Problem 24
Question
Let \(s(t)\) give the number of acres of wetlands in a state in year t. Assuming that the area of wetlands goes down over time, say what the statements tell you about the wetlands. \(\frac{s(20)-s(10)}{20-10}=-520\)
Step-by-Step Solution
Verified Answer
Answer: The average rate of decrease in the area of wetlands per year is 520 acres, which indicates that the wetlands are shrinking and losing area over time, possibly due to human activities, natural causes, or a combination of both.
1Step 1: Identify the given information
The given information is a relationship between the areas of wetlands at years 10 and 20. The relationship is:
\(\frac{s(20)-s(10)}{20-10}=-520\).
2Step 2: Interpret the fraction
The fraction \(\frac{s(20)-s(10)}{20-10}\) represents the average rate of change of the area of wetlands, which indicates how the area decreases over time. Since the numerator represents the change in area and the denominator the change in time, the fraction represents the average rate of change per year.
3Step 3: Interpret the sign
The fact that the average rate of change is negative (-520) indicates that the area of wetlands is decreasing over the given period from year 10 to year 20. If the area was increasing, the value would have been positive.
4Step 4: Find the average rate of decrease per year
By simplifying the given expression, we can find the average rate of decrease in the area of wetlands per year:
\(\frac{s(20)-s(10)}{20-10}=-520\) translates to \(\frac{\Delta s}{10}=-520\).
To find the change in area per year, multiply both sides by 10:
\(\Delta s=-520\times 10 = -5200\).
5Step 5: Interpret the result
The result \(\Delta s=-5200\) indicates that over the 10-year period between year 10 and year 20, the area of wetlands in the state has decreased by 5,200 acres. The average rate of decrease per year is 520 acres per year. This tells us that the wetlands are shrinking, possibly due to human activities, natural causes, or a combination of both.
Key Concepts
Interpretation of Linear FunctionsDecrease in Function ValuesReal-World Applications of Algebra
Interpretation of Linear Functions
Linear functions represent relationships where the change is constant, often modeled by a straight line when graphed. In real-life scenarios, they describe situations where one quantity depends linearly on another. For instance, in the context of wetlands, the function \(s(t)\) relates the year \(t\) to the number of acres of wetlands.
Linear functions are often characterized by their slope, which indicates the rate at which one variable changes in response to another. The formula for the average rate of change, such as \(\frac{s(20)-s(10)}{20-10}\), is a tool used to measure how one quantity varies over a set period.
This particular linear relationship shows the average decrease in wetlands over a decade. If the change is consistent, linear models can be very accurate. However, they also assume that the rate of change does not fluctuate, something to bear in mind.
Linear functions are often characterized by their slope, which indicates the rate at which one variable changes in response to another. The formula for the average rate of change, such as \(\frac{s(20)-s(10)}{20-10}\), is a tool used to measure how one quantity varies over a set period.
This particular linear relationship shows the average decrease in wetlands over a decade. If the change is consistent, linear models can be very accurate. However, they also assume that the rate of change does not fluctuate, something to bear in mind.
Decrease in Function Values
The decrease in function values is a critical concept in understanding how quantities reduce over time or circumstances. When a function's average rate of change, such as \(-520\), is negative, it indicates a decline in the quantity it measures.
In the example of wetlands, this negative change shows that the number of wetlands is reducing every year. The calculated average rate of decrease is \(520\) acres per year. To visualize it, imagine losing \(520\) football field-sized areas of wetlands every year.
Identifying a negative slope helps us pinpoint the factors contributing to this decline, like urban development or climate changes, and understand the speed of these changes. This knowledge is pivotal when making decisions about conservation efforts or urban planning.
In the example of wetlands, this negative change shows that the number of wetlands is reducing every year. The calculated average rate of decrease is \(520\) acres per year. To visualize it, imagine losing \(520\) football field-sized areas of wetlands every year.
Identifying a negative slope helps us pinpoint the factors contributing to this decline, like urban development or climate changes, and understand the speed of these changes. This knowledge is pivotal when making decisions about conservation efforts or urban planning.
Real-World Applications of Algebra
Algebra provides invaluable tools for solving real-world problems by modeling situations with mathematical expressions. These models help us predict and analyze trends and outcomes.
In the case of decreasing wetlands, algebra aids in visualizing and calculating their decline. Policymakers and environmentalists use such analyses to understand and act upon ecological issues. For example, understanding the rate of decline helps allocate resources more effectively to conservation programs.
Outside environmental science, algebra models similar trends in economics, engineering, and social sciences. For example, calculating depreciation rates of assets or predicting future population changes. By using algebra, we can tackle various challenges, making it an essential part of many fields.
In the case of decreasing wetlands, algebra aids in visualizing and calculating their decline. Policymakers and environmentalists use such analyses to understand and act upon ecological issues. For example, understanding the rate of decline helps allocate resources more effectively to conservation programs.
Outside environmental science, algebra models similar trends in economics, engineering, and social sciences. For example, calculating depreciation rates of assets or predicting future population changes. By using algebra, we can tackle various challenges, making it an essential part of many fields.
Other exercises in this chapter
Problem 23
Abby and Leah go on a 5 hour drive for 325 miles at 65 mph. After \(t\) hours, Abby calculates the distance remaining by subtracting \(65 t\) from \(325,\) wher
View solution Problem 23
Let \(f(T)\) be the volume in liters of a balloon at temperature \(T^{\circ} \mathrm{C}\). If \(f(40)=3\) (a) What are the units of the 40 and the 3 ? (b) What
View solution Problem 24
If \(w(x)=0.5-0.25 x\), solve $$ w(0.2 x+1)=0.2 w(1-x) $$.
View solution Problem 24
To calculate the balance after investing \(P\) dollars for two years at \(5 \%\) interest, Sharif adds \(5 \%\) of \(P\) to \(P\), and then adds \(5 \%\) of the
View solution