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DochEvi [55]
2 years ago
11

A rectangular athletic field is twice as long as it is wide. If the perimeter of the athletic field is 90 yards, what are its di

mensions?
Mathematics
1 answer:
Virty [35]2 years ago
6 0

Answer:

L=30 W=15

Step-by-step explanation:

90 divided by 6 is 15. 15 is the Width. Multiply 15 by 2 and you get 30 which is the Length.

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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
PLEASE HELP, MARKING BRAINLIEST!!!
adell [148]
3y = y + 50
2y = 50
y = 25

ANGLE B
3(25)
75

ANGLE G
25 + 50
75
8 0
3 years ago
Does anyone know how to do this? i need help please
Veronika [31]
The answer will be c
5 0
3 years ago
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Three points, A, B, and C exists in space such that B is "between" A and C. It is known that AB=7, BC=4, and AC=9. Are points A,
EleoNora [17]

If the points are collinear, then that would mean that the points lie on the same line when joined to each other. In order to confirm this, the distances between the points must sum up to the length of the line.<span>

AB + BC = AC
7 + 4 ? 9
11 ≠ 9

<span>Since they do not sum up, therefore, the points are not collinear.</span></span>

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2 years ago
Graph the line with the equation FASTTTT PLSSSS
Ahat [919]

Answer:

the graph

Step-by-step explanation:

hi there hope this helps

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