help-36

[novel jackal]

ive did

but how do i get y to be smaller then x

[tired walrus]

break 3^4 into 3^3 * 3

subsequently, $\sqrt[3]{3^3 \cdot 3} = \sqrt[3]{3^3} \cdot \sqrt[3]{3} = 3 \sqrt[3]{3}$

[soft zealot]

Ann

[novel jackal]

Ahh ty took me a sec to get it

.close

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[final saddle]

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[gray acorn]

Is it possible for a propositional function to be a paradox?

[gray acorn]

like a proposition could be one for example " this sentence if false" is one, but idk about propositoinal functions

[valid carbon]

is this actually math.

ok it seems legit

[gray acorn]

discreet math

[tired walrus]

define "propositional function" & define "paradox"

[old quarry]

ong

thought it was english grammar

[gray acorn]

watch me start writing poetry in math server

[gray acorn]

propositional function --> a statement template with free variables like lets say P(x,y) and P is x+y < 5

Paradox uh A set of statement that produces a contradiction

I dont know how to really define it sorry

[tired walrus]

yeah then your question is not really answerable at this stage

[gray acorn]

ok ty

.close

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[final saddle]

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[latent dragon]

@rapid dragon, if you still have a question, please answer in a new channel

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[flint summit]

integral understandnig help

[flint summit]

so im learning about integrals

and theres some concepts i dont get

i get that dy=f'(x)dx

[lethal totem]

Post a picture

[flint summit]

and that f'(x)*dx represents the area under f'(x) which also represents the net change in y of f(x)

[flint summit]

its theory/concept

[flint summit]

but then when we integrate

why do we antiderive backwards if its f'(x)*dx and not f(x)*dx

[vital crag]

f'(x) is also a function

[flint summit]

yes

[vital crag]

it has an area under the curve as well

[flint summit]

yes

isnt that area under the curve the net change in f(x)

[vital crag]

$\int f'(x) dx = f(x) + C$

[soft zealot]

riemann

[flint summit]

that i dont get

[hybrid heath]

Basically the definition of antiderivative

[flint summit]

i dont get how we go from very very small changes in dx*f'(x) (rieman sums)

to how it equals f(x1) - f(x2)

[hybrid heath]

oh, you mean a definite integral

[vital crag]

hand wavy proof https://www.youtube.com/watch?v=pWtt0AvU0KA

[flint summit]

yes

[hybrid heath]

fundamental theorem is not a trivial proof

[flint summit]

i had a lecture on indefinite integrals, but im assuming the logic is the same but with the application of + c to represent infinte solutions of an antiderivative

[vital crag]

more thorough proof https://math.clarku.edu/~ma120/FTCproof.pdf

[flint summit]

thank you ill take a look

doing it is easy but i hate not understanding what im doing

[vital crag]

yea read every line of Theorem 1 FTC and ask questions if you don't understand

[hybrid heath]

Intuitively, I just think that the rate of change of area is just equal to the height of the graph, which makes sense to me

[flint summit]

i mean i can see the relation the more i think about it

how infintely small changes in x * its given slope at that value equals those changes in y

ohhh shit

[flint summit]

dy/dx (slash not divison) is bottled up into f(x) since it rerpesents changes at any given moment

therefore if i want to analyze the amount of change and not the instantaneous change i need to antiderive

because the amount of change is infintely represented by the amoutn of instataneous change

[hybrid heath]

if that helps your intuition, then go for it

[flint summit]

(is that wrong)

probably wrong terminology

[hybrid heath]

i wouldn't say wrong. It's not mathematically rigorous, but that's okay. Mine isn't either

Just a heuristic to help you visualize it

So there's no wrong way to see it, imo

[flint summit]

when it gets mathematically rigorous thats when its confusing as fuck

im a phys major ill take my basic understaning of calculus haha

ty

[hybrid heath]

I majored in physics

welcome, brother

[flint summit]

ty

im in my first term, im liking it so far

[hybrid heath]

It's fun. But, disclaimer: you'll need to be pretty good at math. You won't need all the theory, but physics is very math heavy:

• calculus
• complex analysis
• differential equations
• linear algebra
• statistics
  And, if you want to get into higher level, you'll need some basic topology, and maybe even abstract algebra

[flint summit]

fairfair

physics is whats kicking my ass tho, the way my prof has lectures set up is very ezpirmental/theory heavy and not very example heavy

its basically highschool physics but with dimensional analysis and all the barebones frameworks

alot of algebra

[hybrid heath]

Yeah that can be hard. I'll give you some advice, that text book will be your best friend. Read it, reread it, and read it often

always be asking questions.

[flint summit]

thats what my prof says funny enough

[hybrid heath]

it's the truth of college life

[flint summit]

i find with math i can get all i need from course notes+ the exercises in there

but physics, my profs lectures are basically him brainfarting on the board how to analyze the dimensions that contribute to the suns pressure and shit so he assuems you have textbook knowledge

im a few lectures behind in terms of review, im gonna catch up this weekend, just doing the easy things(calculus and chem) first

[hybrid heath]

in college, the EXPECTATION is that you have read the chapter BEFORE going to class.

I cannot stress that enough

[flint summit]

seems very true

[hybrid heath]

It is

[flint summit]

is there a good spot for practice problems apart tyhe textbook]

any good like websites that have big banks

[hybrid heath]

the book and your hw should really be sufficient

[flint summit]

kk tyvm

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[soft pewter]

!help

An agriculturalist must sow his land with corn and soy. He must sow at least 14 km² of his land and at most half of it with corn. Each square kilometer of corn requires 140 kg of seeds and 4 hours of work, and each square kilometer of soy requires 100 kg of seeds and 8 hours of work. The farmer has at most 1840 kg of seeds and at most 104 hours. One square kilometer of corn yields 25 k$ and requires 20 kg of fertilizer, and one square kilometer of soy yields 50 k$ and requires 25 kg of fertilizer.

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[soft pewter]

Which area should the farmer sow with each seed type, knowing that he wishes to maximize revenue first and then minimize the amount of fertilizer?

[final saddle]

@soft pewter Has your question been resolved?

[crude island]

how much fertilizer does the farmer have?

As of now I can't see why that's even relevant

[soft pewter]

yeah i dont think it is

[crude island]

So you want to maximize profit essentially?

[soft pewter]

yeah i found some stuff

[crude island]

Just spam soy ngl

[soft pewter]

yeah

[crude island]

reasoning is that

Soy is twice as profitable

so you would expect it to take twice as many resources

but it only takes 2x as many hours and actually requires less fertilizer

[soft pewter]

wait u mean corn?

[crude island]

I still don't think this is optimal thought but I do think

you need to use mainly soy

[soft pewter]

soy is 100 kg corn is 140kg

[crude island]

he has to sow 14 km^2

so spam soy as much as possible while still reaching that target seems like a good idea

[soft pewter]

i dont think that part is importasnt

soy has less kg than corn tho

[crude island]

12 soy 2 corn?

I think this might be optimal^

[soft pewter]

wait why do we need more soy

[crude island]

I'm pretty convinced that it is

[crude island]

it yields twice as much money

[soft pewter]

oh

i see

[crude island]

honestly 12 soy 2 corn probably is it

oh wait

at least 14 km^2

so he can do more

lol

nevermind then

it's gonna change

hold up

[soft pewter]

we gotta start with the inequations i think

[crude island]

actually nah

I think it might be the same

there's no way to replace any soy with more than 2 corn

so profit will stay the same

the key is prioritizing hours because that is much more limiting

[soft pewter]

alr

we should do the inequalities

[crude island]

Let c = number of corn km^2

Let s = number of soy km^2

[soft pewter]

alr

[crude island]

140c + 100s <= 1840

[soft pewter]

got that too

[crude island]

4c + 8s <= 104

25c + 50s = MAXIMIZE THIS

[soft pewter]

yup

c + s >= 14

[crude island]

10s 6 c

or 12 s 2c

they work equally as well actually

It's most likely the case that fertilizer

changes things

[final saddle]

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[crude island]

because otherwise there are multiple solutions

[soft pewter]

.reopen

[final saddle]

✅

[soft pewter]

is there any more inequalities

.close

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[final saddle]

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[latent dragon]

hi!

where are ya stuck?

[sterile pebble]

hlelo

so I dont know which ratios to go for

here

cos W is 9/41

so I have a 9, 40, 41 triangle

wait

do I solve for triangle ADC's ratio via triangle ABC's side lengths/

if I assign DC as 9

and AD as 40

because AC is hypoteneuse of triangle ADC

I get 9/40 as my answer

answer key says 40/9 though

<@&286206848099549185>

Proportions look good

[odd seal]

can you find relation between w and z angles?

[sterile pebble]

no

[odd seal]

wdym no?

try

[sterile pebble]

no like

Z belongs to angle A

while W belongs to angle B

[odd seal]

and?

[sterile pebble]

so they form 90 degrees?

[odd seal]

well, that just sounds like a random guess

can you support it with maths?

you should look at angles

the angle BCA is common in the triangles ABC as well as ADC right?

and both are right angled triangles

[sterile pebble]

yes

so AC is = DC?

[odd seal]

do you know the sum of angles for a triangle?

[sterile pebble]

in ratio

that is

[sterile pebble]

180

[odd seal]

yea, use that in both triangles

sum of angles of ABC = sum of angles of ADC

and fill in all the known/labelled angles

[sterile pebble]

what

we dont know any angles besides the 90's

[odd seal]

I mean just name the names

like ABC + ACB + yada yada

[sterile pebble]

ummm

I want to know why my proportions are off

[odd seal]

and you will

trust the process

[sterile pebble]

I do not understand how naming all 7 angles of these 3 triangles is going to do anything

[odd seal]

1. two triangles only have 6 angles
2. one of them is common, so you are left with only 4

Dude, you are throwing darts in the dark

you are claiming stuff out of nowhere

[sterile pebble]

1. angle A is cut across 2 triangles within triangle ABC, the 90 degree angle that is split by angles CAD and BAD counts as an angle

[odd seal]

this is not what I am asking you to do

I asked you to find a relation between the angles w and z

and I am telling you how to approach that

by comparing the triangles they are in

[final saddle]

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[cedar obsidian]

If \alpha, \beta \neq 0, \text{ and } f(n) = \alpha^n + \beta^n \text{ and }

\begin{vmatrix}
3 & 1 + f(1) & 1 + f(2) \
1 + f(1) & 1 + f(2) & 1 + f(3) \
1 + f(2) & 1 + f(3) & 1 + f(4)
\end{vmatrix}
= K (1 - \alpha)^2 (1 - \beta)^2 (\alpha - \beta)^2,

\text{ then K is equal to }

(1) \alpha \beta \quad (2) \tfrac{1}{\alpha \beta} \quad (3) 1 \quad (4) -1

[soft zealot]

BlackidoZΣ
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cedar obsidian]

Im stuck at the last step

[modest fog]

I don't think so you r supposed to do like this

You should split the determinant in 2 parts

Such that product of both is this determinant

Write that 3 as 1+1+1

[cedar obsidian]

can you help me with that what will be the first element in both new determinants which are in product now

[modest fog]

Okk

[cedar obsidian]

is it | 1 ....| x | 1+1 ....|

[modest fog]

Lemme grab my copy

[cedar obsidian]

but it dont make 3

[modest fog]

Wait wait

See this @cedar obsidian

[cedar obsidian]

,rotate

[modest fog]

Now both the determinants r same

So it becomes squared

Now this will be easier to open

Also we can perform row operations too to do it easier

[cedar obsidian]

doing..

i am getting

[(alpha-1)(beta-1) {beta-alpha} ] ^2

[modest fog]

Yes

Now (alpha-1)² is same as (1-alpha)²

Same for beta term too

Now you can compare what they gave

You will get the value of k as 1

[cedar obsidian]

yes

you got correct idea quickly

idk what i was doing

[modest fog]

?

[modest fog]

Nah man

I did the same q in class

I remember

I first tried what u were doing

[cedar obsidian]

oh

[modest fog]

My sir said then do it like this

[cedar obsidian]

ty

[modest fog]

It's a pyq ig ?

[cedar obsidian]

it is in exercise 2

[cedar obsidian]

in my book

[modest fog]

Clicking unknown links isn't a very good idea

[modest fog]

Oh I see

Its a pyq also I remember

Jee loves to trap people

Smth a scammer might say

[cedar obsidian]

lol

i may close this now

.close

[final saddle]

Channel closed

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[tired walrus]

@hearty swift this is not the place to advertise.

[hybrid heath]

<@&268886789983436800>

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[terse folio]

I was trying to find inspiration for my physics project and stumbled upon this expression online, may i ask what is the technical meaning of this?

[terse folio]

i understand R+ := (0,+inf)

but just not C^2([0,L) x R+)

[final saddle]

@terse folio Has your question been resolved?

[final saddle]

@terse folio Has your question been resolved?

[drowsy epoch]

u is a twice differentiable function on the interval [0,L] x (0,οο) but the codomain is unclear. (Presumbaly to R, maybe) Where did you get this online from?

[terse folio]

a university lecture note

[terse folio]

do we get "twice differentiable " from the 2 in C^2?

[drowsy epoch]

Yes

C^n is the function space of functions that are n times continuously differentiable

[terse folio]

i see

[terse folio]

the interval would be 2 dimensional..?

sorry not too well versed in this

[drowsy epoch]

That's usually what it's supposed to mean, but without further context it could be also something else

[drowsy epoch]

Yeah

[terse folio]

I see

thank you for your clarification

[terse folio]

.close

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[exotic shale]

how can we show this for taylor series

[rain lance]

Hey

[desert mantle]

differentiate the series n times

[exotic shale]

hhow

n can be any non negative integer

[rain lance]

series around a point "a". The series is:

f(x) = Σ cₙ (x-a)ⁿ, where cₙ = fⁿ(a)/n!.

Each coefficient cₙ is the nth derivative of the function at "a", divided by n! (factorial). For convergence, the function must be differentiable or analytic in the interval.

[desert mantle]

its a polynomial series

those are easy to differentiate

properly speaking do induction but like who cares

[tired walrus]

i think we're about to get hit with the question of why said differentiation is legitimate

[desert mantle]

its an obvious induction

differentiate it the first fww times and you'll see the pattern

[tired walrus]

esp given how i said in an earlier channel the assumption of f having derivatives of all orders is just bookkeeping

[exotic shale]

no i see the pattern

was asking how to show its true for all n

so we use induction for it

ok

one more thing

idk how to ask this

if we can just integrate/differentiate and compare to 1/1-r why do we need taylor series

[desert mantle]

not everything results from a geometric series...?

[exotic shale]

but everything can be written as taylor series?

wait but why is it important that we can write a function as an infinite series

[desert mantle]

easier to work with in a lot of cases

[exotic shale]

so there is some cases

where writting a function as a taylor series

makes it easier to proceed with what we are doing

idk just seems more complicated as a series to me but ill probably understand as i continue and see some examples/problems

anyways thanks for the help take care

.close

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[magic finch]

How do i proove E is the midpoint of DC??

[magic finch]

(Yes, DC and AB are equal and parallel, and same rules for the other pair of sides)

(AK = 2KE)

[tired walrus]

let $\ora{AB} = \bd{u}$ and $\ora{AD} = \bd{v}$ and express various vectors in the picture in terms of these two

[soft zealot]

Ann

[magic finch]

is it ok if AB is "a" and AD is b?

[tired walrus]

sure yeah use whatever names you like

[magic finch]

anyways ive tried this already and other than knowing BD is b - a, im not really sure how to use it other than that

i cant find a use for AK = 2KE and i also cant come up with a condition for DE being equal to EC

[zenith cedar]

Did you try using similarity

[magic finch]

no?

[zenith cedar]

Alright

Which similar triangles can you find

[magic finch]

ADE with BEC

and the other 2 triangles formed by them, AME and EMB

[zenith cedar]

Uh any other ones

What about ked and abk

[magic finch]

Oh

Well they have parallel sides with DE and AB, AK with KE with known lengths, but what about DB

[zenith cedar]

Are you trying to prove the similarity?

Or finding the lengths

[magic finch]

The only similarity i see is that one triangle has what seems to be 2x smaller sides than the other

[zenith cedar]

What are the other similarity criteria

[magic finch]

Angles? they all match up

[zenith cedar]

So now what can we conclude from this similarity

[magic finch]

Vectors made out of them would have equal directions?

other than that the triangles only have a difference in lengths of sides

[zenith cedar]

Not vectors what can we conclude about the lengths of sides

[magic finch]

Okay so since the angles are equal then to increase the triangle it would be like multiplying all the sides by a number, and since when we know AK = 2KE, wouldnt that mean that all the sides of the bigger triangle ABK are 2x bigger than DKE's?

[zenith cedar]

Yes

[magic finch]

Ohhhhh and by that logic we can find out DE is just AB/2

thanks so much, i always get stumped on these questiosn because i cant pick up the right direction to follow

Okay well

I guess ill be closing this if you dont have anything to tell me

Ty again

.close

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[lone lichen]

can someone explain me the steps for solving this question please?

[ornate tree]

Please claim another channel

[lone lichen]

aight

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[polar owl]

I really don’t get the solution

It says the height is 2 right? but it clearly very obviously isn’t 2. So that’s my confusion. GPT is saying something about it being consistent within its scaling system but it isn’t really because it doesn’t correspond to the 1 unit measurement given above

[crude idol]

the height of what

if you are wondering where the 2 came from then its the distance between the segments AB and CF

so that the height is divided by that ratio

[final saddle]

@polar owl Has your question been resolved?

[polar owl]

but it says hence triangle cpf has height 5/8 x 2

why r they multiplying by 2

[crude idol]

5/8 is the ratio between the height of the triangle and the distance between AB and CF

and 2 is the distance between AB and CF

[polar owl]

i thought 5/8 was the scale factor for the sides of the triangles

since triangle APB is 3 units and triangle CPF is 5 units

[crude idol]

the height also scales

with the same ratio

[polar owl]

yes i agree

but why do they choose the number 2 if the height of neither triangle is 2

[crude idol]

the sum of the heights are 2

so the ratio is used to see which one gets which part

[polar owl]

Guy

❤️

my beloved

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[polar owl]

thank you 👊🏿

[crude idol]

youre welcome

[final saddle]

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[soft pewter]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[ornate tree]

hello, what do you need help with?

[soft pewter]

An agriculturalist must sow his land with corn and soy. He must sow at least 14 km² of his land and at most half of it with corn. Each square kilometer of corn requires 140 kg of seeds and 4 hours of work, and each square kilometer of soy requires 100 kg of seeds and 8 hours of work. The farmer has at most 1840 kg of seeds and at most 104 hours. One square kilometer of corn yields 25 k$ and requires 20 kg of fertilizer, and one square kilometer of soy yields 50 k$ and requires 25 kg of fertilizer.

Which area should the farmer sow with each seed type, knowing that he wishes to maximize revenue first and then minimize the amount of fertilizer?

[soft pewter]

hey

[soft pewter]

x: square kilometer of corn
y: square kilometer of soy

x + y >= 14

140x + 100y =< 1840

104 >= 4x + 8y

[junior token]

yes

[soft pewter]

not sure if theres more inequalities

[junior token]

at most half of his land must have corn

[soft pewter]

x + y >= 1/2x?

[junior token]

not really

you can look at it this way

at least half of the land must have soy

so the area of soy is always at least that of corn

[junior token]

so check this again

[soft pewter]

kk

wait half of his land has to be corn then?

[junior token]

no, at most half can be corn

but not more

wait, I’m just realizing an ambiguity in the question

[soft pewter]

which

[junior token]

he has to sow at least 14 km^2 of his land, but does it have to be with corn?

I guess we can assume the 14 km^2 doesn’t have to be with corn

[soft pewter]

i dont think that part is important

im not sure tho

[junior token]

anyway

[junior token]

how do you write this constraint?

[soft pewter]

trying to figure that

theres definitely a 1/2x in there

[junior token]

it’s not 1/2 x

[soft pewter]

oh wait

[junior token]

because that would be “half of the area of corn”

[soft pewter]

ah

i mean i got x =< 1/2x + y

idk if thats good or not

[junior token]

remember to include parentheses

[soft pewter]

we didnt learn that yet

[junior token]

[junior token]

this would be “area of soy plus half the area of corn”

[soft pewter]

so its wrong then right

[junior token]

if you put brackets then it’d be correct

[soft pewter]

it cant be that then cause we didnt see brackets in constraints yet

[junior token]

well, all I mean is this: x =< 1/2(x+y)

because you halve it after you add the areas of corn and soy

and that would be correct: the area of corn is at most half the total area of both corn and soy

so I assumed that’s what you meant

[soft pewter]

but he only wants half his land to be corn

the rest can be soy

[junior token]

yes

at most half

[soft pewter]

we should find another one to replace that cause i dont think my teacher will allow that

in a real test

[junior token]

on the other hand, you can simplify the inequality

[soft pewter]

which inequality

x =< 1/2(x+y)

[junior token]

yes

[soft pewter]

1/2x + 1/2y

[junior token]

write out the full inequality and continue

collect like terms

[soft pewter]

x =< 1/2x + 1/2y

[junior token]

Or you can apply this reasoning

[junior token]

yes

[soft pewter]

alright

ima write that

[junior token]

well, that works

You can simplify it further though

[soft pewter]

not sure how

[junior token]

Subtract 1/2x from both sides

[soft pewter]

alright

1/2x =< 1/2y

[junior token]

yes

[soft pewter]

is that the fully simplified version

[junior token]

No

[soft pewter]

u sure?

[junior token]

I’m sure

[soft pewter]

i cant see how we can simplify it more

maybe x2

[junior token]

Yes

[soft pewter]

2x =< 2y?

[junior token]

no, it would be $2 \times \frac 12 x \leq 2 \times \frac 12 y$

[soft zealot]

soup_norm

[junior token]

and then the 2 and 1/2 would cancel

so you had the right idea

[soft pewter]

oh so 2x 1/2

makes 1

so x =< y

[junior token]

yes

[soft pewter]

i see

[junior token]

(Earlier I outlined another way you could arrive at this conclusion: The area of corn is at most half the total area, So the area of soy is at least half the total area, therefore at least the area of corn. So the area of soy is at least the area of corn)

and that’s all the inequalities

[soft pewter]

wait is there more because look here

An agriculturalist must sow his land with corn and soy. He must sow at least 14 km² of his land and at most half of it with corn. Each square kilometer of corn requires 140 kg of seeds and 4 hours of work, and each square kilometer of soy requires 100 kg of seeds and 8 hours of work. The farmer has at most 1840 kg of seeds and at most 104 hours. One square kilometer of corn yields 25 k$ and requires 20 kg of fertilizer, and one square kilometer of soy yields 50 k$ and requires 25 kg of fertilizer.

what about ths

[junior token]

the question asks “Which area should the farmer sow with each seed type, knowing that he wishes to maximize revenue first and then minimize the amount of fertilizer?”

so you first need to maximize revenue

can you find revenue in terms of x and y?

[soft pewter]

we cant

we gotta do the graph

we gotta represent graphically the constraints

[junior token]

well you can make the graph first

[soft pewter]

kk

[junior token]

you can write revenue in terms of x and y later, but you need to do this eventually

[soft pewter]

okay

i did the graph

we gotta put x + y >= 14

[junior token]

let’s see, we currently have:
x is the area of corn in km^2
y is the area of soy in km^2
the constraints are:
x + y >= 14
140x + 100y =< 1840
104 >= 4x + 8y
x =< y

[soft pewter]

yup

[junior token]

can you send the graph?

[soft pewter]

its on a paper

[junior token]

then take a picture

[soft pewter]

all i did was just 2 lines one vertically and one horizontal

i didnt put anything on it yet

[junior token]

Oh

[soft pewter]

yeah

[junior token]

Then graph the lines for now

[junior token]

have you done questions like this before?

[soft pewter]

yes but i get stuck at the graph part

[junior token]

oh

[junior token]

do you know how to graph lines?

[soft pewter]

what is that

like the constraints?

[junior token]

like, if I asked you to graph x + y = 14, could you do it?

[soft pewter]

nah thats where i get stuck

[junior token]

Yes, you need to graph those

[junior token]

oh, I see, have you learned how to do that before?

[soft pewter]

yes i did some practices but its still a bit difficult

[junior token]

were you able to do it in those practices?

[soft pewter]

kind of

[junior token]

what do you mean?

[soft pewter]

when i remembered how to do it i was able in class

but now i forgot

[junior token]

well, try to graph these constraints, using the methods you learnt

[soft pewter]

for x + y = 14

i think we have to isolate y

.close

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[soft pewter]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[soft pewter]

how do i graph x + y >= 14

like this?

[latent dragon]

Why did you use the factoid on yourself?

[soft pewter]

huh

[latent dragon]

this

[soft pewter]

what is that

[latent dragon]

sigh nevermind

[soft pewter]

it said that by itself

[latent dragon]

it didn’t, you triggered it by doing !help

[soft pewter]

yeah thats what u say when u ask a question

[latent dragon]

Nevermind. Anway, you need help graphing 𝑥 + 𝑦 ≥ 14

[soft pewter]

i did it already

i don't know if thats right though

[latent dragon]

looks right to me

[soft pewter]

alright ty

.close

[final saddle]

Channel closed

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[final tangle]

graphing utilities usually have no issues with stuff like that

[final saddle]

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[vague anchor]

24

[vague anchor]

,rccw

[pearl solar]

which one?

[light depot]

solving the pyqs huh?

[vague anchor]

Yepp

[vague anchor]

24th

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vague anchor]

2nd step

[vague anchor]

JEE?

[final saddle]

@vague anchor Has your question been resolved?

[vague anchor]

<@&286206848099549185>

[final saddle]

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

[ornate tree]

Alright, show everything you've done so far

[vague anchor]

Ok

@ornate tree
Can u read that

[ornate tree]

A second

hmm, I'm not capable of solving such problem, might pass it to the next helper

[ivory bear]

isme error hai

or maybe i could be wrong

[ivory bear]

ans is d?

or maybe i think @vague anchor that you are making significant blunders; recheck your calculations

if needed ill dm the soln

[vague anchor]

Fr

.close

[final saddle]

Channel closed

Closed by @vague anchor

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[vague anchor]

Yepp

[final saddle]

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[sage stirrup]

Just a conceptual doubt

[vague anchor]

Ohk

[sage stirrup]

How to go from step 1 to 2

[vague anchor]

Most of the numbers can be represented as the factors of 1 or more no

[sage stirrup]

so?

[worldly mesa]

Are you asking why we can factor a polynomial into linear terms?

[vague anchor]

We express polynomial in factors

[sage stirrup]

ehhh whatever ya understand like the derivation I am asking

[vague anchor]

And the no. will be 0 if any of these factors became 0

[sage stirrup]

Ok so we express in terms c = a *b where a and b are factors

so where did leading coefficient come from

[vague anchor]

Let factor be. X-a
So it will be 0 only if x=a

[sage stirrup]

yes

[vague anchor]

So it will make the whole polynomial 0

[worldly mesa]

You want to match the a at the beggining of the x^n

[sage stirrup]

ehhh well the bookish answer is infinite polynomials have same roots

but my doubt is why not take B

Yk like ax^2+bx+c , just take b common

why a?

Why leading coeeficient

[worldly mesa]

You want both to have the same leading coefficent

[sage stirrup]

y?

[worldly mesa]

Otherwise they cant be equal

[sage stirrup]

Ok...... but the same question why not take B common

like we can use the same argument for B

[worldly mesa]

We can't

[sage stirrup]

like we want same coefficient for B as well right

[worldly mesa]

The leading coefficent won't match

[sage stirrup]

how will the polynomial be equal then

[worldly mesa]

The roots manage that

[sage stirrup]

Show how

@worldly mesa

Are ya writing lmao?

If ya are keep on

pretend I said nothing

[worldly mesa]

?

No need for me to answer?

[sage stirrup]

Noooooo

I thought you were

writing

on a copy or smth

[worldly mesa]

I am

[sage stirrup]

and sending

Btw a question

are ya showing for general polynomial or............... quadratic?

[worldly mesa]

if p is a monic polynomial and it has alpha1,...,alphan
then
p(x) = (x-alpha1)(x-alpha2)...(x-alphan)

[worldly mesa]

General

[sage stirrup]

roots?

[worldly mesa]

Yeah

[sage stirrup]

Monic mean what?

Linear?

[worldly mesa]

leading coefficient 1

and n=degp

[sage stirrup]

ok ok

go on

[worldly mesa]

When p is not monic, with leading coefficient a, then p/a is monic so it can factored like before and then
p(x) = a (x-alpha1)(x-alpha2)...(z-alphan)

Note p and p/a have the same roots

[sage stirrup]

assumption or is there a reason behind it?

[worldly mesa]

p(x)=0 iff p(x)/a =0

[sage stirrup]

what........

Ok Ok

I see

[sage stirrup]

agreed

then

you didn't really explain how roots manage terms other than leading coefficient

[worldly mesa]

If you open coefficients here (x-alpha1)(x-alpha2)...(x-alphan) then the coefficients will be the coefficients of p/a

which are just the coefficients of p(x) divided by a

[sage stirrup]

hmmm reverse engineering I seee

How are the coefficintes the sam etho

P(x)/a will have same coefficients as P(X)

that makes no sense

[worldly mesa]

Equality here is not just as functions its as polynomials

[worldly mesa]

No!

if p(x)=ax^2+bx+c
then p(x)/a = x^2 + b/a * x + c/a

[sage stirrup]

exactly

OHHH

understoood

thenk u

.close

[final saddle]

Channel closed

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[native basin]

i found this proposition/statement in an exam online (part of the mathematical logic course)

[native basin]

and it's negation is somehow:

But we cannot have the quantifier ∀ in a proposition P; in this case, P⇒Q.

[lofty sinew]

Why not?

[empty vigil]

In general, the negation of "For All x, P" is "There exists x, negation of P"

[lofty sinew]

Like, some forms of propositional logic don't include quantifiers

But there are absolutely forms of propositional logic that do allow their use

[terse crypt]

I find the notation to be quite weird..

[onyx peak]

that notation is terrible

[lofty sinew]

I'll agree to that

[terse crypt]

It's combining ${\forall x (e^x \geq 1)}$ with ${(x > 0) \implies (e^x \geq 1)}$

[soft zealot]

k

[onyx peak]

Wtf does Ex > 0 -> e^x < 1 mean

[native basin]

well in this case "For all x > 0" is a whole proposition, not defining where x is in the beginning

[onyx peak]

Even if they meant Ex: (x > 0 -> e^x < 1), it would be wrong

[terse crypt]

i think it's saying that the negative of ${\forall x (e^x \geq 1)}$ is ${\exists x (e^x < 1)}$

[soft zealot]

k

[terse crypt]

the latter is the full form of for all

i think the author mixed the shorthand one with the full form

and the result is questionable

[onyx peak]

where'd the -> come from tho

[native basin]

What I know is that the negation of (𝑃⇒𝑄) is (P∧¬Q) .

[onyx peak]

oh wait is that -> used for : (such that)

[lofty sinew]

It might be more meaningful to say the negation of (𝑃⇒𝑄) is (¬𝑄⇒¬𝑃)

[onyx peak]

and that -> just separates the quantifier from formula?

[lofty sinew]

Both are true, but the version I gave keeps the implication around

[native basin]

that ⇒ is weird, shouldn't we annotate it:

∀x>0, e^x ≥ 1

instead of :

∀x>0 ⇒ e^x ≥ 1

[lofty sinew]

I agree

[terse crypt]

Yes

[onyx peak]

The negation of $\forall{x}(x > 0 \implies e^x \ge 1)$ is $\exists{x}(x > 0 \land e^x < 1)$

[soft zealot]

MathIsAlwaysRight

[onyx peak]

the other possibility would be that they wanted to write sth like this

that'd explain the first ->, but make the 2nd -> wrong

[native basin]

yup, that anottation could actually be just wrong

thanks for the help y'all ❤️

[final saddle]

@native basin Has your question been resolved?

[final saddle]

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[hardy coyote]

Can someone explain how to get CB and OC

[plucky rover]

Well what have you done so far

[hardy coyote]

Only AB

[plucky rover]

Good

Now AC = 2 CB

How else are AC and CB related

[hardy coyote]

Im not sure

[plucky rover]

Well give it a guess

[hardy coyote]

C divids the segmenr AB ?

[plucky rover]

Yes

So AC + CB = ?

[hardy coyote]

AB

[plucky rover]

Good

Now can you find CB from this

[hardy coyote]

1/3 AB?

[plucky rover]

Good

Now can you find OC

[hardy coyote]

Its
OB + BC right

[plucky rover]

Yes

Good, you're done then

[hardy coyote]

Whats BC?

[plucky rover]

-CB

Cuz you're just reversing the direction

[hardy coyote]

So - 1/3 AB?

[plucky rover]

Ye

I hope you can do the rest yourself

[hardy coyote]

So its

b+ -1/3a

But answer is suppozed to be 1/3a + 2/3 b

<@&286206848099549185>

<@&286206848099549185>

[muted prairie]

These are not negations

They are equivalent

[lofty sinew]

Oh yeah I guess you're right

Probably shouldn't talk about this in this channel right now tho

[muted prairie]

AB is not a

Also

!15m

[final saddle]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[muted prairie]

So you subtract 1/3 * a, when you should subtract 1/3 * AB instead

[hardy coyote]

Huh

Oh soeey that was a texting mistake

I meant AB

-1/3 (b-a)

So OC= OB+BC
So

b+ -1/3 (b-a)

Is that correct?

[muted prairie]

yes

Simplify and you should get what's on the answer key

[hardy coyote]

That part im confused with only

Is this correct?

[final saddle]

@hardy coyote Has your question been resolved?

[hardy coyote]

.close

[final saddle]

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• Show your work, and if possible, explain where you are stuck.
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[winter lava]

i think you can drop a perpendicular from B to DC

[severe verge]

yea and calculate the short side at the bottom

[winter lava]

yeah

because there will be congruent triangles

[rain lance]

Use coordinate geometry

and Use the Pythagorean theorem on triangle

[golden cave]

oh

[thin cloud]

Have you tried anything?

[golden cave]

ye

[thin cloud]

okay show me

where you got stuck

[golden cave]

idropped a perpendicular from B to DC and named it X and its length is m, same thing wth A to DX and named it m so y-2m=x.

DC

[glossy zephyr]

i think i got a crazy trig idea which tbh im kinda stumped to even start to explain

[golden cave]

lol

[thin cloud]

which segment is your m?

[winter lava]

you don't need to call it m since we know it's x

[golden cave]

dh

[thin cloud]

okay,... DH isn't quite useful here

you should find HC instead

[winter lava]

oh do you mean XC is m?

[golden cave]

ye

[winter lava]

yeah y-2m = x

that's right

[golden cave]

dh=md?

[winter lava]

what is md?

m is a distance and D is a point

[thin cloud]

uhh.. there's no point labeled M

[winter lava]

you should use capital letters for points, it'll be less confusing

[glossy zephyr]

time to simplify the cursed answer i got

[winter lava]

y-2m = x
solve for m

[golden cave]

2sqrt(13)/13

[winter lava]

it's not what i got

[golden cave]

what did u do

[golden cave]

how

[winter lava]

add 2m-x to both sides, then divide both sides by 2

[glossy zephyr]

3/5, mb, at the end

i can show if you wanna

its basically pulling out x as a proportion of y

[winter lava]

i got 3/5

[glossy zephyr]

was it x/y or y/x?

[golden cave]

huh

[thin cloud]

x/y

[golden cave]

lemme do smth

[thin cloud]

i got the same

[glossy zephyr]

oh yeah, mb, did it inverse

[glossy zephyr]

once you find out what "m" is, its pretty easy from there if you know your algebra

[golden cave]

k

i got $5x^2 + 2xy - 3y^2 = 0,$

[soft zealot]

glitched-ronin

[golden cave]

which factors as $(5x - 3y)(x + y) = 0.$

[soft zealot]

glitched-ronin

[glossy zephyr]

Yes, but theres an easier way using difference of squares

[golden cave]

$5x = 3y,$

[soft zealot]

glitched-ronin

[golden cave]

alr

[glossy zephyr]

given the previous formulas, this is basically trivial

[golden cave]

?

[glossy zephyr]

Is this supposed to be the same trapezoid or just another one?

[golden cave]

another

[glossy zephyr]

oh ok, yeah, 8 and 9 as lengths there, lmao

[golden cave]

lol

[thin cloud]

you should try smth first

[glossy zephyr]

okay i already got a solution based on angles

[thin cloud]

stop using trig bro

[glossy zephyr]

just the general idea, just as a hint, use law of sines

[glossy zephyr]

no chance

tbh i dont think theres any other way to pull it off either, idk

[winter lava]

there is another way

[golden cave]

how

[winter lava]

dropping the perpendiculars again and doing a bunch of algebra

[glossy zephyr]

oh yeah i can see how i think

[final saddle]

@golden cave Has your question been resolved?

[winter lava]

an isoceles trapezoid is cyclic so you can get AB with ptolemy's theorem

[final saddle]

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[urban cedar]

i have a question can anyone help me

[proper dagger]

do send the question

[urban cedar]

x^2+1/x^2 = 27 , find x+1/x

[proper dagger]

It is given that $x^2 + \frac{1}{x^2} = 27$. Find $x + \frac{1}{x}$.

[urban cedar]

yes

[soft zealot]

fox's directional derivative

[urban cedar]

yes this is the question

[proper dagger]

retyped for other helpers

[worldly mesa]

Now, what have you tried?

[umbral hamlet]

try working backwards from x + 1/x, see if you can get to an expression similar to the equation

[final saddle]

@urban cedar Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[thin cloud]

Given the prism $ABC.A'B'C'$ with an isosceles triangle ABC as the base where $AB=AC=1$ and $\angle BAC=120^{\circ}$. The projection from A' onto the base plane lies on a line that passes through $A$ and is perpendicular to $AC$, knowing that the angle between $CC'$ and the base plane is $\alpha$ and $\tan(\alpha)=\frac{\sqrt{21}}{7}$. Let M be a point on the segment $CC'$ such that $CC'=3CM$ then $BM$ is perpendicular to $AB'$. Find the volume of the prism

[soft zealot]

Alexis_Fx

[thin cloud]

tbh I help no idea where to start

[abstract bramble]

what the fuck did i just read

id start with a sketch of the prism

[thin cloud]

the fact I also have no clue how to draw a diagram

[abstract bramble]

oh

[thin cloud]

i mean I can draw prism

[abstract bramble]

can you draw the base

[thin cloud]

But other info

[thin cloud]

letme try

[severe verge]

Typical TSA bullshit

[thin cloud]

no it's not my guy

TSA can't be this hard

[sharp vessel]

classsssic problem

[sharp vessel]

draw em first

[thin cloud]

I'm not sure I draw it right

[final saddle]

@thin cloud Has your question been resolved?

[final saddle]

@thin cloud Has your question been resolved?

[thin cloud]

.close

[final saddle]

Channel closed

Closed by @thin cloud

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[exotic shale]

hello

[ornate tree]

hello, do you have a question?

[exotic shale]

so for the taylor series representation of a function we have a theorem

where we can use lim Rn = 0

i was told lagrange found a way to calculate the lim Rn

using some formula

can someone tell me about it cause it isnt covered in my source

[desert mantle]

google taylor series lagrange remainder term

[tired walrus]

https://en.wikipedia.org/wiki/Taylor's_theorem scroll to "Explicit formulas for the remainder"

[exotic shale]

this?

[exotic shale]

these right?

[tired walrus]

yup

[exotic shale]

which the easiest

[tired walrus]

for what purpose

[exotic shale]

i wanna see how these were found

[exotic shale]

proof

[tired walrus]

dunno

well hmm

for proof maybe lagrange

[exotic shale]

only equation i have

[tired walrus]

bc it is basically a generalization of lagrange MVT for higher order derivs

[exotic shale]

is f(x) = Tn(x) + Rn(x)

idk anything else about Rn

so i have no clue how we would even be able to find any other formula for it

i thought it cant be done until i was told otherwise

ill first try to use it ig

ok so for example for e^x

@tired walrus is this formula true for every function ?

[tired walrus]

do you want the direct-and-literal answer or the informative answer

[exotic shale]

i dont understand

so idk

[desert mantle]

you literally screenshotted the theorem statement

it tells you for which functions its true

[exotic shale]

oh where

oh the top

its for all the formulas

ok so e^x i can use

when i use this formula

do i use it with any a

[desert mantle]

how have you defined e^x

[exotic shale]

or do i pick an a ?

[desert mantle]

for any a you want

[exotic shale]

idk why does it make a difference if its the same function

[desert mantle]

the usual taylor series for e^x that you know is a=0

[desert mantle]

because if you defined e^x via its taylor series then this is pointless

[exotic shale]

yea the example says to try to find it about x=0

ok so in the limit

do i put 0 instead of a

or i can keep a to show its true for all a

[tired walrus]

you're being unclear what your proof goal is rn

[exotic shale]

im trying to use the theorem for e^x

this

so i need to calculate lim Rn

but im not sure if i choose a specific a

like a=0

or i just use a

[exotic shale]

im guessing not

but idk what definitions of e^x there are

[desert mantle]

look you cant do proofs if you dont have rigorous definitions

its pointless

if you dont know what e^x is then how do you think you can prove a statement about it

a possible definition that you could use is the following: f(x)=e^x is the unique function satisfying f'(x)=f(x), f(0)=1

[exotic shale]

i thought its just like

the 2.71.... number

on the base

and x on the power

[desert mantle]

but what does something like e^pi mean

[exotic shale]

yea i know that that results onto e^x only

[desert mantle]

it cant mean "e multiplied with itself pi times"

that makes no sense

[exotic shale]

but didnt know we use that as definition

[desert mantle]

you can define it for rational functions and then by continuity for real numbers. but thats also awful to work with

[exotic shale]

then whatt do you do

[exotic shale]

then what does it mean

[desert mantle]

well thats the question

thats my point

[exotic shale]

arent we supposed to all have same defintions

so no problems occur

[desert mantle]

all definitions end up equivalent

but we still have to start with some definition

[exotic shale]

ok