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yawa3891 [41]
3 years ago
8

Sandy is working with a carpenter to frame a house. They are using 8-foot-long boards, but each board must be cut to be 7 feet i

nches long. How much is cut off each board? Note: 1 foot = 12 inches A. 1 1/4 inches B. 1 3/4 inches C. 7 inches D. 3/4 inch
Mathematics
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

5/4in

Step-by-step explanation:

The initial length:

1 ft = 12 in

8 ft = 8 * 12 in    

= 96 in

The final length:

7 ft 10 3/4 in = 7 ft + 10 in + 3/4 in

                     = 7 * 12 in + 10 in + 3/4 in                    

                     = 84 in + 10 in + 3/4 in        

                     = 94 in + 3/4 in

The final length:

96 in - 94 in - 3/4 in = 2 in - 3/4 in = 8/4 in - 3/4 in = 5/4 in

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This is how you solve it..
95/n=n
You could put 5 in and get 95/5=19.
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Lola saved $86.25 in 3 months. She saved $30.75 in the first month and $7.99 in the second month. How much money did Lola save i
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$47.51

Step-by-step explanation:

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2 years ago
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A tree that is 3 feet tall is growing at a rate of 32% each year. How tall will it be in 3 years?
pashok25 [27]

Answer: 5.88 feet tall

Step-by-step explanation:

Using this formula

Interest =principal×rate×time/100

Principal is the actual growth=3ft

Rate is the 32%

Time is 3yrs

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Interest =2.88

To get the value for 3yrs,add the interest to the principal value

Exact value in 3yrs=

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4 0
3 years ago
6x – 2y = 10 2x + 3y = 51 solving the first equation above for y gives:
Masja [62]
The first equation is 6x - 2y = 10. To solve for y, you will use inverse (opposite) operations to undo what is happening to y.  Please see the steps below for the work.

6x - 2y = 10
-6x          -6x
<u>-2y </u>= <u>(-6x + 10)</u>
-2            -2
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6 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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