help-10

[brave bramble]

You CAN construct brand new vector spaces. In later courses you will.

You'll have more tools to do this, though

[obtuse pebble]

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[worn cove]

Could anyone check if i have this question right ( a and b), please?

"A metal sphere with mass is suspended from a horizontal ceiling using a rope (with negligible mass) and swings back and forth. The mass of the sphere is m = 627 g, and the gravitational acceleration is 9.81 m/s². At a given moment, the rope makes an angle θ = 28.0° with the ceiling.

(a) Calculate the magnitude of the force that the rope exerts on the sphere at that moment:
Fs = I Fg ·L I / II L II
Fg is the gravitational force vector, and L is a vector from the beginning to the end of the stretched rope.
Provide an answer with three significant figures.

(b) in which range [a, b] can the outcome of a scalar product GENERALLY be found." I got ]-infinity, + infinity[ Is this also correct?

[obtuse pebble]

@worn cove Has your question been resolved?

[obtuse pebble]

@worn cove Has your question been resolved?

[worn cove]

<@&286206848099549185>

[worn cove]

<@&286206848099549185>

[obtuse pebble]

@worn cove Has your question been resolved?

[unique nebula]

I did not check your gravitational but for part b, does that really make sense? I'd use a function of both a and b instead, rather than a uniform 'all reals' answer (which is generally not informative)

[worn cove]

the multiple choice answers were: 0, 1 ; -1,0 ; -1, 1 ; 0, +infinity ; -infinity, 0 ; -infinity, +infinity

[unique nebula]

Why is it suddenly a multiple choice

[worn cove]

Only (b) is a multiple choice question

[worn cove]

<@&286206848099549185>

[obtuse pebble]

@worn cove Has your question been resolved?

[obtuse pebble]

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[warm canopy]

stop deleting your original message

[timid silo]

Let $m, n \in \mathbb{N}$
Find the number af multiples of $m$ that are contained within the set ${1,2, \ldots n}=: A$
Denote $p(m, n):=#{k m {1, \ldots n}, n \in \mathbb{N}}$

I want to prove that $P(m, n)=\left\lfloor\frac{n}{m}\right\rfloor=\max \left(\left{x \in Z: x \leq \frac{n}{m}\right}\right)$
If $n=0$ then $A={}$ and $p(m, n)=0$
If $m=0$ then $k \cdot m=0$ $\notin$ $A \Rightarrow P(m, n)=0$

If $n, m \neq 0$
I want he prove that $p(m, n)=\left\lfloor\frac{n}{m}\right\rfloor$
To prove this I will first show that $p(m, n) \leq\left\lfloor\frac{n}{m}\right\rfloor$ and then $\left\lfloor\frac{n}{m}\right\rfloor \leq P(m, n)$.
$$
\begin{aligned}
& P(m, n)=#{k m {1,2, \ldots n}, n \leq \mathbb{N}} \leq \
\end{aligned}
$$

[warm shale]

12345

[timid silo]

Sorry about that

Got it properly now

This is where I'm stumped. What kind of upper estimate can we find for p(m,n)? Maybe we can directly say that p(m,n) $\leq$ $\max \left(\left{x \in Z: x\cdot m \leq n\right}\right)$

[warm shale]

12345
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[warm canopy]

yeah but now you need to open a new channel again as per the bot message

[timid silo]

Ah I see. First time on here

[warm canopy]

its ok sorry if this came across strongly

[timid silo]

All g

Bet it happens loads

[warm canopy]

just open a new channel and dont delete first msg 🙂

[timid silo]

On it. Do I just .close this one now?

.close

[obtuse pebble]

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[fallen inlet]

How can I rewrite 21x/2 to something more understandable? I got help earlier for x/3 which turned out to be 1/3x. Is there something that can be done for 21x/2 ?

[timid silo]

you mean 1/3 * x

[subtle sinew]

Same logic can be applied, in general, $\frac{ax}{b} = \frac{a}{b} x$

[timid silo]

not 1/(3x) different things

[warm shale]

CaptainNova22

[subtle sinew]

So $\frac{14x}{3} = \frac{14}{3}x$

[warm shale]

CaptainNova22

[brazen gorge]

21x/2 = (21/2)*x

[fallen inlet]

Yes

[timid silo]

uh if 21/2 can be simplified

[fallen inlet]

Ohh ok

So the same thing can be done

That’s very helpful

Thanks everyone

[subtle sinew]

If you are done with the channel, don't forget to close it using .close

[fallen inlet]

.close

[obtuse pebble]

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[vestal cape]

for c)

I did 8c6 x 0.6^6 x 0.4^2 + 8c6 x 0.5^8

and then divided it by 2

what did I do wrong

[obtuse pebble]

@vestal cape Has your question been resolved?

[vestal cape]

<@&286206848099549185>

[wet nebula]

You have to divide by the probability of having tails on your first throw

You also have*

By the definition of conditional probability, P(E | F) = P(EnF) / P(F)

[vestal cape]

since it's given that the first of the 9 flips lands on tails I thought it was not needed to be calculated

[wet nebula]

Have you learned the definition written just above?

[vestal cape]

yeah

P(A n B)/P(B)

[wet nebula]

And then also to compute P(EnF), you need to be careful as you don't have any conditional event now

[vestal cape]

but then I got 0.02211068611

which is wayyy too small

cause p(B) is 8c1 x 0.4 + 8c1 x 0.5 no?

[wet nebula]

Because P(EnF) should give you

[8C6 × (0.6)^6 × (0.4)^3 + 8C6 (0.5)^9]/2

What did you define as B? Having tails on first throw ?

[vestal cape]

yeah

[wet nebula]

No, this should give you (0.4 + 0.5)/2

[vestal cape]

so wait what is p(f)

[wet nebula]

My F is your B

[vestal cape]

oh ok

so 0.45?

[wet nebula]

Yes

[vestal cape]

okkk thank you so much

[wet nebula]

And P(EnF) is there

[vestal cape]

yeah I messed up thinking that I could just ignore the first flip

[wet nebula]

Be careful the exponents arent the same as you first suggested

[vestal cape]

yep I see it

I appreciate the help

.close

[obtuse pebble]

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[bright arrow]

So the question is f(x,y) = xy constrained by g(x,y) = x^2 + y^2 - xy - 9

Using Lagrange Multipliers to find the maximum and minimum

The values in the 2nd picture is what I got when plugging back into f(x,y) but I wanna check if my work is correct

Cuz online calculator says otherwise but I dont trust myself enough to say that its wrong or right

<@&286206848099549185>

:(

[obtuse pebble]

@bright arrow Has your question been resolved?

[bright arrow]

<@&286206848099549185>

[obtuse pebble]

@bright arrow Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[jagged cargo]

i need to find the general solution of this DE but im lost on how this factors

[forest sinew]

what did you try?

you mean youre not sure how to solve the quadratic?

[jagged cargo]

oops nevermind i figured out what i did wrong

quadratic works yeah i did not realize that, but i was supposed to use the superposition principle using 2 linearly independent solutions which were given

but thanks

.close

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[modern palm]

Is this the correct integral?

(\int^1_{-1}\int^{\sqrt{1-x}}_{\sqrt{1+x}}2x^2-2y^2dy dx)

[warm shale]

hamr

[bronze mica]

unfortunately, no

[late sequoia]

your y between 1 and - 1 and x between y^2 - 1 and 1 - y^2

[bronze mica]

if you pick x=1/2, draw the points (1/2,y) where y lies betwen sqrt(1+1/2) and sqrt(1-1/2)

you're going to see you're integrating over those points, but they're not in the region you want to be integrating over

[modern palm]

.close

[obtuse pebble]

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[wintry hedge]

does anyone know how to solve $ ln(x + 2) = x + 1$

[wintry hedge]

$$ln(x + 2) = x + 1$$

[warm shale]

Hzhou1

[brave bramble]

Impossible

[latent walrus]

theres one solution

[brave bramble]

Well, I guess x = -1

Just by inspection

Is that the only solution?

[wintry hedge]

oh

shoot I was working on a problem and got to this point

yea it should be -1

but can I just say that?

[brave bramble]

Why not? "I plugged in x = -1 and it worked"

[wintry hedge]

lmaooo

[brave bramble]

The greats do that kind of math daily

[wintry hedge]

alr I'll try it

thanks

.close

[obtuse pebble]

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[violet rapids]

T

[violet rapids]

Top graph

ok

so we know it goes up 4

but now it’s kind of strange because at x = 2 it changes

so how would I write that in a equation

[static furnace]

you don't care about the actual equation, the equation is y=f(x). you just need to express each graph as a transformation of f(x).

[violet rapids]

Oh

like this transformation would be going up 4 ?

[static furnace]

y=10f(x) for example, would stretch the graph by a factor of 10

what you really care about is what graph (a) looks like compared to f(x)

[violet rapids]

ohhhh

yea ok I get it

so A would be

compressed

[static furnace]

right

[violet rapids]

but how do we know how much compressed by? or that don’t matter

would be 2 ?

[static furnace]

I first look at the max value of the graph

then compare that to the max value of graph (a)

kinda

it's getting shrunk, so what can u multiply values by to make them shrink. the max value of f(x) and graph (a) are 4 and 2 respectively.

so how do you get 4 to turn into 2. what do you multiply it by?

[violet rapids]

1/2

[static furnace]

so to make sure we're correct, let's look at another point.

[violet rapids]

yes

[static furnace]

the graph flattens out at y=2 in f(x) and y=1 in graph (a)

as well as y=-4 and y=-2, respectively

so that checks out

so then how do you write that it's being compressed by 1/2

[violet rapids]

Idk

I just wrote “compressed by 1/2”

[static furnace]

so you're multiply all the y values by 1/2, that's what we're doing, correct?

and y=f(x)

[violet rapids]

ya

[static furnace]

so how could you represent graph (a) using the term f(x), which is y, knowing that you're multiplying all y values by 1/2?

y equals what

[violet rapids]

1/2

(1/2)x

no

(1/2)y?

lol

[static furnace]

use the term f(x)

[violet rapids]

ok

f(x)=1/2x

[static furnace]

almost. y=(1/2)f(x)

that's the graph of (a)

since the y-values of f(x) all get compressed by a value of 1/2 in graph (a)

[violet rapids]

Ok

Let me try b first

[static furnace]

if you were to stretch it by any factor A, the the resulting transformation is y=Af(x)

okay

[violet rapids]

ok this one is compressed

but

the other way

for the x values

[static furnace]

right

the x values

yes exactly

by how much

[violet rapids]

1

2

[static furnace]

2, yes

so since it's the x-values, then how can we represent that using y=f(x)

[violet rapids]

It can’t be infront of f(x) right

[static furnace]

no, that would mean we're changing the y's

[violet rapids]

so it has to be infront of f(x)

I mean

wait

before f(x) is y values change

Af(x)

[static furnace]

correct

[violet rapids]

then after f(x) is x values change lool

[static furnace]

almost. lol. it's inside the function.

so the numbers are being compressed again yea?

so it's a factor of 2 that it's changing by, but is it 2 or 1/2 that is being multiplied by x?

[violet rapids]

hmmm

1/2

[static furnace]

yes.

[violet rapids]

f(x-1/2)

[static furnace]

no.

[violet rapids]

What the e heck

[static furnace]

it does go inside, but we've decided that it's multiplying every x value by 1/2

not subtracting

[violet rapids]

f(x)1/2

[static furnace]

it goes inside

f(1/2*x)

[violet rapids]

Phhh that’s what u mean

[static furnace]

yeah inside the function, f

the output is f(x), which is y, and the input is x

[violet rapids]

yes

that mean sense

[static furnace]

change x, change inside. change y, change outside

[violet rapids]

Make

[static furnace]

are those the only two questions?

[violet rapids]

There’s one more

But

That one is flipped over the x axis

So is the negative here or here

f(-x)

-f(x)

[static furnace]

it's flipped over the y-axis, remember that bottom flat line was on the left side for f(x)

so if it's flipped over the y-axis, then none of the y-values change

[violet rapids]

-f(x)

No

I lied

[static furnace]

if none of the y-values change, then we're changing the x values

[violet rapids]

F(-x)

[static furnace]

yes perfect

f(-x)

[violet rapids]

Thx wumbo

Appreciated ur time

[static furnace]

no problem friend

[violet rapids]

.close

[obtuse pebble]

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[round knot]

the vector in the opposite direction to u = <5,2> and of half its length is __

[round knot]

<@&286206848099549185>

[daring sorrel]

!status

[obtuse pebble]

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

[round knot]

I have 0 clue how to do this

[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.

[round knot]

okay i apologize

[daring sorrel]

Intuitively, you can think of vectors as arrows. How would you find the opposite direction of where an arrow is pointing

[round knot]

-vector?

[daring sorrel]

yeah

[round knot]

next step?

[daring sorrel]

Do you know how to get the magnitude of a vector?

[round knot]

yes

[daring sorrel]

ok do that then divide by 2

actually

you can just divide the original vector by 2

I was gonna make this more complicated then it had to be

[round knot]

sqrt(29)/2?

[daring sorrel]

yeah nvm what I said about finding the magnitude

[round knot]

okay so
<5/2,1>?

2/2=1

[daring sorrel]

That would be half of the original vector yeah

But it needs to be in the opposite direction so you can now multiply it by -1

Or you could flip it then half it

[round knot]

correct

whats the second question asking

[daring sorrel]

For that question we need to find the magnitude of v.

[daring sorrel]

Yeah

We want the magnitude to be 8 agree?

[round knot]

ok

[daring sorrel]

If we multiply sqrt35 by 8/sqrt35 we would get 8 obviously, but we can also multiply the original vector by 8/sqrt35 and its magnitude will become 8

so multiply each component by 8/sqrt35 to get your answer

[round knot]

ok

damn thank you so much

.close

[obtuse pebble]

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[hexed jungle]

Do i take the derivtive and set it to 0?

[fierce lagoon]

you can

[hexed jungle]

that's what i tried doing and still got it wrong

[fierce lagoon]

show me your f'(x)

[hexed jungle]

then after 3 attempts, i put it into desmos, found the maximum x value, and its still wrong

😭

[hexed jungle]

f'(x) = -(ln(x)-1)/3x^2

[fierce lagoon]

and what did you get for f'(x) = 0?

[hexed jungle]

oh wait

@fierce lagoon wouldn't this just be e

[terse sable]

What is that???

[fierce lagoon]

yeah

[hexed jungle]

wait no bc if i set it to 0, then that'll be the minimum

[terse sable]

Do u guys know how to do this

[hexed jungle]

dont i have to find the double derivative to find the maximum

[fierce lagoon]

get your own cahnnel

[hexed jungle]

channel is already full, go to #❓how-to-get-help

[fierce lagoon]

you could

or just do sign analysis on f'(x)

[hexed jungle]

e was wrong

one last attempt 💀

[fierce lagoon]

it wants you to round to 5 decimal places

so round e to 5 decimal places

[hexed jungle]

oh yeah i j noticed that

i think 6 decimal places bc it says 5 places after the decimal point

so 2.71828

[fierce lagoon]

thats 5 decimal places

5 after the dot

[hexed jungle]

oh

yeah that terminology makes sense

lessgoooo

thanks

ppreciate it

.close

[obtuse pebble]

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[hexed jungle]

why isn't this just 2000/22?

[drifting wraith]

it could be

they don't say what kind of growth it is, but like, bacteria wouldn't do it linearly

[hexed jungle]

how would i set up the equation for exponential growth

[drifting wraith]

it would just be 2^(1/22)

[hexed jungle]

so growth rate would be
2^(1/(insert time))

as the formula?

[drifting wraith]

no the 2 is difference

like you calcuated 2000

[hexed jungle]

ohhh so 4000/2000 = 2 and that's how u get the difference

[drifting wraith]

yes

[hexed jungle]

so difference^(1/(time))

[drifting wraith]

sure

[hexed jungle]

so i plug the difference between starting and end population, add the amount of time

and the answer for growth constant would be 2^(1/22)?

for this question\

and this applies for every other question that's structured like this?

[drifting wraith]

i would think so

but like

maybe it's supposed to be in exponential form

[hexed jungle]

it says give an exact answer

[drifting wraith]

click the reference thing, maybe it explains more

[hexed jungle]

reference is just a math calculator

idk why its even labeled reference lol

[drifting wraith]

yeah so "growth constant" imlpies exponential form, so you also log that

[hexed jungle]

yeah im looking thru the provided pdf rn and its hard to wrap my head around it

[drifting wraith]

ln(2) / 22

[hexed jungle]

ah

so 2^(1/22) would be the wrong form?

[drifting wraith]

yeah, ln(2^(1/22)) = ln(2) / 22 is the right one

[hexed jungle]

i see

like this?

or ln(2/22)

[drifting wraith]

yes like that

[hexed jungle]

aii

lesgoooo

ppreciate it

🙏

.close

[obtuse pebble]

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[obtuse pebble]

@vestal acorn Has your question been resolved?

[vestal acorn]

<@&286206848099549185>

[honest sail]

what have you tried?

[obtuse pebble]

@vestal acorn Has your question been resolved?

[vestal acorn]

none they didnt teach us this

[honest sail]

search for factor theorem online

[vestal acorn]

.close

[obtuse pebble]

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[obtuse pebble]

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[paper scarab]

hi could someone please help review my methods/stats section of my thesis? don't need any work done for me or anything but just to need some advice and for someone to check that my methodology is sound, ty! willing to compensate

[versed pier]

I'm not sure if this is the correct place for this

I can't help you bc I'm a lowly freshman

but good luck

[paper scarab]

oh sorry if it's the wrong place, do you know a more appropriate place?

[versed pier]

@fathom flicker

where would something like this go

[fathom flicker]

Your thesis?

[paper scarab]

hi yes

[fathom flicker]

Can you be more specific what you are asking about

your thesis for what

[paper scarab]

it's a dissertation in the social sciences, it's not sophisticated stats or anything

[versed pier]

I mean you can post if you are willing to

but no promises

[fathom flicker]

I think you should try #advanced-lounge instead

Or one of the advanced statistics channels

[versed pier]

hmm ^^

[fathom flicker]

this is meant more for a single math question you need help with solving

it is unlikely someone will look over your dissertation in a help channel

[paper scarab]

ohh alright, thank you

[versed pier]

do you wanna close this cannel?

[paper scarab]

yeah sure thank you

[versed pier]

type in ".close"

[paper scarab]

.close

[obtuse pebble]

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[merry girder]

hi

[obtuse pebble]

@merry girder Has your question been resolved?

[merry girder]

<@&286206848099549185>

[copper void]

To first start this problem do you know what form the equation is in?

[merry girder]

vertex form

ik how to do a

its (200,30)

idk how to do b

[copper void]

on b?

[merry girder]

cuz it has no x intercepts

[versed pier]

can distance be negative

[merry girder]

no

[versed pier]

can distance be 0

[merry girder]

uhhh

yeah..?

[versed pier]

yep

so [0,

how long can the cable be

[merry girder]

uh heroizontal wise or vertical wise?

horizontal

[copper void]

correct

basically, how far can this rope stretch out horizontally

mathematically focus in on (x-200)^2

[merry girder]

oh

well

idk lol

[merry girder]

do i solve for x..?

[copper void]

give me one minute im trying to think of a better way to explain it to you

[merry girder]

okay lol

thanks

[copper void]

Think back to what nosq said, with distance

if distance cannot be negative then your x-value has to start at 0 right?

[merry girder]

yeah

[copper void]

graphically

if distance cannot be negative, where should the graph stop then?

[merry girder]

80

y-int

80

and x-int 0?

[copper void]

yes for the domain it will start on
[0,

now to find the other side

[merry girder]

idk it kinda continues

[copper void]

do you remember how to find the focal width of a parabola

[merry girder]

whats that

i mean

i guess i could just do 200=0+x/2

[merry girder]

400 = x

[copper void]

yepppppp

[merry girder]

but like

the thing is

that only apploes when the other side is 0

right?

i wanna do it if the other side was not 0

also is c [0,30]

?

[copper void]

that's beyond the scope of the question, you have to set it to be 0 since distance is scalar

[copper void]

your height isn't capped at 30

[merry girder]

huh

[copper void]

the vertex is (200, 30), meaning the absolute minimum is 30

[merry girder]

ohh

[copper void]

( since it's facing upward )

[merry girder]

wait it starts from 30 then

whoops

right?

[copper void]

mhm

[merry girder]

is it[30,80]

[copper void]

i believe so

[merry girder]

for

g is it 0 and 400

wait no

nvm its meant to be y vale

[copper void]

i'm unsure if it is asking for a coordinate or an actual measurment for distance

[merry girder]

on th answer it says 30 for f and 80 for g

[copper void]

then yeah it's the x and y value of the coordinate point

f) the closest distance from the cable to the base of the bridge

would be on the vertex as it would be the lowest point

as for g
g) the greatest distance is at the highest point

be wary though if they gave the bridge a line for example or a point you'd most likely have to use the distance formula to calculate it since you'd have 2 points

[copper void]

does that make sense?

[merry girder]

yeah

[copper void]

[merry girder]

i got another question if thats fine

[copper void]

i can try

[merry girder]

question c

wait nvm

[copper void]

you already plugged it in

right

[merry girder]

its 2 lol

[copper void]

bingo

[merry girder]

trial and error lets go

[copper void]

1/9 = 3^-t

[merry girder]

lol i havent done log yet so

i have to use trial and error

[copper void]

i was abt to bring logs up if u didnt know

but yeah

aight

@merry girder im gonna go sleep, if you're done w/ the channel or still need help keep it open

if not please type .close 😊

[merry girder]

yah

tysm

.close

[obtuse pebble]

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[prime axle]

how would I solve this

x is between 0 and 2pi

I notice sin2x can be rewritten as

2sinxcosx

so I currently have

2sinxcosx+cosx-2sinx-1=0

[upbeat fulcrum]

2sinxcosx + cosx = 1 + 2sinx
cosx(1 + 2sinx) = 1 + 2sinx
cos x = 1
x = 0 ?

[crude coral]

i dont think u have to find x

i think u have prove that all of that is = 0

[upbeat fulcrum]

ooh

[prime axle]

well yes

but that's how you find x

right?

[crude coral]

what is the question

[upbeat fulcrum]

I guess

[crude coral]

is it Prove : OR is it Solve for x

[prime axle]

solve for x

[crude coral]

then that is the answer

give complete information or else it is very hard for helpers to help

[upbeat fulcrum]

agreed

[prime axle]

that is the information though

solve for x

and that's the question

where x is between 0 and 2 pi or equal to 0 or 2 pi

[crude coral]

this is ur answer, which btw is againt the rules of this server

[upbeat fulcrum]

ah?

[crude coral]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upbeat fulcrum]

Am I not supposed to do that?

oof, sorry for that

Kinda new here

[crude coral]

yeah but i think its ok , the guy already knew sin2x = 2sinxcosx

[upbeat fulcrum]

I c

[crude coral]

u helped them with factoring and canceling

[prime axle]

but there's more to it no?

[crude coral]

no

[shut lagoon]

Just quickly, doesn't this assume 1+2sin(x) != 0?

[crude coral]

oh yeh thats wrong then

[prime axle]

Because the answers say 0, 7pi/6, 11pi/6 and 2pi

[crude coral]

the answer the guy gave was wrong

[upbeat fulcrum]

oh right that's true

[crude coral]

2sinxcosx + cosx - 1 - 2sinx = 0
cosx(1+2sinx)-1(1+2sinx)=0
cosx-1 = 0 OR 1+2sinx=0

[upbeat fulcrum]

err... how come?

[crude coral]

solve both the cases

[upbeat fulcrum]

i c

[shut lagoon]

By going to the cos(x) = 1 step, you divided by 1+2sin(x).

[prime axle]

Ohhhhh Ok

thanks

I was just confused with how I could factorise it

so (cosx-1)(2sinx+1)=0

and that's easy to solve for then

ok

thanks

.close

[obtuse pebble]

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[red swallow]

Need help with this

[shut lagoon]

What is 2^0?

[red swallow]

1

[shut lagoon]

What is 2^1?

[red swallow]

wait what is ^

[shut lagoon]

Exponent

You just have to plug it in.

[red swallow]

oh ok

[shut lagoon]

2^0, 2^1, 2^2, 2^-1, ...

[red swallow]

ok

Did I do it right

[crude coral]

all is correct

except e is only 0.5

not -0.5

[red swallow]

oh ok thanks

.close

[obtuse pebble]

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[balmy sand]

pls let me know if any of these are incorrect

[obtuse pebble]

@balmy sand Has your question been resolved?

[balmy sand]

<@&286206848099549185>

[obtuse pebble]

@balmy sand Has your question been resolved?

[obtuse pebble]

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[balmy sand]

pls let me know if any of these are incorrect

[obtuse pebble]

@balmy sand Has your question been resolved?

[obtuse pebble]

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[amber oak]

I think this is the repeated squaring method, but I'm currently stuck because I don't fully understand what my teacher meant with this

[amber oak]

How do you go from 16, 13 to nothing

and why can you use fermat's little method on 9 and 3 to make it 10?

[pseudo swift]

it's not FLT

it's just straight up multiplication

9*3 = 27 = 10 mod 17

[amber oak]

oh wow, ofcourse

[pseudo swift]

and 16*13 = 4 mod 17, that's what they did

it didn't disappear

[amber oak]

thank you, that makes a lot more sense. But how did he go from 16 11, 17 to -11?

[drifting wraith]

16 = −1

[pseudo swift]

and 10*13 = 11 mod 17 it seems

[drifting wraith]

it's solved in two ways at the same time for some reason

[pseudo swift]

¯_(ツ)_/¯

[amber oak]

Yeah, it really confused me

but thank you both, that clears it up a bunch

[drifting wraith]

they get 16×11, and also 4×10

[amber oak]

.close

[obtuse pebble]

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[spice plinth]

Guys help me in this

[raven solar]

use this information

[spice plinth]

So equations with Pythagoras?

DLM and DLO?

[raven solar]

remember this is a circle

AO=DO

so DLO

DLM is a long route since DM length is still unknown

@spice plinth

[spice plinth]

Oh right.....

Thanks

[raven solar]

you need to show O is the center tho

but that is easy

[spice plinth]

Intersection of diameters

One more thing

We don't know ratio of DH to DF

[raven solar]

you dont need it

since O is center

then NO is radius

radius = Diameter/2 = 1.5

NL is 1

so LO is???

[spice plinth]

Yes but we need to find DH

[spice plinth]

0.5

[raven solar]

DO is?

[spice plinth]

DF is 2 root 2

[spice plinth]

3

[raven solar]

DO is 1.5

[spice plinth]

Yep

[raven solar]

then DL is?

[spice plinth]

Sry

Root 2

Now we need ratio?

[raven solar]

(DO)^2 - (LO)^2= (DL)^2

use pythagorean to solve DL

[spice plinth]

It is root 2 I have solved it but we need to find DH

[raven solar]

now EO = HL

by perperndicularity

EO = HL =0.5

[spice plinth]

Ok

Oh right

Got it

.close

[obtuse pebble]

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[obtuse pebble]

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[gleaming crypt]

There is a polygon with n sides, where each side having the length x. There is an ant at each vertex, and each ant will keep walking directly to the ant clockwise of them. Find in terms of n and x, the total distance an ant will walk before reaching the centre.

[gleaming crypt]

How do I continue this, I tried drawing a triangle and square and doing small movements and it creates a spiralling towards the centre

drawing an outline with the ants as the vertexs creates a smaller polygon, repeating this process, you can see its the overall larger polygon getting smaller and smaller and spiraling clockwise to the centre

[timber island]

directly to the ant clockwise of them..doesnt that mean it will move always on the perimeter?

[latent walrus]

the other ant is also moving - so they all kinda curve as they move around - leading to a spiralling effect

[gleaming crypt]

when one moves slightly forward, the one behind it wont be moving straight

[scarlet gale]

No, that would be going along the perimeter rather than directly.

[timber island]

oh right, all of them are moving

[scarlet gale]

They also meet in the center, so they're not staying on the perimeter.

[timber island]

yeah makes sense

well one ant has to be the first

[gleaming crypt]

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

@gleaming crypt Has your question been resolved?

[cyan solar]

I think is x

No matter of n

[gleaming crypt]

its not just x, I thougt that too

[cyan solar]

The vector of velocity (v) is always pointed to the other ant neighbor

[gleaming crypt]

we can tell its not x by thinking as n approaches infinty

it approaches a circle

and the inward spiral will definitely have a length longer than x

as x is apporaching 0 when n approaches infinity

[cyan solar]

Yeah

[gleaming crypt]

so any clue on how to proceed?

[cyan solar]

And the radial velocity = v

[gleaming crypt]

we knwo that the relative angle is always fixed

how do we get the distance out of that tho?

[cyan solar]

Yeah, my bad. Yesterday I solved similar problem but it was with square

[gleaming crypt]

this is similar

but its for any polygon

so the generalisation of a square

[cyan solar]

There the vectors of velocity are perpendicular

[gleaming crypt]

we have the angle since its 180-360/n

<@&286206848099549185>

[cyan solar]

If their velocities are equal the shape will be the same in every time

[gleaming crypt]

yes

but how do I get a distance

[regal tundra]

Can you find the time take for them to meet each other by using relative velocity?

[gleaming crypt]

how?

that's a curve

[regal tundra]

Ok better way:

Find the component of velocity towards the centre of the polygon

[gleaming crypt]

there isnt

[regal tundra]

?

[gleaming crypt]

as a whole or a single ant?

[regal tundra]

Single ant

[gleaming crypt]

the ant moves directly forward

how's there any centripetal force

[covert jackal]

will i get ban if i ask question if the help is not my channel

[regal tundra]

Yea but there's a component of its velocity towards the centrw

[covert jackal]

🤔

[gleaming crypt]

if it helps my question, I dont mind

[covert jackal]

it doesnt but its really quick

the instruction is to list the constant term of the polynomial

if there is no constant term do i write
"none" or "0"🤔😞

[gleaming crypt]

0 is the constant then

[gleaming crypt]

I see, but how do I find out the magnetude of that?

[regal tundra]

You get this?

[gleaming crypt]

no...

wait I see how it's working

ish

[regal tundra]

This is a slice of the polygon

[gleaming crypt]

ye

what's the vcos

[regal tundra]

And the angle at the centre will be 2π/n

[gleaming crypt]

oh u changed it to parallel and perpendicular

[regal tundra]

Yes

Now can you imagine the ants will retain the shape of the polygon while they move towards each other?

[gleaming crypt]

yes got that part

[regal tundra]

And the polygon will always be a regular polygon

[gleaming crypt]

but dont know how to get the overall distance travlled

yes

[regal tundra]

We are getting there

So the component of velocity of an ant towards the centre of the polygon will always be the same ?

[gleaming crypt]

yes

[regal tundra]

The magnitude of it

[gleaming crypt]

oh ye, the magnitude

but focusing on one ant, it should be the same too

[regal tundra]

I am focusing on a single ant

As the ants move the polygon will get smaller while it rotates

[gleaming crypt]

ye

[regal tundra]

So the magnitude of velocity of the ant towards the centre is constant and the other component of the velocity is perpendicular so it doesn't interfere yes?

[gleaming crypt]

yes

perpendicular velocities dont interact in a sense

[regal tundra]

Yes

[twin pasture]

hi

[gleaming crypt]

hi

[twin pasture]

is it the ant question in moving bowl?

[gleaming crypt]

no

a polygon

[twin pasture]

then?

[gleaming crypt]

#help-10 message

[regal tundra]

So isnt the time taken to reach center = distance of centre from the ant divided by the velocity of ant towards the centre?

[gleaming crypt]

no

its a curve not straightline

[regal tundra]

Didn't we just establish the velocity towards the centre is the only thing that matters

Like that's what I've been trying to tell you all along

[gleaming crypt]

oh

but it just doesnt seem like its that

but does make sense sicne its the only force

going towards the centre

and the otehr is perpendicular

[regal tundra]

The other component is perpendicular so it doesn't interfere with us and velocity towards the centre isn't changing

[gleaming crypt]

but how do we find the velocity of the ant?

[regal tundra]

You don't need to

Just let it be as v

It will cancel out when we are finding the distance

[gleaming crypt]

ok

[cyan solar]

ant will travel x/(1+cos(360/x))

i think

[gleaming crypt]

where does that just pop out from

[regal tundra]

So as you can see from the diagram our velocity towards the centre is v sin(π/n)

[gleaming crypt]

yes

[regal tundra]

And can you use a bit of geometry to find the distance of vertex to the centre?

[gleaming crypt]

distance is $\frac{x/2}{sin(\frac{\pi}{n}}$

[warm shale]

龘

[regal tundra]

Yes

[gleaming crypt]

so time is x/2

[regal tundra]

Divide properly bro

[gleaming crypt]

oh v

s/2v

x/2v

[regal tundra]

It's x/2vsin²(π/n)

[gleaming crypt]

... yea

thought the two cancelled, didnt do that right

[regal tundra]

Yes

Now you have time of traveling

And the ant always had net speed as v

Not talking about the centre anymore, im talking about the resultant speed

[gleaming crypt]

yes

[regal tundra]

It was always v towards the other ant

[gleaming crypt]

so distance will just be velocity * time

[regal tundra]

Yes

[gleaming crypt]

$\frac{x}{2sin^2(\pi/n)}$

[warm shale]

龘

[regal tundra]

Yep

[cyan solar]

why

[regal tundra]

Good job bro, i was thinking how was I gonna explain all that to someone but you got it quite easily

[regal tundra]

What why

[gleaming crypt]

knew the stuff needed, but just had no idea how to formulate it all

thxs

.close

[obtuse pebble]

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[plush hill]

Context: the class is called topics in algebra. We have been talking about groups

[plush hill]

Question: a b c and x are elements of a group G. X^2a = bxc^-1 and acx = xac. Express c in terms of a b c and their inverses

[frigid gale]

#help-10 If a, b, and c are consecutive integers such that a < b < c, which of the following must be true?
A) b is even
B) c is odd
C) a + b + c is even
D) b – a is odd
E) b + c is even

[lilac jewel]

x is a variable yes?

[plush hill]

I guess technially, but we don't use typical inverse properties

[twin pasture]

a,b,c,d is answer

[plush hill]

Well your options are either odd even odd or even odd even. Apply both scenarios to see what is true for both

[frigid gale]

thanks

[lilac jewel]

Why are you guys using this channel?

[plush hill]

Idk. I was here first

[twin pasture]

do u need expla?

[pine barn]

Who's question are we answering?

Walrus or soap

[lilac jewel]

No clue

MasterWalrus ig

[plush hill]

Just some hint or clarification. I really don't understand what it is asking

[lilac jewel]

So acx = xca is useless

[plush hill]

It helps provide idea of a abelian ig

[lilac jewel]

What is an abelian

[plush hill]

A group that us commutative

[lilac jewel]

It's literally useless in this question

[plush hill]

Alright then

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dull rune]

wait, are you solving for x or c?

[plush hill]

Solving for x

[lilac jewel]

The question is weird in itself

What in the

b/ac i think

[dull rune]

solving for x makes more sense, yea

[lilac jewel]

Wait why did I give out the answer immediately

[dull rune]

idk man?

X^2a = bxc^-1 i'd try to deal with this equation first

specifically, get rid of the c^-1

[plush hill]

I think I get an idea now. Ty for the help

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[obtuse pebble]

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[vestal crest]

are those 2 things same?

[vestal crest]

my teacher today told us that those 2 are different , and ive been trying both methods on same examples and i always get the same answer

[royal basin]

they look the same as written by you

are you sure that you did not miscopy?

[vestal crest]

no, let me write down how she told us it goes

[royal basin]

so your teacher specifically told you that the expressions $\sqrt[n]{a \sqrt[m]{a}} = \sqrt[n]{a \cdot \sqrt[m]{a}}$ are unequal?

[warm shale]

AnnGhost

[royal basin]

or what

[vestal crest]

and so she gave us an example, did it both ways and the both ways gave same answer and she said that it was coincidence

[vestal crest]

yes

[royal basin]

that is strange

putting things next to each other means multiplication

[timid silo]

ann got de-honorabled?

[royal basin]

yes i did next question

[timid silo]

gg

[vestal crest]

she said that in the left one u can drag a into this

and in right u cant

[timid silo]

pochemy

[royal basin]

bullshit lmao

[vestal crest]

yeah i did like 4-5 examples and they were all same answers

so i was just asking to be sure

because i always thought that if a is next to some sqrt, that it means a*sqrt

[royal basin]

it does

either your teacher is on some bullshit or you misunderstood something

[vestal crest]

ill ask her next class, i am sure she said it like this because she used the different methods

thank you tho

appreciate it

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[green pagoda]

someone help me solve this

[twilit meteor]

use cosine rule?

idk

[obtuse pebble]

@green pagoda Has your question been resolved?

[green pagoda]

<@&286206848099549185>

no helpers???

bad experience omg

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[tribal tide]

How can i solve the elacity of this function?

<@&286206848099549185>

[sand bough]

.close

[tribal tide]

?

what

<@&286206848099549185>

[stark nebula]

Start with the formula you would use

I’m assuming it’s the elastic energy formula

[tribal tide]

its divirident i think

[stark nebula]

the left side looks like an integration factor

[tribal tide]

okay i can try

what will i be starting with Elx f(x)?

[stark nebula]

Sorry multitasking

What’s the original equation

[obtuse pebble]

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[torpid locust]

how can i make a graph doing that? ive been trying for 10 minutes

[brazen gorge]

a graph doing what? these are just inequalities

[torpid locust]

sum like that, the lowest in this example is "a"

[torpid locust]

here the lowest is c

[obtuse pebble]

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[latent frost]

a chord (?) in the unit circle is seen from a point in the circle with the degree of 120, decide the length of the chord

[royal basin]

ok so the circle has radius 1

[latent frost]

so the way I did this originally is most likely not "correct", so I'd like some clarification on if it's ok ro not

[royal basin]

ok show us

[latent frost]

so i'm using c^2 = a^2 + b^2 - 2bccos(alpha)

and the idea I have is that the beta must be 240 degrees since inscribed angle says y = 2x

but I think it's not applicable the way I did it

[royal basin]

wym

yeah that big arc is 240 degrees you're right

thus alpha is 120

[latent frost]

all the images online makes it seems as if the only way I can do it is if it's a triangle within another triangle..

although this one does show that it can be to the left as well

so that approach is ok for calculus then?

[royal basin]

do you mean calculation

[latent frost]

no like that level of math

[royal basin]

uhh

well i dont see why it would be bad in a calculus context so sure

[latent frost]

are there more ways to get the angle of alpha?

[obtuse pebble]

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[latent frost]

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[bright inlet]

i need help with this question, i'll translate it in the next message

[bright inlet]

consider that x is a real number, how many positive integers exist such that they can't be written as floor(x^2) + floor(x) while being smaller than 2023

the part i don't understand is that you can write any real number into the floor function

is there a really really special case that i don't know

[obtuse pebble]

@bright inlet Has your question been resolved?

[bright inlet]

womp womp

[obtuse pebble]

@bright inlet Has your question been resolved?

[bright inlet]

bleh

maybe later

[obtuse pebble]

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[ember oar]

In this graph, why is the interval 0<x<5 considered negative?

[ember oar]

The function is x/x-5

There is an asymptope at 5

Between 0 and asym. 5, the sign of f(x) is considered negative

According to my book

[neon vector]

Yes

The graph is under the x -axis

[ember oar]

Huh ok that makes sense I thought it meant something esle thank you

[neon vector]

Well

Both can't be the output

😄

[neon vector]

The sign of a function just talks about whether the function is positive or negative

It's positive when you get positive values for your inputs (or the graph is above the x -axis), and negative when you get negative values for your inputs (the graph is under the x -axis).