help-42

[austere topaz]

like

1. factor ac by cross multiplying, use trial and error
   2....

[astral mural]

This is the most accurate steps

Although they did make a mistake by not multiplying x

[austere topaz]

oh

wym multiplying x

like adding them

[astral mural]

It's 3x and 1x on the left side

and 2 and 4 on the right side

I told you how the multiplication of the left side should equal the coeff of x^2, right?

Well, 3 * 1 isn't 3x^2

3x * 1x = 3x^2

[austere topaz]

yea

[astral mural]

It's just proper notations

[austere topaz]

ok i see

ima do a few questions

[astral mural]

Practice factoring tonight

[austere topaz]

using ur method

yea

[astral mural]

And you'll be fine for the test tmr

[austere topaz]

yea

if you dont mind

keep this open for like 5 mins

and if im fine ill close it

[astral mural]

?

[austere topaz]

ill let u know if i have a issue tho

[astral mural]

You have to be active in order to make this channel open

[austere topaz]

well i will be active

doing amth

ill emote every 2 mins

does that work

[astral mural]

ig

[austere topaz]

ok

ill ping u if anythin

[astral mural]

Even if it is closed, I will give you permission to ping me

[austere topaz]

ok so i noticed my numbers aren't nesserairly in order

answer says this and i got (2y+1) (3y+8)

@astral mural

should be fine tho either way right

[astral mural]

This way works too

But it doesn't work always

That's the thing

But it's fine

[austere topaz]

ok then

1 more and im outa ur way

ok i see how i can make it right

thank you very much for your help

thank you other person as well

fairy

god bless

hope you both become your dream jobs

stay safe

.close

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[atomic heron]

I want to do a Math AAHL IA (ibdp) which is almost like a large research task done at year 12 of school. I want to do it one the Theoretical limit of the 100m dash. I dont know how to start approaching this, so any ideas on how I can approach it would really help . there is a sample on the theoretical limit for the swimming 100m and they have spilt it into like different parts of the race, and then tried to analyse each part on its own and maximise the time

[atomic heron]

<@&286206848099549185>

[unkempt drift]

Bro why are you asking here

For IA help just ask on the server

[atomic heron]

nobody respnds on the server

[unkempt drift]

Like this really isn't the place to brainstorm ideas like that, sorry no one responded on the server tho

[atomic heron]

plus im just asking for how the maximum speed you run at can be a function of something

[unkempt drift]

Ah okay you want a function right yeah that makes more sense

[atomic heron]

yes

if i want to maximise the speed you run at, how can that be a function of something

[unkempt drift]

https://sportsscientists.com/2008/08/beijing-2008-men-100m-race-analysis/

Approximate the graph in there as a polynomial and you're good

You can integrate it to find the distance travelled after x metres

It would be easy to find the distance travelled after 10, 20 metres etc using SUVAT so you can cross-check your answers and see how well they match

To see the fit of the model

[atomic heron]

yeah but

im not doing it for bolt

im trying to find the maximum speed a human can run at

so like if it was a function of the downward force you apply

then i could find the maximum downward force a human can apply

and then get the horizontal vector from that

and like how can i maximise the acceleration of a human

[unkempt drift]

That's kind of theoretical and not a focus of a maths IA honestly, it isn't bio yk

We have the data already so I was thinking why complicate it

You don't have the knowledge to do cutting edge sports science and biomechanics research anyways

I guess you're interested in how the 100m or other sprint race times have gone down over the years

Have a look at that also, that would be nice

[unkempt drift]

Yeah you'd have to find research papers on this

[atomic heron]

there was a sample on the swimming 100m

[atomic heron]

its not enough for a math ia is it

[unkempt drift]

https://hal.science/hal-00908318/document

Why go through what researchers have done already by yourself tho

[unkempt drift]

The maths IA is about the maths you use, so like converting the problem into a mathematical one is only the first step

You just have to use you know integration techniques, stuff in your course or slightly above

[atomic heron]

ill have a look

thanks

[unkempt drift]

Yeah no worries

You have good quality data there actually, it's a good paper

[atomic heron]

yeah

[unkempt drift]

Take their data points and do some stats on it

[atomic heron]

im going through it briefly

seems alright

[unkempt drift]

Like check their r^2 coefficients and so on

No worries

[atomic heron]

yeah ill see what i can do

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[deft python]

how do i make a phase potrait? I know it tells us about stabiltiy but how do i make one?

[calm coral]

Please don't occupy multiple help channels.

[deft python]

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[remote mural]

We are given the PDE $\ds\laplacian\psi = \psi_{xx} +\psi_{yy} = 0$. I must find the solutions of the form $\psi =x^\gamma \6f\xi$ where $\xi = \5xy$. The PDE falls under the boundary condition $\psi_x - \psi_y = 0$ along the line $x=y$. What are the possible values for $\gamma$ and what are the corresponding functions $\6f\xi$ that solve the PDE?

[remote mural]

plugging the solution in gets you the ODE [
\xi^2(\xi^2 + 1)f''(\xi) +2\xi(\xi^2 +\gamma)f'(\xi) + \gamma(\gamma - 1)f(\xi) =0
]
Which is a Sturm-Liouville ODE, but idk how you evaluate it. Any ideas?

[calm coral]

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[marsh agate]

#odes-and-pdes

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[worn yoke]

I’m doing past papers for ext 1 maths

Yr 11

I’ve read the answer but I stil don’t understand the question

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[brazen elbow]

not sure where to start here

[fathom shuttle]

the projection of S onto ABC is the circumcenter of ABC

you can then figure out the height and volume of ABC

SABC*

then check the ratio of base and height to figure out the ratio of volumes

[brazen elbow]

.close

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[brazen elbow]

thank you btw!

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[austere hazel]

uh

[austere hazel]

small doubt

how did they simplify this

[neon sage]

$3- 2\sqrt{2} = ({\sqrt{2}})^2 + 1^2 - 2(1)(\sqrt{2}) = (\sqrt{2} - 1)^2$

[potent lotus]

nebula40

[austere hazel]

ahh

i see

thank you

.close

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[gray pebble]

isn't here supposed to be $g(x) > 0$

[potent lotus]

wcasa

[gray pebble]

absolute values can't have (-) value coming out of them; we have to restric g(x) being minus

!close

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[eternal shard]

If f(x) ≥ 0 then |f(x)| = f(x) otherwise if f(x) < 0 then |f(x)| = -f(x).

So -f(x) ≤ g(x) is equivalent to f(x) ≥ -g(x)

[gray pebble]

thank you!

but how wasn't this channel closed

[gray pebble]

I already figured it out and closed it ._.

[eternal shard]

ah ok

well it is it just takes some time

[gray pebble]

.close

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[shy light]

Hi, I want to know why 21² = 20² + 20 + 21?

[shy light]

Are there any properties associated with it?

[cinder blaze]

21^2 = (20+1)^2

hm

yeah

20^2+40+1 =20^2+20+21

[shy light]

Oh yeah right I missed that part

[cinder blaze]

💀

[shy light]

Lmao it's not like I forgot the distributive property

ikr, like 22² = 21² + 21 + 22

I thought that there were some other properties associated with it

lmao anyways

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[shy light]

Here's the proof for the above question

.close

[white atlas]

21^2 - 20^2 = (21-20)(21+20)

[shy light]

thanks

I've tried to prove it the other way, but great observation

[white atlas]

what way? without making any modifications to both sides?

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[white atlas]

if so, then you can prove the right hand side:

20^2 + 20 + 21 = 20(20 + 1) + 21 = 20(21) + 21 which is indeed (21)(21)

[white atlas]

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[graceful dust]

[white atlas]

wow layla's always there to help me

[graceful dust]

[shy light]

yes

Like this

[white atlas]

Lol you don't need all of that but sure

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[remote mural]

theres a bit too many variables here

[still marlin]

x = 100 - r

Ur equation is wrong also, left side should be something like A for area

[remote mural]

why though

[still marlin]

The total length of he cable is 100

[remote mural]

oh wait

it's just rearranged?

and where'd the pi and x go

[still marlin]

wym

[remote mural]

so why isn't it

100=x^2 + pir^2

[still marlin]

100 is the length of the wire

right side is the total area of the shapes

[remote mural]

total area of a square is x^2

total area of a circle is pi r^2

[still marlin]

Altho hold on I made a small mistake, if r is the radius of the circle then x = 100 - 2pir

[still marlin]

since when does a length equal an area

[remote mural]

the amount of material

is being used

to make two things

and we're asked

for the area

of those two things

[still marlin]

It's being used to make the perimeter of the things

[remote mural]

but we're asked for the area

to maximize and minimize

[still marlin]

It's 100 cm long

cm not cm^2

[remote mural]

ok

so you use it

to find perimeter

and then area?

[still marlin]

perimeter of circle + perimeter of square = 100 so I made another mistake

[remote mural]

in that case 100=4x + 2pi r

[still marlin]

ye

[remote mural]

$\frac{100-2\pi r}{4}=x$

[potent lotus]

Remlis

[still marlin]

yup

[remote mural]

$25-\tfrac12 \pi r = x$

[potent lotus]

Remlis

[remote mural]

and then we use

area formula

substituting for x

?

[still marlin]

yuup

[remote mural]

ok gotcha

thanks

[still marlin]

Np!

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[uneven summit]

Hi

[uneven summit]

Can someone help me factorise this question

[clear sapphire]

!status

[calm coral]

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

[uneven summit]

I don’t know

Where to begin

[brazen elbow]

$20 \times ... = 60$

[potent lotus]

∮μ²ƞdS

[uneven summit]

What

[brazen elbow]

answer it

[uneven summit]

3

[brazen elbow]

yes

[uneven summit]

Mhmmm

[brazen elbow]

now using this information, rewrite this

[uneven summit]

I got no clue still

[brazen elbow]

you see the 60

[uneven summit]

Yes

[brazen elbow]

you know its = 20x3

[uneven summit]

Yes

[brazen elbow]

rewrite it so

[uneven summit]

Idk how to rewrite it

[brazen elbow]

$(x+4)^{20}-(x+4)^{...}$

[potent lotus]

∮μ²ƞdS

[brazen elbow]

fill in the blank

[uneven summit]

Would that be 3

?

[brazen elbow]

and?

3 what

[uneven summit]

20

3 and 20

[brazen elbow]

yes

so whats the final expression

[uneven summit]

$(x+4)^{20}-(x+4)^{3x20}$

[potent lotus]

gary tate

[uneven summit]

Oops

[brazen elbow]

\times

[uneven summit]

Messed it up

Ye messed up

$(x+4)^{20}-(x+4)^{3/20}$

[brazen elbow]

$(x+4)^{20}-(x+4)^{3\times20}$

[potent lotus]

gary tate

∮μ²ƞdS

[brazen elbow]

ok

[uneven summit]

So is this the answer

[brazen elbow]

no

not yet

now answer the next question

$x^{ab}=(x^a)^b$

[uneven summit]

Gosh I don’t get that

[brazen elbow]

ill give you a fairly big hint

[uneven summit]

Alr

[potent lotus]

∮μ²ƞdS

[brazen elbow]

this is what im referring to

[brazen elbow]

so, with this information

[uneven summit]

Why am I such an idiot

[brazen elbow]

rewrite this

[uneven summit]

I still don’t get it

[brazen elbow]

try harder

actually give it some time to think before speaking

i get your idea

but its not what we wanted here

try a different approach

[uneven summit]

Idk

Wait

[brazen elbow]

could you stop saying idk

you are not stupid, but think before you speak

give it some time, we have plenty

[brazen elbow]

no

[brazen elbow]

again,remember this

in this case, "x" is (x+4)

[uneven summit]

I don’t get that

Ohhh

[brazen elbow]

try again

[uneven summit]

I can’t

Tbh

I can’t think

[brazen elbow]

your deleted message was on the right track

[uneven summit]

Is it fine if u can give the steps as I lean easier and best that way?

[brazen elbow]

so lets say $x^{3\times4}$

[potent lotus]

∮μ²ƞdS

[uneven summit]

Alr

[brazen elbow]

using this formula, what can you infer?

[uneven summit]

Idk

[brazen elbow]

give it some time to think

[uneven summit]

I just don’t get this at all

I gtg thanks for ur time tho 🙂

.close

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[vernal tiger]

(x) ^ 60

is (x^3) ^ 20

idunnu

forgtet

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[hollow ruin]

Hi guys I was just wondering for a circle inscribed in an isocoles triangle and you know the length of the long side of the triangle can you find the radius

[hollow ruin]

Or is there a way to approximate the radius

[fluid flame]

you can arbitrarily squish the two 14.5 sides closer to (or further away from) each other so that’s insufficient

[hollow ruin]

What if that triangle are part of 6 identical triangles that when put together make a hexagon thats inscribed inside the circle

[vagrant jay]

write similation between ODB and ACB

[half sparrow]

The triangle would be equilateral so yes

[hollow ruin]

oh waiiit yeah i overlooked that it makes sense now thanks!

[unkempt drift]

You need to be able to uniquely define a triangle for the inscribed circle

So knowing the length of the base would be enough (SSS)

[hollow ruin]

ok yeah got it

[unkempt drift]

Knowing the angle in between would be enough, or any angle as it's isosceles (SAS etc.)

Npnp

[hollow ruin]

thanks guys 👍

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[clever smelt]

can someone tell me what the ~CN() function is

[clever smelt]

just the name would be fine

[calm coral]

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[clever smelt]

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[vocal sorrel]

Is this O(n^7)

[clear delta]

pain

but I don't see where you get n⁷

start from the inside and work out

[calm coral]

@vocal sorrel Has your question been resolved?

[vocal sorrel]

I got n^7 becayse each while loop is n, and since its nested they multiply. Starting at bar() or plg() u get n^2 and eventually u just leads to n&5?

so was my first ans right? I think I forgot bar() was outside

[clear delta]

is every while loop n?

[vocal sorrel]

yes. because every runtime we assume worst case

[clear delta]

not every while loop is n

[vocal sorrel]

its not?

[clear delta]

no

look at the increment and the starting values

[vocal sorrel]

whgat do you mean starting values?

[vocal sorrel]

uhhh I dont get what its highlighting

[clear delta]

c = n^2

but also notice the other line, the k = k/2

[vocal sorrel]

I can see that but I dont get where this leads

[clear delta]

well...

that k-loop

how many iterations is it?

[vocal sorrel]

Idk since n is arbitrary?

oh

log_2 n?

[clear delta]

that is correct (approximately. if it's not an exact power of 2 it won't be exactly that but close enough)

[vocal sorrel]

just log?

but what about the plg() part?

[clear delta]

again, start from the inside and work your way out

[vocal sorrel]

sorry the bar(0 part

[clear delta]

here's what you should do

take the code

and replace the call to qlg() with O(n)

then replace the entire while-q loop with O(n²) -- see why?

do a similar thing for the other functions

and while loops

[vocal sorrel]

yea I see the for th q loop why its O(n^2) because it terminates aftere first iteration

Ok ill try that

[clear delta]

wait yeah it's only one iteration of O(n) so that loop would also be O(n) my mistake

[calm coral]

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[vocal sorrel]

.reopen

[calm coral]

✅

[vocal sorrel]

would it not be O(1)?

[clear delta]

don't ignore the bot next time

[clear delta]

no, because qlg itself is O(n)

so in total it's O(1) * O(n)

[vocal sorrel]

oh ok

Im getting n^6 log (n)

[hollow totem]

Why multiply steps that are performed one after the other

If I do one thing that takes n^2 time and then I do a second thing that takes n^2 time, it doesn't mean it takes n^4 time

[vocal sorrel]

o icic. so its just n^3?

[clear delta]

you didn't do the thing I asked you to do, which is to replace the text

[vocal sorrel]

I replaced it with the drawings

[clear delta]

the text is still there

[vocal sorrel]

Like do you want me to cross it iyt>

out?

[clear delta]

i want you to do something like this

[vocal sorrel]

Oh ok

[vocal sorrel]

Im getting O(n^4logn)

using ur diagram

[clear delta]

that seems more reasonable

[vocal sorrel]

but it doesnt appear as an option

[calm coral]

@vocal sorrel Has your question been resolved?

[vocal sorrel]

@clear delta yea I cant figure out how to get any other option

<@&286206848099549185>

[calm coral]

@vocal sorrel Has your question been resolved?

[vocal sorrel]

<@&286206848099549185>

[calm coral]

@vocal sorrel Has your question been resolved?

[thick sinew]

hm no help provided k

I'll start from the inside without reading the above first

We know that this innermost part takes Θ(n²) since plg() takes Θ(n²)

now, how often do we perform this loop? Once, since q is directly set to 1 within it. Meaning this block has time complexity 1*Θ(n²)=Θ(n²):

on the same depth, bar() is executed, which takes Θ(n²), meaning our complexity for these two boxes is Θ(n²)+Θ(n²)=Θ(n²) since they are executed one after another:

now how often is this loop executed? n times, since i starts at n and counts down by 1 each iteration. so our complexity for the following box is n*Θ(n²)=Θ(n³):

on the same depth as the box, foo() is executed which takes Θ(n²), so our new runtime is Θ(n²)+Θ(n³)=Θ(n³):

how often is this loop performed? log2(n) times, since k starts at n is divided by 2 each time until it's 1 or less. so our complexity becomes log2(n) * Θ(n³) = Θ(log2(n)*n³):

now, on the same depth multiple things happen:

first there is another loop, which executes adv() n² times and since adv() takes Θ(n²), that loop has a complexity of n² * Θ(n²) = Θ(n^4).

And then there is a single wcr() call which takes Θ(n²)

meaning these three boxes together take Θ(log2(n)n³) + Θ(n^4) + Θ(n²) = Θ(n^4):
[If this summation is unclear to you, note that any logarithmic term is smaller than a polynomial one. meaning log2(n) * n³ is smaller n * n³ which is n^4 (since log2(n) < n).]

how often is this loop executed? n times, as g starts at n and counts down by 1. Meaning our new time complexity is n * Θ(n^4) = Θ(n^5):

@vocal sorrel

hope that clarifies it 🐛

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[calm coral]

@vocal sorrel Has your question been resolved?

[vocal sorrel]

<@&286206848099549185>

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@vocal sorrel Has your question been resolved?

[rustic osprey]

<@&268886789983436800> ^

[whole hinge]

you might get better help in #discrete-math btw, I think graph theory qs tend to receive more attention there than in the help channels

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[vocal sorrel]

alright than

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[unreal crater]

I desperately need someone's help! IT's pretty easy calc-4 stuff, but

[unreal crater]

I have tried so much! I wasn't here for the lecture so I am just lost here

[cloud nexus]

https://tutorial.math.lamar.edu/classes/de/StepFunctions.aspx

found this online

[unreal crater]

I just don't know how to put it in function form

that's kind of helpful but they format it in a very weird way, like u(n)

[cloud nexus]

why is that a weird way

[unreal crater]

because for my function notation its h(t-a), which requires parts of u that are not listed

[cloud nexus]

so translate it between them

[pallid halo]

with your proposed answer, notice that if you plug in a t value >= 3 then you get f(t) = 8

because both h(t-1) and h(t-3) will be 1 in that case

[unreal crater]

srry I just had a dumb adding mistake!I got it

thanks guys!

idk how to close thsi tab haha\

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@rustic abyss Has your question been resolved?

[civic dirge]

Just messing around a bit, I've found the function is odd.

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[gray pebble]

considering 8 + sec... bla bla.. find the value of the tan theta

descriptiors only wrote the

[green haven]

is 6tgtheta 6tantheta?

of 6xtheta?

[pure kayak]

tg is tan yeah

turn sec^2 into tan^2 via an identity, then youll have a quadratic of tans

[green haven]

so maybe break up sec and tan into cos/sin and sin/cos then get the same denominator

[gray pebble]

oh i tried it just now

didn't work out; just got a mess

[green haven]

oof

1+sec^2 is tan^2 if i remember my identities correctly

[pure kayak]

nearly

[green haven]

other way around?

[pure kayak]

1+tan^2=sec^2

[green haven]

1+ tan^2?

mm

[gray pebble]

how do you even remembewr such things

[green haven]

so sec^2 break into tan^2 and 1

[pure kayak]

im just a maths man

its routine to me

[gray pebble]

man of culture I see

[green haven]

can u help solve my help question then pls

after u help this man

gentleman

[gray pebble]

prettyy much solved already

thatnk

.close

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[green haven]

:D

[crystal forge]

!help

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[green haven]

i did...

[modern peak]

i think he meant !noadvert

[green haven]

oh i see

so just gotta wait until someone checks

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[bleak ermine]

how do i find the factor this is impossible

if tried f(everything up to 10)

both positive

negative

and fractions

[pallid halo]

do you know the intermediate value theorem?

[bleak ermine]

no...

maybe this is a topic we didnt do

[pallid halo]

ok well let's reason through it intuitively

suppose i plug in two values of x

[pallid halo]

and i find that one of them gives me a negative value when i plug it into x^3 + 5x - 10

[bleak ermine]

its from one of these

we actually did them all

[pallid halo]

and the other gives me a positive value

then you know x^3 + 5x - 10 needs to be zero somewhere between those values

because it's a continuous function

[bleak ermine]

oh ok

[pallid halo]

it can't jump

[bleak ermine]

we did not learn that at all

which exercise would that be from

from my image above

[pallid halo]

it's related to the bisection method (6K)

[bleak ermine]

oh we havnt done that yet

far out i thought i was screwed ive got a test tomorrow

[bleak ermine]

were doing bisection later

[pallid halo]

also possibly related to 6G

[bleak ermine]

it sounds like it is

is there any way to find a factor

for that polynomial

[pallid halo]

and maybe 6D depending on what they cover there

[bleak ermine]

and graph it

because i was trying to graph it and i could not find the factor

[pallid halo]

no it doesn't have any nice factors

,w x^3 + 5x - 10 = 0

[potent lotus]

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[bleak ermine]

fraction?

[pallid halo]

irrational

[bleak ermine]

is that doable by hand

[pallid halo]

there is a complicated formula for cubic equations

but you probably don't want to know about it

[bleak ermine]

im in highschool

im just trying to figure out if i should know how to do this or not

[pallid halo]

yea then def don't worry about it, it's very nasty

[bleak ermine]

ok

[pallid halo]

just use what i said above

[bleak ermine]

alr thank you

[pallid halo]

plug in the two values they give for each option

and see if you get one positive and one negative result

if so, you know there's a zero solution between them

[bleak ermine]

ok thanks so much

havea good one

.close

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[green belfry]

Redoing questions i got wrong. For b, do i need to solve for x as well? Or is that already considered a complete answer?

[green belfry]

Holy compression! the function is f(x) = 2x^2 + x - 1 , and I have -2x^2-x+1 for the answer to -f(x).

(to be clear, that is not the answer that was marked wrong, this is my new answer)

[green haven]

ah

[green belfry]

@sly geyser To be a bit more clear about my question: Am I expected to solve for x when I am asked to find -f(x) of a function? The videos I watched all seemed to stop at this point. Is that considered the full answer?

[green haven]

find -f(x)'s equation i would assume

so ur answer looks fine to me

[green belfry]

Okay that's what I figured, just wanted to make sure. :) Thank you!

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[green haven]

yea its only 1 mark so u would probably only have 1 line of working

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[hybrid zephyr]

in a triangle sin A =3/8, sin B=3/4, if a=5 find b

[winged smelt]

!status

[calm coral]

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

[winged smelt]

Do you know law of sines?

[hybrid zephyr]

sin = o/h

[winged smelt]

https://www.mathsisfun.com/algebra/trig-sine-law.html

[hybrid zephyr]

yeah i know sine rule

[modern peak]

good, then how do you think you can apply it here

[hybrid zephyr]

i need to make my own triangle then

[modern peak]

mhm

[winged smelt]

That will be better

You should draw a triangle with all the information given.

[hybrid zephyr]

k

[hybrid zephyr]

then i would use the equation right

[winged smelt]

Why is that angle written as a=b

[hybrid zephyr]

its meant to be 5

[winged smelt]

I see

That's not the angle actually

It's the side opposite to angle A

a is the length of the side opposite to angle A

It's also very important to name the triangle

To avoid confusion later on

I suggest you do that as well

[hybrid zephyr]

ok i understand my mistake

i was mainly doing this to see if i did it right on the exam

[winged smelt]

Even so If you do things in a proper manner it won't stress you later on with the confusions that you may have encountered without doing them in a proper manner

I'm just advising you to be neat with your work

[unkempt drift]

Yeah it's like how if i tp lk ths y m stl b abl t unstd m

But it would be very annoying to read

Clear communication is just as important in English as it is in maths

[winged smelt]

True

[hybrid zephyr]

i know

[unkempt drift]

I should have chosen something with really bad spelling but anyways

[hybrid zephyr]

perhaps you should have

[calm coral]

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[vernal tiger]

I did it up to here,
My question is

Why is it bad to simplify the inner brackets

and make the +c into ln k

[chilly stirrup]

how to solve inequality

[modern peak]

!occupied

[calm coral]

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[vernal tiger]

eh nvmm

I got it

I just found it weird

:<

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[vernal tiger]

all good

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[hoary juniper]

,,\lim_{n \to \infty} \bigg(\frac{1^2-2^2+3^2-4^2+....... \ n \ terms}{n^2}\bigg)

[potent lotus]

Intecules ∮

[hoary juniper]

so here how i did it

I first wrote it like $$\lim_{n \to \infty} \bigg(\frac{(1^2+3^2+5^2+.....) -(2^2+4^2+6^2+.....)}{n^2}\bigg)$$

[potent lotus]

Intecules ∮

[hoary juniper]

wait a min i did smth wrong with the latex

yea so just focusing on numerator i did this $$\sum_{r=1}^{\frac{n+1}{2}} (2r-1)^2-\sum_{r=1}^{\frac{n-1}{2}}(2r)^2$$

[potent lotus]

Intecules ∮

[hoary juniper]

which gives

,w summation of (2r-1)^2 from r=1 to (n+1)/2 - (summation of (2r^2) from r=1 to (n-1)/2)

[potent lotus]

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[hoary juniper]

this

which gives me the answer infinty

but the book says its 1/2

also n is an odd number

what am i doing wrong

oh wait

godamn it

wrong brackets

sorry my bad

,w summation of (2r-1)^2 from r=1 to (n+1)/2 - (summation of (2r)^2 from r=1 to (n-1)/2)

[potent lotus]

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[hoary juniper]

sorry

.close

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[zenith shoal]

i don’t understand this question can someone help me ☹️

[wispy plank]

this might be different to what you've learnt bc im in australia but do yk about similarity reasoning?

so like equiangular

[zenith shoal]

nope ☹️

[wispy plank]

can u tell me what you've learnt

in terms of similarity

[zenith shoal]

i only learned the angles theorems like interior and exterior angles

[wispy plank]

what abour co-interior, vertically opposite, alternate and corresponding?

[zenith shoal]

i don’t think i’ve heard of co-interior

[wispy plank]

essentially, co-interior angles are angles on a straight line that add up to 180

that may not be the exact definition

ill show u a real image of it

with those 4 types of angles, you can make "shapes" to figure out which one it is

for example, co interior has a C shape

corresponding has a F shape

alternate has a Z shape

[zenith shoal]

oh yea i’ve seen it but it’s called consecutive for us

[wispy plank]

ohhh alr

well thats fine then

so in this image

we're asked to prove that traingle eab is similar to triangle edc

now i don't want to confuse you here as what my reasoning would be might be different to yours

do you still want to hear it?

[zenith shoal]

yea

[wispy plank]

ok so there are 4 conditions for 2 triangles to be similar

equiproportionality

which are all lengths in the same ratio

equiangularity - 2 corresponding angles equal

equiproportionality-equiangularity - lengths of 2 sides and an angle in between

and equiproportional with right angle which is just a right angle + hypotenuse along with one shorter side

in this case

can you tell me which case in those 4 cases fit this question?

[zenith shoal]

uhh ae and ed?

[wispy plank]

nope

we're only given 3 sides

which means none of those conditions would fit

what's angle bae and angle edc?

[zenith shoal]

right triangles

[wispy plank]

right angles

so we that leaves us with the 2nd condition and the 4th condition

are we given a hypotenuse for any of the triangles?

[zenith shoal]

is it be and ec

[wispy plank]

yep

but is there a definte lenght?

[zenith shoal]

no

[wispy plank]

correct

so that leaves us with the 2nd condition

equiangularity

remember how i told you about those 4 different types of angles

[wispy plank]

^

[zenith shoal]

yea

[wispy plank]

so in these two triangles

excluding the right angle

can you find two angles that fit one of those 4 types?

[zenith shoal]

i’m sorry this is kind of confusing cause it’s not a transversal 😭😭

[wispy plank]

it's okay

take ur time

i'll give u a hint

what's a common point in both triangles

where does one point meet another point

[zenith shoal]

point e

[wispy plank]

good

which means it's vertically opposite angles

and that fits our requirement for the 2nd condition

2 corresponding angles

so you would write, "Therefore, triangle EAB ~ triangle EDC (equiangularity)

now you might have to check with your teacher

bc the reasoning will most likely be different

but it should be equivalent to what i've said

[zenith shoal]

thank you this is very helpful sorry u have to put up with me 😭😭😭

[wispy plank]

no no it's allgs

glad you understand it

[zenith shoal]

thanks again

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[latent fern]

how can i get a cosh function with parameters

[latent fern]

based on this catenary shape

https://www.desmos.com/calculator/qn4nvsflpp
here's the graph and i was trying to get the value for a and b

[honest quiver]

You can create a (nonlinear) system of two equations by substituting the given (x,f(x)) values into the middle of this equation

The point on the y axis should prove especially helpful

[latent fern]

how come

[honest quiver]

x=0

[latent fern]

oh

so is a=-3.1

[honest quiver]

Not quite

e^0 + e^0 = …

[latent fern]

bruh i'm stupid

-1.55?

[honest quiver]

Yes

Although…..hold up

[latent fern]

like i acutally got the equation from desmos by using a slider

but it was more likley to be a trial and error method

[honest quiver]

acosh(bx) cannot be equal to 0 if a ≠ 0

Because that would require e^bx = -e^-bx

That’s odd…

[latent fern]

look how amazing the technology is but i need to write my process of deriving that weird function

[honest quiver]

Your function will have to be acosh(bx) + c

The issue is that’s three unknowns so we probably need three points

[latent fern]

there're bunch of coordinates available

[honest quiver]

Oh nice

Any three should work

[honest quiver]

Hold up it’s not an even function?

[latent fern]

umm

that's a great question

[honest quiver]

The plot thickens

You might need a phase shift term

[latent fern]

let's just around up 8.35 to 8.4

will it work tho

oh actually no it's because the center of my image was a bit off

it is an even function

[honest quiver]

Oh good

So now you have y = acosh(bx) + c

[latent fern]

hold up

i'm still stuck with the variable a

what am i supposed to do with the three coordinates

[honest quiver]

Plug them in to this

You then get a system of three equations

Not linear but should be solvable

[latent fern]

is there an online calculator that can solve system of cosh equations

or i'll just expand them

[honest quiver]

Yeah you’ll have to expand them

Also make sure you only use points on one side of the axis because your image is slightly off and if it’s not even, the system will be unsolvable

[latent fern]

smt weird has happened

i was trying to use those three points

[calm coral]

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[prime wasp]

question

[prime wasp]

would we consider ln(-2)

to be in the domain of ln

[blazing coyote]

Not if you are considering the real valued log function

I mean log (-2) isn't even defined

On R

[prime wasp]

Ok

I am trying to find critical numbers of ln(x-2)

derivative is

1/x-2

0 would make undefined, but 0 I assume is not in domain of ln(x-2) like you said

1/x-2 = 0

can you help me with trying to solve for x here

[blazing coyote]

A critical point is a point where the derivative is either 0 or undefined

[prime wasp]

I already stated that

[blazing coyote]

so where is the derivative not defined?

[prime wasp]

oh

kill me lol

I mistakenly wrote 0, undefined at x = 2

ln(0)

is that in domain

no

[blazing coyote]

the domain is ||(2,infty)||

[calm coral]

@prime wasp Has your question been resolved?

[calm coral]

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[vast violet]

could anyone help me understand how to first approach this question?

[graceful dust]

can you think of any numbers a,b with a > b but a^2 <= b^2?

you want to find an x such that sin(x) > cos(x) but not sin^2(x) > cos^2(x)

[graceful dust]

thinking about this should help

[vast violet]

I used 3pi/4 as an example, but I am worried that just bc that example might prove set A is not a subset of B i still havent proved it enough

bc it is a specific value and not general enough

maybe I am overthinking?

[graceful dust]

yea you are lol

you just need to find one element of A that isn’t in B

[vast violet]

oh okay haha, ig i just needed that reassurance

thank you for your help :)

and reassurance lol

[graceful dust]

[vast violet]

.close

[calm coral]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[tiny steppe]

help

[glacial inlet]

do you know the formula for the area of a trapezoid?

[tiny steppe]

B1 x B2 / 2

[glacial inlet]

times the height

[tiny steppe]

B1 x B2 x h / 2

[glacial inlet]

yep

what’s our b1

[tiny steppe]

7.5

[glacial inlet]

correct and b2?

[tiny steppe]

13.7

[glacial inlet]

close

[shy light]

no, try again!

[glacial inlet]

b2 is the entire lower base

[tiny steppe]

5.3

• 13.7

[glacial inlet]

yep

[tiny steppe]

19

[glacial inlet]

now the height?

[tiny steppe]

6.8

[glacial inlet]

yep

now we just have to plug that into the formula

oh wait also the formula is a little wrong

it should be $A=\frac{1}{2}(b_1+b_2)h$

[potent lotus]

y0shi

[glacial inlet]

not b1 times b2

[tiny steppe]

so 19/2

7.5/2

• eachother

than x 6.8

[glacial inlet]

yep

[tiny steppe]

ok

I have another question

[glacial inlet]

go ahead

[glacial inlet]

so what is the length and width of the outer rectangle

[tiny steppe]

2 x 3

[glacial inlet]

not exactly

if the frame is 2 wide

we would have to add 2 twice to each of the length and the width

since it is covered all around

[tiny steppe]

Im confused

2 x 4

x 2

[glacial inlet]

where are you getting 4 and 2 from?

[tiny steppe]

4 sides all around

2 units wide

[glacial inlet]

the 2 doesn’t signify the length or the width of the outside rectangle

but rather it’s distance from one of the outer edges to the corresponding inner edge

[tiny steppe]

How do i solve it

[glacial inlet]

so since we know that distance is 2

[tiny steppe]

yes

[glacial inlet]

if we look at the inner length 5

[tiny steppe]

yes

[glacial inlet]

there are two extra lengths that needs to be added to get the outer length

and that length is given to us as 2

so we would have to do 5+2+2

for the outside length

[tiny steppe]

9

[glacial inlet]

yep

and how about the outer width

using the same concept we did above

[tiny steppe]

3 + 2 +2

[glacial inlet]

yep exactly

so to get the area of that outer rectangle

we just need to multiply them

[tiny steppe]

9 x 7

[glacial inlet]

yep

[tiny steppe]

63

[glacial inlet]

and to get that surrounding edge that is shaded

we need to do the outer rectangle minus the inner rectangle

so what is the area of the inner rectangle?

[tiny steppe]

I might sound dumb but is it 5x 3

[glacial inlet]

yep it is

[tiny steppe]

48

[glacial inlet]

so we would have to do 63 minus that number

and yep that’s it

[tiny steppe]

is that the awnser

[glacial inlet]

yep

[tiny steppe]

Can i get help with ONE more

[glacial inlet]

ofc go ahead

[glacial inlet]

well we have to break this up to calculate the area

so let’s split it up to a triangle and a rectangle

[tiny steppe]

ok

[glacial inlet]

so let’s find the area of the triangle first

what would be our base and height

[tiny steppe]

4 and 4

i think

[glacial inlet]

yep

so the area would be?

[tiny steppe]

16

[glacial inlet]

well not exactly

the area of a triangle is

[tiny steppe]

8

[glacial inlet]

yep

[tiny steppe]

16/2

[glacial inlet]

exactly

now we have that

we also need the rectangle

what is the length and the width?

[tiny steppe]

9 x 10

[glacial inlet]

yep

[tiny steppe]

90

and not /2 because its a rectangle

[glacial inlet]

exactly