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Leni [432]
3 years ago
5

Which expression is half as large as the expression 345+23?

Mathematics
2 answers:
kipiarov [429]3 years ago
6 0

Final Answer: 184

Steps/Reasons/Explanation:

We will need to divide 345 + 23 by 2 to get the expression that is half as large of the sum of 345 + 23.

Question: Which expression is half as large as the expression 345 + 23?

\frac{345+23}{2}  = \frac{368}{2} = 184

~I hope I helped you :)~

Whitepunk [10]3 years ago
5 0

Answer:

368:2= 184

................

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PLEASE HELP
Usimov [2.4K]

Answer:

r\approx0.03\text{ or about $3\%$}

Step-by-step explanation:

The standard compound interest formula is given by:

\displaystyle A=P(1+\frac{r}{n})^{nt}

Where A is the amount afterwards, P is the principal, r is the rate, n is the times compounded per year, and t is the number of years.

Since we are compounding annually, n=1. Therefore:

\displaystyle  A = P (1+r)^{t}

Lester wants to invest $10,000. So, P=10,000.

He wants to earn $1000 interest. Therefore, our final amount should be 11000. So, A=11000.

And our timeframe is 3.3 years. So, t=3.3. Substituting these values, we get:

11000=10000(1+r)^{3.3}

Let’s solve for our rate r.

Divide both sides by 10000:

1.1=(1+r)^{3.3}

We can raise both sides to 1/3.3. So:

\displaystyle (1.1)^\frac{1}{3.3}=((1+r)^{3.3})^\frac{1}{3.3}

The right side will cancel:

\displaystyle r+1=(1.1)^\frac{1}{3.3}

So:

\displaystyle r=(1.1)^\frac{1}{3.3}-1

Use a calculator:

r\approx0.03

So, the annual rate of interest needs to be about 0.03 or 3% in order for Lester to earn his interest.

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What is 123,499 rounded tot he nearest thousand
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pashok25 [27]

Answer:

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Step-by-step explanation:

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A colony of bacteria triples in link every 10 minutes. It's length is now 1 mm. How long will be in 40 minutes
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If the bacteria is tripling every 10 minutes, that means the "rate of increase" on that period is 200%, so if say the current amount is "c", 200% of "c" is just 2c, so c + 2c is 3c, a tripled amount.

\bf \textit{Periodic Exponential Growth}\\\\
A=I(1 + r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1\\
r=rate\to 2\%\to \frac{200}{100}\to &2.00\\
t=\textit{elapsed time}\to &40\\
p=period\to &10
\end{cases}
\\\\\\
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