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The Mathematics of The World Cup

June19

soccerEvery four years, The World Cup (the second biggest sporting event) forces fans to remember their mathematics lessons. In fact, working out what each team needs from its final match to finish in the top two of a group and advance to the knockout rounds takes some algebra knowledge and powers of prediction.

After Brazil and Mexico played to a scoreless draw last Tuesday, the calculation became clear: both teams just need to draw in their next matches to advance with five points in Group A. Croatia, which beat Cameroon Wednesday, would get to six points by beating Mexico. So a draw with Cameroon would still get Brazil through with five points. If Mexico beats Croatia, Brazil would advance even if it loses. But if Mexico and Croatia draw, and Brazil loses — then it gets complicated with tiebreakers. Got that, or do you need a calculator and a notepad? For most of the other teams in the tournament, the calculations are almost endless.Read more here.

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NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

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