Okay,
This is just one thing it can be used for
http://web2.0calc.com/questions/a-cannonball-is-launched-horizontally-from-the-top-of-an-78-4-meter-high-cliff-how-much-time-will-it-take-for-the-ball-to-reach-the-ground
I'm not sure - maybe
Can someone please tell me if infinity has square of round brackets please
Also can someone tell me how to write [in latex without using it with \left etc
Thanks
$$\\x^2 \ne -2 \\\\ \left[\infty,-2)\; \cup \; (-2,\infty]$$
Will let you know if the online test accepts either answer j have a feeling it will and that I am really weak working with fractions like this. Also, there is a problem with mobile view/normal view on this site showing a large portion of the pagebin white and unable to zoom in to the section that is condesed to the lhs.
OK
I tink I it! What about you Stu?
Heureka and I have interpreter it a little differently.
Heureka's answer is better.
x+3=12 or x+3=-12
You can finish it.
Beats me!
info: enter a more detailed answer..
Well if the width is w then the length is 2w
You have to make an equation for the perimeter and solve it.
I don't understand this problems.
You don't necessarily have 15 on the first one do you?
If you have x on the first one then you have 60-x on all the other 5 combined.
I'd start by squaring both sides and solving it as a quadratic. Hope that helps.
Look here
http://web2.0calc.com/questions/how-do-i-type-in-fractions
In the sticky topics on the right hand side of your page is a thread called
"How do I type in fractions" Look in there.
NO
4(n-4)-1 = 4*n - 4*4 -1 = 4n-16-1 = 4n-17
If you have not heard of complex or imaginary numbers yet then there are no (real) solutions to this.
Google "scientific notation" abd then work through the "mathsisfun" site that it is bound to throw up.
Can you simplfy the following please
$$w=\frac{-1}{-1/2cos(theta)^2+(7/16)} = \dfrac{-1}{-\dfrac{1}{2} \cos{ ( \theta^2 )} +\dfrac{7}{16} } = \dfrac{-2} { \dfrac{7}{8} - \cos{ ( \theta^2 )} } = \dfrac{2} { \cos{ ( \theta^2 )} - \dfrac{7}{8} }$$
$$\\w=\frac{-1}{-1/2cos(\theta^2)+(7/16)}\\\\ =-1\div \left(\frac{-1}{2cos(\theta^2)}+\frac{7}{16}\right)\\\\ =-1\div \left(\frac{-8}{16cos(\theta^2)}+\frac{7cos(\theta^2)}{16cos(\theta^2)}\right)\\\\ =-1\div \left(\frac{7cos(\theta^2)-8}{16cos(\theta^2)}\right)\\\\ =-1\times \left(\frac{16cos(\theta^2)}{7cos(\theta^2)-8}\right)\\\\ =\frac{16cos(\theta^2)}{8-7cos(\theta^2)}\\\\$$
Is that simple enough?
I don't know where this number: ((32!^4)*20!)*66.466304816293425, comes from (and why bother with all those decimal places!) but even if a computer could manage 20! calculations a second the 32!^4 is so large it would take more than the lifetime of the Universe!
arc sine and inverse sine are the same thing.
On the web2 calc it is called atan
Press the 2nd key on the calculator (http://web2.0calc.com/ ) to see tan-1
Use asin(0.3971) or sin-1(0.3971). sin-1 is obtained by first pressing the 2nd button on the calculator here: http://web2.0calc.com/
Just type this in pretty much as you've written it, except change (5root7) to (5*71/2) or (5*sqrt(7))
h
sintheta ^2
