help-27

[woven radish]

parthisjoking

[stone stump]

yes

[dapper fable]

yes

[stone stump]

solving DEs is often much harder than proving statements like this about the function

[pure cedar]

ohhh

i get it

then i find range of

rhs

[restive river]

if they give non linear DE then you should know you are not supposed to solve it 🥀

[pure cedar]

right 🥀

[dapper fable]

just invent new math to solve it

[pure cedar]

im also getting tan-1x

[dapper fable]

yup

bound arctan x for x >= 1

[pure cedar]

ohhh

thats why its + pi/4 and not -pi/4

due to arctan

that was such a smart solution

how do yall come up with these

[dapper fable]

i forgot how to solve des

[pure cedar]

gg

[dapper fable]

i mean when they give you an inequality most likely you arent gonna solve for the function

[stone stump]

this identity is obvious

and you are asked to show f(x) <= something

so the obvious step is to try to show that int (...) <= int(...)

[dapper fable]

most of the time you work around it via laziness

[stone stump]

the integral has f(x) inside it

thats obviously shit

you want to remove that

you can easily see f(x) >= 1

so plug that in

[pure cedar]

right

[stone stump]

and from there is basically autopilot and you are done

[pure cedar]

thank you guys so much

if i clear my exam coz of this server

ill be so happy

.close

[devout snow]

Channel closed

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Use .reopen if this was a mistake.

[dapper fable]

👀

[restive river]

give $1 to everyone in this server

[pure cedar]

in this economy?

but alr ill give nitro or something

[devout snow]

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[pure cedar]

i solved a and c , need help with b

[dapper fable]

prime factorization

you can just find a contradiction actually

[pure cedar]

will that work in a proof based exam

[dapper fable]

yea

[pure cedar]

i was thinking more of a general solution

[dapper fable]

if a statement says "for all ...."
a counterexample would be " ∃ something for which this is not true"

[pure cedar]

sure ig

that will work

thanks

[dapper fable]

if you want that then

write a = product of primes, b = ..l

look what happens to the exponents

you can also simplify a*b to look nicer without the lcm

[pure cedar]

oh right

i can take a , b and c as a general product of primes

and basically just plug values

and ill get that lhs =! rhs

gg

.close

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[devout snow]

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[sinful badge]

Did I do this wrong

[sinful badge]

I'm not confident 💔

[green crypt]

should be y/2

[sinful badge]

That's my x

You me like 2 should be with my y?

Instead of a half

[green crypt]

here you divided the entire question by 2 yes?

[sinful badge]

Yeah isn't 2/2 = 1 and then the y has a 1 and 1/2 = a half

[green crypt]

yes so half of y

you wrote 1/2y instead of y/2

[sinful badge]

I guess I dont understand why it would be written like that

[green crypt]

okay so

when x is multiplied by a half

it should be half of x right?

[sinful badge]

Yeah

[green crypt]

wait did you write y/2 as 1/2 times y?

[sinful badge]

1/2 I mean as 0.5

[green crypt]

💀

okay nvm

i thought you put y in the denominator

[sinful badge]

It's okay, thats why I was lost

[green crypt]

how did you get a 4 here?

[sinful badge]

That's my problem there

Oops

Should have been 3☠️

If I change that to a 3 and everything else I had the 4 and recalculate will all be good

Or did I mess up any other steps

[green crypt]

nah other than that, you're good

[sinful badge]

Haha okay thank you so much

.close

[devout snow]

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[devout snow]

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[pure cedar]

couldnt understand the question even

[candid maple]

do you not understand the whole thing or just certain parts of the question?
if you can understand some part of the question, could you explain what you did and didn't understand?

[stone stump]

you have the biased die and are throwing it twice. how often will you see the number 2

[pure cedar]

what do they mean by mean here

[candid maple]

mean = average.

[restive river]

the question says you have a biased die with P(2) = 3/10 meaning probability the die rolls a 2 is 3/10 , if you had an ubiased with all outcomes equally likely the probability for any number would be the same 1/6 however the condition is different here

[pure cedar]

i get that

[restive river]

so where are you stuck

[pure cedar]

how will one calculate mean of only 1 observation

[candid maple]

but there are two!

the question said the die is thrown twice.

[pure cedar]

right

and its asking to calculate mean of number of times

2 is obtained

in each of the throws

[candid maple]

across both throws, yes.

[pure cedar]

but that can only be 1

in a single throw

2 can only be seen once

[candid maple]

but you have two throws, not one!

the mean concerns both throws taken together, not separately.

[stone stump]

have you never seen a question such as "you throw a normal dice 30 times. how often will you expect to see a 6" ?

[pure cedar]

or is it asking to calculate mean of the probability of seeing 2

[pure cedar]

i have , but what does mean have to do with this

[stone stump]

its the same question

just a different word

"you throw the biased die twice. how often do you expect to see a 2"

if you think its easier, first answer "you throw the biased die 30 times. how often do you expect to see a 2"

[pure cedar]

is that it😭

so i make cases here

both 2s

and a single 2

[stone stump]

how do you answer this

pls dont say cases

[pure cedar]

umm

total cases are 6^30

[stone stump]

overthinking

[restive river]

have you heard of binomial distribution or shit like that

[stone stump]

this is not a trick question

[pure cedar]

not really

[pure cedar]

1/6?

[stone stump]

thats the probability to get a 6 in one throw

so how many 6s do you think you will probably see if you throw it 30 times

[pure cedar]

5

[stone stump]

good

so (number of throws)*(probability to get a 6 in one throw)

[pure cedar]

oh

[stone stump]

can you answer this?

[pure cedar]

9 times

[stone stump]

good

and can you now answer the original question?

[pure cedar]

is it 6/10

[stone stump]

yes

[restive river]

voila

[pure cedar]

so mean just means

hpw often you see something

[stone stump]

mean=average=expected number=...

you arent guaranteed to actually see 5 sixes when you throw your dice 30 times

you could get unlucky and only see 3

or 4

or 7

[pure cedar]

😭😭

so how is probability of getting 4 or 7 calculated

[stone stump]

thats more complicated

not sure if thats relevant for you yet

[stone stump]

(it uses this)

[pure cedar]

actually it was in our book but teacher never taught us

i might have to do it myself

thanks anyway

.close

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[devout snow]

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[solar goblet]

any hint for b?

this is my work so far to a

[dapper fable]

if X>= 20, then after 19 games no one has 10 wins, so you can bound the amount of both their wins

[devout snow]

@solar goblet Has your question been resolved?

[solar goblet]

what do you mean?

[devout snow]

@solar goblet Has your question been resolved?

[devout snow]

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[tawdry meteor]

Can I just…..divide this by 2?????(bottom line)

[dull parrot]

you can but it doesn't help

[tawdry meteor]

It won’t cancel the squares?

[dull parrot]

no it wouldn't cancel

[tawdry meteor]

Yeahhhh

Monkey brain saw 3 2’s and wondered if they could cancel

[dull parrot]

$\frac{x^2}{2} \neq x$

[woven radish]

MxRgD

[tawdry meteor]

Yeah I thought so

[dull parrot]

as an example

[dull parrot]

you can use the trig identity that $1 + \tan^2(x) = \sec^2(x)$ however

[woven radish]

MxRgD

[tawdry meteor]

That’s what I was gonna suggest next

[trail eagle]

What happened in the line before that?

[tawdry meteor]

If I use it on both I just…essentially swap places on tan and secant

[tawdry meteor]

Multiplied the denominator of the fraction on the bottom by the top to cancel the fraction on the bottom

Which leaves a 1+1 and that equals 2

Should I only use the pythag on one of them?

[trail eagle]

If you multiply the denominator by something you gotta multiply the numerator by that thing too though

Am I missing something?

[worldly coyote]

the easiest way would be to convert the whole LHS in terms of sin and cos and simplifying it, youll get sin/cos^2

which can be translated to sin/cos * 1/cos
and then ultimately, tan*sec

[trail eagle]

Oh wait are you claiming that $(\sec(\theta) + \tan(\theta))^2 = \sec^2(\theta) + \tan^2(\theta)?$

[woven radish]

Azyrashacorki

[tawdry meteor]

Ok one sec on this I’m in lunch

Away from my work

[trail eagle]

In any case it would probably be more straightforward to put $\frac{1}{\tan\theta} + \frac{1}{\sec\theta}$ on the same denominator first. What you get from multiplying the whole thing by $\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}$ seems like a mess.

[woven radish]

Azyrashacorki

[trail eagle]

I would maybe understand multiplying by the conjugate of the numerator, but I'm pretty sure what you would get out of that would be more complicated than setting a common denominator for 1/tan(t) + 1/sec(t) in the first place

[restive river]

what i see is two mistakes
1/a + 1/b = (1+1)/ab
and as mentioned (a+b)(a+b) = a^2 +b^2

,, \frac 1a + \frac 1b = \frac{a+b}{ab}

[woven radish]

hail

[restive river]

then you could multiply sectheta.tantheta in numerator but more easier way is convert all to sin and cos then simplify

[devout snow]

@tawdry meteor Has your question been resolved?

[tawdry meteor]

Hold on

Back it up

[tawdry meteor]

Just use x it’s the same and it’s easier to type

[restive river]

di you haveyour lunch

[tawdry meteor]

I did

Was pretty good

[restive river]

wonderful

so @trail eagle

[tawdry meteor]

What part of the problem were you all discussing? The second line(s)?

[restive river]

let's see if they are there

[restive river]

there were multiple people including me suggesting different things

[tawdry meteor]

Interesting

I’ve been told this problem was an issue for others too

[restive river]

but yeah overall second line is an issue

[tawdry meteor]

So…run it back

[restive river]

this one

[tawdry meteor]

Right..?

That’s what I have on there

[restive river]

and then consider that

[tawdry meteor]

1+1/secxtanx

[restive river]

nuh uh

[tawdry meteor]

But that is a+b/ab is it not?

[restive river]

do you mean $\frac{1+1}{secxtanx}$?

[woven radish]

hail

[tawdry meteor]

Yes

Wait- I see my problem

[restive river]

again $\frac 1a + \frac 1b \neq \frac{1+1}{ab}$

[woven radish]

hail

[tawdry meteor]

It went from 1/tanx +
1/secx to 1+1/tanx**•**secx

Yeah I see that now

Common denominator is…secxtanx though yes?

So that gets me to…… secx + tanx /secxtanx

And then just….cancel the top and bottom

[restive river]

yh, when adding fractions you try to make the denominator common by taking lcm ,
for example if you have $\frac 1a + \frac 1b$ and you want to make denominator ab :
for first term , you multiply in denominator and numerator by b $\frac{b}{a\cdot b}$
for second term , you multiply in denominator and numerator by a $\frac{a}{a\cdot b}$
then add them

[woven radish]

hail

[tawdry meteor]

Then it’s tanxsecx

[restive river]

yh

[restive river]

indeed

[tawdry meteor]

Ok perfect

[restive river]

you have proved

[tawdry meteor]

Can I have you on standby while I work on the next one?

[restive river]

forgive me for slow typing tho xD

[restive river]

mm.. yh

[tawdry meteor]

No worries

If I have smth like…. X/X^2 and cancel I’m still left with X from the bottom yes?

[restive river]

yes

[tawdry meteor]

Now to figure out how to cancel…

Here’s where I’m at- I need to cancel a secant- to get….i guess sec/1?

Is making both secants into cosines a good idea here?

[summer summit]

you cant cancel that in the second line

you can cancel once you factor it, since its in the form a^2-b^2

[tawdry meteor]

Wdym?

It’s not fully cancelled

It’s 1-cosx/1-cos**^2**x

[summer summit]

what did you leave on the bottom?

[tawdry meteor]

Which leaves 1- coax

Cosx**

[summer summit]

no

[tawdry meteor]

I forgot the 1 on paper

OH

[summer summit]

you can't cancel those since you have that subtraction

you must factor that into (1+cos)(1-cos)

[tawdry meteor]

I’m not following

[summer summit]

remember the formula, $a^2-b^2 = (a+b)(a-b)$

[woven radish]

Krish

[tawdry meteor]

But they aren’t both squares

[summer summit]

1 is the same as 1^2

[tawdry meteor]

Ohhhhhh I’m thinking of the wrong thing

Ok ok

[summer summit]

so you have $\frac{1-\cos\theta}{1-\cos^2\theta}$

[woven radish]

Krish

[tawdry meteor]

So factor them into (1 + cos) and (1-cos)

[restive river]

you can cancel like that if you have (1-cosx)^2 in denominator whuch is not the case

[summer summit]

that can factor into $\frac{1-\cos\theta}{(1+\cos\theta)(1-\cos\theta)}$

[woven radish]

Krish

[summer summit]

and you can cancel the term from there, being left with 1/(1+cos)

[tawdry meteor]

Gotcha

Let me write that out

[summer summit]

go for it

[tawdry meteor]

Right so that leaves me with secx/ 1+ secx = 1/ 1+cosx

[summer summit]

yes

[tawdry meteor]

Is it a smart idea to convert both secants to cosine- or is it possible on the bottom since there’s a 1 attached

[summer summit]

i didnt read above so im not sure- are you just supposed to verify if they are equal?

[tawdry meteor]

Yes

Trig proofs

[summer summit]

then yeah convert into cosines

[tawdry meteor]

I’m looking for advice- not a full walk through……….this time

Can I on the bottom? Since it’s 1+ secx

[summer summit]

you would have 1 + 1/cosx

remember to keep the 1 there, it doesnt go away

[tawdry meteor]

Yea I know

[summer summit]

and from there you can simplify using common denominator

[tawdry meteor]

I just wasn’t sure if it was one unit or not

[summer summit]

try doing that on your own and see what you get

[tawdry meteor]

They just cancel-

I ended up with cosx/cosx

[summer summit]

can you show what you did?

[tawdry meteor]

Yes let me write it down

[summer summit]

,rotate

[summer summit]

this is what i was talking about, you cant get rid of that 1

you cant multiply by that reciprocal since its not one term in the denominator

[tawdry meteor]

Right- what do I do with the 1? Is my problem

[summer summit]

use a common denominator

[tawdry meteor]

Elaborate

[summer summit]

if you use a common denominator, you can turn that bottom into all one fraction, which then you can multiply by the reciprocal

[tawdry meteor]

How do I do that

Because as of now….it’s all over cosx?

[summer summit]

first of all, what denominator can you use?

lets just look at the bottom part of the fraction for now, $1 + \frac{1}{\cos\theta}}$

[tawdry meteor]

Wait where do that other 1 come from

[woven radish]

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

[summer summit]

there we go

[tawdry meteor]

Oh ok

I see

[summer summit]

now we have a cosine in the denominator of one of the terms, yes?

[tawdry meteor]

Could cosx be used? Multiply the 1 by a value of 1 (the value being cosx/cosx)

[summer summit]

yes!!

[tawdry meteor]

Ok ok

[summer summit]

so now we have $\frac{\cos\theta}{\cos\theta} + \frac{1}{\cos\theta}$

[woven radish]

Krish

[tawdry meteor]

Hold onc

Writing

[summer summit]

sure

[tawdry meteor]

Are the 1 on top and the cosx the same value? Can they be added?

Or keep them separate

[summer summit]

the same value, no but can they be added, yes

[tawdry meteor]

Ok

[summer summit]

you can just combine it into one fraction

[tawdry meteor]

That’s what I was asking lol

[summer summit]

so what is that combined fraction looking like?

[tawdry meteor]

2/cosx

[summer summit]

no, theyre not the same value like i said

$\cos\theta \neq 1$

[woven radish]

Krish

[tawdry meteor]

Ohhhh I read that wrong

1 + cosx/ cosx

[summer summit]

yes

so now lets replace the bottom of the fraction with (1+cosx)/cosx

[tawdry meteor]

Do I need the brackets?

[summer summit]

just replacing, doing nothing else we have $\frac{\frac{1}{\cos\theta}}{\frac{1+\cos\theta}{\cos\theta}}$

[summer summit]

once you write it out no but since we're writing it in text yes

[woven radish]

Krish

[tawdry meteor]

Yes

[summer summit]

now we can multiply by the reciprocal of the entire denominator as a whole

[tawdry meteor]

So 1+cosx/cos • cos/1

[summer summit]

$\frac{1}{\cos\theta} \cdot \frac{\cos\theta}{1 + \cos\theta}$

[tawdry meteor]

Oh I did it backwards

[summer summit]

close, but other way around, the numerator stays the same and the denominator is the one that flips

remember the phrase "keep change flip"

[tawdry meteor]

Yeah

[summer summit]

1. keep the numerator the same
2. change the division into multiplication
3. flip the denominator

[tawdry meteor]

Am I cross multiplying or multiplying in a straight line

[summer summit]

oh wait i wrote this wrong hold on

that second fraction should be flipped

[woven radish]

Krish

[tawdry meteor]

Yeh

The cosx’s cancel

[summer summit]

yes they cancel

and what are you left with?

[tawdry meteor]

And you’re left with 1/1+cosx

[summer summit]

good

[tawdry meteor]

And then multiply by a value of 1-cos/1-cos

[summer summit]

yes

however if you already converted the other one into 1/(1+cosx) i dont think you have to do anything else

[tawdry meteor]

Then it’s 1-cosx/1-cosx = 1-cosx/1-cosx

[summer summit]

what?

[tawdry meteor]

Oh wait

I was looking at the wrong thing

2 more 🙏

[summer summit]

nice almost done

[tawdry meteor]

But I have no more paper…

And that page is FULL

[summer summit]

oh- thats unfortunate

[tawdry meteor]

When I turn this page in I’ll go get more

When my current teacher finishes-

[devout snow]

@tawdry meteor Has your question been resolved?

[devout snow]

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[trail harness]

any math huzz in here

[devout snow]

Please don't occupy multiple help channels.

[worn salmon]

https://giphy.com/gifs/huzz-guzz-the-jtPO6jt0Ckn6qsphgx

[worn salmon]

Also @trail harness please don't occupy multiple helps

[restive river]

!notroll

shit no factoid

for trollers

[uncut crow]

thank goodness

[restive river]

:kek they would have killed that as well

[unique monolith]

<@&268886789983436800> troll, occupying channel

[split stream]

.close

[devout snow]

Channel closed

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[faint zinc]

No trolling help channels please

[green crypt]

based

[devout snow]

Available help channel!

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[mossy berry]

These derivatives?

[ionic jasper]

on not derivates i mean other write

[worn salmon]

Do you mean derivations?

[ionic jasper]

mb

mb

mb

[devout snow]

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[ionic jasper]

wait

[mossy berry]

Oh. Just rewriting the exponents?

[ionic jasper]

ye

are these correct

[restive river]

!msgdel

[devout snow]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[worn salmon]

They are not correct

[ionic jasper]

all?

[worn salmon]

the first one is

[mossy berry]

Only the first one is.

[narrow nexus]

the first one is correct

[mossy berry]

Try that

[restive river]

Correct is one

[narrow nexus]

for the others you are making the same mistake. Do you know what $\frac{2}{5} \cdot \frac{1}{x}$ is equal to?

[woven radish]

gribble19

[worn salmon]

$\frac{a}{bx} = a (bx)^{-1}$

[woven radish]

WeAreIngram

[mossy berry]

^

[lethal wraith]

remember the parentheses

[ionic jasper]

2/5x ?

[narrow nexus]

yes, exactly

now keep this in mind, and try to figure out what went wrong

if you don't see it, try rewriting the answers you got back to a fraction

[worn salmon]

$\frac{2}{5}x^{-1}$

[woven radish]

WeAreIngram

[narrow nexus]

for example, write $10x^{-1}$ as a fraction. What would you get? Is this the same as what you started with?

[woven radish]

gribble19

[ionic jasper]

i need an sec there is much information here rn

oh

[narrow nexus]

yes of course, there is no rush. Take your time to look over and think about things, and if things are still unclear just ask for another nudge in the right direction!

[worn salmon]

This is the key idea

[ionic jasper]

i get now what u were doing here

[ionic jasper]

but i dont get that

[ionic jasper]

i get that

[narrow nexus]

this is a generalization of the idea that my example was also trying to tell you

[worn salmon]

$\frac{a}{bx} = a(bx)^{-1} = a (b)^{-1} x^{-1} = \frac{a}{b} x^{-1}$

[woven radish]

WeAreIngram

[narrow nexus]

if you replace the b with 2, and the a with 5. What do you get?

[ionic jasper]

taht is correct right?

[narrow nexus]

yes exactly!

[ionic jasper]

is is that right too?

does an ^-3 go to positiv?

[devout snow]

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[ionic jasper]

unclose

[ionic jasper]

w

[narrow nexus]

yes exactly! You could also write this as $\frac{1}{x^{-3}}=(x^{-3})^{-1}$

[ionic jasper]

perfect thanks

[woven radish]

gribble19

[ionic jasper]

oh

[narrow nexus]

and as you might know you can then multiply the -1 and the -3

which would be 3

and thus its the same as x^3

[ionic jasper]

makes sense

[restive river]

,, \frac{1}{x^{-a}} = x^{-(-a)}= x^a

[woven radish]

hail

[ionic jasper]

are these middlesteps correct?

[restive river]

.reopen is the command

[ionic jasper]

aight

[worn salmon]

Yes

[narrow nexus]

perfect!

[ionic jasper]

aight thanks

appreciate the help

.close

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[devout snow]

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[tribal mauve]

hello i want someone to just solve this with ni explanation <@&286206848099549185>

[supple knot]

!15m

[devout snow]

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.

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[tribal mauve]

sorry sorrry

[tribal mauve]

wait so they wont give me the solve i already got anumber

number

[tender cobalt]

take a look at EG and DG
they tell you CDFE is a parallelogram
so in a parallelogram, do you think EG and DG are always the same length?

[tribal mauve]

yes they are

[tender cobalt]

now their lengths are 3(2x + 1) and 9x - 6

do you see a way to solve for x

as a reminder, you can place an = between two things when you know they should be equal

[tribal mauve]

i know wait

3 (2x+1)

=

9x minus 6

right?

[tender cobalt]

yep

now you can solve for x

[tribal mauve]

3

its 3

[tender cobalt]

nice, thats x

now for what the question wants: the length of DF

[tribal mauve]

7x2

7x +2

[tender cobalt]

right, then?

[tribal mauve]

man im learning this is fun

[tender cobalt]

thats not something you see everyday, keep going

[tribal mauve]

23

[tender cobalt]

yep thats it

[tribal mauve]

because x is 3

IM SO SMART

[tender cobalt]

because x is 3, yep

[tribal mauve]

and what now

[tender cobalt]

well since the question is done, youd then move on to a different question

[tribal mauve]

thats it?

[tender cobalt]

any time you get stuck, you try and figure out where you can go, what you can do, and what youre capable of

here, look at the question and notice it wants "the length of DF"

now lets see what we found: we found 23 for the length of DF

[tribal mauve]

ooo

thanks so much man

[tender cobalt]

np

[tribal mauve]

🙏🏻

close

[tender cobalt]

.close

[devout snow]

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[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[frozen aurora]

<@&268886789983436800> hell nah 🥀

[woven radish]

Please give me a valid number of messages to delete. See the help for this command for usage.

[devout snow]

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[fair canopy]

Found the avg velocity how do I find the inst velocity?

[fair canopy]

Is it same as finding the slope of tangent line?

[abstract plover]

,rccw

[candid lance]

Im sure you gotta differentiate and set t = 3

Coz ds/dt is v

And instantaneous is basically velocity at an instant when t = 3

[fair canopy]

Ye jst asking is my concept right

[candid lance]

Yep

[fair canopy]

Inst velo is like same as slope of tangent line

And avg velo is same as slope of secant line

?

[candid lance]

1st one is prolly correct but idk the 2nd thing

[fair canopy]

Alr ty

[wind cypress]

differentiate s(t)

[fair canopy]

Another basic q I m new to derivates so if a q is given in this form how r u going to tell it's the derivates of what with respect to what I get the y x

[fair canopy]

Ye I got 8t +2 and ans as 26

[candid maple]

,rccw

[fair canopy]

It's from first principle ig I was able to do with normal x plus delta x etc but got confused with other variables

I got it wrong in my mcqs

[candid maple]

hm?

[fair canopy]

So it's derivates of what with respect to what

?

[candid maple]

I'm just here to rotate the image for others, sorry. I'm not committing to this question at the moment.

[fair canopy]

Oh alr np

[rain summit]

that's the definition of derivatives right?

[fair canopy]

Ye

[rain summit]

what about it?

[fair canopy]

How do u like tell it's the derivates of what with respect to what

Like look at that one

Do u look at what's being subtracted?

[rain summit]

yea

[fair canopy]

F prime of a is same as df/da or dy/da in leibniz notation?

[rain summit]

i don't really know much about that

[fast agate]

df/da

[fair canopy]

Alr

.close

[devout snow]

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[devout snow]

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[plain ibex]

how do i program the median function in google sheets so it doesnt average the 2 medians if its even

[worn salmon]

What do you want it to do instead?

[plain ibex]

lets say the medians are 5 and 6.5
it will output 5.75 instead of [5 , 6.5]

[worn salmon]

Remove the highest element, calculate median, remove lowest element, calculate median

take the pair

[plain ibex]

i have this and i want it to be so that if there are 2 medians both boxes will highlight
its set up in such a manner that if a variable in a column is in a range where it will round to that columns number, only that box will be highlighted.
each row is set to be the "average" of each color in the average chart, and only highlight in the right rows so it can change. i will do the same with median but the median function is garbage

[plain ibex]

i suppose that could work

but it wouldnt work in my current setup i dont think

,av @worn salmon

[woven radish]

weareingram's Avatar

Click here to view the image.

[plain ibex]

awesome

[worn salmon]

I don't have much experience with excel

Have you looked up if there is a function with your desired functionality

[plain ibex]

i have

i did the mode function in a roundabout way, then tried it with the actual mode function upon learning it exists, and it gave incorrect outputs because i needed multiple

the mode function i created had issues but ive since fixed them

thank you for trying to help anyways

[woven vale]

to do this specifically, you can use
=IF(ISEVEN(COUNTA(1:1)), AVERAGE(COUNTA(1:1)/2, (COUNTA(1:1)/2) + 1), MEDIAN(1:1))

• COUNTA(1:1) returns the number of nonempty cells in row 1, so we simply check if it is even.
• if it is even, take the average of the middle two numbers, which will always be n/2 and (n/2)+1 for a list of length n
• if not, compute the median as normal

[plain ibex]

this chart represents all inputs overlayed
if a box in the 5 column says 2, it means that 2 inputs for that color were 5

the mode i set up works by judging if its larger than all other boxes in the row or equal to another largest

[plain ibex]

the average of the 2 is the exact thing im trying to avoid

the preset does that

if there are 2 middle variables i want it to output both not average

0 reds code for example looks like this
its mostly just a lot of nesting "and" functions, and the other parts are to fix errors and to return everything that isnt the mode as grey box

[woven vale]

oh whoops i read what you said wrong

[plain ibex]

the "conditional format" just means the boxes change color if certain aspects are met

deleting the "value is equal to 1" would delete the ruleset elsewhere and i just havent dealt with it but it shouldnt be a roblem

the mode is just a good example of how ive been doing this sorta stuff

oh shit wait

maybe i can do something similar to what you said but instead of averaging i could use an "OR" function

idk if thatl work

i should probably take a break

[woven vale]

modified the formula to get
=IF(ISEVEN(COUNTA(AA6:AU6)), JOIN(", ", INDEX(AA6:AU6, 1, COUNTA(AA6:AU6)/2), INDEX(AA6:AU6, 1, COUNTA(AA6:AU6)/2+1)), MEDIAN(AA6:AU6))
using this formula will output 5,6 in the situation you described (without the square brackets). you could encase the formula in SPLIT(... , ",") to splt 5 and 6 into two separate cells - might be easier to work with that way

INDEX accesses the range at row 1 (since we fed it a single row), then at the middle two indices that we wanted. JOIN puts them together with ,

[plain ibex]

i was working on it now i gotta test your thing ty

[plain ibex]

wait

how would i apply this

what box would that go in if its talking about AA6:AU6

[woven vale]

let's say you write this formula in A1, then 5 will go in A2, and 6 will be in A3

but you can place this formula wherever you'd like, as long as you're referencing the same range it won't matter

[plain ibex]

i mean tysm for writing code but i dont think it worked
oh wait wait
i see

no

we are not medianing the values in the total

if a box in the 6 column says 4, then 4 of the inputs are 6

tysm tho

i might be able to make this work if i just change how it takes in inputs

ima try that

[woven vale]

yeah tbh i wasn't understanding what you wanted the highlighting to do when i wrote that, so that's my bad

[plain ibex]

tf you mean bad you just wrote code for someone for free

[woven vale]

so let's say we have 4 sixes, 2 twos, 1 one for a certain color, then the set we want to take the median of is [1, 2, 2, 6, 6, 6, 6]?

[plain ibex]

yes

also your thing kinda worked when i changed the inputs in the first part, it returned "0,0"
after replacing all of them with the range i got an error but thats bc i put in errors

the inputs look like this and i tried it a different way but it wasnt very useful

the blank ones return errors

[woven vale]

you can handle this by checking if COUNTA = 0, since you are doing very good error handling

[plain ibex]

?

[woven vale]

as in, if COUNTA(AA6:AU6) = 0,then return "", otherwise do the formula

[plain ibex]

what is counta and what do you mean im doing very good error handling

[woven vale]

one idea is to write this range out somewhere else, then reference that "expanded" range when calculating this formula.
perhaps there's a way to handle the original range as a sort of weighted range? but i dont know what that would be

[stone stump]

ok I havent read through the chat but why not just put in each cell =COLUMN()-MEDIAN(...) and then conditional formatting for this value to be between -0.5 and 0.5

[woven vale]

so COUNTA checks how many nonempty cells there are (has either an explicit value or some formula definition in it), it's basically the length of the list for our purposes

[stone stump]

or something along those lines anyway

[plain ibex]

each color row ive had so far has had all the correct value and the conditional format only highlights the column that corresponds to the variable

the median function will not output 2 answers and instead averages them and i fucking hate that

[plain ibex]

ohhh

ima go figure out how to do that while making it ignore errors

[plain ibex]

the average and mode chrts do this

mode looks like this tho

[devout snow]

@plain ibex Has your question been resolved?

[devout snow]

@plain ibex Has your question been resolved?

Channel closed

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[worn salmon]

Had an epiphany it seems

[devout snow]

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[distant torrent]

Did I answer 3 b correctly?

[polar chasm]

,w simplify 1/(x^2 - 1) - 2/(x^2 + 3x + 2) * (x+1) / (4x^2)

[woven radish]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[distant torrent]

Uh did I get it right?

[polar chasm]

hmm werent you supposed to subtract it too?

[distant torrent]

ah

This?

[polar chasm]

yeah

Also, try to avoid using => like that

you should probably keep a column of the simplifications (and use = there) and then another column / space for extra computations (such as x^2 - 1 = (x-1)(x+1), x^2 + 3x + 2 = (x+1)(x+2))

[distant torrent]

Got it

Nvm

.close

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[idle dune]

hik

[polar chasm]

is this a lagrange multipliers problem?

[idle dune]

ye

naw im just curious now

i remember from linear algebra something about quadratic stuff

and that it's usually points like (c,0,0) or (0,c,0) or (0,0,c) that are the maximum or minimums for some constraints

so i just wanna know the theory that i can just plug in values from (0,0, +- 2) instead of all of these points

[polar chasm]

i think its kind of a coincidence here

if it was z^2 = 4xy + 4, the minimum would be different

[idle dune]

oh

wait actually i just realized

i cant solve for that -xy = 4

wait nvm nvm

[polar chasm]

hmm wdym

[idle dune]

naw mb i solved for it mb

[polar chasm]

the closest point would be at (1, -1, 0) and (-1, 1, 0)

[idle dune]

?

[polar chasm]

in here

[idle dune]

oh

[idle dune]

oh is that so.

so i do have to evaluate all of these points

[polar chasm]

probably

if you added some extra x and y, the symmetry would be gone completely (and you would prolly get a non-zero z as well)

[idle dune]

shouldn't be bad since i squared the distance function

kk thanks bro

i do mæth too

.solved

[devout snow]

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[polar chasm]

in this specific case, there is likely some algebraic shortcut, not in general though

yeah, the constraint only fixes xy and x^2 + y^2 is minimized if x = +-y (given fixed product xy)

[devout snow]

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[indigo sail]

Does This Question have a single answer I'm stumped? I concluded that the path the Hare taken is 25 km. This is also meant to be a hard Problem.

[fair juniper]

have you drawn a diagram?

[indigo sail]

Yes?

[fair juniper]

can we see it?

[indigo sail]

I don't have my phone ,but I'll try drawing it on my computer.

[fair juniper]

if its too much of a hassle, you can also try describing it

specifically, highlight the equations you need

[indigo sail]

Imagine doing that on a trackpad pain.

[fair juniper]

the tortoise and hare start at the same point

[indigo sail]

Wait they do?

Ohhh OK I get it. I hate my stupid mistakes.

[fair juniper]

looks like this

[indigo sail]

13km bc i did a quadratic equation?

Which simplyfied into a Linear one.

[fair juniper]

yup

if you notice, this is one of those special pythagorean triangles

5-12-13

[indigo sail]

Thank you.

[fair juniper]

youre welcome

[indigo sail]

.close

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[indigo sail]

Hello. I feel like IM:ML:LA = 1:2:3 ,but I can't prove it. So I need help to prove it's true.

[indigo sail]

I always get stuck in geometry. Is there a way to hack these geometry questions?

[granite island]

ok so how would you find the area of the traingle

[indigo sail]

1/2bh?

[granite island]

i would firs find the area of ABJ then the area of ALJ

[granite island]

yes this is true

[indigo sail]

20

[granite island]

yes that is ABJ

now i would find BL:LJ to finish the problem

do u know how to do this?

[indigo sail]

No?

[granite island]

ok

do u know mass points?

mass points is the easiest way to solve this problem

[indigo sail]

Like centroids?

[granite island]

no

like you assign a points to a specific weigh and then use that to assign other points weights

[lunar harbor]

I saw this question a few days ago

[granite island]

oh no not that

[indigo sail]

It's from a popular competition.

They released their past questions this year.

[granite island]

here you can read this article
https://artofproblemsolving.com/wiki/index.php/Mass_points?srsltid=AfmBOopfV0cd9AgIGhoR6IWg_U4N3xafnFyhfP48nHTo7TqkSic7mnZ0

if u don't understand it there is another more complicated way

that i can explain

[indigo sail]

OK go on.

[granite island]

wait a sec

MB gang

so sorry

it was a typo

😭

[indigo sail]

What?

[granite island]

lol i mistyped wait a sec on acident into something not really nice

anyways give me a sec and ill explain it

ok so extend BJ. Then, draw a line that goes through point A that intersects BJ and is parralel to BC

[indigo sail]

OK

[granite island]

wait ill draw a diagram

ok this might take a while u can think about it while i draw it

ok

so what os AJ/JC

[restive river]

Hello ppl

[granite island]

hi

um

are you here to help

[restive river]

✅

[granite island]

ok

[indigo sail]

OK?

[granite island]

can you do this?

[restive river]

Jsn am looking for a pencil

Alr lemme see

[indigo sail]

1/2

[granite island]

good

so what is NJ/JB

[indigo sail]

IDK

[granite island]

think similar triangles

NJA is similar to BJC

so wht can we say about AJ/JC

[restive river]

To find the ratio \frac{AJ}{JC}, we need to look at the geometric configuration in your image. This setup is a classic application of Menelaus' Theorem or Ceva's Theorem variations.
​Based on the visual construction (which appears to be a standard subdivision of a triangle where I, K, and J are specific points on the sides), we can determine the ratio by looking at how the lines intersect.................... Bro am I saying Enrelervant stuff

Wait

No

Idk

[granite island]

umm

thats a lot of text

[restive river]

Sryyy I type a lot

[granite island]

and a little confusing

[indigo sail]

1/2

so 1/2

[granite island]

good

[indigo sail]

scale factor 1/2

[granite island]

so can you find NL/LB

if we know AL/LI=1/1

[woven vale]

!nogpt

[devout snow]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[granite island]

remember think similar triangles again

[indigo sail]

1/1?

[granite island]

perfect

[indigo sail]

I guessed that idk why its 1/1

[granite island]

oh

[indigo sail]

I'm so bad at this

[granite island]

so notice that triangles ALN and BLI are similar

and because AL/LI is 1, so is NL/LB

[granite island]

its ok i remeber learning this is my early math contest years and it was so frustrating

[indigo sail]

So they're congruent?

[granite island]

yes

so does it make sense now? i can elaborate more if needed

[indigo sail]

It does make sense ,but how did you know I had to extend a side?

[granite island]

like the first step?

where i extended BJ?

[indigo sail]

Yes.

[granite island]

so um basically for most of the problems where you have to find ratios of intersecting lines in a triangle

like two lines intersectig and you want to find the length ration

i usually always extends a line to form parrelel lines which i can then use to make similar triangles, like in this problem

[indigo sail]

OK

[granite island]

ok

so can we continue?

[indigo sail]

yes

[granite island]

ok so let NB equal x. so can we write NJ, JB, BL in terms of x?

[indigo sail]

ok

Yes

we cna

[granite island]

ok

so what will we get?

[indigo sail]

NJ = 1/3x

BL=1/2x

[granite island]

and JB?

[indigo sail]

so JB = 1/6x

[granite island]

i hink you ment JL

yes JL=x/6

[indigo sail]

OH yes

JB = 2/3x

[granite island]

yes thats right

so can we find BL/LJ

[indigo sail]

1/2x/1/6x

[granite island]

ye

which simplifies to...

[indigo sail]

3

[granite island]

ohh

sorry i mean LJ/BL

[indigo sail]

1/3

[granite island]

yes

so what is LJ/LB

wait mb

what is LJ.JB

LJ/JB

[indigo sail]

1/6x/2/3x

1/4

no x

[granite island]

good

[indigo sail]

So 20/4 = 5

[granite island]

perfecty

good

so i would reccomend hat you take a look at mass points

because it makes it 10x simplier

if u want to

but yea this method is also really good for these problems

great job!

you can close the cahnnel now unless you have any furthur questions

i gtg go now. have a nice day!

[light storm]

i can takeover

[granite island]

ok good

[granite island]

so this is the answer?

anyways i gtg

[granite island]

mass points gives the same answer too so prolly correct

[indigo sail]

Who are you?

[indigo sail]

Bye Thanks.

[light storm]

helper

[worn salmon]

Says the non-helper

[indigo sail]

.close

[devout snow]

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[devout snow]

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[acoustic coral]

what is the question even asking

[zealous pewter]

Is there no list?

[acoustic coral]

no

[zealous pewter]

It has to be continuous on 1
And f(1)=1
So lim (x -> 1+) f(x)=1
Meaning a + b+c = 1

[acoustic coral]

how did u go from lim (x -> 1+) f(x)=1 to a+b+c=1

[zealous pewter]

That limit is the limit of f on the plus side of 1
Meaning where f(x)= ax²+bx+c
So for x->1+, the limit is equal a+b+c

When you say x->1+
It means x is approaching 1 from the numbers bigger than 1
And since f=ax²+bx+c when x>1
That is the form we use for the limit x->1+

I hope you understand^^

[acoustic coral]

yeah thx

[zealous pewter]

You are welcome
Do .close if u are done ^^

[acoustic coral]

.closwe

.close

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[rare kernel]

mods are too fast bruh

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[ivory linden]

well, i have found, 28/5, but imma not sure abt that.

[ripe grove]

this mf is working on a whole as tv

[ripe grove]

Looks good to me

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@ivory linden Has your question been resolved?

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[lost laurel]

for <=
As the gcd of $a,b$ is $1$. There are elements $r,s$ sucg that $ar+bs=1$.Let $r' \in R$ be arbitrary. Then $r'ar+r'bs=r'$. $r'ar \in (a); r'bs \in (b)$. Thus $R \subseteq (a)+(b)$ . Subseqeuntly, $R=(a)+(b)$

[woven radish]

wai

[lost laurel]

(a) + (b) subset of R as they are ideals of R and the sum of their elements is too

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[lost laurel]

.close

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[dim ember]

Hello smart people,

I have something that I get, but I don't fully understand. I hope you can shine a light for me 🙂

I am working on my understanding of fractions. And what I found is that they are seen as an ambiguous duality between being a process of division, and an object of a number on the number line. That bit is something that I understand. What I find a lot of different opinions on is the consequence of such a perspective on calculations.

Easy one. 6 : ½
Looking at ½ as a process, I would say I have to read it as 6 : 1 : 2 = 3
Looking at ½ as an object, I would say I have to read it as 6 : (½) = 12

What do the rules say is right, and why?

[wheat pawn]

for a start, they arent ambiguous
however, in the "examples" you've put, you're not dividing by 1/2

6:(1/2) is not the same as 6:1:2, since the division is not associative

thus:
6:1:2 = (6:1):2 which is not equal to 6:(1:2) = 6:(1/2)

your first operation would be (6/1):2 instead

[devout snow]

@dim ember Has your question been resolved?

[rain summit]

usually, $\frac{1}{2}$ means $1 \div 2$, well specifically $(1 \div 2)$

[woven radish]

1 divided by 0 equals Infinity

[rain summit]

so your $6 \div \frac{1}{2}$ would be counted as $6 \div (1 \div 2)$

[woven radish]

1 divided by 0 equals Infinity

[rain summit]

that's usually how people did it

[restive river]

,, \frac{a}{(\frac bc)} = \frac{ac}{b}

[woven radish]

satan

[restive river]

,, \frac{(\frac ab)}{c} = \frac{a}{bc}

[woven radish]

satan

[dim ember]

So am I right to say that a fraction bar comes with parentheses that we just dont show?

[rain summit]

wdym?

[dim ember]

Every case of a/b (i am not too familiar with the scripts here to make it a proper fraction, sorry) should be read as (a/b)
So in my example, 6 : 1/2 is read as 6 : (1/2)

[restive river]

yes

[dim ember]

Thanks for helping me understand it better @restive river @rain summit

.close

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[obsidian plank]

does anyone have any tips for verifying trig identities I

[obsidian plank]

im having trouble reconizing when I should do certain things

like when should I turn everything into sine and cos

or when should I try and get both denominators the same of an expression

like in one example I turned everything into sine and cosine and I still ended up with the answer but my teacher said I over worked massively

[trail eagle]

For most "simple" identities, writing in terms of sine and cosine will eventually work, but like your teacher said it's kind of a brute-force method that may take long and facilitate mistakes.

If you have a sum of fractions it's always a good idea to write them on the same denominator.

<@&268886789983436800>

[nova mountain]

My calc professor told me to remember 3 trig identities:

1. sin^2 (x) + cos^2 (x) = 1
2. sin(2x) = 2sin(x)cos(x)
3. cos(2x) = cos^2 (x) - sin^2 (x)

Other identities come from those 3 identities

[obsidian plank]

thank you I guess Ill first look for anything I could use the pythagorean identities on and if I dont see any then ill go to common denominators and if that doesn't work sine and cos

[nova mountain]

The pythagorean identity comes from the first identity

And remember the unit circle

Learn it and you should be good

[graceful cosmos]

You are turning one side into the other. You take note of what the other side "looks like" and use that to choose what you do. It comes with experience

Did you have an example of a question that tripped you up?

[obsidian plank]

yeah

give me a second

[primal badge]

its just practice ig

[restive river]

experience helps in trig

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@obsidian plank Has your question been resolved?

[devout snow]

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[tall trellis]

can someone explain sin and cosine addation indenites to me

[tall trellis]

i dont get it

[unique canyon]

which ones?

[rain summit]

sin and cosine identity

is that what you meant?

[tall trellis]

yes the one its slike sin(a+b)=sina+cosb+-cosbsina

and then one for cosine as well

[unique canyon]

yeah

like you know the proof right?

[tall trellis]

no

i dont udnerstand it

[unique canyon]

you can search it, proof is kinda weird

like its not the most intuitive

but there are lots of proofs in the internet

[tall trellis]

i dont really get it i searched it up

[rain summit]

i'd just use geometry if i can

[tall trellis]

becuase its like sin(45)+cos(45) to get sin+cos=90 i think

thats all i get

[rain summit]

oh

$\sin \theta = \cos(90 - \theta)$