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2 years ago

Please help will give brainliest

Mathematics

1 answer:

mylen [45]2 years ago

8 0

3 ,4,5 .....................

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I dont get this one someone help me

worty [1.4K]

You have to set up a ratio and solve

5 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

2 years ago

Use the discriminant to determine the number of real solutions to the equation. −4x^2=−8−10x

ICE Princess25 [194]

Answer:

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Fill in the missing digits to find the base 2 representation of 59.

Anna007 [38]

Answer:

111011

Step-by-step explanation:

Following the binary rule we can find the base 2 presentation of the decimal number 59.

To find the binary equivalence of 59 we use the sum of powers of 2.

$2^{0}=1$

$2^{1}=2$

$2^{2}=4$

$2^{3}=8$

$2^{4}=16$

$2^{5}=32$

$2^{6}=64$

Now we take our number and find out what the binary number will by taking our largest number closest to the number first.

59 = 32

We chose the number 32 since 64 will be a larger value than 59.

We then check how much we have to add to 32 to get 59.

59 = 32 + 27

We then look for the closest number to 27 in our powers of 2.

59 = 32 + 16

Now we check again for how much we need left to get a total of 59.

59 = 32 + 16 + 11

Now we repeat the same process of finding which value in the powers of 2 are closest to the number.

59 = 32 + 16 + 8 + 3

59 = 32 + 16 + 8 + 2 + 1

Now since we already have a total of 59, our binary number will be all the numbers present will have a value of 1 and the numbers now used will have a number of 0.

32 16 8 4 2 1

This can also be represented as:

2^5 2^4 2^3 2^1 2^0

Now we have to include the numbers that we skipped to get the total binary number.

32 16 8 4 2 1

1 1 1 0 1 1

This can be represented as:

59 = 32 16 + 8 + 0 + 2 + 1

1 1 1 0 1 1

3 0

2 years ago

Read 2 more answers

What is the y-intercept of the equation y=-x?

Aleks [24]

Answer:

0,0

Step-by-step explanation:

6 0

2 years ago