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Ratling [72]
3 years ago
12

Help me please with this

Mathematics
1 answer:
Sphinxa [80]3 years ago
8 0
They're all $15.99 each
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What is the solution to this equation? 8 (e) 2x + 1
CaHeK987 [17]

Answer:

<u><em>16ex+1</em></u>

Step-by-step explanation:

<u>Remove parentheses: (e) = e</u>

= <em>8e2x+1</em>

<u>Multiply the numbers: 8 x 2 = 16</u>

<em>=16ex+1</em>

4 0
3 years ago
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
Can the sum of two mixed numbers be equal to 2? ​
Sholpan [36]
No it depends on the sum of the two numbers
4 0
3 years ago
Read 2 more answers
HELP????????????????????????????????????????
balandron [24]

Answer:

No

it goes on forever so no

8 0
3 years ago
Determine a and b so that figure ABCD is a parallelogram
ziro4ka [17]

Answer:

C(8,2)

That is the answer for a and b

4 0
4 years ago
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