help-33

[vernal forge]

that... shouldn't matter

(i have no clue what you are talking about)

show an example pls

[jovial mulch]

xD

sorry.
I have to go

Im about to loose it

I need to go for a walk lol

Sorry for that

Ty for the help from before tho

.close

[marsh citrus]

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[dire ermine]

Here only -sqrt3 is correct?

[dire ermine]

Since + is outside of the domain?

Or are both correct

[marsh citrus]

@dire ermine Has your question been resolved?

[marsh citrus]

@dire ermine Has your question been resolved?

[marsh citrus]

@dire ermine Has your question been resolved?

[dire ermine]

.close

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[limber wing]

does somebody can help me with physics ?

[marsh citrus]

@limber wing Has your question been resolved?

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[limber wing]

.reopen

[marsh citrus]

✅

[limber wing]

<@&286206848099549185>

[marsh citrus]

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[marsh citrus]

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[clever mortar]

I’m trying to find the solutions for this but I’m a little bit confused. What’s the difference between the equation above and if it was just sin(t) and not 1/sin(t)?

[clever mortar]

Does the difference between 1/sin(t) or just sin(t) affect the solutions? Would it be 1/Y instead of just Y since it’s 1/sin(t)?

[gleaming pecan]

$\text{remind that: }\text{ }\left| \sin\text{}x \right|\le1$

[elfin berry]

Joanna Angel

[gleaming pecan]

think about what this might mean in your case

[marsh citrus]

@clever mortar Has your question been resolved?

[clever mortar]

Ik that sqrt3/2 is less than one, but idk where to from there

Or what to do with that info

Ik that if it was sin(t)= sqrt3/2, I would have to look for a point on the unit circle that has a y value of sqrt3/2, but does this change if it was 1/sin(t)?

[gleaming pecan]

but you must also thikn how sint = , in your equation

it is not sqr(3) / 2

[gleaming pecan]

$\frac{1}{\sin\text{}t}=\pm\frac{\sqrt{3}}{2}\Leftrightarrow \sin\text{}t=\pm\frac{2}{\sqrt{3}}$

[elfin berry]

Joanna Angel

[gleaming pecan]

analys e it now

what is your conclusion then?

[clever mortar]

Oh wait I understand, I think I messed up, let me show what i have

[gleaming pecan]

ok

[clever mortar]

I accidentally wrote 1/sin for step 5

Otherwise it would have been correct as just sin(t)

Right?

[gleaming pecan]

can you change it in yoru paper and pase it corrwect ?

[gleaming pecan]

now i agree

🙂

$\sin\text{}t=\frac{\sqrt{3}}{2}\text{ }\text{ }\vee \text{ }sin\text{}t=-\frac{\sqrt{3}}{2}$

[elfin berry]

Joanna Angel

[clever mortar]

haha ty

[gleaming pecan]

use this:

[clever mortar]

Is this correct

[gleaming pecan]

yes and you can add 2kPi to every such soution

[clever mortar]

Ok tysm

[gleaming pecan]

yw)

[clever mortar]

.close

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[steady mural]

can someone explain how to do 2a.

how do I find the distinct words?

I used 10! to get all possible arrangements but they arent distinct as repeating As and Ns

[marsh citrus]

@steady mural Has your question been resolved?

[sand fable]

you divide by the factorial of appearing letters

so BANANARAMA has 5 As and 2 Ns

in practice, this would look like: 10! / (5! * 2!)

[vast frigate]

It has different properties

Wow they basically gave it lol

Its 10! / (5! * 2!)

The solution is to find the distinct letters

And count how many there are

FOr example in the word "two"

They are all distinct letters

So its
3! /(1!*1!*1!)

WHich is just 3! Lol

[marsh citrus]

@steady mural Has your question been resolved?

[marsh citrus]

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[lilac fulcrum]

I need help

[lilac fulcrum]

With algebra

<@&286206848099549185>

[thorn tide]

Wassup

Send ques

[still temple]

ask away

[thorn tide]

Don't ask to ask

[lilac fulcrum]

Wait hang on

My teacher messaged me on how to solve it

[thorn tide]

Oke

[lilac fulcrum]

thanks guys

[thorn tide]

So you don't need help anymor?

[still temple]

np

[lilac fulcrum]

nope

[thorn tide]

.close

!close

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[sharp mauve]

.rotate

[sharp mauve]

What do i fill in the inside intergral?

I rearranged r to a circle function

[sharp mauve]

.rotate

.rotate

,rotate

[sharp mauve]

.close

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[wooden storm]

The length of segment AB is 12 centimeters. Point M is taken on this segment. Find the lengths of segments AM and BM if 3AM=2BM

[wooden storm]

<@&286206848099549185>

[atomic ruin]

take a ratio

AM/BM = ?

[wooden storm]

3/2?

[dull parcel]

yea

[atomic ruin]

what

isnt it 2:3

[dull parcel]

hahaha

its 2/3 ofcos

[wooden storm]

Oh yeha my bad

[atomic ruin]

mhm

[dull parcel]

mhm

[atomic ruin]

so now we can express AM as 2k and BM as 3k where k is a positive integer

[dull parcel]

? - ?

[atomic ruin]

AB = AM+BM = 2k+3k = 5k

5k = 12, k = 12/5

substitute k=12/5 into AM = 2k and BM = 3k

[dull parcel]

2k = 4.8, 3k = 7.2

so, AM =4.8cm and BM=7.2cm

[atomic ruin]

yes, that gives you AM and BM

correct

[wooden storm]

Answers say that its 7 and 5

[atomic ruin]

huh

[dull parcel]

wait

[atomic ruin]

that doesnt even make sense

3am = 2bm

[wooden storm]

Idk

[atomic ruin]

am = 7 bm = 5

then 3(7) =/= 2(5)

[elfin berry]

Question mark

Question mark

[wooden storm]

WAIT

[dull parcel]

yes?

[wooden storm]

Sorry 3AM doesn't equal to 2BM

[dull parcel]

.....ok

[wooden storm]

3AM - 2BM=11

[dull parcel]

what is it

[wooden storm]

Damn my bad

[dull parcel]

....

ok

[atomic ruin]

Oof

[dull parcel]

so let AM be x and BM be y

so

[atomic ruin]

simultaneous equation

x+y = 12
3x-2y = 11

[dull parcel]

$x+y=12$ and $3x-2y=11$

waittt

sorry

wrongtype

wait again

[atomic ruin]

2y not 3y

[elfin berry]

Question mark

[dull parcel]

sorry

then $$2x+2y=24$$ and $$5x=35$$ therefore $$x=7$$

[elfin berry]

Question mark

[lavish kernel]

Yeah correct

[dull parcel]

and because $$x+y=12$$

[elfin berry]

Question mark

[dull parcel]

$y=12-7=5$

[elfin berry]

Question mark

[dull parcel]

ok, solved

!status

[marsh citrus]

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

[lavish kernel]

$$y = 12 - x $$ substituting into 3x - 2y = 11 $$x=7$$

[elfin berry]

David K.

[wooden storm]

Thanks guys

[dull parcel]

$$y = 12 - x $$ substituting into $3x - 2y = 11$ $$x=7$$

[elfin berry]

Question mark

[dull parcel]

looks better~

[wooden storm]

Thank you

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

wouldnt both probabilities be higher than 1?

[still temple]

idk maybe im being dumb

[vernal forge]

probability can't be higher than 1

[still temple]

yea ik

but when i calculate the circled ones...

i get higher than 1 idk how

[vernal forge]

show me your calculations

[still temple]

i said P(B|A) = P(B intersect A) / P(A), as A is a subset of B then P(B intersect A) = P(B), so P(B|A) = P(B) / P(A)

as A is a subset of B then P(A) <= P(B)

then P(B|A) => 1 ?

[vernal forge]

no no

$A \subset B \implies A\cap B = A$

[elfin berry]

artemetra

[vernal forge]

not B

[still temple]

ohh wait true lol

[vernal forge]

you are confusing it with $\cup$

[elfin berry]

artemetra

[still temple]

wait but if B happens shouldnt A also happen tho

[vernal forge]

no

if A happens, then B happens
if B happens, then A might happen or not

[still temple]

wait but

[vernal forge]

yes

there are some possible outcomes in B that are not in A

but all outcomes in A are also in B

[still temple]

ohhh and if A happens since it is inside B then B must happen if A happens

[vernal forge]

yes

[still temple]

ok so similarly P(omega | B) = P(omega intersect B) / P(B), as P(omega intersect B) = P(B) then P(omega | B) = 1 ?

[vernal forge]

yep

[still temple]

ayy thxthx

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[marsh citrus]

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[lucid zenith]

A function $f:\mathbb{R}\to\mathbb{R}$ is additive and satisfies $f(x)f(\frac1{x})=1$. I need to find all functions f

[elfin berry]

kheerii

[lucid zenith]

I choose $y=x+\frac1{x}\implies f(y)=f(x+\frac1{x})=f(x)+f(\frac1{x})=f(x)+\frac1{f(x)}$

[elfin berry]

kheerii

[lucid zenith]

so, $|y|\ge 2\implies |f(y)|\ge 2$

[elfin berry]

kheerii

[lucid zenith]

is this a boundedness condition?

if I recall, the last inequality has to be reversed for boundedness to occur

like, |f(y)|<=M for some M in some range of y

[marsh citrus]

@lucid zenith Has your question been resolved?

[lucid zenith]

<@&286206848099549185>

[zenith python]

take $n \in \mathbb{N}^*$, then $f(n) \cdot f \left( \frac{1}{n} \right) = 1 \iff f \left( n^2 \cdot \frac{1}{n} \right) \cdot f \left( \frac{1}{n} \right) =1 \iff n^2 \cdot \left( f \left( \frac{1}{n} \right) \right)^2 = 1$

[elfin berry]

LF

[lucid zenith]

what is N*?

[zenith python]

non zero positive integers

[lucid zenith]

ah ok

[zenith python]

and you can solve for f(n)

[lucid zenith]

so we get $f(\frac1{n})=\pm\frac1{n}$

[elfin berry]

kheerii

[lucid zenith]

but that won't give us any info about irrationals

I am trying to prove that f is bounded on some interval so I can use $f(x)=cx\forall x\in\mathbb{R}$

[elfin berry]

kheerii

[lucid zenith]

which would give us the result directly

[zenith python]

we suppose that f is continuous?

[lucid zenith]

no, if f were continuous then the result would follow directly

https://en.wikipedia.org/wiki/Cauchy's_functional_equation

I need to prove one of these three conditions

[zenith python]

right

actually i think you can show its continuous

it might be easier than using boundedness

especially since you only need continuity at one point

[lucid zenith]

at what point?

[zenith python]

x=1 would go nicely i think

$1 + \epsilon < 1 + \frac{1}{n}$ for some n

[elfin berry]

LF

[lucid zenith]

.close

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[abstract niche]

Is there any way to do this in one step in Wolfram Mathematica?

[abstract niche]

Nvm found it

.close

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[halcyon cradle]

Let R be the set of all realnumbers and 𝑹𝟐 = {(𝒙𝟏, 𝒙𝟐): 𝒙𝟏, 𝒙𝟐 ∈ 𝑹}, then which one of the following is a subspace of 𝑹𝟐 over R?

a){(𝑥1, 𝑥2): 𝑥1 > 0, 𝑥2 > 0}
b){(𝑥1, 𝑥2): 𝑥1 < 0, 𝑥2 < 0}
c){(𝑥1, 𝑥2): 𝑥1 ∈ 𝑅, 𝑥2 > 0}
d){(𝑥1, 0): 𝑥1 ∈ 𝑅}

[void elm]

do you know what properties a subspace must have?

[halcyon cradle]

Addition property and multiplication

And scaler multiplication

[void elm]

could you be more specific?

i think you know what a subspace is, but i just want to be sure lol

[halcyon cradle]

Ahh yes. I understand the properties but I don't know how they applied

[void elm]

so a subspace is closed under addition and scalar multiplication, and also contains the zero vector

[halcyon cradle]

Yes yes

[void elm]

so for each of your candidate subspaces

[halcyon cradle]

B is discarded

[void elm]

you must verify these properties

[void elm]

you're right, but why?

[halcyon cradle]

1st and 2nd doesn't have addition idenity vectors

[void elm]

why does that matter?

[halcyon cradle]

They should have addition idenity vectors

Right?

[void elm]

not necessarily

check if B is closed under scalar multplication

[halcyon cradle]

If i multiply B with any negative numner

[void elm]

i.e. for any (x1,x2) in B, is c(x1,x2) in B?

[halcyon cradle]

It will give positive vector

[void elm]

thats right

and such a positive vector cant be in B

[halcyon cradle]

Same thing go to 1st option

[void elm]

yup

[halcyon cradle]

Let me think about 3rd and 4th now

How to check 3rd 4th?

[void elm]

same thing

[halcyon cradle]

They have real and 0 both vectors

[void elm]

let's start with C

[halcyon cradle]

Yes

[void elm]

check scalar multiplication

again

[halcyon cradle]

Yes if we multiply them by any scaler

It will be closed i guess

[void elm]

why do you think so?

[halcyon cradle]

can I seperate them?

X1 and x2

[void elm]

seperate them?

wdym

(x1,x2) is a single vector

there's nothing to seperate

[halcyon cradle]

if i multiply x2 with a negative number

(-2x1,-2x2)

But here x2>0 should be

Which is failed right?

[void elm]

yes

[halcyon cradle]

I see i see

[void elm]

x1 can be anything, but x2 must be >0

so if you multiply by a negative scalar c

[halcyon cradle]

It will ne negative

[void elm]

then cx2 is negative, yes

[halcyon cradle]

Ohh D will work

Because 0 will remain it same

X1 can be scaled

And 0×R=0

[void elm]

well D is closed under addition and scalar multiplication, yes

so it is a subspace of R^2

[halcyon cradle]

I see i see now i understand how things work and oroperties

[void elm]

(x1,0) + (y1,0) = (x1+y1, 0) and c(x1,0) = (cx1,0) and (0,0) is in D

[halcyon cradle]

Most of question can be checked with scaler multiplication

[void elm]

don't forget addition though

sometimes a subset is closed under scalar multiplication but not addition

[halcyon cradle]

Addition identity is must?

[void elm]

but yes, scalar multiplication is usally faster

[void elm]

the additive identity need not be part of the subset in order for it to be a subspace

but the subset should be closed under addition

so if the subset is called W, then for all x,y in W, x + y should be in W

[halcyon cradle]

Yes yes

Closed property

[void elm]

yes

[halcyon cradle]

I have read definitions but you know things can be clear when we see in examples

[void elm]

indeed

[halcyon cradle]

I understand now the properties application

Thanks

[void elm]

👍

[halcyon cradle]

Let me upload my next question soon

.close

[marsh citrus]

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[echo jolt]

Each of them I got 0, it's impossible? 4b) ODE question

[marsh citrus]

@echo jolt Has your question been resolved?

[echo jolt]

<@&286206848099549185>

[marsh citrus]

@echo jolt Has your question been resolved?

[echo jolt]

Still confused 🤔

[echo jolt]

.close

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[halcyon cradle]

If a/b=11/13
What will be value (a^3-b^3)/(a^3+b^3)

I solved it by expansion of cube a+b and got 0.25

[halcyon cradle]

Is there any simple method

If any, please share

[spark otter]

I would just write a = (11/13)*b

Replace every instance of a with this expression

And it becomes $\frac{\frac{11^3}{13^3}-1}{\frac{11^3}{13^3}+1}$

[elfin berry]

rafilou2003

[halcyon cradle]

Ohh

Interesting

[spark otter]

Oops sorrh

You can rewrite it as $\frac{11^3 - 13^3}{11^3 + 13^3}$

[elfin berry]

rafilou2003

[spark otter]

Then you will get a negative result

It just boils down to calculations, so might be simplified with some cubic formulas

[halcyon cradle]

Do you remember cubic numbers?

I need to calculate 11^3, 13^3 actually

[spark otter]

11³ will be pretty easy

11² is 121

And then 121×11 = 1210 + 121 = ...

1331

Then for 13³ you get 169×13 = 1690 + 507 = ...

2197

And then hard division

So no real need for computing the decimals, but maybe just checking for common factors to simplify

[marsh citrus]

@halcyon cradle Has your question been resolved?

[halcyon cradle]

Yes it is very long

But no other option I guess

[spark otter]

Not very long, only 2 is a common factor

-866/3528 = -433/1764

And here's the (atrocious but in simplest form) answer

I'm just gonna put this comedic gem here

[halcyon cradle]

I guess question makers were drunk

They have given
0.24,0.25 both option

[spark otter]

And a/b = 11/13

First of all if both a and b are positive, you can easily see that a < b and so a³ - b³ < 0 so the result is negative

Very wanky question

[halcyon cradle]

Would you like to see options 🫣

[halcyon cradle]

Agree☺️

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[halcyon cradle]

I can only see the sequence is like

tan9° tan27° tan 45°tan63°

[still temple]

Discord is having media issues. Please transcribe your issue in text so we can see your problem.

[halcyon cradle]

tan(π/20) ×tan(3π/20) ×tan (5π/20)× tan(7π/20)× tan(9π/20) is equal to

A 1
B -1
C 1/2
D none

[next raft]

big sad

[next raft]

U need brackets 😭

[halcyon cradle]

Like how?

Clear now?

[next raft]

on the last two as well

[halcyon cradle]

🙄🙄

[stoic slate]

You can rewrite 7pi/20 as pi/2-3pi/20 and 9pi/20 as pi/2-pi/20

Also 5pi/20 is pi/4

[marsh citrus]

@halcyon cradle Has your question been resolved?

[halcyon cradle]

I see thanks samuel

Let me try

[marsh citrus]

@halcyon cradle Has your question been resolved?

[halcyon cradle]

Omg

You made it so simple

I got tan45°

1

[halcyon cradle]

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[echo jolt]

How do I get from top to bottom? Do I use partial fractions? And K is the constant

[marsh peak]

Multiply top and bottom by K of the left hand side

[echo jolt]

Let me try it

So we can play around the fraction?

[marsh peak]

You can always multiply the numerator and denominator of a fraction by a nonzero number

[echo jolt]

Oh okay

Then the rest I just use the partial fractions to finish off with ln

Thanks

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[still temple]

hi i'm doing the junior maths challenge (first time in intermediate bracket), any tips? thx

[still temple]

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[past charm]

if you can have any amount of positive integers that sum to 2023, how can you pick them such that their product is maximised?

[past charm]

i think it might be something to do with am-gm but i'm not sure

[hasty ruin]

usually when you see some nonsense like this if it's not immediately obvious you probably want to try with smaller numbers first

and see if you can gain some intuition

[past charm]

what sort of things would you need to do to solve this question specifically though

[hasty ruin]

instead of 2023 try smaller numbers

like numbers you can quickly maximize

although there is one idea that if you come up with lets you solve the question in about 3 minutes

[past charm]

hmm so to get the biggest product you have to spread the numbers out as evenly as possible (not sure how to word it)

say the number that it summed to was 10

if you had 2 positive integers you would use 5,5 (can just do 10/2): product = 25

if you had 3 positive integers you would use 3,3,4 (10/3 is a decimal, 3.333, so you would have to use 4s and 3s): product = 36

if you had 4 positive integers you would use 2,2,3,3 (10/4=2.5, using 2s and 3s): product = 36

if you had 5 positive integers you would use 2,2,2,2,2 (10/5=2): product = 32

so the observations are that the product peaked at a point, then started decreasing again

how would you find the point at which the product peaks for an arbitrary number (like 2023 in the original question)

[hasty ruin]

think about the following thing:

what's the biggest number you would ever want to pick for any arbitrarily large number

to get you started, you would never want to pick 1000, because 500 + 500 = 1000 contributes the same amount to the sum but contributes 500*500 = 25000 instead of 1000 in the product

[past charm]

well, you would want the number of integers to be obviously less than the sum (that would give you a product of 1)

and if you number of integers is just 1 then the product would just be same as the sum

not sure how to find the maximum though

[hasty ruin]

you should probably keep trying to lower that biggest number

you can actually make it pretty low

[past charm]

not sure what you mean by that

[hasty ruin]

what's the biggest number you would ever want to pick, in your bunch of numbers, to sum to some target (in this case 2023), while trying to maximise your product

it's smaller than 1000 (because if you ever pick 1000 you should pick 500+500 instead to get a bigger product)

if you keep going by that logic you can make that biggest reasonable number very very low

[past charm]

would it just be 2 then

but then it cant sum to anything odd

[hasty ruin]

no by your own example

[past charm]

meaning you would have to put a 1

[past charm]

why not

im confused

[hasty ruin]

you already showed in your own example that when you want to sum to 10,
2 * 2 * 2 * 2 * 2 is smaller than 2 * 2 * 3 * 3

[past charm]

hmm

is it something to do with turning it into the largest amount of primes

[hasty ruin]

uh, i guess?

[past charm]

so what does that have to do with this

[hasty ruin]

it lets you narrow down very quickly how you should choose numbers once you get the right largest reasonable number

have you tried lowering the bound i gave for you yet
i already gave you a starting point of 1000

[past charm]

umm

i guess you can keep going down until 125

after which it becomes a decimal

[hasty ruin]

you don't have to always divide it by 2..

[past charm]

so it has to be made from 2s and 3s?

[hasty ruin]

yes

if you don't need to show a proof you're basically 99% done, if you need to show a proof you're maybe say 85%

[past charm]

right so im guessing we need to use as much 3s as possible

2022 is a multiple of 3 but that gives only 1 left

2019 is also a multiple of 3, leaving 4

2019/3 = 674

so the maximum would be 674 3s and 2 2s?

[hasty ruin]

yup

[past charm]

oh right, thanks for the help

how would you go about proving it though like you said

[hasty ruin]

you might do the whole lowering largest number and keep going, or through some other method

and eventually you would find that the largest reasonable number is 4 (which is equal to 2 * 2 but doesn't matter)

and then you would want to show that if you pick any number larger x>4 you have a way to pick 2 numbers a, b such that a+b = x and a * b > x

which rigorously shows that the largest number you would pick is 4

then you have to show that picking a bunch of 3s first is better than picking a bunch of 4s

[past charm]

is there any way to do it algebraically

[hasty ruin]

for which part? if you mean picking a,b

pick a = 2, b = x-2

then try showing the result yourself

[marsh citrus]

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[vagrant robin]

$\int^1_{-1} \frac{\ln{(x+1)}}{x}dx = \frac{\pi^2}2$

[elfin berry]

ykj3

[vagrant robin]

I need to prove this but don't know where to start made the subs x <- 1/x

also did x <- x - 1 and tried ibp

[still temple]

Did u try substituting u = ln(1+x)

[vagrant robin]

doesnt do anything

[still temple]

Differentiate it too, to replace dx

With du

du=dx/1+x

dx=du(1+x)

Where x = e^u-1

So dx = (e^u)du

[stoic slate]

This doesn’t have an elemental solution I think

[vagrant robin]

yeah i did that the integral becomes ue^u/(e^u - 1)

[vagrant robin]

for those bounds it does

ok nvm

looks like i just use dilogarithm inversion formula

1/2log(-x) term vanishes at x = -1

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@vagrant robin Has your question been resolved?

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[long cape]

Is this correct? The black line that I drew

And Is this formula somehow tied to Pythagorean theorem?

[still light]

Yes

It's essentially pythagorean theorem applied twice

[long cape]

Wdym by twice?

[still light]

Think of it like finding the length of the diagonal of a rectangular prism

If you apply Pythagorean theorem on one of the faces, you get the diagonal of the face

If you apply it again on that face diagonal and the 3rd dimension, you get the space diagonal

[long cape]

OK

Ty

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[still temple]

having trouble with b

I found g(f(x))

but I'm not getting the answer

this is what I got

$(\frac{-2x^2+24x+2}{x^2-1})(-2)$

[elfin berry]

mj

[still temple]

this is right so far

I just don't know how to get to the answer

[void elm]

how did you get this?

[still temple]

subbing and simplifying

my concern isn't with this

it's correct

[void elm]

ah i see

where is the final (-2) coming from though?

[still temple]

it's b

[void elm]

got it

[still temple]

I assume it's multiplied by -2

[void elm]

oh

no

it's asking you to evaluate g(f(-2))

i think

[still temple]

hmm

that would've been

much simpler

let me try that though

yrp

yep*

getting it now

thanks

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[magic belfry]

How would I work out b?

<@&286206848099549185>

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[viscid edge]

Hi, could I get some help with this weird task?

“Mathematicians are strange sometimes,” thought the inspector.
“We had all of these partially filled glasses lined up on a table one after the other.
Only one of them had poison in it, and we wanted to know which one before we drank the glass
Examined fingerprints.
Our laboratory could have examined the contents of each jar, but these examinations cost money
Time and money, so we wanted to examine as few as possible. We called in the
University and they sent a mathematics professor to help us.
He counted the glasses, smiled and said:
'Take any glass you want, Commissioner, and we'll do that first
investigate.'
'But wouldn't that spoil the whole investigation?' asked the inspector.
'No,' he said, 'this is the beginning of the best selection process. We have to do a glass first
investigate. It doesn't matter which one it is.'«
“How many glasses were there at the beginning?” asked the inspector’s employee.
»I can't remember it exactly. Any odd number between one hundred and
two hundred.”

Determine the number of glasses.
Explain that the math professor's method was not the best.

[viscid edge]

I dont even know how to start

[marsh citrus]

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[viscid edge]

<@&286206848099549185>

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[still temple]

wouldn't the domain of $log_3((2x-1)(x+4))$ be $x>0.5 or x>-4$?

[elfin berry]

!

[still temple]

because apparently not

[quaint elm]

show work

[still temple]

the orignal question

the answers

what I did was I factored the inside of the log

by doing product/sum with -8 and 7

$log_3(2x^2+8x-x-4)$

[elfin berry]

!

[still temple]

$log_3(2x(x+4)-1(x+4))$

[elfin berry]

!

[quaint elm]

oh ok so you need $(2x-1)(x+4) > 0$ right?

[elfin berry]

hayley

[still temple]

yep

wait a min do you have to do interval notation for this

[quaint elm]

no

[still temple]

or i mean number line thing

[quaint elm]

that would help yes

although i tend to just draw a really rough sketch

[still temple]

okay so

doing that

I get a positive sign

from -4 to -inf

and 1/2 to inf

so therefore x>1/2 or x<-4

ah i see now

[quaint elm]

ye

[still temple]

I just took the roots lol

thank you

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[quasi kettle]

i need help

[quasi kettle]

with this

[atomic ruin]

what do you not understand?

[quasi kettle]

wht does 2r mean

i forgot

[atomic ruin]

Where is 2r

[quasi kettle]

i mean 2x

[atomic ruin]

2 times x

or do you mean x^2

[quasi kettle]

oh i mean x^2

[atomic ruin]

that means x multiplied by itself

if x = 3 x^2 = 9

[quasi kettle]

oh

[marsh citrus]

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[tepid hollow]

Any hints on this cant think of a way to do this

[sand fable]

disregard the "largest possible" portion for now

what is the domain of that function?

[novel juniper]

The domain is the inputs a function can take

what values of x can this take as an input

[tepid hollow]

$\mathbb{R}$

[elfin berry]

luke1

[novel juniper]

correct

what about its outputs

that's the range

[tepid hollow]

same thing

its just the wording was a bit confusing for me

[novel juniper]

correct.

[novel juniper]

oh,ok

[marsh citrus]

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[bronze reef]

whats the approach for this question

[sand fable]

would you agree that this problem tells you that he walks such that the required distance is minimized?

[bronze reef]

yes

[sand fable]

alright

for problems like this, you want to reflect one of the points over the line which he's required to touch

in this case, you could reflect point Q over the x-axis

and the straight-line distance between P and Q is the minimum

[bronze reef]

ohh image concept

[bronze reef]

ok I'll give it a try

thank you for helping

[bronze reef]

i got the answer

[sand fable]

nice 🎉

[bronze reef]

thank you

[sand fable]

welcome

[marsh citrus]

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[marsh citrus]

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[main mica]

o so i got a pyramid with a triangle as a base the lateral edges are sqrt70 sqrt99 and sqrt126 the latral edges are also perpendicular to each other i need the vlume of the pyramid and the area of the base

[main mica]

i alr found the sides of the base and the area too i need to find the hight of the pyramid now

<@&286206848099549185>

<@&286206848099549185>

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[undone notch]

can someone tell what to do next to get angle gpb

like what to do to get the measur of pb

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[tardy idol]

im not sure why the angle is wrong

[tardy idol]

i thought the angle should be (pi/2) - arctan(20/5)

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[main mica]

ok so i got a pyramid with a triangle as a base the lateral edges are sqrt70 sqrt99 and sqrt126 the lateral edges are also perpendicular to each other i need the vlume of the pyramid and the area of the base
i alr found the sides of the base and the area too i need to find the hight of the pyramid now
i alr know the area of the base
now i need hight of the pyramid so i can find the volume

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[tardy idol]

i thought the angle should be (pi/2) - arctan(20/5)

[marsh citrus]

@tardy idol Has your question been resolved?

[tardy idol]

<@&286206848099549185>

[marsh citrus]

@tardy idol Has your question been resolved?

[marsh citrus]

@tardy idol Has your question been resolved?

[main mica]

So what woylild it be here

And i think thats wron

G

[marsh citrus]

@tardy idol Has your question been resolved?

[lethal ocean]

How exact does the answer need to be? (pi/2)-arctan(20/5) = 0.244978...

[quiet anvil]

Also worth noting child is moving east of North, and if North is 0 radians, then typically directions clockwise to the 0 direction are considered negative

@tardy idol

[tardy idol]

it was the exactness thanks 😭

[tardy idol]

thank you for this too

.cloae

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[raw saffron]

Is this correct?

[median cloud]

off topic, but you're icelandic?

[main tartan]

isn't -4a+5a=a

[raw saffron]

Yes

[woeful palm]

don’t forget to add brackets after expanding the (a+5)(a-4)

[median cloud]

the answer sholud be|| 2a^2 - a + 20 = 0|| assuming you are just simplifying?

[median cloud]

cool i am too

[raw saffron]

I see

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[marsh citrus]

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[signal wyvern]

Having trouble understanding thjs Consider the pictures below. Each angle
θ
is an integer when measured in radians. Give the radian measure of the angle

[signal wyvern]

Pheta = 150

[marsh citrus]

@signal wyvern Has your question been resolved?

[signal wyvern]

<@&286206848099549185>

[hearty moat]

is this with polar?

should be right

x = rcos(theta), y = rsin(theta)

[marsh citrus]

@signal wyvern Has your question been resolved?

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[deft horizon]

Need help

[deft horizon]

Delta x is 1

And I don't know where to go from there

[shadow plume]

search up the question

and get the eqaution

[marsh citrus]

@deft horizon Has your question been resolved?

[deft horizon]

.close

[marsh citrus]

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[muted cargo]

hi

[muted cargo]

can anyone helps

[nova solstice]

That is hard to read

[muted cargo]

tanx = 1/2

0<x<pi/2

[nova solstice]

Using SOHCAHTOA you know Tan(x) = Opposite/adjacent
cos(x) = adjacent/hypotenuse

[muted cargo]

ok?

@nova solstice and what?

.close

[marsh citrus]

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[muted cargo]

no its not

[dusky dust]

Yes it is the top is

[marsh citrus]

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[marble glacier]

https://gyazo.com/bb42167ff427a8a37e1d2bb6a30b8ab7

[marble glacier]

anyway to represent a semi circle explicifity

like f(x) = ..

instead of (cos(t), sin(t))

[fallow wind]

f(x)=sqrt(1-x^2)

[marble glacier]

thx

i aint integrating that

.close

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[narrow crow]

Hello, I've been working on the question shown in the image and got to the conclusion that after 20000 years of death, there is around 8.89...% or 8.90% of carbon-14 remaining, could anyone double check this? There aren't any answer sheets.

[narrow crow]

I'm really not sure of the method I've used to solve this, so I'm not sure if I got it correct at all

[stoic saddle]

well show us your method

[narrow crow]

alright

you can ignore the finding r part

<@&286206848099549185>

Please ping me if you decide to respond, thx

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[marsh citrus]

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[worldly trail]

4 + 24 + 144 + 864 + ...

this is geometric right?? so you can find the ratio with the formula below

i did the formula below with 24 (2nd term) = (4 (first term) * (r^2 -1)) / r-1 (i did r^2 cuz snd term)

what did i do wrong??

[worldly trail]

r = 5

[stoic saddle]

!xy

[marsh citrus]

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.

[worldly trail]

first one

[stoic saddle]

ok so then why find any sums

you aren't asked for it

also r isn't 5

[worldly trail]

...

[stoic saddle]

24/4 isn't 5 and 144/24 isn't 5 and 864/144 isn't 5

[worldly trail]

ok yeah ik but you should be able to get r from the formula right?

[stoic saddle]

you can also take off your left shoe by sticking your right hand between your legs from the back, and doing a one-handed handstand on your left while doing it

just because you can doesn't mean you should!

(and you messed it up in some way anyway. no, don't ask me to fix it. refer back to the handstand analogy.)

[worldly trail]

alr thx, so just do it simplier

thx

.close

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[marsh citrus]

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[lofty wren]

How do you find the solution to the inequality

[thorn sentinel]

@lofty wren what have you tried so far

[lofty wren]

Trying to see what numbers are on the inequalitys if thats what its called like the numbers on top

But im lost

Can u tell me like the steps without telling me what to do

with the numbers

[thorn sentinel]

so you can do this in proper way or redundent way

you can try each point on both inequalities and see who fits both

or you can build a graph, draw both inequalities and see where each point stands

makes sense @lofty wren ?

what you are trying to do here is see if which x,y combination from the alternatives satisfies both inequalities at the same time

[lofty wren]

uh

Ohhh

I seee

[thorn sentinel]

from what i said, what is confusing ?

[lofty wren]

Okay so

I think i understand

so with each point i put the numbers in them and see which equations are true like but both of them have to be true

[thorn sentinel]

yes

[lofty wren]

so i put 0 next to 1/2?

[thorn sentinel]

you literally try 5=-1/2 * 0 + 5

yes

[lofty wren]

and then 5 at y

[thorn sentinel]

exactly

[lofty wren]

okay

[thorn sentinel]

this is the redundent way tho

[lofty wren]

Im gonna try

[lofty wren]

thats what we do in mt class

my

[thorn sentinel]

it gives you the solution but it's not very effective if you have a lot of points

that's why the next step is to draw both graphs

[lofty wren]

Oh

[thorn sentinel]

and from there you have a visual representation of all points that belong and don't

[lofty wren]

to the shaded area?

[thorn sentinel]

yes

[lofty wren]

that sounds easier

Alr hold on im gonna pull one up and do that

how do i find out the shaded part

[thorn sentinel]

okay so lets start with a xy graph

[lofty wren]

Yes

i was gonna use this one

[thorn sentinel]

this would be -1/2x+5

m is negative so slope is downhill

and c = 5 so y is 5 for x = 0

[lofty wren]

ok

[thorn sentinel]

okay perfect

now what the inequality says is

[lofty wren]

.

[thorn sentinel]

we like y so that y is equal or lower than the graph

so the line itself is a solution

[lofty wren]

okay

[thorn sentinel]

cause we like all y that are equal to y in the graph

and everything that sits lower than the graph

[lofty wren]

yes

[lofty wren]

So now we find the points

Yea

[thorn sentinel]

this is all the points that satisfy the first inequality

lets start with the second one now

[lofty wren]

yea

Ok

-5,-2

[thorn sentinel]

this is the second graph line

4x - 1

positive slope

and c = -1

[lofty wren]

okay

[thorn sentinel]

now the inequality says, y has to be equal or bigger

so whatever is in the line and bigger value than the line

[lofty wren]

okay

[lofty wren]

Can u put them together now

[thorn sentinel]

yes

[lofty wren]

alright

[thorn sentinel]

wherever we see colors colliding

we know that sector is a solution to both inequalities at the same time

[lofty wren]

So those would be right?

[thorn sentinel]

exactly

[lofty wren]

Alright

[thorn sentinel]

each point you have on alternative you check on the graph

[lofty wren]

on alternative?

[thorn sentinel]

options?

idk what you call them

[lofty wren]

yea

options

Lmao

[thorn sentinel]

alternatives/options/answers

xD

[lofty wren]

yes

xD

so do we start puting the points down now

putting

[thorn sentinel]

sure 1 sec

[lofty wren]

kk

0,5
-5,-2
3,1
8,2
-10,5

[lofty wren]

E and B are my anwsers

?

[thorn sentinel]

A also is a solution

[lofty wren]

Yea

Noticed that

[thorn sentinel]

so A E and B are the solutions to both inequalities

[lofty wren]

Alright

Yea

I understand

[thorn sentinel]

:)) if you have any confusion lmk

[lofty wren]

Thank u

[lofty wren]

what do u use for this btw?

[thorn sentinel]

geogebra

[lofty wren]

okay

cus i have like 2 more

[thorn sentinel]

you can use it but i would advise for your own benefit to try and draw them yourself to become faster at solving function and graphing them

[lofty wren]

yea i just looked at it and it looks confusing

[lofty wren]

I tried the first one i think i did it right but how do i graph a fraction?

.close

[marsh citrus]

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[still temple]

can someone explain how the inclusion exclusion plays into here

[restive sun]

for each prime p that divides 210, there are 210/p numbers that are not coprime with 210

but also for each pair of primes p1, p2 that divide 210, we've overcounted the 210/(p1*p2) numbers that share a factor of both p1 and p2

and similar for all triples of primes, etc.

so you're summing up 210/p for prime p dividing 210, minus the sum of 210/(p1 * p2) for p1 < p2 dividing 210, plus the sum of 210/(p1 * p2 * p3) for p1 < p2 < p3 dividing 210, minus ... (by inclusion-exclusion)

[still temple]

so youre saying it would be phi(210) = |[210] \ {X2 U X3 U X5 U X7}|

which is just [210] - |X2 U X3 U X5 U X7|

but then isnt |X2 U X3 U X5 U X7| = |X2| + |X3| + |X5| + |X7| - |X2 U X3| - |X2 U X5| - |X2 U 7| - |X3 U X5| - |X3 U X7| - ...

[restive sun]

I'm unfamiliar with the notation

I can write out the sum though

[still temple]

okay cool

i would appreciate that

[restive sun]

\begin{align*}
\varphi(210)&=n-\left(\sum_{p\in{2,3,5,7}}\frac{n}{p} - \sum_{p_1<p_2\in{2,3,5,7}}\frac{n}{p_1p_2} + \sum_{p_1<p_2<p_3\in{2,3,5,7}}\frac{n}{p_1p_2p_3} - \sum_{p_1<p_2<p_3<p_4\in{2,3,5,7}}\frac{n}{p_1p_2p_3p_4}\right)\
&=n\left(1-\sum_{p\in{2,3,5,7}}\frac{1}{p} - \sum_{p_1<p_2\in{2,3,5,7}}\frac{1}{p_1p_2} + \sum_{p_1<p_2<p_3\in{2,3,5,7}}\frac{1}{p_1p_2p_3} - \sum_{p_1<p_2<p_3<p_4\in{2,3,5,7}}\frac{1}{p_1p_2p_3p_4}\right)\
\end{align*}

[elfin berry]

Desync

[restive sun]

you could just compute it at this point, but you can simplify it using the product form of the totient function

[still temple]

they did it like this which confuses me

[restive sun]

do you know what the X subscripts notate?

[still temple]

I think its multiples of

[restive sun]

oh

well, that's the same thing as this really

the |X2| + |X3| + |X5| + |X7| is the first sum I gave

the next minus ones are all the pairs

etc.

[still temple]

oh so like because 2 times 3 and then 2 times 5 and then 2 times 7

for the sets that have a multiple of a prime

[restive sun]

yeah

[still temple]

why are you adding 30, 42...

[restive sun]

inclusion exclusion

when we subtracted the pairs

we accidentally subtracted all the triples as well

so we need to add them back in

but now we've double counted the factor of all 4, so we subtract that back off

[still temple]

oh like 2 x 3 times 5

[restive sun]

yes

[still temple]

is there a good video youd recommend for this general concept, i think my understanding is kinda weak

[restive sun]

uhh, I have a worked example of a different problem

not sure of any video though

[still temple]

Like i think the problem is i memorized the general formula but then couldnt apply my understanding

[restive sun]

another one is: how many ways are there to partition [n]={1,2,3,...,n} into exactly k non-empty subsets? (much trickier than the previous 2)

have a go at these problems

you can do them with inclusion-exclusion as well

[still temple]

this is sorta like counting stuff right

[restive sun]

yes

inclusion-exclusion is a combinatorial result

[still temple]

thanks, also off topic but can you use P(N) to show something is uncountable since there exists no bijection between P(N) and anything

[restive sun]

no bijection between P(N) and anything countable yes (by cantor's theorem)