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kupik [55]
3 years ago
11

At a soccer tournament 12 teams are wearing red shirts, 6 teams are wearing blue shirts, 444 teams are wearing orange shirts, an

d 2 teams are wearing white shirts
Mathematics
1 answer:
salantis [7]3 years ago
6 0

Answer:

red

Step-by-step explanation:

The complete question is as follows;

At a soccer tournament 12 teams are wearing red shirts, 6 teams are wearing blue shirts, 4 teams are wearing orange shirts, and 2 teams are wearing white shirts. For every 2 teams at the tournament, there is 1 team wearing ____ shirts

Solution;

The total number of teams we have is 12 + 6 + 4 + 2 = 24 teams

Now looking at the number of teams wearing red shirts , we can see 12 of them which is exactly half the total number of teams

So what this means is that for every two teams at the tournament, there is 1 team wearing red shirts

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The radius of a circle can be found by dividing the circumference of the circle by 6.28. What is the radius of a circle that has
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5 0
3 years ago
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)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

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}

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

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

4 0
3 years ago
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