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[zinc haven]

hmm

[quasi vector]

and?

[zinc haven]

so if we apply this to Tv, where v in V

brb

[keen plinth]

no come back

answer the question

[zinc haven]

ok i came back

[keen plinth]

answer this

this is not the complete answer

[zinc haven]

A_jj = A_ii, and every other coefficient A is 0

wait this is really reminiscent of the identity matrix

[keen plinth]

okay so what does that mean

write out T again

[zinc haven]

$Tv = T(x_1e_1 + \dots + x_ne_n) = x_1Te_1 + \dots + x_nTe_n$

[keen plinth]

i meant using this

you only ever need to compute a linear map on a basis

writing out a linear combo is never a good way to write down a linear map

[zinc haven]

oh

[keen plinth]

i keep saying this

you are given lemmas by axler

use them

this is absolutely fundamental

[quasi vector]

writing out a linear map on a basis is so nice

[zinc haven]

ah, so that means $Te_i = A_{ii}e_i$

[keen plinth]

yes exactly

and also, the A_i,i are all the same

[zinc haven]

so that means T simply sends an element to itself times by some constant, namely A_ii

ohhh

[zinc haven]

wait, i don't quite grasp this part. A_i, i are all the same because they all equal to A_j, j, but isn't A_j, j also not fixed?

[keen plinth]

so you have A_1,1 = A_2,2 = A_3,3 = ... = A_n,n

call their common value A or something

then Te_i = Ae_i

for the constant A

[zinc haven]

is A = A_j, j?

[keen plinth]

its the common value of all of them

A_i,i = A_j,j for all i, j

[zinc haven]

ah i see

how does this look?

[keen plinth]

i feel like you should explicitly state that A_i,j = 0 for i /= j

rather than vaguely describing it in english

[keen plinth]

and you should also explicitly state that lambda is the common value of the A_i,i like i've done here

,, \lambda = A_{1, 1} = A_{2, 2} = \dots = A_{n, n}

[zinc haven]

alright, thank you very much

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[keen plinth]

👍

[zinc haven]

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[lament patio]

Hey all

[lament patio]

Can I get some help in here with these?

What the hell

<@&268886789983436800>

[charred jewel]

aint no way people are still cheating in fortnite in 2025

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@lament patio Has your question been resolved?

[lament patio]

No

No it' hasn't been resolved...

[violet jetty]

What have you done so far? These are straightforward questions, if you can't do them its probably a good idea to relearn how to get eigenvalues

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@lament patio Has your question been resolved?

[lament patio]

I'm doing exactly that right now

thanks

[lament patio]

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[alpine sable]

.

[ocean seal]

villegas_bozo

villegas_bozo

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[alpine sable]

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[still ibex]

I was just messing around with different ideas in maths because they're useful when teaching people, and I stumbled across this weird thing and would really appreciate if someone could explain what's going on:

So for the product rule: take x^2*sin(x) for example. Thinking from the perspective of someone learning for the first time, they might think to write this as x^2 + x^2 + x^2 + ... + x^2 (sinx times), making the derivative 2x + 2x + 2x + ... + 2x (sinx times) = 2x*sinx. Of course this isn't right, and function multiplication doesn't work like this (although if it doesn't then I'm not sure what it represents). But there's more to this:

We can also look at it from another perspective: that x^2*sin*(x) can be thought of as sinx + sinx + sinx + ... + sinx (x^2 times), making the derivative cosx + cosx + cosx + ... + cosx (x^2 times), which is x^2*cos(x)

The curious thing about this that I realised, is that when we add together these two incorrect ways of calculating the derivative of a product - you get the right answer! Is there a reason behind this? Or is it merely coincidence? Surely it can't be a coincidence, and it has to have SOME connection to the product rule given that it does technically kinda work

[rustic coral]

Note that if you generalise this to functions f and g, then:

• take the derivative of f and add it g times -> f’g
• take the derivative of f and multiply it g times -> g’f

Meaning that this is the statement of the product rule in all but name. This is kind of related to the geometric illustration of the proof of the product rule, so you may want to check that out.

Also, since you talked about teaching, #math-pedagogy may be of interest to you. (There’s some pretty interesting conversations if you comb through there) Also, math educator’s stack exchange also has some interesting chats.

[still ibex]

I'm familiar with the geometric intuition, but the problem is that for this "intuition", it seems like you should be able to just do f'g and then you're done, but that's not the case - there's a flaw in the logic somewhere

And I'll check out that channel - thank you!

[rustic coral]

The flaw in the logic is that g also depends on x, so you can’t treat it as a constant in the first bullet.

Same idea for the second

The method is basically splitting it up like in the geometric proof, hence why I said what I said

Whether that’s by coincidence or something else, idk

[still ibex]

How am I treating it like a constant? (Sorry that's not supposed to sound rude I'm just trying to figure it out fully so I can explain it to someone lol)

Assigning a arbitrary side length of g(x) to a rectangle is the same as adding up a number g(x) times, except the latter is a strictly discrete version

[rustic coral]

You’re factoring g out of the sum when you do the first bullet

[still ibex]

But I didn't factor anything

I just interpreted multiplication as repeated addition

[rustic coral]

Same idea.

The number of times that you’re adding the derivative isn’t a “pure” constant - it also depends on x.

[still ibex]

So why doesn't the geometric intuition for the product rule fall on its face for the same reason?

[rustic coral]

b/c the geometric proof is just putting the more “proper” proof into rectangle form

aka it’s hiding some of the finer details that are seen in the actual proof

[still ibex]

I just messed around to figure out the actual proof there since it's actually been a while since I've done it lol

I'll definitely have to mess around with it while looking at the geometric interpretation at the same time

But it kinda brings up another unsatisfying thing - it feels so random to randomly add and subtract u(x)*v(x + h) to make the proof work

Is it likely that people just discovered that through messing around?

Or is there a stronger motive for it

[rustic coral]

You mess around to guess what the rule is

Then you mess with the terms to force the terms you want to come out

[still ibex]

Hmm okay

I'll mess around with it

Thanks for your help!

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[bold silo]

can someone help me on (a), i have no clue on where to start

[severe portal]

Notice that's a geometric series (and you can easily show it's true). They do want you to write a inductive proof tho.

[bold silo]

yeah idk how to do an inductive proof

[severe portal]

Ah I see. If I had the time, I'd explain myself but I don't so I'll out source:
https://math.libretexts.org/Courses/Mount_Royal_University/Mathematical_Reasoning/3:_Number_Patterns/3.1:_Proof_by_Induction

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@bold silo Has your question been resolved?

[median chasm]

@

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@bold silo Has your question been resolved?

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[dense lily]

real

[mint orchid]

umm last night i tried to get help and they said that the velocity id the derivative of the acceleration

so i tried to take the derivative of the function which gave me 10 but that was wrong

[nimble sleet]

Other way around, acceleration is the derivative of velocity

[nimble sleet]

You'd instead need to integrate

[mint orchid]

oh okay i can do that

[nimble sleet]

Be mindful of your integration constant; that's your initial velocity

[mint orchid]

so its (40t^2)/2 - 40t +c

[nimble sleet]

1- where did the first 40 come from
2- read the message i sent above

[mint orchid]

1. i got 40 from the function: 10t-40 2) i dont remember what a integration constant is

[nimble sleet]

Your function is 10t + 40 but you made it 40t+ 40

[nimble sleet]

Integration constant is the C here

[junior kraken]

v = ds/dt
a = dv/dt
v = integration a
v = 10t²/2 + 40t + c where c is the constant of integration, it is given at t=0 , v = 50
So you can figure out c

[mint orchid]

oops that was a typo it is supposed to be (10t^2)/2 -40+c

[junior kraken]

Why -40 ?

[nimble sleet]

Well u still have two typo's in there

[mint orchid]

omg sorry + 40t

[nimble sleet]

That's one

There you go

[mint orchid]

okay great so do i need to use the v(0)=50 in this part of the question? or is the whole thing just asking for the integral

[nimble sleet]

Well

[nimble sleet]

Have you figured out what C is meant to be here after all?

Try to substitute in t = 0 for example

[mint orchid]

would it be 0

[nimble sleet]

No

you have v(t) = 5t^2 + 40t + c

v(0) = ?

[mint orchid]

v(0)= 50

[nimble sleet]

Which means c is?

[mint orchid]

c=50?

[nimble sleet]

Indeed

[mint orchid]

okay so i plug that in for c and now my function is: 5t^2+40t+50

[nimble sleet]

Right

[mint orchid]

okay awesome

now i know for the second part i need to do absolute value

[nimble sleet]

Well

You gotta integrate again

[mint orchid]

integrate which part?

[nimble sleet]

.

[nimble sleet]

Or this i guess

[mint orchid]

here is a picture becasue i dont want to type it

[nimble sleet]

Ok yes

Now

Here's the deal

It wants the total distance

Meaning you can't just do x(10) - x(0)

You need to figure out where x(t), what you got, is equal to 0. From there, you need to find the regions where x(t) ia negative

Well in reality it'll be all positive so i guess there's no point in doing the above, but it's good to know that you need to do so

[mint orchid]

okay so i set the above equation to 0 firsto to find where it is 0 so that i can know where to set the bounds?

[nimble sleet]

Yeah but it's ok to realise that since t > 0 it'll be strictly positive anyways

So u dont need to do that

You can just do x(10) - x(0)

[mint orchid]

silly question but i can leave out the +c when im doing this right?

or do i plug 10 and 0 in for c too

[nimble sleet]

Yep

U r subtracting them off so thr Cs cancep

[mint orchid]

okay great

thank u!

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[distant loom]

Can someone please show me step by step how to resolve this

[distant loom]

And the answers i can choose are {-3; -2} , {-3;2} , {-2; 3} and {2; 3}

[jagged cobalt]

what are your ideas

[distant loom]

I think i should multiply the first one with √5 over √5

[jagged cobalt]

you could do that but i wouldnt say it is too helpful

[distant loom]

Well, what do you suggest?

[jagged cobalt]

what else could you do with sqrt(5)

[distant loom]

Uhh, could it work to simplify the √5 with the 5√5?

[jagged cobalt]

how?

[distant loom]

I dont really know man

We just started learning about this and i dont understand anything

[jagged cobalt]

how could i get the left to just be 2x-1

[distant loom]

Get rid of √5

[jagged cobalt]

by doing what?

[distant loom]

Like divide the √5 with √5 so it becomes 1

So 2x-1 over 1 is 2x-1

[jagged cobalt]

thats the idea, so youre saying to multiply by 1/(1/sqrt(5))?

[we would normally say just multiply both sides by sqrt(5) ]

[distant loom]

I dont really understand what 1/(1/sqrt (5)) is supposed to look like, can you draw it?

[ocean seal]

AℤØ

[tacit arch]

https://www.mathsisfun.com/fractions_division.html

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[spring lodge]

I don’t think this is maths

[obtuse berry]

damn tru

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[obtuse berry]

mb gng

[spring lodge]

GL

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[alpine sable]

My lecturer made a mistake, right?

[spring lodge]

Yup

Wait

Yea

Mistakes in the first line of equations

Need the correct steps?

[alpine sable]

In the first line? Not the second line?

Yes please

[spring lodge]

Wish I had my white board… 😦

Anyways

[little drum]

Just replace the 2.5 with 1.5 on top line right index

And you're set

[alpine sable]

oh bruh

indeed that was the mistake

thank you guys! 🙂

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[frigid elbow]

guys if A is a square matrix M3x3 and |A| =/ 0, its invertible what about a Matrix M3x2 with |A| =/ 0

[hushed locust]

you can't take the determinant of a non-square matrix

[frigid elbow]

ohh true

ok thx ahahah

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[naive valley]

and a non-square matrix can't have an inverse

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[raven lava]

.help

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Commands:

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[raven lava]

HELP MEEE

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[raven lava]

2x(squared)+7xy+3y(squared)+9x+7y+4

factoring

plz

someone explain

[fathom grove]

$2x^2 + 9x + 7xy + 7y + 3y^2 + 4$

[ocean seal]

890s

[raven lava]

tx

.end

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.close

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[charred hollow]

In isosceles triangle ABC with base BC, AB = 5.1 cm. Point D is on side AB so that the length of BD = 1.7 cm. Point E is on ray AC so that the area of triangle ABC is equal to the area of triangle ADE. What is the length of segment AE?

[charred hollow]

I tried to diagram it but I don't know how to proceed from there

Maybe I'm drawing it wrong?

<@&286206848099549185>

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@charred hollow Has your question been resolved?

[charred hollow]

<@&286206848099549185>

[vagrant crow]

nah diagram looks right to me

[charred hollow]

How do I keep going?

[vagrant crow]

im having a brain fart rn

but try find the length of BC

[charred hollow]

But you don't have any angles

[vagrant crow]

no but i'm sure that you can do it idk how tho

no you cant

[charred hollow]

A part two says that the length of DE cannot be determined and wants me to find the difference between the longest and shortest lengths of DE

But first I want to find AE

All that's given is that ABC and ADE have the same area and the lengths of AB and BD

And that ABC is isosceles

[charred hollow]

Yeah

How do I proceed from there?

[vagrant crow]

not sure

[charred hollow]

<@&286206848099549185>

[jolly timber]

What is it

[charred hollow]

question is pinned

[charred hollow]

<@&286206848099549185>

In isosceles triangle ABC with base BC, AB = 5.1 cm. Point D is on side AB so that the length of BD = 1.7 cm. Point E is on ray AC so that the area of triangle ABC is equal to the area of triangle ADE. What is the length of segment AE?

[charred hollow]

<@&286206848099549185>

[safe meadow]

rotate you triangle so ac is the base so as ae

your triangle will look like this

except the point e in the question is being replaced by f

and altitue line from d to ac and name it point h

DA/DH = AB/BG

let line DH = x

then BG = 3x/2

name point CF = Y

x(5.1+y)/2 = 3x(5.1)/4

you can find the length of CF

the question asked for AF = AC+CF

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[dawn shard]

Please @ me if you see this

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[potent dust]

@dawn shard

[dawn shard]

Oh hi

[chrome shuttle]

@dawn shard

after you converted to kilowatts

all you do is multiply the killowatts expended in a week of usage by the price

[dawn shard]

Thank you already figured it out

[chrome shuttle]

oh

!done

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[sick crest]

how do i compare # 5?

im confused how to do this

A = O(b) etc..

[buoyant saddle]

use a^b = e^{bln(a)}

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[regal parrot]

how is the intial condition applied?

[tacit arch]

Did you solve the first order linear differential equation

[regal parrot]

yes

this image is my work, the previous ones are from a journal

[tacit arch]

You changed the variable names?

[regal parrot]

phi is given in the journal, whichare the first 2 images, and delta is in the second image which is my work

so \phi is the same as \delta

<@&286206848099549185>

[tacit arch]

And did you use eqn 7

Whatever that is

[regal parrot]

yes i did

what i dont get is how the initial conditoin is applied in the journal

and more specifically how its applied, if applied to Eq 7

as im getting 1/\delta = 0, which is wrong

[tacit arch]

I don't see where you even plugged in initial conditions

[regal parrot]

i havent done that yet, ii was looking at how it was done in the journal, which is what i dont understand

but it would look like this

$\frac{1}{\delta}=\frac{1}{s}\frac{ds}{dt}\
\frac{1}{\delta_0}=\frac{1}{s_0} \frac{ds_0}{dt}\$

[ocean seal]

beesaH

[regal parrot]

but then ds_0\dt=0 and 1/delta_0 \neq 0

[tacit arch]

You don't plug in initial conditions to the differential equations

[tacit arch]

Plug it into 32

[tacit arch]

Oh you did it here

[tacit arch]

The limits on the right side of 31 should be S_0 to S

[regal parrot]

yeah

[regal parrot]

right, to eliminate the integration constant

[tacit arch]

Not sure where beta * I_0 comes in though

[regal parrot]

was also confused about that

[tacit arch]

Probably from other information given you're not showing

[regal parrot]

ythis is from the journal

[tacit arch]

Yea you left out way too much information at the start to answer your question

[regal parrot]

yeah sorry about that

does it make better sense now?

[lone heart]

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[regal parrot]

.close

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[sterile trench]

anyone can help me with composition of inverse trig functions

[sterile trench]

T inverse T for eg. arcsin (sin 100))

[wind cloak]

you need to utilise the periodicity of sine

[sterile trench]

yup ik how to solve that

i was just giving example tho

[wind cloak]

!xy

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sterile trench]

i want to solve it quicker tbh

no like i have a doubt over a concept

not a particular question

[wind cloak]

what's the problem?

[sterile trench]

sometimes i get confused

uhm wait lemme show you how

for eg. there is arc cos ( cos 1000))

[wind cloak]

,calc 318*pi

[ocean seal]

Result:

999.02646384155

[sterile trench]

so we know 1000 rad would be somewhere between 318pi

and 319 pi

[wind cloak]

yeah

[sterile trench]

then how will we like see in which quad does it lie

[wind cloak]

doesn't matter here though

[sterile trench]

i think seeing quad will make it easier to easier to solve?

[wind cloak]

1000 - 318π is in between 0 and π

that's all we care about

[sterile trench]

oh

so like if we have sane qye with tan

same que*

arc tan tan 1k

[wind cloak]

,calc 1000 - 318*pi

[ocean seal]

Result:

0.97353615844577

[wind cloak]

,calc pi/2

[ocean seal]

Result:

1.5707963267949

[wind cloak]

yeah it will still work

[sterile trench]

so it's in first quad so its pos

[wind cloak]

since the value is in between -π/2 and π/2

[sterile trench]

and ans will be 1000 - 318pi

[wind cloak]

yeah

[sterile trench]

idk why i am not able to express my doubt

[wind cloak]

,calc tan(1000)

[ocean seal]

Result:

1.4703241557027

[sterile trench]

gimme a sec

alr so
arc cos (cos(12)) - arc sin (sin(14))

everything in rad

so 12 rad is around 688 deg

and 14 rad is around 799 deg

then we see quad?

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[celest rapids]

Hello

[zinc haven]

[little drum]

:hi:

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@celest rapids Has your question been resolved?

[raw jetty]

do you have a question?

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[thorny helm]

How do i find my y-intercept and slope when given the correlation coefficient?

[thorny helm]

I have a correlation efficient that I obtained from the partial correlation coefficient formula, and I don't have a data table to give me the slope and y-intercept

is it possible to work backwards from the correlation coefficient to get my slope and y-intercept

[hushed locust]

the correlation coefficient only tells you how much the data matches the trend line, not what the trendline actually looks like (except the sign of the slope)

[weary path]

can anyone explin like the vol representation for double intregrals

[thorny helm]

so it wouldnt be possible to work backwards when given the correlation coefficient?

[hushed locust]

not if that's the only information given

you can have the same correlation coefficient for completely different trendlines

[thorny helm]

ah

does it help that I have a data table for the coefficients that i used to get the partial correlation coefficient?

[hushed locust]

i would recommend you post your original problem and any information you were given

[thorny helm]

its more of a paper

i just wanted to know if i could find the slope and y-intercept when im only given the correlation coefficient

if i cant its alright

[hushed locust]

for a trivial example, any data that lies on a straight line with positive slope will have a correlation coefficient of 1, regardless of the slope or y-intercept

[thorny helm]

👍

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[thin zinc]

i can say we have 3 numbers of 3k type
2 of 3k+1 type
and 2 of 3k-1 type
to get sum of 4 consecutive to be divisible by 3 it can be
3k 3k+1 3k-1 3k
3k+1 3k+1 3k-1 3k-1

using the box method
i listed down the possible permutations and im getting 240 as the answer ( there are 3!2!2! ways to interchange numbers of each type)

which i believe is wrong

[prime badge]

it has to be like 1 2 0 1 2 0 1

in this case 0 1 2 0 1 2 0

3!2!2!

sounds good to me

wait i thought 3 consecutive

[thin zinc]

yea its 4

[prime badge]

so e.g 0 1 2 0 0 1 2

[thin zinc]

going through all perms would be difficult like this ig

[prime badge]

1 2 0 0 1 2 0

the first three must have 1 zero

[thin zinc]

i think yes

[prime badge]

well it's 3!

times 3!2!2!

middle is 0, left is free, right = left

[thin zinc]

???

i dont understand it

oh yes middle has to be 0

and 3! ways to fix the left sidey

the right side is fixed automaticall

[prime badge]

0 1 2 0 0 1 2
0 2 1 0 0 2 1
1 0 2 0 1 0 2
1 2 0 0 1 2 0
2 0 1 0 2 0 1
2 1 0 0 2 1 0

you got it

[thin zinc]

yepp

so its 144

tysm

.close

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[prime badge]

eyah

[fair kelp]

Does anyone have the picture of the banner of this server to send me? I love it and would love to use it as my phone wallpaper

That pic

[thin zinc]

bro jus search math wallpaper on google

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[slow pine]

can anyone help me understand functions pls

[north rover]

give a question, and be more specific, pls

[slow pine]

i can’t , i have an exam on it and don’t understand any of it

graphic readings , etc

[still prism]

what did you learn in class?

[lone heart]

@slow pine Has your question been resolved?

[grand grotto]

functions? highschools

[lone heart]

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[radiant hedge]

Why am I wrong?

[pseudo ice]

The sign function has a primitive of the absolute value function

[radiant hedge]

idk russian

[radiant hedge]

wdym sign function?

[tardy stag]

please keep conversations to english unless otherwise specified...

[radiant hedge]

he's trolling

[pseudo ice]

Gives you the sign of a number, +1 if strictly positive, -1 if strictly negative

[radiant hedge]

so?

[tardy stag]

so the sign function has a jump discontinuity at 0 and has a primitive near 0

[radiant hedge]

oh I see

[radiant hedge]

Also the answer key says the answer is D

Why is D false?

[tardy stag]

think of some relatively simple rational functions

[radiant hedge]

1/(1-x^2)

Are trigonometric function not rational?

[tardy stag]

trigonometric functions are not rational, no

[radiant hedge]

Oh so that's why

thx

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[sturdy crypt]

HELLO

[sturdy crypt]

hey hey

hey

[lone heart]

@sturdy crypt Has your question been resolved?

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[tight pier]

Probably trivial but can somebody still check it please

[little drum]

If you have learnt to show that, for any prime p, there exists a field of order p^n, n natural. Then you should show the same for 7^3 in the same way. That is what was probably intended by this question

Otherwise this is equivalent to a one line sol. Since 343 = 7^3, hence there exists a field K with o(K) = 343

[tight pier]

good point

I will have to see then, but I actually believe one sol line, regardless thanks!

.solved

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[eternal badge]

can somebody help me understand the circular segment and circular crown area? im 14 btw and i dont speak english very well

[little drum]

inappropriate ._>

[eternal badge]

why'

?

OH

now?

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[stable palm]

hii i need help on a wuestion i got. its translated from swedish but says “the lines y=(-x/2)+2 and y=kx+m both have the same x intercept. together with the y axis, they shape a triangle with an area of 12. what is the x intercept” i have no idea what to do

[stable palm]

plz help :(

<@&286206848099549185>

<@&286206848099549185> i dont even know where to start

[quick fable]

do you have a drawing of this?

[stable palm]

nope :(

[quick fable]

x incept means similar point for both?

[stable palm]

they both have same x intercept meaning they cross x axis at same point

[quick fable]

okay

thanks

lemme draw it

this one is the first line

-x/2 +2

[stable palm]

ok yes

[quick fable]

are we told what k and m are?

[stable palm]

nope

[quick fable]

then it's with parameter

we have to solve every possible case then

[stable palm]

the answer sheet says (4,0) but i have no idea how they got that

or how to get it

[quick fable]

might be truth

if so

imgine line

going up as x increases

and it's point which touches y axis is (0;4)

then we have trinagle

and y axis is base

[quick fable]

then m is equal 4

[stable palm]

im still confused bc weare supposed to work the x intercept out algebraicly

[quick fable]

oh

maybe

try

this thing

and

if we have to find "together point"

we can do it like y=y

so

[quick fable]

@stable palm .

[stable palm]

ok i will try

[quick fable]

and then solve x maybe

[quick fable]

i guess the point might be (-2;3) then'

not 100% truth yet

[stable palm]

Found teh answer! it is 4

Thank u for helpign

.close

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[quick fable]

if something's still wrong you may

dm me

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[young steppe]

can someone pls explain this qs

i have no idea how to solve it

i got the common side between the 2 triangles which is 6 but idk what to do after that

[gusty yew]

a

[cedar hawk]

Correct bu5 how would you explaon that to them

[gusty yew]

Pythagoras

[cedar hawk]

Ik but do they

[gusty yew]

Ok

I'm not good at explaining

But I'll explain to them

[gusty yew]

So basically

We know that we have a right angle triangle

which mean we need to use pythagoras

10^2 - 8^2

Square root the answer

6

then we use tan

6/8

4/3

[little drum]

what is the other angle in the right angled triangle, in terms of theta

[cyan saffron]

if you try and get some angles in terms of theta, it just so happens that one of them is also theta!

[gusty yew]

💀

[cedar hawk]

You really judt said 6/8
4/3

[gusty yew]

uh

4/3

thats the answer

[young steppe]

hey guys i'm so sorry i'm back

[little drum]

the question asks for tan theta

[young steppe]

but why with 8

[little drum]

do you understand the figure?

[gusty yew]

wait a second

c?

[cedar hawk]

?

[young steppe]

the answer's a but idk how

[gusty yew]

Ok

I was correct

[cedar hawk]

It is 4/3

[gusty yew]

Yeah

[young steppe]

so the triangles are similar?

[little drum]

Yes

but that's besides the point

[young steppe]

how did you know that, like how did you get that the triangles are similar

[little drum]

Think you can figure the altitude?

[little drum]

do the angle chasing, I did not know. I found it

[young steppe]

ik what to do after this

but i'm just confused on how you found that the 2 triangles are similar

would the big triangle also be similar to the other 2 small ones

[little drum]

what is the angle on the left vertex, in terms of theta

[young steppe]

wdym

[gusty yew]

Why you doing this?

[young steppe]

like where the triangle splits???

[young steppe]

yea what does the altitude have anything to do with this

[little drum]

this

[young steppe]

ik all the rules, i just didn't get how you figured out they were similar

they're similar by which rule??

[gusty yew]

I don't get why you would need all of this

I found out the answer in a few steps

You can explain what you do though

I'm out

[unreal jolt]

angle angle

[little drum]

1. You found the altitude here
2. You used angle chasing to figure 8/6 in the smaller triangle is the ratio we require
3. All you did was look at the problem, flex your answer and left. Idts any of that random rambling really helped the OP

[young steppe]

which 2 angles?

the right and the acute?

[gusty yew]

I wasn't flexing?

[little drum]

the big triangle is a right angled triangle

[unreal jolt]

yes, the shared angle and the right angle

[little drum]

what is angle ACD

Idk why people can't just label their figures

what's angle ACD

[young steppe]

DAB

[little drum]

in terms of theta

[young steppe]

like it's the same as DAB right

[little drum]

but that does not help us in this situation.... I can say I had oreos for breakfast but how does that help here?

[gusty yew]

Okay

[little drum]

what's ACD in terms of theta

[young steppe]

what's 90-theta gonna do for me

[little drum]

what's CAD then

[young steppe]

i'm not getting any info, idk what the value of theta is

[little drum]

what is CAD in terms of theta

[young steppe]

cad is theta

[little drum]

that's how I drew this figure

[little drum]

ring a bell now?

[young steppe]

no

[little drum]

wow... so what about this figure did you not understand

[young steppe]

what did i benefit from it

i asked you how they were similar

by which rule were they similar

[little drum]

when did I mention they were similar..

and why would you require similarity

[young steppe]

.

[gusty yew]

He said its besides the point

[young steppe]

the answer in my solution pdf mentions similarity

[little drum]

Alr. ABC = CAD, BCA = DCA => triangles DCA and ACB are similar

but why would that help

[young steppe]

i just didn't understand how the solutions pdf explained it

[little drum]

you require tan theta

from here you clearly get tan theta = 8/6 = 4/3

why similarity

[young steppe]

ignore what you drew

you see where theta is

[little drum]

yes

[young steppe]

how it's in it's own separate triangle

[little drum]

yes

[young steppe]

triangle ADC and triangle ABD are similar

thats why theta is angle CAD

wdym it's not important

[little drum]

ADC ~ BDA

[young steppe]

yea that

[little drum]

HUHHHH... we got CAD = theta without any need for similarity tho?

[young steppe]

i typed it wrong

[little drum]

Also, they used CAD = theta for their similarity condition probably

nevertheless, we're straying from the Q

[little drum]

.

you were asked tan theta, it's on you whether you calculate all the remaining sides and angles of the triangle to answer the Q, or get to the point and find tan theta

I've explained to the best of my abilities. Tyvm

[young steppe]

okay well thank you

.close

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[brittle lake]

Can someone help me how I would evaluate this?

[brittle lake]

Im confused on how I would solve this

i know tan^x = sec^x -1

but that doesnt help me

here

i think

[north rover]

so $\tan^2(x)=\sec^2(x)-1$.

[ocean seal]

;(

[north rover]

you can just realize that this is a splittable fraction

i.e., it turns into $\int\cos^2(x)-\tan^2(x)\cos^2(x)dx$

[ocean seal]

;(

[north rover]

and this is pretty simplifiable

recall that tan=sin/cos

[brittle lake]

icic

.close

thanks

[lone heart]

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[brittle lake]

.reopen

[lone heart]

✅

[brittle lake]

nvm

wait

[north rover]

wat

[brittle lake]

How do i know what methods to use for integration?

[north rover]

for?

[brittle lake]

by just lookinga t it

because youd want to use integration by parts if you want to solve the integral for

$\int\csc^3(x)$

[ocean seal]

Light

[brittle lake]

and not

trig sub power reduction whatever

[north rover]

yes, since power reduction isn't really nice here.

[tight pier]

dx

yes

[spiral lily]

you dont

its mostly hugely trial and error and recognizing little things little by little

kinda like geometry

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@brittle lake Has your question been resolved?

[north rover]

"nah i'd be the notationless one"

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[indigo hound]

How is he getting the 30 degree angle here on the right side?

[north rover]

he just rotated the image

[hushed locust]

there is a theorem saying "vertical angles are congruent"

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@indigo hound Has your question been resolved?

[lone heart]

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[proven yoke]

how do you know when you can factor to solve quadratics vs using the square root method?

[placid zinc]

ax² + c = 0
vs
ax² + bx + c = 0

The top can be solved just by isolating for x. I think that's what you're calling the "square root method"

The bottom one is a bit more complex and needs better methods

[proven yoke]

ok, but specifically what properties of an equation make factoring possible instead of solving using square roots?

[tardy stag]

i mean technically they can all be factored

if you want to factor them over the reals then b^2 - 4ac has to be ≥ 0

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[wicked kettle]

how do i do this

[wicked kettle]

nvm

how do i do this

plz

[safe meadow]

Draw a line from O to C

Now you have a isosceles triangle AOC and isosceles triangle COB

Angle CAO = angle OCA ( since its isosceles)

[tardy stag]

a+c = 180

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[thin meadow]

lmao

[zenith flax]

damn gg

[thin meadow]

gg indeed

[north rover]

what is RREF form for you?

[thin meadow]

i have no clue

1. In each row which has any non-zero entries, the first nonzero entry is a 1.
   We call this 1 the leading one for that row.
2. In each column which contains the leading 1 for some row,
   all other entries are 0.
3. The leading ones for the rows move from left to right as you
   read down the matrix.
   i.e. each is further to the right than those above it.
4. If there are any rows that only contain zeroes, they are at
   the bottom of the matrix

.close

[lone heart]

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[lone heart]

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[thin meadow]

I am super lost

[tight pier]

each row tells you an equation

and the entries of that row is the coefficient of the variables, I wrote above

[thin meadow]

then whats the 7 2 and 3

and why are the t,r,s in the options?

[tight pier]

that what comes after =

[tight pier]

usually you redefine so called "free variables"

The 3rd row tells you for example what x_5 is

that information alone should suffice what answer to pick

[thin meadow]

option 3 right?

[tight pier]

yep

[thin meadow]

thx

i got 2 more question if you dont mind?

[tight pier]

These are difficult to tell instantly, so try to write the system into a matrix and bring it into row echelon form

[solid kettle]

Dababy

[thin meadow]

whats row echelon form?

[lone heart]

@thin meadow Has your question been resolved?

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[solar glade]

someone help me?

[zinc haven]

!sseei

[lone heart]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[solar glade]

oh

my fualt

[broken pivot]

which ones do you need help with

[solar glade]

like all tbh

i dont really understand it

i can do 6 and 7

and 13

but i dont really understand all of it

[zinc haven]

then do it

!1q

[lone heart]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[solar glade]

i did them

its just not on the paper

cuz thoe are the questions

i got 5-18t for 6

2x-6x^2 for 7

and 13 accelartion is c

velocitiy is A

and position is B

i think

but the others idrk

[broken pivot]

13 is wrong (i think)

[solar glade]

yeah

[broken pivot]

I was never good at identifying that stuff so dont take my word for it

[solar glade]

A and B might be switched

but C is def accelration

i think

[broken pivot]

$$g(x) = \sqrt{9 - x}$$

[ocean seal]

King Leo

[broken pivot]

First, i recommend expressing the square root as an exponent

[solar glade]

that one i got $$-1/2sqrt{9-x}$$

[ocean seal]

Ryanqwazwsx

[broken pivot]

$$-\frac 1{2 \sqrt{9 - x}}$$

[ocean seal]

King Leo

[broken pivot]

Thats right

[solar glade]

yeah

i used the cahin rule

chain*

or sum

[broken pivot]

9. $G(t) = \frac{1 - 2t}{3 + t}$

[ocean seal]

King Leo

[broken pivot]

Quotient rule:
$$\dv{x} \qty(\frac{y_1}{y_2}) = \frac{y_2 \dv{y_1}{x} - y_1 \dv{y_2}{x}}{\left.y_2 \right.^2}$$

[ocean seal]

King Leo

[broken pivot]

"Low D High Minus High D Low over Low Low"

[solar glade]

what does that mean

😭

[broken pivot]

Its a "poem" to memeorize the quotient rule of differentiation

[solar glade]

oh

so uh

how am i supposed to use that

to solve 9

[broken pivot]

Quotient rule:
$$\dv{t} \qty(\frac{y_1}{y_2}) = \frac{y_2 \dv{y_1}{t} - y_1 \dv{y_2}{t}}{\left.y_2 \right.^2}$$

[ocean seal]

King Leo

[broken pivot]

Assuming $\frac{y_1}{y_2} = \frac{1 - 2t}{3 + t}$, what are $y_1$ and $y_2$?

[ocean seal]

King Leo

[solar glade]

y1 is 1-2t

and y2 is 3+t

right

[broken pivot]

Now can you try to use the formula

[solar glade]

yeah hold on

Is it that

[broken pivot]

• you need parentheses in your numerator
• you must actually solve for d(...)/dt

[solar glade]

wdym by solve for d.../dt

like

for d(1-2t)

d-2td?

[broken pivot]

$$\dv{t} \qty(1 - 2t)$$
$$\dv{t} \qty(3 + t)$$

[ocean seal]

King Leo

[solar glade]

wait what

how does that work

when the formula has the y2 and y1 on the numerator of the fraction

[broken pivot]

$\dv{y_1}{t} = \dv{t} \qty(y_1)$

[ocean seal]

King Leo

[solar glade]

oh

ok

hold on

so its

like that?

[broken pivot]

Now you just need to solve for the derivatives

[solar glade]

idk how to do that ngml

ngl

[broken pivot]

You dont know the derivaitve of (1 - 2t)?

[solar glade]

oh

i do

mb

its -2 right

or am i tripping

[broken pivot]

Now whats the derivative of 3 + t

[solar glade]

1

?

[broken pivot]

So now, whats $\dv{G}{t}$

[ocean seal]

King Leo

[solar glade]

-2?

or what

idk

wait

do i just substitute

-2 and 1

[broken pivot]

Can you substitute in the derivatives of 1 - 2t and 3 + t

[solar glade]

for d/dt(1-2t) and d/dt(3+t)

ohhh

ok

let me solve rq

Like that?

[broken pivot]

You messed up here

[solar glade]

oh

i didnt distribute negative right

[broken pivot]

Also, American AP Calculus Ab will allow you to just keep the first line (without further simplification)

[solar glade]

yeah im taking that next year

i think

or bc

[solar glade]

wdym first line

[broken pivot]

You can just give your answer as
$$\dv{G}{t} = \frac{-2(3 + t) - (1 - 2t)}{(3 + t)^2}$$

[ocean seal]

King Leo

[solar glade]

oh

ok

ill simplify tho

[broken pivot]

-# assuming youre studying american ap calculus ab

[solar glade]

like that?

or did i mess up something again

[broken pivot]

You will lose points for omitting your denominator, even if you put it back later. Other than that, looks good

[solar glade]

ok ok

[broken pivot]

,w d/dx (1 - 2x)/(3 + x)

[solar glade]

can you explain the questions starting from question 1

from after this question

so i can write it down in order

@broken pivot

?

uh

@broken pivot ?

uh

<@&286206848099549185>

can one of you guys help me

the other guy just kinda left

[heavy delta]

what question u asking?

can u repost again

[solar glade]

^

those

kinda all of them

i alr did 6-9

the others i kinda lost

[heavy delta]

just reposting so i can see it easily

[solar glade]

ok

[heavy delta]

so for q1

uk what gradient mean right

[solar glade]

uh

no

lol

we like js started learning this stuff

and limits

[heavy delta]

basically gradient mean how steep the tangent line is

[solar glade]

oh

i think we talked about it in class

[heavy delta]

steep line = high gradient
horizontal line = low gradient

[solar glade]

ok

[heavy delta]

more steep, higher gradient

so derivative of a function f(x) is a function that represent the gradient of tangent point of f(x)

so like for (a)

eh wait its harder to say in words

give me a min

[solar glade]

ok ok

[heavy delta]

see from left to right, the line go from neg slope and when the line kinda like "U -turn" it become positive slope

[solar glade]

yeah

ik that

[heavy delta]

so the derivative function should be increasing from negative value and pass "slope of zero" and become positive slope

and we know that the derivative function should be neg -> 0 -> positive from left to right

so it's II

[solar glade]

wait what