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2 years ago

PLS HELP ILL GIVE BRAINLIEST

Mathematics

1 answer:

inysia [295]2 years ago

6 0

<h3>Answer: 14x - 8</h3>

=======================================================

Explanation:

I'll use the quadratic formula to find the roots or x intercepts. This slight detour allows us to factor without having to use guess-and-check methods.

The equation is of the form ax^2+bx+c = 0

a = 12
b = -11
c = -5

This leads to...

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-11)\pm\sqrt{(-11)^2-4(12)(-5)}}{2(12)}\\\\x = \frac{11\pm\sqrt{361}}{24}\\\\x = \frac{11\pm19}{24}\\\\x = \frac{11+19}{24} \ \text{ or } \ x = \frac{11-19}{24}\\\\x = \frac{30}{24} \ \text{ or } \ x = \frac{-8}{24}\\\\x = \frac{5}{4} \ \text{ or } \ x = -\frac{1}{3}$

Now use those roots to form these steps

$x = \frac{5}{4} \ \text{ or } \ x = -\frac{1}{3}\\\\4x = 5 \ \text{ or } \ 3x = -1\\\\4x - 5 =0 \ \text{ or } \ 3x+1 = 0\\\\(4x-5)(3x+1) = 0$

Refer to the zero product property for more info.

Therefore, the original expression factors fully to (4x-5)(3x+1)

Use the FOIL rule to expand it out and you should get 12x^2-11x-5 again.

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We did that factoring so we could find the side lengths of the rectangle.

I'm using the fact that area = length*width

L = length = 4x-5
W = width = 3x+1

The order of length and width doesn't matter.

From here, we can then compute the perimeter of the rectangle

P = 2(L+W)

P = 2(4x-5+3x+1)

P = 2(7x-4)

P = 14x - 8

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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:

$\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\$