Guys i need help with exponential growth / decay and functions.

1 answer:

7 0

Exponential growth functions show an increase. Whereas, exponential decay functions show a decrease. For example. If you are losing money it would be an exponential decay

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If an initial amount A0 of money is invested at an interest rate i compounded times a year, the value of the investment after t

seropon [69]

Answer:

Following are the solution to the given point:

Step-by-step explanation:

Please find the comp[lete question in the attached file.

Given:

$\bold{ \lim_{n \to \ \infty} (1+ \frac{r}{n})^{nt} =e^{rt}}$

In point 1:

$\to y = (1+ \frac{r}{n})^{nt}$

In point 2:

$\to \ln (y)= nt \ln(1+ \frac{r}{n})$

In point 3:

Its key thing to understand, which would be that you consider the limit n to $\infty$ , in which r and t were constants!

$=lim_{n \to \ \infty} \ln (y) = lim_{n \to \ \infty} nt \ln(1+\frac{r}{n})\\\\= lim_{n \to \ \infty} \frac{\ln(1+\frac{r}{n})}{\frac{1}{nt}}\\\\= lim_{n \to \ \infty} \frac{\frac{-r}{\frac{n^2}{(1+\frac{r}{n})}}}{- \frac{1}{n^2t}}\\\\= lim_{n \to \ \infty} \frac{\frac{rn^2t}{n^2}}{(1+\frac{r}{n})}\\\\= lim_{n \to \ \infty} \frac{rt}{(1+\frac{r}{n})}\\\\= \frac{rt}{(1+\frac{r}{0})}\\\\=rt$

In point 4:

$\to \lim_{n \to \ \infty} = (1+\frac{r}{n})^{nt}$ and

$\to \lim_{n \to \ \infty} = A_0e^{rt}$

7 0

3 years ago

Derrecks dog sitting $12 plus $5 per hour

vova2212 [387]

A.
5x+12=3x+18
2x+12=18
2x=6
x=3
you would get paid the same amount if you worked 3 hours at each job

B.
5(5)+12
10+12
22

3(5)+18
15+18
33

you would get paid more at the second job if you worked for 5 hours

8 0

3 years ago

Simplify 7/the square root of 2 (as a fraction)

ioda

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