help-13

[potent fiber]

yup

[kindred storm]

And then you can use the Pythagorean theorem to get the length of a side of the octagon.

And then the middle square will have that as a side length.

[potent fiber]

that comes to like 1.414

[kindred storm]

Right, or sqrt(2).

[potent fiber]

yea

[kindred storm]

So, both red parts are sqrt(2).

[potent fiber]

yea

got that

then is this right?

[kindred storm]

Yes, that's right.

That's like the top side of the octagon.

Or one of the sides.

[potent fiber]

yup

so the bottom side is sqrt(2)+2?

[kindred storm]

Now what you want to do is to draw in yellow maybe one side of the strip on the octagon picture.

[kindred storm]

Like take the side of the strip that forms the top of the octagon.

That side of the strip then goes straight down from the leftmost point of the top side.

Does that make sense?

[potent fiber]

?

[leaden otter]

Something like this I suppose

[potent fiber]

ok

[kindred storm]

Yeah, starting like this:

[potent fiber]

ok

[kindred storm]

So, you know two of the sides of the red rectangle and you can find the other two.

[potent fiber]

so that would be 2(sqrt(2))+2?

[kindred storm]

I think that's not exactly right.

What's the length of the long side of the rectangle?

[potent fiber]

is it like this?

[leaden otter]

That's half of the complete answer

[kindred storm]

The short sides are sqrt(2) each.

What are each of the long sides?

[potent fiber]

sqrt(2)+2?

[kindred storm]

Right.

So, 2 times that plus 2 times a short side.

[potent fiber]

so this?

[kindred storm]

Oh, sorry, I didn't know that was your answer.

Yes, that's right.

[potent fiber]

ok

is this correct then?

[kindred storm]

No, notice the 1 is longer in that than the sqrt(2).

[potent fiber]

o wait yea

[kindred storm]

The ends are both the same length.

And they're the same size as the middle section when they're put together.

[potent fiber]

its like this right?

[kindred storm]

Right, so half the height of the rectangle.

[potent fiber]

uhh

?

[kindred storm]

Right.

Then, you can check by adding together the parts and seeing if it's 4 sqrt(2) + 4.

[kindred storm]

So your individual segment lengths are correct.

[cedar kiln]

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[boreal trail]

how would you do this

[boreal trail]

for

i get Pr(HH) = 5/8

(type in img)

but how do numerator intersection

is there a way to use these or smth

[exotic furnace]

The numerator is basically just saying “what’s the probability that the coin is the fair coin AND flips heads”

[boreal trail]

yep

but if i think of it mentally, ik that getting a fair coin is 1/2

[exotic furnace]

Yes

[boreal trail]

and flipping heads is 5/8

[exotic furnace]

Why is it 5/8?

[boreal trail]

denominator

basicly

since arent we accounting flipping heads weather its fair coin or also a heads coin

so 1/1 * 1/2 + 1/2 * 1/2

[exotic furnace]

Yes

Which is…?

[boreal trail]

ye

.

[exotic furnace]

Where did you get 1/8 from

[boreal trail]

3/4

[exotic furnace]

Yeah, 3/4 is the total probability of heads

Not sure where you got 5/8 or 1/8 from

[boreal trail]

but then sure 3/4 then what

[exotic furnace]

Do you understand why it’s 3/4 tho or is it not clear

[boreal trail]

i get it

[exotic furnace]

Ok

[boreal trail]

so thats just Pr(HH)

[exotic furnace]

Yes

[boreal trail]

then whats this, where does 3/4 come into this

[exotic furnace]

This solution doesn’t match up with the question

The question shows only one toss

[boreal trail]

how would you do

[exotic furnace]

Yet the solution is solving for 2 tosses

[boreal trail]

you toss once to determine weather its a fair or unfair

[exotic furnace]

Yeah, so see, the solution you’re showing is for b, not a

[boreal trail]

second is tossing for heads/tales with the chosen

[exotic furnace]

For a, we’re only doing one toss

[boreal trail]

oh ye 'b'

mb'

[exotic furnace]

If it’s b then the solution’s correct

So you’re asking about solving b or solving a

[boreal trail]

b

mb

[exotic furnace]

Ok.

[boreal trail]

i was confused how they do Pr(F intersection HH)

[exotic furnace]

Let’s start by doing fair and double heads

[exotic furnace]

You have a formula here for independent events

[boreal trail]

ye

and its dependent

[exotic furnace]

The choosing of the coin and the flip of the fair coin is actually independent

Choosing either coin will not change the weight of the fair coin suddenly

[boreal trail]

isnt landing a head dependednt on if the coin is a fair or headed coin ?

[exotic furnace]

When we’re flipping a coin, the other coin won’t affect its probability

[boreal trail]

?

[exotic furnace]

It’s just like this

[boreal trail]

as in marbles is dependent because you lose a marble

but in this, its infinite and nothing is lost?

[exotic furnace]

Imagine me having a six-sided die and a ten-sided die

If I pick the six-sided die, will the ten-sided die affect the six-sided die’s rolls?

[boreal trail]

as in rolling one after another

[exotic furnace]

No, just one roll

[boreal trail]

wording is scuffed, brain isnt braining

'will the ten-sided die affect the six-sided die’s rolls?

[exotic furnace]

I mean, if I roll the six-sided die

Will the ten-sided die change anything

[boreal trail]

no

[exotic furnace]

Yeah exactly

So it’s the same thing here

For the probability that the fair coin lands heads

[boreal trail]

so how do that

[exotic furnace]

The two-headed coin’s probability doesn’t matter

Yeah hold on

We’re doing fair and two heads

We’ll do rigged and two heads next

So, if the coin is fair, what does that mean

[boreal trail]

1/2 head

tail

headed is 1/1 head

ye

[exotic furnace]

Ok.

Focus on just the fair coin

What’s the probability that the fair coin lands heads twice

[boreal trail]

1/4

i get this

[exotic furnace]

Ok

So what’s the probability that the random coin is fair and lands heads twice?

[boreal trail]

1/8

1/2 * 1/4

may you just explain what they did at the top

[exotic furnace]

yeah we just did that

1/2 * 1/2 * 1/2

1/8

[boreal trail]

xd'

[exotic furnace]

Now we go to the bottom

To find the total probability of landing heads twice

Let’s do the easy one now

[boreal trail]

is there a formula for the numerator, or is it just mental

[exotic furnace]

What’s the probability that the rigged coin lands heads twice?

[exotic furnace]

It’s case-by-case

[boreal trail]

1

[exotic furnace]

Yeah, then what’s the probability of the chosen coin being rigged?

[boreal trail]

1/2

[exotic furnace]

Now what’s the probability of the chosen coin being rigged and landing heads twice?

[boreal trail]

1/2

[exotic furnace]

Yep

[boreal trail]

ah yep and pr(hh|f) = 1/8

[exotic furnace]

Now, you have $P(R\cap HH)$ and $P(F\cap HH)$

[wraith dagger]

CST

[boreal trail]

• 1/2

[exotic furnace]

Yeah exactly

That gets the 5/8 for the total probability of landing heads

Then you can just apply conditional probability

[boreal trail]

ah k ty ag

[exotic furnace]

No problem

[boreal trail]

.close

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[ruby topaz]

not sure how to go about this

[vagrant elbow]

Putnam

[ruby topaz]

question was in a textbook I was using, and I can't even find the solution :C

[mighty drift]

Taylor expand each term and combine them
You probably get something interesting

[vagrant elbow]

It's weird that you can't find the solution to a Putnam problem lmao

[mighty drift]

All be it not that simple to expand

[ruby topaz]

is there a different way to do it. I currently have no idea about taylor series

[vagrant elbow]

It's interesting to note that $\prod_{k \geq 0} (1 + x^{2^k} ) = \sum_{k \geq 0} x^k$ btw

[ruby topaz]

the problem is under Sequences & Series chapter

[wraith dagger]

NEON

[mighty drift]

That's the kind of thing I had in mind yes

Hoping something like this appears

[vagrant elbow]

The bottom can be factorized with difference of squares I suppose

Or you could divide by the numerator and see where that leades you

Stuff might get cleaner

$\frac{1}{x^{-2^n} - x^{2^n}}$

[wraith dagger]

NEON

[ruby topaz]

ah

y'all bursted the problem open fast. i'll try those ideas and see where I get, thanks!

.close

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[eager pilot]

What is the coefficient of 𝑥3𝑦14
in the expansion of (2𝑥+3𝑦)17
?

[eager pilot]

its actually What is the coefficient of 𝑥^3𝑦^14
in the expansion of (2𝑥+3𝑦)^ 17
?

lol

[surreal cave]

are you familiar with the binomial theorem? :)

[radiant fjord]

do you know binomial expansion theorem

lol

[eager pilot]

yes i am

i did this: C(17, 3) =

680

then i set equal the first and second expression to get 27y^3

680 * 2^14 * 𝑥^14 * 27𝑦^3

does tht look right?

11,059,840 ?

680 * (2𝑥)^14 * (3𝑦)^3

[surreal cave]

,w 6802^1427

[wraith dagger]

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Requested by XxMrFancyu2xX#6050

[eager pilot]

then its 680 times 16384

times 𝑥^14 * 27𝑦^3

but its wrong yeah

[surreal cave]

nope 300 million is right

,w expand (2x+3y)^17

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by XxMrFancyu2xX#6050

[surreal cave]

there we go compare the coefficient for x^14y^3 and you answer :)

[eager pilot]

so its 131072 plus 300810240?

times ***

[surreal cave]

your answer of C(17,3)*2^14*3^3 is the coefficient

[eager pilot]

oh

lol ill study this a little bit more after doing the homework

thank you

wait it marked it as wrong

so its not 6802^143^3 = 300810240

[surreal cave]

can you show the question?

[eager pilot]

11,059,840 didnt work either

sure

[surreal cave]

ahhh I see what went wrong

the question is asking about x^3*y^14

[surreal cave]

not x^14*y^3

[eager pilot]

oh

woops

[surreal cave]

just rinse a repeat the same way you did earlier :)

[eager pilot]

oh

C(17,3)3^32^14

now?

the times in the middle isnt showing oh

C(17,3) times 3^3 times 2^14

[surreal cave]

C(17,3)*3^14*2^3 the 3 is with the y so we raise that to the 14th

[eager pilot]

so its 680 times 2^3 times 3 ^ 14

okay

which is = 32491591040

not 27339240960 either

3^14 = 4,782,969.

2^ 3 = 8

680 * 4,782,969 * 8 = 27,339,240,960. is wrong

[surreal cave]

,w 6802^33^14

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by XxMrFancyu2xX#6050

[surreal cave]

always the damn arithmetic that gets you :)

try typing that instead

[eager pilot]

😭 yeah

it works

the simple things ugh

thank you

i have 2 questions left and im lost on the union part

its asking for the union of b and h so isnt it just 111 + 196 - 17 = 290.

290 is wrong, 324 is wrong

[cedar kiln]

@eager pilot Has your question been resolved?

[eager pilot]

.close

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[turbid spoke]

someonme can help me with a inntegral

[turbid spoke]

idk why is that

[surreal cave]

do you need help with the integral or the logarithm?

[turbid spoke]

inntegral

2pi/3 sin is sqrt3/2

not 1/2

[surreal cave]

be careful when doing your bounds

pi/3 becomes pi/6 not 2pi/3 :)

[surreal cave]

huh that is a method I would've just done u=x/2

[hearty arch]

yea same, when you substitute pi/3 into x, sin(x/2) becomes sin((pi/3)/2)

[turbid spoke]

I saw a video of khan academy that you can let the bounds?

https://www.youtube.com/watch?v=etpzxWBulN4 finnal

it is necesary change bound?

[hearty arch]

when using a substitution, yes. the bounds change as a consequence of the substitution

but if you're not substituting, then you do not change the bounds

[surreal cave]

a u-substitution is scaling/manipulating the number line so you need to change bounds to keep the area the same

[hearty arch]

yea that's a good way to explain it

[surreal cave]

nice! made that up on the spot

[hearty arch]

yea that was really clever. i've never thought about it like that

that's exactly what it is though

[surreal cave]

that's how my friend explained the Jacobian to me, he's really good at explaining stuff :)

[slate lintel]

twisting it around sometimes

[silent finch]

It’s because schools fail to teach intuitively

[slate lintel]

warping the very fabric of space

[slate lintel]

i think it takes some practice and a lot of introspection to really understand something, i can't really blame high school teachers there

[surreal cave]

I still don't know how a determinate of partials changes space tho

[slate lintel]

i just know that determinant measures stretchiness

[surreal cave]

not teachers fault just the institution itself

[hearty arch]

this may be a better conversation for the discussion channel

[silent finch]

Yeah you’re right

[hearty arch]

is there anything else we can help you through?

[turbid spoke]

my last procces is this

i need change bounnd then?

[cedar kiln]

@turbid spoke Has your question been resolved?

[hearty arch]

the sine of pi/6 is not sqrt(3)/2

wait

yea no, it's not

after plugging in the upper and lower bounds you should have

$2ln(1)-2ln|sin(\frac{\pi}{6})|$

[turbid spoke]

pi/2/3 = 3pi/2?

[hearty arch]

no

[hearty arch]

(pi/2)/3=(pi/2)*(1/3)=(pi/(2*3)=(pi/6)

however, pi/(2/3) would equal 3pi/2

incorrect usage of parentheses highlights its importance here

so you started with (pi/3) and then plugged it into x/2, so you get (pi/3)/2

[wraith dagger]

Wumbo

[hearty arch]

did that make sense, or i can write that in latex to make it more legible

[turbid spoke]

okay i get

[turbid spoke]

thx

[cedar kiln]

@turbid spoke Has your question been resolved?

[cedar kiln]

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[solid juniper]

my friend asked me about but i don’t understand it either

[solid juniper]

and before you ask no she cannot come ask herself bc she's in china lol

what is "their" in their union here?

it's from lang algebra

[flint plinth]

why such a rigamarole to prove that sigma has to be a 3 cycle or the identity?

all you need to know is that 2 cycles aren't in A_n ah wait no that's wrong thinking

[solid juniper]

maybe if i just typed this 5 seconds later it would have doubled as an answer to your question

[civic eagle]

union of the orbits?

[solid juniper]

it means some two orbits that exist by the previous sentence right

[civic eagle]

at least two, but yeah, as far as english goes that is what i would assume.

[solid juniper]

we are assuming all the orbits have 2 elements

on their union can't be referring to all the orbits right

[flint plinth]

(i had to go and look up what the hell J_n means ,it's just {1,2,...,n} haha)

[solid juniper]

just picking some 2 that exist?

[flint plinth]

yea he's just saying take any two orbits of size 2

[solid juniper]

ok that's what i thought but we were still struggling

ig that's all i wanted to confirm, thanks @flint plinth

.close

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[hybrid wren]

A force of 1150 N acts parallel to a ramp to push a 250-kg gun safe into a moving van. The ramp is frictionless and inclined at 30o 17°. (a) What is the acceleration of the safe up the ramp? (b) If we consider friction in this problem, with a friction force of 120 N, what is the acceleration of the safe?

[hybrid wren]

Am I doing this right?

It’s sin because the angle is between the opposite and the hypotenuse right?

i get a negative force

is that because the angle is so steep?

[zenith sail]

" inclined at 30o 17° " - what did you mean by this?

did you just mean to say 30 degrees?

if so, that seems right, but let me double check

@hybrid wren

[hybrid wren]

oh weird

its only supposed to be 30

[hybrid wren]

yeah its supposed to be 30

let me run it through real quick

because it keep getting -75

[hybrid wren]

is that because the angle is so steep?

for my net force

[zenith sail]

Yeah that's what I get too

The gravitational force in the direction down the ramp is greater than the applied force

In other words, the force they describe as "pushing it up the ramp" is not actually strong enough to overcome gravity

kinda seems like a mistake, judging by the wording of the problem

[hybrid wren]

so im doing it right pretty much its just an odd ball?

[zenith sail]

Yes, I agree with your answer

[hybrid wren]

so now just apply friction?

.close

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[hybrid wren]

.reopen

[cedar kiln]

✅

[hybrid wren]

wait

whoa

so if the object is sliding down

at a negative force

the friction would act in a positive force right?

[cedar kiln]

@hybrid wren Has your question been resolved?

[hybrid wren]

<@&286206848099549185>

[patent elbow]

Math

[hybrid wren]

yes

Since I have a negative force

Wouldn’t the friction be positive?

.close

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[jolly sable]

178

[jolly sable]

Ohh repeated can be only at 0

And at 0 it is positive means value c>0 so can't be roots repeated

[tropic oxide]

this is word salad.

we don't give out answers here.

[nimble veldt]

dont call her dude.

[dusty hazel]

Oh my god, are you so lonely that you need attention like this?

" Argument "
It was you crying. You don't know how this server works. Maybe read rules if time is all you have.

[grand kernel]

[cedar kiln]

@jolly sable Has your question been resolved?

[cedar kiln]

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[jolly sable]

What happened here?

[jolly sable]

What?? I know the answer and have a solution with it

I explained my vision how i came to the option A

[gritty galleon]

What was ur question

[tropic oxide]

ah, this was me responding to somebody else who eventually got banned

[gritty galleon]

Some bitch was shitposting

[jolly sable]

Why this equation cannot have repeated roots at x=0 means c>0

[cedar kiln]

@jolly sable Has your question been resolved?

[vestal wolf]

It can have a repeated root

Example: f(x)=x²+2x+1

f(x)=0 => x=-1 (repeated)

What does this have to do with c>0

What even do you mean by "c"

The answer c?

[cedar kiln]

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[jolly sable]

Means above axis

[vestal wolf]

Above axis what

I don't even understand what you want to ask here

[fallen moat]

hi guys

[vestal wolf]

Hello

[fallen moat]

if there's no question,we better close this

[vestal wolf]

The question is this

[fallen moat]

i see

[vestal wolf]

He said 178 so I'm sure he wants to ask about question 178

But I don't understand at all what he wants to ask

[vestal wolf]

Like, what does this even mean

[jolly sable]

I guess my approach is wrong

So the given answer is wrong

[vestal wolf]

I don't know what even your approach is

Please could you explain

[fallen moat]

[jolly sable]

I thought it will be repeated at 0 only

Abd function doesn't pass through 0

[vestal wolf]

What will be repeated at 0 though

[jolly sable]

Roots if it

But it is wrong

[vestal wolf]

Roots of WHAT

[jolly sable]

I'm mixing things badly so I'm wrong

[jolly sable]

Nothing

Leave it

[vestal wolf]

Okay what do you want to ask here

On question 178

[jolly sable]

Option a correct?

As you explained it can have repeated roots

[vestal wolf]

Alright finally

Has your answers been solved?

[jolly sable]

Not yet

[vestal wolf]

Okay

What do you want to ask

[jolly sable]

The given answer of the question 178 is a which is correct or not?

[vestal wolf]

No, answer a is not correct

[jolly sable]

Then??

[vestal wolf]

Do you know how to approach each solutions

Do you know what a polynomial is

[jolly sable]

Other 3 options are fine

I know it can happen

[jolly sable]

Bruh

I guess no completely

[vestal wolf]

Okay so you do know which is the correct answer, you said you know why b, c, d is correct/incorrect

[vestal wolf]

And I told you that option a is incorrect because of the following example

[vestal wolf]

https://en.m.wikipedia.org/wiki/Polynomial

[jolly sable]

Bro i can google too and i can read all the Wikipedia

Even everyone can

But they still ask questions here

[gritty galleon]

Don't be rude

[jolly sable]

They are correct

[jolly sable]

What about them?

[gritty galleon]

Hindi mein likh yaha

[jolly sable]

Did I start first?

.close

[cedar kiln]

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[vestal wolf]

You said you don't know what a polynomial is

[gritty galleon]

He was expecting u to explain it to him

[vestal wolf]

So I posted a link on the definition of a polynomial

[jolly sable]

Don't worry

No need to talk right now I'll skip the question

[cedar kiln]

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[dim tiger]

@jolly sable

[dim tiger]

i will try to help

so do you want to solve the question or no ?

nvm then

anyways if you want to solve the question post it and if i am online i will answer

@jolly sable

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[merry prairie]

Someone help pls

[sinful ocean]

Triangle AHC and ABC are similar

[cedar kiln]

@merry prairie Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[pearl raptor]

i need help understanding why the constant before the integral becomes 1/sqroot(3)

[mighty shuttle]

let $\frac{x}{\sqrt{3}}= u ,so dx= \sqrt{3} du$.

[wraith dagger]

physicsrocks

[pearl raptor]

but what about the 1/3 that was there previously?

[deep oriole]

oh wait

im talking shit

you find dx in terms of ds

and then you take out sqrt3 not 1/sqrt3

[mighty shuttle]

the integral becomes $\int \frac{\sqrt{3}ds}{3(1+s^2}}$

[deep oriole]

so you have 1/3 * sqrt(3) which becomes 1/sqrt(3)

[wraith dagger]

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

[pearl raptor]

oh i didnt know this

So basically

why is this?

[obsidian coral]

Rationalize the fraction

Take 1/sqrt(3) multply by sqrt(3)/sqrt(3)

[deep oriole]

because 3 = (sqrt(3))^2

x/x^2 = 1/x

[pearl raptor]

ooohhh ok got it

thank you sm :D

.close

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[turbid mesa]

Let G be a finite set with an associative binary operation satisfying these two conditions:

1. ab=ac implies b=c
   2.ba=ca implies b=c

Prove that G is a group and give an counterexample to show that this fails when G is infinite

[turbid mesa]

Now see, ab=ac implies b=c simply means that if the elements are say a1,...,an, then ai•aj where j is from 1 to n will contain all elements of G exactly once, hence there will be some aj such that ai•aj is ai. Now how do I show that this aj will work as identity for every ai in G?

I feel like I am missing something trivial

<@&286206848099549185>

[crimson delta]

you can write every element in the group as something*a_i

[turbid mesa]

Okay, how does it help?

[crimson delta]

so x = y a_i for any x and some fitting y. then what is x a_j

[turbid mesa]

That would be y

[crimson delta]

no

[mighty drift]

use the bijection Denascite showed

[turbid mesa]

We had ai•aj to be 1, right?

[crimson delta]

no

you wrote it equals ai

[mighty drift]

for every a, b in G, there is a c in G such that b = c a

[turbid mesa]

Ah, yes

We haven't shown yet that 1 exists

[turbid mesa]

So x a_j is y a_i, which was x. So x a_j is again x. Since x was arbitrary this holds for all x. For the left identity, we can just do the same with property 2

[mighty drift]

yes

[turbid mesa]

...

I need some sleep

Anyways, thanks for helping!

.close

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[mental crag]

Can someone explain why we subtract by 0.5000? Also, do we always subtract by 0.5000 or does it depend on the question? And if it depends on the question, how do we know what to subtract it by? Because the 0.5000 wasn't given in the question

[royal rampart]

ok

so

let me reak this

read

hmm

[mental crag]

<@&268886789983436800> This is the same guy that was trolling me yesterday. Surprised he hasn't been banned on the channel

[royal rampart]

this question is little difficult

but i can help

i know the answer

[mental crag]

Please ban this guy @royal rampart

[royal rampart]

<@&268886789983436800> i am helping this man/women and this man/women is saying to bann me

for no reason

[royal rampart]

do you want to know the ans ?

and how it is the ans ?

[mental crag]

<@&268886789983436800> Ask Eric (forgot his Discord user) who is a Helper and he'll tell you that this dude was trolling me all day yesterday

[royal rampart]

wth

i ma trying to help you

[woven herald]

I'm not sure how saying that he's reading it warrants a mod ping

[crimson sedge]

@royal rampart Just go away dude

they don't want to talk to you

[royal rampart]

ok yes

i was just trying to help

cause i know this question

nvm

[mental crag]

@wanton sail Hey Eric, sorry to bother you! This guy @royal rampart is the same person that kept trolling me yesterday. Can you please ask the mods to ban him?

[woven herald]

??

I'm right here and i read yesterdays exchange as well

[royal rampart]

yesterday i was just joking

i know it was my bad

and i accepted that

[woven herald]

Eric decided that yesterdays chat did not warrant moderator action and now you pinged mods just because he said he's reading the question

[crimson sedge]

I'm 100% sure mods aren't gonna ban you. Just stop talking in this channel and you'll be good

[mental crag]

LOOL Deleted User changed his user to @royal rampart. You ain't slick

[royal rampart]

true

[royal rampart]

i was agent47 from starting

[woven herald]

We use discord IDs anyway so this does not do anything

[royal rampart]

i just changed to deleted user for just yesterday

[mental crag]

Can you guys talk somewhere else? Like, I actually need help..

[woven herald]

I've read everything and i'm telling you to just drop it and continue with your question

[umbral dew]

wdym by discord id's?

[woven herald]

if Agent47 happens to type something actually troll like feel free to ping mods but don't do it when he just says he's reading the question

[woven herald]

not relevant for this channel, let's not make this off-topic

[mental crag]

<@&286206848099549185>

Can someone explain why we subtract by 0.5000? Also, do we always subtract by 0.5000 or does it depend on the question? And if it depends on the question, how do we know what to subtract it by? Because the 0.5000 wasn't given in the question

[crimson sedge]

300 is the middle right?

[mental crag]

Yeah, the mean is 300

[crimson sedge]

so the area under the right side of the cruve is 0.5

[mental crag]

Wdym

[crimson sedge]

the area under a normally distributed curve is one right?

[mental crag]

Sorry I'm a visual person

[mental crag]

Where's the 0.5

[crimson sedge]

If you were to compute the area under the curve (the poorly filled in red)

the area would be 1

so

if you were to split the area in half, ( left and right)

the right side of the curve would have half the area which is 0.5

[mental crag]

Ohhh

So the entire area in red is 1

[crimson sedge]

yeah

and the right side(half) is 0.5

[mental crag]

I see

[crimson sedge]

and you have a range on the right side of the curve that goes from 300 to 330

[mental crag]

For this question, the mean is in the middle where 0 is, right

[crimson sedge]

yeah

[mental crag]

From the middle to the right side would be from 300 to 330

[crimson sedge]

no?

[mental crag]

300 to 300+ I mean?

[crimson sedge]

yeah

[mental crag]

It goes forever on the right

[crimson sedge]

everything above 300

[mental crag]

Yes

[crimson sedge]

the 0.4332 number you got

was the area under the curve from 300-330

[mental crag]

Yes, I looked at the z table

[crimson sedge]

and the rest of the area to the right of the mean(center) is 0.5 - 0.4332

[mental crag]

So not all questions, you have to subtract by 0.5000

It just depends on the question, right

[crimson sedge]

yeah

because this question was asking you for the area to the right of that range

[mental crag]

Could you ask me a question where we wouldn't subtract by 0.5000 and subtract by a diff number

[crimson sedge]

I know absolutely nothing about statistics so no

[mental crag]

So if it's right range, you subtract by 0.5000?

And if a question was asking for the left range?

[crimson sedge]

also you weren't subtracting by 0.5

you were subtracting 0.5 by 0.4332

[mental crag]

Yes

[crimson sedge]

it would be the exact same thing

[mental crag]

What if they were asking for the left range

[crimson sedge]

because the are to the left of the (center of the )curve is also 0.5

[mental crag]

Yeah

Would it be negative if it was left range?

[crimson sedge]

area can't be negative

[mental crag]

Oh okay

Thank you

[crimson sedge]

this question would make more sense if you knew integral calculus

[mental crag]

I wish I did

Do you have any other pointers to know

[crimson sedge]

watch nothing besides learning calculus.

[royal rampart]

24karat if your doubt is not clear so you can also scan that image it google lens then maybe your doubt will be clear

[cedar kiln]

@mental crag Has your question been resolved?

[cedar kiln]

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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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• 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.

[near wing]

like am i doing something wrong here

why arent the numbers matching up

im doing exactly how they did this in the example video

[crimson sedge]

Are you comfortable in radian notations?

[near wing]

no not really sadly

[crimson sedge]

Uhmm

[crimson sedge]

Okay here

You know the quadrant thing

Like where all trigo functions are negative and positive

[near wing]

yes

[crimson sedge]

Like sin is positive from 0-180° right?

1st and 2nd quadrant

[near wing]

yes

[crimson sedge]

And √3/2 is value of cos 30 and sin 60

Right?

[near wing]

yes

[crimson sedge]

And that trigo fns change at angles that are multiples of 90° when added or subtracted

And do not change at 180° multiples

Like sin(180+ x) = -sinx

[near wing]

mhmmm

[crimson sedge]

• sign cuz third quadrant and x as it is added to 180

You know this stuff?

Or should I write it all as points

[near wing]

i do but everyone and everything was just confusing me or making it harder than it actually is

but yeah please do that

[crimson sedge]

Okay sure

So here

Starting with basics

It'll take 5 min extra just sit and relax and do write it doe somewhere in a concise form

And revise it everyday to remember

Okay

So

[near wing]

okay thats fine

thank you

[crimson sedge]

Anytime you have to convert functions they do so as

Sin into cos and vice versa

Tan into cot and vice versa

Cosec into sec and vice versa

Right?

[near wing]

yup

wait csc into sec?

[crimson sedge]

Yeah

Now

[near wing]

oh okay

[crimson sedge]

Over angles that are multiples of 90°

Like, 90,270

Odd multiples

Not 180 360

Of the form (2n+1) 90°

[near wing]

mmm okay

[crimson sedge]

That is n belongs to natural numbers

Angles such that 90,270,(360+90)

[crimson sedge]

These conversions take place

If you get sin (90±theta) it is always as an answer of cos (theta)

That is

The function changes

Tan will change into cot and like so for sec and cosec

Understood?

[near wing]

yup your explaining it better than most on here

[crimson sedge]

Yeah

Now when you go to the even multiples of 90

Of the form (2n) 90

That is 180,360,540,720

These functions don't change

Sin will remain sin

Cos will remain cos

Tan will remain tan

And like so all else

So cos(180±theta) is of the form of cos(theta)

Understood?

Now the main part

[near wing]

yup

[crimson sedge]

What you did wrong was you didn't take the - sign into account

Wait Lemme draw the quad for you

[near wing]

okay

[crimson sedge]

Okay so what this basically telling is

At which angles. , Which functions are positive

The pneumonic is - All Students Take Coffee

So can you tell me where is cos function positive?

[near wing]

oh i know it as As Students Take Calc

[crimson sedge]

So you know it already good

[crimson sedge]

Yet just

Like the range of angle

[near wing]

Quadrants 1 and 4

[crimson sedge]

Yeah that's

In angle?

0-90 and 270-360 right

Can you do so for sin ??

[near wing]

yeah 0-90 in Q1 and 270-360 in Q4 okay im kinda getting it more

[crimson sedge]

Yep

[crimson sedge]

Try it

Like after our discussion is over , write it all yk

[near wing]

umm 0-90 in Q1 and 90-180 in Q2?

[crimson sedge]

Make four coloumns 0-90 90-180 180-270 270-360 and write for all

[crimson sedge]

Yep correct

So you agree that in Q 3 and Q4 sin will be negative?

[near wing]

yup

[crimson sedge]

Yeah

So now to your question

[crimson sedge]

Sin( something ) = - √3/2

Now follow these steps ALWAYS

Step 1 : What does the sign tell you ?

In minus root whatever, what does the minus tell you?

Okay

See

Starting again you there?

[near wing]

yes im still here

[crimson sedge]

Yeah

So

If the angle in sin was in Q3 or Q4

Was the answer negative?

[near wing]

yes

[crimson sedge]

So if the answer is negative

Must the angle be in Q3 and Q4?

[near wing]

yup

[crimson sedge]

Now

You know 2 theta should be in Q3 or Q4

Right

[near wing]

yes

[crimson sedge]

(I'll just write theta as x okay?)

So now

You would also know √3/2 is equal to sin 60 and cos 30 , after comparing

Right?

[near wing]

yessir

[crimson sedge]

Step 2: Trynna make these using multiples of 90

So what have we got now?

Angle should be in Q3 or Q4

For sin it should be 60
For cos it should be 30

Okay?

[near wing]

yup

[crimson sedge]

Yeah

Now

Is sin(180 + x) = sin x

Trynna answer

[near wing]

no would'nt it be a negative

[crimson sedge]

Yessss good

It isn't

Trick question

😂😂

[near wing]

shit i was overthinking fr

lol

[crimson sedge]

Sin (180+x) = -sin x

Now

We have on the RHS -sin 60 agree??

[near wing]

yup

[crimson sedge]

Now we want the LHS to be the same

So

We can't just add a minus so we will take the angle to be in Q3 or Q4 therefore

2x must be in 180-360 range

Sure?

(don't start comparing answer rn)

[near wing]

yes

[crimson sedge]

Yeah

Now

What will sin(180+60) give you?

[near wing]

240

[crimson sedge]

Try again

[crimson sedge]

..

[near wing]

120

[crimson sedge]

Oh wait I ddint tell you?

Nah nah nah nah

Wait

[near wing]

huuuh

[crimson sedge]

Okay so what I meant by functions don't change is

Sin/cos/tan (180 + x) will give x

Like

Sin (180+x) = sin x with appropriate sign

Cos (180+x) = cos x with appropriate sign

While sin (90+x) = cosx

See

The 90 multiples disappear and just leave x

Understood?

[near wing]

hmm

[crimson sedge]

You understood kinda

Okay now

Feel free to ask

[near wing]

na im sorry could you explain it again lol

[crimson sedge]

Yeah again

So

If you wanna get an angle

Example. 210

Can you write it as 180+30 ?

Or 270-60

??

That's what it adds upto right??

180+30=210

Innit?

[near wing]

yes

[crimson sedge]

Yeah

Now just the things is you have to remeber this

That at even multiples of 90 , functions don't change

Odd multiples of 90 fn change

And giving any angle by adding or subtracting any angle with multiples of 90 gives you the angle itself

Like

Where f is a trigonometric function of your choice

So you add or subtract any angle x , from multiple of 90

You get some function of x with appropriate sign

You got this?

I'm bad at explaining

😂😂

[near wing]

na your good im just a bit slow lol

[crimson sedge]

Okay

Listen

Just remeber this

Whenever there is a angle added to some multiple of 90, answer will always be that angle and multiple of 90 will disappear

Okay?

One question btw which class you in

[near wing]

okay i got that

[crimson sedge]

Aye aye

Then im not talkign about graphs

Cuz that'll be too much for this question 😂

[near wing]

like which class imn taking

im*

[crimson sedge]

Okay so what would be sin(180 + x) be?

[crimson sedge]

Yeah standard

Division

Whatever you call it

Secondary higher secondary

[near wing]

trig

[crimson sedge]

Umhmm

[near wing]

ummm

[crimson sedge]

Okay this

Okay now?

What would be sin(180+x)

[near wing]

bro i feel like im overthinking so hard omg

[crimson sedge]

Just answe it out

Don't matter of you wrong

Okay waittt

Here steps

1 : Find which quadrnt angle belongs to

Tell me

180+x would belong to which quadrant?

If x is between 0 to 90

[near wing]

Q1

[crimson sedge]

No

[near wing]

man'

[crimson sedge]

I'm asking for 180+x

Like there's a angle 180+45

Which quadrant would it be in?

[near wing]

3??

[crimson sedge]

Yes

Okay you need practice in this fr

270-73

[near wing]

shit i do

[crimson sedge]

I'm just writing some question you tell me the quadrants

Dotn overthink

270+6

360-1

90+67.5

Andd

180-65

You get 5 minutes

[crimson sedge]

This one ??

OKAY here

[near wing]

3

[crimson sedge]

@near wing

[crimson sedge]

Yeah

[near wing]

4

[crimson sedge]

See

Here

[near wing]

4

[crimson sedge]

Those axes I made on the qudrant these are based on those

Good

[near wing]

2

[near wing]

2

[crimson sedge]

Yeah good

Now

If instead of 67.5

Hadi written 90+x

What answer you would have given

[near wing]

so that 67.5 is the x

[crimson sedge]

Yep

Which quadrant then?

[near wing]

157.5 right?

[crimson sedge]

Yeah

Which quadrant?

[near wing]

2

[crimson sedge]

180-x

[near wing]

still using 67.5 right?

[crimson sedge]

Doesn't matter

[near wing]

112.5?

[crimson sedge]

Any number between 0-90 will give you the same result

Yep

Now if I ask you

180+x

?

Can x be 60?

Can x be 30?

[near wing]

yes

[crimson sedge]

Will you still get the qudrant to be Q 3??

[near wing]

yeah

[crimson sedge]

Can x be 45 , and here you still get Q3?

[near wing]

yes

[crimson sedge]

Yeah so any number x between 0-90 will give you Q3 when it is 180+x okay?

Now after 45 mins of conversation

[near wing]

ohhhh okay so as long as is 180+ any number 0-90 its always Q3 then

yeah my bad i know im being difficult to teach

[crimson sedge]

You want the angle 2x to be in Q3 and Q4 right

[crimson sedge]

Nah man I don't know how to teach either 😂😂

[near wing]

i appreciate you for taking time out of your day to help my dumbass lol

[crimson sedge]

It's night over here

Lmao

1 am

😂😂

Nvm I have to sleep and wake up in 5 hrs

[near wing]

almost 1pm here

oh damn

[crimson sedge]

So now back to question

[near wing]

mhm

[crimson sedge]

.

[near wing]

yup

[crimson sedge]

And to get sin 60 or cos 30

Sin (180+2x) = - sin 60

Agreed??

[near wing]

mhm

[crimson sedge]

Sin (180+2x) = -sin 2x

Agreed?

[near wing]

yes

[crimson sedge]

Yeah

Don't worry like my man said x was between 0 to 90 now it's over that quadrant would change

No.

Take 2x as something new theta

It'll stay in Q3

Rules of math

So yeha

[crimson sedge]

Putting this in

[crimson sedge]

This we get

-sin (2x) = - sin 60

Correct?

[near wing]

yup

[crimson sedge]

2x = 60