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Gnesinka [82]
2 years ago
6

Given :

Mathematics
1 answer:
pashok25 [27]2 years ago
8 0

\\ \sf\longmapsto A=(1-4)-(2+1)

\\ \sf\longmapsto A=-3-3

\\ \sf\longmapsto A=-6

\\ \sf\longmapsto B=(3-\dfrac{1}{2})-(2-4)

\\ \sf\longmapsto B=\left(\dfrac{5}{2}\right)+2

\\ \sf\longmapsto B=\dfrac{5+4}{2}

\\ \sf\longmapsto B=\dfrac{9}{2}

Now

\\ \sf\longmapsto AB=-6\times \dfrac{9}{2}

\\ \sf\longmapsto AB=-3(9)

\\ \sf\longmapsto AB=-27

BA

\\ \sf\longmapsto \dfrac{9}{2}\times (-6)

\\ \sf\longmapsto 9(-3)

\\ \sf\longmapsto -27

Observed:-

\\ \sf\longmapsto AB=BA

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y = \frac{3}{4} + x ----> Swap x and y

x = \frac{3}{4} + x -----> Subtract 3/4 from both sides.

x - \frac{3}{4} = y ----> rearrange to appropriate order

y = x - \frac{3}{4}


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3 years ago
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Answer:

Step-by-step explanation:

\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\

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