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3 years ago

What is 67,396 divided by 123

Mathematics

2 answers:

melomori [17]3 years ago

8 0

When you round it up you'll find 548

Vikki [24]3 years ago

7 0

547.935 I calculated, so I think and sure this is correct answer...

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BASED ON THE GRAPH, WHICH STATEMENT IS CORRECT?

abruzzese [7]

Answer:

i would go for a but im not really sure. sorry if incorrect

Step-by-step explanation:

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2 years ago

The vertex angle of an isosceles triangle measure 40.whats the measure of the base angle?

Eddi Din [679]

I don't know if my answer is completely correct, but I think it's 100. All triangles add up to 180, and an isosceles triangle has 2 equal sides. 40 would be the 2 sides, and 180 subtracted from 80 equals 100.

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3 years ago

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Do we have to know all the answers to the problems

musickatia [10]

Answer: Not quite

Step-by-step explanation:

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3 years ago

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mamaluj [8]

y = mx+b where m is the slope and b is the y intercept

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3 years ago

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

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3 years ago