Compare 5/6 and 1/3 using the benchmark fraction 1/2
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The functions I(x) and E(x) of the theatre's income and expenses are illustrations of linear functions.
<h3>The equation of the theatre's income</h3>The theatre charges $4.50 per student.
Assume the number of students is x, the equation of the theatre's income would be:
I(x) = 4.5x
<h3>The equation of the theatre's expenses</h3>The theatre expense per student is $1.25, and the operating cost on the staff is $130
The equation of the theatre's expenses would be:
E(x) = 1.25x + 130
<h3>Complete the table</h3>Using the formulas I(x) = 4.5x and E(x) = 1.25x + 130, the complete table is:
Students, x 0 10 20 30 40 50 60 70
Income, I 0 45 90 135 180 225 270 315
Expenses, E 130 142.5 155 167.5 180 192.5 205 217.5
<h3>The graph of the theatre's income and expenses</h3>See attachment
<h3>The pattern by which theatre's income and expenses increase</h3>The functions I(x) and E(x) are linear functions.
So, the pattern with which the functions increase is a linear pattern.
<h3>The number of students when the theatre's income and expenses are equal</h3>
This means that:
I(x) = E(x)
So, we have:
4.5x = 1.25x + 130
Subtract 1.25 from both sides
3.25x = 130
Divide both sides by 3.25
x = 40
Hence, the number of students is 40
<h3>The theatre profit</h3>This is the difference between the theatre expenses and their income.
So, we have:
P(x) = E(x) - I(x)
This gives
P(x) = 1.25x + 130 - 4.5x
Simplify
P(x) = 130 - 3.25x
<h3>Solution to the inequalities</h3>We have:
E(x) < 255
This gives
1.25x + 130 < 255
Subtract 130 from both sides and divide by 1.25
x < 100 students
Also, we have:
I(x) > 675
This gives
4.5x > 657
Solve for x
x > 146 students
Hence, the number of students for the inequalities are less than 100 and greater than 146
Read more about linear equations and inequalities at:
