All Answers 358083 Answers

I need to find a way to give you more likes...

Not sure, my best guess is that it is related to how many like one gets with that answer & how popular the question is viewed, you can ask the staffs when they are online about that :)

How do I make someone 'best answer?' 

Oh...I get it now.  Kind of like internet-version Terminator...

The robots won't come for only the purpose to advertising, but used aswell to clog a server's internet connection by putting immense useless data on it, making the connection taking ages to created (i.e. DDoS attack), or to spam personal political messages like who to support in a presidential election etc., also adblock is used to block the ads you see on webpages on the blank spaces left or right to it (i.e. Sponsored content, that's how websites mostly get their funds and money from), not to block comments with ads encrusted in them.

Oh. Thanks for answering again. 

But wouldn't using adblock be better than the captcha?

It's called a captcha, and unfortunatly no, you can't disable it, if you can, then it would defy the purpose that it is created in the first place, since people who want to spam can just disable it and do whatever they want to do on websites.

It might be annoying but hey, atleast you won't see spam advertisements in this forum with their protection :D

Lol...I'll read it now. Thanks :)

I studied particle physics for three years, I might aswell be classified as a "scientist" if you want so.

(How dare you skip over my explanation XD)

I wish I could give you more likes...

Thank you! You are a genius! You should be a scientist (or are you one already?). But I have to say... I skipped most of your explanation and looked at your conclusion... 

And are you related to Albert Einstein? 

Good job for your answer! No wonder those two functions look alike.

Einstein's famous equation \(E=mc^2\) tells us that matter is equivalent to energy, and so does the other way around, in this case, fire is rising air with oxidizing chemicals in them, since it is made out of gas molecules (i.e. matter), it is an alternative form of energy.

It is easy to misintepret that fire is a kind of plasma because of its similarities and uniqueness, but in reality, although fire does ionize atoms, due to the lack of temperature and energy intensity unlike welding arcs and lightnings, most of the atoms are just in the form of heated gas rather than a plasma.

Conclusion: It is both matter, energy, and partial plasma but mostly gaseous.

So...fire is matter?

Because plasma is sometimes considered as the fourth state of matter.

But isn't plasma matter? 

No, frasinscotland.

What you did is compound interest. The question clearly says simple interest.

Finance Company A: $15,000 x 0.125 =$1,875.00 simple interest for 1 year.

$1,875.00 x 5 years =$9,375.00 Interest cost for 5 years.

Finance Company B: $15,000 x 0.102 =$1,530.00 simple interest for 1 year.

$1,530.00 x 6 years =$9,180.00 Interest cost for 6 years.

As you can see, Finance Company B is slightly cheaper by $195. This is on the assumption that no principal is paid back until maturity of 5 and 6 years respectively.

It is?  Thanks.

It is neither!. It is "plasma".

Sorry. Couldn't see the picture.

Well...if the coordinates of the point can be plugged into the equation of the line, and it works, then the point is on the line.

For example, point (1,6) is on the line y=3x+3.

Plug the coordinates in:

6=3(1)+3

6=3+3

6=6

So point (1,6) is on the line y=3x +3

Or you could locate the point on the line in a graph...

Is this image too big? I used https://www.desmos.com/calculator to graph.

Correct me if my answer is wrong. 

All of them

Company A : Principal sum  X (1.125)^5

Company B : Principlal sum X (1.102)^6

Notice that you don't need to actually involve the principal sum in the calculation.  You just need to evaluate the exponentials.

(1.125)^5      compared with  (1.102)^6      and see which is the least for the best deal.

What problems? 

2. -2 A^-2(A^-2+A^-3)^-1+2A³(A²+A³)^-1+ 3/4 A(A^-1+I)^-1+ 3/4 A³(A+I)^-1

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} = \ ?\\ \hline \end{array}\)

1.
\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\\\ && & (A^{-2}+A^{-3})^{-1} = [(A^{-1}+I)A^{-2}]^{-1} \\ && & (A^{2}+A^{3})^{-1} = [(A+I)A^{2}]^{-1} \\\\ &=& -2 A^{-2}[(A^{-1}+I)A^{-2}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2} = A^{-1}A^{-1} = (AA)^{-1} = (A^2)^{-1} \\ &=& -2 A^{-2}[(A^{-1}+I)(A^2)^{-1}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A^{-1}+I)(A^2)^{-1}]^{-1} = [A^2(A^{-1}+I)^{-1}] \\ &=& -2 A^{-2}[A^2(A^{-1}+I)^{-1}]+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 A^{-2}A^2(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2}A^2 = A^{-1}A^{-1}AA = A^{-1}(A^{-1}A)A = A^{-1}(I)A = A^{-1}A = I \\ &=& -2 I(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A+I)A^{2}]^{-1} = [(A^2)^{-1}(A+I)^{-1} ] \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = A^{-2} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3A^{-2}(A+I)^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2AA^2A^{-2}(A+I)^{-1} \quad & | \quad A^2A^{-2} = I \\ &=& -2 (A^{-1}+I)^{-1}+2AI(A+I)^{-1} \\ &=&\color{red}{ -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} } \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2[(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(AA^{-1}+IA^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(I+A^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(A^{-1}+I )^{-1} \\ &=& \mathbf{0} \\\\ && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } = 0 \\ \hline \end{array} }\)

2.
\(\begin{array}{|rcrcl|} \hline && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\ &=& \color{red}{-2 (A^{-1}+I)^{-1}+2A(A+I)^{-1}} & =& 0 \\ && -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} & =& 0 \\ && 2 (A^{-1}+I)^{-1} &=& 2A(A+I)^{-1} \\ && \mathbf{(A^{-1}+I)^{-1}} &\mathbf{=}& \mathbf{ A(A+I)^{-1}} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& 0 + \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A\underbrace{(A^{-1}+I)^{-1}}_{=A(A+I)^{-1}}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 AA(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A^2(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 ( A^2 + A^3)(A+I)^{-1} \\ &=& \frac34 A^2(I+A)(A+I)^{-1} \\ &=& \frac34 A^2\underbrace{(I+A)(I+A)^{-1}}_{=I} \\ &=& \frac34 A^2 I \\ &\mathbf{=}& \mathbf{\frac34 A^2 } \\ \hline \end{array}\)