$? Convert each fraction to a decimal, write an absolute-difference expression for each distance from $0.41$, and compare the three distances.

Hints

- Convert all three candidates before measuring their distances from the target. - Represent each distance with an absolute difference, not an unsigned subtraction chosen by order. - The closest value has the least nonnegative distance.

Solution

1. Convert: $\frac{3}{8}=0.375$, $\frac{2}{5}=0.4$, and $\frac{4}{9}=0.444\ldots$. 2. Distances are $|0.41-0.375|=0.035$, $|0.41-0.4|=0.01$, and $|0.41-0.444\ldots|\approx0.0344$. 3. The smallest distance is $0.01$, so $\frac{2}{5}$ is closest.

Answer

$|0.41-0.375|=0.035$ $|0.41-0.4|=0.01$ $|0.41-0.444\ldots|\approx0.0344$ Therefore, $\frac{2}{5}$ is closest to $0.41$.

5113627

Without calculating the final value, explain why the two expressions must be equal. Expression A: $\left(-\frac{3}{4}+0.5\right)\cdot(-2)$ Expression B: $\left|-\frac{3}{4}+0.5\right|\cdot2$

Hints

- First determine the sign of the quantity inside the parentheses. - What does multiplying by $-2$ do to a negative value? - What does absolute value do to a negative value? - Compare the resulting forms rather than calculating the final number.

Solution

1. Let $x=-\frac{3}{4}+0.5$. Since $0.5=\frac{1}{2}<\frac{3}{4}$, $x<0$. 2. Expression A is $x\cdot(-2)=(-x)\cdot2$. 3. Because $x<0$, $|x|=-x$. Therefore, Expression B is also $(-x)\cdot2$. 4. Both expressions have the same form, so they must have the same value.

Answer

Let $x=-\frac{3}{4}+0.5$. Since $x<0$, $|x|=-x$. Both expressions therefore equal $(-x)\cdot2$, so they have the same value.

5117997

The segment on a number line from $-1.2$ to $2.4$ is divided into three equal parts. Find the two division points.

Hints

- Find the distance between the two endpoints. - Divide the total distance into three equal lengths. - Starting at the left endpoint, add one section length at a time.

Solution

1. Find the total length: $2.4-(-1.2)=3.6$. 2. Divide by $3$ to find each part’s length: $3.6\div3=1.2$. 3. Starting at $-1.2$, the first division point is $-1.2+1.2=0$. 4. The second division point is $0+1.2=1.2$.

Answer

$0$ and $1.2$

5121717

An elevator connects parking levels below ground with office floors above ground. Ground level is $0$ feet, and each floor is $11.5$ feet high. The elevator travels from the floor of Parking Level $-4$ directly to the floor of Level $6$. What vertical distance does the elevator travel?

Hints

- Represent floors below ground with negative numbers and floors above ground with positive numbers. - Find the height of each endpoint relative to ground level. - The vertical distance is the difference between the two signed heights.

Solution

1. Parking Level $-4$ is at $-4\cdot11.5=-46$ feet. 2. Level $6$ is at $6\cdot11.5=69$ feet. 3. The vertical distance is $69-(-46)=115$ feet. 4. Equivalently, the elevator moves across $6-(-4)=10$ floor intervals, and $10\cdot11.5=115$ feet.

Answer

$115$ feet

5122847

A point is at $-3.4$ on a number line. A second point at coordinate $x$ is exactly $7.2$ units away. a) Write an absolute-value equation that models the distance condition. b) Solve the equation by considering the two possible signed differences.

Hints

- Model “distance from $-3.4$” with an absolute difference involving $x$. - An absolute value equal to a positive number produces two signed cases. - Check that both solutions are exactly $7.2$ units from $-3.4$.

Solution

1. The distance equation is $|x-(-3.4)|=7.2$, or $|x+3.4|=7.2$. 2. Thus, $x+3.4=7.2$ or $x+3.4=-7.2$. 3. The solutions are $x=3.8$ and $x=-10.6$.

Answer

a) $|x+3.4|=7.2$ b) $x=3.8$ or $x=-10.6$

5122857

For each pair, write an absolute-difference expression and evaluate the distance. Give each answer as a decimal or a fraction in simplest form. a) $-\frac{3}{4}$ and $0.2$ b) $1\frac{1}{2}$ and $-\frac{2}{5}$

Hints

- Use an absolute difference before converting to a convenient common representation. - The absolute value ensures the distance is nonnegative regardless of subtraction order. - Keep the absolute-value expression in the final response.

Solution

1. a) $0.2=\frac{1}{5}$, so $\left|\frac{1}{5}-\left(-\frac{3}{4}\right)\right|=\frac{19}{20}=0.95$. 2. b) $1\frac{1}{2}=\frac{3}{2}$, so $\left|\frac{3}{2}-\left(-\frac{2}{5}\right)\right|=\frac{19}{10}=1.9$.

Answer

a) $\left|0.2-\left(-\frac{3}{4}\right)\right|=0.95=\frac{19}{20}$ b) $\left|1\frac{1}{2}-\left(-\frac{2}{5}\right)\right|=1.9=\frac{19}{10}$

5122987

On a number line drawing, $1\,\text{cm}$ represents $0.2$ unit. a) Describe where to place $-1.4$, $0.6$, $-0.8$, and $1.2$ relative to $0$. b) Find the distance between the least and greatest of the four numbers. Give the numerical distance and the corresponding length on the drawing.

Hints

- Identify the least and greatest numbers. - Find their distance by subtracting the lesser value from the greater value. - Divide by $0.2$ to convert a numerical distance to centimeters on the drawing.

Solution

1. Since $1\,\text{cm}$ represents $0.2$, divide each absolute value by $0.2$. Thus, $-1.4$ is $7\,\text{cm}$ left of $0$, $0.6$ is $3\,\text{cm}$ right, $-0.8$ is $4\,\text{cm}$ left, and $1.2$ is $6\,\text{cm}$ right. 2. The least number is $-1.4$, and the greatest is $1.2$. 3. Their numerical distance is $1.2-(-1.4)=2.6$. 4. On the drawing, the length is $2.6\div0.2=13\,\text{cm}$.

Answer

a) $-1.4$: $7\,\text{cm}$ left; $0.6$: $3\,\text{cm}$ right; $-0.8$: $4\,\text{cm}$ left; $1.2$: $6\,\text{cm}$ right. b) The numerical distance is $2.6$, which is $13\,\text{cm}$ on the drawing.

5123007

Points $A=-2$ and $B=3$ are marked on a number line. The segment from $A$ to $B$ is divided into $10$ equal intervals. a) What value does one interval represent? b) What number is at the fourth tick mark to the right of $A$? c) How many intervals lie between $-0.5$ and $1.5$? Explain.

Hints

- Find the total numerical distance from $A$ to $B$. - Divide that distance by the number of equal intervals. - Use the interval value to move from one point to another.

Solution

1. The total numerical distance is $3-(-2)=5$. 2. One interval represents $5\div10=0.5$. 3. Four intervals to the right of $A$ gives $-2+4\cdot0.5=0$. 4. The distance from $-0.5$ to $1.5$ is $1.5-(-0.5)=2$. Since each interval represents $0.5$, the number of intervals is $2\div0.5=4$.

Answer

a) $0.5$ b) $0$ c) $4$ intervals

5139707

Consider these three expressions: $A=-3\cdot4+10$ $B=15\div(-3)+4$ $C=0.5\cdot(-6)+2.8$ Which expression has a value closest to $0$? Justify your answer with calculations.

Hints

- Evaluate each expression separately before comparing them. - Think of each result as a point on a number line. - The distance of a number from $0$ is its absolute value.

Solution

1. Evaluate $A$: $-3\cdot4+10=-12+10=-2$. Its distance from $0$ is $2$. 2. Evaluate $B$: $15\div(-3)+4=-5+4=-1$. Its distance from $0$ is $1$. 3. Evaluate $C$: $0.5\cdot(-6)+2.8=-3+2.8=-0.2$. Its distance from $0$ is $0.2$. 4. Since $0.2$ is the smallest distance, expression $C$ has the value closest to $0$.

Answer

Expression $C$, with value $-0.2$, is closest to $0$.

5181577

Write an expression for each description and evaluate it. a) Add the opposite of $-350$ to the absolute value of $-120$. b) Subtract $450$ from the sum of $-200$ and $800$.

Hints

- Find the opposite by changing the sign. - Evaluate absolute value as distance from zero. - Follow the stated operation order.

Solution

1. The opposite of $-350$ is $350$, and $|-120|=120$. Thus, $|-120|+350=120+350=470$. 2. The sum is $-200+800=600$. Then $600-450=150$.

Answer

a) $|-120|+350=470$ b) $(-200+800)-450=150$

5181607

Match each verbal instruction to an expression, and then evaluate it. (1) Add the absolute values of $-18$ and $-12$. (2) Add $-12$ to the opposite of $-18$. (3) Find the sum of $-18$ and $-12$. (4) Find the opposite of the sum of $-18$ and $-12$. Expressions: A: $-18+(-12)$ B: $|-18|+|-12|$ C: $-[-18+(-12)]$ D: $-(-18)+(-12)$

Hints

- Translate “absolute value” and “opposite” carefully. - Determine which operation happens first. - Evaluate only after matching each instruction.

Solution

1. Instruction (1) matches B and gives $18+12=30$. 2. Instruction (2) matches D and gives $18+(-12)=6$. 3. Instruction (3) matches A and gives $-18+(-12)=-30$. 4. Instruction (4) matches C and gives $-[-30]=30$.

Answer

(1) B; $30$ (2) D; $6$ (3) A; $-30$ (4) C; $30$

5181617

Write and evaluate an expression for each instruction. a) Add the opposite of $24$ to $-16$. b) Add $-15$ to the absolute value of $-35$. c) Add the absolute values of $-12$ and $18$. d) Find the opposite of the sum of $-10$ and $10$.

Hints

- Translate each instruction into symbols first. - Absolute value is nonnegative. - A number and its opposite sum to zero.

Solution

1. The opposite of $24$ is $-24$, so $-16+(-24)=-40$. 2. $|-35|+(-15)=35-15=20$. 3. $|-12|+|18|=12+18=30$. 4. $-[-10+10]=-[0]=0$.

Answer

a) $-16+(-24)=-40$ b) $|-35|+(-15)=20$ c) $|-12|+|18|=30$ d) $-[-10+10]=0$

5181627

Decide whether each statement is true or false. Justify with calculations. a) The sum of the absolute values of $-7$ and $-3$ equals the opposite of their sum. b) Adding $-5$ to the opposite of $-5$ gives $0$. c) The absolute value of the sum of $-15$ and $5$ equals the sum of their absolute values.

Hints

- Calculate both sides of each claim separately. - Distinguish the sum of absolute values from the absolute value of a sum. - A number and its opposite add to zero.

Solution

1. $|-7|+|-3|=7+3=10$, and $-[-7+(-3)]=-(-10)=10$, so a) is true. 2. The opposite of $-5$ is $5$, and $5+(-5)=0$, so b) is true. 3. $|-15+5|=|-10|=10$, but $|-15|+|5|=15+5=20$, so c) is false.

Answer

a) True b) True c) False

5181987

Evaluate this claim: “Subtracting the absolute value of an integer from the integer always gives $0$.” Decide whether the claim is true or false and justify your answer with examples.

Hints

- Test both a positive and a negative integer. - Remember that absolute value is always nonnegative. - An “always” claim is false if one counterexample exists.

Solution

1. For a positive integer such as $7$, $7-|7|=7-7=0$. 2. For a negative integer such as $-4$, $-4-|-4|=-4-4=-8$, not $0$. 3. Because one counterexample is enough to disprove an “always” claim, the claim is false.

Answer

The claim is false. For example, $-5-|-5|=-5-5=-10$, which is not $0$.

5183047

A laboratory uses several cooling areas. The room temperature is $22\,^\circ\text{C}$, a refrigerator is $4\,^\circ\text{C}$, a freezer is $-18\,^\circ\text{C}$, and a dry-ice chamber is $-78\,^\circ\text{C}$. a) Write and evaluate an absolute-difference expression for the room-to-freezer temperature difference. b) Write and evaluate an absolute-difference expression for the refrigerator-to-dry-ice-chamber difference. c) Write absolute-difference expressions for the refrigerator-to-freezer and freezer-to-dry-ice-chamber differences. Which is greater?

Hints

- Use the same absolute-difference structure for every temperature separation. - Keep the signed temperatures inside the expression before taking absolute value. - Compare the resulting nonnegative distances in part c).

Solution

1. a) $|22-(-18)|=40\,^\circ\text{C}$. 2. b) $|4-(-78)|=82\,^\circ\text{C}$. 3. c) $|4-(-18)|=22\,^\circ\text{C}$ and $|-18-(-78)|=60\,^\circ\text{C}$. The second difference is greater.

Answer

a) $|22-(-18)|=40\,^\circ\text{C}$ b) $|4-(-78)|=82\,^\circ\text{C}$ c) $|4-(-18)|=22\,^\circ\text{C}$ and $|-18-(-78)|=60\,^\circ\text{C}$; the freezer-to-dry-ice-chamber difference is greater.

5184467

Write an expression and evaluate it. Subtract the opposite of $-18$ from the absolute value of the sum of $-54$ and $21$.

Hints

- Evaluate the sum inside the absolute-value bars first. - Find the opposite of $-18$. - Pay attention to which quantity is being subtracted.

Solution

1. The sum is $-54+21=-33$. 2. Its absolute value is $|-33|=33$. 3. The opposite of $-18$ is $18$. 4. Therefore, the expression is $|-54+21|-18$, and its value is $33-18=15$.

Answer

$|-54+21|-18=15$

5184477

For the numbers $-125$ and $75$, add the absolute value of their difference to the sum of their opposites.

Hints

- Find the difference in the order the numbers are given. - Then find and add the two opposites. - Combine the two intermediate results last.

Solution

1. Their difference in the given order is $-125-75=-200$, so its absolute value is $200$. 2. Their opposites are $125$ and $-75$, whose sum is $125+(-75)=50$. 3. Adding the results gives $200+50=250$.

Answer

$250$

5184487

Determine whether the two expressions have the same value. Show your calculations. Expression A: the absolute value of the difference of $15$ and $40$ Expression B: the difference of the absolute values of $15$ and $40$

Hints

- Translate each verbal description into symbols separately. - Follow the order of operations in each expression. - Compare the two final values.

Solution

1. Expression A is $|15-40|=|-25|=25$. 2. Expression B is $|15|-|40|=15-40=-25$. 3. Since $25\ne-25$, the expressions do not have the same value.

Answer

No. Expression A equals $25$, and Expression B equals $-25$.

5185477

Evaluate each expression step by step. a) $|-42|+58-(-100)$ b) $|15-40|+(12-30)$

Hints

- Evaluate inside absolute value bars and parentheses first. - Absolute value is a number’s distance from zero, so it is never negative. - Subtracting a negative number is the same as adding a positive number.

Solution

1. In a), $|-42|=42$, so $42+58-(-100)=42+58+100=200$. 2. In b), $15-40=-25$, so $|15-40|=25$. Also, $12-30=-18$. Therefore, $25+(-18)=7$.

Answer

a) $200$ b) $7$

5185487

Evaluate each expression. a) $|150-210|-(45-90)$ b) $|-25-15|+[-30-(-50)]$

Hints

- Treat absolute value bars like grouping symbols and evaluate inside them first. - Keep careful track of signs when subtracting a negative number. - Write intermediate values before combining them.

Solution

1. In a), $150-210=-60$, so $|150-210|=60$. Also, $45-90=-45$. Therefore, $60-(-45)=105$. 2. In b), $-25-15=-40$, so $|-25-15|=40$. Also, $-30-(-50)=20$. Therefore, $40+20=60$.

Answer

a) $105$ b) $60$

5185497

Evaluate each expression. a) $|120-450+130|-[-100-(250-600)]$ b) $|-55|+(-125)-|200-350|-(75-200)$

Hints

- Work from the innermost grouping symbols outward. - Record the value of each absolute value and parenthetical expression. - Check every sign before combining the intermediate results.

Solution

1. In a), $120-450+130=-200$, so the absolute value is $200$. Also, $250-600=-350$, and $-100-(-350)=250$. Therefore, $200-250=-50$. 2. In b), $|-55|=55$, $|200-350|=150$, and $75-200=-125$. Therefore, $55-125-150-(-125)=-95$.

Answer

a) $-50$ b) $-95$

5185597

A research team measures elevations relative to sea level. <table> <tr><td>Location</td><td>Elevation</td></tr> <tr><td>A (peak)</td><td>$+340\,\text{m}$</td></tr> <tr><td>B (meadow)</td><td>$+115\,\text{m}$</td></tr> <tr><td>C (village)</td><td>$-25\,\text{m}$</td></tr> <tr><td>D (lakeshore)</td><td>$-142\,\text{m}$</td></tr> <tr><td>E (cave floor)</td><td>$-205\,\text{m}$</td></tr> </table> For each part, write and evaluate an absolute-difference expression. a) Find the elevation difference between A and C. b) Find the vertical separation between D and E. c) Find the elevation difference between the highest and lowest locations.

Hints

- Use absolute difference even when you already know which elevation is higher. - Keep negative elevations signed inside the subtraction. - For part c), identify the extreme elevations before writing the expression.

Solution

1. a) $|340-(-25)|=365\,\text{m}$. 2. b) $|-142-(-205)|=63\,\text{m}$. 3. c) The highest is A and lowest is E, so $|340-(-205)|=545\,\text{m}$.

Answer

a) $|340-(-25)|=365\,\text{m}$ b) $|-142-(-205)|=63\,\text{m}$ c) $|340-(-205)|=545\,\text{m}$

5185607

Ms. Berger records her checking account balance at the end of each weekday. <table> <tr><td>Monday</td><td>$+\$12$</td></tr> <tr><td>Tuesday</td><td>$-\$4$</td></tr> <tr><td>Wednesday</td><td>$-\$18$</td></tr> <tr><td>Thursday</td><td>$-\$9$</td></tr> <tr><td>Friday</td><td>$+\$22$</td></tr> </table> a) Find the signed change from Monday to Tuesday and state whether the balance increased or decreased. b) Identify the lowest balance and write an absolute-difference expression for its distance from Friday's balance. c) Write the four absolute daily changes between consecutive days and identify the greatest one.

Hints

- Keep signed change and magnitude of change distinct. - Use absolute difference when comparing how large two balances or daily changes are. - In part c), calculate all four consecutive-day magnitudes before choosing the greatest.

Solution

1. a) $-4-12=-16$, so the balance decreased by $\$16$. 2. b) The lowest balance is $-\$18$ on Wednesday. Its distance from Friday is $|22-(-18)|=40$, or $\$40$. 3. c) The absolute daily changes are $|-4-12|=16$, $|-18-(-4)|=14$, $|-9-(-18)|=9$, and $|22-(-9)|=31$. The greatest is $\$31$ from Thursday to Friday.

Answer

a) $-4-12=-16$; the balance decreased by $\$16$. b) Wednesday; $|22-(-18)|=\$40$. c) Daily changes: $\$16$, $\$14$, $\$9$, $\$31$; greatest from Thursday to Friday.

5185617

A laboratory stores samples in refrigerated compartments set to the temperatures shown. <table> <tr><td>Compartment 1</td><td>$+6\,^\circ\text{C}$</td></tr> <tr><td>Compartment 2</td><td>$-2\,^\circ\text{C}$</td></tr> <tr><td>Compartment 3</td><td>$-15\,^\circ\text{C}$</td></tr> <tr><td>Compartment 4</td><td>$-28\,^\circ\text{C}$</td></tr> </table> a) Find the temperature difference between Compartments 1 and 3. b) A sample is moved from Compartment 4 to Compartment 2. By how many degrees does its temperature increase? c) Compartment 5 will be set halfway between the temperatures of Compartments 1 and 4. What should its temperature be?

Hints

- Treat a temperature difference as a distance on a number line. - For part b), subtract the starting temperature from the ending temperature. - To find the midpoint, add the two endpoint values and divide by $2$.

Solution

1. The difference between Compartments 1 and 3 is $6-(-15)=21\,^\circ\text{C}$. 2. The increase from Compartment 4 to Compartment 2 is $-2-(-28)=26\,^\circ\text{C}$. 3. The midpoint is the average of the two temperatures: $\frac{6+(-28)}{2}=\frac{-22}{2}=-11\,^\circ\text{C}$.

Answer

a) $21\,^\circ\text{C}$ b) $26\,^\circ\text{C}$ c) $-11\,^\circ\text{C}$

5185787

Which tasks have a value of $30$? (A) $|-18|+|-12|$ (B) $18+12$ (C) Add the absolute values of $-18$ and $-12$. (D) Subtract $-12$ from $18$. (E) $18-(-12)$ (F) Find the distance between $18$ and $-12$ on a number line. (G) $-30+60$ (H) $-18-12$

Hints

- Absolute value gives distance from zero. - The distance between two numbers is the absolute value of their difference. - Subtracting a negative number is the same as adding a positive number.

Solution

1. A, B, and C each equal $18+12=30$. 2. D and E each equal $18-(-12)=30$. 3. The distance in F is $|18-(-12)|=30$. 4. G equals $-30+60=30$, while H equals $-30$.

Answer

A, B, C, D, E, F, and G

5317787

Use the number line with points $A$, $B$, $C$, and $D$. a) Write the decimal represented by each point. b) Which of the four numbers has the greatest absolute value? c) Write the value of $B$ as a fraction in simplest form. d) Find the number exactly halfway between the values of $A$ and $C$.

Hints

- Determine the value of one small interval. - Absolute value is distance from $0$. - A decimal in tenths can be written over $10$. - The midpoint is the average of two values.

Solution

1. Each small interval represents $0.1$. Reading the marked points gives $A=-1.2$, $B=-0.3$, $C=0.7$, and $D=1.4$. 2. Their absolute values are $1.2$, $0.3$, $0.7$, and $1.4$. Therefore, $D=1.4$ has the greatest absolute value. 3. Since $B=-0.3$, $B=-\frac{3}{10}$. 4. The midpoint of $A$ and $C$ is $\frac{-1.2+0.7}{2}=\frac{-0.5}{2}=-0.25$.

Answer

a) $A=-1.2$, $B=-0.3$, $C=0.7$, $D=1.4$ b) $D=1.4$ c) $-\frac{3}{10}$ d) $-0.25$

5351867

Use the number line with points $A$, $B$, $C$, and $D$. a) Write the decimal represented by each point. b) Find the distance between points $B$ and $C$.

Hints

- Determine the value of one small interval. - Count from a nearby labeled value. - Distance is the absolute value of the difference between two coordinates.

Solution

1. The interval from $0$ to $0.5$ is divided into $5$ equal parts, so each small interval represents $0.1$. 2. Point $A$ is two intervals right of $-1.5$, so $A=-1.3$. 3. Point $B$ is two intervals right of $-1$, so $B=-0.8$. 4. Point $C$ is three intervals right of $-0.5$, so $C=-0.2$. 5. Point $D$ is four intervals right of $0$, so $D=0.4$. 6. The distance from $B$ to $C$ is $|-0.2-(-0.8)|=0.6$.

Answer

a) $A=-1.3$, $B=-0.8$, $C=-0.2$, $D=0.4$ b) $0.6$

5510317

Points $A$ and $B$ are shown on the number line. a) Read the coordinates of $A$ and $B$. b) Find $A-B$ and $B-A$. c) Find $|A-B|$ and $|B-A|$. d) What is the distance between $A$ and $B$? Explain why the two signed differences are opposites but their absolute values are equal.

Hints

- Use the spacing between labeled whole numbers to determine the value of each small interval. - Subtract in the order requested and compare the signs of the two differences. - A distance is nonnegative, so connect each signed difference to its absolute value.

Solution

1. The number line shows $A=-1.25$ and $B=0.75$. 2. $A-B=-1.25-0.75=-2$, while $B-A=0.75-(-1.25)=2$. 3. Therefore, $|A-B|=|-2|=2$ and $|B-A|=|2|=2$. 4. The distance between the points is $2$ units. Reversing the subtraction reverses the sign of the difference, but absolute value removes direction and keeps the same distance.

Answer

a) $A=-1.25$, $B=0.75$ b) $A-B=-2$, $B-A=2$ c) $|A-B|=2$, $|B-A|=2$ d) The distance is $2$ units. The signed differences are opposites, but both absolute values represent the same distance.

5117977

The number $0.5$ is exactly halfway between $-3.5$ and an unknown number $b$. Find $b$.

Hints

- Find the distance from the known endpoint to the midpoint. - The unknown endpoint is the same distance from the midpoint on the other side. - Check by averaging the two endpoints.

Solution

1. The distance from $-3.5$ to the midpoint $0.5$ is $0.5-(-3.5)=4$. 2. Move the same distance to the other side of the midpoint: $0.5+4=4.5$. 3. Therefore, $b=4.5$. This also satisfies $\frac{-3.5+4.5}{2}=0.5$.

Answer

$b=4.5$

5510327

The number line shows a target $T$ and three machine readings $A$, $B$, and $C$. A reading is acceptable when its distance from the target is at most $1.25$ units. a) Read the values of $T$, $A$, $B$, and $C$. b) For each reading, compute its absolute difference from $T$ and decide whether it is acceptable. c) The target is recalibrated $0.5$ unit to the right. The three readings do not move. Determine which readings are now acceptable, and explain which acceptance decisions change.

Hints

- Read each marked value by using the quarter-unit tick spacing. - The distance from a reading to the target can be represented by the absolute value of their difference. - “At most” includes a distance equal to the stated tolerance. - For the recalibrated case, change only the target value; keep the three reading values fixed.

Solution

1. The number line shows $T=11$, $A=9.75$, $B=10.25$, and $C=12.5$. 2. For $A$, $|9.75-11|=1.25$, so $A$ is acceptable. For $B$, $|10.25-11|=0.75$, so $B$ is acceptable. For $C$, $|12.5-11|=1.5$, so $C$ is not acceptable. 3. Moving the target $0.5$ unit right gives a new target of $11.5$. Then $|9.75-11.5|=1.75$, so $A$ is not acceptable; $|10.25-11.5|=1.25$, so $B$ is acceptable; and $|12.5-11.5|=1$, so $C$ is acceptable. 4. The status of $A$ changes from acceptable to not acceptable, the status of $C$ changes from not acceptable to acceptable, and $B$ remains acceptable.

Answer

a) $T=11$, $A=9.75$, $B=10.25$, $C=12.5$ b) $A$: distance $1.25$, acceptable; $B$: distance $0.75$, acceptable; $C$: distance $1.5$, not acceptable. c) New target: $11.5$. $A$: distance $1.75$, not acceptable; $B$: distance $1.25$, acceptable; $C$: distance $1$, acceptable. $A$ and $C$ change status; $B$ does not.

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