help-23

[lone crescent]

no 2

it was 2f(0)=f(0)^2

[obsidian oracle]

Not in what you wrote

[lone crescent]

oh wait a minute let me check

no it looks right

it is origin

I know that there is solution f(x)=-x but I already deduced that f(x)=0

then how it is possible

oh wait

I just found mistake...

But now I have no idea how to prove anything at all

Just some thoughts P(x,0)then
f(x^2)=xf(x)
then P(x,-x)
f(x^2)+f(-x^2)=f(x)*f(-x)-xf(x)+xf(0)
xf(x)-xf(-x)=f(x)*f(-x)-xf(x)+xf(0)

2xf(x)-xf(-x)-xf(0)=f(x)f(-x) and then it completely unclear what to do

[safe radish]

@lone crescent Has your question been resolved?

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[forest perch]

hi

[safe radish]

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[buoyant nexus]

How do i approach this?

[buoyant nexus]

(a is not the correct answer btw)

[main mural]

u substitution

[buoyant nexus]

i did that and got: (-1/2x+1) +c

but thats not even an answer

[main mural]

you did it incorrectly, then

show your steps

[buoyant nexus]

i tried doing:
(2x+1)^-2

then:
2(2x+1)^-1/1 (deriving whats in the brackets then integrating)

ill show my steps one sec ill get an pic

[torpid condor]

I think its (-1/2(2x+1)) + C

is what you should of gotten if im right ?

[main mural]

this is correct

[fierce quiver]

Thats what answer says

[torpid condor]

I think if you write it as (2x+1)^-2 it can help you see it better ?

[main mural]

you might've forgotten something from this step

[buoyant nexus]

(its the working out under the line)

[main mural]

u = (2x+1)
what's du?

[fierce quiver]

I did the question too and working went

1/(2x+1)^2

(2x+1)^-2

(2x+1)^-1/-1

Idk how to go further than that

[buoyant nexus]

where is du meant to go?

[main mural]

wdym

du=<something>dx

[buoyant nexus]

du = 2, but idk where to sub it in

[main mural]

du=2dx

thus, dx=du/2

therefore, you substitute dx to be du/2, not du

and since 1/2 is a constant, you can factor it out of the integral

[main mural]

so your answer is correct, is just that you need to also multiply it by 1/2

[lavish gazelle]

Yo can anyone help me

[safe radish]

@buoyant nexus Has your question been resolved?

[safe radish]

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[narrow vault]

The angle theta at the left top is given. Can I assume the other angle theta is indeed theta? And if so, why?

[grim plover]

is this related to pendulums?

[narrow vault]

No

[grim plover]

ah K 😐

can u tell what other angle is? the bottom one marked with theta?

[narrow vault]

It’s not given

The angles with the dot are 90 deg.

[grim plover]

well if i am thinking about the correct angle then yes

[narrow vault]

But why

[grim plover]

by triangle angle sum property in triangle ABC

what is angle ACB?

[safe radish]

@narrow vault Has your question been resolved?

[safe radish]

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[coarse imp]

Factor a^{3}(b-4a)-27b+108a

[coarse imp]

I need to factor that

$a^{3}(b-4a)-27b+108a$

[flat frigate]

Slade

[icy lance]

factor the right pair

[coarse imp]

A³(b-4a)

?

How do i do that?

[icy lance]

this is fine

factorise -27b+108a

[coarse imp]

Uh

Is it 4(b+27a)?

[icy lance]

how would that work

thats 4b+108a

[coarse imp]

Wait right

Wait what

Oh my bad its 27(b+4a)

Right

?

[icy lance]

no

[coarse imp]

Would you please explain

[icy lance]

thats 27b+108a

[icy lance]

we had this

[coarse imp]

So its 27(-b+4a)

[icy lance]

yeah but pull out -1 too

[coarse imp]

-27(b-4a)

[main mural]

yep

[coarse imp]

Oh

Yeahhhh

Hold on lemme see if i can move on from that

I cant

😭😭

[icy lance]

you have a^3(b-4a)-27(b-4a)

[coarse imp]

We have a³(b-4a

Yeah

[icy lance]

factor out b-4a

[coarse imp]

What do i do next

(b-4a)(a³-27)?

[icy lance]

done

[coarse imp]

Wait i need to expand further. We can type a³-27 as a³-3³

Yeah thats easy after that. Thank you so much @icy lance

[icy lance]

this isnt true

if you have a^3-b^3 thats the same as (a-b)(a^2+ab+b^2)

but i dont see the need for that

[coarse imp]

There is no option like this

[icy lance]

show me

[icy lance]

oh

[icy lance]

apply this then

[coarse imp]

(b-4a)(a-3)(a²+3a+9)?

So E?

[icy lance]

uh huh

,w factorise a^3(b-4a)-27b+108a

[flat frigate]

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[icy lance]

same thing

[coarse imp]

Yeah thank you

$x^{2}-9x-y^{2}+9y$

[flat frigate]

Slade

[coarse imp]

When factoring this how do you know you should add xy and -xy? Like here

<@&286206848099549185>

[nocturne heron]

they used -xy and +xy bc they cancel out, to know which one you want to use look at the symbol of the y^2

look at which symbol combination will give you the same brackets

[safe radish]

@coarse imp Has your question been resolved?

[safe radish]

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[stray axle]

is my unit normal vector right?

im getting something different from the textbook

im not sure where i am going wrong

also from the 4 to 5 line, when finding the derivative, we dont find the derivative of the |R'(t)| but just the vector

[safe radish]

@stray axle Has your question been resolved?

[stray axle]

.close

[safe radish]

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[boreal grotto]

Greetings! I have a problem but not sure what should I apply to solve it. On the board there are natural numbers n > m. John divided n by m with remainder and got an incomplete part q1 and a remainder r1. Steve divided n - 1 by m with remainder and got an incomplete part q2 and the remainder r2. It turned out that q1 + q2 = r1 + r2. Prove that 2n is a square of a natural number.

[quasi bison]

$n = mq_1 + r_1$ and $n-1 = mq_2 + r_2$

[flat frigate]

Ann

[quasi bison]

this is what you should begin with

just writing out those divisions with rmainder

[safe radish]

@boreal grotto Has your question been resolved?

[boreal grotto]

Well that is clear

But just doubling the first equation didn't give me anything useful for the proof

[tiny wraith]

What happens if q1 = q2?

[boreal grotto]

r2 = r1 + 1 if I'm not mistaken

[tiny wraith]

Yeah, so try plugging this in, into q1 + q2 = r1 + r2

*it's the other way around actually r2 + 1 = r1

[boreal grotto]

Well that means 2q1 = 2r2 + 1

[tiny wraith]

Do you see why that gives a contradiction?

[boreal grotto]

Perhaps because q1 = r2 + 1/2 but 1/2 can't be a remainder?

[tiny wraith]

Yeah, basically left side is even, right side is odd

So, q1 and q2 aren't equal

[boreal grotto]

That's understandable, how should I use this for 2n though?

[tiny wraith]

q1 and q2 have to be close to each other though, right?

n - 1 is just 1 less then n

[safe radish]

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[boreal grotto]

Right

[tiny wraith]

use .reopen

[boreal grotto]

.reopen

[safe radish]

✅

[tiny wraith]

So, what would that mean?

[boreal grotto]

q2 = q1 - 1 I believe

[tiny wraith]

Yeah, the difference can't be more (if you want to be complete prove this with inequalities 0 <= r1 < m and 0 <= r2 < m)

try subsituting that again

[boreal grotto]

2q1 - 1 = r1 + r2 if I substitute it into q1 + q2 = r1 + r2

[tiny wraith]

and into these equations?

[boreal grotto]

n - 1 = mq1 + r2 - m
n = mq1 + r1

[tiny wraith]

Yeah, it might actually be easier to reason about this. The incomplete part of n when divided is 1 more then the incomplete part of n - 1 when divided, so what is the remainder of n when divided?

[boreal grotto]

Remainder of n division is equal to the reminder of n - 1 division then

[tiny wraith]

No, it doesn't have to be

Try n = 8, and m = 4

[boreal grotto]

The remainder for n division is zero and for n - 1 division is three. So the remainder in the second case seems to be m - 1

[tiny wraith]

Yeah, and in the first case 0. There fit exactly q1 m's into n

otherwise the incomplete part wouldn't decrease for n - 1

[boreal grotto]

So now I should subtitute m - 1 instead of r2 in n = m(q1 - 1) + r2 + 1 I suppose

And r1 is equal to 0

[tiny wraith]

Yeah, I forgot the exact substitions, but you should be able to get there with just algebra now

Dont forget r1 + r2 = q1 + q2 btw

[boreal grotto]

I got to 2n = 2mq1 + 2m - 1, so 2n = 2m(q1 + 1) - 1, that doesn't seem like a square

How can I get at least one square in the expression so I could factor it?

If I use 2q1 - 1 = r1 + r2, then 2q1 = m, so 2n = m^2 + 2m - 1, but this way 2n = (m + 1)^2 - 2

[tiny wraith]

I think something went wrong there

2n = q1 * m + r1 = q1 * m, because r1 is 0

@boreal grotto

[boreal grotto]

It seems that 2n is equal to 2mq1 + r1 though, then 2n = 2 * m * (m / 2), twos cancel out and 2n = m^2

Not sure if it's correct

[tiny wraith]

Yeah, that's correct

[boreal grotto]

I guess that was it. I'll try to compose it into a full solution and notify if something goes wrong. Big thanks for your help!

.close

[safe radish]

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[buoyant sphinx]

How would I go about proving this? I know the formal def of limit, and my friend gave me a hint to let epsilon=1/n, but how did they know to do that?

[quasi bison]

1/n can be replaced with any other sequence of positive numbers that approach 0

[buoyant sphinx]

gotcha thank you

why would we want any sequence that approach 0?

[quasi bison]

well, you should know that for any ε > 0 there exists a point in S that is greater than sup(S) - ε

i.e. a point that lies strictly between sup(S)-ε and sup(S)

when you then take values of ε along a sequence that goes to 0, you can get a sequence of points in S to which the squeeze theorem is readily applicable

[buoyant sphinx]

gotcha

thank you for explaining that I appreciate it

.close

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[native epoch]

I need to solve this system of equations, is this correct?

[native epoch]

If so how do I continue?

[quasi bison]

this feels like a big overecomplication

[native epoch]

right, so do I need to use matrices?

[quasi bison]

you don't.

do you know how to solve systems like this where the coefficients are real?

[native epoch]

yeah but I dont have em real here

[quasi bison]

well the good news is that doesn't matter. all techniques that you learned for real systems still apply here.

you have two equations:

1. (1+i)z_1 - z_2 = i
2. (1-i)z_1 + (1+i)z_2 = 1

you can multiply both sides by i in the first equation, which will make the coefficient on z1 into -1+i

and then add the second equation in, which will cancel out the z1 terms and so eliminate z1.

[native epoch]

should I multiplicate this first equation by the conjugate or just to make it i^2 ?

[quasi bison]

i told you exactly what to do

i said to multiply the first equation by i

not by a nebulous "conjugate" (conjugate of what???)

but by i

and i also told you exactly why i wanted you to do that.

[native epoch]

I still dont get it how multiply by i can solve it

[quasi bison]

and then add the second equation in

you've gotten the first equation into
(-1 + i)z_1 - iz_2 = -1
and the second equation is
(1 - i)z_1 + (1 + i)z_2 = 1

to repeat myself, i want you to add these two equations together

[native epoch]

ok so I got 0z1 +1z2 =0

[quasi bison]

indeed you did

[native epoch]

so there is no solution?

[quasi bison]

why would there be no solution?

[native epoch]

well if I move the variables over, I dont get any value

like 0z1 = -1z2

[quasi bison]

??

...you understand that your equation is literally just $z_2 = 0$, right?

[flat frigate]

AnnGhost

[native epoch]

so thats the only solution?

[quasi bison]

we're not done yet

we found only one of the variables: z_2

now you need to also find the other variable, z_1

do not overcomplicate this!

[native epoch]

ok let me plug it back into it

I got 0z1 = 0

[quasi bison]

did you try to plug the equation z_2 = 0 into itself?

[native epoch]

yes

[quasi bison]

why

there's no reason to do that

you would not do this if the system was real

it won't give you anything

it's GUARANTEED not to give you anything

[native epoch]

i*0 = 0

[quasi bison]

(1+i)z_1 - z_2 = i

you were supposed to plug it into this

the first equation from your system

or the second if you prefer

but not into itself

[native epoch]

oh so If I do the multiplication I would just get 0 tho

[quasi bison]

what multiplication...

[native epoch]

if I plub back into the first one

I get (i-1)z1 = -1

[quasi bison]

yes...

so then z1 is what

[native epoch]

dang thats true I dont need to plug in the second one as well

[quasi bison]

you need to review systems of linear equations

after this

[native epoch]

yep

so should I leave z1 = -1/(i-1) as answer?

.close

[safe radish]

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[quasi bison]

.reopen

[safe radish]

✅

[quasi bison]

if you're not going to wait for me to answer, then at least say "nevermind" or something... that was kind of rude of you.

but no, -1/(-1+i) can and should be simplified.

[native epoch]

oh ok

sorry I thought u were done

0.5+0.5i ?

[quasi bison]

decimals meh

fractions are better.

[native epoch]

ok

do u know any online calculator for double check? since I have a bigger one to do as well

[plucky elk]

,calc 1/(1-i)

[flat frigate]

Result:

0.5 + 0.5i

[native epoch]

?

[plucky elk]

What

[plucky elk]

I input this into a calculator

[plucky elk]

As you requested here

[native epoch]

no I meant for another system of equations

is this how I can start this one?

.close

[safe radish]

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[jade charm]

i am trying to solve this differential equation but i cant figure out why my answer is wrong. I am trying to use undetermined coefficients

[jade charm]

i then combined the general colution with yp that i found at the end

im wondering if its just some algebra error for finding yp? but im double checking it and it looks fine

[raven heart]

,w solve y''+6y = -294x^2e^(6x)

[flat frigate]

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[raven heart]

,calc 168/42

[flat frigate]

Result:

4

[raven heart]

yeah your B is definitely not 7

@jade charm

[jade charm]

Ooof, i see

I will look again at my algebra

[raven heart]

seems like your only mistake

ofc you have to see if it influences C

@jade charm

[jade charm]

It will. I will check those and update

i got it now. thank you!

.close

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[fickle abyss]

anyone know where i did wrong here?

ans is 2x-8

[celest ember]

you wrote x-3 as 2x-3

6th line

[fickle abyss]

ahh

thanks

.close

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[mossy nova]

hi

[mossy nova]

how would I find the deriative of that ?

would it just be 2e^-x

[peak estuary]

why would it be that

[mossy nova]

i rewrote it as e^-x^-2

oh wait would it be -2e^-x

[peak estuary]

no

[mossy nova]

so how would I do this

I thought deriv of e^x is e^x

[peak estuary]

yes

[mossy nova]

so why would the answer I get be wrong?

wouldnt I use exponent rule to bring the exponent down and subtract 1

[peak estuary]

no

well, not directly

exponent rule is about x^number, not number^x

you have to use chain rule here

[mossy nova]

so is my first step of doing e^-x^-2

correct?

like is that legal move

[peak estuary]

well ignoring all the missing parentheses, yes

[mossy nova]

ok so how would I do chain rule for this

[peak estuary]

do you know how to use chain rule in general?

[mossy nova]

yea

wait can let me try it

would it be e^-x^-2 * e^2x^-3

[peak estuary]

no

but closer

why does the second term have the e^ in it

[mossy nova]

so it would just be e^-x^-2 * 2x^-3

[peak estuary]

yes

[safe radish]

@mossy nova Has your question been resolved?

[safe radish]

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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'm trying to see if there's a pattern of some sort so I can generalize, but I can't see any

even in R^3, I'm not seeing anything :/

[devout shale]

if W is a 2d dimensional subspace of the vector space R3 and suppose that x is a vector in R3, but x is not in W, then what is the smallest subspace of R3 that contains both W and x?

[rustic goblet]

R^3 itself

?

[devout shale]

why?

[rustic goblet]

because the vector x would have to not lie in the 2d plane for it not to be in W

that would form a basis for R^3 now

[devout shale]

because we need to contain the d dimensions of W, and we need to add in the smallest subspace of V that contains x. At least adding this subspace will add one dimension, because x is not in W. So since d was 2, 2+1=3 and that is all 3 dimensions

if we were in R4 instead

what would dim(U) be?

[rustic goblet]

I want to say 4?

but I'm not convinced

[devout shale]

hmm let us think about this

[rustic goblet]

my issue in R^3 is that 1D lines are subspaces too

so then dimU would be 2

in R^n you could have 1D lines, 2D planes, etc

[devout shale]

sure

let us say again that W is a 2 dimensional subspace of R4

[rustic goblet]

alright

[devout shale]

and let us say that a vector x is in V, but is not in W

then x is not in the span of W

[rustic goblet]

mhm

[devout shale]

what is the smallest subspace of R4 that contains x?

how many dimensions large is it?

[rustic goblet]

3?

[devout shale]

No

[rustic goblet]

oh wait

[devout shale]

if we are looking for just 1 vector in R4, we only need 1 dimension

so a subspace that contains x is 1D

this subspace is not overlapping at all with the W subspace

from earlier W was 2D

the smallest subspace that contains x is 1D

these don't overlap

so what is the smallest dimension that contains both W and x?

[rustic goblet]

wait, i'm confused

W is a plane in R^4, and x lies on a line that is a subspace, or am i mistaken?

[devout shale]

yes

[rustic goblet]

wait but then why could they not overlap

[devout shale]

x doesn't overlap with W

since x not in W

[rustic goblet]

but the line could, no?

[devout shale]

at a point

but that point would not be x

[rustic goblet]

ok

oh wait

you could take an element from the line and add it to one from W, and it would no longer be on the line or W

then it has to be R^4 itself

wait but what if W were just a line

an element x not on W would just form a plane with W, no?

and the plane itself would be a valid subspace i think?

[devout shale]

no

a line is a 1D subspace

a plane is a 2D subspace

if the line is not contained in the span of the plane

then both of them together span 3D

not 4D

[rustic goblet]

didnt i say this though?

my first intuitive thought was that dim(U) = dim(W) + 1, but idk how to justify it

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@rustic goblet Has your question been resolved?

[rustic goblet]

<@&286206848099549185>

[rustic goblet]

is this right, or is something wrong?

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@rustic goblet Has your question been resolved?

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[rustic goblet]

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[supple valve]

where does the -1/2 come from

[bold ferry]

du = -2 dx
- 1/2 du = dx

[supple valve]

oh ok ty

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[reef estuary]

using product rule, what is p'(x) in terms of f and g

[frigid plinth]

Uhg

[frigid plinth]

f(x)g'(x) + f'(x)g(x)

Oh I see

So uh

6(0) + -1(3) = -3
-1

f(2) = 7
f'(2) = 1
g(2) = 3.5
g'(2) = -0.5

3.5(1) - 7(-0.5) = 3.5 + 3.5 = 7/3.5^2

3.51 - 7-0.5 = 3.5 + 3.5 / 7

= 7

4/7

@reef estuary I'm having trouble with the quotient rule one

I'm getting 4/7

That seems correct

The machine is saying it's not

Nvm uts coreect

Sorry

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[tardy mango]

yeah

[buoyant shadow]

yeah

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[lean otter]

Hi, how do I solve this using the comparison test?

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<@&286206848099549185>

[frosty oriole]

i cant help you @lean otter :(

[lean otter]

2 = a

👍

[lean otter]

Huh?

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[idle cave]

how would i go about solving this

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[idle cave]

you probably didnt type that out right now, but if you ddi thanks a lot :)

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[lyric island]

A little help doesn't do anything, haha

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[quartz torrent]

I am not sure on how to do trig sub for this

[quartz torrent]

I get 1/cos(theta) and then I tryto like,, do the silly triangle

andthen its 1/cos(sqrt(2*x^2)/x^2)

idk if that is , right

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[wheat pine]

help

[elfin moon]

again

what's up

which one now

[wheat pine]

help me with q 1

it due today

[elfin moon]

didn't we just do this 1

okay

this is my last question today

[wheat pine]

ok

it due in 10 min 💀

[elfin moon]

first we change 3 * sqrt(12) to 3 * 2 * sqrt(3)

do you agree with this step

[wheat pine]

idk i know nothinh

nothing

[elfin moon]

bro

honest opinion

mathaway it and then ask the question tommorow

bc you aren't going to learn anything in 10 min

[wheat pine]

💀

i am dead

[elfin moon]

Im sorry

it is the truth tho

[wheat pine]

bro

grade 8 hard asf ngl

[elfin moon]

aint no way

[wheat pine]

ok bye

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[coarse kelp]

Beethoven wrote 9 symphonies and Mozart wrote 27 piano concertos. If a university radio station announcer wishes to play first a Beethoven symphony and then a Mozart concerto, in how many ways can this be done?

Why 9 * 27 * 26 = 6318 is wrong?

[still charm]

Where do you get 26

[coarse kelp]

nvm I'm wrong

[still charm]

9 choices for #1, 27 choices for #2

[coarse kelp]

i thought they play 3 songs

thanks

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[magic pasture]

Would the answer be B since 1 is in the interval -2,3

[charred slate]

hey

yeah i said to put in a different help channel bc that one got reserved by scriptedeli already

anyway

let's see

yeah i think you got it

you just need to check the endpoints

(1)^2 - 3(1) = -2

and 6(1) - 2 / 1 - 3 = -2

[magic pasture]

Thank you

I have another one

[charred slate]

go for it

[magic pasture]

I’m guessing it’s E x/-x = -1

[charred slate]

yep

[magic pasture]

Thank you again

[charred slate]

no problem

make sure to do .close

[magic pasture]

.close

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[grim jackal]

Would the correct answer be that “The function has a zero in the interval because the 5 sqrt(5) is between f(-1) and…”

[dull sequoia]

What does 10 have to do with the question

[grim jackal]

The question was asking for a zero, which is a root, so I thought that when you have an x intercept, y/f(x) would be zero

I set the function to 0 and then subtracted 10 on both sides

[dull sequoia]

But you didn’t have to know explicitly the function f to apply the IVT

The 10 means nothing

IVT requires only that the function is continuous on the interval and the value of 2 end points

[dull sequoia]

If you aren’t too certain about this, try to state the IVT

[grim jackal]

I need to show if the function has a zero in the interval so I'm not sure how to do that

[dull sequoia]

What does the IVT say?

[grim jackal]

I don't understand sorry

[dull sequoia]

State the intermediate value theorem.

[grim jackal]

For IVT, don't I need to know the range of the interval and the y of the function

[dull sequoia]

Well state it

First state the theorem then use it

But that means we need to be able to state the theorem

So let’s start there, state the IVT

[grim jackal]

If f is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in [a, b] such that f(c) = k

[dull sequoia]

Ok

f is a polynomial continuous on (-inf, inf)

This implies that it is also continuous on [-1, 2]

Do you agree?

[grim jackal]

yeah

[dull sequoia]

Let us choose k = 0

[dull sequoia]

Since 0 is between f(-1) = 12 and f(2) = -54, this k fits the requirement for IVT

Now what does the IVT say

[dull sequoia]

Can you stop

[grim jackal]

k is between f(-1) and f(2), which would satify the requirement

I see why i'm wrong. 0 is f(x). It doesn't matter what the equation is if the x-int is going to be y=0

[dull sequoia]

And thus ???? By IVT

Yep

[frozen ledge]

Intermediate value theorem that is the meaning of ivt

[dull sequoia]

It matters what the (sign of the) endpoints are

And that it’s continuous

[grim jackal]

Yeah I get it now. Thank you

[dull sequoia]

Try complete this sentence

It’s good practice to articulate the conclusions

[grim jackal]

IVT applies to the function because it is continuous and 0 is between f(-1) and f(2)

[dull sequoia]

Then?

[grim jackal]

then f(x) = 10-2x^5 has a 0 in the interval? I'm not sure. My response is usually accepted

[dull sequoia]

Thus there exists a c between -1 and 2 such that f(c) = 0, and c is a zero of the function, by IVT.

[grim jackal]

oh. Thank you

[dull sequoia]

The whole idea of these theorems is like

If (conditions) then (some conclusion)

Here it’s if (f cont, know end points) then (there’s an input in the middle bla bla bla)

So you show the “if” part holds

Then say this implies the “then” section by the theorem

[grim jackal]

Okay got it.

.close

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[rare owl]

f(x)=int(2x)

[rare owl]

What does int mean?

[knotty sigil]

integral or integer

depends on the context

[rare owl]

I’m supposed to find the domain and intercepts

And graph

How do I start?

[quasi bison]

uhhh

can you show a picture of the problem

because i don't think any of us can tell you what int() is supposed to mean

[quasi bison]

does your book give you a definition anywhere for this int function?

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[storm herald]

what did they here?

[storm herald]

I understand the rest but what steps did they do in the highlighted bit

this is a proof question, ik about the sin^2+cos^2=1 but what did they do to the second bracket

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@storm herald Has your question been resolved?

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@storm herald Has your question been resolved?

[storm herald]

<@&286206848099549185>

[willow gust]

If 2sin(A)cos(A) = sin(2A)
Then, 2sin(2theta)cos(2theta) = sin(?)

[storm herald]

.close

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[zinc pebble]

uhhh

[zinc pebble]

<@&286206848099549185>

Idk

[high mica]

1. ping after 15 minutes

secondly do you know sum of infinite GP formula

[zinc pebble]

?

i dont

[high mica]

a = base number, r = common ratio

[zinc pebble]

so i'll apply that formula?

[high mica]

yes

[zinc pebble]

oh okay

[high mica]

you apply that on question 2

for question 1 you need the infinite sum formula

[zinc pebble]

oh

[whole acorn]

why do you need infinite sum formula for 1?

[zinc pebble]

a is the..

[high mica]

wait

[whole acorn]

there is only 8 terms no?

[high mica]

sorry i thought it was to infity

infinity

💀

my bad

[zinc pebble]

so whats gonna be the formula?

[whole acorn]

this one ^^

[zinc pebble]

oh okay thanks

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[little niche]

a question regarding the existance of $\int_0^1 e^{dx}-1$

[flat frigate]

skittle

[little niche]

please explain why doesn't this work

and don't come at me that dx doesn't work like that, that's just a new way of expressing it

which no one ever saw before

https://www.youtube.com/watch?v=KqmIGr68RCo&ab_channel=FlammableMaths
there are a few videos that solve it like that

<@&286206848099549185>

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[little niche]

<@&286206848099549185>

[open tulip]

yes?

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[little niche]

.

@open tulip

[open tulip]

im actually in grade eight so basically i cant solve this, but someday i can

[little niche]

alr

[open tulip]

srry for my childish state

[little niche]

why are you talking like that

[open tulip]

*i mean my age

[little niche]

@plucky elk

[little niche]

.

[safe radish]

Please do not ping individual helpers unprompted.

@little niche Has your question been resolved?

[little niche]

<@&286206848099549185>

[lean otter]

No pain. No gain. Do it on your own

[little niche]

huh

I used my knowledge, and I am asking for help

[plucky elk]

You need to stop pinging me

Or any individual helper for that matter

[little niche]

right

im sorry

[little niche]

can you please answer this?

[safe radish]

@little niche Has your question been resolved?

[kind fractal]

I guess this would be ok with respect to recreative math I guess (?. I mean, unless you define properly what dx and integral would mean in this expression, then it doesn't really make sense. It's just the result of intuitively working with differentials.
For example, is the sense of integral still the same here as we work through? 🤔
Try comparing it with the Riemann sum. Is this integral sign in this problem also linear? I mean can we separate it like
integral e^dx - integral 1 ?
Would that make sense? 🤔

What I mean is that it's ok for recreative maths (I guess (? ), but it's not really logical. Try starting from a Riemann integral. You will see that you can't do that limit just like that

[little niche]

I think what is shown is a product integral

[kind fractal]

Yes but we use the same logic as Riemann integral

well it looks like it works well in a Riemann sum 😵‍💫

I don't have a clue on why we can put the limit inside the sum

https://www.desmos.com/calculator/24n6mywz3j

I guess it's ok, but I can't tell rigorously why

still, it's concentually incomplete too. I guess it's possible to define all the properties of this kind of operation 🤔

[safe radish]

@little niche Has your question been resolved?

[celest ember]

This is fine I don't see where the problem is

[little niche]

cool

[celest ember]

What are you asking then?

[little niche]

because people dont often see the differential in the exponent

so almost everyone denies it

[celest ember]

Are you putting x=e in this?

[little niche]

mhm

[celest ember]

But x is a variable

You can replace it with a constant

[little niche]

yeah but its the proccess is the same

[celest ember]

No

[little niche]

.

you transfer it into a limit

the limit with x equals ln(x)

and the limit with e equals 1

[celest ember]

The integration won't give you the same result

[little niche]

right, because they are different integral

s

[celest ember]

Yes

[little niche]

normally this substitution doesnt work

but here youre just converting into a limit

both ways work

[celest ember]

huh?

What are you trying to say?

[little niche]

look

,w lim x->0 (e^x-1)/x

[flat frigate]

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Requested by skittle266

[little niche]

,w lim x->0 (a^x-1)/x

[flat frigate]

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Requested by skittle266

[little niche]

see?

if we put a=e here, we get ln(e)

which is 1

so if you see that the integral $\int_0^1 x^{dx}-1$ works, then $\int_0^1 e^{dx}-1$ also works

[flat frigate]

skittle

[celest ember]

Hm

[little niche]

yhyh

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[boreal sparrow]

I was hoping someone could help with my understanding for this question:
For any sets A, B, and C, prove by contraposition: if A ⊆ B and B ⊆ C, then C' ⊆ A'

My work so far:
contrapositive: if C' ⊄ A', then A ⊄ B or B ⊄ C

Then, given the premise, we know there exist an element that is contained within C', but not in A'. Let this element be x. So x ∈ C', but x ∉ A'. If x ∉ A', then by the definition of the complement, x ∈ A.

Now, to prove the conclusion, we must show that either A ⊄ B or B ⊄ C.

Kind of stuck here now, because then would this mean I use a proof by cases approach for each element of the conclusion?

[safe radish]

@boreal sparrow Has your question been resolved?

[boreal sparrow]

so we are allowed to use a different proof method within this proof by contrapositive?

Alrighty then, I had written my proof down via that, perhaps I will write it here once it is proper and done

[safe radish]

@boreal sparrow Has your question been resolved?

[boreal sparrow]

For any sets A, B, and C, prove by contraposition: if A ⊆ B and B ⊆ C, then C' ⊆ A'

My work so far:
contrapositive: if C' ⊄ A', then A ⊄ B or B ⊄ C

Then, given the premise, we know there exist an element that is contained within C', but not in A'. Let this element be x. So x ∈ C', but x ∉ A'. If x ∉ A', then by the definition of the complement, x ∈ A. Similarly, this would mean that since x ∈ C', then x ∉ C.

Now, to prove the conclusion, we must show that either A ⊄ B or B ⊄ C.

Case a) - B ⊄ C: Suppose A ⊆ B. Then, by the definition of set inclusion, every element of A is also an element of B. Thus, or every y ∈ A, it implies y ∈ B. If it were the case B ⊆ C, then every element of B is also an element of C. From our hypothesis, we had found that there exists an element, namely x, that is not contained in C (ie. x ∉ C). Then, this would mean that not all elements of B is in C, since we had that x ∈ B due to A ⊆ B (where ∀x(x ∈ A --> x ∈ B)). In other words, B ⊄ C. As such, since B ⊄ C is true, then so is the conclusion (A ⊄ B) or (B ⊄ C).

Case b) - A ⊄ B: Suppose B ⊆ C. Then, by the definition of set inclusion, every element of B is also an element of C. Thus, or every y ∈ B, it implies y ∈ C. If it were the case A ⊆ B, then every element of A is also an element of B. If so, then by transitivity, we have that A ⊆ B and B ⊆ C --> A ⊆ C, but our premise stated that there exists an element in A that is not in C, namely x. Then it cannot be that A ⊆ B, rather A ⊄ B is true. As such, since A ⊄ B is true, then so is the conclusion (A ⊄ B) or (B ⊄ C).

As a result, by cases (a) and (b), we see that for any sets A, B, and C, if C' ⊄ A', then either A ⊄ B or B ⊄ C. Now that the contrapositive has been proven, this indicates that the claim in question is also proven.

what do you think @lean otter ?

damn, I guess you are right

(now to decide which is better to keep)

(now you're just killing me)

thank you again @lean otter for the help

https://tenor.com/view/spider-man-gif-26192774

of course (woohoo new friend)

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[sinful crown]

i have the casio cg50 and when i add in a text file to the program folder it uploads but i cant open them

[sinful crown]

all the files are in .txt and then it converts it itself to the g3m format

lowkey im just gonna turn my txt into images and figure this out this weekend

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[gritty heart]

Hi, I am confused as to how I should start this problem:roblox

[gritty heart]

I know how to find derivatives using product/power/quotient/chain rule, but I'm unfamiliar with the dy/du notation.

[plush isle]

whats happens if you take d/du of y. what is whole equation

[plush isle]

in other words dy/du is the derivative of y with respect to u.

[urban dune]

For this case, since u and y are functions of x, take dy/dx and du/dx. Then, dy/du = (dy/dx)/(du/dx)

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[gritty heart]

dy/dx is 2x while du/dx is 2, so dy/du=x?

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[ruby glen]

I got everything right except for the last 4 boxes

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[upper ginkgo]

hi i need help from the synthetic division 2x^3 + x^2 + 3/2x^2 + 1

[lean otter]

what have u tried

[upper ginkgo]

im just getting confused for a while hahaha

[lean otter]

do you know how to do it

[upper ginkgo]

how to do it yeah

[lean otter]

okay and ur working

[ruby glen]

.reopen

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[thin helm]

theres a typo here

it shouldn't be x^4

but also, in the final equation, why is x^2? Shouldn't it be x?

there is a typo here right? am i missing something

oh nvm the original equation is a typo it should be x^4 it just isn't in the first term

smh

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[tiny heart]

hi

[tiny heart]

how do i solve the equation

[upper mulch]

This is just a question finding the domain and range.

Any ideas of how you find domain?

[tiny heart]

36-x^2 > or equal to 0

how do i solve that

-x^2> -36?

but dont i switch sides

cuz negative

so

x^2<36

[upper mulch]

This is a way, but you'll find that the answer to that solution is x<6 and x<-6, which only gives x<-6.

[tiny heart]

how do i find that

how i find its x<6 and x<-6

?

[upper mulch]

If you put sqrt on both side, then you'll get those two solutions.

[tiny heart]

oh right +- 6

[upper mulch]

But that is a bit confusing. Instead, solve for this equation 36-x^2=0.

[tiny heart]

x = +- 6

[upper mulch]

So x = 6 and x = -6 are zero points, based on that, what is the domain?

[tiny heart]

-6<x<6

?

[upper mulch]

What if x = 6 or x = -6? Would that still give a valid value?

[tiny heart]

yea

its equal ot less right

greter

greater

[upper mulch]

Yeah, so the domain is -6 <= x <= 6.

Now you got domain, then it's just finding the range.

[tiny heart]

how i find the range

[upper mulch]

What is range?

[tiny heart]

y vslues

values

[upper mulch]

What do you mean by y-values?

[tiny heart]

permissible

[upper mulch]

Is there prerequisites for that?

[tiny heart]

??

[upper mulch]

Is there limitations on how x should be?

Hmm, this explanation is a bit confusing.

In other words, if range means valid values of y, then based on that, x should also have a "range" of valid values, which is domain.

Then it's just finding the values that the domain "allows", meaning the smallest and biggest x-values.

[tiny heart]

ok then

[upper mulch]

Do we know domain to the function?

[tiny heart]

-6 <= x <= 6.

[upper mulch]

Apologies, that wasn't the case.

[tiny heart]

oop

[upper mulch]

A value is one of the x endpoints. And the other is when x^2 = 0

Do you got the two values?

[tiny heart]

uhh for y its 0 and -6

[upper mulch]

So you know the domain and the range, then it's finding the identical option.

Do you got your final answer?

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[placid root]

How should I start with this?

It's easier to show B=DA has a unique solution

how do i go about the opposite?

[quiet juniper]

the problem says theres not a unique solution tho

Hmmm

[placid root]

yea, how should this be done? i need a nudge T_T

Should I use this theorem ?

Theorem 1. Let A be a square n × n matrix. Then Ax = b has a unique solution if and only if the only
solution of Ax = 0 is x = 0. Let A = [A1, A2, . . . , An]. A rephrasing of this is (in the square case) Ax = b
has a unique solution exactly when {A1, A2, . . . , An} is a linearly independent set.

And should go for the opposite?

so Ax=b will have no unique solution if the solution of Ax=0 (so Da=0 in this case) is a is not equal to 0?

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[lofty trail]

for simialrity in maths rihgt

[lofty trail]

wahts does the 3 staright vertical liens mean

and what does the 2 vertical liens mena

[hasty wagon]

do you mean by horizontal lines instead?

$=$v.s.$\equiv$

[flat frigate]

Biscuity

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[cyan vector]

I am taking a class in linear algebra. I do not understand what is happening here. How do we turn the (1,0,0) into (9,0,0), and what step do we need to take in order to get from the first step to the second?

[rocky ferry]

(9/9)<1, 0, 0> = (1/9)<9, 0, 0>

[cyan vector]

Thank you!

[rocky ferry]

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[near oak]

help

[safe radish]

Please don't occupy multiple help channels.

[split ether]

!onechannel

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Please stick to your channel.

[quasi bison]

!noclopen

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Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[quasi bison]

seriously what the fuck are you doing

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[fathom lake]

What are these small “o”s and big “O”s found sometimes in Taylor series?

[mortal ingot]

f(x) = o(g(x)) means f(x) / g(x) -> 0 as x goes infinity.

Meanwhile f(x) = O(g(x)) means f(x) / g(x) is bounded as x goes infinity.

[peak estuary]

well not always x->infty

[cosmic grove]

^

[mortal ingot]

But you should have learned the definitions in the class if you are encountering them in notation of Taylor series

[peak estuary]

in general for x->c where c is some fixed point

[mortal ingot]

Ah, right. Yea

May apply to any convergent filters

[peak estuary]

especially here its about x->c where c is the point around which we write the taylor series

[mortal ingot]

..idk how I missed that.

[fathom lake]

so O can mean the same thing o does sometimes?

[mortal ingot]

Well they are still different

But if f(x) = o(g(x)), then f(x) = O(g(x)), if that is what you mean.

[fathom lake]

yeah

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[burnt obsidian]

HOW

[burnt obsidian]

The

Fuq

Do you calculate arg(z)

If

You know that z= 1+i *sqrt(3)

When you have to keep in mind that you can't use calculators

In the exams

Sorry for typing so slow im on tabletvimma switch to phone now

[icy lance]

arctan(sqrt(3)), which is something you should know of if you know the unit circle, or the trig triangles

[burnt obsidian]

I understand how you got the whole triangle sides

But not how you got the angles

Can you explain a little bit more how you calculated that?

[quasi bison]

these are half of an equilateral triangle and half of a square resp.

[burnt obsidian]

Now that you mention it yea I can see it. How do I use that information to get to the answer I’m looking for?

Ah

Now I get it I think

But wait aren’t both of them half of a equilateral triangle and half of a square?

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@burnt obsidian Has your question been resolved?

[burnt obsidian]

No I don’t understand how you’re supposed to use the hints I’ve been blessed with :(

[burnt obsidian]

I’ve looked at this video specifically excersises D and E. EVERY OTHER EXCERSISES I UNDERSTAND how he does it. But not E and D how is it so obvious that arg(2-2i) is equal to -3pi/4??????????
https://youtu.be/bdyyPjw2Deo?si=m5xen6pY3Sc7-F71

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@burnt obsidian Has your question been resolved?

[burnt obsidian]

<@&286206848099549185>

Is this given
As in is it supposed to be known?

[vague frost]

wdym given? those are values u should know for triangles

[burnt obsidian]

Ok now I understand how it was solved

You already had info I thought you calculated that stuff without a calculator

My bad

[burnt obsidian]

Thank you so much I hope you have a great rest of your week

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[jagged carbon]

I need to identify the vertex, axis of symmetry and also sketch the graph, I’m kind of lost on this kind of math

[solid shell]

Start with the y-intercept, do you remember how to get it from standard form?

[jagged carbon]

isn’t it the C value? If so wouldn’t it be -29 and -5

[solid shell]

Yup

Do you rememeber how to find the vertex from standard form?

[jagged carbon]

is that -b/2a? For the formula

[solid shell]

Yup, that will give you the x value of your vertex

[jagged carbon]

Okay thanks

[solid shell]

Which you can then plug into the equation to get the y value