All Answers 358514 Answers

Well, the only key top answerer who is unaccounted for is Rom.

It is nice to 'see' all you others here!  Hopefully private messaging will be restored soon.

When I tried to sign in on my smart phone I got the most severe warning that I have ever seen.  It did not trust the sign in security on this new site.  I decided to take it seriously and I did not sign in.  My lap top, which also has google chrome is not complaining at all.  Computers are a mystery to me! 

Enjoy the rest of your day.  

If it's a regular polygon, then each interior angle is 180°-360°/11 so the sum of 11 of them is 11*(180°-360°/11) = 11*180°-360° = 1620°

Each of the 6 teams plays 5 others so at first sight it looks as though there are 6*5 = 30 matches.  However, this counts each match twice, so, in fact, there are 30/2 or 15 matches in total.  

Hmm!  Math(Input=Result) doesn't display properly after posting!

Ok, it does now!  Weird, it didn't at first, it showed an icon!

Now back to an icon!!!!!

Looks like the old maths stuff doesn't display.  I note there are two new maths buttons. Let's see if they work:

$${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{8}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)} = {\mathtt{495}}$$  This is Math(Input=Result)

Can't see how Math(Input) works!

Wow! the whole thing changed! Wicked!

$$\begin{pmatrix}
a&w&e&s&o&m&e \\
:)&:)&:)&:)&:)&:)&:)
\end{pmatrix}$$

Happy * -1

Melody here

I just estimated my answer

It is 44.8 or 45 to the nearest whole number.

I believe it is correct.

Let's rearrange things slightly...we have...

225x² + 60x + 2 = 0

I can tell right away that this won't factor.

Using the on-site solver, we have

 $${\mathtt{225}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{0}} = \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}{{\mathtt{15}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{15}}}}\\
\end{array} \right\}$$

The 'quadratic formula" would give us this answer, too. It's pretty messy, no matter which way you go......

thanks thata the anwser and youre right its not exaclty 45, im sorry if that was missleading.

One more problem.....no "preview" function/button ??

I liked that one A LOT !!!

I think Melody made a slight error towards the end....I'm pretty sure the answer isn't EXACTLY 45, either.

(T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 ???

When I stick "45" in for "T,"  we get

$$\left({\mathtt{45}}{\mathtt{\,\times\,}}\left({\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}}\right)\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{45}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,-\,}}{\mathtt{50}}\right) = {\mathtt{0.190\: \!978\: \!224\: \!309\: \!896\: \!5}}$$

Let's see what we can do....taking Melody's work from this point......

50√(6) / (1 + √(3))

Let's multiply by the whole fraction by the conjugate of (1 + √(3)) on "top" and "bottom"

We have

50√(6) / (1 + √(3))   *   (1 - √(3)) / (1 - √(3))

This gives us

50√(6) * (1 - √(3)) / (-2)

Sinplifying the top, we have

[50√(6)- 50√(18)] / (-2)

Let's multiply top and bottom by (-1) to get rid of that pesky (-2).....Nore that the things in the numerator just switch places when we do this...so we have

[50√(18)- 50√(6)] / (2)....and dividing everything by 2, we get

25 (√(18)- √(6)).....we could simplify the "18" under the square root, but we would still have two square roots, so let's just leave it alone.

So "T" = 25 (√(18)- √(6)) .......BTW this is close to 45 !!!!!

And plugging in for "T" we have

25*(√(18)-√(6))*(cos(45)/cos(30))*sin(30)+(25*(√(18)-√(6))*sin(45)-50)

If you evaluate this answer inyour calculator (or the calculator on this site), you'll get something that ≈ 0

(Sorry....they've made some changes to this site and I haven't figured out how to evaluate things, just yet   )

Anyway.....I feel pretty confident that this is the correct answer

Hi Chris,

I think I mentioned that problem in the email that I have already sent to Andre.  The whole 'member' section is missing.

Actually I am hoping that Andre is reading this thread.  There has been at least one 'admin' response to a question this morning.  

Hi Kitty,  

It is nice to know that you are around.  And yes, we all hope that aAdre can 'spray' the bugs soon.  A good does of Pyrethrum might do it.  I hope that are not SUPER bugs that have become genetically immune. (There is no laughing icon!)

Another problem?  Also you can't copy and paste in here.  

Funny; I have the same log-in problem as Chris:(

Seems like Mr. Massow is experimenting again, which is alright.  Guess we'll just have to wait and see how it all turns out.

<= This emoticon is somewhat cute though.

Cheers, ~Kitty<3

Hi,

I have changed my answer a little bit but I don't think that it is 45.

Other people will look at this thread so you will get more input.  

I think mine is correct.  

We will see what the others think.

Thanks for posting and for the feedback. 

Melody

PS There was a bracket missing in the original question - is my interpretation of what you wanted correct?

One more problem I've noticed.....I can't tell if someone is online or not......am I missing something???

Im so sorry that im bothering you guys on a thread that doesnt apply to my question but i dont know how to contact you guys privately, anyways melany i was wondering if you could have a look at the problem you had helped me with a bit back,

$$\mathrm{\ }$$

(T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0

The response you gave was well worked but the solution wasnt right (at least i dont think it was) anyways the correct anwser should be 45. Think you could help me 

more info please

Idk how to reply to anwser but i think you did it wrong melody, thanks for the help anyways but the anwser is 45 not 75*square root(2) - 50*square root (6). If you figure out where you went wrong though i would be glad to hear you explanation, you work was really helpful so thanks

Mmmm Yes a few ants!

Thanks for letting us know Chris.   Hopefully Andre will help us get you back up with your original username. 

I just used the built in Latex.

I couldn't get align to work and it is not particularly friendly for long displays. (Which mine was)

It may be quite nice for little displays.  We will see.

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

I have already written to Andre regarding the lack of member section and the missing messages.

I will write another regarding your lock out problem and other issues but i might let them build up a little first.

Hopefully it will all be sorted out quite quickly.

I was looking at the emicons but I don't think I like the new set very much.   (that one is ok)

$$T\times\frac{cos45}{cos30}\times sin30+Tsin45-50&=0\\\\
T\left(\frac{1}{\sqrt{2}}\div\frac{\sqrt3}{2}\right)\frac{1}{2}+T\times\frac{1}{\sqrt2}&=50\\\\
T\left(\frac{1}{\sqrt{2}}\times\frac{2}{\sqrt3}\times\frac{1}{2}\right)+\frac{T}{\sqrt2}&=50\\\\
\frac{T}{\sqrt{6}}+\frac{T}{\sqrt2}=50\\\\
\sqrt{6}\times\left(\frac{T}{\sqrt{6}}+\frac{T}{\sqrt2}\right)=50\times\sqrt{6}\\\\
T+\sqrt{3}T=50\sqrt{6}\\\\
T(1+\sqrt{3})=50\sqrt{6}\\\\
T=\frac{50\sqrt{6}}{1+\sqrt{3}}\\\\
T=\frac{50\sqrt{6}}{1+\sqrt{3}}\times\frac{1-\sqrt3}{1-\sqrt{3}}\\\\
T=\frac{50\sqrt{6}-50\sqrt{18}}{1-3}\\\\
T=75\sqrt{2}-25\sqrt6$$

It may be just me (Cphill), but I tried to log in at my local library.....unfortunately, I had forgotten my password (my home computer stores it automatically) and after a few attempts at guessing it unsuccessfully, I got "locked" out. (I'm an idiot - what can I say ??)

Well, when I got back home, I still couldn't log in, so I just created a new user name - namely, mine...so now, "Cphill" is just "Chris"......

Melody is correct......no private messages.....(and no access to them ???)

Mmmmm.......looks like a few "bugs" have shown up at the picnic...  

We have tanθ = h/2(miles)

And so, the height of the balloon at any time is given by

h = (2miles)* tanθ

What we're looking for is dh/dt (the change in ht. per min)

So we have

dh/dt = dh/dθ * dθ/dt

Note that dh/dθ = (2miles)*sec²θ and dθ/dt = .1 rad /min

(Also, note that the 2 miles is a constant...i.e., the "x" part of the tanget isn't changing, only the "y" part  - the height - is.)

Then dh/dt = (2miles)*sec²θ * .1 rad /min

And when θ = π/6, we have

dh/dt = (2miles)*sec²(π/6) * .1 rad /min

dh/dt = (2 miles) * (2/√(3))²  *  .1 rad / min

dh / dt = (2 miles) * (4/3) * .1 rad / min

dh /dt = (8 / 3) * (1 /10) miles / min

dh /dt = (8/30) miles / min = (4/15) miles / min

(Since rads are a "dimensionless quantity" they "fall out" of the answer)

I believe that's it...............