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4 years ago

Lot the points (0, −2) and (4, 1) on the coordinate plane.

Mathematics

1 answer:

S_A_V [24]4 years ago

6 0

$(0,\ -2)\to x=0,\ y=-2$

from orgin (0, 0):

0 units left-right

2 units down

$(4,\ 1)\to x=4,\ y=1$

4 units right

1 unit up

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What’s the answer to this

madam [21]

B is the correct answer I think

4 0

3 years ago

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Find a third degree polynomial function of the lowest degree that has the zeros below and whose leading coefficient is one.

Tanzania [10]

Answer:

The polynomial function of the lowest degree that has zeroes at -1, 0 and 6 and with a leading coefficient of one is $p(x) = x^{3}-5\cdot x^{2}-6\cdot x$ .

Step-by-step explanation:

From Fundamental Theorem of Algebra, we remember that the degree of the polynomials determine the number of roots within. Since we know three roots, then the factorized form of the polynomial function with the lowest degree is:

$p(x) = (x-r_{1})\cdot (x-r_{2})\cdot (x-r_{3}) = 0$ (1)

Where $r_{1}$ , $r_{2}$ and $r_{3}$ are the roots of the polynomial.

If we know that $r_{1} = -1$ , $r_{2} = 0$ and $r_{3} = 6$ , then the polynomial function in factorized form is:

$p(x) = (x+1)\cdot x \cdot (x-6)$ (2)

And by Algebra we get the standard form of the function:

$p(x) = x\cdot (x+1)\cdot (x-6)$

$p(x) = x\cdot (x^{2}-5\cdot x -6)$

$p(x) = x^{3}-5\cdot x^{2}-6\cdot x$ (3)

The polynomial function of the lowest degree that has zeroes at -1, 0 and 6 and with a leading coefficient of one is $p(x) = x^{3}-5\cdot x^{2}-6\cdot x$ .

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3 years ago

What is the value of fraction 1 over 3x2 + 5.2y when x = 3 and y = 2?

LenaWriter [7]

$\dfrac{1}{3}x^2+5.2y\\\\for\ x=3\ and\ y=2\\\\\dfrac{1}{3}\cdot3^2+5.2\cdot2=\dfrac{1}{3}\cdot9+10.4=\dfrac{9}{3}+10.4=3+10.4=13.4$

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3 years ago

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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:

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