All Answers 358387 Answers

almost anon

In [Math Formula]   or  [Calculation]      in the ribbon above it is    sqrt( )

In [LaTex]   in  the ribbon above   it is               \sqrt{   }

Sun 15/3/15

Thanks albert 

I think it would be really good if we had "an event of the day" entry more often. 

Hang on, you are a day early !!!

I'll put this in my wrap tomorrow........

Yes, I will second that  

harrypotter, please include your working because otherwise no one can learn anything 

Yea I wouldn't mind getting a couple of those emails either.  LOL

I like your answer, your logic is good  :)

The question really does not make sense so that is probably why anon thinks you are wrong 

Hi  TSC,

Welcome to web2 forum :)

I guess you have already worked out how to go back and forth between the English and the Spanish forum?

$${{\mathtt{600}}}^{{\mathtt{13}}} = {\mathtt{1\,306\,069\,401\,600\,000\,000\,000\,000\,000\,000\,000\,000}}$$

How cot inverse (x) is (pie - tan inverse (1÷ x) ) and sec inverse x is simply cos inverse (1÷x) when x

Lets see if I can make sense of your question :)

How come         $$cot^{-1}(x)\;\; is\;\;\; \pi - tan^{-1}\left(\frac{1}{x}\right)$$

and           $$sec^{-1}(x)=cos^{-1}\left(\frac{1}{x}\right) \qquad when\;\; x <0 ?$$

from my graphs I can see that

See Anonymous's answer below.

log₅(625) = 4  (because 5⁴ = 625)

³√-125 = -5

so 

$$\frac{\log_5{625}}{17+^3\sqrt{-125}}=\frac{4}{17-5}=\frac{1}{3}$$

.

sin-1  is     asin

I do hope that you have not passed on your CDD infection to other members of the forum Nauseated!

If any of us get CDD we will know exactly who to blame!

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On a more serious note I am going to give your 3 by 3 solution more thought. 

I have not fathomed Nauseated's answer (perhaps he can better explain)

but I did it a different way and got the same answer.

This is how I did it.

I thought of it as three ladders placed side by side.

There is the ground level - L1

the first rung                     L2

the second rung                L3

and the top rung               L4

So you have to be on each of the ladders and you can not go down a rung between ladders or you will not get to the end point in 6 units.

So here are your 'rung' possibilities as you go across the ladders.

L1    L1       L1,2,3 or 4                     4 possibilities

L1    L2       L2,3 or 4                         3 possibilities

L1    L3        L3 or 4                           2 possibilities

L1    L4        L4                                   1 possibilities

L2    L2        L2,3,or 4                         3 possibilities

L2    L3        L3,   or 4                         2 possibilities

L2     L4         L4                                 1 possibilities

L3     L3         L3 or 4                          2 possibilities

L3     L4           L4                               1 possibilities

L4     L4            L4                              1  possibilities

Total                                              = 20 possibilities

NOW Nauseated, can you explain why our 2 answers are the same.

Do you actually understand the logic behind your answer?  If so, please share  

i could've broken the say.....only if i hadnt said it!

I get that a lot!

thats too short.......something more is needed!

Same thing, just one is a positive way of putting it and the other is negative.

lets see if another genius comes.......im hoping this will be cracked soon!

I'm afraid you've stumped me, rosala.....!!!!!

I've got no clue........

We can't solve for x, because there is no equation, here.

It's like saying

3x

"Solve for x"   ......  x could be anything!!!!

I am going to "steal" some of Melody's work, here.....

Look at the following pic.....

Let's suppose that the border IS 2 ft wide.  But notice, we can't say WHAT the specific side length of the border might be because, as you see, L₂ in the picture is longer than L_1.

And as Melody has indicated using similar triangles, L₂ = L₁ (r + 2) / r

Since ABQP forms an isosceles trapezoid......we could give an "average" side length = [L₁ + L₂ ]/ 2  = [r*L₁ + L₁(r + 2)] / [2 r] = [r*L₁ + L₁(r + 2)] / d    which is just the length of the median of the trapezoid.

Thus......no general answer as to side length is possible....

log_.6 .36      we can solve this with the change of base rule

The answer is given by     log . 36  / log .6   = 2

Similarly

log₈  .0039    =   log .0039 / log 8 = about  -2.667

28/512  = 14/256  = 7/128

-8x^2 + 4x + 3 = 0   this won't factor....using the onsite solver and the Quadratic Formula, we have

 $${\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.411\: \!437\: \!827\: \!766\: \!147\: \!6}}\\
{\mathtt{x}} = {\mathtt{0.911\: \!437\: \!827\: \!766\: \!147\: \!6}}\\
\end{array} \right\}$$

-4x^2 + 8x + 5  = 0      factor

(-2x -1)(2x -5)  = 0       and setting each factor to 0, we have that x = -0.5 and x = 2.5

-9x^2 + 7x + 6 = 0    same as the first......no factorization is possible

$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{265}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{265}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.515\: \!490\: \!033\: \!116\: \!650\: \!4}}\\
{\mathtt{x}} = {\mathtt{1.293\: \!267\: \!810\: \!894\: \!428\: \!1}}\\
\end{array} \right\}$$

So.....x = -0.52 and x = 1.29