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Stels [109]
3 years ago
8

In order to go to college, Chris goes from working full-time making $30,000 per year to working part-time at half the salary for

two years. The cost of his education will be $5,000. If Chris makes $35,000 per year after getting his degree, approximately how many years will it take him to recover his investment?​
Mathematics
1 answer:
Korolek [52]3 years ago
6 0

Answer:

Step-by-steIn order to go to college, Chris goes from working full-time making $30,000 per year to working part-time at half the salary for two years. The cost of his education will be $5,000. If Chris makes $35,000 per year after getting his degree, approximately how many years will it take him to recover his investment? *

it would take him 7 years to recover the 35,000 he invested

p explanation:

he was in college two years and it cost him a total of 5000 and he lost 30000 from the two years he worked part time

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The time is 135 min.

Step-by-step explanation:

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t = time

T_0 = initial temp of the heated object

k = constant

From the information given we know that:

  • Initial temp of the cake is 310 °F.
  • The surrounding temp is 72 °F.
  • After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}

T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}

Next, we find the time at which the cake's temperature will be 100°F.

100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1

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