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4 years ago

5

A factory employs skilled laborers at$25 per hour and unskilled laborers at $12 per hour. The total hourly cost of the factory’s

24 workers is $405. How many of each type of laborer works at the factory.

1 answer:

yarga [219]4 years ago

6 0

For this case what you should do is the following system of equations:
x: skilled laborers
y: unskilled laborers
By writing the system we have:
x + y = 24
25x + 12y = 405
Solving the system:
Step 1:
-25x + -25y = -600
25x + 12y = 405
Step 2:
-13y = -195
y = 195/13
y = 15
step 3:
x = 24-y
x = 24-15
x = 9
Answer:
skilled laborers = 9
unskilled laborers = 15

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Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

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<u>Step 1: Define</u>

$\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}$

t = 1

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<u>Step 2: Integrate</u>

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[Right Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt$
[Right Integrals] Reverse Power Rule: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C$
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<u>Step 3: Solve</u>

Substitute in variables: $\displaystyle 8 = \frac{5(1)^4}{4} + \frac{9}{1} + C$
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[Subtraction Property of Equality] Isolate <em>C</em>: $\displaystyle \frac{-9}{4} = C$
Rewrite: $\displaystyle C = \frac{-9}{4}$

Particular Solution: $\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Differentials Equations and Slope Fields

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