help-0

[naive valley]

a bit gnarly

there may be a short cut that i'm overlooking, but i guess i would start by writing the most general form for g and h:
g(x) = ax + b
h(x) = cx^2 + dx + e

that's five unknowns

you have 2 constraints g(0) = 6 and h'(0) = 5
plus the formula for f'(x) gives you three more constraints

so that should be enough to solve for the 5 unknowns

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.

[alpine sable]

I have the answers but I dont understand what this question is asking me. I haven't touched my books in months so It might be something basic.#

Firstly how am i supposed to know how they went from 2.1 to 26.901

Am I supposed to know that?

And how should I have known that F represents the gradient of the chord. Is that a formula that I should be familiar with?

Maybe it would be more useful if someone could tell me exactly what topic this is covering

[lusty hearth]

this seems to be an introduction to derivatives

f(x+h)-f(x) all over h gets closer and closer for the slope at a singular point depending on the value of h

the smaller h gets, the better the approximation

[alpine sable]

So is this a formula I'm meant to remember? And why are we trying to zero in on a specific point?

[alpine sable]

Yes this i can understand as h gets closer to 0 limits

[lusty hearth]

the difference quotient is good to remember yes

[alpine sable]

okay i remember the quotient rule and i know how to find the derivative dy/dx but how does that link?

Oh is this from first principles?

[lusty hearth]

heres an example function

The question is asking what column F represents, which is the derivative or another simpler word which is?

[alpine sable]

gradient?

[lusty hearth]

yes

So column F represents the gradient at the point (2,24)

[alpine sable]

so its trying to find the gradient at P by making H smaller do that i can make an approximation of the gradient?

[lusty hearth]

yes

[alpine sable]

But im confused because f(x+h) is roughly 24 with the smallest value of h and f(x) is 24 so 24-24 is 0 no?

[modern vessel]

hi

[alpine sable]

So then why does F show as roughly 28 on the table?

[alpine sable]

Hi

[lusty hearth]

its just a bit higher than 24

[modern vessel]

i'm a new student

[modern vessel]

no

[alpine sable]

Thats cool but im in the middle of a question so you should go to discussions not here

[lusty hearth]

f(x+h) is always at least somewhat bigger than f(x) (at least in this example since its a positive slope), so the top never equals 0

[modern vessel]

ok i'm sorry

[alpine sable]

Yeah but i mean wouldn't it still be close to 0? I dont see how they get 28

[lusty hearth]

the thing is you divide by h, which is also very small

its .0028001/.0001 in the best approximation on the figure

which still becomes approximately 28

[alpine sable]

Oh I see - So i would just have to accept it as being 28 without needeing to do the calculation myself.

I think thats answered my question thanks a lot

[lusty hearth]

the main thing the chart shows is that they get very close to 28

[alpine sable]

Yeah I see that now I completely ignored the F column when i first looked at the question

[lusty hearth]

you can’t approximate better than the chart since it never gives you the function, but the chart itself is enough to see that the gradient approaches 28 when h approaches 0

which is what ii) is all about

[alpine sable]

Ah okay would i be able to make f(x+h) with this information btw? Like if i got another point could i recreate a table like his or would i need more infromation?

Because Im also a bit confused as to how they managed to calculate f(x+h)

[lusty hearth]

if you knew the function you could make another table like this yes

f(x) in this example is just an arbitrary function as they don’t give it to you

[nova pond]

hi, does anyone know a good book on combinatorics? i want it to be higher level as i’m good at intermediate stuff like mostly sums

[lusty hearth]

#math-discussion or a help channel when it opens @nova pond

[alpine sable]

So what would a good example for a function be? I just want to get an idea

[lusty hearth]

like take for example the function f(x)=x^2

[alpine sable]

ohhhhhhhhh

[lusty hearth]

lets find the gradient at (2,4)

for h=.1, itd be f(2.1)-f(2)/.1 or (2.1)^2-4/.1

[alpine sable]

you would sub in (x+h) into x^2 right?

[lusty hearth]

yes

[alpine sable]

I see its coming back to me now

[lusty hearth]

which ends up being 4.1

[alpine sable]

wait wouldnt it be f(4)?

[lusty hearth]

f(x) is for the x coord

(2,4)

f(2)=4

[alpine sable]

oh oh oh i see

yes i see that

[lusty hearth]

if you let h be .001, the difference quotient becomes 4.001

[alpine sable]

i read h =1 not 0.1

so the gradient would be 4 then?

[lusty hearth]

yes

[alpine sable]

right becaise it gets closer to 4

[ocean seal]

Ruby

[alpine sable]

yes i see thank you i understand now

.close

[lone heart]

Channel closed

Closed by @rough ravine

Use .reopen if this was a mistake.

[dusk birch]

Hey

[lone heart]

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.

[supple quiver]

.close

[lone heart]

Channel closed

Closed by @supple quiver

Use .reopen if this was a mistake.

[lone heart]

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.

[ornate tundra]

Can someone verify that this is correct please

[vocal lagoon]

your green sector should be extending outwards forever but otherwise this seems to look alright

i would probably erase the hard line on the outside edge of that sector

[lone heart]

@ornate tundra Has your question been resolved?

[ornate tundra]

i drew arrows

[vocal lagoon]

oh i see that now

yeah looks good then

[ornate tundra]

oh thats true i wont make the hard line connect. it was meant to represent a constant angle

[lone heart]

Channel closed

Closed by @ornate tundra

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.

[pearl wave]

How can we solve part b?

[naive valley]

i can think of a fairly simple example with f=g

[pearl wave]

I try to solve it like that is it true?

[naive valley]

,rotate

[naive valley]

oh b sorry, i was reading c

one observation, f(0+0) = f(0)+f(0), so immediately you know that f(0)=0

also notice that f(x) = ax is a solution, for any real a

see if you can prove that these are the only solutions

(if you don't assume continuity then there are other solutions)

[pearl wave]

How?

We don’t study yet the continuty of the function

[pearl wave]

How can i prove it?

[naive valley]

you need the axiom of choice, see here https://math.stackexchange.com/questions/2377638/additive-function-t-mathbbr-rightarrow-mathbbr-that-is-not-linear

[naive valley]

show that it satisfies the two criteria

[dreamy olive]

someone PLEASE

PLEASE HELP ME

IM DYINGG

[lone heart]

@pearl wave Has your question been resolved?

[dreamy olive]

mb might've sent it in wrong chat

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.

[blazing dragon]

The medians of triangle ABC drawn from vertices B and C intersect at right angles. Find the length of the median of the triangle drawn from vertex A if BC = 42 cm.

[fallen berry]

Do you know what a centroid is

[lone heart]

@blazing dragon Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

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.

[hybrid ravine]

why did they say the answer is x<=-3 for part ii) when that means it goes into the negative direction. Doesn't that mean its a smaller domain?

[hybrid ravine]

oh wait nvm i just realised

also for completing the square to find the vertex is the answer y=(x-3)^2-19

[fickle sandal]

no

not -19

[hybrid ravine]

oh wait its 1 right

[fickle sandal]

yes

[hybrid ravine]

oops alright thx

.close

[lone heart]

Channel closed

Closed by @hybrid ravine

Use .reopen if this was a mistake.

[lone heart]

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.

[hoary nimbus]

I have no idea how to approach this question

Tried triple simultaneous eqns but couldn’t solve em

Tried an approach with (x+1)(y+1)(z+1) compared to xyz but i got nowhere with that

[lavish cave]

(a + 1)(b + 1)(c + 1) = abc + (ab + bc + ca) + (a + b + c) + 1

Yeah and each term compares with something in the info you are given

[hoary nimbus]

Yeah i did this

And then?

[lavish cave]

Like the surface area for example: what's that?

[lavish cave]

And then you should be able to plug everything in

[hoary nimbus]

2xy+2xh+2yh

[lavish cave]

Exactly

So that compares with xy + xh + yh

[hoary nimbus]

Hmm

a+b+c is half of sum of edges right?

No wait

1/4th?

@lavish cave

[lavish cave]

Yes, 1/4th

[hoary nimbus]

Ah damn

The solution was right there in front of me

I just had to connect the dots

[lavish cave]

Exactly

[hoary nimbus]

@lavish cave Thanks a lot dude

[lavish cave]

No worries

[hoary nimbus]

.close

[lone heart]

Channel closed

Closed by @hoary nimbus

Use .reopen if this was a mistake.

[lone heart]

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.

[craggy reef]

yo

[craggy reef]

i got 7735.9 for lateral surface area

can someone check if its right

[jagged cobalt]

,calc (pi*7.6)10.830

[ocean seal]

Result:

7735.8577501995

[jagged cobalt]

sure

[craggy reef]

thanks

.close

[lone heart]

Channel closed

Closed by @craggy reef

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.

[minor anchor]

can someone help me explain the calculation for P(R|M,N)?

[torpid dirge]

its the same as the previous part, but replace M with M, N

[minor anchor]

cant really understand why P(M,N) = p1(1-p0)^n-1

[lone heart]

@minor anchor Has your question been resolved?

[lone heart]

@minor anchor Has your question been resolved?

[lone heart]

@minor anchor Has your question been resolved?

Channel closed

Closed by @minor anchor

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.

[chrome jungle]

well just need the correct answer

[median oar]

We don’t do that here

[chrome jungle]

what you mean ?

[valid quail]

We are not solving tasks

[chrome jungle]

ok

[gray isle]

if you just want the answer, use a bot to simplify

[chrome jungle]

tried that

using chatgpt but i can t quite type this in chat

[valid quail]

Explain why you cannot solve it

And we may help with that

[gray isle]

don't use gpt

[chrome jungle]

what should i use ?

they arleady told me that i know

[gray isle]

if you want to use mathematical calculator, use something that actually knows what it's doing, like Wolfram

[chrome jungle]

oh ok

thanks

[valid quail]

Providing the answer is quite useless, especially in this context

[chrome jungle]

.close

[lone heart]

Channel closed

Closed by @chrome jungle

Use .reopen if this was a mistake.

[lone heart]

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.

[past shadow]

I'm trying to figure out whether my set of vectors spans R^3 but honestly I'm confused on how to do that

[past shadow]

.closed

.close

[lone heart]

Channel closed

Closed by @past shadow

Use .reopen if this was a mistake.

[lone heart]

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.

[rare pollen]

Determine the equation of a plane passing through the point M(2,0,5) and parallel to the planes (alpha): x + y + z + 3 = 0 and (beta): 2x + 3y - z + 10 = 0.

[rare pollen]

This is what we did in class but i don't understand it

[lone heart]

@rare pollen Has your question been resolved?

[young quarry]

I need help

Give me a radical expression using 3 different operations. Must consist of numbers and variables inside and outside the radicals. Its final answer should be negative.

[lone heart]

@rare pollen Has your question been resolved?

[strong prism]

can someone help me find the length of l6

ill pay

[dreamy plover]

Hi!! Could someone please help me with this I’m doing an assignment and I’m a bit lost, math is not my best subject.

[lone heart]

@rare pollen Has your question been resolved?

[lone heart]

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

[lone heart]

Channel closed

Closed due to timeout

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.

[autumn quarry]

I have a question about how to solve the integral J of sin^2(x) from -pi to pi
I know that people use 1-cos (2x). However, my method was to do the integration by parts for the first time giving us (-sincos) from -pi to pi + integral of cos(x)^2
Then again integration by parts for cos^2
Then you say that 2J = 2(-sinxcosx) from -pi to pi
It gives 0, which is wrong, so why is my reasoning wrong ?

[royal grail]

when you said you're integrating $\sin^2{x}$ by parts, what exactly are you differentiating and what are you integrating here?

[ocean seal]

lgkoo

[autumn quarry]

sin(x) and sin(x)

sin(x)'=cos
and sinx=-cos

this is what I did

[royal grail]

what did you get for integration by parts for cos^2

[autumn quarry]

-sinx*cosx (from -pi to pi) - integral of sin^2

[royal grail]

I got sinx*cosx (from -pi to pi) + integral of sin^2

which sin cos cancelled with the previous and thus not helpful

[autumn quarry]

I don't see why
Before that, you got + intergral of cos^2, correct ?

[royal grail]

yes

$\frac{d}{dx}\cos{x} = -\sin{x}$ and $\int \cos{x} dx = \sin{x}$ so by parts gives you $\cos{x}\sin{x}$ from -pi to pi $ -\int -\sin^2{x} dx = \int \sin^2{x} dx$ from -pi to pi

[ocean seal]

lgkoo

[autumn quarry]

Damn that was a silly mistake from me to mess up the integral of cos

Thanks

[royal grail]

no worries

[autumn quarry]

.close

[lone heart]

Channel closed

Closed by @autumn quarry

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.

[random crag]

I can help you

Ok so look for A just substitute y=3x-5 in the other equation

Then expand and simplify

Get a and i will explain b

Yeah

That’s fine

Just lemme make sure again

Ok now for b

Look a tangent means only one solution

Yeah

But equate to zero

It’s a tangent

Means it’s touching at one point only

So only one solution 😊

Just a name

Get it and equate to zero it will give you one solution

Ok

Umm lemme explain something to you

When it says value(s)

That means the s is optional

So it could be value or values

Dont worry

I will show you rn

Just writing it down

Actually am sorry it could be two tangents so yeah one solution for every tangent

Since he didn’t say k>0

So take both plus minus 360

Oh forgot the root lmao

Plus minus 6root10

Look

Just replace

And you took the formula

axsquared +bx+ c

So a is gonna be 36

Since k is squared it’s coefficient is 36

Oh right

Look the descrimenant formula

Is b^2-4ac

Look

Check dms

I sent a full explanation lol

[lone heart]

@abstract socket Has your question been resolved?

Channel closed

Closed by @abstract socket

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.

[compact garnet]

I need to get the coordinates for N

Using vectors

Everything I have is on the left side and I can use it, I know it shows the coordinates for N but how do I get it

AN is the height of the triangle

[charred flint]

vectors can get you the height by going through the area formula

[compact garnet]

I know I need to use this formula somehow

[charred flint]

nah it's cross product, area is 1/2|a x b|

and you can relate it to 1/2*base*height

[compact garnet]

ok but how does that get me the coordinates for N

I tried Cosine's theorem but it doesn't work because the solutions are imaginary numbers.

[charred flint]

oh I guess using area is really roundabout

the problem is more that you want the point on line BC closest to A

[compact garnet]

Yes thats true

[charred flint]

and you do that by writing that the vector along BC dot product the vector from A to a point along BC is 0, right angles

the formula for BC is B + x*a

[charred flint]

yea

[compact garnet]

I did this is what i got using vectors

then I tried using this

and I got this but it has 2 solutions idk which one to take

[charred flint]

hmm that xn=-yn-2 is just from making the vector 0i+0j which doesn't make sense

I'd try writing that left term with (B+t*a)-A

like a one dimensional equation for a point on line BC, to the point at A

t is a parameter along BC

o wait I misread that looks right actually

clearly y has to be positive so you take the positive solution

[compact garnet]

wolfram alpha calculated this

so the positive one

is the one i take

[charred flint]

yea

[compact garnet]

but why

there has to be like a reason and idk what reason

idk i will just write because Xb and Xc are positive so thats why

.close

[lone heart]

Channel closed

Closed by @compact garnet

Use .reopen if this was a mistake.

[lone heart]

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.

[weary epoch]

can someone just read it and tell me i did it right thx

[lavish cave]

Yes, this is correct

[weary epoch]

south ur everywhere omg

okay ty

😄

.close

[lone heart]

Channel closed

Closed by @weary epoch

Use .reopen if this was a mistake.

[lone heart]

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.

[sour basin]

you r proving for divisibility so probably start with induction

[amber atlas]

do not understand the hint, why is it enough to prove that either 2 or 3 divides it?

[sour basin]

if n^2+2^n divides by 2 or 3, it has 2 or more factors at least

hence not prime

hence composite

[amber atlas]

you mean I have to use induction to prove 2 | n^2+2^n ?

[sour basin]

yeah

[amber atlas]

how about division algorithm?

cause it is saying that n is not of the form 6k+1

[sour basin]

that, im not too sure

my approach would be induction but if u can prove it through a different approach, u should give it a go

[alpine sable]

do you know modular arithmetic

you can consider each of the possible cases individually. So suppose n = 6k, 6k+1, 6k+2, 6k +4 or 6k+5, then you can consider the divisibility of each one of those

[amber atlas]

just replace n in n^2+2^n and see if it's divisible by 2 or 3?

[amber atlas]

not into that yet

[alpine sable]

yeah put them in n^2 and just like figure out the last digit of 2^n

then u can figure out if they are odd or even

[amber atlas]

how do I figure out its last digit?

[alpine sable]

of course that tells you only the divisibility by 2

[amber atlas]

2^6k is even cause 6k is even

[alpine sable]

well i guess my way isn't quite necessary. If you have n = 6k + j then if j is even, n is even, and n^2 + 2^n entirely is even

that should be pretty direct

then you can check the n = 6k + 1 and 6k+5 cases directly

substitute in for the expression with that

the n^2 term will get you something divisible by 3, so you are left to prove the leftover terms formed by 2^n will be divisible by 3 as well

[lone heart]

@amber atlas Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

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.

[ocean shard]

Hello

[royal grail]

what is your thought/attempt so far

[ocean shard]

idk

[royal grail]

no thought at all?

[ocean shard]

nope

[royal grail]

hmm, ok, let's try this

do you understand the question?

[ocean shard]

yes

x = ?

[royal grail]

yup

so the goal is to isolate x

here x is only in the RHS (right hand side)

so try to get everything to the left

[ocean shard]

idk

[royal grail]

say if I want to make x the subject in:
$a = \frac{x}{10}$

[ocean seal]

lgkoo

[royal grail]

do you know what to do here?

[ocean shard]

yes

x = a x 10

[royal grail]

good, now apply that to this

[ocean shard]

how

x = square root of gk - a squared / 10

idk

[royal grail]

no, try to do this step by step and not rush in one go

[royal grail]

first, multiply both side by 10 like in the simpler case

so you will get 10*gk = (x+a)^2

do you understand this?

[ocean shard]

yes

[royal grail]

good, now that you see there's a square on the RHS, what would you try to do to get rid of the square?

[ocean shard]

square root

[royal grail]

yup

so what do you get when square root both sides?

[ocean shard]

idk

[royal grail]

what is $\sqrt{(x+a)^2}$ ?

[ocean seal]

lgkoo

[ocean shard]

x + a

[royal grail]

yup, and you do the same to the left, but there's nothing to simplify on the LHS so you just leave it as

$\sqrt{10gk}$

[ocean seal]

lgkoo

[royal grail]

so $\sqrt{10gk} = x+a$

[ocean seal]

lgkoo

[royal grail]

now do you know how to isolate x @ocean shard

oh also

did you learn about +- when taking a square root?

[ocean shard]

yes

i just forgot how to do these weird subject questions

[royal grail]

ok, in that case you need to write
$\pm \sqrt{10gk} = x+a$

[ocean seal]

lgkoo

[royal grail]

and now you re-arrange for x

[ocean shard]

so 10gk - a

[royal grail]

+-sqrt(10gk) - a

but yh

[ocean shard]

thanks

[royal grail]

np

[ocean shard]

.close

[lone heart]

Channel closed

Closed by @ocean shard

Use .reopen if this was a mistake.

[lone heart]

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.

[remote fulcrum]

14 pls

cause ik i most likely need to simplify the fraction is just need to know why z1z2+z1z3+z2z3 = 0

ik what’s given and z1z2z3=1 cause unit circle

[round geyser]

if |z| = 1 (which in this case, it is), then 1/z = z* (complex conjugate of z)

try using that

[remote fulcrum]

so just the conjugate of z1 z2 and z3? @round geyser

how does that equal 0?

[round geyser]

if z1 + z2 + z3 = 0, then what does that tell you about the sum of the conjugates

[remote fulcrum]

does it just imply that it’s also 0?

@round geyser

[round geyser]

why though

what's your reasoning

[remote fulcrum]

idk

that’s what i’m asking

[round geyser]

try writing the complex numbers as a + ib, c + id, x + iy

and then see what you can notice

[remote fulcrum]

it’s not the same? @round geyser

[round geyser]

let a + c + x = p
and b + d + y = q
So now the first eqn tells you p + qi = 0

if p + qi = 0 then what can you say about p and q

(think of 0 as 0 + 0i)

[remote fulcrum]

p=0

q=0

[round geyser]

yep

[remote fulcrum]

ahhh

ok thankyou

[round geyser]

np

[remote fulcrum]

.close

[lone heart]

Channel closed

Closed by @remote fulcrum

Use .reopen if this was a mistake.

[lone heart]

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.

[dense lily]

Logs

[dense lily]

just had spring break and i just came back to school and day one i forgot everything

need some assistance

i don't know whether im supposed to do log base 5 of 9 - log base 5 of 1/3 or +?

[royal grail]

,rotate ccw

[royal grail]

you can either do log base 5 of 9 - log base 5 of 3, or log base 5 of 9 + log base 5 of 1/3. Because - log 3 = log 1/3

[dense lily]

right ok

ill do the second version

so, wait a second, after I do that, I would have to...

wait

one second, ill rewrite it

is this correct?

if it is, do i then multiply 9 * 1/3 * 36?

or no

[royal grail]

yes

[dense lily]

ok

that would get me 108

so then 108 * 3?

[royal grail]

108 is correct, but why multiply by 3

[dense lily]

idk how to simplify further, i now have

3(log base 5 of 108)

do i put the 108 cubed?

i wouldn';t think so...

[royal grail]

yes

but it depends what the form of answer your school wants

[dense lily]

probably as simple as physically possible

[royal grail]

cuz 3 log 108 and log 108^3 is equivalent

[dense lily]

ill PROBABLY do that, but have a base 5 obv

[royal grail]

I mean, I think most ppl here prefer 3 log 108, maybe even taking some factors inside the log to the coeffiecient, cuz 108^3 is huge number

[dense lily]

ok

i have another question, one sec

is this corre t?

or am i tripping

[royal grail]

that is correct

I just dont know how much expansion your school wants, it really depends on the teacher's demand

[dense lily]

it's prolly right, ill ask tomorrow

ln is just log of base e right?

[royal grail]

yes

[dense lily]

like, how

i dont even get this

[royal grail]

so ln(2x+5) = log of (2x+5) to the base e, and this = 0

can you convert this in the form of exponential/power?

[dense lily]

hmm

switch x and y?

so basically just make 0 a y and plug it in?

[royal grail]

no

ok lemme ask you this

log of x to base 2 = 4, what is the value of x?

[dense lily]

2

[royal grail]

and how did you solve it

[dense lily]

because

2^2 = 4

[royal grail]

yup

that's what I meant

apply the same logic to the question

[dense lily]

so to what does e need to be raised to to become 2x + 5?

e = 2.718 right?

[royal grail]

well here you're told e is raised to 0, but yh the question is basically what x satisfies e^0 = 2x+5

[royal grail]

yes

[dense lily]

so

e^x = 2y + 5?

solve for y?

[royal grail]

no

you're given the eqn already

[dense lily]

well ok but should it or should it not be the inverse?

[royal grail]

$\ln({2x+5}) = 0$ means $e^0 = 2x+5$, and just rearrange for x

[ocean seal]

lgkoo

[dense lily]

well ok, i got (e^0 -5)/2

how does that sound

[royal grail]

do you know what e^0 is

[dense lily]

well

1

[royal grail]

yh

[dense lily]

so... -2?

[royal grail]

yh

[dense lily]

X

=

-2?

[royal grail]

yh

[dense lily]

WHHHHHAAAAAT

[royal grail]

you can check by substituting it back to ln(2x+5)

[dense lily]

NAHH BUDDY IM GOOD THAT WAS ENOUGH FOR ME

TRUST

that would just equal 1 though right

it would be ln (1)

[royal grail]

yh, which is?

[dense lily]

log base e of 1

?

[royal grail]

yh what is that

[dense lily]

0

OMG

[royal grail]

yh

[dense lily]

DUDE MATTTTHHHHHH

gimme a sec

This correct?

[royal grail]

I couldn't see the question due to reflection of light on your screen

[dense lily]

it's just logbase 3

oh wait

sorry

that's log base 3 (2x - 5)

there we go

[royal grail]

ok

where does the (2x-5) on the RHS come from?

[dense lily]

that's just multiplying both sides by 2x - 5

[royal grail]

but you didn't do that to the left as well

[dense lily]

sorry forgot to write it in

my bad

[royal grail]

ok

so you get x^2 = 18x-45

then you divided both side by x?

[dense lily]

uhh yep

then

18 - 45

= -27

[royal grail]

why isn't your 45 divided by x then

[dense lily]

uhh

do i have to do that..?

[royal grail]

if you gonna divide both side by x (or anything) you have to do it to all terms

[dense lily]

so - 27 aint the answer then

[royal grail]

no

[dense lily]

so

wait

what do i do then

[royal grail]

x^2 = 18x-45 can be rearranged as x^2-18x+45 = 0

do you recognise what type of equation this is

[dense lily]

oh

yeahhhh

i do

[royal grail]

what is it

[dense lily]

x -

and x -

so

-15 and -3

so x = 3, 15

[royal grail]

yup

[dense lily]

so here, first i thought,

x^2 - x -2

= 4

then that would become, theoretically

$x^2 - x - 6$

[ocean seal]

GMDennis

[royal grail]

yup

[dense lily]

correct?

ok

so

I can't figure out how to factor that

my brain aint braining

[royal grail]

so the question to ask yourself is

[dense lily]

( x + __) (x - _)

[royal grail]

which two numbers multiply to give -6, but sum to be -1

[dense lily]

-3 and + 2?

YOOOOOOOO

[royal grail]

yeah

[dense lily]

thanks man

[royal grail]

you're welcome

[dense lily]

that's it

have a wonderful eclipse, wherever you live

.close

[lone heart]

Channel closed

Closed by @dense lily

Use .reopen if this was a mistake.

[royal grail]

thanks, you too

[lone heart]

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.

[summer void]

I need help this question asks for the y intercept of a quadratic graph but the y intercept is not shown how do I get it

[true trout]

solve for x=0

[opaque monolith]

This is parabola. Find it's equation first if not given to you. Then put x = 0 and get the value of y. That'll be your y intercept.

[summer void]

How would I get the equation? I just started learning this part Friday on our own and we hadn’t been shown how to do that yet

[last thorn]

just by inspection, no?

[true trout]

i dont remember the formulae for that lol, i always go visually :p

[last thorn]

u can see the intercepts are 1 and 7

[summer void]

Dang

[summer void]

That’s x I’m looking for the y intercept

[formal ginkgo]

i think you can like create the equation of this parabola then solve it

[summer void]

Well this isn’t good

[last thorn]

ah

[lament forge]

i mean any three points on the graph determine it, but yeah it's easier to find special points if you have them

[lament forge]

well no, the x intercept is enough

[opaque monolith]

y = ax² + bx + c is the general form.
See for three points that lie on the parabola and put them into equation one by one. Now you'll end up with three linear equations in a, b and c. Solve them and get the values of a, b, c

[lament forge]

now you know the roots of the quadratic

...wait no is that enough?

ok well it's useful

[summer void]

Idk I’ve barely started learning

[lament forge]

ok yeah this makes more sense actually

just pick any three points, you get three equations, then solve them for the coefficients

[summer void]

I can try

Thanks
Figured out what it was the teacher just wanted us to put none I wasted like an hour on that

.close

[lone heart]

Channel closed

Closed by @summer void

Use .reopen if this was a mistake.

[lone heart]

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.

[viscid trench]

Which one of these is the odd one out

[royal grail]

what does ^ and upside down ^ and ¬ mean?

[viscid trench]

By the way, my answer is that ג is odd. My cousin says א so that's why I'm asking here ^^

^ means and upside down ^ means or

And the last one means not

[royal grail]

ok gimme a sec to think

I think the answer is ג like you said, cuz the middle one should mean that: there exists x real, such that (x>5 and x not element of A) which is what the first one is. The last one means that for all x real, (x>5 and x not element of A). The last one is for all x and the first two are only there exists an x

[viscid trench]

Yeah

[lone heart]

@viscid trench Has your question been resolved?

Channel closed

Closed by @viscid trench

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.

[eternal star]

Differentiation - Change of variables

[eternal star]

Perform derivatives dx/dr, dx/ds, dy/dr, dy/ds and dz/dx, dz/dy. Finally, calculate dz/dr and dz/ds

[lone heart]

@eternal star Has your question been resolved?

[lone heart]

@eternal star Has your question been resolved?

Channel closed

Closed by @eternal star

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.

[north hemlock]

I'm trying to do this and I'm a bit confused, there is no rotation in the Klein-4 group sooo........ what do I do?

[north hemlock]

Maybe just finding the left and right cosets would be easier but then it'd be longer

[lone heart]

@north hemlock Has your question been resolved?

[north hemlock]

<@&286206848099549185> 😔🙏

.close

[lone heart]

Channel closed

Closed by @north hemlock

Use .reopen if this was a mistake.

[lone heart]

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.

[misty hearth]

Using Comparison Test or Limit Comparison Test, determine whether the series converges or diverges. I'm assuming this is convergent with LCT, but with CT it's either one...

[lone heart]

@misty hearth Has your question been resolved?

[misty hearth]

<@&286206848099549185>

[thorny copper]

do you consider the p-test a comparison test?

because if you use that then you can show that your sum is smaller than 1/n^1.2 which does converge by p-series

[lone heart]

@misty hearth Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

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.

[left topaz]

Any chance someone can help me start this?

Sorry😅

[tiny fable]

don't yoink someone's help channel, there are more available

[sturdy lava]

When I sent the question, her questions was sent, sorry :'v

[left topaz]

It was like same second it's alright

[tiny fable]

all good :p

[royal grail]

they sent the questions pretty much at the same time, so saying they yoink is a bit mean ngl

it's just unfortunate timing

[tiny fable]

just tryna keep things organized, people steal help channels all the time and I didn't see what happened lol

[royal grail]

yh I get that, that does happen

[left topaz]

That's fair, anyone any good at statistics and probability?

[tiny fable]

well recognize that 2.18 is where the z score would usually be

[subtle light]

i would say try and use a normal distribution table to find the tail probability

[tiny fable]

so 2.18 is the z-score. you just need to find the confidence level corresponding to that z score

[subtle light]

that’ll give us the alpha value

we can get the exact confidence level using that

[left topaz]

So how do I find the confidence level without the rest of the information though?

[subtle light]

you really only need the z score

you don’t need anything else

do you have a normal distribution table?

[left topaz]

This right?

[subtle light]

yeah

so what is the tail probability of a z score of -2.18

since this gives us the left tail, we use the negative version

[left topaz]

.0146

[subtle light]

yep

and that’ll be our alpha value

do you know how to get the confidence level from there?

[left topaz]

Not really

[subtle light]

have you learned that $\alpha=\frac{1-C}{2}$

[ocean seal]

y0shi

[subtle light]

where C is the confidence level

and alpha is the tail probability

if not we can just use the table again

[left topaz]

I don't believe we learned that

[subtle light]

alright

so the confidence level is the middle probability, or the area between the positive and the negative z score yes?

[left topaz]

Yeah

[subtle light]

so using the table

we really just need the probability to the left hand side of the positive z score and then we subtract that by the tail probability of the negative z score

so that we can get the area of the middle bit

so what’s the probability of the positive z score?

[left topaz]

Is that still using the same table?

[subtle light]

yeah

[left topaz]

.9854

[subtle light]

yep

so we essentially just do .9854-.0146

and that’ll be our confidence level

[left topaz]

Oh ok that makes sense

[subtle light]

alright good

[left topaz]

So when I subtract that that should be my answer right?

[subtle light]

yep

[left topaz]

Thank youuu

[subtle light]

yw!

[left topaz]

.close

[lone heart]

Channel closed

Closed by @left topaz

Use .reopen if this was a mistake.

[lone heart]

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.

[dreamy bramble]

<@&286206848099549185>

[dreamy bramble]

help with these 3 please

[slow steppe]

What's your problem @dreamy bramble

[dreamy bramble]

idk how to do these

[slow steppe]

What don't you know? Give me a bit more info please. What's the goal of the problem

[dreamy bramble]

it says express as a product, im not sure how to get it in the right form, u can only use multiplication

and for the 3rd one i cant simplify further

im sorry, it's just that i've asked about 5 times and no one helps

[slow steppe]

By product, do they mean a multiplication or $\Pi$?

[ocean seal]

Tittom_123

[dreamy bramble]

multiplication

[slow steppe]

Ok

@dreamy bramble

[dreamy bramble]

u can't use +

@slow steppe

it should be like 2*sin(x)*cos(x)

something strict like that

[slow steppe]

You can't

[dreamy bramble]

yeah

no its wrong

[royal grail]

I mean you can, but not without making it uglier, by having some sin(alpha+pi/4)

[dreamy bramble]

yeah u need that

ugly is fine

@royal grail

And I don’t think u can use 1/cos(x)

@slow steppe

[slow steppe]

Just say sec(x) then, lol

You should use trig identities

[lone heart]

@dreamy bramble Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.

[tawdry urchin]

Let $a,$ $b,$ $c,$ $d,$ and $e$ be the distinct roots of the equation $x^5 + 7x^4 - 2 = 0.$ Find
\begin{align*}
&\frac{a^3}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^3}{(b - a)(b - c)(b - d)(b - e)} \
&\quad + \frac{c^3}{(c - a)(c - b)(c - d)(c - e)} + \frac{d^3}{(d - a)(d - b)(d - c)(d - e)} \
&\quad + \frac{e^3}{(e - a)(e - b)(e - c)(e - d)}.
\end{align*}

[ocean seal]

Dork9399

[tawdry urchin]

How would anyone even begin this question

This is from alcumus btw

Topic is polynomial interpolation

[main path]

Find the roots?

By the fund. theory of algebra there should be 5

[tawdry urchin]

my only concern is that some of them may be complex

or irrational

in which case it would be very difficult to find them

[main path]

Any chance of graphing?

[tawdry urchin]

calcs would be cheating

rather we are supposed to do without calc

[main path]

Yeah idk, looks like geometric series from a quick googling

[tawdry urchin]

hmm

so there are def no rational roots

i used rational root thm

<@&286206848099549185>

[dreamy bramble]

alr thanks

but how would u do 1-cot(a)

i simplfiied to csc(a)(sin(a)-cos(a))

but idk if there is a formula for sin(a)-cos(a)

<@&286206848099549185>

[tawdry urchin]

!occupied

[lone heart]

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

[tawdry urchin]

open a new channel for ur question

[slow steppe]

Well, in his defence, it used to be his channel and he probably didn't know that the channel timeout

[tawdry urchin]

yea

ik

[slow steppe]

Open a new channel and I will help you

[tawdry urchin]

<@&286206848099549185>

[slow steppe]

Probably start by finding the roots

[tawdry urchin]

um

[dreamy bramble]

oops

[tawdry urchin]

how would i do that

[dreamy bramble]

alr

[tawdry urchin]

they arent rational

they would be complex or irrational

so I have no way of guessing them

[slow steppe]

Idk

[tawdry urchin]

<@&286206848099549185>

please

fine

.close

[lone heart]

Channel closed

Closed by @tawdry urchin

Use .reopen if this was a mistake.

[lone heart]

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.

[worldly scaffold]

The only think I can think of is AA^T = In

but ig im asking can you take the determinant of both sides

like an equation

it seems weird

unless im on the wrong trajectory

[placid zinc]

That's exactly what you should do

[worldly scaffold]

oh!

[placid zinc]

Then keep in mind determinant properties

[worldly scaffold]

ik that det(A) = det(A^T)

because they have to be square right

[placid zinc]

It's true that det(A^T) = det(A)

Any matrix you can take the determinant of needs to be square

[worldly scaffold]

so would i j show that det(AA^T) = det(In)

and det(AA^T) = det(A)det(A^T)

so that would also mean

ok

there is where i get lost

i mean det(A) = det(In)/det(A^T)

oh

wait

okay

because (det(A))^2 = det(In)

and that means

det (A)

could either be 1

or -1

bec det (In) = 1

is that right @placid zinc

[placid zinc]

Yeah!

[worldly scaffold]

wow ok tysm

[lone heart]

@worldly scaffold Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

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.

[manic raft]

i am looking for some help with these questions, im confused how do i approximate the equations when theres no value other than infinity. not sure how to plug everything into a calculator

[livid sage]

you could just plug in an extremely large number

[manic raft]

i tried in desmos and the answer is still wrong

[main path]

Why does this kind of look like the equasion from which e was derived

[livid sage]

then you use the fact that these limits have well-known values

or plug into calculator that can give you exact answer (such as wolfram alpha)

in particular

[manic raft]

thanks lemme try

[manic raft]

how would you find the answer if it was 3x on the bottom instead of just x

[livid sage]

by property of fractions, you can notice 5 divided by 3x is the same as 5/3 divided by x

[manic raft]

thanks for the help

i was able to solve everything

.close

[lone heart]

Channel closed

Closed by @manic raft

Use .reopen if this was a mistake.

[livid sage]

yw 🎉

[lone heart]

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.

[hexed glacier]

suppose 2nd derivative of A(x,y) with respect to either variable is a constant.

then it stands to reason that A(x,y) is some polynomial of x of degree 2, yes?
and we know that the constant is twice the coefficient of that of the leading term

suppose 2nd derivatives of A(x,y) wrt x + 2nd derivative of A(x,y) wrt y = 4c
since we know, 4c is a sum of 2 terms, and we know that each 2nd derivative contributes twice the power of the coefficient of the leading term(2nd power)

[hexed glacier]

hmmm...actually, no.

2 x coefficient of x^2 might not be 2, same for that for y^2

yeah, my question is how to find the ratio of contribution x^2 term vs y^2 to term towards the laplace equation

[lone heart]

@hexed glacier Has your question been resolved?

[lone heart]

@hexed glacier Has your question been resolved?

[lone heart]

@hexed glacier Has your question been resolved?

[lone heart]

@hexed glacier Has your question been resolved?

[south otter]

I couldn't get this right, help me😅

[lone heart]

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

[unborn herald]

Allen module?

[south otter]

Nah! It's from a course(BBS) text book

[lone heart]

@hexed glacier Has your question been resolved?

[chrome glen]

wsg

[hexed glacier]

.close

[lone heart]

Channel closed

Closed by @hexed glacier

Use .reopen if this was a mistake.

[hexed glacier]

i will ask again another time

[lone heart]

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.

[cobalt token]

Looking for some help with this question... I'm pretty bad at Linear Algebra so I'm not exactly sure where to start. I figure that this is asking me to cross check it with the following:

[cobalt token]

but I'm not positive on how to check these

.close

[lone heart]

Channel closed

Closed by @cobalt token

Use .reopen if this was a mistake.

[lone heart]

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.

[valid tendon]

Hello just a continuation of what I was discussing in #help-7｜zen1thxyz (you have to scroll up a bit) I remember there's a rule about you can only have an equal to and greater than on one side of the domain

[valid tendon]

like this

Is this valid?

you get exactly the same thing with this

[lone heart]

@valid tendon Has your question been resolved?

[valid tendon]

<@&286206848099549185>

just a basic question it's not gonna be a big long discussion

cause apprently you've got to have one of the previously discussed things on each side of the domain

if it has multiple sides

it's busy today isn't it

ASODME ODMSD PS

c'mon lads

am I not explaining it cleary enough?

nvm

.close

[lone heart]

Channel closed

Closed by @valid tendon

Use .reopen if this was a mistake.

[lone heart]

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.

[restive swan]

Anyone know how to get started after setting (c+is)^4 = to cos4θ + isin4θ?

tried expanding the bracket fully but doesnt seem to get anywhere

[past musk]

try multiplying the cos^4(theta) on the right side denominator to the left

[restive swan]

but dont you have to start with (cos + isinθ)^4 = cos4θ + isin4θ

then somehow get to what is shown in the q

[past musk]

not really

[livid sage]

uhh yes???

your first suggestion is fine; helps see what is going on in the expression

but your final goal should be to move forwards from the given expression to achieve the desired result

[past musk]

I would start off with saying something like "consider cos^4(theta) times the left hand side"

the goal is then to show that it equals 2 cos(4 theta)

then you just divide at the end to get to that result

i mightve made it more confusing but yeah

[restive swan]

these are the first couple steps apparently but idrg what is going on here

i understand theyve flipped the sign although idk why u would think to do that but idk where the 3rd line has come from lol

[livid sage]

do you know why the 2nd equation is true

[livid sage]

u think to do that because that's exactly what's in the thing you're trying to prove lol

[past musk]

the third line comes from adding the previous two lines together

[restive swan]

oh yeah

makes more sense now haha

thank you

.close

[lone heart]

Channel closed

Closed by @restive swan

Use .reopen if this was a mistake.

[lone heart]

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.

[vivid badge]

this should give me -2/3 instead of -4/3 where is my error

[vivid badge]

my previous steps

[coral flower]

Last step from here

Where did the 2 in denominator go

[vivid badge]

i added it later

[wraith stratus]

you also integrated $4\int u^{1/2}$ wrong, there's an extra 1/2

[ocean seal]

chebyshev's infinite pee norm

[coral flower]

Also the 1/2 is multiplied with the entire thing starting from 3rd step

[vivid badge]

idk

its weird

i think i will try it again

[coral flower]

Tbh it would be better if u rewrite step 4

More clearly

[vivid badge]

i also tried sub by parts

but doesnt work

[coral flower]

U sub works why go for by parts

Just go the easy way

[vivid badge]

ill try it again and ill draw clearlier

if i have the same error ill ping you

[coral flower]

Ok

[vivid badge]

the same

first part finding u, du, x^3

changing variables

and expanding

simplifications/ replacing u, and end of the problem