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3 years ago

7

Solve. If there are X teams in a sports league and all the teams play each other twice, a total of N(x) gamers are played, where

N(x)=x^2-x. A soccer league has 10 teams, each of which plays the others twice. If the league pays $46 per game for the field and officials, how much will it cost to play the entire schedule?

2 answers:

s2008m [1.1K]3 years ago

6 0

Answer:

$\$4140$

Step-by-step explanation:

According to the problem, <em>x </em>refers to teams, and $N(x)=x^{2}-x$ represents numbers of games played.

So, if there are 10 teams, we just have to replace this value for the <em>x</em> and calculate the number of games. Once we have it, we multiply all games by $46, which is the cost per game.

$N(x)=x^{2} -x\\N(10)=(10)^{2}-10=100-10=90$

This means that in the entire schedule there are 90 games. If each one costs $46, the whole schedule will cost:

$90(\$46)=\$4140$

WINSTONCH [101]3 years ago

4 0

number of games is 90, so cost is

<span>90×46=4140</span>

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