help-10

[valid oxide]

I see

was wondering about that

but regarding the substitution

if I did n as infinity, the fraction would approach 0 no?

[tame python]

thats not what I meant by substitution

[valid oxide]

oh

[tame python]

do not try to plug in n=infinity in there

it doesnt work

that is why you are doing a limit

[gloomy vector]

$\lim_{n\to\infty}\frac{\qty(1+\frac{-3}{n+5})^n(1+\frac{-3}{n+5})^5}{(1+\frac{-3}{n+5})^5}$

[warm shale]

Skill_Issue

[gloomy vector]

is this technichally possibke here

you get e^-3/the bottom lart

yea ok i should shut up

[slim cove]

Skill_Issue I think you're confusing rihito

I think mtt's got it

[gloomy vector]

mk

[valid oxide]

appreciate the help though!

[gloomy vector]

tbh im morr asking mtt cause i also wanna know

[gloomy vector]

yea sorry

[valid oxide]

nothing to be sorry about!

[tame python]

Im back

[tame python]

this is why you think through your thoughts before you unveil them at someone without warning

[gloomy vector]

huh

[tame python]

@valid oxide we'll need to do a few more steps until we get to here - itll allow us to do the substitution

keep that fraction in mind

[valid oxide]

gotcha

[tame python]

first, let me show you what a substitution looks like

[valid oxide]

ok

[tame python]

you know from earlier that n + 5 also -> inf

[valid oxide]

yes

[tame python]

so that would mean:

,,\lim_{n\to\infty}\qty(1+\frac r{n+5})^{s(n+5)}=e^{rs}

[valid oxide]

I see

cause n = n+5

[tame python]

not really

[valid oxide]

oh

[warm shale]

mtt

[tame python]

the idea is more of

n -> inf means n+5 -> inf

and if n+5 -> inf, then this is just like the limit we had before but with a different-looking "n"

can you see how that works

the n+5 "acts like" the n in the limit we know baout

[valid oxide]

since they both approach infinity

[tame python]

yep

[valid oxide]

ok

[tame python]

now this is very close to what we have

we currently have $\lim_{n\to\infty}\qty(1+\frac{-3}{n+5})^n$

[warm shale]

mtt

[tame python]

now off the bat, can you take a guess at what r and s would be

remember that r and s are both constants

[valid oxide]

r being -3

s being 1?

[tame python]

thats correct

[valid oxide]

cool cool

[tame python]

,,\lim_{n\to\infty}\qty(1+\frac{-3}{n+5})^{n+5}=e^{-3}

[warm shale]

mtt

[tame python]

that +5 there is all thats left

we'll need a +5 somehow

[tame python]

this is where this image comes in

look at it closely and youll see its another change to the (1 + -3/(n+5))^n limit

[tame python]

can you see that

[valid oxide]

is it when you multiply them, you'd get the plus 5?

[tame python]

thats not what I specifically asked but thats the intention this has

Ill ask again,

how do you get from here

to here

[valid oxide]

you just multiply the top and bottom by (1+ -3/n+5)^5

which is the same as saying 1

[tame python]

yea thats all

but you can see there its a very specific choice to multiply it by that

you see for why, right

[valid oxide]

yes

[tame python]

thats good

this is a common trick, youll need to keep it in mind

[valid oxide]

will do

[tame python]

after this, you have $\lim_{n\to\infty}\frac{\qty(1+\frac{-3}{n+5})^{n+5}}{\qty(1+\frac{-3}{n+5})^5}$

[warm shale]

mtt

[valid oxide]

yep

[tame python]

do you think you can simplify the limit from here

it takes a bit

[valid oxide]

I don't think so, i'll be honest

i'm learning this from you rn, so this is all new

[tame python]

alr heres a hint

you then have $\frac{\lim_{n\to\infty}\qty(1+\frac{-3}{n+5})^{n+5}}{\lim_{n\to\infty}\qty(1+\frac{-3}{n+5})^5}$

[warm shale]

mtt

[tame python]

thats the first step

you can take the limit of the numerator and of the denominator separately

do you know the numerator's limit?

[valid oxide]

it should be infinity?

[tame python]

we've seen that exact limit earlier

you can scroll up and refresh your memory

[valid oxide]

oh

[tame python]

what did we end up figuring out that numerator limit to be?

[valid oxide]

e^-3

[tame python]

thats it

you then have $\frac{e^{-3}}{\lim_{n\to\infty}\qty(1+\frac{-3}{n+5})^5}$

[warm shale]

mtt

[tame python]

denominator limit, you can try putting in n = infinity first

[valid oxide]

should be 1

[tame python]

correct

[valid oxide]

so then the final answer would be e^-3?

[tame python]

yep

[valid oxide]

oh

just wondering

whoops

wrong one

couldn't have we just plugged in -3 and gotten the answer

using the hint?

[tame python]

you needed to prove that the bottom being n+5 didnt change the result

[valid oxide]

ah

so this would've been a lucky thing if I had just done that

[tame python]

yep

sometimes youll notice that a minor change doesnt affect anything

this is one of them

[valid oxide]

mhm

thank you very much

[tame python]

np

[valid oxide]

I appreciate your patience and teaching me

have a wonderful night!

[tame python]

you too

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[woeful fulcrum]

why is this wrong???

[woeful fulcrum]

heres my work

[slim cove]

Your general process looks right to me, so it must be some silly calculation mistake somewhere. Could you send another picture with just the calculations of Mx and My? Sorry, the picture you sent is kinda blurry.

[woeful fulcrum]

this photo is more close up

I checked it several times and a tutor did too and didnt find any error

[slim cove]

I think the error is in the calculation of Mx somewhere

[woeful fulcrum]

hmmmmmmm

[slim cove]

I think you forgot the factor of 4/3 in this step

when you were simplifying 4/3 (3-x)^3

[woeful fulcrum]

ohhh

okay let me fix that!

[woeful fulcrum]

i fixed it and got it right thank u so much!

[slim cove]

you're welcome!

glad it worked out

[woeful fulcrum]

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[earnest knot]

I need to find the area of a circle with radius r using this limit

[earnest knot]

I feel like I´ve tried everything.

[brazen viper]

Ok. So

[earnest knot]

Sorry for my english.

[brazen viper]

What did you get for the area of the triangle?

[earnest knot]

I tried using law of cosines

[brazen viper]

Too complicated

[earnest knot]

I know the lenght of 2 sides is R.

[brazen viper]

Yup

Drop an altitude from the center perpendicular down to one of the sides of the 2^n-gon

[earnest knot]

Ujum

[brazen viper]

Now you have two right triangles

[earnest knot]

That would be the apotema?

[brazen viper]

Yes

[earnest knot]

Okay.

[brazen viper]

Apothem in English

[earnest knot]

Sorry.

[brazen viper]

It's ok, it's clear we're talking about the same thing.

Just wanted to be helpful

Your English is perfectly fine, for the record.

[earnest knot]

Thanks.

But.

[earnest knot]

I don´t know how to work with that?

[brazen viper]

The apothem bisects the center angle. And now you have a right triangle.

Use sine and cosine to find the height and base using the hypotenuse

[earnest knot]

Okay.

[brazen viper]

It will also be useful to remember that (lim x->0 of sin(x) = x), and (lim x->0 of cos(x) = 1).

[earnest knot]

Okay.

Mmmmm.

I don´t think I got it.

Hahahha.

[brazen viper]

!show

[obtuse pebble]

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

[earnest knot]

base: h*sqrt(1-cos(pi/2n))

2 to the nth power.

Okay.

[brazen viper]

Where h is hypotenuse?

[earnest knot]

Yeah,

[brazen viper]

Ok, maybe for consistency use r?

[earnest knot]

Ooo..

yeyeyee

[brazen viper]

As h, at least in English typically means "height"

As in area of a triangle = bh/2

😉

So what did you get for height, speaking of?

[earnest knot]

Is it clear?

[brazen viper]

Wait wait sorry

I am not paying enough attention

[brazen viper]

This is not correct, my b

It would be correct if the cos^2

But it's way way easier to use sine

Remember sine = opposite/hypotenuse

Cosine = adjacent/hypotenuse

[earnest knot]

Omg.

I´m dumb.

[brazen viper]

In English we have a mnemonic, SOHCAHTOA

[brazen viper]

Not dumb, just made a mistake.

So b = r sin(pi/2^n), h = r cos(pi/2^n)

So what is the area of the triangle

[earnest knot]

Yes yes.

[brazen viper]

(keep in mind that the base you found is half of the whole base)

[earnest knot]

So.

Are you paying enough attention?

Sorry.

Jajajaja

[brazen viper]

Yes

[earnest knot]

Jk.

Sorry sorry,

[brazen viper]

Np

[earnest knot]

And thats it?

I mean.

[brazen viper]

Well, now you have to multiply by 2^n and take the limit

[earnest knot]

O gosh,

[brazen viper]

Don't forget this

[earnest knot]

I find trig limits quite hard.

Bro.

Thank you so much for your help.

[brazen viper]

Do you see the answer now?

Like x-ray vision just everything cancels.

[earnest knot]

Wait.

[brazen viper]

Yes?

[earnest knot]

Wait wait.

[brazen viper]

What's up?

[earnest knot]

I think.

I can´t do it.

Hahahaha.

[brazen viper]

So you have

lim n->inf 2^n r^2 sin(pi/2^n) cos(pi/2^n)

[earnest knot]

I tried using a variable.

[earnest knot]

Yep.

[brazen viper]

Now as n gets large, what can we say about pi/2^n?

[earnest knot]

0

[brazen viper]

Ok, and what did I say about sin and cos as lim->0

[earnest knot]

Yeah.

But.

I need to do a "change of variable".

Like.

To do it right, don´t i?

[brazen viper]

Sure.

[earnest knot]

Ooooooo.

Teah.

[brazen viper]

Use the fact that lim x->x0 a(x) b(x) = (lim x->x0 a(x)) (lim x->x0 b(x))

Then perform the change of variables in only the limits where it makes sense

So in this case you can say u = 1/2^n and you have a limit as u->0

If you want to be careful.

[earnest knot]

Jummmm.

Well.

I can´t.

But.

I will try again later.

Thank you so so much bro.

[brazen viper]

No problem

[earnest knot]

Why are you that good?

Like.

you got the idea how to solve it in 1 second.

[brazen viper]

I've done this problem before

A handful of times

I didn't have to figure it out, so much as remember it

[earnest knot]

Okay okay.

And.

What did you study in college?

[brazen viper]

Nuclear engineering

[earnest knot]

One more time.

Thank you so much.

Have a great life bor.

Bro.

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[distant ridge]

im not really sure where to start

[distant ridge]

.close

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[ripe wadi]

I have 2 more questions
what is the predicted current investment for an individual who had a college debt of 5000
and what is the proportion of the variation in the current investment is explained by college debt
for the first one I assume I put it into the regression formula in the place of X (I just need confirmation that this is the right course of action and I don't know how to do the proportion part)

[obtuse pebble]

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[ripe wadi]

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[limpid oak]

why on f'(x) there's (3) on behind?

[civic socket]

in 3rd line?

[limpid oak]

yes

[civic socket]

derivative of (6+3x)

becomes 3

[limpid oak]

how does the algebra look like?

[civic socket]

wdym algebra

do you know how the power rule works

[limpid oak]

I mean like (a+b*c)^-1/2

[civic socket]

what about it

first thing they did rewrite the function to make it easier for derivative

but the transcript looks weird

should be -½-1

it becomes -3/2

then they simplified the function

[obtuse pebble]

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[zinc heron]

Is this method correct?

[obtuse pebble]

@zinc heron Has your question been resolved?

[gray kiln]

looks fine

nothing wrong

[zinc heron]

thankssss!!!

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[modern lagoon]

How do i derive the heisenberg's uncertainty principle?

[modern lagoon]

(please ping)

[obtuse pebble]

@modern lagoon Has your question been resolved?

[modern lagoon]

<@&286206848099549185>

[void holly]

It's complicated

They will teach in uni

[modern lagoon]

but how

[slim cove]

Well, it's questionable how a physical principle can be said to derived. Ultimately, it comes from experimental observations, not mathematical proof. To get more intuition behind what uncertainty principles are though, I recommend this video, which approaches things from a mathematical perspective. https://www.youtube.com/watch?v=MBnnXbOM5S4

[modern lagoon]

I understand the uncertainty principle thoroughly

[slim cove]

Okay

Do you have a specific confusion about where the Heisenberg uncertainty principle comes from then

[modern lagoon]

But its like how e=mc^2 is derived

As of my information, it is derived via another equation

[dark stirrup]

(This implies you are maximally uncertain about something else)

[modern lagoon]

e=(lorentz factor)mc^2

[modern lagoon]

genius moment

[slim cove]

Anyways the simplest explanation is probably that there's two key steps here

1. Position and momentum are a Fourier-transform pair
2. Fourier-transform pairs obey an uncertainty principle

Which part are you confused about?

[modern lagoon]

...how to derive it?

[slim cove]

How to derive which part of this

I just gave you two statements, do you understand the words in all of them?

Particularly, what a Fourier-transform pair is?

[modern lagoon]

not really

i will research that topic

thank u

[slim cove]

Check out the video I just sent then

It explains it

[obtuse pebble]

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[modern lagoon]

.close

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[terse sparrow]

Hello, how can I change the case expression to not be indented to the center of the page?

[wicked crown]

it probably comes from the \[\] environment. Inline math should solve it though it might be harder to write this whole thing down that way. You should be better off asking in #latex-help

[warm shale]

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

[obtuse pebble]

@terse sparrow Has your question been resolved?

[terse sparrow]

What do you mean with "inline math" should solve it?
What is the "\[\]" environment? I just found the code on codestack where someone asked how to make cases. I mean it does more or less do what I want, but I'd like to be able to make documents and latex code that displays like I intend it to.

No one answers in latex help im afraid 😦

[wicked crown]

well maybe you need to wait a bit for people to answer there

but short answer is dollar signs are inline math

the brackets make up a math environment that makes such a centered math block

try chatgpt it works very well in explaining tex

[eager niche]

Single dollar signs are inline double dollar signs are separated

Oh that’s just with the latex bot I guess

[eager niche]

Does this solve it?
https://tex.stackexchange.com/questions/145657/align-equation-left

[obtuse pebble]

@terse sparrow Has your question been resolved?

[terse sparrow]

yes it does, thanks a lot!

[terse sparrow]

Cool, thanks for the suggestions!

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[slate crater]

we are draggining this homogeneus cylinder of mass M which is rotating without sliding. What is the acceleration of the object? Draw all forces acting on the cylinder

[slate crater]

idk how to do this

[drifting wraith]

same

wait is that your drawing?

[slate crater]

yes

[drifting wraith]

maybe F is somewhere else

[slate crater]

no wait

that is the drawing that is given

[drifting wraith]

ok

[slate crater]

F is there

i just know

v = w*r
and
a = alpha * r

[drifting wraith]

i'm pretty sure those are the other forces

but it doesn't help much

[slate crater]

yeah and also gravitational right?

[drifting wraith]

oh and gravity also

[weary reef]

I think I understand

Do you agree that the resultant force on the cylinder only due to force F?

[slate crater]

I don'T understand

[weary reef]

How do you calculate torque in general (given a force)

[slate crater]

r * F

[weary reef]

Right. Now, consider the torque due to the weight and the reaction force

[slate crater]

reaction force?

Ok so gravitation is 0 because the r = 0, and if you mean the force of the surface is also 0 because its perpendicular to its r

[weary reef]

yes that's what I meant

[slate crater]

what about this force?

[weary reef]

ahhh I forgot about that tbh

[slate crater]

but idk i just drew this maybe it doesnt exist

[weary reef]

It should because it's rolling without slipping

there should be a frictional force

were you given any other information?

[drowsy cargo]

what linear algebra ?

i need help

with that

[slate crater]

No thats it. I mean im not supposed to solve anything just express the acceleration of the cylinder

[weary reef]

there were free help channels you can claim

[drowsy cargo]

oh

[drifting wraith]

it's gotta be about moment of inertia and stuff

i don't know anything about it though, can't solve it

[weary reef]

Oh alright then I'm at a loss in that case. I believe that force does exist and to find it, you would need the coefficient of friction with the ground.

That said

If we assume it's not acting on the cylinder

Recall the formula T = I a

Moment of inertia x Angular acceleration

You can equate F r = I a to find the (resultant) angular acceleration. The resultant linear acceleration is just a x r

Actually, do you know what the correct answer is?

[slate crater]

what?

[weary reef]

Like if you had a solution in the textbook idk

[slate crater]

No

Like i have video tutorial fro problems like this

so i know how to solve this

But in that case there is the force acting at the top

so im lost

[slate crater]

ok yeah this makes sense

[weary reef]

That might be correct. I just realized I don't understand rolling without slipping

[slate crater]

ill ask a friend if he knows the answer but for now this is good enough ty

[weary reef]

you're welcome. Sorry I couldn't be more help

[slate crater]

wait

if I do this i get
$$a(mr-\frac{1}{2}mr)=0$$

[warm shale]

OHHELLNAH

[slate crater]

so ig im cooked

@weary reef

[obtuse pebble]

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[elfin sleet]

Can someone help

[elfin sleet]

<@&286206848099549185>

help

[timid silo]

b c d are positive constants or negative?

[elfin sleet]

so is it A

[timid silo]

no

[elfin sleet]

esiy

wait

sorry

Is it D

[timid silo]

ig its a or d

[elfin sleet]

whicih one

[timid silo]

idk ngl lol sure it has sm tricky methods

[elfin sleet]

Even I am not sure

[timid silo]

is therea ques before it ?

[half silo]

x-> infinity it should tend to +2

x-> - infinity it should tend to +2

[elfin sleet]

Is it A then

[timid silo]

what

[half silo]

?

[timid silo]

how it tends to 2?

no

[half silo]

divide numerator and denominator by x^5

[timid silo]

yhyhyhyh right

but what if the constants are infinity?

[elfin sleet]

so D

[timid silo]

nah b

[timid silo]

@half silo

[half silo]

Huh?

[timid silo]

wdym huh what if b c d = infinity

[half silo]

they are constants

[timid silo]

nvm

[half silo]

You can have x=b*2000000000000000992829291819191018191^c^d

if go far enough

[elfin sleet]

I am confused

what is the answer at this point

[half silo]

b

do you know limits?

[timid silo]

yh

its b tho

[half silo]

yeh

[elfin sleet]

ok

thank you

[timid silo]

tnx man n srr cuz of that dump ques

[obtuse pebble]

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[west geyser]

In this step they simplify down to |x^2|, it is to my understanding that infinity divided by infinity is undefined and therefore indeterminate? If so how does this simplify into |x^2|

[unique kettle]

derive top and derive the bottom in respect to n

[west geyser]

won't that make x^2 go away?

I'm not familiar with taking derivaties when there is more than one variable present

[unreal musk]

(The x^2 is constant with respect to n, so really, you want to work out the limit of (2n + 1)/(2n + 3) as n goes to infinity, which you could do by lopitals (suggested here), or alternatively, divide top and bottom by n)

[west geyser]

wait wait

are we splitting it into 2 limits?

[unreal musk]

More pulling a constant out of the limit, I would say

[west geyser]

so this wouldn't work?

I understand what you are saying though you are doing:

[high lily]

it would

[west geyser]

yes of course, all makes sense

thank you

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[shell skiff]

can someone help with this problem?

[obtuse pebble]

Please don't occupy multiple help channels.

[shell skiff]

i thought P(none get car) = P(Jeff doesn't get car) • P(Brita doesn't get car)

= 0.61 • 0.27

[gloomy vector]

but isnt there abit of an overlap

lemme draw it

i think this is correct?

basically theres an overlap in there

[timid silo]

yup

[gloomy vector]

im pretty sure its asking for $(J \cup B )'$

[warm shale]

Skill_Issue

[shell skiff]

what does that mean

[timid silo]

https://tenor.com/view/awkward-introuble-ohno-shocked-whoops-gif-9056781886829673082

[gloomy vector]

it basically wants you to find the thing outside of the circle

the outside of ths circle would basically be 1-the combined thing of the circle

[foggy quail]

This would be correct if the events were independent

[shell skiff]

i see

[timid silo]

(J U B)' = U - (J U B)

[shell skiff]

why aren't they independent?

[foggy quail]

because the probability that they both get cars is not equal to the probability that one gets a car times the other getting a car

so if J and B are events that they get cars, P(J)P(B) != P(J and B)

what you tried doing is $(1-P(J))(1-P(B))=1-P(J)-P(B)+P(B)P(J)$ but what you want is $1-P(J)-P(B)+P(B \text{ and } J)$

[warm shale]

Bair

[foggy quail]

as krill issue said, try to find the area of the union of J and B as in the picture

[shell skiff]

i see

[shell skiff]

so if P(J and B) = 0.2847 it would indicate that J and B are independent and my solution would have worked, right?

[foggy quail]

right

[shell skiff]

i think i understand now

P(J or B) is the complement of P(neither J nor B) so that's why you're doing 1 - P(J or B) = 1 - (P(J) + P(B) - P(J and B)) = 1 - P(J) - P(B) + P(J and B)

thanks

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[timid silo]

deep mathsmatical quotes

[obtuse pebble]

@timid silo Has your question been resolved?

[gilded needle]

do you have a question?

[timid silo]

nah i just need mathsmatical quotes

[gilded needle]

"No-one shall expel us from the paradise Cantor has created."

• David Hilbert

[obtuse pebble]

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[thin pawn]

I‘m probably missing some simple thing, but I cannot for the life of me figure out how to do $\int_{0}^{\frac\pi 2}u\cot(u) \dd u$. It looks so innocent tho 😭

[warm shale]

𝓲𝓶𝓣𝔂𝓹𝓞

[thin pawn]

Oh ffs

Rubber duck programming is real

And applies to other things

.solved

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[inland portal]

ok so like how would I divide something from one side, and do the same to the other. I cant remember if I divide both sides entirely or just the one thing youre doing on one side and then the entire opposite side

[bitter terrace]

if you have 3 + x = 4 + y and want to divide both sides by 2 you would do 3/2 + x/2 = 4/2 + 7/2, which is the same as multiplying by the reciprocal (1/2)(3+X) = (1/2)(4+y) if that was what you were asking

your dividing each term in the equation by 2

[inland portal]

like 2x+4=4x+2 if I divided x what do I do

like systems of equations for example when equalizing both sides

[bitter terrace]

you would get (2x/x) + 4/x = 4x/x + 2/x

[inland portal]

alr ty

[bitter terrace]

if you wanted to solve that equation

[inland portal]

so like

this ffor example

not that equation

[bitter terrace]

you would subtract both sides by 2x

[inland portal]

my equation that im trying to solve is for quadratic functions

so

f(x)=2x^2+4x+1

find the vertex

the intercept and coordinates

and coordinates of the two x intercepts

just like

I subtract 1 right

and then when dividing

the 2 from x

I divide it from 4

so
y-1=2x^2+4x
y/2-.5=x^2+2x

[bitter terrace]

to find the intercepts you set y equal to 0 and the vertex is x = -b/2a and y = y(-b/2a) I believe

not sure if im understanding your question, why did you subtract 1

[inland portal]

its how my professor is wanting us to solve for the vortex in equations not in vortex form

ok

just help me solve this

im so confused

.close

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[lucid snow]

I am flipping a coin, for that flip i'm allowed to flip two coins instead and use one of those results as that coin flip. (1 - (1/(2^2^1))) = 75% of calling the flip right.

I can continue flipping a coin using this rule until I fail, so my odds of getting three in a row are 0.75 * 0.75 * 0.75 = 42.18%

If I'm allowed to have two instances of flipping this way, what are my odds of getting at least three flips in a row with both instance's probability combined (42.18%).

How is that expressed?

[obtuse pebble]

@lucid snow Has your question been resolved?

[lucid snow]

<@&286206848099549185>

Let me rephrase it:
I am flipping a coin, for that flip i'm allowed to flip two coins instead and use one of those results as that coin flip. (1 - (1/(2^2^1))) = 75% of calling the flip right.

I can continue flipping a coin using this rule until I fail, so my odds of getting three in a row are 0.75 * 0.75 * 0.75 = 42.18%

I'm allowed to keep flipping until I fail the call, then when I fail I can start flipping again until I fail the call again. What are my odds of getting at least three flips in a row with both instances of flipping.

How is that expressed?

[obtuse pebble]

@lucid snow Has your question been resolved?

[drifting wraith]

it's the same thing

0.4218 × 0.4218

[obtuse pebble]

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[lucid snow]

.reopen

[obtuse pebble]

✅

[lucid snow]

@drifting wraith Yeah your right, thats for three in a row. My bad I phrased that wrong. What's the odds of getting at least a combined total of three successful calls over both instances.

[drifting wraith]

i think it's like making 4 calls and having 3+ successes

,calc 0.4219^4 + 4(0.4219^3)(0.5781)

[warm shale]

Result:

0.20534055835554

[drifting wraith]

no wait 0.75

,calc 0.75^4 + 4(0.75^3)(0.25)

[warm shale]

Result:

0.73828125

[obtuse pebble]

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[timid silo]

Yes

There is no c because after differantiate the function, constant disappears

Its equals to samething with integral

Then put f(x)+c/dy

Is that means wolfram alpha confirms its integral?

[tame python]

are you assuming y = f(x)

@timid silo you here?

[obtuse pebble]

@timid silo Has your question been resolved?

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[wet sphinx]

I am stuck

[wet sphinx]

Since both the triangles are congruent

And we know the combined perimeter

Why cant I just divide it by 2 to get the individual triangle perimeter and hence find the Unkown value

[past sand]

The perimeter of DEFHG does not include FG

[wet sphinx]

Huh

I'm confused

[past sand]

?

DEFHG is a (non regular) pentagon

[wet sphinx]

Wdym does not include FG

[past sand]

FG is not a side of that pentagon

[wet sphinx]

Oh

But it's inside a Pentagon

Ok but how do I solve it then

Cuz I can't figure out how to

[past sand]

That's the whole point of the exercise

[wet sphinx]

I understand now

But I still don't know how to solve it

FG = EF WE DONT KNOW EITHER

<@&286206848099549185>

[timid silo]

Yes

[wet sphinx]

Sry but that question closed

[tame python]

you can go ask the question again in another channel

[crude solar]

Consider EF=GF= x, what would you get for GD then

[wet sphinx]

Hmm

[wet sphinx]

Idk

Like are you talking about the perimeter

Or what

[crude solar]

DF is 15, and GF is x, so DG?

Length

[wet sphinx]

15-x

[crude solar]

Yeah

Perimeter is given, add all sides of the shape

[wet sphinx]

Wait

The perimeter is not including the line FG

[crude solar]

No, it won’t

[strong sentinel]

EF=GF

[wet sphinx]

Then how can I add it

[crude solar]

Only GD

[wet sphinx]

Kk

[crude solar]

We found found GD = 15-x, use that

[wet sphinx]

Ok

46=50

[strong sentinel]

It doesn't make sense tho coz then you get DG+EF=19 but if EF=GF, it should be 15

[wet sphinx]

X+8+15-x+8+15=50

[strong sentinel]

We're missing a 4 for some reason

[wet sphinx]

Yep

[crude solar]

Yeah I just realised it, but the two triangles are congruent

[strong sentinel]

Yes that's why it doesn't make sense

Maybe wrong question?

[crude solar]

Feels like it

[wet sphinx]

The answers are

[strong sentinel]

Coz if they are congruent the sides as the mentioned in the question needs to be equal

[wet sphinx]

DG=7

EF=10

[crude solar]

But the perimeter won’t be same then, and the triangles wouldn’t be congruent

[strong sentinel]

Yes only48

[wet sphinx]

ok ima ask my teacher tmr

thanks anyways guys

[strong sentinel]

Alright

[wet sphinx]

.close

[strong sentinel]

Goodluck

[obtuse pebble]

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[dusky arrow]

how

[dusky arrow]

shouldnt it be delta a/delta l = 2l + delta l squared?

[turbid comet]

it's just simple algebraic manipulation

[dusky arrow]

whats being done

[turbid comet]

factor ΔL out

[dusky arrow]

?

[turbid comet]

2LΔL + ΔL^2 = ΔL(2L + ΔL)

[dusky arrow]

ohhh\

omg im so dumb

.close

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[turbid comet]

it is just basic algebra

[obtuse pebble]

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[dusky arrow]

i got the answer as a

can anyone verify?

[obtuse pebble]

@dusky arrow Has your question been resolved?

[brazen viper]

Let r be the number of red balls initially in the bin. Notice in both cases, we are removing a ball from the bin, and in the first case that is a red ball. So in both cases we have (n-1) balls remaining in the bin. And in the first case we have (r-1) red balls remaining in the bin, in the second case we still have r balls.

So we can set up a system of equations, fill in the missing parts.

(r-1)/???? = 1/7
????/???? = 1/6

Then we solve the system of equations for r and n

[obtuse pebble]

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[wary bobcat]

is anyone able to spot my mistake?

so a topological vector space (tvs) is a vector space (say over R or C) with a topology such that every singleton is closed, and addition and scalar multiplication are continuous

if V is a tvs then X subset V is bounded if, for every open neighbourhood N of the origin, there is an s in R s.t. t>s implies tN supset X

now consider eg V=R² with the discrete topology (so every set is open). note that every function on V has to be continuous and every singleton must be closed, so this is a tvs. then N = {(0,0),(0,1)} is an open neighbourhood of 0, but there is no real s such that sN contains {(1,0)}, so in particular the latter set is unbounded, which seems very absurd (i remember reading compact implies bounded somewhere)

[safe meteor]

there is no hack for (2^2+2^3+2^4+2^5) ?
or im forced to calculate each of them and then add them up

[wary bobcat]

thats just a geometric sum, have you encountered the formula for that?

[safe meteor]

no

[obtuse pebble]

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[dim galleon]

hi how do u answer this

i have never encountered smt like this bc this has never been discussed 😭 im confused as to how u find the probability for 2 dice .. at the same time

[timid silo]

i tried using kirchoff's voltage law, but im getting negative current, is my equation incorrect or does that mean somthing directionally?

[timid silo]

you should get another help channel

[dim galleon]

oh okay thank u 🥲

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[fossil crag]

negative current only means the current is flowing the opposite way from the reference you chose

so you can use voltage law, but current law as well, so that if you name i the current that flows from E1 to A, i1 the one that flows from a to E3 and i2 the one that flows from a to b

you get i = i1+i2

etc

in any case don't be scared because of negative current, it just means it's flowing the opposite way

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[stoic yacht]

looks right to me

,w integral from 0 to 1 of (x^2 - 2) / sqrt(x^3 - 6x + 7) dx

[warm shale]

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Upgrade to WolframAlpha Pro!

Requested by nightshade5107

[stoic yacht]

,w 2/3 (sqrt(2) - sqrt(7))

[warm shale]

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[obtuse pebble]

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[near kraken]

can anyone explain this algebraically?

[plucky flax]

(4-x) = (-1)(x-4)

[obtuse pebble]

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[near kraken]

oh

[obtuse pebble]

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[trim dove]

i have a headache

So, my thought process was that I'd need to get the derivative of this before trying to get the limit

so I got ((cosx-cospi)(x-pi)-(1)(sinx-sinpi))/(x-pi)^2

which can be factored and simplified into ((cosx-cospi)(x-pi)-(sinx-sinpi))/(x-pi)(x+pi)

should i try to separate it into (cosx-cospi)(x-pi)/(x-pi)(x+pi) - (sinx-sinpi)/(x-pi)(x+pi)

and simplify into (cosx-cospi)/x+pi - (sinx-sinpi)/(x-pi)(x+pi)?

i don't know where to go from here, honestly

it seems so simple, but the teacher's explanation for an example like it was cut short because she forgot to continue it after getting f'(c)

she left that as the answer and moved on to the next question

[ivory spoke]

sin pi = 0

sin x - sin pi = sinx - 0 = sin x

then l'hopital, no?

(sin x)'/(x - pi)'

cos x/ 1

evaluate

= cos pi

thus, second option?

$$\frac{\sin \left(x\right):-:\sin \left(\pi \right)}{x-\pi }=\frac{\sin \left(x\right)-0}{x-\pi }=\frac{\sin \left(x\right)'}{\left(x-\pi :\right)'}=\frac{\cos \left(x\right)}{1}\rightarrow \cos \left(\pi \right)$$

[warm shale]

Shinutsi

[trim dove]

oh

l'hopital, fair enough

wait, so in what contexts would we need to use the quotient rule instead of l'hopital's?

[ivory spoke]

if we get 0 as the denominator and numerator

x - pi

= pi - pi = 0

but idk, are you guys allowed to use that

[trim dove]

we would get 0 as the numerator AND demoninator if we didn't immediately simplify sin(pi), so could you show me what i did wrong when trying to apply the quotient rule?

[trim dove]

we are, i guess i just blanked and didn't see it

thank you.

.close

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[sacred elk]

How would i do #21

[keen quarry]

you can factor by grouping

i.e: 3x^3 + 9x^2 - 15x + 25 -> 3x^2(x+3) - 5(3x+5)

[sacred elk]

But when i try to group i get 3x(x+16)-5(x^2+16)

[keen quarry]

you should almost always group the x^3 and x^2 terms together

as that will always result in a linear factor

[sacred elk]

ahhh

Lemme try to do dat

So i shiuld get

+-4i

And 5/3?

Well if i write it correctly into what its asking it should be f(x) = (x-4i)(x+4i)(3x-5)?

Or do i keep the answer (x^2 + 16)(3x-5)

Sorry, @keen quarry

[keen quarry]

yea that’s correct

[sacred elk]

thanks

.close

[obtuse pebble]

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[sleek mural]

Is this a sine graph?

[sleek mural]

Shouldn’t it originate at (0,0)?

[obtuse pebble]

@sleek mural Has your question been resolved?

[obtuse pebble]

@sleek mural Has your question been resolved?

[weary cypress]

yes, unless there is a phase shift

[sleek mural]

Oh, I see

So would this still be considered a sine graph?

[weary cypress]

yeah theoretically

[sleek mural]

Got it

Thanks

[weary cypress]

although it may be unlikely to have a phase shift if a homework asks to determine if it is or isnt

i actually think yours is a cosine graph @sleek mural

[sleek mural]

Oh, I see

I get what you mean

Thanks

[weary cypress]

notice how it "originates" from the "bottom" of the curve rather than halfway

so its a cosine

[sleek mural]

Got it

[weary cypress]

Np

[sleek mural]

Appreciate it

[obtuse pebble]

@sleek mural Has your question been resolved?

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[tribal cloud]

how do I solve the question
Calculate the displacement change (magnitude and direction) of the ball when it is moved from the initial to the final position given as follows.

Initial position: (xi, yi) = (-10.0 m, 10.0 m);
Final position: (xf, yf ) = (15.0 m, -20.0 m).

[gloomy vector]

is magnitude like the regular distance and direction is the angle?

[tribal cloud]

magnititude is just the distance

[tepid sapphire]

I would use vectors to do this. Starting by identifying my vectors and finding the relation between these, wanting to solve for the displacement vector and then its magnitude.

[tribal cloud]

if you dont mind awnsering how would I do that

would I just add the potiitions as vectors

[tepid sapphire]

Not add, it would give the black vector in the image, which is clearly not the same as the red vector, which is the one you want.

[tribal cloud]

alr

[tepid sapphire]

Correction, still wrong if adding.

[tribal cloud]

ok

[obtuse pebble]

@tribal cloud Has your question been resolved?

[obtuse pebble]

@tribal cloud Has your question been resolved?

[obtuse pebble]

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[surreal cosmos]

Can someone tell me the meaning

[gloomy vector]

a cyclic quadilateral is a quadilateral where all the vertex are on the circumfrence of a circle

[obtuse pebble]

@surreal cosmos Has your question been resolved?

[surreal cosmos]

all of them are tho

<@&286206848099549185>

oh

[grave herald]

I ain't helping no one helps me

[surreal cosmos]

Ill help you

What do you need

[pulsar cloud]

b

check the vertices

vertice C of the quadrilateral is not on the circumference of the given circle

theres a difference right

[gloomy vector]

dont tell answers lmao

[obtuse pebble]

@surreal cosmos Has your question been resolved?

[obtuse pebble]

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[gloomy vector]

find $\frac{300^2}{253^2-247^2}$

[warm shale]

Skill_Issue

[gloomy vector]

denominator is (500)(6)

[iron edge]

a^2-b^2 identity

[gloomy vector]

so 30^2/6

wait no

oh is it 30

[high lily]

depends

[iron edge]

ig you had some mistakes on your calculation

That’s why you’re asking

[high lily]

is what 30

[gloomy vector]

i thought it was 900/6 and there wasnt a choice for it

so i was confused

[iron edge]

nah, incorrect calculation

Just double check again, you’re good

[iron edge]

Correct

Didn’t notice it

[gloomy vector]

find a+b if a^2+b^2+3ab=719 and a<=b

a,b are positive integers

(a+b)^2+ab=719

[timid silo]

Everyone I wanna improve my maths level but the matter is that I don't know where to start

[iron edge]

Real question 😆

[iron edge]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[gloomy vector]

what

[obtuse pebble]

@gloomy vector Has your question been resolved?

[gloomy vector]

im gonna make a new channel cause i already asked here

.close

[obtuse pebble]

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[slim mortar]

Left: Question
Right: Answer

In this exercise I don’t understand why the value of the expression increases as it moves away from (2,1). How you reach that conclusion?

The rest I understand it

Btw, x,y are POSITIVE

[short spire]

The expression is of a quadratic form, so see if you can complete the square to rewrite it in a more familiar fashion

This is probably how they drew that conclusion so directly

[slim mortar]

Maybe a circle?

Ahh okok, (x-2)^2 +(y-1)^2 -5

[short spire]

Yeah how you’d usually rewrite the equation of a circle yes

[slim mortar]

Oh cool so the minimum is when (2,1)

[short spire]

Yup! Try to see why

[slim mortar]

Oh right thank you!!

[slim mortar]

It is squared, so minimum is 0

Oh wait, so why it needs to be proportional?

[short spire]

I think you have the right idea but I’m not parsing your message, so for example x^2 is the smallest at x=0 since x^2 >= 0 for all x

[short spire]

What has to be?

[slim mortar]

Yeah I catch it

[short spire]

Oh I see, so the original question has a domain which is restricted, your point lies outside

So what is the closest point to (2,1)?

[slim mortar]

I mean it says that in order to find the minimum value with x,y from Q we have to find the closest point to (2,1).

[slim mortar]

How do we get that conclusion?

[short spire]

So first off, we know (2,1) is the minimum if the domain was not restricted

[slim mortar]

Yes

[short spire]

So since we’re working with a quadratic form

When we move away from this point we will naturally only strictly increase, do you see why?

[slim mortar]

Ohh I see

[short spire]

And to really nail down this point

[short spire]

If we look at this as a circle for a moment, you’ll see that it’s offset from the origin

So let’s forget all specific numbers and shift it back to something of the form x^2 + y^2

Then we notice that there’s a rotational symmetry

[slim mortar]

Ah yes, I was confused by so many variables

[short spire]

So if we graph this as z = x^2 + y^2

[slim mortar]

The farther x,y from 0 the larger the value, so we want the closest

[short spire]

Yup!

[slim mortar]

Ok!!! I completely got it

Thank youuu

[short spire]

You’re welcome!

I guess finding the closest point is not the issue here then?

[slim mortar]

Yes, no issue

[short spire]

Alright awesome!

[slim mortar]

We get the tangent and the it’s perpendicular

And substitute it to (2,1)

I will close it Thanks again!!

.close

[obtuse pebble]

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[obtuse pebble]

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[weary garden]

howdy hey

[weary garden]

<@&286206848099549185>

[timber hull]

do you need help

[misty hearth]

yo

[weary garden]

yep

[misty hearth]

i can help

[weary garden]

alright

thanks mate

one sec

[misty hearth]

k

[timber hull]

i think the folder got you

[weary garden]

aigh thanks creator

[misty hearth]

yeah u can go

[timber hull]

alr

[weary garden]

for 4 a

i got 2(x+1)^2-8

can you check that for me

[misty hearth]

yeah this is right

[weary garden]

alright

and can you also

go through 4b

[misty hearth]

yup

[weary garden]

thanks

go ahead

[violet vine]

have you learnt derivatives

[weary garden]

(i made these sausage bites in the airfryer and i blew them up 😭

)

[hushed garnet]

yeah just use the formula for minima

[violet vine]

^or this

[weary garden]

ill say no for the purpose of this

[hushed garnet]

or you could use the graphical method and figure out the vertex

but that's a longer method

[misty hearth]

can i be completely honest with you?

[weary garden]

but kinda easier

ngl

[weary garden]

go ahead

[misty hearth]

i dont actually know how to do math

sorry

[hushed garnet]

🔥

[weary garden]

bro what 😭

[misty hearth]

https://tenor.com/view/saitama-one-punch-man-gif-22096485

[weary garden]

wait can someone check my answer for 4a

then

[misty hearth]

nope

[weary garden]

someone who knows math 😭

[hushed garnet]

wait i'll solve it using minima and i'll share the answer here

[misty hearth]

i dont even know what x means

or b

[hushed garnet]

4b i mean

[misty hearth]

or what ever the fuck any of it even means