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2 years ago

Write an expression in factored form to represent the total area of each rectangle

2 answers:

7 0

The area of any rectangle is

(the rectangle's length) times (the rectangle's width) .

If you have one or more specific rectangles to be discussed,
you'll have to reveal them.

natali 33 [55]2 years ago

4 0

L time W
L= length
W=width so LxW

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Which two integers is 42 between?<br> -8 and -7<br> -5 and -4<br> -7 and -6<br> -9 and -8

Karo-lina-s [1.5K]

Answer:

6 and 7

Step-by-step explanation:

we know that 42 will lie between 36 and 49

And 36=6; 49=7

so we can be sure that 42 lies between 6 and 7

3 0

2 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

2 years ago

PLEASE HELP!!! Jake cleaned up his room. He organized his toys into boxes. He put 7 toys in each of 3 bins.If t represents the t

timama [110]

Ok well..
7 (3) = x
meaning 7 times 3= 21 toys altogether! :)

7 0

2 years ago

Read 2 more answers

Which pair of numbers, if included in this set, would not change the median?

sattari [20]

Answer:

Step-by-step explanation:

Order the set first

27, 28, 35, 37, 43, 47

Median now is: 36 (which is the mean between 35 and 37, the two central numbers )

If you add A) 35,50 then you will still have to compute the mean between 35 and 37 in order to find the median, because you have:

27,28,35,35,37,43,47,50

7 0

3 years ago

Read 2 more answers

√24 is between which 2 numbers? Question 6 options: 1 and 2 2 and 3 3 and 4 4 and 5

Goryan [66]

The answer is: between 4 and 5

The way I figured this answer was by simply finding the square of the numbers

$1^{2} = 1$
$2^{2} =4$
$3^{2} =9$
$4^{2} =16$
$5^{2} =25$

Since 24 is between 16 and 25 then that has to be your answer.

6 0

2 years ago

Read 2 more answers