help-0

[balmy plank]

how do we know pr+s>ps+s

is it because s<r

[merry depot]

b=pr+s, 0≤s<r.

[oak perch]

Yeah

[merry depot]

you'd still need to justify that p is at least 1

you're only given that p is an integer

[balmy plank]

ah would I just say since a and b are greater than 0 then p also has to be greater than 0 but i don't think that works

[merry depot]

that's a good start

if p was negative then b would be negative (given the restraints on r and s)

you need a little more than that for excluding 0

why can't b = s?

[balmy plank]

because s<b

[merry depot]

by transitive property, yes

[balmy plank]

okie thank you so much!

.close

[lone heart]

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[solar oasis]

Can someone explain the first part to me

[solar oasis]

How does x^3 -xy + y^3 differentiate to 3x^2-3y with respect to x

[limpid spade]

if u diff wtt x

consider y a constant like 2,4,pi etc and ull see

[solar oasis]

Wouldnt it just be 3x^2 - y thi

Tho*

[limpid spade]

yes

[solar oasis]

Why is it 3y then

[limpid spade]

error maybe

[lone heart]

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[lone heart]

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[alpine sable]

by 16 years, 99% of patients have at least a 2-year period of remission

[alpine sable]

can this be solved?

[tacit arch]

• Show your work, and if possible, explain where you are stuck.

[alpine sable]

99% of the patients will have a crisis 2 out of 16 years

what percantage of time will one patient stay in crisis

[lone heart]

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[lofty tartan]

will somebody help

[alpine sable]

brother ive never seen this shit in my life

[lone heart]

@lofty tartan Has your question been resolved?

[placid zinc]

So the line you've got there says
"The battery starts at 25% and drops to 0% one minute later"

Can you see why your line is saying that?

[lone heart]

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[blazing rose]

I have no idea what to do here

i know theres 2 properties that determine if a function is a homomorphism

f(X + Y) = f(X) + f(Y)
and
f(c * X) = c * f(X)

where X and Y are vectors

but how do i check these properties

[obsidian chasm]

For the first property, choose two arbitrary vectors in the domain, in this case quadratics, say ax^2 + bx +c and dx^2 + ex + f. Plug these into the function, and see if you can manipulate one side of the property to get the other.

[placid zinc]

Less stylish but usually easier, simplify both f(x + y) and f(x) + f(y), show you get the same thing

Might not matter here though.

[blazing rose]

can you explain

[placid zinc]

rzrshr is right on the money with his suggestion

Mine is just a style difference

[blazing rose]

im a little slow in lin alg tho so bear with me

we have f(x) = ax^2 + bx + c

right

thats the function they give us

[placid zinc]

Ah I see the problem

Yeah we should be consistent here

f(x) = ax² + bx + c
We shouldn't really think of this as a function, it's a polynomial. In this context, it doesn't take inputs/outputs, it's just an algebraic element in its own right

h(x) is the function they gave you, which takes quadratics, and returns members of R²

[blazing rose]

oh yeah so h requires quadratic inputs and gives back a matrix in 2 dimensional space?

[placid zinc]

In order to complete the question, you want to prove that
h(u + v) = h(u) + h(v)
h(au) = ah(u)

h gives back a vector with two real entries, yes

So as per the suggestion, compute h(u+v) for some general quadratics u and v.

Show that this is the same as h(u) + h(v)

[blazing rose]

ok ill give it a try right now

thank you

@placid zinc im confused

i am not doing this right

[obsidian chasm]

try collecting like terms on that last line

[blazing rose]

there are none ?

are there

[obsidian chasm]

ax^2 + dx^2 = ?

[blazing rose]

shoot

ok so thats (a+d)x^2?

[obsidian chasm]

yep

[blazing rose]

OH

wait

do i group c + f

then i have a coeff for x^2, x, and 1

so then i can put that in the form of a matrix

right

[obsidian chasm]

a vector yes

[blazing rose]

a vector?

[obsidian chasm]

it looks like a matrix, but the rows in the example in the question are a+b, a+c, not a b, a c.

[blazing rose]

OH

RIGHT

righhtttt

ok so then i do the right hand side too

and see if they are equal

[obsidian chasm]

yes

[blazing rose]

brb

gonna try

]89

so this is the left hand side

BOOM

so since these are equal to each other, the function h is a homomorphism

[obsidian chasm]

perfect

[blazing rose]

no wait

i need to do the other check now

[obsidian chasm]

right

[blazing rose]

thanks so much

linear algebra is so confusing

i feel like theres so many gaps that havent been closed

in my knowledge of this subject

[obsidian chasm]

it gets easier with practice

next time you have to prove something is a homomorphism, the argument will be very similiar to the one you just did

[blazing rose]

did i do this right?

i dont think so

[obsidian chasm]

q isn't being plugged into the quadratic

its just a scalar out the front

[blazing rose]

oh

OH

[obsidian chasm]

remember we aren't really treating the quadratics like functions currently

[blazing rose]

so with h(qv) its aqx^2 + bqx + c

and with q*h(v) its q(ax^2 + bx + c)

?

[obsidian chasm]

yes

[blazing rose]

wait

[obsidian chasm]

just don't forget its +cq

not +c

[blazing rose]

yeah thats what i forgot

alright

thank you for all the help

[lone heart]

@blazing rose Has your question been resolved?

[lone heart]

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[hidden hazel]

Hey I need some help with a calculus problem I don't know how to come up for more than one answer and I don't know how I am supposed to use non identity functions to solve the function operation
I keep getting 9-x for f(x) and square root of x for g(x) but that's apparently incorrect

[grim delta]

fog(x) means you’re passing g(x) in as the “x” of f(x)

like f(g(x))

[hidden hazel]

doesn't it mean f composed with G

yeah but I've done that and I keep getting an incorrect answer and I don't know what they mean by non identity funcionts

*functions

[grim delta]

a good example

[grim delta]

an identity function is like y=x
where nothing changes the output from the input https://en.m.wikipedia.org/wiki/Identity_function

no identity functions is basically just saying don’t cop out and do something like f(g(x))=sqrt(9-g(x)) where g(x) is just x

it’s not what you did wrong

[hidden hazel]

okay so let me try it again...

[lone heart]

@hidden hazel Has your question been resolved?

[hidden hazel]

almost there

so if F composed with G= H(x) which is square root of 9-x then I need to find the values of f(x) and g(x) that will equal h(x) when multiplied right?

[grim delta]

not necessarily multiplied

[hidden hazel]

or do they just want me to find the two functions

[grim delta]

think of f(x) as an outer function and g(x) as an inner function

[hidden hazel]

oh. wait...Yeah that helps a lot thanks

[grim delta]

it might be helpful to rewrite it as
$\f(g(x)) = (9-x)^{\frac{1}{2}}$

[hidden hazel]

got it so rewrite it and then solve. I kept trying to derive the answer from H(x) but that's not going to help so I'll rewrite it and try again

[ocean seal]

Dogecode

[hidden hazel]

okay I'm officially stuck.

[buoyant zephyr]

What r u trying to do

[hidden hazel]

I'm trying to solve this problem but I can't figure out the two solutions or how to use the correct values

[grim delta]

ill tell you those answers would be correct if it was asking for gof(x) [aka g(f(x))]

[hidden hazel]

but it's not if Fog. so what do I need to substitute] differently

[grim delta]

think about what you can separate out as the inner function that can be plugged into an outer function

this is a really good couple of examples of how this works

[hidden hazel]

okay

do you have an answer? I think I know how it works but I want to be surer

*sure

also so I can prevent getting stuck like this on the SAT

yeah I'm really stuck

sorry

[noble sinew]

huh? You were just told it was correct if it asked for g(f(x))?

so try flipping definition of f and g?

[lone heart]

@hidden hazel Has your question been resolved?

[lone heart]

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[native temple]

This rule of alturnating interior angles was mentioned nowhere in my textbook

why did they mention it here

also my textbook is algebra and trignometry by openstax

sorry

taken

[native brook]

i it took a while to upload sorry

[native temple]

okay

[lone heart]

@native temple Has your question been resolved?

[native temple]

<@&286206848099549185>

[swift shore]

@native temple what’s the question exactly

Do you not know what “alternating interior angles” means

[native temple]

yes

I have no clue

[swift shore]

Ah

It’s from geometry

Give me a sec

“Alternating interior angles” is just a fancy name for this Z shape

The thing about Z shapes is their angles are always equal. This is because a Z shape is symmetrical: it looks exactly the same if you rotate it a half turn around its center point. That means the angle at the top must be the same as the angle at the bottom. Does that make sense?

@native temple

[native temple]

I forgot geometry class

[swift shore]

Don’t worry, just read that explanation and think about it for a second

[native temple]

yes it does

[swift shore]

Ok cool

That’s really all there is to it

In your problem the Z is a little hard to see, but it’s there

[native temple]

waitI'm having trouble in my trig class who's final lectures are about problem solving
does this mean I just have to practice more math problems

[swift shore]

You’re having trouble with problem-solving, and you’re wondering what to do

?

[native temple]

yes....

[swift shore]

Yeah so practice is really all there is to it, but there are a couple caveats.

First, you need to understand what you’re doing and why. If you’ve only learned how to mechanically solve a type of problem, but don’t really know why the method works, you’ll have a lot of trouble when the problem switches up a little bit.

Second, you need to have some good problems on hand. You can’t just keep doing the same type of problem over and over again. The whole vibe of problem-solving is being able to take a fundamental set of ideas and tools and apply them to a wide range of problems; having some good problems is very important for this.

[native temple]

openstax textbook?

[swift shore]

What

Wym

What about it

[native temple]

it has problems

you know what openstax is right?

[swift shore]

No, I may have heard of it tho

[native temple]

free textbooks

[swift shore]

Ah

[native temple]

https://openstax.org/details/books/algebra-and-trigonometry

my book

https://openstax.org/

all of them

[swift shore]

Well I don’t know it, so I can’t really recommend it

But

[native temple]

…what?

[swift shore]

Ok I just checked it out

This is my worst nightmare

It’s like they copied and pasted the same problems over and over again

Ok gimme a sec

I think I found a place with some good problems

[native temple]

how?

[native temple]

They just do the same category?

this textbook is bad???

[swift shore]

It’s not bad per se

They have some good stuff

But the problems look dreadful

It’s like the students are chimpanzees, trained to do one task over and over again

It’s just sad

[native temple]

huh

[native temple]

I don't get it

it's just repeating the same problems

right?

[swift shore]

Yes, that’s the issue

If you know how to do one of them, you know how to do all of them

There’s no new skill being attained; you’re just getting faster at doing one very specific type of problem

[native temple]

I think it's practice

in case you still struggle with one type of issue

besides I don't do all of those exercises

because that's just ehhh practice

[swift shore]

Yeah

Well in any case

We need to find some trig problems that vary

I’m trying to look online

https://artofproblemsolving.com/wiki/index.php/Category:Introductory_Trigonometry_Problems

These are very good problems, pretty difficult

Do the first three (do not look at the solution until you have a final answer)

And let me know how it goes

I have to leave for 20 minutes now sorry

[native temple]

those don't they have different categories

I mean those are problems for a unit

[swift shore]

Wdym by different categories

Like this is “law of sines” problem, this is a “Pythagorean identity” problem

[native temple]

hea

that’ what it does define vary

[swift shore]

So this might sound foreign to you, if you’ve never done problem-solving stuff

But that’s rubbish

Good problems are ones that arise naturally, and that you don’t know how to solve yet

They may use a variety of different ideas and techniques

[native temple]

how is that part of a unit?

Like if I say law of sines you would want to bring in law of cosines

and vectors

or polar coords

[swift shore]

That’s what I’m saying

Mathematicians don’t invent problems in a box

[native temple]

I suck at that with math

every time a lecture bring in someht8ng I didn’t know I get frustrated and fail

[swift shore]

There’s no such thing as a “law of cosines” problem. You can have problems that use the law of cosines in the solution, but not the entire problem is based off of it

[native temple]

2 years ago I was crying about this literally not kidding

[swift shore]

I totally get it

I know all this might sound ridiculous but the school system sucks at teaching math

[native temple]

my dad said it’s how school works I never fully accepted it

[native temple]

IDK i can’t judge the school system

[swift shore]

Every mathematician would agree with this

[native temple]

I’ll only judge it when I’m out

[swift shore]

You can absolutely judge it, you are the buyer, the consumer

[native temple]

nahh

I don’ believe that

are you a mathmatician?

[vagrant phoenix]

Can someone help me with linear relations

[swift shore]

#❓how-to-get-help

[swift shore]

I like to think I am, but I’m far from a professional one lmao

You probably have an idea that criticizing education as a student is “not your place”

[native temple]

it isn’t

[swift shore]

It absolutely is your place. You’re the one receiving the product; you have the right to have an opinion about it

You just told me you were crying about this two years ago, and now you’re saying you as an individual have no right to criticize it

[native temple]

Yes

[swift shore]

You don’t see the obvious problem here?

You were pushed to tears because you couldn’t do some math problems. You’re rejecting the evidence of your eyes and ears; the system is terrible and you know it

[native temple]

alright

[swift shore]

Anyway

[native temple]

anyway

Uhh practice right?

Idk what you’re talking about there but practice problems that come a bit out of left field right?

[swift shore]

I’m trying to help

But I can try to find some practice problems

You said you had that book

What was wrong with those problems?

@native temple

[native temple]

nothing

Really

Just I need to do them I guess

[swift shore]

Ok

[native temple]

you said something was wrong

that i wasn’t varied

[swift shore]

The problems don’t vary

But that’s irrelevant now

Im sure the problems in the book are fine for your class

[native temple]

So just do multiple problems from multiple units

so I can master problem solving?

[swift shore]

Yeah, I think that’s a good idea

Definitely go across multiple units

Make sure you don’t forget anything

[native temple]

that’s probably the intention

tbh

[swift shore]

Yeah

Is this unit all problem solving?

[native temple]

yep

[swift shore]

Sorry I mean

All trigonometry

?

[native temple]

prelude to finale exam

yea this was my trig

[swift shore]

Ok cool

So I think it would be good to review a bit of geometry

As you saw in the problem you posted, you need some tools from geo

[native temple]

I have baron’s regents geometry 2020

That was from a textbook not my course

the problem it was a similar one though

[swift shore]

That’s probably fine

But it might be hard searching through a big textbook looking for review material

For example you won’t need to know about quadrilaterals

[native temple]

I can do it

[swift shore]

Ok

[native temple]

uhhhh…right

Page 1 I guess

or that khan academy course

probably both

[swift shore]

Khan academy is great

You could skim through their geometry course and stop whenever you see something you don’t know

[native temple]

ahh

a textbook i a course in book form right?

[swift shore]

Not exactly

Textbooks are mostly reference materials

They’re not really pedagogical, meaning they’re not good for learning something the first time through

But if it’s review then you’ll be fine

[native temple]

so what happends if you buy a textbook to learn soemthing and go thri it covor to covor

Someone did this right?

[swift shore]

Yeah I did that once

[native temple]

I swear someone did this

What happends?

you’ll probably need a tutor if your school doesn’t have/isn’t at this class yet

[swift shore]

Well the thing is, you need to buy the right textbook

The one I bought was meant for self-study

[native temple]

but what happenns?

[swift shore]

Wdym

I read the lesson, did the problems

[native temple]

do you not learn it?

[swift shore]

No it was a great experience

[native temple]

That’s why I said it was like a course in book form as a good one has the same infomation

[swift shore]

I learned everything very thoroughly and it was fun

Yeah

You gotta get the right textbook tho

I bought a textbook once that was not pedagogical, and it was really painful trying to self-study it

And of course @native temple if you have trouble with any problem feel free to post it here

[native temple]

Oh the openstax is is definitely pedalogical(literally just looked up what that meant)

[swift shore]

Oh that’s great

[native temple]

see the units

how it gives prerequisites

[swift shore]

Are you comfortable reading their explanations? Some textbooks are quite dry

[native temple]

and then teaches you this explains the lack of variance

for the most part

[swift shore]

Ah

That makes sense

[native temple]

there was one explaination of radians

that was confusing but

everything else is fine

[swift shore]

Ok that’s good

[native temple]

do you think that would be good to pick up for you one of those

it’s free

[swift shore]

what

[native temple]

….do you think those textbooks are good

they are for college students

[swift shore]

Openstax?

[native temple]

yes

[swift shore]

From what I’ve seen they’re super mid

But the fact that they’re free kinda makes up for it I guess

[native temple]

wdym mid?

[swift shore]

Mediocre

Run of the mill

[native temple]

oh

[swift shore]

I may be judging too harshly

[native temple]

probably mathmaticiand made this

[swift shore]

It’s a standard textbook, I’m sure any course could use it and it would be great

[swift shore]

Ehh

The mathematician may have wrote the book

but they probably didn’t come up with what to teach, or any of the problems, or any of that

Often it’s school boards and whatnot that decide those things

Which is why what you learn from year to year is pretty disconnected and random

[native temple]

uhhhh Did you expect above avergae IDK what that even looks like

I mean a textbook if it’s good snd pedatricsl contains the same info as a course

[swift shore]

I mean, when the average is garbage, then what can you expect? yknow

[native temple]

sometimes right down to assignments(science textbooks’

[native temple]

This is garbage?

[swift shore]

No, I’m exaggerating a bit lol

It’s not garbage it’s very okay

Again maybe I’m judging too harshly, this is an okay textbook for a high school or university course

[native temple]

You’re deep into a phd or something right

not learning the basics like me…..

[swift shore]

Nah lmao

I’ve been exposed to math outside the school system, so I have an ounce of perspective

Here

https://www.maa.org/external_archive/devlin/LockhartsLament.pdf

[native temple]

you’re a college student

[swift shore]

I will leave this here, maybe you’ll understand my strong views on this subject

[swift shore]

This is a fantastic essay (written by a mathematician, no less) criticizing the K-12 math education system

Even if you disagree with his argument, I think it’s worthwhile to read because his writing is excellent

Anyway, I have to go

Let me know if you have any trouble with any problems

See ya

[lone heart]

@native temple Has your question been resolved?

[lone heart]

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[calm bone]

If you buy 6 pens and one pencil, you get 1$ change back from your 10$ bill, but if you buy 4 pens and 2 oencils, you get 2$ change, how much does each cost?

[short cobalt]

have u tried setting up a system of equation?

[calm bone]

the thing is idk how

we got this in the homework and we havent done it

i tried something like 6x+y=9 and 4x+2y+8, is that what you meant?

[short cobalt]

something like that. idk if that eq is 100% for sure, but definitely the thought process I was thinking

[calm bone]

alright, so where do i go from there?

should i just substitute numbers??

[short cobalt]

I would rearrange the first eq u proposed so it equalled y. then sub it in

[calm bone]

oh word alr

so 6x-9=y?

[short cobalt]

the 6 would be negative and the 9 positive

[calm bone]

@short cobalt ive tried numbers but i cant find one that makes sense in the other equation

[short cobalt]

what do u mean?

[calm bone]

like i cant sub in anything that makes sense

[short cobalt]

ah

you sub in the y=9-6x into the other eq

so it would be 4x +2(9-6x)=2

[lone heart]

@calm bone Has your question been resolved?

[lone heart]

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[stone hill]

Can someone explain to me this? I get how it could be pi/2, and 3pi/2, but not the other two

[storm grotto]

can you post what you've done so far?

[storm grotto]

this isn't right

[ocean seal]

sadkid

[stone hill]

d/dx sin^2(x) = 2sin(x)*cos(x)

[storm grotto]

hmmmm

nvm

[stone hill]

lol

[storm grotto]

double angle

[stone hill]

ye

[storm grotto]

when you set cos(x)=0 you get pi/2 and 3pi/2, but there should be another factor aswell

did you simplify further?

[stone hill]

I emailed my prof, i think I got it now

[storm grotto]

👍

[stone hill]

.close

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[warm latch]

hi

[warm latch]

how do u do this?

[tepid mauve]

Find angle V first

[warm latch]

ok

is it all just adding?

[atomic stirrup]

Yes

[warm latch]

so

[atomic stirrup]

You need to find WVU first so you can then find WVX

[warm latch]

its 102

the V?

[atomic stirrup]

No

[warm latch]

What

[atomic stirrup]

So first we know that the angles of a triangle make 180

[warm latch]

yes

[wary stream]

For angle V? No

[warm latch]

they said add

[atomic stirrup]

So W + U + WVU = 180

[wary stream]

[atomic stirrup]

Then you would substitute the angles you know for each of the values

[warm latch]

So W + U then add W again?

[wary stream]

No

<W + <U + <V = 180

[warm latch]

oh

[atomic stirrup]

Its sometimes important to differentiate angles, so thats why i put <WVU

[warm latch]

WVU = 80

180\

wvx

[atomic stirrup]

No

[warm latch]

WVU=WVX?

[atomic stirrup]

No, thats the thing

The letters show the points that create the angle, the letter in the middle showing where the angle is

[warm latch]

The point is in X?

[atomic stirrup]

Points/vertices*

[warm latch]

points

[atomic stirrup]

Lets just move on

[warm latch]

ok

[atomic stirrup]

So W + U + V = 180

And we know the values of W and U

[warm latch]

so the value of V is 78

wait

[atomic stirrup]

No, slow down for a sec

[warm latch]

74

V=74?

[atomic stirrup]

Ye

[warm latch]

ok

[atomic stirrup]

But we still dont have our answer

[warm latch]

ye

What do we do next

now we got the value of V

[atomic stirrup]

We then see that WVX and <V share a line

We know a line is 180

[warm latch]

ye

[atomic stirrup]

So to complete the angle from 74degrees

[warm latch]

do we add

[atomic stirrup]

No

[warm latch]

subtract?

[atomic stirrup]

Ye

To find the missing angle that completes 180

[warm latch]

74 -

[atomic stirrup]

180-74

[warm latch]

180-74

106

[atomic stirrup]

Thats the answer

[warm latch]

106?

m/WVX=106

thanks

[atomic stirrup]

Ye, dont forget degrees symbol

.close

[warm latch]

I am gonna right it on the paper

thanks

.close

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[elfin snow]

is there such thing as an element of a matrix

[elfin snow]

like could I say $a \in \begin{bmatrix} a & b \ c & d \end{bmatrix}$?

[ocean seal]

guh mode

[elfin snow]

or am I being dumb

[quick field]

help

Use the graph to write a linear function that relates y to x

[simple turtle]

y value is same for every x.

[elfin snow]

please get your own channel

[quick field]

sorry

[simple turtle]

What do you mean?

[elfin snow]

this is my channel

read #❓how-to-get-help

ugh whatever

.close

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[sterile vale]

Just adding the intermediate step so no calculations are needed to be made

.close

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[daring kindle]

how do you show that a point. say |X| at 0,0 is not differetiable? I know if their derivitives arn't equal its not differentiable.

[daring kindle]

Eg at 0,0 the derivitive on the right is 1 and the left is -1

is that all I need to say?

[tight locust]

yes

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[zenith compass]

I stuck at here

Idk how to deal with the f(tx,ty) and f(x,y) since the question just said f

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[zenith compass]

.close

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[zenith compass]

.reopen

[lone heart]

✅

[zenith compass]

This is my question

[zenith compass]

This is my work

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@zenith compass Has your question been resolved?

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[tacit arch]

.reopen

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✅

[tacit arch]

f'(x,y) doesn't make sense

The prime notation only makes sense for functions of one variable

f is a function of two

[swift shore]

Such power

You can reopen channels at will

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[wild pebble]

Two connected pipes have inner diameters 8cm and 10 cm. Theese pipes want to be replaced with one pipe so, that the flow speed will stay same. How big should be the pipes inner diameter?

[wild pebble]

How do I do this

I have no logic for this at all

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[wild pebble]

<@&286206848099549185>

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[limpid spade]

sum?

iteration?

[knotty spire]

product?
recursion?

[limpid spade]

integral?

[lone heart]

@dark schooner Has your question been resolved?

[knotty spire]

I mean you can obviously also integrate a discrete function.

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[static basin]

Can anyone vouch that the best way to learn math is by practicing?

[limpid spade]

yes

[knotty spire]

I mean you still need to be able to understand and learn the theory before being able to practice on it

But yes, practice is the most important component to learning math in my opinion

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[static basin]

Thank you!

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[valid tundra]

7x3+3/3=8
what is the truth value of this? do we consider pemdas? if i use pemdas, it will be always false

[limpid spade]

truth value?

,w truth value

[valid tundra]

if the mathematical sentence is true or false

[limpid spade]

$$\frac{73+3}{3}$$ or $$(73)+\frac{3}{3}$$

?

[ocean seal]

The Fractalogist

[valid tundra]

that's the problem, the sentence is ambiguous

[limpid spade]

?

[valid tundra]

it is written this way 7x3+3/3=8

[limpid spade]

show screenshot

then

7x3+1

22!=8

[valid tundra]

ahh I see, thanks!!

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[desert compass]

i kind of want a full explanation on solving 5.

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[alpine sable]

Please help, lesson title is : Polygon and Other Related Terms

[upbeat gorge]

What parts do you know?

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[sturdy kayak]

Question 14. Im stuck on it

[spark creek]

So you have the interest, rate, and time
You can calculate the principal

[sturdy kayak]

Oh okay cheers 😁

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[neat apex]

how do i get the parametric representation for a line out of something like this?

[neat apex]

or carthesian, doesn't matter

just a way to find the direction vector?

a more efficient way than just finding 2 random points

[vale wigeon]

well you could take one of the variables (doesnt matter which) as your parameter

and express everything else in terms of it

[neat apex]

That's the best way?

well fair enough

Just wanted to make sure

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[grave oyster]

If there is a horizontal tangent at point (c,f(c)), does that mean f'(c) = 0

[rigid smelt]

yes

[grave oyster]

Kk Thanks.

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[raven rover]

Read this.

[misty eagle]

Huh

[raven rover]

Not that hard to see/read

[misty eagle]

I’m trying to calculate t so I added 5 to u and left v w alone and it’s wrong

[raven rover]

You're close with t

Think about directions

And sign

positive vs negative

[misty eagle]

I don’t think anything is negative

If it is idk why

[raven rover]

Try using u + t = v

[misty eagle]

So U is negative because that is t’s origin

Or is there a different reasoning for it

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[violet panther]

Could someone help me with this please

[violet panther]

Could someone show me step by step explanation? I have no idea how to do it

<@&286206848099549185>

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@violet panther Has your question been resolved?

[spiral aspen]

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

I think

Set

f(x) = -(x^4) -3

You add the minus 3 part because you’re going down 3 units on the y axis

Then you also need to make the x^4 negative because you are reflecting on the x axis

Now u need to find f(x/2 - 3)

For translations on the x plane, you do the opposite (if you want to move three right, you need to subtract three)

Same for dilation on the x axis. If you want to dilate by a factor of 2 about the y axis (which spreads it wider horizontally on the x axis), you need to instead divide by 2

So your equation is

[spiral aspen]

^

[spiral aspen]

^

[violet panther]

Thank you for the detailed explanation

I got it

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[sharp storm]

Can anyone explain me how i know in which corner is the angle alpha, beta or gamma?

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[slim idol]

Hi, problem is a student is waiting for a bus to school, there are 2, bus A and B, and the time they appear are 2 random variables of exponential distribution, for which I have 2 parameters, λa = 1/10 and λb = 1/20. Let T be the time the student is waiting, I need to demonstrate that T is a exponential variable of parameter 3/20.
At the top of my head I could think I just have to sum both parameters, but what's the explanation for that?

[strange urchin]

How do you prove that the solution set to the system of linear equations Ax=b is not a vector space?

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[slim idol]

@mellow saffron

oops, wrong ping, sorry

[slim idol]

<@&286206848099549185>

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[tacit arch]

Google poisson distribution bus problems

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[alpine sable]

someone help please

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[alpine sable]

<@&286206848099549185>

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

[last ether]

@alpine sable use your chain rule

[last ether]

To find what F'(x) would be

[last ether]

Buddy someone else reserved this

Read rules

[river sluice]

oh sorry

[last ether]

Note how the channel has a name next to it

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[cunning raven]

@worldly skiff

[worldly skiff]

Coolio

This is another property that ideally you wanna prove to use to save time for that question

So you don't have to keep repeating it

[cunning raven]

I see

[worldly skiff]

The online proof also uses it to say r|a and r|b implies r|s

since s is the gcd(a,b)

[cunning raven]

I'll post the full proof here again

yeah so why are you allowed to just assume a|b a|c and what frame of mind leads you do decide to do that?

I think for proofs the most important part to get down is the frame of mind

Like for example for regular induction equality proofs my frame of mind is to
"try to match IS with IH by separating k+1 terms"

[worldly skiff]

Person online is just stating the property to use

They actually don't do anything with that 1st statement

[cunning raven]

OHHH

[worldly skiff]

They mention it as reasoning for why r|b and r|d tells us that r|d-bc

And why s|a and s|b implies s|a+bc

[cunning raven]

how exactly does r|d-bc work?

[worldly skiff]

And then they use algebraic manipulation with d=a+bc to get r|a and s|d

I can post that snippet as I finished that part

[cunning raven]

but yeah here's my (so far) frame of mind for euclidean proofs

1. Take gcd(b,c) = a and turn it into a|b, a|c

[worldly skiff]

Yes

This is 1st part explained in more detail

[cunning raven]

I think I'm missing s|a+bc part but I'm kind of confused how to incorporate that

[worldly skiff]

I took out the property the online proof used to show how long it is to prove this

When you can just prove the property or mention it (if allowed) to save time

So yeah. That's why that person did so

similar reasoning for s|a+bc part

[cunning raven]

why were you aiming for r|(d-bc) ?

[worldly skiff]

the reason why was that we know d=a+bc

we wanna say r|a

and from d=a+bc we get a=d-bc

it comes up later in the proof, but we want r|a

So that we can say r|a and r|b, so r|gcd(a,b) aka r|s must be true

Are you familiar with Bezout's lemma by any chance

[cunning raven]

ok then why do you know that

r|b, r|b means r|d-bc or r|a?

[cunning raven]

absolutely not

[worldly skiff]

Ahhhhhhhhhhhhhhhh okay then proving r|a and r|b implying r|gcd(a,b) will be a bit trickier

so r|d-bc and r|a are equivalent

because a=d-bc

[cunning raven]

yeah I get that a = d-bc

[worldly skiff]

We know that r|b and r|d means r|d-bc by what I posted up a bit earlier

https://cdn.discordapp.com/attachments/490557019623915520/958454584106422282/Screen_Shot_2022-03-29_at_12.55.35_PM.png

A lot of divisibility proofs starts with the definition and then algebraic manipulation

[cunning raven]

damn this is really tricky please give me like 2 minutes

[worldly skiff]

For sure

I'm gonna get food rn, so I might be a bit before I respond

Or I'll be responding a bit slower cause phone

[cunning raven]

alright

thanks

[worldly skiff]

Once you’re done looking at that. Take a look at this

This shows why s|a and s|b implies s|a+bc

[cunning raven]

I'm just wondering from how you did r|d-bc

Is it True to also say that

For some x,y in whole numbers
r|dx-by
using the same frame of logic?

You are basically applying
a|bx+cy frame of logic but subtraction so it is valid right?

(I'll also go get food will still be on discord)

[worldly skiff]

Yep. All statements you said are correct

I'll eat for a bit and then can explain the r|a and r|b implying r|gcd(a,b) part

[cunning raven]

I'll also go eat.

I guess a more general formula for this would be

$gcd(b,c) = a \implies a|b \land a|c \implies a|bx \pm cy \land a|cy \pm bx$, Note that x and y can be 1.

[ocean seal]

Salt

[worldly skiff]

yep. you got it

😎

[cunning raven]

I'll ask about the later steps after I get food then

[worldly skiff]

for sure

I'll be in class from 2 to 2:50 PM PST

But after that I'm free to help

[cunning raven]

tysm

[worldly skiff]

No problem. Def work on what else you need or take a look at other questions in the mean while

I'll @ when I get back to my dorm room after class

[lone heart]

@cunning raven Has your question been resolved?

[lone heart]

@cunning raven Has your question been resolved?

[lone heart]

@cunning raven Has your question been resolved?

[worldly skiff]

@cunning raven

[cunning raven]

hihi

[worldly skiff]

I can't help too much for longer as I gotta work on some hw with a friend, but I got like 30 minutes to go over stuff atm

[cunning raven]

aight

[worldly skiff]

So we got through most of it. We just have to do the r|a and r|b stuff onwards

I'm gonna post this theorem called Bezout's Lemma. I'm not sure if your professor has gone over it yet or not. But I can post the proof if you want it. Unsure though what you learned in Discrete rn though, so proof may not make the most sense

This is Bezout's lemma

We need to use this theorem to prove that if something divides a and b

Then it also divides it's gcd

And here is that

I just picked random variable names. This has no relation to the problem. It's just showing why the property works

So that's why from r|a and r|b, they could say r|gcd(a,b). And since gcd(a,b)=s, r|s

And then they do the same thing for s|b and s|d. So s|gcd(b,d). gcd(b,d)=r. So s|r

[cunning raven]

a|b, a|c, a|bx+cy is bezout's lemma?

[worldly skiff]

a|b, a|c, a|bx+cy is not bezout's lemma

Bezout's Lemma is that for any integers a,b

There exists some other integers x,y that we can multiply against them

Such that their sum equals the gcd

So ax+by=gcd(a,b)

I can try to explain the proof, but it might take a bit. Do you know about well ordering principle and division algorithm? If not then for now, just accept that Bezout's Lemma is true and that we can use it to show that d|a and d|b implies d|gcd(a,b)

Ofc go back and take a look at it when you have more time

[cunning raven]

I kind of know division algorithm for euclidean but idk what well ordering principle is

[worldly skiff]

Ah okay then no problem. Well Ordering Principle is simple enough. It says that every non-empty set of positive integers contains a least element

But for now then we'll not worry about it. Let's accept Bezout's lemma at face value and using that we get d|a and d|b implies d|gcd(a,b)

[cunning raven]

wait so with bezout you can do
$a|b \land a|c \implies a|gcd(b,c)$

[worldly skiff]

So using that same property and logic. r|a and r|b implies r|s. s|b and s|d implies s|r

[ocean seal]

Salt

[worldly skiff]

So a|b and a|c getting the whole a|bx+cy is a different thing than Bezout's Lemma

$a|b \land a|c \implies a|bx+cy$ is a property of divisibility. doesn't have any fancy name I am aware of

[ocean seal]

3K

[cunning raven]

so altogether you have
$a|b \land a|c$
$\implies a|bx \pm cy \land a|gcd(b,c)$

[ocean seal]

Salt

[worldly skiff]

Bezout's Lemma says that there exist some x,y such that you can make a linear combination with a and b, so that ax+by=gcd(a,b)

So for Bezout's Lemma all you need to know is are a,b integers

$\forall a,b \in \mathbb{Z}, \exists x,y \in \mathbb{Z} \implies ax+by=gcd(a,b)$

[ocean seal]

3K

[worldly skiff]

So all you need to know is that a,b are integers. Then boom Bezout's Lemma applies

I can post the proof at the end for you to look at after we sort out through the current proof

[cunning raven]

is it possible to form a relationship between property of divisibility and bezout's lemma?

[worldly skiff]

Won't be able to explain it today

You can def do so

[cunning raven]

since I noticed that they have the same conditions of $a|b \land a|c; a,b,c \in \mathbb{Z}$

[ocean seal]

Salt

[worldly skiff]

But there's multiple properties of divisibilities, so not one good way to generalize it

And ah so here's a big distinction

a|b and a|c, where a,b,c, are integers, are the necessary assumptions to say that a|bx+cy

bezout only requires that a,b are integers

so that divisibility property requires 3 conditions. bezout only requires the one

you use a lot of them in conjunction with each other. you can def try to generalize some stuff using both together

but be careful to not mix up conditions

[cunning raven]

This is all the theorems I have so far

[worldly skiff]

Oh shoot I should have specified. That's my bad

First three are all good and yes. All divisibility proofs you work on should have everything be an integer

The one you have below it is not Bezout's lemma. That's just another property of divisibility and gcd

[cunning raven]

1 is actually just more simplified version of 3

[worldly skiff]

That theorem is correct. It's just not Bezout's Lemma

Here is Bezout's Lemma

Yeah. There's a lot of overlap between them

This is Bezout's Lemma

Kinda tiny to see though

Lemme see if I can make it bigger

[cunning raven]

is there much importance in deliberately stating the x and y part in proofs?

I thought it's main use is just get c|gcd(a,b)

[worldly skiff]

You use Bezout's Lemma actually for a lot of other things, so it is still important

[cunning raven]

ok

[worldly skiff]

the c|gcd(a,b) is just one of the things you can prove using Bezout's Lemma to help

[cunning raven]

ax+by looks really similar to divisibility

[worldly skiff]

Yeaaaaaaah. Divisibility, Bezout, Division + Euclidean Algorithm, gcd, lcm stuff are all really similar

I kinda group them all together in my head as they rely on divisibility definition a lot and algebraic manipulation

[cunning raven]

I'm kind of confused how the separate divisibility preconditions apply for bezout

c|a and c|b for example

how does it satisfy ax+by

wait

I'm messing up lettering

fixed?

[worldly skiff]

Notation looks all fine with what you have. The separate divisibility preconditions don't matter for Bezout at all

Maybe it would have been better if I picked different variable names for Bezout

But the c|ax+by from c|a and c|b

Is not necessarily the same for ax+by=gcd(a,b)

c|ax+by NEEDS the preconditions of c|a and c|b to be true

ax+by=gcd(a,b) does NOT need the precondition of c|a and c|b to be true

it only needs that a,b are integers

[cunning raven]

oh wait

c|ax+by is essentially the same as c|gcd(a,b)

so bezout uses same process as divisibility before it reaches c|ax+by

bezout is just the conversion of ax+by into gcd(a,b)

[worldly skiff]

That's one way to think of it. I'm sure your teacher will cover it in more detail when the time arrives

How long has the discrete class been going

Are you guys on a quarter, trimester, or semester system

[cunning raven]

I watch another prof's lectures online but like i said he's sick for divisibility and euclidean

uh

we're on semester system

my current prof's lectures confusing af

[worldly skiff]

Ah gotcha. Our discrete classes are structured differently then. I'm on quarter system and we learned divisibility first then sets when I took the class

[worldly skiff]

And yeah it's rough. Idk if your teacher is super old or not

[cunning raven]

quick question does ax-by or by-ax still work for gcd

[cunning raven]

he's new to teaching this course this is his first year

[worldly skiff]

But our math department has a lot of really nice professors, but they're all really old

And with COVID their technological skills leave room for improvement

Ah gotcha

[worldly skiff]

do you mean like does bezout's lemma still apply for ax-by=gcd(a,b)?

and such

[cunning raven]

yeah

since it's so linked to divisibility

[worldly skiff]

yeah that's also fine

x,y can be positive or negative

same thing a,b

so signs and order don't matter too much

since addition is commutative

aka yes what you said is correct

😎

Alright let's try to finish off the proof. So we have r|s and s|r because we proved that d|a and d|b implies d|gcd(a,b) part

https://cdn.discordapp.com/attachments/490557019623915520/958453505696362526/Screenshot_2022-03-28_225930.png

So now we just have the last two lines to explain. The last one is self-explanatory. It's just restating what we wanted to prove, so I won't go over that

The last really important part is understanding why r|s and s|r means that r=s

[cunning raven]

yeah that's the core of the frame of mind when doing this proof imo

your target is to match the two

but I'm still confused

[worldly skiff]

Exactly

So, a key implication from r|s

That you might wanna write down with the definition of divisibility is that r|s implies r<=s

$r|s \implies r \leq s$

[ocean seal]

3K

[worldly skiff]

Let's think of this intuitively. We know that r|s is saying that r is a factor of s

[cunning raven]

yeah that part makes sense

[worldly skiff]

Let's call s some random number. I'll pick 20

Oh perfect then

[cunning raven]

yeah

[worldly skiff]

Then we can finish up the proof

So we know r<=s from r|s

Then we also know that

$s|r \implies s \leq r$

[ocean seal]

3K

[cunning raven]

I see

[worldly skiff]

So we have these two inequalities that we have to satisfy and BOTH have to be true

Because we know that r|s and s|r

So the only way for both inequalities to be true is if s=r

Thus, from the two inequalities s=r

And those were set equal to the gcd stuff

So then you just rewrite what the question wanted you to prove

And you're done

[cunning raven]

so how do you know right away that you are trying to match the two just from the question itself

[worldly skiff]

If there's any other last clarifications you need lemme know, but I think that about covers it. I'll post Bezout's lemma here too for you to look at

But you don't need to fully understand it

So, that's part of some proof strategies

With the information that you're given, you should first assume that the proof is divisibility based

[cunning raven]

like for example this one I didn't have to match the two

[worldly skiff]

This is because of the definition of gcd

so let me get that for you

Again a bit small unfortunately

but the gcd definition is based off of divisibility

and this just comes with practice

but a common way to prove something is equal to one another in divisibility

is by getting a|b and b|a

So the person realized "Hey. This is a gcd problem where they want me to get something equal. I know that gcd definition involves divisibility, so if I can get r|s and s|r. Then I know r=s"

[cunning raven]

Did I mess up writing 4 then?

[worldly skiff]

Let me take a look

[cunning raven]

was it supposed to be
$(ng)|h \land h|(ng)$