help-17

[lyric relic]

np

[grand linden]

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[wheat falcon]

Hello !
how to find derivative of this function at x=0:
Thanks !

[lyric relic]

this function is even

so if the derivative exists then it must be zero

[wheat falcon]

woooow
so how to prove that it exists ?

[lyric relic]

if by any luck the exercise is assuming that it is differentiable ..then you hav already done

[wheat falcon]

it doesnt 😦
and i couldnt reach any conclusion using derivative definition

[vast shale]

you can watch the derivative at x=0- and x=0+

[rough patrol]

okay um use l'hospitals rule to calculate derivative of $e^{-\frac{1}{x^2}}$

[twin meteor]

Arnavutköy

[wheat falcon]

It just goes around a loop

[rough patrol]

oh whoops

oh wait l'hospital rule helps you find limit

not derivative im trolling

[lyric relic]

no beacuse you do knothe derivative for x not equal zero

so you can do the limit

[rough patrol]

simply calculate the derivative of e^{-1/x^2} via regular derivation rules

[wheat falcon]

mhmmm

i will think about it thanks !

i will close this for now

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[lethal heart]

Consider an orthonormal reference system in space ( S ). Two planes are given, respectively:

[
\pi_h : hx - hy - (2h + 1)z + 2 = 0
]
[
\pi' : 2x - y + 3z + 1 = 0
]

where ( h \in \mathbb{R} ) is a parameter.

(a) Determine for which values of ( h ) the two planes are perpendicular.

(b) Given the value of ( h ) from the previous point, let ( r ) be the line of intersection of the two planes. Determine a skew line ( s ) with respect to the line ( r ).

[twin meteor]

jandro0103

[lethal heart]

I don't know how to do the point b)

a) h=-1

Pls help

[vocal sleet]

@lethal heart Has your question been resolved?

[lethal heart]

<@&286206848099549185>

[vast shale]

wdy m :"Determine a skew line ( s ) with respect to the line ( r )" ?

[hexed needle]

hint: any line perpendicular to either plane that doesn't intersect r is a skew line of r

[lethal heart]

Can we go through it?

[hexed needle]

no, skew lines never touch

[grand linden]

i think he means to do the point with you

[hexed needle]

oh lmao you're probably right

[lethal heart]

Yes

[hexed needle]

yeah sure

what did you get for r?

[grand linden]

i think to find r you have to solve this

[twin meteor]

jandro0103

[grand linden]

but im not sure

[lethal heart]

Mmm

@hexed needle I don't know how to get it

[hexed needle]

it's what jandro said, just solve the system of equations

[grand linden]

you can sum the equation !

[hexed needle]

since it has 3 unknowns and 2 Equations, you'll end up with a line

[lethal heart]

Okay !

Lemme see

[twin meteor]

Task Bot

[grand linden]

so x = -4z - 3

[lethal heart]

$x=-4z-3$

[grand linden]

now you sub this into (1) or (2)

[twin meteor]

Task Bot

Task Bot

[lethal heart]

@hexed needle

[hexed needle]

do you understand how to get x and y here?

[twin meteor]

jandro0103

[hexed needle]

jandro seems to want to help so I'm gonna let them take over so as not to crowd the channel. if either of you get stuck you can ask me lol

[lethal heart]

@hexed needle But what should I do now?

[grand linden]

im not sure how to continue here ... maybe z = t, but i dont know how to explain why

[hexed needle]

oh, alright. well that would be correct. the set of Equations jandro wrote down do indeed form a line, but it's easier to see if you set z = t. then,

x = -4t - 3
y = -5t - 5
z = t

that's your solution to the system

so that's the line that intersects the two planes

and t is some number. if you plug anything into t, you get a point on the line

[lethal heart]

And what should I do now?

[hexed needle]

you need to find a point on one of the planes that doesn't touch r

how would you go about doing that?

[lethal heart]

Mmm

[hexed needle]

isn't that what I said?

[lethal heart]

But do I have to try things out?

[hexed needle]

yeah trial and error basically. if you look at the first equation in the problem for example, (0,0,-2) is in that plane. is that point in r?

[lethal heart]

What

I got 8=3, 10=6 and t=-2

The point belongs to the plane

@hexed needle

[hexed needle]

what did you get these from?

[grand linden]

but does the point have to belong to a single plane, not both right @hexed needle ?

[lethal heart]

In r I substituted coordinates for x,y,z

[hexed needle]

which point did you substitute?

[lethal heart]

(0,0,-2)

[hexed needle]

okay, so you should have

0 = -4t - 3
0 = -5t - 5
-2 = t

[lethal heart]

Yes

[hexed needle]

and if you solve for t you get different stuff

so clearly that point doesn't work

[lethal heart]

Oh so I don't have to put -2 in the other equations?

[hexed needle]

the plane Equations?

[lethal heart]

Here

t=-2

[hexed needle]

oh you can sure

but you'll get nonsense lol, which it seems like you got

[lethal heart]

I got 8=3, 10=6 and t=-2

[hexed needle]

that means (0,0,-2) is not part of the line, since you get nonsense if you plug it in

[twin meteor]

jandro0103

jandro0103

[grand linden]

Right?

[hexed needle]

as long as s is not parallel with r, yes

[lethal heart]

And now ?

What should I do?

[hexed needle]

well now you have a point on one of the planes that doesn't touch r.

you need to make a line going through that point that isn't parallel with r

then it is a skew line with respect to r

[grand linden]

so we just need to take a different direction

[hexed needle]

the easiest way to do that is to make it normal to the plane

[grand linden]

d = (1,1,1)

[hexed needle]

do you know how to do that @lethal heart ?

[lethal heart]

I can't use vectors

[hexed needle]

why not?

[lethal heart]

I didn't do them

[hexed needle]

like you haven't learned them yet?

[lethal heart]

Yes

[hexed needle]

oh, well that's fine. you just need to make sure that whatever line you create doesn't intersect r

if you take the line

x = -4t - 2
y = -5t + 1
z = t + 3

for example, that is parallel to r

since the "slope" is the same for each component

so you just need to change the "slope", then make sure it doesn't intersect r

does that make sense?

[lethal heart]

Yes

[hexed needle]

this is ignoring the whole "point on the plane" thing, i was doing that bc i thought this involved vectors

what i just said is actually easier anyway

[grand linden]

but since the direction of r is d = (-4,-5,1), isnt enough to take a different direction like d = (3,3,3) ???

[hexed needle]

that's what I'm saying, yes...

[hexed needle]

but you need to ensure they don't intersect since they could still intersect at a single point

even if the direction is different

[lethal heart]

Mmm

And how is this line found?

[hexed needle]

well easiest first step is to change the slope.

take your line r and make all of the t coefficient different

[lethal heart]

Ok

[hexed needle]

that changes the direction of the line so it isn't parallel to r

[grand linden]

[lethal heart]

x=-5t-3, y=-7t-6, z=2t

[hexed needle]

sure, that works

[lethal heart]

I finished ?

[hexed needle]

not yet

how do you make sure the line doesn't touch r

[lethal heart]

I have to equalize the parameters and a false relationship must emerge

[hexed needle]

👍

[lethal heart]

Mmm

2t=t -> 2=0

[hexed needle]

there ya go lol

so that does work

well no

that says t = 0

so plug t = 0 into the x and y to see what happens

[lethal heart]

But t is simplified

If t =0 -> 0=0

[hexed needle]

if 2*t = t, the only way that can happen is if t = 0

[lethal heart]

.

[hexed needle]

so t must be 0

[lethal heart]

But in this way a true equality emerges

[hexed needle]

that's only for the z

if x or y fail to be equal, it's still a problem

[lethal heart]

Okay

t=0

All true equalities

So no

[hexed needle]

that's not true

[lethal heart]

Mmm

Why

[hexed needle]

if you plug t = 0 into r, what do you get

for x

[lethal heart]

-3=-3

[twin meteor]

jandro0103

[hexed needle]

well it works for x, I meant y sorry

[grand linden]

you have to use this to build s

[lethal heart]

-6=-6

[hexed needle]

the line r has y = -5t - 5

[lethal heart]

-5t-6=-7t-6

Plug t=0

-6=-6

Maybe I misunderstood

[hexed needle]

.

look at the Equations for r again

[lethal heart]

I saw it

[hexed needle]

why did you use -5t - 6 for y?

[lethal heart]

Ops

Sorry

-5=-5

x=-5t-3, y=-7t-5, z=2t

Then it was wrong here too, now I've corrected it

I think

[hexed needle]

oh I see

well then if that's your line, it wouldn't work since they intersect at t = 0

so you need to change the intercept too

[lethal heart]

Ah

but of all the components?

[hexed needle]

you only need one of them to be different

then they won't intersect at t = 0

just change the x intercept for example

then you're done

[lethal heart]

r:{ x=-4t-3 , y=-5t-5, z=t . r':{x=-5t-5, y=-7t-5, z=2t

[hexed needle]

there you go, that works

[lethal heart]

Now to see if they intersect

I didn't understand how to do it

Maybe

t=2t -> t=0

[hexed needle]

in order to intersect, the x, y, and z coords all need to be equal for some t

[hexed needle]

yes, true

[lethal heart]

-5=-5, -3=-5

For y and x

So they don't intersect?

[hexed needle]

exactly

and they're not parallel since they have different directions

so it is a skew line with respect to r

[lethal heart]

r':{x=-5t-5, y=-7t-5, z=2t

This right ?

(s)

Is this ?

[lethal heart]

Oh okay

Nice !!!

[hexed needle]

👍

[lethal heart]

Thanks very much

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[lethal heart]

.reopen

[vocal sleet]

✅

[lethal heart]

@hexed needle But what was the point of finding the point P(0,0,-2)?

[hexed needle]

that was only done bc I thought you were allowed to use vectors lol. it's an entirely different method

[lethal heart]

Aaaaaah

Clear

Thanks again

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[sacred frigate]

I have this problem, would someone be able to help?

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[fervent wasp]

bro, it's the same idea

[pliant pivot]

I already tried 7/3x

isnt that right

[fervent wasp]

no, how did you arrive at that?

[pliant pivot]

in class we learned just input the y= into desmos and count based off slope

and I got 7 over 3

[fervent wasp]

Can you elaborate the details?

[untold stone]

you can just tell from the equation

[pliant pivot]

how

[pliant pivot]

wdym

[untold stone]

the x coefficient in a linear equation is the slope

if you have y = mx + b

m is the slope

[pliant pivot]

IM SO FUCKING STUPID GODDAMN

[fervent wasp]

#help-26 message

[pliant pivot]

so its -6/7x?

[untold stone]

yes

no

[pliant pivot]

my bad

oh

[untold stone]

the slope is a number

not the whole thing

[pliant pivot]

lemme solve this rq

[vast shale]

Hi

[pliant pivot]

ok im good thanks.

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[green crane]

A 3.1 kg ball moving west to east struck a 2.0 kg stationary ball. If the 3.1 kg ball slowed to
1.2 m/s and was deflected to 25º north of east and the 2.0 kg ball was deflected to 48º south of
east with a velocity of 1.06 m/s. What was the original velocity of the 3.1 kg ball? (4 marks)

[green crane]

so i know momentum before = momentum after

momentum of ball + stationarry ball = 1st ball + 2nd ball

v = inital speed of 3.1kg ball
3.1xv+2x0= before

and the first ball is went out in like a triangle like this

so the momentum of the first ball after is 3.72NS horizontally

oh i got my trig wrong somehow

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[green crane]

i forgot to use

degress

instead of rads

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[fathom gyro]

anyone

<@&286206848099549185>

[flat whale]

You should just show your work

[fathom gyro]

ok

look

3 1/4 is 3.25 meters per sceond

x 60

is 195

then 2 3/4 is 2.75 x60 is 165

and in one lap they cross and 195/165 = 1.18

wait nvm

ok look

you do 195 x24 to find total distance ran

then divide by 180

to find laps ran

which is

26

and then (165 x24) / 180 = 22

then they passed 4 times

<@&286206848099549185>

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[fathom gyro]

bro

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[karmic glen]

So I get that D and E are wrong since those are not concave down

Just not sure how to use f'(3) = 2

I used MVT, but it doesnt tell us the value of c where the derivative is 2

We just know the average rate of change is 2

<@&286206848099549185>

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[low haven]

show that x^3 - x + 1 is irreducible in Z/3Z[x] (this part i already did) and find all primitive elements of the finite field Z/3Z[x]/(x^3-x+1), which i need help

[half imp]

primitive element is one with multiplicative order 26?

[low haven]

yes

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[warped osprey]

Hi for sketching graphs, how do you know if there is a vertical tangent at a certain x value?

[trim trout]

do you have a specific graph in mind you're talking about

or a specific type of function

[warped osprey]

17c is the part i got stuck on

and this was what the graph looked like

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[warped osprey]

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[cursive turret]

so you need help with "show the function is odd"?

[vast shale]

a)f'(x)=x^2/sqrt(x^2-1)+sqrt(x^2-1)

[vast shale]

c)sqrt(x^2-1) is even function bc x^2 and sqrt function are even
but x is odd and odd×even is odd

[warped osprey]

Nah I need help with determining the shape of the graph in general

Mostly how do I know if there is a vertical tangent at 1 and -1

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[final magnet]

which one is the one corresponding to a?, the (-1)n is throwing me off,

[final magnet]

oh wait

it's -x4

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[patent crag]

I don't understand this example and its solution, could someone explain the problem for me? How can -3>0?

[marsh charm]

where do you see -3 > 0

i could just be blind

[twin horizon]

-3x<0 is what I saw.

[vast shale]

its a rule in inequalites basically if x > -y then -x <y

so 3x> 0 ==> -3x<0

[patent crag]

Oh shit its -3x<0

[vast shale]

f(x) = 2-3x from R->R+
now y = 2-3x ==> x = (2-y)/3 (Rearrange)

Now x>0 ==> 2-y/3>0
2-y>0
2>y
Hence range = (-inf,2)

when you want to find range that the function as y

and bring x in terms of y

and then solve like you do for domain of a function

[patent crag]

oh so ur transferring the negative to the lhs? I didn't learn the quadratic inequalities

[vast shale]

no

when you change signs just flip the inequality

eg: 2>1

-2<-1

[patent crag]

ok so the signs change all together?

[vast shale]

yes

just change the sign

and make sure to change the inequality sign

[patent crag]

Understood, to find range, we need to bring it in terms of x and to find domain, it needs to be in terms of y right?

[patent crag]

alright thank you

[vast shale]

yes

[patent crag]

wait I have another question

Sorry to take ur time, could you help me understand what f(x) = |x-3| means? I'm not sure what these brackets || mean

[vast shale]

ok so

|any value| is the modulus of that value

y = |x| is defined as y = { x when x>0, 0,x=0 , -x when x<0

hence it is similar to absolute value

another way to define |x| is the distance of x from the origin

[vast shale]

we have f(x) = |x-3|

f(x) = {x-3 when (x-3)>0, 0 when x-3 = 0 , -(x-3) when (x-3)<0

[patent crag]

wait what is modulus? Is it just a positive number of a number, like modulus of 5 is 5 and modulus of -5 is 5?

[vast shale]

not exactly

[vast shale]

so ur function can be written as f(x) = {x-3, x>-3, 0, x=-3, -(x-3), x<-3

i hope its clear now

[patent crag]

Ohh OK I understand now thank you!

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[bitter pilot]

Let $G$ be an abelian group with neutral element $e.\$
Let $a,b \in G$ with orders $\operatorname{ord}(a) = m$, $\operatorname{ord}(b) = n.\$
Show [ [a] \cap [b] = {e} \Rightarrow \abs{U} = mn. ]
$\textbf{Proof.}\$
Let $u_1,u_2 \in U.\$
Then for $i_1,i_2,j_1,j_2 \in \mathbb{Z}$ with $0 \leq i_1,i_2 \leq m-1, : 0 \leq j_1,j_2 \leq n-1$ there is
[ u_1 = a^{i_1}b^{j_1}, u_2 = a^{i_2}b^{j_2}. ]
Consider
[ u_1 = u_2 \Rightarrow a^{i_1}b^{j_1} = a^{i_2}b^{j_2} \Rightarrow a^{i_1-i_2}b^{j_1-j_2} = e. ]
Because of the assumption $[a] \cap [b] = {e}$ it is $i_1 = i_2$ and $j_1 = j_2$ which implies every element of U is unique.$\$
Therefore $\operatorname{ord}(U) = mn.$

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

Is the proof correct

[vocal sleet]

@bitter pilot Has your question been resolved?

[hard atlas]

what does that assumption have to do with the line before

[bitter pilot]

[ u_1 = u_2 \Rightarrow a^{i_1}b^{j_1} = a^{i_2}b^{j_2} \Rightarrow a^{i_1-i_2}b^{j_1-j_2} = e \Rightarrow a^{i_1-i_2}=b^{j_2-j_1}. ]
They are only equal if both are the neutral element because of $[a] \cap [b] = {e}$.

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

i_1 - i_2 = 0 and j_2 - j_1 = 0

[hard atlas]

why is that the only way for them to be equal?

and why does a^(i1-i2)=e imply i1-i2=0 ?

[bitter pilot]

Because [a] and [b] are distinct sets without e and it is a^0 equals the neutral element

[hard atlas]

I would have liked to hear something like "the left side is an element in [a], the right side is an element of [b], so in total that element is in [a] cap [b]"

is 0 the only solution to a^k= e?

[bitter pilot]

Isn't that obvious

How far does one go into explaining

[bitter pilot]

No

k is a multiple of m

[hard atlas]

no offense but there are some other obvious things you havent seen. so I wasnt sure whether you just claimed what you wanted or whether that was actually the argument you have in mind

[bitter pilot]

So what do I change

I will add this

the left side is an element in [a], the right side is an element of [b], so in total that element is in [a] cap [b]

i_1 - i_2 = 0 mod m
j_1 - j_2 = 0 mod n

[hard atlas]

you havent used 0 <= i1,i2 <= m-1

(with that condition it is obvious to say that the only option is 0)

[bitter pilot]

i_1 = m-1 = i_2 then i_1 + i_2 = 2m-2

What I am trying to say is with the sum

Won't get out of that interval

No

you are right

If i_1 and i_2 are in that interval

i_1 = i_2 = 0

[hard atlas]

yes

then the rest is fine

[bitter pilot]

Then I showed every element of U is unique

Can you give me a hint on
[ \gcd(m,n) = 1 \Rightarrow \abs{U} = mn ]
I started with $V := [a] \cap [b]$ is after some theorem a group and therefore a subgroup of $G.\$
Then $V \subseteq [a], [b]$ it is $V$ is also a subgroup of $[a]$ and $[b]$

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

The intersection of arbitrary many groups is a group

[hard atlas]

let c in V, so c=a^k and c=b^ell and then some playing around with that. and bezout probably also helps

[vocal sleet]

@bitter pilot Has your question been resolved?

[vocal sleet]

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[placid crescent]

I have a question

[placid crescent]

I wanna know if you can reverse the derivatives ?

Like how to reverse a derivative?

[scenic ravine]

integrate

That's an operation

[placid crescent]

I mean like dy/dx converted to dx/dy

Let they equal a/b

[mellow void]

that happens to equal 1/[dy/dx]

dx/dy that is

[placid crescent]

That just doesn't seem right

[mellow void]

well, you know implicit differentiation?

[scenic ravine]

,rotate

[scenic ravine]

!xy

[vocal sleet]

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.

[placid crescent]

I understand now

I got it

[vocal sleet]

@placid crescent Has your question been resolved?

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[elfin sorrel]

i dont know how to do this

[peak matrix]

think about the angles, maybe you will spot some similar triangles...

[graceful peak]

Triangle OMN and OQP are similar

[graceful peak]

just take care of order and implement their properties

[elfin sorrel]

so MO = 19.4 and PO = 6.5?

[elfin sorrel]

is that correct?

[elfin sorrel]

^

[graceful peak]

yeah it is correct

i guess you have approximated PO

[elfin sorrel]

thank u

[vocal sleet]

@elfin sorrel Has your question been resolved?

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[vast shale]

I just asked this question but actually I’m still lost

I’m not understanding when you flip the sign of the answer and when you don’t

[graceful peak]

are u asking when to flip sign of x^2

[vast shale]

For the answer of -6 and 5

[outer warren]

don't reduce it down to sign flipping

first focus on the process

[graceful peak]

you need to find values of x using factorisation
you need to make factors
for eg (x + a).(x + b) = 0

now first consider x + a = 0 which implies x = -a
and next consider x + b = 0 thus x = -b
and x = -a, -b

[outer warren]

the first part is the factorisation / simplification
factoring out the -1 leading to
x^2 + x - 30 = 0
regardless of whether you have that = 0 there or not, the factorisation of the bolded part will still be
(x+6)(x-5)

[vast shale]

Where did the -1 come from

[outer warren]

the original equation they gave had a leading coefficient of -1

its usually easier to work with a positive leading coefficient which is why they factored that out

[vast shale]

-x^2-x+30=0

Implies a -1

[outer warren]

what?

[graceful peak]

see you don't necessarily need to remove that negative sign of x^2

but it is generally done to ease factorisation

simply consider it multiplying LHS and RHS by negative 1

[vast shale]

Okay

So I could leave the problem negative if I wanted

[outer warren]

wdym

[graceful peak]

if you could do factorisation with the negative sign infront of x^2
you can proceed with it

[vast shale]

I wouldn’t have to change the problem to
X^2+x-30=0

I could leave it as

-x^2-x+30=0

Right? Or wrong

[outer warren]

you can if you really want

its just less comfortable to factorise that

[vast shale]

So if I see -x^2 I should reverse the signs at the beginning of the problem to make it easier?

I’ll write down my thought process and take a pic and you tell me where i go wrong.

[graceful peak]

if there is a RHS you can multiply LHS and RHS by -1 and proceed with it

if there is no RHS then you take -1 out from equation like say
-ax^2 + bx + c
-1(ax^2 - bx - c)
and then factorise it but -1 will remain like
-1.(x + alpha)(x + beta)

[vast shale]

Wait

Let me redue that

[graceful peak]

you have made wrong factors

the factors come out to be (x + 6)(x - 5)

[vast shale]

I see the problem

[thick bridge]

yes, bc x² + x - 30 = 0 has the same solutions of x as -(x² + x - 30) = 0

[vast shale]

Now

-6 and 5 don’t get to positive 30

And 2 negatives don’t get to -x

How would you do the problem without factoring -1?

[graceful peak]

like as in -x^2 - x + 30

[vast shale]

Yeah

I don’t see how you do it without -1

[thick bridge]

if x² + x - 30 = 0 in factorised form is (x - 5)(x + 6) = 0,

then -x² -x + 30 = 0 in factorised form is -(x - 5)(x + 6) = 0 or (5 - x)(x + 6) = 0, you're getting mixed with the signs

but the -1 in front technically doesn't matter

[graceful peak]

consider m and n to be factors of the equation

then m + n = -(-1)/(-1)
and m.n = 30/(-1)

[thick bridge]

ur just finding values of x that satisfy the equation to be 0

[vast shale]

Thanks

[vast shale]

Yeah I get it now

[graceful peak]

yeah this is right

[vast shale]

The part that didn’t make sense to me was factoring out -1. I also see you do this if you had something like -3x^2….

.close

[vocal sleet]

Channel closed

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[vocal sleet]

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[craggy hollow]

How do i do that

[civic schooner]

Factor 12d

[craggy hollow]

What does that meaj

[distant shoal]

Do u know how to factor?

[lilac pebble]

what can multiply together to make 12, for example?

[distant shoal]

Wronskian! 👻

[craggy hollow]

6x2

[lilac pebble]

yeah

and what about 24?

[craggy hollow]

12x2

[lilac pebble]

do they share anything in common?

[craggy hollow]

They can be multipled by 2

[lilac pebble]

so you can take 2 out to the front

[craggy hollow]

Wym

[lilac pebble]

2(12d^2-6d)

[craggy hollow]

Ohhh

Wait so

Why do u have ^2 ur doing distrubutive property righ

[lilac pebble]

yeah

$2(12d^2-6d)$

[twin meteor]

Axe

[silk osprey]

brother why are we factoring 2

kevin does 12 divide 24

[lilac pebble]

we can factor more stuff

[craggy hollow]

Yoo i got it

Appreciate ut

So thats the answer

[lilac pebble]

that's not the final answer

[craggy hollow]

Oh shit

[lilac pebble]

they share more factors

[craggy hollow]

Thhen what i do

[silk osprey]

^

[craggy hollow]

Im getting it

[craggy hollow]

Why

[silk osprey]

in other words is 24 divisible by 12

[craggy hollow]

Thas 2

[silk osprey]

because if it is then we can factor it

[silk osprey]

i’m aware but i think it’s clear that you can factor more from the beginning

instead of factoring each prime one by one

[craggy hollow]

Wait wym by that

[lilac pebble]

yeah 24 is 12*2 and 12 is 12*1
so they share 12

[craggy hollow]

Yup

Ik

That

[silk osprey]

so factor out 12

[lilac pebble]

so you can take 12 out to the front

[silk osprey]

then deal with the d after if you can’t spot that immediately

[silk osprey]

nah just go from the start

[lilac pebble]

why is there c?

are you doing a different problem?

[silk osprey]

24d^2 - 12d

[craggy hollow]

Oh shit i was teyna makdena brwxket

[craggy hollow]

Wtf does ^ dis mean

[silk osprey]

to the power of

[craggy hollow]

Ohh

[silk osprey]

d squared

[craggy hollow]

Ok

Oh yes ik whay that is

Ai so

Factor out 12 then what we do after this

[lilac pebble]

what did you get after factoring out 12?

let's make sure it's right

[craggy hollow]

Wdym by factoring out

Like

Dividing?

[lilac pebble]

moving to the front

like i did with the 2

[craggy hollow]

Why did we factor 12

[lilac pebble]

to get a polynomial multiplication expression

[lilac pebble]

yeah that's right

👍

[craggy hollow]

Thats the answer?

[lilac pebble]

there's more we can do

[craggy hollow]

Like would my teach take did

When u mean more like simifly?

[lilac pebble]

my guess is your teacher would want you to do more

we're done with the numbers but

now the d's

d^2 means d*d

so you can see they actually share a d

[craggy hollow]

Yup

Ik wym by tha

[lilac pebble]

and you can move d to the front

there's a d in both 2d^2 and 1d

[craggy hollow]

How would do that tho one is d^2 and one is d^1

[lilac pebble]

because d^2 is d*d

[craggy hollow]

Show the equation

[craggy hollow]

Oh yea

So 12d?

In the beging

[lilac pebble]

12d in front, yeah

$12d(2d-1)$

[twin meteor]

Axe

[craggy hollow]

Whys the quality so ass

[lilac pebble]

good job

[craggy hollow]

So thats final answer?

[lilac pebble]

yeah

[craggy hollow]

Aii appreciate it bro i got this retest

Tmr

It fucked up my grade so badly

96 to a 92

Can i dm u if i need help?

[lilac pebble]

[craggy hollow]

Wha tha mean

[silk osprey]

it means no

!vol

[vocal sleet]

Helpers are just people volunteering their time to help you. Be polite and patient.

[craggy hollow]

Oh shii

Aii thx tho

How do i close if

!close

.close

[vocal sleet]

Channel closed

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[vocal sleet]

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[jagged cargo]

why is it 2.5m/s to the right?

[jagged cargo]

i got 0.5m/s to the right

using conservation of momentum, i have m_1v_1f = m_2v_2f

so thats 1.5 * 0.5 = 1.5 * v_2f

[bronze osprey]

No, 1.5 × 3.0 = 1.5 × 0.5 + 1.5x

Technically there are 4 terms

But 1.5 × 3.0 + 1.5 × 0 on the left hand side

Before and after, two pins

[jagged cargo]

oh

so it's m_1v_1i + m_2v_2i = m_1v_1f + m_2v_2f

oh yeah that does output 2.5m/s

thanks

.close

[vocal sleet]

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[brisk monolith]

can someone help?

[dense eagle]

well we gave you some pointers last time, where did u get up to with those?

[brisk monolith]

i use them

but didn t got anything

[dense eagle]

well let's take mod 3 for instance

what does our thing become?

[brisk monolith]

k^2==2mod3

if n is odd

k^2==0mod3

if n is even

[dense eagle]

and go on

k^2 = 2 (mod 3)

what's the issue here?

[brisk monolith]

7^n=1mod3

[dense eagle]

well as in you should always remember that squares can only take a limited value mod n

so if k is a multiple of 3, what can you say about k^2 mod 3

similarly if k is 1 more than a multiple of 3 etc.

[brisk monolith]

k^2=0mod3 if k is multiple of 3, k^2=1mod3 if k is multiple of 3 +1 and k^2=1mod3 if k is multiple of 3 +2

[dense eagle]

yep

so that tells us k^2 = 2 (mod 3) is impossible

so what can we say about n?

[brisk monolith]

7^n-2^n= multiple of 3 or multiple of 3 +1

7^n is a multiple of 3 +1 for any n

2^n=2^n(mod3)

[dense eagle]

i think ur a little bit confused

we've shown that 7^n - 2^n - 9 = k^2 reduces to k^2 = 2 (mod 3) if n is odd, and k^2 = 0 (mod 3) if n is even

we've said up there that k^2 = 2 (mod 3) is impossible

so what can we say about n?

[brisk monolith]

ohh

n is even

[dense eagle]

yep

i'm going to make the substitution n=2m for now

we now want to know when $7^{2m} - 2^{2m} - 9$ is square

[twin meteor]

LY

[dense eagle]

now i was saying yesterday, we know $7^{2m}$ is square right?

[twin meteor]

LY

[brisk monolith]

yes

(7^m+2^n)(7^m-2^n)=x^2+3^2

[dense eagle]

you could try that, but then we can't factor x^2 + 3^2

(well u could work in some different ring but that's uni and let's try to avoid that)

[brisk monolith]

ok

[dense eagle]

what i was saying last time was this thing is intuitively a little bit smaller than 7^2m

so we expect it to be bigger than (7^m-1)^2 and smaller than (7^m)^2

what's the issue now?

[brisk monolith]

m-1>0 so n>2

?

[dense eagle]

no as in let's suppose we prove that most of the time

(7^m - 1)^2 < k^2 < (7^m)^2

what's the issue?

remember that (7^m - 1) and 7^m are consecutive

so like if i have a number between 7^2 and 8^2, can it be square?

[brisk monolith]

oh

yes, so k^2 is a perfect square between two consecutive perfect squares

so it doesn t exist for m>1

[dense eagle]

so that means there's only one possible value of n which might work, n=2

does n=2 work?

[brisk monolith]

yes

[dense eagle]

so that's the answer

[brisk monolith]

thank you very much!!

[dense eagle]

nw!

[brisk monolith]

^^

[dense eagle]

btw if ur done, type .close or .solved

[brisk monolith]

i reread the solution

and then close it, it is ok?

[dense eagle]

yh

[vocal sleet]

@brisk monolith Has your question been resolved?

Channel closed

Closed by @brisk monolith

Use .reopen if this was a mistake.

[brisk monolith]

.reopen

[vocal sleet]

✅

[brisk monolith]

@dense eagle

i have a question

can we proof that (7^m)^2-2^2m-9 is bigger then (7^m-1)^2 ??

[dense eagle]

wait i thought u proved it

uh you just expand

then stuff happens

[brisk monolith]

nvm

yeah

i just did that

:))

ty

.close

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[vast shale]

i need help with experiment number 2. I don't understand how to answer number 2-4.

[rough patrol]

okay

so basically, when they mean the "existence of a limit" for question number 2

this means that as we plug in x and x gets closer and closer to 1

the value gets closer and closer to - infinity or positive infiinity

[vast shale]

yes

but there's a hole at 1

so the limit can't be 0

i'm very confused with all of this

sorry 😓

[rough patrol]

okay so there is a difference in the "calculation" of a limit, which can be done at any point, and the "existence" of a limit, which implies an asymptote

specifically, the asymptote is the line this function is approaching

when this line is vertical

we say that the x-coordinate represents the limit

so why would you think there might be a limit at x=1?

well, the denominator goes to 0

and the numerator over a very small number might be very big right?

so like, if we got 1 for .9 and then like 100 for .99 and then like 10000 for .999, we see that the number is growing bigger and bigger

this would indicate that there is an asymptote at x=1

because the value just keeps on increasing more and more

but since there is simply a hole, there is no limit in this case

or wait

in this case, sorry there is a limit

[vast shale]

so the limit DNE?

[rough patrol]

i recommend just looking at the graph

the limit simply is x=1

[vast shale]

but why?

[rough patrol]

okay basically, the "slope" of the line as x approaches one is inifnity

think of a limit as in, if we look at the graph of the function when x=1 locally, it basically resembles a "vertical" line

[vast shale]

right

[rough patrol]

so we see in your graph that for a minimal change in x (like .009 or .09, we are seeing a 'lot' of relative change in the y-coordinate

this indicates that the function is nearly vertical at that point

which indicates that it is a limit

does that make sense?

[vast shale]

umm

kinda?

if there's an asymptote there is it just hard to see the change in the graph because it's minute?

like usually when there's an asymptote it's easy to see so because the graph doesn't touch it

but if there's a asymptote at x=1 then why is the graph going through that point?

why not just stop there?

[rough patrol]

it is possible for a graph to go through in asymptote

[vast shale]

it is???

[rough patrol]

yes, although most examples you see don't do that

[vast shale]

okay

[rough patrol]

the point is that the function is almost "tangent" with that asymptote

[vast shale]

right

okay

one more question

so when the question asks for the existence of a limit at x=1 what does that exactly mean?

[civic otter]

You have to make sure that left-sided and right-sided limits at 1 exist and they are equal

[vast shale]

but at one it's undefined

they're aproaching 1 but don't equal it

[civic otter]

If this happens, then the limit at 1 exist

[vast shale]

i'm aware of that

[civic otter]

The function might be undefined at 1, yes

[civic otter]

The limit can't approach something, maybe you mean the function?

[vast shale]

i mean the left and the right hand side

if you look at the data it's approaching 1 but doesn't equal it.

[civic otter]

Yeah that's totally fine

[vast shale]

does that mean the lim x-> 1 = 0?

[civic otter]

Mmh no?!

Could you send the graph/exercise you're referring to?

[vast shale]

.

it's the second one

@civic otter

[civic otter]

Oh then yes

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

Channel closed

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[vast shale]

.reopen

[vocal sleet]

✅

[vast shale]

And using the graph makes more sense to figure this out than a table value right ?

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

[civic otter]

Usually yes, but in this server I've seen people struggling with graphs and preferring tables 🤷‍♂️

[vocal sleet]

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[worldly otter]

<@&286206848099549185> please and thank you 😭 💀

[hot tree]

Please tell me Lexi

[worldly otter]

yes?

[hot tree]

How can I help

[worldly otter]

explaination

for a) specifically

[hot tree]

Ok

[worldly otter]

because my friends say its 8x2 which i think isnt correct

cuz its not a vertical line its slightly slanted towards the side

and yes diameter of a circle inside a square is equal to the side length of square but the circle is outside of the square

[hot tree]

As per the diagram and as per the question just given that circle inscribed inside the square having the radius 8 cm
Which means the total length of one side of a square is 8 x 2 means 16

Actually it is outside but you can see the dots

Mathematics and in engineering if we see those dots it means we just imagine that's structure but don't work on that

[worldly otter]

but the circle is outside the square

i dont think u can use that

[hot tree]

Yes Circle is outside but the margin is negligible that it doesn't matter we can neglect that value

[worldly otter]

what in the physics

[glacial osprey]

uhm

wat

[worldly otter]

scrollup

[hot tree]

Means

[worldly otter]

dude u fr?

🗿

[glacial osprey]

give me a moment

[worldly otter]

i dont think it should be neglected ngl 😭

[drowsy cedar]

we cannot neglect that value 😭

[hot tree]

Yes you can

[drowsy cedar]

howw

and why

[worldly otter]

how?

[glacial osprey]

you really just cannot

[worldly otter]

yea

[glacial osprey]

Calculators are allowed?

[hot tree]

we can neglect and if you want to calculate that value it doesn't even matter because it's so small that when we add this it will just change a fraction of that value

[worldly otter]

duh this isnt 1980s

imagine using a log book counting trigo

[stark berry]

no you cannot neglect it

[worldly otter]

hell

[glacial osprey]

ok but usually questions ask for the exact answer 😭

[worldly otter]

yes

exactly

😭

[hot tree]

Ok then assum it and add .
That won't affect anything

[stark berry]

yes it will?

what are you assuming

[worldly otter]

my answer is 14.78

[stark berry]

why are you assuming anything

[hot tree]

Like pie =3.14 and so on

[worldly otter]

which is like 1.22 off

[worldly otter]

in my country we use 3.142

🗿

[hot tree]

But in my world we take 8 digits after decimal

[glacial osprey]

what

first you say that this smidge of a circle is negligible

but obviously 8 digits after the decimal isnt

[worldly otter]

well this isnt physics

this is maths

and maths wants precision

😭

[stark berry]

show your working so we can check

[hot tree]

I know

Please don't cry 🙏🏻 .
Let me upload

[hot tree]

Wait a moment

Ma'am

[worldly otter]

im male jose

💀

[stark berry]

sin 45/2 is not that

[worldly otter]

why

circle is divided into 8 sectors

360/8=45

[stark berry]

im confused

walk me through your process

what are you trting to do

[worldly otter]

i just use formula thats abt it

ignore what i did and what would u do

🗿

[stark berry]

i would find the perpendicular distance from O to the square

[stark berry]

focussing on this right triangle

[worldly otter]

yea thats what i did

🗿

because idh the vertical line

and bottom line as well

the formula i used was formula for chord in circle

[stark berry]

where did uou get 3.062 from

[vocal sleet]

@worldly otter Has your question been resolved?

[worldly otter]

6.123/2

4sf at least 2dp

[stark berry]

where did 6.123 come from?

range of sin is -1 to 1

[worldly otter]

2(8) sin 45/2

[vocal sleet]

@worldly otter Has your question been resolved?

[mossy atlas]

I got 8*sqrt(2-sqrt(2))*(1+sqrt(2)), which is indeed about 14.78

[vocal sleet]

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[smoky pebble]

perpendicular distance from a eqn is zero to center

[smoky pebble]

then that line is diameter right

is this rigorous enough?

[marsh charm]

are you ai

[smoky pebble]

no?

[outer warren]

!original

[vocal sleet]

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.

[vast shale]

exactly what an AI would say

[vocal sleet]

@smoky pebble Has your question been resolved?

[vocal sleet]

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[young plank]

<@&286206848099549185> find the shortest distance between two points and it should touch a line and let the points be a,b and the let the line be ax+by+c=0.

[smoky pebble]

what this is a general question

not related to question

[outer warren]

the question isn't clear

which is normally an issue when questions are paraphrased which is why i asked

[vocal sleet]

@young plank Has your question been resolved?

[young plank]

Not yet

[stark berry]

not sure what this is tring to say

could you be clearer

what should touch what line

if youre looking for shortest distance its just rhe distance formula

[young plank]

That's why this is the question that is amazing everyone

[stark berry]

$d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$

[twin meteor]

ashy!

[vocal sleet]

@young plank Has your question been resolved?

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
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[pseudo wave]

I need assistance with this.

[calm light]

have you found the components of v already?

[topaz eagle]

you can visualize this with @twin meteor

,tikz
\draw[->] (0,0) -- (220:2);

[twin meteor]

vin100

[topaz eagle]

In the above command, \draw[->] means to draw an arrow. In (220:2), 220 is the angle that the arrow made with the positive x-axis, and 2 is the length of the arrow.

to scale this vector, you can simply mutiply the 2 in (220:2) by a constant, like

[pseudo wave]

1 moment

[topaz eagle]

,tikz
\draw[->] (0,0) -- (220:2);
\draw[->,yellow] (0,0) -- (220:2*2);

[twin meteor]

vin100

[pseudo wave]

I'm just having trouble finding c.

I know that the magnitude of vector v is 2 and since it's magnitude changed to 1/2, c must be a scalar of +/- 1/4.

I just don't know which.

I know that the angle changing from 200 degrees to 40 has something to do with the positive/negative factor.

[topaz eagle]

if you have difficulty in one way of thinking, transforming your doubt into another way of thinking can often provide you a solution
it's like when you're stuck in one pathway, you choose another pathway

if you're stuck at numerical calculations, you can think of drawing them in graphs

you can test with @twin meteor, which will reveal the answer

you can also draw grid quickly to measure the lengths

,tikz
\draw[opacity=0.5] (-3,-3) grid (3,3);
\draw[->] (0,0) -- (220:2);
\draw[->,yellow] (0,0) -- (220:2*2);

[twin meteor]

vin100

[pseudo wave]

.close

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[thick niche]

why is he allowed to flip from o to inf to from inf to 0

afer the u sub?

nvm

its because

when x is 0, u apraoches inf

and when x is inf, u approaches 0 right

[rocky wyvern]

i am pretty sure he just took out the negative and that allowed him to flip bounds

[thick niche]

i meant this step

[rocky wyvern]

yeah

[thick niche]

its just replacing the x values with u right

[rocky wyvern]

no

[thick niche]

then here you just take the negative and flip bounds

[rocky wyvern]

well yes but that would still be infinity to 0

[thick niche]

ok

thanks