help-13

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[crimson sedge]

for this question do u just draw a straight line by hand not using any points for ref

[crimson sedge]

and would u just two points from that line u drew for part b ?

[west stag]

just draw a tangent line

prolly should label the coordinates

of the point that is touching

[cedar kiln]

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[shy pagoda]

$\frac{\partial F}{\partial y} = {\ln(x)}\frac{\partial f}{\partial y}$

[wraith dagger]

rach4

[shy pagoda]

$\frac{\partial F}{\partial x} = \frac{y}{x}\frac{\partial f}{\partial x}$

[wraith dagger]

rach4

[neon moon]

$f$ is a single function of one variable (let’s say it is "$t$"), so you don’t need to use the partial to denote its derivatives. 🙂

[wraith dagger]

jimmy1234

[shy pagoda]

so LHS should be ${y}\frac{\partial f}{\partial x} + {y}{\ln(x)}\frac{\partial f}{\partial y}$

[wraith dagger]

rach4

[shy pagoda]

oh right

so LHS is ${y}\frac{df}{dx} + {y}{\ln(x)}\frac{df}{dy}$

[wraith dagger]

rach4

[shy pagoda]

i think at least

$\frac{\partial^2 F}{{\partial x}{\partial y}} = \frac{1}{x}\frac{df}{dx} + \frac{y}{x}\frac{d^2 f}{{dx}{dy}}$

[wraith dagger]

rach4

[shy pagoda]

this is where i expect ive got something wrong

[neon moon]

What i mean by "no partial" is that, there are no separate "dx" or "dy" for f.

There is just a single "dt" for f(t) as a function of a single variable t.

[shy pagoda]

oh

so i should let t = y * ln(x) and then deal with derivatives of f with respect to t?

[neon moon]

Let $t = y \ln x$, then $\frac{\partial F}{\partial x} = \frac{df}{dt} \frac{\partial t}{\partial x}$

[wraith dagger]

jimmy1234

[shy pagoda]

ok

that makes sense

so $\frac{\partial F}{\partial x} = \frac{y}{x}\frac{df}{dt}$ and $\frac{\partial F}{\partial y} = {\ln(x)}\frac{df}{dt}$

[wraith dagger]

rach4

[cedar kiln]

@shy pagoda Has your question been resolved?

[shy pagoda]

sorted

.close

[cedar kiln]

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[weak mountain]

What's the general idea to show that a space is not hausdorff?

[weak mountain]

Show that for every neighborhood around a point x, y must be in it?

(open nbh)

Trying to show this for the co-countable topology on an uncountable set X

nvm i was dumb

.close

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[deep patrol]

Hello

So i looked at this answer key and started wondering

So if you look at the bottom right

There are two answers

And they are answers in which they repeat every 180 degrees

But how about we put them into just one answer

And repeat every 90 degrees instead?

Like, starting from 30

So

Pi/6+k*pi/2

How about this?

Would this still be the same as the bottom right of the image?

[karmic cedar]

The difference between the two is 2pi/3

120 deg

You would get 120 degrees instead of 5pi/6(150)

[deep patrol]

Ooh

Nvm

I thought both of them together would be every 90

But i was wrongp

Wrong

[karmic cedar]

Yeah that happens

Radians r hard

For me at least

[deep patrol]

Yeah

They are not all the same

Angles

Only two of them are

Alright i solved the problem

All done

Much appreciated

@karmic cedar

.close

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[drifting solar]

Can someone double check me on this?

my work is sending lol

Honestly just trying to make sure my bounds are correct, the actual setting up the integral s where I think I make the most mistakes

@ me when someone gets this pls

[cedar kiln]

@drifting solar Has your question been resolved?

[cedar kiln]

@drifting solar Has your question been resolved?

[hoary spoke]

I have a question

[small dust]

#❓how-to-get-help

[hoary spoke]

How many 3 digit numbers exist such that their first digit equals to the Sum of of other two

[small dust]

get ur own help channel

[hoary spoke]

I edited it

There was a mistake

Oh....o

Is this occupied?

Sorry then

[drifting solar]

All good

[cedar kiln]

@drifting solar Has your question been resolved?

[cedar kiln]

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[crimson sedge]

you’ve picked 3 cards out that arent hearts

how many cards are there, how many are hearts?

[grave radish]

then thered be 39 cards left

13 would be hearts

[crimson sedge]

not 39

but yeah

[grave radish]

wait sorry why wouldnt it be 39

[heavy sorrel]

bcoz u have already taken out 3 non heart cards

[grave radish]

oh right

then would it be like

P(heart on fourth draw | first three not hearts) = (39/52) * (38/51) * (37/50) * (13/49) / (39/52) * (38/51) * (37/50) = 13/49

[cedar kiln]

@grave radish Has your question been resolved?

[grave radish]

<@&286206848099549185>

[fair geyser]

sure

[grave radish]

so it'd be 13/49 and not 13/52 right

[fair geyser]

yeah

[grave radish]

okok thank u

[grave radish]

.close

[cedar kiln]

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[cobalt wyvern]

How can I solve for ii and iii

[cobalt wyvern]

I did i and got -8 + 8 * root(3)

[cedar kiln]

@cobalt wyvern Has your question been resolved?

[cobalt wyvern]

<@&286206848099549185>

[lethal jackal]

what's the square root of $e^9$?

[wraith dagger]

Saccharine

[cedar kiln]

@cobalt wyvern Has your question been resolved?

[gritty galleon]

e^4.5 ig?

[cedar kiln]

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[coral vortex]

hey can i get some help please

[coral vortex]

can you help with IB AA HL?

[lethal jackal]

maybe

[coral vortex]

its just some section A questions

this is the question

[gritty galleon]

since its rate of change..Derivatives ig?

[coral vortex]

yes

i dont understand how on earth to do ti

it

[gritty galleon]

ah'

lemme see

Okay

Let f(x) = pix^2

[coral vortex]

okkkk

[gritty galleon]

Now, Can you rewrite the expression for V in terms of the function f(x)?

[cedar kiln]

@coral vortex Has your question been resolved?

[cedar kiln]

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[covert island]

when i solve using fraction decomp

i kind of get stuck on a value

[lunar plaza]

have you tried integration by parts

[covert island]

uhhhh

isnt fraction decomposition

integration by parts

[lunar plaza]

my bad

i didnt know

[covert island]

if i break it up you get

A(y+2)(y+3) + B(y)(y+3)+ C(y+2)y

2 of the values overlap and can cancel to 0

so B = 5 and C = -45

but i cant figure out A

wait

im dumb

[lunar plaza]

wait wa

wont you get ln

[covert island]

i figured it out

i think

[lunar plaza]

try using partial fraction

[covert island]

.close

[cedar kiln]

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[elfin otter]

Anyone know how to do this with basic trigo? (meaning no sine/cosine rule)

[lunar plaza]

can you take a better picutre please

a closer one

are you malaysian btw

[gritty galleon]

A translation would be really helpful

[lunar plaza]

theres english at the bottom too

[elfin otter]

Ye

[gritty galleon]

OH

sorry 😅

[elfin otter]

,rotate

[gritty galleon]

Perimeter of the shaded region...??

Isnt that just the circumference of thec circle? OH NO OFC NOT

nthing

Do you know tangents drawn to circle from the same point theorem?

It means PS = PQ

[lunar plaza]

soq

should be double of 53

and minus

360

[elfin otter]

Find
1) x
2) length of SU
3) perimeter of shaded region
4) area of shaded region

[olive lance]

sorry, which ones do you not know?

and what have you tried so far

[elfin otter]

Yes

[lunar plaza]

x should be 127

[elfin otter]

I saw WY=2 then I blanked out

[gritty galleon]

@lunar plaza first explain

[lunar plaza]

because ps pq

[elfin otter]

This is can do AM but cannot do normal Maths syndrome

[lunar plaza]

is tangent

and soq is inside of the circle

[olive lance]

so here's a couple of useful facts

[lunar plaza]

so by taking 53 times 2

you will get the soq smaller angle 106

and 360 minutes 106 will give you the bigger angle

and the big angle is 254

and finding x

you just have to simply divide it by 2

and it iwll get you 127

[elfin otter]

I forgor that rule exists.

[limber bay]

Hi, I wanna know if I did the question right, q7

[olive lance]

#❓how-to-get-help

[lunar plaza]

anything else?

[elfin otter]

Gonna do it from here but will come back if I blank out again

[olive lance]

just to be clear

i think zong is wrong here

[lunar plaza]

wa

[olive lance]

x is acute here

it cant be 127

[lunar plaza]

true

[olive lance]

i think the mistake you make is that angle SOQ is 53*2

[lunar plaza]

im sorry

[olive lance]

i dont think that is the case

[lunar plaza]

and the point p isnt in the circle too

[olive lance]

its fine

[lunar plaza]

but does 116.5 make sense

[olive lance]

for what?

[lunar plaza]

the angle x

[olive lance]

it should be less than 90 degrees

there's a right angle in the triangle

[lunar plaza]

@elfin otter you are not leaving until you tell me the answer

[lunar plaza]

🥲

[olive lance]

so like i said, here's a couple of useful facts and a hint:

• if a radius is perpendicular to a chord, then it bisects it.
• You already know this but it's important to restate, OQ, and OW are both radii, so you know their lengths.
• ||The quad PQOS is cyclic, do you see why?||

[lunar plaza]

because all points is touching the edge ?

[gritty galleon]

OS = 2r - 2, I just realized

and the chords gon be half like Shell said

So that gives you the sides

OY = r-2

You have all the sides

[lunar plaza]

like theres a butterfly effect

[gritty galleon]

Do inverse ratio (asin / acos / atan)

[olive lance]

here's why the last claim is true:
||PQR, and PST are tangents to the circle, hence the angles PQO and PSO are 90 degrees each. So, the quad is necessarily cyclic.||

[lunar plaza]

.

[gritty galleon]

Wait.. thats necessary.!?!?

[olive lance]

so you get the QOS angle, yes

[lunar plaza]

this is holding me back no asin acos

[elfin otter]

K u lost me here

[olive lance]

SW = SY + YW

[elfin otter]

Yes

[olive lance]

whats SW?

[elfin otter]

10

[olive lance]

and YW?

[elfin otter]

2

[lunar plaza]

shell are you good at alevel mechanics do you mind helping

[gritty galleon]

Its SY, not OS

[olive lance]

so then SY is?

[elfin otter]

8

[olive lance]

thats what they were saying

because r = 5

let me put things this way

[elfin otter]

Ohhh

[olive lance]

if you want x, you need SY and UY

SY can be gotten this way

UY is not so simple

[olive lance]

recommend looking back at this

i also recommend working backwards

if you want UY, what are the different ways of getting UY?

[elfin otter]

Last one I don't know
We nvr learnt abt it

[olive lance]

thats just fancy terminology

let me put it this way

what is the sum of all angles of a quad?

[elfin otter]

I wld hv struggled with angles n lengths then Pythagorean

[elfin otter]

360

[olive lance]

and you know one of the angles, QPS right?

[elfin otter]

Ye

53

[olive lance]

but what are PQO and PSO (angles)?

given that PQ and PS are tangents

[gritty galleon]

May I answer ✋

[elfin otter]

180-

[olive lance]

let them think abt it

[olive lance]

whats 180?

[elfin otter]

Opposite sides + tgt = 180 ?

For QOS

[olive lance]

wdym

[elfin otter]

My angle for QOS would be 180-53

[olive lance]

ah

yes

[elfin otter]

180-53) /2

[olive lance]

why /2?

[elfin otter]

For PQO n PSO

Oh no

Fk

Dumb

[gritty galleon]

Why...??

[olive lance]

nah

again, remember that PQ and PS are tangents

and that OQ, OS are radii

[gritty galleon]

SOQ is 127..?

the obtuse one

[elfin otter]

Ye

[gritty galleon]

and the inside one is 233

[olive lance]

just to be clear

the angles PQO and PSO are both 90 degrees

do you see that?

[elfin otter]

Ah.

[gritty galleon]

Cus tangent-from-same-point-radius theorem thing right?

[elfin otter]

Yes

[olive lance]

so now that you have that

its just that

90 + 90 + QOS + 53 = 360

and you get the angle from there

[elfin otter]

Aight I think I got it

Can't afford to be this brain dead tmr...

[olive lance]

relatable

[gritty galleon]

Well with SOQ, what next? YOQ is 233 - some quantity, but what?

[elfin otter]

National exam for Maths tmr wish me luck :)

[olive lance]

good luck!

[gritty galleon]

May the lord be with you 🙏

[elfin otter]

N with ur spirit

Thanks guys 🫂

.close

[cedar kiln]

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[reef lynx]

Hi I know this is a math server but I was wondering if anyone could help still, I'm sort of unfamiliar with this type of circuit, mainly the circle with the arrow in it and the resistance that has a current. The question asks to find the value of the Current and Voltage on R3

[lusty grotto]

Circle with an arrow is a current source

[reef lynx]

do you know how to solve for r3?

[lethal jackal]

nodal analysis is your friend

[reef lynx]

elaborate please

[lethal jackal]

https://en.wikipedia.org/wiki/Nodal_analysis

[reef lynx]

could you help me implement it?

[cedar kiln]

@reef lynx Has your question been resolved?

[reef lynx]

@lethal jackal

[cedar kiln]

@reef lynx Has your question been resolved?

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[cedar kiln]

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[tough karma]

I don't understand how they used tan(X/2)= u to write cos

This is what I've reached at

I understand the partial fractions part

<@&286206848099549185>

[ancient lodge]

https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution

[tough karma]

Oh

Alright thank you

.close

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[stark bronze]

[\frac{30(4x-3)^{2}}{(2x+1)^2} * \frac{1}{(4x-3)^{2}}]
[\frac{30(4x-3)^{2}}{(2x+1)^4}]

[wraith dagger]

dopediscorduser

[stark bronze]

How does this first step equal the second?

I don't understand how you can just add the exponents of the denominators when they're two different terms?

[tranquil oracle]

yeah i don't think it is

perhaps there's further context?

[stark bronze]

This is the question

I solved by using the extended power rule, k(g(x))^k-1 * g'(x)
Using the product rule on the derivative g'(x)
And then simplifying the result

[tranquil oracle]

yeah i don't see where you get that step

[stark bronze]

Give me a sec and I'll type out my work

[-3(\frac{2x+1}{4x-3})^-2 * \frac{d}{dx}(\frac{2x+1}{4x-3})^2 \text{ Extended power} ]
[-3(\frac{2x+1}{4x-3})^-2 * \frac{-10}{(4x-3)^2} \text{ Product}]
[30(\frac{2x+1}{4x-3})^-2 * \frac{1}{(4x-3)^{2}}]
[\frac{30(4x-3)^{2}}{(2x+1)^2} * \frac{1}{(4x-3)^{2}}]
[\frac{30(4x-3)^{2}}{(2x+1)^4}]

[wraith dagger]

dopediscorduser

[stark bronze]

Sorry I don't know why the negative exponents aren't displaying correctly

[tranquil oracle]

you need to $a^{-2}$

[wraith dagger]

Element118

[tranquil oracle]

not sure what you mean by extended power, but shouldn't the exponent decrease? @stark bronze

[stark bronze]

https://slideplayer.com/slide/5834731/19/images/4/Extended+Power+Rule+(power+rule)+(extended+(or+general)+power+rule).jpg

[stark bronze]

Oh that's my bad, the exponent should be -4 not -2, right?

[-3(\frac{2x+1}{4x-3})^{-4} * \frac{d}{dx}(\frac{2x+1}{4x-3})^2]

[wraith dagger]

dopediscorduser

[stark bronze]

Like this?

[tranquil oracle]

yeah that seems to be it that part's fixed but not yet

but i'm not sure why your d/dx

you should get this?
$-3\left(\frac{2x+1}{4x-3}\right)^{-4}\times\frac{d}{dx}\frac{2x+1}{4x-3}$

[wraith dagger]

Element118

[tranquil oracle]

@stark bronze not sure why there's the squared for the part you differentiate

[stark bronze]

I didn't mean for that to be there

Sorry my latex is really a mess

[cinder venture]

@stark bronze

[cinder venture]

This is correct

What’s the situation

[stark bronze]

I know the answer is right

I just don't understand why

[cinder venture]

For what part

[stark bronze]

Give me just a sec

I'm typing something out in latex

[cinder venture]

Okay

[stark bronze]

For the last step here right before the answer, I don't understand how the denominator gets a 4 in the exponent

[cinder venture]

I didn’t use the method you used but would you like to see it? Or you have to use the extended power thing

[stark bronze]

I'm typing my work out rn

[cinder venture]

This is not fully correct either

The working

[stark bronze]

[-3\left(\frac{2x+1}{4x-3}\right)^{-4}\times\frac{d}{dx}\frac{2x+1}{4x-3}]
[-3\left(\frac{2x+1}{4x-3}\right)^{-4}\times\frac{-10}{(4x-3)^{2}}]
[\frac{30(2x+1)^{-4}}{(4x-3)^{-4}} \times \frac{1}{(4x-3)^2}]
[\frac{30(2x+1)^{-4}}{(4x-3)^{-2}}]
[\frac{30(4x-3)^{2}}{(2x+1)^{4}}]

[wraith dagger]

dopediscorduser

[stark bronze]

Does this look correct?

[stark bronze]

Where was my mistake?

[cinder venture]

The exponents

[stark bronze]

Are the exponents still wrong in this?

[cinder venture]

Yes

Well wait

[stark bronze]

Where am I going wrong?

[cinder venture]

This is correct

No issue

[stark bronze]

Awesome

Thanks for the help!

[cinder venture]

No problem

[stark bronze]

.close

[cedar kiln]

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[supple forge]

I really need help with this question asap and I don't know the basics either cause I wasn't well and I didn't attend the lectures. My friends are no help either. pls help 😦

[sullen saffron]

Why do you need help so quickly?

[supple forge]

well beacuse the prof wants the submission by 12am IST. and i have like 2

2.15 hrs

[sullen saffron]

We don't give out answers and you said you don't know the basics so I'm not sure the server can be much help for you.

[supple forge]

oh okayy. thanks then

.close

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[elder kestrel]

does the modulus of derivative equal the derivative of modulus?

[crimson delta]

general advice, dont upload random files

also no, chain rule and lets not ignore that absolute values are not differentiable everywhere

[south tundra]

Yup, consider the modulus of the derivative of x at 0 and the derivative of the modulus of x at 0

[cedar kiln]

@elder kestrel Has your question been resolved?

[cedar kiln]

@elder kestrel Has your question been resolved?

[cedar kiln]

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[crimson sedge]

How can I add these two together

Then being in a fraction form is throwing me off

[sage forge]

First try to cancel the two in the left denominator

[crimson sedge]

I don’t follow?

So the left side jsut divide it ?

[sage forge]

OK, it might be faster if we do it in a different order

For the right side of the + you can use that sqrt(a/b) = sqrt(a)/sqrt(b)

Can you try to apply it?

[crimson sedge]

Uh

[wraith dagger]

Stephen

[sage forge]

Thx

[crimson sedge]

I see

@sage forge

[sage forge]

,rotate

[sage forge]

Yes?

[crimson sedge]

I was just pinging u so u could see that I applied it lol

[sage forge]

Yeah

[crimson sedge]

What would my next step Be? @sage forge

[sage forge]

Can you simplify sqrt (112)?

[crimson sedge]

Yea into 16 x rt 7

So it would just be 28

4 x 7 = 28

[sage forge]

Why 16 sqrt(7)? It is sqrt(16*7), which is sqrt(16)*sqrt (7)

OK?

[crimson sedge]

Oh yea

But then it would 4sqrt7 tho right

[sage forge]

Yes

Can you do the same for sqrt (28)?

[crimson sedge]

4 sqrt 7

Sqrt 4 sqrt 7

Then 2 sqrt 7

[sage forge]

Very good

Now to the sqrt(x^2)

Do you have an idea how to simplify this?

[crimson sedge]

Uh

No

[sage forge]

sqrt(x^2) = |x|

Okay?

[crimson sedge]

Uh

I get the x part

Not the abs values part

[sage forge]

Because sqrt ((-4)^2) = sqrt(16) = 4

sqrt does always give you the positive solution to the equation x^2 = a

Did it become clearer?

[crimson sedge]

Yes

[sage forge]

OK, now we can say that |x| = sign(x) *x. Sign of x is defined as 1, if x is positive and -1 if it is negative. So this is basically a fancy way of writing +- x

Okay?

[crimson sedge]

Ok thank you for ur time

.close

[cedar kiln]

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[harsh rover]

i have no clue how to go about this so someone please help/hint

[cedar kiln]

Please don't occupy multiple help channels.

[harsh rover]

sorry

.close

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[cedar kiln]

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[brittle field]

sqrt 9-40x becomes (9-40x)^1/2 ? and thats my f

how do i find g?

[red pumice]

$\sqrt{9-40x} = (9 - 40x)^{\frac{1}{2}}$ by definition

[wraith dagger]

heavy

[red pumice]

what do you mean by find f and g?

do you mean for the chain rule

[brittle field]

yes

[red pumice]

9-40x should be one of your functions

and sqrt(x) the other

such that f(g(x)) = sqrt(9-40x)

[brittle field]

wont it be ((9-40x)^1/2)(g(x)) ?

[red pumice]

your function is (9-40x)^(1/2)

which you want to differentiate

using the chain rule

[brittle field]

but chain rule needs f and g right?

[red pumice]

yeh

[red pumice]

look here

[brittle field]

and i multiply that with g ? what will g be?

[red pumice]

$\dv{x} f(g(x)) = f'(g(x))g'(x)$

[wraith dagger]

heavy

[brittle field]

yes, so its 9x-40(g(x))

[red pumice]

no

[brittle field]

oh, do you know how it should be?

[red pumice]

like this

[brittle field]

so i need f and g right

[red pumice]

yes

[brittle field]

how would i get g

[red pumice]

did you get f?

[brittle field]

yes you said its 9x-40 right?

[red pumice]

choosing g(x) = 9 - 40x makes more sense

[red pumice]

otherwise you would have to switch f and g here

[brittle field]

ok so f(9x-40)

[red pumice]

should be equal to ...

[brittle field]

f'(9-40x) * g'(9-40x)

[red pumice]

no

you need to do the first part

find f(x) and g(x) such that f(g(x)) = sqrt(9 - 40x)

[brittle field]

how would i do that

[brittle field]

.

.close

[cedar kiln]

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[wary anchor]

answer for question 6)a) is wrong right?

beucase its cos(x) should it be 360-ans then ±360

[wary anchor]

what they did was 180-18.4 which is for sin(x)

https://media.discordapp.net/attachments/935900823525916705/1079446939449446430/19446A2C-24F6-4788-A8D8-D616BF45489C.gif

<@&286206848099549185>

[tardy storm]

hmm, let me try help

so, i should give only one solution to number 6? @wary anchor

[wary anchor]

well

in the interval -180< x <180

[cedar kiln]

@wary anchor Has your question been resolved?

[wary anchor]

any progress?

https://tenor.com/view/lonely-gif-18978208

[tardy storm]

no :(

kinda busy tbh

do it yourself :) the hard way

[wary anchor]

<@&286206848099549185> anyone else?

[crimson sedge]

Did you use Trig Identities?

[wary anchor]

no

[wary anchor]

i got until the blue lines

and also +_ 18.4

[crimson sedge]

Solved here, but I didn´t understand your question exactly

For me it fits also for 161,6 degrees

[nimble veldt]

if you think 161.6 is wrong, just calc 10*cos² (161.6) and look at the result.

[crimson sedge]

That´s what I did.

[wary anchor]

i got -9.488

wait

[crimson sedge]

it´s 9.

[wary anchor]

i got 9

yeh

[crimson sedge]

9.00365

[nimble veldt]

10 times a quadratic can be negative?

[crimson sedge]

haha

[wary anchor]

but what about the intervals

because how i was taught, you do arccos(answer) to get one angle

then you do 360- answer to get the next

eg

arccos(x) = 18.4

360-18.4 =341.6

then 341.6-360 = -18.4

[crimson sedge]

it´s +_ sqrt 9/10

you have to check for both signals

[wary anchor]

ahh

thats where i done wrong

well

not complete

actually makes sense now

[crimson sedge]

Indeed!

[wary anchor]

Thanks

💀 1 hour for 1 question

[nimble veldt]

if you have a solution x in [0,360], and you calc 360 - x then you get another solution in [0,360], but you should find solutions in [-180,180].

[crimson sedge]

He did right actually, it's commonly used method. The interval [-180;180] represents which of the thetas will be accepted.

so if for instance he gets 193,4 degrees it'll not be accepted as an answer for this particular question.

[nimble veldt]

maybe its a commnly used method. but usesless in such cases of examples. if i have a solutionx in [0,180], why should someone calc 360-x which leads to another solution outside the accepted intervall?

[crimson sedge]

You got a point but it's not because an interval doesn't allow some numbers that means it doesn't exist such numbers.

But yeah, in some intervals it's completely useless

[nimble veldt]

@native sapphire get your own channel.

[wary anchor]

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

this is I guess Physics

[proven kelp]

any idea on how to solve?

[cedar kiln]

@proven kelp Has your question been resolved?

[cedar kiln]

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[torpid urchin]

my prof said that this question is solvable through partial fractions, as it is an implicit function so we cant go about finding a solution normally. could anyone help me out with how to do this via partial fractions?

sort of confused on where to pull fractions from and quite honestly where to start lol

[cosmic steppe]

Wdym implicit function

[torpid urchin]

like

[cosmic steppe]

I'm pretty sure your primitive is gonna be an equation not a function

[torpid urchin]

y+e^y

cannot isolate for y here

[cosmic steppe]

It's fine as is

It's an implicit equation

[torpid urchin]

right

weve never done these before

so i am confused as to where to go

[cosmic steppe]

Yeah no that's fine

What you have should work

[torpid urchin]

ok

so to isolate for y i was thinking implicit differentiation or something but im not sure that provides anything?

[cosmic steppe]

although

Hm

Well the equation is symmetric along y=x

Lemme graph it

Because the thing is the relationship for y is the exact same for x

Like if I let p(n) = n + e^n

p(x) = p(y)

[torpid urchin]

right

[cosmic steppe]

Lemme graph this

I'm thinking it simplifies down to y = x

[torpid urchin]

but supposedly partial fractions have to make an appearance here

but im not sure where that would be relevant

[cosmic steppe]

Yeah same, I'm just using analysis

Turns out I was right

p(y) = p(x) implies a linear relationship

[torpid urchin]

yeah it logically makes sense there

y=x

but why would i ever need partial fractions 😭

[cosmic steppe]

No idea

[torpid urchin]

so without this analysis

is there a way to go about isolating for y

[cosmic steppe]

Lambert W

[torpid urchin]

ok so not something weve learned

is this what youre referring to O_O

[cosmic steppe]

No

Well yeah

That's the

Uh

Shit I'm blanking

But yes that's Lambert W visualized in a weird way

Forgor the word 💀 but yeah

[torpid urchin]

woah it has a real and imaginary plane

ok yeah

definetly not something weve covered

[cosmic steppe]

But lemme ask if

[cosmic steppe]

If this implies y = x

[torpid urchin]

then y=x 🤩

im just stuck on the fact he mentioned partial fractions

could it have to do with manipulating the integrands before i even integrate so the implicit stuff doesnt appear?

[cosmic steppe]

Possibly

I'm gonna figure out something hold on

Hopefully in time because I do have to go soon

[torpid urchin]

no problem

symbolab is mentioning lambert w as well

[cosmic steppe]

Apparently p(x) = p(y) works because p(n) is injective

Which means the function is "one-to-one"

1 + e^n is one-to-one because e^n is monotonically increasing

So that's the analysis side of things lol

[cedar kiln]

@torpid urchin Has your question been resolved?

[torpid urchin]

interestingly enough learned about this today so cool to see them actually being applied

appreciate the help

[royal loom]

$$ y+e^y=x+e^x $$only when $$ y=x $$ $$ y+e^y=x+e^x \implies y-x+e^y-e^x=0 \implies 1(y-x)+1(e^y-e^x)=0 \implies y=x or y-x=e^y-e^x \implies y=x $$ again

[wraith dagger]

AustinU

[torpid urchin]

right

so no need for partial fractions?

or even possible application

[royal loom]

I am hesitant to say for sure either way

[torpid urchin]

i personally cant see a use for them

but i understand this question from that perspective

just not how to use partial fractions

im going to shoot my prof an email for how he did this but thankfully i can get the answer to this for now

thanks

.close

[cedar kiln]

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[wide cove]

I need help finishing this trig substitution. I don't know if I started this right

And then I just need anyone to help me walk through this problem

[shut panther]

,rcw

[cedar kiln]

@wide cove Has your question been resolved?

[wide cove]

<@&286206848099549185>

[wide cove]

.close

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[humble sleet]

Hello

[humble sleet]

is it possible to verify a solution set to an inequality

inequation

[dire geode]

yes

[cedar kiln]

@humble sleet Has your question been resolved?

[cedar kiln]

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[hollow rock]

I think the answer should be (729/8000)/(504/6840), but its not right

[hollow rock]

man i got a goofy ah question compared to everyone else's help

[cedar kiln]

@hollow rock Has your question been resolved?

[hollow rock]

<@&286206848099549185>

.close

[cedar kiln]

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[bronze briar]

I need to find gcd(100, -35) and write it in the form of ax + by, where x,y are integers. Im not really so sure what to do

[cedar kiln]

Please don't occupy multiple help channels.

[bronze briar]

i know that the gcd is 5

sorry, just closed the other channel. forgot about that one

or do i just divide both numbers by 5 and say that it's -7x + 20y?

<@&286206848099549185>

[devout galleon]

can you show us the text of the problem?

[bronze briar]

yeah sorry:

[devout galleon]

if gcd(100, -35) = 5 then x = -1, y = -3 and you can write -100x+35y = -100 + 3*35 = -100 + 105 = 5, etc.

[bronze briar]

im so confused. Where did you gegt x = -1, and y = -3? Also where does the number 5 play a role in the equation?

[devout galleon]

your question in other words: for every pair of a and b, find x, y pairs that stisfyies this equation gcd(a, b) = ax + by

x, y in Z

[bronze briar]

isn't that this?

[devout galleon]

check it )

you see

[bronze briar]

so the linear combination should add up to 5, right?

[devout galleon]

yeah, (should up to gcd(a, b) )

[bronze briar]

i mean, when x is 2 and y is 5 it works out

20(2) - 7(5) = 5

[devout galleon]

hey no

in the 23) a = 100, b = -35 and you calculated gcd(a, b) = gcd(100, -35) = 5.

[bronze briar]

yes

[devout galleon]

then you neet to solve this ax+by=gcd(a, b)

100x-35y=5

[bronze briar]

ohh so that's it?

i apologize for my extreme stupidity lol

that was so obvious

[devout galleon]

yes

[bronze briar]

thank you, i appreciate it!

.close

[cedar kiln]

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[crimson sedge]

I don't understand

[crimson sedge]

Why when taking the x out of the sqrt

The sqrt becomes - sqrt

wtf

[fervent locust]

Oh

Mb

[crimson sedge]

/help

[wooden quarry]

Did you know Beemo supports Slash Commands? They are the new, recommended way to discover and run commands! Type / to see them all.

[crimson sedge]

help

😦

<@&286206848099549185>

[crimson sedge]

.

Why does the sqrt turn negative?

Wdym

Im in my second semester of uni

Nice troll

<@&268886789983436800>

[crimson sedge]

. <@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[devout galleon]

lol

[cedar kiln]

@crimson sedge Has your question been resolved?

[harsh pelican]

hint consider (x + x^2)^{1/2} - x

and how u coulf factor out x from thsi expression

[cedar kiln]

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[dusk finch]

I have diff. eq. $\frac{d^{2}s}{dt^{2}}=\frac{1}{s^{2}}\\$
Assuming that $v=\frac{ds}{dt}\\\$
It can be said that $\frac{d^{2}s}{dt^{2}}=\frac{dv}{dt}=\frac{dv}{ds}\cdot\frac{ds}{dt}=\frac{dv}{ds}v\\\$
Therefore $v\cdot\frac{dv}{ds}=\frac{1}{s^{2}}$ and $v\cdot dv=\frac{1}{s^{2}}ds\\\$
So $\int v\cdot dv=\int \frac{1}{s^{2}}ds$ ‏‏‎ and $\frac{v^{2}}{2}=-\frac{1}{s}\\\$
Substituting for v $\left(\frac{ds}{dt}\right)^{2}=-\frac{2}{s}\\\$
Therefore $\frac{dt}{ds}=\left(-\frac{2}{s}\right)^{-\frac{1}{2}}$ and $\int dt=\int \left(-\frac{2}{s}\right)^{-\frac{1}{2}}ds\\\$
Evaluating $t=\frac{\sqrt{2}s}{3\sqrt{-s^{-1}}}\\\$
Solving for s $s=-\sqrt[3]{\frac{9t^{2}}{2}}$

[dusk finch]

Except for forgeting conditions and +C, what's wrong here?

[wraith dagger]

MathIsAlwaysRight

[dusk finch]

I mean even tho I forget conditions, the second derivative of the expression should be 1/s^2

but it's not

No way

it actually works

there are more solutions than 1

[long swan]

why would the second derivative be 1/s^2

[dusk finch]

By the initial diff equation

[long swan]

keep track of what is your dependent and independent variable

so i mean yes it would be 1/s^2 where s = what you got on the final line

but maintain that distinction

[dusk finch]

What does it mean? Sorry I am not very experienced at diff equations

[long swan]

let's try verifying:

,w d^2/dt^2 -cbrt(9t^2/2)

[wraith dagger]

Results provided by WolframAlpha

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Requested by Mr. Gamer#6658

[dusk finch]

,w 1/(-cbrt(9t^2/2))^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by MathIsAlwaysRight#4079

[long swan]

that negative sign!!!

,w int (-2/s)^(-1/2) ds

[wraith dagger]

Results provided by WolframAlpha

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Requested by Mr. Gamer#6658

[dusk finch]

That means that s is always negative?

Does this have multiple solutions or have I done something incorrectly?

[cedar kiln]

@dusk finch Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[hoary spoke]

I have a q

[hoary spoke]

If 7k+5 is divisible by 9

And k is natural

What does it tell about k?

What all can k be?

I am certain that k cannot be a multiple of 9

[cedar kiln]

@hoary spoke Has your question been resolved?

[hoary spoke]

No

Fine I have another one

[fair geyser]

i think it tells us that k+2 is divisible by 9

[hoary spoke]

hm....

How?

[fair geyser]

somehow

i got it by looking at all k

[hoary spoke]

hm....

[hoary spoke]

Can u show me how u deduced this?

[fair geyser]

7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 112 : multiples of 7
7 5 3 1 8 6 4 2 0 7 5 3 1 8 6 4 : multiples of 7 mod 9
↑ ↑the fours
so k = 7, k = 16 are the first two solutions

and it holds forever

[hoary spoke]

Sorry but I don't know mod operations

Can u tell me what does multiple of 7 mod 9 mean?

Isn't 7 mod 9, 2?

[fair geyser]

the first line is the multiples of 7

[hoary spoke]

Yup

[fair geyser]

the second line is the first line

mod 9

7 mod 9 is 7

[hoary spoke]

You mean remainder after dividing by 9?

[fair geyser]

yes

[hoary spoke]

Oh

[fair geyser]

every time it is 4

it will become divisible once you add 5

[hoary spoke]

Wow

That's interesting

And also amazing

hm....

So I have to find all the numbers where I got 4 after mod 9

[fair geyser]

now there' must be a clever way to do it too, but i don't know it, this was caveman math

[hoary spoke]

Okay....

[hoary spoke]

Can u help me with this?

[fair geyser]

hm

[hoary spoke]

And don't get distracted by the 5 in the last line

It is irrelevant

[fair geyser]

well divisibility by 36 is divisibility by 4 and 9

[hoary spoke]

Yup

I deduced that the last number should be 2 or 7

Sorry

I mean 2 or 6

[fair geyser]

no problem

i'm stuck at this point

[hoary spoke]

hmm ..

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[hoary spoke]

I am not done yet

[fair geyser]

.reopen

[cedar kiln]

✅

[fair geyser]

it would understand if you clicked the ❌ back then, no problem

[hoary spoke]

Oh okay

I have never used discord 😅

I just had this account

Anyone?

[fair geyser]

oh wait, 5 is given

i didn't notice

[hoary spoke]

Ya

3rd digit

[hoary spoke]

It happens

[fair geyser]

well we can bruteforce that too then

if it ends in 52, the first 2 digits must add up to 2 or 11

[hoary spoke]

what

😮🤔

wdym?

[fair geyser]

the trick for divisibility by 9 is that digits added up are divisible by 9

so 2 + 7 or 11+ 7

[hoary spoke]

Sorry i am confused

Why ?

[fair geyser]

i don't know, it's a trick

[hoary spoke]

Really?

Give me an example please

Well I can see for 2 digits tho

0+9

1+8

2+7

3+6

4+5

And so on...

But I am not sure for 3 or 4 digits

[fair geyser]

,calc 351/9

[wraith dagger]

Result:

39

[fair geyser]

,calc 2358/9

[wraith dagger]

Result:

262

[fair geyser]

,calc 90909/9

[wraith dagger]

Result:

10101

[hoary spoke]

hmm.....

So you mean that sum of all digits of multiple of 9 are also multiplies of 9?

[fair geyser]

yes

[hoary spoke]

Wtf

When did that happen

[fair geyser]

probably more than 800 years

[hoary spoke]

💀

[fair geyser]

many number theory results are ancient history

[hoary spoke]

Do we have proofs for these?

[fair geyser]

uh you can probably see it let me think

[hoary spoke]

Okay

[fair geyser]

2358 = 2×1000 + 3×100 +5×10 +8
= 2×999 + 2 + 3 × 99 +3 + 5×9 +5 +8

so those parts are divisible by 9 by themselves,
2×999 and 5×9 etc.
we can ignore them, they don't affect the divisibility of the sum

[hoary spoke]

Isn't 2358 not divisible by 9?

[fair geyser]

it's divisible by 9

[hoary spoke]

Oh it is

Sorry my bad

hm...

Now I got it.

The proof

For the sum of digits of multiple of 9 must also divisible by 9

Thank you

[fair geyser]

okay

[hoary spoke]

So how are we gonna solve this by this fact?

[fair geyser]

as you said, there are 2 cases, it's __52 or __56

[hoary spoke]

Yes....

[fair geyser]

if it's 52, the first digits must add up to 2 or 11

[hoary spoke]

hm...

Ph

[fair geyser]

then the entire sum will be 9 or 18

[hoary spoke]

*oh

So that the sum is multiple of 9!

I mean 9

Not factorial

[fair geyser]

we can only make it 2, if it's 1152

because we can't use the digit 2 again

[hoary spoke]

But also, we can't use 1 twice

[fair geyser]

oh

we can't

[hoary spoke]

Ya

No repetitions allowed

It is written in the question

[fair geyser]

you're right

83
74
65 can't
56 can't
47
38

so those 4 should work

[hoary spoke]

Okay.....

I think I can solve it now

[fair geyser]

if it ends in 56, the other two digit must add up to 7 or 16

[hoary spoke]

Yeah....

So they can be.....

16

34

61

97

43

79

So I have to just check one by one?

If they satisfy the question's condition or not?

Thank you very much

You are my messiah

Bot, i am done

,calc 99/9

[wraith dagger]

Result:

11

[hoary spoke]

. close

[fair geyser]

you don;t have to check they all should work

[hoary spoke]

Oh

[fair geyser]

just not the 61 or 16

[hoary spoke]

Oh ya

6

[hoary spoke]

Why?

[fair geyser]

why not

they are divisible by 36

that was the condition

they are because they are divisible by 4 and 9, which is equivalent

[hoary spoke]

Wait wait I think i got it

hm...

Thanks

.close

[cedar kiln]

Channel closed

Closed by @hoary spoke

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.