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lianna [129]
3 years ago
13

PLEASE HELP IVE BEEN WORKING ON THIS FOR 76 HOURS I WILL MARK BRAINIEST PLEASE HELP I AM SO DESPERATE:A student is doing a readi

ng assignment. After 41 minutes of reading, she skims the pages ahead and estimates that she still has 53% of the reading to do. According to the girl's estimate, how long is the total reading assignment? Round to the nearest minute.
Mathematics
2 answers:
RUDIKE [14]3 years ago
7 0

Answer:

87 minutes.

Step-by-step explanation:

to get the percentage of the read part, do 100% - 53% = 47%

41min = 47%

  ?        = 100%

cross multiply;

41 × 100 ÷ 47 = 87.2340...

round off to get 87 min

notka56 [123]3 years ago
5 0

Answer:

87 mins

Step-by-step explanation:

The percentage of pages she read = 100-53= 47%

47%------------41 mins

53% ------------- x

x= (41×53)/47

x= 46.23 mins

nearest minute = 46 mins

Total reading assignments is 41+46= 87 mins

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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
Please help i know this is kind of easy but I am having a hard time with it unfortunately​
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C

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y= 60(2)+10

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y=65(2)

y=130

8 0
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Whats the question here
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Answer:

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Step-by-step explanation:

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A is the area of the surface perpendicular to the force

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So 53% are tigers.

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