Let's assume
amount invested is P
Since, we are given
Nolan invested equation as
![V_n=P(1.02)^x](https://tex.z-dn.net/?f=V_n%3DP%281.02%29%5Ex)
Anias invested equation as
![V_a=100(1.02)^{x-3}](https://tex.z-dn.net/?f=V_a%3D100%281.02%29%5E%7Bx-3%7D)
now, we can set them equal
![P(1.02)^x=100(1.02)^{x-3}](https://tex.z-dn.net/?f=P%281.02%29%5Ex%3D100%281.02%29%5E%7Bx-3%7D)
now, we can solve for P
![P=\frac{100(1.02)^{x-3}}{(1.02)^x}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B100%281.02%29%5E%7Bx-3%7D%7D%7B%281.02%29%5Ex%7D)
now, we can simplify it
![P=100(1.02)^{x-3-x}](https://tex.z-dn.net/?f=P%3D100%281.02%29%5E%7Bx-3-x%7D)
![P=100(1.02)^{-3}](https://tex.z-dn.net/?f=P%3D100%281.02%29%5E%7B-3%7D)
![P=94.2322](https://tex.z-dn.net/?f=P%3D94.2322)
So, she needed to invest $94.2322 to have the same amount of money that she has three years later........Answer
Answer: You Multiply
Step-by-step explanation:
Answer:
78.5 square units
Step-by-step explanation:
![d = 10 \implies \: r = \frac{10}{2} = 5 \\ area \: of \: circle \\= \pi {r}^{2} \\ = 3.14 \times {5}^{2} \\ = 3.14 \times 25 \\ = 78.5 \: {units}^{2} \\](https://tex.z-dn.net/?f=d%20%3D%2010%20%5Cimplies%20%5C%3A%20r%20%3D%20%20%5Cfrac%7B10%7D%7B2%7D%20%3D%205%20%5C%5C%20area%20%5C%3A%20of%20%5C%3A%20circle%20%5C%5C%3D%20%5Cpi%20%7Br%7D%5E%7B2%7D%20%20%20%5C%5C%20%20%3D%203.14%20%5Ctimes%20%20%7B5%7D%5E%7B2%7D%20%20%5C%5C%20%20%3D%203.14%20%5Ctimes%2025%20%5C%5C%20%20%3D%2078.5%20%20%5C%3A%20%20%7Bunits%7D%5E%7B2%7D%20%5C%5C%20)