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lions [1.4K]
3 years ago
5

The eatery restaurant has 200 tables on the reason even if there's no there's no reservation for one half of the tables were

Mathematics
1 answer:
a_sh-v [17]3 years ago
4 0
What is a true<span> statement about muscles?

</span>
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Which function does the graph represent?<br> (click on picture)
Crank

Answer:

B is correct

Step-by-step explanation:

7 0
3 years ago
In a recent survey of 700 people, 15% said that red was their favorite color. How many people said that red was their favorite c
Alex17521 [72]

Answer:

Answer: 105

Step-by-step explanation:

In a recent survey of 700 people, 15% said that red was their favorite colour. 105 people said read was their favorite colour.

6 0
3 years ago
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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}&#10;\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\&#10;&&\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)}\\\\\\&#10;\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}\\\\\\&#10;\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\&#10;\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)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\&#10;=\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
3 years ago
Please help fast i need help ​
Nutka1998 [239]

Answer:

omg

I

1. 3×3×3×3 = 81

2. 2²a³

3. (-5)³ (-5×-5×-5) = -125

4. 1

5. a^m-n

II

1. 1.6807× 10⁴

2. (7×7×7) × (2×2×2×2×2×2) = 21952

3. 32²

4. 2187

5. 803025

3 0
3 years ago
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James Works in a flower shop he will put 27 tulips in vases for a wedding he must use the same number of tulips in each vase the
Svetllana [295]
Umm..can you help  i dont get it.
3 0
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