help-23

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@jade lotus Has your question been resolved?

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[mystic fox]

I'll send everything in a minute

[mystic fox]

Idk if ive done the right method but i got the correct answer

[toxic shoal]

you can write 2^x as e^{ln(2) * x}

[mystic fox]

ik but its a show that question so i need to be specific on what steps i make

[toxic shoal]

well you would say note that y = e^{ln(2) * x} (and maybe you can argue this)

[frank glen]

Does it say you need to use implicit differentiation?

[toxic shoal]

then take the derivative via chain rule

if you can

[mystic fox]

it is an implicit differentiation question and ive realised iove done no implicit differentiation

[frank glen]

Looks implicit to me

Even though it was hard to follow the steps

The LN isn't written that way btw

You write it like this

$y = 2^x \ \
\ln(y) = x\ln(2)$

[flat frigate]

VulcanOne

[mystic fox]

yeah i kind of have a habiot of doing it in capital

[frank glen]

Small letters and you put brackets to make it clear you are using natural log

Anyways you did that then you went and divided by ln(2)

Which doesn't matter anyways since it is a constant

Then you did

[mystic fox]

i differentiated but i forgot dy/dx i think

[frank glen]

$\frac{1}{y\ln(2)} \cdot \dv{y}{x} = 1$

[flat frigate]

VulcanOne

[mystic fox]

yeah forgot that part

[frank glen]

Yeah this is implicit

Btw you can do this

$\frac{1}{y\ln(2)} \cdot \dd y= 1\dd x$

[flat frigate]

VulcanOne

[frank glen]

Still works

Then at the end when you want dy/dx you divide dy by dx

On one side

Which is the part we all usually skip

[mystic fox]

oh i see

i could divide by (dy/dx) then (dx/dy)^-1=Dy/Dx so i just do ((yLn(2))^-1)^-1 and substitues y=2^x

thanks i really appreciate it 🙂

.close

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[frank glen]

Be careful because this might confuse you

Keep things simple

[mystic fox]

yeah but it makes more sense now

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[dense tiger]

just show your work

chain rule

[slate drift]

First one is right

[dense tiger]

Oh it’s in the exponent

and it’s not multiplied

yes

[slate drift]

Correct

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[dense tiger]

right skewed

[lean otter]

yeah right skewed seems right

[mortal sandal]

well also bimodal quite possibly

not the smooth curve but the histogram

I would go bimodal

hard to say for sure though

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[mortal sandal]

ur good np

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[vital junco]

wait nvm I just realized it must be x = c

would I be correct on that assumption?

ye it has to be

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[crisp holly]

hello

[crisp holly]

i need help with a simplification of a formula i found in a research paper

First equation z¨ = (α / m) · (vw - z˙)^2 - g

Second equation: g = (α / m) · v^2eq

And it comes out....

one moment

filling in g in the first equation gives
z¨ = (α / m) · (vw - z˙)^2 - ((α / m) · v^2eq)

but from then on i qouldnt know what to do

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@crisp holly Has your question been resolved?

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@crisp holly Has your question been resolved?

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[rustic goblet]

!status

[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[rustic goblet]

1

I've been considering a lot of sets and functions, but for any X, f, and g I think of, I can't seem to get g o f to be bijective

When I draw the domains/codomains/maps out, I usually find that the composition isn't injective

[lean otter]

Here's a nice thread that might help https://math.stackexchange.com/questions/198379/example-where-f-circ-g-is-bijective-but-neither-f-nor-g-is-bijective

[rustic goblet]

Let me have a look, thanks for the link!

[broken yew]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[broken yew]

surely its much better to hint them towards an answer 😒

[lean otter]

I generally agree with that but that post provides a ton of perspectives

Usually I'm the one doing !nosols xD

[broken yew]

regardless, the original format really does indicate it wants the person to find it

example hunting is a v important skill to develop

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[safe radish]

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[cold tide]

.close

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[wicked basin]

hi so i’m supposed to prove sqrt 3 is irrational

i did so in the picture i sent, but the part that throws me off is the last part mentioning 3j, 3j+1 and 3j+2? how do i prove it using that?

[bold aurora]

a can't be 3K, you said a is prime

Well in actuality you just need a and b coprime, so it's fine

[wicked basin]

wait so how should i rewrite that?

[bold aurora]

where a,b are COPRIME instead of PRIME

[wicked basin]

👍🏾👍🏾👍🏾thank you

[bold aurora]

We just need a and b not to share any factors

[wicked basin]

👍🏾and the rest is good?

[bold aurora]

Nope

[wicked basin]

o

[bold aurora]

You get that b^2 = 3k^2, so 3 must be a factor of b (by the fundamental theorem of arithmetic), but we said a is also a multiple of 3. CONTRADICTION, a and b were supposed to be coprime

[wicked basin]

ahhhh yes i had to fix that one too

thank you very much

i might just go with this method cuz i have no clue what the 3j,3j+1,3j+2 thing is

[bold aurora]

The 3j, 3j + 1, 3j + 2 stuff idk what they are on about, don't think I've seen a proof like that, is j the imaginary unit?

That is by far the most convoluted way to do the same thing you did

I would also submit the classical proof which is the one you did

After fixing it though, it's key that they are coprime, e.g. share no factors, not prime

[wicked basin]

and thanks again for fixing that comprimé mistake

coprime* french keyboard autocorrect

[plucky elk]

FYI this is bannable

[wicked basin]

oh snap i did not know my bad

i will revise the rules rn

[plucky elk]

Yea if you can't explain the answer yourself just don't help

[bold aurora]

It's his own question btw, he asked the ai

[wicked basin]

wait i’m the one asking the question 😭

[plucky elk]

My bad

[wicked basin]

i was discussing an answer i got from it again tho my bad i have now revised the rules

[plucky elk]

It's still bannable though

[wicked basin]

and will no longer do so

.close

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[bronze dust]

Where did my work go wrong? Work below

[high aspen]

I'd recommend using the Quotient Rule rather than h substitution.

Which is f'g-g'f/g^2

[bronze dust]

oh i dont know how to do that yet so i assumed im not allowed to use it in this homework

[high aspen]

I see

[fathom goblet]

It says to use "the definition of derivative"

[high aspen]

I'm aware

But I always ignored that and used the optimal rule when I was taking Cal 1

Some professors are different ig

[bronze dust]

okay, so can you guys explain what I did wrong then please?

[high aspen]

Well the answer is supposed to be a fraction

Your numbers aren't necessarily incorrect, but somewhere along the way it looks like you disregarded that you're working with a denominator.

Try 9/v^2 + 81 or 9/(v+9)^2

[bronze dust]

how did the 9 get up there

[bronze dust]

the first one is wrong

and i assume the second would also be

[high aspen]

Probably just wants you to not solve the bottom part

And leave it as (v+9)^2

[bronze dust]

@high aspen

[high aspen]

You're missing a comma

[bronze dust]

oh you right

.close

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[marble lake]

I am having a hard time with this for some reason,
so we use the chain rule for this where the outer (f(x)) is (x^3) and the inner (g(x)) is the quotient 1/5t +1
derivatives being 3x^2
i am drawing an awful blank on how to use the quotient rule, could someone help me get the derivative of the quotient

[marble lake]

not trying to just look up the answer for it so wanted to actually work it out

or am I wrong to think we use the quotient rule to find the derivative?...

[plucky elk]

Unnecessary

,tex .exp rules

[flat frigate]

riemann

[marble lake]

i dont see how those properly apply to this to find the derivative... maybe im just stupid idk...

[plucky elk]

Negative exponent

[marble lake]

im still having a hard time seeing how thats used here...

maybe im just not understanding something

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[marble lake]

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[rough matrix]

I need help

[rough matrix]

Question 3

What do u get for the answer

@devout shale

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@rough matrix Has your question been resolved?

[devout shale]

We don’t give answers here

[rough matrix]

Wdym?

I got 40

Is that right?

Can i ask that

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[deft pivot]

how to factorize this :
(x+1)(x+3)(x+5)(x+7)+15

long division not allowed

[quasi bison]

why is long division not allowed?

plus you could circumvent it anyway once you find a root

you could write (x+1)(x+3)(x+5)(x+7)+15 = (x-c)(Ax^3 + Bx^2 + Cx + D), where c is a known root, and equate coefficients

and in doing so you will accomplish the same thing as you would with long division

so why exactly is it prohibited? @deft pivot

[deft pivot]

because it is low grade math problem

[quasi bison]

it's high-grade enough for factorization to be a thing we know of, but it's low-grade enough that we have no access to long division?

i mean fuck i guess maybe write the thing as ((x+4)^2 - 1)((x+4)^2 - 9) + 15

then youll have an almost quadratic

[deft pivot]

something like this ?

[quasi bison]

no

[deft pivot]

oh its not even tru

[quasi bison]

x+3 = (x+4)-1, x+5 = (x+4)-1, x+1 = (x+4)-3, x+7 = (x+4)+3

[deft pivot]

yes yes

[quasi bison]

do you at least have access to (a-b)(a+b) = a^2 - b^2?

or is this also forbidden?

[deft pivot]

that i think is correct way

[quasi bison]

maybe it's also forbidden.

maybe it's too high grade.

[deft pivot]

no do not be sarcastic

[deft pivot]

so that next?

[quasi bison]

then group these into pairs

((x+4)^2 - 1)((x+4)^2 - 9) + 15

and from then on, treat (x+4)^2 as a single indivisible unit

maybe give it a new name like t

unless that's also forbidden

in which case don't

you'll have (t-1)(t-9) + 15, a quadratic

[deft pivot]

aha

i see

i will do that now

[deft pivot]

this is good this is not forbidden

is thia correct ?

yes it is

tnx Ann

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[waxen moon]

Hi, is this correct?

[quasi bison]

the "x=r" bit should probably be removed

looks fine otherwise

[waxen moon]

-1 < x < 1 refers to the common ratio value

Could I just replace x with r then?

[quasi bison]

yes

you call it r

so let it be r

the letter x is already in use

[waxen moon]

Alr then

Thx

[waxen moon]

Btw I have another question. Is this correct?

I got 1 / -(x-1) instead

[west zinc]

You did a calculation mistake

[quasi bison]

1-(2x-1) ≠ -2x

1/(1-x) is correct actually

i thin?

i think*

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@waxen moon Has your question been resolved?

[waxen moon]

Alr then, thx

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[ripe gyro]

how did the power magically get out

[devout shale]

it didn't get out of anything

it is the same way of writing the same thing

[ripe gyro]

so its the same thing?

ahhhh well

thanks

[devout shale]

yes

[ripe gyro]

.close

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[pliant elk]

Hello

[pliant elk]

I'm having trouble finding out where to start this set proof

If $A, B$ and $C$ are sets, then $A - (B \cap C) = (A - B) \cap (A-C)$

[flat frigate]

Ziggy

[pliant elk]

How do I start the proof?

$x \in A, B,,\text{and},C$

[flat frigate]

Ziggy

[pliant elk]

hm

🥺

[gentle herald]

Venn diagrams ?

[pliant elk]

no

venn diagrams does not equal proof

[final halo]

Take an element in the LHS and show its an element of the RHS

And then vice versa

[pliant elk]

so any element in the left hand side?

so i can start with

$x\in C$ or $x \in B$ or $x \in A$

[flat frigate]

Ziggy

[final halo]

No in the whole left side

So start with something in A - (B n C)

[pliant elk]

okay

then just move through the definitions?

but i dont really undrestand

what tha means?

[pliant elk]

$x \in (A-(B \cup C))$

where is x contained?

[flat frigate]

Ziggy

[final halo]

That is where you start yes

[pliant elk]

so this means

$x \in A $ and $x \notin (B \cap C)$

[flat frigate]

Ziggy

[pliant elk]

that should be intersection not union sorry

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[lean lantern]

pls help

[boreal kernel]

easy

b is 180 - 21.5 - 65.5

because those 2 are parallel

[amber shore]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lean lantern]

you subtract 180?

[stuck crest]

look for a pair of alternate angles

[quasi bison]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[boreal kernel]

sorry guys

i didn't know that i can't help someone

[lean lantern]

thx victor

[stuck crest]

helping someone isnt the same as giving out the answer tho

[lean lantern]

it is

[boreal kernel]

It is i did the same one like 3 years ago

maybe 4

Can someone here help me ??? (help-2)

[lean lantern]

i did hte same thing victor did here but dosent work @boreal kernel

[boreal kernel]

what did you get for b

[stuck crest]

again you can look for a pair of alternate angles

[lean lantern]

thats what im saying how do i solve for b what doi i do

[boreal kernel]

look what taro said

[lean lantern]

ahhh

[stuck crest]

are you familiar with angle relationships?

[lean lantern]

i was going to subtract 180-125-30

[lean lantern]

not that much

?\

[lean lantern]

oh ok

[stuck crest]

the one your focusing on is the alternate interior angles

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@lean lantern Has your question been resolved?

[nimble shadow]

How to solve this problem?

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[pulsar pecan]

Hey

[pulsar pecan]

I ask about this one a few times but I am still super confused

1. Draw the amounts in the complex plan (i did)
2. Calculate A invert_b B

I dont know how to do the 2nd one

I just know that they touch each other in 1+i and 1-i from my drawing but i seriously do not know how to calculate it

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[fleet ivy]

can someone help me figure out how to do this? so far ive made the new ln x and ln y table i even have the slope of 1/2 i did all this on desmos graphing but i can not figure it out after that point.

[fleet ivy]

this is a practice one if you want to see the format of what is needed

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[haughty nexus]

can anyone give me a clue on how to prove that lnx< ( or equal ) x-1. f(x)=xe^-x. i found that f is increasing from 0,1 and increasing from 1 to + ♾️ but now i’m stuck

[safe radish]

@haughty nexus Has your question been resolved?

[safe radish]

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[lean otter]

Can I have some help, I need someone to check my work

I got u_t = k(u_rr - u_r/r) instead of with plus

My steps were that u_x = du/dr * dr/dx + du/dt dt/dx = du/dr * dr/dx = u_r * x/r, and similarly for u_y.

Then I got that u_xx = d/dr (u_r *x/r) * dr/dx + d/dt(u_r x/r) * dt/dx =
d/dr(u_r x/r) dr/dx = (u_rr * x/r + u_r(-x/r^2)) * x/r = (u_rr)(x^2/r^2) - (u_r)(x^2/r^3), and similarly u_yy = (u_rr)(y^2/r^2) - (u_r)(y^2/r^3), then adding together with the fact u_zz = 0:

u_t = k(u_xx + u_yy + u_zz) = k((u_rr)(x^2 + y^2)/r^2 - (u_r)(x^2 + y^2)/r^3) = k(u_rr - u_r/r).

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185> sorry for double ping but need help with this today

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[quartz ledge]

can you help me calculate the mass of the mcdonalds big mac please

[quartz ledge]

need it for a friend 🙂

https://tenor.com/view/wait-benjamin-engvid-hold-on-wait-a-minute-gif-25784741

[oak wraith]

look it up

[quartz ledge]

bruh

[oak wraith]

https://letmegooglethat.com/?q=mass+of+the+mcdonald's+big+mac

[quartz ledge]

i need to show how i work it out 😉

May 21, 2018 The un-cooked weight of a Big Mac® is 240g and is broken down as follows: Bun - 74g, Beef patty - 45g x 2 = 90g, Cheese -14g, Lettuce - 28g, Pickle - 2 x 3.5g =7g, Onion - 7g, Sauce - 20g.

so that's the uncooked formula

i need a cooked one

[quartz ledge]

cocked*

anyways

so i need to know how the cocking affects the formula

@oak wraith

[split fulcrum]

Cocking effects

🤔

[oak wraith]

how is anybody realistically supposed to know that

[quartz ledge]

maths

[plush spruce]

‘The weight of the final product will change due to the cooking process therefore we cannot declare a served weight’

[quartz ledge]

next

[split fulcrum]

stop throwing shade go buy some maccas and test yourself

[plush spruce]

It’s from google 💀

[quartz ledge]

https://tenor.com/view/cringe-gif-22399496

[safe radish]

@quartz ledge Has your question been resolved?

[safe radish]

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[wheat island]

the first one is correct
the second is incorrect and i am not sure why

if repsonding, pls ping, i am continuing after this question

[safe radish]

@wheat island Has your question been resolved?

[safe radish]

@wheat island Has your question been resolved?

[wheat island]

still no

@versed bronze

sorry, meant
<@&286206848099549185>

[wheat island]

nvm .close

.close

[safe radish]

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[thick fog]

what does this want me to do I'm confused

[thin bridge]

which specific part/phrase there confuses you

[thick fog]

the endpoints...

it gives the endpoints

then ask to find?

[thin bridge]

it gives the endpoints of a radius

they want the endpoints of a diameter

[thick fog]

oh...

[thick fog]

thanks

got it

.close

[safe radish]

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[void steeple]

may I have help on this?

[void steeple]

I know the first one is wrong

[icy lance]

why

[twilit spindle]

why would it be wrong?

oh actually

it isn't the negative, the negtive isn't distributed

[void steeple]

im not sure it said it was wrong on my first submission

[hazy pumice]

It is wrong, because of the negative not being distributed in the -f(x)

[icy lance]

ah yeah, the 5

and the 3x

[void steeple]

so it would be the last one?

[thin bridge]

no

[twilit spindle]

what is f(x+h)?

[void steeple]

oh-

f(x=h)=4x^2+3x-5 right?

where should I begin?

[hazy pumice]

Hint: ||Your substituting (x+h) for x||

[void steeple]

is that not what i did when i said f(x+h)=4x^2+3x-5?

i just gotta solve that?

[thin bridge]

f(x) = 4x^2 + 3x - 5
f(x + h) is not the exact same thing

[hazy pumice]

Your not substituting x+h for x, for example say we have 6x^2 +5x+9
We would substitute x+h for x, which would get us 6(x+h)^2 + 5(x+h) + 9

[void steeple]

ohhhhh so it would be 4(x+h)^2+3(x+h)-5?

okay so where do i go from there now?

[hazy pumice]

You've got f(x+h) now we need -f(x) which we already know

If you recall you were given f(x) = 4x^2 +3x - 5

[void steeple]

yes yes

[hazy pumice]

So if you connect the pieces together... what does a negative do to a number?

[void steeple]

sorry i got lost is ti multiplying?

it*

so we know its not a or d correct?

[hazy pumice]

Well yeah but it can also be distributing, lets go back to our earlier example
We have f(x) = 6x^2 =5x+9, since we have a negative in front of f(x) we will get -(6x^2+5x+9)

[hazy pumice]

Yeah its not a or d

[void steeple]

ahhhh i see

sooooo

it should be c?

[hazy pumice]

If we go back here...

[void steeple]

cause b doesnt make sense

[hazy pumice]

(4x+4h)^2 does not equal to 4(x+h)^2

[void steeple]

okay so then how on b did they get the 2 with the 2xh?

[hazy pumice]

What is (x+h)^2

[void steeple]

ohhhhhh

I see

[hazy pumice]

yeah so it will be B

[void steeple]

2xh

okay i have 2 more questions if thats fine?

thank youuuu

[hazy pumice]

Sure

[void steeple]

okay one sec let me screen shot it

[hazy pumice]

Ok

[void steeple]

this is the next one

[hazy pumice]

Hmm ok

[void steeple]

i had variations with 5 and 4 instead of three but i stil got them incorrect

its the same concept no?

[hazy pumice]

Yeah it is the same concept, but I have never experienced these kinds of problems.

[void steeple]

ohhhh i see

[hazy pumice]

I have only been taught the surface of the product quotient

[void steeple]

would you like to see the other problem first?

[hazy pumice]

Not the other variations

[hazy pumice]

Sure

[void steeple]

okay

[hazy pumice]

Ok, I know how to do these types

[void steeple]

Okayyy

[hazy pumice]

Same concept except we aren't given the x variable in f(2-h), we can just substitute 2 for x

[void steeple]

okayyy

[hazy pumice]

actually thats wrong

[void steeple]

yeah i was a bit confused looking at that

its okay

[hazy pumice]

wait a minute let me try it out

[void steeple]

okay!

[hazy pumice]

I haven't done a problem that replaces the x variable in f(x+h) before

now that I think about it

[void steeple]

ohhhh i see

[thin bridge]

the same principle applies

[void steeple]

so where should i start exactly then?

[thin bridge]

exact same principle of functions

given f(x) = stuff
how do you get
f(thing)

replace all x in that equation with (thing)

[void steeple]

soooo f(2+h)=5(2+h)-6?

[thin bridge]

just like to get f(x+h), you replaced all x with (x+h)

yes

[void steeple]

okay one sec

f(2)=5(2)-6 and f(2+h)=5(2+h)-6

correct?

[thin bridge]

yes

[void steeple]

okay so what now?

[thin bridge]

use those expressions in your difference quotient and simplify

[void steeple]

soooo f(2)=5(2)-6 -> 10-6 -> 4

f(2+h)=5(2+h)-6 -> 10+5h-6 -> 4+5h

sooooo 4+5h+4/h?

[hazy pumice]

Remember -f(x)

[void steeple]

sooooo 4+5h-4/h?

[hazy pumice]

yup

[void steeple]

soooo the answer is 5?

[hazy pumice]

yeah

[void steeple]

okay did you figure out how to do the other one by chance?

this one

[hazy pumice]

No, I was following along with the other one

[void steeple]

ohhhh i see

[hazy pumice]

But then again same principles are still applied here

[void steeple]

i feel it would be the same

yeah

sooo 3x-1+h-3x-1/h?

[hazy pumice]

I've kind of figured it out

[void steeple]

okay so how exactly

[hazy pumice]

Think about f(x) = 4x^2 if we did f(x) = (x+h)
We would do 4(x+h)^2 same thing here, we have j(x) = 3^(x-1)

If you look at it 3^(x-1) is similar to 4x^2

[void steeple]

okay okay i kinda see it

soooo 3^(x-1+h)-3^x-1?

[hazy pumice]

,tex j(x) = $ 3^{x-1}

[flat frigate]

Bingbing123

j(x) = $ 3^{x-1}
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.59 \end{document}
                   
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[void steeple]

hmmm okay

[hazy pumice]

I had 3(x+h)^x-1 - (3^x-1) I maybe wrong though

Im working and improvising on the go

[void steeple]

i see it just seems the answers given are weird

[hazy pumice]

Yeah but they will start to make sense if we continue to go deeper

[void steeple]

sort of with the first part now its just the second part

with the 3^h

i dont think its the one with just the 3^h but I could be wrong

due to the fact thats the type of answer I put the last two attempts

[hazy pumice]

,tex it maybe $ 3^{x+h}{-1} $

[flat frigate]

Bingbing123

it maybe $ 3^{x+h}{-1} $

[hazy pumice]

no

thats not it

[void steeple]

you mean b?

[hazy pumice]

,tex it maybe $ 3^{(x+h)(-1)} $

[flat frigate]

Bingbing123

[hazy pumice]

there we go

It might be this for j(x+h)

[void steeple]

ohhhh okay okay

but for j(x) its 3^x-1

meaning maybe it is B

cause 3^h-1

[hazy pumice]

,tex I'm assuming that it is $ 3^{(x+h)(-1)} -3^{(x-1)} $

[flat frigate]

Bingbing123

[hazy pumice]

for f(x+h) - f(x)

[void steeple]

yeah i think so too but how do they get the h for the second part in the answers

[hazy pumice]

Well we have to simplify this first to figure it out

I would rather not guess or critique the answer choices

[void steeple]

so the first part would be 3^x-1

fair enough

[hazy pumice]

For j(x) ?

[void steeple]

for j(x+h) i think

[hazy pumice]

I got this for j(x+h)

[void steeple]

ohhhh i thought that was both

[hazy pumice]

This is for both f(x+h) -f(x)

[void steeple]

ohhhh okay okay

sooo j(x+h) = 3^(x+h)-1 for sure

[hazy pumice]

yeah

[void steeple]

soooo like you said with the equation 3^(x+h)-1 -3^(x-1)/h

[hazy pumice]

yeah

[void steeple]

would 3 be distributed into h and make 3h?

[hazy pumice]

wait where

In the f(x+h)

[void steeple]

yes

[hazy pumice]

I got 3^h

[void steeple]

for the second part?

[hazy pumice]

Then again I maybe wrong

[hazy pumice]

f(x+h)?

[void steeple]

no f(x)

i think its either b or c

cause it cant be a cause of the 3^h by itself

and then not d because of the first part

[hazy pumice]

ok

[void steeple]

so i would think b because of the negative

[hazy pumice]

Yeah thats what im thinking

[void steeple]

okay so were on the same page

[hazy pumice]

I gotta go and eat dinner

[void steeple]

okayy thank youuu

[hazy pumice]

Your welcome! You should use the helper ping

[void steeple]

i got a 100 😄

[hazy pumice]

Oh nice

[void steeple]

i appreciate your help

i understand now

[hazy pumice]

Ok, I'm happy I've been of some help

.close is the command for closing the channel by the way

[void steeple]

okayyy

.close

[safe radish]

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[unreal prism]

For this claim, I think I know how to prove the set part of it but not sure how to really start

[unreal prism]

I know "Let r be any factor of q" is for the end

So I don't need that right now

but how do I use "Let p and q be positive integers that are relatively prime"

Should I just start with the $x\in\mathbb{Z}$ stuff

[flat frigate]

42

[sterile apex]

is r . p multiply?

[unreal prism]

yea

[sterile apex]

sorry no idea how to solve that, I understood, but idk how

[unreal prism]

yeah all g

i think i know how to do all that but the relatively prime thing is confusing me

ill try proving it normally and see how to add the prime thing

[sterile apex]

they said that to not create a contradiction, cuz p | x , and q | x , it would be kp = k'q , ((k,k') from Z), so making them relatively prime prevent the equqlity of k and k', thus the eqhqlity of p and q.

[sterile apex]

okay let me see how if u make it

I'm curious

[unreal prism]

ah i see why they did that

[unreal prism]

i got to this

where pk = qf where l,f are ints

[sterile apex]

good

bro that's hard

🥲

[unreal prism]

yeah its literally a first year course man

its so hard

its our 2nd problem set

wait

but if i get there right

pk = qf

wait nvm idk

they dont have any common factors

so that has to come in this step i think

[safe radish]

@unreal prism Has your question been resolved?

[sterile apex]

yes exactly

[unreal prism]

this is waht my ta said

[sterile apex]

hmmm

I don't know how to use real numbers in that question

they are ints

[unreal prism]

yeah exactly

let me ask again

[sterile apex]

okay

wait

ARE U A 42 SCHOOL STUDENT

[unreal prism]

no, i wouldnt be this stupid

i actually didnt know what that was until i searched it up lol

[sterile apex]

LMAO

XDDD

I just saw ur name and thought u could be in that school

[unreal prism]

lmfao haha

nah idk

i like 42 as a number

just looks nice

yeha im still stuck at the same spot

oh not real numbers

is ok

[sterile apex]

😔

[unreal prism]

yeah i dont know

ill save it for tomorrow

nvm i need to finish it now

<@&286206848099549185> i need some extra help

nvm ig

.close

[safe radish]

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[lean otter]

👋 Hey, I was wondering if someone would be able to help me create an calculation?

I'm currently attempting to create an easy calculation for sizing different blocks in a 2d space.

So lets say there is a floor, the dimensions of that floor are 10 units length wise

Lets say also there are 5 pillars, now I need to create a way to calculate how large the pillars should be length wise so that the pillars would be evenly spaced apart by n.

So for example:

Lets just use the 5 pillars, on a 10 unit long floor and I want 1 (our n value) to be the space between them. So how would I calculate how long the 5 pillars should be?

[azure delta]

Do the pillars touch the edges, or do you have equal spacing from the pillars to the edges too?

[lean otter]

The pillars touch the edges

[azure delta]

So you have 5 pillars and 4 gaps

[hardy lion]

It would help if you drew a diagram both for us to see and to help you work it out

[azure delta]

so with pillar length p and gap length g, you would have 5p + 4g = 10

[lean otter]

so these are the original lines

then i need to figure out

how large they should be

so that the gap would be n units

for example 1 unit

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

.close

[safe radish]

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[safe radish]

@lean otter Has your question been resolved?

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[safe radish]

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[jade berry]

I need help with both A and B

[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[dense prism]

What did you get

[jade berry]

9:1

[dense prism]

And how did you arrive at that

[dense prism]

Hmm it should be correct, did they say anything about the order?

[jade berry]

nope

[dense prism]

For surface area right?

[jade berry]

yes

[dense prism]

Try 1:9 perhaps they want it the other way

[jade berry]

alright

[haughty nexus]

no

[jade berry]

is the answer for the volume one is also 1:9???

nvm

it's 1:27 right?

[dense prism]

Yes

1:27

Because radius is cubed

In volume

[jade berry]

oh ye!!!

for both question iii and iv, which ratio are they talking about?

1:9 to find the surface area and 1:27 to find the volume, right?

[safe radish]

@jade berry Has your question been resolved?

[jade berry]

.close

[safe radish]

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[safe radish]

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[civic elm]

is this correct

[safe radish]

@civic elm Has your question been resolved?

[thin bridge]

show your reasoning

[civic elm]

.close

[safe radish]

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[pulsar pecan]

hii

[pulsar pecan]

We look at the line in the complex system which is between z1 and z2. The line is now deformed in the following way: The line is scaled with a factor of 8, and rotated against the click around origo with an angle of 0.5. At last the deformed piece is paralellel offset with the vector (3,2)

Give the complex numbers w1 and 21 which make up the ends of the deformed line. The complex number closest to origo is w1

[safe radish]

@pulsar pecan Has your question been resolved?

[safe radish]

@pulsar pecan Has your question been resolved?

[safe radish]

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[mystic vale]

if this is the formula for nth root of complex numbers

if i want to find the roots of 1 , i should be able to plug in for tetha 0, 2pi, 4pi.... right?

[mystic vale]

if i use tetha zero, the first root with k=0 would be 1.
but if i use tetha 2pi i would get a different result...

[mortal sandal]

Any of them would work

You'd get the same roots just in a different order

Usually you go with the smallest non-negative

If you use all of the possibilities (0, 2π, 4π, etc) instead of just one, you don't even need the +2kπ part to find all the roots

[mystic vale]

you're right i was making an error later and i got wrong values ;D

ok thanks a lot!

.close

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[mystic vale]

true, that's the point of having +2pi XD

[mortal sandal]

np

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[shrewd topaz]

how do i solve this

[karmic falcon]

figure out where the function slopes up

and give the x values that define the range where the function move upwards

[shrewd topaz]

-1 and 1

?

[karmic falcon]

At x = -1, we see the function is at y = 1

At x = 1, we see the function is at y = -1

y goes from 1 to -1 so its decreasing there

that interval is not what we are looking for

the function should go up (y should increase proportionally to x)

[shrewd topaz]

how do i display that

[karmic falcon]

what do you mean

you find where it goes up then input the x values in interval notation

should I break it down more?

[shrewd topaz]

well the highest it goes is 1 and the lowest is -1

otherwise idk how to find where it goes

[karmic falcon]

I mean locally

when x is between -3 and -1: the slope is positive, the y value increases, our line goes up

when x is between -1 and 1: the slope is negative, the y value decreases, our line goes down

when x is between 1 and 3: the slope is positive, the y value increases, our line goes up

What do you not understand exactly?

[shrewd topaz]

so how do i communicate in interval notation that this is whats happening

[karmic falcon]

do you know how to do interval notation?

[shrewd topaz]

kind of. (a,b). i understand that a is lower bound and b is upper but i dont really understand what bound means

[karmic falcon]

interval notation is a way to describe a value that can be anywhere between two numbers

for example, an interval notation of (a, b) for value x means

that x can be anything between a and b except for a and b themselves

x for (1, 4) means x can be 1.0001, 1.002, 2.383, 3.5848, 3.9999 and all the values in between

using square brackets instead of parenthesis - [3, 4] instead of (3, 4) includes those "bound" values as possible x values

x for [1, 4] means x can be exactly 1, 1.009, 2.113, 3.8, 3.9999, exactly 4 and all the values in between

do you get it or should I explain more?

[shrewd topaz]

okay i understand that.

so i dont use the inf sign in this case

[karmic falcon]

no- we dont assume the function is increasing into infinity unless the question states so

past -3 and the function might curve down and past 3 the function might also curve down

so we only give intervals based on what we see on the graph

[shrewd topaz]

so we..input the x values?

the x values are -1 and 1

[karmic falcon]

those are the y values

the y values range between -1 and 1

the x values' domain is between -3 and 3

[shrewd topaz]

oh the x ranges between that

ahh

[karmic falcon]

so we see the function increase when x is between the numbers -3 and -1

we also see the increase between 1 and 3

how do we describe that with an interval?

[shrewd topaz]

(-3,-1) and (1,3)?

[karmic falcon]

do we include these values as part of the intervals though?

is the function increasing at x=-3, x=-1 x=1, and x=3?

[shrewd topaz]

it increases and then decreases

[karmic falcon]

yes

its turns at these points

therefore we keep the parenthesis

[karmic falcon]

so this answer is correct

[shrewd topaz]

its not

[karmic falcon]

did your teacher ask for you to use the union symbol?

You might need to use it

[shrewd topaz]

i separated it with commas and it told me it was the wrong answer

[karmic falcon]

(-3, -1), (1, 3) is different from (-3, -1) ∪ (1, 3)

(-3, -1) ∪ (1, 3) means to include all values in either bound

[shrewd topaz]

wrong

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[ionic blaze]

If x+y=4 and xy+6z=z^2+13, find x+y+z

[ionic blaze]

so like

no specifics if they're integers

otherwise z would've immediately been equal to 3

but

how to find answer '<'

the thing is my only answer is 7 because I assumed z to be an integer(ikr, doesn't make sense) and (answer keys haven't been given to us because of data privacy thing)

so I can't check

Maybe it could be integer

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[violet lion]

The question is asking to sketch the limit and I am confused on how it would look like.

[violet lion]

This is what I have currently, I don't know if it is correct

I'm mainly confused about lim x approaching infinity f(x) = infinity as well as lim x approaching negative infinity f(x) = negative infinity

[cold aurora]

You can't exactly go to infinity on a graph. x goes to infinity means that x is somewhere on the rightmost part of x-axis on graph. y being infinity means that y is somewhere on the topmost part of the y-axis on the graph. So, show a line going to topmost tight corner.

Similarly for negative infinity part.

[violet lion]

oh so would it look this then?

would that also affect the curved line in the middle?

[cold aurora]

As long as that last curve in the rightmost part is going upward.

[cold aurora]

No.

[violet lion]

oh, so the first picture was wrong because the rightmost part was going downwards and the seconds picture is correct because the rightmost part is now going upwards ?

[cold aurora]

Yes

[violet lion]

Can i ask one more question

[cold aurora]

Of course.

[violet lion]

what does it mean by "f is odd"?

[chilly lance]

where the graph of an odd function has rotational symmetry with respect to the origin, meaning that its graph remains unchanged after a rotation of 180 degrees (or pi radians) about the origin.

[violet lion]

so if this is my original graph:

then it will change to.

is that what you mean? Also what would the function being even, mean then?

[cold aurora]

Is the original graph just this much?

How are you getting the middle part then?

[violet lion]

For this picture I forgot the point at (0,0), and I didn't draw the up because I didn't know what "f is odd" meant.

The original was only this much, without "f is odd"

then Prime said : where the graph of an odd function has rotational symmetry with respect to the origin, meaning that its graph remains unchanged after a rotation of 180 degrees (or pi radians) about the origin.

then I got this image

[cold aurora]

Yeah. That's correct.

[cold aurora]

An even function means that f(x) = f(-x).
Which means that it is symmetric about y-axis.

[violet lion]

so something like this?

[cold aurora]

Symmetric about y axis.

Not x axis.

[violet lion]

oh so something like this? then how would it be different an odd function, I'm confused now.

[cold aurora]

Yes

If it was an odd function, your line would have continued like a straight line in 3rd quadrant.

See. It's important that you use these definitions to understand this properly.

Odd functions are such functions where f(-x) = -f(x).
Even functions are such functions that f(-x) = f(x)

[violet lion]

sorry about this can i ask one more question, if you have the time

[cold aurora]

Sure.

You don't have to worry about asking me more questions. Ask as many as you need.

[violet lion]

ok, so I'm confused about these 3 parts?

so I understand that there will be a hole at (3,4)

[cold aurora]

No, there won't be.

f(3) = 4

[violet lion]

but how would i graphically be drawing 3 from the right and left?

oh, it's a dot

[cold aurora]

Yes.

[cold aurora]

You draw graph in such a manner that on the left side of 3, its y values are reaching 1.

[violet lion]

would it be something like this?

or do we not use the dot?

[cold aurora]

No.

Come on. y values are reaching 1 when x reaches 3 from the left side.

Your y is reaching 4 when x reaches 3.

[cold aurora]

Yeah. This works.

It doesn't have to be a straight line.

Should just be reaching (3,1).

[violet lion]

ok since I got that down, if I was given something like (it going to take a some time to draw this)

would 10 be something like this?

[violet lion]

if it doesn't have to be a straight line, then what does that imply?

[cold aurora]

You drew an almost parallel to x-axis line for that one. I was saying that it could have taken any route.

[cold aurora]

This isn't even a function.

See for yourself. Does it pass the vertical line test?

[violet lion]

no it doesn't

[cold aurora]

So, it's clearly wrong.

[violet lion]

I don't know how to express this properly is there any suggestion?

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[violet lion]

?

.close

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[glass gust]

how can i make this limit work, can someone break it down for me?

[stray socket]

Expand everything

Collect like terms

When it comes to derivative limits, you just gotta expand everything

Alternatively, you can do the limit for x^2, -2x, and 1

And add them all up if that makes it nicer

[glass gust]

we havent covered derivatives yet

[glass gust]

and wdym by this

[stray socket]

I mean the limit you're doing is the literal mathematical defintion for the derivative. I'm just calling it the "derivative limit"

[stray socket]

You're familiar with the algebraic limit theorem, right?

$\lim (f + g + h + ...) = \lim f + \lim g + \lim h + ...$

[flat frigate]

Umbraleviathan

[stray socket]

At least for addition

So when you have to do the derivative limit for x^2 - 2x + 1

You can just take the derivative limit for x^2, -2x, and 1 separately and then add them up

So Gimmie a sec

$\lim_{\Delta x \to 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x} + \lim_{\Delta x \to 0}\frac{-2(x+\Delta x) + 2x}{\Delta x} + \lim_{\Delta x \to 0} \frac{1 - 1}{\Delta x}$

[flat frigate]

Umbraleviathan

[stray socket]

And now you got 3 bite-sized limits

[glass gust]

yeah i never learned this

this there any other way to solve this?

[stray socket]

Expand it and collect like terms

You'll end up doing the mathematically same thing

[stray socket]

Are you sure this isn't familiar?

[glass gust]

no ive never seen it

iirc he said he werent allowed to use derivatives or l'hopitals since we hadnt learned them

[stray socket]

They're not L'Hopitals or derivative tricks

It's just algebraic properties of limits

[glass gust]

this is the closest one i see

[stray socket]

Oh I mean if the limit for f(x) is L and limit for g(x) is K

Then yeah

Same thing

[glass gust]

so its the same idea as what you showed above?

[stray socket]

Yeah

[glass gust]

going back to this you would be left with delx^2-2delx+2, right?

[stray socket]

There shouldn't be any deltas

[glass gust]

1 sec

when expanding (x+delx)^2, would x*delx just be delx^2?

[stray socket]

No

Δx is a different variable on its own

[glass gust]

so is it xdelx

[stray socket]

Yes

[glass gust]

so i end up with x^2+delx^2+2xdelx-2x-2delx+1-x^2+2x+1

combining like terms i get delx^2+2xdelx-2delx+2

all i end up with is deltas

[stray socket]

Split the fraction up into different terms

$\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$

[flat frigate]

Umbraleviathan

[stray socket]

So after combining terms, just break it all up

[glass gust]

i feel like that doesnt change the fact that im left with is deltas which you said shouldnt happen

[stray socket]

Well delta x should cancel out

The numerator should only have terms with a delta X in it

[glass gust]

so its ok to deltas in numerator after expanding and combining like terms?

[stray socket]

Well yeah in fact, if you did it correctly, all terms in the numerator should have delta x as a factor

Show me what you did for the numerator

[glass gust]

i dont know how to use the commands

[stray socket]

The 1's should cancel out

1 - 1 = 0

[glass gust]

oh ur right

yeah so its all deltas

[stray socket]

Mmhm

[glass gust]

i forgot to distribute the negative to the 1

[glass gust]

wait what about the 2xdelxx

[stray socket]

That stays

[glass gust]

alri

oh and the limit is as delx approaches 0

[stray socket]

Yes

[glass gust]

so im cancelling out the delx in the denominator using the delx^2 in num, which puts me at delx+2xdelx-2delx = 2xdelx-delx

[stray socket]

That doesn't sound right

[glass gust]

yeah i thought so

[stray socket]

Show me the work again

[glass gust]

i was gonna try cancelling out the denominator

is that a bad idea, im just tryna avoid /0

[stray socket]

Cancel out delta x

[glass gust]

using the -2delx or the delx^2

[stray socket]

Division

Every single term in the numerator has delta x as a factor

[glass gust]

oh

wait

[stray socket]

You legit just cancel out common factors