Question

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The expression 1000(1+r)^2 represents the balance after 2 years where r is the annual interest rate in decimal form. Write a polynomial in standard form.

The interest rate is 3%. What is the balance of your account after 2 years.
The guitar costs $1100. Do you have enough money in your account after 3 years?

Answer 1

The balance of your account after 2 years is $1,060.90. You do not have enough money in your account after 3 years to buy the $1100 guitar.

Step-by-step explanation:

To write a polynomial in standard form for the balance after 2 years, we substitute the given interest rate of 3% (0.03 in decimal form) into the expression $1000(1+r)^2. This gives us $1000(1+0.03)^2. Simplifying this equation, we have:

$1000(1+0.03)^2
= $1000(1.03)^2
= $1000(1.0609)
= $1,060.90

Therefore, the balance of your account after 2 years is $1,060.90.

To determine if you have enough money in your account after 3 years to buy the $1100 guitar, we can substitute the values into the expression $1000(1+r)^3. Using the same interest rate of 3% (0.03 in decimal form) and solving the equation, we have:

$1000(1+0.03)^3
= $1000(1.03)^3
= $1000(1.0927)
= $1,092.70

Since $1,092.70 is less than the cost of the guitar, you do not have enough money in your account after 3 years to buy the guitar.

Answer 2

let's move like a crab, backwards some.

after 2 years?


`\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$1000\\ r=rate\to 3\%\to (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+(0.03)/(1)\right)^(1\cdot 2)\implies A=1000(1.03)^2`

after 3 years?


`\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$1000\\ r=rate\to 3\%\to (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+(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.