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I get your point, but nothing justifies mocking someone. I read the questions he answered, and i dont think its nice to do that. I think the question you gave an example of is legitimate. But you think its not. Suppose its not. So what? what if that user made a mistake, or forgot something, what if he's an idiot? it doesnt justify mocking him in front of everyone.

I know those moments. Sometimes im REALLY frausterated, sometimes i think someone asks a really stupid question, and i feel like i just NEED to tell him. How could someone be that stupid? we all have those moments. But unlike him, i calm myself down and remind myself its just a question. No need to write those things.

And yeah, im not going to lie, what im doing is very embarrassing for ginger ale. But that is the only way to stop him, to let him know he's rude. Im doing it because i dont want him to swear more people. And i know some people, some of the people he made fun of, agree with me. Its not my "punishment". im doing it because there is no other way, and i want him to stop hurting other people.ginger ale, Im sorry it has come down to this.

I cant message him. He wont listen, and no one else can be there to tell him to stop. a conversation of 2 people cant end. trust me. reporting him wont help either. It will delete the posts, but it wont stop him, and wont sweet the bitter taste in the mouths of the people he made fun of.

again, im really sorry it had to end like that.

Why, are you going through my profile?

No comment.

1. Divide the red numbers and multiply the yellow numbers

\({\color{red}22\div 5}+{\color{yellow}21\times 2}\)

2. Add the red numbers

\(\color{red}4.4+42\)

Answer

\(46.4\)

I don't think you can do that on this calculator. I could be wrong.

1. Add up the positive numbers

\({\color{red}5+5+5+5}+5x{\color{red}+3}\)

Answer

\(23+5x\)

Lol, but the reCAPTCHA system is by google, I'm pretty sure you can report problems here: https://groups.google.com/forum/#!forum/recaptcha

Here are a few ways to indicate an angle.

The first two seem the most clear, to me.

The “\degree” command does not work in Tex, but the explicit ° sign renders properly.

\(\begin{array}{|rcl|} \hline \angle {ABC} &=& 27° \\ \measuredangle{ABC} &=& 27° \\ \stackrel {\; \frown} {ABC} &=& 27° \\ \stackrel { \hspace{.1em} \wedge} {AC} &=& 27° \\ \stackrel {\, \hspace{.1em} \frown} {AC} &=& 27° \\ \hline \end{array}\\\ \hspace{1.5em} \overset{\; \frown}{AC} = 27° \\ \)

The third and the last option look identical but use different commands. The “\overset” command does not work inside an array for some reason.

Here are the Tex commands in text form.

\begin{array}{|rcll|}

\hline

\angle {ABC} &=& 27° \\

\measuredangle{ABC} &=& 27° \\

\stackrel {\; \frown} {ABC} &=& 27° \\

\stackrel { \hspace{.1em} \wedge} {AC} &=& 27° \\

\stackrel {\, \hspace{.1em}  \frown} {AC} &=& 27° \\

\hline \end{array}\\\

\overset{\; \frown}{AC} = 27° \\

Im going to have a liver transplant surgery, but you're right! i should be proud, not insecure! im beautiful on the outside AND inside! thanks Arya!

a=(b1+b2)/2 h

h=16in

b1=32in

b2=24in

\(a=\frac{32+24}{2}in \cdot 16in=448in^2\)

or

\(a=\frac{[32+24)in}{2\cdot 16in}=1\frac{3}{4}\)

  !

In this case it is the same thing but the calculator is using correct order of operation.

X/100*59/118   

(X/100)*(59/118)

assuming that the distribution is normal and the mean is 100 then the prob will be 88%

Haha thank you Melody!

I obviously didn't make this though! Just thought I'd clarify that 

I love your illustrated thought process Hectictar :)

I used this site:

What the question is apparently saying is that y prime = y double-prime . The only place I've heard of primes is in geometry, so good luck! Sorry I couldn't help more. DX

Arr, Hectictar was answering :)))

Yes, our guest is correct.   :)  

(I though Hectictar was answering ://)

Anyway...

I always struggle with these myself.  I just spent quite a bit of time working it out.

2, 4 and 5 start with a positive or 0 gradient so 3 cannot be a derivative of any of them.

so 

g    must be  4.4.3    (which I will call 3)

Now 3 has a zero, or close to zero gradient at  x=0,  and x=2.8    (they are turning points)

This is where its derivative must cross the x axis and it cannot cross any other place.

Only 5 does this.

g'   is graph 5

5 has a zero gradient at about    x=0,x=1.8 and maybe x=3.8.

So its derivative must be 2

g'' is graph 2

2 has a zero gradient at about  x=0,  x=1.2 and x=2.7

So its derivative must be  4

g''' is graph 4

It is easier to see what is happening if you line the graphs up one directly below the other.

There are other features you can use to but loking at whether the gradient is positive or negative or 0 is the main one. Looking at changes in concavity can also be used.

I think the answer is:

Figure 4.4.3 = g

Figure 4.4.5 = g'

Figure 4.4.2 = g''

Figure 4.4.4 = g'''

Because:

Look at figure 4.4.3.... the "peaks of the hills" are where the slope = 0

The peaks of the hills on this graph occur around x = 0 and x = 2.9

The slope = 0 around x = 0 and x = 2.9

The derivative = 0 around x = 0 and x = 2.9

The graph of the derivative crosses the x axis around x = 0 and x = 2.9

And figure 4.4.5 does just that! So figure 4.4.5 must be the derivative of figure 4.4.3.

Figure 4.4.5 has peaks around x = 0, x = 1.8, and x = 3.8

The derivative of this graph should cross the x-axis around x = 0, x = 1.8, and x = 3.8

Figure 4.4.2 does this, so figure 4.4.2 must be the derivative of figure 4.4.5.

Then just continue with that train of thought.. That's just how I did it! 

What i hate seeing that word sin or it is from the calculator i am in the wrong site. BYEEEEE!!!!!!!!!

Sorry we're all dead at the moment