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EastWind [94]
3 years ago
13

Does anyone know how to do these word problems?

Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0

What is your grade level and which curriculum are you using?



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If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha
castortr0y [4]
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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3 years ago
Write an inequality that represents the situation below:
stellarik [79]

Answer:

4 ( x - 2) > 5

Step-by-step explanation:

I think that's the right answer

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F(x+h)-f(x)/h<br> f(x) = 2x 2 + 7x
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174992774 there you go chicken
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Expand "6y" in algebra​
fenix001 [56]

Answer:

6 x y

Step-by-step explanation:

Expanded

Or add y 6 times

y + y + y + y + y + y

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HELP ASAP 30 POINTS WORTH !!!!! OωO
Len [333]

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A because (0,-6) is the y-intercept so you start at the point you know. Then because slope is a rise over run fraction the slope can also be written as 2/1 which is rise 2 over 1.

Step-by-step explanation:

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