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Sagot :
Let's solve this problem step by step:
### Step 1: Determine the linear function [tex]\(C(x)\)[/tex]
We start with the given data points:
- In 1990, the cost of tuition was [tex]$101 per credit hour. - In 1997, the cost of tuition was $[/tex]206 per credit hour.
Let's define the variables:
- Let [tex]\( x \)[/tex] be the number of years since 1990.
- Let's denote 1990 as [tex]\( x = 0 \)[/tex].
- In 1997, [tex]\( x = 1997 - 1990 = 7 \)[/tex].
Hence, we have two points [tex]\((0, 101)\)[/tex] and [tex]\((7, 206)\)[/tex].
We need to find the slope ([tex]\(m\)[/tex]) of the linear function first. The slope formula for a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Plugging in the values:
[tex]\[ m = \frac{206 - 101}{7 - 0} = \frac{105}{7} = 15 \][/tex]
Now, we know the slope [tex]\( m = 15 \)[/tex].
The general form of a linear equation is:
[tex]\[ C(x) = mx + b \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
To find the y-intercept ([tex]\( b \)[/tex]), we use one of the points on the line. Let's use point [tex]\((0, 101)\)[/tex]:
[tex]\[ C(0) = 15(0) + b \rightarrow 101 = 15(0) + b \rightarrow b = 101 \][/tex]
Hence, the linear function [tex]\(C(x)\)[/tex] is:
[tex]\[ C(x) = 15x + 101 \][/tex]
### Step 2: Determine the tuition in the year 2000
To find the tuition cost in the year 2000, we need to determine [tex]\( x \)[/tex] for the year 2000:
[tex]\[ x = 2000 - 1990 = 10 \][/tex]
Using the linear function [tex]\( C(x) = 15x + 101 \)[/tex]:
[tex]\[ C(10) = 15(10) + 101 = 150 + 101 = 251 \][/tex]
So, the tuition in the year 2000 is [tex]$251 per credit hour. ### Step 3: Determine the year when tuition will be $[/tex]386 per credit hour
We need to find the year when [tex]\( C(x) = 386 \)[/tex]. Using the linear function:
[tex]\[ 386 = 15x + 101 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 15x + 101 = 386 \][/tex]
[tex]\[ 15x = 386 - 101 \][/tex]
[tex]\[ 15x = 285 \][/tex]
[tex]\[ x = \frac{285}{15} \][/tex]
[tex]\[ x = 19 \][/tex]
Therefore, [tex]\( x = 19 \)[/tex], and the year is:
[tex]\[ 1990 + 19 = 2009 \][/tex]
So, tuition will be [tex]$386 per credit hour in the year 2009. ### Conclusion The linear function \( C(x) \) is: \[ C(x) = 15x + 101 \] In the year 2000, the tuition will be $[/tex]251 per credit hour.
In the year 2009, the tuition will be $386 per credit hour.
### Step 1: Determine the linear function [tex]\(C(x)\)[/tex]
We start with the given data points:
- In 1990, the cost of tuition was [tex]$101 per credit hour. - In 1997, the cost of tuition was $[/tex]206 per credit hour.
Let's define the variables:
- Let [tex]\( x \)[/tex] be the number of years since 1990.
- Let's denote 1990 as [tex]\( x = 0 \)[/tex].
- In 1997, [tex]\( x = 1997 - 1990 = 7 \)[/tex].
Hence, we have two points [tex]\((0, 101)\)[/tex] and [tex]\((7, 206)\)[/tex].
We need to find the slope ([tex]\(m\)[/tex]) of the linear function first. The slope formula for a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Plugging in the values:
[tex]\[ m = \frac{206 - 101}{7 - 0} = \frac{105}{7} = 15 \][/tex]
Now, we know the slope [tex]\( m = 15 \)[/tex].
The general form of a linear equation is:
[tex]\[ C(x) = mx + b \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
To find the y-intercept ([tex]\( b \)[/tex]), we use one of the points on the line. Let's use point [tex]\((0, 101)\)[/tex]:
[tex]\[ C(0) = 15(0) + b \rightarrow 101 = 15(0) + b \rightarrow b = 101 \][/tex]
Hence, the linear function [tex]\(C(x)\)[/tex] is:
[tex]\[ C(x) = 15x + 101 \][/tex]
### Step 2: Determine the tuition in the year 2000
To find the tuition cost in the year 2000, we need to determine [tex]\( x \)[/tex] for the year 2000:
[tex]\[ x = 2000 - 1990 = 10 \][/tex]
Using the linear function [tex]\( C(x) = 15x + 101 \)[/tex]:
[tex]\[ C(10) = 15(10) + 101 = 150 + 101 = 251 \][/tex]
So, the tuition in the year 2000 is [tex]$251 per credit hour. ### Step 3: Determine the year when tuition will be $[/tex]386 per credit hour
We need to find the year when [tex]\( C(x) = 386 \)[/tex]. Using the linear function:
[tex]\[ 386 = 15x + 101 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 15x + 101 = 386 \][/tex]
[tex]\[ 15x = 386 - 101 \][/tex]
[tex]\[ 15x = 285 \][/tex]
[tex]\[ x = \frac{285}{15} \][/tex]
[tex]\[ x = 19 \][/tex]
Therefore, [tex]\( x = 19 \)[/tex], and the year is:
[tex]\[ 1990 + 19 = 2009 \][/tex]
So, tuition will be [tex]$386 per credit hour in the year 2009. ### Conclusion The linear function \( C(x) \) is: \[ C(x) = 15x + 101 \] In the year 2000, the tuition will be $[/tex]251 per credit hour.
In the year 2009, the tuition will be $386 per credit hour.
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