Respuesta :
let's move like a crab, backwards some.
after 2 years?
[tex]\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 &\$1000\\ r=rate\to 3\%\to \frac{3}{100}\to &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &2 \end{cases} \\\\\\ A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2[/tex]
after 3 years?
[tex]\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 &\$1000\\ r=rate\to 3\%\to \frac{3}{100}\to &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &3 \end{cases} \\\\\\ A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3[/tex]
is that enough to pay the $1100?
now, let's write 1000(1+r)² in standard form
1000( 1² + 2r + r²)
1000(1 + 2r + r²)
1000 + 2000r + 1000r²
1000r² + 2000r + 1000 <---- standard form.
after 2 years?
[tex]\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 &\$1000\\ r=rate\to 3\%\to \frac{3}{100}\to &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &2 \end{cases} \\\\\\ A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2[/tex]
after 3 years?
[tex]\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 &\$1000\\ r=rate\to 3\%\to \frac{3}{100}\to &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &3 \end{cases} \\\\\\ A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3[/tex]
is that enough to pay the $1100?
now, let's write 1000(1+r)² in standard form
1000( 1² + 2r + r²)
1000(1 + 2r + r²)
1000 + 2000r + 1000r²
1000r² + 2000r + 1000 <---- standard form.
