Answer:
Short-Term investment: 5000$
Intermediate-Term investment: 65000$
Long-Term investment: 30000$
Step-by-step explanation:
To construct our first equation lets define sort-term bond investment as x, long-term investment as y.
So the equation is:
From the equation it is found that:
Instead of y, if we put 25000+x the equation will be as following:
From the equation it is found that:
Short-Term investment is 5000$
Long-Term investment is 30000$
Rest of the money is Intermediate-Term investment 65000$
<span>The Volume of a Cylinder = </span><span>π <span>• r² • height<span>
</span></span></span>radius = 13.5 feet
volume = PI * 13.5^2 * 4 feet (the problem states it is to be filled to a depth of 4 feet)
volume = PI * 182.25 * 4
volume =
<span>
<span>
<span>
2,290 cubic feet</span></span></span>
The hose delivers 80 <span>cubic feet per hour so it will take:
</span>2,290 / 80 =
<span>
<span>
<span>
28.625
</span>
</span>
</span>
hours
**************************************************************************
The volume of the pool is:
volume = PI * 182.25 * 4.5
=
<span>
<span>
<span>
2,576.</span></span></span>5 cubic feet
Answer:
Step-by-step explanation:
If an exponential function is in the form of y = a(b)ˣ,
a = Initial quantity
b = Growth factor
x = Duration
Condition for exponential growth → b > 1
Condition for exponential decay → 0 < b < 1
Now we ca apply this condition in the given functions,
1). 
Here, (1 + 0.45) = 1.45 > 1
Therefore, It's an exponential growth.
2). 
Here, (0.85) is between 0 and 1,
Therefore, it's an exponential decay.
3). y = (1 - 0.03)ˣ + 4
Here, (1 - 0.03) = 0.97
And 0 < 0.97 < 1
Therefore, It's an exponential decay.
4). y = 0.5(1.2)ˣ + 2
Here, 1.2 > 1
Therefore, it's an exponential growth.
You need to multiply the two numbers. Or you can use long addition.
Hmm if I'm not mistaken, is just an "ordinary" annuity, thus
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![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)
