Melody

Huh?  Why don't you change it back to a mixed numeral Chris ?

maybe t=6/2

No :)

I am Australian,

CPhill (Chris) is from the US

Alan is from the United Kingdom

and

Radix is from Germany.

There are actually 5 seperate sites here, English, German, French, Spanish and Russian.

There is a gateway into the other forums at the bottom of most pages.  You will be automaticallly logged out but you can log back in again.  Why don't you take a look :))

The message system, and the Top Users board  is the same in each forum so it is not unusual for me to get messasges in other languages and they do not necessarily realize that I only speak English.     

That is excellent forensic maths Chris :))  5 points from me :))

Well aren't you the clever one!!!

Come on Parker, this is a calculator site! 

Well,the exercise needs a solution that has to do with bolzano's equations or with continuous functions or the equations that are the aftermath of bolzano's theories,such as m<=f(Xo)<=M

If you knew that is what you needed why didn't you say so.  !!

I/We did not know whether you were year 9 or whether you were in University. 

When you said you did not want a graph I thought you meant that you did not want a computer generated graph that you did not understand.

Please express your questions better!

sqrt[     ( 4[(75kg)(9.81m/s^2)(sin37)] - [(76kg)(9.81m/s^2)(0.060)(cos37)]      ) /    (1.80m x 0.500m)     ]

sqrt[     [    [  (4*75*9.81*sin37)   -   (76*9.81*0.060*cos37) ] kgm/s^2    ) /    (1.80 x 0.500) m^2    ]

((4*75*9.81*sin(37))-(76*9.81*0.060*cos(37)))/(1.8*0.5) = -2142.41383611526428

No solution - you cannot find the square root of a negative number in the real number plane.

Both those answers look great to me :)

In this case the -1 does not mean flip (find reciprocal) 

It mean find the angle whose sin is x ..     which is also asin x

\(If \qquad sin^{-1}(x)=\theta \qquad then \qquad sin \theta =x\\ and \qquad sin^{-1}(x)=asin\;(x)\)

I think the asin notation is much better because the -1 notation is very confusing.

If you want the reciprocation you need to write

\((sin\theta)^{-1} = \frac{1}{sin(\theta)}=Cosec(\theta)\)