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3 years ago

On a school bus, the teacher sitting in front of me asked if I passed gas -- why did she ask me this and why did she WHISPER it

2 answers:

4 0

She probably asked you if you passed gas because she smelled something that stink and thought they since u were clos3r to her that it was you. she could hav3 done it, or another student, it all depends. this question was do funny lol. and sh3 whispered it becaus3 she didnt want to possibly embarrass u. anyways ciao

lawyer [7]3 years ago

3 0

She probably did not want to embarrass you in front of the rest of the students.

You might be interested in

Which expressions have the same solution as –4(6)?

gogolik [260]

Answer:

-24

Step-by-step explanation:

8 0

3 years ago

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

PLEASE HELP!!<br><br> Order the following fractions from least to greatest: 3/8, 3/4, 5/16, 1/2

Lynna [10]

The answer is 5/16, 3/8, 1/2, 3/4

3 0

3 years ago

Read 2 more answers

I will give u brainliest

Ray Of Light [21]

Answer:

17.549

is the answer to this

5 0

3 years ago

Read 2 more answers

Please help I will mark brainliest!

Ymorist [56]

Answer:

Step-by-step explanation:

You know that the negative number with the largest magnitude is farthest to the left on the number line.

3/4 = 9/12

2/3 = 8/12

Then 3/4 > 2/3, so ...

-3/4 < -2/3

7 0

3 years ago