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3 years ago

PLEASE HELP!!!

1 answer:

4 0

EXPLANATION: By definition two lines are said to be parallel when the distance between two lines is same when measured from any point on one line.

However the criteria for intersection is that the distance between two lines should change and eventually should change to zero at intersection point.

Step-by-step explanation:

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Does anybody know the answer

denis-greek [22]

Answer:

hypotenuse aka the longest line on a right angle

7 0

3 years ago

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the weight of a nickel is 80% of the weight of a quarter.if a nickel weighs 5grams how much does a quarter weigh?

Rasek [7]

Just divide 100% by 80% and multiply that answer by 5. I'm pretty sure that's how you do it.

8 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

3 years ago

Determine if the two triangles are congruent. if they are, state how you know.

kicyunya [14]

Yea they are cuz they have the same side and angle measurements

8 0

2 years ago

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What is the sum of -6 and 9

katrin2010 [14]

Positive 3, If You're starting at -6 and you add the first 6 it's at 0, now you have 3 more to add to make 9. And that gives you positive 3. or if you do 9 - 6 you can get your answer that way too.

7 0

3 years ago