28. Suppose an investment doubles in value every 5 years. This is year the

Mathematics

1 answer:

dybincka [34]2 years ago

6 0

I think the answer is in 10 years from now the investment will be worth 24,960 and five years ago it was worth

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Emaan charges a fixed amount for a bracelet, with an additional charge for each charm a customer adds to the bracelet. Write a l

DIA [1.3K]

Answer:

f(x) = y + (x*p)

Step-by-step explanation:

Since we are not given actual values we will need to make the function with only variables. Each variable will represent the following...

Fixed Cost: y

Cost per charm: p

Number of charms: x

Therefore, using the variables mentioned above we can combine them into the following linear function using the number of charms as our main input for our function...

f(x) = y + (x*p)

6 0

3 years ago

Is negative 17/24th bigger or smaller than negative 11/12

Triss [41]

It would be smaller because 11/12 = 22/24

5 0

4 years ago

How much more would $1000 earn in 5 years in an account compounded continuously than an account compounded quarterly if the inte

choli [55]

$\bf ~~~~~~ \textit{Compounding Continuously Interest Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
t=years\to &5
\end{cases}
\\\\\\
A=1000e^{0.037\cdot 5}\implies A=1000e^{0.185}\\\\
-------------------------------\\\\$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &5
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.037}{4}\right)^{4\cdot 5}\implies A=1000(1.00925)^{20}$

compare both amounts.

8 0

3 years ago

19 tens + 11 tens equal

Nata [24]

19*10= 190 11*10=110 190+110=300

7 0

4 years ago

Read 2 more answers

Simplify <br><br>(a^m/a^n)^m+n×(a^n/a^p)^n+p8×(a^p/a^m)^p+m<br><br>step by step

Rasek [7]

Answer:

m+a-1mn+m2+a-1np+n2n+a-mp+p2p8