help-17

[swift bane]

you could also drop bisectors

that's what u mean right

[astral pilot]

angle bisectors

[swift bane]

yea

so

if you connect a line to the incentre

it can be a bisector?

just like you did here

[astral pilot]

yeah

[swift bane]

I see

[astral pilot]

its an angle bisector

it divides the angle into two parts

[swift bane]

mhm

[astral pilot]

Didn't they teach this?

[swift bane]

yea but

kind of hard to see in this fashion

did you also

drop a height here?

[astral pilot]

yeah

H divides into AC into two parts

[swift bane]

wait how?

oh

isosceles right

[astral pilot]

O is centre of the circle and AC is the chord

[swift bane]

kind of like this

🥲

[astral pilot]

😭

[swift bane]

wait

[astral pilot]

But anyways

[swift bane]

the radius property right

[astral pilot]

<BCI = C/2

[astral pilot]

Idk what they call it

[swift bane]

its like this

[astral pilot]

yeah

[swift bane]

yeah i see

[swift bane]

alright

what's next?

[astral pilot]

<COH = B

[swift bane]

how is that 🥲

[astral pilot]

angle subtended at the centre is twice the angle subtended on the circle

[swift bane]

oh yea

inscribed angle and center angle right

[astral pilot]

I'm sorry if I'm talking nonsense, Its 1:30 am in India rn

[swift bane]

I see

yeah its alright

I just forgot the rules 🤣

[astral pilot]

<COH = B

then what is <OCA?

[swift bane]

is A the top point

.

[astral pilot]

yeah

[swift bane]

yeah I see

[swift bane]

yeah I have no idea how you find OCA 🤣

[astral pilot]

Hint: look at triangle COH

[swift bane]

isn't OC a angle bisector

[astral pilot]

no it isn't

[swift bane]

or is that IC

[astral pilot]

IC

[swift bane]

only the incenter right

[astral pilot]

IC is angle bisector

yeah

[swift bane]

🥲

well <H = 90; <O = b

[astral pilot]

we know <COH

[swift bane]

no idea how can you find <C

or rather oca

[astral pilot]

angle sum property

<OCA = 90 - B

[swift bane]

yeah I see

🤣

[astral pilot]

<BCI = C/2

now what is <OCI

Hint: <C = < BCI + <OCI + <OCA

[swift bane]

you have to just use algebra for this right

[astral pilot]

yeah

[swift bane]

yeah well

you could just get BCi and OCA on the left hand side

I guess?

[astral pilot]

<OCI = B + C/2 - 90

I think this is the moment we use Cosine rule

[swift bane]

yeah I see

[astral pilot]

Look at triangle OIC

we know IC = r (inradius), OC= R (circumradius)

[swift bane]

yea

[astral pilot]

so, what was that formula I forgot it

[swift bane]

.

you meant

the law of cosines

:rofkl

man I love that formula above

[swift bane]

$c = sqrt(a^2 + b^2 - 2ab cos \alpha)$

[twin meteor]

asm

[astral pilot]

yeah

that one

now plug some stuff in that

[swift bane]

that's kind of the hard part :rofkl

oh yea

we know the radiuses

IC = 3; OC = 6.25

<OCI = B + C/2 - 90

[swift bane]

what do we do about this though

[astral pilot]

law of cosines to find d

[swift bane]

yea but

how would i put <OCI = B + C/2 - 90 in cos

[astral pilot]

Hmm, just put it ig, we'll see about it later

[swift bane]

,w d^2 = sqrt(3^2 + 6.25^2 - 2(3)(6.25) * cos(B + C/2 - 90))

[twin meteor]

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[swift bane]

uh oh

[astral pilot]

I think there's a way

to make all this go away

[swift bane]

I messed the formula up

,w d = sqrt(3^2 + 6.25^2 - 2(3)(6.25) * cos(B + C/2 - 90))

[twin meteor]

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[swift bane]

i'm starting to appreciate the creator of that formula more now

🤣

I think i'll ask my teacher on doing this manually

seems kind of tedious

I'll have to move on since I have like 4 more word problems

thanks for the help yet again 👍

[astral pilot]

<ICO can be simplified to (B-A)/2

[swift bane]

do we try plugging it in?

[astral pilot]

wait, I am forgetting something

nvm

You do you, use the formula if you want to, I'll dm you the derivation tomorrow, I'm sleepy asf

[swift bane]

you could just forget about it

my teacher will probably teach us the manual way

[astral pilot]

,w \sqrt{6.25^2-12.5(3)}

[twin meteor]

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[swift bane]

yea

that's the correct answer

[astral pilot]

there you go

[swift bane]

that formula is too nice

🤣

thank you 👍

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[halcyon gulch]

would appreciate if anyone could send solutions i could check my workings against

[delicate nimbus]

what is the question at all

ha ha

[dense eagle]

i think it'll be easier if u just sent ur soln and then someone can check it

[halcyon gulch]

calm

ill send it

now

the working isnt too clear ...

will try explain

basically formed a differntial equation

using the schrodinger equation

then solved it

and plugged in boundary conditions

[delicate nimbus]

write it proparly it is immposible to read

[halcyon gulch]

ye mb

gimme like 5 mins

to write it up

[delicate nimbus]

ok ok take time

and click the image from close

[dense eagle]

ok so u've done part b but have u done part a?

[halcyon gulch]

i skipped part a

bc i think im good with that one

[dense eagle]

also i don't think u needed to find the constant, it seems like the q only wanted u to give the form of the schrodinger eqn

[dense eagle]

ah fair

[dense eagle]

why do you have ur wavefunction as A+Be^(kx)?

[delicate nimbus]

send the image again

[halcyon gulch]

almost done writing it up

shoot

it sent rotated

hold on

[flat whale]

,rotate

[halcyon gulch]

oh thanks

[dense eagle]

i mean, d^2y/dx^2 + stuff*y = 0 gives y = e^(k1x) + e^(k2x)

so how did u end up with it as A+Be^kx?

[halcyon gulch]

so for one of them

k = 0

so it just become a constant

[delicate nimbus]

if A:B=3:4 and B:C=2:3 then what is the answer ofA:B:C.

pls help me guys

[dense eagle]

um... how did u get that?

[halcyon gulch]

when u solve the dfe

u factorise

the alpha

[dense eagle]

d^2y/dx^2 - y = 0

[halcyon gulch]

and so a solution is alpha = 0

ohhh

shiiiit

mb

[dense eagle]

lol nw

[halcyon gulch]

just clocked

would the rest of the method be right otherwise?

[dense eagle]

ngl i'm not sure what ur doing for the rest of it

we'd need to solve the S.E. in the region 0<x<L to get the form of the wavefunction in the region 0<x<L

then that gives us b.c.s at 0 & L since the wavefunction is cts at 0 & L, and has a cts derivative at L

[halcyon gulch]

basically just subbed in U(x) = 0, for x = 0 and U(x) = U0 at x = L, and solve for the unknown constants from that

[dense eagle]

but how does knowing U(x) give u information about the wavefunction?

[halcyon gulch]

because the schrodinger equation has a U(x) term in it

next to Psi(x)

[dense eagle]

oh right so did u literally just sub it into S.E.?

oh wait no i think i'm understanding what ur doing now

[halcyon gulch]

nah, because its a constant i thought that you could sove the dfe as u would normally then sub it in to solve for constants

[dense eagle]

well ur substitution doesn't work for x=0 cus the wavefunction doesn't look like that near 0

since it'd have a different form for 0 < x < L

[halcyon gulch]

in the box tho the potential is 0 as well as at x = 0 no?

[dense eagle]

yeah the potential is 0 as well but that's not how it works

if i just the potentials

U(x) = 1 for x > 0

U(x) = 4 for x < 0

let's take E = 0

[halcyon gulch]

but the question is designed so that the box has zero potential?

between x = 0 and L

[dense eagle]

yh but you can't plug in boundary conditions for x=0

you've found out that the wavefunction has to look like Aexp(kx) for x > L

but you can't figure out the value of A by plugging in x = 0

[halcyon gulch]

so is that equation only valid for x > L?

Keeping A = A for now

without subbing the boundary for x = 0

[dense eagle]

so that's what we get

but we've solved the equation for x > L

[dense eagle]

yh

[halcyon gulch]

ahhh

[dense eagle]

it's like if someone tells you their function is

[halcyon gulch]

makes sense

but i have a quesiton

how can we solve for A?

[dense eagle]

f(x) = Ax if x>0
f(x) = e^x if x < 0
f(-1) = e^-1
then you can't solve for A by plugging in x = -1

[halcyon gulch]

ye ye

i get that

[dense eagle]

you'd need to solve for the middle bit

[halcyon gulch]

now

[dense eagle]

so for 0 < x < L, you end up with a different equation

let's sps E > 0 so that gives us like

$\chi(x) = A\sin(\kappa x) + B\cos(\kappa x)$

[twin meteor]

LY

[halcyon gulch]

ye ye

[dense eagle]

then we use our b.c.s

[halcyon gulch]

i know how to solve that one

i get it now

i appreciate the help

[dense eagle]

we now $\chi(0) = 0 \implies B = 0$

[twin meteor]

LY

[dense eagle]

nw

i'll quickly finish this off

[halcyon gulch]

ye thanks

[dense eagle]

then you always have that the wavefunction and its derivative are continuous (at least for "normal" functions)

so that gives you 2 boundary conditions for x=L

^ is normally the best way to do problems like these

in general, it's like normally impossible to solve the S.E. for an arbitrary function as our potential

so a lot of questions will just have like constants as the potentials

so the general method is you solve to end up with sin/cos or exp

and then use b.c.s asw as continuity of the wavefn and its derivative

(the derivative isn't cts at 0 cus U(x) = infinity is not a normal function)

hope that helped!

[halcyon gulch]

ye that really helped

gonna review this stuff

now

how do i close the channel after

[dense eagle]

do .close or .solved

[halcyon gulch]

ok thanks again for the help

[dense eagle]

nw!

[halcyon gulch]

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[trim pecan]

is it because of the 2k

it has to be in one these two forms

[viral copper]

yeah no alternating series test for this one

since (-1)^(2k+1) is always -1

[trim pecan]

so it does not alternative meaning we cannot use the alternating serious test?

[viral copper]

yeah

you need (-1)^n*a_n where a_n is either always positive or always negative

[trim pecan]

ok I understand thanks

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[trim pecan]

how did they get negative

[mental falcon]

there's a (-1)^n in the numerator, seems like those calculations in the 2nd image are just ignoring that?

[trim pecan]

yup silly mistake

https://tenor.com/view/facepalm-really-stressed-mad-angry-gif-16109475

its been a long day

wait I still dont get it

there answer

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[true mango]

is the closure {y,z,w}? i had this wrong in my exam. and how is it not dense 🥹

[hybrid flicker]

you mean {x,z,w}?

[true mango]

ohhh i see omg from the closed subsets huhu

[hybrid flicker]

the closure of {x,z} is the smallest closed set that contains it

[true mango]

okie thanks 🫶

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[odd sigil]

Show that the solutions of | k 𝓧-1 | = | 𝓧-k | are independent of k

[odd sigil]

idk how to start

i tried doing the pos and neg vers of the equation but i dont know how to get rid of k

[undone herald]

Try to isolate x first

[odd sigil]

Ok

oh i got it

thank you 🙏

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[finite hatch]

whats it mean by convert a + bi onto form rcistheta

[vale frigate]

Convert it to polar form

[finite hatch]

ik that if it equal a + bi

it equals

[vale frigate]

Rectangular to polar

[finite hatch]

rcostheta + (rsintheta)i

[finite hatch]

how do they go from r(costheta + isintheta) to cis

oh nvm

i think i get it

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[vale frigate]

Suppose we have a perfect cylinder that has a water that is continuously draining in it. When will the radius of the water exhibit a rate of change

[ornate sun]

As it's getting poured into the cylinder

Assuming that it is poured as a stream and not at a radius matching the radius of the cylinder

[vale frigate]

Water is getting drained, not poured

[mellow oyster]

unless there's specific numbers associated with this question, it doesn't seem very specific but intuitively the radius of the water stays constant until the end when it fully drains out of the bottom

this could also be interpreted semantically, technically it always has a rate of change, that rate of change could just be 0

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[lime kraken]

here if p = -1 would alos have no solution no?

[lime kraken]

row 2 would then be inconsistent

wait no nvm

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[grizzled vault]

hmm

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[spring meteor]

what is this method of factorizing called in the denominator?

[trim trout]

factor by grouping

[nocturne sorrel]

I'm pretty sure it's called grouping

[spring meteor]

another thing i dont get is how my teacher know what to add in order to group the common term
example 2x^2 to x^2 + x^2 and 6x to x + 5x

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@spring meteor Has your question been resolved?

[shadow girder]

You mean you don't understand why we're adding those?

[shadow girder]

If that's the case, let's look at it this way

Why does x^2+x^2=2x^2
For simplicity, let's imagine x^2 is a bun, now we have one bun (x^2) + another bun (also x^2), you get =two same buns (2x^2)

I also struggled with that at one point, this simplified way helped me understand

[shadow girder]

I'm assuming you mean here why does 6x=x+5x

We can see in x+5x both of the numbers have something in common: x
This is why we are allowed to add them in the first place (again, because they have something in common, in this case, it is x; in our previous case it was x^2)

Now we know that there is 5x, we can see the 5 in front of the x, but why when we do 5x+x we get 6x?

When there is nothing standing in front of the x, we have to remember that there's a 1 in front of it, which we do not write as in most cases we either remember it or know that multiplying anything by 1 gives us the same outcome
(This point is just something you have to remember, when there's nothing in front of the x, there is actually a 1 there)

So 5x+x could be rewritten as 5x+1x which gives us 6x

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[oblique juniper]

Is this correct?

[oblique juniper]

Ping me

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@oblique juniper Has your question been resolved?

[oblique juniper]

<@&286206848099549185>

[astral pilot]

,rccw

[oblique juniper]

a² - b² = (a+b)(a-b) ?

Thats what i did

[astral pilot]

but you forgot about 1/2

you have include that in a^2

[oblique juniper]

Oh yeahh

But

Its seperated from ²

Oh nvm i understand now

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[brittle basin]

Hey, I have a question about vector calculus.

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@brittle basin Has your question been resolved?

[brittle basin]

<@&286206848099549185>

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@brittle basin Has your question been resolved?

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@brittle basin Has your question been resolved?

[flat whale]

just show the solution of the past paper

[brittle basin]

But is this not wrong since F is not defined for x=0, y=0?

[flat whale]

show their definition of F

[flat whale]

the corollary is correct. the converse is true if F is defined on all R^3, but that's not necessary. F just needs to be defined on a simply open and connected set.
https://en.wikipedia.org/wiki/Conservative_vector_field#Irrotational_vector_fields
and
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/16%3A_Vector_Calculus/16.03%3A_Conservative_Vector_Fields

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[lofty orchid]

is it a or c

[lilac pebble]

it looks like student II is trying to integrate along the y-axis

but won't that include the little wedge in the bottom-left?

[lofty orchid]

ye thats why i was thinking so A?

[lilac pebble]

sounds good to me

[lofty orchid]

Axe so if you are free, I just did my practice test but it doesn't have solutions, can you go over my answers rq?

[lilac pebble]

[lofty orchid]

its only a few qusetions

like 4

this is D right?

[mighty nacelle]

show your work

[lofty orchid]

you plug in sec as x right

then sec^2(x) - 1 = tan^2(x)

so it should be d but gemini says its B

so idk

[mighty nacelle]

first off, dont ask an AI

[cerulean drift]

Fr

Um hold up

[lofty orchid]

how else would i check

this one doesn't solutions

[mighty nacelle]

how are you checking right now

[lofty orchid]

gemini

[mighty nacelle]

ok bye

[lofty orchid]

wait

how should i do it

how should i check?

[cerulean drift]

I think this would help

[cerulean drift]

!noAi

[lofty orchid]

mb

but how should i check my answer tho 😭

[cerulean drift]

I just showed you the trig functions

Find which one looks like it

[lofty orchid]

ik but what about other questions

there are different types

[cerulean drift]

Is this your homework?

[lofty orchid]

its a practice exam for my final exam today

my classes endded lol

[cerulean drift]

You jsut have to recognize their derivatives

It sucks

Memorize

Only choice

[lofty orchid]

i have these memorized but this is trig sub righit?

so its different?

[cerulean drift]

You can tell it is a trig sub because the first value under the square root is squared and the second is also a square

There this is more accurate

[lofty orchid]

oh ok so it is D then?

[cerulean drift]

I dont know

I didn’t do it

[lofty orchid]

oh alr but i got it

does my work look correct

[pine crow]

I have a problem

[cerulean drift]

!help

[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[pine crow]

yeah

[cerulean drift]

Visit #❓how-to-get-help

[pine crow]

I want your help

[lofty orchid]

lol

[lofty orchid]

help

[lofty orchid]

the work i did

[vocal sleet]

@lofty orchid Has your question been resolved?

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[viral walrus]

what therom would i need

to solve thi

s

[lilac pebble]

SAS similarity

UC/UV = UB/UT and angle CUB is congruent to angle VUT
therefore triangle CUB is similar to triangle VUT

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[heavy palm]

Can someone explain this problem to me

[heavy palm]

I mean

not problem

rather the second step

how does x^-2/3 -8x^1/3 = x^-2/3 (1-8x)

[flat whale]

Factored out x^(-2/3)

1/3 -(-2/3) = 1

[heavy palm]

ic

thank u

.thaknk

..close

close

AAA
A
A

a

AA

.clsoe

.clsoe

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[hexed ivy]

Does a go right of y or left?

[jagged cargo]

right, away from Y

[hexed ivy]

So is my mark wrong?

[trim trout]

yes

A' is twice the distance from Y as A is from Y

so it should be on the right of Y similar to how A is to the right of Y

[hexed ivy]

Ah ok

Thanks 🙂

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[radiant lake]

Can you teach yourself math? im not doing too good in the brain department as a 19 year old I was wondering do you all teach yourself?

[past hare]

<@&268886789983436800> Should this question be solved in another channel?

[autumn trail]

I don't think asking for general advice is totally against the rules.

[radiant lake]

What's the channel called

[past hare]

Nah, just double checking if it should be answered here, that's all

[autumn trail]

Kinda uncommon but so long as things stay productive/not toxic it should be fine.

[past hare]

Copy 👍🏾

[past hare]

What do you mean by teaching yourself math?

As in reading books by yourself?

[radiant lake]

Like, learn on own from books and online without any help

Your

[past hare]

It's possible, it's just that it requires discipline

I relearned linear algebra from scratch at 21

Using some good books

About the help part, I don't think it's advisable at all to discard it

[brittle cipher]

khan academy can get you pretty far on your own if you don't cheat yourself I guess

[past hare]

If one gets stuck, one should try to talk to like-minded people

[brittle cipher]

you have these help channels for free too so no reason not to use them

[past hare]

Exactly my point ☝🏾

[radiant lake]

Thanks both of you👍

[past hare]

The thing is to try to put effort

Try to push yourself

Take your time solving a problem

Once you're utterly exhausted or blocked, try reaching for help

But make the effort first.

It will enable you to start seeing patterns and make further learning easier every time

Like a snowballing effect

[vocal sleet]

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[thorn spindle]

why did they ln both sides?

how did lnx^1/x become lnx/x

[stark berry]

by properties of logs

you can bring down the exponent

[thorn spindle]

oh i see

so it becomes lnx (1/x)

which is lnx/x

i thought the ln went away after

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[thorn spindle]

for the second step

how did they get the 4e^-t (-1)

i know its chain rule but shouldnt it just be 4e^-t

[outer warren]

chain rule again

[heavy yoke]

the derivative of e^(-t) is -e^(-t)

[thorn spindle]

wait so chain rule is f'(g(x)) g'(x) so wouldnt it be -1(1+4e^-t)^-2 (4e^-t)

[thorn spindle]

so 4e^-t is just -4?

[outer warren]

no

[lilac pebble]

the derivative of e^(-t) is -e^(-t)

[heavy yoke]

to find the derivative of 1 + 4e^(-t) you have to apply the chain rule

[lilac pebble]

e^(-t) is f(t)=e^t composed with g(t)=-t

[outer warren]

first level of chain rule
d/dt (1 + 4e^-t)^2 = 2(1+4e^-t) * d/dt(1 + 4e^-t)
and then apply chain rule again for the bolded part

[thorn spindle]

oh

okay i see so

d/dt(1 + 4e^-t)

is -4e^-t?

[outer warren]

yes

[thorn spindle]

and then they took out the -1

so -1 times 4e^-t

is there a reason as to why they did that

or it just looks better

[deft vapor]

its basically differentiating f(g(x)) where f(x) = 1 + x and g(x) = 4e^-t

when u differentiate 4e^-t you get -4e^-t

[thorn spindle]

yeah i see that now

[deft vapor]

oh in this case they did f(x) = 1+4x and g(x) = e^-t

instead

thats why you have the -1

same thing

it doesnt matter

[thorn spindle]

yeah but there a reason why

if its helpful ill do that in the future

[deft vapor]

it is

[thorn spindle]

like if it helps with siimplifing

[deft vapor]

so you dont make silly mistakes

for me I like representing everything as exponents when differentiating

[lilac pebble]

you can do the chain rule like
d/dx f(g(h(x))) = f'(g(h(x)) g'(h(x)) h'(x)

[deft vapor]

yeah

or just make k(x) = g(h(x))

cus in this scenario u dont have to split everything

to

separate functions

[thorn spindle]

i see

alright thanks

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[safe cloak]

yall is it even possible to use squeeze theorem for this at (0,0)

[desert hornet]

it is

[safe cloak]

could you help me figure out the upper and lower bound

i was thinking about this

but idk if it's allowed to do abs(xy)

[desert hornet]

why not

the center thing is |f(x,y)| so you have 0<=|f(x,y)|<=|xy| or -|xy|<=f(x,y)<=|xy|

[safe cloak]

i didn't know if it's valid to take out xy for the inequality

[safe cloak]

oh thats a relief

[safe cloak]

is this a valid move to make

[desert hornet]

why wouldn't it be?

[safe cloak]

in all the textbook problems and in my prof's lessons i've never seen a situation like this where i had to take out both an x and a y

so i unconsciously thought it was odd aand therefore not allowed ig

thanks for opening my eyes LMAO

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[hexed umbra]

help

[hexed umbra]

140 - 3x completes the triangle

for the angle next to the p

but idk what it should equal to

or should i just use the equation 140 - 3x + 40 + 3x = 180

and would the x in 3x be the x in this

[hexed umbra]

but if you combine like terms in this equation there would be no x

[hexed umbra]

^^ (wrong reply before)

or wait

140 + 90 + x

nvm

140 - 3x + 90 + x = 180

and 3x + y = 180

man im confused

okay i did not know how i did that

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[real zenith]

hi can some one halp me pls The natural number N with exactly three divisors and the sum of its divisors equal to 183 has the sum of its digits:
A) 3
B) 4
C) 5
D) 12
E) 16

[hexed umbra]

n/3 = 183?

so to find n what would you need to do

wait nvm

[stark berry]

you need to find 3 prime numbers addjmg to 183

[real zenith]

if it has 3 divisors then it is n=p^2 where p is prime

[stark berry]

wait nvm im stupid

[real zenith]

so 1+p+p^2=183

right?

p+p^2=182

[queen root]

yes

[stark berry]

yws

[queen root]

whyd u do that

[real zenith]

??

[queen root]

whats ur idea afterwards?

[hexed umbra]

whyd u square cuz idk what the question is talking abt 😭

[stark berry]

irs correct, solve for p and fimd p^2

[queen root]

oh wait yeah its correct

[real zenith]

p=13?

[stark berry]

yup

[real zenith]

and 13+169=

yea

[stark berry]

no

the original number is just p^2

[hexed umbra]

what grade r u

[real zenith]

yes

so n=169

and sum of digits is 16

[real zenith]

6th

and im preparing for national olympiad

[hexed umbra]

damn im cooked 💀

[stark berry]

all the best

[real zenith]

thxx

byee

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[hexed umbra]

in order to figure out the blue angle, do i subtract 50 - 3x?

[hexed umbra]

50 - 3x = 180?

or is 50 irrelevant

[bronze osprey]

50 is irrelevant

so yeah it's just 180 - 3x

[solar cape]

By adjacent angles on a straight line

[hexed umbra]

ok

thx

ill try 180 - 3x first

uhh where did i mess up

3(60) is 180 which couldnt be true

[bronze osprey]

can you show the full diagram?

[bronze osprey]

ah so that means angle ABE is a right angle

so you should focus your attention on triangle ADC

find angle CAD and find angle ACD

then set up an equation in terms of x = 180 degrees

[hexed umbra]

ok

ill try

[bronze osprey]

(btw this follows by alternate angles in parallel lines)

[hexed umbra]

would this equation work

[bronze osprey]

no, angle ACD is not x

[bronze osprey]

you were pretty close here, so angle ACD + x = 180 - 3x

[hexed umbra]

ohh ok

so i got angle acb

60

and got x

3(40) + 60 = 180

so therefore ADC is 120

[bronze osprey]

how?

[hexed umbra]

oh wait

oh

so 180 - (50 + 90) = angle A

toh

ohhh

oopsies i read angle ABC as 90

[bronze osprey]

well I was looking for 40 + 3x + (180 - 4x) = 180

so x = 40 yes

it just feels to me that you guessed the answer somehow

[hexed umbra]

but mistaking angle ABE is not how to properly get to x = 40

lemme try again

[bronze osprey]

oh it's very possible to do it another way

[hexed umbra]

50 + 3x + CBE = 180?

then subtract ABC by CBE

[bronze osprey]

angle CBD = ||180 - x - (180 - 3x) = 2x||
angle EBC = ||90 - 2x||

||90 - 2x + 3x + 50 = 180 means x = 40||

[bronze osprey]

yes!

[hexed umbra]

how do i get CBE tho

[hexed umbra]

lemme try this first

[bronze osprey]

first find angle CDB

then you can find angle CBD

then finally 90 - angle CBD = angle CBE

[hexed umbra]

do i divide cbe by 3 since u have to divide 3 on both sides

[bronze osprey]

yes

but 180 - 50 = 130

[hexed umbra]

oh right

im pretty bad when there are 2 variables

how do you isolate CBE

[bronze osprey]

my point is

you can find CBE through finding other angles first

instead of isolating

you need angle CDB then angle CBD

[hexed umbra]

ok lemme try finding the other angles first

damn im lost

you still need to isolate to get ACD

[bronze osprey]

well my point is that you should write angle ACD in terms of x

so that would be ACD + x + 3x = 180

[hexed umbra]

how can i know when i need to use x or another variable (y)

cuz earlier you said i cant use x for CBE in 3x + 50 + x = 180

[bronze osprey]

oh you don't know CBE yet

cause the only other angle that can give you CBE is angle CBD, and you don't know CBD either yet

[bronze osprey]

yeah you need another variable y here

cause you are saying that angle CBE = x, which isn't true

but you know angle ACD directly, from angles in a straight line

[hexed umbra]

but isnt ACD and CBE kinda similar?

3x + 50 + CBE = 180
3x + 40 + ACD = 180

[bronze osprey]

yeah those equations are true

[hexed umbra]

so why i do need to know a prior angle for CBE while i can use x for ACD

[bronze osprey]

.

[hexed umbra]

oh i see now

3x + x + ACD = 180

180 - 4x = ACD

okay got it

properly this time

all i had to do was subtract 180 - 4x

thanks

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[next grove]

What does linear and vertical mean

[rough patrol]

vertical angles are the pair of opposite angles made by the intersection of two lines

so in the bottom diagram for example, 1 and 2 are vertical angles

similarly 3 and 4 are also vertical angles with each other in the bottom diagram

[toxic pivot]

Number 4

[meager shoal]

open ur channel

[next grove]

The top diagram is confusing me

.close

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[plain sand]

How do I draw g(x)?

[stark berry]

,rotate

[stark berry]

find the roots, turning point, and axes intercepts

[plain sand]

When I set g(x)=0 I get negative root

[stark berry]

then there are no real roots

find the turning point

and y intercept

[plain sand]

@stark berry like this?

[stark berry]

,rotate

[stark berry]

yup looks like it

[plain sand]

Tnx for help!

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[dusk grove]

hi

[dusk grove]

i need help

with unit circles

never mind it’s ok nvm

wait

no i do im lying

someone help PLS

[vast shale]

What

[dusk grove]

but like idk how to ask

like i have no clue where to start

it’s fine nvm

byeeee

[flat whale]

!done

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If you are done with this channel, please mark your problem as solved by typing .close

[flat whale]

.close

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[tulip lance]

i have a square in a square, sq1 a = 12cm, sq2 a = 30cm. how can i calculate the distance between them between two different sides

[tulip lance]

example:

[stark berry]

half the difference

[tulip lance]

so a2 - a1 / 2

[stark berry]

yup

[tulip lance]

okay, thanks alot

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[polar ice]

how do i solve
sin2x = 2sin^x

[lilac pebble]

use the double angle formula

sin(2x)=2sin(x)cos(x)

[polar ice]

This is what ive done so far but im not sure

Oh wait

I dint have to switch out 2sin^2x do i

Is this right

[lilac pebble]

yeah but you will also have solutions when sin(x)=0

since you divided by sin(x)

[polar ice]

Got it

Thanks

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[fluid bough]

I don'

[fluid bough]

understand how I got this incorrect

& this?

[trim trout]

what line did you reflect across here?

draw it

[vocal pagoda]

this wasn't in line x=3.

it was in y=0 🙂

[fluid bough]

when i got this correct.

[vocal pagoda]

you drew x=3 correctly (the line)

[fluid bough]

how come? It's passing through 3 right

[vocal pagoda]

see in your example you did correct.

reflection is on other side of line.

[fluid bough]

oh

[vocal pagoda]

image is above y=1, reflection is under

you reflect on other side of line.

[signal pendant]

reflect some shape in the line L means treat the line L like a mirror and see where the virtual image of the shape is

[vocal pagoda]

here: A is left side of x=3. reflection should be on right side of x=3.

[fluid bough]

so in the other box?

OH

i understand!

but theres no grid there

[vocal pagoda]

there's just enough space in both of them

to draw the reflection.

Try one of them, then send here 🙂

[fluid bough]

Hm oka. reflection should be behind x = 3?

[vocal pagoda]

yes; object and reflection cannot be on same side of line.

the line should separate them.

[fluid bough]

oof, that makes a lot of sense

[fluid bough]

ot

should it be in the same box? @vocal pagoda

[vocal pagoda]

see here?

it's like in a mirror

[fluid bough]

oh

[vast shale]

i need help with maths

[vocal pagoda]

the exercise with y=3 should be very similar

[vocal sleet]

Please stick to your channel.

[vocal pagoda]

open a new channel... and ask your questions there @vast shale

[vast shale]

how

[vocal pagoda]

on left side; look for math help (available)

go in help 19, 23, or 35 (these are there)

click one of them.

ask your question.

[vast shale]

thx

[vocal pagoda]

someone might help you if they know how 🙂

[vocal pagoda]

you're welcome.

[vocal pagoda]

I'm afraid not 😦

[fluid bough]

ik.. idk wh i keep getting it wrong..

[vocal pagoda]

https://www.mathsisfun.com/geometry/reflection.html I noticed this nice site. Maybe it helps you see them better.

if you go a little down, it has a nice application where you move the line, and it shows the reflection of the triangle.

[fluid bough]

should mv answer be the opposite wa around?

[vocal pagoda]

right now, it's on correct side, but not right place...

[vocal pagoda]

I see the method you use is reflecting the corners of the object, then connecting them.

[vocal pagoda]

wrong channel again.

[fluid bough]

ohh so next to it?

[vocal pagoda]

yes, in this case, they're 'connected'.

like someone else suggested earlier in this channel. you can see it as the reflection in a mirror.

[fluid bough]

the fact I was about to do it that wa too 😆 .

[vocal pagoda]

so it will be 'behind the line', but also 'mirrored'.

[fluid bough]

it's easier for me

[vocal pagoda]

yess

🙂

[fluid bough]

thanks so much 🥰 .

[vocal pagoda]

You're welcome.

[fluid bough]

I will do the other question reall quick before I close the channel

I have a question

[vocal pagoda]

yes?

[fluid bough]

for a question like this, I had to draw it behind the line, es, but in another box. Is this because there was not enough space to do it for example like this:

[vocal pagoda]

Ah, yes; but you don't need to worry about the 'boxes'; all that matters is the red line and the object you reflect.

[fluid bough]

I get it now, if there is enough space to do it in a certain place I do it. I alwas thought it had to be done in a different axi box.

[vocal pagoda]

no, also it might sometimes cover 2 different boxes at same time, or more.

I'll draw something.

[fluid bough]

oh.. so it should just be behind the line basicall. Sorr for confusing things

[vocal pagoda]

ah, I didn't draw all numbers 🙂 wait a sec

[fluid bough]

haha

[vocal pagoda]

I didn' draw all the squares; just an approximate figure. If you were to reflect the blue triangle in the line y=1, you'll get another triangle that's in 2 boxes.

[fluid bough]

that makes sense wh u told me not to worr about boxes then haha

[fluid bough]

this is a good example, thanks

.close

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[hexed kestrel]

Parallel side of parallelogram have the same length

[hexed kestrel]

No

3c -77 and 4c-74 are not parallel side

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[shadow dirge]

Hi I don't get reference angle, why are these given specifically subtracted to these numbers?

[thick shard]

I think the idea is to find an angle that is less than or equal to 90 degrees.

draw pictures.

if you go 310 degrees counter clockwise, its the same as going 50 degrees clockwise.

similarly, you go 180 degrees counter clockwise, then 10 degrees clockwise, it's the same as going 170 degrees counter clockwise.

[shadow dirge]

oh

thank you!

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[quasi nova]

Composite Areas Using Addition

i do not know how to do this for sheet

[pale perch]

split it into shapes you know the formulae for

i like parallelograms, just on a random note (subtle)

[quasi nova]

rule but i dont understand it

[pale perch]

theyre both just base * height really

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[exotic ravine]

pls explain me the 2nd last property

[exotic ravine]

nvm i got it

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[junior summit]

Given the following Markov Chain. What is P(A|E)? will this be 0 or 0.6?

[flat whale]

There are no arrows from E to A

0.6 is the arrow from A to itself

[junior summit]

so is it correct to say the self-referential probability is the "default". I am assuming that was no past state when I read P(A|E).

P(A|E) mean "A given E" right. I was confused if I could use the self-referential probability or because E doesnt have an arrow it will be 0 because its an absorbing node?

[junior summit]

So are you saying its more correct to say its 0?

[flat whale]

E is absorbing yes

[junior summit]

So 0, right.

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[wild meteor]

For 9b idk what I did wrong

Okay so to get to said lines wouldn't it be 180-45

And 180+45

[flat whale]

,calc cos(225 deg)

[twin meteor]

Result:

-0.70710678118655

[flat whale]

,calc cos(135 deg)

[twin meteor]

Result:

-0.70710678118655

[flat whale]

Yes? What part do you think is wrong

[wild meteor]

Oh nvm

Mb 💀

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[warped osprey]

Hi for part a i got everything except for drawing the integral. Like I got that as N -> infinity, the integral -> 1 but how am I meant to know what the area under the new graph should look like

[stark berry]

its the same graph, just sketch it starting from 1

[warped osprey]

Wait now I don't really get the fact that the integral approaches 1

Also in the second part it says it 'diverges to infinity' what does that mean

[stark berry]

to evalute improper integrals

you tKe the limit as N approaches infinity