help-28

[trail barn]

oh wait

I accidentally generalised

derp

[tired oak]

lol

[trail barn]

Anyway, each element of D8 is of the form r^n or sr^n.

[tired oak]

oh shoot oops

[trail barn]

now it's enough to check whether $sr^n=r^ns$ and if $rr^n = r^nr$, i.e. if any of the $r^n$ commute with the everything

[tired oak]

why?

[glossy valve]

Edward II

[tired oak]

oh yes sry

[trail barn]

and if $ssr^n=sr^ns$ and $rsr^n=sr^nr$

[glossy valve]

Edward II

[trail barn]

because every element is a product of $s$ and $r$, so commuting with the generators is enough to commute with every element

[glossy valve]

Edward II

[tired oak]

@trail barn so i check

uh

what exactly

[trail barn]

Check each element $x\in D_8$ if $sx=xs$ and if $rx=xr$. If both hold, then $x\in Z(D_8)$. Split into cases where $x=sr^n$ and $x=r^n$

[glossy valve]

Edward II

[trail barn]

you can also do $r^ns$ instead of $sr^n$ is that's the way you're used to writing the default elements, there's no difference

[glossy valve]

Edward II

[tired oak]

surely i dont have to check every single element

[trail barn]

well you can do all the possible n at once

[tired oak]

can u do it man

lol

[trail barn]

no

[tired oak]

: (

pls

[trail barn]

I'll do the $sr^n$ case but not the other

[glossy valve]

Edward II

[tired oak]

the other case being r^n?

[trail barn]

yes

$ssr^n=r^n$ but $sr^ns = r^{-n}ss = r^{-n}$, these are not always equal (when are they?)\
$rsr^n=sr^{-1}r^n=sr^{n-1}$ but $sr^nr=sr^{n+1}$ which are never equal.\\
So for any $n$, $sr^n$ does not commute with all elements and $sr^n\notin Z(D_8)$.

[glossy valve]

Edward II

[trail barn]

this really isn't that long as you can see

[tired oak]

is $r^k$ ever equal to $s^{-1}r^ks$? @trail barn

[glossy valve]

jsidind810

[tired oak]

only for k=0?

[trail barn]

there is another

[tired oak]

8

[trail barn]

ig but r^8=r^0

there is k between 1 and 7 that this holds for

[tired oak]

how

how can i see that clearly

[trail barn]

Note $s^{-1}r^ks=ssr^{-k}$. So we want $r^k=r^{-k}$, which is the same as $r^{2k}=e$ where I'm using $e$ for the identity

[glossy valve]

Edward II

[trail barn]

using ss=e to cancel out

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[sharp fable]

!status

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

[sharp fable]

1

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@sharp fable Has your question been resolved?

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[atomic vortex]

how does one simplify the following expression

1 + (n - 3) + (n - 4)*(ceil(log_3(n) + 1) + ceil(log_3(n)) + ceil(n/2 + 1) + ceil(n/2) + 3)

[onyx glen]

$1+n-3+(n-4)(\ceil{\log_3(n)+1} + \ceil{\log_3(n)} + \ceil{n/2+1}+\ceil{n/2} + 3)$

[glossy valve]

AnnGhost

[onyx glen]

and for what purpose do you need this simplified?

[atomic vortex]

i want to find the determinant

wait

im reyatded

i meant

dominant term

mb

[onyx glen]

ok first off xyproblem

second don't use the R slur

[atomic vortex]

ai mb

[atomic vortex]

what does that mean

[onyx glen]

http://xyproblem.info/

and third, you have some linear terms, some n log(n) terms, and some n^2 terms

(n-4) ceil(n/2) is Θ(n^2)

same goes for (n-4) ceil(n/2+1)

[atomic vortex]

so the dominant term would simply be n^2 right

[onyx glen]

yes

!original tho.

[full forum]

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

[atomic vortex]

oh alr sure one sec

[onyx glen]

weh ok

[atomic vortex]

ty doe

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[rough ruin]

howdy can anyone explain the wording of this question to me? (Vector projection)

heres a clearer pic

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[signal tangle]

how to do?

[minor crater]

notice that there's a common term in both the bracket and under the square root

[tiny flame]

maybe rewrite it with the exponents for each term

probably will help you see what you need to do more clearly

[minor crater]

hint: $\sqrt{x}\sqrt{x} = x$

[glossy valve]

artemetra

[signal tangle]

thx for the help 🙂

.close

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[signal tangle]

ow to do 3e

[royal hare]

I think you are expected to write it first into common denominator

And then simplify the numerator

[tiny flame]

put the terms into exponenets again

hint: $\frac{1}{x} = (x)^{-1}$

[glossy valve]

chef

[tiny flame]

try replacing x with $\sqrt{x-1}$

[glossy valve]

chef

[tiny flame]

and putting all the terms into exponents

[royal hare]

And what's next?

[tiny flame]

my bad i thought that was multiplying not adding

so yeah, make the second term have the same denominator as the first, then simplify

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[kind wind]

I am so bad at functions. I didn't understand a single thing said in our video lessons. May someone explain to me how to solve this? Like where do I start?

[scarlet oracle]

x(v(t)) ?

basically just let x=88t - 2t^3

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[ruby vale]

need help with d)

[torn jolt]

Alright so let’s start with a simple geometry question: what defines a rectangle ?

[ruby vale]

:/

[torn jolt]

That is important because you need to interpret those results into vectors after

From the properties of a rectangle you deduce the coordinates

[ruby vale]

I solved harder vector questions than these but when it comes to geometry my brain just disfunctions

[ruby vale]

can u just give the answer with an explanation and let my brain interpret it?

[torn jolt]

Alright can you draw a rectangle then ?

[ruby vale]

yes?

[torn jolt]

No because then you’ll never have the right creativity/steps in mind when doing it

[torn jolt]

Right draw an ABCD rectangle

And accurately (so with the equal sides and angles on it)

[ruby vale]

tf no thats a waste of time

[torn jolt]

It ain’t because if you see that AB=DC and BC=AD you can have an idea on how to solve your damn thing

[ruby vale]

anyone else wanna help?

[torn jolt]

LMAO

WHAT AN ||ass||

youre quite ungreatful

[ruby vale]

you cant even spell ungrateful

[torn jolt]

Because my keyboard isn’t in English so yeah autocorrect does f me over

[ruby vale]

what ur telling me to do is done in like 3rd grade here

and that is NOT needed

if you dont wanna help, leave

[torn jolt]

Yeah and you can’t solve a 9th grade question because you’re not using the basics from 3rd grade

I do want to help

Im just giving you the steps

[ruby vale]

actually im in 8th grade

[ruby vale]

I dont need those steps

[torn jolt]

Now if my method doesn’t correspond to you just say it nicely and I’ll leave

[ruby vale]

Give me the direct answer with an explanation

[torn jolt]

No need to be rude and say “anyone else wanna help”

[queen crater]

Let's not overcomplicate things
If BC = (4,-8) then what's AD?

[ruby vale]

(4, -8)?

answer is 0, -9 btw

[torn jolt]

Thats not how the server help channels are suppose to work tho

[queen crater]

Yes

[ruby vale]

yes anddd?

[queen crater]

I'm not sure what is blocking you, do you not know that you can convert points into vectors and back or do you not know that you can add vectors?

[ruby vale]

i do know, i just cant figure out the marking scheme method since i forgot lol

[torn jolt]

Well since you e got coordinates from A and B and the length of AB and comsequently the opposite side you therefore can have the coordinates of the needed points

Just imagine the points as translation of the original points along the vectors

[ruby vale]

yes i am imagining them as translations, but that doesnt really help much

[torn jolt]

Alright so

[ruby vale]

i feel like im overlooking this

[queen crater]

You know AD = (4,-8) and you know A = (-4,-1)

[ruby vale]

actually ABvec = 6, 3

[queen crater]

Sorry misread

[ruby vale]

np

[ruby vale]

yes

OHHH I GET IT NOW

[torn jolt]

Nice

[ruby vale]

OD = OA + AD

damn i figured it out on my own, but thanks for the help anyway

[torn jolt]

Yes that’s chasle

LMAO the arrogance you hold is insane my guy but sure anyways you got it that’s the point

[ruby vale]

Im sorry I was getting impatient

[torn jolt]

Its alright just close the channel when you’re done ! ^^

[ruby vale]

yeah idk how to do that

[torn jolt]

.close

[ruby vale]

.close

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[ruby vale]

okay

[torn jolt]

Have a good one

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[warm vault]

Anyone have a mnemonic for csc = 1/sin and sec = 1/cos?

[warm vault]

I always get confused cause csc looks more similar to cos and sin looks more similar to sec

[torn jolt]

All I can think of is that you can change "sin" to "sec" and drop/add "co"

cosine => cosecant
cosine => cosecant

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[gaunt hull]

Can anyone help me? I can't figure out how to do efgh

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@gaunt hull Has your question been resolved?

[gaunt hull]

<@&286206848099549185>

[mossy loom]

a: 11^(1/2) c: 2*3^(1/2)

b, d are the same thing as a, c but the exponents are negative

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[clever cargo]

!nosols

[full forum]

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.

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[gleaming steeple]

Is it possible to apply a linear map T to an entire subspace U? Is the output a subspace?

[limpid moat]

yes because is linear

[next sail]

yeah why not

[limpid moat]

but attention to this, if T:V--> V then T|U:U-->V not T|U:U-->U

[next sail]

applying T to a subspace is like, just apply T to all the points in the subspace and see what you get

[gleaming steeple]

what's the bar?

restricted to U?

[limpid moat]

yes

its a commun notation

[gleaming steeple]

i see

[gleaming steeple]

in this case I'm dealing with invariant subspace U, so writing the second should be okay right?

[limpid moat]

yes

[gleaming steeple]

thanks

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[pseudo cape]

Let n in IN with n >= 3.
Show that there exists atleast one prime p with n < p < n!

[pseudo cape]

Hint: look at the prime decomposition of n! - 1

How would we use that hint?

[onyx glen]

n! - 1 is divisible by some prime p, as all natural numbers are.

can this p be ≤n?

[pseudo cape]

No, because n! = 1 * 2 * ... * n and since we subtract 1 from that, we "destroy" all those factors

It has to be > n

[onyx glen]

C tier

better to say that n! is divisible by all numbers from 2 to n and n! - 1 is thus not

(because it is congruent to -1 modulo them all)

[pseudo cape]

Ah, so we should be done

[pseudo cape]

How would you reason that p < n!?

"p has to be less than or equal to n! - 1, otherwise, the second factor in the prime decomposition of it would need to be less than 1, but the smallest prime is 2."?

[onyx glen]

any factor of a number is leq that number

so p ≤ n! - 1

[pseudo cape]

shouldn't we specify this to "prime factor", or is that already implied by that?

[onyx glen]

do you claim it's false without said specification

[pseudo cape]

Well, 5 = 10 * 0.5.
Or we need to specify that the factors are in IN

[onyx glen]

N

please don't write it as IN it is painful to look at

also of course we are working in N

[pseudo cape]

Alright, lol

[onyx glen]

we always have been

unless you want me in N to remind you in N that we are in N always working in N and all in N our in N numbers in N are in N

[pseudo cape]

Thank you

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[strong pawn]

Prove by contradiction that, if A ∩ B ⊆ C and x ∈ B, then x isn't in A \ C.

[random canopy]

well by contradiction if $$P\Rightarrow Q$$ then $$P\Rightarrow\neg Q$$

[glossy valve]

42

[random canopy]

so by contradiction $$A\cap B\subseteq C$$ and $$x\in B$$ then $$x\in A-C$$

[glossy valve]

42

[random canopy]

ill give you another hint

if $$x\in A-C$$ then $$x\in A\land x\notin C$$

[glossy valve]

42

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@strong pawn Has your question been resolved?

[strong pawn]

Here's my proof. What do you think?

We can say x is in A \ C instead. So that means x is in A but not in C. But if we add the other requirement, x should be in B too. This means x is in A and B. But A and B have to be subsets of C (they are both inside C, making x inside C too) and we know that x is not supposed to be in C.

[random canopy]

yeah

thats good

you have ur contradiction

seems fine

i am convinced 👍

[strong pawn]

.close

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[frigid phoenix]

hey, asking here again

[frigid phoenix]

if i use l'hopital, and diff top and bottom, i get
sin(x^2)/0

but it's wrong. how am i supposed to use lhopital here?

[sharp vine]

d/dh of the top should be

sin((h+2)^2)

and of the bottom 1

[frigid phoenix]

right dh !

and when i differentiate a def integral, why do i take the upper limit only?

[sharp vine]

not only

just sin(2^2) is constant

derivative of constant = 0

this is why I've missed it

[limpid moat]

foundamental theorem of calculus

sin(4)

[frigid phoenix]

so when i have a def integral and i take the anti derivative i should do
F(b) -F(a) right?

[limpid moat]

in this case yes, since h is only in the domain of integration...if h is inside the integrand function then F(b) notation couldn't be useful

[frigid phoenix]

the "method" should be to rewrite integrate what's inside the integral substituting the bounds as per FTC

take the antiderivative in this case, for each of the 2 expressions

correct?

so i would have
integral (sin((2+h)^2) - integral(sin(2^2))

and i woudl take the derivative of each...?

no doesnt work

need to sleep

thanks good night

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[worthy minnow]

i just need help with the whole question step by step, my test is on monday

[glacial pasture]

what have you done thus far

[worthy minnow]

i tried completing the square

[glacial pasture]

on what?

have you done the 'show that' yet?

[worthy minnow]

no

ill do it now

with what ive got

not closed yet though

[glacial pasture]

thats very wrong

[worthy minnow]

I realised

what should I do

Im here btw (silent typing plugin)

[glacial pasture]

completing the square isnt really the way id go about it but if thats what you want to do.\newline
The second line should be:
$$2\left(x^2-\frac{5}{2}x\right)-18=0$$
$$2\left[\left(x-\frac{5}{4}\right)^2-\frac{25}{16}\right]-18=0$$
$$2\left(x-\frac{5}{4}\right)^2-\frac{25}{8}-18=0$$
$$2\left(x-\frac{5}{4}\right)^2-\frac{169}{8}=0$$

[glossy valve]

AℤØ

[worthy minnow]

what would be the first line

ill try that right now

[glacial pasture]

the first line is just the equation 2x^2-5x-18=0

[worthy minnow]

ohh

alright

ill lyk if it works

[glacial pasture]

id just do this by factorisation though

[worthy minnow]

with that i got

x-2 = 0

2x+9 = 0

when i tried factorisng

ill try yours now

[glacial pasture]

think you have those signs wrong

2x^2+4x-9x-18=0
2x(x+2)-9(x+2)=0
(2x-9)(x+2)=0

2x=9 or x=-2

x cant be negative

[worthy minnow]

I figured

[glacial pasture]

or less than 4

[worthy minnow]

Ill try that right now

it says its incorrect

[glacial pasture]

,w solve (2x^2-5x-18=0)

[glossy valve]

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[glacial pasture]

perhaps its because you never did the show that

[worthy minnow]

should i do both?

[glacial pasture]

mathswatch wont let you off so easily

wdym both?

you need to actually do the derivation of 2x^2-5x-18=0

[worthy minnow]

sorry if im taking the piss

wdym by derivation

[glacial pasture]

derivation just means show how they got to that equation

which they just did by getting the area of that shape

[worthy minnow]

alr

i just got boosted from 86 - 94

thanks

.close

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[plush egret]

do you understand why this is a system of 2 equations for 2 unknowns?

or, alternatively, what kind of function is the derivative of a quadratic?

can you give a generic quadratic function?

[rapid rain]

3 unknowns if you include the constant

[plush egret]

lets say that like

$f(x) = ax^2 + bx + c$

is more what i meant

[glossy valve]

jan Niku

[plush egret]

okay

whats f'?

sure, but what is it equal to?

just take the derivative of f here

use the power rule

its just like this

$\dv{f}{x} = \dv x \qty( ax^2 + bx + c)$

[glossy valve]

jan Niku

[plush egret]

so $\dv{f}{x} = \dv x (ax^2) + \dv x (bx) + \dv x (c)$

[glossy valve]

jan Niku

[plush egret]

okay

so what is f'?

given f(x) = ax^2 + bx + c

what about a and b?

sure

so $\dv x (ax^2) = a \dv x (x^2) \dots$

[glossy valve]

jan Niku

[plush egret]

so isnt $f'(x) = 2ax + b$?

[glossy valve]

jan Niku

[plush egret]

okay

this is the derivative of a quadratic function

they tell you its equal to 3 when x = -1

can you write an equation using that information?

yes

or $-2a + b = 3$

[glossy valve]

jan Niku

[plush egret]

alright, the other piece of information was that f' is 6 when x=3

whats that equation?

[grizzled yarrow]

its

2a(3)+b = 6

[plush egret]

\begin{align*}
-2a+b &= 3 \
6a+b &= 6
\end{align*}

[glossy valve]

jan Niku

[plush egret]

can you solve a system of equations?

[grizzled yarrow]

I used to but its been a long time

I cant remember

[plush egret]

its good to practice

[grizzled yarrow]

Do i addition them?

[plush egret]

nothing will cancel out as theyre written here

[grizzled yarrow]

Yeah

[plush egret]

you could maybe subtract them

then the b's will cancel

[grizzled yarrow]

So it becomes

4a

[plush egret]

no

[grizzled yarrow]

No

6a

[plush egret]

no

[grizzled yarrow]

8 a

I mean

8a

[plush egret]

okay, sure

[grizzled yarrow]

Because it becomes positive

[plush egret]

you did 6a-(-2a)

[grizzled yarrow]

Yes

[plush egret]

and b - b

[grizzled yarrow]

8a

[plush egret]

so that leaves what on the other side of the equals?

[grizzled yarrow]

8a = 3

[plush egret]

seems good

now what?

[grizzled yarrow]

So a= 3/8

[plush egret]

seems right

now how do you get b?

[grizzled yarrow]

Now

We use the point (-1,3)

Plug back in the y’

To solve for b

[plush egret]

sure

you mean use $-2a+b=3$

[grizzled yarrow]

Ohhh

[glossy valve]

jan Niku

[grizzled yarrow]

Yes

[plush egret]

okay

[grizzled yarrow]

And use the point

And then ill get the answer

[plush egret]

alright, so we gotta solve $-2 \frac 38 + b = 3$

[glossy valve]

jan Niku

[grizzled yarrow]

Ok let me do this

3.75

[plush egret]

yup

[grizzled yarrow]

And that is 30/8

[plush egret]

okay so $a=\frac38$ and $b=\frac{15}{4}$

[glossy valve]

jan Niku

[grizzled yarrow]

Oh yeah

And for the b in the quadratic equation

[plush egret]

Okay, we solved the system.

[grizzled yarrow]

Do i pit 15/4x so when it becomes derivative

its 15/4 only

[plush egret]

all we gotta do is go back to this

$f(x) = ax^2+bx+c$

[glossy valve]

jan Niku

[grizzled yarrow]

Oh yeah

Woah you’re a very good teacher

[plush egret]

we solved a system, so we can just write $$f(x) = \frac38 x^2 + \frac{15}{4} x + c$$

[glossy valve]

jan Niku

[grizzled yarrow]

Yeah thats the answer

Ur a genius

[plush egret]

https://www.desmos.com/calculator/nsk9yxzftq

because i made it lol

if you care

i was anticipating confusion about c

[grizzled yarrow]

Thank you so much

What would happen though

If the b’s dont cancel out

[plush egret]

you can run into all the usual problems you get with systems of equations

like, you can have not enough information

so all you can do is express a in terms of b

or you could be given contradictory information, so there isnt any solution at all

[grizzled yarrow]

I see

Yeah

Thanks so much for your help

[plush egret]

no problem

[grizzled yarrow]

Have a good evening

[plush egret]

you too

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[woeful sparrow]

The answer is (A) and (D)

But I dunno how

I can't get it

Right

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<@&286206848099549185>

[woeful sparrow]

...

<@&286206848099549185>

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@woeful sparrow Has your question been resolved?

[woeful sparrow]

<@&286206848099549185>

.close

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[plush egret]

well

lets answer the i and ii

and see what the problem is

maybe we can make an adjustment

whats your thought for i and ii

sure

okay

it looks like

it looks like the problem is that iii is 5 too

so maybe we can make an adjustment

my first thought would be

you could have h(x) = x-5

or you could have it be like

h(x) = (x-5)(x-1)(x-2)(x-3)....

you could add as many factors as you want, right?

that first one is still gonna make it be 0

what if

okay heres an issue

if we have one factor of x-5 in both g and h

then there will be a factor of x-5 left in h^2

that will make iii be 0

what if g(x) = (x-5)^2

and say

h(x) = (x-5)(x-2)

this isnt the answer but its my thought

that makes it $\lim _{x \to 5} (x-2)^2$

[glossy valve]

jan Niku

[plush egret]

so, you could make some adjustments here

this would be where id start, probably

see if you can generalize this factor

idk maybe ax+b?

its a nice problem solving exercise

so call $h(x) = (x-5)(ax+b)$ knowing the ultimate problem is to decide a and b such that $\lim _{x \to 5} (ax+b)^2 = 2$

[glossy valve]

jan Niku

[plush egret]

my thought is you dont need a

[plush egret]

think this is a root solving problem

so algebra

can you use latex

the math bot

do you mean $\frac{ (x-5)^2 }{ \frac{1}{10} (x-5)^2 x}$?

[glossy valve]

jan Niku

[plush egret]

new preamble broke my xfrac

i guess, yea

its less complicated than what i was suggesting though

its not guessing, its problem solving

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@grizzled yarrow Has your question been resolved?

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[tawdry grove]

sort of a general question but I do have examples, uhhh.

when calculating derivatives when do you if you are meant to do product rule or chain rule. also, when can you do the exponent rule where the exponents number jus drops infront of the function/terms

[plush egret]

product rule comes when you have product of functions

chain rule is for composition

so you just have to be able to identify those

the exponent rule are you talking about the power rule?

[rough tundra]

there's two $\frac{d}{dx}[n^x]=n^x\ln(n)$ and $\frac{d}{dx}[x^n]=nx^{n-1}$

[glossy valve]

PajamaMamaLlama

[rough tundra]

for exponents identify whether the variable of differentiation is the power or base :)

[tawdry grove]

for example, what rule is this.

[plush egret]

can you identify any function composition here?

lol ur good pajama

i stepped on ur toes

[rough tundra]

😅

nah kinda gave it there, go ahead, all yours chief!

[plush egret]

[plush egret]

you gotta look if there are functions inside of other functions

if you can see that, then you need the chain rule

[rough tundra]

note: $\cos^{5}(4x)=\cos(4x)^{5}$

[glossy valve]

PajamaMamaLlama

[tawdry grove]

wellll

[plush egret]

?

[tawdry grove]

i saw this in multiple ways, I saw it as chain rule because therre is a inner function and a outter function.

i also saw this as product rule because the 4x is being multiplied by the COS^5

i also say this as power rule, maybe. Can the power of 5 just drop infront of the cos and thats it, (without putting a new exponet of 4)

[plush egret]

what you're calling the exponent rule is the power rule i think

and you do use it here, but you've described it incorrect

also the 4x isnt being multiplied by the cos^5, its like, an input to the function

cos^5 doesnt mean anything on its own

[rough tundra]

start with the composition

I think your getting overwhelmed with how much is happening here, back to basics, what is being composed? :)

[rough tundra]

[tawdry grove]

but I don't like that

here let me ask the question a differnt way

true or false

[rough tundra]

that is false

[tawdry grove]

sooooooooooooooo, when can you use power rule

[rough tundra]

when you have x^n

only when x is linear and n is constant wrt to x

[tawdry grove]

like this?

[rough tundra]

that's not true either

[dense cape]

the original power is reduced by 1

[rough tundra]

so 3x^2

[tawdry grove]

yes i agree

[rough tundra]

would be your derivative

[tawdry grove]

n-1

[dense cape]

yeah

[tawdry grove]

but

[tawdry grove]

then what does this mean?

[dense cape]

the original is x^n

n x ^ n-1 is the formula for the derivative of x^n with respect to x

[tawdry grove]

well yes but what does this mean?

[rough tundra]

$\frac{\dd}{\dd x}[x^3]=3x^{3-1}=3x^2\neq3x$

[glossy valve]

PajamaMamaLlama

[rough tundra]

x^2, x^5,x^(69) are all of the form x^n

[tawdry grove]

soo when do we use power rule

becasue x^5 is 5x^4

not

5x

in what case do you just drop it infront

[rough tundra]

well dropping the exponent in front is part the power rule

it's not the power rule itself

the second part is subtracting one from the exponent

[tawdry grove]

in waht case do we not do this

or do we always do it

[rough tundra]

always do it

it's a rule

[tawdry grove]

okay, allow me to ask question in differnt way again

[rough tundra]

also false

[dense cape]

thats isnt differentiation

[rough tundra]

oh yeah

[dense cape]

that is a property of the log function

[rough tundra]

$\log_{2}(\sqrt[3]{x})=\underbrace{\log_{2}(x^{\frac{1}{3}})=\tfrac{1}{3}\log_{2}(x)}_{\text{By }\log(a^b)=b\log(a)}$

[tawdry grove]

so in this case, you can do this

[glossy valve]

PajamaMamaLlama

[rough tundra]

because it's a property of the log function

has nothing to do with derivatives :)

[tawdry grove]

i seeeeee, any other cases like that I should know of?

[rough tundra]

not really with derivatives, there aren't a lot of things that go wonky with them

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

<@&286206848099549185>

[sonic orchid]

post your question first

[pine frigate]

hold up il send some pictures in a bit

[hot herald]

do NOT ping helpers before 15m, especially if you haven't even asked a question yet

[pine frigate]

ur life wont end if someone pings u wont it

[reef dust]

ur comparing two completely diff things there 🥱

[hot herald]

it wont, but its best that you try and follow the guidelines and not double down

[pine frigate]

ok'

[hot herald]

and not piss people off

[sonic orchid]

yours will for sure change for better if you act more nicer to people

[pine frigate]

ur just making assumptions

[sonic orchid]

buddy you acted rude for no reason, calm down & post your question if you have any

[pine frigate]

i wasnt rude i was just honest

[hot herald]

tripling down...

[sonic orchid]

lol

[pine frigate]

anyway

How do I turn this into a root

[torn jolt]

The top number is the exponent of the number in the root, and the bottom number is the exponent of the root

x^1/2 is the same as sqrtx

[sonic orchid]

there is exponential rule
this

[pine frigate]

so what would it look like in teh end

5Vx v2y?

[torn jolt]

Turn them both in the sqrt flrm then multiply

[pine frigate]

idk what a sqrt flrm is man im horrible in math can u just give me the result?

[torn jolt]

Also is 1/2 negative? The quality of the picture is really bad

[pine frigate]

its just 1 under 2 no negative

[torn jolt]

Just turn the exponents into a square root like in the picture above

It was a typo sorry

[torn jolt]

Use this

[pine frigate]

So V1X V2Y?

[torn jolt]

Not really

[torn jolt]

Let's say over here m=1 and n=5

So x^1/5

Convert it like in the picture

What do you get?

[pine frigate]

i dont understand shit sir

[torn jolt]

Do you understand what's in the picture?

[pine frigate]

no i just need the end result

[sonic orchid]

just try to follow along with the steps @pine frigate

[pine frigate]

english not my first language so its hard

i just need end result for homework thats it

[torn jolt]

You need to get to the end result yourself, i can't just tell you the anwser

[sonic orchid]

then you are in wrong server, we dont give the direct answers, people here will guide you to get there not actually do it for you

[pine frigate]

why would we waste 40 mins of time when u can give me the answer in 2 mins

[sonic orchid]

whats your mother tongue (also its not my first lang either)

[sonic orchid]

so you wont have to come here next time u have same type of problem with changed numbers

and you will save your time if you actually learn it

[pine frigate]

i wont have same type of problem for sure

[pine frigate]

it wont

[torn jolt]

No you need to understand it so that you can solve it yourself, i am helping you understand not solve for you

[pine frigate]

i just need teh answer

pls

not any understandings

or guids

[torn jolt]

Well we'll get the anwser once you solve it

[sonic orchid]

and we are trying to help you to get there

[pine frigate]

help me by just sending me it

[torn jolt]

What other languages do you speak? I might know those as well i know a few

[pine frigate]

by now we would ahve been done

[torn jolt]

Go to another server then

[pine frigate]

i will

[torn jolt]

I'm just trying to help you

[sonic orchid]

yea true, if you would have just co-operated with purpar

you would have the answer figured out by now

[pine frigate]

yea bro cooperate i dont know shit about math and u want me to solve some shit which i dont even know about

[sonic orchid]

again we are trying to help you so you know your maths

[pine frigate]

meanwhile only reason i joined is to get help

[sonic orchid]

there is reason why you dont know cuz you arent trying

[pine frigate]

and only help ur giving me is no help

[torn jolt]

That is exactly why we're helping you understand

[pine frigate]

BRO I DONT WANA UNDERSTAND

[sonic orchid]

doubling down on that, can you tell us what other lang you are comfortable with

[pine frigate]

JUST GIVE ME THE GOD DAMN ANSWER

IM NOT GOING FOR MATH DEGREE

ITS JUST STUPID ANSWER I NEED

[sonic orchid]

well then good luck with getting any help here

[pine frigate]

goodnight im going to suicide.

[sonic orchid]

cuz you wont get any direct answer, without you understanding anything

[torn jolt]

If you don't know math doesn't mean you can't do it. Charmy showed you a picture on how to do it, i tried to explain to you yet you aren't even trying

[ocean cedar]

I what school year are you?

[pine frigate]

high schol

2nd year

[ocean cedar]

Alright, you will still have a 3rd year to go through right?

[pine frigate]

yea but no math there

thank God

[ocean cedar]

I see, which is why you are losing patience

Well now that I think about it

There is a whole year with Maths in it

I promise we don't try to make you lose your time at all. There is a purpose behind why we are approaching the problem with you this way

If we give you the answer right away, a problem of lack of autonomy will appear, and this is a big problem because in a situation where you are in a test or an exam, you will be on your own

I know it's annoying, but Maths don't have much of a way around

[pine frigate]

i dont think u understand

[torn jolt]

@pine frigate could you please at least try to solve the problem? It's really easy. I can explain again if you didn't understand it well

[pine frigate]

i just need the answer show it to the teacher and i get a passing grade

[pine frigate]

ok explain

[torn jolt]

Do you understand this?

[torn jolt]

This is the exact same thing i said as an image

[pine frigate]

top number 1 bottom number 2 all i understood.

[torn jolt]

Ok

[ocean cedar]

What I am trying to explain is that what you are trying to get is an easy way through it, but by taking the easy way, you are being dependant on people. And... I'm afraid nobody wants to be at the service of someone for a long time, mostly when it's done voluntarily by people, when it's not paid

[torn jolt]

In the picture the top number(m) and the bottom number(n). Do you see where they are?

[pine frigate]

m top number n bottom number

[pine frigate]

i see

[torn jolt]

The top number (m) is right of the number and the bottom number (n) is at the left

m is near the number and n is near the root

[pine frigate]

so 1 and 2?

[torn jolt]

In the picture you've sent if we look at x, 1 is the top number and 5 is the bottom number. And in y 1 is the top number and 2 is the bottom number

[torn jolt]

Yes!

[pine frigate]

alright

[torn jolt]

So can you replace m and n with the numbers of the pictire you've sent?

Do once with x (1 and 5) and once with y (1 and 2)

[pine frigate]

x1/2

y1/5

correct?

[torn jolt]

@pine frigate

Here is an example on how to do it

I used different numbers

[pine frigate]

its solved here but different numbers right?

[torn jolt]

Yes

These are your numbers that you've sent

This is not solved

[torn jolt]

Can you solve it yourself by looking the this example?

[sonic orchid]

yea what purpar showed you is no different from the rule they were explaining. You just need to plug in the values in place of the variables depending on your question, the rule stays same regardless

[pine frigate]

vx=5
vy2

[sonic orchid]

can you write down on a paper & send the pic, cuz its really hard to guess what you mean by the way you are typing, in your answers

[torn jolt]

Can you send a picture of what you mean?

Just solve it on a paper and send a picture

[pine frigate]

https://prnt.sc/wLrLvUOdTv7s

[sonic orchid]

no

here let me write it down for you again

[pine frigate]

https://prnt.sc/YFI0CSwQS3Ag

[torn jolt]

Pay close attention at what i did here, i can do another one if you want

[sonic orchid]

here, what will go in place of these question marks ?

[sonic orchid]

You can refer back to this image if you need clue

[torn jolt]

@pine frigate here is another example with different numbers

[pine frigate]

https://prnt.sc/lcuSQgcmaHfP

[torn jolt]

YES

[pine frigate]

so thats the result

thank fuck

[torn jolt]

That is the first one

Now do with the second number

[pine frigate]

what 2nd one

[sonic orchid]

there is y term in your question, which is raised to power 1/2

[sonic orchid]

repeat the same thing as you did here for y^1/2

[pine frigate]

so now i do the 1/2

[sonic orchid]

yup

[torn jolt]

With y

[pine frigate]

https://prnt.sc/pIEHxxq3GglQ

[sonic orchid]

LESGOOO

Yup you got it right

[pine frigate]

whats next now

[torn jolt]

Do you have to multiply those or is that it?

Is there anything else in the question?

[pine frigate]

well the quyestion was to turn it as a root

[torn jolt]

Then you're done!

That's it

[pine frigate]

so end result will be

[torn jolt]

This

[torn jolt]

And this

[pine frigate]

alr leme get the paper rq can u not delete this chat yet

[torn jolt]

From now on if you have any questions you know that you can solve to find the anwsers

[sonic orchid]

Its not deleted, you can always find back your chats by search function on top corner (though idk where it is on other devices, I am on pc)

[pine frigate]

alr

[sonic orchid]

just make sure to close the channel

once u done with writing it down, so others can use it

[pine frigate]

k

[torn jolt]

@pine frigate type .close if you don't have anymore questions

[full forum]

@pine frigate Has your question been resolved?

[pine frigate]

i got new question now

[sonic orchid]

alright, ask then

[sonic orchid]

and react to the red emote there if you still have question

[pine frigate]

"Root it"

"tap the roots"

yk i got the answer from a friend but i need to uh know how to do it my self

bcz teacher will ask different numbers

then its over for me

[pine frigate]

u here

https://prnt.sc/KO_YrjlxU396

tap the roots

https://prnt.sc/7ZUDlcv0GjnF

this is the answer

but how do i get it?

[full forum]

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[pine frigate]

.reoopen

.reopen

[full forum]

✅

[pine frigate]

<@&286206848099549185>

[tough marsh]

Yup, not gonna answer a rude student.

[viscid pumice]

whats the problem

[tough marsh]

OP starts the question with a helper ping, and does not apologize when being reprimanded

[pine frigate]

tap the roots

[pine frigate]

ive been waiting for mroe then 15 minutes

no need for me to apologise

matter of fact it should be u apologising

because u are late and did not arrive in time

[viscid pumice]

i dont understand the question

[pine frigate]

i sent u the answer

idk how to get it though

[viscid pumice]

how tf am i supposed to read that

[pine frigate]

https://prnt.sc/SrZ9Nqcfcs2v

[viscid pumice]

what is b)²

[pine frigate]

?

[viscid pumice]

thats what you wrote

[pine frigate]

u should know

[viscid pumice]

²√a = b)²

[pine frigate]

https://prnt.sc/7ZUDlcv0GjnF

this is the answer

but how do i get it

what do u do

[viscid pumice]

idk

what even is b)²

a graph for b²?

[pine frigate]

theres just b2

nothing else

[viscid pumice]

ik but in the first one

[pine frigate]

wym

[viscid pumice]

lemme say it clearly

what the fuck is b)²

[pine frigate]

its just b2

[tacit siren]

damn this is a long channel

[viscid pumice]

here

what does b)² mean there

[pine frigate]

i dont know

[tacit siren]

Kek it's free help u don't get to demand people be on time

[viscid pumice]

then why did you write it

[pine frigate]

because its there

[viscid pumice]

its where

[pine frigate]

on the paper

[viscid pumice]

show me a picture of the paper

[pine frigate]

its teh same as this

[sonic orchid]

bro chill, you were rude to others not once but multiple times now. You really shouldnt be asking others to apologise to you while they are helping you here for free

[viscid pumice]

just send a pic

[pine frigate]

i cant

[viscid pumice]

why

[pine frigate]

pc

[viscid pumice]

get your phone then

[pine frigate]

dead

[sonic orchid]

how did u send pics before ?

[viscid pumice]

also theres a camera on laptops

look for camera app

[pine frigate]

https://prnt.sc/GnI9el0Rr1jd

[pine frigate]

when the phone wasnt dead?

[viscid pumice]

fucking hell that looks nothing like what you sent before

[pine frigate]

lol

but thats solved the issue is i dont know how to solve it

[viscid pumice]

also the equasion (√a² = b)² = 3√(a4)b² isnt mathematically correct

it implies that √a² = b)²

and b)² doesnt exist

[pine frigate]

so how would it be

Lol

[viscid pumice]

the little = in (√a² = b)² is simply wrong

that isnt on your paper

[pine frigate]

anyway then can we move on ton ext one

[viscid pumice]

?????

JUST FUCKING SEND EXACTLY WHATS ON YOUR PAPER

[pine frigate]

.

[viscid pumice]

ITS NOT THAT HARD

[pine frigate]

rlea.

relax.

[viscid pumice]

where did you get the first = from

[viscid pumice]

thats not whats on here

[pine frigate]

bro

fuck that one

we cant solve that

we move to this one

[viscid pumice]

whats "this one"

[pine frigate]

.

https://prnt.sc/uOm0zwcyT3-_

[viscid pumice]

learn how to write

i cant read that

[pine frigate]

<@&268886789983436800> this person is being very rude to me and is not helpful at all i am starting to have suicide thoughts because of him.

[viscid pumice]

yk what me too

how tf do u expect me to help when i cant read what you write

also use a calculator it looks like there are no letters here

[pine frigate]

https://prnt.sc/1ThPlEimgD_b

[marsh vault]

I haven't even gotten to the part where he enters the conversation, and you're looking pretty rude in the scrollback already.

[viscid pumice]

i entered the first thing into my calculator and it says around 7,49

[marsh vault]

@viscid pumice if the helpee this obviously doesn't want you around, I'd strongly suggest disengaging no matter who is right or wrong.

[pine frigate]

thats clearly not teh answer

[viscid pumice]

3√(3^4) . √3 = 7.49

[pine frigate]

it should be 3x4

[viscid pumice]

heres proof

[viscid pumice]

then why is there no indication that youre multiplying

it says 3^4

[pine frigate]

i dont think u know what ur doing

[viscid pumice]

same for you

literally anything that doesnt have letters involved can be solved with a calculator

just use it

[pine frigate]

i figured it oout on my own

[marsh vault]

Okay, halt it. I see several suicide threats from @pine frigate in the scrollback; that kind of emotional blackmail is not something we have to accept. You get a two-day timeout to re-evaluate your approach to interacting with other people, and please behave better when you come back.

.close

[full forum]

Channel closed

Closed by @marsh vault

Use .reopen if this was a mistake.

[viscid pumice]

wow you learned how to use a calculator

[wind willow]

hello gentlemen.

[viscid pumice]

hi

[wind willow]

whats the problem here.

[viscid pumice]

whatever this astrocity means

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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• 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.

[slender onyx]

.close

[full forum]

Channel closed

Closed by @slender onyx

Use .reopen if this was a mistake.

[slender onyx]

just move on yall

[full forum]

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 ridge Has your question been resolved?

[hoary ridge]

I reviewd the problem and I missed a +1, so the parcial result would be -1 = dx/dy (would love to verify if that's correct)

[full forum]

@hoary ridge Has your question been resolved?

[full forum]

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.

[wicked edge]

I am integrating this fraction. Yet for some reason I can't seem to get 2x-2 in the end for my natural log. Am i splitting up the fractions wrong in the first step?

[hot herald]

you didn't factor out 1/2 correctly

[cinder vale]

^

[wicked edge]

Was it supposed to be 1/2 x-1

[cinder vale]

$$
\int \frac{1}{2x-2}dx = \frac{1}{2}\int
\frac{1}{x-1}dx$$

[glossy valve]

_Kookie

[wicked edge]

when i did it that way i didnt end up with 2x-2 for the end natural log

[cinder vale]

okay, let's say you did

I'd write it like this right
$$\frac{1}{2}\ln(2x-2) + C$$

[glossy valve]

_Kookie

[cinder vale]

your final answer

[wicked edge]

is what you did basically a u sub?

[cinder vale]

you can do a u sub

but

right now I'm assuming you've somehow arrived at the above answer

[cinder vale]

this thing here

[wicked edge]

no this is where i have trouble getting to

[cinder vale]

I'm going to do something sneaky

$$\frac{1}{2}\ln(2x-2) = \frac{1}{2}(\ln(2) + \ln(x - 1)) + C = \frac{1}{2}\ln(x - 1) + \frac{1}{2}\ln(2) + C$$

[glossy valve]

_Kookie

[cinder vale]

Now the (1/2)ln(2) is just a constant, like C

so I can group these two together to form a new constant

and the answer becomes
$$\frac{1}{2}\ln(x - 1) + C$$

[glossy valve]

_Kookie

[wicked edge]

thats what i got

[cinder vale]

which is what you got initially

[cinder vale]

now as I was doing the sneaky stuff I think I figured how how to get to this directly

try setting u = 2x - 2

and then integrating as usual

either way, your original answer and the given answer are equivalent

[wicked edge]

i know that works but why does it work, because here wouldnt i just take the x-1 as my u?

[cinder vale]

on the right hand side you'd take x - 1 as your u

on the left hand side you can take u = 2x - 2

[wicked edge]

yes

but wouldnt pulling 1/2 not matter

[cinder vale]

depends on what you mean. It matters because you will arrive at different representations of the final answer

but it also doesn't matter because the different representations are equivalent

they mean the same thing

[wicked edge]

yes that is what i meant

[cinder vale]

but they are different only by a constant

[wicked edge]

so this