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[Sealda]

The Clapeyron equation,
log10 P = A + B/T

Needs to be converted into Y=AX+B style linear function. The problem is that both P and T are column vectors, full of data.

 

[close to the edge]

Try using sin(2t) = 2cos(t)sin(t).

Thanks (and thanks to the others who responded), that did the trick.

 

[zoku88]

I'm trying to figure this out:



Thanks

If you calculate 2^32, you get that large number, which is 4 * 1024 * 1024 * 1024

 

[Lonely1]

The Clapeyron equation,
log10 P = A + B/T

Needs to be converted into Y=AX+B style linear function. The problem is that both P and T are column vectors, full of data.

Let me see if I understand what you are saying. You want to reduce the dimension of the vectors to 1?

 

[Haly]

Posting for a friend:

u[x] + u[y] + u = e^(x+2y)

Where u[x] is the partial derivative of u with respect to x

u[y] is the partial derivative with respect to y

And the initial condition is u(x, 0) = 0

Way beyond my understanding but maybe GAF can help?

 

[TimesEunuch]

Posting for a friend:

u[x] + u[y] + u = e^(x+2y)

Where u[x] is the partial derivative of u with respect to x

u[y] is the partial derivative with respect to y

And the initial condition is u(x, 0) = 0

Way beyond my understanding but maybe GAF can help?

I got u(x, y) = 1/2 e^x sinh(2y). Seems to work.

If you're want my method of solution: the general solution is given by u(x,y) = u_h(x, y)+u_p(x,y), where u_h is the general solution of the homogeneous equations u[x]+u[y]+u=0, and u_p is *any* particular solution to the original (inhomogeneous) equation. I more or less guessed a particular solution u_p(x,y) = 1/4 e^(x+2y). Using separation of variables, you can show that the general solution to the homogeneous equation is
u_h(x, y) = sum_lambda c(lambda) e^{lambda(x-y)-y} (actually a continuous sum, so really an integral). Now use the boundary condition u(x,0) = u_h(x, 0)+u_p(x, 0)=0 to deduce that all the c(lambda) are zero except for c(1), which is -1/4. So
u(x,y) = -1/4 e^(x-2y)+1/4 e^(x+2y) = 1/4 e^x sinh(2y).

There's probably a more direct way to do this, though.

 

[peakish]

I got u(x, y) = 1/2 e^x sinh(2y). Seems to work.

If you're want my method of solution: the general solution is given by u(x,y) = u_h(x, y)+u_p(x,y), where u_h is the general solution of the homogeneous equations u[x]+u[y]+u=0, and u_p is *any* particular solution to the original (inhomogeneous) equation. I more or less guessed a particular solution u_p(x,y) = 1/4 e^(x+2y). Using separation of variables, you can show that the general solution to the homogeneous equation is
u_h(x, y) = sum_lambda c(lambda) e^{lambda(x-y)-y} (actually a continuous sum, so really an integral). Now use the boundary condition u(x,0) = u_h(x, 0)+u_p(x, 0)=0 to deduce that all the c(lambda) are zero except for c(1), which is -1/4. So
u(x,y) = -1/4 e^(x-2y)+1/4 e^(x+2y) = 1/4 e^x sinh(2y).

There's probably a more direct way to do this, though.

I'd try an easier ansatz, I'm too bad at math to handle sums and stuff

Derivatives of and the function itself must include an exponential term as on the right hand side, so assume that u(x,y) = Ae^(x + 2y), where A is some constant. Then solve the equation for A and check that the function fulfills the DE. I just love expontential functions.

Btw I got the answer to A = 1/4, so u = (1/4)e^(x + 2y).

Edit: Wait, totally missed the boundary condition stuff, I'm waay off mark, blagh.

Edit 2: Borrowing your answer I instead ansatz:ed u(x,y) = A e^x ( e^2y - e^(-2y) ), or 2A e^x sinh(2y) as you wrote. Vanishes at y = 0, can then be solved by derivating the function as I did earlier, A = 1/4 and shown to fulfill the DE. Dunno if that's the correct way to do it though. I really should know this stuff better, hawkward.

 

[alveus]

Thanks for the help, peakish and TimesEunuch (I am the friend of Halycon's who was too shy to post myself). I'm not sure what's going on with the continuous sum/separation of variables that TimesEunuch mentioned? But I did solve for the homogenous and then appended the inhomogenous solution to get the same answers you guys did, although I'm not sure if my method was right.

What I did for the homeogenous equation:
u[x] + u[y] + u = 0
divide everything by u: u[x]/u + u[y]/u = -1
then take some v(x, y) = ln(u) (such that v[x] = u[x]/u, etc) and then solve for v, whereupon I got that v = f(x - y) - y for some arbitrary function f
and then the solution for u follows

What I'm not sure about in the above is whether I'm actually allowed to divide by u like that or if I'm just massively fudging something here.

 

[Gouty]

Solve the following systems of equations by graphing:

X-3y=6
x-y=4

My solution is coming out to (3,0) but the answer on my teacher's key is (3, -1)

Once again, thanks for any help.

 

[Therion]

Solve the following systems of equations by graphing:

X-3y=6
x-y=4

My solution is coming out to (3,0) but the answer on my teacher's key is (3, -1)

Once again, thanks for any help.

You can plug the answers into the equations to see that the teacher's answer is right. Your answer isn't actually on either line.

When it says solve by graphing, is that on a calculator? If so, what exactly are you entering? If you're doing it by hand it's tough to find the problem without seeing the graphs.

 

[Gouty]

I'm doing it by hand.

For the first problem I wind up with m = 1/3 and b= (0, -2)

For the second problem I get m = 1 and b = (0, -4)

Are these right or wrong?

 

[Gouty]

Wow, I'm dumb. I was just looking at the wrong line. It is -1. Hur dur. Thank you.

 

[TimesEunuch]

Edit 2: Borrowing your answer I instead ansatz:ed u(x,y) = A e^x ( e^2y - e^(-2y) ), or 2A e^x sinh(2y) as you wrote. Vanishes at y = 0, can then be solved by derivating the function as I did earlier, A = 1/4 and shown to fulfill the DE. Dunno if that's the correct way to do it though. I really should know this stuff better, hawkward.

This is certainly correct, but it would probably take me some time playing around to come up with this ansatz.

Thanks for the help, peakish and TimesEunuch (I am the friend of Halycon's who was too shy to post myself). I'm not sure what's going on with the continuous sum/separation of variables that TimesEunuch mentioned? But I did solve for the homogenous and then appended the inhomogeneous solution to get the same answers you guys did, although I'm not sure if my method was right.

What I did for the homogeneous equation:
u[x] + u[y] + u = 0
divide everything by u: u[x]/u + u[y]/u = -1
then take some v(x, y) = ln(u) (such that v[x] = u[x]/u, etc) and then solve for v, whereupon I got that v = f(x - y) - y for some arbitrary function f
and then the solution for u follows

What I'm not sure about in the above is whether I'm actually allowed to divide by u like that or if I'm just massively fudging something here.

Honestly, most of the time you're trying to solve PDEs in the standard ways, it kind of amounts to a fudge, so I think you're fine.

Guess I'm not sure exactly how you would go about matching the boundary condition using your method: you'll end up trying to find a function f such that e^{f(x)}+1/4 e^x=0. There's no such real function. This is a consequence of assuming that (the homogeneous solution) u is positive when you take v=ln(u). Though you could get around this a posteriori by taking a branch cut for ln along the positive real axis, say, so then v=w+i*pi, where w is real. That is, f=g +i*pi, where g is real.

Anyways, point is, I think you should always imagine these methods as being fudges that just aid you in getting the final answer. Once you've got that, you can just verify directly that it satisfies the original differential equation plus boundary conditions, and forget about how you figured it out in the first place.

 

[Le-mo]

Ok gaf, I have a math question for you all.

In some cases, the populations of predator and prey oscillate, but the amplitude of the oscillations gets smaller and smaller, so that eventually both populations stabilize near a constant value. Sketch a rough graph that illustrates how the populations of predator and prey might behave in this case.

I thought about it, but still couldn't figure it out. Does it involve some kind of disease to keep the predator and prey population stabilize?

 

[Sdawg36]

Ok gaf, I have a math question for you all.

In some cases, the populations of predator and prey oscillate, but the amplitude of the oscillations gets smaller and smaller, so that eventually both populations stabilize near a constant value. Sketch a rough graph that illustrates how the populations of predator and prey might behave in this case.

I thought about it, but still couldn't figure it out. Does it involve some kind of disease to keep the predator and prey population stabilize?

I'll take a stab at it:
I think what happens is that since there is an unbalanced proportion of predator and prey, so an excess in prey leads to an increase in population growth of predators which in turn lowers the population of prey. When the predators reach a max value, there is not enough prey to support them and their population decreases while the prey's goes back up. I'm guessing that each time the populations rise and fall, they never reach the same max point again and continue to oscillate until they both equal a constant and exist in equilibrium where there is not enough prey to increase the rate of growth of predators, so they need a constant amount of prey, so the prey's rate of growth also stays constant. Basically, each time the population oscillates, the max is lower, so the resulting growth/decline is lower each time until they both reach a point were there is just enough of the other to keep the populations in check without allowing them to decrease or increase. Idk if you needed help with the graph or just why it stabalizes, but the graph would just be an oscillating curve that gets closer to the line its oscillating on each time.

 

[hemtae]

help on an applied integration problem

 

[HamPster PamPster]

What is the actual math behind the excel RATE() function

I have the number of payments, the amount of the payments, and the present value. What is the formula that excel uses to get the rate?

Google isn't helping so I have turned to gaf

Is it just cycling through possible answers like an IRR?

 

[Father_Mike]

What is the gradient of a distance funtion? so

grad(sqrt((x1-x2)^2 + (y1-y2)^2))

 

[Lonely1]

The gradient in R^3 is the vector defined as, if f is your function:

(df/dx,df/dy,df/dy)

so, you only need to derivate:

d((X^2+Y^2+Z^2)^(1/2))/d[x|y|z]

 

[Jonsoncao]

What is the gradient of a distance funtion? so

grad(sqrt((x1-x2)^2 + (y1-y2)^2))

if you define dist function of an arbitrary (x,y) to (x_0, y_0) as d(x,y) = \sqrt{(x-x_0)^2 + (y-y_0)^2 }

then the gradient of this distance function is a vector: grad(d(x,y)) = <x-x_0, y-y_0>/d(x,y)

this is also true for any finite n dimensional linear normed space:
if dist is d(X) = |X - X_0|, then grad(d(X)) = (X - X_0)/|X - X_0|

 

[Junior Asparagus]

I have a question that was on a 5th grade worksheet of a kid I was helping out. All that was given was this:

A box has a surface area of 184 square in. Give dimensions for length, width and height. Hint: find three areas that add up to 92 sq in.

Now, I was able to find answers (there are several) using a computer's help, but I'm still not sure how a 5th grader is supposed to get the answer aside from educated guessing. Someone please refresh my memory.

 

[Feep]

I have a question that was on a 5th grade worksheet of a kid I was helping out. All that was given was this:

A box has a surface area of 184 square in. Give dimensions for length, width and height. Hint: find three areas that add up to 92 sq in.

Now, I was able to find answers (there are several) using a computer's help, but I'm still not sure how a 5th grader is supposed to get the answer aside from educated guessing. Someone please refresh my memory.

To my knowledge, there are an infinite number of solutions (the teacher meant integers?). This question is bullshit.

 

[Junior Asparagus]

Yeah, even with integers there are still like 5 or 6 valid sets of answers.

Just wanted to make sure I wasn't missing anything that the 5th grader should have known.

 

[orion434]

Need some help:

sin (pi*x/a) + sin (pi*2x/a) = ?

edit, trying to simplify it more... but this is actually the best it can be.

 

[Deleted member 10571]

help on an applied integration problem

If you found the answer: Could it be W=1,176MJ?

 

[hemtae]

If you found the answer: Could it be W=1,176MJ?

The answer key says 2,352,000 J

 

[dabig2]

The answer key says 2,352,000 J

Just need to break the basic formula down as much as possible:

Basically, W = &#8747; Fdx where F represents the force.

The hydrostatic force for this problem is also equal to Pressure * Area.
P = m*g*h. For water, the m and g will be 1000 and 9.8 respectively. The h in your problem requires a little more work (pun intended) to figure out. That height refers to the distance the force acts upon the water during its travel out the tank. Well, that distance will be from the bottom of your tank till the very top of the sprout. This means at all times the force is acting upon the distance (a+d) -x = (4+2)-x = 6-x, where x represents the actual spot in the tank the water is currently at where x=0 represents the bottom and x = 6 represents the top of the sprout. Hope that makes sense.

Alright, so now at the moment you have this:
W = &#8747; (9.8 * 1000 * 6-x) * Area

Next part is find your area. That is simply 9xdx in this case. So now you have this integral to work out:
W = (9.8 * 1000) &#8747;(6-x)*(9x) dx
W = (9.8 * 1000) &#8747;(54x - 9x^2) dx

And this integral will operate from x = 0 to x = b, which is 4.

So basically this will be your end equations after you integrate:
W = (9.8 * 1000) * ((54/2)*(4^2) - (9/3)*(4^3))
W = 2,352,000

And that's it.

 

[Deleted member 10571]

Thanks guys

 

[Scuba Steve]

Pretty simple one here.

I need to find the derivative of f(x) = sq.rt. of (1/x^3)

I made it to f(x) = (x^-3)^1/2

edit: I probably won't have help til it's too late

The answer I got is

f '(x) = -1.5x^-0.5

 

[phalestine]

Pretty simple one here.

I need to find the derivative of f(x) = sq.rt. of (1/x^3)

I made it to f(x) = (x^-3)^1/2

ok good, you are on the right track. from here you are going to use the chain rule.

basically if you had (x)^2 wouldnt it just be 2x? really its 2x(1), the (1) comes from taking the d' of whats inside. so just do the same here.

pretend (x^-3)^(1/2) is just (x)^(1/2) you are going to get 1/2(x)^(-1/2)*(1). but its not x, its x^-3, so whats the dervative of that? (-3x^-4) so just replace the 1 with that.
1/2(x^-3)^(-1/2)*(-3x^-4).

double check my answer with wolframalpha.com though

click on show steps http://www.wolframalpha.com/input/?i=derivative+sqrt(x^-3)

edit - fixed.

 

[Scuba Steve]

ok good, you are on the right track. from here you are going to use the chain rule.

basically if you had (x)^2 wouldnt it just be 2x? really its 2x(1), the (1) comes from taking the d' of whats inside. so just do the same here.

pretend (x^-3)^(1/2) is just (x)^(1/2) you are going to get 1/2(x)^(-1/2)*(1). but its not x, its x^-3, so whats the dervative of that? (-3x^-4) so just replace the 1 with that.
1/2(x)^(-1/2)*(-3x^-4).

double check my answer with wolframalpha.com though

click on show steps http://www.wolframalpha.com/input/?i=derivative+sqrt(x^-3)

Thanks for breaking it down, I liked how you substituted the x^3 with x to explain the fundamentals of what's going on.

So 1/2(x^-3)^(-1/2)

becomes

1/2(-3x^-4)^(-1/2)

or is it

1/2(x^-3)^(1/2)(-3x^-4)

I appreciate it, thanks alot!

 

[phalestine]

Thanks for breaking it down, I liked how you substituted the x^3 with x to explain the fundamentals of what's going on.

So 1/2(x^-3)^(-1/2)

becomes

1/2(-3x^-4)^(-1/2)

or is it

1/2(x^-3)^(1/2)(-3x^-4)

I appreciate it, thanks alot!

yeah sorry its

1/2(x^-3)^(-1/2)(-3x^-4)

good catch.

nicer way of writing this is,

(-3/x^4) / [2(x^-3)^(1/2)]
OR
(-3/x^4) / [2(1/x^3)^(1/2)]

 

[Scuba Steve]

yeah sorry its

1/2(x^-3)^(-1/2)(-3x^-4)

good catch.

Thank you kindly <3

 

[big ander]

Those are both, personally, pretty messy ways of writing it. And while your method works, there's a lot simpler way to do it.

Since you have
(x^-3)^1/2
you can multiply the exponents to obtain
x^(-3/2)
And from there the derivation is fairly easy.
(d/dx)(x^(-3/2))
(-3/2)(x^(-5/2))

The way you're doing it also simplifies to this answer.
Quick Edit: ACTUALLY, the answer of

(-3/x^4) / [2(x^-3)^(1/2)]

isn't correct. That should be a negative four in the exponent on top. THEN it simplifies to the correct answer.

 

[Scuba Steve]

Those are both, personally, pretty messy ways of writing it. And while your method works, there's a lot simpler way to do it.

Very nice, thank you too!

 

[phalestine]

Those are both, personally, pretty messy ways of writing it. And while your method works, there's a lot simpler way to do it.

Since you have
(x^-3)^1/2
you can multiply the exponents to obtain
x^(-3/2)
And from there the derivation is fairly easy.
(d/dx)(x^(-3/2))
(-3/2)(x^(-5/2))

The way you're doing it also simplifies to this answer.
Quick Edit: ACTUALLY, the answer of

isn't correct. That should be a negative four in the exponent on top. THEN it simplifies to the correct answer.

Double check that, it should be correct.

From wolfram ___ (-3/x^4)/(2 sqrt(1/x^3)) ___

but yes, that way seems more like the logical way of doing it .

 

[big ander]

Ah you know what, I missed

(-3/x^4)/(2 sqrt(1/x^3))

I thought it was written
(-3(x^4))/(2sqrt(1/x^3))

It really should be much easier to type math on computers.

Spoiler

and I should really work on my OWN homework

 

[Kimosabae]

subscribing

 

[movie_club]

i invest $50 a week for 10 years and i leave the money in the bank for an additional ten years. I make no additional deposits in the second ten years. The account pays at 5% interest compound continuously during these 20 years. What is my balance after 20 years?

I know the answer is $55,617...anyone know why?

 

[Beaner]

I know this isn't the usual kind of math dealt with in this thread, but there isn't a Geography help thread.

I have the latitude and longitude of a point (in degrees, minutes, and seconds), and I also have some northings and eastings of it's distance from a reference point. I'm trying to find where this reference point is, so that I can find the northings/eastings and properly map some other points which I only have the coordinates of (no northings or eastings).

The reference point is pretty far away (about 6,000,000meters) so it's been hard to figure it out using google maps or anything like that. Is there some conversion that can be done or what have you?

 

[TimesEunuch]

I know this isn't the usual kind of math dealt with in this thread, but there isn't a Geography help thread.

I have the latitude and longitude of a point (in degrees, minutes, and seconds), and I also have some northings and eastings of it's distance from a reference point. I'm trying to find where this reference point is, so that I can find the northings/eastings and properly map some other points which I only have the coordinates of (no northings or eastings).

The reference point is pretty far away (about 6,000,000meters) so it's been hard to figure it out using google maps or anything like that. Is there some conversion that can be done or what have you?

Bear in mind I know nothing about this topic, and am just going by what I read here, but: do you know that your reference point is one of the standard UTM zones? If so, you could use this to convert longitude/latitude for each of the 60 zones, and see which matches your northings/eastings value best. That will tell you which UTM zone you're in, and you can then use that to calculate northings/eastings for all your other points. Otherwise, you're just going to have to solve the equations at the bottom of the above linked Wikipedia page for the reference meridian latitude lambda_0. Probably want to use a numerical solver for that (e.g. Mathematica, Matlab).

No idea if this even helps.

 

[Seda]

How can I prove these two field extensions are equal?

Q(&#8730;3, -&#8730;3, i, -i) = Q(&#8730;3+i) where Q is the field of the rational numbers.

I got Q(&#8730;3+i) &#8838; Q(&#8730;3, -&#8730;3, i, -i), that direction is easy.

How can I prove Q(&#8730;3, -&#8730;3, i, -i) &#8838; Q(&#8730;3+i)?

Let b &#8712; Q(&#8730;3, -&#8730;3, i, -i)

Therefore b = q + p&#8730;3 + ri where q,p,r &#8712; Q

The fact that Q is a field and has additive inverses takes care of the -&#8730;3 and -i.

But I have to prove b = q + p&#8730;3 + ri = q + (SOMETHING IN Q)* (&#8730;3+i)

Any ideas?

 

[Ave22]

I need assistance in solving what looks like a simple system of equations.

2A + B = 9
C + B = 9
2A + D = 18
C + D = 18

I've figured out A=3 B=3 C=6 D=12, but I need to prove this step by step, and I keep going in circles without being able to isolate any variables.

 

[TimesEunuch]

But I have to prove b = q + p&#8730;3 + ri = q + (SOMETHING IN Q)* (&#8730;3+i)

You made a mistake at this point. You're forgetting that Q(&#8730;3+i) doesn't just contain &#8730;3+i, but also (&#8730;3+i)^2 and (&#8730;3+i)^3. If you work these two out, it should be clear why Q(&#8730;3+i) contains i and &#8730;3.

I need assistance in solving what looks like a simple system of equations.

2A + B = 9
C + B = 9
2A + D = 18
C + D = 18

I've figured out A=3 B=3 C=6 D=12, but I need to prove this step by step, and I keep going in circles without being able to isolate any variables.

There are actually an infinite number of solutions to this set of equations, which is probably what's throwing you. The best way to see this depends on how strong your linear algebra is, but here's a straightforward way: elimate D by subtracting the third and fourth equations - you get 2A=C. Now using this in the second equation, you'll get the first. So your four equations aren't independent.

What you'll need to is write three of the variables in terms of the remaining one. Might as well pick A as the independent variable. Then (as above) C=2A, and B=9-C=9-2A (from the second equation), while D=18-C=18-2A (from the fourth equation). Your general solution is (A, B, C, D)=(A, 9-2A, 2A, 18-2A). You can check this satisfies the original set of equations. A=3 corresponds to the original solution you found.

 

[Ave22]

Problem solved.

 

[hemtae]

Ah, that makes sense. I was focused in on A=3 B=3 C=6 D=12 because this is part of a geometry problem. But since there are infinitely many solutions to the system of equations, this doesn't help me. I'm still stuck without a clue as to what to do.

Given an isosceles trapezoid ABCD, with bases BC and AD, legs AB and CD, where AB=BC=CD, and angles ABD and DCA are right angles, I need to prove that angle A=D=30 and B=C=120.

Problems like these usually involve some clever extensions of lines, but after tossing this around for two days now, I still can't see how I could prove this.

I don't think you can prove that since those angles just add up to 300 while a trapezoid is suppose to have a sum of 360 for its interior angles.

 

[Therion]

But I have to prove b = q + p&#8730;3 + ri = q + (SOMETHING IN Q)* (&#8730;3+i)

Do you really have to prove that?
I would think b = q + p&#8730;3 + ri + s&#8730;3i = w + x(&#8730;3+i) + y(&#8730;3+i)^2 + z(&#8730;3+i)^3. But (&#8730;3+i)^2=2&#8730;3i and (&#8730;3+i)^3=8i, so you should have all the pieces you need to equate the coefficients.

I think that should do it, but I haven't worked out the full proof.

Edit: Bah, I didn't see that this was already answered.

 

[McKeeverFever]

Problem Solved

 

[Ave22]

Ugh, I solved my trapezoid problem - turns out it was the easy one, and the one that I thought was easy turns out to be the hard one.

Given isosceles triangle ABC, with AB = BC, and given a point D (along with line AD) on BC such that AC = BD, with angle B = 20 and angle C = 80, find and prove the angle of ADC = 30.

I thought I had this one figured out until I started writing out my proof, then it all fell apart. I created another isosceles triangle with base EB, where ED=BD, and the base angles were both 20; but it didn't prove anything.

 

[big ander]

My question:
A building consists of 2 floors. The 1st floor is attached rigidly to the ground, and the second floor is of mass 1000 slugs (pounds*seconds squared all over feet) and weighs 16 tons (32,000 lbs). The elastic frame of the building behaves as a spring that resists horizontal displacements of the second floor; it requires a horizontal force of 5 tons to displace the second floor a distance of 1 foot. Assume that in an earthquake the ground oscillates horizontally with amplitude A° [how I'll write A-naught] and circular frequency &#969;, resulting in an external force F(t)=mA°(&#969;^2)sin[&#969;t] on the second floor.
a) What is the natural frequency &#969;° of oscillations of the second floor? Answer in hertz.
b) If the ground undergoes one oscillation ever 2.25s with an amplitude of 3 inches, what is the amplitude of the resulting forced oscillations of the second floor?

Ugh, I solved my trapezoid problem - turns out it was the easy one, and the one that I thought was easy turns out to be the hard one.

Given isosceles trapezoid ABC, with AB = BC, and given a point D (along with line AD) on BC such that AC = BD, with angle B = 20 and angle C = 80, find and prove the angle of ADC = 30.

I thought I had this one figured out until I started writing out my proof, then it all fell apart. I created another isosceles trapezoid with base EB, where ED=BD, and the base angles were both 20; but it didn't prove anything.

Do you mean triangle or trapezoid? Because with only 3 vertices you can't have a trapezoid.