7/9 Has how many 1/3s in it?

Mathematics

1 answer:

Yuliya22 [10]3 years ago

4 0

7/9 only has 2 thirds in it

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A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest

faltersainse [42]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73$

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396$

add both amounts, and that's her investment for the 11 years.

7 0

3 years ago

Someone pls help im so close

Sliva [168]

Answer: A) 126.6

<u>Step-by-step explanation:</u>

Since we know ∠B = 85° and ∠C = 53°, we can use the Triangle Sum Theorem (angles of a triangle = 180°) to calculate ∠A.

∠A + ∠B + ∠C = 180

∠A + 85 + 53 = 180

∠A + 138 = 180

∠A = 42

Now we have: