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marishachu [46]
3 years ago
14

A student got a summer job at an electronics store. the student is given two different compensation plans. plan A pays a weekly

salary $250 plus commission of $25 for each phone sold. Plan B offers no salary but pays $50 commission on each cellphone sold. how many cellphones will the student have to sell to make the same amount of money on both plans? when is plan B a better plan? when is plan A the better plan?
Solve using "System of Equations"
Mathematics
1 answer:
kykrilka [37]3 years ago
4 0
They will make the same amount after 10 phones sold. Before that Plan A is the better plan. After 10, Plan B is the better plan. 

You can find the point of intersection by modelling the two equations and setting them equal to each other. If x is equal to phones sold:

250 + 25x = 50x
250 = 25x
10 = x

Therefore, you make the same at 10 phones. 
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40/3, 172/3, respectively

Step-by-step explanation:

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lys-0071 [83]

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\dfrac{-1}{x(x+h)}, h\ne 0

Step-by-step explanation:

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Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

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5 0
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example:

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7 0
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Answer:

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7 0
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