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lukranit [14]
3 years ago
11

A large waterfall is managed so that a minimum of 110,000 cubic feet of water per second sec) fow over the fals. Find the minimu

m amount of water that will tow over the falils in a 24-hour period during tourist season. Write your answer in scientific notation Aminimum of?cubic feet ofwater flow over thefallsina 24-lourperiod Use scientific notation. Use the multüiplication symbol in the math palette as needed)
Mathematics
1 answer:
Liula [17]3 years ago
5 0

Answer:

9.504×10⁹ ft³

Step-by-step explanation:

We are told the amount of water in cubic feet per second.  We need it in cubic feet per hour, then multiply that amount by 24 to get the cubic feet of water that falls in a 24-hr period.  Converting the seconds to hours:

\frac{110,000ft^3}{sec}*\frac{3600sec}{1hr}

The seconds label cancels out leaving us with a label in cubic feet per hour.  That number, NOT in scientific notation (we'll do that at the end) is:

396000000 cubic feet per hour.

Multiply that by 24 and you get:

9504000000 cubic feet or

9.504×10⁹ cubic feet

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scoundrel [369]
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Since the sides are 10 ft you multiply that by 2 to get the length of both sides which gives you 20 feet. 

Then to find the circle, since both are half circles. You can just do the normal circumference equation and you will end up with the same answer.

For example in this problem: this problem gives you the diameter but you need the radius of the circle. C=2(π)(1.5)

C=9.42 < this is the circumference of the circle. 

Then you add the sides of the table which are 20 feet. 

So the final answer is 29.42 feet. 
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4 years ago
Can someone answer this question please (picture)
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31.5 units^2

Step-by-step explanation:

Rectangle area (RA) = l × w

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RA = 3 × 7

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TA = 10.5

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Add areas together

21 + 10.5 = 31.5

7 0
3 years ago
and a list of numbers the pattern increases by 0.001 as you move to the right if the third number list is 0.0 64 what is the fir
Ksivusya [100]

Answer:

  0.062

Step-by-step explanation:

The numbers will decrease by 0.001 as you move the the left, so the list of numbers can be found as ...

  3rd number: 0.064

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7 0
4 years ago
Sanjay borrowed 7,000 at a simple interest rate of 3% per year. After a certain number of years he had paid $840 in interest alt
alexgriva [62]

Answer:

It was 4 years.

Step-by-step explanation:

To solve this problem we need to use the appropriate formula for simple interest, which is shown below:

M = C*r*t

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7 0
3 years ago
find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration
Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]

now for given values:

\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\

\to  (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to  (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} -  \frac{1}{4}] =0 \\\\\to  (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\

\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\

       = max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}

In point b:

Its  

spectral radius is less than 1 hence matrix is convergent.

In point c:

\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) =   \left(\begin{array}{c}3&1&2\end{array}\right)  , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\  \to x^{(k+1)} =  \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right]  \\\\

after solving the value the answer is

:

\lim_{k \to \infty} x^k=o  = \left[\begin{array}{c}0&0&0\end{array}\right]

4 0
3 years ago
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