help-33

[vital yew]

so 64 * 49 no?

[main idol]

you counted the pair twice. your "corresponding square" also counts the first square choice

[vital yew]

which I don't understand

help me in visualization

I mean when I choose 1 mango and 1 apple from 2 apples and 2 mangoes , i just do 2 * 2 right ?

why would I not divide by 2 here

I mean what is the difference b/w these 2 quest.

[oak oxide]

let say you have 2 choose 2 square e4 and h5

• When you pick e4 first then you have 49 ways to choose the next square which h5 is one of them
• When you choose h5 first then you have 49 ways to choose the next one and e4 is one of them
  In both case you're choosing e4 and h5, just different in order

[vital yew]

Ok........

[vital yew]

so why is this any different/

[oak oxide]

And you count both case

As when you do 64, you have 64 cases for every square

[oak oxide]

And for every 2 cases there is a over count case which is what i said here

[vital yew]

Well my point is why is this quest different from the example I gave

I mean I get u r right

it's just if I look superficially it doesn't seem any different (The questions I mean)

[oak oxide]

Oh mb, letmme read those

[oak oxide]

These right?

[vital yew]

ya ya

[oak oxide]

Well the difference is in the board problem you choose every single square ( 64 squares) but here you only do 2 right?

[vital yew]

Elaborate

[twin pike]

diffrent colors ??

[oak oxide]

Let say, toi choose in order
e4-h5 and h5-e4 are 2 cases in your board problem

[twin pike]

are the boxes of diffrent colors

like one black on white

[vital yew]

yes standard chess board looks like that

[vital yew]

yea

and that problem I don't see in mango apple case

[oak oxide]

But we can't choose A1-M2 and M2-A1 cuz using this argument you are counting A1 or M1 as your first steo

[twin pike]

no bro like do the boxes that we have to choose are of the same color any color or diffrent color

[vital yew]

Well nothing about color is mentioned

I copied an pasted

understand wha tyou want

[vital yew]

Yes

I got it

but as someone

[oak oxide]

So in the apple-mango you don't have over count

[vital yew]

So becoz the the squares are the same thing so we over count?

[oak oxide]

If you choose one of the fruit ( don't matter mango or apple) first

We will have 4•2

[vital yew]

yea I def don't understand

Ok......

ok this is tough to understand

[oak oxide]

Yeah you see
First step: e4
Second: h5
And in your argument we also have the second case
First step: h5
Second step: e4

But in the mango- apple

We dom't have these

[vital yew]

yea

but my fundamental question is why

[oak oxide]

First step: A1
Second: M1

That is it

2•2 here right?

[vital yew]

yea

[oak oxide]

The first 2 mean 2 case for Apple which mean choosing apple have to be the first step

And the second 2 is for choosing mango

By this we're avoid overcount

By removing the case:
First step:M1
Second step: A1

[oak oxide]

That cause over count for this cases

[vital yew]

what......

[vital yew]

explain this

I understood upto that

[oak oxide]

Basically, mango can't be choose as the first step like I said

[vital yew]

Why not.......

[oak oxide]

So we won't have this case

[oak oxide]

.

2(cases for choosing apple)•2 (cases for choosing mango)

[vital yew]

Okay

I think I understand

thnks

.close

[marsh citrus]

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[marsh citrus]

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[fickle shell]

Just so that I understand, the way to solve this:

First I check with my calculator if [1, 4] are within the domain between g(5) and g(7)

Then I solve y = (x + 1)/(-2x + 1) for x

Then I swap those x's for y's

And then sub that new fraction into g(x) in the original equation

And solve and that is my proof?

[knotty trellis]

can u show what exactly u did?

i dont follow ur procedure

[fickle shell]

I just check g(5) and see if its less than 1 and then check if g(7) is greater than 4, if yes its in the domain

In my head I think about if theres any edge cases

If not then I know its surjective

And hi Chartbit ❤️

Why is because it needs to be within its domain to be surjective no?

Why sad nodding kitty hehe

Theres a better way?

[glass silo]

Cause you need to be a bit careful

[knotty trellis]

what if it wasnt continuous between x=5 and x=7? What if there was some sudden jump?

[fickle shell]

That would be an edge case I guess 😅

What way would you guys do it instead

[glass silo]

You also could have the function such that the endpoints aren't "as you stated", but it still happens to be surjective (e.g. if I gave you f : [0, 2pi] -> [-1, 1], f(x) = sin(x), testing the endpoints wouldn't give you enough information so see it's surjective!)

[fickle shell]

I just wanna do something quick cause its not gonna be in my proof anyways

But if yall have a better method id love to know

[knotty trellis]

the general method would be solving for x in terms of y, and then checking if there are solutions for y in [1,4] with x in [5,7]

your method is fine, as long as u really know what you're doing

the reason why your method works is that the function is continuous (the only "discontinuity" would be at x = 8, but thats out of the domain) and so you can apply IVT. g(5) = 1, g(7) = 4 and so by intermediate value thm, for each y in [1, 4], there is x in [5,7] with g(x) = y

but as me and chart pointed out, it doesnt work if:

1. the function isnt continous (IVT doesnt hold in this case)
2. the function grows / falls beyond the values at endpoints such as the sin(x) on [0, 2pi] example (in this case, the range would be larger than you'd expect just by applying IVT)

[fickle shell]

So you get it in terms of x, then replace the y's with x's, then sub in [1, 4] into those x's?

[knotty trellis]

roughly

the slight issue is that you cant sub in every number from that interval

and that replace y's with x's step just makes it confusing

there is no reason to do that

just "sub [1,4]" into y directly

the point is that if you have some y in [5,7] then choosing $x=-\frac{9}{2y+1}+8$ should make g(x) = y, proving y is in the range

[elfin berry]

MathIsAlwaysRight

[knotty trellis]

so now you only have to prove that the x is within the specified domain

and thats just some inequality bashing

[fickle shell]

Ahhh ok I get it

I will try that for this next one

Thank you so much for the info!!

❤️

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[marsh citrus]

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[raw whale]

Can someone help me understand why these vector graphs match the way they do? I get why Equation 1 lines up with graph D but none of the others

[elfin cairn]

The best way to do this distiguishing is to find features of each function and understand how these features reflect on the graph

[raw whale]

okay, so for equation 2, for example

What distinguishing features could I extract from that?

I fear my problem may be that I'm trying to look at it as F = <f(x), f(y)>, and trying to view it as equations of x and y

[elfin cairn]

F(x,y) = (f(x,y),g(x,y))

[raw whale]

Okay, that's a little confusing for me, but I'll try my best

do you think you could explain what the movement of (2x+2, y) would look like?

[elfin cairn]

well in this case its just increasing linearly with x and y

a good example of features is just values at points

plug in x=0,y=0 and see what matches

or really any pair of value for x,y

In this case if x=y=0 then the result is (2,0)

So we are looking for a vector at (0,0) pointing right

[raw whale]

So then equation 2 would be C?

[elfin cairn]

Yes

[raw whale]

So we're plugging in coordinates on the graph, and they tell us where they're pointing

[elfin cairn]

F(x,y)=(a,b) means that at (x,y) there is the vector (a,b)

Usually drawn as an arrow starting at (x,y) with some direction and magnitude

[raw whale]

so then is (a,b) the magnitude?

[elfin cairn]

No

Its the vector at (x,y)

[elfin cairn]

This is only how its drawn

And usually when drawing graphs like this they scale the magnitude down by a lot

So its readable

[raw whale]

Okay, so plugging in points is the way to go

So for equation 4, plugging in (2,2) for x and y gives us (4,0) so the vector at (2,2) has to point toward the right?

[marsh citrus]

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[rigid trellis]

I'm trying to obtain https://arxiv.org/pdf/1804.03275 Eq. (4.69) starting from the Eq. (5.3.1) here: https://arxiv.org/pdf/1510.04430 . The definition of psi is exaclty the same definition as the definition of Z_V in the first paper, and x and z in both notations coincide. It's then just a problem of eliminating the first derivative term in the ODE of the second paper. However, doing that we are doing a rescaling in psi so we lose our original definition and both definitions (Z_V and psi) would not match. The other option is doing a change of variable in x, but again we are changing the definition so it's not recommended. Then, we just can impose the first derivative coeficcient to be 0 in D? Isn't this a very concrete case?
And how can be psi solution of both equations, one with first derivative term and other without?

[marsh citrus]

@rigid trellis Has your question been resolved?

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@rigid trellis Has your question been resolved?

[marsh citrus]

@rigid trellis Has your question been resolved?

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@rigid trellis Has your question been resolved?

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[solemn nymph]

Hi for c the lim should be 1 correct?

[golden frost]

is there a general

[tardy meteor]

not quite

[solemn nymph]

if it was f(-1) it would be 1 then im assuming?

[tardy meteor]

it is asking you when limit ask approach -1 what will the y approach

[solemn nymph]

got it mixed up alr

[tardy meteor]

not asking you what is f(-1)

[solemn nymph]

alright thank you 🙏

[tardy meteor]

so it is 2.

[solemn nymph]

yep

d should be und

e*

[tardy meteor]

yeah

[solemn nymph]

what about d?

[tardy meteor]

DNE

[solemn nymph]

because there isnt a -4+ connection?

is that why a and b are 2

[tardy meteor]

cuz the limit from the both side aren’t equal

[solemn nymph]

okay i get it

thanks lots

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[low scaffold]

just a question, how do we know what variable we need to differentiate in terms of (question c)

[vital oracle]

look at the question and see which variable appears the least constant

[novel juniper]

hint: Think about what population change is normally measured agaisnt

[low scaffold]

against time

[novel juniper]

yes

[low scaffold]

least constant?

[vital oracle]

dw, thats confusing to read, go with what wai is saying

[low scaffold]

and also is my answer correct, I wrote this as my final answer but in the answer key, they wrote "92.4 insects per week"

[indigo nest]

Rates of changes can be fractional

[low scaffold]

they never mentioned increasing or decreasing, because previously I was advised to conclude it

But I am assuming theres more than one answers

[odd agate]

hold up. I hate being taken out of context.

what is your current problem?

[low scaffold]

this

[odd agate]

what's wrong with 92.4 insects per week and what does that have to do with what I said?

[low scaffold]

in the answer key, they never mentioned increasing or decreasing

thats what Im confused about

[odd agate]

the answer key doesn't need to mention that, because the question asks for the rate of increase anyway

of course, you can mention it if you want to, and I know you mark your own work, so maybe consider letting others mark your work to check out different interpretations

[low scaffold]

alr got it, my other question is do you think my working out is correct?

it feels a bit odd that I changed the A into Q_0 on the RHS

[stoic saddle]

this work looks mighty messy but i also dont know how to suggest any sort of fix without simply starting out with a different form of the exponential decay equation.

[low scaffold]

the answer is correct tho

nvm I get it

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[wicked furnace]

little bit of chem but mostly math

am confused on how to set this up

i believe ur meant to use p1v1=p2v2

@vernal fractal

[amber birch]

you mean c1 * v1 = c2 * v2, but yes

you can assume that v1 = 3 and v2 = 3 + 12 here

then yes, I also get 2.522 this way

[marsh citrus]

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[faint grove]

What is the length of AP when line AC is 2 * sqrt(21) long and a diameter of circle,line BC is 10 long, line AB is 8 long and point P meets with circle and line BC?

[odd agate]

does the original question have a diagram? if it does, show it please.

[amber birch]

do you see how there's a right angle?

[faint grove]

Probably the angle APC

[lean pebble]

Do you know how to prove that though

[amber birch]

yes, that's correct

[faint grove]

Hold on I measured it with a protractor and it wasn't right?

[amber birch]

so here's another idea: let BP = x such that PC = 10 - x

[amber birch]

diagram isn't to scale

[lean pebble]

The diagram is not to scale I assume but it is very clear APC is a right angle

[amber birch]

it's not drawn accurately

[lean pebble]

One thing

If this is supposed to be a basic geometry problem

There are multiple ways to solve it

[amber birch]

it's geometry combined with algebra

[lean pebble]

However one uses trig

[amber birch]

you don't need trig

[lean pebble]

Am a little new to helping, do we ask whether to use

[lean pebble]

Yeah but it does make it significantly easier

[amber birch]

did you forget ||Pythagoras||?

[lean pebble]

Lmao I guess I did

[oak oxide]

well it depends, you can't tell 9th grade to use calculus

[faint grove]

Does it create sqrt(64 - x * x) == AP, AP == sqrt(84 - (x * x - 20 * x + 100))?

[amber birch]

sometimes there are 9th grades who ask about calculus, that's the other thing

[oak oxide]

so it's best to ask what tools they are allowed

[lean pebble]

Thanks this was the answer I needed

[amber birch]

yep! that's correct

[lean pebble]

Well I’m taking multivariable rn and it is terrible

Having that experience in 10th grade would fry me

[faint grove]

When it is sqrt(64 - x * x) == sqrt(84 - (x * x - 20 * x + 100))
Would 64 - x * x == 84 - (x * x - 20 * x + 100) also hold true?

[amber birch]

yep!

you can square both sides given those are equal

[faint grove]

I think I solved it

[marsh citrus]

@faint grove Has your question been resolved?

[amber birch]

축하해요!

.close

[marsh citrus]

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[amber birch]

you should get AP = sqrt(48) = 4 sqrt(3) if you want to check

[marsh citrus]

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[jagged compass]

Hello, I require help with Q3 a.iv which seems to be elemental but my brain isn’t braining, I get VX to be (4b-a)/2 which is obviously wrong as VX is parallel to a and must be in terms of a. The answer key is also attached, but I do not understand the reasoning behind it. Any help is appreciated, thank you!

[twin pike]

just extend P and T to form a triangle then use simmilarity conditions to find VX

[jagged compass]

I’m sorry but I can’t visualize the triangle 😭 is that triangle PTX?

[twin pike]

extend sp on the side of p

and extend ut on the side of t

[jagged compass]

Thanks alot!!

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[sweet sparrow]

given the following quartic equation where none of the coefficients are 0 and the equation has 4 distinct roots. prove 3 (a1)^2 > 8 a2

[sweet sparrow]

vieta does work but the problem asks us to solve this using Cauchy's theorem

[stoic saddle]

cauchy's thm?

[sweet sparrow]

yes

not the inequality

[stoic saddle]

uhhhh can i have a refresher

[sweet sparrow]

Cauchy's theorem mentioned in the book is this one

[stoic saddle]

oh THAT cauchy's theorem

uhhhhhhhhhhh

ok right so you've got your shit factored as $(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ ig

[sweet sparrow]

I don't really have an idea on how to approach this using that theorem

[elfin berry]

Ann

[stoic saddle]

now the real question is what should we take as our f and g

[twin pike]

there's more than one 💀

actually one's an inequality one's a theorum so nvm

[marsh citrus]

@sweet sparrow Has your question been resolved?

[marsh citrus]

@sweet sparrow Has your question been resolved?

[sweet sparrow]

solved it

god that took so long

[lethal lodge]

i had a good idea

i was just writing it down 🙁

[sweet sparrow]

just take f(x) = x^3 . (a1)^2 and g(x) = 4x^2 . a2

let the range be [1; +inf]

[lethal lodge]

i didn't see it like that

the inequality holds

oh you said without viete's

that's so boring lol

[sweet sparrow]

from this, divide both sides by c so the left hand side > 3/2 . (a1)^2/(4a2)

wait no

[lethal lodge]

but you would need to proove it with induction

you maybe just got lucky

[sweet sparrow]

ugh thought i had it

i screwed up hard

[sweet sparrow]

looks like gpt

[marsh citrus]

@sweet sparrow Has your question been resolved?

[queen sun]

.solved

[marsh citrus]

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[sweet sparrow]

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[sweet sparrow]

i havent solved it, i made a mistake

[queen sun]

Oh sorry

[marsh citrus]

@sweet sparrow Has your question been resolved?

[inland frigate]

$f(x)$

[elfin berry]

NotYourAverage_3CL1P53

[marsh citrus]

@sweet sparrow Has your question been resolved?

[sweet sparrow]

<@&286206848099549185>

[cunning fiber]

I think the strat is to do this where a,b are roots

This gives you ||P’ and P’’ have three and two real roots respectively||

[sweet sparrow]

alr I'll try later

really my main problem is how do I crank out (a1)^2

[sweet sparrow]

it worked lol

thanks

absolutely disgusting problem

.solved

[marsh citrus]

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[hollow apex]

Does P(a or b) include P(a and b)

[odd agate]

yes, unless the "or" is exclusive.

[lime moth]

what’s 1+1

[odd agate]

!occupied, and please don't troll in help channels.

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hollow apex]

Hmm

Ok thanks

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[buoyant jetty]

need help with ii)

[brazen wraith]

Try to get a particular solutions first

That's easy to get it

[buoyant jetty]

care to elaborate?

[brazen wraith]

For example
a+b=0
(-1,1) ; (0,0) ... Are particular solutions

For the equation a+b=0

[buoyant jetty]

but my equation is
20a + 16b = 36

@brazen wraith

[brazen wraith]

I give you just an example for understand

For 20a+16b=36
We can get (1,1) is a solution
Right ?

[buoyant jetty]

i think I need to use bezouts lemma

[buoyant jetty]

what about it?

[buoyant jetty]

you here @brazen wraith ?

[brazen wraith]

Pgcd(20,16)=4
So we need to transform the equation to
5a+4b=9

[brazen wraith]

I mean that we can use the diophanten formula

[buoyant jetty]

care to elaborate?

[brazen wraith]

We can also use Bezout identity but he take long time

[buoyant jetty]

potato tomato same thing

you here @brazen wraith ?

[brazen wraith]

ax+by=c
If (a_0 ; b_0 ) particular solution
And pgcd(x,y)=1
a=a_0+ky with k is an integer

And b=b_0 +kx

[buoyant jetty]

what if the gcd is not 1

this is a particular scenario

in our case the gcd is not even 1 dawg

@brazen wraith

you here?

[glossy flint]

!noping

[marsh citrus]

Please do not ping individual helpers unprompted.

[buoyant jetty]

gcd(X,Y) is not 1 for this case dawg

[glossy flint]

It's very annoying telling this again and again

[buoyant jetty]

gcd(20,16) = gcd(4,16) = gcd (4,8) = gcd(4,4) = gcd(4,0) = 4

[buoyant jetty]

well he was explaining his way is better but now he left

and the diophantine formula he gave was not applicable for this case that the gcd is not 1

[glossy flint]

@buoyant jetty

[buoyant jetty]

or we can use bezouts, but surely

4 and 5 are coprime and we can use the formula he gave

but... idk if its solving the same equation?

[buoyant jetty]

also im not sure if this trick will apply to iv)

[buoyant jetty]

since 11 is prime, we would need to divide 1555 by 11 and it doesnt divide evenly

you know what I mean?

whatever

forget it

i will figure out myself or open a channel later if I am still stuck

i need to learn more about diophantines because this is the easiest exercise on the homework

so I am already behind on the coursework

.close

[marsh citrus]

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[marsh citrus]

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[humble osprey]

can you not manipulate it so that gcd=1?

[humble osprey]

shit sorry

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[sweet sparrow]

given x_i, y_i ∈ (a;b). x_i > y_i and i ∈ {1, 2, 3, ..., n}. Prove that if f is differentiable on (a;b) then there exists c ∈ (a;b) so that the following is true

[sweet sparrow]

I have solved this beforehand but ended up having to use Darboux's theorem. Wonder if there's any simpler method.

[marsh citrus]

@sweet sparrow Has your question been resolved?

[marsh citrus]

@sweet sparrow Has your question been resolved?

[marsh citrus]

@sweet sparrow Has your question been resolved?

[sweet sparrow]

<@&286206848099549185>

[sturdy carbon]

Is this something like numerical analysis?

not seen Darboux's theorem before, but looks like it's exactly what you'd be expected to use since it's clearly the property that's needed.

any other proof likely just reconstructs something like the proof of Darboux's theorem

[sweet sparrow]

??????

[sturdy carbon]

If you solved it using Darboux's theorem, that'll almost certainly be the expected method (if you've seen Darboux's theorem in the course). Since the property being shown in this question is so similar to the statement of Darboux, you'll likely need to reconstruct most of the Darboux's theorem proof to be able to prove it without using it.

So I doubt there's any simpler methods

[sweet sparrow]

nope, darboux isnt taught in the course

[sturdy carbon]

huh, what do you know Darboux from then?

[sweet sparrow]

i just found it while trying to prove a special case of it

[sturdy carbon]

ah

in that case, they're probably just looking for you to do that proof yourself. presumably you've looked through the proof and can understand it?

[sweet sparrow]

kinda

[sturdy carbon]

well, I wouldn't worry about finding a simpler method, since the proofs of darboux look sufficiently simple that they're probably what the question is looking for you to understand

and it's clearly the property that's important for that question

[sweet sparrow]

This is the special case in an earlier part of the book

I used Rolle's and IVT to did it iirc

[sturdy carbon]

is this labelled in a way you're expected to be able to make use of? if so, you can probably just rescale something on this sort of interval

btw, what language is this?

[sweet sparrow]

vietnamese

[sturdy carbon]

yay I can recognise vietnamese :). Anyway, I was mostly just responding quickly since you seemed to have been stuck for a while. I don't think there's really much to say for this beyond just "understand this proof and how the proofs relate to each other"

[sweet sparrow]

oh well

i might go ask my professor on this later

[sturdy carbon]

that seems reasonable. I doubt they'll have much more to say about it either though.

[sweet sparrow]

neither the book nor the lecture slides mention about Darboux

so yeah

[sturdy carbon]

if you can understand a proof of Darboux, then you should be able to reconstruct it yourself and trying to figure out exactly what part of it is needed might be a good exercise

[sweet sparrow]

alrighty

thank you for the help

.close

[marsh citrus]

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[unkempt herald]

did i make calc error

,rotate

[unkempt herald]

or do i have to find the value of x

.close

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[marsh citrus]

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[onyx geyser]

Do i do both examples 2 and 3 correctly?

[onyx geyser]

Wait i just re read the questions and the examples were poorly placed

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[onyx geyser]

Did i do example 6 correctly? Something tells me i didnt

[rancid geode]

firstly

why do u think (a) is undefined; i believe its incorrect

(b) is incorrect in the substitution of p(4)

[onyx geyser]

For a theres no -1 in the table

[rancid geode]

are u finding p(-1)?

[onyx geyser]

No

[rancid geode]

what input for p are using in h(-1)

[onyx geyser]

I dont know

[rancid geode]

(c) is also incorrect. subsitution also

ur set ups are correct

[onyx geyser]

I dont know the values for h(x)

[rancid geode]

ok

lets go slowly

do similar thing u do with (b), (c) with (a)

try setting up slowly

what do u notice

[onyx geyser]

Substitution?

[rancid geode]

like

for (b), 3p(4) - 1 = 3(3) - 1

but u did 3(4)

substituting p(4) with 4 not 3

using the table

[onyx geyser]

But x=4?

Ohhh

Im actually slow

You substituted p(x)

not x

[rancid geode]

yes

cuz p(4) = 3

[onyx geyser]

Yes

[rancid geode]

so what is 3p(4) - 1

[onyx geyser]

I got 3p(2)(3))-1

[rancid geode]

i think there is some confusion

[onyx geyser]

Maybw

[rancid geode]

p(4) means the value of p where the input is 4

its like

one value

its not p multiplied by 4

[onyx geyser]

Oh

So how would i write it if h((x)=3p(2x)-1

Do i substitute for p

Instead of x

Im just confused, the lesson my teacher sent us didnt cover this 😞

[rancid geode]

thats fine

alright

so u have this expression, right

h(x) = 3p(2x) - 1

[onyx geyser]

Yes

[rancid geode]

now

they are asking u to substitute x with some number

[onyx geyser]

Yes

[rancid geode]

alright

lets do (b) as an example

[onyx geyser]

Ok

[rancid geode]

so

[onyx geyser]

H(2)

[rancid geode]

yup

h(2) = 3p(2(2)) - 1

now notice

p(2(2)) is just one number

[onyx geyser]

Oh

[rancid geode]

it is equal to p(4)

so

the expression reduces to h(2) = 3p(4) - 1

like u did

now

do u know p(4)

[onyx geyser]

But p(4) isnt on the table?

Is it undefined?

[rancid geode]

p(4) is p(x) when x = 4, right?

[onyx geyser]

oh

Yeah p(4)=3

[rancid geode]

wonderful

so

knowing that p(4) = 3

[brittle solstice]

Is this power set

[rancid geode]

we can substituting this back into h(2) = 3p(4) - 1

thus

h(2) = 3p(4) - 1 = 3(3) - 1

[rancid geode]

absolutely no

[brittle solstice]

Oh alr

[onyx geyser]

So h(2(=8

What about the 3 that was infront of p earlier

Oh wait nvm

[rancid geode]

u got it?

[onyx geyser]

We‘ve already mutiplied

Yeah ill try a

[rancid geode]

could u elaborate

[onyx geyser]

3(3)-1

The 3 infront of (3)

After getting p(x)

[rancid geode]

its the same 3 as the 3 in front of p(4)

[onyx geyser]

Yes

[rancid geode]

alright

u're not confused, right?

[onyx geyser]

No

[rancid geode]

nice

alright

can u walk me through (c)

[onyx geyser]

I just finished a

Sorry

I got

[cinder bear]

guys can u help in my maths

[onyx geyser]

h(-1)=2

[rancid geode]

another channel plz

[rancid geode]

nice

👍

[onyx geyser]

h(-1)=3p((2)(-1)-1
=3p(-2)-1
=3(-1)-1

Which us 2

[rancid geode]

p(-2) is not -1

its 1

but 2 is correct

[onyx geyser]

Typo

Mb

[rancid geode]

be careful with the brackets

[rancid geode]

h(-1)=3p((2)(-1))-1
=3p(-2)-1
=3(1)-1
=2

but ur answer and idea are correct

nicely done

[onyx geyser]

Yeah i accidently put double on 2

[rancid geode]

👍

[onyx geyser]

For c i got h(0)=5

[rancid geode]

hm

[rancid geode]

u sure?

[onyx geyser]

h(0)=3p(2)(0)-1
=3p(0)-1
=3(2)-1
=6-1
=5

?

Oh wait

p(0) is -1

Not 2

[rancid geode]

yup

u got the mistake

[onyx geyser]

I get confused with the top graph and bottom one

The x and y

[rancid geode]

ye

understandable

[onyx geyser]

But h(0)=-4

Thanks

You have a good day

.close

[marsh citrus]

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[waxen marten]

@maiden edge

[maiden edge]

damn u really did

where are u stuck in it

[waxen marten]

I can’t even start

If the whole thing is a^2)/2

Then the smallest is (a/4^2)/2

But that’s all I got

[maiden edge]

what do we have to find

?

[waxen marten]

What percent of the whole area is yellow

[waxen marten]

Did I accidentally ask for the answer

[maiden edge]

no 😭

[waxen marten]

U don’t wanna do it

[maiden edge]

meaning?

wait gimme a min

[waxen marten]

It’s too easy and waste ur time

[maiden edge]

ur not wasting my time

i was joking

i really don't get a lot of basic things in starting too

[maiden edge]

well is anything else given wirh the question?

can u post the whole question for clarity

[waxen marten]

It’s in Swedish but I think it says that both of the legs are the same length

It’s right angle

Two of the triangles sides are divided in four equal parts

[maiden edge]

start by naming the sides

and writing what is given

[waxen marten]

a and a

[maiden edge]

i meant A B C

like the vertices

[waxen marten]

What

[maiden edge]

wait

[waxen marten]

Sorry

I solved it

Sorry for time waste

[maiden edge]

im crying i was making the diagram

[waxen marten]

You divided the triangle into 4 different triangles

[maiden edge]

lmao i feel more good that u solved it yourself

keep it upp

[waxen marten]

Then you take the third area - the second

[maiden edge]

yes

[waxen marten]

Thank you bye

[maiden edge]

the area is just base x height x 0.5

[waxen marten]

Wait

[maiden edge]

hm?

[waxen marten]

The top numbers

It’s a on the first one

The denominator is 4

It’s the legs of each of the smaller ones

[maiden edge]

what is ur approach

can u explain in steps

but i think only a variable is used

[waxen marten]

Well the leg is divided into 4 equal parts

[maiden edge]

yes

[waxen marten]

So the leg of the smallest one is a/4

so on and so forth until we get to the forth with is just a

a being the whole leg

So now I take the area of the third triangle and negative the second triangle

That just leaves the yellow part

Then I divide by the whole thing to get percentage

[maiden edge]

thats a good approach

[waxen marten]

I hope

[maiden edge]

does ur answer match

[waxen marten]

I will check give me 3 mins I will ping you if it doesn’t

[maiden edge]

fosho

[waxen marten]

Ok I was wrong but I was close to being right

Do you want the answer or can you try to tell me where it went wrong

[maiden edge]

whats ur answer

[waxen marten]

I got the answer 5/32

[maiden edge]

?

shouldn't it get it in terms of a or is value of a given

[waxen marten]

It became 5a^2 divided by 32a^2

[maiden edge]

should your work lemme see where u might have missed it

[waxen marten]

Which it just 5/32

[waxen marten]

Is messy but ye

Tell me where you need explanation

[maiden edge]

i got 5/16?

5a^2/16

[waxen marten]

Yes I don’t know where it went wrong

5/32 is close to 5/16

5/16 is correct

[maiden edge]

2a^2m

??

[waxen marten]

Where

[maiden edge]

i think that is wrong

[waxen marten]

I did not write that

[maiden edge]

in the area of smaller traingles

u did

(2a)^2/4

the side is a/2

[waxen marten]

Huh

I did ehh

I’m lost

If I hadn’t taken the denominator times two i would’ve gotten the right answer

But don’t I need to do that

Keep flip change

[maiden edge]

try it again taking the side of smaller triangle as a/2

[waxen marten]

Sorry which one

The smallest or the half point

Oh yeah hold on

[maiden edge]

we are only taking 2 triangles right

[waxen marten]

Basically yes

But wait

The third triangle

The denominator becomes 32 every time

My leg of the third triangle is 3a/4

[maiden edge]

ur actually correct wait

oh

bruh

they asked percentage

[waxen marten]

It doesn’t matter

5/16 is the percentage

I just don’t know how to get to 5/16

[maiden edge]

lmao

[maiden edge]

why did u divide by a^2?

[maiden edge]

i forgot about the percentage tf

thats why i got confused

[waxen marten]

Uh

The wait

[maiden edge]

the whole area is something else right

[waxen marten]

I need to translate to English give me 20 sec

[maiden edge]

r we missing something

[waxen marten]

The proportion is the part divided by the whole.

That’s why I divide by

Wait

[maiden edge]

is the whole area a^2?

[waxen marten]

No

Haha wait but I don’t think that solves it

Hold on

[maiden edge]

it does

what is rhe whole area

[waxen marten]

A^2 divided by two

[maiden edge]

divide it by 5/32

[waxen marten]

Haha

I solved it

I just missed the crucial part that the whole area is a^2)/2 not a^2

[waxen marten]

It’s correct right

The image I sent now I solved it

Thanks for help

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[marsh citrus]

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[green gull]

hello, can someone give me a hint on this? If a,b are positive and real and ab>=1 then prove (a+2b+2/(a+1))(b+2a+2/(b+1))>=16

[oak oxide]

$\left( a+2b + \frac{2}{a+1} \right) \cdot \left(b+2a +\frac{2}{b+1} \right) \geq 16$

This?

[green gull]

no, ((a+2b)+(2/(a+1))

[green gull]

and the second one also like that

[elfin berry]

Fionna The Unemployed

[green gull]

yes

[oak oxide]

Wait $ab \geq 1$?

[elfin berry]

Fionna The Unemployed

[green gull]

yes

[analog elm]

have u tried am-gm?

[green gull]

yes, i tried adding 1 and subtracting it in each bracket to group and use am gm to get rid of the denominators, but i didnt get anything from it.

i think we should use am gm at some point in the solution, but i dont know where

.close

[marsh citrus]

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[keen saddle]

is there an easy way to solve this?

[keen saddle]

i know that it can be solved using chain rule and product rule, but there has to be some other way. when i try to solve it, i end up creating a huge equation and i make mistakes

[quick kindle]

simplify the log

[red nimbus]

Have you tried applying some log rules?

[keen saddle]

what do you mean?

do you mean the d/dx(ln(x)) = 1/x?

[quick kindle]

ln(ab) = ln(a)+ln(b)

[keen saddle]

ohh

[quick kindle]

ln(a^n) = nln(a)

[keen saddle]

lemme try, i forgot about that rule

[red nimbus]

-# If n was an even number then you would use absolute values

[keen saddle]

$\frac{d}{dx}\left(\ln\left(\frac{x\left(x^{2}+1\right)^{6}}{\left(2x-1\right)^{\frac{1}{3}}}\right)\right)!=\frac{d}{dx}\left(\ln\left(x\left(x^{2}+1\right)^{6}\right)+\ln\left(\left(2x-1\right)^{\frac{-1}{3}}\right)\right)$

[elfin berry]

a pretty dope cat

[keen saddle]

@quick kindle do you know why both of these are not equal to each other?

[proud ice]

$\ne$

[elfin berry]

SWR

[keen saddle]

wait, i think i was wrong, i think they are equal to each other

but when i plugged it into desmos, the position half of the equation matched up, but the negative half did not

what did i do wrong here?

[vital oracle]

try placing absolute value bars around x and around 2x - 1

keep in mind in the original expression that if x and 2x - 1 were both negative, their negatives would cancel out and so its the ln of a positive number
however when its expanded, this no longer happens
you get around this by placing absolute value bars around all of them so that positive/negative is not a concern

[marsh citrus]

@keen saddle Has your question been resolved?

[winged delta]

correct, would be better with absolute value bars yeah

[marsh citrus]

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[fossil pebble]

Ello

[fossil pebble]

No question in particular, I kinda just wanna know if there is a way to prove that d/dx a^x = a^x(ln(a))

like can you prove it with chain rule?

[cunning fiber]

https://proofwiki.org/wiki/Derivative_of_General_Exponential_Function

[shadow warren]

yes

try and rewrite a^x in terms of e raised to some power

[fossil pebble]

perfect thank you

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[loud stratus]

infinite series (n*ln(n))/(n+1)^3 converges or diverges?

[loud stratus]

i know it must be comparison test

[past thicket]

Starting at 1 im assuming?

[loud stratus]

yes

i tried limit comparison test and got infinity, but that doesn't tell me anything does it

with 1/n^2

[vagrant raft]

maybe try with another :)

[past thicket]

1/𝑛² does convege.

[vagrant raft]

but it's not enough for the comparison test

[past thicket]

$\sum_{n=1}^{\infty}\frac{n\log(n)}{(n+1)^3}$ converges or diverges?

[elfin berry]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[past thicket]

writing in latex for simplicity

[loud stratus]

thanks

i thought since we had n in the numerator and n^3 in denominator u could compare it to n/n^3 = 1/n^2

that ln n is tripping me up though

[blazing pulsar]

You would need to compare it with something like ln(n)/n^2 to get the ln(n)'s to cancel using that strategy.

[vagrant raft]

can you think of another series that converges, but decays slower than 1/n^2?

[loud stratus]

ln n/n^2?

[vagrant raft]

if you know it converges, sure

[loud stratus]

yeah not sure how to show it

[vagrant raft]

then you should pick another one

keep in mind that ln n grows slower than n to any power

(positive)

[loud stratus]

do i need to show that?

[vagrant raft]

it's useful to know that yeah

[loud stratus]

is it sqrt(n)/ n^2?

which converges cuz p > 1

not sure how that would help me though

[vagrant raft]

that works yeah

[loud stratus]

lemme try

i got that it converges

but i'm not sure if i did it right

i compared the original series to ln n/n^2 and then to sqrt(n)/n^2

[vagrant raft]

Yeah that's good

[loud stratus]

alright thank u

.close

[marsh citrus]

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[marsh citrus]

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[wary bluff]

any idea what im doing wrong here

[wary bluff]

Im trying to create the same buffer from the left image

but its not working

when i have pmos on top and nmos on bottom it turns into an inverter

[marsh citrus]

@wary bluff Has your question been resolved?

[wary bluff]

im alsop confused by this or gate circuit

and the one they provided in that image doesnt work

[jovial otter]

this is a math server....

[wary bluff]

ye ik

but ive seen ppl ask these type of questions here before

and this is prolly the most active server im in for help

[tropic cloak]

might get more help in the electrical engineering server in #old-network ?

[marsh citrus]

@wary bluff Has your question been resolved?

[wary bluff]

alr

.close

[marsh citrus]

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[keen saddle]

i dont understand

[keen saddle]

where do i have to put the bars?

[desert dirge]

the first x and the 2x-1 in the third fraction

since they have to be positive in the initial log

[keen saddle]

you mean like this?

[desert dirge]

sure

[keen saddle]

but the graphs still don't match up

it could be something with how i came up with the answer probably? do these steps look right to you?

[desert dirge]

ln(x)+6ln(x^2+1)-1/3 ln(2x-1)
1/|x|+12x/(x^2+1)-(2/3)/|2x-1|

ah i see

we need to restrict the domain i think

give me a second

x>1/2

yeah its fine

desmos is just plotting things it shouldnt be

[keen saddle]

ohh okay

[keen saddle]

hmm, it still says its wrong

sorry for being super annoying haha

[desert dirge]

hm, even though i know its fine ill let the wolf check

[past thicket]

Ohhhh wolfram!!! thousand pound beast walks in

[desert dirge]

Im putting them in a time out since they combined fractions

[desert dirge]

honestly, what you put here should be completely fine

im unsure why its acting up

with t of course

[keen saddle]

hold on, lemme double check that answer

[desert dirge]

maybe it doesnt like how youre dealing with the coefficients

[paper raptor]

yeah I was gonna say maybe its having a freak out over 6*1/(t^2+1)2t

[keen saddle]

yeah its wrong. also can't submit any more solutions 😭

so i guess we never know ...

[desert dirge]

i know its just squished in the box (I hope) but that hurts my eyes, and my soul for that matter

[keen saddle]

yeah mine too ..

[paper raptor]

well, that time you wrote your answer in terms of x

[keen saddle]

anyways, i was doing another probleme that i can't figure out

[keen saddle]

oh fuck!!

😭

youre right

[desert dirge]

honestly drifted right by me

you can use other variables than x in desmos by the way and itll still graph

might maintain more clarity

[keen saddle]

but when i try to use variables like t in desmos, it shows an exclamation mark and asks me to add a slider

[desert dirge]

put y=...
or any other letter

if its an equation of two variables itll graph it no matter what they are

[keen saddle]

oh youre right

nvm then

[marsh citrus]

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[mortal panther]

need help

[topaz narwhal]

do you have an attack strategy here?

[mortal panther]

yes

pi x r x 60/360

[topaz narwhal]

good. you're almost there with this.

can you express the area of the shaded region in terms of the area of the small/big sectors?

[mortal panther]

no i cant ngl

[topaz narwhal]

then please consider Moosey's diagram.

[mortal panther]

so should i do

3.142 x 12 x 60/360

[topaz narwhal]

that only gives you the area of the smaller sector.

(or in Moosey's diagram, the red sector.)

also, plesae ensure you're allowed to take pi as 3.142.

[frail orbit]

shouldn't it be r^2?

[mortal panther]

idk

[topaz narwhal]

oh, right. I didn't catch that, my apologies.

[frail orbit]

the area of a circle is pi*r^2

[topaz narwhal]

do you know the formula for the area of a circle? the formula for the area of a sector is just a scaled/proportionate version of it.

[frail orbit]

so the area of a sector would be ((angle in degrees)/360) * pi*r^2

[frail orbit]

so what you have here, but r^2 instead

[frail orbit]

now notice the green circle, what is the radius of the green circle portion/sector

[frail orbit]

i would not round pi until the end

[odd agate]

message to OP

[mortal panther]

yes

i am

[topaz narwhal]

in that case, please consider this question next.

[mortal panther]

4

[topaz narwhal]

incorrect.

[mortal panther]

16

[topaz narwhal]

do note that the green sector is the entire diagram, so it must be way longer than 4.

now, what is the radius of the red sector?

[mortal panther]

12

[topaz narwhal]

now, do you agree that if we take away the red sector from the green sector, we will be left with the shaded region?

[mortal panther]

yes

[topaz narwhal]

can you now continue on your own?

[mortal panther]

so its

3.141 x 4 x 4 x 60/360

[topaz narwhal]

where is this 4 x 4 coming from, and why has pi suddenly become 3.141 instead of 3.142?

there should be a subtraction somewhere in your working.

[mortal panther]

well its r^2

so i did 4x4

[topaz narwhal]

but neither of the two radii we have is 4.

we have a sector of radius 12, and another of radius 16. we cannot just take the difference in their radii alone.

[mortal panther]

bruh

[topaz narwhal]

you agreed with this, meaning I presume you should be able to deduce that the area of the shaded region = area of green - area of red.

do exactly that, and don't subtract the two radii. (you'll need to use the sector area formula twice, once for the green and once more for the red sector.)

[mortal panther]

bro i hate this topic

ok

i got

58.6

mm

[topaz narwhal]

,w (pi * 1/6) * (16^2 - 12^2)

[elfin berry]

Results provided by WolframAlpha

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Requested by alwayshyalping

[topaz narwhal]

correct.

[mortal panther]

ok hthanks

why did i have to subtract tho

[topaz narwhal]

you're taking away a smaller area from a bigger one.

[marsh citrus]

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[marsh citrus]

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[keen saddle]

how is the answer to this not -10?

[keen saddle]

nevermind, its that i shouldn't have used the negative sign

.close

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[novel juniper]

backward direcction

so here's what I did

$\lambda(v+t)+ (1- \lambda)(w+t)-t$

[elfin berry]

wai

[novel juniper]

which is useless

.close

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[rare owl]

can someone help me with part a? I don’t need the full answer, js a lil nudge abt how to even begin this problem 😭 I js know looking at part b that I can probably develop an explicit expression for f(n) maybe because it’s kinda impossible to count for 40 integers :p ig I also see an easier way to think of it might be a+d=c+b

[marsh citrus]

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[odd agate]

either provide context, or don't repeat what someone else said.

if you have no questions, please head to #discussion to chat.

pinging for attention

[calm harbor]

.solved

[marsh citrus]

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