Question 6

Passing through (-2,-6) and parallel to the line whose equation is y=-3x+4

Licemer1 [7]

Answer:

y = -3x - 12

Step-by-step explanation:

Lines that are parallel have the same gradient.

Therefore using the point (-2,-6) you can make your own equation.

-6 will be y, -2 will be x

All you have to do is find the y-intercept which is C in y=mx+c

-6 = -3(-2) + c

-6 = 6 + c

-12 = c

Therefore;

y = -3x - 12

6 0

3 years ago

Question 1 Which choice shows a 90° clockwise rotation of the figure about the point P?

Arada [10]

The third one is correct.....so go for it without hesitation

6 0

3 years ago

What is one and one fourth times 30

n200080 [17]

1 1/4*30=37.5 because, first to make this easier to work with I like to turn the fraction into a decimal like so, 1/4=.25 because 1/4=.25 so 1/4=.25, so now we have 1.25*30=? What this is pretty much saying is that 30+(25% of 30)=? Which to solve this we just need to find how much 25% is of 30 which is .25*30=7.5 because 7.5*4=30 making it true that this is 25% of 30. Now what we have to do is take 30+7.5 (because we have a 1 infront of the 25% we add this instead of subtract it, if we had no whole number infront of the 30 and where solving this that way we would have to subtract instead of add) which adds up to be are end answer of 37.5.

Hope that this helps and makes sense, Enjoy!=)

6 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:

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