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2 years ago

Question 3

Mathematics

1 answer:

horrorfan [7]2 years ago

8 0

Answer:

Second one:

AAS

Step-by-step explanation:

searched it

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Steven is buying a new mountain bike on layaway for $324. If he pays $36 each week how many weeks will it take steven to pay for

Reptile [31]

Answer:

9 weeks

Step-by-step explanation:

324/36 = 9

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2 years ago

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Which of the following gives a product of x^3 - 3x ?

Vilka [71]

Answer:

Option a

Step-by-step explanation:

Lets verify

RHS

$\\ \rm\longmapsto x(x^2-3)$

Use distributive law
a(b-c)=ab-bc

$\\ \rm\longmapsto x(x^2)-3x$

$\\ \rm\longmapsto x^3-3x$

$\\ \rm\longmapsto LHS$

Hence verifed

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2 years ago

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Find a coterminal angle between 0° and 360°.<br> 780

erastovalidia [21]

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2 years ago

Can someone please help

ASHA 777 [7]

Answer:

They cost 730 all together.

Step-by-step explanation:

One the surface area of the hall is 310m^2.

so, 310 divided by 25 is 12.4, which means they need 13 tin of paint.

The cost for it is 13*10 which is 130.

For the area of the floor, it is 40*15, which is 600m^2.

Which costs 600.

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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