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marin [14]
3 years ago
5

B(n+5)=(n+p)3 what does p=​

Mathematics
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

I think it -5

Step-by-step explanation:

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B=3/4a<br> A+b= 21<br> Solve for a and b
maxonik [38]

Answer:

A=12 B= 9

Step-by-step explanation:

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A net of a square pyramid is shown. The total area of the pyramid's triangular faces is 80 cm'. What is the area of the pyramid'
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5 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
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faltersainse [42]

Concept

Quotient the number obtained by dividing one number by another Dividing 10 by 5 gives a quotient of 2.

6 0
1 year ago
PLEASE HELP ASAAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER
Nutka1998 [239]

Let i = sqrt(-1) which is the conventional notation to set up an imaginary number

The idea is to break up the radicand, aka stuff under the square root, to simplify

sqrt(-8) = sqrt(-1*4*2)

sqrt(-8) = sqrt(-1)*sqrt(4)*sqrt(2)

sqrt(-8) = i*2*sqrt(2)

sqrt(-8) = 2i*sqrt(2)

<h3>Answer is choice A</h3>
5 0
3 years ago
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