Problem 64
Question
Temperature On Monday the temperature reached a high of \(28^{\circ}\) above \(0 .\) That night it dropped to \(16^{\circ}\) below 0. What is the difference between the high and the low temperatures for Monday?
Step-by-Step Solution
Verified Answer
The temperature difference is \(44^{\circ}\).
1Step 1: Understanding the Problem
We need to find the difference between the highest temperature and the lowest temperature on Monday. The high was \(28^{\circ}\) and the low was \(-16^{\circ}\).
2Step 2: Setting Up the Calculation
To find the temperature difference, subtract the lowest temperature from the highest temperature. This means calculating \(28 - (-16)\).
3Step 3: Applying Arithmetic Operations
When you subtract a negative number, it is the same as adding the opposite. Therefore, \(28 - (-16) = 28 + 16\).
4Step 4: Calculating the Result
Now, perform the addition: \(28 + 16 = 44\).
5Step 5: Conclusion
The temperature difference between the high and the low on Monday was \(44^{\circ}\).
Key Concepts
Understanding Arithmetic OperationsInteger Subtraction SimplifiedAbsolute Value Makes It Clear
Understanding Arithmetic Operations
Arithmetic operations are the foundation of mathematical calculations and are essential in solving everyday problems. In this example, we focus on subtraction, which is one of four basic operations. When solving problems like temperature differences, arithmetic helps us find solutions quickly and accurately.
Subtraction is used when we need to find how much one number differs from another. It involves taking one integer away from another, often used in temperature comparisons and other similar situations. Knowing how to handle these basic operations simplifies our approach when dealing with numbers.
Subtraction is used when we need to find how much one number differs from another. It involves taking one integer away from another, often used in temperature comparisons and other similar situations. Knowing how to handle these basic operations simplifies our approach when dealing with numbers.
Integer Subtraction Simplified
When subtracting integers, especially when one of the numbers is negative, it can seem a bit tricky at first. Essentially, subtracting a negative number is the same thing as adding a positive. This can make calculations easy once you understand this rule.
Consider our exercise: The difference between two temperatures is found by subtraction, i.e., given by the formula: \(28 - (-16)\). Here, when we subtract \(-16\), it is equivalent to adding \(16\). Thus, \(28 - (-16) = 28 + 16\). Recognizing these patterns can aid in visualizing and solving typically difficult problems with ease.
Consider our exercise: The difference between two temperatures is found by subtraction, i.e., given by the formula: \(28 - (-16)\). Here, when we subtract \(-16\), it is equivalent to adding \(16\). Thus, \(28 - (-16) = 28 + 16\). Recognizing these patterns can aid in visualizing and solving typically difficult problems with ease.
Absolute Value Makes It Clear
Absolute value represents the magnitude of a number, disregarding its sign. It is denoted by two vertical lines around the number, like \(|x|\). The importance of absolute value is to help determine how far a number is from zero on a number line, regardless of whether it is positive or negative.
In the context of temperature, this relates to how much different one temperature is from another, ignoring whether one is above or below zero. When evaluating the difference from \(28\) to \(-16\), using absolute value gives \(|28 - (-16)|\), which simplifies the calculation to reflect a true magnitude of \(44\) units, as the negative sign is omitted in distance calculations.
In the context of temperature, this relates to how much different one temperature is from another, ignoring whether one is above or below zero. When evaluating the difference from \(28\) to \(-16\), using absolute value gives \(|28 - (-16)|\), which simplifies the calculation to reflect a true magnitude of \(44\) units, as the negative sign is omitted in distance calculations.
Other exercises in this chapter
Problem 63
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