Absolute value applications

Learning goals

Interprets absolute value as distance from zero.
Evaluates absolute-value expressions.
Uses absolute value to represent magnitude in context.
Distinguishes magnitude from signed direction or change.

Difficulty: 1 – 5

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Absolute value applications 1

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.

Absolute value applications 2

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 $0$.

Absolute value applications 3

Evaluate $\left|-\frac{3}{5}\right|$. What does your answer tell you about the number's distance from zero?

Hints

- Ignore direction when finding distance from zero. - Distance is never negative. - The fraction's size does not change when only its sign changes.

Solution

1. Absolute value is distance from zero, which cannot be negative. 2. $\left|-\frac{3}{5}\right|=\frac{3}{5}$. The point is $\frac{3}{5}$ unit from zero.

Answer

$\left|-\frac{3}{5}\right|=\frac{3}{5}$; distance $\frac{3}{5}$ unit

Absolute value applications 4

A point is exactly at $0$ on a number line. What is its distance from $0$? Write an absolute-value expression and explain.

Hints

- Distance measures separation between two points. - Can a point be any distance away from itself? - Use absolute-value bars around the coordinate.

Solution

1. The point has no separation from zero. 2. Its distance is $|0|=0$.

Answer

$|0|=0$. The point is already at zero, so its distance is zero.

Absolute value applications 5

a) Read the coordinates of $A$ and $B$. b) Find $|A|$ and $|B|$. c) Explain why the absolute values match.

Hints

- Work out how much each small tick represents. - Read the sign from which side of zero the marker lies on. - Absolute value records the distance, not the side.

Solution

1. The markers are $A=-0.6$ and $B=0.6$. 2. Both are $0.6$ unit from zero, so $|A|=|B|=0.6$. 3. They have opposite signs but equal distances from zero.

Answer

a) $A=-0.6$, $B=0.6$ b) $|A|=0.6$, $|B|=0.6$ c) The two points lie the same distance from zero on opposite sides.

Absolute value applications 6

Which is farther from zero, $-0.8$ or $0.3$? Show the absolute values and explain your choice.

Hints

- The larger signed number is not always farther from zero. - Find both nonnegative distances. - Compare those distances rather than the original signed coordinates.

Solution

1. $|-0.8|=0.8$ and $|0.3|=0.3$. 2. Since $0.8>0.3$, $-0.8$ is farther from zero, even though it is the smaller signed number.

Answer

$-0.8$ is farther from zero because $|-0.8|=0.8>0.3=|0.3|$.

Absolute value applications 7

Read the coordinate of the marked point $P$. How far is it from zero? Write and evaluate an absolute-value expression.

Hints

- Start at a labeled tick and count the equal intervals. - Keep the sign when giving the coordinate. - Remove direction when giving the distance.

Solution

1. The ticks are $0.5$ unit apart. Point $P$ is at $-1.5$. 2. Its distance from zero is $|-1.5|=1.5$ units.

Answer

$P=-1.5$; $|-1.5|=1.5$ units

Absolute value applications 8

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.

Absolute value applications 9

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 $-112$.

Absolute value applications 10

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

Hints

- Work out the difference inside the absolute-value bars first. - The absolute value of a negative number is its positive distance from zero. - Then carry out the remaining operations in order.

Solution

1. Inside the absolute-value bars, $30-75=-45$. 2. Take the absolute value: $|-45|=45$. 3. Finish the calculation: $45-20+15=40$.

Answer

$40$

Absolute value applications 11

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.

Absolute value applications 12

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

Absolute value applications 13

A temperature drops from $-1.5\,^{\circ}\text{C}$ to $-4.5\,^{\circ}\text{C}$. a) What is the signed temperature change? b) What is the size of the temperature change? Write an absolute-difference expression. c) Why do the answers have different signs?

Hints

- For a signed change, subtract the starting reading from the final reading. - Put absolute-value bars around that signed difference to find its size. - Explain what information the sign carries.

Solution

1. The signed change is final minus initial: $-4.5-(-1.5)=-3\,^{\circ}\text{C}$. 2. The size of the change is $|-4.5-(-1.5)|=|-3|=3\,^{\circ}\text{C}$. 3. The signed change says the temperature fell; the absolute difference records only how much.

Answer

a) $-3\,^{\circ}\text{C}$ b) $|-4.5-(-1.5)|=3\,^{\circ}\text{C}$ c) The negative sign shows a drop, while the distance has no direction.

Absolute value applications 14

Two points on a number line have coordinates $-\frac{3}{4}$ and $-\frac{1}{8}$. Write an absolute-difference expression for their distance, and give the result as a fraction in simplest form.

Hints

- Keep both negative coordinates signed when you subtract. - Rewrite quarters as eighths to compare the positions. - Distance is never negative, whichever point you subtract first.

Solution

1. Use absolute difference: $\left|-\frac{1}{8}-\left(-\frac{3}{4}\right)\right|$. 2. Rewrite in eighths: $\left|-\frac{1}{8}+\frac{6}{8}\right|=\left|\frac{5}{8}\right|=\frac{5}{8}$.

Answer

$\left|-\frac{1}{8}-\left(-\frac{3}{4}\right)\right|=\frac{5}{8}$

Absolute value applications 15

Two different points are each $1.25$ units from $0$ on a number line. Find both coordinates and check each one with absolute value.

Hints

- Where could you move from zero to travel the given distance? - Can two different signed coordinates be the same distance from zero? - Check each candidate using its absolute value.

Solution

1. Move $1.25$ units right from zero to reach $1.25$. 2. Move $1.25$ units left from zero to reach $-1.25$. 3. Check: $|1.25|=1.25$ and $|-1.25|=1.25$.

Answer

$-1.25$ and $1.25$; $|-1.25|=|1.25|=1.25$.

Absolute value applications 16

a) Write an absolute-difference expression for the distance between $-2.4$ and $5.6$, and find the distance. b) Find the distance between $-\frac{3}{4}$ and $\frac{5}{4}$. Show an absolute-difference expression. c) Point $P$ is at $-2$. Point $Q$ is to the right of $0$ and is twice as far from $0$ as $P$. What coordinate does $Q$ have? Use absolute value to explain.

Hints

- A distance is the absolute value of the difference between the coordinates. - Keep the negative signs when you write each subtraction. - For the last part, use distance from zero and the side of zero given in the question.

Solution

1. The distance in a) is $|5.6-(-2.4)|=|8|=8$. 2. In b), $\left|\frac{5}{4}-\left(-\frac{3}{4}\right)\right|=\left|\frac{8}{4}\right|=2$. 3. In c), $|-2|=2$. Twice that distance is $2\cdot2=4$. Since $Q$ is to the right of $0$, its coordinate is $4$.

Answer

a) $|5.6-(-2.4)|=8$ b) $\left|\frac{5}{4}-\left(-\frac{3}{4}\right)\right|=2$ c) $Q=4$, because $|-2|=2$ and $|4|=4$.

Absolute value applications 17

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 $0.41$.

Absolute value applications 18

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.

Absolute value applications 19

Point $A$ is at $-3.5$ and point $M$ is at $0.5$ on a number line. Point $B$ is to the right of $M$, at the same distance from $M$ as $A$. a) Find $|0.5-(-3.5)|$. b) Find the coordinate of $B$. Check it by calculating $|B-0.5|$.

Hints

- The distance between two coordinates is the absolute value of their difference. - Which side of $M$ must contain $B$? - Check that your chosen coordinate has the same distance from $M$ as $A$.

Solution

1. The distance from $A$ to $M$ is $|0.5-(-3.5)|=|4|=4$. 2. Since $B$ is $4$ units to the right of $0.5$, its coordinate is $B=0.5+4=4.5$. 3. Check the distance: $|4.5-0.5|=|4|=4$, which agrees with part a).

Answer

a) $|0.5-(-3.5)|=4$ b) $B=4.5$, since $|4.5-0.5|=4$

Absolute value applications 20

A lift serves three stops at these heights relative to ground: $P=-46\,\text{ft}$, $Q=69\,\text{ft}$, and $R=-62\,\text{ft}$. a) Find the vertical distances from $P$ to $Q$ and from $P$ to $R$. Show an absolute-difference expression for each. b) Which trip is shorter, and by how many feet?

Hints

- A trip's length is not the same as its signed height change. - Subtract the two heights in each trip, then take the distance. - Compare the two positive distances to answer b).

Solution

1. The distance from $P$ to $Q$ is $|69-(-46)|=115\,\text{ft}$. 2. The distance from $P$ to $R$ is $|-62-(-46)|=16\,\text{ft}$. 3. The trip from $P$ to $R$ is shorter by $115-16=99\,\text{ft}$.

Answer

a) $|69-(-46)|=115\,\text{ft}$; $|-62-(-46)|=16\,\text{ft}$ b) $P$ to $R$, by $99\,\text{ft}$

Absolute value applications 21

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 $|-10| > |2|$.

Absolute value applications 22

A point is at $-3.4$ on a number line. Two other points are each $7.2$ units from it: one is to its left, and one is to its right. a) Find both coordinates. b) Write an absolute-difference expression to check each distance.

Hints

- One point lies the given distance to the left and the other to the right. - Moving left decreases the coordinate; moving right increases it. - Check both positions by taking an absolute difference from the starting coordinate.

Solution

1. Move left by $7.2$: $-3.4-7.2=-10.6$. 2. Move right by $7.2$: $-3.4+7.2=3.8$. 3. Check: $|-10.6-(-3.4)|=|-7.2|=7.2$ and $|3.8-(-3.4)|=|7.2|=7.2$.

Answer

a) $-10.6$ and $3.8$ b) $|-10.6-(-3.4)|=7.2$ and $|3.8-(-3.4)|=7.2$

Absolute value applications 23

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) $\left|1\frac{1}{2}-\left(-\frac{2}{5}\right)\right|=1.9=\frac{19}{10}$

Absolute value applications 24

Two pairs of points have coordinates $A=-2$, $B=3$ and $C=-5.5$, $D=1.5$. a) Find the distance within each pair using an absolute difference. Which pair is farther apart? b) Find both points that are exactly $2$ units from $A$. Check both answers using absolute differences.

Hints

- Use a positive distance to compare each pair of coordinates. - A point at a fixed distance may lie to the left or right of the starting point. - Check the distances for both candidates, not just their coordinates.

Solution

1. The first distance is $|3-(-2)|=5$ units. 2. The second distance is $|1.5-(-5.5)|=7$ units, so $C$ and $D$ are farther apart. 3. From $A=-2$, move $2$ units left or right, giving $-4$ and $0$. 4. Check: $|-4-(-2)|=2$ and $|0-(-2)|=2$.

Answer

a) $|3-(-2)|=5$; $|1.5-(-5.5)|=7$, so $C,D$ are farther apart. b) $-4$ and $0$, since $|-4-(-2)|=|0-(-2)|=2$.

Absolute value applications 25

Find the value of each expression. Which value is closest to $0$? Show the absolute values to explain. $A=-3\cdot4+10$ $B=15\div(-3)+6$ $C=0.5\cdot(-6)+2.8$

Hints

- Finish each expression before comparing the results. - Where would the three results fall relative to zero on a number line? - Could a negative value be closer to zero than a positive one?

Solution

1. $A=-3\cdot4+10=-2$, so $|A|=2$. 2. $B=15\div(-3)+6=1$, so $|B|=1$. 3. $C=0.5\cdot(-6)+2.8=-0.2$, so $|C|=0.2$. 4. Since $0.2<1<2$, expression $C$ is closest to $0$.

Answer

$A=-2$, $|A|=2$; $B=1$, $|B|=1$; $C=-0.2$, $|C|=0.2$. Expression $C$ is closest to $0$.

Absolute value applications 26

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) $(-200+800)-450=150$

Absolute value applications 27

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; $30$

Absolute value applications 28

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) $-[-10+10]=0$

Absolute value applications 29

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) The distance between $-5$ and $5$ equals $|-5|+|5|$. c) The absolute value of the sum of $-15$ and $5$ equals the sum of their absolute values.

Hints

- Translate each claim into the two expressions it compares. - Compare the two sides of each claim; do not assume the same rule works for every pair. - Show both sides before deciding whether they match.

Solution

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

Answer

a) True: $|-7|+|-3|=10=-[-7+(-3)]$. b) True: $|5-(-5)|=10=|-5|+|5|$. c) False: $|-15+5|=10$, but $|-15|+|5|=20$.

Absolute value applications 30

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 $0$.

Absolute value applications 31

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.

Absolute value applications 32

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

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

Absolute value applications 33

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

$250$

Absolute value applications 34

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 the two descriptions separately before you calculate. - Check whether the absolute-value bars enclose the difference or each number. - Compare the resulting 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 differ.

Answer

No. $A=|15-40|=25$, while $B=|15|-|40|=-25$.

Absolute value applications 35

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) $7$

Absolute value applications 36

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) $60$

Absolute value applications 37

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) $-95$

Absolute value applications 38

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) $|340-(-205)|=545\,\text{m}$

Absolute value applications 39

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

- Signed change and size of change are different quantities. - For a distance, keep both signed balances inside an absolute difference. - Calculate each consecutive-day distance before selecting the largest.

Solution

1. $-4-12=-16$, so the balance decreased by $\$16$. 2. The lowest balance is $-\$18$ on Wednesday. Its distance from Friday is $|22-(-18)|=40$, or $\$40$. 3. 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's balance was $-\$18$. Its distance from Friday's balance was $|22-(-18)|=40$, or $\$40$. c) $|-4-12|=16$, $|-18-(-4)|=14$, $|-9-(-18)|=9$, $|22-(-9)|=31$; greatest from Thursday to Friday ($\$31$).

Absolute value applications 40

Which expressions or descriptions have a value of $30$? Select all that apply and give the value of every choice. A) $|-18|+|-12|$ B) $|18+(-12)|$ C) The distance between $-12$ and $18$ on a number line D) $|18|-|-13|$ E) $|-18-(-9)|$ F) $|-42|-|-12|$ G) $|-21+3|$ H) The distance between $-7$ and $-27$ on a number line

Hints

- For each absolute-value expression, first evaluate what is inside the bars. - Distance between two positions is the absolute value of their difference. - A sum of absolute values and an absolute value of a sum need not be equal. - Work out every value, not just the choices that look likely to work.

Solution

1. A equals $18+12=30$; B equals $|6|=6$. 2. C is $|18-(-12)|=30$, while D equals $18-13=5$. 3. E is $|-18+9|=9$; F is $42-12=30$. 4. G equals $|-18|=18$; H equals $|-27-(-7)|=20$. 5. Only A, C, and F equal $30$.

Answer

A) $30$ B) $6$ C) $30$ D) $5$ E) $9$ F) $30$ G) $18$ H) $20$ Correct choices: A, C, F.

Absolute value applications 41

a) Write the decimal represented by each marked point. b) Which point has the greatest absolute value? Give that value. c) Which point has the smallest absolute value? Give that value. d) Write an absolute-difference expression for the distance from $A$ to $C$, and find the distance.

Hints

- Use the numbered ticks to work out the size of each small interval. - Absolute value tells how far a point is from zero, regardless of its side. - For the last part, use the difference of the two marked coordinates.

Solution

1. Each small interval represents $0.1$. The points are $A=-1.2$, $B=-0.3$, $C=0.7$, and $D=1.4$. 2. Their absolute values are $1.2$, $0.3$, $0.7$, and $1.4$. Point $D$ has the greatest magnitude, $1.4$, and $B$ has the least magnitude, $0.3$. 3. The distance from $A$ to $C$ is $|0.7-(-1.2)|=1.9$.

Answer

a) $A=-1.2$, $B=-0.3$, $C=0.7$, $D=1.4$ b) $D$; $|1.4|=1.4$ c) $B$; $|-0.3|=0.3$ d) $|0.7-(-1.2)|=1.9$

Absolute value applications 42

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) $0.6$

Absolute value applications 43

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.

Absolute value applications 44

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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Absolute value applications

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