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gayaneshka [121]
2 years ago
6

E) Over a season a football team won, drew and lost matches in the ratio 2:1:5.

Mathematics
1 answer:
IRISSAK [1]2 years ago
8 0

Answer:  4 ties or draws

8 wins, 4 ties, 20 losses

===========================================================

Explanation:

The ratio 2:1:5 is of the format wins:ties:losses where getting a draw is the same as getting a tie.

So one possibility is that they won 2 matches, tied once, and lost 5 matches. The only problem is that it doesn't fit with the statement that "The team lost 12 more matches than they won."

We could guess-and-check our way to the answer, but I think using algebra is the most efficient route.

----------

Let x = number of ties

2x = number of wins and 5x = number of losses

Note the ratio 2x:x:5x reduces to 2:1:5 after dividing everything by x.

Since the team lost 12 more matches than they won, this means,

loss count = (win count) + 12

5x = (2x) + 12

5x-2x = 12

3x = 12

x = 12/3

x = 4

The team tied 4 times and has 2x = 2*4 = 8 wins and 5x = 5*4 = 20 losses.

The ratio 8:4:20 reduces to 2:1:5 when you divide all parts by the GCF 4.

Also, the gap between 8 wins and 20 losses is 20-8 = 12 to help fully confirm we have the correct answer.

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<em>Check: </em>

(1) |2(7) + 3| - 6 = 11                     (2) |2(-10) + 3| - 6 = 11

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What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

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so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

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Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

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