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VladimirAG [237]
3 years ago
7

Daniel used to collect baseball cards as a kid. He purchased a particular

Mathematics
1 answer:
Illusion [34]3 years ago
7 0

Answer:

y < 85 \div  \div 9614 = 88

Ты лоххх

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Katena32 [7]
Let's move like a crab, backwards some.

after 2 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$1000\\&#10;r=rate\to 3\%\to \frac{3}{100}\to &0.03\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annually, thus once}&#10;\end{array}\to &1\\&#10;t=years\to &2&#10;\end{cases}&#10;\\\\\\&#10;A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2

after 3 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to &\$1000\\&#10;r=rate\to 3\%\to \frac{3}{100}\to &0.03\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annually, thus once}&#10;\end{array}\to &1\\&#10;t=years\to &3&#10;\end{cases}&#10;\\\\\\&#10;A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3

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.
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