help-17

[civic oracle]

it is

[tribal sapphire]

ye ur right

how come they dont give the same answer though

like sine60 isnt sin120

wait it is

just tried it on my calculator

google is messing with me

💀

[civic oracle]

check if your calculator is in the right mode

[earnest rampart]

google uses radians by default

[civic oracle]

you can also look at the sin graph and see that pi - x = sin(x)

put in y=sinx and see for yourself

maybe organic chemistry tutor has a video on this

[tribal sapphire]

alright

I think I got it now

thank you guys

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[deep bough]

Can someone check my answer?

[obsidian stream]

That looks fine, but I'd add a few more lines to explain why that restriction property holds

[deep bough]

can you explain a bit more
where would I want to add more information?

[obsidian stream]

Well, if you suppose v is in W, then you need to show that theta(v) = alpha(v)

[distant eagle]

You also should prove that θ is well defined

[deep bough]

if v is in W doesn't mean that theta(v) = alpha(v) by how we define theta?

I thought that's what the line "the restriction of which W is equal to alpha"

[obsidian stream]

Well, the way you defined theta depends on having an element in W and an element in W`

So you need to justify why that definition allows for that simplification

Namely that for every element in W, you can write it as the sum of an element in W and an element in W`

[distant eagle]

And also that θ of that element does not depend on the way you write it

[deep bough]

would I just need to say that since W' is a subspace, it contains 0v. Thus w+w' is in W since w' can be zero. So every element in W can be written as a sum of an element in W and in W'. And same argument for W'

[obsidian stream]

Basically yeah

theta(v+0) = theta(v) where v is in W and 0 is clearly in W'

[deep bough]

gotcha

as for showing that the function is well defined

would I do that with basis

?

[distant eagle]

I don't know it that would work, but I don't think it would be a simple way

You have to use the hypothesis about the intersection

[deep bough]

Yeah I've been confused on how I would use the intersection property that the question gave me
So I would use the property to show that it is well defined then? somehow

[distant eagle]

Exactly
You have to show that, if a vector v+v' is also w+w', then α(v)+β(v')=α(w)+β(w')

[deep bough]

wait
how will me proving that show that the function is well defined

[distant eagle]

Well, to show that the function is well defined you just have to prove that v=w implies θ(v)=θ(w), right?

[deep bough]

...

yeah

I'm sleepy man

XD

thank you

[distant eagle]

But this is not obvious given how θ was defined
Perhaps I can write the same vector as v+v' and w+w' with v and w different, and v' and w' different too

For example, if $V$ is $\mathbb{R}^3$, $W={x_3=0}$ and $W'={x_1=0}$, then the vector $(1,1,1)$ is $(1,1,0)+(0,0,1)$ but also $(1,0,0)+(0,1,1)$, and you would have to prove that $\alpha(1,1,0)+\beta(0,0,1)=\alpha(1,0,0)+\beta(0,1,1)$

Actually I wrote W and W' wrong

[twin meteor]

d

[deep bough]

👍
is there any hint you can give me of what I should be thinking of to better define the function theta than what I have?

[distant eagle]

The definition you gave is ok
What I meant is that you have to prove that it is a good definition, not to change it

[deep bough]

oh I see

[distant eagle]

And to prove it you have to show this

[distant eagle]

Like in this example

[distant eagle]

Hint: if $v+v'=w+w'$, then $v-w = w'-v'$. Let $x = v-w = w'-v'$. Now $x\in W\cap W'$

[twin meteor]

d

[deep bough]

okay I'll give it try
thank you for all your help
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[civic nimbus]

just use double angle or half angle formula for the denominator

[sly sierra]

or use the taylor series for cos

[civic nimbus]

$\cos 2x = 1 - 2\sin^{2}x$

[twin meteor]

Gowtham

[civic nimbus]

$2\sin^{2} \dfrac{x}{2} = 1 - \cos x$

[twin meteor]

Gowtham

[civic nimbus]

we get it to

then solve

@fluid solar

[fluid solar]

That would be better

x/(√2 sinx)

BUT IT WILL BE +-?

[civic nimbus]

seperate it

1/√2 and x/sin x

and apply limits seperately

[fluid solar]

Sinx/x is 1

So we get 1/√2 and -1/√2

[sly sierra]

for the left and right limits, respectively

[fluid solar]

By taylor series it will be 0/0?

We can see it directly no

Cos(x)=cos(-x) it will depend on numerator

[sly sierra]

a picture says it all:

[fluid solar]

We are not allowed to use any gadgets in the exams

So all i have to learn by pen and paper. Btw thanks for 3d

[vocal sleet]

@fluid solar Has your question been resolved?

[tribal moss]

$\text{Hint for you:}\\\frac{x}{\sqrt{1-\cos\text{}x}}=\frac{x}{\sqrt{1-\cos\text{}x}\cdot \sqrt{1+\cos\text{}x}}\cdot \sqrt{1+\cos\text{}x}=\\=\frac{x}{\sqrt{1-\cos^{2}x}}\cdot \sqrt{1+\cos\text{}x}=\frac{x}{\sqrt{\sin^{2}x}}\cdot \sqrt{1+\cos\text{}x}=\\=\frac{x}{\left| \sin\text{}x \right|}\cdot \sqrt{1+\cos\text{}x}$

[twin meteor]

Joanna Angel

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[desert hornet]

Given $a,b,c\in \mathbb{R}^+$ such that $16\sum a\ge \sum \frac1{a}$, prove that $\sum_{cyclic} \frac1{a+b+\sqrt{2(a+c)}} \le \frac{8}{9}$

[twin meteor]

kheerii

[desert hornet]

I have gotten the inequality down to proving that $\frac{\sum a}{(a+b)(b+c)(c+a)} \le 6$

[twin meteor]

kheerii

[desert hornet]

we have, $(a+b)+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}}\ge 3\sqrt[3]{\frac{(a+b)(a+c)}{2}}$, so the required sum $$\sum_{cyclic} \frac1{a+b+\sqrt{2(a+c)}}\le \frac{4(a+b+c)}{27(a+b)(b+c)(c+a)}$$

[twin meteor]

kheerii

[desert hornet]

oh nvm I got it.

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[river kettle]

The circle $x^{2}+4x+y^{2}-6y+12=0$ is transformed by a vertical translation 3 units down and a horizontal translation 5 units right. Find the equation of the transformed circle

[twin meteor]

water beam

[river kettle]

I dont understand I tried adding a "-5" to the x terms

hold on

wait nevermind

i just did it lol

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[river kettle]

.reopen

[vocal sleet]

✅

[river kettle]

nevermind i dont got it

i get this as the transformed

but how come the constant changes the size???

[paper depot]

eh hold up

that's not the right transformation

for an implicit equation you translate down not by just subtracting 3 from the lhs

but by replacing y with (y+3)

[river kettle]

oh?

so i leave the 12 alone?

[paper depot]

yes

[river kettle]

oh okay

$\left(x-5\right)^{2}+4\left(x-5\right)+\left(y+3\right)^{2}-6\left(y+3\right)+12=0$

[twin meteor]

water beam

[paper depot]

yeah that's about right

[river kettle]

looks like its good now i plotted it in desmos

thanks

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[icy spruce]

Can someone teach me on how to answer this trigonometry question. Thanks

[viral copper]

factorize it

[halcyon turtle]

take lcm

[icy spruce]

Huh, I'm sorry, what's lcm?

@viral copper and how do I factorise it?

[outer warren]

identify a common factor,
factor that out

[icy spruce]

Is it smtg like this?

[viral copper]

what is the common factor among the two terms

[icy spruce]

Ouh

Oww can someone help me .... I'm still confused

[lapis lodge]

factorize with cosx

cos x = 0 or 1/sinx + 3 = 0

[icy spruce]

Like this

[lapis lodge]

yeah

and just solve

[icy spruce]

Ouhh thanks! ❤️

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[icy spruce]

.reopen

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✅

[icy spruce]

I'm sorry, but I was wondering, why can't I do it this way?

[outer warren]

can't read you first step

can you write out that attempt more clearly

[icy spruce]

Ok

[lapis lodge]

because you created your own math there

how did you go to cot (x)^2

it is just cot(x)

[icy spruce]

Ouh yeah, just noticed that haha

[lapis lodge]

just factor you will be fine

[icy spruce]

Ok, thanks

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[chrome ridge]

Shouldn't the answer to (ii) and (iii) be false?

because it hasn't stated whether 90 > θ > 0

[outer warren]

yes, they are false

[chrome ridge]

😕

guess ill have to accept it

[outer warren]

is human going to be marking this

[chrome ridge]

yes

[desert hornet]

what are these questions bro

[chrome ridge]

Oh I could say "Assuming 90 > θ > 0, the answer is true."

[outer warren]

then ignore what the key says and justify your result

justify why its false, don't accommodate to problematic answer keys

[desert hornet]

i have gotten into way too many fights with my teacher about problematic answer keys

[outer warren]

or write a whole essay

[desert hornet]

we were taught log limits before we were taught log :/

[outer warren]

about when its true / when its false

[chrome ridge]

gotcha, thanks

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[earnest shoal]

Trig question which I wanted to use law of sine but I forgot how to use law of sine :/

Two snorkelers, Bob and Carl, are directly 20 m apart swimming on the surface of calm water. They both
spot a shark that is below them. The angle of depression from Bob to the shark is 55°. The angle of
depression from Carl to the shark is 40°. How far are each of the divers from the shark to the nearest tenth of
a metre?

just did the angle of the shark but I don't know what to do with the info.

[earnest shoal]

do I use basic functions first with the angle and the distance between B and C given to me?

[outer warren]

start with drawing a diagram

[earnest shoal]

sure let me draw one digitally

[outer warren]

do you know what the sine law states?

[earnest shoal]

length of opposite side over sin(angle)

ah ok wait

so basically I do

for both bob and carl

and I solve for the missing one

[outer warren]

yeh pretty much

[earnest shoal]

thanks!

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[junior plover]

nd help in this exer

[paper depot]

subtract 2pi/3 from everything

[junior plover]

this is my work

[junior plover]

cn u talk french?

[paper depot]

un peu

[junior plover]

ok mrc

[paper depot]

ça c'est la bonne solution, quel est ton problème ?

t'as divisé tout par pi puis soustrait 2/3 de tout

[junior plover]

la deuxieme question

[junior plover]

oui

[paper depot]

pourquoi n'as-tu pas dit au début que tu veux de l'aide avec le 2 et non avec le 1 ?

[junior plover]

srry

[paper depot]

je ne comprends pas ce que tu as fait pour le 2a)

as-tu essayé de faire une sorte d'induction ??

[junior plover]

j'ai essayer de la resoudre par soustraction

puis par reccurence

mais ca ne marche pas

[atomic cypress]

Das ist nicht facile

[paper depot]

la récurrence n'est pas nécessaire

P(n) se factorise en (n+3)(n+4) évidemmeent

[junior plover]

oui

oui

[paper depot]

(n+3)(n+3) < (n+3)(n+4) < (n+4)(n+4)

et c'est tout

sinon qu'on attend de toi une preuve en 17 pages que n+3 < n+4 pour tout n naturel ...

[junior plover]

oui mrc

[junior plover]

pardant je ne suis pas trop meileure au maths

[paper depot]

pardant ?

[junior plover]

mais j'essaie de comprendre

[paper depot]

que signifie "pardant" ?

[junior plover]

srry

[paper depot]

ah, c'était un "pardon" mal écrit ...

je me suis confuse

[junior plover]

i mean im noob in maths

[paper depot]

this is not a problem of math, this is a problem of orthography

i thought "pardant" was some kind of verb

[junior plover]

yh

[junior plover]

but im talking ab the exer

[paper depot]

qu'est-ce qui reste de l'exo ?

on a fini avec le 2a non ?

[junior plover]

2.b

[junior plover]

oui

[paper depot]

ok mais c'est pas trop difficile non ?

sqrt(P(n)) est entre n+3 et n+4 ...

comment peut-elle être naturelle ?

[junior plover]

idk

lets try afk some minutes

[paper depot]

je viens de te raconter presque la solution entière

[junior plover]

mais par absurde

on doit dabord supposer que racine P(n) app a N

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[obsidian bison]

Where I have done mistake?

[vast shale]

wait

oops

[obsidian bison]

On wolfram alpha it shows plus instead of minus in the second term in brackets

The solution says the integral coverages

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@obsidian bison Has your question been resolved?

[obsidian bison]

<@&286206848099549185>

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[restive salmon]

Hi! I don't understand how you go from one to the other. Here is the first function

[restive salmon]

and this is the second

So i understand for the beginning but not for "40/3v5r2"

[regal slate]

just algebra

[loud island]

In the second you substitute r with the 3rd root of 40/pi

Using the second row from the first picture

[tender tartan]

this was an interesting one, i believe it's this way:

[restive salmon]

Oh i see thanks you !

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[high rapids]

Uhh

[elfin surge]

x=y

[high rapids]

Oh

[elfin surge]

think why

[high rapids]

Then

[desert hornet]

!nosols

[vocal sleet]

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.

[high rapids]

Whats x

[elfin surge]

Ok post your q

I just assumed

[high rapids]

I have no idea how and why 💀

[high rapids]

What

[elfin surge]

What is your q?

[high rapids]

Solve for x and y

Oh shi

Y = 41?

Or nah

💀

Im dumb asf

So x = 41 too

82/2

Dam

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[lilac rain]

anyone know how to do part b?

[elfin surge]

name the points a,b,c,d

now just read

the q

recall part

[lilac rain]

do i need prove that point a,b is convex, a,c is convex and a,d is convex?

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[brazen isle]

what's the easiest way to do b)

i don't wanna just guess n check sadly

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[proud cave]

integrate 1 dV = integrate (g * sin theta)/(1 + (kappa ^ 2)/(R ^ 2)) dt

[proud cave]

Where K is the radius of gyration

[scenic ravine]

uh, are you sure it's dt and not dr?

[proud cave]

IT IS dt

It is the acceleration of a round object rolling down a inclined plane

[next ocean]

Then it's just...

[proud cave]

I know we can use kinematic equations to solve it

[next ocean]

(gsin(theta)/(1+K^2/R^2))*t

Because if you're integrating in terms of t and there's no t in there

That's all just one large constant

[proud cave]

No it isn't because the answer is

√2gh/(1 + k²/R²) using the kinematic equation

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[jade stone]

I need guidance

a person told me to get ln to ln((n+1))/n)

then use log(a/b)=log(a)-log(b)

in means to find b_n

where we use p-series to find all α for which the series are congruent

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@jade stone Has your question been resolved?

[jade stone]

<@&286206848099549185>

[vocal sleet]

@jade stone Has your question been resolved?

[jade stone]

<@&286206848099549185>

[jade stone]

<@&286206848099549185> please, it's been over an hour

[vast shale]

gimme 5 mins

[vast shale]

isnt it cauchy condensation

the special case of it

apply the role then youll get a condensed series

[jade stone]

might be, but for the previous two tasks, our lecturer used an/bn

for lim n->inf

if either is congruent, then both are

vice-versa

but if it works with cauchy, let's see

[vast shale]

just after getting the comdensed series youll find that a = a constant and use the ratio test

or u can do the root test too whatever u prefer

if u couldn't do it tell me

[jade stone]

well yeah

I'm burned out

sorry

[vast shale]

no it's fine i already did it but i cant do latex so wait until i type it

[jade stone]

ty, ping me

I'm sorry for taking from your time

I would've done it myself an hour earlier but I'm very tired

[vast shale]

this is The condensed ∑_(n=0)^∞ 2^n (ln(1+1/(2^n)))^(2^n), i used an online latex editor i hope it's typed right

[vast shale]

don't worry people for people

and when using the ratio test it shows that it's approaching a=1

so the original one shown to converge for a = 1 too

is it clear?

[vast shale]

.

[jade stone]

I gotta go sorry

tysm though I'll look through it later

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[surreal bone]

So I can't figure out what I'm doing wrong here in this problem.

[surreal bone]

I've done hanging mass problems a while ago in AP physics but this one popped up in my introduction to calc 3 and I'm failing to solve it

I also don't understand how the tension and magnitude are going to be different values

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[opal sleet]

<@&286206848099549185>

Мое решение

[vast shale]

send english translation

very few people would know russian here, or whatever language that is

[opal sleet]

Given the set of all five-digit natural numbers M that do not contain a single digit zero. Let the set A and B be the sets of all numbers from M that satisfy the specified conditions. A - the number contains exactly one digit seven and exactly one digit eight, B - the product of the digits of the number is divisible by ten. Find the intersection and union of sets A and B.

can you check my solution?

it's a matter of life and death

[tranquil trellis]

do you mean how many elements are in the intersection/union?

[inner osprey]

yes

you see the cardinality symbols in the original paper

[tranquil trellis]

good point

[opal sleet]

yes

элементы, которые обеспечивают данные условия

[tranquil trellis]

I don't think your answer for |A| is correct

[opal sleet]

elements that provide these conditions

[tranquil trellis]

your solution doesn't account for the fact that 7 and 8 can be any of the 5 digits in the number I think

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by woshiari

[tranquil trellis]

nvm that's incorrect

[inner osprey]

for B?

[tranquil trellis]

,wolf 910^4-89^4-(5^5-4^5)

[twin meteor]

Results provided by WolframAlpha

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

Requested by woshiari

[tranquil trellis]

That's my answer for |B|

[opal sleet]

The omega set is a five-digit natural number that does not contain zero.

the set of all five-digit natural numbers not containing equal to zero is 59049 in theory

[tranquil trellis]

wait

you're right

For B then you really just need to have at least one 5 as a digit and one even numbered digit

[opal sleet]

From the omega dimension I need to find A and B and their intersection and union

[inner osprey]

i'm getting a substantially different answer for B

[opal sleet]

The teacher wrote to me that my conditions did not take into account provisions 7 and 8.

[tranquil trellis]

Yeah as I said your answer for |A| is wrong because you didn't take into account that 7 and 8 can be any of the digits

not necessarily the 1st two in that order

[opal sleet]

I don't understand this, can you describe it?

[tranquil trellis]

Can I ask what you got

[tranquil trellis]

Your answer was 7^3 = 343 right?

[opal sleet]

yes

[tranquil trellis]

The answer is 20 times that

Because you have 5 choices for which digit is the 8, 4 choices for the 7, and then 7^3 for the remaining 3 digits

[inner osprey]

90000 - 8^5 - 2100 = 55132

[tranquil trellis]

Uh I got 9^5-8^5-5^5+4^5

9^5 - 8^5 is number of 5-digit natural numbers that contain no 0 and at least one 5

if I'm not mistaken

[inner osprey]

oops i used 90000 instead of 9^5

in that case our answers are 1 off 🤔

[tranquil trellis]

wait really

[inner osprey]

ah i know why

yes you're right then

[vocal sleet]

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[soft halo]

hi

[soft halo]

i need help with this q

im sending it now

<@&286206848099549185>

I need help quicly

since I have an

exam tmrw

It would be much appreciated

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[inner sonnet]

Three numbers a, b, and c form a geometric series such that a+b+c = 35 and abc = 1000. Determine a, b, and c.
Can someone help me with this?

[vocal sleet]

@inner sonnet Has your question been resolved?

[sage gazelle]

Two solutions I think..

@inner sonnet

So

I'm assuming a,b,c = a, ak and ak^2 respectively? @inner sonnet

in that case, we get:

a*ak*ak^2=1000
<-> a^3k^3=1000
<-> ak = 10
<-> b = 10.

Then

a+b+c=35
<-> a+ak+ak^2=35
<-> ak/k + ak + ak*k = 35
<-> 10/k + 10 + 10k = 35
<-> 10/k + 10k = 35
<-> 10 + 10k^2 = 35k
<-> k = 0.5 or 2

ah wait

Yeah there's only one solution i think

[heady ibex]

yes

[sage gazelle]

both those k values should yield the same 3 numbers but in different sort of order

[heady ibex]

i looked it up on wolfram

[sage gazelle]

yeah so look

(a,b,c) = 10/0.5, 10, 10*0.5 = 20, 10, 5

[heady ibex]

there's one solution if you don't care about which one a is or b or c

[sage gazelle]

but

exactly what I mean

(a,b,c) = 10/2, 10, 10*2 = 5, 10, 20

so it's the same. Not sure why I didn't realize that first since 0.5 = 1/2 lol

[inner sonnet]

oh

alr

thanks

[sage gazelle]

Do u understand how I got to the solution?

[inner sonnet]

yeah

[sage gazelle]

ok good 👍 I left out a bit of things here:

...
<-> 10 + 10k^2 = 35k
<-> k = 0.5 or 2

but u get what happens

[inner sonnet]

ty

/close

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[tardy temple]

if i want to get a position by time graph

and for that i would get the areas

what would i do in the negative part?

how would i calculate the areas?

[flat whale]

!msgdel

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[crimson sequoia]

I would like someone to try and explain what the cross ratio actually is

I looked into projective geometry

I'm a bit confused about the topic and I got no idea about how I can prove these

[glad python]

its a ratio that is invariant under projection

theres a nice handout lemme find it

https://alexanderrem.weebly.com/uploads/7/2/5/6/72566533/projectivegeometry.pdf

[crimson sequoia]

soo

there has to be a point O

that projected onto my points A,B,C,D

it results that (A,B,C,D) = (A'B'C'D')

but A = A' since it's an invariant point

[glad python]

Yeah

[crimson sequoia]

but

I'm dumb so bear with me

how is this proving that they are concurrent or parallel

visual representing it makes 101% sense

[glad python]

O is where they intersect

[crimson sequoia]

alright and in case they are parallel

doesn't that mean they never intersect?

[glad python]

O is the point at infinity

If they are parallel

[crimson sequoia]

and if they are concurrent it means that B',C',D' are sitting on top of B,C,D right?

[glad python]

What

[crimson sequoia]

wait nvm

it means they intersect in a point

my bad

[glad python]

Yeah

the best way to learn projective is through doing problems, there’s a lot of intuition to develop that’s hard to explain

[crimson sequoia]

so to be explained step by step

if two cross ratios are the same

then there is a point O

at which they intersect

or vice versa

[glad python]

They have to share a point

[crimson sequoia]

a point O that can project the points of a line onto another line

[crimson sequoia]

alright

and for something like this

[glad python]

It’s weird to learn with prism lemma first

[crimson sequoia]

a,b,c,d are concurrent therefore there is a point O1

[glad python]

I would recommend doing simpler problems first

Yeah

[crimson sequoia]

there are no simpler ones here xD

these are the simplest ones given by our professor

as homework

[glad python]

Is this for Olympiad geometry?

[crimson sequoia]

no

basic homework

[glad python]

Oh what class?

[crimson sequoia]

uhhh, wdym by "what class"

I'm not familiar with some terms in English

that's why I'm asking

[glad python]

Like what are you learning?

[crimson sequoia]

studying Computer Science

[glad python]

Oh nvm then, you can learn how you’d like, I assumed this for Olympiad

[crimson sequoia]

nope, first time I heard of this topic

so can you stay a bit with me while I try to explain this exercise

and you can correct me where I'm wrong

[glad python]

Yeah

[crimson sequoia]

if you've got the time and if you want to

thank you!

so there are two points O1,O2

let's say they're different for now

O1 intersects the first 4 points, a,b,c,d

and O2 intersects the other 4 points

a,b', c', d'

(a, b | c, d) = (a, b' | c', d') means that their cross ratios are equal

therefore they intersect in a point O3 (For now)

the intersection points of b and b'

should probably be the points drawn down from our point O3

right?

[glad python]

What intersects at o3?

[crimson sequoia]

a, b, c, d

a, b', c', d'

no

the lines drawn from these two lines towards O3

I figure there is a term I'm missing

right?

[glad python]

Im getting confused, do you have a drawing?

[crimson sequoia]

nope, let me make one

this is what I figured

isn't perfect but I hope it's alright

[glad python]

There’s no A’

A’ shoulf equal A

[crimson sequoia]

oh yeah true

it's invariant

so something like this?

[glad python]

Draw AD’ and the lines intersections

[crimson sequoia]

oh that

since I didn't name them accordinly

[crimson sequoia]

I'll just rename them like that

[glad python]

So what’s O1?

[crimson sequoia]

O1 should be the intersection point for A,B,C,D

right?

[glad python]

Wdym?

The lines?

[crimson sequoia]

uh actually

O1 (what I was thinking)

is that it was the point from where the projection came onto the points A,B,C,D

that's why I said the intersection points for A,B,C,D

in our case O1 is the only O written in the drawing

[glad python]

Yeah

[crimson sequoia]

O2 I was thinking

that is the intersection point

basically the same projection

that came onto the points

A, B", C", D'

now, since their cross ratios are the same

then the same O1 should be equal to O2

so we know B", C", D' are just the projections of A,B,C,D from our point O1

now I got no idea how to go about showing they are collinear

only if

there isn't another point

that if project upon A,B,C,D

we get the same A, B", C", D'

because (in my head)

if we find two such points, then their cross ratios must be equal

therefore collinear

I got no idea what I said so please correct my thoughts xDDD

or actually can we somehow show that the points A, B", C", D' intersect in the same O

as our initial A,B,C,D points

therefore (A,B | C,D) = (A, B" | C", D') => collinear

?

[vocal sleet]

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@crimson sequoia Has your question been resolved?

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[outer warren]

Hi if the derivative of an inverse is the reciprocal of the functions derivative
where are you getting that?

[tribal moss]

I was showing it to you, yesterday too 🙂

it's ok )

[outer warren]

show exactly what you're reading

what you typed is not an accurate description of that

you're conflating stuff with its inverse or something like that

read carefully

you're supposed to have
$$(f^{-1})'(x) = \frac{1}{f'(\blue{f^{-1}(x)})}$$
but you're describing
$$(f^{-1})'(x) = \frac{1}{f'(\red{x})}$$

[twin meteor]

ℝαμΩℕωⅤ

[outer warren]

how?

why do you think $\blue{f^{-1}(x)}$ is the same as $\red{x}$

[twin meteor]

ℝαμΩℕωⅤ

[outer warren]

what doesn't mean that a=b

wdym

yes

what?

[flat whale]

y = f(x), so f inverse of y = x

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

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[plain aurora]

by constant, do they mean integer a?

otherwise, why is $O(\sqrt n)$ not considered polynomial time?

[twin meteor]

Kalgar

[flat whale]

#old-network for CS server

[plain aurora]

this is from a number theory course

feelsbad

thanks doe

[flat whale]

whats EBO

[plain aurora]

elementary binary operations

i.e., one additiion of bits between two k-bit numbers

[silk comet]

well $O(\sqrt{n}) \subset O(n)$

[twin meteor]

Tushar

[silk comet]

so anything that is O(sqrt(n)) is trivially also polynomial

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[granite solar]

It's crazy hard because of all the variables, but I want to have the general formula or method to solve it, I found some limitations in respect to the graphs distance due to the limited segment length, basically, they can only be in a certain area

[granite solar]

I tried to look for the relation between pairs of equations with constant slope, but I still don't understand how they relate to each other when they are curves because they change every increment and consequently all their relationship.

[vocal sleet]

@granite solar Has your question been resolved?

[granite solar]

No

<@&286206848099549185>

I know it has a lot of possibilities but that's the point

I just wanna know how to kinda search for the theme or something that relates to that,

[vocal sleet]

@granite solar Has your question been resolved?

[granite solar]

No

[vocal sleet]

@granite solar Has your question been resolved?

[granite solar]

Nope

Can someone just tell me if they have any idea of what branch of the maths study that ?

[idle temple]

this might be one of those mess around and find out moments, though I don't have too much outside of that (sorry)

[granite solar]

Oh, thanks you anyways

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[frail breach]

does x=3 or x=-2/3 make sense? thanks

[vocal sleet]

Please don't occupy multiple help channels.

[frail breach]

<@&286206848099549185>

.close

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[orchid pier]

how do i solve this easy with 4 uknowns?

[orchid pier]

.close

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[final bridge]

Prove by contradiction, if x is irrational then 7 + x is irrational

[final bridge]

i said assume x is rational then 7 + x is irrational

is this the right way to start?

[ancient knoll]

no

[final bridge]

fuck

what is the right way to start

[paper depot]

to prove "if A then B" by contradiction, you start with "A and not B"

[final bridge]

so assume x is irrational such that 7 + x is rational

@paper depot

[desert hornet]

correct

[final bridge]

.clos

.close

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[fluid swallow]

kx+4=3x-3r/2r-5 make r the subject

[fluid swallow]

pwease welp

[paper depot]

$kx+4=3x-\frac{3r}{2r} -5$

is this your equation?

[twin meteor]

Ann

[fluid swallow]

no

[paper depot]

then write it properly?

[fluid swallow]

3x and 5 r also in the fraction

[paper depot]

also show progress

[fluid swallow]

idk how to

[paper depot]

(3x-3r)/(2r-5)

[fluid swallow]

(kx + 4 = \frac{3x - 3r}{2r - 5})

[twin meteor]

Mohaimin

[fluid swallow]

this

[fluid swallow]

yesir

[sudden matrix]

multiply both sides by (2r-5)

and expand the brackets on the LHS

[fluid swallow]

lemme try

[fluid swallow]

how do I multiply 2r-5 with kx+4

[sudden matrix]

do you know foil?

did your teacher teach you both foil and factoring?

[fluid swallow]

yes

[sudden matrix]

So use foil to expand the brackets

[fluid swallow]

im juss dumb

yes dumb me

[paper depot]

please do not call me "sir"

[fluid swallow]

the answer was close but I didnt get it

in book the ans is
( r = \frac{5kx + 2x + 20}{2kx + 11} )

[twin meteor]

Mohaimin

[sudden matrix]

Let me try work it out step by step

[fluid swallow]

tyy!

[sudden matrix]

$$(2r-5)(kx+4) = 3x - 3r$$
Expansion $\rightarrow$
$$2rkx + 8r - 5kx - 20 = 3x - 3r$$
Now note that we have to isolate all the terms containing r so we can factor:
$$2rkx + 8r + 3r = 20 + 3x + 5kx$$
$$r(2kx + 8 + 3) = 20 + 3x + 5kx$$
Now we isolate for r
$$r = \frac{20+3x + 5kx}{2kx+11}$$

[twin meteor]

0_0

[sudden matrix]

you mean 3x right?

@fluid swallow

[vocal sleet]

@fluid swallow Has your question been resolved?

[vocal sleet]

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[fluid swallow]

no sadly

[fluid swallow]

.reopen

[fluid swallow]

I like got the same answer

but it was 3x not 2x

well anyways ill juss ask my teacher

btw if any1 has a d2 he can cross check

it was d2 workbook pg 61 Q26 part g

omg I wrote the question wrong

im soooo sorry

its (kx + 4 = \frac{2x - 3r}{2r - 5})

(kx + 4 = \frac{2x - 3r}{2r - 5})

[twin meteor]

Mohaimin

[fluid swallow]

well bro thank u sooo much I got the answer right using ur method

tysmmm

means a lot

.close

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[fast warren]

part b) since it stated hence i need to use the ans from part a) right?

[outer warren]

yes

[silk comet]

yes

[fast warren]

part a) answer is (2x+1)(4x²+1)+6

how do i integrate this

[paper depot]

you integrate that divided by 4x^2+1

so 2x + 1 + 6/(4x^2+1)

[fast warren]

huh

since 8x³+4x²+2x+7 / (4x²+1) = (2x+1)(4x²+1)+6

[paper depot]

no

bad notation

[vast shale]

dividend = quotient*divisor + remainder

you should remember this

[paper depot]

missing parentheses.

and also, it is 8x^3+4x^2+2x+7 itself that is equal to (2x+1)(4x^2+1)+6.

[fast warren]

oh ya

man i have trouble remembering names

dividend quotient divisor

i only know remainder 💀

[vast shale]

3/2 the dividend is 3 and the divisor is 2

[left jay]

dividend is the number being divided by the divisor

[fast warren]

and quotient?

[left jay]

the result

[fast warren]

right

[left jay]

have you come up with the solution for part a)?

[fast warren]

yes solved it

thx u guys

[left jay]

did you do b)?

[vast shale]

represent your answer using this

[fast warren]

yea someone helped

ann

[left jay]

:/

ok

[fast warren]

.close

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[sinful bay]

for part a), the quotient is 2x +1

and remiander is 6

[vocal sleet]

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[thin root]

guys if I got 3 unique vectors in a set, is it enough to say they are not linearly independent because they are not scalar multiples of each other

[thin root]

or is that only for set with 2 vectors?

[paper depot]

only for sets of two vectors.

[thin root]

ok thanks

[paper depot]

otherwise you can have something like
{(1,1), (1,-1), (-2,0)}

no two of which are parallel

but their sum is zero

[thin root]

okay

are all symmetrical matrices always diagonalizable?

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[terse fiber]

An equilateral triangle has height H and sides S.

The point P lies inside the triangle. We drop the normals A, B and C from P to each of the sides of the triangle.

Use the information, show that H = A + B + C.

How can this be done?

[elfin surge]

@terse fiber

[terse fiber]

Hey

[elfin surge]

Look you will be able to see three triangles

addition of their areas will be equal

to the area of the big triangle

if the side of the square is 'a' inches

try to equate em

[terse fiber]

These 3 correct?

[elfin surge]

no

That would be the half

of total area of the big triangle

[terse fiber]

Hm i have no idea to be honest

Because the lines of A,B,C should be as long as the middle line H right?

[elfin surge]

You have to think

Which three triangles

when their respective areas

are added

is equal to area of big triangle

Coz this is easy

you have to think

@terse fiber

you there?/

[terse fiber]

Yeah im just thinking

I'd say A,B,C

if we take the full triangles of all 3

[elfin surge]

I didnt get that

could you draw it

like you did?

[terse fiber]

maybe a wild guess but like so?

[elfin surge]

Nice

You have height and base for each triangle

so add their areas up

and its equal to the area of big triangle

[terse fiber]

Oh i see yes

So

We can show that H = A,B,C by adding those up

[elfin surge]

Yes

do it

[terse fiber]

But might be me, but with H do they just mean the line H or is H also the total area of the triangle

[elfin surge]

H=a+b+c

[elfin surge]

H

[terse fiber]

Ah so the middle line

[elfin surge]

yeah

[terse fiber]

So the next step is to count all the areas up?

[elfin surge]

yeah

add em up

and they are equal

to area of big triangle

[terse fiber]

Alright, so I don't have any numbers that I could use to find the area so I assume you'd have to measure them?

[elfin surge]

no

[elfin surge]

for the blue triangle area will be 1/2 * s * a

formula used

1/2 * base * height

[terse fiber]

ah alright, we use area = s*b^2

[elfin surge]

b^2?

whyy

[terse fiber]

i mean

no my bad

s*b/2

[elfin surge]

yes

add them all up

[terse fiber]

So

A = s x a/2
B = s x b/2
C = s x c/2

x = *

[elfin surge]

A+B+C = ABC (BIG TRIANGLE AREA)

[terse fiber]

So that's the end? That's how we can explain that H = A + B + C

[elfin surge]

BRO

form the equation

[terse fiber]

A+B+C / 2

[elfin surge]

huh?

https://en.wikipedia.org/wiki/Viviani's_theorem

Give this a read

I gtg

I thought you could do this

[terse fiber]

alright, thanks for the help

I'm new to this lol

[elfin surge]

ITs fine

no problem

[terse fiber]

.close

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[zealous surge]

can you explain ratio for me

[zealous surge]

https://discordapp.com/channels/268882317391429632/903463353417072641/1196498788152508468

[foggy cloak]

a ratio is a way to represent a difference in the amount of something.

for example if there are 5 bananas and 6 apples, then we can say there is a ratio of 5 bananas to 6 apples.

sometimes this will be written with a colon between them as in 5:6 or as the two being divided like 5/6

you can simplify some ratios if you can divide both the top and bottom by the same number.

[twin meteor]

river

[foggy cloak]

.close

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@zealous surge Has your question been resolved?

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[finite hatch]

hi

[finite hatch]

can someone explain how to do this pls?

so basically

i think that i means the x intercept

and j means the y intercept

so i just did this

oh...

hi

yes

oh i see

ye

subtract to get AB

3i-5j-(i+2j)

3i-5j-1-2j

= 2i -7j

so to draw diagram

i can draw it in any direction i want

ok! thank u

is it me or is vectors reallt hard

really

ive been working on them for 3 hours

over 3 days

):

[vast shale]

ig you could represent this in a plane to make it easier?

[finite hatch]

did i just draw mt diagram wrong

i thought i stood for

0i

...

but i completely forgot it always means 1

[finite hatch]

how come u drew 2 triangles?

[vast shale]

the triangle with the dotted lines is just to show the hright and width of the other triangle

[finite hatch]

excuse me

[vast shale]

ig representing vectors as arrows on a cartesian plane is more intuitive.

hm?

[finite hatch]

for this question

would BP also be 5i?

[vast shale]

i think it would be -5i if we are going to the left.

[finite hatch]

ah ok thanks

vectors is finally getting easier :D

[vast shale]

glad i could help!

[finite hatch]

.close

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[strong olive]

how can i find the value of E

[paper depot]

x**^2** - 5x - 1 = 0?

[strong olive]

yes

[paper depot]

are you sure?