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The manager at a restaurant found that the cost to produce 150 cups of coffee is $21, while the cost to produce 300 cups of coffee is $36. assume the relationship between the cost y to produce x cups of coffee is linear.
a.) write a linear equation that expresses the cost,y, in terms of the number of cups of coffee,x.
b.) how many cups of coffee are produced if the cost of production is $56?

Sagot :

MrRoyal

Answer:

[tex]y = \frac{1}{10}x + 6[/tex]

500 cups cost $56

Step-by-step explanation:

Given

[tex]x = cups[/tex]

[tex]y = cost[/tex]

[tex](x_1,y_1) = (150,21)[/tex]

[tex](x_2,y_2) = (300,36)[/tex]

Solving (a): Linear Equation

First, we calculate the slope (m)

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Substitute right values

[tex]m = \frac{36 - 21}{300- 150}[/tex]

[tex]m = \frac{15}{150}[/tex]

[tex]m = \frac{1}{10}[/tex]

The equation is then calculated as:

[tex]y = m(x - x_1) + y_1[/tex]

Where

[tex](x_1,y_1) = (150,21)[/tex]

[tex]m = \frac{1}{10}[/tex]

This gives:

[tex]y = \frac{1}{10}(x - 150) + 21[/tex]

Open bracket

[tex]y = \frac{1}{10}x - \frac{1}{10}*150 + 21[/tex]

[tex]y = \frac{1}{10}x - 15 + 21[/tex]

[tex]y = \frac{1}{10}x + 6[/tex]

Solving (b): Cups of coffees for $56

Substitute 56 for y in [tex]y = \frac{1}{10}x + 6[/tex]

[tex]56 = \frac{1}{10}x + 6[/tex]

Subtract 6 from both sides

[tex]-6+56 = \frac{1}{10}x + 6-6[/tex]

[tex]50 = \frac{1}{10}x[/tex]

Multiply both sides by 10

[tex]10 * 50 = \frac{1}{10}x * 10[/tex]

[tex]10 * 50 = x[/tex]

[tex]500 = x[/tex]

[tex]x = 500[/tex]

Hence: 500 cups cost $56

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