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Stop thinking that the world circles around you. You seem like a lazy person who just wants to sit in front of your computer and play games. Btw, this is to IhaveHairystuff

My family actually has a Tesla and actually has a house, unlike most people. 

True

I took a look at it.  Yeah, I remember that problem.  I can't solve it.  Sorry, wish I could help.    

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I've seen many of the questions on here a million times.  I've got to where I save answers I'd previously made and copy/paste them in the the new iteration of the math problem.  I don't think it's one person posting the same thing over and over and over.  Consider a math problem, take this one for example.  I think it's part of a math course – whether classroom, or online, or whatever – and students encounter it at different times.  Maybe they enroll at different times, or work at their own speed, but for whatever reason, they come to the question at different times.  Then they  bring it here.  

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Doesn't that just show that the guest is just a troll and spams a question that has already been asked and answered, while also making no sense at all. 

It's this weirdo again.

True, but you do not know the amount I like the name history. 

To rude guest: please stop! This is getting out of hand!

By the way Bosco (and anyone else interested) this question (https://web2.0calc.com/questions/interesting-question_7) is actually quite interesting and I would like to see it solved. Do any of you guys have ideas.

@Hairyweirdo We are not your subjects. Your self-proclaimed titles are invalid and have no effect on reality. Your screenshot sounds like a projection of your own reality.

I'm sorry that the answer offended you so much you need to declare yourself "a literal genius" when over half of your answers are going to get deleted; and after all no one was talking to you. If you are such a genius, why don't you go solve this question: https://web2.0calc.com/questions/interesting-question_7

In physics, Ohm's law says that current through a wire, I, is directly proportional to voltage, V, and inversely proportional to resistance, R:
I = V/R  
It's also true that resistance is directly proportional to the length of the wire.  We have a piece of wire.  We pass 60 volts through this wire and measure 200 milliamps of current.  If I cut the wire in half and pass 400 volts through it, how many milliamps of current will I measure?   

We begin with the basic V = IR from which R = V / I and I = V / R    

V is in Volts.   I is in Amperes.   R is in Ohms.   

V = 60 Volts   

I  = 200 mA = 0.2 Amp  

                                                                R  =  V / I  

                                                                R  =  60 / 0.2  

                                                                R  =  300 Ohms   

When you cut the wire in half,   

the resistance of one piece of   

it is half of what the total was.  

New V = 400 Volts   

New R = 150 Ohms  

                                                                I  =  V / R    

                                                                I  =  400 / 150  

                                                                I  =  2.6667 Amps  

There are 1000 mA per Amp, thus          I  =  2666.7 mA  

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This is my "negative money" answer. 

I answered this problem a week ago. 

This is a copy / paste of that answer.    

Alice and Bob each have a certain amount of money. If Alice receives n dollars from Bob, then she will have 4 times as much money as Bob. If, on the other hand, she gives n dollars to Bob, then she will have 8 times as much money as Bob. If neither gives the other any money, what is the ratio of the amount of money Alice has to the amount Bob has?    

                                        (A + n)  =  4 • (B – n)   

                                         A + n    =  4B – 4n    

                                         A – 4B  =  –5n                   (1)    

                                         (A – n)  =  8 • (B + n)   

                                          A – n    =  8B + 8n   

                                          A – 8B  =  9n                    (2)   

Multiply both sides  

of (1) by 9                        9A – 36B  =  –45n              (3)    

Multiply both sides  

of (2) by 5                        5A – 40B  =    45n              (4)      

Add (3) and (4)               14A – 76B  =  0   

Add 76B to both sides               14A  =  76B   

Divide both sides by 76B   

                                                  14A          1    

                                                  ––––   =   ––    

                                                  76B           1                       

Multiply both sides by 76/14    

                                                      A          76                         38

                                                    –––   =   –––   reduces to   –––   

                                                      B          14                          7   

It works if you accept the  

concept of negative money.  

That is, n must equal –2 dollars.  

Say Alice has 38 and Bob has 7        

                                                     (38 + n)  =  4 • (7 – n)     

                                                      38 + n   =  28 – 4n    

                                                             5n  =  –10     

                                                               n  =  –2       

and               

                                                     (38 – n)  =  8 • (7 + n)    

                                                      38 – n   =  56 + 8n      

                                                          –9n   =  18    

                                                               n  =  –2       

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>>>> probably to boost their ego through imaginary internet points <<<<   

Maybe you'd get more satisfaction by commenting on things you actually know something about, if anything fits that description.  To the contrary, solving problems here is a humbling experience for me, when I see the number of math problems that I not only (1) don't know how to solve, but also (2) don't even recognize some of the math terminology they use. 

Some of these probably are students bringing their homework here.  But even if I do solve a problem for them, if somebody hadn't shown them, then they wouldn't have seen how to do it at all.  It's to their own advantage to try to learn something from the responders' efforts.  Responders who try to help, that is; not responders like the scolds hiding behind "Guest" postings who imagine themselves the self-appointed conscience of the site.  

_.

To answer this question, we use Ohm's Law. We know that current, I, is equal to voltage, V, divided by resistance, R, so we can rearrange the equation to solve for R:

R = V/I

Using the values given, we can calculate the resistance of the wire:

R = 60V/200mA = 0.3 Ω

Now that we have the resistance, we can use Ohm's Law to calculate the new current when the voltage is increased to 400V:

I = V/R = 400V/0.3Ω = 1333.3mA

Since we cut the wire in half, the resistance of the wire is doubled, so the new resistance is now 0.6Ω. We can use Ohm's Law to calculate the new current:

I = V/R = 400V/0.6Ω = 666.7mA

Therefore, the total current passing through the two halves of the wire is 1333.3mA + 666.7mA = 2000mA, or 2A. Since this is in milliamps, the answer is 2666.7mA.

To find the area of triangle PQR, we need to calculate the lengths of PX and YQ.

Using the Pythagorean theorem, we can calculate the length of PX: 

PX^2 = PR^2 - PQ^2 = 1296 - 1296 = 0

PX = 0 

Using the Law of Cosines, we can calculate the length of YQ: 

YQ^2 = QR^2 + MY^2 - 2*QR*MY*cos(QPR)

YQ^2 = 256 + 2500 - 2*16*50*cos(QPR)

YQ^2 = 2808

YQ = 53.08

Now, we can calculate the area of triangle PQR using Heron's formula:

A = sqrt(s*(s-PQ)*(s-PR)*(s-QR))

where s is the semiperimeter of the triangle

s = (PQ + PR + QR)/2

s = (36 + 36 + 16)/2 = 88/2 = 44

A = sqrt(44*(44-36)*(44-36)*(44-16))

A = sqrt(44*8*8*28)

A = sqrt(2808*28)

A = 280.8

You sound bitter because I’m a literal genius and you’re a nobody.

Well, if I didn't care about myself I'd be a loser. That's a valid hypothesis, but incorrect. I'm spending only a few seconds answering these questions in order to increase my public perception in the eye of my subjects. I'm actually gaining time and power by answering these questions.

Evaluate:   (2013^3 - 2×2013^2×2014 + 3×2013×2014^2 - 2014^3 + 1)/(2013×2014)

Why don't you use your calculator, or just paste it in the calculator here on this site:

Answer = 2013

If you were cheating on another website, I don't think you would use the same username as ao ps.

You are wasting your own time by answering their questions. I think you care about yourself.

This problem has already been asked. Also, this question makes no sense, and this question I've already seen multiple times. Just stop spamming. 

Honestly, who cares about homework cheaters? They are just hurting themselves.