help-36

[shell condor]

I agree

[native sluice]

Are R O1 O2 collinear

[shell condor]

I don't think we can say that

[woven ledge]

if the common tangent is l then both O1R and O2R are perpendicular to it

thats why they are collinear

[shell condor]

Oh yeah point

Then finding O1O2 could help too

[native sluice]

It seems difficult to find O1O2

[woven ledge]

damn why is this so difficult

[native sluice]

Yeah

[shut minnow]

Can you tell what the angle PO2O1 should be

[native sluice]

I can’t tell

[shut minnow]

What's angle O2PT

[native sluice]

90 degrees?

[shut minnow]

Now the other one

It's 90 deg right

[native sluice]

Yeah

[shut minnow]

The angle in down is angle inscribed in semi circle

So it should be 90 too

[native sluice]

Yeah so it’s a square

[shut minnow]

Now can you find O2O1

[native sluice]

How?

Pythagoras theorem using r (radius of O2?)

I can’t get the result for that one

[shut minnow]

Internal tangent

[native sluice]

What do you mean, I am sorry I can’t get it

[shut minnow]

A common theorem

Is that the centre of touching circles

And point of contact are collinear

[native sluice]

So what can we do about them

[shut minnow]

O2O1R is a line

[native sluice]

Yeah?

We still need O2O1 right

[shut minnow]

So find O2O1 in terms of r2 and r1

[native sluice]

I used pyth theorem can’t get the result

[shut minnow]

Hmm

[native sluice]

O1O2 = sqrt( (r2-0.5a)^2 + (0.5b-r2)^2 ) ?

[shut minnow]

The( a+b-c)/2

[shut minnow]

Is something you get for incircle.of right traingle

[shut minnow]

try writing in radius

[native sluice]

O1O2 + r = 0.5c ?

Why is it incircle

[native sluice]

I don’t think this is correct tho

[final saddle]

@native sluice Has your question been resolved?

[final saddle]

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[radiant igloo]

Wanted to confirm if this question is correct or not?

If it helps, this question was taken from the UNSW Integration Bee 2022 Knockout Round-3 Question 2

https://assets.ctfassets.net/xv1hr0zszyo1/1Bui41TER2cQ6aUqTIuHbO/cd46c2114f0b5083102d5cdfef6b0536/Integration_Bee_Questions_and_Answers.pdf

[ionic venture]

Is the question missing its third right parenthesis?

[radiant igloo]

Oh

But still even if we consider the 2nd term to come out 2 ways; x^(k²+2k) or x^(k²+k) I still didn't get the right answer which was 3ln2

I tried Wolfram alpha after an hour but it couldn't even interpret the whole question

[bold turtle]

What do you mean "correct" here?

[radiant igloo]

By correct here I mean that the question gives the intended solution of 3ln2

[bold turtle]

ah

Well, as Erebus mentioned, yh it's not done the parenthesis correctly; it's still missing a closed bracket

Since it deals with a k in the summand, I'd surmise the remaining one should be before the dx

Try swapping the Sigma with the Integral and see if that leads you to anything

(There are conditions that have to be met to allow you to do that, but let's just assume they are met)

[radiant igloo]

Hmmm

Gonna try

Ouch that's not gonna work the first term to apply sigma is 1/n which diverges

(or sigma k=0 to inf 1/(1+k) to be precise here)

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@radiant igloo Has your question been resolved?

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[mystic agate]

Solve: 7(2x-1)-4(x+3)=5(3x-1)

[mystic agate]

I am so lost trying to solve this

but this is what I did so so far I did 14x-7-4x-12=15x-5

So far is it good?

[rancid zenith]

thats a good start

yes

well now what you want to do is get something that looks like x = ??

[mystic agate]

yeah yeah

[mystic agate]

Then from that we choose the right side 15x is bigger than 14x

good?

[rancid zenith]

well the left side also has -4x so actually 15x is "bigger than" 10x but yes thats also correct

[mystic agate]

oh

you need to add them up or minus them

okay thanks I need to watch out for that when doing other solves

let's contuine

[rancid zenith]

in general, you can move everything involving x to one side, move everything involving the numbers to the other without worrying which side is bigger but you can definitely continue like this for now

[mystic agate]

then I think we do 15x-5-14x+4x

[mystic agate]

Oh okay I mean the teacher told us to choose the bigger side

[rancid zenith]

yes thats completely fine

[rancid zenith]

lets continue with this

[mystic agate]

then we need to serprate the odd one out the -5

so then the left side becomes 5-7-12?

okay so that's -14

then we need to do 15x-14x+4x

so then

-15 divison 5

?

[rancid zenith]

why -15?

[rancid zenith]

you said -14 here

[mystic agate]

Oh my bad

My brain not working lol

-14 divison 5

so then the answer is -2.8?

[rancid zenith]

always a good idea if you have the time is to plug back x = -2.8 to be sure

[mystic agate]

I am still so not confident with this subject man I know how to solve these but I take ages

[mystic agate]

Yeah the teacher told us that too

[rancid zenith]

its fine, just some practice will do

[mystic agate]

But like is there something wrong with me

like

[rancid zenith]

not at all

[mystic agate]

I am in math class

and

I pause for 5 seconds or even more

just to do 13-7

.......

[rancid zenith]

well its fine if you take time but get the correct answer

ofcourse if you are writing a test that is timed, it may cause an issue

[mystic agate]

I am lacking so far behind class everyone already done I am the only one not done and before I am done the teacher is correcting the classwork I didn't finish

[mystic agate]

yeah

That's the problem

[rancid zenith]

once you practice solving the equations a lot you will be able to do the subtractions/additions mentally

[mystic agate]

I feel like I might be stupid or mentally slow

[rancid zenith]

i dont think so at all

you basically did this problem fully on your own (i just gave some commentary)

slowness will fix itself with practice and almost everyone will be slow when they start

[mystic agate]

So then I guess I doubt myself to much

[rancid zenith]

yes just practice

[mystic agate]

I will go check my textbook questions that I have for homework

Oh my I will take ages to finish these

I will close this ticket as I don't have doubts anymore what to do.

[mystic agate]

Thanks alot though for the help

.close

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[tranquil charm]

Compute
[
\int_C \frac{1}{z},dz,
]
where (C = C_1 + C_2), (C_1) is the line segment with initial point (-1) and endpoint (-2 + i), and (C_2) is the line segment with initial point (-2 + i) and endpoint (-\sqrt{8} - i\sqrt{8}).\\

So far i have this parametrization
[C_1\cdots\varphi_1(t)=-1+t(-1+i), t\in[0,1]]
[C_2\cdots\varphi_2(t)=-2+i+t(-\sqrt{8}+2+i(\sqrt{8}-1)), t\in[0,1]]

[tranquil charm]

how should i continue?

[soft zealot]

Slowaq

[tranquil charm]

<@&286206848099549185>

[cursive bough]

Firstly I think your second parameterization is off by a little, when t = 1 it gives you -sqrt(8) + i*sqrt(8)

[cursive bough]

use the formula

$\int_{t_0}^{t_1} f(z(t)) z'(t)dt$

[soft zealot]

Doaby

[final saddle]

@tranquil charm Has your question been resolved?

[final saddle]

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[heady dagger]

I am looking for a formula to calculate this, I cant seem to find any patterns to simplify this

[tired walrus]

that looks like some harmonic number bullshit.

!xy

[final saddle]

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.

[heady dagger]

I dont have an origional question unfortunatly so give me one sec to try and get the details around

c is capacity
d is devices

z is powered devices
y is batteries

Capacity is a constant (example 50)

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[velvet pivot]

Can I get an explanation for what they did? My answer (the long digits) isn't the same

[vital crag]

you left out a lot of information

[velvet pivot]

Do you recall any of this in grade school?

[dense coral]

what’s part B?

[velvet pivot]

it's this part

[vital crag]

yea buddy, that's why i asked you to provide information necessary to answer your question

[velvet pivot]

I have, I believe.

[vital crag]

yet you could have just answered my question

[velvet pivot]

I wasn't trying to be sarcastic or anything, i meant to like build upon that

i was wondering what you meant

because i didn't provide any formulas or steps, so maybe you needed those

[vital crag]

yea don't keep doing that

anything being referenced in the screenshot you should provide

[velvet pivot]

don't keep doing what

[vital crag]

this

[velvet pivot]

i didn't though

[vital crag]

you also didn't show your answer and your work getting your ansewr

[velvet pivot]

Well I only used a calculator

[vital crag]

^

only you know what you input into the calculator

[velvet pivot]

i assumed it was known, my bad, because it wasn't necessary because it was already in the picture

[velvet pivot]

ah

[vital crag]

how is it possibly known what "part B" refers to from this image

[velvet pivot]

i put 2.71 because earlier this was mentioned

[vital crag]

[velvet pivot]

no you can't

[vital crag]

,calc e

[soft zealot]

Result:

2.718281828459

[vital crag]

2.71 isn't even rounded correctly

[velvet pivot]

This was never shown in my class

So i assume that 2.71 is what i'm supposed to use

because they used it for other examples

i'm really confused right now

[crimson bronze]

We want to see part A and B because there's not enough context in the screenshot you've given. What's f(x) and g(x) for example?

[velvet pivot]

Okay, sure!

[crimson bronze]

And how did you end up with your answer? We want to see the steps in order to tell you what you did wrong

[velvet pivot]

i will show screenshots of part a and b, i hope this helps

[velvet pivot]

i used a calculator in windows because i felt it was okay to use (the history is from bottom to top), first i put 4 in place of the x in the exponent and calculated that it is e^-0.16, and replaced e with 2.71 and then multiplied the result by 2.5 and subtracted -0.2(4) from the result an then subtracted by -1.2

And i did the same but with 5 for the "upper bound (5)" one, and for "average of the bounds (4.5)" i took the results from both and added them together and divided them by two

this is why

im sorry, i didn't know what someone who'd want to help would know or wouldn't, i just assumed because i've used this channel before for very basic algebra things that maybe it'd also be similar

[crimson bronze]

I see, thanks for explaining I can't double check the computations right now, but I'm sure someone else will come along to help. I'm guessing it's just a simple arithmetic error or typo somewhere

[velvet pivot]

Ty!

[velvet pivot]

I'm not sure what happened, because my first answer for the lower bound is correct to the hundredths place, idk why they don't have long numbers just like me, maybe they rounded or something

[velvet pivot]

Tysm! ❤️

[crimson bronze]

Whenever there's a question that is part of a bigger problem it's nice to see the entire problem, so we know what variables are defined etc

[velvet pivot]

-0.012 here is the lower bound and upper bound added and divided by 2 and rounded because the actual answer is - 0.0115

[velvet pivot]

Right!

usually i've had problems where they don't need context

unlike this one

[crimson bronze]

Can you try the calculation again, but with a more precise value of e? As provided here

[velvet pivot]

Okay

i assume that they just rounded it then? because this is with the constant e in my calculator

[crimson bronze]

Yeah of course they rounded it to 0.13, I thought the problem was with the other values? I mean 4.5 and 5, which are not the same as what you found, even after rounding

[velvet pivot]

Yep

i'm not sure why

i'll try e^-0.04(5) now

i always wanted to use e, i saw it on my calculator as a kid and wondered what it was

now im an adult and im finally using it myself

[crimson bronze]

There's something wrong with your calculation, can you use anything other than windows calc? It's pretty error prone when you have to copy the results repeatedly

[velvet pivot]

Okay!

[velvet pivot]

Omg! it's correct now, tysm!

im gonna ditch windows calculator it sucks

Tysmmmm ❤️

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[vocal iris]

Is it necessary to graph this?

[vocal iris]

İn Cartesian plane as illustration

[stoic temple]

The same question?

[stoic temple]

If it is then graphing isn't necessary but it's helpful

[vocal iris]

Which one here doesn’t need to be graphed

[vocal iris]

Yeah omg sorri

[stoic temple]

!xy

[final saddle]

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.

[stoic temple]

Also, what step are you stuck on?

[vocal iris]

Im done with akl solutions

I just need illustrations

I need help

[stoic temple]

Draw the x and y axes then locate the points A and B

[final saddle]

@vocal iris Has your question been resolved?

[final saddle]

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[next thorn]

guys... whats the perimeter of this shape?

[next thorn]

is it 4x + n + m

or is it just 4x

[severe verge]

which shape in particular?

[hasty mist]

It looks like m and n are interior line segments?

[ivory vessel]

imo it's 4x, perimeter shouldn't include interior segments

[icy current]

Is this figure to scale?

[hasty mist]

Since perimetre only counts the outer boundary, it should be 4x I reckon

[next thorn]

abcd

[next thorn]

so you're telling me that only after 9+ years of math i only know NOW that perimeter doesnt include interior segments..

[severe verge]

why do you need the perimeter of the shape?

[icy current]

Is it a rectangle or just a quadrilateral?

[next thorn]

it was a rhombus in a national test and i calculated the perimeter INCLUDING the diagonal

like i didnt just add the 4 sides

i even added the diagonal in the middle

[severe verge]

then it's just 4x?

[hasty mist]

Yeah, I'm pretty sure the perimeter of a 2D shape is the total length of its outer boundary.

[next thorn]

i guess 😭

[next thorn]

lovely...

im gonna be in college next year btw ❤️

[hasty mist]

[icy current]

Well depends on whether BC = x or (the point at which n is intersecting with BC)C =x

[icy current]

Hey man, happens to the best of us.

[next thorn]

x's are the lengths of the sides

[icy current]

Yea then it's just 4x

[next thorn]

https://tenor.com/view/choke-choke-emote-choke-myself-gif-14544625584098674544

ty everyone

.close

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[severe verge]

🐧

I have no comments on this

[next thorn]

https://tenor.com/view/triggered-stan-twitter-screaming-meltdown-sad-gif-6570216357328685463

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[timber plume]

"thats the cardinality of your final answer set, not the probability. thats the 10/36 possible outcomes" what does he mean

[timber plume]

I thought the question asked for proability

[loud sundial]

I see the numerator of 10 circled

this comment seems to refer to that

there's 10 outcomes in the set of favorable outcomes (giving the cardinality of that set to be 10)

"10/36 possible outcomes" meaning that there's a total of 10 favorable outcomes out of 36 possible ones

[timber plume]

so its still refering to probability

because he said its not probability so I was confused

ohhh ok I get it now

he is indeed reffering to 10

👍 thanks

.close

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[loud sundial]

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[sonic spindle]

How does this 1-y turn into y-1

[sonic spindle]

Is it because we already know y is 2 ? And it's modulus

[thick veldt]

the absolute value would make it positive regardless so if y=2 the absolute value makes it positive 1 so changing the form only takes the absolute value into consideration

[sonic spindle]

If we didn't know y was 2, what would we have done instead?

[supple viper]

I think they did that because it was already given. Because it’s absolute value, the only way to make the denominator positive would be to flip it

[sonic spindle]

Yeah but what if we needed the differential equation in the form y=f(x) and we didn't have any given X or y and this couldn't work out lnk. (So the general solution and not particular) How would we do that given that we wouldn't know whether to flip the modulus or not given we don't know whether it's positive or negative

[thick veldt]

did the problem give you any restrictions for ex: x>1?

[sonic spindle]

Sorry, let me show you the question.

Number 7

From the Edexcel year 2 pure maths a level textbook

[thick veldt]

i’d say usually they’d never give you a problem like that without giving you a specific restriction or boundary

[sonic spindle]

Alright, thank you for your help. I was just curious (lol). Cheers mate.

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[left trail]

So I couldn't think of any way of proving this from the axioms so I assumed it is true and came up with the first photo. It seems true and from there I could get to what I want to proof. Is it obvious enough or do I need to prove it too?https://cdn.discordapp.com/attachments/486841106298961930/1440871169690435756/20251119_200028.jpg?ex=69206418&is=691f1298&hm=8b0cbf1483bf67f0ac4b16521d57f8c52d52c4c8964102e41699bfb82c2f4d6e& https://cdn.discordapp.com/attachments/486841106298961930/1440871170093350952/Screenshot_20251119_200204_WhatsApp.jpg?ex=69206419&is=691f1299&hm=03eef1be3adb4b2bfd8046e70a9c48aa3cfcc96efe52aea2ef37cdc76d0f136e&

[pliant shore]

do you know inclusion-exclusion for 3 events?

[dusk lake]

im looking for a study patner in michigan

[pliant shore]

this isn't your channel

#study-discussion

[left trail]

Yes that is what I used to get what I sent

[pliant shore]

ah, you only need to prove $P(A \cap B) \ge P(A) + P(B) - 1$

cause that implies $P((A \cap B) \cap C) \ge P(A \cap B) + P(C) - 1$

[soft zealot]

south

[left trail]

Wait what?

[pliant shore]

yes

[left trail]

One sec let me write it out

$p(A \cup B \cup C) = p(A) + p(B) + p(C) - p(A \cap B \cap C) - p(A^c \cap B \cap C) - p(A \cap B^c \cap C) - p(A \cap B \cap C^c)$

wait how do you latex again here?

[pliant shore]

more likely the bot is just slow

[left trail]

ok

[pliant shore]

you should really understand how this works

disregard my previous comment about inclusion-exclusion

[left trail]

ye I misread what that was

ok so for the first part of what you said all I could see was that we move the one over and write it as the union of a and b and the complimenet of that union which has to be greater than p of a and p of b

[pliant shore]

yes, the first part comes from 1 >= P(A union B), which is true for all A, B

[tired walrus]

nah it's bc op put a space before the ending dollar

[pliant shore]

I have faith that you know how to prove it

[pliant shore]

that's evil.

[soft zealot]

BigBen

[tired walrus]

btw there is a nice visual way to do this imo

[left trail]

Ok but this was because I misread what you wrote

[left trail]

What is it?

[tired walrus]

rewrite your goal as $$P(A)+P(B)+P(C) - P(A \cap B \cap C) \overset?\leq 2$$

[soft zealot]

Ann

[tired walrus]

make a venn diagram something like this

[left trail]

No I see it cannot be possibly greater than 2

[tired walrus]

i was not done

P(A) counts the probability of each piece inside A once, so put a 1 in each of these regions

P(B) adds 1 to every piece inside B, so P(A)+P(B) looks like this [where each red number is the coefficient with which that region's probability is counted]

this is P(A)+P(B)+P(C)

and then subtracting P(A \cap B \cap C) puts the red number in the centermost bit back to 2

heres what the LHS looks like.

each region is counted no more than twice

[pliant shore]

I KNEW THERE WAS A COUNTING ARGUMENT

[candid pulsar]

o

thats smart

[pliant shore]

honestly both of our approaches have merits

I think you can also generalise Ann's logic to N events too

[left trail]

Ok so I followed all this but I don't see how that can show that the total area less than or equal 2

[pliant shore]

the maximum number of times any region is counted is the centre region
which is counted n - 1 times

[tired walrus]

my logic generalizes to also proving inclusion-exclusion

[pliant shore]

so it follows that the equivalent expression is <= n - 1

[tired walrus]

if you are thinking of probabilities as areas

then the "area" of the entire picture is 1

like, if you were to put a "1" in each region, incl the outside one, then the total is 1.0

[left trail]

But for what you did we ignore the e outside region

[tired walrus]

then it's going to be less!

let me cap off my argument

[left trail]

Also for what you said earlier you were showing that each single one cannot be more than 2

[tired walrus]

you misunderstood me

each region is counted no more than twice

[left trail]

Ok. so then the logic is because we need all components to be 2 but in our scenario the components are all less than equal to 2 it must also then be less than or equal to 2. And here we say less than equal because if something is less than we can rewrite it as less than or equal to right? But why would we do that? What's the benefit?

[tired walrus]

urghururhhr.h.

you're KINDA hitting the right-ISH notes but like

it's a bit hard to tease apart the correct understanding from the misconceptions.

[left trail]

What misconception?

[tired walrus]

And here we say less than equal because if something is less than we can rewrite it as less than or equal to right?
very roughly this but i cannot possibly point at it in any more precise way and would like you to not ask me to.

i can put it in more formal/bureaucratic terms maybe.

[left trail]

Ok

[tired walrus]

ok so im just gonna label the probabilities of each region on the venn diagram as p_0 through p_7

[left trail]

Ok

[tired walrus]

in this fashion

so for example $p_1$ will refer to $P(A \cap B' \cap C')$, $p_5 = P(A \cap B' \cap C)$ etc.

[soft zealot]

Ann

[tired walrus]

the red number stuff i showed above shows that $$P(A)+P(B)+P(C) - P(A \cap B \cap C) = p_1+p_2+2p_3+p_4+2p_5+2p_6+2p_7$$

[soft zealot]

Ann

[tired walrus]

and we also know that $\sum_{i=0}^7 p_i = 1$

[soft zealot]

Ann

[tired walrus]

and that each p_i ≥ 0

[left trail]

Ok I'm following

[tired walrus]

$p_i \geq 0$ implies that

$p_1+p_2+2p_3+p_4+2p_5+2p_6+2p_7 \ \leq {\color{red}2p_0} + {\color{red}2p_1} + {\color{red}2p_2} + 2p_3 + {\color{red}2p_4} + 2p_5 + 2p_6 + 2p_7 = 2$

[soft zealot]

Ann

[tired walrus]

(messed-with terms highlighted)

[left trail]

I'm not seeing the implication

So each term is greater or equal to zero so the sum will also be

[tired walrus]

$p_1 \leq 2p_1$ and $p_2 \leq 2p_2$ and $p_4 \leq 2p_4$ and $0 \leq 2p_0$

[soft zealot]

Ann

[tired walrus]

each term is greater or equal to zero
...which means that making the coefficient on it bigger will also make the sum (nonstrictly) bigger!

[left trail]

Ok I see this for the individual term but when we make the comparison between our two thing I lose it

[tired walrus]

do you understand that inequalities can be added

i.e. do you understand that if a ≤ a' and b ≤ b' then a+b ≤ a'+b'

[left trail]

Yes

[tired walrus]

thats what im doing here

[left trail]

But we have 0≤ to our first thing and then 0≤ second thing = 2

[tired walrus]

0 ≤ 2p_0
p_1 ≤ 2p_1
p_2 ≤ 2p_2
p_4 ≤ 2p_4
-# 2p_3 + 2p_5 + 2p_6 + 2p_7 ≤ 2p_3 + 2p_5 + 2p_6 + 2p_7

........

what you just said made no sense at all

ok

you're getting hung up on the fact that both sides of my ineq are nonnegative and i make no mention of that fact? or what.

im trying to make this as followable as i possibly can

[left trail]

Let me clarify this. By first one I mean the blue underlined and by second one I mean the non underlined

Both are greater or equal to 0

[tired walrus]

yes

sure they are

so what

[left trail]

0 ≤The second is = 2

I don't see how we can add these two statements as you said to get the inequality we want

[tired walrus]

we arent adding THOSE statements.

[tired walrus]

i listed EXACTLY whats being added here

[left trail]

Ok I think I see for the terms that don't match you have them as inequalities and add them all up. Then for the terms that do match you add them to both sides of the inequality

[tired walrus]

basically yes.

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[left trail]

Thank you

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[sand schooner]

Help with 2b please

[sand schooner]

This is my working but I think I have an extra term compared to my tutors solutions

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@sand schooner Has your question been resolved?

[sand schooner]

<@&286206848099549185>

[jovial grotto]

The last step is wrong

the 2 terms are cancelling each other as one is positive and the other negative

so they both disappear

[sand schooner]

Which 2 terms ?

[jovial grotto]

You have not eliminated the second one

[jovial grotto]

The last 2

oh wait nvmd

theyre different

srry

[sand schooner]

Yes

[jovial grotto]

btw, which term is extra in your solution?

[sand schooner]

This is the solution so I think it's the last -pi/4n term

[jovial grotto]

In your 3rd last step, I think the application of FTC is wrong

the signs are wrong on the sin and cos terms on the second line

This would be correct if you added them, but you are subtracting the whole thing in the bracket

@sand schooner ?

[sand schooner]

Ftc?

[jovial grotto]

Fundamental THeorem of Calculus

integral a to b f(x) dx = F(b)-F(a)

Get it?

[sand schooner]

Which line sorry

[jovial grotto]

The one where you substitute the values x=pi/2 and x=pi

[sand schooner]

I used sin(-x)=-Sin(x) when substiting

[jovial grotto]

you didnt use it on the first sin term

Its positive, should be negative

and thats the term thats extra in the solution

@sand schooner ?

[sand schooner]

This term you mean?

[jovial grotto]

the second one underlined

[sand schooner]

I cancelled both the terms out that I underlined so instead I would get 2pi²/4n ? And then only subtract one of them so the term would still be there unless I made a mistake elsewhere too

[jovial grotto]

wait a minute

in your last step, how do you simplify the sins?

you do $-\pi^2/4n(sin(n\pi/2)+\pi^2/4n(sin(n\pi) = -\pi^2/4n(sin(n\pi/2)$, which is very incorrect

[soft zealot]

The Foolish Almond

[jovial grotto]

try to simplify again from the mistake step

[sand schooner]

I cancelled the sin (npi as it equals 0

[jovial grotto]

is n an integer?

[sand schooner]

Ok thanks I will keep looking at it

[jovial grotto]

If n is provided to be an integer, then you can cancel it

oh wait its fourier series nevermind

[sand schooner]

Yes

It's ok I will just skip it for now, thanks for your help

.close

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[wanton geode]

Oh god.

[vital crag]

don't open a help channel without a math question, thanks

.close

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[radiant carbon]

how do i do this

[radiant carbon]

this is like an ellipse instead of a circle

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[radiant carbon]

.close

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[tranquil pine]

Am I supposed to do this by letting z be a+bi for example, or are there different ways, like geometrically as the first equation of just a circle radius root 2 centre (1 + i) ?

[drowsy epoch]

radius sqrt2 yea

for the other one you can indeed let z be that and apply the def of |z|

[tranquil pine]

Ok thank you so much !

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[shrewd wren]

A surface of revolution is obtained by rotating the graph of a function 360 degrees about the y-axis. Find the area of the surface when the function is given by $$f(x) = x^2-1 \quad\text{and}\quad D_f = [\sqrt{2}, \sqrt{6}]$$

[soft zealot]

dream

[shrewd wren]

i did this

but, its doesnt work..

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[zealous storm]

shouldn't it be the derivative of invers(f)

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[obtuse river]

How do I do 2a 😭😭

[green ibex]

this is not what the curves will look like

the left one is correct but the right is not

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@obtuse river Has your question been resolved?

[obtuse river]

Fixed it

How do I find h(y) 😭😭

[rocky leaf]

think so , linearly method like trigonometric as well and use the condition reflexed on the equation .

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[timber plume]

how is x_34=13 and x_35=13

[proper dagger]

you have 17 copies of 11 and 15 copies of 12, so they total to 32 scores, as in the CF column

[timber plume]

yep

[proper dagger]

then, you know you have 13 copies of 13, so scores number 33-45 (inclusive) are all 13

that's how

[timber plume]

13, so scores number 33-45 (inclusive) are all 13

I dont understand this part

[proper dagger]

the first 32 scores are all 11s and 12s. agreed?

[timber plume]

agree

[proper dagger]

good. and the 13 scores after that would be all 13s. agreed?

[timber plume]

agree

[proper dagger]

so let's take a look at this example

or rather this way of looking at it

so score #32 is 12, and score #33 is 13. we know that the next 13 scores after 32 are score 13, so we expect scores #33 - #45 to be all 13s

and look what we have: score #34 and score #35 are well within the range of the score 13 group

[timber plume]

how do we expect scores #33 - #45 to be all 13s?

[proper dagger]

because it's stated there are 13 of them?

look at the frequency column for score 13

[timber plume]

score 13?

[proper dagger]

?

[timber plume]

theres no score 13

oh

[proper dagger]

then what's that circled number??

[timber plume]

i meant in this

[proper dagger]

then what are those 13s I wrote in the second column?

[timber plume]

is that a differnet example?

[proper dagger]

no

it's based off the same table

the first column is the current count or number of scores we've tracked so far, and the second is the actual score at that position

it's... not that complicated. (hopefully.)

<@&268886789983436800>

[timber plume]

ohh ok I get it now

ty

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[soft zealot]

Renato

[whole halo]

this is too scattered to be readable

theres also a mistake on the second image

[gentle zephyr]

where?

[whole halo]

nvm, the image is correct

[gentle zephyr]

a mistake?

[whole halo]

I misread the image
there is no mistake in the second image

[gentle zephyr]

well is a hard exercise ofc it's going to be messy

[whole halo]

renato you could afford to organize your work before you spill them all over

but yes this work is correct

[gentle zephyr]

wdym? how would you make the proof perfect?

[whole halo]

I didnt say to make it perfect

I said to make it readable, or approachable

[gentle zephyr]

i am all ears on how to improve my proofs to make them look more professional

[whole halo]

is this proof a rough sketch or your final work?

[gentle zephyr]

final

[whole halo]

really?

[gentle zephyr]

well idk what else can i add

[whole halo]

thats not correct

think of this like writing

you have completed the "content" part of writing

which is all the work you do in the actual content leading up to the first draft

you now need to "revise" your writing

which is to simplify and refine your draft until it is shorter and easier to read

for example, take a look at this line

nvm, you ultimately wanted to get to this line

[tired walrus]

wait isn't this the very same question we did yesterday

[whole halo]

this isnt your work is it, ann

[tired walrus]

no

the handwriting is not neat enough for it to have been mine

[whole halo]

and also pages 3 and 4 are very scattered

it went back on itself once, technically

[whole halo]

anyways, Im getting tripped up over your work here trying to show whats happening, I deleted an image where I screenshotted the wrong line

[whole halo]

now the easier approach would be to expand out both ∑ 2/(n+1+i) and ∑ 2/(n+i), then show what terms needed to be added or subtracted

[gentle zephyr]

fair enough yes

[whole halo]

you can see here the reason is because you didnt know where you were going at the time

you were still attempting to figure it out

that alone means this is a sketch, not the final proof

[pliant shore]

I'm not going to try and confuse Renato

[whole halo]

you can simplify the steps in your sketch to make it easier to read, or easier to follow

there are shorter ways to explain or show what you are doing, they may not line up with the initial ways you did things

[pliant shore]

but I'd do it by bounding the sum below by an integral, the integral of that from i = 1 to n + 1

and then it's not bad from there

turns out you just need to check n = 2, 3 individually

[tired walrus]

i suggested that yday

he screamed

[whole halo]

wouldnt you then have to bound the sum above too

[gentle zephyr]

this is for my algebra class i can't use anal machinery, walled garden etc @tired walrus

[whole halo]

or is that not necessary for the integral test

[pliant shore]

oh you suggested that lmao

[gentle zephyr]

yes

[whole halo]

we're not allowed the integral test either for this or the previous real analysis class

so bounding it with an integral, even as a rough approach, would mean we would need to prove those techniques first

[pliant shore]

knew it, thanks for clarifying matt

[whole halo]

those riemann sums keep popping up when I dont expect them

[pliant shore]

I mean the integral is a bound below cause that is a decreasing function (of i)

there's some nice picture representations of Riemann sums

[pliant shore]

(so then y = f(1) for 1 <= x < 2 is an upper bound, and likewise)

[whole halo]

I didnt see the riemann sum when I said this, I had the wrong idea

wait bounded below by the integral? this is a right riemann sum

[gentle zephyr]

fair enough

I'm still new with proofs, shit is hard

[pliant shore]

I mean, you can choose whether to use the left or the right Riemann sum

here we use the left

so the bounds are 1 and n + 1

[whole halo]

now youre phrasing it like a left riemann sum uses bounds 1 to n instead of from 0 to n - 1

you can see in the picture itd be 0 to n for our P(n+1) case

Im looking at this and the sum would be bounded above by the corresponding integral

[pliant shore]

with the integral approach you can avoid induction entirely

[whole halo]

I didnt say to use induction

[pliant shore]

shift each of those those rectangles 1 to the right

[whole halo]

you need to say that bro

and then additionally show the rectangles dont subtract enough area

[pliant shore]

fair enough, I was picturing it in my head and it seemed obvious to me

I did clarify it was a left Riemann sum later

[whole halo]

and you said to begin at 1 too?

[pliant shore]

yeah

[whole halo]

for a left riemann sum?

that breaks convention all by itself

[whole halo]

you at least need to say this before you just drop that on someone

its not even a clean example of a riemann sum

[frail moat]

Where is op

[gentle zephyr]

here

[whole halo]

do you see any areas to revise yet

[gentle zephyr]

say my name

[frail moat]

Renato

[whole halo]

youre goddamn right

[whole halo]

this is mostly a do-it-yourself kind of problem since usually you know how you phrase things better than anyone does

it still looks like induction is the cleanest

doing a riemann sum then doing a perpendicular hack to fix it just tastes disgusting

if you phrase it right, you can get the number of lines youre writing down

to be honest, I dont see much for revision given youre showing all the work required
you should definitely shorten pages 2, 3, 4 in first showing that this is true

when you see this line, place a ≥ before it

[gentle zephyr]

one time i tried using polynomial division and rrt on a analysis question and i got 0 points out of that iirc

[tired walrus]

that is such utter bullshit

[whole halo]

I mean you couldve just gotten it wrong, its entirely plausible

[gentle zephyr]

well

[tired walrus]

your profs are incompetent and overstrict

[whole halo]

I wouldve just figured I got it wrong, theres just no other way a professor has to force a proof out of rational root theorem

[gentle zephyr]

well it was preuni at the time but yeah

[whole halo]

what

preuni analysis question?

[gentle zephyr]

well "analysis", just finding the range/image of a single variable function

[whole halo]

oh also when youre solving this, simplify the left to be 1/(n+1)
then subtract 1/(n+1) - 2/(n+1)
shortens how much youre seeing

[gentle zephyr]

oh, nice catch

[whole halo]

heres how I wouldve written it:

but I still recommend you try coming up with this stuff yourself

[gentle zephyr]

ohh way better

i see

[whole halo]

youve clicked on the spoiler already

[gentle zephyr]

yes

i appreciate the help provided

[gentle zephyr]

prolly, but i like to think it's for the greater good of everyone, university is big, is not like we are a class of 20 people, anal class had like 700 or 800 first year students, if everyone does things differently it will be way harder to grade the exams, maybe i just did a mistake in the polynomial division + rrt, but don't think so

anyways i appreciate the help from everyone as always

[tired walrus]

"for everyone's good" ≠ "for the grader's convenience"

[gentle zephyr]

yeah maybe they are just lazy

i can certainly agree on dat

thanks for the help

.solved

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[smoky copper]

Hello, is this correct?

[smoky copper]

Can I cancel out the 2 I think it belongs to the identity

[tired walrus]

aside from the many many missing brackets, yes

[smoky copper]

Im in a hurry😭

[tired walrus]

and missing equals signs tbh

[tired walrus]

why are you doing math while hurried

[smoky copper]

Ill do it properly during the exam

[smoky copper]

I have my math exam tmrw

Im far behinf

[frail moat]

gl

[digital wave]

Relax

You'll do well

[smoky copper]

Thankyou so much🤍 I needed that

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[radiant carbon]

i divide both sides by cos(3t) to make tan(3t) = ...

so i lose solutions at cos(3t) = 0, right

[rocky tusk]

you can do that or divide both sides by 2 then use sin(u - v) = sin(u)cos(v) - sin(v)cos(u) and take u = 3theta and v = pi/3

[radiant carbon]

so why is there no solutions to cos(3t) = 0 in the list

only solutions to tan(3t) = sqrt(3)

[frail moat]

nvm my bad

[radiant carbon]

undefined

[vital crag]

you only lose solutions if they were solutions in the first place

[rocky tusk]

do way 3

so you don't have to think about losing solutions

[radiant carbon]

and how do i tell that before diving by cos(3t)

[rocky tusk]

🤔

[vital crag]

many ways. pythagorean theorem. solve for t in cos(3t) = 0 and plug it into the starting eqn.

[rocky tusk]

i mean its quite obvious that if cos is zero then sin cant also be zero since they have the same argument so it wont be a solution here

[ember peak]

anyone can help me with measure theory in easy way ?

[vital crag]

!occupied

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[radiant carbon]

solve for t in cos(3t) = 0 and plug it into the starting eqn.
this works

[radiant carbon]

where is sin = 0 here?

[rocky tusk]

did you read what i said

[radiant carbon]

i tried to

[frail moat]

btw you avoid losing soln this way

[rocky tusk]

^

[frail moat]

oo didnt read dat

[radiant carbon]

im aware that there are better ways to do this, but i'm asking this question for future reference where the equation cannot be rewritten using sin(u-v) for example, and where i have to divide by a trig ratio

[rocky tusk]

doubt it

if you see a combination of sin and cos like this its pretty much always possible to exploit angle addition/subtraction

just scale properly

[radiant carbon]

okay

thanks

[final saddle]

@radiant carbon Has your question been resolved?

[final saddle]

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[tranquil pine]

Let f(x) is differentiable functional and satisfy f(x+y)=f(x)f(y) for all x and y belonging to R and f(2)=9 then find f(x)

[tired walrus]

!status

[final saddle]

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

[tranquil pine]

2. I need to find a differiantial equation

[muted prairie]

I don't think you do

Well, maybe if you have to prove this from scratch

But I would just inspect it

[tranquil pine]

I think thats the approach, integrate the equation thats the approach for almost every question of this topic

[noble salmon]

f(x+y) = f(x)f(y) is a very interesting property for a function to have

[tranquil pine]

I haven't hoped onto the solution yet

[muted prairie]

There's a rather natural guess for such a function

I suppose informally you could set y=dx

[potent cypress]

It's a functional identity for f(x) = a^x where a in an arbitrary constant the question has data to find a(f(2)= 9)

[muted prairie]

how to formalize... ig take the limit as y -> 0?

[tranquil pine]

its in my class notes, my friend got me the notes, we need to partially differentiate both sides

[potent cypress]

So f(x) = 3^x

[potent cypress]

This this correct?

[tranquil pine]

yeahh

youre right

can you write the solution in detail please

[potent cypress]

Are u preparing for jee by can chance?

[tranquil pine]

yeah I am

[muted prairie]

!nosols

[final saddle]

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.

[potent cypress]

So in jee there are basically 5 or 6 functional identities which you have to remember in other to these questions

[tranquil pine]

isee, thats gotta be really helpful

[potent cypress]

Like here I used if f(x+y)=f(x)f(y) then f(x) = a^x

Question will have data to find a

[tranquil pine]

I didn't understand

[tranquil pine]

that's the question

[rocky tusk]

💀

you can use an easy induction argument to show f(2n) = 9^n

[potent cypress]

It's an identity if a function follows f(x+y)=f(x)f(y) then f(x) must be the type of a^x

[rocky tusk]

this is the question mate

[potent cypress]

Bro what are u talking about

[potent cypress]

The question is that a function follows this

[tranquil pine]

@potent cypress send me those identities in my dms

[potent cypress]

And we have to find f(x)

[rocky tusk]

it’s not very helpful to assert the conclusion of the thing we’re trying to arrive at

that accomplishes nothing

show those identities hold

that’s the point

[tranquil pine]

I didn't even understand the identity

whats the proof?

[potent cypress]

U can look it up

[tranquil pine]

how did you come up with a=3?

[potent cypress]

Coz the question says

[rocky tusk]

https://math.stackexchange.com/questions/1548249/a-function-with-a-property-fxy-fxfy

[potent cypress]

F(2)=9

[rocky tusk]

you can use my induction argument to prove the statement on the integers

[potent cypress]

And by the identity say f(x) is a^x

Then a is 3

[tranquil pine]

thanks bud

[rocky tusk]

they outline the proof for extending it to Q then you use continuity

bud is crazy

[tranquil pine]

ohhh got it got it

[muted prairie]

this probably isn't a proof question

[potent cypress]

And the proof isn't important trust me

[tranquil pine]

I was looking for an alternative rather solving by partial differentiation

[tranquil pine]

okay okay, ill note that definitely

[muted prairie]

if it were the y->0 way would probably be the way to go

[rocky tusk]

you should at least be able to get this

y = 0 gives f(x) = f(x)f(0) then since f(2) = 9 you can divide by f(x) to get f(0) = 1. from there you can use induction to show f(2n) = 9^n then n --> n/2

to get f(n) = 3^n

f(x + 2) = 9f(x)

[worldly mesa]

You need to extend this to the rationals

[rocky tusk]

yes i said this sir

[rocky tusk]

^

[worldly mesa]

sir

[rocky tusk]

^

[tranquil pine]

alright i got the hang of it thanks @potent cypress @rocky tusk

.close

[final saddle]

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[soft zealot]

Renato

[worldly mesa]

Induction of course

[final saddle]

@gentle zephyr Has your question been resolved?

[gentle zephyr]

where did i mess up?

[crimson bronze]

Seems like a good start. The inequality you found is not true for n = 1 or 2, but I think it's true for all n > 2. So one way to solve that is to just add another base case

[gentle zephyr]

what?

[crimson bronze]

what

do you have a question?

[gentle zephyr]

how?

[crimson bronze]

how what? I can't be bothered if you're just gonna write one-word sentences

[gentle zephyr]

i mean how is p(n) true for n in N but this quadratic in inequality i get be true for n > 3 onwards

TA is saying the exercise is not wrong but i have to figure out why

[gentle zephyr]

i appreciate the fix, i just want to know because it seems like i found a glitch in the induction matrix

[crimson bronze]

I'm not sure, but as far as I can see, you never actually prove that P(n) implies the inequality that only holds for n > 2, you just conclude that it's sufficient to prove this inequality

[gentle zephyr]

what does this mean in simple words, my English is bad

[crimson bronze]

like, you know that a > b, and your goal is to prove a > c. In order to do that, you try to prove b > c, which implies a > c by transitivity. But you never actually prove that a > b implies b > c

[gentle zephyr]

we assume > is a total order relation

[crimson bronze]

The thing is that you've kinda written your proof backwards - you've started with your goal, then reached an inequality you want to prove to reach that goal. Which is a good strategy, but it shouldn't be how you write the proof down for presentation

I know > is a total order relation, but my point is that you're doing things backwards - you haven't proved b > c, you have only proved that proving it is sufficient for proving a > c. I recommend writing your proof the correct way around, so it's clearer to understand

[gentle zephyr]

care to give an example and elaborate

im not sure i follow

[crimson bronze]

hmm, first of all, do you see that you've written "I want to prove:" then a series of inequalities? Did you start with something you know is true there, or something you we're trying to prove is true?

like, you know how a proof starts with something true, then deduces something else? For example, I can't say that 1 = 2 implies 0 * 1 = 0 * 2 implies 0 = 0 therefore 1 = 2 is true. I need to start with a truth

[wary nymph]

you can start with something you are trying to prove

if you chain the statements via equivalences

[sth i want to prove] iff sth1 iff sth2 iff … iff [sthn we know is true]

[terse dagger]

Why would you do this? it seems rather more difficult to prove bidirectionals the whole way

[crimson bronze]

I know, I'm just trying to explain that he hasn't actually proved a falsehood from a truth, because he didn't start with a truth

[wary nymph]

comes often handy

especially w inequalities

[terse dagger]

I'm not sure i see which inequality isn't true @crimson bronze

[crimson bronze]

the final inequality here is only true for n > 2, while the thing he's trying to prove should be true for all n. But as I mentioned, it's not a big deal, since you can just add a base case

[terse dagger]

This one doesn't hold for n = 1

i think

as the right side is 6

and in order for it to hold 15/x > 6

but that makes no sense with integers

natural numbers

2x

the only solution is not in natural numbers

isn't it

[crimson bronze]

where do you see an x? I agree, it doesn't hold for n = 1, the RHS is 6 and the LHS is 3.75 I think

[terse dagger]

n+1 = x

i'

i'm just saying the only solution is n = 0 if even that

just ignore it it's how my brain solves it

[crimson bronze]

hmm, wdym? I'm pretty sure the inequality is true for n > 2. I mean, we're not trying to solve it, we're just trying to prove it holds

[terse dagger]

just ignore it

i plugged in n = 1 for everything but the fraction

and then i see what the solution for n+1 would give > 6

but n = 1 is not valid

it would have to be n = 0

but therefore it doesn't work

[crimson bronze]

ah, I see. A lot of work to avoid plugging in some extra values 😅

[terse dagger]

human brain not even once

anyway @gentle zephyr your induction step doesn't work if your base case is n = 1

[gentle zephyr]

but we already proved P(1) holds

[terse dagger]

sure

but the implication P(n) ⇒ P(n+1) is not true

so no matter what P(n) we've proven

in this specific case

lemme restate

P(n) ⇒ P(n+1), n ≥ 1 is not true

and so you cannot use P(1) as your base case

as the implication is not true

which means your induction doesn't cascade

it doesn't go to P(2)

and then P(3)

etc.

So even though P(1) is true

your induction won't work with your current base case

I hope that makes sense

If however you calculate P(1), P(2), and P(3) and continue from there to prove P(n) ⇒ P(n + 1) it will work

[gentle zephyr]

why do we need to prove P(3)

[terse dagger]

input n = 3 into this

and it will be true

therefore for n = 3 P(n) ⇒ P(n + 1) is true

[final saddle]

@gentle zephyr Has your question been resolved?

[gentle zephyr]

.close

[final saddle]

Channel closed

Closed by @gentle zephyr

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[final saddle]

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[south pulsar]

can someone help me with this question

[south pulsar]

my plan is to show the cardinality of Im(phi) divides both m and n and because m and n are co prime and so the cardinality of Im(phi) is 1 and so phi is a trivial homorphism but I dont really know how to do it

or there is a better way of doing it

just any help is good

[plucky rover]

Have you done first isomorphism theorem

[south pulsar]

yes

G/ker(phi) is isomorphic with Im(phi)

I think

[static fractal]

you could also use that the two groups are cyclic

[plucky rover]

Yes

[plucky rover]

That's a deeper argument I'd say

[versed crater]

Oh that’s clever

[south pulsar]

how nwould I use that?

[plucky rover]

Now attach numbers to those two and see what happens

[versed crater]

Group isomorphisms preserve the order of elements

[plucky rover]

Homomorphism here

[south pulsar]

ohhh shittt cause pigeonhole principle

thats so smart

[plucky rover]

Homomorphisms don't preserve order though

[south pulsar]

now I just gotta figure out what Z/Zm/ker(phi) is

[versed crater]

I thought so

Something was off aboit that edit

[plucky rover]

They can decrease order

[south pulsar]

cause they arent bijective

[plucky rover]

The new order must divide the old order

[south pulsar]

but if I show Z/Zm/ker(phi) has an order of 1

[plucky rover]

Anyway this is getting too complicated

[south pulsar]

then I win

[plucky rover]

You don't even have to do that

[south pulsar]

do I not 🙁

[plucky rover]

Let's say the cardinality of the kernel is k

[south pulsar]

yep

[plucky rover]

And the cardinality of the image is p

[south pulsar]

yep

[plucky rover]

What does first isomorphism theorem give you

[south pulsar]

m/k=p?

[plucky rover]

Yes

Now p divides n

Cuz the image is a subgroup

[south pulsar]

ohhh ye cause the image is a subgroup of z/zn

[plucky rover]

Yes

[south pulsar]

damnnnn

[plucky rover]

Now you're done

[south pulsar]

thank you man

.close

[final saddle]

Channel closed

Closed by @south pulsar

Use .reopen if this was a mistake.

[plucky rover]

Do you see why, just to confirm

[south pulsar]

do I see why what?

[plucky rover]

Like do you see what happens now

.reopen

[final saddle]

✅ Original question: #help-36 message

[south pulsar]

ye im(phi) divides both m and n

[plucky rover]

Just wanted to confirm that you actually understood

[plucky rover]

Yes

[south pulsar]

and we know m and n are coprime

so the only thing that can divide both m and n is 1

so p=1

[plucky rover]

Yup

All right, thanks for the confirmation lol

[south pulsar]

and if p is 1 then that means all the elements in Z/Zm only get mapped to 1 thing by the homorphism

[plucky rover]

Feel free to close it again

[south pulsar]

so phi is a trivial homorphism

[plucky rover]

Ye

Nice

[south pulsar]

yeee

thank you bro

.close

[final saddle]

Channel closed

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[silk fjord]

$y^2=18x$

[soft zealot]

UCYT5040

[silk fjord]

What am I missing here? How can the 6 and 8 be used here?

We aren't really given angles so I don't see how the triangle sides are related

[latent dragon]

Why’d you just ghost ping mods?

[steep hatch]

[please don't delete mod pings even if they've been handled, it can be confusing for us]

[silk fjord]

Wait mb shouldn't have deleted it sorry

Someone sent spam

[half raptor]

do you know the tangent chord theorem?

[silk fjord]

Like this?

Have it in my calc

I don't see how this is applied here though, perhaps I'm thinking of the wrong thing

[half raptor]

Nah this one: the angle between a tangent line and a chord through the point of contact is equal to the angle in the alternate segment of the circle

[silk fjord]

Is angle ADC the first angle in that?

[half raptor]

wait i don't get what your asking

[silk fjord]

Is ADC the angle between a tangent line and a chord through the point of contact?

[half raptor]

oh yea

[silk fjord]

Then what's the other angle?

The alternate segment one

[half raptor]

DCB is congruent to ADC

[silk fjord]

How is that useful though?

I try to find the two triangles similar but i can't seem to

[half raptor]

Hmm idk

its just additional info you might need XD

[final saddle]

@silk fjord Has your question been resolved?

[half raptor]

hey I gotta go but I think I worked the problem out

the answer should be 81/8 i think

i know theres a no solution policy but I gotta go sleep so ye

[final saddle]

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[tranquil pine]

what do they mean by “standard definition “