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3 years ago

Explain the process of finding the product of two integers

2 answers:

4 0

If you have two negatives they cancel each other out and make a positive. The same if you have to positives the product will be positive. If you have a negative and a positive you multiply the numbers and bring down the negative. Same if you have a positive and a negative. Only two negatives cancel each other out, same with positives. For example -4(-5)(-1) you would use order of operations, 4 and the 5 cancel each other out and you multiply 20 by -1 which is -20 since you bring down the negative.

slava [35]3 years ago

4 0

If the integers have two different signs, the answer will be negative. If they have the same signs, the answer is positive. And a example being: -5(3)= -15. -5(-6)=30 8(7)=56.

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Rina8888 [55]

Answer:

$\sf 1092 \: dm^2$

Refer to the attachment for explication of steps. :)

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2 years ago

What is the circumference of a circle whose area is 121 pi square meters?

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3 years ago

What is the slope of a line that is perpendicular to the graph of y= 5x + 5?

AVprozaik [17]

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3 years ago

A line passes through (−1, 7) and (2, 10).

Pachacha [2.7K]

The slope m of a line through points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by :

$\displaystyle{m = \frac{y_2-y_1}{x_2-x_1}
$

Thus, the slope of the line passing through points (−1, 7) and (2, 10) is

 $\displaystyle{m= \frac{10-7}{2-(-1)}= \frac{3}{3}=1$


The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).

We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):

y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8

Answer: C. −x+y=8

7 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