help-10

[solemn mason]

hm

okay

x² - 2x - 8 = 0?

correct?

[devout sable]

yes

now you can use quadratic formula or just factorise

[solemn mason]

so A = 1, B = -2, C = -8?

[devout sable]

yes

[solemn mason]

Yeah but how do I find the nature or root or whatever?

[stable rain]

,w what is the quadratic formula

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Circle#0025

[stable rain]

that bit in the sqrt

thats the discrimiinant

so if its <0 then thr r no real roots

[devout sable]

nature of roots is determined by discriminant

[solemn mason]

okay lemme solve real quick, check my answer k?

[devout sable]

okay

[solemn mason]

how do i use this

[devout sable]

umm what?

[stable rain]

k?

[solemn mason]

how do i use bot

[devout sable]

oh the bot

well for now leave it

lets solve this first

then i can guide you

i mean well there’s a quick start guide in #resources last message

[solemn mason]

okay here

X = -2 ± ✅️ (-2)² - 4(1)(-8)

---

2(1)

okay so after calculating

It's X = -2 ± ✅️ 4 + 32

---

2

so X = -2 ± ✅️ 36

---

2

correct??

...

That 36 is a perfect square no?

so do I like... find the square root?

So like... X = 2 ± 6

---

2??

@devout sable @stable rain

Why did you guys leave me

[stable rain]

[solemn mason]

why

Okay so

It becomes Like X = 2 + 6 and X = 2 - 6 right?

divided by 2??

[stable rain]

hi

[solemn mason]

hi

[stable rain]

,w solve x^2 - 2x - 8 = 0

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Circle#0025

[solemn mason]

wow

[next dagger]

you can (and should) check your answer by substituting it back into the original equation

[solemn mason]

i want to die

[stable rain]

no

urs is wrong

HAHHAA

[solemn mason]

how do i substitute

[stable rain]

i throughly checked it

[next dagger]

no, you had them right

[solemn mason]

[stable rain]

wait it is

[next dagger]

the "divided by two"

[stable rain]

why is there a /2

so weird

but ok u got this rome

[next dagger]

(2+6)/2) = 4, and (2-6)/2 = -2

[solemn mason]

wow

Wait so I don't need to find the roots or what

?

[next dagger]

because the quadratic fomula is "over 2a" and here a is 1, so yhou divide by 2

[solemn mason]

ok

[next dagger]

you did find the roots

[solemn mason]

oh

So... that's it right?

@next dagger do i need to find the sum of roots or something?

or thats just it

[next dagger]

only if you're asked to by the problem, i didn't see the original problem

[solemn mason]

What if I was?

Here :

x² + 2x = 4x + 8

hm

probably not

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

Suppose that f and f' are continuous on t = [0, ∞) and of exponential order as t → ∞. Use integration by parts to show the following:
F(s) = \scriptL{f(t)} → lim_(s → ∞) F(s) = 0

[kind hawk]

What have you tried

What's the relevant integral you need to consider?

[timid silo]

I have gathered the following info so far

By the definition of "exponential order", \exists c \in \doubleR, M \in \doubleR^+, T \in \doubleR^+ \ni \forall t > T, abs(f(t) <= Me^(ct)

[kind hawk]

Put all of that in $..$ so the bot renders it as latex

[timid silo]

The same applies for f', where
d \in \doubleR, N \in \doubleR^+, S \in \doubleR^+
And the following holds:
\forall t > S, abs(f'(t)) <= Ne^(ct)

$\exists c \in \doubleR, M \in \doubleR^+, T \in \doubleR^+ \ni \forall t > T, abs(f(t) <= Me^{ct}$

[warm shale]

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

[kind hawk]

For the bot it's \bR

[timid silo]

$\exists c \in \bR, M \in \bR^+, T \in \bR^+ \ni \forall t > T, |f(t)| \leq Me^{ct}$

[warm shale]

CrimsonDevil_Rias

[timid silo]

This is the definition of "exponential order"

Since f and f' are continuous, $\scriptL{f(t)}$ and $\scriptL{f'(t)}$ exist

[warm shale]

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

[timid silo]

Meant to write as Laplace transform symbol, which is a script letter L

[kind hawk]

\mathcal{L}

Not sure if the bot has a macro for those

You can set up your own preamble so it knows your own commands

If you wanna bother to do that

[timid silo]

Anyways, by integration by parts, I need to show that if F(s) is the Laplace transform of f(t), then the limit of F(s) is 0 as s goes to ∞

[kind hawk]

Yes. So what is the relevant integral you need to take the limit of

[timid silo]

I think it's $\int_0^\infty e^{-st}f(t) dt$

[warm shale]

CrimsonDevil_Rias

[kind hawk]

Ok. Next they already give the tip to use ibp

Which function should you probably integrate and which should you differentiate

[timid silo]

So, I assign the following:
$u = f(t), dv = e^{-st} dt, du = f'(t) dt, v = -e^{-st}/s$

[warm shale]

CrimsonDevil_Rias

[timid silo]

I meant to write what I assigned v as $\frac{-e^{-st}}{s}$

[warm shale]

CrimsonDevil_Rias

[timid silo]

By integration by parts, we have:
$\int_0^\infty e^{-st} f(t)dt - \int_0^\infty \frac{e^{-st}}{s} f'(t) dt = -\frac{e^{-st}}{s}f(t)$

[warm shale]

CrimsonDevil_Rias

[timid silo]

In this TeX code, how to I write in a right line bracket with the limits (t = 0) and ∞?

[kind hawk]

Right line bracket?

$\Big|_{t=0}^\infty$

This?

[warm shale]

Denascite

[kind hawk]

Usually I see ibp written as (integral)=(stuff to evaluate)-(other integra)

And I think that's also generally more useful

[timid silo]

ibp is derived from the product rule of derivatives, which is why I put it that way

[kind hawk]

Yes. But we use it to calculate the first integral in terms of the second most of the time

So it's more helpful to have the two integrals on two different sides of the equality sign

Anyway now we did the ibp, next the limit

Instead of showing that the limit F(s)=0, show that the limit of |F(s)|=0

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

What is the meaning of an expression being invariant? I kind of understand what it means but I am curious why specifically it is called "invariant". Invariant to what?

This example relates to the scaling of a vector to probabilities

[timid silo]

Assume my task is to constrain a vector so that

the elements of the vector (i.e. $w_i \in \mathbb{R}$) represent probabilities $p_i \in [0, 1]$ and $\Sigma_i P_i = 1$

[warm shale]

LongBear

[timid silo]

I could do $p_i = \frac{w_i}{\displaystyle \sum_{i = 1}^{n} w_i}$

[warm shale]

LongBear

[timid silo]

But apparently using a softmax function is better because it is invariant to... something

[kind hawk]

Well like you said, invariance is always with respect to something

So without that something it doesn't make sense

And there are probably different softmax functions depending on that something

[timid silo]

Apparently the issue that we are trying to avoid with softmax (which would be the case if we used the above function) is this:

So I guess it would be "invariant to the size of the numbers". Maybe there is a more accurate way to describe it.

Instead he proposes:

@kind hawk Is there an answer to what "invariant" means or should I close this question?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[kind hawk]

Well first you have to make very clear what the issue is

100, 101 and 102 are close numbers so without further context it's not unreasonable that they should all correspond to about 1/3 probability

If you use that other formula you end up with roughly 10%, 24% and 66%, whether that's more reasonable depends on the context

I wouldn't really call it invariance tho, it's just context depending on what you want

[obtuse pebble]

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[short grove]

hello can someone slove this

[slim leaf]

a bit lengthy but apply quotient rule with product rule and chain rule

[high lily]

hint, let:
$$p(u) = \frac{u^2+5u+1}{u^3-2u-1}$$
$$y = (u-3)^2p(u)$$

[warm shale]

ℝamonov

[obtuse pebble]

@short grove Has your question been resolved?

[obtuse pebble]

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[serene furnace]

Hi

[serene furnace]

If I want to express the half of an x number

I could express it like this

$\frac{1}{2}x$

[warm shale]

Raid

[serene furnace]

But can I also express it like this?
$\frac{x}{2}$

?

[warm shale]

Raid

[tidal hawk]

yesh it’s the samesies

[serene furnace]

ty ❤️

Can I also as

ask

what does it mean to have like the third part of a number?

[tidal hawk]

oh wdym?

[serene furnace]

I mean If I would express the third part of 9

what is it?

[tidal hawk]

third part of 9?

[serene furnace]

9/3 ?

yep

[tidal hawk]

oh a third of 9?

yesh

[serene furnace]

for example

but could be any number

[tidal hawk]

yes

[serene furnace]

would it be the number / 3?

[tidal hawk]

$\frac{1}{3}$ (that number)

[serene furnace]

wdym

XD

oh

yea

same thing

ty ❤️

.close

[obtuse pebble]

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[warm shale]

swaggofishballs

[serene furnace]

.reopne

[tidal hawk]

HAHAHHA welcc!

[serene furnace]

.reopnqwdinqi0ndqwinbdqibodq

XD

[obtuse pebble]

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[timid silo]

How do i know which type of integration factor to use?

[timid silo]

Because there is 5 types

Do i just try one by one until i get something i can work with

[abstract flame]

you mean integration techniques?

you just have to do a lot of problems

until you can recognize which one to use

that's the only way to get better

[random depot]

lhopital cheese

[slim leaf]

@timid silo do you mean integrating factors for differential equations?

if so, depends on the differential equation. for linear differential equation, it is straightforward

but for exact differential equations, there are 3 rules if i remember correctly

Or probably 4

[timid silo]

Yes yes

Differential equations

[slim leaf]

just look at the differential equation and the rules and you'll know what to do

[timid silo]

Okay bro thanks

.close

[obtuse pebble]

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[timid silo]

I need help with the 27th question

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Nah but its ight

[obtuse pebble]

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[obtuse pebble]

@burnt slate Has your question been resolved?

[burnt slate]

<@&286206848099549185>

[obtuse pebble]

@burnt slate Has your question been resolved?

[bold bane]

Disregarding the highlighted area, which of these look like independent events.

@burnt slate

[obtuse pebble]

@burnt slate Has your question been resolved?

[burnt slate]

The first one?

[bold bane]

The reason I told you to disregard the highlighted area because each line is asserting a true statement, that for the LHS the RHS is true. All you have to do is determine if the RHS of each equation is dependent or independent.

Is P(E) an independent probability?

[burnt slate]

Yes?

[bold bane]

Yes.

That same applies for P(F). It is an independent event.

P(E) x P(F) however is not.

[burnt slate]

So which one do I choose

[bold bane]

The question asked which not an independent event, in other words, which is a dependent event?

[burnt slate]

I chose P(e)xp(f) and it's wrong

I have 3 more tries

[bold bane]

Hmm, then I don't know. I tried reasoning it out myself but I was wrong.

I want to say its the fourth option if the first option was wrong.

[burnt slate]

Yeah

It's 4th one

[bold bane]

Which I could explain why. I need to go over this stuff again. 😉

[obtuse pebble]

@burnt slate Has your question been resolved?

[midnight scaffold]

6/10 * 9/10

[burnt slate]

tysm

[obtuse pebble]

@burnt slate Has your question been resolved?

[humble wind]

ayo close the channel

.close

@burnt slate

[stable rain]

.close

[obtuse pebble]

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[stable rain]

💕

[obtuse pebble]

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[dire fox]

what are the steps to this?

[weary delta]

You write the definition of the limit. Then it's easy to show a bound on delta in terms of epsilon

[dire fox]

how do i do that?

I'm completely lost

[obtuse pebble]

@dire fox Has your question been resolved?

[dire fox]

.close

[obtuse pebble]

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[young dagger]

Halp

[dense moon]

is x the distance/displacement equation?

[young dagger]

Yes

[dense moon]

ok then do dx/dt

to get velocity

[young dagger]

Can you explain further, sorry im just new to this physic cal thing

[distant moth]

I dont think u need to do dx/dt here

[next dagger]

the equation given is the cyclist's position at time t

yhou do not need calculus for this

just the forumat for average velocity, which is just distance traveled dvided by time

[young dagger]

Oh also, the topic is called MOTION IN ONE DIMENSION

Thats the only question remaining and still cant solve it

[next dagger]

so the cyclist's position at $t=0$ is $x(0)$ and at $t=5$ it's $x(5)$, the distance traveled is $x(5) - x(0)$, and the time elapsed is $5-0$, so the answer to a is $\frac{x(5)-x(0)}{5-0}$

[keen swift]

Average Velocity = Displacement / Change in Time

[tranquil quiver]

smh

nah

[next dagger]

there, sorry about the missing diggle

[tranquil quiver]

swap 5 and 0 on one of the

[next dagger]

oh, boogers

[young dagger]

Wait so im just going to replace the "t" with the given numbers like 0,5?

[warm shale]

rome of oxtrot

[next dagger]

yes

[unborn valley]

yes

[young dagger]

Wait, no derivative thingy needed ?

[tranquil quiver]

nah thats for instataneous

[next dagger]

no, this is average velocity, not instantaneous

[young dagger]

Ohhhhh i mean i did that already and thought it was wrong

Cuz imagine a bicycle travelling 300+ m/s

[next dagger]

lol

[dense moon]

rip

[next dagger]

the equation given is silly physically

[young dagger]

So, all this time thats the easiest question

[unborn valley]

if your calculation is right, don't worry about the values, these are usually not based on practical applications or real life.

[next dagger]

after a relatively short time, the bicyclist will have pedaled his way into a black hole

[unborn valley]

lol

[young dagger]

lol

[next dagger]

created by his ridiculous;ly high mass-energy as his velocity approaches the speed of light

[young dagger]

Guys you dont know hmuch time that question took away from me

[stable rain]

HAHHAA

[young dagger]

Anyways, thanks alot to everyon who helped

How to end this thing?

[tranquil quiver]

.close

[young dagger]

.close

[obtuse pebble]

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[wild chasm]

why are the options in decimals?

[wild chasm]

the 10th percentile would be 20 no?

50th would be 80 and 100th is 105

why are they in decimals?

[obtuse pebble]

@wild chasm Has your question been resolved?

[stable rain]

um so

the 50%

wld be the

middle of 5th n 6th person

aka average of them

like if u throw down all the ppl in order n give them a percentile

ull c the stuff dont fit nicely

n when they dont u just take the avg of the 2 closest

as usual let 0% be the smallest

100% rhe largest

the the rest just fills in

but idk

so like

11.11% will be the 2nd person

so ig 10% is avg of smallest n 2nd smallest?

nvr rly done this b4 tho

bit weird to me

[obtuse pebble]

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[wild chasm]

so 83 in this case

got it

[stable rain]

yay

[obtuse pebble]

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[signal snow]

yoyoyi

[signal snow]

<@&286206848099549185>

bro this question sucks

[stable rain]

!15min

[obtuse pebble]

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.

[signal snow]

yo @stable rain help me

someone bro

ive been at this for an hour

[random depot]

Simplify it ez

[stable rain]

idk

thrs a

r^r

which seems sus

[random depot]

Susssyyy

[signal snow]

this entire question is sus bro

dont tell me i wasted an hour

[brisk grove]

that's some determination

a whole hour on this

[signal snow]

bro ols

some math god descend

<@&286206848099549185>

[brisk grove]

You are not gonna get constants for a, b, and c I'm pretty sure

that's they goddess said it's sus

[high lily]

the sqrt(qr) probably wasn't intended to be in superscript

[signal snow]

bruhh

BRUH

ONE HOUR

[stable rain]

HAHAHHAA

[devout sable]

one hour spent well

[stable rain]

well said

deep words

[obtuse pebble]

@signal snow Has your question been resolved?

[obtuse pebble]

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[timid silo]

In this video
https://www.youtube.com/watch?v=gMlf1ELvRzc&t=937s
At 1.:40 to 2:23
why is he dividing ?
I don't understand that part ?

[timid silo]

he expands it

the ^2

@slim needle this channal is taken by me

okay

In this video
https://www.youtube.com/watch?v=gMlf1ELvRzc&t=937s
At 1.:40 to 2:23
why is he dividing ?
I don't understand that part ?

he's dividing the hexagon to calculate its perimeter

so that he can compare it with the circle's perimeter

but perimeter of hexagon is 6a
why is he dividing it?

[sleek wyvern]

what do u mean by dividing it?

in the calculations or in the drawing first?

he divides it into 6 triangles so the far out part, hexagon sides can be visualized as a circles perimeter

[timid silo]

how dividing the diameter with perimeter of hexagon
help with perimeter of circle ?

[sleek wyvern]

yes he divides the hexagon into 6 pieces but only for the sides and divides it with the diameter of the hexagon which also happens to be the diameter of the circle to calculate the perimeter of the cirlce

[timid silo]

okay

thanks ,, understand

[trim thicket]

hello guys i have a question what do u use usually to write math with?

[sleek wyvern]

np

#❓how-to-get-help

[timid silo]

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

I have a question on how I'd generally approach this type of problem.

I take the lengths and dot product

set them equal and solve?

is that literally all there is to this? this seems like a really elementary example but it wasn't on the homework at all

[kind hawk]

That's all there is

The numbers here are chosen so that the equation is nice

But that's the general approach that always works

[timid silo]

appreciate it, just wanted to make sure

.close

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[kind hawk]

Sometimes exam questions are easier than homework

[timid silo]

this is undoubtably the case

the homework gives me a skin rash every time I look at it

[kind hawk]

Because they just want to test if you know how to approach a problem

And not additionally give you hard/annoying numbers to make stupid mistakes

Well good exam questions anyway

[obtuse pebble]

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[keen turret]

how do i find the minimum value of this function without derivatives?

[keen turret]

is it possible?

[keen turret]

<@&286206848099549185>

[brisk grove]

Why would you want to do that?

<@&268886789983436800>

Thanks

[keen turret]

to find the minimun time

[brisk grove]

Why without derivatives?

[keen turret]

because i don't know calculus yet and this problem showed up in my exercise list

its a physics problem

[brisk grove]

What topics are you learning?

Oh

Maybe they assume you already learned calculus?

[keen turret]

probably

its a list for a test

idk what theyre called in english

like jee

he wants to get to B in the least time

velocity in AC is 2m/s
AB is 1m/s

BC = 40, AC = 30

he starts at A

[brisk grove]

You must go either on AB or on AC?

[keen turret]

no

you go to a point D, that is between a and c

and then go to B

[brisk grove]

How would you know the velocity on your way to D?

Based on the angle with AC or something?

I could try helping if you give me enough info on this problem. I haven't studied that much physics after all

and this is a math server

[keen turret]

wait

im drawing

[brisk grove]

and I'm not sure this has a simple solution

[keen turret]

Velocity in DB is also 2

[brisk grove]

Oh I see

[keen turret]

1*

[keen turret]

(x^2 + 40^2)^1/2

[brisk grove]

Ok I see where you got your question from

Well, can't really see how you would do this

You might want to make sure they aren't assuming to know calculus.

[keen turret]

yeah

i will learn the basics

just so i can do those problems

[brisk grove]

Calculus isn't a small topic

[keen turret]

yeah but basic physics problems dont use advanced calculus

just derivatives and integrals

like if want to go to position function to velocity funcion

[brisk grove]

"just"

ok

[keen turret]

and the just the basics of it like chain rule and that trick to subtrtact 1 from the exponent

and multiply

at least for my test

not being cocky

[brisk grove]

well seems like you are familiar with it already

yeah all good

if you feel like you can get it quickly then thats great

good luck!

[keen turret]

thanks

.clsoe

.clsoe

.clsoeee

.clsoe

man

.close

[obtuse pebble]

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[timid silo]

help me

[timid silo]

PLEASE

i got 1/12

somehow its wrong

PLWaAAAAAASE#

[dim patio]

What's wrong

[royal shard]

try to get common denominators

18/7=36/14

3/14 equal one bracelet. we have 36/14

in other words how often can 3/14 go into 36/14

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

why did you double it?

[royal shard]

if you have a fraction like a/b then we can multiply nominator and denominator by the same number and the grmmfraction wont change
so
a/b=2a/2b

[obtuse pebble]

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[obtuse pebble]

@burnt slate Has your question been resolved?

[wary magnet]

(8/13)*(8/13)

(64/169)

@burnt slate

For the first problem

[burnt slate]

ok

and for the second i use @wary magnet 5/13*5/13?

[wary magnet]

Yes

[burnt slate]

ty

[wary magnet]

Np

[obtuse pebble]

@burnt slate Has your question been resolved?

[humble wind]

@burnt slate close the channel

[obtuse pebble]

@burnt slate Has your question been resolved?

[burnt slate]

.close

[obtuse pebble]

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[thick shore]

sin(L) = 0.502
what's going to happened to 0.502?

[warm canopy]

wdym whats going to happen

[thick shore]

like

is it sin(0.502)?

or something else?

[warm canopy]

is what?

[royal shard]

you want L?

[thick shore]

well yes

[warm canopy]

then tell us😭

[thick shore]

here's the thing

[royal shard]

sin(L)=0.502
L=sin^-1(0.502)

[thick shore]

ooh

[royal shard]

dunno if sine^-1 is the same as arcsin

[thick shore]

^ is root right?

[royal shard]

expoment

^-1 denotes inverse

[thick shore]

so 3^2 is 9?

[royal shard]

better call it arcsin

L=arcsin0.502

[thick shore]

i feel so embarassed honestly

[royal shard]

,arcsin(0.502)

,w arcsin(0.502)

[warm shale]

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[royal shard]

lets test it

,w sin(0.52591)

[warm shale]

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[royal shard]

nice

[thick shore]

so i was following this site and

https://www.mathsisfun.com/algebra/trig-solving-ssa-triangles.html

it turns to 30.136

I'm just confused honestly

[royal shard]

yeah

they use degrees

not radians

[wild swallow]

,w arcsin(0.502) in degrees

[warm shale]

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[thick shore]

so i had to convert this from deg to radian?

[wild swallow]

bruh

wolfie pls

,w arcsin(0.502) degrees

[warm shale]

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[royal shard]

to get radians we multiply by 360/2pi i think

,w arcsin(0.502)×(360/(2pi))

[warm shale]

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[royal shard]

nice

[thick shore]

:0

so do arcsin first then convert?

[royal shard]

order doesnt matter i think

i think

[thick shore]

oh yay i got it

[royal shard]

if you convert first you have to put ° though

i think

i never use degrees haha

[thick shore]

this mistake messed up everything in my calculation

let's see if it works well or not

i think it might've went negative

welp gotta debug then

.clos

.close

[obtuse pebble]

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[solemn crater]

would someone be able to take a look at this for me?

I got to the indeterminate form through direct substitution, now I am supposed to simplify it so that it can be evaluated. Something does not feel right here though

(since you can't really evaluate 5/0 anyways)

Did I do something wrong, or is it possible that this function just cant be evaluated at x=0?

[timid silo]

use L hopitals rule

[warm canopy]

$\sqrt{x^2+9} \neq x+3$

[warm shale]

ΣAC

[solemn crater]

is it not? That would probably be what I am doing wrong then

[wraith mountain]

Yes, but its also true that x do not = 0

[warm canopy]

you also cannot cancel like this: $\color{red} \frac{6-\cancel{x}}{\cancel{x}x}$

[warm shale]

ΣAC

[solemn crater]

lol, yeah that one felt wrong, but I was not sure where to go at that point

[warm canopy]

i suggest rationalising the numberator, by multiplying the top and bottom by the same thing

[solemn crater]

That sounds like a promising plan, I'll give that a go

[obtuse pebble]

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[tawdry crane]

Hello, can I please receive any guidance with this question? Thanks.

[lilac glade]

where are you stuck?

[obtuse pebble]

@tawdry crane Has your question been resolved?

[obtuse pebble]

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[timid silo]

need help on these two please

[random ocean]

standard form would be 2.5*10^7

[timid silo]

omf im so stupid

i lit done that but not in standard form.

[random ocean]

10^n is a square number when n is even

[timid silo]

didnt read the question properly lol

thankss

.close

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[indigo steeple]

n.n - 2n = 15

[teal turret]

n*n ?

If yes, set up as a quadratic in ax^2 + bx + c = 0 form, and then use quadratic formula to solve. There are other methods but they take a lil bit more explaining

[obtuse pebble]

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[obtuse pebble]

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[somber spade]

help

[somber spade]

I have no clue how to do this

its ok

[royal shard]

a^0=1

so 2×a^0=2 for all a

so 2×a^x works

or easier
f=2+a×x

where a is a random constant

x could also have exppnents if you want

[somber spade]

ok ty

.close

[obtuse pebble]

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[faint cave]

I’m really struggling with this question

This is my second attempt

But I keep getting a 0

And before I derives it again but secx goes to secxtanx and tan0 is 0

So it won’t work either

[obtuse pebble]

@faint cave Has your question been resolved?

[faint cave]

Never mind solved!!!!

Resolved

Solved

How to end thsi chat

.close

[obtuse pebble]

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[obtuse pebble]

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[sleek mural]

For a false conjecture’s counterexample, do I need the angle A is equal to 55 degrees part? Or do I just need the drawing?

[civic zealot]

you've cut off what you're trying to counter....

so we have no idea.

[surreal pecan]

so I'm stuck on which one to choose and i cant pass if i keep getting them wrong can someone help the answer is 61

[nocturne minnow]

#❓how-to-get-help

[sleek mural]

“Show the conjecture is false by finding a counterexample: “If angle A can be bisected, then angle A must be obtuse”

[civic zealot]

so you're saying mA = 55 just to show that it's not obtuse?
that's fine. I don't know that's necessary if the drawing is clearly of an acute angle.

[obtuse pebble]

@sleek mural Has your question been resolved?

[sleek mural]

Ty

Also

@civic zealot

The teacher proved this differently, but is mine still correct?

[obtuse pebble]

@sleek mural Has your question been resolved?

[obtuse pebble]

@sleek mural Has your question been resolved?

[obtuse pebble]

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[split wraith]

I need to prove that a function is a subgroup of the real numbers. The function is bijective and has an inverse.

[split wraith]

<@&286206848099549185>

<@&286206848099549185>

plz DX

@timid silo

[devout sable]

!15m

[obtuse pebble]

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.

[split wraith]

oh sorry, i wanted to show them where I was

i dont think anyone is going to come help tho.

.close

[obtuse pebble]

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[mental siren]

Could anyone tell me how to get the answer on a and b

[tardy epoch]

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

[mental siren]

I don't know how to start or apply the formula

[obtuse pebble]

@mental siren Has your question been resolved?

[mental siren]

.close

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[minor belfry]

this is a past exam question

i have very little clue how to do this

ive worked out one path

if y = x

and u take the limit of (x,x) -> (0,0)

u simplify it to a 0 on the numerator with values still on the denominator, so it can equal 0

but i need a 2nd path

which i cant find

i assume u cant do x = y?

since they are literally the same thing

wait i think i applied l'hop incorrectly when i tried smth earlier

idk if thats the correct approach?

could smn let me know

<@&286206848099549185>

<@&286206848099549185> hate to double ping, but would appreciate some help

[tardy epoch]

use y=kx for some constant k

and k not equal to 1 obviously

[minor belfry]

was my y = x acceptable?

[tardy epoch]

no idea

[minor belfry]

and is doing y = kx considering a different path?

oh right

[tardy epoch]

if k is different from 1, yes

[minor belfry]

if i do y = kx, and x = my, are those different?

the very top 3 lines are my steps for y = x

does that look fine?

[obtuse pebble]

@minor belfry Has your question been resolved?

[obtuse pebble]

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[rough dirge]

anyone know why this is wrong

heres my work

i changed the beginning to 6/n

[obtuse pebble]

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[obtuse pebble]

@rough dirge Has your question been resolved?

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[echo steppe]

sda

[echo steppe]

ok i thought tangents were diagonal lines

why is this straight

[final thunder]

tan(x) = 2

[echo steppe]

that still makes it straight

[final thunder]

Yeah that makes it straight

Because Tanx is 2 for certain values of x

[high lily]

plotting tan(x)=2
gives the values of x where that equation is true

[echo steppe]

so what does that make the slope?

[high lily]

well the equation tan(x)=2 by itself
has no relation to your parabolic function

[echo steppe]

how would i make it represent a slope

for the parabola

[high lily]

have you started derivatives?

[echo steppe]

barely

[high lily]

would you be able to find the derivative of f(x)

[echo steppe]

f'(x)= -2x+2

[high lily]

yes

now, do you have a clear description of what you're trying to do?

are you trying to find the tangent to the curve at x=2?

or finding the tangent with a slope of 2
or something else

[echo steppe]

finding it at the curve of x=2

[high lily]

ok, so sub x=2 into

f'(x)= -2x+2
to get the slope at x=2

[echo steppe]

so that would make it -2

correct?

-2x2+2?

[high lily]

don't use x for multiplication

-2 * 2 + 2 = -2

that will be your slope at x=2,

sub x=2 into the original function to get your y-coordinate and point

[echo steppe]

so -2^2 +2 * 2 + 3?

[high lily]

yes

[echo steppe]

-1?

[high lily]

no

also why did you edit

[echo steppe]

is * times or power?

[high lily]

-2^2 +2 * 2 + 3

times

^ is for power

[echo steppe]

3

[high lily]

yes

so at x=2, you have
slope of 2
and the point (2,3)
then apply stuff like point slope formula from coordinate geometry to get the equation of your line

[echo steppe]

okay thankyou for all the help ill probs be back later

.close

[obtuse pebble]

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[placid totem]

Im tryna do my calculus homework, I got the b section doen but Im not sure how to do a

[placid totem]

actually wait

does chain rule apply here?

[plain stag]

Indeed

[royal basin]

does it apply in any nontrivial way? idts

oh the question was about part a mb

...what's with the $\ln \cdot x$ in the parentheses \thonk

[warm shale]

Ann

[placid totem]

oh shit

💀

happens

[plain stag]

l, n \in R

Obviously

[royal basin]

flashbacks to when i saw someone type natural logarithm as In all over the page

[devout sable]

wut

lol

[placid totem]

💀

[plain stag]

Love it

[devout sable]

Iove it

[alpine raven]

i snort coke

[placid totem]

I snort coke zero

but wait, If im applying chain rule then its d/dx(ln(ln(x)))*d/dx(ln(x) ) or?

Oh wait yeah I think I got it now

.close

[obtuse pebble]

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[sacred cove]

Is anyone able to help me with how to go about this? Wouldnt i just have to define these values in matlab to plot them or is there some additional steps?

[sacred cove]

<@&286206848099549185>

[obtuse pebble]

@sacred cove Has your question been resolved?

[kind hawk]

just have to define them and then use the plot command

[sacred cove]

ok yeah i think i overthought the question that makes sense thanks

[sacred cove]

do you know how i can find the tangent vector to (x,y)

[kind hawk]

well you hopefully should have done that in your course

or do you mean with matlab?

not sure if matlab has a command for that

[sacred cove]

nah i understand what tangent vectors are im just not sure how to apply it here

[sacred cove]

would i just differentiate both x and y?

[obtuse pebble]

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[sacred cove]

.reopen

[kind hawk]

with respect to xi, yes

[obtuse pebble]

✅

[sacred cove]

and then im left with two tangent vectors that are with respect to xi?

[kind hawk]

its only one tangent, yes

[sacred cove]

right

so itd be like
[x
y] and thats one tangent?

[sacred cove]

ok sure that makes a lot of sense thanks, sry im a bit rusty

[kind hawk]

you may have to move that vector around so it actually touches the curve

[sacred cove]

what do you mean?

[kind hawk]

well like this it points in the correct direction but starts at the origin

[sacred cove]

are you talking about the original vector or tangent vector?

[kind hawk]

the tangent vector

[sacred cove]

hmm im still a bit confused how would i do that?

[kind hawk]

how about you just plot it and see

[sacred cove]

ok yeah, also how do i differentiate this in terms of xi if xi is a range of values?

[kind hawk]

you just compute the derivative by hand as always?

don't differentiate numerically

well I guess you could but why

[sacred cove]

right and then sub in 0 and 1 as xi?

after deriving

[kind hawk]

xi is a real number between 0 and 1

[sacred cove]

yeah but im confused what you mean by derive in terms of xi

oh u mean juust derive normally?

[kind hawk]

yes

[sacred cove]

😂

that makes sense sorry

thanks for ur help

[sacred cove]

i tried deriving the equations normally and got -1 and 1

[sacred cove]

if thats right is this what u mean?

[sacred cove]

i plotted it but im not sure what you mean by this

[kind hawk]

you made a mistake with the derivative

those sin terms don't just vanish

[sacred cove]

really i checked it with a calculator as well

hmm

oh right

[sacred cove]

i got this now but doesnt the vector already touch the curve?

[kind hawk]

really not sure what you are plotting there

but I wouldn't call that a tangent

[sacred cove]

right but if you derived x and y wouldnt u get a cos graph?

[sacred cove]

because theyre both defined in terms of sine

[kind hawk]

well ok those are indeed both cos curves

but they are not tangent to the original curve we care about

[sacred cove]

right

so im not meant to derive the original equations?

[kind hawk]

you are. just plot the result differently

[sacred cove]

could you give me a hint on what u mean?

[sacred cove]

because surely those graphs cant be tangent if they go directly through the parametric plot

[kind hawk]

ok. lets take a specific point. lets say xi=0.5

plot the tangent vector as a line

[sacred cove]

ok i got these by plotting at xi=0.5

[kind hawk]

what is that other line?

plot a line from (0,0) to (x'(0.5), y'(0.5))

and also plot your original curve

[sacred cove]

right im not sure how to do this in matlab yet, but would this be the expression for the tangent vector then?

[sacred cove]

why did you choose the line to go to x'(0.5),y'(0.5) instead ot x'(1),y'(1)

[kind hawk]

because I chose xi=0.5 arbitrarily

[sacred cove]

but thats not what im meant to do right?

[kind hawk]

no but we might get there. maybe. I actually have to go soon

[sacred cove]

are u able to tell me before u go what im meant to do?

[kind hawk]

I mean the original question just asks about the curve itself. that part you did

I don't know the exact phrasing of the question about the tangent

[sacred cove]

ill bring it up

[sacred cove]

this is what it is

and this is the initial values

[kind hawk]

oh you don't even have to plot it

[sacred cove]

wait im confused how have we already done this?

[kind hawk]

I mean you don't have to plot the tangent

and you found the expression already by differentiating etc

[sacred cove]

wait so can i just clairfy my understanding

[sacred cove]

when i found the derivative i got a cos graph, but when i used a calculator and found it at xi=0.5 i got a tangent line

[sacred cove]

which one is the expression that ur talking about?

[sacred cove]

are u saying the expression is just the two cos equations that i got?

[kind hawk]

yes

[sacred cove]

oh right thanks a lot im not sure why i got so confused lol

[obtuse pebble]

@sacred cove Has your question been resolved?

[obtuse pebble]

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[north cove]

In this problem, how do we know that the range of possible values for [x] will be [0,6]?

[north cove]

GIF is the floor function, btw

[royal basin]

do you mean {0,1,2,3,4,5,6} maybe?

[north cove]

Oh yes

my bad

I meant integral values

[north cove]

But how do we know that

[royal basin]

well i can come up with something a posteriori to justify why there do not exist any solutions for our equation outside of [0, 7)

coming up with the range of floor(x) just from the equation is tricky though

[north cove]

Oh yes that would be great

I asked this question earlier

and I got wonderful answers

but I just realized I didn't quite understand this part

[fervent cradle]

that's not obvious to me

we could go like x^2 - 7(x - e) + 5 = x^2 - 6x + (7e + 5) = 0, where e is a number between 0 and 1

then quadratic formula it

x = 0.5(6 +- sqrt(36 - 4(7e+5))) = 3 +- sqrt(4 - 7e)

which is [1, 5]

[north cove]

Quadratic formula won't work here, will it?

[fervent cradle]

why not

[north cove]

we have x^2 and [x]

[fervent cradle]

look closely at what i did

so i let [x] = x - e

[north cove]

oh right

e being the fractional part

[obtuse pebble]

@north cove Has your question been resolved?

[obtuse pebble]

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[smoky vault]

Im doing a constrained optimization problem but I keep getting the FOC as constants so I dont know how to make progress

[obtuse pebble]

@smoky vault Has your question been resolved?

[smoky vault]

<@&286206848099549185>

[obtuse pebble]

@smoky vault Has your question been resolved?

[obtuse pebble]

@smoky vault Has your question been resolved?

[obtuse pebble]

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[umbral wren]

Hi, I don't know what I'm doing wrong here and I don't really understand the proposed calculation very well, could someone pls explain? I've been trying to figure out this question since yesterday

[umbral wren]

combinations I've tried:
c&d, b&d, a&c, a,b,&c.

[viral blade]

well this is a pretty dumb question I think

but there's only 1 answer choice that states true information

the others are all wrong so it's process of elimination

[umbral wren]

I tried a, b, & d and it's not the answer either

[obtuse pebble]

@umbral wren Has your question been resolved?

[viral blade]

is (1-1) = (-1+1)?

[obtuse pebble]

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[summer stone]

how do i go this?

do i make my functions, f(x) = sqrt(x^2 + y^2 +z^2) and g(x) = y^2-xz = 25?

[slim leaf]

f(x,y,z) = sqrt(x^2 + y^2 + z^2) and g(x,y,z) = y^2 - xz = 25

[summer stone]

okay, awesome! just wanted to confirm

.close

[obtuse pebble]

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[obtuse pebble]

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[cold flower]

i'm stuck at b ;-;

i'm not sure how to do ln

[obtuse pebble]

@cold flower Has your question been resolved?

[obtuse pebble]

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[hazy junco]

Do I need to find the equation of the graphs first

[teal turret]

No

That’d be tough lol

[hazy junco]

idk what to do lol

[teal turret]

So for the 1st one

(f+g)(0)