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Katen [24]
3 years ago
9

55 is 40% of what number?

Mathematics
2 answers:
Vika [28.1K]3 years ago
6 0
Assume that 40 percent of X is 55 
<span>This means X (40 / 100) = 55 </span>
<span>0.4 X = 55 </span>
<span>X = 55 / 0.4 </span>
<span>X = 137.5 </span>
<span>55 is 40 percent of 137.5</span>
galina1969 [7]3 years ago
4 0
Simple just do 55/0.40 and that gets you 137.5
so 40% of 137.5=55 here is the proof 137.5x0.40=55

Answer is 137.5
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Answer:

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Step-by-step explanation:

Area is found by mulitplying width and height.

A=7.5(5)\\

A=37.5\\in

The ratio is 0.5in : 1ft, so if you multiply that ratio to the answer, you should get the answer in feet.

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2 years ago
A pony keg (quarter barrel) of beer contains 62 pints.  If the local drinking establishment sells draft beer in 300 mL pilsner g
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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)

\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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