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natulia [17]
4 years ago
12

The point (0,-5) lies on which quadrant.

Mathematics
1 answer:
Umnica [9.8K]4 years ago
6 0

Answer:

Quadrant II

Step-by-step explanation:

(0,-5) on graph would lies on II quadrant

hope it helps!

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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}
\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
3 years ago
An automobile has a price of $16,000 after being discounted 20%. what is the original price?
stepan [7]
 <span>$16,000 / (100% - 20%) = $16,000 / .8 = $20,000 
to verify, work the answer backwards. 
$20,000 - $20,000 * 20% = $20,000 - $4,000 = $16,000 
</span>
6 0
3 years ago
Find the following <br> (u•w)(2)=<br> (w•u)(2)=
Stells [14]

Step-by-step explanation:

substitute 2 into x

1. (u ○ w)(2)

u(w(x))

= 4(-5x+1)+2

= 4(-5.2+1)+2

= 4(-9)+2

=-34

2. (w ○ u)(2)

w(u(x))

= -5(4x+2)+1

= -5(4.2+2)+1

= -5(10)+1

= -49

8 0
3 years ago
What is the Value<br> out in the equation<br> 3/4 (z+4)=9
alisha [4.7K]

Answer:

z = 8

Step-by-step explanation:

3/4 (z+4)=9

3/4z + 3 = 9

3/4z = 6

z = 8

8 0
3 years ago
Write the inverse of g(x) = e x<br><br> g -1(x) = _____.
mezya [45]

Answer:

g^{-1} (x) = \ln x

Step-by-step explanation:

From the definition if inverse function we can say if y = f(x) and x = h(y) then h(x) is the inverse function of f(x).

Here the given function is g(x) = e^{x} and we have to find the g^{-1} (x) i.e. the inverse function of g(x).

Now, let us assume y = g(x) = e^{x}

Now, taking ln both sides we get, \ln y = \ln e^{x} = x \ln e = x  

{Since \ln e = 1}

⇒ x = \ln y

Therefore, g^{-1} (x) = \ln x (Answer)

6 0
3 years ago
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