help-36

[pliant shore]

no, the x-coordinate of B is not going to be the same as P

correct! nice observation

[thin cloud]

No i meant

Hold on

[silk fjord]

pretty easy from here then, i can just solve for the missing values and then sum the distances

[pliant shore]

yes you get an equation with the line BQ and a

solve for a then find the distances

I got 17 this way

[silk fjord]

i'll try that myself now

[thin cloud]

the smallest distance should be P'Q' , no?

[pliant shore]

yeah that is a nice method

I had it in the back of my mind, but I wasn't able to articulate it

@silk fjord completely forgot about this method above ^

[silk fjord]

honestly im kinda stuck on the algebra for the first method

$y_{PA} = \frac{5-a}{4} x + a\ b=\frac{-12}{5-a}+11$

[soft zealot]

UCYT5040

[silk fjord]

these are the only meaninful equations i've found

where point be B is at (b,0)

[pliant shore]

honestly forget about what I said

try to understand the new diagram

AP = AP' and BQ' = BQ

[silk fjord]

ok, ill look at that

oh yeah that makes sense

perfect i got 17

thank you both!

.close

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[late laurel]

yo how would yall approach this i lowkey don't really know where to start

i tried letting c=a+bi but it didn't really get anywhere

[flat mason]

consider: what is z+z*

[late laurel]

2Re(z)

2Re[(r-c*)z]=is a fixed value

is it a vertical line with real part r^2 - |c|^2

wait no

[flat mason]

i dont think it's vertical, no

[late laurel]

what else could it be tho if the real part of z is a fixed value

[flat mason]

the real part of (r-c*)z is a fixed value
sorry i misread jn

[ivory bear]

maybe i obtained a circle-like equation

[flat mason]

it's definitely a line

[ivory bear]

lemme type it out on LaTeX and send here

[late laurel]

circle?

[ivory bear]

wait maybe i misidentified it then

lemme send it

maybe i commited an error then

[late laurel]

oh wait i think i got an equation

[flat mason]

when all else fails we can expand and simply use the real part instead of solving by inspection

[ivory bear]

mine?....

[late laurel]

i got something like this

[flat mason]

where did tge squares come from

[small patio]

guys here it says How many irreducible proper fractions are there whose denominator is 76?
(idk if I translated that right)

[ivory bear]

i expanded the terms and made perfect squares

[ivory bear]

!occupied

[final saddle]

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

[ivory bear]

#❓how-to-get-help

[flat mason]

i... suppose that works?

[flat mason]

wellllllll you can draw the locus from this

[ivory bear]

what about |c|^{2}?

[late laurel]

what about it

[ivory bear]

ok mates i am bad at locuses this year

i read it last year and did hard ques too

[flat mason]

please send a photo of the locus when youre done haha

just wanna see it

[ivory bear]

but forgot everything

[small patio]

@everyone who's the best at math, can I dm pls

[ivory bear]

anyone works.

[late laurel]

yeah this chapter is kinda buns

its a hit or miss

[ivory bear]

kind of

but i did absolutely brilliant probs last year for this chapter i remember

fine @late laurel close this if your doubt is done

[late laurel]

its a straight line with vertical intercept -[r^2-|c|^2]/2?

aite thanks

.close

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[pulsar axle]

can someone explain

[pulsar axle]

why the curve looks like that?

usually what i would do is like the "theta" is the angle its at and the "r" is the measure of the line (with a ruler)

js a approximate sketch

but why does it look like that

[bold zenith]

For theta=0 you have r=1 so it’s far from the origin, as you increase in theta, you go anti-clockwise, but at the same time r is decreasing so it ends up at r=0.3

[pulsar axle]

Im askin like

why is it

under

why isnt it above

usually its like this

[bold zenith]

What’s under?

That arrow in the mark scheme is just to show which direction theta=0 should be

The origin is the dot curled up under the line

[pulsar axle]

oh

but it says theta = 0 on the line

ohhh

so its js weirdly drawn

@bold zenith also also

one more thing

[bold zenith]

!noping

[final saddle]

Please do not ping individual helpers unprompted.

[pulsar axle]

mb

anyway

on b

i dont get the

-7 -2 -5

how do i know its a common point

[bold zenith]

Common points are points that are on both planes

Tbh I forget how to find them

Oh wait

Get the equation of the planes and see if you can find something that satisfies both

[pulsar axle]

but lik

ehow am i supposed to find it

lol

idek

[bold zenith]

Can you find the equation for the 2nd plane?

[pulsar axle]

yeah

2x-5y+5z=5

[final saddle]

@pulsar axle Has your question been resolved?

[pulsar axle]

<@&286206848099549185>

[bold zenith]

You need to find x y and z that satisfy both equations

Try putting the first into the second one

[bold zenith]

I don’t think this is correct

[pulsar axle]

oh no

its

-5x+3y+5z

ik how to

find the common point

i just dont knio

which equations to use

for the first one its x+y=-7

for the second one its r(-5i+3j+5k)=4

is the

parametric form for

plane 2

-5x+3y+5z=4

got it

i think

[pulsar axle]

the equation is 2x-5y+5z=0

a no

i got z=-7

bro this shit is so confusing inma crashout

[final saddle]

@pulsar axle Has your question been resolved?

[bold zenith]

So
y+z=-7
-5x+3y+5z=4

Just pick a variable that is in both equations and set it to whatever value you like, then solve the other two for the point

Then use the cross product to find a vector that is parallel to both planes. Finally, use that to find the line:
$$\vec{r}=\vec{a}+\lambda*\vec{n}$$

[soft zealot]

BBMaths

[pulsar axle]

yeye

ty

.close

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[bold zenith]

Sorry for delay, I didn’t realise until now the longest waiting are at the bottom of the list

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[safe venture]

Hey everyone.

[safe venture]

I'm in PreCalc doing zeros on polynomials.

One of the questions is "Degree 3, x = -4, x = 1 + 5i"

I know you have to do some FOIL stuff here, but not sure what the variable is. Book doesn't say much of it.

[proven panther]

so for problems like these

you gotta use the fact that the coefficients to the polynomial have to be real

and for that to happen, the imaginary parts need to cancel out

you already know x=-4 and x=1+5i

so that means 2 of the factors are (x+4)(x-(1+5i))

but since its degree 3, theres a 3rd factor, or a 3rd zero

and this 3rd zero has to cancel out the imaginary coefficients brought about by the other one

this turns out to be the conjugate

the conjugate of a complex number a+bi, is a-bi

because (a+bi)(a-bi) = a^2 + b^2

see if you can find the factor using that

[final saddle]

@safe venture Has your question been resolved?

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[wanton yarrow]

I've figured out that a, b and c work, but d and e do not. However I'm unsure what the consistent pattern is for determining this (for the activity when they say all tetrominoes they mean the 5 distinct ones -- straight, square, s/z, t and l/j)

[opal plinth]

Can you show how you tiled the ones you could, and the reason you couldn't tile the other ones?

[wanton yarrow]

So at first I thought it was a matter of having the number of squares be 20

That ended up not being true

Then I thought maybe it was based on whether the areas were coprime values (both don't share a factor) such as how a, b and c are technically 3 x 7 - 1 square

[opal plinth]

Ok and what are the b and w about

[wanton yarrow]

I was playing around with the idea of chess board layout

So then I was like "oh the first 3 have 9 white and 11 black" but 3 has the same and it still doesn't work

[opal plinth]

3?

[wanton yarrow]

Parts a, b and c

[opal plinth]

but 3 has the same and it still doesn't work
what 3

[wanton yarrow]

Sorry my keyboard didn't register

I meant e

[opal plinth]

Ah, alright

But it's still a good argument for d

[wanton yarrow]

It is, but now it's a matter of what the argument for e is

Or if there's something more consistent

Well

Actually e is technically a 5 by 6

Like how d was a 4 by 5

And the others were a prime by prime area

Versus d and e were a prime by composite area

That's my only thought as to something consistent

[opal plinth]

ok I need to go do something but I'll come back to it in a few minutes; maybe play around a bit more trying to actually tile e

[wanton yarrow]

Well I have that was why I thought to come here

[opal plinth]

Okay I'm back

Try to tile e with the square in the bottom-left corner and the line in the top-right corner

[wanton yarrow]

I see I see

[opal plinth]

Let me know if you find anything

[wanton yarrow]

Unfortunately I end up coming across remaining spots that are duplicates of the other pieces

I really don't think it's possible

[opal plinth]

Can you show me an example?

[wanton yarrow]

A square is left when I need an s

[opal plinth]

Right, what if you replaced that L with the S

[wanton yarrow]

Doesn't fit if the t stays

[opal plinth]

Yes it does, the top three squares stay, the bottom square moves up-right

[wanton yarrow]

HOLY

That's it

Wait okay so

Then it's gotta be the distribution of black and white spaces like a chess board

Because there's 10:10 in d but 9:11 in the others

[opal plinth]

Yes, again, that's a good argument for d being untileable

It's not really an argument for any grid being tileable, but you provided a tiling for the other grids, so...

[wanton yarrow]

Very true

[opal plinth]

Here's a different, poorly drawn tiling for e

[wanton yarrow]

Man I was really not thinking

[opal plinth]

That's the kind of problem where it's very easy to think you're stuck while you almost have a solution

[wanton yarrow]

Very true

[opal plinth]

There's a systematic way (like an algorithm a computer can use) to find tilings like this but it's complicated

Anyway, if you're done you can .close the channel

[wanton yarrow]

Much appreciated thank you ^^

.close

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[final maple]

can someone tell me where i went wrong?

[plucky rover]

Well showing us your work would help

[final maple]

mb

i thought it sent w the screenshit

most of it was mental math

i switched s to x make it easier for me

but i found a GCF

GCF in first group is 3x^3

9x^4 / 3x^3 = 3

-6x^3 / 3x^3 = -2

GCF in second group is -4

and u can see the math i did

the system says I need to foil it

@plucky rover theres my work

[plucky rover]

-4.2 is -8

[plucky rover]

You want +8

[final maple]

AH

I see

Stupid mistake

[edgy mauve]

This channel is occupied
Go to #❓how-to-get-help ,to see how u can claim urself a channel

[final maple]

Thank you so much

.close

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[timber plume]

im a bit confused in this line, why are they multiplying it by x?

[formal trail]

they are multiplying the gradient of the line by x to get the equation of the line

[timber plume]

isnt the equation of a line is in the form of y =mb + b

[formal trail]

the premise of the question is that the line passes through the origin (because they said it is in the form y = mx)

[timber plume]

ah I see

so are they trying to find the equation of the tangent at P since they multiplied it by x

[formal trail]

yes

[timber plume]

why did they times it by x tho? what is x

[formal trail]

just the input variable for the line equation

[tired walrus]

y=mx + b, you mean, surely?

[formal trail]

like we already knew that y = mx is the form of the line equation we want, really they just substituted in an appropriate value of m

[timber plume]

yes

[tired walrus]

so where is the question of "why did they times [sic!] the slope by x" coming from, exactly.

[timber plume]

they didnt ask I was curious

about this line

Im just having difficulty understanding this slide overall

[tired walrus]

the question's premise is that some line of the form y=mx is tangent to the graph.

do you understand this or not

don't try to think more or jump ahead

focus only and exclusively on this one tidbit.

and tell me: do you understand or not

[timber plume]

yes

[tired walrus]

ok

the key thing about a tangent to a graph is where it's tangent at. the point of contact.
on this slide, the x-coordinate of this point of contact is called x_0.

don't try to think more or jump ahead.
focus only and exclusively on this one tidbit.
and tell me: do you understand or not

[timber plume]

I understand

[tired walrus]

ok

so we write down the slope of the tangent at x = x_0

and we say, this must be the m in the problem's premise of y = mx

and it's \textbf{from this} that we say \ that the tangent line has equation $y = \underbrace{2e^{2x_0}}_m x$.

[soft zealot]

Ann

[tired walrus]

don't try to think more or jump ahead.
focus only and exclusively on this one tidbit.
and tell me: do you understand or not

[timber plume]

Im having some difficulty understanding

[tired walrus]

then reread this last bit like 2 more times

[timber plume]

ok I understand now

[tired walrus]

cool\

finally, we say, there's one point this tangent line ought to pass through

that point is the point of contact -- the point with x = x_0 on our graph.

the point is $(x_0, e^{2x_0})$. it must lie on this line $y = 2e^{2x_0}x$.

[soft zealot]

Ann

[tired walrus]

so we plug its $x$ and $y$ coordinates in, \ and we get the equation $e^{2x_0} = 2e^{2x_0}x_0$.

[soft zealot]

Ann

[tired walrus]

don't try to think more or jump ahead.
focus only and exclusively on this one tidbit.
and tell me: do you understand or not

[timber plume]

plug it in where

oh nvm I get it

[tired walrus]

and then finally, we solve this equation for x_0.

doing so gets us x_0 = 1/2.

do you understand how x_0 = 1/2 comes to be? yes or no.

[timber plume]

Ill come back to this later, thanks

.close

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[high pine]

Is this a valid proof?

[high pine]

"If A and B are mxn matrices, the A and B are row equivalent if and only if A and B have the same RREF"

[storm crystal]

What’s your go-to caffeine hack for late-night study sessions? Coffee? Energy drinks? Something else?

[bold zenith]

!occupied?

[final saddle]

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

[high pine]

In the first part I (think I) proved that some set of elementary row operations can be applied to A to produce B. I then solve for A so I have A as a product of elementary row operations on B. I then apply some separate set of elementary row operations to A to produce the RREF form R. I can similarly apply that set of elementary row operations to the product I found before.

In the second part, I assumed that there was a set of row ops on A to produce R, and some separate set of row ops on B to produce R. If I solve for A and solve for B, I get A and B as two separate products, thus proving that A is row equivalent to R, and B is row equivalent to R.

[final saddle]

@high pine Has your question been resolved?

[final saddle]

@high pine Has your question been resolved?

[high pine]

Anyone?

[loud sundial]

Should be fine as long as you don’t need pedantic levels of detail

[final saddle]

@high pine Has your question been resolved?

[final saddle]

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[dense palm]

Hello people i need help with 1.62 to 1.64 i only know basic formulas that sin²alpha + cos²alpha = 1 and that tgalpha is sin alpha / cos alpha

[willow tiger]

!show

[final saddle]

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

[dense palm]

Well i am stucked at the beggining i did 1.64 with chatgpt trying to explain me but i am still confused

[willow tiger]

ok lets start with 1.62

why is there 2 equations

[dense palm]

I think that its a typo but im not sure

[willow tiger]

ok so lets look at 1.63 then

[dense palm]

Yeah okay

[willow tiger]

so you can turn the left side into something

try it

or what is the question to solve it or to prove it

[dense palm]

To prove it

Also this is the one i did with chatgpt

[willow tiger]

ok so lets try turn whats there in the left side to the right side

on the left side you can recognise a²-b²

[dense palm]

Yeah i did that

[willow tiger]

k what'd you get

[dense palm]

Now i can do the formula right?

[willow tiger]

yea

[dense palm]

This is it for now

[timber cove]

TRY DOING WITH TRADITIONAL RAW METHOD

[willow tiger]

huhhh

oh yea its right so far

[timber cove]

LIKE WITH REAL VALUES UK

[willow tiger]

ill leave him to you i have to go

[timber cove]

LIKE SIN ALPHA = P/H AND SO ON BY FAR U R DOING RYT IN CASE U R LOOKING FOR ANY OTHER METHOD

[late rose]

Do you recognise a difference of squares identity?

[willow tiger]

turn of caps lock bro

[willow tiger]

i just told him

[timber cove]

yeah sorry for that

[dense palm]

I did this one is it goo

[dapper hull]

nobody accepts that in identities/proofs

[timber cove]

its nowhere mentioned idc maths is when u feel its foundation

[open pike]

https://tenor.com/view/monster-versus-vs-alien-monster-vs-alien-gif-16037253809360562033

[dense palm]

What is that bruh

[timber cove]

the raw values like perpendicular/hypo , base/hypo... and so on

[dense palm]

Can u write it or do you have any pictures? Or can i just do my metod that we did in class?

[timber cove]

yeah u can do ur method np in it i just recommended a substitute like in place where u write sin alpha write it just as it is in raw as we know it is ratio of perpendicular and hypo of triangle

[dense palm]

Yeah i understand what do you mean but i dont think that i need to do that

[timber cove]

it can be bit long

[dense palm]

Can you help me with the rest

[timber cove]

where r they

[dense palm]

Here

[timber cove]

that 1.63 one ryt?

[dense palm]

Yeah i did that one

[timber cove]

so which one?

1.64?

[dense palm]

Yeah lets do that one

[timber cove]

okh lemme solve it first

[dense palm]

Ok

[dapper hull]

yk sec(x) and csc(x) ?

[dense palm]

What

[dapper hull]

1/cos(x) , 1/sin(x)

[dense palm]

I know the basic formula sin²l + cos²l = 1

Like if sin²l = 1-cos²l?

[dapper hull]

ok I want you to divide this formula by cos^2l

[valid carbon]

guys use brackets pleaes this looks so weird

[dapper hull]

what if I don't :3

[dense palm]

Is the answer sin²l?

[dapper hull]

the whole formula by cos^2(l)

[timber cove]

i got ur answer of 1.63

[dense palm]

Yeah i got that aswell its sin²l - cos2l now i am doing 1.64

[dense palm]

I got sin2l because cos cancels out right?

[dapper hull]

you still have the other side of the equality

[dapper hull]

canceling in addition is 1 not 0

[timber cove]

sorry i got answer of 1.64

i thought it was 1.63 but i solved 1.64

proceed it by making them raw except rhs

[dapper hull]

$\frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}$

[soft zealot]

qimmah

[dense palm]

Do you mean to use the formula on this one or?

[dense palm]

What do i do with this

[dapper hull]

you get an identity realting
tg^2(x) and 1/cos^2(x)
which is tg^2(x)+1=1/cos^2(x)

[dapper hull]

,rccw

[dapper hull]

and take a guess what the equation have

[valid carbon]

good

[timber cove]

bro proceed by making it raw then in (p/b)^2 -( b/h)^2 apply a2-b2 identity and beside them canceling out will yeild (p/h)^2 and now put their real names as goes by tan alpha or cos alpha or sin alpha according to ratios and u will get it

[dense palm]

@timber cove do i transform tg into sin alfpha / cos alpha or?

[dapper hull]

:3

[dapper hull]

,rccw

[dapper hull]

...

[dense palm]

What do i do now do i transfrom tg into sinl/cosl

[timber cove]

no need my method was short and conscise

now listen carefully

[dense palm]

Your words are confusing me its so hard to keep on with both of you at the same time

[timber cove]

transform cos^2 alpha as base / hypo u can write it b /h and do this for all trignometric functions exceppt rhs

[timber cove]

i will go one by one

[timber cove]

do this first

[dense palm]

Can i just do it like i did 1.63 please this is confusing me

[dense palm]

Can i do it like this

[timber cove]

like do tg alpha as b/p

[dense palm]

Or just send me how u did it and explain it because i cannot see it in my mind

[timber cove]

i m on my pc lemme see if i can send

i cant my phone is discharged

[dense palm]

Are you free to come to call and screenshare paint ir something

[timber cove]

u can follow as i said

[dense palm]

And explain it like that

[timber cove]

i m very insecure kinda i can go frm very basic here

so ig u r aware that trignometric functions are ratios of a right angle triangle

??

[dense palm]

Yeah but i dont understand your words bro like do it as b/p but how how will i write 1-cos2 alpha like that

[timber cove]

yeah thats wt i m explaining

[dense palm]

Yeah how will i ever write that

As p/h

Because i dont know the ratio i just need to prove it

[timber cove]

so now since u know it we define sin alpha (fact here we call commonly as sin theta) as ratio of perpendicular and hypotenuse of a right triangle??

[dense palm]

That (1-cos² alpha) (1+tg² alpha) is = to tg² aplha

[dense palm]

But i do not know the ratios or anything i just need to prove it

[timber cove]

ok lemme get it by ur method

wait

i got it

[dapper hull]

this existed for 12 minutes :3

[timber cove]

by ur method too

proceed by making tg^2 alpha as sin^2 alpha/cos^2alpha

[dense palm]

I do not understand it :3

[dense palm]

Okay

What now

[timber cove]

now multiply (1-cos^2 alpha )(1+sin^2 alpha/cos^2 alpha)

[dense palm]

Look how i did

[timber cove]

u will get 1+sin^2 alpha/cos^2 alpha-cos^2 alpha-sin^2alpha

[dense palm]

What do to now because first thing i transformed 1-cos²alpha inso sin²alpha

And did yours

Now what do i do

[timber cove]

u r going ryt

[dense palm]

What do i need to do

Nezt

[timber cove]

no waitt

dont transform 1-cos^2 alpha as sin ^2 alpha let it be as it is

[dense palm]

Oh okay

[timber cove]

now multiply

[timber cove]

u should get

[dense palm]

How will i get that

[timber cove]

by multiplying bruh

[dense palm]

Bruh by each day that passes i dont think that i can go into electrican engineering

[timber cove]

heyy wait u r engineer?

i m in 10th grade lol

[dense palm]

No i want to be an engineer

I am in high school

[timber cove]

ohh gud

tommorow i have exams

of my science

nd lol i m here

[ornate canyon]

Guys out of topic but…Anyone good at chem? Or know an active chem server?

[timber cove]

on dc

[ornate canyon]

Which subject?

[timber cove]

chem like

[ornate canyon]

Chemistry, organic, inorganic, physical

[timber cove]

all combined physics chem bio

[timber cove]

u r frm?

[dense palm]

Bruh i cant get it my brain is on fire

[ornate canyon]

Ooooh best of luck, need any help with concepts feel free to ask

[ornate canyon]

India

[timber cove]

same u must be older than me

[dense palm]

Bro i have chem exam in a week could you maybe help m

[ornate canyon]

100% I will love to help for exam

[timber cove]

bruh its just simple calculations

[ornate canyon]

Yes I guess, because u still have combined science

[dense palm]

Its intro to biochem

[timber cove]

yeah lol

[dense palm]

I have been doing math for 7 hours today

I cant do this

[ornate canyon]

Where is the original question?

[timber cove]

damn u r french?

[dense palm]

Here is the question

[dense palm]

Im bosnian

[ornate canyon]

You have to solve for alpha?

[timber cove]

ah ok

no prove it

[dense palm]

No i have to prove that left side is the same as right side

[timber cove]

thats too easyy

[ornate canyon]

Bro u know 1 + tan square identity?

[dense palm]

Okay so i multiplied it (i just rewrote what u said lol/

[dense palm]

Tan alpha = sin alpha / cos alpha

[timber cove]

bro u gottt titt

[ornate canyon]

No I + tan square alpha is sec square alpha

Mensur, make a list of identities on the side that will help you

[timber cove]

mensur u did it ryt now just take minus common and cos2alpha + sin2alpha will become 1 and all will be cancelled out and wuill left tg

[ornate canyon]

Otherwise just convert everything to sin and cos, you will get the answer

You can solve it even without identity

[timber cove]

yeah with raw method but he dont even know ratios

[ornate canyon]

But to satisfy your school teacher use identity 😅

[timber cove]

true asf in asia

[ornate canyon]

Oh

[timber cove]

its ryt bro

[timber cove]

now just do this

[dense palm]

Bruh what minus common

Bruh i get confused by the language xd

[ornate canyon]

Open your book and look at identities

[timber cove]

minus "-" this sign u know

[ornate canyon]

What textbook are u using?

[timber cove]

substraction sign

[dense palm]

Yeah i know that

[timber cove]

yeah so just do it and congo u will made it

make*

[ornate canyon]

Drixs u have exam?

[timber cove]

yeahh

[dense palm]

So cos² alpha - sin² alpha is 1 right?

Or -1?

[ornate canyon]

Mfker

You didn’t study

Open book and study identities

[timber cove]

bro when u will take common it will look like this
1+tg2alpha -(cos2alpha+sin2alpha)

and frm here u will get
1 + tg2alpha -(1)

[dense palm]

Yeah i got that

[timber cove]

congo

[ornate canyon]

Good Job

[dense palm]

Thank God i did it in like 2 hours

[ornate canyon]

Mensur full name Mensuration 😂

[timber cove]

lmaoo

[dense palm]

I have one more to do

Im practicing for my test

[timber cove]

i gotta go i have exams today havent studied anything

[dense palm]

Its in 2 days

[ornate canyon]

Good luck, if u need concept clarity ask me

[ornate canyon]

Study hard

[timber cove]

surely i will

bye guys gn tc

[dense palm]

@ornate canyon can you help me with the new task

[ornate canyon]

Yeah send q

btw actually I came to ask if anyone knows any chemistry server

I got distracted

Send ur q I will help

[dense palm]

I just need to prove it

So im thinking to change cos² alpha into 1- sin² alpha

[ornate canyon]

Now you know what 1 + tan square is right?

[ornate canyon]

No

You don’t change simple to complicated

Change complicated to simple

[dense palm]

What it is

[ornate canyon]

It’s an identity

1 + tan square = sec square

Sec square is the inverse of cos square

So at last all will cancel and answer will be 1

[dense palm]

Can we just use sin, cos, tg and ctg

Because we dont use sec in school

[ornate canyon]

Yes but convert tan

Don’t covert cos

Cos will cancel out at the end

Covert tan to sin and cos and solve

No need to use sec

[dense palm]

Ok lemme do the first part

Okay now what do i have to do

[bold turtle]

That's not what tangent is equal to

[ornate canyon]

Solve that

[ornate canyon]

Yikes I ignored that 😂

He is right

Look for the real value of tan mensur

[dense palm]

Oh yeah fuck

This has to be right

[bold turtle]

,rccw

[bold turtle]

ye*

[ornate canyon]

Yes correct

Your answer is also an identity

[bold turtle]

But if you're trying to write a proof, you shouldn't write it out like that

[ornate canyon]

Which means for any value of alpha, it holds true

[bold turtle]

LHS = (1 + tg^2 alpha) (cos^2 alpha)
= ...
= cos^2 alpha + sin^2 alpha
= 1 = RHS

[dense palm]

Guys can you give me a task simmilar to this one to do it right now

[bold turtle]

I mean you could google "trig proof questions"

[dense palm]

Yeah ur right i guess

Nah

They use

Angles

Do you have any perchance like in your head or sum

How to do this one

@ornate canyon @bold turtle

[bold turtle]

That's not an identity

The LHS there is not always equivalent to the RHS

[ornate canyon]

What is the question?

Original q

[dense palm]

This is the question

[bold turtle]

and I mean you can spot that right here since that fraction on the left equals ONE

https://tenor.com/view/hahaha-one-snapcube-sonic-the-hedgehog-gif-27436474

-# obligatory GIF is obligatory

[bold turtle]

Right but then this isn't a "proof" question

[dense palm]

How would i get an indentity

[bold turtle]

You're likely being asked to solve for alpha

[ornate canyon]

No

He is not even allowed to use sec

He is just wanting LHS = RHS

but the question is faulty

[bold turtle]

1. "cos x = 1/2" is a similar question; who tf uses secant for this
2. I've literally just pointed out that this isn't an identity, so it's illogical for this to be a proof question

[ornate canyon]

I think you are really stupid with all due respect

You are trying hard to be smart 😅 u will be one day

[bold turtle]

My name's green for a reason, you know

[ornate canyon]

Don’t waste my time

[bold turtle]

No, you stop wasting mine

@dense palm Where was the original question from?

[dense palm]

I got it from chatgpt

Is this one an indentity

[bold turtle]

...Ah

[bold turtle]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[bold turtle]

We have this factoid for a reason

ChatGPT and similar tools are language learning models, or LLMs

Their main function is to mimic human language

That means they try to "look" as correct as possible

That does not mean they are correct

,rcw

[dense palm]

Yeah i get it

[dense palm]

Is that good

[bold turtle]

This one is an identity though

[bold turtle]

Though again, for a proof, this structure is more solid

[dense palm]

I have these 1.112 - 1.114

Lemme try to do one and can you say is it good or bad

[bold turtle]

oh no, Croatian

I am out of my depth here /hj

[dense palm]

Its bosnian but yeah same

Language

[bold turtle]

yh

It's one of those "politically they're different"-type situations

[dense palm]

So like cos660⁰ = cos(360 + 300⁰) = cos300⁰?

Is that correct

[bold turtle]

yes, you can keep going

[dense palm]

calculate the values ​​of expressions

This is the question translated

[bold turtle]

The idea behind these questions in particular is, you should be able to evaluate these by hand

(meaning you can then check these with a calculator)

cos(x) = cos(360⁰- x) is another rule you should be familiar with

[dense palm]

Yeah i did that

[bold turtle]

No you've done the
cos(x) = cos(360⁰ + x)
one

The one with the plus

I'm talking about the one with a minus

[dense palm]

I did 1.112

Is it good?

[bold turtle]

No I mean, we're not done with the first one

cos(300⁰) can be further simplified

[dense palm]

Yeah but hoe

How

[bold turtle]

via this

All of these are evaluable to get a nice closed-form number

So as an example, cos(240⁰) = cos(360⁰ - 240⁰) = cos(120⁰)

[dense palm]

Oh i see so its gonna be cos 300 = 360 - 300

[bold turtle]

ye (though you're missing the cos in the RHS there)

[dense palm]

And its gonna be 60⁰ and its gonna be 0,5

[bold turtle]

There we go

All the others there have similar rules and goals

(btw this is slightly wrong - what does 3 times 180 give you?)

[dense palm]

How to do it with sin870

[bold turtle]

sin has the rule sin(x) = sin(180⁰ - x)

[dense palm]

Its not sin

Its tan

[bold turtle]

?

[bold turtle]

This is for the fourth one there

[dense palm]

Yeah

Its tg or tan585⁰

[bold turtle]

This being the second one, I gave you the relevant sine rule

[bold turtle]

yh but my point is that 585 isn't equal to 3 x 180 + 35

[dense palm]

Oh yeah

Its 45

[bold turtle]

yh

[dense palm]

I got sin150⁰

But the right answer is sin30⁰

How to get that

[bold turtle]

And then

[dense palm]

Oh now 180 okay

[bold turtle]

I gotta go, but the rest of these involve the same sorta tricks; here's a list of them I've found online

[dense palm]

Thank you

I did the sin

[bold turtle]

(remember that ctg is the reciprocal of tg, so no further tricks needed)

[dense palm]

Can u just check

Im sending it

[tepid narwhal]

ASTC

[bold turtle]

(and then sin(30⁰) equals...?)

[tepid narwhal]

sin 30° you gotta know is 1/2

[dense palm]

1/2

[tepid narwhal]

You gotta know that
(90°, 270°, 540°) Will change whatever the trig function that's in front.

E.x. sin (270°-X) it changes into cosx as for signs depends on the quadrant

[bold turtle]

@tepid narwhal Can you stop pinging me for this?

[tepid narwhal]

Oh sorry

[dense palm]

Anyways that enough for today my brain is fried

.close

[final saddle]

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[final saddle]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[tropic void]

Hello , how to write well in mathematics to have a complet mark in an exam ? or to solve well problems by a very schematic system ?

[cosmic warren]

Can you ask a more specific question than that? Like, do you have a specific problem to work on?

[bleak granite]

And if you mean in general, I would assume just show every logical step, ensure clarity, and keep handwriting neat?

also hi phoenixperson

[fallen geyser]

This question already has an open thread: https://discord.com/channels/268882317391429632/1419819119859077212

Please don't open several help threads or channels on the same question. One is enough.

I'm going to close this help channel now.

.close

[final saddle]

Channel closed

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Use .reopen if this was a mistake.

[tropic void]

I mean Because any answer to a question has three spaces :
the space of introduction , middle , end
Introduction is to prepare what there's and for what , the middle is execution
the end is concluding the solution
So how to increase the probability to 100/100
that I will have a complet mark ?

[fallen geyser]

Why not put this in your thread instead: https://discord.com/channels/268882317391429632/1419819119859077212

[tropic void]

@fallen geyser

They were helping me

But now nobody .

[final saddle]

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

i need help understanding cumulative distributive function for discrete random variables, there's a question I can send

[vital crag]

yes just do that in the beginning

[bold zenith]

Send it! :)

[errant root]

how do I find P(1.4<T<6) for this?

[vital crag]

Use F(x) = P(t <= x)

You should also have a list of properties of F

like F(b) - F(a) = P(a <= t <= b)

[errant root]

gotcha thanks, for cdf Im trying to lock in an upper boundary right?

Like for P(T>1.4), I would do 1-P(T<=1.4)?

[final saddle]

@errant root Has your question been resolved?

[final saddle]

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[lucid nymph]

So you have a function f(x)

If you found the area under the graph you take the definite integral, which is like splitting it into many tiny rectangles right

For revolutions its the same but its the volume of a bunch of disks

If you take one value of x and the tiny change dx as the width of a disk then the area of that disk is pi(f(x)^2)dx

Since f(x) is essentially the radius

And you just take the integral for that new function

[final saddle]

@clear saffron Has your question been resolved?

[final saddle]

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[foggy whale]

Can someone help me with the factoring part

[ornate tree]

What have you tried in advance?

[proper dagger]

please don't jump into another help channel to ask for help. you may ping helpers after 15 minutes if no one has noticed your channel.

[foggy whale]

i cant im not good at math

[dapper cloak]

okay so first you want to try make a binomial

then get out

[proper dagger]

age please?

[ornate tree]

bruh

[proper dagger]

<@&268886789983436800>

sorry, potential ToS violation, can't risk that here.

[dapper cloak]

@south sage u in big trouble

[proper dagger]

ok ignoring all that

[proper dagger]

what have you tried?

[foggy whale]

subtracted the two equations for ax-bx+b-a=0

then did (a-b)x+(b-a)=0

[dapper cloak]

what?

[proper dagger]

hm

[dapper cloak]

hm

[proper dagger]

there's one more factoring to make here

[dapper cloak]

yah

[proper dagger]

need a hint?

[foggy whale]

yes

[proper dagger]

or can you figure it out?

[foggy whale]

hint

[dapper cloak]

okay

[proper dagger]

so, you have (b-a), or more specifically, +(b-a), as the second term

do you see a way to manipulate this to get the (b-a) into an (a-b)?

[foggy whale]

😭 i've only done the easy kind of factoring before

is this algebra 1 or 2

[dapper cloak]

algebra 3

so basically you need to multiply by something

[proper dagger]

i have no idea what algebra it is

i don't share your education system

[foggy whale]

oh u are japanese?

[umbral igloo]

i got an interesting question if someone wants to help me out

[proper dagger]

!occupied at the moment, sorry.

[dapper cloak]

create your own channel

[final saddle]

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

[umbral igloo]

how

[umbral igloo]

can i dm u after youre free

[proper dagger]

#help-49 is open atm.

[proper dagger]

read my bio. (tl;dr, no. not for math help)

[proper dagger]

half, but it matters not

[foggy whale]

i dont think this is algebra 1 bc i havent seen this type of it

[proper dagger]

well, it's time for you to see some magic then

suppose i have x - y

i want to turn this into a y - x

note that the x in the first expression is positive, and the y is negative; exactly the opposite of our target, where y is positive and x negative

so we need to somehow flip the signs of both x and y

and the best way to do that is to multiply the expression by -1

or, factor out a -1 from the expression

so (x-y) is the same as -(y-x) (note the negative sign in front)

[proper dagger]

apply that here.

[foggy whale]

umm (a-b^2)(x-1)=0? idk

[proper dagger]

the square should not be there

but otherwise correct

it should just be (a-b)(x-1) = 0

there's nothing to square the b by anyway

[foggy whale]

ohh

[proper dagger]

now

what are the two solutions to this equation?

[foggy whale]

plus minus 1

[proper dagger]

not correct

[foggy whale]

oh

[proper dagger]

ok so from the equation i mentioned

let either factor be 0

then solve

[foggy whale]

wait thanks for the help so far but i have to go

it did help but im still confused kinda but tysm

cya

[proper dagger]

mkay

.close (OP gtg)

[final saddle]

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[quiet garden]

bruh what i dont get how x(t)=2cost and y(t)=2sint

[robust mulch]

do you know how to describe a point in polar coordinates?

[quiet garden]

i forgot

how

[robust mulch]

the sin and cos come from trig

[quiet garden]

so x=cos/r

and same for y

[robust mulch]

not quite

you want to multiply by r to cancel the denominator

so cos(theta)*r = x

[quiet garden]

oh yeah

so r is radius?

[robust mulch]

yes

[quiet garden]

how do u do b

[robust mulch]

have you considered the hint

[quiet garden]

i cant understand it bruh

for a i got 2cost,2sint, 2cost+3

[robust mulch]

you have (-1,sqrt3,2)=(2cost,2sint,2cost+3)

consider each coordinate separately

[quiet garden]

so i just plug those into t?

[robust mulch]

not plug in

solve for t

for example

-1=2cost

this will give you a two values for t

to pick the right value for t, you look at the second coordinate

[quiet garden]

ok i got cos=-1/2, sin=sqrt3/2 and cos=-1/2

[robust mulch]

keep t in there

because you need to solve for t

e.g. cos(t)=-1/2

[quiet garden]

yeah

so do i inverse of cos and sin on both side?

[robust mulch]

yeah do the appropriate inverse trig function

[quiet garden]

ok so t=2pi/3 and t=pi/3

[robust mulch]

remember, inverse trig functions (usually) have two different solutions on the interval [0,2pi), make sure t matches for both coordinates

[quiet garden]

they both work

[robust mulch]

if you plug in 2pi/3 and pi/3 into cos(t), you get -1/2?

for both

[quiet garden]

pi/3 is for arcsin

[robust mulch]

there is another value on the interval such that sin(t)=sqrt(3)/2

[quiet garden]

for cos its 2pi/3 and 4pi/3 and sin is pi/3 and 2pi/3

[robust mulch]

which value of t satisfies both?

[quiet garden]

2pi/3

i thought range for arcsin is only first and fourth quadrant tho

[robust mulch]

if youre looking at the function of arcsin, we limit it so it passes the vertical line test

however, there are an infinite amount of values that satisfy sin(t)=-1/2

in reality there are two families

sorry, cos(t)=-1/2

we want to pick a value from the right family that matches with sin(t)=sqrt(3)/2

[quiet garden]

so the range doesnt mater thne

?

[robust mulch]

when you are solving trig equations, things are different

[quiet garden]

k what i do now

[robust mulch]

find the derivate of each coordinate

then plug in t to find the direction of the tangent line

[quiet garden]

i plug in 2pi/3 into t?

[robust mulch]

no, find the derivatives first

sorry i gotta go, but youre very close to the end here

[final saddle]

@quiet garden Has your question been resolved?

[final saddle]

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[onyx smelt]

hows 18 ≡ 11p mod 26 same as 11p ≡ 18 mod26 ?

[tired walrus]

symmetry of modular congruence

if a ≡ b (mod m) then b ≡ a (mod m)

agree or disagree w/ this?

[onyx smelt]

because m divides both a-b and b-a ?

[final saddle]

@onyx smelt Has your question been resolved?

[onyx smelt]

@tired walrus

[tired walrus]

... yes, but i'm unhappy that you didn't give me a direct answer to my question

[onyx smelt]

sorry i wasnt sure

is hard for me i j started this class

but i got it now so tq

.close

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[onyx smelt]

how we wrote step 4 ?

[onyx smelt]

sorry 5

de =1+k one

[final saddle]

@onyx smelt Has your question been resolved?

[fiery bluff]

on the side:

1. handwriting issues: phi is not emptyset, and typically people use lowercase phi
2. to solve the equation de congruent to 1 mod phi(n), you can use extended euclidean algorithm between e and phi(n)

[onyx smelt]

can you pls elaborate 2?

[final saddle]

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[final saddle]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[olive burrow]

can someone please explain how to do 11

[olive burrow]

i don’t understand the reasoning

[ivory bear]

first of all find the relation b/w powers of 3 and 81

then try again

[timber cove]

CAN I DO THE 11 OR ANYELSE WT U WANT

[ivory bear]

!noans

[final saddle]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[paper warren]

take the hcf and you'll get the minimum side

[timber cove]

OKI

[paper warren]

*hcf

sorry

[olive burrow]

hmm

[olive burrow]

huh wdym

[olive burrow]

but like am i supposed to be knowng that

or is there a reason behind it

[ivory bear]

like 3^1=2, 3^2=9, 3^3=27 and so on

[timber cove]

THERES A TRICK I CAN SHARE

[ivory bear]

that is pretty basic uk

[olive burrow]

yes

[timber cove]

ALWAYS TRYNA PREDICT THAT IS UR ANSWER THAT WILL COME IS SHORTER THAN GIVEN VALUES IF YES THEN ALWAYS FIND HCF AND IF U PREDICT ITS GONNA BE HIGH THAN THE GIVEN VALUES FIND LCM MAKE SURE U DECIDE THIS BY JUDGING STATEMENT OF QUESTION

[olive burrow]

hmmm

[olive burrow]

i see ok thnx

.close

[final saddle]

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[craggy plank]

Why all caps? 💀

[timber cove]

by mistake turned on

[final saddle]

Available help channel!

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[craggy plank]

I see

[craggy plank]

.solved

[final saddle]

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[formal trail]

!da2a

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[pliant shore]

!15m

[final saddle]

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.

[plucky rover]

!xy

[final saddle]

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.

[dapper hull]

!status

[final saddle]

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

[spring haven]

!nosols

[final saddle]

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.

[odd seal]

!1q

[final saddle]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[spring haven]

!noans

[final saddle]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[edgy mauve]

Factoids practice

[plucky rover]

!noint soon (hopefully)

[spring haven]

!noint

shame

[edgy mauve]

!noint

(Rooting)

[plucky rover]

Root on the suggestion thread

[edgy mauve]

My version of rooting " "

[plucky rover]

Fair enough

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[leaden crypt]

Can someone help me with these and explain the laws used to convert them

[leaden crypt]

I've tried it with truth tables but I need to do it with replacement method

[final saddle]

@leaden crypt Has your question been resolved?

[final saddle]

@leaden crypt Has your question been resolved?

[final saddle]

@leaden crypt Has your question been resolved?

[junior token]

what are PDNF and PCNF?

[junior token]

what is the replacement method that you’re being asked to use

[final saddle]

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[leaden crypt]

.reopen

[final saddle]

✅

[leaden crypt]

principal disjunction and conjuctive normal form

[leaden crypt]

we have to convert it to pdnf and pcnf without using truth tables

using laws

[junior token]

so can you just use the distributive laws

[leaden crypt]

its one of the laws but you need to make it such that its conjunction of max terms

for pcnf

for pdnf disjunction of min terms