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IrinaK [193]
2 years ago
12

A printer can print 12 pages in 9 seconds. What is the closest estimate of the

Mathematics
2 answers:
vivado [14]2 years ago
8 0

Answer:

80

Step-by-step explanation:

12/9 is the amount of pages it can print in one second, which equals 1 1/3.

multiply that number by 60 (60 seconds in a minute) and there is your answer.

kolbaska11 [484]2 years ago
7 0

Answer:

72

Step-by-step explanation:

there are 60 seconds in a minute so about 6 9 seconds in a minute. then there is 12 pages per 9 seconds and since there is 6 9 seconds then you can just multiply 6 by 9 to get the answer 72

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Answer: B. 15

Step-by-step explanation: The number on the bottom of the fraction is always the denominator, therefore 15 is the denominator in the fraction.

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SYSTEM OF EQUATION WITH SUBSTITUTION
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Step-by-step explanation:

X=7

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. If 3n + 8 = 44, what is the value of n?​
algol13

Answer:

3n + 8 = 44

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2 years ago
Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.
andriy [413]

Answer:

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Step-by-step explanation:

We have been given a indefinite integral \int \frac{2}{3x\left(3x-5\right)}dx. We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}

\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx

\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx

Using difference rule of integrals, we will get:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)

Now, we need to use u-substitution as:

Let u=3x-5.

\frac{du}{dx}=3

dx=\frac{1}{3}du

\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|

\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|

Substitute back these values:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)

Let us add a constant C.

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Therefore, our required integral would be \frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C.

5 0
3 years ago
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