You invest $9000 with a 6% interest rate compounded semiannually. After 9 yrs, how much money is in your account?

1 answer:

7 0

$\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 &\$9000\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\

t=years\to &9
\end{cases}
\\\\\\
A=9000\left(1+\frac{0.06}{2}\right)^{2\cdot 9}$

You might be interested in

4 is a solution of x < -5 true or false -3 is a solution of y > -2 true or false

san4es73 [151]

the first one is true and the second one is false

3 0

2 years ago

karas jug contains 2 quarts of apple cider.she fills 8 mugs with cider.each mug can hold six three fourths of fluid ounces.how m

bazaltina [42]

So first of all you need to know that in 1 quart there is 31.9995 ounces so since shes says there is 2 quarts lets do the math:)
31.9995 x 2 = 63.999 ounces
now she says she took 8 mugs and filled them up each with 6 and 3/4 ounces. lets do the math:)
6 and 3/4 x 8 = 30 ounces
now lets subtract that from our 2 quarts or 63.999 ounces
63.999 - 30 = 33.999
so we have 33.999 fluid ounces left in the jar:)
hope this helps lol

8 0

2 years ago

Plz write this on paper help me and send it❤️

vovangra [49]

Answer:

1. $27^{\frac{2}{3} } =9$

2. $\sqrt{36^{3} } =216$

3. $(-243)^{\frac{3}{5} } =-27$

4. $40^{\frac{2}{3}}=4\sqrt[3]{25} =4325$

5. Step 4: $(\frac{343}{27}) ^{-1} =\frac{27}{343}$

6. $D. -72cd^{7}$

Step-by-step explanation:

Use the following properties:

$a^{\frac{x}{y} } =\sqrt[x]{a^{y} }$

$\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}$

$a^{-n} =\frac{1}{a^{n} }$

$(xy)^{z} =x^{z} y^{z} \\\\$

$(x^{y}) ^{z} =x^{yz}$

$x^{y} x^{z} =x^{y+z}$

So:

1. $27^{\frac{2}{3} } =\sqrt[3]{27^{2}} =\sqrt[3]{729} }=9$

2. $\sqrt{36^{3} } =\sqrt{36*36*36} =\sqrt{36} \sqrt{36} \sqrt{36} =6*6*6=216$

3. $(-243)^{\frac{3}{5} } =\sqrt[5]{-243^{3} } =\sqrt[5]{-14348907} =-27$

4. $40^{\frac{2}{3}}=\sqrt[3]{40^{2} } =\sqrt[3]{2^{6} 5^{2} } =\sqrt[3]{2^{6} } \sqrt[3]{5^{2} } =2^{\frac{6}{3} } 5^{\frac{2}{3} } =4 *5^{\frac{2}{3} } =4\sqrt[3]{5^{2} } =4\sqrt[3]{25}=4325$