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Romashka [77]
3 years ago
8

Last one anyone that can help me out?

Mathematics
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:

Part a. t = 7.29 years.

Part b. t = 27.73 years.

Part c. p = $3894.00

Step-by-step explanation:

The formula for continuous compounding is: A = p*e^(rt); where A is the amount after compounding, p is the principle, e is the mathematical constant (2.718281), r is the rate of interest, and t is the time in years.

Part a. It is given that p = $2000, r = 2.5%, and A = $2400. In this part, t is unknown. Therefore: 2400 = 2000*e^(2.5t). This implies 1.2 = e^(0.025t). Taking natural logarithm on both sides yields ln(1.2) = ln(e^(0.025t)). A logarithmic property is that the power of the logarithmic expression can be shifted on the left side of the whole expression, thus multiplying it with the expression. Therefore, ln(1.2) = 0.025t*ln(e). Since ln(e) = 1, and making t the subject gives t = ln(1.2)/0.025. This means that t = 7.29 years (rounded to the nearest 2 decimal places)!!!

Part b. It is given that p = $2000, r = 2.5%, and A = $4000. In this part, t is unknown. Therefore: 4000 = 2000*e^(2.5t). This implies 2 = e^(0.025t). Taking natural logarithm on both sides yields ln(2) = ln(e^(0.025t)). A logarithmic property is that the power of the logarithmic expression can be shifted on the left side of the whole expression, thus multiplying it with the expression. Therefore, ln(2) = 0.025t*ln(e). Since ln(e) = 1, and making t the subject gives t = ln(2)/0.025. This means that t = 27.73 years (rounded to the nearest 2 decimal places)!!!

Part c. It is given that A = $5000, r = 2.5%, and t = 10 years. In this part, p is unknown. Therefore 5000 = p*e^(0.025*10). This implies 5000 = p*e^(0.25). Making p the subject gives p = 5000/e^0.25. This means that p = $3894.00(rounded to the nearest 2 decimal places)!!!

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Hello!

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Step-by-step explanation:

Because it is a equilateral triangle, it means all sides have the same length.

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AB = 5/4 y - 1  = 5/4 (12/13) -1 = 2/13

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Please check my other articles in my site, just copy and paste this word to google search engine: learningandassignments diy4pro.

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