A land conservation trust has an initial area of 720 acres. The growth plan will add 84 acres of land to the trust each year. If a function is written in the form f(x) = ax + b, where f(x) represents the number of acres of land in the trust and x repre-sents the number of months that have passed, what is the value of x when f(x) = 1, 126 ?
The correct answer and explanation is:
To solve this problem, we need to write the equation for the growth of the land conservation trust in terms of the number of acres of land at any given time, based on the number of months that have passed. The function for this problem is given as: f(x)=ax+bf(x) = ax + b
Where:
- f(x)f(x) represents the number of acres of land in the trust.
- xx represents the number of months that have passed.
- aa is the rate of growth, which is 84 acres per year.
- bb is the initial amount of land, which is 720 acres.
Step 1: Convert the yearly growth rate to a monthly growth rate
Since the function is in terms of months and the growth rate is provided per year, we need to convert the yearly rate to a monthly rate. There are 12 months in a year, so the monthly growth rate is: 84 acres12 months=7 acres per month\frac{84 \text{ acres}}{12 \text{ months}} = 7 \text{ acres per month}
So, the equation for the function becomes: f(x)=7x+720f(x) = 7x + 720
Step 2: Set up the equation to find xx
We are asked to find the value of xx when the number of acres in the trust is 1,126. So, we set f(x)=1,126f(x) = 1,126 and solve for xx: 1,126=7x+7201,126 = 7x + 720
Step 3: Solve for xx
To solve for xx, first subtract 720 from both sides: 1,126−720=7×1,126 – 720 = 7x 406=7×406 = 7x
Now divide both sides by 7: x=4067=58x = \frac{406}{7} = 58
Final Answer:
The value of xx when f(x)=1,126f(x) = 1,126 is 58 months.
Explanation:
This problem involves understanding how growth occurs over time. The land is increasing by a fixed amount each month, and the equation reflects this linear growth. The number of months it takes for the trust to reach 1,126 acres is found by setting up the equation and solving for xx.