help-13

[whole wigeon]

Final velocity - intial velocity upon time right

Use the formula for vector subtraction

you have theta

the magntiude will be same, 80 kmh

[inland solar]

so 80-80?

[cedar kiln]

@inland solar Has your question been resolved?

[cedar kiln]

@inland solar Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Will this proof suffice to show bezout's coefficients for integers a,b are relatively prime, in both ways (<= and =>)?

[crimson sedge]

I made a typo for s' in (<=), it should be s' = lcm(a,b)/|a| sorry

[cedar kiln]

@crimson sedge Has your question been resolved?

[solar vector]

In the second half you have gcd(x,y)=1 and you apply bezout to claim xs+ys'=1 for some integers s and s'? If so, then how can you assign them values in terms of a,b and the lcm? You don't know which particular integers s and s' are?

I might be misreading you of course.

[crimson sedge]

Then how about saying let's try to solve:

gcd(a,b)s = |a|

And if we get an integer s, then such s must exist

Will this work?

[solar vector]

How do you know s satisfies this equation at all?

[crimson sedge]

Ah actually it will be better to directly show from the identity:

gcd(a,b)(lcm(a,b)/|a|) = |b| and
gcd(a,b)(lcm(a,b)/|b|) = |a|

Then

x gcd(a,b)(lcm(a,b)/|b|) + y gcd(a,b)(lcm(a,b)/|a|)
= gcd(a,b) (x (lcm(a,b)/|b|) + y (lcm(a,b)/|a|))
=gcd(a,b) (x (a/gcd(a,b)) + y(b/gcd(a,b)) ) = gcd(a,b)

[solar vector]

I don't see how this shows it.

I'm actually a little lost on what you are trying to prove in the second half? Is it gcd(x,y)=1 implies gcd(a,b)=ax+by?

[crimson sedge]

Yes

[solar vector]

Is this true if we take x=2, y=3, a=2, b=4?

gcd(x,y)=1. gcd(a,b)=2. But 2(2)+3(4) > 2?

Not sure if I am failing arithmetic today lol.

[crimson sedge]

😭 I read the question wrong again from the book it's only => and not <=>

No no you're fine

[solar vector]

Ah okay so you got it right then and discovered a counterex to the false converse. This is good.

[crimson sedge]

Thanks for the help

.close

[cedar kiln]

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[cedar kiln]

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[hidden mural]

!status

[cedar kiln]

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

[abstract pollen]

got stuck where

[maiden violet]

I substituted the powers of x^2 as x-1, but I'm stuck simplifying x+x^2022

[abstract pollen]

show your working

[hidden mural]

you'd treat 2022 as any other power

[abstract pollen]

this is just a geometric series though

[hidden mural]

for all you can write x^n = (a_n)x + (b_n) for some a_n and b_n

[hidden mural]

true

[maiden violet]

Excuse my writing

[opaque root]

[abstract pollen]

that is a mess

[hidden mural]

,rotate

[maiden violet]

🙏

[opaque root]

Rotate make it worse

[hidden mural]

what

no one else rotate

i got this

oh nvm

are u saying the whole thing is x + x^2022

[maiden violet]

Yeah I'm like 99 percent sure

I did the working

Unless there's somewhere I went wrong that's where I'm stuck at

[pastel grove]

wait i think this is insolvent

[dusk finch]

it's correct

[pastel grove]

if x^2 - x + 1 = 0

[dusk finch]

at least according to my similarly messy calculations

[pastel grove]

then there is no real variable for x

[maiden violet]

We have to choose from one of the multiple choice options

[dusk finch]

but there is complex one

[pastel grove]

imaginary numbers ?

[pastel grove]

ye but if x doesnt exist then how can the solution exist

[dusk finch]

complex numbers.

[pastel grove]

but it must be unreal , no?

[maiden violet]

X can be a complex number but the multiple choice options has X in them, although its value isn't explicitly stated

[dusk finch]

btw, if you want to simplify x^2022, I'd suggest you to first simplify 1, x, x^2, x^3, x^4, x^5, x^6... and youll probably notice a pattern

[pastel grove]

oh

true

thanks

[dusk finch]

yes, it will be unreal. However power of an "unreal" (complex) number can be real, so its possible that it would actually sum up back to a real number

[maiden violet]

I want to kill myself because I've been stuck in the problem for 1 hr but this was in my last year test and somehow I had solved it in my mind back then but now no matter how hard I try now I can't solve it 🙏

[pastel grove]

😭

[dusk finch]

It's not that bad, just simplify those few powers and youll see the pattern

its quite nice pattern

im actually quite impressed that you managed to get x + x^2022 without doing that in prior

[maiden violet]

It was mostly a hit and try method, I have more pages of my clean calculation if you want to see

I tried factoring, substituting, grouping but those didn't work

[dusk finch]

i don't need to, ik it's correct

[abstract pollen]

i swear this comes out as zero

[dusk finch]

1 = 1
x = x
x^2 = x-1
x^3 = -1
x^4 = x^3 * x = -1 * x = -x
x^5 = -1 * x^2 = -x^2
x^6 = 1

now x^7 will be x again

x^8 will be x-1 again

etc

it's periodic basically

[abstract pollen]

the geometric series is $\frac{1-x^{24}}{1-x}$ no?

[wraith dagger]

Xetrov

[dusk finch]

x^2022 = (x^6)^337

[abstract pollen]

oh

[maiden violet]

FU-

[abstract pollen]

i kept writing 24

[maiden violet]

I DIDNT THINK ABOUT THAT

[abstract pollen]

23

not 2023

gotcha

[dusk finch]

in these questions, its always a good idea to just explore it

just write out the small terms

simplify them

[maiden violet]

Bro I didn't even think x^3 = x^2 + x which can be written as x-1 +x

[dusk finch]

x^3 = x^2 * x

[maiden violet]

Idk I was impatient or if I forgot

[dusk finch]

= (x-1) * x = x^2 - x = (x-1) - x = -1

then x^6 = (x^3)^2 = (-1)^2 = 1

and so x^2022 = (x^6)^337 = 1^337 = 1

[maiden violet]

I see now

[dusk finch]

btw there is one even neater solution using this pattern

[maiden violet]

The final answer then is 1+x

[maiden violet]

Sure

[dusk finch]

x^0 = 1
x = x
x^2 = x-1
x^3 = -1
x^4 = x^3 * x = -1 * x = -x
x^5 = -1 * x^2 = -x^2

Note that if we sum all of these up, we get 0

1 + x + x^2 + x^3 + x^4 + x^5 = 0

that's because x^3 = -x^0, x^4 = -x, x^5 = -x^2

that means, that we can literally remove any 6 consecutive terms, without changing the sum

so 1 + x + x^2 + x^3 + ... + x^2023 = 1 + x + x^2 + ... + x^2017
Because we can simply remove the last 6 terms

and we can repeat this

1 + x + x^2 + ... + x^2017 = 1 + x + x^2 + ... + x^2011

=1 + x + x^2 + ... + x^(2011-6)

and we can essentially subtract any multiple of 6 from the final exponent

but 2022 is a multiple of x

so 1 + x + x^2 + x^3 + ... + x^2023 = 1 + x^(2023 - 2022)

= 1 + x

[maiden violet]

I didn't think about that

Are there any reference books that contain questions like these

[dusk finch]

you can only get that if you play with the problem instead of directly trying to solve it

instead of solving it, I just wrote out first few terms and tried to simplify them

[maiden violet]

Because my school syllabus doesn't have them, and id like to practice more problems like this

[dusk finch]

hmm

this looks like olympiad kind of a question

tbh idk many books with such questions

but you can try asking in #book-recommendations

[maiden violet]

True, it's from the Cathedral Maths Competition, for 9-10th graders

[maiden violet]

I asked

But what to do if these questions come up in an exam

Where there's a time limit

It'll just be practise right?

[dusk finch]

yeah, by practising these questions you can definitely get faster

[pastel grove]

tbh the best option in my opinion is to give up and look back for easier questions =))

[dusk finch]

Bruh

[pastel grove]

because hard questions vary, some you can do mostly u cant

[dusk finch]

easy question can boost ego, but if you already know how to solve them, they dont improve your "math skills"

[pastel grove]

i mean whats the point of doing all the hard questiosn if ur not sure if the easy ones are correct are not ;-;

the same situation in my country: 2 extremely hard questions at the back

so most ppl chose to give up and make sure all the easy questions are correct

[dusk finch]

but ultimately, the difference between winners and successful solvers are those 2 extremely hard questions

i think one should at least give them a try

[pastel grove]

true but ... it is life determining factor TOT

cannot risk it all

[dusk finch]

oh are you talking about some entrance exam?

[pastel grove]

every exam has the same form

[dusk finch]

strategy in that might be different than the one in olympiads

[abstract pollen]

my only advice would be: do more questions

[pastel grove]

=}}}}}

https://tenor.com/view/funny-gif-23169978

[maiden violet]

Tysm guys

Y'alls deserve every good thing in life

My question is resolved now

!close

[pastel grove]

dang no one evrey said that to me blush

[maiden violet]

I love this dog meme lol

.close

[cedar kiln]

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[drifting peak]

Use midpoint formula for y coordinate

-12/2 =-6

[twilit escarp]

he simplify badly

[drifting peak]

No

[twilit escarp]

Look what he wrote on step 3

[drifting peak]

Step3 is wrong

Step 4 is correct

[drifting peak]

Yes

[charred magnet]

so step 3 was wrong

[drifting peak]

Yes

[twilit escarp]

Ye

[charred magnet]

thank you

.close

[cedar kiln]

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[zealous solar]

how would i solve this?
find the smallest integer k such that 26 divides 2^378910 - k

[fading jewel]

Plug and chug baby

Playing

[lost gyro]

k is even

[fading jewel]

This looks similar to an olympian problem I've seen

[lost gyro]

mod 13 should be a cycle

2^n cycle (mod 13): 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1

[fading jewel]

I love mod cycles

[lost gyro]

find which 2^378910 is and choose the smallest k that is even to get the divisble by 13

[abstract pollen]

,w prime factor decomposition of 378910

[wraith dagger]

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Requested by xettt

[fading jewel]

W

Solution

💀💀

[zealous solar]

thanks

[fading jewel]

There it is

[abstract pollen]

,w prime factor decomposition of 378909

[wraith dagger]

Results provided by WolframAlpha

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

Requested by xettt

[lost gyro]

this doesn't help 💀

unless im stupid

[fading jewel]

Its in there

Idk I thought it did something

[abstract pollen]

oh seeing if Fermat's little theorem did something

[fading jewel]

I thought he knew what he was doing

💀💀

I know the answer

[lost gyro]

it's a 12 cycle. 12 is def not a factor of 378910

[abstract pollen]

I'm not looking at cycles lol

[fading jewel]

You can find what 2^378910 is congruent to

Which should be the answer

So its a mod problem

[lost gyro]

"p" is not a prime in this case even if you're using fermat's little

[abstract pollen]

i could have tried 13

[lost gyro]

but it's not 2^13...

it's 2^378910

[abstract pollen]

bro that's why i checked

chill

[fading jewel]

💀

[twilit escarp]

Why not using eulers one ?

[lost gyro]

nah ur stupid cause 378910 is clearly even

[fading jewel]

Chill

[abstract pollen]

1. no need for the ad hominem
2. 2^12*(x) gives 1^x

[zealous solar]

so to be clear would i just need to do this 278910 times because i'm not really familiar with what you guys are talking about

[twilit escarp]

We use the fact that phi(26) = 12 so with the euler theorem , 2^378910 = 2^10 = 1024 = 16 [26]

The number is 16

[zealous solar]

ok

thanks

[abstract pollen]

i didn't even think of this tbh

[twilit escarp]

Fermat would work but i dont know how

[abstract pollen]

at this point I'd just do 378910 mod 13 and search through the group

or 12 actually

,w 378910 mod 12

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by xettt

[zealous solar]

can i close the channel?

[abstract pollen]

,w 378910 div 12

[wraith dagger]

Results provided by WolframAlpha

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

Requested by xettt

[abstract pollen]

yeah go ahead

[zealous solar]

.close

[cedar kiln]

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

help

[neat marsh]

<@&286206848099549185> Grade 11 Pre-Calc Radicals finding the area

idk what im doing

[dire jay]

nah you got this

[neat marsh]

I dont

actually

[dire jay]

reminder anything multiplied by itself is squared

so...

[neat marsh]

no its negative

if u look

[dire jay]

uhm

[neat marsh]

why is it negative

what did I do wrog

which step did I mess up

[dire jay]

ok double checking it rn

[neat marsh]

ok

[dire jay]

so anything multiplied

-32

• -32

is?

[neat marsh]

huh

im so confused

can u draw it out

[dire jay]

if you square everything

then -32 times -32

doesnt it make a positive?

[neat marsh]

wait can we start frm A

[dire jay]

A is correct tho?

[neat marsh]

yk how I wrote A

it's correct

so then step 2 is where I messed up

[dire jay]

yea

when you multiplied

-32^2

you wrote -1024

not 1024

it should be positive

[neat marsh]

ohh

[dire jay]

see

you got this

double check it

[neat marsh]

wait why am I doing that

[neat marsh]

why am I doing that

im confused on why I did that

[dire jay]

sorry which part are you confused on?

[neat marsh]

what am I supposed to do

like

yk how 10 rooted 2

is the height

why am I doing 32 and that together

[dire jay]

oh wait yeah

[neat marsh]

why am I doing that

😭

[dire jay]

its fine just square root the answer

so it makes it

[neat marsh]

im so confused

holy hsat

[dire jay]

ok

so your idea

is correct

you should square everything

if you dont want to deal

with square root

but you didnt square everything

[fading jewel]

why ping

[neat marsh]

should I solve overall @dire jay

[neat marsh]

wdym

[dire jay]

yea try solving it now

[neat marsh]

like

should i do 12 rooted 2

times 10 rooted 2

[fading jewel]

i got pinged

[neat marsh]

yes like long long time ago?

[dire jay]

no

[fading jewel]

oh it was the first msg

[dire jay]

you cant solve overall right now

because you wouldnt be able to seperate it

[neat marsh]

no so I could subtract it?

idk

[dire jay]

you have to solve it step by step

you were doing so good

[neat marsh]

idk what im doibg

why would I do that height times that

[dire jay]

10^2 * sqrt/2^2 - 32^2

[neat marsh]

can we call

im like stressing

[fading jewel]

whats happening

[neat marsh]

idk why I'm doing that

[dire jay]

im on my bed and got like no mic

[neat marsh]

ok

[fading jewel]

i got a mic

[neat marsh]

it's ok

[fading jewel]

🗣️

[neat marsh]

ok why am I doing the 2nd step

[neat marsh]

can u call

[fading jewel]

yeah

[neat marsh]

I call u

[dire jay]

hey

[fading jewel]

my fault gang

[dire jay]

you could always

just subtract 10*sqrt2 by 4sqrt2

[fading jewel]

yes

[neat marsh]

idk what I doing

[fading jewel]

thats what i was gonna tell her

[neat marsh]

wait y

[fading jewel]

💀

[fading jewel]

i can explain here or on call

[neat marsh]

can u like

answer

yes

clal

[fading jewel]

wdym

[neat marsh]

😢

call

[fading jewel]

i dont see anything

[neat marsh]

oh

[fading jewel]

$10\sqrt{2} - 4\sqrt{2}$

[wraith dagger]

quevivamexico

[fading jewel]

This is the height for B

[fading jewel]

10x-4x

[drifting peak]

Yes

[fading jewel]

we on call dw

nemesis

[drifting peak]

Yup

Take √2 common

√2(10-4)

[fading jewel]

she got it

dont over complicate 💀

[neat marsh]

wow

yippe

did it

[dire jay]

👍

ty vivo

[neat marsh]

ok

I need help again

it's grade 11 trig

I forgot how to do it

question B

2**

<@&286206848099549185>

can smo pls help me😢

[civic sand]

heeeey

can I help?

[neat marsh]

omg

yes

wow

after like 30 mins

🥰

[civic sand]

I have a magazine to run

but let's solve this problem!!

[neat marsh]

ok

[civic sand]

so, what does solving a triangle mean? is it just finding all the values??

[neat marsh]

huh

[civic sand]

I'm not from an english speaking country

[civic sand]

Just makin sure, what is the objective?

Finding all lenghts and amplitudes?

@neat marsh heeey I'm trying to help!

[neat marsh]

is this right

[civic sand]

you tried solving it?

[neat marsh]

idk y it's not sending

omg my wifi

is so ass

😡

[civic sand]

yes! that's the sine law

[neat marsh]

yes

what am I supposed to do

[civic sand]

the ratio between the sine of an angle and the oposite side's lenght is always the same

so let's start with it

[neat marsh]

do I find a

[civic sand]

you cannot yet

you see, the oposite angle to a is yet not known

what info do we have?

[neat marsh]

is this right

[civic sand]

no.. it's not 60 its sine of 60

which is sqrt3/2

[neat marsh]

huh

[civic sand]

you wrote 60 in the equation

it's sine of 60

[neat marsh]

ok

[civic sand]

you cannot find anything related to a or A yet

[neat marsh]

ok

now what

[civic sand]

what info do you have?

[neat marsh]

thag

that*

on the paper

that's the info I got

[civic sand]

I know, I can see, I'm asking you to write it to see if you can figure out what the next step could be

no-one solves problems by just pluging numbers in formulas

you need to know what you can do and what is not yet available

[neat marsh]

so what to do

[civic sand]

find sine C

can you do it?

[neat marsh]

how

[civic sand]

let me just try latex rl quick

ok so

[neat marsh]

yea

[civic sand]

$\frac{a}{b}=\frac{c}{d}$ $\rightarrow$ $a=\frac{b\times c}{d}$

[wraith dagger]

Max-Cat

[neat marsh]

huh

why is there d

[civic sand]

this is the rule of three simple

[neat marsh]

ohh

yea

[civic sand]

when you have too ratios you can find any value missing using the other three

so, to find SinC we just need to use it

want to try doing it?

[neat marsh]

now what do I do

[civic sand]

no, wait, you're solving for SinC

and remove the SinA/a you can't do anything with it yet

[neat marsh]

huh

ok

[civic sand]

... :/

$Sin(C) = \frac{Sin(60)\times 27.8}{40}$

[wraith dagger]

Max-Cat

[neat marsh]

im so confused

[civic sand]

$Sin(C) = \frac{\frac{\sqrt{3}}{2} \times 27.8}{40}$

[wraith dagger]

Max-Cat

[civic sand]

@neat marsh do you know how the rule of three simple works?

[neat marsh]

im like

very confused now

[civic sand]

can you elaborate?

[neat marsh]

like

it's confusing

[civic sand]

what part?

[cedar kiln]

@neat marsh Has your question been resolved?

[cedar kiln]

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• 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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[split cosmos]

Hey can someone help me understand what the problem is asking? English isnt my first language so i dont understand this question well

[civic sand]

what language do you speak?

[split cosmos]

hindi but i woukld prefer if you explained in english

[civic sand]

okay

[drifting peak]

Send all sub questions

[split cosmos]

thats all

[lone plaza]

It wants you to first create a new model based off of that model I believe

dont just give out answers lol

[drifting peak]

Do you have the answer?

[lone plaza]

since you are working one hour less after 1988

[lone plaza]

I haven't solved it he just wanted it to be explained

[drifting peak]

Sorry about that

[lone plaza]

Its fine just remember lol

[split cosmos]

Would a good model be W/(t-1988)

yes

daily wage / the amount of hours worked each day

wait

im sumb

i figured it out

thank you for help

.close

[cedar kiln]

Channel closed

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[drifting peak]

What's the answer to a?

[drifting peak]

.

[cedar kiln]

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[rapid pond]

For each positive integer $n$, let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$. Find the largest value of $n$ for which $f(n) \le 300$.

i rewrote this problem as the sum of the number of digits of each kn for that k = 1-100, and that sum is <= 400. which means the average number has 4 digits and k > 100. now what?

the solution of this problem says "it doesn't take too long to check," wtf?

[wraith dagger]

Aurora

[rapid pond]

the answer is 109, and i have no idea how im supposed to efficiently check all numbers from 101 to 109 manually

(problem source 2010 AIME I P14)

[modern compass]

log(kn) = log(k) + log(n)
So you get the sum of log(k), which you can calculate and is constant plus 100log(n)

[rapid pond]

there is a floor function

plus, no way im able to calculate by hand without calculator what log(k) and log(n) is

[noble granite]

Same, $\lfloor\log kn\rfloor=\lfloor\log k+\log n\rfloor$

[wraith dagger]

nameless individual

[cedar kiln]

@rapid pond Has your question been resolved?

[rapid pond]

that doesnt help solve the problem

[cedar kiln]

@rapid pond Has your question been resolved?

[cedar kiln]

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[pliant oxide]

https://www.youtube.com/watch?v=0jZlx7GeUKg&t=66s

[buoyant latch]

Do you have a question?

[cedar kiln]

@pliant oxide Has your question been resolved?

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[fossil marten]

hey could someone help walk me through this?

integration by parts i believe

[drifting marlin]

Seems likely

[fossil marten]

i get stuck ill show you my work

first should i make u=(5x^2+7x+3), or u=e^x, and why?

isnt there LIATE?

[drifting marlin]

That is an acronym that exists, sure

You want to choose whichever u leaves you with a simpler integral in the next step

[cedar kiln]

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[spring wadi]

hey dawg

[spring wadi]

i need help

idk how to solve this'

<@&286206848099549185>

[drifting marlin]

!15m

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[spring wadi]

i have less than 15 mins thats the thing

kinda struggling here

[drifting marlin]

Irrelevant

[spring wadi]

wth man

[nimble veldt]

respect the rules!

[spring wadi]

okkkk!

[paper drift]

whats 1 + 1

[spring wadi]

2

[paper drift]

thx

[spring wadi]

ur welcome

[icy shard]

Solve what (I'm blind)

[spring wadi]

this whole thing

its not even mines like

its a friends but

if i solve it i get something out of it

its smth with airplanes

[cedar kiln]

@spring wadi Has your question been resolved?

[cedar kiln]

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[crimson sedge]

i got 62 1_2 % as a fraction

[crimson sedge]

how do i do it

[misty furnace]

What do you mean

I am not understanding what you want to say

[crimson sedge]

@misty furnace

[misty furnace]

Basically

Whatever the percentage is

Divide it by 100

[crimson sedge]

ok

[misty furnace]

Yea

[crimson sedge]

i got 0.62

[lone plaza]

0.625*

dont forget the 1/2

[crimson sedge]

@lone plaza thank you

what do i do now

[lone plaza]

you basically have 0.625/1

you need to make that 0.625 an integer without any decimals

[crimson sedge]

i don't follow

@lone plaza how do i make a integer

just learned how to do

so i remove the percentage

divide 62 by 100

can someone help me

on how to do actually divide percentages

[vast pilot]

they mean without any decimals, so youd end up making it 625 by multiplying 0.625 by 1000, since there are 3 decimal places -> 3 0's (1000)

[crimson sedge]

thank you

[vast pilot]

over here, since you multiplied 0.625 by 1000, that means youd have to multiply the bottom (denominator of the fraction) by 1000 too. so then it'd be 625/1000

you can ask chatgpt for how to do long division or watch a video

get some practice in

[crimson sedge]

thank you

i can't watch lecture videos

i tried

@vast pilot

i'm gonna use chatgpt like you said

[crimson sedge]

how do i make a fraction from this

that is my last question

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

no

[teal token]

alr

wwhats the problem

redbeans

.

lol

@crimson sedge

close this

[crimson sedge]

no

[teal token]

oh

[crimson sedge]

i need help

[teal token]

youre here

ok

so

whats the problem

sned pic

[crimson sedge]

i need help with turning this into a fraction

that is my last question

[teal token]

can u send pic

625/1000

that is fraction

@crimson sedge

ts a fraction already

[crimson sedge]

i mean to a fraction with 2 numbers

[teal token]

ok

[crimson sedge]

like 1_2 or 7_4

thank you

[teal token]

1_2?

[crimson sedge]

it was just a example

[teal token]

oh ok

[crimson sedge]

i can't put 1 over the 2

[teal token]

divide each by 5

you get 125/200

[crimson sedge]

so divide this by 5 each

[teal token]

yes

[crimson sedge]

ok thank you

[teal token]

alwasy multiply or divide with same number

with 125/200 diide by 5 again

you get 25/40

then simplest form is 58

5/8

yeah np

dont forget to close

[crimson sedge]

thank you

@teal token i got one last quesiton

how do i simply 25/40

then i will close

[teal token]

divide by again

do u know hcf and lcm

@crimson sedge

[crimson sedge]

no

[teal token]

oh..

hcf basically means

highest common divisor

so like

until u cant further divide it

in 25/40

if we divide both sides by 5

we get 5/8

but now

in 5/8 we cannot divide it fursther into a simpler fraction because we get decimals

so highest commin divisor means like the largest number we can divide on both numbers to get simplest form

[crimson sedge]

thank you

i will close

@cedar kiln

close

[teal token]

@crimson sedge

do .close

@crimson sedge

[crimson sedge]

.close

[cedar kiln]

Channel closed

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[teal token]

yeah

[cedar kiln]

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[static grail]

Factor the denominator and resolve into partial fractions?

[lyric narwhal]

Yes

[static grail]

is there gonna be Bt + C

the numerator of the second partial fraction

?

[uncut ridge]

t(t^2+4) so the other partial fraction would be At+B, yes

[static grail]

OK, but im getting the values A=1

and B=-1, C=0

[uncut ridge]

yeah, should be correct. i haven't solved

[raven shard]

the terms are in the correct ratio but…

[static grail]

woops

i wrote down the wrong question

theres a four in the numerator

do i still have to apply partial fractions

[uncut ridge]

its okay just multiply your final answer by 4

[static grail]

here are the fractions

but they dont equal the original fraction

[uncut ridge]

did you solve this or is it the solution?

seems incorrect

[static grail]

this is what i got

when i solved the partial fracts

A=1, B=-1, C=0

[carmine dock]

Is B the coefficient of t?

[static grail]

yea

its Bt + C

,w partial fractions of 1/(t^3 +4t)

[wraith dagger]

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[carmine dock]

So the numerator should be -t

[uncut ridge]

yeah you made a mistake calculating the partial fractions

[vestal quarry]

Plz

[static grail]

,w partial fractions of 4/(t^3 +4t)

[wraith dagger]

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[static grail]

@uncut ridge

look theyre correct

[uncut ridge]

you didn't have t in the numerator of the second partial

[static grail]

wait

oh

so its gonna be -t

ok ok

mb

[opaque root]

No

[static grail]

now ima take the integral

[opaque root]

1/4 and -t/4

[static grail]

any idea on how to integrate this

[uncut ridge]

u = t^2 + 4

[static grail]

2t

oh ok

[uncut ridge]

yes so take half of it

[static grail]

what if it was

1/ t^2 +4

arctan???

[lyric narwhal]

Yeah

[uncut ridge]

yeah (1/2)arctan(t/2)

[static grail]

correct integral?

@uncut ridge

[opaque root]

||

[static grail]

?

[uncut ridge]

ln|t^2+4|, as samuel said and it's correct

[static grail]

1/2?

theres a 1/2 cuz of the u sub

[opaque root]

Need absolute value in logx

[static grail]

the derivative is 2t

[static grail]

i put the 1/2 out of the log

oh ok

[uncut ridge]

yeah yeah, -1/2

[opaque root]

Ln(|t|)

[static grail]

i dont know how to type this lmao

looks like i dont have a key for it

[opaque root]

Also

I am not sure it is correct

[static grail]

how

[opaque root]

Oh it is t^3

[static grail]

yup

i got thecorrect ans

thanks

.c;pse

.close

[cedar kiln]

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[opaque root]

Yes only absolute value over ln(|x|)

[static grail]

yup

[opaque root]

In the second one is not needed

[static grail]

the integral is definite

and its from 1 to 2

[opaque root]

That also sorry

[static grail]

nw

.reopen

[cedar kiln]

✅

[static grail]

@opaque root

yo wait

what about the constant of integration

i didnt put any value for it and still got the correct answer

@lyric narwhal

.close

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[jagged viper]

hi i wrote a proof and im not sure if it's correct (im not even sure if the conclusion is correct) and i was just wondering if it looks good

Theorem: Given some injective linear transformation T: V -> W. If the list v1, ..., vn is linearly independent in V then the list Tv1, ..., Tvn is linearly independent in W

Proof: Every v in V can be written as v = c1v1 + ... + cnvn. then, by the linearity of T,
T(c1v1 + ... + cnvn) = c1 Tv1 + ... + cn Tvn
Suppose that Tv = 0. because null T = {0}, we have c1 Tv1 + ... + cn Tvn = 0 if and only if c1v1 + ... + cnvn = 0, implying that c1 = ... = cn = 0 (because v1, ..., vn is linearly independent). therefore, Tv1, ..., Tvn is linearly independent in W

[jagged viper]

oh wait shoot not every v in V can be written as a linear combination of that list

does v1, ..., vn have to be a basis instead of just a linearly independent list for this to hold?

i guess as long as T is finite dimensional v1, ..., vn could be extended to a basis v1, ..., vn, w1, ..., wm and then we can show that Tv1, ..., Tvn, Tw1, ..., Twn is linearly independent. then of course Tv1, ..., Tvn is a sublist of that so it's also linearly independent

agh now im just confusing myself 😭

okay i think i fitgured it out

.close

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[static grail]

Does anyone have Wolfram Alpha pro??

[static grail]

I want to check the full solution to this integral

,w integrate (5t^2 -3t +18)/(t(9-t^2))

[wraith dagger]

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[static grail]

,w definite integral of (5t^2 -3t +18)/(t(9-t^2)) from t=1 to t=2

[wraith dagger]

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[static grail]

this one

<@&286206848099549185>

[static whale]

you need to use partial fractions

[static grail]

Ik

But the answer in my book is 4.35

And here it's different

[static grail]

.

[static whale]

are you certain that you put the exact same thing into WA as the book

[static grail]

100% yes

[static whale]

then the book is wrong i suppose, not uncommon

[static grail]

Not uncommon!

Wdym

[static whale]

often the answers in textbooks are wrong

[static grail]

@static whale

some guy on youtube did it like this

the final asnwer is same as wolfram alpha

.close

[cedar kiln]

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[static grail]

,w integrate (e^2 - e^x)/2

[wraith dagger]

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[static grail]

,w integrate lnx

[wraith dagger]

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[cedar kiln]

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[potent flower]

The number of ways to form a number that is divisible by 3 consisting of 5 different digits of different numbers from the numbers 0, 1, 2, 3, 4, 5 is equal to...

[potent flower]

how do I even start with this

[eager sparrow]

use the divisibility of 3 rules

[potent flower]

the hard bit is with numbers like 54321 that are divisible by 3 somehow

[potent flower]

what are they

[eager sparrow]

for example a random number

1239234

I just typed that randomly

to get if it is divisible by 3 you add each digit together

1+2+3+9+2+3+4

then check if the sum is divisible by 3

you can do this repeatedly if you are still not sure until you get to 1 digit only

the sum here is 24

I know that 24 is divisible by 3

so the random number i just typed is divisible by 3

if you are still not sure just add 2 and 4 together

you get 6

and 6 is divisible by 3

so the original number is still divisible by 3

[potent flower]

aha

does this apply to any number or just 3

[eager sparrow]

for multiples of 3 and 9

[potent flower]

thnx

[eager sparrow]

I believe it applies to any powers of 3

but I'm not entirely sure

[potent flower]

.close

[cedar kiln]

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[placid jacinth]

So what's your answer

[potent flower]

.reopen

[cedar kiln]

✅

[potent flower]

I've just realised that I still can't do it

@eager sparrow how to go further with this

[placid jacinth]

Since you have to test every combination you might need to use factorial

[potent flower]

yes

the problem is how to put the divisibitly by 3 in the calculation

[placid jacinth]

I mean the sequence 1,2,3,4,5 is already divisible by 3 isn't it?

So no matter where the digits are placed it will always be divisible by 3

[potent flower]

ah right

oh wait let me try this

[potent flower]

thnx

I solved it

.close

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[dusk finch]

.Just to clarify this, it doesn't. It breaks at 27 already. Try e.g. 9891
@potent flower @eager sparrow .
It works only for 3 and 9, mainly because 10 is equivalent to 1 modulo these 2.

[eager sparrow]

Aaah

[eager sparrow]

ok ok tnx bro

.close

[cedar kiln]

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[placid jacinth]

Can i quote you on this?

[dusk finch]

Why? lol

[noble granite]

for what?

[dusk finch]

maybe I forgot 1

but that one is quite trivial

[placid jacinth]

Because i might have said it before in some channel and i just want to show how i discovered this

[dusk finch]

I dont exactlyy get what you mean, but sure. Do it

[placid jacinth]

Thanks

[cedar kiln]

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[timid cloak]

I'm having a hard time figuring how to approach this question. I got so far as to figure out that combined they polish 44/3 gems / minute. Then divide that by 8 and times it by 5 to get robot A's rate. But my answer is incorrect.

[livid hound]

where's 44/3 coming from

[timid cloak]

uhm 88/6

44/3 gems / minute. I made a mistake in my initial message

[livid hound]

the way you're multiplying/dividing afterwards gives B's rate, not A

[timid cloak]

How come?

a = 3/5b -> 5a = 3b

Ah, wait ratios are different from the the algebraic expressions

if a = 3 then b = 5

[livid hound]

yes

[timid cloak]

Ah yikes its good that I know this now, lol

[timid cloak]

Thank you

easy msitake to make if you're not paying attention. Thank you

[timid cloak]

How did you figure out so quickly that it was the other way round? It was not immediately apparent to me.

[livid hound]

from my knowledge of ratios

and considering the base case of what happens when there are 8 parts

[timid cloak]

if a = 3/5b means if there are 8 parts a would be 3 of them?

Initially, I assumed it would be 5 parts and the breakdown would be 2:3

.close

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[neat raven]

if tan11 deg = a
express tan1 deg in terms of a.

[neat raven]

@gleaming path

[lyric narwhal]

!original

[cedar kiln]

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.

[haughty osprey]

Try to write tan(1) in the form tan (a+b) with 11 as one of the variables, then apply the formula

Also reduction cannot be used here

[neat raven]

got it. first calculated tan(11 * 2), then (22 * 2). Then did tan(45-22*2)
Thank.ss

.close

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[mild narwhal]

How come im wrong in taking x = 5sectheta

[lyric narwhal]

You are not wrong

Oh wait

You are

[lyric narwhal]

How did $4x^2-25$ turn to $4(25\sec^2\theta-25)$?

That’s not right

[wraith dagger]

kheerii

[mild narwhal]

yes i assumed that i could leave the co-efficient

[lyric narwhal]

What

?

The 25 magically turned into a 100

The last 25

[mild narwhal]

well i multiplied the 4 into 25sec^2

my brackets there is a mistake

[lyric narwhal]

$4(25\sec^2\theta)-25\ne 4(25\sec^2\theta-25)$

[mild narwhal]

well this whole thing is wrong in general

[wraith dagger]

kheerii

[mild narwhal]

yes ik that

my brackets placing was incorrect

but its wrong anyways

[lyric narwhal]

Yes but you’re close

[mild narwhal]

$4(25\sec^2\theta)-25

its supposed to be this

[lyric narwhal]

Instead use 2x=sec theta

[lyric narwhal]

It is

[mild narwhal]

yes

can you explain to me why i have to use 2x

thats the part that im not understanding

is it because the root of the co-efficient is 2

@lyric narwhal

so im practically rooting what ever is (ax)

and taking that as my substitution correct?

[lyric narwhal]

You can think of it like that yeah

[mild narwhal]

i see thanks

my vision is clear

.close

[cedar kiln]

Channel closed

Closed by @mild narwhal

Use .reopen if this was a mistake.

[cedar kiln]

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[fathom zealot]

<@&286206848099549185> may I know how to do this question?

[dark birch]

What do you know about range and domain ?

[fathom zealot]

range is the value i can get using the domains?

[dark birch]

exactly

So find the largest and smallest possible value you can get

Those are the limits of the range

[fathom zealot]

I see but I have already used that but the range is not correct

like I got 2 and 2.0001

but the answer is 2 and 4

[dark birch]

That sounds like you are using degrees instead of radians.

Ancient meme

[fathom zealot]

oh yeah

true

Thanks tho 🙂

[dark birch]

I guess you just plugged in the limits of the domain given?

[fathom zealot]

yes

now i got the correct answer

[dark birch]

Do note that that does not always work, since cos could be decreasing or increasing depending on what the limits are

[fathom zealot]

I see

[dark birch]

So for example the max value (upper limit of range) could be in the middle of the domain

But here it works out 🙂

[fathom zealot]

welp, I didnt know that

[fathom zealot]

yes 🙂

.close

[cedar kiln]

Channel closed

Closed by @fathom zealot

Use .reopen if this was a mistake.

[dark birch]

If it would be defined within this domain for example, you see that the max value and the min value are in the middle of the domain

[fathom zealot]

Oh I see

thanks 🙂

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[crimson sedge]

I need to show that for odd integer a,

a² + (a+2)² + (a+4)² + 1 is divisible by 12.

What I thought of was splitting into cases for a = 3k_1, 3k_2 + 1 and 3k_3 + 2 for odd k_1, even k_2 and odd k_3 respectively; and it worked

But is there a shorter way to show this? I tried messing around with triangular numbers but they don't seem to help much

[brisk nova]

use induction

[crimson sedge]

Oh damn yeah

[hot crag]

uh

you don't really need to idt

it's rather simple

@crimson sedge what is the expansion of the expression

[crimson sedge]

Doing it 2 min

[hot crag]

use WA

it's like 3 steps lol

[lyric narwhal]

,w expand a^2+(a+2)^2+(a+4)^2+1