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Question
- rodney is performing a science experiment. the table shows the temperature of two liquids he is using. graph the integers that represent the temperatures on a number line. which beaker’s liquid is closer to 0°c? explain.
(table: beaker a -4°c, beaker b 2°c)
(number line: -5 -4 -3 -2 -1 0 1 2 3)
- sydney owes her mother $5 and her brother owes her mother $7. graph the integers that represent the amount they owe their mother as a negative integer on a number line. how much more will her brother have to repay their mother than sydney? explain.
(number line: -8 -7 -6 -5 -4 -3 -2 -1 0)
- mp use math tools explain how to find the distance between 1 and −3 on a number line.
- at midnight, the outside temperature was 0°f.
a. by 6:00 a.m., the temperature had dropped 4°f, and then the temperature raised 10°f by noon. what is the temperature at noon?
b. what represents zero in this situation? explain.
- create describe a real-world situation that can be represented by a negative integer. then write the integer.
- mp justify conclusions craig has $28 in his checking account. he wants to make a withdrawal of $30. will his checking account balance be represented by a positive or negative integer after the withdrawal? justify your conclusion.
Question 15
Step1: Find absolute values
The temperature of Beaker A is \(-4^\circ C\), its absolute value is \(|-4| = 4\). The temperature of Beaker B is \(2^\circ C\), its absolute value is \(|2| = 2\).
Step2: Compare absolute values
Since \(2 < 4\), the distance of \(2\) from \(0\) is less than the distance of \(-4\) from \(0\).
Step1: Represent owed amounts as integers
Sydney owes \(\$5\), so we represent it as \(-5\). Her brother owes \(\$7\), so we represent it as \(-7\).
Step2: Calculate the difference
To find how much more her brother has to repay, we calculate \(|-7| - |-5| = 7 - 5 = 2\).
Step1: Recall distance formula on number line
The distance between two numbers \(a\) and \(b\) on a number line is \(|a - b|\) (or \(|b - a|\)).
Step2: Apply the formula
For \(a = 1\) and \(b = -3\), the distance is \(|1 - (-3)| = |1 + 3| = |4| = 4\). Alternatively, we can count the units between \(1\) and \(-3\) on the number line: from \(-3\) to \(0\) is \(3\) units, and from \(0\) to \(1\) is \(1\) unit, so total \(3 + 1 = 4\) units.
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Beaker B's liquid is closer to \(0^\circ C\) because the absolute value of \(2\) (which is \(2\)) is less than the absolute value of \(-4\) (which is \(4\)), meaning \(2^\circ C\) is closer to \(0^\circ C\) than \(-4^\circ C\).