Ask question

Ask question

All categories

3 years ago

15

Can someone answer this??

1 answer:

Yuri [45]3 years ago

8 0

Answer:

B. -0.4 ÷ 0.25

I hope it helps..

You might be interested in

Ms Wilson draws a model

Korolek [52]

I need the model and the rest of the question to help you ;-;

7 0

3 years ago

.divide the problem.

quester [9]

Answer:

$x - 2x - 15 \: divided \: by \: x - 3$

according to the above x²-2x-15 divided by the number below x-3 first I cancel out the x
$x - 2x - 15 \: divided \: by \: - 3$
the we solve the next x-5 divided by x²-9 then as usual rule out the x
$- 5 \: divided \: by \: x - 9$
now we only have x-2x-15 divided by-3 divided by-5÷x-9 now our next step is to again single the equations out
$x - 2x - 15 \: divided \: by \: - 3 \: lets \: rue \\s \: rule \: out \: - 3 \: and \: 15 \: giving \: us \: x - 2x - 5$
now the next -5 divided by x-9 here there's nothing to rule out therefore the sign "÷" is changed to "×" cancel out the -5 and therefore we have
$x - 2x = x - 9 \: as \: final \: answer$

6 0

2 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

Will make brainily Help please due soon

aksik [14]

Answer:

my wattapp number

is 7667587493

Step-by-step explanation:

I think the answer is 8

7 0

3 years ago

Find the equivalent fraction of 1/3 equals

kolezko [41]

Equivalent Fractions for 1/3:

1/3, 2/6, 3/9, 4/12, 5/15, 6/18, 7/21, 8/24, 9/27, 10/30,

6 0

4 years ago

Read 2 more answers