Set \(\displaystyle P = 5,000\) and solve for \(\displaystyle t\) :

\(\displaystyle 5,000 = 253 \cdot 1.04^{t}\)

\(\displaystyle 1.04 ^{t} = \frac{5,000 }{253 }\)

\(\displaystyle 1.04^{t} \approx 19.7628\)

\(\displaystyle \ln 1.04^{t} \approx \ln 19.7628\)

\(\displaystyle t \ln 1.04 \approx \ln 19.7628\)

\(\displaystyle t \approx \frac{\ln 19.7628}{\ln 1.04 } \approx \frac{2.9838}{0.0392} \approx 76.1\)

January and February have 59 days total; add March, and this is 90 days. The 76th day is in March.

Set \(\displaystyle P = 100\) and solve for \(\displaystyle t\):

\(\displaystyle 100 = 318\cdot 1.07 ^{t}\)

\(\displaystyle \frac{100}{318} = 1.07 ^{t}\)

\(\displaystyle 1.07 ^{t} \approx 0.3145\)

\(\displaystyle \ln 1.07 ^{t} \approx \ln 0.3145\)

\(\displaystyle t \ln 1.07 \approx \ln 0.3145\)

\(\displaystyle t \approx \frac{\ln 0.3145}{ \ln 1.07} \approx \frac{-1.1568}{0.0677} \approx -17.1\)

17 days before January 1 was in December of 2014.

We set final principal \(\displaystyle A = 60,000\) original principal \(\displaystyle A_{0} = 40,000\), and interest rate \(\displaystyle r = 0.055\). We solve for \(\displaystyle t\) in the  continuous compound interest formula:

\(\displaystyle A = A_{0} e ^{rt}\)

\(\displaystyle 60,000 =40,000 e ^{0.055t}\)

\(\displaystyle \frac{60,000 }{40,000 }=\frac{40,000 e ^{0.055t}}{40,000 }\)

\(\displaystyle 1.5 = e ^{0.055t}\)

\(\displaystyle 0.055t = \ln 1.5\)

\(\displaystyle t = \frac{\ln 1.5}{0.055} \approx \frac{0.4055}{0.055} \approx 7.37\)

The correct response is therefore between 7 and 8 years.

Apply the compound interest formula:

\(\displaystyle A = A_{0} \left ( 1 + \frac{r}{n}\right ) ^{nt}\).

We set final principal \(\displaystyle A = 50,000\) original principal \(\displaystyle A_{0} = 30,000\), interest rate \(\displaystyle r = 0.045\), number of periods per year \(\displaystyle n = 4\) (quarterly). We solve for \(\displaystyle t\) in the equation

\(\displaystyle 50,000=30,000 \left ( 1 + \frac{0.045}{4}\right ) ^{4t}\)

\(\displaystyle 50,000=30,000 \left ( 1 .01125 \right ) ^{4t}\)

\(\displaystyle \frac{50,000}{30,000}=\frac{30,000 \left ( 1 .01125 \right ) ^{4t}}{30,000}\)

\(\displaystyle 1.6667= \left ( 1 .01125 \right ) ^{4t}\)

\(\displaystyle \log 1.6667= \log \left ( 1 .01125 \right ) ^{4t}\)

\(\displaystyle \log 1.6667= 4t \log 1 .01125\)

\(\displaystyle t = \frac{\log 1.6667}{4\log 1 .01125} \approx \frac{0.2219}{4 \cdot 0.004859} \approx 11.4\)

This is rounded up to  the next quarter of a year, so the correct response is 11.5 years, or 11 years 6 months.

The formula to find area is:

\(\displaystyle = length\cdot width=area\) is correct.

In our case our length is equal to our width which is \(\displaystyle y\).

Substituting our values into our equation we get:

\(\displaystyle y\cdot y= A\)

\(\displaystyle y^2=A\)