help-4

[proper pike]

You have your slope

Plug in the coordinates you know and isolate the y variable

what did ya get

[toxic narwhal]

im lost

linear line with the slope of -4 that has points (k,13) (k+7,-15) and y intercept of (k-5,b)

so doesnt k = 5 since the y intercept has to be 0,b

[proper pike]

K doesn’t necessarily equal 5

[toxic narwhal]

but doesnt it have to since the y intercept has to be (0,b)

[proper pike]

You can’t grasp that from the information shown at least

[proper pike]

Not really

Why does X have to be 0?

Oh right I’m confusing myself

[toxic narwhal]

because thats where the line would intercept the y axis

[proper pike]

But you don’t need to know what K is necessarily to solve this is my point

You can plug in (K,13) to point slope

[toxic narwhal]

yeah and the k cancelled out

[proper pike]

Right

I actually did not catch that icl

That also might be a way of solving

But usually for problems like this you have to do it through solving the line equation, that is a good catch though lmao

[proper pike]

I think you have it more figured out than I do at this point

You should close if no further questions

[toxic narwhal]

ok thanks for your help!

.close

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[gusty wind]

I need help with this numbers, I'm still kinda confused as to how and why

[fossil mulch]

11 or 12

[gusty wind]

Both

[fossil mulch]

Welp ion man

Good luck

From Google you just sum mr^2 for both disks

For 11a.)

[chrome beacon]

ok first thing is to make a sketch

[gusty wind]

Does it look like the wheel from under a car?

[chrome beacon]

kinda yeah

[gusty wind]

Like that yeh?

[chrome beacon]

they're concentric

[gusty wind]

What does that mean? sorry

[chrome beacon]

they have the same centre

[gusty wind]

Ohh

[gusty wind]

Okay so do we just do mr squared the two just like what dezion said?

[chrome beacon]

i think so for moment of inertia

[gusty wind]

Oh wait, is it not 1/2? because of this thingy

[chrome beacon]

oh right yeah

but yes

[gusty wind]

Okay

so we'll use 1/2?

so it'll be like

[gusty wind]

1/2(.90)(.028^2)+1/2(1.80)(.056^2) = Moment of inertia

[chrome beacon]

yea

keep units around

[gusty wind]

what does that mean?

[chrome beacon]

that means .90 kg not just .90

[gusty wind]

Ohh right right mb

1/2(.90kg)(.028^2m)+1/2(1.80kg)(.056^2m) = 0.0031374kgm^2

[gusty wind]

Is this right?

[chrome beacon]

sure seems good

mainly the units help you make sure your numbers make sense on the next one

this is one of those problems where you can kinda multiply everything together

[gusty wind]

Ohh

So is A good now?

[chrome beacon]

yea

[gusty wind]

Okay, how do we do b?

[chrome beacon]

thats the 'just multiply it all together' thing hehe

hmm

actually no thats too simple

you'll probably need to use actual torque on the wheel assemblage

[gusty wind]

How do we do that

Im too slow to comprehend such

[chrome beacon]

wjat torque is the block applying? force * distance from origin

[gusty wind]

None?

[gusty wind]

what I dont understand

sorry

[chrome beacon]

see how the rope is attached to the wheels in the diagram

[gusty wind]

Yes

[chrome beacon]

do you see how that would pull down on the wheel

and start it spinning

[gusty wind]

Okay, I see how that would happen

[chrome beacon]

okay. how much force is the block applying?

(the rope)

[gusty wind]

9.81 times 1?

due to gravity?

[chrome beacon]

units

but yes

[gusty wind]

mb

9.81m/s times 1kg

[chrome beacon]

yep so 9.8 N

and how far away from the centre is that force being applied?

[gusty wind]

2m?... idrk

[chrome beacon]

it'll be the radius of the smaller circle

[gusty wind]

ohhh

Ok what's next?

[chrome beacon]

so what's the torque?

[gusty wind]

so it would be force times distance

9.81N times .028m

[gusty wind]

.27468Nm

[chrome beacon]

yes good

ugh this is going to be a bit annoying

The relationship is defined by the rotational equivalent of Newton's second law: Torque (τ) = Moment of Inertia (I) × Angular Acceleration (α) [τ = Iα].

so we can figure out the angular acceleration. go ahead and do that.

[gusty wind]

Okay, let me just write it

[chrome beacon]

here's a walkthrough actually https://www.rose-hulman.edu/Class/phao/PH/JOW/Ph113 Labs/Physics_lab_manual/chapter_3/angular/angular.html

theres a very annoying step where you have to consider the inertia of your wheel and of the brick at the same time

[gusty wind]

87?

but the unit is rad/s^2

and the two is Nm and Kgm^2

[chrome beacon]

yeah thats fine they cancel right

[gusty wind]

yes

they leave /s^2

[chrome beacon]

no it's rad / s^2

yea

so... how many radians does the wheel turn in total? remember the block drops 2m

[gusty wind]

87.55rads/s^2

what do I do with the height?

[gusty wind]

Do I multiply by 2?

[chrome beacon]

it'll involve the circumference

think about the string on a yoyo unwinding

[gusty wind]

.028x2 is the circumference right

[chrome beacon]

gonna guess pi is involved somewhere

[gusty wind]

okay so .56xpi

.1759m

[gusty wind]

What do I do next

[chrome beacon]

ok heres a better idea

find a new moment of inertia including the point mass of 1kg at distance .028m

then use K = ½ I ω² to find the final angular velocity

[gusty wind]

Ill just add its weight?

[chrome beacon]

to what

[gusty wind]

to the weight of the small wheel?

[chrome beacon]

no, because it's not a wheel

it's a point mass

https://en.wikipedia.org/wiki/List_of_moments_of_inertia

[gusty wind]

ML^2?

[chrome beacon]

yes

[gusty wind]

1kg(.056m)^2?

[chrome beacon]

where did 0.56m come from

[gusty wind]

Diameter or do we use radius?

[gusty wind]

Okay so 1kg(.028m)^2 = 0.000784 kgm^2

[chrome beacon]

yes so now whats the total moment of inertia of our system?

[gusty wind]

0.0031374kgm^2 + 0.000784kgm^2 = 0.0039214kgm^2

[chrome beacon]

cool, now how much potential energy is in the system due to the brick?

[gusty wind]

(1kg)9.81m/s^2(2m) = 19.62kgm^2/s^2?

[chrome beacon]

looks good

you could call that 19.6 J if you wanted but eh

ok! so now

[chrome beacon]

[gusty wind]

what is the w

[chrome beacon]

angular velocity

[gusty wind]

Ok so 19.62J = .5(0.0039214kgm^2)ω²

19.62J/0.0019607kgm^2 = ω²

10006.6302851 /s^2

[gusty wind]

100.033145932 rads/s^2

[chrome beacon]

hmm

velocity should be in rad / s

what happened?

[gusty wind]

but joules is kgm^2/s^2 then kgm^2

okay so square root

100.033145932 rads/s so it become this

[gusty wind]

what's next?

[chrome beacon]

ok so now you know how fast the wheel is spinning when the brick hits the ground

now how fast is the brick moving? refer to the diagram

[gusty wind]

9.81m/s times 2m?

[chrome beacon]

no. think about a yoyo

the rope is on the outside of that small wheel

[gusty wind]

It moves directly proportional to how the wheel spins?

[chrome beacon]

yes

it moves at the linear velocity of the edge of that wheel

[gusty wind]

So what do we do next

[chrome beacon]

convert angular velocity to linear velocity

[vale dock]

@gusty wind Has your question been resolved?

[gusty wind]

How

[chrome beacon]

multiply by radius, pretty simple

[gusty wind]

Alright

100.033145932 rads/s x .028m = 2.8009280861m/s

[gusty wind]

is it correct?

[chrome beacon]

does it feel like roughly the right amount? like if you imagine the scenario irl

[gusty wind]

Not really sure I guess so

[chrome beacon]

yeah not off by like thousands of times

ok cool

now do (c)

[gusty wind]

alright ty for your patience

c and b is just the same but the big wheel instead right?

[gusty wind]

okay ty

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[austere grove]

I need help with combinatorics, can't really understand the task

[austere grove]

3 women and 2 men are sitting on a bench together. the women are all sitting next to each other and so are the men. prove that there are 24 ways they can sit together like that

[light marlin]

Ayo finally unmuted

Finally unmuted after 6 months

Who even muted me in the first place 😭💔

[austere grove]

well no it says that there are 24 ways and I have to prove that lol

[errant gyro]

how many ways are there for the women to be arranged amongst themselves?

[austere grove]

3!

[errant gyro]

good

what about the men?

[austere grove]

oh

3!*2! in total then?

[errant gyro]

that gives us only 12

[austere grove]

oh right

[errant gyro]

there's one more thing you need to consider

the women can overall be to either side of the men

[austere grove]

oh okay

so then 2x3!x2!

[errant gyro]

there you go

end of story

[austere grove]

alright thank you so much!

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[shut osprey]

wow very funny

[brittle anvil]

im being genuine

i cant get past this question in the game

i want to play more of it :(

[stark wedge]

<@&268886789983436800> more of this bondi baldi troll shit

[river shale]

You should really stop

[river shale]

Baldi*

:))

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[gloomy harness]

repeat offender? 😭

[chrome beacon]

come back in 2027

[stark wedge]

dunno if repeat, im just seeing more of the same crap from potentially multiple ppl

[gloomy harness]

i see

[shut osprey]

do u just ping the moderators role for reporting trolling and other incidents?

[river shale]

Yes

[shut osprey]

alright thanks

[river shale]

But in some cases, you can negotiate with them.

[stark wedge]

that or modmail

[river shale]

Not this one, cuz it’s a mass spam

[shut osprey]

wait ahhh

ok now ic

[shut osprey]

hmmm

[river shale]

Bruh 😭

[chrome beacon]

(it is a year; the bot auto refreshes it)

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[tame stream]

10 light bulbs are shining in a row. If a lightning strikes, then some (or all or none) of the light bulbs may go out (all possibilities being equally likely). What is the chance that after a lightning at least two consecutive light bulbs are still shining.

[errant gyro]

is the probability for any individual bulb given?

[tame stream]

equally likely

[errant gyro]

oh right

[tame stream]

\Why is N_i = that

(ofc other method suggestions are welcome)

[tidal swift]

idk why, but stars and bars immediately come to mind

[jovial edge]

it is star and bar i think

[tame stream]

how?

[jovial edge]

so 10 bulbs have 11 gaps minus i bulbs so 10-i bulbs has 11-i gaps, put i bulbs into those gaps give us the equation

[tame stream]

Ooo I think I got it

thanks

[jovial edge]

-# how the hell you get helpful role in a week, how many hours/day you spend in this sever

[tame stream]

(They live on the server ig, or maybe a alt)

anw I'mma close

.close

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[glass elm]

idk what borders of integration should be

-pi/2 pi/2 ?

[flint phoenix]

$$4\pi R \int_{-r}^{r}\sqrt{r^2 - y^2}\mathrm dy $$

[rocky lotus]

This is sad 😢

[cinder creek]

Do you have a clue how $$\sqrt{r^2 - y^2}$$ looks like?

[rocky lotus]

Good

[candid bolt]

yeah

[glass elm]

i figured it out

like this

[candid bolt]

yeah

[glass elm]

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[opaque ingot]

I forgot how to do this can someone help me

[flint phoenix]

,rcw

[opaque ingot]

gradient of a is 2 and h intervals is -4

[opaque ingot]

M = 2 y = -4

guys can someone please help me

[fringe lava]

just draw a straight line that has a slope of 2 and y-intercept of 4 and label the axis intercepts

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[coral lagoon]

For which values of $n$ and $k$ is the following statement true?

Given $n$ lines in (Euclidean) 3D space, no $k$ of which are coplanar, the lines must be concurrent.

[rocky lotus]

chloe

[coral lagoon]

This is always true when $k=3$

[rocky lotus]

chloe

[coral lagoon]

I believe the first instance where this fails is $n=6$ and $k=4$: take the edges of a tetrahedron, for instance

[rocky lotus]

chloe

[coral lagoon]

also, is there any information available on sets of n pairwise intersecting lines in 3d space? i cant seem to think of a set of more than 6 pairwise intersecting lines that arent coplanar or concurrent

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[worthy nova]

Hi, can somebody please explain to me, how does this expansion work?

I have a $$\mathbf{r}(t,\tau) = \mathbf{R}(t) + \varepsilon\boldsymbol{\varrho}(t,\tau)$$

Now I want to do Taylor expansion of

$$\mathbf{B}(t,\mathbf{r}) = \mathbf{B}(t, \mathbf{R} + \varepsilon\boldsymbol{\varrho})$$

which is

$$\mathbf{B}(t,\mathbf{r}) = \mathbf{B}(t,\mathbf{R}) + \varepsilon(\boldsymbol{\varrho}\cdot\nabla)\mathbf{B}|_{\mathbf{R}}$$

I think I just dont get, how this work, could somebody please explain it to me? Thanks.

[rocky lotus]

konxmok

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[floral haven]

Can anyone check why got 12b wrong?

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[mortal wagon]

Someone help me im not sure what to do for h( ) and how to solve the inequality

<@&286206848099549185>

[marble furnace]

That's the interval where h(x) is below g(x) on the graph

[marble furnace]

You don't have to find individual zeroes for that

[mortal wagon]

I don’t have to solve it?

But how do I know because it’s not a exact number on the graph

[marble furnace]

You have to solve when h(x) = g(x)

[marble furnace]

Just look at the parts where h(x) is below g(x)

[mortal wagon]

Yes I see the graph

[mortal wagon]

So I put them equal to eachother and find x?

[marble furnace]

That's one way yes you'll get 2 values for that

[mortal wagon]

Can you help me with the steps to do that

What is the fastest way

[marble furnace]

h(x) -g(x) < 0

So that'll be x +2 +(3x/x-2)

Less than or equal zero

Just solve the quadratic then

[mortal wagon]

Oh ok I just solve for x rigjt

[marble furnace]

Yes

[mortal wagon]

I’ll get 2 answers tho?

[marble furnace]

Yes that's the point

You need anything interval where h(x) is less than g(x)

And you can see on the graph

[mortal wagon]

How do I know when to find the zeros or when to do this in a problem

[marble furnace]

Anytime there's an inequality

[mortal wagon]

Inequalities I do this and to write interval notation I find zeros?

[marble furnace]

There's some stuff after you find the zeroes too

Like if the zeroes are a and b then will the interval be (a,b) or R- (a,b)

And will they be included?

R is the set of real numbers

[mortal wagon]

Oh Yes Ik that, Ik how to solve this now, thank you very much

[marble furnace]

Alright

[mortal wagon]

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[midnight pier]

Hello, I managed to do the entire exercise correctly however the last part, z1/z2, i just couldnt figure out. The answer is provided to, i tried to solve this using reciprocal, so * denominator/denominator but then i get a weird fraction which doesnt really help

[midnight pier]

or does someone maybe know how z1/z2 is derived in such a nicem anner that the angles are just subtracted of each other? it doesnt really make sense to me

[quasi valve]

do you know the exponential notation $e^{i \theta} = \cos(\theta) + i\sin(\theta)$?

[rocky lotus]

Bungo

[normal hollow]

By this do u mean that u tried to solve it by multiplying both numerator and denominator with the conjugate of denominator?

[vale dock]

@midnight pier Has your question been resolved?

[midnight pier]

By my own knowledge, yes, but it was not covered in my course

[midnight pier]

Yes sorry thats what i meant

[quasi valve]

ok so prove it in terms of cosines and sines

$$\frac{\cos(\theta) + i \sin(\theta)}{\cos(\phi) + i \sin(\phi)} = (\cos(\theta) + i\sin(\theta))(\cos(\phi) - i\sin(\phi)) = \cdots$$

[rocky lotus]

Bungo

[quasi valve]

multiply out the right hand side and use trig identities on the real and imaginary parts

[midnight pier]

How did u go from left to right

[quasi valve]

multiply num and denom by the conjugate of the denom

[midnight pier]

Ahhh and then youll get that?

[quasi valve]

the denom becomes 1 since cos^2 + sin^2 = 1

[midnight pier]

Thanks ill try it out

[quasi valve]

yw

[midnight pier]

Bruhh of course lol

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[midnight pier]

How abts to start this?

[keen tundra]

calculate the right side

its just a number

[light saddle]

||it's also a double angle formula||

[midnight pier]

without calc tho

[midnight pier]

yea trying to do this

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[alpine bear]

What equation do I find the speed from a graph?

[keen tundra]

slope

assuming the graph is a line

[flint phoenix]

yep, slope

[dire cloud]

of a distance- time graph specifically tho, right?

[alpine bear]

Yeah time 🙂‍↕️

It is a line , thanks for the help

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[cosmic terrace]

How does it go from that to that, help

[cosmic terrace]

I still don't understand knowing this

[keen tundra]

just open (4+sqrt(7))^2 with this formula

[dire cloud]

try to split 23 + 8rt7 into a^2+b^2+2ab

[red tulip]

2ab should be 8 sqrt7

[dire cloud]

yep

and a^2+b^2 should be 23

[cosmic terrace]

Can someone write it down on paper I still don't get it

[keen tundra]

$(4+\sqrt{7})^2 = 4^2 + 2 \cdot 4 \cdot \sqrt{7} + (\sqrt{7})^2$

[rocky lotus]

ExpertSqueeSQUEE

[cosmic terrace]

OHH okay I get it now

[keen tundra]

$=16 + 8\sqrt{7} + 8=23+8\sqrt{7}$

[rocky lotus]

ExpertSqueeSQUEE

[cosmic terrace]

Thanks

.close

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[coral ocean]

I'd like to close the sum : $\sum_{k=0}^{n} \binom{2n}{2k} 2^k$. It's relatively easy to solve $\sum_{k=0}^{n} \binom{n}{2k} 2^k$, but I can't even solve the sum $\sum_{k=0}^{n} \binom{2n}{2k}$

[rocky lotus]

Médicis

[broken condor]

Umm first thing that comes to my mind

$$(\sqrt{2} + 1)^{2n} $$

[rocky lotus]

casework

[coral ocean]

that's where i come from

[broken condor]

Wait what...

Whats the original problem?

Also looking at $$(\sqrt{2} - 1)^{2n}$$ also helps

[rocky lotus]

casework

[broken condor]

I mean your thing is basically AM of those 2

[coral ocean]

the original problem was to prove that $(4+\sqrt{8})^{4n+2} + (4-\sqrt{8})^{4n+2}$ was an integer, and then to get a neat result out of it. I managed to prove that this relation was equal to $2^{6n+4} \sum_{k=0}^{n} \binom{2n}{2k} 2^k$.

[rocky lotus]

Médicis

[broken condor]

Why are you trying to calculate that

Its obviously an integer

You are done

[coral ocean]

i'm helping a friend out with a Zeta function problem

he got somewhere the result $(4+\sqrt{8})^{4n+2} + (4-\sqrt{8})^{4n+2}$ but he can't go much further

[rocky lotus]

Médicis

[broken condor]

Im confused..the problem is to prove that is an integer right?

[coral ocean]

no.

1. prove it's an integer
2. simplify it

i succedded in 1)

not in 2)

i can't close the sum

[broken condor]

I would simplify it as

$$\frac{(\sqrt{2} + 1)^{2n} + (\sqrt{2} - 1)^{2n}}{2}$$

[rocky lotus]

casework

[broken condor]

I dont think that you can get it in a more simple form

[coral ocean]

ok nevermind I actually got the right answer

i don't know why I thought i was wrong

😭

$close

[flint phoenix]

.close

[river shale]

.close

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[soft vine]

hi

[soft vine]

i have a question

yx-2x squared divided like over line thats the top line under that is -4y + x squared (x=2 / y = -5)

this

is very confusing

pls helppp

[broken condor]

Can you send the original question

[red tulip]

!xy

[vale dock]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[fleet burrow]

Evaluate $\frac{(yx - 2x)^2}{-4y + x}$ where $x = 2$ and $y = -5$ is that your question?

[red tulip]

i think its $\frac{yx-2x^2}{-4y+x^2}$

[rocky lotus]

ImOakley

[vale dock]

@soft vine Has your question been resolved?

[soft vine]

yes

thats the one

but no

its not solved

[fleet burrow]

substitute in the values of y and x given in the equation

[vale dock]

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[sour crown]

.reopen

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[sour crown]

@soft vine

[vale dock]

Please don't occupy multiple help channels.

[sour crown]

.close

[vale dock]

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[soft vine]

how

this is what i got and i asked ai

simplyfiy -18 / 24

is -3/4 i belive

[sour crown]

yeah it’s -3/4 i did the last part wrong sorry

i was going to fast

[soft vine]

oh its okay tysm

[soft vine]

yeahh it happenes trustttttttttt

thanks for your helppp

[vale dock]

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[humble pagoda]

I have no clue what to do for b, c, and d

[steady charm]

for b), the "smallest value we could get as a lower bound" would translate to the largest interval in that form

[humble pagoda]

Like the thingy from a?

(I don’t know what those are I just picked up that it’s just between discontinues and to put u and or bracket, never seen them before, pretty sure they were from pre calc which I didn’t take)

[steady charm]

well i don't think they actually wanted you to union them, just list them

[humble pagoda]

So remove the u

?

[steady charm]

yes, the U is when you want to combine two intervals into a larger set

but they didn't ask "what's the set of points on which it's continuous", they just wanted a list of intervals

[humble pagoda]

So what do I do about the k inclusive

Does it have anything to do with x at -5

[steady charm]

well the interval can't be longer than (2, infinity)

[humble pagoda]

So how do i find the smallest value

From the interval

And how is it [k,inf), there is no k inclusive (I think) that leads to infinity (I also don’t really know how any of this works)

[flint phoenix]

[steady charm]

well there are definitely values of k for which it's continuous on [k, infinity), for example k = 3

but we do have to ask whether there is a smallest possible k

[humble pagoda]

Wait what does k mean

Start of interval?

[steady charm]

k is just a number. in context it is the left endpoint of an interval

[humble pagoda]

So b would be -5, but how is there a k inclusive in the same interval of inf

Because (k,inf) is fine after the first hole, but the only inclusive is at the end of something coming from negative infinity

[steady charm]

if in b) k = -5, that means you are claiming it is continuous on (-5, infinity)

[humble pagoda]

So it was 2

I was right for the wrong reason

-# Took 2 from the distance between -5 and -3

[vale dock]

@humble pagoda Has your question been resolved?

[flint phoenix]

[vale dock]

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[molten blade]

Can someone walk me through answering this question?

[wintry oxide]

Look at the denominator inside the logarithm. What value will be problematic ?

[molten blade]

ah, 0 would make it undefined

[wintry oxide]

So you’re almost sure there’s a vertical asymptote there

[molten blade]

shouldn't x = 5 also be an answer considering how it would make the denominator of the first fraction undef too?

[wintry oxide]

Apart from that (and -5) I don’t see any other problem

[wintry oxide]

I don’t think so

The function is clearly undefined there

But it’s not gonna shoot up or down to infinity. It’s just gonna « break »

[quasi valve]

the function will change signs as you cross x=5, that's a jump though, not an asymptote

[wintry oxide]

It’s gonna switch signs

[molten blade]

I'm not sure I get it 😅

[broken condor]

|x|/x is just sgn(x) (well almost)

Its not really defined for 0 in the |x|/x case

[wintry oxide]

You can replace the first fraction by -1 before 5. And 1 after 5

So it is gonna cause a discontinuity at 5

But no infinities

And the precise value for 5 won’t exist. It’ll be like a jump from a value on the right to its opposite

[molten blade]

oh ok, so I can disregard it as a hole because x-5/x-5 creates a discontinuity but since the numerator of the fraction is abs value the sign also switches when graphing it?

[wintry oxide]

You know the graph of x^2 right

[molten blade]

yeah the parabola

[molten blade]

what the

[wintry oxide]

Look what happens to it if I multiply it by that fraction

[molten blade]

it gets reflected off the x axis

but why?

[wintry oxide]

Before 5, it’s like multiplying by -1.
After 5, it’s like multiplying by 1
At 5, it’s not defined (no value, no infinity)

That’s because before 5, |x-5| = -(x-5)

So the fraction is -1

[molten blade]

ohh I see so only the part of the parabola that considers values under 5 gets reflected

[wintry oxide]

After 5, the fraction simplifies into 1

In 5, the fraction is not defined (but no infinities either

[molten blade]

I see now, yeah alright so it's a jump like you said

[wintry oxide]

So that fraction is gonna have the same effect on your function. It’s gonna break and flip it before 5. But it won’t cause infinities

[molten blade]

I understand it now, thanks!

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[golden elk]

how do I do this?

[steady charm]

can you show the question it's asking?

[golden elk]

its asking for all 5 x values

[steady charm]

all 5 x values that do what?

[golden elk]

well 2 of them are the boundaries and the rest are in between

[golden elk]

this is what its asking

[steady charm]

that doesn't tell us what the x-values are supposed to mean. can you show a full screenshot of all context?

[vale dock]

@golden elk Has your question been resolved?

[cloud coral]

yeah really looks like something is missing there

[golden elk]

this is literally all the context that i can possibly show

theres nothing else they gave

[cloud coral]

who assigned this to you? you might want to contact them

[golden elk]

what am i supposed to ask?

[cloud coral]

"what do the x_i correspond to here"

you can say something like
"It seems to me that these boxes refer to this question, but I can't tell what x2, x3, x4 are supposed to be. Also, the length of the period is not rational."

[golden elk]

the lop is 2pi no?

[cloud coral]

yeah the period is 2pi and 2pi is not rational

[wintry oxide]

I think the most « relevant » 3 values within a period would be the max, the min and the root

But your teacher should definitely be slapped for asking shit ass questions like this

[golden elk]

does it have anything to do with intervals?

as in multiplying the lop by 1/4th and then adding on and on to get the values from the lb

[cloud coral]

but max, min would be the y values

yeah the question is poorly designed

[wintry oxide]

You can find the corresponding x values. Like choose a period and provide all these values

[golden elk]

.close

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[vale canopy]

The title of this page is "Practice Solutions of an Equation in Two Variables". Problem is " Write a Solution of the equation y = 4x - 0.5". I forgot how to do it😭

[tawny moon]

your job is to write a solution?

as in, any solution?

[vale canopy]

I assume so, like to use for a line that goes at a constant rate in a graph.

[tawny moon]

in that case, pick some random x-value, and then substitute it into the equation and solve for y.
(hint: technically, you don't have to if you remember what the last term / rightmost term means.)

[hollow rune]

quick recap: Two Variable Equations have infinite answers made up of pairs of (x,y)
💔

[west cloud]

it has infinite answers if there is one or no equation, and one answer if there are two or more equations
equations as in, distinct equations

not "generally"

but anyway, in OP's case, there are infinite solutions (as you said)

[hollow rune]

that would be a system, but yeah

I suppose i was thinking of a few equations which have a specific "domain", but that doesnt make the answers non-infinite, mb

[tawny moon]

if an equation is linear and not vertical or horizontal, it has infinite solutions assuming no other restrictions

[west cloud]

I was being a bit nitpicky there sorry about that

[strong shale]

Could I ask for assistance with a physics problem in here?

[tawny moon]

but all this discussion aside, is OP still here

[tawny moon]

yes, but please do so in an available channel, like #help-45

[strong shale]

Okay, thanks

[vale dock]

@vale canopy Has your question been resolved?

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[glacial raven]

im doing some khan academy stuff and this was the solution for a question apparently? where did the -36x come from lol im so confused

[shut osprey]

as in the 2ab of (a-b)^2 identity

-2.9x.2

=> -36x

[glacial raven]

ohhhh

.close

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[ionic drum]

hello for this question, do i cube the 1 gal/231^3 inch again? or can i just keep it like that since its already cubed

[tawny moon]

gallon is already a unit of volume

cubing the gallon unit gives gallon^3, which is less a unit of volume and more a unit of whatever 9-dimensional exotic material scientists are working on nowadays

so no, don't cube

[ionic drum]

haha okay thank you :)

also for #18, do i find the area of the classroom in meters first and then convert it into square inches

and if i do find the area in meters first, would the units be meters^2

[tawny moon]

there are two ways for it

either find the area in sq. meters first, then convert to sq. inches (a bit dangerous, reason later)
or convert both lengths to inches first, then find the area in sq. inches

the reason the first option is dangerous is that if you forget to square the conversion factor, you'll be off by orders of magnitude

but if you are confident in your conversion, the first option is faster, because it involves only one conversion instead of two

[ionic drum]

it’s all good, so if i used the first option would it look something like this?

[tawny moon]

yup

[ionic drum]

got it thank u so much :))

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[sturdy oak]

hey so im stuck with this limit thing pls help

[sudden forum]

i think u multiple by 1/infinite / 1 /infinity or something

why this wrong @sturdy oak

--

[sturdy oak]

for the first step u muliply denominator and nominator

by

x+ sr x2-3

[sudden forum]

oh wait i think u multiply by (2/x)/ (2/x) looking at my notes

am i right @pearl nacelle

[pearl nacelle]

Multiply top and bottom by x+sqrt(x^2-3)

[sturdy oak]

yea thats what i did

but at the last step i got stuck

wait ima do it again

and send u a pic of where i get stuck

[pearl nacelle]

I don’t think sp

[pearl nacelle]

Okay

[sudden forum]

i think im rite and ur wrong

[sturdy oak]

yall did calculus right?

[pearl nacelle]

Yeah

[sudden forum]

im in calc

[pearl nacelle]

The denominator will end up being

3

So itll be infinity

[sudden forum]

ok why this wrong @pearl nacelle

[sturdy oak]

answer is supposed to be -infinity

but yh

[pearl nacelle]

Oh yeah

[sturdy oak]

cus i have the answer to the exercise

[pearl nacelle]

Limit is to negative infinitt

I didn’t see

But same principle applies

[sturdy oak]

yes

[tawdry minnow]

+1

[sudden forum]

why my wrong

[pearl nacelle]

Do you understand what you have to do

[tawdry minnow]

what did you even do

[pearl nacelle]

What

[sudden forum]

bro the lim equation is literally the 2nd one

[tawdry minnow]

also i'd suggest open another channel for your doubt instead of mixing with aym.sna's

[tawdry minnow]

was supporting your arguement

[sudden forum]

but mine isn't complicateed

[pearl nacelle]

Oh right

[sturdy oak]

im to the step where i get 3 as the denominator

[pearl nacelle]

Sure

[tawdry minnow]

now just substitute the limit, as all the terms are in numerator only, there's no indeterminancy, and you'll get your ans

[tawdry minnow]

what is up with the image? you have a doubt?

[sudden forum]

yah my answer wronging

[sturdy oak]

oh alr, i wanna jus trynna get the whole approach

but thanks

[tawdry minnow]

thats fair

[tawdry minnow]

wdym? what did you get as 'a' and 'b'?

[sudden forum]

what i need to solve for a and b?

[tawdry minnow]

given the function is continous at x=1,3 , you need to equate lim x tends to 1- with 1+ and 3- with 3+ as defined in the ques

put x=1 in x+4 and the given quad and equate them and then do the same at x=3

[sudden forum]

idk

[tawdry minnow]

do you know how a continous function is defined at any point?

[sudden forum]

yes as long as it's closed

the answer is 1

so it didn't want the function

ax2+bx+2

ok u can close now

@sturdy oak

[sturdy oak]

ight thanks

idk how to close lel

[sudden forum]

type .close

@sturdy oak

[sturdy oak]

wait one last question

[sudden forum]

ya

[sturdy oak]

-infinite times 0

is 0?

[noble coral]

Indeterminate form

[sturdy oak]

ohh

k

alr thanks bye

.close

[vale dock]

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[sudden forum]

good boy

[sturdy oak]

lol

[vale dock]

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[fleet rover]

i have two question, i understand why x is taken as 0.1, becausw from the graph it is 10 cm, but why would the radius be taken as 10 cm, the radius should be of the circle, whereas 10 cm is the distance from the circular phase to the starting which is the lenght of electric field line.

2nd question: while caculating flux 1, we do not give it a negative value, as in acc to my notes and general thesis if an electeic field is going into a surface, the electric flux must be negative ( have attached the notes for the samw), so even though we did get negative pi as the answer, shouldnt we again multiply it with a negative symbol whcih makes it plus pi

@normal hollow

[normal hollow]

Is this a given info?

[fleet rover]

YOU ARE BACK

YAYAYAY

[fleet rover]

YES

[normal hollow]

Which means that the radius is given as 10 cm

[fleet rover]

NO THE DISTANCE OF ELECTRIC FIELD IS GIVEN AS 10 CM

ok

NVM

IM SLOW

I READ THAT AS X

MY FIRST QUESTION HAS BEEN WELL CLEARED

its not my book, so i dony understand the handwriting

WHAT ABOUT GIVING THE ELECTRIC FLUX NEGATIVE THOUGH, BECAUSE IT IS GOING INTO THE SURFACE

[normal hollow]

No i dont think u need to multiply it with a negative again

[fleet rover]

so is it like, if its a flux going into the surface, it automatically becomes negative?

[normal hollow]

The negative sign represents that the electric flux is going into the surface

[normal hollow]

Yes u could say that

[fleet rover]

BUT LIKE SEE THE ELECTRIC FIELD VALUE GIVEN HERE WAS 10 X I CAP, WHAT IF IT WAS -10 I CAP, then the value would have come out to be positive pi

it would still be goinf into the surface

left over all remains same

just that

the values becomes positive whcih doesnt make sense as per the rule

[normal hollow]

I would suggest tagging helpers for this cuz igtg for now

[fleet rover]

WOKAY

[normal hollow]

But there must be a restriction to that

[fleet rover]

that makes sense

ill get it cleared though just incase

THANK YOU SO MUCH

[fleet rover]

<@&286206848099549185>

lmao somehow ur back again

[normal hollow]

Ik this wont clear the concept but the value of E which u use in the formula of electric flux is the magnitude of electric field it must be positive

[normal hollow]

Got 1 min ig

[fleet rover]

okay NO WAIT THAT DOES CLEAR THE TOPIC

BECAUSE IF ITS ALWAYS POSITIVE

THEN THE RULE REMAINS INTAXT

[normal hollow]

If the E is -10 i cap then,it would mean that the direction of the field it towards negative x axis

Otherwise magnitude remains same

[fleet rover]

YOU ARE THANK YOUED

[fleet rover]

THANKS A LOT KIND PERSON

[normal hollow]

Alr np and igtg
others can help if u have confusions

[fleet rover]

YES

WOKAY

.CLOSE

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[sullen pagoda]

can somebody guide me on how to solve Laplace transforms?

[vale dock]

@sullen pagoda Has your question been resolved?

[sullen pagoda]

Im not sure how to continue from here

And im not sure if what i did so far is right or not lol

<@&286206848099549185>

[vale dock]

@sullen pagoda Has your question been resolved?

[past wigeon]

Are you using any textbook?

[sullen pagoda]

nope, i have some exercises from previous exams, and im trying to get ready for the exam 😅

[past wigeon]

I was hoping that there would be similar questions from a textbook you could use

[sullen pagoda]

i did manage to get to an asnwer, now i'm not sure if it's right or not 😂

[past wigeon]

Test your answer via substitution

[sullen pagoda]

yeah i did test and y(0) and y'(0) give the same so seems like it's right

now all i have left to do is figure out how to solve a Fourier series or something like this lol

this would be an example for fourier, any thoughts? lol

[sullen pagoda]

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[trail aspen]

Helpp

[trail aspen]

Is symmetry and congruency same

[keen tundra]

In what context

And probably no

[fossil mulch]

I guess if a shape were symmetrical, the two shapes on both sides of the line of symmetry would be congruent

If that's what you mean

[trail aspen]

So it's congruent by ss

Ass*

Sss*

[fossil mulch]

Can we see the question for context

But if we're thinking of the same thing yea SSS

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[azure osprey]

hi, prof explained this concept really poorly and i got 0 clue where to start. i want to figure out how to actually solve it and approach problems like this vs just ask chat gpt first so anything helps

[azure osprey]

calc 3^

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@azure osprey Has your question been resolved?

[safe fulcrum]

https://tutorial.math.lamar.edu/classes/calcII/vectorarclength.aspx

[azure osprey]

awesome, this is what i needed

thank you

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[neat heath]

is the condition for the bottom graph 0<a<1 ? i guess the graph doesnt make sense if a becomes -ve ?

[distant galleon]

Yes. a should be positive.

[neat heath]

thanks dude

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[honest stone]

welp

[limber turret]

12x2+12+4x(-6)

[honest stone]

open a new channel

plz

[limber turret]

is -160 right?

[limber turret]

mb

[cinder sonnet]

Wrong brotha

Try again

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[hybrid comet]

Help pls

[flint phoenix]

what have you tried in advance?

[normal hollow]

Try using similar triangles

[hybrid comet]

I just calculated the lengths of AM, MC, and AB

Couldn't get further

[normal hollow]

These r the given values

Notice triangle AQM and ABC

[hybrid comet]

Yes,,,

[normal hollow]

Do u see any equal angles

[hybrid comet]

Yes

BAC

[normal hollow]

And?

[hybrid comet]

Uhh

AQM?

90⁰

[normal hollow]

Yes exactly AQM and ABC r equal

So can u say that these two triangles r similar to eo

[hybrid comet]

Yes

But not congruent right?

[normal hollow]

And do u know the property of similar triangles

[hybrid comet]

Yes

[normal hollow]

Not congruent

[normal hollow]

Which is?

[hybrid comet]

Corresponding angles are congruent and similar sides are proportional?

[normal hollow]

Yea

[hybrid comet]

So I do the same for the second triangle too!

[normal hollow]

So the ratio of sides opposite to equal angles of two traingles r equal to eo

[hybrid comet]

Hmm

Right

[normal hollow]

Aha ig

So can u identify the opposite sides of 90° angles of rhose two triangles

[hybrid comet]

AM and MC?

These are 7.5 cm each

Because M is the midpont

*i

[normal hollow]

AM is the side opposite to <AQM

But CM isnt-

[hybrid comet]

CM is the side opp to MPC

[normal hollow]

AC is the side opposite to <ABC

[hybrid comet]

Right

Sorry I thought we were looking at ∆AQM and ∆MPC

[normal hollow]

Got u ,nw

[hybrid comet]

So AC is 2AM right?

[normal hollow]

Can u find the sides opposite to the other 2 equal angles

[normal hollow]

Yes

[hybrid comet]

Yes

[normal hollow]

They r?

[hybrid comet]

So MQ and BC are in a fixed ratio,

What about AQ but?

[normal hollow]

Dont use MQ and BC here

Since u dont know the value of either of thwm

[hybrid comet]

But the value of BC is given

BC is given 5 cm

[normal hollow]

That's AB

And yes u can find BC basing on AB and AC

But y increase steps when u can do it at a go

[hybrid comet]

Hmmm

Right

[normal hollow]

Notice that if two corresponding angles of two triangles r equal then the other angles r equal as well

[hybrid comet]

Right

[normal hollow]

So u can say that <AMQ=<ACB

[hybrid comet]

Due to the angle sum property

Yes correct

[normal hollow]

Yes

[normal hollow]

And from that u can say that the ratio of sides opposite to those angles is?

[hybrid comet]

2:1?

AB=2AQ?

[normal hollow]

Yes

[hybrid comet]

Are you able to find the question, or should I resend the image?

[normal hollow]

Take time to scroll ,lemme send it rq

[hybrid comet]

Ok

More convenient

[normal hollow]

Yes now as u can see u were given AB and now u can find AQ

[hybrid comet]

Right, AQ should be 2.5 cm

And then using Pythagoras theorem, I can find the remaining sides

Thank you so much!

I can take it from here

.close

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[normal hollow]

Np

[hybrid comet]

.close

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[ruby ingot]

The only things i can think of is that it will have an x factored out and the fact that f(1) = 1 and f(-1) = -1

[distant galleon]

Hint: Observe that $f(x)=x$ for $x=-1,0,1$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

This implies that ||f(x)-x|| has ||roots of -1,0,1||

[flint phoenix]

[ruby ingot]

wait hold onn

how did we come to f(x) -x

[river sapphire]

If f(x) = x then f(x) - x = 0

[ruby ingot]

yea but for only certain values of x no?

how i sthat tru for all values

[umbral elbow]

what does which polynomial mean tho?

like what context

[ruby ingot]

its asking for the equation

[umbral elbow]

ah

see

[river sapphire]

You need 4 points to represent a cubic anyways

I don't see how that's helpful either

[ruby ingot]

this i dont understand since this is only true for x = -1, 0 , 1

[umbral elbow]

general equation of a cubic polynomial ax^3 +bx^2 + cx + d = f(x)

[ruby ingot]

i mean i noted i had those values, i assumed it would be some simulatneous eq

yea

but this one

d = 0

and

[umbral elbow]

now at 1 and -1 f(x) is 1

[ruby ingot]

we kow 2 other points

[ruby ingot]

no

[umbral elbow]

so substitute in the general equation

[river sapphire]

We need one more point

[ruby ingot]

at 1 f(x) = 1 at -1 f(x) = -1

[river sapphire]

So it doesn't really matter anyways haha

[umbral elbow]

ah yes sorry

so now

[ruby ingot]

yea exactly

[umbral elbow]

take x=0

[river sapphire]

N+1 points to represent an Nth degree polynomial

[umbral elbow]

at this point f(x) =0

which means d is 0

[ruby ingot]

yea

[umbral elbow]

now the equation is now

ax^3 +bx^2 +cx

[ruby ingot]

yh

[umbral elbow]

now this is reducible to a quadratic form

[ruby ingot]

yh

x(ax^2 + bx + c)

[umbral elbow]

so x(ax^2 +bx +c) = f(x)

yes

[distant galleon]

hold on what's happening

[umbral elbow]

now plug in the values of x = -1 ,0 ,1

in the new equation

[ruby ingot]

this wont help

[distant galleon]

My intention here was that $$f(x)-x=k(x-1)x(x+1)$$ for some constant $k$. You know that the local minimum is $-1$, so derivative bash to compute $k$ and you're done.

[umbral elbow]

let me finish.

[ruby ingot]

alr

eyeball 😭

surely theres a definite wya

[umbral elbow]

so now we get equations in terms of a , b c

[ruby ingot]

yh im following

[umbral elbow]

then solve ab c

[ruby ingot]

u cant

u only have 2 equations in terms of a b c

[umbral elbow]

hmm

[rocky lotus]

Civil Service Pigeon

[distant galleon]

this was what I was going for

[ruby ingot]

i hada question on this

can we assume the min and max are at -1/2 and 1/2 ?

[distant galleon]

mm?

[ruby ingot]

since it says it just fits

[distant galleon]

The stationary points are always halfway between the roots of cubics, so you don't even need to assume this - it must be true.

[ruby ingot]

ahh shit

lemme cook this up rq

[distant galleon]

verifying this is straightforward by derivative bashing

proof omitted because I'm walking

[ruby ingot]

lol alr

[silver birch]

if you have done derivatives then you can assume that -1/2 and 1/2 are stationary points

[ruby ingot]

calm

i hope i wouldve done derivatives

[umbral elbow]

ok wait

[ruby ingot]

going into undergrad without derivate knowledge 😭

[umbral elbow]

we can use