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lakkis [162]
3 years ago
8

Corinne has a job selling magazines. She earns $7.50 per hour plus 20% of the total amount of her sales. She also gets an allowa

nce of $40 per week for gas. She knows her weekly earnings can be shown using the following expression: 7.50h + 0.20s + 40 Part A: Identify a coefficient, a variable, and a constant in this expression. (3 points) Part B: If Corinne works for 25 hours and sells $300 in magazines, how much does she earn for the week? Show your work to receive full credit. (4 points) Part C: If Corinne gets a raise and begins earning $9 per hour, would the coefficient, variable, or constant in the equation change? Why? (3 points)
Mathematics
1 answer:
Lynna [10]3 years ago
4 0
Part A:  Coefficient:  either 7.50 or 0.20
Part B:

7.50 (25) + 0.20 (300) + 40
187.50 + 60 + 40 = $287.50

Part C: After her raise, the first term will change from 7.50h to 9h
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Solve the system. <br> x+3y=1 <br> -2x+7=-23
Viefleur [7K]
X+ 3y = 1  -----(first equation)
-2x+7 = -23 ---(second equation)
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Now, Substitute the value of x in first equation,
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In short, Final Answers are: x = 15 & y = -14/3

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5 0
3 years ago
Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution
Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\

for y=e^{6x}:

y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\

Now we replace in the differential equation  y'' − y' − 30y = 0

for y=e^{-5x}:

25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0

for y=e^{6x}:

36e^{6x}-6e^{6x}-30=0\\36-6+30=0

Now, to know if both function are linearly independent we calculate the Wronskian

W(f,g)=fg'-f'g

W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0

I hope this is useful for you

Best regard

7 0
3 years ago
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