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Novay_Z [31]
3 years ago
8

Which of the following is 10 times what the 5 represents in 8,145

Mathematics
2 answers:
Pepsi [2]3 years ago
8 0
A. 50

5 in ones place should be multiply by 10
5x10
50
Alexxx [7]3 years ago
7 0

Answer:

A

Step-by-step explanation:

5 in 8145 is in the ones place. Hence 10 times of that

= 10 × 5

= 50

Hence answer is <u>A</u><u>.</u>

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Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t
Anika [276]

Answer:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

Step-by-step explanation:

Given

\int\limits {\frac{x}{x^4 + 16}} \, dx

Required

Solve

Let

u = \frac{x^2}{4}

Differentiate

du = 2 * \frac{x^{2-1}}{4}\ dx

du = 2 * \frac{x}{4}\ dx

du = \frac{x}{2}\ dx

Make dx the subject

dx = \frac{2}{x}\ du

The given integral becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

Recall that: u = \frac{x^2}{4}

Make x^2 the subject

x^2= 4u

Square both sides

x^4= (4u)^2

x^4= 16u^2

Substitute 16u^2 for x^4 in \int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du

Simplify

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

In standard integration

\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)

So, the expression becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)

Recall that: u = \frac{x^2}{4}

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

4 0
3 years ago
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