All Answers 358089 Answers

ahhh. yes thats wat i needed. :) *snatches it away and goes in a corner

why do i even bother. -_- *slaps face

It is completely possible to have 100%+ of something. For example, 300% of 6 means 3 times more than 6.

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so harsh, Z! LOL

close ur ears chris and melody! O.O

Melody and Chris love me so much! they would never do such a thing! LOL

(dont even think about it Chris and melody! ) -.-

I wont bite you but ill sure as heck get melody and Cphill to bite you. Cphill has bony teeth and melody has well sharp teeth....So pick your poison!

We're not always "boring" and "responsible"  Z-Man.....!!!!

well excuse me math police but i have a RIGHT to speak to my fellow minions! thank u very much!

bite me.

Sorry mam but you are obstructing learning and im going to have to remove your post from this-oh s***w it being  boring and responsible is no fun.....How does melody and Cphill do it

welcome to the club, anon

Thanks Alan,

I had to think about that one, but yes of course any x where sin x =0 will be a solution.

This is why it is best to use facorization to make sure you get ALL the solutions.

$$\\2sinxcosx=2sinx\\
2sinxcosx-2sinx=0\\
2sinx(cosx-1)=0\\
sinx=0\;\;OR\;\;cosx=1\\
x=0,\pi\;\; OR\;\;x=0\\
$The 2 solution are $x=0\:\;and\;\;x=\pi$$

Thanks Alan, I did it exactly the same way BUT

can you see any way of getting the total number of paths possible?

I suppose you could do it backwards.

Any individual path has the probability of   $${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{6}}}}}$$      

So

$${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{6}}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{1}} \Rightarrow {\mathtt{x}} = {\mathtt{64}}$$

So there must be 64 possible paths taken, 22 of these are favourable outcomes

so   P(ending at startpoint) =      $${\frac{{\mathtt{22}}}{{\mathtt{64}}}} = {\frac{{\mathtt{11}}}{{\mathtt{32}}}} = {\mathtt{0.343\: \!75}}$$

This is interesting -

I cannot remember ever using probability backwards like this to work out the number of possible outcomes.  

0^∞  and  1^∞  are "indeterminate" expressions  --- which means that their value is determined by the problem that created those expressions and has no specific value for all cases.

Also, 0^-∞  and  1^-∞  are also indeterminate expressions.

Expressions, such as (½)^-∞ can be reduced to (2)^∞ so it won't approach 0. --- Therefore, fractions between 0 and 1 won't approach 0.

2401  =  (2 + 4 + 1 + 0)⁴

y = 2(x + 1) (x +2) ......let's expand this to get this form ... y = ax^2 + bx + c

y = 2x^2 + 6x + 4  

The x coordinate of the vertex is given by  -b/2a  =  -6/(2*2)  = -6/4 =  -3/2  = -1.5

To find the y coordinate of the vertex, just plug this value into the function and we get

2(-1.5)^2 + 6(-1.5) + 4 = 

2(2.25) - 9  + 4 =

4.5 - 9 + 4 = -0.5

So the vertex is (-1.5, -0,5)

Yep....it is.....

I like using anonymous since no one would hate me if I accidentally gave a wrong answer.

"It does not equal zero, but it gets closer as it aproches -infinity"

Here are some links you can look at:

you just type it in.

for some you may need to put the x  (or whatever) in brackets.