And you want me to believe all that garbage you spew with that language? Learn some manners, and some math!

On Tuesday, September 19, 2023 at 3:49:45 PM UTC-4, Eram semper recta wrote:
> On Tuesday, 19 September 2023 at 13:12:27 UTC-4, Mathin3D wrote:
> > Look f ckt rd, you really do not know much mathematics. Try learning it first before you can discuss it.
> >
> > Look at this dude:
> >
> > https://www.youtube.com/watch?v=63HpaUFEtXY
> >
> > he knows mathematics.
> > On Tuesday, September 19, 2023 at 10:22:44 AM UTC-4, Eram semper recta wrote:
> > > You cannot understand calculus until you understand the Holy Grail of Calculus which is derived from my Historic Geometric Theorem of January 2020.
> > >
> > > To understand the simplicity, elegance and genius of the Holy Grail of Calculus, follow the steps as outline:
> > >
> > > Use Online Geogebra at: https://www.geogebra.org/classic
> > >
> > > Step 1: Graph f(x)
> > > Step 2: Place anchor point (x,f(x)) on graph of f(x)
> > > Step 3: Draw tangent line at anchor point
> > > Step 4: Place another point (x+h, f(x+h)) on f(x) to form a non-parallel secant line
> > > Both points in step 2 and step 4 must be movable.
> > > Step 5: Draw a horizontal through anchor point
> > > Step 6: Draw a vertical through the other end point of the non-parallel secant line
> > > Step 8: Intersect tangent line with vertical from step 6
> > > Step 9: Intersect horizontal from step 5 with vertical from step 6
> > > Join points of intersection to arrive at f1, f2 and h.
> > >
> > > f'(x)= f2/h
> > > Q(x,h)=f1/h
> > >
> > > To test with other functions, simply change f(x). You may have to rescale axes to see all of the curve and diagram.
> > >
> > > As a final exercise:
> > >
> > > you can find the area under the graph of f'(x) from x to x+h by h[ f2/h + f1/h ].Be careful! The graph you are using is f(x), not f'(x). So the area will be for f'(x) and not f(x).
> > >
> > > As a follow-up exercise, you can find the area under the graph of f'(x) from x to x+h by h*[ f2/h + f1/h ].Be careful! The graph you are using is f(x), not f'(x). So the area will be for f'(x) and not f(x).
> > >
> > > You can check your answer by evaluating \int_a^b f'(x) dx because
> > >
> > > \int_a^b f'(x) dx = h*[ f2/h + f1/h ]
> > >
> > > Holy Grail (f(x+h)-f(x))/h = f'(x) + Q(x,h) described in detail here:
> > >
> > > https://www.academia.edu/105576431/The_Holy_Grail_of_Calculus
> > >
> > > and in far more detail with proofs here:
> > >
> > > https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020
> Hey moron.
> The OP clearly states that it's for human consumption, not fuckwads like you.
>
> Impale yourself on a blunt pole, you fucking vile piece of shit!
>
> FUCK OFF, MORON!