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mixer [17]
3 years ago
13

If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha

Mathematics
1 answer:
castortr0y [4]3 years ago
7 0
6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}


use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2

6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}
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Answer:

You can do this by rolling dice 4 times

Step-by-step explanation:

If you roll a fair dice 4 times and record the findings and their orders, you can generate 6^4= 1296 possible outcomes. You can pick 1-1000 as numbers and then re-roll all the dice if the number generate 1001-1296  

To translate the dice result into a number you have to multiply the dice result as following:

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Lorico [155]

Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

<em>Is the transformation a rigid motion?</em>

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

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Answer:

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Step-by-step explanation:

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