Answer: The 2 answers are inequality form and interval notation: n>200, and (200, ∞).
Step-by-step explanation: To solve for n, you’ll need to simplify the both sides of the inequality, and then isolating the variable.
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$10000\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &12 \end{cases} \\\\\\ A=10000\left(1+\frac{0.04}{1}\right)^{1\cdot 12}\implies A=1000(1.04)^{12}\\\\\\ A\approx 16010.32](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%0A%5C%5C%5C%5C%0AA%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%0A%5Cquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%20%26%5C%2410000%5C%5C%0Ar%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cto%20%260.04%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20thus%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0At%3Dyears%5Cto%20%2612%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D10000%5Cleft%281%2B%5Cfrac%7B0.04%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%2012%7D%5Cimplies%20A%3D1000%281.04%29%5E%7B12%7D%5C%5C%5C%5C%5C%5C%20A%5Capprox%2016010.32)
he then turns around and grabs that money and sticks it for another 9 years,
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} ~~ \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$16010.32\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semi-annually, thus twice} \end{array}\to &2\\ t=years\to &9 \end{cases} \\\\\\ A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18} \\\\\\ A\approx 24970.64](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%0A%5C%5C%5C%5C%0AA%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%0A~~%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%20%26%5C%2416010.32%5C%5C%0Ar%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cto%20%260.05%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bsemi-annually%2C%20thus%20twice%7D%0A%5Cend%7Barray%7D%5Cto%20%262%5C%5C%0At%3Dyears%5Cto%20%269%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D16010.32%5Cleft%281%2B%5Cfrac%7B0.05%7D%7B2%7D%5Cright%29%5E%7B2%5Ccdot%209%7D%5Cimplies%20A%3D16010.32%281.025%29%5E%7B18%7D%0A%5C%5C%5C%5C%5C%5C%0AA%5Capprox%2024970.64)
add both amounts, and that's how much is for the whole 21 years.
![\sf \: 12x^{4} ( - 56x^{3} {y}^{2} )](https://tex.z-dn.net/?f=%20%5Csf%20%5C%3A%2012x%5E%7B4%7D%20%28%20-%2056x%5E%7B3%7D%20%20%7By%7D%5E%7B2%7D%20%29)
Multiplying a positive and a negative equals a negative: (+)×(-)=(-)
![\sf \: { - 12x}^{4} \times 56 {x}^{3} {y}^{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5C%3A%20%20%7B%20-%2012x%7D%5E%7B4%7D%20%20%5Ctimes%2056%20%7Bx%7D%5E%7B3%7D%20%20%7By%7D%5E%7B2%7D%20)
Calculate the product
![\boxed{ \tt - 672x^{7} {y}^{2} }](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Ctt%20-%20672x%5E%7B7%7D%20%20%7By%7D%5E%7B2%7D%20%7D)
Answer: 1/10
Step-by-step explanation: -2/5+1/2=0.1
If we convert 0.1 into a fraction, it becomes 1/10.