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3 years ago

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A rectangle of perimeter 100 units has the dimensions shown. Its area is given by the function A = w(50 - w). What is the GREATE

ST area such a rectangle can have?

1 answer:

Inessa05 [86]3 years ago

7 0

Using a Graph we can determine the greatest area the rectangle can have using the midpoint between the two w-intercepts.

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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

What is another way to wright the same equation, for example 1/4x-5

Finger [1]

Answer:
Y= -5+ 1/4x
There are other ways to write too

6 0

3 years ago

It takes 15 seconds for danial to play a card game. if the probability of winning the game is 0.1 and he has two minutes before

ycow [4]

If Daniel has two minutes and each game lasts for 15 seconds then he has eight tries to win. ( 120 seconds divided by 15 seconds = 8) And if his odds are .1 that means he has a 10% chance each game. 10% x 8 tries = 80% chance. or .8

4 0

4 years ago

A quiz has 5 questions with answers that are either true or false. How many different combinations of answers are possible?

Liula [17]

512..please mark bianliest or at least 5 stars

5 0

3 years ago

The area of a triangle is 13 square feet. If the base of the triangle measures 4 feet, what is its height?

VashaNatasha [74]

The area of a triangle is 1/2 base X height. So, 1/2 4 X h = 2h
h= 7.5

4 0

3 years ago

Read 2 more answers