Ask question

Ask question

All categories

bagirrra123 [75]

2 years ago

7

How far is the Water Slide from the Restrooms?

2 answers:

snow_tiger [21]2 years ago

4 0

Answer:

4,-3

Step-by-step explanation:

anygoal [31]2 years ago

3 0

Answer:

i think it is 7 bc from the restroom its 7

Step-by-step explanation:

You might be interested in

Can the sides of a triangle have length 1,3 and 5

Katen [24]

Answer:

No.

Step-by-step explanation:

The two smallest sides must be greater than the biggest side when added up in order to form a complete triangle.

1+3 is 4, which is less than five, so the triangle wouldn't be connected.

Essentially, imagine a triangle where one line is too short to reach the other two, which is what was just explained--the smallest sides must add up to be greater than the biggest side in order to form a complete triangle.

8 0

2 years ago

at a local soccer game, 6/7 of the people were fans. one-half of those fans were wearing hats. what fraction of the people was w

Brut [27]

Answer: $\frac{3}{7}$

You simply need to multiply $\frac{6}{7}*\frac{1}{2}$ .
Multiply the numerators.
$6*1=6$
Multiply the denominators.
$7*2=14$
Simplify. $\frac{6}{14}*\frac{3}{7}$

7 0

3 years ago

Who wants free gram.m.arl.y pre.mi.um acco.unt

Damm [24]

Don’t fall for this it’s fake

7 0

2 years ago

Plssss help meeee!!!

pogonyaev

The answer is c b/c you multiply

3 0

2 years ago

Read 2 more answers

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

2 years ago