help-43

[icy nymph]

well

when u differentiate,

you multiply by the inside

because of chain rule, right?

when you integrate,

you divide by the inside

because of reverse chain rule

when you divide by half

that's the same thing as multiplying by 2

[mystic crater]

okay makes sense

and then 2 * -3 = -6 okay

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[quartz yoke]

Not too sure where ive gone wrong

[unkempt epoch]

,rcw

wait wrong thing

[quartz yoke]

K

[unkempt epoch]

how exactly did you get sin38 * 14= BC

[thorny urchin]

ABC isn't a right triangle, you can't use some like that

[quartz yoke]

Oh shit thats true

I cant use sine or cosine rule either

Or pythagoras

[unkempt epoch]

why not?

[quartz yoke]

oh i see it isnt right sngle

[quartz yoke]

Becausr i only have one side and one angle

[unkempt epoch]

you can use it

[quartz yoke]

how?

[unkempt epoch]

use pythagoras twice in BAD and CAD, then cosine rule in ABC

[quartz yoke]

Hiw do i use it in CAD?

[unkempt epoch]

AC^2-CD^2=64

(BC+CD)^2=196-64

and cosine rule in ABC

[quartz yoke]

kk ill try it

[unkempt epoch]

you have a calculator available right?

[quartz yoke]

yee

[unkempt epoch]

or are you supposed to estimate?

oh ok

[quartz yoke]

I dont have AC so how do i hse pythagoras flr that, because i need CD aswell

[unkempt epoch]

look what im saying is

[unkempt epoch]

you have 3 equations from this

and 3 variables (AC, CD, BC)

so its just a bunch of algebra at this point

[quartz yoke]

Yea and the variable i want is cd

Which i dont have

I have BD

[pure solar]

You have a calculator right?

[quartz yoke]

ye

[pure solar]

Why not get the angles..

By like arcsin or arccos

[unkempt epoch]

yes, so you get that by solving these equations

[quartz yoke]

wym?

[pure solar]

I will see that respond as you didnt learn it yet and cannot use it

[quartz yoke]

So is tht how i find AC? Then i use CD

[thorny urchin]

approach seems complicated
consider finding angle BAD first

[quartz yoke]

So its 90 + BAD + DBA = 180

[thorny urchin]

using inverse trig

[quartz yoke]

yh so like sin-1 nd studd but my question is how do i do that

[unkempt epoch]

you dont find just AC, you find all of AC, CD (which you want) and BC

[pure solar]

Oh did you learn inverses of trig functions?

[thorny urchin]

identify what information you're given
the angle involved

[unkempt epoch]

sin-1 is the same as arcsin

[quartz yoke]

k

[thorny urchin]

and from sohcahtoa
identify which trig function and it's inverse you should use

[unkempt epoch]

sohcahtoa?

what the hell is that?

[quartz yoke]

For triangle CAD? We need two different sides tho?

[thorny urchin]

the angle you want to use atm is BAD

[quartz yoke]

Sin cos tan

[unkempt epoch]

what?

[thorny urchin]

as that's part of a right triangle

[quartz yoke]

But what happens to the line AC then

[unkempt epoch]

how do you interpret sin cos tan from "sohcahtoa"?

[thorny urchin]

you don't worry about AC for now

[quartz yoke]

Sine = opposite/hyp cosine = adjacent/hyp and tan = opp/adj

[quartz yoke]

k

[unkempt epoch]

ohhh

[thorny urchin]

don't try jumping straight to the end goal, you could work backwards to see what would be useful
AC is part of the smaller right triangle
determining angle CAD will help you find it
and to get that angle you can first get angle BAD

(since BAD is the sum of CAD and the given angle)

[quartz yoke]

38 + 90?

[thorny urchin]

no

wheres that coming from

[quartz yoke]

Given angle 38 and the right angle 90

[thorny urchin]

sorry I wasn't clear
the given angle I was referring to was just the 38° angle

note that the 38° angle together with CAD form BAD

[quartz yoke]

the 38 is CAD tho??

[thorny urchin]

it's not,

38 is BAC

[quartz yoke]

Ohh kk

Soo i do 180-90 which is 90 and CAD + ACD = 90?

[thorny urchin]

true but no

you're not reading what I'm saying

[quartz yoke]

So then what are you saying I’m confused?

[thorny urchin]

focus on the angles around A,

note that the 38° angle together with CAD form BAD

[quartz yoke]

yep so how does tbat help?

[thorny urchin]

don't try jumping straight to the end goal, you could work backwards to see what would be useful
AC is part of the smaller right triangle
determining angle CAD will help you find it
and to get that angle you can first get angle BAD

[quartz yoke]

Cus im not sure what to do with this, as i dont know how to get BAD

[thorny urchin]

identify the angle you want to use (currently BAD)
then identify the relative positions of the sides you want to use

what is the position of the 8cm side AD relative to that angle?
is it the adj,opp or hyp?

[quartz yoke]

Adj

AD right?

[thorny urchin]

yes

similarly is the 14cm side the adj, opp or hyp

[quartz yoke]

Hyp

[thorny urchin]

yes

now referring to sohcahtoa from earlier
which trig function relates the angle, adj and hyp?

[quartz yoke]

cos

[thorny urchin]

yes

and can you set up an equation using that

[quartz yoke]

Cos x = 8/14 so x = cos-1 8/14

[thorny urchin]

I'd use BAD explicitly instead of x, but yes

[cerulean bramble]

-# don't mind me, just gonna repost the original question but rotated so that no one else has to keep contorting their neck

[quartz yoke]

Shit ty i didnt even notie😭😭

[quartz yoke]

kk so that comes out tk 55.15

[thorny urchin]

,w arccos(8/14) * 180/pi

[boreal girder]

Results provided by WolframAlpha

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Requested by ramonov

[thorny urchin]

yes

[quartz yoke]

So BAD = 93.15

[thorny urchin]

no

what we just got was BAD

[quartz yoke]

ohhhh so CAD = 55.14-38

[thorny urchin]

yes

55.15 - 38

[quartz yoke]

so its 17.15

[thorny urchin]

that will be your angle CAD

now you have the info needed to determine AC using the smaller right triangle

use the same approach again
identify the position of sides relative to the angle you're using

[quartz yoke]

AD = adj and we want the opposite so we use tan

[thorny urchin]

yes

[quartz yoke]

So its tan 17.14 x 8

[thorny urchin]

.15 assuming that's a typo again?

[quartz yoke]

Yeahhh mb

Thats abit big…

Its 137.2..

[thorny urchin]

make sure your calculator is in degrees mode

if your angle is in degrees

[quartz yoke]

Yeah it is

Wait i retyped it

Its 2.4687

Which seems much better lol

Awesome thank you so much for the help and patience @thorny urchin 🙏🙏

[thorny urchin]

np

[quartz yoke]

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[dreamy topaz]

Is this correct?

[keen granite]

,rccw

[keen granite]

[dreamy topaz]

[keen granite]

Mathematically, you've done:

[\frac{1}{60}\frac{hr}{mi}=\frac{0.1}{6}\frac{hr}{mi}]

then:

[\frac{6}{0.1}=\frac{mi}{hr}\frac{60}{1}\frac{mi}{hr}=60\frac{mi}{hr}]

[boreal girder]

PajamaMamaLlama

[keen granite]

[dreamy topaz]

Thank you very much

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[icy radish]

basically, uhm

simplify

[peak talon]

what have you tried?

[icy radish]

I've tried to see if I can factorize, but I didn't see anything like that. Then I multiplied both sides by -1, but it led to nothing

I'm just stuck. I'm just staring at the question. I don't know what to do.

Maybe I could get some hints or something like that?

But like the chapter is about dividing and multiplying different rational things, so maybe it's something with dividing and multiplying. I think

Like this?

Yeah, I did this before as well, but like, what do I do after? I'm like so stuck

Oh, okay

[cerulean bramble]

(Do you notice that 1 is another square number?)

[icy radish]

Wait, what?

[cerulean bramble]

(So now the numerator is the difference of two squares)

This thing - it's a difference of two squares

[south sage]

Perhaps more simply, let c = a+b

[icy radish]

Ooh, okay. Wait, let me try that

[icy radish]

?

[south sage]

Perfect

[icy radish]

Oh, okay, so you have to use like substitution, but when do you know when to use substitution?

[cerulean bramble]

Almost

[south sage]

Wait

[icy radish]

I find it easy to use, but I can never guess when I should use it, and I don't know the method. Oh fuck

[cerulean bramble]

c = a + b

[south sage]

1 - (a+b)

[cerulean bramble]

So -c = ...?

[icy radish]

Oh, so it's 1 - A - B, right?

Or am I onto the wrong track?

[south sage]

Yeah you got it

Regarding your earlier question

[icy radish]

Do you just do a lot of practice questions and then you just naturally get it

[south sage]

You never HAVE to substitute

But it makes things easier to see

[cerulean bramble]

It ultimately comes from practice

[icy radish]

How do you guys practice?

[south sage]

It helps to study basic identities

[cerulean bramble]

Just by doing

[icy radish]

I don't have time! 😭

Fuck math!

Okay, but thank you so much!

[south sage]

Like you should instantly recognize a difference of squares

And also the expansion of a squared binomial

(a+b)^2 = ?

[icy radish]

That's like A with the 2 on top of the A + 2 * A + B + B with the 2 on top of the B, you know?

Okay, sorry. I'm using a voice-to-speech thing, it's making a weird

ok my brain is fried thanks

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[thorn saffron]

I’m confused about #2 and #3 here

[thorn saffron]

,rotate ccw

[rugged stump]

What are you confused abt

[thorn saffron]

all of it tbh

Actually ik this

so I’d wanna find the lim, which I did for 2

but then how do I do step 2?

[strange pendant]

did you found the limit yet?

[thorn saffron]

here

wait I think I’m just dumb

isn’t h(9) =9+2 = 11 bc you just want to plug in x into x+2

and since the limit also =11

we conclude its 11 since f(a)=lim x->a f(x)

[strange pendant]

yes

so what would k be?

[thorn saffron]

11

[strange pendant]

yep

[thorn saffron]

and just to clarify a is the hole in terms of x?

[strange pendant]

you talking for 3) now?

[thorn saffron]

okay I’m also very lost on 3

[thorn saffron]

in 2, the value of 9 just comes from setting the canceled factor = 0

x-9=0
x=9?

[strange pendant]

for 2) is good that you did x^2 - 7x - 18 = (x-9)(x+2) and cancelled the (x-9)s

you concluded the limit is 11, right?

[thorn saffron]

yep

I see how that problem works

3 however… I’ve got no clue what it wants me to do

paper is filled out but teacher didnt explain it lol

[strange pendant]

you need to use the definition of continuity you wrote earlier

f(a)=lim x->a f(x)

[feral parcel]

hi guys idk how to do that sorry

[strange pendant]

for example, f(a)=lim x->a f(x) , when a = 0

[thorn saffron]

hmm okay

wait so are you finding all one sided limits for 0 and 5?

[strange pendant]

evaluate f(0) and compare that to lim x-> 0^- f(x)

[strange pendant]

yes since we assuming that this function is continous for some a and b, we need to equate f(0) with the lateral limits of x approaching 0 from the left and x approaching 0 from the right

aswell as 5 from the left and 5 from the right, for f(5)

when x = 0 , f(x) = ?

[thorn saffron]

b

[strange pendant]

ye

now use the limit of when x is approaching 0 from the left

[thorn saffron]

2

[strange pendant]

and you know from the definition of continuity that what?

[thorn saffron]

the left and right must be =

so b=2??

[strange pendant]

yes

now find a

when x = 5, f(x) = ?

[thorn saffron]

okay so as x ->5 from the left, you have ax+2

then u need give x to be some value

but idk how to find that

[strange pendant]

first, find when x = 5, f(x) = ?

[strange pendant]

its, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} ax + 2$

[boreal girder]

Renato

[thorn saffron]

yeah

[strange pendant]

that limit is ?

[thorn saffron]

a(5)+2

[strange pendant]

yeah, now compare with the limit as x is approaching 5 from the right

[thorn saffron]

-3(5)+7

So -8

[strange pendant]

yep\

now equate them together (because we are assuming continuity)

[thorn saffron]

so a=-2

!??!?

[strange pendant]

yes

[thorn saffron]

alright Im starting to see how it works

I’ll do some practice problems to make sure I got it

[thorn saffron]

I really appreciate the help

[strange pendant]

good luck, have fun

[thorn saffron]

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[royal spindle]

haiia i got these graphs and data but idk what the mean , median and mode would be ngl.. insight would be much appreciated! i also cannot tell how the graph is skewed

[twin idol]

https://cdn.discordapp.com/attachments/948777486018285638/1415550888663322655/20250910_231102.jpg?ex=68c39dff&is=68c24c7f&hm=d56d0b83b4624eedd435f058c395b2f5db1cd4b17fb1d105aad480d91f7d03f7&

,rccw

[twin idol]

The frequency polygon at dice score 2 and 3 should be 0

Same at dice score 11

[royal spindle]

sorry j do not understand wym

[twin idol]

Like so
Yellow line = line that the question wants

Anyways

Mean = arithmetic average

It's where you add all your data and divide the result by the number of data you have

@royal spindle

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[royal spindle]

thank u!!

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[boreal umbra]

im stuck on part b (actually i did part a on a whim a few days ago so i dont really get what i wrote)

thanks!

‘if f(x) is divided by x, remainder is 24’ does this mean that f(x)=Q(x)(x)+24? because f(x)=quotient x divisor + remainder

[cerulean bramble]

Almost - who said the quotient's the same as when you divided by x+3?

[boreal umbra]

ah right

[cerulean bramble]

If I said 35 is divisible by 7; and when I divide it by 4 I get a remainder of 3, I couldn't conclude for instance
35 = 5 x 7;
35 = 5 x 4 + 3

(you can see why that's not true lol)

[boreal umbra]

yepp

how do i solve it without knowing the quotients?

[cerulean bramble]

You've got the same issue going on here btw

[boreal umbra]

riight

[cerulean bramble]

You could start with this line though

[good handwriting btw that's always a plus]

We're told this function's divisible by x - 1

What other thing here can you spot that's divisible by x-1?

[boreal umbra]

{haha thanks! you can see im frustrated in b part lol}

[boreal umbra]

uh

no idea sorry

[cerulean bramble]

Could we factorise the (x^2 -1) somehow?

[boreal umbra]

ohh (x+1)(x-1)

[cerulean bramble]

ye

So in fact that whole Q(x) (x^2 -1) has a factor of (x-1) lurking in there

And the whole function is divisible by (x-1) [we're told as much]

[boreal umbra]

yep

[cerulean bramble]

So then surely the other part should be divisible by x-1?

That is, x-1 is a factor of kx + 8

[boreal umbra]

wait so since f(x) is divisible by x-1, the other part is also divisible by x-1?

[cerulean bramble]

ye

I'll give a numbers example

[boreal umbra]

sorry whats 'the other part'? like q(x)?

[boreal umbra]

thank you :D

[cerulean bramble]

If 999999 = 1407 + 998592 (it is), and you can spot that 7 divides 1407, and you're told that 999 999 is divisible by 7 as well

We can conclude that 998592 is divisible by 7

[boreal umbra]

:0 right i never thought of that

so kx+8 is divisible by x-1

do we factorize it..?

[cerulean bramble]

I mean we can try to

But you can probably spot it?

I'll give you a hint

The factor theorem says
if (x-a) is a factor of a function f(x), then f(a) = 0

[boreal umbra]

💀 right

[cerulean bramble]

(any function f, dw about it not being the same as the f() in the question)

[boreal umbra]

omd sorry that was kind of stupid

[cerulean bramble]

So if we've got "x - 1 is a factor of kx + 8", what value of x could we try?

[boreal umbra]

1

right i got -8

[cerulean bramble]

ye

YE

[boreal umbra]

thank you :D

but now part b is even more confusing

[cerulean bramble]

So now we've got k (in a way that's a bit more plausible-sounding lol)

[boreal umbra]

yep yep

so f(x) is divided by x, remainder is 24

i cant use remainder theorem cause x=0 right?

[cerulean bramble]

Admittedly (what you call) the remainder theorem is less so a theorem, more so a definition of remainder

[boreal umbra]

ah

[cerulean bramble]

There's some quotient function P(x) such that f(x) = x P(x) + 24

That's sorta what a remainder means

(I'm using "P" so as to not confuse ourselves with the Q from part a)

[boreal umbra]

yep

wait

do i just x P(x)+24=0

[cerulean bramble]

Not quite...

We want to find the roots, sure

But we want to be a little strategic

[boreal umbra]

hmm yeah

[cerulean bramble]

using the other bits of information we have

[boreal umbra]

x+3 is a factor of f(x)?

[cerulean bramble]

For instance, yh

Here's a thought - f is a cubic function. If I divided it by (x^2 - 1), what function should my quotient be?

[boreal umbra]

mx+r?

[cerulean bramble]

ye

(I'll work with those constants lol; I'd've gone with a and b but that still works lmao)

[boreal umbra]

thanks :')

hmm

we can use a and b

idm

[cerulean bramble]

Aight

Right, so we have that Q(x) as some (ax + b)

[boreal umbra]

yep

[cerulean bramble]

How can we rewrite this line then?

(noting also that we now have k as well)

[boreal umbra]

f(x) = (ax+b)(x+1)(x-1)-8x+8

[cerulean bramble]

ye

Okay, let's use the factor+remainder theorem to see what else we can find

Firstly - we're told x + 3 is a factor of f(x)

[boreal umbra]

yep

[cerulean bramble]

So what value can we put for x to make f(x) = 0?

[boreal umbra]

-3

[cerulean bramble]

ye

[cerulean bramble]

plonk that in, see what equation you get

[boreal umbra]

wait but x-3 is a factor does it mean it is a factor with (x^2-1)

[cerulean bramble]

Nah

To refer to my numbers example...

[boreal umbra]

this right

so can i use them tgt?

[cerulean bramble]

999999 is divisible by 9, but that doesn't allow us to then say "does 9 divide 1407?"

[boreal umbra]

yep

[cerulean bramble]

[spoilers: 1407 is not div. by 9 lol]

[boreal umbra]

so why am i subbing x=-3 into this? (sorry TT)

[boreal umbra]

lol

[cerulean bramble]

Because we know that f(-3) = 0, from the factor theorem

But hey, we have a way to express f(x)

Meaning we can write some expression for f(-3)

And we can claim that this is equal to 0

[boreal umbra]

yeah

[cerulean bramble]

So we get an equation in a and b

[boreal umbra]

that would mean another quotient that isnt (x^2-1) right?

[cerulean bramble]

I mean, yes, but we don't need to worry about that

[boreal umbra]

ah

[cerulean bramble]

We're just substituting x = -3 into this expression

[boreal umbra]

24a-8b=32

[boreal umbra]

ah

[boreal umbra]

wait f(x) is divisible by x-1 so can i just sub x=1 into here?

[cerulean bramble]

You could, but that's going to remove the a and b in the resultant equation

So it's not really a meaningful substitution

[boreal umbra]

rightt

[cerulean bramble]

You did mention some other substitution we could use tho...

[boreal umbra]

x=0?

[cerulean bramble]

yee

[boreal umbra]

oh that works rightt

[cerulean bramble]

Because dividing f(x) by x gives us a remainder of 24, we then have that f(0) = 24

[boreal umbra]

riight

b=-16

a=-4?

[cerulean bramble]

I'll take your word for it, i have yet to check the working out lol

But now we can find f(x)

And since we have some factors, we can then just attempt to factorise it

[boreal umbra]

so i just set f(x)=0 and solve it right

[boreal umbra]

f(x) = (-4x-16)(x+1)(x-1)-8x+8=0 ?

[cerulean bramble]

Slightly different; factorising is a way to find the roots (and finding the roots is synonymous with solving f(x) = 0)

[boreal umbra]

rightt

[cerulean bramble]

But since the ultimate aim is precisely to find the roots, I won't discourage that

[boreal umbra]

can i factorize out (x-1) first

[cerulean bramble]

Yh

[boreal umbra]

all are integers

so claim is correct

[cerulean bramble]

ye

[boreal umbra]

wow

thank you sm!!

[cerulean bramble]

no worries

!done

[compact pewter]

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

[boreal umbra]

.close

[compact pewter]

Channel closed

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Use .reopen if this was a mistake.

[boreal umbra]

tysm :D

[compact pewter]

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[carmine garden]

for every integer more than 2, n^2-1 isn't prime

so U(n) doesn't form a group, much less a subgroup

[mild sky]

What is U(n)?

[carmine garden]

Group of units

[mild sky]

Of?

Roots of unity?

[kind viper]

yeah actually what's the definition of U(n) @carmine garden

the FULL definition

[carmine garden]

one minute

[carmine garden]

multiplicative group of integers modulo n

[kind viper]

ok right so what's this about U(n) and U(n^2-1)?

[carmine garden]

I'm saying n^2-1 is not prime for n>2

[kind viper]

sure it isn't

but what does this mean for U(n^2-1)?

[carmine garden]

and U(n) forms a group iff n is prime

[carmine garden]

It doesn't form a group

[kind viper]

wrong

[hushed magnet]

U(n) by definition is a group

[hushed magnet]

group

[kind viper]

U(n^2-1) is never empty and it is in fact a group

it just does not consist of EVERY SINGLE integer between 1 and n^2-2 incl.

[hushed magnet]

if the question says

show that the group U(n^2-1) is not cyclic
then your answer really shouldnt be
its not a group

[compact pewter]

@carmine garden Has your question been resolved?

[mild sky]

Notice n^2 - 1 = (n-1)(n+1)
and gcd(n-1,n+1)=1 for n >= 4

[carmine garden]

Ah

Really sorry for disappearing

I fell asleep

[mild sky]

Its fine

[carmine garden]

Lemme try again, I'll reopen the channel if I fail

.close

[compact pewter]

Channel closed

Closed by @carmine garden

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[compact pewter]

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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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[gilded lark]

5th and 3rd

[vale star]

R u allowed to use calculator

[gilded lark]

No

Even if yes I wont

[peak talon]

that is blurry, could you take the pic again?

[gilded lark]

Ohk

[gilded lark]

Wait

[peak talon]

Take your time

[vale star]

Notice that sin40=sin(60-20) and sin80=sin(60+20)

[pine thicket]

Write sin80 aa sin10

The the gp formula

[vale star]

For 3rd

[peak talon]

wsp

[pine thicket]

<@&268886789983436800>

[peak talon]

just go to #discussion or #chill

[pine thicket]

@gilded lark

[gilded lark]

aa?

[pine thicket]

as

[gilded lark]

Sin 10?
Or cos 10

[pine thicket]

Yeah cos10 mb

[gilded lark]

What?

[gilded lark]

Done

Now put 2SC formula right.

[peak talon]

it looks much clearer, sorry for disturbing

Just continue and ignore me

[pine thicket]

Have you learned the formula x, 60+x, 60+x?

[graceful kelp]

jee moment

[gilded lark]

😬

[graceful kelp]

oh wait yeah

u should know it as you are doing JEE

[pine thicket]

No?

[gilded lark]

It isn't formally told but ik

[pine thicket]

Tell me what it is

[graceful kelp]

?

[graceful kelp]

are u asking me?

[pine thicket]

No, I was asking him

[gilded lark]

Answer is 2sqrt(3)

[gilded lark]

1/4S3x

[graceful kelp]

ye

[pine thicket]

Nice

[gilded lark]

So now let's go to 5th

[pine thicket]

1/4 S3x

Space is important

[gilded lark]

Yea

[pine thicket]

What have you tried?

[graceful kelp]

oo classic technique

[gilded lark]

It takes so much time to send an image

[pine thicket]

Lol

[gilded lark]

#discord

Hell my writing

[pine thicket]

Reacting to your own message🥀

[gilded lark]

Expanded the terms and added both equations

[graceful kelp]

while that works, there is another trick

[graceful kelp]

2a = (a-b)+(a+b)

[gilded lark]

Pls elaborate

[graceful kelp]

$\tan(2 \alpha) = \tan([\alpha - \beta] + [\alpha + \beta])$

[boreal girder]

mersenne primes

[graceful kelp]

and u can easily find alpha-beta and alpha+beta

[gilded lark]

Ohk alpha + beta

But Alpha - beta👎

[graceful kelp]

no like

not finding those

but the tangents of them

[gilded lark]

R u indian or JEE Aspirant

I'm an 11th standard guy

[graceful kelp]

cos(alpha+beta) = 3/5
=> tan(alpha+beta) = 4/3

[gilded lark]

So yea

[gilded lark]

Oh that worthy

Fr

[graceful kelp]

indian and preparing for isi

[graceful kelp]

same 💀

[gilded lark]

My dream😭

[graceful kelp]

Solve for tan(alpha-beta) similarly and then apply the direct formula

[graceful kelp]

gl

[gilded lark]

But ig i would have to take BSDS only😭

Then if not via JEE Mains then CUET will be easy

[gilded lark]

12th?

[graceful kelp]

dam gl

[graceful kelp]

11th

[gilded lark]

B stat or Math?

[graceful kelp]

math

[gilded lark]

Done ty

[gilded lark]

Can I dm u?

[graceful kelp]

Sure

[gilded lark]

. close

[graceful kelp]

.close

[gilded lark]

.close

[compact pewter]

Channel closed

Closed by @gilded lark

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[hardy cedar]

$3+2\lga-1/2\lg(4b^2)$ write as a single logarithm to the base 10 where a and b are both positive

[kind viper]

$3 + 2 \lg(a) - \frac{1}{2} \lg(4b^2)$

[boreal girder]

Ann

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

[kind viper]

did you mean this?

[hardy cedar]

yes

[kind viper]

ok, progress? @hardy cedar

[hardy cedar]

used the power rule

3+lga^2-lg4b

3 can be written as

[kind viper]

stop

[hardy cedar]

3lg

[kind viper]

stop

$\frac12 \lg(4b^2) \neq \lg(4b)$

[boreal girder]

Ann

[hardy cedar]

2b?

cuz u take square root of 4 too?

[kind viper]

yes

$\lg(\sqrt{4b^2}) = \lg(2b)$

[boreal girder]

Ann

[hardy cedar]

so 3+lga^2+lg(2b)

3 can be written as

3lg

[kind viper]

3lg?

[hardy cedar]

or lg1000

[hardy cedar]

yea

[kind viper]

it has to be lg(something), you can't just put a bare lg without a number inside of it.

lg is a function.

it is not a number, and 3*lg does not make sense.

[hardy cedar]

how do you write subscripts in latex

3lg10

[kind viper]

_

but you are not looking for a subscript

[hardy cedar]

$3lg_10$

[boreal girder]

Tan

[kind viper]

$\lg(x)$ means $\log_{10}(x)$

[boreal girder]

Ann

[kind viper]

it's $\lg(10)$ and not $\lg_{10}$.

[boreal girder]

Ann

[hardy cedar]

but its

lg 10 to the base 10

cuz anything to its same base is 1

$lg10_10$

[boreal girder]

Tan

[kind viper]

no

[hardy cedar]

$\log_10(10)$

[boreal girder]

Tan

[kind viper]

lg, without the o, ALREADY means base ten.

you should NOT be indicating base ten a second time.

[hardy cedar]

oh

[kind viper]

also {} for superscripts or subscripts longer than one character

e.g. $x^{12}$

[boreal girder]

Ann

[kind viper]

anyway, with this notational mishap taken care of,

yes, $3 = \lg(1000)$.

[boreal girder]

Ann

[hardy cedar]

$x^12$

[boreal girder]

Tan

[hardy cedar]

i see

yeah

so i got

lg1000+lga^2-lg2b

using the order of operations

you do whichever sign comes first

so its more like

(lg1000+lga^2)-lg2b

right??

[kind viper]

if you wanna put those brackets nobody can stop you

but they are not obligatory

[hardy cedar]

no im just putting them to show you

what i mean

[kind viper]

there's no obligation to handle the addition first, but if you want to do it that way, that's perfectly ok.

[hardy cedar]

really?

i heard about it once

like

if its 3-5+2

you thik cuz of pemdas you do addition first

[kind viper]

3-5+2 is NOT 3-(5+2).

[hardy cedar]

but u actually do 3-5 first

[kind viper]

addition and subtraction have the SAME priority.

[hardy cedar]

yes yes

thats what im saying

so if subtraction comes first you still just subtract first

and in the same sense

[lime juniper]

I've only seen lg to mean log_2

[hardy cedar]

if multiplication comes first

you multiply first

[plain quail]

Ive only seen lg to mean log_10

[lime juniper]

¯_(ツ)_/¯

[plain quail]

Countries be like that fr

[kind viper]

i think you're kinda overthinking this, tan

[lime juniper]

okay , glad we have clarity now

[kind viper]

youve got 3 terms here that youre GONNA COMBINE in one order or another\

[plain quail]

Owh

[hardy cedar]

so yeah my final answer is $\frac{lg1000a^2}{2b}$

[boreal girder]

Tan

[hardy cedar]

just

its really..

fishy

[kind viper]

why is the 2b outside of the log?

[hardy cedar]

i wanted to put it in log

[kind viper]

ok then dont shove the log into the fraction mate

[hardy cedar]

but im not good at using latex

[kind viper]

$\lg \left(\frac{1000a^2}{2b}\right)$

[boreal girder]

Ann

[hardy cedar]

yeah

[kind viper]

or if you're a bracket-hater, $\lg \frac{1000a^2}{2b}$

[boreal girder]

Ann

[kind viper]

either way you can and should simplify 1000/2

[hardy cedar]

i heard you cant

one time i got something like

x^2-10\x-5

i thought

oh

thats easy

x+2

but it was not

[kind viper]

you misunderstood

[kind viper]

very badly written fraction

[hardy cedar]

$\frac{x^2-10}{x-5}$

[kind viper]

the thing about fractions is that you CAN cancel when the top AND bottom are PRODUCTS. and if they AREN'T then you CAN'T cancel.

[boreal girder]

Tan

[kind viper]

right, plaintext for that is (x^2 - 10)/(x - 5).

brackets and forwardslash obligatory.

[hardy cedar]

oh

[kind viper]

the thing about fractions is that you CAN cancel when the top AND bottom are PRODUCTS. and if they AREN'T then you CAN'T cancel.

[hardy cedar]

what about the b

do i just leave it alone

[kind viper]

$\frac{1000 \times a^2}{2 \times b}$ is precisely the good'' kind of fraction that \emph{does} allow you to cancel. so do \textbf{not} attempt to walk away with a blanket i can't cancel in fractions or i will get yelled at'' thought pattern.

[boreal girder]

Ann

[kind viper]

yes of course you do.

[hardy cedar]

okay

$\lg(\frac{500a^2}{b})$

[boreal girder]

Tan

[strange ermine]

$\lg\left(\frac{500a^2}{b}\right)$

[hardy cedar]

why are the parenthesises not big

[boreal girder]

Alberto Z.

[hardy cedar]

wth man

$/lg/left1\right$

[boreal girder]

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

[hardy cedar]

latex is so hard

okay way

[hardy cedar]

this is correct, right Ann?

oh yeah i have another question

its like

so they asked to solve some stuff and give answers in exact form

i got 3^1/4

should i just keep it as that

or make it

[kind viper]

yes this is correct

3^(1/4) is fine to leave as-is for exact form

[hardy cedar]

4√3^1

[strange ermine]

Backslashes, not (forward) slashes

[strange ermine]

?

[hardy cedar]

okay thank you

[low canyon]

Help

[strange ermine]

!occupied

[hardy cedar]

,rccw

[compact pewter]

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).

[hardy cedar]

there use this its kinda more understandable

.close

[compact pewter]

Channel closed

Closed by @hardy cedar

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[low canyon]

Sorry

[compact pewter]

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[forest token]

not able to start

[kind viper]

can you find $\int_0^1 x^n \ln(x) \dd{x}$

[boreal girder]

Ann

[kind viper]

for natural n

[forest token]

yes

integration by parts

[kind viper]

ok do it

then consider that $\frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n}$ and use that in your integral (and dismiss any concerns about convergence)

[boreal girder]

Ann

[forest token]

-1/(n+1)^2

[forest token]

could u explain this..how is 1/(1-x^2) = summation (stuff)

[kind viper]

bad brackets.

but it is just infinite GP summation.

[forest token]

oh right

not sure how to proceed

do i use ibp on my integral?

oh wait

[kind viper]

$\int_0^1 \frac{\ln(x)}{1-x^2} \dd{x} = \int_0^1 \sum_{n=0}^{\infty} x^{2n}\ln(x) \dd{x} \overset{\text{MAGIC}}{=} \sum_{n=0}^{\infty} \int_0^1 x^{2n} \ln(x) \dd{x}$

[boreal girder]

Ann

[forest token]

yeah i just got that

okay but how did u see that 😭

[kind viper]

what, like $\frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n}$?

[boreal girder]

Ann

[forest token]

yea like that was tough to notice

[kind viper]

dunno

ig it's like, experience with series expansions?

[forest token]

oh alr alr

thank you

.close

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[compact pewter]

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[copper sandal]

how can i proove there is a triangle with sides equal to the medians of another???
And how am i even supposed to come up with solution on my own for something like this

[copper sandal]

it really feels impossible

[weak cobalt]

Just take an equilateral triangle?

!original

[compact pewter]

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.

[copper sandal]

bulgarian

oh wait i missed out a bit

[weak cobalt]

Translate as accurately as possible

[copper sandal]

okay

Proove that a there exists a triangle, with sides equal and parallel to the medians of said triangle (said underlined as i dont understand it myself clearly)

[weak cobalt]

of any given triangle?

[copper sandal]

i think

im pretty sure im supposed to explain this but i genuiely dont know how

[weak cobalt]

Vectors?

[copper sandal]

Might work, havent tried though

im not really sure how it would work

[weak cobalt]

If I give you three random vectors u,v,w, can you arrange them to form a triangle, and if so, how?

[copper sandal]

yes but thats what im trying to figure out how to do

[weak cobalt]

... just take two random vectors and find a condition for the third vector

[copper sandal]

oh so like

i have 2 vectors with a equal point

and i try to proove the third vector can fit the gap

[weak cobalt]

Right

[copper sandal]

ohhhhh

okay thanks dude

ill be closing this because i actually have to go do something but i think i can do it on my own now

.close

[compact pewter]

Channel closed

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[compact pewter]

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[tender pasture]

how you meant to find this without a calc

[compact pewter]

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

[tender pasture]

2

[cerulean bramble]

CSP, I presume he means "how do you answer b?"

[rotund sphinx]

ik

[cerulean bramble]

Given that it supposes the answer from a should be enough

[rotund sphinx]

I'm asking to see if they noticed things like ||360-30=330|| or ||simplfying cos(360-x)||

[tender pasture]

wht

[cerulean bramble]

which it is, but unless you knew that second spoiler, it wouldn't be so clear-cut

[tender pasture]

i need help with b

bro pls any help

[cerulean bramble]

https://www.highermathematics.co.uk/cast-diagram/ (otherwise the search term is "CAST diagram")

[tender pasture]

what do i click

idk

[cerulean bramble]

..."click"?

Scroll odwn a bit

There's a diagram like this

But essentially, for this particular question the relevant result of this is that cos(x) = cos(360° - x)

[tender pasture]

aaaa

i dont understand any of this

whats the method for this

[cerulean bramble]

brackets here - I've given the search term for you

[tender pasture]

what is a quadrant

[cerulean bramble]

,w quadrant

[boreal girder]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by waes_wires

[cerulean bramble]

oh come on

,w define quadrant

[boreal girder]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by waes_wires

[cerulean bramble]

Sense 1

[tender pasture]

i see

[cerulean bramble]

You draw an angle going from the positive x-axis direction, anticlockwise until you reach your angle

Where you stop is the quadrant of that angle

So 30 degrees is in the first quadrant, for instance

https://youtu.be/pqqmPh2ISRQ, but I recommend searching for videos yourself to see what clicks and what doesn't

[tender pasture]

should i watch tht?

[cerulean bramble]

I mean why not

It explains CAST diagrams in a way I won't be able to easily convey through text

[tender pasture]

why this alvel im doing gcse what had my teacher set me up for 😭

[cerulean bramble]

geometry questions are always like that

Tbh, A Level Year 1 has a lot of bleed-over from GCSE

[tender pasture]

i see ok let me finish watching the vid hopefully i have a better understanding by the end of it

[cerulean bramble]

dw, I do have to specify that to my own students sometimes

[tender pasture]

wait by the end of it i should be able to ans the question right?

[cerulean bramble]

In theory yh

[tender pasture]

because it went from drawing cos graphs to drawing angles

[cerulean bramble]

Though in fairness, I have given you the main hint for this particular question here:

[further hint - what happens if I let x = 330 degrees?]

[tender pasture]

cosx=330?

x= cos-1(330)

[cerulean bramble]

(no, replace x with 330)

[tender pasture]

root3/2

i used a calc

[cerulean bramble]

The point is you don't NEED a calc

(which I thought you got from the first message you sent)

[tender pasture]

cos330=

idk

[cerulean bramble]

cos x = cos (360 degrees - x)
cos 330 deg = cos (360 deg - 330 deg)

All I've done here is replace x with 330 degrees

[tender pasture]

yes

[cerulean bramble]

Is this clear so far?

[tender pasture]

yes

[cerulean bramble]

Okay, look inside the brackets on the right

[tender pasture]

cos330=cos60

[cerulean bramble]

...no

[tender pasture]

yes

[cerulean bramble]

We've got 360 degrees minus 330 degrees

What's that equal to?

[tender pasture]

30

cos330=cos30

[cerulean bramble]

30 degrees (important to keep that in mind)

[tender pasture]

cos30= root3/2

yes

[cerulean bramble]

Right, and part a already gave us what cos(30 deg) was

So we're basically done

[tender pasture]

yes wow

nice

sickk

that was awesome ty

[cerulean bramble]

np

[tender pasture]

.close

[compact pewter]

Channel closed

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[compact pewter]

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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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[hidden walrus]

he substitues E for sigma / epsilon not, and the reason given is gauss law e value for infinite plane sheet of charge, but according to gauss law, e value is sigma / 2 epsilon not, not sigma/epsilon not

[rigid perch]

but there are two plates so the electric field adds up to that

each individual plate contributing half

[hidden walrus]

oh

that makes a lot of sense

[hidden walrus]

thank you

i will be back with more questions very soon

.close

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[compact pewter]

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

Can someone help me understand the question? I thought $\mathbb{R}^{[0,1]}$ is the set of real valued functions on 0 to 1

[boreal girder]

BOSS

[vivid breach]

not every function is continuous

[delicate musk]

wait, so even non-continuous functions make a vector space over R?

like what exactly does $R^{[0,1]}$ contain

[boreal girder]

BOSS

[vivid breach]

what about continuity is exactly important in making a vector space

you shouldnt be that surprised by this

R^[0,1] is the set of all real-valued functions on 0 to 1

[delicate musk]

Do the functions go from [0,1] -> R or [0,1] -> [0,1]

[vivid breach]

it says R^[0,1] instead of [0,1]^[0,1], what do you think

[delicate musk]

im just trying to understand i saw the notation for the first time today lmao

ok but yeah to R then

Ok so its all functions from [0,1] to R, so something like f(0.5)=12 would be in here?

[vivid breach]

you cant define a function by just a single point

[delicate musk]

f(x)=12

[vivid breach]

yea

[delicate musk]

ok cool

[vivid breach]

also, a function like this

this is a real-valued from [0, 1] to R

[delicate musk]

yeah non-countinuous

makes sense

[vivid breach]

so youre already given that R^[0, 1] is a vector space

what you now need to confirm is that the continuous functions within R^[0, 1] form a subspace

[delicate musk]

Yeah ok so

to show its not empoty we can say that

f(x)=0 is in U (calling the new set U)

[vivid breach]

yep

[delicate musk]

sk we have the additive identity

[vivid breach]

we have that, yea

[delicate musk]

Then for additive closure

asssume that $g,f \in U$

[boreal girder]

BOSS

[delicate musk]

I mean the sum of two continuous functions is always continuous

do i need to prove that out?

[vivid breach]

to be honest, I dont know

when I learned about continuous function, it was taken as a given that their sum and scaled versions are also continuous

[delicate musk]

yeah same... lmao

[vivid breach]

if you want to prove it with epsilon delta, maybe you can look something up and use that

[delicate musk]

like any real number times a function would just be that functions output scaled

not in the mood for real analysis rn tho

but yeah

it makes sense

[vivid breach]

just seems enough to take it as a given

[delicate musk]

[vivid breach]

its more important you know what it is youre proving

and we've already done that now

we are good

[delicate musk]

w yeah it makes snese

sence

the function notation was throwing me off a little

[vivid breach]

massive W for the "skipping 90% of the work" community

[delicate musk]

so wait for like

$R^n$ I can just think of it as a sec of functions from the list of n numbers to some real number

[boreal girder]

BOSS

[delicate musk]

like

[vivid breach]

yea this is what I use as a reference

[delicate musk]

$\mathbb{R}^3$ is $\mathbb{R}^{(1,2,3)}$

ok cool i think i understand what yoy were talking about last night

[boreal girder]

BOSS

[delicate musk]

like where for any x in R^n; (x_1,x_2,x_3) could just be taking 1,2,3 to their respective real number values, which is the vector space

tyty @vivid breach

.close

[compact pewter]

Channel closed

Closed by @delicate musk

Use .reopen if this was a mistake.

[compact pewter]

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[last willow]

Alright here it goes. I am trying to understand how -2 times 2 divided by -3 gets me -3 when i do the steps that are asked of me i get -1.33 repeating.

[icy nymph]

$-2 \times 2 \div -3$

[boreal girder]

Flatus

[icy nymph]

ok so this is what you want to evaluate, right?

this does not become -3 btw

[last willow]

it is a fraction -2/-3x= 2

[icy nymph]

uh?

$\frac{-2}{-3x} = 2$

this?

[boreal girder]

Flatus

[icy nymph]

and you're solving for x?

[last willow]

the x is beside the fraction not the 3 but yes

[icy nymph]

ah

$\frac{-2}{-3}x = 2$

[boreal girder]

Flatus

[last willow]

I feel like im going crazy with this

[icy nymph]

ok so first

we can immediately cancel out the negative

at the top of the fraction (numerator) and the bottom of the fraction (denominator)

so we get

$\frac{2}{3}x = 2$

[boreal girder]

Flatus

[icy nymph]

do you get this?

[last willow]

Right. following you so far. then it would be 2x2 div 3

[icy nymph]

no

multiply both sides by 3

to get rid of th edenominator

[last willow]

Ok so ive been having it explained to me incorrectly

[icy nymph]

$\frac{2}{3}x \times 3 = 2 \times 3$

[boreal girder]

Flatus

[icy nymph]

$2x = 6$

[boreal girder]

Flatus

[icy nymph]

then, divide both sides by 2

$\frac{2x}{2} = \frac{6}{2}$