{{
`${userDetail.user.displayName}${userDetail.user.cachedBungieGlobalDisplayNameCode ? `#${standardizeBungieGlobalCode(userDetail.user.cachedBungieGlobalDisplayNameCode)}` : ""}`}}
{{getUserClans()}}
10/1/2016 4:56:03 AM
12
Could you clarify your meaning by single variable and multi variable calc
English
-
If you're unsure of the meaning, you're probably in single variable calculus. You can't have taken one without the other, and you would know if you were working with more than one variable. Calculus is the study of change, thus, single variable calculus considers functions which change with respect to one value, while multivariable calculus considers functions which change with respect to more than one value. So like, a linear function you consider in single variable would be: F(x)=Ax+By+C Where as a linear function in multi variable would be: F(x,y)=Ax+By+Cz+D
-
[quote]F(x,y)=Ax+By+Cz+D[/quote] I think that is incorrect. F (x,y)=P (x,y),Q (x,y) Or Ax+By+Cz=D No z since it was not F (x,y,z) = etc......
-
[quote][quote]F(x,y)=Ax+By+Cz+D[/quote] I think that is incorrect. F (x,y)=P (x,y),Q (x,y) Or Ax+By+Cz=D No z since it was not F (x,y,z) = etc......[/quote] Pretty sure you were just complaining about z, not D.... Also, Ax+By+Cz=D And Ax+By+Cz+D Are the same exact thing. D is some constant. Maybe if you understood what was presented in that you would know this.
-
..dbl post
-
There isn't anything wrong with F(x,y)=Ax+By+Cz+D. Remember, F(x)=Ax+By+C? Like, this is standard notation lol.
-
Show me where your formula is on this page..... http://www.google.com/url?sa=t&source=web&cd=3&ved=0ahUKEwjv7o6U6rnPAhVTxCYKHdK0A3oQFggfMAI&url=http%3A%2F%2Fwww.math.harvard.edu%2F~knill%2Fteaching%2Fmath21a2011%2F21a_fall_2011.pdf&usg=AFQjCNHMQ4wWHRiaCGgY2zRE2mdaOMwDyw&sig2=4Bm9yzYN5XXwjCBSJYaQMg It's no where on there. But you will see the ones I posted. Odd isn't it? I guess the people at Harvard are wrong as well........ So, mind elaborating more on your erroneous formula?
-
Like, I legitimately don't understand how you don't understand this lmao F(x,y)=Ax+By+Cz+D Cz=F(x,y)-Ax-By-D If z=0 F(x,y)=Ax+By+D If z=/=0 F(x,y)=Ax+By+(Cz+D) Does that make it easier? F isn't a function of z. That doesn't mean z cannot exist. You've obviously never actually dealt with multivariable functions if you found fault with that lmao
-
Edited by Unicorn Goo: 10/1/2016 7:36:01 PMI'm not talking about z. I know z exists. I'm talking about D. It does not belong in your equation. As I've said, the equation is; Ax+By+Cz=D Not ........+D Again I ask, elaborate....... And I ask [i]again[/i], show me your equation on the link I posted, from Harvard. Not there. Funny though........mine are.......
-
[quote][quote]F(x,y)=Ax+By+Cz+D[/quote] I think that is incorrect. F (x,y)=P (x,y),Q (x,y) Or Ax+By+Cz=D No z since it was not F (x,y,z) = etc......[/quote] Pretty sure you were just complaining about z, not D.... Also, Ax+By+Cz=D And Ax+By+Cz+D Are the same exact thing. D is some constant. Maybe if you understood what was presented in that you would know this.
-
[quote][quote][quote]F(x,y)=Ax+By+Cz+D[/quote] I think that is incorrect. F (x,y)=P (x,y),Q (x,y) Or Ax+By+Cz=D No z since it was not F (x,y,z) = etc......[/quote] [quote]Pretty sure you were just complaining about z, not D....[/quote] Nope. Always been D [quote]Also, Ax+By+Cz=D And Ax+By+Cz+D Are the same exact thing.[/quote] Nope, it would be Ax+By+Cz[b]-[/b]D not +D [quote]D is some constant. Maybe if you understood what was presented in that you would know this.[/quote] And D would not be a constant.
-
[quote]Nope, it would be Ax+By+Cz[b]-[/b]D not +D [/quote] [quote] And D would not be a constant.[/quote] LMFAO YOU DONT KNOW WHAT A CONSTANT IS
-
Oh I'm in single only a junior
