help-49

[twilit field]

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[twilit field]

thanks everyone

[midnight plank]

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[slim portal]

3^(2x+2) + 8*3^(x) -1 = 0

[slim portal]

$3^{2x+2} + 8*3^x -1 = 0$

[grand pond]

カザミ

[rain holly]

what is the issue

[slim portal]

How to solve it?

[rain holly]

substitute u = 3^x

then it reduces into a quadratic

[slim portal]

So $3^{2x+2} + 8*u-1 = 0$?

[grand pond]

カザミ

[rain holly]

rewrite the first term $3^{2x+2}$ as $3^2 \cdot (3^x)^2$

[grand pond]

Deftioon

[rain holly]

remember though you should check your answer for u to make sure its in the range of 3^x since 3^x > 0 for all x

[slim portal]

Thank you ♥️

[slim portal]

So 3^x = -1 not included since theres no exponent numbers < 0?

I got x = -2

Thanks

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[rain holly]

yeah

if you get a u = negative number solution that should be rejected if you're solving for real x

[slim portal]

How about 3^(-999999999)?

Is it = 0

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[floral grail]

what went wrong for me? the typed pic has the correct answer

[lyric charm]

$\frac{1}{4\sqrt{y}} = \frac14 y^{-1/2}$ not $4y^{-1/2}$

[grand pond]

ann.in.a.teacup

[floral grail]

ohh ok thx

.close

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[subtle peak]

How do I solve this?

[subtle peak]

Well I think I'm solving it correctly, I just can't get it into standard linear form

[midnight plank]

@subtle peak Has your question been resolved?

[sharp wave]

could you send a better picture of the question

and your solution

[subtle peak]

Certainly!

[sharp wave]

after your first step why dont you divide by y^4/3 first then proceed

i think you have made a calculation error afterward

[sharp wave]

your substitution is correct

[subtle peak]

Because that wouldn't match Bernoulli's equation

[sharp wave]

i dont think it matters, since you get rid of the dy/dx term straight away

let me write and show you one sec

[subtle peak]

Why not? Using bernoulli's equation is to then make the v substitution right afterwards

It's the whole point of it

See how I'm putting it into the bernoulli equation form to then find v?

[subtle peak]

Ok, but how do you solve from there? It's not separable

[sharp wave]

Oh, idk ive been taught to divide by y^n first then sub 1/y^n-1 as u or t

[sharp wave]

its a linear differential equation

[subtle peak]

I think that's called an extra unnecessary step

[sharp wave]

Well, whatever works for you

[subtle peak]

I think I see what you're saying

[sharp wave]

Your method is not wrong, im just not accustomed to it let me try though

[subtle peak]

Yeah, and you seem to have gotten a more correct answer, so I don't even have the right to judge

[sharp wave]

i think you should differentiate this with respect to x

[subtle peak]

I see my mistake

[sharp wave]

not the equation written below it

[subtle peak]

No it's at where I plug it in

Because I didn't divide by 3

I multipled by 3*4

[sharp wave]

oops

[subtle peak]

Getting v^-12

[sharp wave]

great you got your mistake

[subtle peak]

Am I tripping or do we still not have the same answer though

[subtle peak]

You forgot x on the rhs of the first step

[sharp wave]

i divided by x

the whole equation

[subtle peak]

You're right, that's my second mistake

I still have something different than you

[sharp wave]

how have you differentiated?

dy/dx = -3v^-4dv/dx

[subtle peak]

Oh right. My math work is a festival of disappointments

[sharp wave]

you can refer this if it seems easier to you

i just feel its a lot less work

again what you feel is best

[subtle peak]

I'll stick with what the teacher taught us because I don't have time to afford learning a new technique

[sharp wave]

sure!

[subtle peak]

I will send this to myself and save it for later on though if I can

[sharp wave]

great

[subtle peak]

Dude, this semester has been brutal. It's been a year since I've touched calculus

[sharp wave]

i hope you can carry on its like 12am for me and i really need to sleep

[sharp wave]

all the best!

[subtle peak]

Thank you for your help. I'll refer to your work

Have a great night! Rest well

!done

[midnight plank]

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[subtle peak]

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[sharp coral]

factoring is part of it, the other part is multiplying by the conjugate

[gritty rune]

I see nothing wrong in your solution (its 12am so maybe im wrong too) but you can solve this using hopital it will be faster

[lyric charm]

lhopital is a nuke

is somebody forcing you to do it that way??

hrgh

[gritty rune]

Bruh why its so good🫡

[lyric charm]

well ig you could pull a factorization of x-2 as x+2-4 and then as (sqrt(x+2)-2)(sqrt(x+2)+2) out of thin air

[sharp coral]

you could factor [ x - 2 = (x+2) - 4 = (\sqrt{x+2} + 2)(\sqrt{x+2} - 2) ] but that's baroque enough that I don't think you could really think of it ahead of time

[grand pond]

cloud

[lyric charm]

it is a nuke in that it is such heavy machinery that it's very very easy to misuse

[sharp coral]

i think that the method you used is fine for that prompt

yes

[midnight plank]

@jolly roost Has your question been resolved?

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[unique hornet]

I need help finding the area of the shaded region, i have no other angles or values that i can work with, can somebody help me continue?

[shy marsh]

Calculate x using pythagoras

[unique hornet]

ty, i got its equal to 6 and the totaal area is 54, but how do i find the areas of the other two sectors?

[shy marsh]

Put them together

like see them as one sector

cuz they have same radius

[unique hornet]

ohh so their angles combined are 90?

[midnight plank]

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[dense holly]

anyone good at circuits and physics

[dense holly]

?

like kirchoffs law and stuff

need help with this question

[carmine sigil]

The two nodes immediately above and below the circled A are at the same potential. So R2 and R1 make a parallel circuit, and so do R3 and R4.

The current is divided through the two sets of resistors independently, so the current through A depends on how much they differ from each other.

I would approach this problem in the following way:

1. Find the effective resistances of R1 || R2 and R3 || R4
2. Use those effective resistances to find the total current, and the voltage at A.
3. Using the known voltage at A, find the current through each of the four resistors
4. Using the known currents find how the currents differ, and how much needs to flow through A to make the circuit happy

@dense holly

[dense holly]

i will read it thank you

[dense holly]

so you are saying I should use KVL?

to find U/V for the ammeter

after i have the total resistance

[carmine sigil]

You use Kirchoff's laws in several different ways

But you can't use U/V for the ammeter, because there is no voltage change

There is also no resistance

[dense holly]

oh alr as for the

adding the resistances

its gonna become a series circuit right?

with Rp -- ammeter -- Rp smth like this?

[carmine sigil]

R1 and R2 are in parallel

Ah, yes

Exactly

[dense holly]

ok and i add both Resistances

[carmine sigil]

Ignore the ammeter though, it's 0 resistance

[dense holly]

i can get the total I

1.5/R total

then get the I individually for all the Rs?

and just look around till i get a value?

[carmine sigil]

Use voltage divider

[dense holly]

whats that

[carmine sigil]

To get voltage at A

V = IR for individual circuit elements as well

[dense holly]

i dont understand the current flow tho

is it from the bottom or up to the ammeter

it depends on the sign of th e value?

[carmine sigil]

You don't need to know immediately

[dense holly]

aör

alr

ima give it a try

see what. ican do

thank you

[carmine sigil]

Though with a little thinking qualitatively you can determine direction simply in this case

We know that lower resistances get more current in parallel

So we know more current goes through R1 than R2 and more current goes through R3 than R4

[dense holly]

didnt know that wow but its true

[dense holly]

so i see

more current from bottom

leads to more current upwards

through the ammeter node

[carmine sigil]

Exactly, we can qualitatively determine that the ammeter is seeing current flowing upwards

[dense holly]

i see

i got total I as 0.0063amp

6.3 mA

[carmine sigil]

Then for the voltage divider, we have V_tot/R_tot = V_part/R_part as long as the part is in series with the rest of the circuit

[dense holly]

sounds wrong

[dense holly]

oh

[dense holly]

i did this

[carmine sigil]

This is equivalent to observing that the total current is constant

[dense holly]

1/100 + 1/200 = 3/200 so Rp = 200/3

did for both and in the end added as a series

i then did 1.5/the value

i got 0.0063 mA

[carmine sigil]

That sounds reasonable

[dense holly]

however

I1 is 15mA

I2 is 7.5mA

or no wait

forget that

those were old values

from previous arttempts

how stupid can i be

[carmine sigil]

Not stupid

Just learning

[dense holly]

actually

that does make sense

so look

R1 for example is 100 Ohms

i do 1.5/100 ohms

that gives me 15mA

thats more than the total?

[carmine sigil]

Not quite!

You haven't found the voltage at A

[dense holly]

isnt the voltage at A 1.5?+

[carmine sigil]

The entire 1.5 volts isn't spent across that resistor

[dense holly]

oh

[carmine sigil]

It's less than 1.5

[dense holly]

i forgot about that

U = RI

so

uh

do i use

supplied voltage = total voltage drop?

to figure that out

or

[carmine sigil]

V_tot/R_tot = V_part/R_part

Using R1 || R2 as the part

[dense holly]

uh

i dont get it

what is meant by a part

[carmine sigil]

Using R1 and R2 in parallel as the part of the circuit that were examining to see the voltage drop across it

[dense holly]

wait i found it i think

for R2 and R1

the parallel thingies

isnt it just

U = R * 0.0063

cause the I total still applies there

[carmine sigil]

Yes, that is another way to find this value

[dense holly]

that gives

0.42V

[carmine sigil]

That seems reasonable

[dense holly]

you said the ammeter has its own voltage right

[carmine sigil]

So the voltage at A is what?

[dense holly]

1.5-0.42?

or

am i wrong

wait

let me say smth

the current is constant rn right?

cause its technically a series circuit

[carmine sigil]

The current though each pair is constant

The current through the individuals can be different

If we let I1 be the current through R1 and so on, then I1 + I2 = I3 + I4

But none of I1 through I4 are necessarily equal

[dense holly]

oh ok

so what current do i use

for R3+R4 ones

not the same I?

[carmine sigil]

You know the Voltage and you know the resistance, so you find the current

[dense holly]

1.08/R?

[carmine sigil]

0.42/R1 = I1

Sorry

Fixed

[dense holly]

oh

[carmine sigil]

Then 1.08/R3 = I3

And so it goes

[dense holly]

ok

so

4.2 mA for I1

2.1 mA for I2

3.6 mA for I3

2.7 mA for I4

[carmine sigil]

Those look reasonable

[dense holly]

i total was 6.3 mA

tho

doesnt add up to that

idk

[carmine sigil]

It does actually

[dense holly]

oh ur right

cause its for I2 and I1

only

[carmine sigil]

I1 + I2 = I3 + I4 = I_tot

[dense holly]

damn

now

i can use

I total + I1 + I2 + I3 +I4 + Ia?

or will it not work

yeah nvm i sound dumb

[carmine sigil]

Use the node laws, total current in = total current out

[dense holly]

ok

I total = I2 + I1

[carmine sigil]

Focus on the node below A

[dense holly]

ok

[carmine sigil]

How much current is going in?

[dense holly]

I1 = Ia + I4

[carmine sigil]

Ok!

[dense holly]

I1 - I4

it gives

11.25

mA

[carmine sigil]

Huh?

[dense holly]

no wait

im stupid

i used the old values again

[carmine sigil]

Not stupid

Just learning

[dense holly]

4.2 - 2.7

1.5

wow

thats an A question for me

difficult

[carmine sigil]

A because amps? 😄

[dense holly]

lmao

thanks a lot

you helped me i will remember this on the day i take that test

means a lot

gonna close this chat thank you again

.close

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[carmine sigil]

Yw! Have a good one

[dense holly]

ty

.close

.close

[carmine sigil]

It's closed

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[languid remnant]

16/36 simplify to 16

[languid remnant]

How to used squeeze theorem in this problem

[subtle blaze]

What does the squeeze theorem state?

[languid remnant]

f(x)<=g(x)<=h(x) and limit equal at one point

Then the limit of f(x) is also equal

[subtle blaze]

So we need to find 2 functions f and h

We are given a hint that -1 <= sinx <= 1

But the function we want to “squeeze” is sinx/sqrtx

(May be important but x is positive since we will take the limit)

[languid remnant]

are we using substitute ?

[buoyant yoke]

find your squeezing functions f(x) and h(x)

[languid remnant]

alright how to solve it?

[buoyant yoke]

which functions did you find

[languid remnant]

1/squareroot(x)

[buoyant yoke]

is that f(x) or h(x)

[languid remnant]

h(x)

[buoyant yoke]

and then whats f(x)?

[languid remnant]

-1/squareroot(x)

[subtle blaze]

Perfect

At this point you just need to write it up nicely

[languid remnant]

oh ok

.close

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[radiant roost]

how do we know that if r and s are coprime and both divide n, then rs divides n?

[dawn dagger]

did u try to prove it?

[radiant roost]

hmm

i just thought about it a little

[midnight plank]

@radiant roost Has your question been resolved?

[dawn dagger]

try to prove it

[radiant roost]

i'm trying

i haven't found anything worth posting

[dawn dagger]

what do you have so far

[radiant roost]

i just know by bezout's identity we have rx+sy=1 for some x,y

[dawn dagger]

ok good

[radiant roost]

we have rp=sq=n for some p,q

[dawn dagger]

ok good

now you could multiply the equation by n

[radiant roost]

rxn+syn=n

oh

rxsq+syrp=n
rs(xq+yp)=n

[dawn dagger]

welcome in Algebra

[radiant roost]

thanks 🙏

.solved

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[olive yew]

could you explain the question

[orchid grove]

What do you mean?

I didn't understand it either

[olive yew]

same bro

what does it mean in each subject there be necessary 1 bell

[orchid grove]

ABCDE

[olive yew]

okay

what we do here

is first we give 1 particular bell to a class

there is 6 bell and 5 class

then we just give the rest of it to each class

because each bell is distinct we can get 5!

then try to solve it by yourself

[midnight plank]

@orchid grove Has your question been resolved?

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[wintry plover]

So im using a hyperbola to model a water bottle cap

[kindred dagger]

Hyperbola for a bottle cap?

[wintry plover]

Yes

[kindred dagger]

Interesting

[wintry plover]

I wanna rearrange this so itll equate to y=

Which I did here but now im confused

[kindred dagger]

This isnt a one-to-one function, so you cant get it to be of the form y = ...

[wintry plover]

If i should hse the positive or negative value

[wintry plover]

Oh wdym By that

[kindred dagger]

Oh ig +/- works

[wintry plover]

Cz Like i wanna put this function

Into the surface area of revolution

Formula

So idk if should use the positive or negative value of the B

[kindred dagger]

I recommend using the positive value cause its eaiser to work with after integrating

[wintry plover]

I did, and I input the positive integral into chat gpt but it gave a negative value

Or should I use a different program

[kindred dagger]

Where did this come from

[wintry plover]

Differentiation

Of the function

[kindred dagger]

Are you trying to find the surface area or the volume

Also is tihs supposed to be your bottle ?

[wintry plover]

Nah this

I used the hyperbola equation

But i limited the domain

To only fit that part

[midnight plank]

@wintry plover Has your question been resolved?

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[orchid grove]

How can I do this?

I tried to look into 3d

One lies on xy plane and other one over yz

[blissful talon]

This is a concept question. Do you understand what it means to be perpendicular to two vectors at once?

[orchid grove]

If it is perpendicular then they form determinant non zero

LI

[blissful talon]

true, but I mean visually. What does it look like?

[orchid grove]

If I join both vectors and made a line

Then it would perpendicular to that line

So it is 2?

One point is upside

[blissful talon]

what is this saying?

[orchid grove]

Other one downside

A line join both vectors

And if I make perpendicular to its line joining

[blissful talon]

what is a line that joins vectors?

[orchid grove]

Actually i don't know its name

But i can imagine

[blissful talon]

it is 2, but you have no argument for it so far

[orchid grove]

a-b

it is line equation

[blissful talon]

thinking about the line that joins the vectors is not the way to go

[orchid grove]

So by fixed distance 5

Only 2 possible

[blissful talon]

a-b is not perpendicular to a or b

[orchid grove]

Nooo

I am saying perpendicular to a-b

which is 5

Distance away

One is upside

Other one is downside

[blissful talon]

there are infinitely many vectors perpendicular to a-b

[orchid grove]

But 5

Distance

Soooo?((

[blissful talon]

there are infinitely many vectors of length 5 that are perpendicular to a-b

[orchid grove]

Ohh

My bad please guide me

How can I solve it

[blissful talon]

when you have a vector in 3d, there are many perpendicular directions

think about a pen standing up on a table. Any pen lying on the table is perpendicular to it.

There's an entire plane of perpendicular

[orchid grove]

Yes sir

I can feel it

[blissful talon]

But if you must be perpendicular to two vectors, then you must be in both of those planes

[orchid grove]

So if i join these vectors

[blissful talon]

and those two planes intersect in a single line

[orchid grove]

And made a plane

So i can make infinite points?

[blissful talon]

Yes, one good way to approach this is to see that if you are perpendicular to a and b, then you are perpendicular to every combination of a and b, and thus to the entire plan that they span

So if you're perpendicular to a plane, now you either go to one side of it or the other

[orchid grove]

I didn't understand it

Yeah two plane

But here we have vectors

a×b?

[blissful talon]

a cross b is another approach to this question

[orchid grove]

Can you explain visually?

I did not understand what planes they are making

[blissful talon]

the set of vectors perpendicular to vector a is a plane

[orchid grove]

You meant yz plane and xy plane?

Which is y=0?

[blissful talon]

This part actually makes no sense

[blissful talon]

we're looking at vectors a and b. Each one lies in a line. It doesn't make sense to talk about what plane they line in because each one lies in infinitely many planes.

[orchid grove]

I see

So how i do next?

[blissful talon]

Do you understand the set of vectors perpendicular to a given vector?

[orchid grove]

Actually no

I can't imagine

😇😇😇

[blissful talon]

this is one way

[orchid grove]

I will be back

Yes this way i made those vectors where they lie

a vector lies in xy plane

B lies in yz

[blissful talon]

those are technically true, but totally useless

(0,1,0) also lies in xy plane and in yz plane

[orchid grove]

These planes they lie

[blissful talon]

choosing a single plane that your given vector lies in is not useful

[orchid grove]

Can you make that plane in desmos 3d?

So that i can understand better because it is language barrier

[blissful talon]

You keep saying that a vector lies in a plane, but that's useless.

[orchid grove]

Okay

[blissful talon]

The vector (1,1,0) lies in infinitely many planes.

[orchid grove]

Okay sooo

What should I do quickly

This is the plane at the start i was saying you which joins both of them

And perpendicular

5

Not exaclty same plane but it will look like this

[blissful talon]

that plane contains infinitely many vectors of length 5

[orchid grove]

a-b=(1,0,-1)

[orchid grove]

I see so this is final answer?

[blissful talon]

no, because this plane that you wrote down doesn't come from anywhere

Try to name 3 vectors in different directions that are perpendicular to (1,1,0)

Alternatively, you could try to find a single plane that contains both a and b

[midnight plank]

@orchid grove Has your question been resolved?

[orchid grove]

I didn't understand how i can name

[blissful talon]

state

Say what the vectors are

Is (1,2,3) perpendicular to (1,1,0)?

[orchid grove]

I need to check it with determinant

[blissful talon]

How does that work?

[orchid grove]

Ohh you asking me two

I will cross product

Or dot product

[blissful talon]

Those two are very different

I think you are missing a lot of information here.

[orchid grove]

I am confused actually

[blissful talon]

You should look up what not just what dot and cross product formulas are, but what they mean conceptually

[orchid grove]

If cross product is 1

I read about and try to watch in 3D about dot product in cross product so in doubt product if we have prependicular then what's the time will be zero so if don't product is zero then we can asume that those two vectors are perpendicular

@blissful talon

(1+2+0)=3

Not perpendicular

I can't use cross product for perpendicular year if two vectors are given

[blissful talon]

Now find 5 different vectors that are perpendicular to (1,1,0), and graph them all on desmos at the same time

[orchid grove]

I can't graph them

Actually

[orchid grove]

Could you give another hint besides desmos

The question is already too time

[blissful talon]

Now find 5 different vectors that are perpendicular to (1,1,0), and graph them all on paper at the same time

[orchid grove]

(-1,1,2),(-2,2,5)(4,-4,6)

@blissful talon

I can make many

Dot product is zero

.close

[midnight plank]

Channel closed

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Use .reopen if this was a mistake.

[orchid grove]

.reopen

[midnight plank]

✅

[orchid grove]

Cross product of two vectors gives a vector that is perpendicular to both of vector and it can be two types one is upset one is down side

@blissful talon

And the length is 5 so it can be two option

That's it and it was very easy and now I am closing it

.close

[midnight plank]

Channel closed

Closed by @orchid grove

Use .reopen if this was a mistake.

[blissful talon]

that tells you that it's AT LEAST 2, but you don't know that it isn't more. 🤷

[orchid grove]

I am sorry but i am not understanding your way of explaining

Thanks for helping

[blissful talon]

Fair enough. Good luck.

[midnight plank]

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[woven lion]

can someone help for this? I am really confused, dont know where to begin

[white gate]

add both equations together

[woven lion]

what equatons? the e^xsinx + e^xcosx and e^xcosx -e^xsinx?

How do I use the results? Why are we adding?

[white gate]

we want to only obtain e^x cos(x)

[woven lion]

so we need to integrate e^xsinx + e^xcosx?

[white gate]

$\frac{d}{dx}\left( e^x\sin x+e^x\cos x \right)=2e^x\cos x$

[grand pond]

TargetVN

[white gate]

it's just (1) + (2)

[woven lion]

?

idk

[white gate]

i mean you just add those 2 equations together, then integrate both sides

that's it

[woven lion]

where did the sin(x) go?

[white gate]

it vanishes

the sum of those is 0

[woven lion]

ah

so yea 2e^xcosx

then I integrate 2e^x cosx?

[white gate]

more specifically, you integrate both sides

[woven lion]

wdym by both sides?

[white gate]

at the LHS, d/dx and integral sign will vanish

[white gate]

i do mean literally

[woven lion]

so i integrate e^x sinx and e^xcosx?

[white gate]

No

like this

and i do mean literally

straight up integrate both sides

[woven lion]

ah

where did you get 1/2 from?

[white gate]

divide both sides by 2

[woven lion]

ah

[white gate]

the coefficient doesnt affect differentiation and integration

[woven lion]

but 2x^3, dy/dx is 6x^2 the coefficient changes?

[white gate]

i mean, the initial coefficient

[woven lion]

ah

[white gate]

the derivative of x^3 just add more coefficient, but the 2 remains

[woven lion]

i see

[woven lion]

would this be the final answer? or

[white gate]

ofc not

[woven lion]

had a feeling

[white gate]

d/dx and integral sign cancels out, then u have the answer

[woven lion]

does it matter the order of the integraL sign and d/dx? do they still cancel out?

[white gate]

cancel out as normal

[woven lion]

ah i think i get it, so I just add the two equations, then integrate both sides

[woven lion]

how does adding them gives me the d/dx even though they are two separate equations?

[white gate]

2 d/dx of separate functions = d/dx 2 functions

$\frac{d }{dx}\left( f+g \right)=\frac{d }{dx}f+\frac{d }{dx}g$

[grand pond]

TargetVN

[midnight plank]

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[prime garden]

Find a point X inside a triangle ABC such that area(AXB) : area(BXC) : area(CXA) = k : l : m
where k, l and m are given constants.

[last slate]

best to draw this out

(if you haven't already that is)

[lyric charm]

let him share what he's got so far with us, people

[prime garden]

I ask: Is it allowed to divide a line segment in a desired ratio?

[chrome vessel]

pretty sure you are.

[prime garden]

Sol.:
Drop the perpendicular from A onto BC.
area(AXB) = kx, area(BXC) = lx and area(CXA) = mx.
area(ABC) = area(AXB) + area(BXC) + area(CXA) = x(l+k+m)
area(AXB) : area(ABC) = kx:[x(l+m+k)] = k:(k+l+m)
Similarly, area(BXC) : area(ABC) = l:(l+m+k).
area(BXC) = XE * BC/2
area(ABC) = AE * BC/2
area(BXC) : area(ABC) = XE:AE = l:(l+m+k).
So find a point X on AE such that AE:XE = (l+m+k):l
Draw a line parallel to BC such that it goes through X. All the triangles that have the base BC and an area equal to (area(ABC) * l)/(l+m+k) will have the third vertex on the line GF (since triangles having an area equal to one another and on the same base lie between the same parallel lines).
We can do the same for the other sides. The point of intersection of the parallel lines will the actual 'X'.
Is this correct?

Shouldn't I also have to prove that the three parallel lines formed will be concurrent?

[chrome vessel]

Is AE the perpendicular from A onto BC?

[prime garden]

Yes.

[chrome vessel]

"area(BXC)=XE*BC/2" assumes X lies on AE

which might note be the case

[prime garden]

Yes, I should have labelled that point with another letter. The actual X (as I told in the last) can be anywhere on GF (and that's why I drew a parallel line).

[chrome vessel]

yeah to write a clear solution it is better to declare what your variables are before using them (like E, X, which you used without defining)

[chrome vessel]

yes

but two parallel lines are sufficient: Let them intersect at X, and show that the area of the third shape has the desired area

[prime garden]

yup

We can force the third line to intersect at the same point

Thank you, is everything else correct?

[chrome vessel]

the line "All the triangles that have the base BC and an area equal to (area(ABC) * l)/(l+m+k) will have the third vertex on the line GF (since triangles having an area equal to one another and on the same base lie between the same parallel lines)." needs some modification

third point lies on parallel line -> same area, but not the other way around.

you'd have to show (fixed area+ third point lies inside the triangle -> third point lies on parallel line).

but the modification shouldn't be too difficult

[prime garden]

<@&268886789983436800>

[chrome vessel]

Another way to solve it, if you are interested, is to let Y be a random point in the triangle, extend AY to a point D on BC, and relate the ratio area(AYB):area(AYC) to BD:DC

[lyric charm]

no, you don't have the specialized permit for it. you need to submit a request to your local math board and attach exactly 4 forms of identification (5 or more = rejection) each one valid for at least 8 months beyond the date of application, with the form filled out in blue ink if it's an even day or black if it's an odd day (wrong color = rejection), then wait 7 to 12 business days to collect the physical copy of your permit. it is valid for exactly 4 months and 11 days.

(massive joke)

[midnight plank]

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[midnight plank]

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[graceful ferry]

There's a theorem that says if f' is bounded then f is uniformly continuois,
Can I say if f' is not bounded then f is not uniformly continuous ?

[dawn dagger]

are u trying to formulate the contraposition?

[visual tiger]

no, there is no equivalence

[dreamy lichen]

sqrt(x) on the interval (0, inf) could be a counterexample

[visual tiger]

sqrt(x)

[graceful ferry]

Contraposition?

[visual tiger]

A => B is equivalent to notB => notA

[graceful ferry]

Ok nvm sqrt(x) is counterexmaple yea

[visual tiger]

even worse counterexample!

[graceful ferry]

Yes, that's what i thought, how come it doesn't work here?

[visual tiger]

because you asserted notA => notB

[graceful ferry]

Ohh

[dreamy lichen]

But A => B is not necessarily equivalent to ~A => ~B, nor is it equivalent to B => A

[visual tiger]

yeah, also even worse counterexample:

weierstrass function on [-1,1]

uniformly continuous

but derivative doesn't even exist

anywhere

[graceful ferry]

Huh haven't heard of his function yet

[dreamy lichen]

It's basically a fractal-ish function which is continuous everywhere but differentiable nowhere

[graceful ferry]

Omg I've thought what function could be like this

There it is

[graceful ferry]

Wtf

[unique juniper]

math

[graceful ferry]

Why does it look like that help😭

[unique juniper]

it's made to be continouous but indifferentiable

[graceful ferry]

It reminds me of a video I've watched long ago ugh

[dreamy lichen]

https://www.desmos.com/calculator/w0pylg4tas

If you're interested about how you can construct such weird functions, you can do it by summing bunch of (as in infinitely many) cosines which are scaled down by half every term (that guarantees that the sum converges)

[graceful ferry]

That's interesting !

I can't find the video I got a flashback of :(

.close

[midnight plank]

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[midnight plank]

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[prime garden]

Prove that 2p^2 is never a perfect square where p is an integer.

[midnight plank]

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

[prime garden]

2p^2 = j^2
j = root(2) * p
p is an integer. This implies j is irrational. So 2p^2 is not the square of an integer, making it a non perfect-square.
Is this proof correct? Lol
p is an integer

[lyric charm]

what is p?

[fresh sparrow]

assuming you already know that root(2) is irrational, yes

[lyric charm]

ok the fact that p ∈ Z should have been added to the question itself not to your proof

but other than that,

• is the irrationality of sqrt(2) known already?
• if it is, can you explain why sqrt(2)*p is always irrational?

also you want to exclude p=0, yes?

@prime garden

[prime garden]

yes

I should stop labelling everything a 'proof'.

I can prove those two points.

thank oui

.close

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[still tartan]

I need help with that

[still tartan]

I read the previous examples and even watched the video under it...I still couldn't get it

[lyric charm]

ok so notice how all the lines here go through one corner of one triangle and one corner of another

ok well actually no the ones in A and E go through two corners of the smaller one

but that actually lets you reject them automatically bc if they really were passing through the center of enlargement they'd also go through 2 corners on the big one

and they don't

do you understand this point? (i am not done yet)

[still tartan]

Hang on

[lyric charm]

oh actually no i have to go now so im just gonna drop the rest of my explanation

for the other lines here you want to make sure they go through a pair of CORRESPONDING corners of the two triangles.

like for example C won't do because it goes through the right angle in the big triangle but not through the right angle in the small one

[still tartan]

So it's B,D,F

I got it, thanks

[midnight plank]

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[lyric scroll]

3x-4y+2z=5
2x-y-z=0
4x-2y-2z=12

[lyric scroll]

yo so

i did III * 2

and did I-III‘

(III*2 is III‘)

so then

we have IV = -5x+6z= -19

and then we do the II * 4

and then we do I-II'

then we have -5x-2z=5

but now

z=-3

but the solution says no solutions exisz

what did i do wrong?

[alpine gyro]

If you multiply the second equation by 2, you’ll get something identical on lhs to one of your equations.

[lyric scroll]

lhs?

[drifting birch]

left hand side

[lyric scroll]

but what did i do wrong

i eliminated a variable

like always

[drifting birch]

Can you show what you actually did, reading this is hard to understand

😢

[lyric scroll]

yeah but

my handwritins is gruesome

and i didnt properly write everything

ok but wait

let me rewrite it

wait

why?

ok wait

wait it should be correct tho

[drifting birch]

z

[lyric scroll]

z?

[drifting birch]

2z-(-4z)

[lyric scroll]

-2-(-4)

thats -2 right

[drifting birch]

yeah mb

[drifting birch]

no

[lyric scroll]

yo

wtf

😭

[drifting birch]

-2+4

🙏🏻

[lyric scroll]

man i keep doing stupid mistakes

[drifting birch]

oh wait why did he write -2-(-4)

I thought he was right lmao

[lyric scroll]

wait no

its 2-4

3x-4y+2z=5
-8x-(-4y)-4z = 0

wait

im so fucking confused

[grand pond]

lin.fei ‧₊˚❀༉‧₊˚.

[drifting birch]

-(8x-4y-4z)

[lyric scroll]

what?

wdym

oj

yeah

why minus 4 tho

i did times plus 4

wait listen

OH my god

its -z not +z

YO

FUCK YEAH

-5x+6z = 0

and -5x +6z = -19

so now i do -5x+6z = 0 minus -5x+6z = 0

0 is not -19

so no solution

okay 1 more

wait here

so u mean eliminate z

first?

oh

i see

the II and III

ok wait dont go yo

so

4x-5y=-4
5x+3y=32

i did I times 5

and II times 4

and did I'-II'

but idk also doesnt make sense

yo

[fallow scarab]

why not

show the result of the difference

[lyric scroll]

its 37x = 128

[fallow scarab]

what eqn is I and what eqn is II

[lyric scroll]

oh

okay

20x-25y=-20
20x+12y=128

[fallow scarab]

you just subtract them, no differencing needed

20x-20x = 0

[lyric scroll]

wait wdym

[lyric scroll]

so i do 20x-25y=-20 minus 20x+12y=128 or?

[fallow scarab]

yes

[lyric scroll]

yes so then we have

yeah but i think

it doesnt make senes

[fallow scarab]

why

[lyric scroll]

cuz on wolfram it says no solution

[fallow scarab]

show

[lyric scroll]

ok wait

oh

its 4 and 4

ok mb

[midnight plank]

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[grand pond]

waffle

[kindred dagger]

i think you messed up

can you show your work

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by musicalwaffle

[kindred dagger]

^ should be negative

but you said positive 1

[grand pond]

waffle

waffle

waffle

[kindred dagger]

,w true or false arccsc(sec(x)) = arcsin(cos(x))

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by .king.leo.

[kindred dagger]

,w arcsin(cos(x)) = pi/2 - arccos(cos(x))

[grand pond]

Results provided by WolframAlpha

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

Requested by .king.leo.

[kindred dagger]

✅ @earnest forge

[midnight plank]

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[kindred dagger]

lol

[midnight plank]

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[serene rose]

How do I find X

[drifting birch]

Did you not ask this an hour ago

😭

They are the same angle

[drifting birch]

You can find it from supplementary angles, 180-94

[serene rose]

I still don’t get how their supplementary and how to write the equation

[drifting birch]

since they are the same you equate

180-94=13x-5

[serene rose]

Ok

.close

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[vocal briar]

i was told to show more reflection in my math lab

and one of the things I was talking about is how i had a leaf and assumed that it was flat and had no z-axis protrusions

would this be considered a non-euclidean object in that regard and why would it be wise to assume that the leaf is flat?

i'm assuming that the work of implementing multi-variable calculus and non-euclidean geometry concepts would overcomplicate the problem for a minor change in surface area

but how could i speak on it

[humble steeple]

i mean ideally u shouldve flattened out the leaf urself but if u didnt, u dont need to talk about non euclidian geometry at all. u can just say that dealing with multiple variables is futile for negligible changes in results

[vocal briar]

right but i need more thinking about the problems and a proper justification for the reasons

if i js say that the change in SA would be negligible regardless or that I didn't have the technology to take a 3D scan

like I js want to be able to speak on more if I can

since js saying that i didn't want to cuz i couldn't or that its not worth it is not really that strong of a justification in my opinion

another assumption i've made is that this was a perfect ellipsoid

is there something that I can talk about here as to why? The real reason I did this was because if I assumed it was a circle then I would literally plug it into 4pir^2 and get the surface area but assuming it was an ellipse allowed for me to use calc

[humble steeple]

what is ur research question

[vocal briar]

To what extent can calculus be applied to calculate the surface area of leaves?

[humble steeple]

and what is that in the image

[vocal briar]

that is a string-of-pearls leaf

[humble steeple]

i see

[vocal briar]

which is unique in that it appears spherical

[humble steeple]

so u r taking different types of leaves?

[vocal briar]

yeah and finding the surface area using different methods

[humble steeple]

since u r asking this, u would need a close-to-exact value so u can compare how accurate ur assumptions and calc methods are

[vocal briar]

so with the flat banana leaf i took polynomial interpolation and matrices to explain it, with this i explaiend the riemans sum and disk integration and with my last example it was more logic based since it had overlapping domains of piecewise functions

[vocal briar]

yeah

[humble steeple]

perhaps there r values on the internet

[vocal briar]

but there is certain assumptions I have to make for the hs level

[humble steeple]

is this IB?

[vocal briar]

what are R values

[vocal briar]

yeah

math ia

[humble steeple]

perhaps there ARE values

[humble steeple]

aa hl?

[vocal briar]

aa sl

[humble steeple]

or sl

[vocal briar]

oh

[humble steeple]

u should be fine by justifying that the topic gets too difficult with the inclusion of more variables for example

u arent required too much rigor at aa sl

[vocal briar]

yeah so basically my teacher said to include more student voice and reflection of the process

[vocal briar]

right

ok

what about for the second exmaple?

[humble steeple]

reflection could mean some other methods that could be explored

like ur ellipsoid leaf

u could talk about a simpler way being assuming its spherical

does this leaf more accurately resemble an ellipsoid btw?

[vocal briar]

yeah

[humble steeple]

u can justify it through that

[vocal briar]

ok fair enough

also

since i alr asked for the first wto

can you explain for the english ivy example i have

basically, I was doing this the night it was due so after calculating it for half the leaf I sadi we can just assume bilateral symetry so i wouldn't have to do the whole process for the bottom half

but is there a justification i can say as to why

[humble steeple]

thats a bad assumption

u should probably calculate it properly

[vocal briar]

okay fair enough

if I did then what would i talk about then

for reflection criterion

i wouldn't necessarily be assuming anything in that example

[humble steeple]

u can talk about ur previous methods used

so u can talk about how u wanted to assume bilateral symmetry but realized it would cause a large margin of error

[vocal briar]

if i overlay the shaded area on the bottom half as an image and show that this wouldn't be the most accurate assumption then would that be a really good point in terms of PE and reflection?

like are diagrams better for this stuff

or would it not matter

[humble steeple]

yes diagrams r the best

they boost personal engagement

[vocal briar]

ok W

i have a bunch

also

can you see if there is any topics in particular I should be touching on here

I was talking about rieman's sum and after deriving the formula, i said that this just appears in the arc length formula

but I didn't really explain how, is there certain key concepts I should talk about if anything comes to mind?

I feel like this might increase the reflection as well

[humble steeple]

first of, replace the delta by d in ur arclength derivation

its important to emphasize that its a differential, not a discrete change

oh u used it in a discrete way

did u evaluate the integral at the end?

[vocal briar]

yeah this is what I've said

I feel like I can expand on the first sentence maybe?

[humble steeple]

u could actually compute the integral using numerical methods

it will boost engagement

u can use the riemann sums but with n = 10^8 for example

and use a code that iterates those many times (if u have programming background, or just ask gpt to code a riemann sum calculator)

[midnight plank]

@vocal briar Has your question been resolved?

[midnight plank]

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[midnight plank]

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[viral citrus]

This question i need second opinion on the answer

[viral citrus]

The black and red on the top left is what someone i know and their tutor did

And the work i did was the blue on the bottom right

The part i was most skeptical about for my answer was the formula for projected length

The answer the tutor of my friend gave didnt really make sense to me given that the surface area of the ellipse would decrease as theta increases

[viral citrus]

If i used his formula

But SA should increase as the cup tilts

[west iron]

yes, pi*r^2*cos(theta) isn't right for that reason

[viral citrus]

Ok im glad im not just missing something there

I dont think the math i did was wrong but the only thing that really bothers me is the formula for the projected length

[west iron]

the height goes from h to hcos(theta)

the cross sectional area needs to increase to compensate

the new area should be pi*r^2/cos(theta)

[viral citrus]

Ok perfect so i got that right

Now i have a concern and it may be that the question just doesnt specify what happens in the situation but…

At some point shouldnt the water not reach the middle of the bottle when tilting? Its hard to explain with words but lets say there is a water bottle with some water in it. As you tilt the water doesnt reach the radius/center of the bottle which then means the other axis that used to be constant, now isnt.

Sorry if i explained it poorly.

Maybe i can draw diagrams to explain better

[west iron]

ye idk what u mean

[midnight plank]

@viral citrus Has your question been resolved?

[viral citrus]

God my diagram made it more confusing

Honestly i seriously doubt that they’ll consider this since it would make the question nearly unsolvable with the info provided

[viral citrus]

Thank you for the help here :)

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[tired ferry]

Why is the relation clearly reflexive?

[tired ferry]

just by definition of partition?

[dawn dagger]

well an element is in the same partition with itself, if that makes sense

[tired ferry]

Ok, so its like by definition of the partition

The reason im asking because its suposed to be a proof proving that partitions imply equivilance relations

so idk if thats the right way to look at it because it should be something we are trying to prove

if that makes sense

[radiant roost]

he's right

[tired ferry]

whos?

[radiant roost]

this statement ^

i agree with it

[tired ferry]

Ok

my question is

is that not something we are trying to prove?

[radiant roost]

they defined a relation, by saying that a is related to b if there exists X_i with a,b both in X_i