help-19

[fervent hawk]

what have you tried?

[valid swan]

i understand that

But when distributing terms

It ends up becoming 12(3x-4y)^3 - 16(3x-4y)^3 y’ = 9y^2y’

[fervent hawk]

let's see

[valid swan]

Thats where i dont understand on the left side

Thanks

I used chain rule on the left side to differentiate it

[fervent hawk]

i think you should move all the terms with y' to one side, and without y' to the other side

that usually works

[valid swan]

Yeah how do u do it from

Here

how can i move the one on the left

to the right

[fervent hawk]

it's just expanding brackets

I'll try and type it out

[valid swan]

Thank you

Im not sure i understand the expanding because

It goes from this

[valid swan]

to thos

this*

[fervent hawk]

ohhh i think i saw a mistake

[valid swan]

What is it?

[fervent hawk]

d(4y)/dx is 4y' not 4yy'

[valid swan]

Oh yeah

True

[fervent hawk]

i think that's the main problem that makes you not understand

[valid swan]

oh but also

if it ends up being 4(3x-4y)^3 * (3-4y’)=9y^2 y’

[fervent hawk]

4(3x-4y)^3 (3-4y’)=9y^2 y’
then
3(4(3x-4y)³)-4(4(3x-4y)³)y'=9y²y'

[fervent hawk]

and then to this

distribution

on (3-4y')

oh, it's like
a(b+c)=ab+ac

where a=4(3x-4y)³, b=3, c=-4y'

oh, that

3×4=12 and 4×4=16

？

a can be any expression

yea

it's just any expression

you can do sub any expression into a b and c

all good?

[odd edge]

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[mystic saffron]

help i cant find m

[echo ginkgo]

sounds impossible to find such an m yes

@mystic saffron

[mystic saffron]

yup

like i get that 5mi + 7mj = k

idk how to get m

[echo ginkgo]

well yeah it's impossible

if we write it in vector form

$$\begin{bmatrix}5m \ 7m \ 0\end{bmatrix} = \begin{bmatrix}0 \ 0 \ 1\end{bmatrix}$$

[clever fjord]

aPlatypus

[echo ginkgo]

that's what we have

[mystic saffron]

yup

[echo ginkgo]

0=1

sounds like a problem to me

[mystic saffron]

then its fault by my teacher

[echo ginkgo]

well no

they never said it existed

"is there a scalar m..."

[mystic saffron]

yupppp

[echo ginkgo]

the answer can be no

[mystic saffron]

i seee

[mystic saffron]

u know in every question i know that there exist 😂

[echo ginkgo]

if it doesn't exist, i'll ragequit the question

take that, math problem

[mystic saffron]

yupp hahahaha 😂

[mystic saffron]

thank you for helping me alot

[echo ginkgo]

@mystic saffron you got other questions? if not you should close the channel

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@mystic saffron Has your question been resolved?

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[idle elbow]

I’m trying my hand at algebra. Is my proof alright?

[lofty spade]

are you proving the first line?

[crisp wadi]

That's not what surjectivity means

"For all a in A there exists y in B with f(x) = y" is true for any function f: A -> B

[idle elbow]

True

Are there any other problems?

[crisp wadi]

Looks good

[idle elbow]

Thank you so much

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[vocal dawn]

Is quadratic a function? My confusion stems from the idea that functions are supposed to for any given input map to exactly one output, whereas the quadratic has multiple solutions.

[copper quarry]

a quadratic is a function

[vocal dawn]

But then how can it have multiple solutions, and still be defined as a function?

[copper quarry]

have you heard of the vertical line test

[vocal dawn]

It seems familiar

Right, for a vertical, any x, youll get a unique y

[copper quarry]

if you can draw a vertical line at any point on the graph and have the vertical line intersect at most ONE time then it is a function.

If at any point the vertical line intersects the graph twice then it is not a function but a relation

!occupied

[odd edge]

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

[odd crypt]

Yes

[vocal dawn]

So i've def heard of the test, and that makes total sense, but then how can we have two solutions, then when exactly is a relation a function, without mapping it on a graph

for any (x,y) for a relation to be a function you can't have two x mapping to different answers

[odd crypt]

If it's + or - then you can

[vocal dawn]

Well then by definition it wouldn't be a function

xD

oh

f(x) = x^2

Right

[copper quarry]

a quadratic is a many to one

and is still a function

[vocal dawn]

so thats an x^2 term that maps to two different ys

not the x

[odd crypt]

Bro there will be two outputs let's say x^2=2

To solve this take root on both sides to get

X=+ or -root 2

[copper quarry]

do you know what a many to one function is

[vocal dawn]

Sure many to one, but one to many is impossible

[copper quarry]

yeah but a quadratic isnt a one to many

its a many to one

[vocal dawn]

Right

we are finding the x, by using the y

instead of what we usually do in functions

[thin kelp]

dont u find the y by using the x?

[vocal dawn]

You are looking for x

solutions to x

in quadratic it has to have 2 solutions

[thin kelp]

yes

[vocal dawn]

those two solutions might be equal though

[thin kelp]

and?

[vocal dawn]

Im studying relations right now, and the definition of the function confused me

It makes sense, you cant have more than one pair starting with the same term

[thin kelp]

simplest way to determine a function is with the vertical line test, if u can draw a vertical line on any point of the function and only have 1 intercept then u must have a function

[copper quarry]

a many to one is a function where several values of x can correspond to the same y value

[vocal dawn]

Yes, what confused me is that in a quadratic is that we derive the solutions from the y

unlike in normal functions where we plug the x and you get a y

Thanks for the help guys

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[warped phoenix]

someone please explain this

[warped phoenix]

didnt not understand a single thing

how did we prove that it has exactly one solution!?

[odd crypt]

IVT theorem states that a function has to be continuous on an interval [a,b] such that there is a value c between them

By finding f(a) and f(b) you will get answer

[lofty spade]

By contradiction

[lofty spade]

They showed there is at least one, and there can’t be two or more than two

so there must be exactly one

[warped phoenix]

i got f(a) and f(b)

f(-1) = -3 and f(0) = 1,

now how did we say we only have one root?

numerically

aside from looking at the graph

[odd crypt]

In you plug these values in he derivative

You will get 1 answer as zero

Wait*

[lofty spade]

do you know the Intermediate Value Theorem?

[odd crypt]

Because -3<0<1

There is a value c in which f of c can be zero

[warped phoenix]

yes

its like the MVT

[odd crypt]

According to IVT

[warped phoenix]

but doesnt need to be differentiable

only continous

[odd crypt]

Bro Differentiability implies continuity

[warped phoenix]

if i plug f(-1) = -3 and f(0) = 1,
in 3x^2 + 3?

[odd crypt]

Non

[warped phoenix]

u mean this : if i plug f(-1) = -3 and f(0) = 1,
in 3x^2 + 3?

[odd crypt]

No

[warped phoenix]

then what

[odd crypt]

That was a mistake

Do you know Rolle's theorem

[warped phoenix]

yes

[odd crypt]

Read this

[warped phoenix]

thats rolles theorm

this

i just do not understand in the question , how did we prove there is only 1root

[odd crypt]

https://tenor.com/view/who-knows-marques-brownlee-i-don't-know-why-not-it-could-be-gif-17859827335515836671

[warped phoenix]

u do not know?

[echo ginkgo]

Can you pinpoint what part of the argument you don't understand ?

@warped phoenix

[shrewd adder]

imagine a random graph

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[keen rain]

It want me to find sina and I have to use law of sine but I couldnt find it

[keen rain]

Blues is the answer but I still dont get it

[swift rock]

,rotate

[swift rock]

that isnt law of sine

[keen rain]

It is

[swift rock]

law of sine is
a/sina = b/sinb = c/sinc

[keen rain]

Its also 1/2 a.b.sinC in triangle

[swift rock]

thats for area

do u have area of triangle given?

[keen rain]

BAD is x and ADC is 3x I think

[swift rock]

what are these numbers ?

they are not x

how did u get them

[summer trail]

8 and 35?

[keen rain]

I wrote it from my friend I realy dk

She wrote it like this

[swift rock]

so basically , as u said BAD is x and ADC is 3x

formula for area of triangle is 1/2 ab sinC

in BAD the formula becomes
1/2 * 8 * h sin30 = x

in ADC it becomes
1/2 * 12 * h sinA = 3x

divide both equations together
1/2 * 12 * h sinA = 3x
1/2 * 8 * h sin30 = x
and u get ur answer

@keen rain understand?

[keen rain]

Why dividing them together?

[swift rock]

cuz there isnt any other way to find sinA

or u could divide the area of ADC with 3 so u find x and then take that equal to area of BAD

[keen rain]

Thanks so much

[swift rock]

.close

to close the channel

[keen rain]

.close

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[tardy lagoon]

are you familiar with the limit definition of e?

try to use the definition: $\lim_{x\to \infty}(1+\frac{1}{x})^x=e$

[clever fjord]

WhereWolf(ping if needed)

[summer trail]

if u cube the left side

then u cube the right

why 3e?

wouldnt that be 3 * the limit?

ill be honest, im not too sure where the 3/x^2 term comes from

no clue lol

ask whereswolf

he might be better suited to help

oh interesting

[tardy lagoon]

you should be able to apply l'Hopital's rule after taking natural log

exponent is hard to deal with, so take natural log to turn it to multiplication

you can apply continuous function in limits

[summer trail]

yea

and before that

you can even cube root both sides

to get rid of the 3

and the question becomes quite neat after u take the limit

yep

wait no

1 sec

hmm, im not too sure actually what the best way is

maybe u just use the limit to break it up

like lim(a+b) = lima + limb

i could be wrong

but perhaps u could put the limit inside ln (i think you can maybe do that) then if u take the lim of 3/x^2 by itself, it goes to 0, then u get the traditional form of e as a limit

oh wow, take a look at this

https://socratic.org/questions/what-is-the-limit-of-1-1-x-x-as-x-approaches-infinity-2

thats why the mention lhopital

[tardy lagoon]

bruh I told you

[summer trail]

oops i didnt see ahaha

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[gilded crag]

This makes no sense to me. If it rains, shouldn't be get rained on? Therefore, the chance he gets rained on is the chance that it rains..

This logic isn't logicing for me.

That 10% should not affect if he gets rained on tomorrow afternoon considering that he knows that there is a 60% chance of rain..

[brisk urchin]

P(A) -> Event A: It rains in the afternoon.
P(B) -> Event B: Brian gets caught in the rain.

Consider this scenario,
It is raining in the afternoon, and Brian is outside, so he will get caught in the rain,
But if Brian is inside the house while its raining in the afternoon, then Brain won't get wet,

The two events are related only when we know that Brain got wet from the rain, meaning it is raining in the afternoon,

Still the problem statement is faulty because "Brain can get wet from the rain" when its "raining and its not afternoon" but we will ignore all other possibilities for raning in a different time for now and focus on the task at hand.

[brisk urchin]

the 10% in P(A & B) (Intersection can be denoted as & in logical sense) states both events happened, where it was raning and Brain was outside and got caught in the afternoon rain.

P(A & B) = P(A) x P(B | A) is a formula, you can check its proof online.
it just states that the probability of both events A and B happening at the time = the probability of event A happening, and the conditional event of B happening as well (in this case its Brain getting out of the house in the afternoon while it is raining).
Not good with explanations but you can read the actual proof online.

[odd edge]

@gilded crag Has your question been resolved?

[livid skiff]

wait i am gonna do it in paper

[gilded crag]

that formula only makes sense in my mind if brian's chance of getting rained on was 10% if it rains outside and the chance it is going to rain is 60%. Do you understand my logic?

[brisk urchin]

yea

the chance of raining and brain getting wet in it is
the chance of raining and brain getting wet due to the rain

its a bit confusing but its there

[gilded crag]

but doesnt everyone always get wet if it's raining outside and they're outside?? im getting autism from this question..

😂

[livid skiff]

there is a difference between the intersection and the given

in one ocasion we know for a fact that it is raining

it is denoted as p(A|B)

So it does not matter if its raining or not

only if bryan is outside

[gilded crag]

so the chances of getting rained on given the chance of it raining is 100% i my mind.. not 10%.. y'know?

[gilded crag]

but brian **will **be outside, according to the question, because he's driving home. no?

[livid skiff]

here

[livid skiff]

the probability of rain is 0.60, the probability that ryan is outside is 0.10

while its raining

[gilded crag]

i understand.. thank you..

<3

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[sour geyser]

I am doing a task for hypothesis testing. My study aims to look at which age group gains the most benefit from an adrenaline shot. I did this by recording their initial time then giving them an injection and recording a time difference between the post injection run and the initial run. My null hypothesis: there is not a statically significant difference in the mean change in 100m sprint times among the different age groups. My alternative hypothesis is: There is one group which has a statistically significant different in the mean change of a 100m. I need help putting my null and alternative hypothesis into mathematical notation.

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<@&286206848099549185>

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[buoyant solstice]

why is Z_6 X Z_3 isomorphic to Z_3 X Z_3 X Z_2?

[buoyant solstice]

does a theorem tell us this? i know neither are isomorphic to Z_18 and that Z_18 is isomprhic to Z_9 X Z_2, i just dont know why Z_6 X Z_3 is isomorphic to Z_3 X Z_3 X Z_2

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[west apex]

help idk how to start

[thin kelp]

yo

ok so we want to try and find a base that is comparable between the two

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[desert marlin]

"the above" was a proof that for tree the number of edges is exactly v-1

So I am nearly done with this proof but I am making a small mistake somewhere and I'm not sure what it is.

Here is my proof:

For a collection of disjoint trees with k=1 connected components, then this is just one tree, and by the above, e=v-1=v-k and we have done our base case.

Suppose for some n a natural number that for all k<=n we have that a collection of k disjoint trees with k conneceted components satisfies e=v-k. We will show this implies a collection of k+1 disjoint trees w/ k+1 connected components satisfies e=v-(k+1).

If we remove any 1 disjoint tree from our collection, we remove v vertices and v-1 edges, by the above.

Then we have k disjoint trees left, which by our induction hypothesis satisfy e=v-k

adding back the 1 disjoint tree we removed, we get
e+(v-1)=v+v-k
e=v-k+1

but this is not what I wanted.... because I wanted to show that we would get
e=v-k-1

so I am not sure what I am doing to pick up this wrong sign, because I feel like I am approaching the proof in the right way. Can anyone point out my mistake?

<@&286206848099549185>

@zinc glacier

[zinc glacier]

the main thing thats a little screwy in the proof for me is not using a different variable for the number of vertices in the removed graph

it does not change the sign afaik tho

[desert marlin]

hmph

[zinc glacier]

Another catch: k increases becuase you have one more connected graph

[desert marlin]

it seems like this is the right approach

[zinc glacier]

It does

[desert marlin]

e+(v-1)=v+v-k+1

e=v-k

QED

it is because k increases

[zinc glacier]

Yes

[desert marlin]

and I just need to be more careful with the rest of my labeling

that is satisfactory

damnit waste so much time on the tiniest little detail lol

so annoying

but ty for point that out

.close

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[zinc glacier]

Yw

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[quick falcon]

1. How to set this question up

2. I don’t know if the -2.05 is a z score or a temperature

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[civic isle]

Can i square both to get rid of the square roots at the bottom

Thats a given function

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[fiery ermine]

Prove that $\prod_{d|n} \Phi_d(x)=x^n-1}$ where n is a positive integer.

[clever fjord]

bigpufik

Prove that $\prod_{d|n} \Phi_d(x)=x^n-1}$ where n is a positive integer.
```Compilation error:```! Extra }, or forgotten $.
l.57 Prove that $\prod_{d|n} \Phi_d(x)=x^n-1}
                                             $ where n is a positive integer.
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```

[fiery ermine]

Any ideas, I’ve just started cyclotomic polynomials, I understand that $deg \Phi_n(x) = \varphi(n)$

[clever fjord]

bigpufik

[fiery ermine]

<@&286206848099549185>

Can I state that since $e^{\pi\mathi}=1$ then the primitive root is a zero of x^n-1?

[clever fjord]

bigpufik
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

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[distant spoke]

how to do this

[odd edge]

@distant spoke Has your question been resolved?

[distant spoke]

<@&286206848099549185>

[hollow crane]

@tidal glacier

<@&286206848099549185>

[slender venture]

They are all quardratic eqns

[distant spoke]

.close

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[blissful depot]

Prove that if x,y and z are nonzero rational numbers for which 1/x + 1/y + 1/z = 1 then the number A = (xy/z +1)*(yz/x+1)(zx/y+1) is nonnegative and sqrt(a) is a rational number

[blissful depot]

Still didn't solve it.

here's the actual exercise

<@&286206848099549185>

[shy smelt]

I think I solved it

@blissful depot

if x + y + z is rational at least

Try proving these 2 identities first:

[blissful depot]

x+y+z should be rational

[clever fjord]

Jelle

Jelle

[blissful depot]

the first one I know

but the second one is wierd

Alright, so how do I get to use them?

Do I spread out the whole thing?

[shy smelt]

That's basically what I did, yeah. There might be a quicker way. I also let u = xyz and v = x + y + z

[blissful depot]

Alright so I just spread it out and solve

I'll try that.

I thought it's a pretty unorthodox approach

[shy smelt]

you can rewrite the above identities with u and v as well

try to get rid of all x's, y's and z's

[blissful depot]

Should help.

Thank you

I'll come back to you after I figure it out

[odd edge]

@blissful depot Has your question been resolved?

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@blissful depot Has your question been resolved?

[odd edge]

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[copper quarry]

I need help determining the shape of the graph, how do I do that?

by the way there should be a - sign at the y axis for the 5pi/4

ive plotted the stationary points

but not sure what to do next

[odd edge]

@copper quarry Has your question been resolved?

[copper quarry]

<@&286206848099549185>

wddddddd

[zenith tartan]

water beam

are you sure your derivative of e^-1 sinx is correct?

[copper quarry]

its not e^-1

and yes my stationary points are correct.

[zenith tartan]

yea mb

@copper quarry cant you just plot it like this? (sorry if im being unhelpful im just trying)

[copper quarry]

well i guess but how can I tell the shape?

[zenith tartan]

is it asking for it?

[copper quarry]

i mean i do have to graph f(x) lol.....

how do I graph f(x) if i dont know the shape

[zenith tartan]

like shape of what?, just plot it like you have done here, finding the stationary points and then ploting it from 0 to 2pi?

[copper quarry]

the shape of f(x)... what else lol

[zenith tartan]

also when you put 2pi in f(x) it gives you zero, so why not just join them alltogether and show it, like this is what the question is saying you to do, sketch it from 0 to 2pi, why are you worried about the shape lol

or i didnt get the question

[copper quarry]

dud

if i dont have the shape

how can i draw the graphh.....

you do realize it can be this

or

or

this

or this

like.....

i need the shape of the graph???

what else am i asking lol

[zenith tartan]

OHH

okay

[copper quarry]

bruh

💀

[zenith tartan]

just differentiate it again and see if the value of >0 or <0 ?, does this help?

[copper quarry]

noh

because im not allowed to

i cant use any more calculus than i already have used to find the stationary points

[zenith tartan]

but you did thier derivative = 0 that means at these point thier slope is zero how can you do this then

[copper quarry]

you arent understanding

i cant use anymore calculus

[zenith tartan]

yea i know

[copper quarry]

i was allowed to do it because that was part A

[zenith tartan]

you dont have to use calculus anymore

[copper quarry]

part B i cant use any more

[zenith tartan]

yea dont use it, but its normal that in part A you differentitated it and made them equal to 0, that means at thesse point thier slope is zero, that means this point is either maxxima or minima and the second point is downward you so will take your graph down, and then make its slope also equal to zero, YOU ARE NOT USING ANY MORE CALCULUS, you are just using the readings done in part A

you can definitely use calculations of part A

cant you?

[copper quarry]

yes

[zenith tartan]

then?

[zenith tartan]

did you get this?

[copper quarry]

ok

i did it

[zenith tartan]

btw what day is it?

[copper quarry]

its tuesday

[zenith tartan]

[copper quarry]

😂

[zenith tartan]

i have 0 knowledge about calculus yk, i only studied little of it because it is used in physics🤣

[copper quarry]

lmao

what level of maths u doin

[zenith tartan]

like idk what are you asking but im in grade 11, it is in our syllabus but it is in the end

derivatives in 11 and integrals in 12

[copper quarry]

ohhh i see

anyways ima close this

cyas

.close

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[boreal bay]

How do I solve sqrt(-4) using polar form? What is the r here, and where do i even start? With stuff like sqrt(8), where r=sqrt(8) its intuitive.

[echo ginkgo]

wdym solve sqrt(-4) ?

you want to find the square roots of -4 ?

@boreal bay

[boreal bay]

yes, z=sqrt(-4)

but using polar form

[echo ginkgo]

well this means you're looking for the complex numbers $z$, for which $z^2=-4$

[clever fjord]

aPlatypus

[echo ginkgo]

you ok with that ?

[boreal bay]

im i know that, i put it into de moivres theorm, get r^2(cos(2θ) + isin(2θ))

But what is that r

[echo ginkgo]

re, my internet crashed

it's the r for z

if you suppose z = re^iθ

then z^2 = r^2 e^(i2θ)

[boreal bay]

im i know that, i put it into de moivres theorm, get r^2(cos(2θ) + isin(2θ))
But what is that r

typed that before, dunno if u saw that

[echo ginkgo]

yea I saw it

it's just that, "what is that r" is ambiguous as a question

[boreal bay]

i mean how do i calculate it, how do i get |z| of sqrt(-4)

[echo ginkgo]

but ok now I got you

[echo ginkgo]

and you want this = -4, so r^2 e^(i2θ) = -4

it's the r and theta of -4 we want to find

and then we can equal the r's and the θ's on both sides

[boreal bay]

could u elaborate more? What would theta and r be equal to

[echo ginkgo]

the drawing should help you

r = what's the distance from -4 to the center

theta is the angle in red

[boreal bay]

so r=4, theta is pi

[echo ginkgo]

yeah exactly

so we have r^2e^(i2θ) = 4e^(i*pi)

those two complex numbers are equal, so their modulus and argument should be the same

[boreal bay]

alright

now i got u

it was unintuitive for me, cause when we get like cuberoot(-27), we can easily get z=-3, therefore r=3

and its easier to put onto plane

[echo ginkgo]

so r^2 = 4, 2θ = pi mod 2pi

[echo ginkgo]

well sure z=-3 is a solution there

it's not the only one

[boreal bay]

I know

but that was the way i was grabbing that r

[echo ginkgo]

ah yeah right

[boreal bay]

now i know how to grab it generally, ty

.close

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[civic isle]

How do i find a and b

[nimble blaze]

start with multiplying by conjugate/conjugate

[civic isle]

Right

[nimble blaze]

divide numerator and denominator by x

[nimble blaze]

apply x = sqrt(x^2) for x>0

[civic isle]

Wait i don't t get that part

[nimble blaze]

which part

[civic isle]

This

[nimble blaze]

the property i'm mentioning is for the manipulation of the denominator

what's your issue with that property

[civic isle]

Oh its a property?

Ohhh

Ohhh

Nvm nvm

Yeah i get ot

I can like

Actually nvm idk what to do

Ahh i got it

Becomes this?

[nimble blaze]

split those fractions under the roots

[civic isle]

Oh cool

Thats a lazy 1 but yeah

Do i conjugate it again?

I mean rationalize

[nimble blaze]

no, that would just undo what you did earlier

[civic isle]

I rationalized it

Did nothing?

Yeah

[nimble blaze]

now consider what happens as x→inf

[civic isle]

Correct me if im wronh but since the fraction has a square root only the numerator matters?

Like if the value of numerator is 3 then thats the limit

I mean i said that cuz the limit is 3

So i can just do
ax+2-x +b = 3

And system of equations shiz

Or no

Actually nvm im taking shit

I dont have a value x for system of equations lok

[nimble blaze]

you can't completely ignore the denominator

[civic isle]

Idk what to do

[nimble blaze]

now consider what happens as x→inf

[civic isle]

I get the limit even with the a and b?

[nimble blaze]

limits can be expressed in terms of a,b
yes
but what's the first thing you see happening

it would also help to group / factor the x terms on the numerator

[civic isle]

a and b are constants?

So the limit is 2+b?

Cuz the ax-x will like cancel out

[nimble blaze]

for what value of a will that cancel out

[civic isle]

1

[nimble blaze]

yes

[civic isle]

Ohhhhhh

Shit

Cool

[nimble blaze]

a needs to be 1 for that limit to exist

[civic isle]

Yeah

Ok i celebrated too early

[nimble blaze]

what will the denominator of the fraction be?

as x→inf and when a=1

[civic isle]

It'll be 0?

Or wait

[nimble blaze]

how are you getting 0

[civic isle]

No

1

[nimble blaze]

yes

[civic isle]

Wait i prolly messed up my signs

Its gonna be 1-1 on mine?

The 2/x will be ignored also the -(b/x)

But my denom is 1-1?

[nimble blaze]

1 sec, misread the handwriting

don't know why i didn't see this earlier.

but your denominator at the start was wrong

should've been sqrt() + sqrt()

[civic isle]

Oh

Yeah2

Wait

I fucked up the numerator or no?

[nimble blaze]

numerator was fine

[civic isle]

Mb lol

[nimble blaze]

a still needs to be 1,

denominator will be 2 under those conditions

reducing what's left to finding the value of b where
(b+2)/2 = 3

[civic isle]

b=4

Thanks!!!!!!

The goat

Ill close this now

Lol

Thanks!!

.close

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[shy berry]

Help me please

[shy berry]

Help

[upper charm]

can u show the full question

the weights of the shape

[shy berry]

Yeah

[upper charm]

10+2t

[shy berry]

OOHH

i did 10 + t it wouldnt work

[upper charm]

oh

[shy berry]

Thanks man

How did u know?

[upper charm]

know what

[shy berry]

That its 10+2b

I mean 2t

[upper charm]

in previous one u solved the weight as c+2t

and in next one it gave us a condition where c=10

so just put c=10 in c+2t

[shy berry]

Now that makes sense

I got another one

Im stumped at this

Oh nvm

I solved it

[upper charm]

first put t=3

in 10+3t

replce t with 3

[shy berry]

Its 19 right?

Cuz 10+3+3+3

Okay so im in a new problem and i dont see why i am wrong

Help

Is anyone here?

[upper charm]

ye

[shy berry]

Nice

[upper charm]

what we have to do?

[shy berry]

Write the equation

[upper charm]

3s=15+2s

?

[shy berry]

Ill check

ITS CORRECT

[upper charm]

ok so both sides are balanced

we have 3 cubes on left

one cube weights s so 3 will weight 3s

[shy berry]

Ooohh

Was wondering why my answer isnt correct

.close

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[swift rock]

just find inverse of f(x)

and then let that equal to 1 in Q27 and 2 in Q28

[shy smelt]

Or apply f to both sides of the equation

so f(f^-1(x)) = f(1)

and then simplify this

[swift rock]

oh yeah thats simpler

[wooden python]

jelle beat me to the punch but yeah jtp your suggestion was very circuitous

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[mystic saffron]

is there a way to simplify this?

length - spacing - (spacing * i) + (j * spacing + spacing)
yes its from programming

[wooden mango]

length - spacing * (i - j)

[mystic saffron]

woah thanks, works

[nimble delta]

(j-i)

[mystic saffron]

hm let me try with this

[nimble delta]

i mean, the above answer should be with (j-i) at the end

[mystic saffron]

yeah i know what you meant but it doesnt work

the i - j worked

[nimble delta]

oh nvm, he did minus the whole part

[weak flare]

i - j is correct

[merry finch]

It’s just algebra you can do normal things to it

[weak flare]

yea lol

[merry finch]

a - b - (b * i) + (j * b + b)

[mystic saffron]

👍

thanks

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[lone zinc]

I need to find the equation for the tangent line at the point (c, f(c)) what am i doing wrong for question 31?

[lone zinc]

.close

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[forest shale]

What do these each look like when they're expanded into the summation formula(e) that actually are the expected value function?

[forest shale]

e.g. I know that $E(X)=\int_{1}^{i}x_{i}\cdot L(X=x_{i})\cdot dx$

[clever fjord]

AlexanderJ

[forest shale]

Or, with discrete outcomes, that $E(X)=\sum_{1}^{i}x_{i}\cdot P(X=x_{i})$

[clever fjord]

AlexanderJ

[forest shale]

But that's just $E(X)$

[clever fjord]

AlexanderJ

[forest shale]

What do the expansions in the screencap (above) expand to?

[odd edge]

@forest shale Has your question been resolved?

[forest shale]

pls halp

[odd edge]

@forest shale Has your question been resolved?

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@forest shale Has your question been resolved?

[odd edge]

@forest shale Has your question been resolved?

[weary pelican]

If $X$ is a random variable and $f$ is a continuous function, then $E[f(X)]=\int f(x_{i})\cdot L(X=x_{i})\cdot dx$ or $E[f(X)]=\sum_if(x_{i})\cdot P(X=x_{i})$

[clever fjord]

rafilou2003

[weary pelican]

This is called the transfer theorem @forest shale

[pastel dew]

i only think that f must be mesurable (?)

[weary pelican]

yes but I don't think they're at the level of understanding measurability

[pastel dew]

ah ok

rafilou do you know algebraic geometry ?

[weary pelican]

a bit

[pastel dew]

do you know stuff like morphism between agebraic varieties ?

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[austere rover]

SOLVED

[austere rover]

If I had u = (u1, u2) and v = (v1,v2)

would $u \oplus v$ be $(u_{1}v_{1}, u_{2}v_{2})$?

[clever fjord]

LeGM

LeGM

[austere rover]

or would it be like matrix multiplication?

OH NVM

i should just treat them as numbers

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[austere rover]

how to say let $u \in R$ and $v \in R$ and $w \in R$ without using too many "and"

[clever fjord]

LeGM

[low locust]

u,v,w in R

[austere rover]

thanks

.close

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[karmic bloom]

how can I prove that the following equality is true, no matter what values I put inside of it which are rational:

max(x, max(y, z)) = max(max(x, y), z)

[karmic bloom]

max(x,max(y,z)) = max(x,y,z)
max(z,max(x,y)) = max(x,y,z)

could I write something like this and just say that they are equal because of this?

:))

[jagged maple]

I’d say check for x >= y >= z, x >= z >= y, …

[jagged maple]

Then prove that every equation is equal to max(x,y,z)

[south plume]

you can go from the fact that inequality is transitive

y > z, z > x, therefore y > x

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[karmic bloom]

thanks man

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[plain sun]

can someone explain this using trig rules?

.close

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[zenith parcel]

why is this wrong?

[noble forge]

show your working

[zenith parcel]

oh nevermind i figured it out lol sorry

thanks anyway

.close

[odd edge]

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[odd edge]

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[topaz forum]

Given is the function 𝑓 with prescription 𝑓(𝑥) = arctan( 𝑥/a)
Determine 𝑎 such that the tangent to the graph of 𝑓 at the point 𝑃 (1/2,f(1/2)) is parallel with the line 𝑙 -> 2𝑦 − 7𝑥 + 17 = 0.
this is what I already have, but the thing is that the discriminant is negative, so yeah idk what to do now

[topaz forum]

<@&286206848099549185> Someone that can help with a cyclometric question ?

[odd edge]

@topaz forum Has your question been resolved?

[odd edge]

@topaz forum Has your question been resolved?

[odd edge]

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[tall ridge]

Help me to solve these problems.

[tall ridge]

<@&286206848099549185> save my life plz

@errant bronze Bro help me

[nimble blaze]

!patience

[odd edge]

Please wait patiently, and do not interrupt other channels with your question. Helpers in this server are volunteers, and the server cannot guarantee that someone will be able to help you. By being impatient or begging, you will only turn potential helpers away.

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[nimble blaze]

!noadvert

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[nimble blaze]

don't ping individuals without permission, especially mods for help

[tall ridge]

ok

[odd edge]

@tall ridge Has your question been resolved?

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@tall ridge Has your question been resolved?

[tall ridge]

<@&286206848099549185>

[odd edge]

@tall ridge Has your question been resolved?

[raw wyvern]

what is the definition of free? @tall ridge

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@tall ridge Has your question been resolved?

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Available help channel!

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[next bobcat]

Hey, I need some help with this question, I don’t really quite understand it other than the rest of the problems on this one. Thank!

[cold sage]

how many gallons is 1 litre

[next bobcat]

How would we find that?

[cold sage]

treat it like 1*g=3.78541*l, rearrange for l

same idea as if a=5b then b=a/5

[next bobcat]

I’m Ngl, I’m stuck 😭

Wait so

1* 3.78541*l?

That would just give me 18.927

[cold sage]

1 gallon is 3.78541 litres
hence it stands that
1 litre is (1/3.78541) gallons

[next bobcat]

Oh

0.264 * 5 to find the litres

Which would just be 1.3 or 1.320

So it’s 1.3g

[cold sage]

id go with 1.32 gallons but yeah

if you write g people will think you mean grams

[next bobcat]

Oh, alright now it makes sense.

Yeah

Thanks!

.close

[odd edge]

Channel closed

Closed by @next bobcat

Use .reopen if this was a mistake.

[next bobcat]

.reopen

[odd edge]

✅

[next bobcat]

Also I wanted to ask, are either of these correct?

[cold sage]

rounded wrong in the first one
as for the second, you havent answered it yet

[next bobcat]

10,582.2

For the 2nd

Oh and the first one, 35.27392 or ig 35.27

[cold sage]

no

[cold sage]

yeah

[next bobcat]

?

[cold sage]

why are you dividing anything

you know what 1g is in oz

youre converting 30000g to oz

[next bobcat]

So I would have to multiply, but I just did and it was wrong

[cold sage]

because you did the multiplication wrong

you wrote 0.35274, thats not what it says

[next bobcat]

Oh

1,058.22

[cold sage]

yeah

[next bobcat]

Alr, for this one. I am confused..

For the 3000c I did 3000/91.44 but that wouldn’t really help I think

Wait would it be 91.44/3000 giving me 0.03048

[cold sage]

you were right the first time

[next bobcat]

For the meter one would it be .9144*40?

[cold sage]

no

[next bobcat]

Oh

[cold sage]

how many yd is 1 m

[next bobcat]

1.09?

[cold sage]

actually

[next bobcat]

1/.9144

[cold sage]

yeah

so 40m is

[next bobcat]

43.74?

I did 1.09*40

[cold sage]

thats the one

[next bobcat]

Alright

What about this one?

[cold sage]

theres not enough information to do that question

is there an overarching main question this is part of?

[cold sage]

ah nvm, i see

[next bobcat]

Alright uh

Yeah

You would sadly have to go back and forth

Here let me like

[cold sage]

well, first you need the area of the floor in feet^2

[next bobcat]

Where do we even start first off

9.33 yards?

[cold sage]

get the area in yd^2 then convert to ft^2

[next bobcat]

This is gonna sound dumb but like… I forgot what ‘^’means

[cold sage]

exponent

$$a^2$$

[clever fjord]

AℤØ

[cold sage]

is what a^2 would mean

[next bobcat]

Ohhh

To the third

Or to the second

In that case

so the area of yrd^2 then convert it to feet and ^2?

[cold sage]

you cant convert yd^2 to ft, not possible

you can convert yd^2 to ft^2

[next bobcat]

I’m stuck.

So wait

We would first have to find the 10sq to yrds

Then when we have the yrds we would have to ^2?

[cold sage]

i mean, you could? Id literally just directly convert the yd^2 to ft^2 though

1yd=3ft, 1yd^2=9ft^2

[next bobcat]

OHHHH

See

I was right

81

But I got 81 on the dot this time, when I got the 9.3 all I got was 81 and more decimals. It was 87 nvm.

So short answer it’s 81?

[cold sage]

,calc ((((3000/91.44)*(40/0.9144))*9)/10)/60

[clever fjord]

Result:

21.527820833419

[next bobcat]

Oh nah

[cold sage]

whered you get your values from

whats the area in yd^2

[next bobcat]

Alright ima be honest how did we find 21.5

[cold sage]

^

[next bobcat]

Wait but

Wait so

3^2 then 9^2 81

[cold sage]

what

[next bobcat]

Was that not

[cold sage]

where is 81 even coming from

[cold sage]

all i said was this

you still need the area in yd^2 to even use that

[next bobcat]

Yeah I’m lost

[cold sage]

^

[next bobcat]

Wait do you mean by that

[cold sage]

the area of the floor

what is it in yd^2

[next bobcat]

So the area is 10

[cold sage]

?

[next bobcat]

Ok

[cold sage]

not even close, how did you get 10

[next bobcat]

The area of the floor is 3000c

[cold sage]

no, it is not

[next bobcat]

But also by 40m

Oh

[cold sage]

cm is not a unit of area

[next bobcat]

32.80?

[cold sage]

how did you get that

[next bobcat]

I am actually like lost

Where do we even find the area first off

[cold sage]

the floor is 3000cm by 40m, you found both of those values in yd already

[next bobcat]

Oh

[cold sage]

multiply them

[next bobcat]

1,434.672

[cold sage]

more specifically 1435.188056

which is why you shouldnt use rounded results in calculations

but moving on

what is 1435.188056yd^2 in ft^2

[next bobcat]

So now we do 1435.1880^2?

[cold sage]

Why would you do that?

[cold sage]

Youre doing this

[next bobcat]

Wait so wdym by that..?

[cold sage]

Thus

[next bobcat]

Isn’t 1yd^2 just one tho?

[cold sage]

Idk what youre talking about

[cold sage]

We have this many yd^2

So convert to ft^2

[next bobcat]

So convert that to ft?

[cold sage]

Thats not possible

[cold sage]

I gave you this

Just use it

[next bobcat]

Ok

[cold sage]

Convert yd^2 to ft^2

[next bobcat]

One yard = 3ft

So 1yrd^2

=

9ft^2

[cold sage]

Yeah

[next bobcat]

So 81

[cold sage]

How are you doing that

[next bobcat]

9^2?

[cold sage]

Why are you doing 9^2

[next bobcat]

Cuz 9ft^2

[cold sage]

My man

No

[next bobcat]

😭

[cold sage]

You just told me this right

[next bobcat]

Yes

[cold sage]

1yd^2=9ft^2

[cold sage]

You have 1435.188056yd^2

Convert to ft^2 using what you just told ne

[next bobcat]

Wait first off wdym by like yd^2

Do you mean like power of 2

Or are we talking about something different

[cold sage]

Yards squared

[cold sage]

You literally are just multiplying this by 9 thats it

[next bobcat]

12,916.692

?

[cold sage]

Thats the area in ft^2 yeah