<h2>
<u>Sol</u><u>ution</u><u>:</u></h2>
Equation: x² + 10x + 21
<u>Step</u><u> </u><u>1</u><u>:</u> Find two numbers that can add up to 10 and be multiplied to 21. We have: 7 & 3, in the sense that 7+3=10, and 7×3=21. Replacing 10 with 7+3, the equation is now → x² + 7x + 3x + 21
<u>Step</u><u> </u><u>2</u><u>:</u> Get the new equation bracketed → (x² + 7x) (+3x + 21)
<u>Step</u><u> </u><u>3</u><u>:</u> Use 'x' in the equation. For the first part, we have 'x'. x² = x × x so, bring out one x out side the bracket, divide 7x by = 7 → x (x +7). Do the same for the second part by dividing 21 by 3 = 7, and then bringing out 3 from the bracket → 3 (x + 7).
Bringing everything together, we have: x(x+7) +3(x+7) → (x+3) (x+7)
<h3>
<u>Final</u><u> </u><u>ans</u><u>wer</u><u>:</u></h3>
(x+3) (x+7)
<h3 />
Answer:
x = 8.69
Step-by-step explanation:
we know that the perimeter of the dodecagon is 54, so each edge will be 54/12
54/12 = 4.5 cm
if we draw the lines to remove 6 vertices and form a hexagon, 6 triangles with 2 sides of 4.5 cm are formed.
we know that the angle of each vertex is 150 ° because it is a dodecagon
if we apply the law of cosines we can take the other side of the triangle, since we only need 2 side and the opposite angle to the side we want to know
a would be our x
b = 4.5
c = 4.5
A = 150°
a^2 = b^2 + c^2 - 2bc * cos (A)
x^2 = 4.5^2 + 4.5^2 - 2 * 4.5 * 4.5 * cos (150)
x^2 = 20.25 + 20.25 - 40.50 * -0.866
x^2 = 40.50 + 35.07
x = √ 75.57
x = 8.69
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]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D)
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 \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%20-%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%3D0%20%5C%5C%5C%5C)
![\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\\\\](https://tex.z-dn.net/?f=%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B-%5Clambda%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%20-0%5D%20-%200%20-%20%5Cfrac%7B1%7D%7B4%7D%5B0-%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B%28%5Cfrac%7B%5Clambda%7D%7B2%7D%20%2B%20%5Clambda%5E2%20%29%5D%20-%20%5Cfrac%7B1%7D%7B4%7D%5B%5Cfrac%7B%5Clambda%7D%7B2%7D%20-%20%20%5Cfrac%7B1%7D%7B4%7D%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B8%7D%5Clambda%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%5Clambda%5E2%20-%20%5Cfrac%7B%5Clambda%5E2%7D%7B2%7D%20-%20%5Clambda%5E3%20-%20%5Cfrac%7B%5Clambda%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B16%7D%3D0%20%5C%5C%5C%5C%5Cto%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%28%5Clambda%20-%5Cfrac%7B1%7D%7B4%7D%29%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%3D0%5C%5C%5C%5C)


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] \\\\](https://tex.z-dn.net/?f=%5Cto%20c%5E%7B%28k%2B1%29%7D%20%3D%20A%20x%5E%7Bk%7D%2BC%20%5C%5C%5C%5C%5Cto%20x%280%29%20%3D%20%20%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D3%261%262%5Cend%7Barray%7D%5Cright%29%20%20%2C%20c%20%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%29%5C%5C%5C%5C%20%20%5Cto%20x%5E%7B%28k%2B1%29%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D%20x%5Ek%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%5C%5C)
after solving the value the answer is
:
![\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Clim_%7Bk%20%5Cto%20%5Cinfty%7D%20x%5Ek%3Do%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%260%260%5Cend%7Barray%7D%5Cright%5D)
Answer:
<u>k = 220</u>
Step-by-step explanation:
<u>Proportion of y and x</u>
<u>Given</u> :
<u>Solving</u> :
- 55 = k/4
- k = 55 x 4
- <u>k = 220</u>