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2 years ago

A rectangular patio has the length of 12 1/2 feet in an area of 103 and 1/8 ft.what is the width of the patio

Mathematics

1 answer:

Scrat [10]2 years ago

7 0

A=LW
103 1/8=12 1/2W
103 1/8 divided by 12 1/2
8 1/4

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Xan and Katie are twins. Xan weighs 66 6 6 pounds 99 9 9 ounces, and Katie weighs 88 8 8 pounds 11 1 1 ounce. What is the differ

gtnhenbr [62]

Answer:

264 ounces

Step-by-step explanation:

#Convert Xan weight into ounces:

$1lb= 16 \ oz\\\\66\ lb=66\times 16=1056 \ oz\\\\\therefore Xan's=1056 \ oz+ 99 \ oz=1155 \ oz$

Convert Katie's 88 pounds 11 ounce to ounces:

$1lb= 16 \ oz\\\\88\ lb=88\times 16=1056 \ oz\\\\\therefore Katie's=1408 \ oz+ 11 \ oz=1419\ oz$

The difference is calculated by subtracting Xan's from Katie's:

$=1419 \ oz-1155 \ oz\\\\=264\ oz$

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2 years ago

Which of the following is the linear parent function?A.F(x) = 2xB.F(x) = |x|C.F(x) = x2D.F(x) = x

Daniel [21]

The linear parent function is f(x)=x

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2 years ago

A picture representing 24÷6

Licemer1 [7]

Answer:

Step-by-step explanation:

24÷6

6 x ? = 24

6 goes into 24 four times

24÷6=4

7 0

2 years ago

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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

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

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2 years ago

Excluding any outliers, what is the mean of the data set {21, 14, 11, 26, 33, 61, 29}?

sashaice [31]

Answer: 22

Step-by-step explanation:

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2 years ago

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