Yes that is true - I was assuming that the question asker was refering only to cardinal numbers. (Positive integers)

And that 'are divisable by 7 or 3' really meant 'are multiples of 7 or 3'

@@ End of Day Wrap - Tuesday 20/5/14   Sydney, Australia Time 22:20

Hi everyone,

Another extremely busy day.  Great answers were given by reinout-g, Zegroes, Alan, heureka, DavidQD, admin, chilledz3non, CPhill, GoldenLeaf, Kitty<3, davanshbablaq1.  Thanks all.
There were also a few nice shows of gratitude but i don't have the names, sorry.

Our message centre has been redesigned from scratch.  There are some features that are not properly implemented yet but some of the new ones will be really good.

We can now view our messages from oldest to newest or vice versa.  Received messages and sent messages are now together in a converstion view (they are when you go into send message view anyway).  Sent messages are left justified and received messages are right justified.  Plus sent messages can  be edited or deleted right up until they are read by the recipient.  I used to love this feature (it was in the previous version of the forum) We cannot delete our messages at present but I am sure we will be able to soon.  A message export system is now being developed and should be up and running in a week or two.  The feature for choosing a member from a list definitely needs improvement but I think Andre Massow is repairing features and introducing new features in extremely good time.  I commend him on his efforts and the results he has achieved!  Thank you Andre! 

There are a great many questions.  Primary answerers are having a tough time getting through them all but we do do our best.  I have always encouraged people to get involved and post answers were they are able.  Unfortunately there are some people posting who do not have the knowledge base that they need to answer the question properly.  These answers can be very misleading to the question asker which is definitely not good.  They can also be time wasting for primary answerers.  If you are one of these over-confident people who is posting answers outside your knowledge base please STOP!  (I may start deleting point for wrong answers given by repeat offenders)  If you lack confidence but you really believe you can help I do not want to discourage you.  There have been some fantastic answers from casual answerers, today and everyday.  

Here are just a couple of todays posts.  I don't remember any really standing out.

Forum Interest Question - Web2.0calculator

Probability that nbr1 is a boy is p1 = 6/10.

Probability that nbr2 is a boy, given 5 boys left is 5/9

Probability that nbr3 is a boy given 4 boys left is 4/8

Probability that nbr 4 is a boy given 3 boys left is 3/7

$${\mathtt{Overallprobability}} = {\frac{{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{3}}}{\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)}} \Rightarrow {\mathtt{Overallprobability}} = {\mathtt{0.071\: \!428\: \!571\: \!428\: \!571\: \!4}}$$

Approximately 0.071 or 7.1% 

We can see from the graph below that there is no real number solution here.

6x + 8/x -3 has no values in the range -16.856 (the value at x = -2√3/3) to 10.856 (the value at x = 2√3/3)

Every real number is divisible 7 or 4.

100/7=14+

100/4=25

numbers that are divisable by 7 and 4=28,56,84 number =3

So I think the answer is 14+25-3=36

So we have:

$$\\\mbox{start} = a_1=4\\
\mbox{increase}=d=3\\
\mbox{term number}=n=50\\
\mbox{term 50}=a_n=a_{50}?\\
\boxed{a_n=a_1+(n-1)*d}\\
a_{50}=4+(50-1)*3\\
a_{50}=4+49*3\\
\boxed{a_{50}=151}$$

Latex code:

} = a_1=4\\
\mbox{increase}=d=3\\
\mbox{term number}=n=50\\
\mbox{term 50}=a_n=a_{50}?\\
\boxed{a_n=a_1+(n-1)*d}\\
a_{50}=4+(50-1)*3\\
a_{50}=4+49*3\\
\boxed{a_{50}=151}

Term 50 is 151

$$4<6x+\frac{8}{x}-3<5$$    

Start by multiplying everything by x²

This will get rid of the fraction.  And, since x² is positive you don't need to turn any signs around.

Is this the question that you meant or was it actually different.  If you want more help make another post asking for more help.

\theta= \tan^{-1}{(\frac{\Delta{y}}{\Delta{x}})} =

$$\theta= \tan^{-1}{(\frac{\Delta{y}}{\Delta{x}})} =$$

By Heureka.

$$\Delta{x} =$$ $${\mathtt{98\,200}}{\mathtt{\,-\,}}{\mathtt{94\,700}} = {\mathtt{3\,500}}$$

\Delta{x} =

$$\Delta{y}=$$ $${\mathtt{29\,500}}{\mathtt{\,-\,}}{\mathtt{28\,200}} = {\mathtt{1\,300}}$$

\Delta{y}=

$$r=\sqrt{\Delta{x}^2+\Delta{y}^2}=$$  $${\sqrt{{{\mathtt{3\,500}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1\,300}}}^{{\mathtt{2}}}}} = {\mathtt{3\,733.630\: \!940\: \!518\: \!893\: \!867\: \!9}}$$

r=\sqrt{\Delta{x}^2+\Delta{y}^2}=

$$\theta= \tan^{-1}{(\frac{\Delta{y}}{\Delta{x}})} =$$  $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1\,300}}}{{\mathtt{3\,500}}}}\right)} = {\mathtt{20.376\: \!435\: \!213\: \!836^{\circ}}}$$

\theta= \tan^{-1}{(\frac{\Delta{y}}{\Delta{x}})} =

so

$$r = 3733.63 \mbox{ units}$$

r = 3733.63  \mbox{ units}

and

$$\theta = 20.3764 \ensuremath{^\circ}$$

\theta = 20.3764 \ensuremath{^\circ}

Thank you anonymous for you good manners. 

Thanks anonymous.

I assume you just want to expand and then simplify this.

(x+7)(x+9)

=x(x+9)+7(x+9)

=x²+9x+7x+63

=x²+16x+63

 x² + 7x + 19 

$$If \qquad x^2+7x+19=0 \mbox{ which it doesn't - then}\\\\
x=\frac{-7\pm \sqrt{49-76}}{2}\\\\
x=\frac{-7\pm(\sqrt{17})i}{2}\\\\$$

SO

$$x^2+7x+19=\left(x+\frac{7+(\sqrt{17})i}{2}\right)
\left(x+\frac{7-(\sqrt{17})i}{2}\right)$$

I think that is ok.

You are ofcourse correct Bertie.  Thank you.

I gave myself -3 for my lack of diligence!  That'll teach me - maybe???  

Are you looking to turn rectangular coordinates into polar coordinates?

If so, pol(x,y) becomes (r, θ) where θ = tan^-1(y/x) and r = √(x² + y²)

However, I don't know if the values you've supplied are meant to indicate the end values of a range, or if the numbers are to be subtracted or what!  

Thanks Admin, and btw the ! is a factorial sign, it means that number multiplied by all the numbers lower than it up to 1 e.g.

3!=3*2*1=6

Thanks Alan,

I didn't think Chris's was wrong.  Your's is neater though!   

Chris might think i can't have his points but I've still got his sunnies!!!   And, he can't have them back. 

You are right heureka;  sin is also positive in the second quadrant, where cos is negative.

Do you mean find p?  If so then p+10 = 90 - 49, so p = 31°.  Another possibility is in the second quadrant where p + 10 = 180 - 41 so p = 129°  (The 41 comes from the first p+10, i.e. 31+10 = 41), 

Check:

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{49}}^\circ\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{31}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{129}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

750 + ('X% of 750 free') = 900.

X% of 750 = 150.

X = 20.

Hi Alan,

$$\\\cos{(\alpha)}=\textcolor[rgb]{1,0,0}{\pm}\frac{3}{5}\\
\text{if }{+\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(0.6) = 53.1301023542\ensuremath{^\circ}\\
\text{if }{-\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(-0.6) = 126.869897646\ensuremath{^\circ}\\
\sin{(53.1301023542\ensuremath{^\circ})}=\sin{(126.869897646\ensuremath{^\circ})}=0.8=\frac{4}{5}$$

Latex code:

\text{if }{+\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(0.6) = 53.1301023542\ensuremath{^\circ}\\
\text{if }{-\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(-0.6) = 126.869897646\ensuremath{^\circ}\\
\sin{(53.1301023542\ensuremath{^\circ})}=\sin{(126.869897646\ensuremath{^\circ})}=0.8=\frac{4}{5}

Could also note that  cos(2θ) = 2cos²(θ)-1 so 1-2cos²(60) = -cos(120)

$${\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{120}}^\circ\right)} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0.5}}$$

No need to think of removing Chris's points!  There are 10 possible choices for the first wheel.  For each of these 10 there are 9 possible choices for the next wheel.  For each of these 10*9 there are 8 possible choices for the next wheel.  For each of these 10*9*8 there are 7 possible choices for the last wheel.  

$${\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{5\,040}}$$

What, even with those shades!!!