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2 years ago

What expression is equivalent to \root(4)(x^(10))

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

3 0

If this is your expression → $\sqrt{4x^{10} }$
Then your answer is → $2|x|^5$ lower math courses will accept $2x^5$

You might be interested in

. I. The graphs of all linear relationships are a line.

aksik [14]

Answer:

Step-by-step explanation:

The graph of a linear relationship consists of the x and y values. The y values are plotted on the vertical axis against the x values on the horizontal axis. A change in the values of x lead to a corresponding change in the value of y. The correct options are

. I. The graphs of all linear relationships are a line. They are sometimes called straight line graphs

III. The tables of all linear relationships have a constant rate of change. This represents the slope

The true statement is

I and III only

5 0

2 years ago

Which of the following equations has a solution that is negative?

Serga [27]

Answer:

-10(-h-9)=18

Step-by-step explanation:

-10(-h-9)=18

10h+90=18

10h=-72

h=-7.2

5 0

2 years ago

Read 2 more answers

Find $\int \dfrac{x}{\sqrt{1-x^4}}$ Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

6 0

2 years ago

What is the quotient a-3/7÷3-a/21

Airida [17]

I wish you were using parentheses here. I must assume (without knowing for sure) that y ou mean

a - 3/7
----------
3 - a/21

The LCD here is 21, since 7 divides into 21 without a remainder.

Thus, multiply numerator and denominator by 21:

21a - 9
-----------
63 - a

You could leave this result as is, or you could factor out 3:

7a-3
3 ---------
63-a

3 0

3 years ago

Read 2 more answers

Please solve as fast as possible...<br>No spams please......

lisabon 2012 [21]

Answer:

In ∆ABD and ∆ACD

<BAD =<CAD ( each are half of <A )

<D=<D ( each equal to 90°)

AD = AD ( common)

So ∆ ABD is congruent to ∆ ACD.

Then AB =AC (by C.P.C.T)

Hence,∆ABC is an iso - sceles triangle.

8 0

2 years ago