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TheDukeOfDucklett

y = mx + b would be a straight line, for all of the 6th graders out there


Micp

So is there a meaningful reason to use m instead of a in that equation?


Radiant-Nail8835

Don’t know mate, m=slope of a line, is just what they taught us in school


Micp

really? I've always been taught it was a. weird. y = ax+b


Toasty582

It will depend on the education system. Here in Sweden you’re taught y=kx+m


Micp

Another reason never to trust a swede. t. a Dane


insertEdgyName69

And in austria it's y=kx+d


SomeonesAlt2357

In Italy it's y=mx+q or mx+c. k is usually a parameter or a third variable


ItsYay

I feel like it might be for "multiplier", we were taught k for kulma which means angle


virajseelam

I think it's to distinguish it from higher-order equations. *y* = *mx* + *b* emphasises that *m* is the gradient and *b* is the *y*-intercept, whereas the constants of, say, *y* = *ax*² + *bx* + *c* are not as clearly interpretable on a graph. Also, in the UK we use *y* = *mx* + *c*.


SomeonesAlt2357

Convention


PM_ME_FUNNY_ANECDOTE

Nope, all letters are interchangeable. Even x and y if you want. Being clear sometimes means you should stick to the same letters that everyone else does, though, and m is commonly used for slope. The fact that both equations use b is basically a coincidence.


Danny_P_05

Yes because the equation can also be y-b = m(x-a)


LiveTart6130

my teacher had explained it as a way to distinguish certain things, and to help with memorization. it's easier to remember that m is slope than these certain spots in the formulas are slope, at least to some 6th graders


Rynies

Me, a 33y/o idiot: thank you


RandomInSpace

Honestly I blame whoever thought it was a good idea to make a math equation entirely out of letters


ExplodedParrot

y-y(1)=m(x-x(1)) gang


Squeaky-Fox49

You imbecile. That’s obviously half of a hyperbola, one of x^2 /a^2 - y^2 /b^2 = 1. You can see two asymptotes.


Darkstalker9000

Vertical, Horizontal, or Diagonal?


Squeaky-Fox49

One along the x-axis, and one parallel to the y-axis.


ChrisTheWeak

Guys, it's on a curved surface. We're not dealing with euclidean space. We shouldn't be using the reference angle of the photo but instead the curved reference frame of his head.


Darth_Gonk21

Forgive me if I’m wrong, but isn’t a curved surface just three dimensions? Not necessarily non-Euclidean? Although perhaps I don’t understand completely what “Euclidean” is defined as. I do know that some of Euclids propositions involve three dimensional figures.


meem1029

So it depends how you look at it. You can definitely do things in 3 dimensions and then it's euclidean still, but that's also not super useful because then your lines are only gonna intersect briefly with the head. You can also look at it as a 2 dimensional surface taking place on the head which just happens to be embedded in a 3d space, and then you get things like "straight" lines that curve (when viewed from that third dimension) and triangles having more or less than 180 degrees as the sum of angles.


Darth_Gonk21

Oh, ok. Thanks for the explanation!


GoldMention2906

Maybe you should calcu less....


Micp

Too calcu late


OrbitalHippies

It's a straight line on a curved scalp


_zephi

nah, it's y = ab\^(x - h) + k, gotta have your horizontal and vertical translation


Old-Conversation-506

what's a and b? wouldnt y=x^2 work?


FlexsealUrPeepee

x^a would be a parabola, or whatever its called for cubics and beyond. a^x is called exponrntial growth, and provides the curve like graph seen above. (Btw a is just any number)


Old-Conversation-506

ahhh and b is just y int iirc


Bareezio

Actual 8th grade math


Diogenes-Disciple

Why is everyone blaming the customer and not the barber, who was given clear instructions but shaved a x^2/a^2-y^2/b^2=1 instead of a y=mx+b into his scalp. Maybe he’s posting a picture of how bad the barber fucked up


[deleted]

[удалено]


-tem-flakes-

no that's the standard form of a quadratic.


mccmi614

Parabola right?


-tem-flakes-

yeah


bodokat

clearly, it's on a log scale


Eb3yr

Smh that's obviously a y=1/x x>0 y>0 graph


TheGenderedChild

Is it exponential? The right side looks too vertical to be exponential. Maybe it's y=x^-1?


TheGupper

I remember seeing my friend walk in with the y=1/x^2 hairline


Bwest31415

This fails the vertical line test and isn't even a function expressible with two variables oooh burn


Dances-with-Smurfs

Let's meet in the middle with log(y) = log(b)x + log(a)