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Sign in and say that. LOL

WEAK!

For you anon, and try thinking.
http://nrich.maths.org/2473

Does stu stand for stupid?

I can't agree with that. it would be the only number to not have a fixed root since from what I know infinity is changing either approaching on 1, or exapnding further away. As well, only 1 has a sqrt of its self, but it is fixed and not moving.So rather wouldn't it be better to make the sqare root infinity -1? 

As well sqrt(infinity) is not equal to infinity. As the definition of square root is  (the result of multiplying the number by itself, or y × y) which defuses a belief it can be its self since sqrt(infinity) equals infinity, just made infinity an infinity times bigger. But that would also be another first, the only number to grow as a result of a square root. Just can't agree with it.

f(0.5) = 16^0.5 = √(16) = 4

If n = 9 then (-1/3)^n-1 = (-1/3)⁸ = (-1)⁸/3⁸ = 1/3⁸ = 1⁸/3⁸ = (1/3)⁸

ok.  So plot the graph of cos(x) where x is is in degrees.  This time we have to home in on a small portion of the graph to see it clearly:

You can see they are still the same.

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(-{\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

However, your original post stated that the correct answer to the integral was -3.25.  This is only true if x is in radians.

If you mean (1/3)⁸ = (-1/3)^n-1 then n = 9 is an integer solution.

Please explain where the - sign in (-1/3) comes from to make lhs = rhs.

I assume r is a scalar and I is the identity matrix.

im not sure, but only one question is in radian on the exam sample, which specifies radians and gives the sign of radians. All others I have taken to be in degrees. 

It is normal to assume the angle is in radians unless it is accompanied by an explicit degree sign.

To see that cos(-3) is the same as cos(3) look at the graph below:

You can see that cos(-3) has exactly the same value as cos(3).  They are both ≈ -0.99.  

$$f(x)=e^{5x}\\\\
f'(x)=5e^{5x}$$

You can memorize this outcome but if you understand it that would be even better.

-----------------------------------------

you know that if  $$y=e^x$$  then  $$\frac{dy}{dx}=e^x$$

Now

consider $$y=e^{5x}$$

Let u=5x  then   $$y=e^u\quad \mbox{and}\quad \frac{du}{dx}=5$$

$$y=e^u$$

Now use the chain rule:

$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=e^u\times 5 = 5e^{5x}$$

Merveilleux! À partir de maintenant, je vais vous poser la questions de la physique en français

@@ End of Day Wrap - Sunday 1/6/14   Sydney, Australia   Time 12:15 am (Monday really)

Zegroes is right - I didn't write a wrap last night.  It is just as well that he is here to help! 

Great answers were given over the last 2 days by GoldenLeaf, all-them-marvel-feels, CPhill, Bertie, Alan, Rosala, Zegroes, Chilledz3non, DavidQD, NBlueBird, Rom Admin and Problem. Thank you all! 

GoldenLeaf has now replaced Scrutinizer on the Top Answerer list.  Congratulations GoldenLeaf!

NBlueBird has only just joined up.  Welcome to web2.0calc forum.  We hope you like it here. 

@ Answerers must always remember that many/most people who use this forum have never heard of imaginary or complex numbers!  This poster has been trying to figure out our answer to this question for quite some time!

Aha, spot on. That would be what Im looking for. Thanks. 

It was mentioned earlier but i didn't understand why or see why. I get it now, a part of the product rule, and totally overlooked. Cheers.

Why is this in radians and not degrees? It is not specified so I assume it is degree standard. Also, I do not understand how you take the /3 out side the bracked and devide the 5 to get the same result as having it inside. Also, why is it + when there is 3 - sign in the equation. I was told that the two intial minu make positive but then the - on the degree to the cos (-3) comes out side to make the + and - a negative. As shown in my example. SOme further explaination would assist me. As well, this is not the equation from last night.

Do you mean (1/3)⁸ = (-1/3)ⁿ - 1 or  (1/3)⁸ = (-1/3)^n-1? 

If you mean (1/3)⁸ = (-1/3)^n-1 then n = 9 is an integer solution.

For other non-integer solutions you need to involve complex numbers. 

thanks so much