- Locate \(-6\) relative to \(0\). - Absolute value records distance, not direction. - Use the original sign when describing whether the diver is above or below sea level.

1. Absolute value is distance from \(0\), so \(\lvert-6\rvert=6\). 2. The diver is \(6\,\text{m}\) from sea level; the negative sign shows that the diver is below sea level.

a) \(\lvert-6\rvert=6\) b) The diver is \(6\,\text{m}\) below sea level.

- Picture the reading on a thermometer marked at zero. - Count how many degrees you would move to reach zero. - Your expression should use the temperature reading, not just your counted distance.

1. The temperature is eight degrees below zero. 2. Distance is nonnegative, so \(\lvert-8\rvert=8\).

\(8\,^\circ\text{C}\); \(\lvert-8\rvert=8\).

- Think about how you find absolute value on a number line. - What happens to that distance as a point gets closer to zero? - Does zero need to be positive or negative for the distance idea to work?

1. Absolute value measures distance from zero on the number line. 2. Zero is \(0\) units from itself, so \(|0|=0\). 3. Sofia is incorrect. Absolute value is defined for zero.

No. \(|0|=0\) because zero is \(0\) units from itself on the number line.

- Imagine a vertical number line through sea level. - Put the diver and bird on that line before writing the elevations. - Count the gap between each position and the water surface.

1. The diver is at \(-1.25\,\text{m}\), and the bird is at \(+1.25\,\text{m}\). 2. \(\lvert-1.25\rvert=1.25\) and \(\lvert1.25\rvert=1.25\). 3. Their distances from sea level are equal.

The elevations are \(-1.25\,\text{m}\) and \(+1.25\,\text{m}\). Their absolute values are both \(1.25\), so neither is farther.

- Absolute value gives a temperature’s distance from zero. - Compare the two distances. - Use \(>\) when the first distance is greater.

1. The distances from zero are \(|-12|=12\) and \(|10|=10\). 2. Since \(12>10\), the comparison is \(|-12|>|10|\).

\(|-12|>|10|\)

- The equator is the zero point in this model. - Compare each station’s distance from zero, not the signed values themselves. - Points the same distance from zero on opposite sides have opposite coordinates.

1. Distance from the equator is distance from \(0\), so compare absolute values. 2. \(\lvert80\rvert=80\) and \(\lvert-10\rvert=10\). Therefore, station \(A\) is farther from the equator. 3. A point the same distance from \(0\) as \(+80\) but on the opposite side has coordinate \(-80\).

a) Station \(A\), because \(\lvert80\rvert=80>10=\lvert-10\rvert\) b) \(-80^\circ\)

- Treat the lobby as the zero point. - Find each location’s distance from zero before comparing. - Subtract the smaller distance from the larger distance to find how much farther.

1. The parking level is \(\lvert-4\rvert=4\) floors from the lobby. 2. The rooftop deck is \(\lvert9\rvert=9\) floors from the lobby. 3. Since \(9>4\), the rooftop deck is farther from the lobby. 4. Its distance is \(9-4=5\) floors greater.

The rooftop deck is farther from the lobby, by \(5\) floors.

- A deviation’s sign gives direction, while its absolute value gives distance from the target. - Compare the two distances after removing the signs. - Find the difference between those distances.

1. Sensor \(A\)'s distance from the target is \(\lvert-7.5\rvert=7.5^\circ\). 2. Sensor \(B\)'s distance from the target is \(\lvert6.2\rvert=6.2^\circ\). 3. Since \(7.5>6.2\), sensor \(A\) is farther from the target. 4. The difference in distances is \(7.5-6.2=1.3^\circ\).

Sensor \(A\) is farther from the target by \(1.3^\circ\).

- Treat each balance as a signed number. - Absolute value gives its distance from a zero balance. - Include a balance whose distance is exactly \(\$25\).

1. The distances from a zero balance are \(\lvert-18\rvert=18\), \(\lvert27\rvert=27\), \(\lvert-25\rvert=25\), and \(\lvert40\rvert=40\). 2. The distances \(18\) and \(25\) are at most \(25\). 3. Therefore, the balances \(-\$18\) and \(-\$25\) are within \(\$25\) of zero.

The balances \(-\$18\) and \(-\$25\) are within \(\$25\) of zero.

- Interpret absolute value as distance from \(0\). - Consider whether a nonzero value can be zero units from \(0\). - Translate the coordinate back to elevation relative to sea level.

1. Absolute value gives distance from \(0\). 2. The only number at distance \(0\) from \(0\) is \(0\), so \(x=0\). 3. An elevation of \(0\) is exactly at sea level.

a) \(x=0\) b) Zero is the only number whose distance from \(0\) is \(0\). c) The elevation is exactly at sea level.

- Treat \(0^\circ\text{C}\) as the reference point. - Absolute value gives the distance from that reference point. - Separate the amount of the difference from its direction.

1. The distance of \(-\frac{3}{4}\) from \(0\) is \(\frac{3}{4}\). 2. Thus, \(\left\lvert-\frac{3}{4}\right\rvert=\frac{3}{4}\), so the temperature is \(\frac{3}{4}^\circ\text{C}\) from \(0^\circ\text{C}\). 3. The absolute value does not indicate that the temperature is below \(0^\circ\text{C}\).

a) \(\frac{3}{4}\) b) The temperature is \(\frac{3}{4}^\circ\text{C}\) from \(0^\circ\text{C}\). c) It does not include whether the temperature is above or below \(0^\circ\text{C}\).

- Find each number's distance from \(0\). - Compare each distance with \(1\). - Keep each original sign in the final list.

1. The absolute values are \(\frac{5}{4}\), \(\frac{3}{4}\), \(\frac{1}{2}\), \(0.9\), and \(\frac{6}{5}\). 2. The values less than \(1\) are \(\frac{3}{4}\), \(\frac{1}{2}\), and \(0.9\). 3. Therefore, the original numbers less than \(1\) unit from \(0\) are \(-\frac{3}{4}\), \(\frac{1}{2}\), and \(0.9\).

\(-\frac{3}{4}, \frac{1}{2}, 0.9\)

- Use the directional word to choose the sign. - Use absolute value to check only the size of the difference. - Compare the meanings of the positive and negative values.

1. Below the target is negative, so the signed level difference is \(-2.6\,\text{cm}\). 2. The magnitude check is \(\lvert-2.6\rvert=2.6\). 3. The value \(+2.6\) has the same magnitude but represents a level above the target.

a) \(-2.6\,\text{cm}\) b) \(\lvert-2.6\rvert=2.6\) c) \(+2.6\) represents a level above the target.

- Which whole-number distance lies between the two limits? - Which two directions can the elevator travel from the lobby? - Absolute value tells how far a floor is from the lobby, not the direction.

1. The only whole-number distance greater than \(2\) but less than \(4\) is \(3\) floors. 2. The elevator could show \(-3\) or \(3\). 3. \(\lvert-3\rvert=3\) and \(\lvert3\rvert=3\). Each floor is three floors from the lobby, in opposite directions.

a) \(-3\) and \(3\). b) \(\lvert-3\rvert=3\) and \(\lvert3\rvert=3\). Both floors are three floors from the lobby.

- Sketch the track and mark the starting point as zero. - Place the robot six feet from the start on the stated side. - How many feet would the robot travel to return to zero?

1. Being on the negative side means its signed position is \(-6\). 2. Its distance is \(6\), so \(\lvert-6\rvert=6\).

Position \(-6\,\text{ft}\); distance \(\lvert-6\rvert=6\,\text{ft}\).

- Put the four scores in order on a number line. - Count each score's number of steps from zero. - Compare the results of those two approaches before explaining them.

1. The greatest signed score is \(5\), since it lies farthest right. 2. The distances from zero are \(8\), \(2\), \(3\), and \(5\). 3. The smallest distance is \(2\), so \(2\) is closest to zero.

Greatest: \(5\). Closest to zero: \(2\). One compares signed size; the other compares distance.

- The absolute value of a score is its distance from zero. - Find all five absolute values before ordering the players. - Ignore the signs when comparing absolute values.

1. The absolute values of the scores are \(25,18,12,22,3\) for Players A through E. 2. The least absolute value is \(3\), so Player E answers part a). 3. From greatest to least absolute value, the order is Player A, Player D, Player B, Player C, Player E.

a) Player E b) Player A, Player D, Player B, Player C, Player E

- The vertical distance to the x-axis depends on the y-coordinate. - Distance is nonnegative, even for a point below the x-axis. - Compare the absolute values of the y-coordinates.

1. The coordinates are \(P(5, 1)\), \(Q(-4, -3)\), \(R(1, 4)\), \(S(-3, 1)\), and \(T(0, -4)\). 2. A point’s distance from the x-axis is the absolute value of its y-coordinate. 3. The distances are \(1, 3, 4, 1,\) and \(4\) units, respectively. 4. Points \(R\) and \(T\) are each \(4\) units from the x-axis, so they are tied for the greatest distance.

The coordinates are \(P(5, 1)\), \(Q(-4, -3)\), \(R(1, 4)\), \(S(-3, 1)\), and \(T(0, -4)\). Points \(R\) and \(T\) are farthest from the x-axis, at a distance of \(4\) units.

- Read each marked temperature from the number line. - Absolute value gives distance from \(0\). - Compare those distances to identify the closest and farthest readings.

1. The graph shows \(A=-2.5\), \(B=1.75\), \(C=-0.75\), and \(D=3.25\). 2. Their absolute values are \(\lvert-2.5\rvert=2.5\), \(\lvert1.75\rvert=1.75\), \(\lvert-0.75\rvert=0.75\), and \(\lvert3.25\rvert=3.25\). 3. The least absolute value is \(0.75\), so \(C\), at \(-0.75^\circ\text{C}\), is closest to \(0^\circ\text{C}\). 4. The greatest absolute value is \(3.25\), so \(D\), at \(3.25^\circ\text{C}\), is farthest from \(0^\circ\text{C}\).

a) \(C\), at \(-0.75^\circ\text{C}\) b) \(D\), at \(3.25^\circ\text{C}\) c) \(\lvert-0.75\rvert=0.75\) and \(\lvert3.25\rvert=3.25\)

- Interpret each signed floor as a position above or below the lobby. - Compare distances from floor \(0\), not ordinary signed order. - Restore each original sign after ordering the distances.

1. The distances from the lobby are \(\lvert-6\rvert=6\), \(\lvert2\rvert=2\), \(\lvert-9\rvert=9\), and \(\lvert4\rvert=4\). 2. Ordering the distances gives \(2<4<6<9\). 3. Matching those distances to the original signed positions gives \(+2, +4, -6, -9\).

\(+2, +4, -6, -9\)

- Mark the trailhead as \(0\) and consider both directions. - Use one variable for the signed position. - Identify the directional information that absolute value does not preserve.

1. The hiker could be \(3.2\,\text{mi}\) east of the trailhead, at \(+3.2\), or \(3.2\,\text{mi}\) west, at \(-3.2\). 2. Both positions satisfy \(\lvert x\rvert=3.2\). 3. Absolute value gives distance from \(0\) but does not identify the side of \(0\).

a) \(+3.2\,\text{mi}\) and \(-3.2\,\text{mi}\) b) \(\lvert x\rvert=3.2\) c) The equation gives the distance from the trailhead but not the direction.

- Use left-to-right number-line order for the signed comparison. - Use distance from \(0\) for the absolute-value comparison. - Interpret sign and magnitude as different features of position.

1. As signed positions, \(-0.7<+0.7\). 2. Their distances from the trailhead are equal: \(\lvert-0.7\rvert=\lvert+0.7\rvert=0.7\). 3. The hikers are on opposite sides of the trailhead but equally far from it.

a) \(-0.7<+0.7\) b) \(\lvert-0.7\rvert=\lvert+0.7\rvert\) c) They are on opposite sides of the trailhead and the same distance from it.

- Begin with the two numbers that are \(12\) units from \(0\). - Use the inequality to select one of them. - Translate the selected sign into elevation relative to sea level.

1. A magnitude of \(12\) gives \(\lvert m\rvert=12\). 2. The two numbers with that magnitude are \(12\) and \(-12\). 3. Since \(m<0\), \(m=-12\), which represents an elevation \(12\) units below sea level.

a) \(\lvert m\rvert=12\) b) \(m=-12\) c) The elevation is \(12\) units below sea level.

- State the definition of each notation separately. - Distinguish a distance from a new signed position. - Explain the shared result without claiming the operations always mean the same thing.

1. The expression \(\lvert-9\rvert\) gives the distance of \(-9\) from \(0\), which is \(9\). 2. The expression \(-(-9)\) gives the opposite of \(-9\), which is the signed number \(9\). 3. One operation finds magnitude, while the other reverses direction.

Both expressions equal \(9\), but \(\lvert-9\rvert\) means distance from \(0\), while \(-(-9)\) means the opposite of \(-9\).

- First compare the signed values on a number line. - Then find each value's distance from \(0\). - Interpret signed order and absolute value as answers to different questions.

1. As signed elevations, \(-10<+3\), so \(-10\,\text{m}\) is lower. 2. Their distances from sea level are \(\lvert-10\rvert=10\) and \(\lvert+3\rvert=3\), so \(\lvert-10\rvert>\lvert+3\rvert\). 3. Signed order describes vertical position, while absolute value describes distance from sea level.

a) \(-10<+3\) b) \(\lvert-10\rvert>\lvert+3\rvert\) c) The elevation \(-10\,\text{m}\) is \(10\,\text{m}\) from sea level, while \(+3\,\text{m}\) is \(3\,\text{m}\) from sea level.