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:

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

\(\displaystyle y^2=A\)

Put the negative exponent on the bottom so that you have \(\displaystyle \frac{x^{12}}{x^{6}}\) which simplifies further to \(\displaystyle x^{6}\).

\(\displaystyle (5^{0})^{2}=(1)^{2}=1\)

We can also solve this problem using a different apporach

\(\displaystyle (5^{0})^{2}=5^{0\times2}=5^{0}=1\)

Remember that any number raised to the 0th power equals 1

\(\displaystyle ((yz)^{2})^{3}=(yz)^{6}=y^{6}z^{6}\)

Alternatively,

\(\displaystyle ((yz)^{2})^{3}=(y^{2}z^{2})^3=y^{2\times3}z^{2\times3}=y^{6}z^{6}\)

\(\displaystyle ((6^2)^{0})^{3}=6^{2\times 0\times 3}=6^0=1\)

Remember that any number raised to the 0th power equals 1

\(\displaystyle \left (\frac{10^{3}}{10^{0}} \right )^{2}=\frac{10^{3\times2}}{10^{0\times2}}=\frac{10^{6}}{10^{0}}=\frac{1000000}{1}=1,000,000\)

Alternatively,

\(\displaystyle \left (\frac{10^{3}}{10^{0}} \right )^{2}=\left (10^{3-0} \right )^{2}=\left (10^{3} \right )^{2}=10^{3\times 2}=10^{6}=1,000,000\)

\(\displaystyle \frac{4a^{-5}b^{3}c^{2}}{12a^{-7}b^{-5}c^{-2}}\)

Step 1: Simplify the exponents using the division of exponents rule (subtract exponents in demoninator from exponents in numerator).

\(\displaystyle \frac{4a^{-5-(-7)}b^{3-(-5)}c^{2-(-2)}}{12}\)

\(\displaystyle \frac{4a^{2}b^{8}c^{4}}{12}\)

Step 2: Reduce the fraction

\(\displaystyle \frac{a^{2}b^{8}c^{4}}{3}\)

Follow the division of exponents rule.  Subract the exponents in the denominator from the exponents in the numerator.