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lord [1]
3 years ago
9

If $396 is invested at an interest rate of 13% per year and is compounded continuously, how much will the investment be worth in

3 years?
Use the continuous compound interest formula: A = Pert
Mathematics
2 answers:
sineoko [7]3 years ago
5 0

Answer:

$584.88

Step-by-step explanation:

If $396 is invested at an interest rate of 13% per year and is compounded continuously in 3 years.

Formula of continuous compound interest : A=Pe^{rt}

Where

A = Future Amount

P = Principal amount ( $396.00 )

r = rate of interest 13% ( 0.13 )

t = time in years (3 years)

Now put the values in the formula

A=396e^{0.13\times 3}

A=396(2.718282^{0.39})

= 396 (1.476981)

= 584.884394 ≈ 584.88

The amount after 3 years would be $584.88

           

Ipatiy [6.2K]3 years ago
3 0

Answer:

A=\$584.88  

Step-by-step explanation:

we know that

The formula to calculate continuously compounded interest is equal to

A=P(e)^{rt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

t=3\ years\\ P=\$396\\ r=0.13  

substitute in the formula above  

A=\$396(e)^{0.13*3}=\$584.88  

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It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

We have a rational function:

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Given:

The equation is,

2\log _3x-\log _3(x-2)=2

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Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

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\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

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