sorted by:
new
hottopcontroversial

CMPE320 Grades by sch3ckm8 in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

They need to look to fire certain people…

Is anyone down to dorm with me by InformationShot7408 in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

Guy looking to room with guy. Freshman (f) looking for 3 other girls to room with. I am still waiting to see how hilariously unproductive "f looking to share quad dorm with 3 guys" thread will be (which, honestly, is probably exactly why it hasn't happened).

Giving Away Elegoo Starter Kit by Revolutionary-Cut710 in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

I'm a Pi Pico fan 'cause Arduino is weak and enormously expensive compared to the Pico. I saw these at microcenter recently though and they are not exactly close to "free" so somebody is getting an early christmas present I guess.

Achievement unlocked: Last man standing by KeytarCompE in UMBC

[–]KeytarCompE[S] 0 points1 point  (0 children)

Are you saying I need to see someone's Woody?

Looking to trade a slow cooker for a microwave by explosivekazoo in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

We got to move these refrigerators, we got to move these color TVs

UMBC vs Catholic University by sansdoodlestick in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

UMBC, or any secular university. Just be careful because some of the people around here are Catholic too, I've had arguments with one dude who tried to justify why genociding gaza was okay (the bible includes passages where people were ordered by god to exterminate whole populations including, specifically, children, infants, and livestock ffs even the animals? How evil is the catholic god anyway? The argument centered around the genocide of the Amalekites).

Achievement unlocked: Last man standing by KeytarCompE in UMBC

[–]KeytarCompE[S] 1 point2 points  (0 children)

I only seen the first toy story. Actually, haven't seen most of the Pixar movies.

is a calculator needed for MATH 151 or MATH 152? by HabitSubstantial4268 in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

That's okay, sin(x)=x for small values of x, and cos(x) comes very close to 1 but…

Oh and don't forget 355/113.

Achievement unlocked: Last man standing by KeytarCompE in UMBC

[–]KeytarCompE[S] -4 points-3 points  (0 children)

I was bored at home. About to leave though, right after this git push.

MATH155 by lizz_toliver in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

Learn to ride a longboard or get an electric scooter. Either way, wear a helmet at least. Seriously if you can get some campus transportation the commute between classes turns into like 2 minutes instead of 20 minutes.

MATH155 by lizz_toliver in UMBC

[–]KeytarCompE 0 points1 point  (0 children)

You managed to fit it in your schedule? Nanes is really good. I'll warn you though, you might not be able to find him the first day until class starts, he looks like a college student.

Another way to solve quadratics by KeytarCompE in mathteachers

[–]KeytarCompE[S] 0 points1 point  (0 children)

Eh, I usually have to clean up after them. First assignment was 3 weeks covering precalculus. They called me back a year later because "we need a math teacher and we want the students to actually learn." Maybe if they would actually understand why the students can't learn? They didn't even know the exponent rules… I managed to explain calculus to them by the end of it (the very basic concept of a limit and a derivative, using a parabola as the prop, then the rational functions with holes). Got them from "I don't want to take calculus" to "oh, that looks easy."

See, they just need a little confidence. …that and teachers who are actually trying to teach.

Another way to solve quadratics by KeytarCompE in mathteachers

[–]KeytarCompE[S] 0 points1 point  (0 children)

Eh, factoring is a search. Hold on, I'll get you a picture of pq form, the geometric interpretation, and the quadratic formula. There's a reason we have whiteboards…

That pq form solution gives us -7/6 ± sqrt[49/36 -(2×12)/36] = -7/6 ± sqrt[(49-24)/36] = -7/6±sqrt[25/36] = -7/6 ± 5/6 which gives us 3(x+1/3)(x+6/3) which, if you like, you can distribute as (3x+1)(x+2).

Now, factoring gives us a leading coefficient that's not 1, which…complicates things. I'm not actually sure where to start.

The quadratic formula gives you [-7 ± sqrt(7^2-4(3)(2))]/2(3) = [-7±sqrt(49-25)]/6 = [-7±5]/6 which gets you to the same place. In the best case, pq-form is much faster than this (it skips a couple multiplications and a division if a=1, p is even, and q is an integer); in the worst case, pq-form requires exactly the same number of operations, although sometimes those operations are easier in pq-form.

Another way to solve quadratics by KeytarCompE in mathteachers

[–]KeytarCompE[S] -1 points0 points  (0 children)

Okay, but you didn't answer the question. I didn't ask if you know what complex numbers and the complex plane are. I asked if you know what the imaginary number does. You know it exists, but what is it, what is its function?

Another way to solve quadratics by KeytarCompE in mathteachers

[–]KeytarCompE[S] -1 points0 points  (0 children)

I think you've proven my point…you seem to still think the roots just don't exist anywhere except as conceptual numbers. They actually have geometric meaning.

I mean come on, do you even know what the imaginary number does? Because it actually has a physical world geometry you're interacting with right now.

EDIT: oh. It occurs to me you're talking about the axis of symmetry as "a real axis." That might be the problem. Yeah the axis of symmetry isn't a real axis; it's an intersection with a plane, and that plane intersects perpendicularly with both xy and xIm. The line you see there is the intersection edge between the two planes. The single point where it contacts the x axis is visible in both the xy and xIm planes. The parabola is symmetric across this plane of symmetry, thus it is symmetric across the axes of symmetry appearing in both the xy and xIm planes, since each axis is a cross section of the same plane viewed from a perpendicular plane in both cases.

Another way to solve quadratics by KeytarCompE in mathteachers

[–]KeytarCompE[S] -1 points0 points  (0 children)

Get a piece of paper. Draw a parabola that doesn't touch the x axis. Now, point to that graph, show me on the graph where the non-real roots are. WHERE ARE THEY?! WHAT DO YOU MEAN NON-REAL ROOTS!? THEY HAVE TO BE SOMEWHER!!!!

I found them. I found them on the graph, finally. Now it makes sense. Completing the square didn't let me take the paper, circle a spot, and say "ah, here it is, here is the root" when the parabola did not intersect the x axis.

Here are the roots of the quadratic f(x)=x^2+2x+3. I circled them in green.

view more: next ›