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3 years ago

Given that f(x) = 8x + 3 and g(x) = x + 9, calculate. f(g(16) = g(f(16) =

Mathematics

1 answer:

Ludmilka [50]3 years ago

4 0

203,140

Step-by-step explanation:

f(g(16))=f(16+9)=f(25)=8×25+3=203

g(f(16))=g(8×16+3)=g(131)=131+9=140

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Find the value of $ 15,000 at the end of one year if it is invested in an account that has an interest rate of 4.95 % and is com

lora16 [44]

A)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{twelve months, thus}
\end{array}\to &12\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{12}\right)^{12\cdot 1}$

b)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{365 days, thus}
\end{array}\to &365\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{365}\right)^{365\cdot 1}$

c)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{four quarters, thus}
\end{array}\to &4\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{4}\right)^{4\cdot 1}$

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Ierofanga [76]

Answer:

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Step-by-step explanation:

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