help-23

[brave wolf]

no, not really

it's fine, i was just asking if you could explain it if you were asked about it

you dont need to go into details in the proof, if you are confident that you could explain it in few simple sentences

[raven veldt]

yeah I believe I would!

Oh I did have this question : ) I think I said it above but dk. Is it necessary to include all these steps (circled in red) or can I just observe it:

[brave wolf]

I think you can do that in one step

in the "other direction", (g + h)^3 = g^3 + 3g^2h + 3gh^2 + h^3 it follows just by simple expanding

so you can just claim that

[brave wolf]

here you are showing the thought process probably, but for a proof it suffices to just claim that it = (g+h)^3 (because that's apparent by expanding it)

[raven veldt]

Okie got it!

I also have this question.

I was mostly not sure about this closure proof but I wouldn't mind if the other ones could be checked as well : )

[winter whale]

Meth

[brave wolf]

What if the result of some a / b was 0? Would it still be closed?

you have showed that the result will always be real

but what if the result was 0?

[raven veldt]

The only way that you would get 0 is if 0 is a dividend but since this set excludes the 0 from the real, you would never have that

I think

[brave wolf]

Yeah, I feel like that's worth mentioning, given that to the fact that a / b ∈ R you have dedicated 2 whole paragraphs

[raven veldt]

I attempted to in the first line of my proof but now I realized I oly discussed when 0 is the divisor (i.e. to make it undefined). Perhaps I can combine these two ideas together somehow

[brave wolf]

you could say something like "Furthermore, the result of dividing 2 reals will be 0, only if the dividend is 0."

Then the sentence that starts with "0 is excluded from the set..." can be left essentially unchanged, except for the "would be a real number" part, which should be changed to "would be a non-zero real number"

[raven veldt]

Okay makes sense!

[raven veldt]

For the second part (i.e. where I have to show identity) I was not as confident in this one

[brave wolf]

I think you have a typo here

(exists a instead of exists e)

are you sure that identity is defined as e * a = a * e = a in your book btw? Because it asks you to prove that it has an identity, while from that definition it's obvious that it doesnt haveo ne

[raven veldt]

Oh yup I see that now : )

[raven veldt]

i'll check

These is all I have

[brave wolf]

I see, so I guess that you are supposed to show just the existence of right identity here

[raven veldt]

Yup my prof clarified that in class

[brave wolf]

Okay, a / 1= a, that's true and that shows that 1 is a right identity

but for the left identity, you only showed that 1 is not a left identity. You can't conclude that there is no other left identity

what you could conclude is that 1 is not a full identity, since it's not a left identity

[safe radish]

@raven veldt Has your question been resolved?

[raven veldt]

Okay so the way I showed the right identity is correct?

but left identity needs more evidence?

[brave wolf]

If you wish to show that there is no left identity, then yes. You need a new proof

[raven veldt]

Okay makes sense

[raven veldt]

I am a bit confused on how I would go about this

[brave wolf]

Okay, so your proof only shows that 1 is not a left identity

[raven veldt]

mhm

[brave wolf]

you need to show that any e isn't a left identity

[raven veldt]

should I show this with a contradiction?

[brave wolf]

Similar technique will work

Assume that e is a left identity, then for all r (in R \ {0}), e / r = r.

Try substituting in different values of r

[raven veldt]

Okay thank you!

[brave wolf]

by substituting in r = 1, you get that if such e exists, then e = 1. By substituing r = 2, you get e = 4. And that's already a contradiction

[raven veldt]

How's this? I wasn't sure about the last few lines

[brave wolf]

the contradiction arises from a different fact actually

note that e = 1 and at the same time e = 4

that's the contradiction

You assumed that there is a left identity, and called it e. Then you showed that e must be equal to 1, and then you showed that e must be equal to 4. And that's the contradiction

[raven veldt]

Ahh okay that makes sense. How's this

[brave wolf]

I would just write it like this

identity is supposed to be respected by all elements, there is not one identity per element of R \ {0}

if any element does not obey the identity, then it's not an identity.

what im trying to say that the value of e doesnt depend on a at all

[raven veldt]

Okay yeah I see that now thanks!

Here are the last two I need to verify

For commutativity I think I may need to provide more, but wasn't for sure

[brave wolf]

I think that you are using e = 1, because 1 is the right identity. Not because e must be constant

if that was the only condition, we could even take e = 2.718 or sth, which we obviously cant

[raven veldt]

oh yes true! thanks ; )

[brave wolf]

For commutativity, yes, providing a counterexample would probably be good

oh, they were just asking you to explain your intution. It would probably be sufficient to say that division is not commutative

and possibly provide an example

[raven veldt]

Okay do you think the part when I said "The only case.. . " that whole line is necessary

[brave wolf]

It's certainly not necessary

but it's not a mistake either, I see it as a line that's supposed to provide additional insight, although it's not directly related to the proof

For the shortest possible proof, you can just say 1/2 != 2/1, and so it's not commutative

[raven veldt]

I think I am happy with it this way if that works

[brave wolf]

is this a typo?

you are first saying "for all a, b" and after that "there exist a, b"

[raven veldt]

Is it okay to just say this

or should I keep the for all..

[brave wolf]

this looks okay

Do you use black text to write claims that you will be proving?

[raven veldt]

Oh no I just had to write some of this stuff onto LaTex. The red indicates the adjustments I am making

[brave wolf]

I see

[raven veldt]

that I have to change on the LaTex version. since I started typing in LaTex a bit before I checked it over thoroughly, which I probably shouldn't have done but oh well learn for next time

[brave wolf]

You could probably somehow signify that "For some a, b.... a / b != b / a" is a statement you are gonna be proving, or at least visually distinguish it from the rest of the proof.

[raven veldt]

What about this?

[brave wolf]

This is a last little detail I would change, we either say that operation is commutative, or it isn't. We don't say " / is commutative for 1, because 1 / 1 = 1 / 1", so it doesnt make much sense to say that something is commutative for all R \ {0}

[brave wolf]

oh, and probably change it to "for all a, b in R \ {0}"

[raven veldt]

this should be good now : )

Do I really need that comma?

[brave wolf]

the one after forall?

[raven veldt]

I feel like maybe but this is just the nitty gritty now xD

[raven veldt]

yeah the one before the equation

[brave wolf]

you dont

you would need it if you placed the forall behind the equation

[raven veldt]

Ahh I see okay got it !

Thank you : )

[brave wolf]

np

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[viral grove]

119 please help😭

Would rate of change just be the derivative and plug in 3?

[lean otter]

yep

that would give you the rate of change at x=3

[viral grove]

And then do I double it

To make it c?

Bc none of my answers are the multiple choice

[lean otter]

you double it and then equate it to rate of change at x=c

[viral grove]

OHHH

Wait okay gimme a min let me do this

Wait so I got 0.72 for rate of change at 3 and I doubled it 1.44 and set it equal to derivative

[lean otter]

yeah

i would have kept the fractions rather than simplifyinf it

[viral grove]

Ohhh bc I didn’t get the choices

But I got 0.9 something

Oh shit wait

Okay wait nvm I got 3/4

Lol yay Ty I get it now

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[heady dock]

I need to detect if the Green arc is overlapping the Red and\or Blue arcs
All of the arcs have the same radius
Green = 2.35 -> -4.8
Red = -2.35 -> -2.2
Blue = 3 -> 3.4
The result should be that the Green arc is overlapping both the Red and the Blue arcs

[calm coyote]

colors by loona is answer

[pine horizon]

@calm coyote can you stop trolling

[calm coyote]

sorry ill stop.

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@heady dock Has your question been resolved?

[heady dock]

I had an I idea, I can check with the Green arc normally it should get me the Blue arc, and then subtract 2Pi from the Green arc and check again which should give me the Red arc
If the Green arc cross the angular point it should subtract 2Pi if it doesn't it should add 2Pi

<@&286206848099549185> I need your help please 🙏🏻

[toxic torrent]

Hello !

[heady dock]

Hi

[sharp mango]

I need help with a question

please help me

[valid olive]

!occupied

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@heady dock Has your question been resolved?

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@heady dock Has your question been resolved?

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@heady dock Has your question been resolved?

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@heady dock Has your question been resolved?

[severe berry]

┬─┬ノ( º _ ºノ)

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@heady dock Has your question been resolved?

[steep lily]

You might get more help if you clarify what you want

Like, yes the green arc overlaps the red and blue

What else is there to say?

clearly the points in the blue arc are also points in the green, since the radius is the same and the angles of points in blue [3,3.4] are a subset of the angles in green [2.35, 4.8]

and similarly for red except you shift by 2pi

[untold raft]

Are you trying to come up with an algorithm to see if the green ark overlaps the others?

[heady dock]

yes but is there another way?

[heady dock]

yes

[twin chasm]

Need to prove the equation the first one is enough

[twin chasm]

<@&286206848099549185> need to prove the equation

[analog quartz]

which equation?

[hasty marsh]

is tg tangent?

[dapper star]

π/12 rad is 15 deg, we can use tan (A-B) formula for calculating π/12 and same with cot

π/12 = (π/4)-(π/6) (45deg - 30deg)

[twin chasm]

Thanks

[untold raft]

Here's how I would think about it:
If you are familiar with polar coordinates (rather than cartesian- x and y), they are parametrised by r (radius) and θ (angle).

Given that all possible points lie on the same circle, the radius is constant, so each point can be parametrised by just θ (it is the only 'variable')

The range of values which the green arc occupies can be parametrised by its initial and final angle (2.35 and 4.8 radians)

any arc occupies this range if the initial or final value of theta of the arc falls within the range of θ_green_lower < x < θ_green_upper) (if neither the start nor the end of an arc is contained in the bound, it is not overlapped)

---

To deal with any values of theta (angles) that are not principal (not within 0 <= x <= 2pi, such as the negative radian values of the red arc), you can use the modulo (%) operator against each value of theta first before assigning

(if not familliar, read up- it is essentially clock math- 13 modulo 12 = 1, as 13 = 1(12) + 1- essentially where the hand points after a whole number of cycles round the clock

---

in summary:

• each arc can be declared by just its initial and final angle (θ)
• To convert all angles (pos, neg, obtuse) to 0 <= x <= 2pi (principal values), do angle = θ % 2pi (pretty sure this will work, but I havent tested it myself, just thought about it)
• if red or blue's start or finish angle lies between interval (θ_green_lower, θ_green_upper) it is overlapped

Please make sure it makes sense, and let me know if it works. I hope you have done some googling in the meantime; the struggle sucks (not knowing how to solve something and thinking about it forever), but the struggle is how you learn (with some nudges)

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[sterile rivet]

N

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[keen crag]

Yi

[ionic venture]

! occupied

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[severe berry]

bro

u cant plus

or what

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[kind fractal]

Hello. First write tg and ctg in terms of sin and cos

[alpine stone]

.close

[left gyro]

are you still stuck on this

if so, you need to explain in more detail the problem youre stuck with
are you looking for a computer program or a formula that takes in the three arcs and gives out which ones overlap
what are you using this for

[heady dock]

I think that I got a solution, its an algo that I'm working on for path system
just haven't implement it yet tho I think that it will work

.close

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[lean otter]

<@&268886789983436800> this channel has been left open after being closed for a good 9 hours lol

[left gyro]

.reopen

[safe radish]

✅

[left gyro]

.close this keeps happening, have to reset the timers when a post stays open for too long

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[pallid mulch]

I'm having trouble on why we add a test point thats 0, what does it do exactly

[digital sparrow]

So now you know that any points that gives negative too
Will lie on the same side as (0, 0)

Because (0, 0) does not satisfy
No points in that side will

So the area side is automatically the answer

[pallid mulch]

can you make it a bit more clearer?

[solar dragon]

its just a random point that shows you which side of the line satisfies the inequality

[digital sparrow]

I don't know what you mean by cleaner

[pallid mulch]

ah

[pallid mulch]

What i meant is that I still don't understand but thanks

[digital sparrow]

Should I draw a diagram and send?

[digital sparrow]

Now either A1 or A2 will satisfy this eqn

[pallid mulch]

Yeah

[digital sparrow]

(0, 0) does not satisfy

So A2 does not satisfy

You can check any point

But (0, 0) is the easiest to do because you know it's location on the graph (whether it's in A1 or A2)

[pallid mulch]

That makes sense thanks

🫂

[digital sparrow]

!done?

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[kind brook]

isnt it both b and d?

[craggy sedge]

not really

[kind brook]

how though?

because for a union b

since a is a subset of b

b already contains all elements in a

so a union b is b

and as for a intersection b

if a is a subset of b, a contains all the common elements

im so confused

[hazy elbow]

is that from a test?

[kind brook]

assignment

but my explanation makes sense right

it is both b and d

[lime dust]

What is that | thing?

[kind brook]

it's my cursor

mb

[hazy elbow]

well you are correct

[kind brook]

so going by my logic

this has to be d right?

[rare fern]

Yes

[kind brook]

okk ty

.close

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[fossil trail]

Help what am i not getting

A is 6

B is 7

But why

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<@&286206848099549185>

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[buoyant shadow]

The bottom of the graph is y=2

Not 0

So b is 7 yeah

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[quasi hatch]

what did I do wrong 😭

[obsidian oracle]

mmmmmh....

[quasi hatch]

multiply the whole thing by sin x? 😢

idk

[cedar void]

missed that sinx

[quasi hatch]

why would there be a sin x there tho or am I dumb

if i multiply it removes the sin x and makes it sin^2 x right

[cedar void]

to combine both of them the denominator of both must be equal

[quasi hatch]

oh so sinx/1

ohhh

[cedar void]

so $\frac{1}{\sin{x}}-\frac{\sin^2{x}}{\sin{x}}$

[quasi hatch]

ahh okahy

thankss

i keep making these mistakes 😢

.close

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[wraith swift]

how do i proceed?

[wispy monolith]

You can change cos²x to 1-sin²x then treat it as a quadratic equation. But the question is what are they asking you to find? Value of n? Or anything else

[wraith swift]

its more of a trigo equation

the only possibility for the equation to be valid is if it satisifies to be 1 which simplifies to this

[wispy monolith]

Better to post original question

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[cosmic epoch]

can someone please give me an example when i is false?

[cosmic epoch]

I thought it should be true

[brave wolf]

0

[cosmic epoch]

0 isn't positive or negative?

[brave wolf]

0 is neither

positive means > 0

negative means < 0

[cosmic epoch]

oh

oH

Why is 0 not positive or negative

is there

a proof behind it

[brave wolf]

Well, 0 isn't greater than 0

so it's not positive

0 also isn't smaller than 0

so it's not negative

[cosmic epoch]

then what would 0 be considered

[brave wolf]

zero.

[cosmic epoch]

and would i be considered a real number?

[brave wolf]

Every real number is either positive, negative or 0.

[cosmic epoch]

ok that makes sense

[brave wolf]

i as in sqrt(-1)?

[cosmic epoch]

yeah

[brave wolf]

no

i isn't a real number

[wide kindle]

0 is the absence of number

like space

[cosmic epoch]

are those the only examples of non-real numbers?

[wide kindle]

there is nothing

[brave wolf]

[wide kindle]

and we refer to this nothingness as "0"

[brave wolf]

there are other number systems

e.g. quaternions

[wide kindle]

it is but a symbol to refer to the absence

[brave wolf]

so those would be other numbers that aren't real

the thing is, "number" doesnt have an exact mathematical definition

[cosmic epoch]

I have never

heard of those

[wide kindle]

you can use greek numbers

or arabic

[brave wolf]

yeah, they are not used commonly

[wide kindle]

all same

[brave wolf]

what?

[cosmic epoch]

is 0 not considered an integer in higher level math?

[brave wolf]

No, 0 is considered to be an integer

and actually, in most of "higher" math, it's even considered to be a natural number like 1, 2, 3, ...

[cosmic epoch]

that's pretty cool

I'lll close this tab thing so other ppl can use it

thanks for all the help

.close

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[lean otter]

easy

[digital sparrow]

!15m

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[lean otter]

the last two terms of each are negligible

this is the hint

[stable flint]

2024^2025 >>>> 1
so you can neglect one

same for denominato r

okay

[forest wigeon]

wassup?

P < Q < S

that is the order

oh

whats the original question?

first send the original question

then I shall work it out

give me sometime

lemme solve it and will share the answer once done

[digital sparrow]

Hey I have an idea

S and P have 3 terms

Let's ignore the last one

Pand Q*

Both the terms of Q are greater than that of P

2024²⁰²⁵ > 2024²⁰²⁴

And 2024^-2026 is less than 2024^-2025

But there is a minus with it

So it becomes greater

What?

Yes Q is greater

No then we will have to check

Cause it's 1/2024 ^2024

That would have at least 4*2024 digits

If this was in Kg

Then mass of electron is 10^-34

See the scale?

Basically nothing

Yes

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[wary bear]

the answer choices were 6 and 7

no rounding

[plucky elk]

What does "each lateral side was 42" mean?

Area 42 or length 42?

[wild dragon]

area i think

[wild dragon]

So first you just note your system of equations

[wary bear]

surface area

sorry for being vague

[wild dragon]

The surface area of the prism is: 4 * (lateral side area) + 2 * base

so 168 + 2*base = 252

so 2*base = 84

base area = 42

you got a cube

._.

[wary bear]

thats what i thought too

but no

[wild dragon]

Yeah either am doing something hella wrong

Or the data is wrong

[wild dragon]

we can try the variants i guess tho

so if the height is 6 -> base = 42/6 = 7

7^2 * 2 + 42 * 4 = 266

which is wrong...

and if the height is 7 -> base length is 6

and 6^2 * 2 + 42 * 4 = 240

[wary bear]

which is also wrong

damn idk man

[wild dragon]

yup

yeah id just bet on the fact that the problem is wrong

:<

[wary bear]

its an sat question so i doubt it is

maybe i missed a given

[wild dragon]

huh yeah then its unlikely

[safe radish]

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[sleek saffron]

hi I'm clueless on how to do this question

[sleek saffron]

can someone give me a pointer?

here I think V is a constant vector but idk how does the other vector (R) work

[median vigil]

i think $\vb r$ would be something like [ \vb r = x \vb i + y \vb j + z \vb k ]

[flat frigate]

cloud

[sleek saffron]

yea i tried this

i separated v and r into x,y,z component then do the dot product

then after that when i put the partial derivative sign on it im lost

Like this

mb the commas should be "+" but still

[median vigil]

i think that the components of r are specifically r_x = x, r_y = y, r_z = z

[sleek saffron]

sorry I don't quite understand what you mean

[median vigil]

as in this could be written as [ \phi(x,y,z) = \cos(\vb v \cdot (x,y,z)) ]

[flat frigate]

cloud

[sleek saffron]

Oh

ok i think i got the solution

Is this right (or in the right direction)?

[median vigil]

seems right

[sleek saffron]

thanks m8

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[cloud kelp]

for (i) why do we only take -π/6 ?

[safe radish]

@cloud kelp Has your question been resolved?

[plucky elk]

need the entire original question

[plucky elk]

have you learned definition of stationary point

[safe radish]

@cloud kelp Has your question been resolved?

[cloud kelp]

when dy/dx = 0 ?

[cloud kelp]

the gradient is. 0

[plucky elk]

Yes

So find when sin(x)=-1/2

[cloud kelp]

so doesn't that give π/6 and -π/6

[plucky elk]

Wot no

,calc sin(pi/6)

[flat frigate]

Result:

0.5

[cloud kelp]

sorry it confused me a little because my teacher was like if u have like sin x = -1/2 then ignore negative sign and find first positive and then. egative angles

[plucky elk]

That's wrong

Or at least incomplete

sine is positive in the first quadrant

[cloud kelp]

so we're only supposed to get the negative angle ?

[cloud kelp]

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[wind oyster]

Hello guys anyone here who can speak french ? I study math in french and need help with some questions

[wind oyster]

Just the first question

[lean otter]

hhhhhhhhhhhhhhhh hadi lepreuve li dwzt f 2023 💓💓

1-a ghir b TAF

[wind oyster]

Hhh sbhanlah jiti f waqtk

Hoa xof hadxi baqi maqrina mno walo hir bghit xof wsh kijawbo o kfx o sf

[lean otter]

mazal ma9rito theoreme des accroissements finis?

[wind oyster]

Nn 7na baqi f suites numeriques

[lean otter]

aaah iwa sf

ah aslan maynknch tkml had lexo db khss hta t9raw les suites de rieman

yak sm b3da?

[wind oyster]

Aji hadxi d sc maths s3ib awla xwia ?

[lean otter]

nta pc?

[wind oyster]

La sm

[lean otter]

sara7a sahel rah 2023 kay9olo as3ab watani w jbt 17,25 f math bla man9ra tahaja

[wind oyster]

Zaema Xi nasi7a ida kan momkin

[lean otter]

khss ghi dbt lcours wsf

lcours aham haja

[wind oyster]

17.25 bla matqra walo ??

Kont kanxof nas kikhrjo kibkiw

[full marten]

Je parle Français mais je suis en Terminale 💀

[wind oyster]

Sf chokran

[lean otter]

c'est une epreuve du bac marocain 🙄

[wind oyster]

O exercices khsni nkhdm xi hja li s3iba awla nmxi b xwia ?

[lean otter]

khdm ghi lwataniat

[wind oyster]

Mn daba nbda fihom ?

[full marten]

J'ai fini par comprendre que c'était de l'arabe et j'ai commencé a lire 😭

puisque je suis Tunisien

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[safe radish]

@light crypt Has your question been resolved?

[light crypt]

<@&286206848099549185>

[prime locust]

basically both answers are correct. if you consider the entire line x = 5 - 3t, y = -3 + 2t, z = t, then t takes on all values from negative infinity to infinity

so t takes on every positive and every negative value

if you replace t with negative t,
(1) -t also takes on every value from negative infinity to positive infinity
(2) those parametric equations become x = 5 + 3t, y = -3 - 2t, z = -t

that t is called the parameter, and there's multiple different ways to create this paremeter

in this case, using -t as the parameter instead will still give you the same line

[light crypt]

I know, just wasn't sure if there was a convention I should be using but I guess it's not super duper important

Thanks anyways!

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[static hedge]

could anybody give me a hint?

[static hedge]

i know the top would be 0 + 1/2 + 1/3 ... 1/2027

but i dont know how to do the bottom part

3+ 3 + 4 + 5 ... 2015 + 2017?

im not sure

[swift hare]

try to write the numerator of a_n in terms of the denominator of a_n

ie can you think of a funtion that maps 1 to 2, 2 to 5, 3 to 10...

[static hedge]

wait so if i write out he first three terms it would be like

0/3 + 1/2 / 3 + 1/3/4

so if i exclude 0/3

would it be 1/(n+1) / (n+2)

oh wait do i have to use partial fraction decomp

[swift hare]

what does this mean

like is 1/3/4 = 1/(3/4)

[static hedge]

(1/3) / 4

bcz the top is the fraction part

and the bottom is the whole part right

[swift hare]

ok i see what you did

i meant considering a_n and not the new sequence with {x}

but your way also works

[swift hare]

yes

[static hedge]

OKAY thank you s m

lemme try that

so 504/1009?

OMG THANK YOU SO MUCH 😭 🙏

[swift hare]

np

[static hedge]

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[opal oasis]

howdy! I'm having a minor issue with determining a polynomial given the intercepts/factors. I'm mostly there, but I think I'm getting the stretch factor incorrect (it could be something else I'm missing.

[opal oasis]

my solution to try to find a was to plug in (0,30) and solve which ended up with 30=-45a

i then solved for a and got -30/45 which simplifies to -1/3, but when i graph the polynomial the y-intercept isn't correct

[rigid inlet]

-30/45 which simplifies to -1/3
no it doesn't

[opal oasis]

oh lol

oh my goodness

2/3

l m a o

[rigid inlet]

-2/3 indeed

[opal oasis]

HAH, thanks

that was the issue 😅

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[alpine grove]

Quick asymptotic analysis question:

[alpine grove]

Can I do just that to establish this?

Or do I need to do something more?

[mortal sandal]

Capital theta right

Θ

$\Theta$ vs $\theta$

[flat frigate]

mommymorphism aficionado

[mortal sandal]

That's probably good enough yeah were it my hw I wouldn't even bother explaining why ceil(n/10) is Θ(n)

[alpine grove]

Alright thank you, I'll remove that then.

Last question

[mortal sandal]

also technically you can make the second ≤ a <

[alpine grove]

Would it be correct to write this:

Where Theta(1) is the best case scenario

and Theta(n) the worst scenario

Ooh damn I need to correct the Theta vs. theta.

Thank you.

[mortal sandal]

I've never seen that before

[alpine grove]

Cworst <= C(n) <= Cbest

Never seen that?

[mortal sandal]

That means C isn't a well-defined function?

[alpine grove]

It's the compelxity of an algorithm.

In function of n.,

[mortal sandal]

idk it's weird but I'm not a huge CS buff

[alpine grove]

Alright not a big deal.

Thank again!

.close

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[stoic dune]

How do you know they aren't calculated using that?

Why the "= 0" at the end? That's not the partial derivative

[modern bloom]

That's a typo

[safe radish]

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[nocturne cipher]

By Substitution

[nocturne cipher]

Anyone?

[rich elm]

what is the actual question?

[pine horizon]

u have to draw the lines?

[nocturne cipher]

You have to graph by substitution

[rich elm]

graph what

[nocturne cipher]

The equation

or the answer of it

[rich elm]

i think they are asking to graph the point of intersection

[nocturne cipher]

No no

[rich elm]

otherwise substitution doesn't really make sense in this context

[nocturne cipher]

The top eqation is the y value

[pine horizon]

i think its just number substitution

[rich elm]

oh i see

[nocturne cipher]

yea

[pine horizon]

put x=1, 2, 3 and get y points

[rich elm]

well with each line

[pine horizon]

then make the line

[rich elm]

u only need to substitute 2 points for each equation

[nocturne cipher]

huh

[pine horizon]

basically

put x=0 in first line, you will get a y value

then put y=0, you will get a x value

plot these points

and join them

thats your line

[nocturne cipher]

My brain

god

[pine horizon]

do the same with second line

[rich elm]

ok ok lets break it down to things one part at a time

[nocturne cipher]

mhm

[rich elm]

lets graph the first line, y = 2x-1

[nocturne cipher]

Nooo

The first line is what you substitute

I think

idk man

[pine horizon]

what

[sage wren]

what is the question

is it just to graph two lines?

[pine horizon]

yes

[nocturne cipher]

yes two lines

[lean otter]

sustitution makes no sense here, you will just end up with the "values" of x and y

[nocturne cipher]

Hold on

[sage wren]

yeah graph both lines and find the intersection

[nocturne cipher]

let me send in a different equation

[pine horizon]

not any values tho, its the intersection point

[nocturne cipher]

This one

[lean otter]

youre right my friend

[sage wren]

substitution should only be used to find intersection

[nocturne cipher]

Omg

im so sorry

i meant elimination

..

[sage wren]

so are you trying to just graph the lines?

[rich elm]

elimination is just another way to solve simultaneous equations

it doesnt help u graph the lines

[sage wren]

because you dont use elimination for graphing

[nocturne cipher]

Im supposed to use elimination mto get my answer and graph

[rich elm]

these 2 different 'techniques' only really make sense for solving their points of intersection

[sage wren]

your answer being the point of intersection?

[nocturne cipher]

I had my director do this equation for me and he got parallel lines?

[sage wren]

for the recent one you sent

yeah theyre parallel

[nocturne cipher]

The first one

I sent in

[sage wren]

the first one?

[nocturne cipher]

mhm

[sage wren]

this one?

[nocturne cipher]

yes

[sage wren]

they shouldnt be parallel

[lean otter]

they cannot be parallel...

[nocturne cipher]

Wait sorry

not that one

[sage wren]

this?

they are parallel

if you rearrange for y, youll get that they have the same slope

[nocturne cipher]

this one

[sage wren]

yes this one is parallel

rearrange for y

[nocturne cipher]

Thats impressive

[sage wren]

in the second equation

[nocturne cipher]

how you can take one look at the problem

wow

[rich elm]

the coefficients of x are identical

[nocturne cipher]

But

I want to know how to solve this equation by elimination

[sage wren]

so when it comes to solving for the point of intersection for two parallel lines, you cant

[rich elm]

i am 99% sure the question is asking for you to find their point of intersection

not the graph of each different lines

but yes

[sage wren]

parallel lines will never intersect

[nocturne cipher]

I know

[sage wren]

you can check this by subtracting the two equations

y - y = 6x - 6x + 4 - 1

0 = 3 which doesnt make sense

[nocturne cipher]

man

Just tell me how to solve it

[sage wren]

solve which one

[nocturne cipher]

this

[sage wren]

rearrange the second equation for y

then subtract the two equations

[nocturne cipher]

Demenstrate

[sage wren]

this

[lean otter]

watch a video

i already answered your question like 20 mins back, you can simply watch a 15~20 mins video to clear your concepts before you solve maths

[safe radish]

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[proud zealot]

Im not sure if i wrote the correct multiplicitys

[thin bridge]

multiplicities seem fine

[thin bridge]

you've only considered the roots,
you also need to determine the leading coefficient by using another point on the graph

i.e. with the roots you would only know
y(x)=a(x+4)(x-3)^2(x+2)(x-1)
determine the value of a
also you don't need to explicitly write those powers of 1

[proud zealot]

how u determine the value of a?

[thin bridge]

by using another point on the graph

[proud zealot]

any?

[thin bridge]

yes

[proud zealot]

but the zeros?

k

[thin bridge]

use whatever point you find most convenient to you

theres only really one option here

[proud zealot]

(-3,4)

[thin bridge]

wouldn't have been what i'd choose but that'd work

it's much better to use a clearly labelled point, in this case the y-intercept

instead of something that requires a ruler/eyeballing which may not even be exact coordinates

[proud zealot]

uh now what?

[thin bridge]

sub that point into
y(x)=a(x+4)(x-3)^2(x+2)(x-1)

[thin bridge]

don't use a calculator to get decimal approximations

you want exact value, leave as a fraction and simplify that

[proud zealot]

so

[thin bridge]

yes

[austere magnet]

Checks out

Has the question been resolved?

[proud zealot]

https://tenor.com/view/взгляд-2000-ярдов-война-war-soldier-gif-3632617944134077161

[austere magnet]

You still have x in the submitted expression

That would make 0 a root

[proud zealot]

oh. so now x by the 1/36

no*

[austere magnet]

That's the multiply sign???

[proud zealot]

yeah

[austere magnet]

o7, happens to the best of us

[proud zealot]

.close

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[shut valley]

can someone help me with this? my exam is in 45 minss

[languid quest]

How can you rewrite in terms of sin and theta

(Instead of arcsin as it is now)

[shut valley]

yeah i need to find the exact value but im not sure how

how

[languid quest]

I’m asking you to rewrite in terms of sin and theta

[shut valley]

from the formula right

ill do that

[languid quest]

$\arcsin(\frac{\sqrt{3}}{2})=\theta\\
\frac{\sqrt{3}}{2}=\sin\theta\\
\text{Refer to unit circle. What value must angle }\theta \text{ be to } = \frac{\sqrt{3}}{2} \text?$```

[flat frigate]

mari

[safe radish]

@shut valley Has your question been resolved?

[shut valley]

no but ill figure it out

thank you

[plucky elk]

!done

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.close

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[tranquil brook]

According to Intermediate Value theorem shouldn't this be always true?

[safe radish]

@tranquil brook Has your question been resolved?

[plucky elk]

Yes

[tranquil brook]

Yea thanks I was so confused it's legit the definition of the theorem I'm not sure why it's been marked as wrong

[sick scarab]

i think its because of the [ ] and between a and b

there exists a real number c between a and b such that N=f(c)

if N is in ] f(a) , f(b) [ actually :p

[sick scarab]

even in here, its < and not <=

i think thats why..

[tranquil brook]

Would it then be sometimes true?

[sick scarab]

yes, if f(a) < N < f(b), its true, you can always find a < c < b such that f(c)=N

but if N=f(a), then c must be a, doesnt it?

but a < c < b

c cant be a

hence, wrong, or... not always true

same thing can be said for N=f(b)

[tranquil brook]

.close

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[dire ingot]

Hi I have a very random question. So when determining whether a stochastic matrix is regular, is there any other way to tell than to take the power again and again?

[safe radish]

@dire ingot Has your question been resolved?

[dire ingot]

No

[median vigil]

for an n x n matrix you only have to examine up to the (n-1)^2 + 1 power

[dire ingot]

Oh wow

Any proof for this? I'd def want to read

[median vigil]

i found the theorem here (presented without proof): https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom)/10%3A_Markov_Chains/10.03%3A_Regular_Markov_Chains

this stackexchange answer discusses proofs: https://math.stackexchange.com/questions/450090/if-p-is-a-regular-transition-probability-matrix-then-pn2-has-no-zero-ele/450135#450135

here is the original proof: https://people.math.wisc.edu/hans/paper_archive/scanned_papers/hs149.pdf

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@dire ingot Has your question been resolved?

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[median vigil]

try using the angle sum/difference identities

[safe radish]

@somber dust Has your question been resolved?

[safe radish]

@somber dust Has your question been resolved?

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[safe radish]

@lean otter Has your question been resolved?

[wet wigeon]

Huh?

There is only one x value which satisfies this, not multiple

How would you even get those values

Next time, try to input the values you find into the equation and see if they hold, yourself

"x=1,2,7" is usually used to represent those values of x satisfy

$$1 \frac{a}{b} = 1 + \frac{a}{b}$$

[flat frigate]

Cyrenux

[wet wigeon]

Yeah

No its not

[flat frigate]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[safe radish]

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[fallow vale]

what is this set equal to?

[fallow vale]

is it just (-2, 2)?

[lean otter]

ye

[safe radish]

@fallow vale Has your question been resolved?

[fallow vale]

thanks

[safe radish]

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[wind fog]

find all integer solutions to $1+2^{x}+2^{2x+1}=y^2$

[flat frigate]

Anik

[short topaz]

did u mean 2^(2x+1) instead of 2^(x+1)?

or r u trying out a variant of IMO2006 P4?

[wind fog]

yes

that's 2006 imo question

[short topaz]

oh wait i see u've edited it

[wind fog]

no i need to know the original one

[short topaz]

u'll need to type it again cus it's been too long so the bot won't rewrite ur message

[wind fog]

i see

[short topaz]

find all integer solutions to $1+2^{x}+2^{2x+1}=y^2$

[flat frigate]

LY

[raven heart]

it's not, idk why that other guy said yes, it's an intersection not a union you've got here

[short topaz]

right so what have u done so far for this q?

[short topaz]

probably thought it was union rather than intersection

yeah defo not (-2, 2) lol

[wind fog]

I tried to find out a pattern or formula

[short topaz]

that's a start

did u find any solutions?

[wind fog]

nope

[short topaz]

ah k

[short topaz]

well i'd start of by recommending that u rearrange this equation

[fallow vale]

then what is it? is it like {0} or sum?

[wind fog]

how?

[short topaz]

we're trying to find integer solutions (x,y) to our equation here, so trying to rearrange our equation and factorise it sounds like a good idea

[gloomy scaffold]

input ur equation there

[short topaz]

!noadvert

[safe radish]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[short topaz]

<@&268886789983436800>

[gloomy scaffold]

where is advertising in this?

I just want to help

[wind fog]

then?

[short topaz]

do u spot a good way we can rearrange our equation?

hint: notice that x^2 and 1 are both squares

[polar lynx]

Please don't advertise your website here.

[hard crest]

getting a computer to solve it for them will not help them

[short topaz]

posting an online equation solver to help solve an equation is not help

[raven heart]

{0} is too extreme in the other way
might wanna reopen another channel don't wanna bother ly and others much more

[fallow vale]

okayokay thank you

[wind fog]

not yet

[short topaz]

this person wants a method to be able to rigorously find all the solutions

posting a link to ur website doesn't help with that

[short topaz]

so what happens if i write

$2^x + 2^{2x+1} = y^2 - 1$

[flat frigate]

LY

[short topaz]

can u see a way we can factor both the LHS and RHS?

[wind fog]

yep

taking the 2^x out

[short topaz]

good

what about the RHS tho?

the y^2 - 1?

[wind fog]

(Y+1)(y-1)

[short topaz]

nice!

now we have

$2^x(1+2^{x+1}) = (y-1)(y+1)$

[flat frigate]

LY

[short topaz]

can u spot a way to progress?

[wind fog]

no

wait i found a pattern

for n=3 its 2 1 3 or 3 1 2 and for n=6 the pattern is 6 2 4 5 1 3. the first 3 term of n=6 is just n=3 doubled and next three terms is doubled minus one. its true for all n and 2n

@short topaz

[short topaz]

wdym by n=3?

[safe radish]

@wind fog Has your question been resolved?

[safe radish]

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[safe radish]

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[lean otter]

I need help, please
I read that you cannot directly substitute in a limit to find continuity because it assumes that the function is continuous. Direct substitution is only possible for continuous functions, so we must use limit properties to prove continuity.

For example:

lim (x → k) (2x + 3x) = 2 lim (x → k) (x) + 3 lim (x → k) (x) = 2k + 3k

However, I've realized that even if we use limit properties, at one point we still need to directly substitute k into x, which is the same as direct substitution. So, if we think about it this way, it's just not possible to find the limit. Why do they say that we should use limit properties?

[glacial cairn]

Technically, to find the limit at a specific point, you can use an epsilon-delta proof

[lean otter]

I am asking about the use of limit properties to find continuity which also includes direct substitution at a point so why is it said to use the limit properties it made no sense to me

[glacial cairn]

define "limit properties"

[lean otter]

Can you please read this question to understand me more well https://math.stackexchange.com/questions/3769908/why-use-limit-laws-to-verify-continuity-instead-of-direct-substitution#:~:text=In any case - limit laws,other examples - not so much.

[glacial cairn]

However, why would the text explicitly state to use the limit laws to verify continuity? I'm assuming this implies that direct substitution shouldn't be used

You just said:

the use of limit properties to find continuity which also includes direct substitution at a point

That's you

Direct substitution isn't a limit property

It's not really a property at all anyway

[lean otter]

you didn't understand me

Wait

If I apply any limit property, then a point will come where I have to directly substitute c into x right?
Like:
lim (x → c) (2x + 3x) = 2 lim (x → c) (x) + 3 lim (x → k) (x) = 2c + 3c
Look

[glacial cairn]

You use limit properties to break up a limit into easier ones, where you can assume continuity as a shortcut

You can use that shortcut because you've already proven, via epsilon-delta proofs for instance, that these easier expressions have some continuity

[lean otter]

Oh alright

Now I understood

So if everything is just proven already, like polynomial functions are continuous, exponential functions are continuous trig functions are continuous why the online notes still Suggests to use limit laws to make me more confused

[glacial cairn]

For something like 2x + 3x, first off you should really just write it as 5x (basic algebra, nothing to do with limits), and then either you know that kx is continuous for any k, or you prove it some other way

[lean otter]

Alright, thank you

🙏

[glacial cairn]

[lean otter]

Can you answer this please

[glacial cairn]

You'll have to give me an actual example

[lean otter]

Okay

Wait a minute

[lean otter]

Are you here sir, sorry for pinging you

[glacial cairn]

Kinda difficult to read this

[lean otter]

Do I give you the link

[glacial cairn]

It just looks like they said lim_{x -> -1} x = -1

So that's their assumption

But it's easy to prove it with epsilon-delta

[lean otter]

Okay okay thank you, thank you maybe they had not teached that every polynomial and rational function are continuous in their notes yet

Thanks again

🙏

[glacial cairn]

[lean otter]

.closw

.closw

.close

[safe radish]

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[radiant canopy]

Feri invests some money.
The rate of interest for the first year is 2.5%
At the end of the second year the overall percentage increase of feri's investment is 6.6%
find the rate of interest for second year

[radiant canopy]

@cerulean crane

<@&286206848099549185>

[craggy sedge]

!15mins

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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.

[radiant canopy]

oh shoot mb

[lean otter]

So the final percentage should be 106.6% since the start of the first year right?

And the first year IR should be 2.5%

So the interest rate would take the amount after the interest rate of year 1 (102.5%) as 100%.

Then from there, it increases a certain x % to reach 106.6% since the start.

[radiant canopy]

ty

.close

[safe radish]

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[bleak stag]

5d and 7d

[sinful glen]

can you show us how you did the other ones and also where you are stuck exactly?

because 5b and 7d are not that different from the others

[bleak stag]

Shit i meant 5d

[safe radish]

@bleak stag Has your question been resolved?

[night tusk]

just set up the long division like this no?

for 7d its just (divisor)(quotient)+(remainder)

in this case $(x^2+7x-2)(x^4+x^3-11x+4)+(x^2-x+5)$

[flat frigate]

NatTaylorsV

[safe radish]

@bleak stag Has your question been resolved?

[lean otter]

,tex \polylongdiv{x^4 + 3x^3 - 2x^2 + 5x -1}{x^2 + 7}`

Ok so this still works

[flat frigate]

Edmund Cloudsley

[lean otter]

Noice it worked

[safe radish]

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[delicate axle]

hey, I got a 4/12 on this one assignment and I want to resubmit it but I don't understand where I went wrong. Almost all the information was from the video, but put into my own words. I don't get it.
(There are 2 videos and then my short essay from those videos. If anyone has the time and effort to read it, please help me out.)

[delicate axle]

The Equations that Describe Nature
Everything in our universe is moving, often in complicated manners. To conquer the
difficulty of analyzing these motions, the greatest scientific minds in history created, analyzed,
and solved various differential equations, thus opening mankind's physical understanding of the
intricacies of the realm in which we live.
Differential equations are the equations which describe the nature of life at its core,
random (yet predictable) sequences of events that occur many different ways over a certain
interval in time which has no end. Life, like differential equations, is blessed with many
Intensifying yet quantifiable patterns. These equations are crucial to being able to visualize and
quantify the way in which a flower blooms, the Three Body Problem, and the growth or decay of
a given population no matter the size. As a physics major, I will be using differential equations to
describe the Three Body Problem and to solve the Schrödinger equation, as well as to describe
change within a system over a given interval.
The Three Body Problem is In physics, specifically classical mechanics, the three-body
problem is to take the initial positions and velocities of three point masses that orbit each other in
space and calculate their trajectories using Newton's laws of motion and Newton's law of
universal gravitation. The growth or decay of a population is modeled through graphing a
population using a phase diagram coupled with an initial condition and it can be analyzed using
limits to see the population cap and how fast the population will take to reach maximum
capacity. The Schrödinger equation is a linear partial differential equation, and has been solved
using the technique of separation of variables. It models the evolution over time of the wave
function in an isolated system.

The essence of differential equations is boundless and lively. With them we are able to
visualize the phenomena around us. Key figures like Isaac Newton and James Clerk Maxwell
played foundational roles in the development of differential equations, laying the groundwork for
modern physics and engineering.

||https://www.youtube.com/watch?v=eSty7oo09ZI||, ||https://www.youtube.com/watch?v=ifbaAqfqpc4||

Requirements:

Rubric:

[safe radish]

@delicate axle Has your question been resolved?

[safe radish]

@delicate axle Has your question been resolved?

[safe radish]

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[fringe dock]

help;

[fringe dock]

how do u work out the x intercept of the graph

[radiant isle]

Set y = 0

[fringe dock]

thank you!

I have another question

i’m not sure how to start

[radiant isle]

I would start by labeling the x-intercepts

[fringe dock]

(x+3)(x+1)(x-2)
x = -3, -1,2

[radiant isle]

Yep

But for part a notice that there is an offset. So you shift the graph 3 units to the right.

[fringe dock]

yes

how will I do that sir

hmm wait

[fringe dock]

would the new coordinates be

x = 0, x = 2, x = 5

i added 3 to each

[radiant isle]

Yeah exactly

[fringe dock]

ah I see