help-39

[floral tendon]

;o

[stoic imp]

well, the channel closed

[floral tendon]

Hi again

[stoic imp]

I slept I think

,w nullspace {{1,-1,0,-1},{1,-3,-1,0},{2,0,1,-3}}

[jolly parrot]

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[stoic imp]

T = <(-1,-1,2,0),(3,1,0,2)>

S = <(1,1,0,1),(0,-1,1,0)>

S n T = <(1,0,1,1)>

v4 = ?

B = {(-1,-1,2,0),(1,1,0,1),(1,0,1,1), v4}

[stoic imp]

whats with your bochi pfp

no, I think I get it know

[toxic lichen]

v4 can be anything not in the span of the first three

you just need to complete a basis of R^4

fuck, you can even try picking some standard-basis vector as v4. see which ones work if any

[stoic imp]

T = <(-1,-1,2,0),(3,1,0,2)> = <(1,0,1,1),(3,1,0,2)>
S = <(1,1,0,1),(0,-1,1,0)> = <(1,0,1,1),(1,1,0,1)>
S n T = <(1,0,1,1)>
S + T = <(1,0,1,1),(1,1,0,1),(3,1,0,2)>

[stoic imp]

yeah

B = {(1,0,0,0),(1,0,1,1),(1,1,0,1),(3,1,0,2)}

[stoic imp]

the key Idea is to notice that S + T is formed out of extending the basis of SnT

you see what I did here?

[toxic lichen]

hold on

[toxic lichen]

this is wrong for sure

[stoic imp]

let me check

[toxic lichen]

well that or i am not following your calculations at all

[stoic imp]

,w rank {{1,0,0,0},{1,0,1,1},{1,1,0,1},{3,1,0,2}}

[jolly parrot]

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[stoic imp]

the vector in SnT can be used for the basis of S aswell as the basis for T

then I extended the basis of S + T with e1

[stoic imp]

why? there is nothing funky in here

T = <(-1,-1,2,0),(3,1,0,2)> = <(1,0,1,1),(3,1,0,2)>
S = <(1,1,0,1),(0,-1,1,0)> = <(1,0,1,1),(1,1,0,1)>
S n T = <(1,0,1,1)>

you agree with this part?

[toxic lichen]

OH ok no wait i see it now sorry

the fact you put (1,0,0,0) at the front threw me off a bit but thats the v4 in my notation

yes ok thats good then

[stoic imp]

what about d)?

[toxic lichen]

v2 belongs to S but not to T
v3 belongs to T but not to S

think about how you can combine these

[stoic imp]

I think I got it

T = <(-1,-1,2,0),(3,1,0,2)> = <t1,t2>
S = <(1,1,0,1),(0,-1,1,0)> = <s1,s2>

t1 + s1 € S + T

(-1,-1,2,0) + (1,1,0,1) = (0,0,2,1)

but (0,0,2,1) not in T not in S

[toxic lichen]

ok yeah

[stoic imp]

you follow ?

ok

,w rank {{0,0,2,1},{-1,-1,2,0},{3,1,0,2}}

[jolly parrot]

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[stoic imp]

,w rank {{0,0,2,1},{1,1,0,1},{0,-1,1,0}}

[jolly parrot]

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[stoic imp]

yeah I think we did it

I appreciate the help, the S+T made out of SnT basis extension was crazy

Havent had that much fun since like yesterday

.solved

[pearl pond]

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[pearl pond]

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[feral quartz]

!help

[pearl pond]

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

[feral quartz]

wait im sharing i much have i reached in latex

[jolly parrot]

BVB999

[feral quartz]

i ifgured its c

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[eager jewel]

can someone tell me where i went wrong

[pearl pond]

@eager jewel Has your question been resolved?

[eager jewel]

<@&286206848099549185>

[zinc beacon]

can you write that in a way easier to understand @eager jewel

[eager jewel]

yeah

so {x} is always between 0 to 1

sec {x} will always be between 1 and 2

so [sec {x}] will always be 1

so f(x) = 1

g(f(x)) = g(1)<0

and i just did calculations for that

[zinc beacon]

and what do you really wanna know?

@eager jewel

[eager jewel]

why my answer is wrong

[zinc beacon]

but do you know the answer right?

[eager jewel]

yeah

[zinc beacon]

what is it?

[eager jewel]

1

and as u can see i am getting 0

[zinc beacon]

how do you know g(f(x))<0?

[eager jewel]

from the question?

[zinc beacon]

i dont know what is the question

[eager jewel]

^^

[zinc beacon]

you just showed your calculations

soory

sorry*

didnt saw that

my bad

give me 5 minutes to make it by my own

[eager jewel]

yeah ok

[zinc beacon]

the equation is
g(x)= 2x*squared - 3x(k+1) + k(3k+1)
right?

[eager jewel]

yes

[simple trench]

your calculations seem to be correct

[eager jewel]

oh ok then whys my answer mismatching?

[simple trench]

maybe the answer in the book is wrong

i dont see any discrepancy

{x} belongs to [0,1)

[eager jewel]

oh

[simple trench]

sec{x} belongs to [1, sec1)

sec1 approximate 1.8

take gif

[sec{x}] is 1 always

[eager jewel]

yeah

[simple trench]

and your wavy curve is also correct

i checked ur algebra too

[eager jewel]

they are getting 0 but the answer is 1???

[simple trench]

uh

it isnt?

examside right?

[eager jewel]

oh ok thanks a lot then the answer is given wrong

[eager jewel]

hmm?

[simple trench]

bro im jee aspirant ik these sites a lot

[simple trench]

np

close

[eager jewel]

oh no its tarigrade

[eager jewel]

ur giving next year or this year?

[simple trench]

next

bro i wouldnt be here

tmrw is adv

[eager jewel]

xd

[simple trench]

if i was giving this year i wouldnt be here

[eager jewel]

yeah true

well ok

ty anyway

[simple trench]

kk close

[eager jewel]

.close

[pearl pond]

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[simple trench]

@eager jewel

ur solution has a mistake

srry i didnt see

while finding the critical points of the wavy curve

[eager jewel]

.reopen

[pearl pond]

✅

[simple trench]

3k+1 = 0 => k = -1/3

u put k = 1/3

so u got (1/3,1)

[eager jewel]

oh shit

[simple trench]

it actually should be (-1/3,1)

with one integral val (k=0)

[eager jewel]

yeah 💀💀 my bad

thanks

[simple trench]

happens

[eager jewel]

.close

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[slate tide]

Show that if $(A_n)$ is a sequence of non-negative random variables, then for $\lambda \geq 0$,
[
\mathbb{P}\left( \sum_{i=1}^\infty {A_i} \geq \lambda\right) \leq \sum_{i=1}^{\infty} \mathbb{P}\left(A_i \geq \lambda \x t_i \right)
.
] for $t_i > 0, \sum_{i=1}^{\infty} t_i = 1$

[west sapphire]

can you argue that if $\sum A_i \geq \lambda$, then it must be true that $A_i \geq \lambda t_i$ for at least one $i$?

[jolly parrot]

Bungo

[slate tide]

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[west sapphire]

i guess not

[slate tide]

no i got it

thanks

[west sapphire]

yw

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[craggy crypt]

Oh sorry

[craggy crypt]

why does the answer sheet not do continuity correction?

[plush bramble]

show the original question

[craggy crypt]

Ok

[craggy crypt]

Sorry , here they did it

But

Here they didn’t

I was asking why they would not do it here

[craggy crypt]

It’s above

[plush bramble]

where

[craggy crypt]

Exercise 3

[pearl pond]

@craggy crypt Has your question been resolved?

[craggy crypt]

<@&286206848099549185>

[pearl pond]

@craggy crypt Has your question been resolved?

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[midnight haven]

1-1-2sin^2 = -2sin^2

where did the sign go

[azure kite]

1-(1-2sin^2theta) = 1+ (-1 + 2 sin^2theta)

its in the bracket

remeber you have to distribute it to both

[midnight haven]

ohh

[azure kite]

-1(a+b) = (-a-b)

[midnight haven]

yea makes sense

thanks

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[urban briar]

Consider the two functions $f(x) = 2x^2 -1$ and $g(x) = 1-3x$

[jolly parrot]

bagelguy3

[urban briar]

Find the values for 'x' s.t. $f(x) = g(x)$

[jolly parrot]

bagelguy3

[placid geyser]

!status

[pearl pond]

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

[hazy pilot]

where are you struggling?

[urban briar]

Im learning about sets and relations

si

so

I basically make it equal to each other

[sharp smelt]

yes

[urban briar]

$2x^2 - 1 = 1-3x$

[jolly parrot]

bagelguy3

[urban briar]

$2x^2 + 3x - 2 = 0$

[jolly parrot]

bagelguy3

[urban briar]

When I solve this quadratic I keep getting -8 and 2

but the answers are

-2 and 1/2

[placid geyser]

U messed up the quadratic somewhere

[hazy pilot]

could you show the steps in solving the quadratic?

[urban briar]

probably

ok

[placid geyser]

Did u divide by 2a?

[urban briar]

OHH YEAH

I FORGOT

2a

mb

[hazy pilot]

lmao

[placid geyser]

Yeah lol

[urban briar]

thx

idk how i missed that

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[pearl pond]

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[faint mica]

Prove that $\sum_{n=1}^\infty \frac{1}{2^{2n}}
,\sech^2!\Bigl(\tfrac{\pi}{2^{n+1}}\Bigr)
;=;
\frac{4}{\pi^2}
;-;
\csch^2!\Bigl(\tfrac{\pi}{2}\Bigr).$

[jolly parrot]

Orc

[faint mica]

My ideas are limited with this one. I have tried using mean value theorem and observing the integral of csch^2(x), but it's not fruitful

i want to also try see if i can convert the LHS into an integral

however idk how to!!!

[pearl pond]

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[faint mica]

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[rough forge]

Maybe you can find a taylor series if you differentiate/integrate:
[ f\left(\f{\pi}{2}\right) = \f{4}{\pi^2}-\csch^2\left (\f{\pi}{2}\right ) \quad \ri \quad f(x) = \f{1}{x^2}-\csch^2(x) ]

[rough forge]

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[royal galleon]

For number 7 how can I do a base case

[royal galleon]

Because for the base case I will use n of 3 but I x can be any positive integer so do I just choose any or is there another method?

[autumn fossil]

keep x variable

you are supposed to show that (1+x)^3 >= 1 + 3x + 3x^2 for x > 0 (for the base case)

try expanding (1+x)^3

[royal galleon]

Ok thx

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[craggy python]

yo guys! i started yesterday on using real math. how im doing?

[pearl pond]

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[craggy python]

br

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[midnight haven]

lets say you have a spinner with 3 groups (red = 90degree, blue 120 deg, yellow 150 deg)
and each group has 3 fishes with their own points (red: 35,, 40, 45, blue: 20, 25, 30, yellow: 5, 10, 15)
you put 3 fish from the group you got from the spinner. in another spinner but this time it is equally divided into 3 parts. after spinning, you collect the fish (setting it aside) and do the whole process 2 more times. whats the probability of getting 50 or more points?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[midnight haven]

<@&286206848099549185>

pls

[polar dune]

guys i need help with integration

[plush bramble]

! occupied

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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).

@midnight haven Has your question been resolved?

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@midnight haven Has your question been resolved?

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@midnight haven Has your question been resolved?

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[sharp smelt]

Prove that if Sup(A), sup (B) exist, then sup(A+B) exists
\
Proof sup(A)> a, forall a \in A
sup(B)> b \froall b \in B
sup(A)+sup(B)> a+b, forall a in A forall b in B
But a+b, forall a in A forall b in B is the defn of A+B
There thus is an element greater than all elements of A,B

[west sapphire]

the >'s should all be >=

but seems fine otherwise

[sharp smelt]

okie

thanks

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[safe prairie]

about matrices

[safe prairie]

square matrices specifically

all invertible square matrices share the same rref right

[toxic lichen]

yes, the identity

[safe prairie]

so when i read(i mightve read it wrong) that if matrices have the same rref theyre equal, is it for non squares?

[toxic lichen]

you read it wrong

[sharp vigil]

if two matrices have the rref they are "row equivalent", not equal

[safe prairie]

so amm invertible square matrices are row equivalent

so adding a zero vector to the right and do some elimination then theyre all the same?

is it because they all have the same unique solution set the zero vector

[pearl pond]

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[white atlas]

How to find range of rational functions.( quadratic by quadratic or quadratic by linear)

[white atlas]

if i use discriminant method i get something wrong

[sour iris]

the range would just exclude the horizontal aymptoptes i think

[white atlas]

like what is the exact method

whats that

[sour iris]

https://www.desmos.com/calculator/aibuh8vcfh

you see that y=1 is the value that f gets close to but doesn't reach

so f's range excludes 1

[white atlas]

how do u get to know that it gets close to 1

[sour iris]

look at the graph i sent

and as x gets really large, the numerator becomes close to the same value as the denominator

so it is close to 1 but never reaches it because of the + 3

[white atlas]

how will i get to know that it gets close to 1 without making graph

[sour iris]

so what happens when x is really large in $\frac{x}{x+3}$?

[jolly parrot]

hello

[white atlas]

a large number upon a large number

[sour iris]

yes and those large numbers are really close to each other. the only difference is that the denominator is + 3

the value is very close to 1.

but it is not 1

it will never be 1 no matter how large the x gets

because of the +3 on the bottom

so you know that the function doesn't include 1 in its range

$\frac{100000000}{100000000+3}=0.99999997$

[jolly parrot]

hello

[white atlas]

so i will have to just use this kind of logic to get range everytime

[sour iris]

well let me give you the way to find the horziontal aymsptotes:

[eager jewel]

Why are u getting it wrong

[white atlas]

the answer always has some value removed

[eager jewel]

Maybe send an example

Cause that is a really easy method

[white atlas]

(x^2 - x - 6 ) / (x-3)

if i apply discriminant method range is all real numbers

but 5 is excluded from the answer

what does asymptote mean

[eager jewel]

Can u show ur work

[white atlas]

i cross multiplied

.close

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[fossil ruin]

Could somebody answer this for me? There aren't any answers and I am not sure if I am correct. Thanks in advance.

[blissful cloak]

notice that each corner and side piece will have the same area

then a corner can be constructed from a quarter circle and semicircle

and a side from a quarter circle

[compact ridge]

yes as the wording of the question suggests, spam A = pi r^2 * theta/360

[fossil ruin]

Ok, thanks.

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[fossil ruin]

I just need to know the answer to this question to check. You are supposed to find the shaded area.

[hazy pilot]

well you know the area of the full semicircle

[compact ridge]

it's assumed that each circle is tangent to the other circles right

then you have half of a regular hexagon (think of the equilateral triangles on the diagram)

the shaded area is (circular sector with radius 6) - (2 semicircles) + (2 circular sectors with radius 2)

[compact ridge]

what did you do then?

to get the answer

it's much easier to check (and if needed, fix) your working rather than an answer

[pearl pond]

@fossil ruin Has your question been resolved?

[fossil ruin]

Your supposed to answer it in pi so i got 2 pi plus 4 pi over 3

[compact ridge]

!

[compact ridge]

what did you do to get that answer?

[pearl pond]

@fossil ruin Has your question been resolved?

[pearl pond]

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[vague plover]

How to solve this system of equation

[autumn fossil]

for integers?

[hazy bear]

This feels like impossible

[vague plover]

I think so, the problem tells me to solve for x,y & z

[autumn fossil]

one obvious solution is just 2025, 0, 0

[verbal whale]

This doesn't mean at all that you're working with integers 😅

[vague plover]

😭

I have absolutely no idea on how to do it

[autumn fossil]

I feel like permutations of those are actually gonna be the only sols

though i cant prove it rn

|x|, |y|, |z| are sure < 2025

<= i mean

otherwise the first equation cant be satisfied

[meager iron]

If it can be x = y = z, we can be sure a set of solution is like what @autumn fossil said

[autumn fossil]

x = y = z doesnt work

3^2 isnt 3^3

[meager iron]

I mean like

Two variables can be equal to each other

If all variables have to have a unique solution, then perhaps it might not be possible

[autumn fossil]

$x^{3}+2y^{3}=2025^{3} \ x^{2}+2y^{2}=2025^{2}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

this?

[meager iron]

Yeah could be

[autumn fossil]

this should result in x = 2025, y = 0

$\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}<x^{3}+y^{3}+z^{3}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

can we prove this

<= with equality iff 2 of them are 0

oh the RHS can be negative, nvm then

oh

its supposed to be >

$\left(x^{2}+y^{2}+z^{2}\right)^{3}>\left(x^{3}+y^{3}+z^{3}\right)^{2}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

it simplifies to this

this should be trivial

[pearl pond]

@vague plover Has your question been resolved?

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[royal galleon]

so for question 8. My inductive hypothesis is a_n<= a1c^(n-1) implies a_n+1 <= a1c^n. So from what I understand geometric sequences is that each number in the sequence increases by a constant ratio. So the way I want to prove my inductive hyopethesis to to mulltipy both sides by c so then I have a_nc <= a1c^n. Then I asssume that c is the right constant for the sequence so a_nc become a_n+1 <= a1c^n but how can I assume that c is the right multiple

[fluid axle]

wdym by right constant ?

a_n doesn't have to be geometric at all

[royal galleon]

so what can it be then?

[royal galleon]

by right constant I mean a constant that would move term a_n to term a_n+1

how do I know that c is the right constant because that is the only way I see of proving the inductive hyopthesis

[fluid axle]

ok well just use the condition for a_n in the question

you've shown c*a_n <= a1*c^n

[royal galleon]

yes

[fluid axle]

and we also know that a_(n+1) <= c*a_n cause of the hypothesis given in the beginning of the question

combine these two

[royal galleon]

oh so your pluggin in n+1 to the given

so start with the given plug in n+1 a_(n+1) <= a1c^n. go to the inductive hypothesis a_n<= a1c^(n-1). mulltiply by both sides. ca_n<= a1c^n. Combine both inequalities a_(n+1) <= a1c<=a1c^n since n is positive

[fluid axle]

start with the given plug in n+1 a_(n+1) <= a1c^n.
that's not the given, that's your goal

or I don't understand what you mean by 'given'

[royal galleon]

the given is a_n <=ca_n-1

plug in n+1 for the given then we have a_n+1 <=c*a_n+1-1

[fluid axle]

yes

that's not what you wrote above tho

[royal galleon]

sorry that is what I meant but with and the inductive hypothesis which is A(n) = a_n <= a1c^(n-1). A(n+1) = a_n+1 <= a1c^n. A(n) implies A(n+1). multiply A(n) by c so ca_n <= a1c^n. Going back to the given we know that a_n+1<= ca^n <= a1c^n

[fluid axle]

alright

[royal galleon]

so this is sound logic. And more thing why are the bounds >=2 and >=1 for the given and the thing we are trying to prove

[fluid axle]

cause the sequence starts at index 1 : a1, a2, ....

if you're using a_(n-1) somewhere, you need n-1 to be at least 1

so n>=2

[royal galleon]

but for both of them it is a_n so why is one start at 1 and the other 2 shouldln't they both start at the same thing

[fluid axle]

a_n <= c a_(n-1)

a_n-1 is involved in that condition

so you need n>=2

otherwise it doesn't make sense at all

[royal galleon]

oh I see and the other one only has a_n so it can be 1

[fluid axle]

if you had n=1, you'd be talking about a0

which doesn't exist

[royal galleon]

yes makes sense

ok thanks so much

.close

[pearl pond]

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[pearl pond]

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[fallen cosmos]

For question 13 part a i feel like theres a simpler way of doung it then the way i did it considering its only worth 3 marks

[blissful salmon]

What you did was right, and at the beginning you equated the coefficients of x^2 and x^3 correctly. To save time, I guess you could at the beginning divide both sides by (2k)^(n-3) only assuming that k is not 0. Then you get C(n,2) * 2k = C(n,3). Write C(n,2) as 3n(n-1)/6 and C(n,3) as n(n-1)(n-2)/6 (having 6 as our common denominator makes it easy for us).Then we can cancel a lot, and get the equation as 3 * 2k = n-2 or n = 6k+2

[pearl pond]

@fallen cosmos Has your question been resolved?

[pearl pond]

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[fallen cosmos]

Ah I see okay thank you

[blissful salmon]

All good, you can close this channel now if you don’t have any further questions

[fallen cosmos]

.close

[pearl pond]

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[tiny yacht]

kosans

imean questions

[toxic lichen]

i beg your pardon?

[tiny yacht]

i have tried questions

[toxic lichen]

... and what's that supposed to mean

[tiny yacht]

you asked what have you tried?

so i tried questions

[toxic lichen]

ok so you have not tried anything for this question in particular

[shut loom]

first, you need to see that a_i is the coefficient of x^i in a binomial expansion

[tiny yacht]

yes

[toxic lichen]

and you replied in an obtuse way

on purpose

as if you dont actually want to cooperate

dont do that

[tiny yacht]

nope, i didnt understand your question

[toxic lichen]

my question was: did you make any progress and if so what

ok

anyway

here is my recommendation:

[tiny yacht]

yes i can answer this question, no

[toxic lichen]

work out $\frac{a_{i+1}}{a_i}$

[jolly parrot]

Ann

[tiny yacht]

hmmm

ok

[toxic lichen]

think about how this can be used to tell when the sequence a_i goes up and when it goes down

[tiny yacht]

1 + ax + a2x2 + ... + a23x23

should i break the binomial expansion?

[toxic lichen]

the expansion is $1 + a_1 x + a_2 x^2 + \dots + a_{23}x^{23}$ yes

[tiny yacht]

then maybe i can compare the two equations

[jolly parrot]

Ann

[toxic lichen]

i will repeat my suggestion

[tiny yacht]

master yoda has come

[eager jewel]

From yesterday's advanced paper 😬

[tiny yacht]

[tiny yacht]

yes, i did all conics

trying my hands at this

[toxic lichen]

work out the general expression for $a_i$ and hence work out $\frac{a_{i+1}}{a_i}$

[jolly parrot]

Ann

[eager jewel]

👍🏿

[tiny yacht]

did you give the paper?

[eager jewel]

No g mines next year

[tiny yacht]

us bro us

[tiny yacht]

can you explain it in detail?

[eager jewel]

Do u know that formula for numerically greatest term

[tiny yacht]

ya ya ya you divide t_r+1 / t_r and smaller than equal to 1

[eager jewel]

Yeah then use that

[tiny yacht]

binomial theorem 101

doing it

got it the answer's coming round 20

nah ther's a mistake

6

yeah 6 is coming the greatst

6!

.close

[pearl pond]

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[pearl pond]

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[pearl pond]

@signal forge Has your question been resolved?

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[glossy wave]

$\int \sqrt{1+sinx} dx$

Let $u = sinx$, $x=arcsinu$, $dx=\frac{1}{\sqrt{1-u^2}}du$

Then $$\int \sqrt{1+sinx} dx = \int \sqrt{1+u} \frac{1}{\sqrt{1-u^2}}du = \int \frac{1}{\sqrt{1-u}}du=-2\sqrt{1-sinx}+C$$

[jolly parrot]

Lambda

[glossy wave]

Why doesn't it work?

[frank goblet]

You can try plotting all the steps and see that way which step is the culprit

[glossy wave]

yeah good idea

[blissful cloak]

,w int 1/sqrt(1-x)

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by xxmrfancyu2xx

[ruby cargo]

I put x/sqrt(1 - x), what's wrong with me

I'm sorry

[random ermine]

where did u put it

[ruby cargo]

wa

[glossy wave]

I'm wondering if it's because arcsin has a smaller domain, but shouldn't they at least be the same at some interval?

[random ermine]

this feels like a 1=2 proof

there's prolly some domain issue going on

like on the surface it seems fine

,w int \sqrt{1 + sin(x)}

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by mayervietorus

[random ermine]

maybe weierstrass substitution

[autumn fossil]

it works over -pi/2, pi/2 i think

[glossy wave]

when I graph it it doesn't

that's why it's weird for me

[plush bramble]

did you mess up

if i leave out the 0.03 they overlap perfectly

[autumn fossil]

this is because of that u-sub i think

you can easily extend the function though

[glossy wave]

idk maybe I put the integral in desmos wrong somehow

[glossy wave]

.

[autumn fossil]

you rendered just -2sqrt(1-sin(x))

you need to render
-2sqrt(1-sin(x)) - (-2sqrt(1-sin(0)))

to get that integral from 0 to x

[glossy wave]

okay nvm it does overlap, thanks

alright, thanks guys I'm closing for now

.close

[pearl pond]

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[cobalt hinge]

If you want to actually solve it, use the half-angle identities

You have to convert the sine to a cosine first though

[pearl pond]

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[brazen skiff]

Can anyone explain circle therom like the working part I know the reasoning and stuff but not the working

[bitter herald]

What circle theorem

[brazen skiff]

Just them in general

Like for example find the angle bxc

[cobalt hinge]

There's too many

[cobalt hinge]

!xy

[pearl pond]

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.

[brazen skiff]

Let me try find an example

This for example

I don’t get how to work it iut

But if it comes to reasonings it’s easy

[cobalt hinge]

You can use central angles here, no?

And inscribed angles

[brazen skiff]

Pretty sure ya

[pearl pond]

@brazen skiff Has your question been resolved?

[pearl pond]

@brazen skiff Has your question been resolved?

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[magic ruin]

How they did it...please elaborate

[pearl pond]

@magic ruin Has your question been resolved?

[nocturne jetty]

Have you started by calculating the number?

[cobalt hinge]

Notice that 0.85(1.15n)=782

That is essentially what they started with

[magic ruin]

what does the first line means

[cobalt hinge]

Then they did 0.8(1.25(0.75n))) to get the result

[cobalt hinge]

15%=3/20

[magic ruin]

oh ok

fine

got it

thanks

.close

[pearl pond]

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[surreal kraken]

idk if this server helps with stats but

[surreal kraken]

i ran a chi squared goodness of fit test for color distributions in a jolly rancher bag

the observed counts came from a sample of 360 jolly ranchers

the expected counts just followed a uniform distribution

[surreal kraken]

now, i want to show one color is greater than every other color using this data

i understand i can just use two-proportion z-tests, but im not sure what the conditions for that test are

[junior loom]

so you want to test that the distribution is not uniform?

[surreal kraken]

because im using a multinomial sample

[surreal kraken]

no i already did that

i now want to show one color is greater than everything else using pairwise z-tests

[junior loom]

but that has to be true if the distribution is not uniform

no?

[surreal kraken]

no, the chi square test results show that any one proportion differs from another

now, i am trying to show that one specific category count is greater than everything else

i understand how 2-proportion z-tests work

[shut loom]

i was never the brightest at statistics, but i'm pretty sure the conditions are:

• each observation independent
• random sampling
• there must be enough expected successes and failures in both groups
• data are counts or categories, not measurements

[surreal kraken]

but idk how to satisfy the conditions for it since i have one sample

[surreal kraken]

yes these are the conditions

but see how its each observation independent

[junior loom]

oh, out of n colors, probability of color 1 is greater than probability of color j for all j>1?

[surreal kraken]

it fails that, so the alternative way to satisfy it is to show the sample size is 10% of the population

but i just realized that this is one singular sample

i n a two-proportion z-test, youd sample from two independent populations for the colors

[surreal kraken]

yes

actually im being stupid sorry

the conditions dcan be satisfied

thanks though

.close

[pearl pond]

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[surreal kraken]

.reopen

[pearl pond]

✅

[surreal kraken]

its cuz

[shut loom]

lmao

[surreal kraken]

i can just use the counts for one color as one n1

and the counts for another as n2

and proceed

that should be valid right

[junior loom]

but the only way it can be false that the underlying probability of a color is not greater than all the others

is if the distribution is uniform?

[surreal kraken]

no because

[junior loom]

this is just a counting argument

[shut loom]

i would approach it treating the problem as a test within a multinomial sample (McNemar's test)

[surreal kraken]

a chi squared test has Ha: p1 != p2 != p3 != p4 != p5

it doesnt show p1 > p2, p1 > p3, p1 > p4, p1 > p5

[surreal kraken]

hmm i never learned about that

i will look into it

[shut loom]

i think it's exactly your case

[surreal kraken]

that's good

[junior loom]

do you mean for a particular p1, or are you testing for the existence of any p1

[surreal kraken]

i'll give you my data just to verify

[surreal kraken]

particular p1

[junior loom]

oh, okay, that's why I wasnt following

[surreal kraken]

i want to show blue > red, blue > green, blue > pink, ...

ill check if the mcnemar test works for this

@shut loom i looked into it, i dont think it applies

[shut loom]

yeah mcnemar would not work

i was thinking it differently

[surreal kraken]

okay

i just do two-proportion z-tests right

and i determine level of significance with some correction mechanism

since this is a multinomial sample

[shut loom]

yes

or a chi-squared

[surreal kraken]

i have done chi-squared already

this just shows pblue != pgreen != pred ...

[junior loom]

why not just do

[surreal kraken]

i now want to show pblue > pred, pblue > pgreen, ...

[junior loom]

test if p_blue > p_green, then test if p_blue > p_purple...

[surreal kraken]

yes that's what im gonna do

[junior loom]

each one is a one sided proportion test

great

[shut loom]

yeah

[surreal kraken]

so lets say i do a two-prop z-test for blue against red

is n_blue = 159 and n_red = 21

[shut loom]

import numpy as np
n = 360
p_blue = 159 / n
p_green = 53 / n

diff = p_blue - p_green
std_error = np.sqrt((p_blue + p_green - diff**2) / n)
z = diff / std_error
print('z-statistic:', round(z, 2))

can't you do this?

like for every case

or it'll have overlap

[surreal kraken]

yes but i need to check the conditions to ensure z-tests are valid

one being the 10% condition, that the sample sizes are 10% of the population

the problem is that this is a multinomial sample

[surreal kraken]

n_blue and n_red are from the same sample

so im not sure if i can even do a two-prop z-test here

[shut loom]

if they are from the same multinomial sample, i believe you cant right?

[surreal kraken]

i'm not sure, that's why i am asking

there aren't many places online i can verify this sadly

[junior loom]

https://en.wikipedia.org/wiki/Bonferroni_correction this looks relevant

[shut loom]

https://en.wikipedia.org/wiki/Two-proportion_Z-test

[surreal kraken]

yeah this is what you'd do for a two-prop z-test i think

so which test should i do if i cannot do 2-prop z-test

[junior loom]

you can do 2-prop z test but you need to adjust your confidence accordingly, if I understand the Bonferroni correction correctly

[surreal kraken]

i recall reading a textbook saying that post-hoc tests for chi squared are pairwise two-prop z-tests

and that youd correct it due to the higher probability of a type 1 error

[surreal kraken]

so i am allowed to run two-prop z-tests if the random sample condition is met, 10% condition is met, and n1p1 n2p2 n1(1-p1) etc are met?

[junior loom]

yes, but, you need to make your confidence level smaller accordingly

[surreal kraken]

okay

[junior loom]

the significance level

needs to be divided by the number of null hypotheses you're testing

[shut loom]

i think in my textbook there was something like forcing mutual exclusive categories

but this is as far as i can go, i'm just an undergrad

[junior loom]

btw

from xkcd

[shut loom]

but applying to your case, was something like,
if this one is blue, it cannot be green, red, pink or purple

so we assume they are one dimensional

it excludes a bunch of samples, but it makes them viable for the 2-prop z-test

i couldnt find it tho

[surreal kraken]

haha yeah

id need way more samples to actually have meaningful results

this is for a class project for ap statistics though, so i'm sure it doesn't matter that much (:

[junior loom]

not necessarily, you have a very large potion of blue in your sample

[shut loom]

omg you're going really deep

[surreal kraken]

it is definitely true that jolly rancher distributions do not follow a uniform distribution

[shut loom]

i didnt do that much

[surreal kraken]

i'm also a little bored

[surreal kraken]

ik many people who complain about the asymmetric distribution of color in candies

[shut loom]

i think what you have is completely fine for AP Stats

if you want to complete it a bit

[shut loom]

do this for every case

[surreal kraken]

yes

[shut loom]

and ignore the edge cases

idk

[surreal kraken]

this is kinda a shitty project idea though lmao i didnt want to choose something hard

so i chose something a 6th grader could sample

im making up for it by running 5 statistical tests 😂

[shut loom]

i did also choose jelly beans lmao

[surreal kraken]

everyone else in my class is choosing weird complicated topics

idek how they will procure sample data for those topics

maybe their ideas are just too lofty

[shut loom]

one group chose some stuff with people with different ages

i dont remember exactly

i just remember that ages 70+ they had any data to work with

so they couldnt finish

[surreal kraken]

one person is trying to show a correlation of temperature with football performance

i have absolutely zero clue how they will get sample data for that

[shut loom]

hahahahahahah

good luck

[surreal kraken]

especially considering that we have to collect our own sample data

[shut loom]

yeah same here

i just bought a bag of candy

for like 5$ converting to usd

it's the default stats project

[surreal kraken]

😂

anyways thank you both for your help

appreciate it 🙂

[shut loom]

no problem dude

🙏

[surreal kraken]

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[pearl pond]

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[wide dagger]

why do i have to change the limits when i do a problem

[wide dagger]

integral

no the bounds haha

[late spindle]

can you give some examples

alr gotta go sorry, if your talking abt u substitution, think of it like streching and compressing the axes, you must compensate for that by adjusting the bounds @wide dagger

[pearl pond]

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[silver pasture]

Would it be domain for f is -4,4 range is -6,2 domain of g is -4,4 range is -2,6?

g(0) undefined
-3+4 = 1
-4/3
-2*6=-12
-4

[edgy wren]

Ranges of f and g are wrong
g(0) is defined (there should be a value)
g o f (-4) is not -12

[silver pasture]

I thought that when there's an open circle and closed one for a value it's undefined
so would g be 4, f(-4) is -2, g(-2) is 4

[edgy wren]

this means only that the function is not continuous but defined at the closed circle
g o f (-4) is indeed 4
check your mistake on the other parts

[silver pasture]

I think range for f is (-6,-2) and not sure about g because of the open circle

[edgy wren]

range is the y values defined for the function check again

[silver pasture]

g(0)=2
f(-2) + g(-2) = -3+4
g(-2) = 4, f(-2)= 3 -4/3
4
f * f, f(0) = 2. 2x2 =4?

[edgy wren]

you didn't have mistakes here you did it correctly I meant for the ranges you still have them wrong
f o f is still -2 like you said earlier

[silver pasture]

This is what I'm looking at

[edgy wren]

what interval notation do you use
like the square bracket [ ) or just brackets ?

[silver pasture]

The [ bracket sorry

[edgy wren]

ok for f [-6,-2]
but for g the range is still defined like (-3,6]
do you know why ?

[silver pasture]

I'm really not sure I forgot everything about the piecewise graphs

I'm only look at the points with the closed circles and trying to disregard the open circle (white one)

(2,6]U(-2,1)

[edgy wren]

just checked (half asleep sorry)
the range should be (-3,1] U [2,6]
simply -3 isn't in our range but every thing bigger than -3 up to 1 is valid in our range
1>=y>-3
and for the second set
2>=y>=6

[silver pasture]

Thank you, this is for g right

So for part b it's
g(0)= 2
-3 + 4 = 1
g(-2) = 4, f(-2)= 3 -4/3
4
I checked the last one. f of 0 is -4, and f(-4) is -2.

And this would be for part a

[edgy wren]

Yep you should revise the interval notation and domains and range
But yep the answers look correct

[silver pasture]

Thank you so much bro, I'll fix the notations

[edgy wren]

I recommend watching videos about function and notation
To strengthen the understanding

[silver pasture]

Thank you I will

Have a good night

[edgy wren]

it's morning lol

[edgy wren]

but yea if you are finished close the room

[silver pasture]

Lol I'm on east coast, maybe different countries

.close

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[silver pasture]

Wait can you tell me what I would use for the brackets? For everyone one except range for g I would use )?

So (-4,4), (4,4) (6,-2) and then (3,1]U[2,6]

[edgy wren]

you could check interval notation on youtube
and also
[-4,4] , [-4,4] [-6,-2] and that last is correct

[silver pasture]

Thank you again

Oh I see, ( indicates it's excluded so that's why you used (. But ] means it's included (closed circle/black circles)

[edgy wren]

.reopen

[pearl pond]

✅

[edgy wren]

yep

.close

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[silver pasture]

.reopen

[pearl pond]

✅

[silver pasture]

I don't think g is defined

[pearl pond]

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[inland laurel]

In question34 do we take log base 10?

[oak ivy]

If it's not specified

I think you take base e

[indigo fox]

no, that would be ln

[inland laurel]

Nothing is specified their

[shrewd mist]

log usually means either log base e or log base 10, we can't predict what convention your teacher is using

[inland laurel]

Im practicing from book

If i take log base 10 is it legal

[oak ivy]

Here nothing is mentioned

[shrewd mist]

we also can't predict what convention your book is using

[oak ivy]

So do whatever

The book must have answer key so you can check

If it does not match try base e

[inland laurel]

Alr

[oak ivy]

Here I think it's 10

[inland laurel]

Whenever the base is not given why do we have to consider log base 10 or log base e accordingly

[oak ivy]

ln is used for base e

[inland laurel]

Yws

Natural log

[shrewd mist]

its just convention

[inland laurel]

Alr

Will it go if I don't take any base

[oak ivy]

How can you do it without any base

[shrewd mist]

what

[inland laurel]

We can't

[toxic lichen]

if OP has just started learning about logarithms, i'd sooner assume log without a base means decimal log.

solve it as if it were decimal logs throughout, then check your answer here or back of the book.

[pearl pond]

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[inland laurel]

how log² (x-3) base 10 can be written in the simplified form

[toxic lichen]

it can't, and your second step is wrong

why did the -3 end up under the logarithm?

$\log^2(x) - 3 \neq \log^2(x-3)$.

[jolly parrot]

Ann

[toxic lichen]

luckily this mistake is (more or less) fixable.

when we do that, we get: $$\log_{10}(\sqrt{x}) - \log_{10}(x) + \log_{10}^2(x) - 3 = 0$$

[jolly parrot]

Ann

[toxic lichen]

@inland laurel would you like me to suggest you a next step? yes/no

[inland laurel]

No

[toxic lichen]

ok

continue on your own then

[inland laurel]

log(√x/x) base 10 + log²(x) base 10 = 3

log (1/√x) base 10 + log²(x) base 10 = 3

Suggest the next step @toxic lichen

[toxic lichen]

substitute $y := \log_{10}(x)$. note that $\log_{10}\paren{\frac{1}{\sqrt{x}}} = \log_{10} (x^{-1/2}) = -\frac12 y$.

[jolly parrot]

Ann

[inland laurel]

Got it

how log²(x) base 10 can be converted into form log x base 10

[verbal whale]

It stays as it is

You can't do anything

[inland laurel]

Let y = log(x) base 10

it may be represented in terms of y

[toxic lichen]

the idea behind a substitution is to represent everything in terms of y.

in particular $(\log_{10}(x))^2 = y^2$...

[jolly parrot]

Ann

[inland laurel]

Yes

Is the answe 100

Answers aren't given in the book

x = 100

[toxic lichen]

the equation becomes a quadratic in y -- it feels like there should be two answers.

[pearl pond]

@inland laurel Has your question been resolved?

[inland laurel]

Not yet

[toxic lichen]

show what you get immediately after substitution

[jade tangle]

Can u help

#help-5

[toxic lichen]

!noadvert

[pearl pond]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[jade tangle]

:3

[pearl pond]

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[cosmic crag]

Hello

[cosmic crag]

Is this channel free?

How did they arrive at the second step?

I know these relations

[hazy pilot]

they used the fact that lim x->a f(x) = lim x-> e^ln(f(x))

[cosmic crag]

This is the expression I am having trouble with at the moment

[hazy pilot]

and then they used that as x -> 0, ln(1 + x) is approx x

stupid they did that in 1 step but alright

[toxic lichen]

@cosmic crag

[inland laurel]

Another answer is not possible due to fraction

[toxic lichen]

well now you were an hour late so you have to take a new channel again

repost your question and all your work there

[pearl pond]

@cosmic crag Has your question been resolved?

[pearl pond]

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[sick mural]

Is there any way to prove that f2 = x using b2 + b3 = 180 - 2x

[sick mural]

I need the mark for my test

[compact ridge]

where is f2?

[blissful salmon]

Probably the 2 at the F is the angle

[compact ridge]

oh that one I see

[blissful salmon]

Uh can you use cyclic quadrilaterals?

[pearl pond]

@sick mural Has your question been resolved?

[blissful salmon]

What does angle CFA and angle CBA imply?

[sick mural]

I need to prove it can be done using the 180 - 2x

Bc then I can get 1 mark lol

[blissful salmon]

Oh ok, I just found it using another way

[blissful salmon]

You should probably X this

[blissful salmon]

Ok maybe first try find what angle C2 is

[blissful salmon]

I don’t get the markings though. How come both A1 and angle A are both 180-2x? Doesn’t make sense

[pearl pond]

@sick mural Has your question been resolved?

[sick mural]

Oh that was just my teacher angle A isnt that

[pearl pond]

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[tall garnet]

When $x_{1}, x_{2}, \dots x_{n} > 0$, prove
$$\frac{n^2}{(x_{1}+x_{2}+\dots+x_{n})^{2}} \leq \frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\dots+\frac{1}{x_{n}^{2}}$$

[jolly parrot]

polar

[tall garnet]

let all $a=\frac{1}{x}$ ir $b=x$, then (cauchy)

$$
n^{2}\leq \left( \frac{1}{x_{1}^{2}} +\frac{1}{x_{2}^{2}}+\dots+\frac{1}{x_{n}^{2}}\right)(x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}) \iff \frac{n^{2}}{x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}}\leq \frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\dots+\frac{1}{x_{n}^{2}}
$$

[jolly parrot]

polar

[tall garnet]

furthest i manage to get

[cyan bronze]

[
\sum_{i=1}^n x_i^2 \leq \left( \sum_{i=1}^n x_i \right)^2
]

[jolly parrot]

Radiation 𝕏

[cyan bronze]

when x_i are positive

[tall garnet]

omg i somehow messed these up

like

the sign

no

the inequality

so i got >=

thank you lol

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[dapper kraken]

how did they achieve the fact that its either 5 or 8, i get why the answer can be 5 but why 8 in the note

[cyan bronze]

ig i can narrow it down to 5,7, or 8

not sure how they got rid of 7 in the note

it must be divisible by 2 otherwise the child count would drop by 4
it must be divisible by 3 because sum is divisible by 3
it must be divisible by 4 otherwise 4 and 8 would get eliminated, dropping the count by 2 which is also not allowed
it must be divisible by 6 because already divisible by 2 and 3

so 5,7,8 are only ones that dont have a 'must' on them

but again what i said kinda relies on the assumption that not being a kid would imply that your age is not dividing the number

but we only know the opposite, which is being a kid implies your age divides the number

[cyan bronze]

tho ig this could be proved

[tropic saddle]

if 4 is missing, then 8 is not missing, so the number is divisible by 8, but then also by 4

etc

same idea but phrased differently

but I also dont see how to eliminate 7. 5 and 7 are both prime so with respect to divisibility its the same number. and there are no other multiples to consider

[midnight haven]

if b=4, the only number is 5544. It is divisible by 4,6 and 7. Why 7 would be a answer?

[dapper kraken]

the note suggests just thinking about the question you can narrow it down to 5 or 8 without doing the heavy lifting (thats what im getting atleast)

[tropic saddle]

btw am I stupid or why did they just ignore b=5,6,7,8. they removed the symmetry on a,b, by now for example having numbrs bbaa etc

[dapper kraken]

im guessing its checking from b=1 to the top and b=4 just happens to work and they stop there?

[tropic saddle]

yeah but that is incomplete

there could be another solution

also they didnt even use that the last two digits are the age of the dad. however you are supposed to use that

[dapper kraken]

honestly this alternate solution looks better :p

my only complaint is that it randomly uses 0bb0

[tropic saddle]

well number plates in that country might start with 0s

but yeah thats a nicer solution

although you of course have to guess correctly to start with assuming 5 is among the kids

[dapper kraken]

ok

thanks ill close this now

.solved

[pearl pond]

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[midnight haven]

Can someone tell me how to do the c part

I did a and b

[dapper kraken]

to find the normal surface area of a cone, you find the area of the base and add it with the lateral surface area right?

you can see the surface area of the fructum is the base of the big cone+base of the small cone+lateral of the big cone-lateral of the small cone

[forest sigil]

Lateral surface+top surface +down surface

[midnight haven]

Whats the lateral surface?

[dapper kraken]

slanted area

[midnight haven]

Slanted height?

Ohh wait do u mean the curved surface area?

[dapper kraken]

yeah

[forest sigil]

Yes

[midnight haven]

Oki wait ill see if i get the answer

OK i git the answe thank u very much.have a nice day

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[rough forge]

. @rugged hollow

[stuck vapor]

hey!! im learning abt loans and stuff and had a question abt someone loaning 400000 kr (swedish currency) in a bank to start a business. the bank took 6.5% in interest. it then asked abt the interest kosts within the first quarter and first month. i was wondering if i could divide 0.065 by 4 and if that tells me the interests costs per quarter? or is that not right? cause rn i js know to times 400,000 by 0.065 to find the first year and then divide that by 4.

[short sparrow]

no, not quite

6.5% is the interest after a year right?

[stuck vapor]

yes

i was thinkinh of multiplying the 0.065 by 400000 nezt