match each function described on the left with an ordered pair that is a solution to the function. round to…

Explanation:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A = Pe^{rt}$, where $P$ is the principal amount (initial investment), $r$ is the annual interest rate (in decimal form), $t$ is the time in years, and $A$ is the amount of money after $t$ years.

Step2: For the first investment ($P = 500$, $r=0.03$)

Let's assume $t = 10$. Then $A = 500e^{0.03\times10}=500e^{0.3}\approx500\times1.34986 = 674.93$. Let's assume $t = 5$, then $A = 500e^{0.03\times5}=500e^{0.15}\approx500\times1.16183 = 580.92$. Let's assume $t = 8$, then $A = 500e^{0.03\times8}=500e^{0.24}\approx500\times1.27125 = 635.63$.

Step3: For the second investment ($P = 300$, $r = 0.05$)

Let $t = 10$. Then $A=300e^{0.05\times10}=300e^{0.5}\approx300\times1.64872 = 494.62$. Let $t = 8$. Then $A = 300e^{0.05\times8}=300e^{0.4}\approx300\times1.49182 = 447.55$. Let $t = 5$. Then $A=300e^{0.05\times5}=300e^{0.25}\approx300\times1.28403 = 385.21$.

Step4: For the third investment ($P = 600$, $r = 0.02$)

Let $t = 10$. Then $A = 600e^{0.02\times10}=600e^{0.2}\approx600\times1.22140 = 732.84$.

Step5: For the fourth investment ($P = 200$, $r = 0.06$)

Let $t = 0$, then $A = 200e^{0.06\times0}=200\times1 = 200$. So the ordered - pair $(0,200)$ is a solution for the investment of $P = 200$ (when $t = 0$, the amount is just the initial investment).

Answer:

An initial investment of $500$ earning $3%$ interest compounded continuously: No match in given options. An initial investment of $300$ earning $5%$ interest compounded continuously: No match in given options. An initial investment of $600$ earning $2%$ interest compounded continuously: $(10,732.84)$ An initial investment of $200$ earning $6%$ interest compounded continuously: $(0,200)$