The cost T, in hundreds of dollars, of tuition and fees at many community colleges can be approximately by T=1/4c+6,where c is the number of credits for which a student registers. Graph the equation and use the graph to estimate the cost of tuition and fees when a student registers for a 4 two-credit courses.
To solve algebraically, a 4 two-credit courses has total of 4 x 2 = 8 credits. Substituting to the equation,
T = (1/4)c + 6
T = (1/4)(8) + 6
T = 2 + 6
T = 8
Since the units of T is in hundreds of dollars, the tuition is $800.
I'll leave the graphing to you since graph cannot be shown here. But just to give an idea how to graph, let c = x-coordinate and T = y-coordinate. Take 2 values of c and substitute it to the equation to get T. You have already have a point which we found above - point (8, 8).
For instance, c = 4.
T = (1/4)(4) + 6
T = 1 + 6
T = 7
thus another point on the line is (4, 7).
You have two points (8,8) and (4,7). Two points make a line so you can now draw the graph.
Hope this helps~ `u`
To graph the equation T=1/4c+6, we can create a table of values and then plot the points on a graph.
First, let's start with the given equation T=1/4c+6.
We can create a table with values for c and corresponding values for T:
| c | T |
|---|------|
| 2 | 6.5 |
| 4 | 7 |
| 6 | 7.5 |
| 8 | 8 |
|10 | 8.5 |
|12 | 9 |
|14 | 9.5 |
|16 | 10 |
Now, let's plot these points on a graph:
The x-axis represents the number of credits (c), and the y-axis represents the cost of tuition and fees (T).
Mark the points (2, 6.5), (4, 7), (6, 7.5), (8, 8), (10, 8.5), (12, 9), (14, 9.5), and (16, 10) on the graph.
Connect these points with a straight line, and extend the line on both ends.
Now, let's estimate the cost of tuition and fees when a student registers for 4 two-credit courses.
In this case, c = 4 * 2 = 8 (since there are 4 courses, and each course is 2 credits).
From the graph, we can see that when c = 8, the corresponding value for T is approximately 8.
Therefore, when a student registers for 4 two-credit courses, the estimated cost of tuition and fees is approximately $800.
Note: The given equation is in terms of hundreds of dollars, so we need to multiply the result by 100.
To graph the equation T = (1/4)c + 6, we can first set up a coordinate plane with T as the y-axis and c as the x-axis. Then, we can plot points on the graph by choosing different values for c and calculating the corresponding values for T.
Let's calculate the cost of tuition and fees for a few different values of c:
For c = 0 (which means no credits), T = (1/4)(0) + 6 = 6. Therefore, the point (0,6) is on the graph.
For c = 1 (one credit), T = (1/4)(1) + 6 = 6.25. So, the point (1, 6.25) is on the graph.
For c = 2 (two credits), T = (1/4)(2) + 6 = 6.50. Therefore, the point (2, 6.50) is on the graph.
Now let's plot these points on the coordinate plane.
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7 |
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| |
| |
| |
| |
______|_________________________________
0 1 2 3 4 5 6 7
After plotting the points, you can see that they lie on a straight line. So, we can connect the points to get the graph of the equation.
Once we have the graph, we can estimate the cost of tuition and fees when a student registers for four two-credit courses. This means c = 4 * 2 = 8. By looking at the graph, we can estimate that when c = 8, T is approximately 7.50 (or 750 dollars in hundreds).
Therefore, the estimated cost of tuition and fees when a student registers for four two-credit courses is 7.50 hundred dollars or 750 dollars.