help-13

[crimson sedge]

ok

got it

[frozen rock]

let me rephrase z > 0

[crimson sedge]

ok , thanks

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[cedar kiln]

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[fast verge]

Pls help!

[floral bridge]

ok let's start 🙂 there's a couple of parts to this 🙂

[fast verge]

Ok!!

[vapid sand]

can you close your other one yoshi

[floral bridge]

you know the relationships of the gradients of the tangents and normal?

[vapid sand]

thanks

[fast verge]

Yes

And I know that you have to differentiate y=kx^2/3

[floral bridge]

yups, and what do you get?

[fast verge]

So you should get dy/dx=3/2kx^1/2

[floral bridge]

nice, let's leave this here first, and head to the straight line equation

[fast verge]

Yup ok

[floral bridge]

we wanna to get the gradient of this straight line so that we can find the gradient of the curve at Point P.

can you get the gradient for this straight line?

[fast verge]

Is it -2/3

[floral bridge]

yups, now that we know this gradient is -2/3 (which is the normal)

so what should be the value of dy/dx of the curve at point P?

[fast verge]

Umm 3/2?

[floral bridge]

yes!

[floral bridge]

replace this value of 3/2 back into here

[fast verge]

Ohhhhhh so then does this equal 3/2

[floral bridge]

bingo

[fast verge]

Ok hang on

Ohhh

I see ok then that equals a half!!! Ty

[floral bridge]

yay!

you know the rest?

the circle stuff

[fast verge]

Hmmm I think so but can you check with me?

[floral bridge]

sure what do you have?

[fast verge]

So if x=4 then y=4 in the equation

Then do you have to make a new equation at point Q by using the same gradient of p

So for Q: y=-2/3x + b

[fast verge]

Then using this you can find b

[floral bridge]

what you wrote is not exactly correct, but yes you are getting the idea

yups then use the equation that you get to solve for point Q.

[fast verge]

I got the equation y=-2/3x +20/3 for Q but from there how do i find x and y now?

[floral bridge]

nicely done on the equation

there's clues in the questions about point Q hahah

[fast verge]

OHHHHH

so x =0

[floral bridge]

nope

[fast verge]

No sorry

I mean y

So x = 10

[floral bridge]

voila

[fast verge]

Okkk I think I can work out the rest, thank you :))

[floral bridge]

awesome! gl!

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[cedar kiln]

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[stark lily]

can someone check this proof for me pls

[stark lily]

https://i.imgur.com/mmWJ0Fw.png

pf:
Set $w \in W, a \in U_1, b \in U_2$. Since $U_1 \oplus W$ is a direct sum, let $w + a$ represent the unique sum resulting in $(0)$.
$w + a = (w_1 + a_1, \ldots, w_n + a_n) = (0_1, \ldots, 0_n)$
$w = (w_1, \ldots, w_n) = (0 - a_1, \ldots, 0 - a_n)$
But $W + U_2$ is also a direct sum of two subspaces with a unique $w + b = (0)$.
So $w + b = (0 - a_1 + b_1, \ldots, 0 - a_n + b_n) = (0_1, \ldots, 0_n)$.
Hence $(a_1, \ldots a_n) = (b_1, \ldots b_n), and U_1 = U_2$

[wraith dagger]

Suro

[cedar kiln]

@stark lily Has your question been resolved?

[crimson delta]

if w+a=0 then w=a=0 (because clearly that choice works and uniqueness) and similarly b=0. yes in that particular case a=b

but what about it

you cant pick w and a first arbitrarily and then assume that they have to sum to 0

thats not an arbitrary choice of a

also just because V is a vector space does not mean that it is finite dimensional and that you can write its elements as tuples

[cedar kiln]

@stark lily Has your question been resolved?

[stark lily]

ah, yeah true

I think my entire line of reasoning was based on picturing this in $\mathbb{R}^3$

[wraith dagger]

Suro

[stark lily]

and assuming that each element in a vector is either 0 or some arbitrary real number

and then that lead to the conclusion that the $w$ I took as one that sums to the 0 vector for $U_1$ has to be the same "shape" as the w that sums to the 0 vector for $U_2$

[wraith dagger]

Suro

[stark lily]

but yeah obviously all of that is pretty bad

@crimson delta I'm struggling to picture this outside of the finite real numbers though :/ any tips on being able to consider this more abstractly?

ok its 2am I'm gonna give this another go tomorrow lol

thanks for the help

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[visual mirage]

can someone check my homework plz?

[gritty galleon]

Okay, hold up

[rustic siren]

https://www.calculator.net/triangle-calculator.html

put them in here

and it gives you the solution

[gritty galleon]

yeah

also, can u show ur working for these questions? @visual mirage

[visual mirage]

whats this

[visual mirage]

wym

[rustic siren]

u give it the sides and angles(the ones you know)

[gritty galleon]

U must have done some sort of work for calculating the answers right

[rustic siren]

and it gives you the rest

[gritty galleon]

Not helping

[rustic siren]

i do

wdym

[gritty galleon]

Nthing, sry

[visual mirage]

i did this assignment a while ago, i will prolly have to look for my notebook

ill look for it now if u rlly need

[gritty galleon]

Its okay then, you can try using @rustic siren's suggestion

[gritty galleon]

I can still check it if you want tho

[visual mirage]

ya ofc

[rustic siren]

for example the second picture has incorrect answer

[gritty galleon]

whats k?

theres a vertex called K

but no length/angle mentioned

[rustic siren]

but use this site only for checking the answers, it wont help much, if you use only this and forget the formulas and stuff

opposite to "uppercase letter" is a side "lowercase letter"

A->a B->b etc.

[gritty galleon]

Ah thats what it meant

[visual mirage]

is this right?

180°-60°-66°=54°

x/sin66° = 280/sin54°

x = 280sin66°/sin54°

= 316.177 = 316

y/sin60° = 280/sin54°

x = 280sin60°/sin54°

= 299.730 = 300

Length of wires: 316m, 300m

[rustic siren]

yes

[visual mirage]

i need help w this

[hallow quartz]

soh cah toa

sine rule

cosine rule

[cedar kiln]

@visual mirage Has your question been resolved?

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[shell hearth]

Wouldnt that make x = 2u?

how is it 4u then?

[hallow quartz]

x is equal to 2u so when squared it’s equals 4u^2

[hallow quartz]

where does it say 4u

[shell hearth]

in the last line

when we usub dont we just replace the x with the u ?

like replace all the "x" ?

[jaunty mural]

f(x) = x^2

f(1+1) = 1+1^2 = 2 !?

f(2t) = 2t^2 !?

[shell hearth]

oh

ye the last one is correct

ig

i still dont get how was it done in the question

where is the f(x) in the question?

[jaunty mural]

i made up f out of thin air

[jaunty mural]

my point is brackets are required to make a substitution

[jaunty mural]

.

[shell hearth]

wait lemme get it right
now we have (4-x^2)^1/2
and x = 2u

Then what should i do

to substitute

[jaunty mural]

not this.

f(x) = x^2
f(2t) = (2t)^2

[shell hearth]

how if there is a "-" ?

[jaunty mural]

it doesnt make a difference

If you substitute with brackets in place, the substitution will be correct

[jaunty mural]

You "replace" x with (2u)

[shell hearth]

yes so (4-(2u)^2)^1/2

?

[jaunty mural]

ok.

[shell hearth]

correct?

<@&286206848099549185>

[cedar kiln]

@shell hearth Has your question been resolved?

[cedar kiln]

@shell hearth Has your question been resolved?

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[worthy ridge]

any tips on how to get started?

[cedar kiln]

Please don't occupy multiple help channels.

[worthy ridge]

I don't even understand how these are linear transformations

aren't these SLEs?

to check whether or not something is a linear transformation

I am told to check these axioms:

for $\Theta: V \rightarrow W$, that $\Theta(\lambda v)=\lambda\Theta(v)$

$\Theta(v+w)=\Theta(v)+\Theta(w)$

[wraith dagger]

normalAtmosphericPa=101,325

normalAtmosphericPa=101,325

[worthy ridge]

so respects vector addition and scalar multiplication

[autumn fox]

A transformation f : ℝⁿ → ℝᵅ is linear if
L1: f(x + y) = f(x) + f(y) for x,y ∈ ℝⁿ
L2: f(ξx) = ξf(x) for x∈ ℝⁿ and ξ ∈ ℝ

You already wrote that

Damn

[worthy ridge]

holy

[autumn fox]

Sorry, was in my own little mind there

[worthy ridge]

imagine typing that without LaTex

seems tiring just thinking about it hahaha

[autumn fox]

Try testing A. for the two requirements

[worthy ridge]

I am so confused lol

I have been given two equations right?

and 4 variables /unknowns

$y_1=x_1+5$

[wraith dagger]

normalAtmosphericPa=101,325

[worthy ridge]

and $y_2=x_2$

[wraith dagger]

normalAtmosphericPa=101,325

[tropic oxide]

i think these are linear maps written in a format reminiscent of y=f(x)

[worthy ridge]

I see

so I've been given sets of two maps?

[autumn fox]

If x = (x₁, x₂) and z = (z₁, z₂), is L1 satisfied? That is, is f(x + z) = f(x) + f(z)?

[tropic oxide]

no

the x's are input coordinates and the y's are output coordinates

the way i understand it

[worthy ridge]

wat

this notation is completely new to me lol

[worthy ridge]

so how do I use the given notation and check with these axioms?

[autumn fox]

Let f be the name of the transformation. For A. we have

f(x + z) = [(x₁ + z₁) + 5, (x₂ + z₂)]

Try checking for f(x) and f(z) now

[tropic oxide]

T_1 [x1; x2] = [x1 + 5; x2]
T_2 [x1; x2] = [4x1; 3]
T_3 [x1; x2] = [0; x1*x2]
T_4 [x1; x2; x3] = [8x2; -10x3; -4x1]
T_5 [x1; x2] = [2x1 + x2; -x1]
T_6 [x1; x2; x3?] = [-7x1; 9x1; 10x1]

these are your maps written in a more commonplace manner

[worthy ridge]

ohhh

like vectors / matrices

wait

[worthy ridge]

so something like $\begin{bmatrix}x_1 \ x_2 \end{bmatrix} = \begin{bmatrix}x_1+5 \ x_2\end{bmatrix}?$

[wraith dagger]

normalAtmosphericPa=101,325

[tropic oxide]

no

have you like... worked with linear maps before

[worthy ridge]

sort of (maybe not)

[autumn fox]

$T_{A}(\mathbf{x}) = T_{A}\begin{bmatrix}x_1 \ x_2 \end{bmatrix} = \begin{bmatrix}x_1+5 \ x_2\end{bmatrix}$

[worthy ridge]

Is that left multiplication by the linear map?

or the matrix representation thereof

[autumn fox]

Just a name for the transformation

[wraith dagger]

Mikkel

[autumn fox]

Just like writing f(x) like you usually do for a transformation with name f of 1 variable

[worthy ridge]

I see

so this is basically an SLE isn't it

written as a matrix

[autumn fox]

I'm not sure what SLE means, to be honest

[worthy ridge]

system of linear equations

sorry my bad

[autumn fox]

Well, if they are linear or not is what we want to find out

[worthy ridge]

system of equations then

lol

wat

wait

that is so unintuitive

becuase normally when I see an equation written in the form "y=ax+b"... that to me is linear

in the elementary algebra sense

or is this one of those things in math where "linear" can mean different things

[autumn fox]

That's because there are two different definitions of what is linear

Depending on context

[worthy ridge]

is it one of these moments

or is it like

[autumn fox]

First we are introduced as linearity being things that "look like lines", later on we meet the definition usually introduced in linear algebra

[worthy ridge]

are they related

or completely different concepts

whereby mathematicians have decided to just recycle the same word

[autumn fox]

Well, linear transformation looks like lines (in the dimensions we can graphically comprehend)

They just go through the origin

In linear algebra, something on the form ax + b is called affine

[worthy ridge]

isn't a linear transformation fundamentally a "function" though?

[autumn fox]

And can be looked on as something linear, but where the "origin is shifted"

[worthy ridge]

I don't understand how a "function" can be visualised"

[autumn fox]

Transformation and function are synonymous

[worthy ridge]

like

[autumn fox]

A graph

[worthy ridge]

isn't the graph the visualisation of the image thopugh

i.e. the parabola's curve in the cartesian plan is the set of all images of the function

[autumn fox]

A function is a set of ordered pairs, which you can visualize in a coordinate-system

[worthy ridge]

oh true

set of all ouputs as well as inputs

[autumn fox]

Yes

Well, if you want to get back to it, you need to check for linearity of A by evaluating f(x + z) and f(x) and f(z) for x,z being vectors in the domain, to see if f(x + z) = f(x) + f(z)

[worthy ridge]

wait wait wait

what if

the input and output of teh function are in completely different spaces

i.e. f: X --> Y

X is the cartesian plane

Y is the complex plane

how would the geometric representation of such a function (or analogous) work?

i.e. the set of all inputs and outputs

[autumn fox]

f : ℝ → ℂ with f(x) = x² + xi can probably be modelled by using a 3-dimensional coordinate system

Haven't tried it though

[cedar kiln]

@worthy ridge Has your question been resolved?

[worthy ridge]

wow chatgpt is surprisingly useful here

[cedar kiln]

@worthy ridge Has your question been resolved?

[worthy ridge]

halppp

I'm still so stuck

Do i do this?

and conclude that A. is not a linear transformation

[autumn fox]

You can collect the final vectors in the second line into 1 vector to make it more apparent

But yes

[worthy ridge]

so I was correct to introduce another vector $\begin{bmatrix}x_3\x_4\end{bmatrix}$?

[wraith dagger]

normalAtmosphericPa=101,325

[worthy ridge]

because in context of that question, I basically just made it up

[autumn fox]

Yes. It is a bit unorthodox to name it like that, when checking for linearity, but not less correct nontheless

[worthy ridge]

Is the second part correct?

second aximos

[autumn fox]

We usually name the separate vectors with names like x, z, w, v and so forth, and their respective entries with (x1, x2, x3, ..., xn)

[worthy ridge]

I see

so conventionally, I should've done like y1 y2 for the second vector

or w1 w2

[autumn fox]

Yes. But for it to be linear, both requirements of the definition have to be satisfied

So if the first requirement is false, you don't have to check for the second

[worthy ridge]

wait I don't understand how this works thoughj

[autumn fox]

Yeah

[autumn fox]

Wait

[autumn fox]

That is wrong, yes.

It should be λx1 + 5 at the top

Not λ(x1 + 5)

[worthy ridge]

why though

[autumn fox]

Just like when you have a transformation from one variable, say f(x) = x² + 5, you would evaluate f(λx) by replacing x with λx, that is f(λx) = (λx)² + 5

[worthy ridge]

OOPS

hmmm

because it scales all inputs right

and constants are not affected

[autumn fox]

With f(λx) we scale each entry of x by λ and transform this new vector by the transformation f. With λf(x) we transform x by f and scale this transformed vector by λ.

[worthy ridge]

yeah

thanx

[worthy ridge]

Just to confirm: is this correct then?

the part that I am concerned about is the constant

because for B., the component 3 is always mapped to 3 right?

[autumn fox]

x2 is always mapped to 3

[worthy ridge]

so is $3\lambda$ mapped to $3\lambda$?

[wraith dagger]

normalAtmosphericPa=101,325

[autumn fox]

B. says that $\begin{bmatrix} y_{1} \ y_{2} \end{bmatrix} = \Theta \begin{bmatrix} x_{1} \ x_{2} \end{bmatrix} = \begin{bmatrix} 4x_{1} \ 3 \end{bmatrix}$

[wraith dagger]

Mikkel

[worthy ridge]

OOPS

[autumn fox]

$\begin{bmatrix} y_{1} \ y_{2} \end{bmatrix} = \Theta \begin{bmatrix} x_{1} \ x_{2} \end{bmatrix} = \begin{bmatrix} 4x_{1} \ 3 \end{bmatrix}$

[wraith dagger]

Mikkel

[autumn fox]

Hmmm

Pretend those are columns

And not rows

[cerulean sail]

$\pmqty{y_{1} \ y_{2}}$

[wraith dagger]

@cerulean sail

[cerulean sail]

They've got shortcuts like that to make your life easier btw

[autumn fox]

LaTeX is bullying me on a daily basis

[cerulean sail]

Trust me, I was so infuriated to learn they had those already built in here did them the long way only to see those and be like

[worthy ridge]

like this?

[cerulean sail]

equal are they?

[worthy ridge]

so as a general observation, whenever the rule of the linear transformation contains a constant in the image, the second axiom will always fail?

[autumn fox]

Also a side-note, I wouldn't really call these axioms. More like requirements for the definition of linearity to be met

I don't know if you call them axioms in the US, though

[worthy ridge]

I'm not from the US

and you're right

I think it's just in the definition

[cerulean sail]

[except for zero of course]

[worthy ridge]

ah becuase it always preserves (vectors on the) lines through the origin

[autumn fox]

Generally for a linear transformation, the 0-vector always have to map to a 0-vector.

t(0) = t(0 + 0) = t(0 + (-1) ∙ 0) = t(0 + (-0)) = t(0) + t(-0) = t(0) + t((-1) ∙ 0) = t(0) + (-1) ∙ t(0) = t(0) + (-t(0)) = 0

Okay, that may have used a bit too many steps than necessary. But it's always good to practice your discrete math

[worthy ridge]

Oh yeah right

lol

[autumn fox]

But yeah, this can sometimes be used to check for linearity instead

For B you have that B(0, 0) = (0, 3)

Which is equivalent to it being non-linear

[worthy ridge]

rightttt

[autumn fox]

Be careful with this though. T(0, 0) = (0, 0) does not necessarily make T linear. Only T(0, 0) ≠ (0, 0) tells you something about the linearity of T

If I remember correctly

chartbit is my source now

[cerulean sail]

counterexample e.g. T(x1, x2) = (x1^2, x2^2)

[autumn fox]

And I trust that source

[cerulean sail]

It's like the "divergence test" for series, where having the limit of the sequence being zero is necessary but not sufficient for the series to converge

[autumn fox]

I'm excited for my second Analysis course, where we're gonna have about sequences. I feel like I'm not part of the cool-kids club everytime they are talked about

[worthy ridge]

I vaguely rememeber it being covered in calculus

doing real analysis next sem

rlly scared

will probs bash through Rudins in the sem break before thoe

[cerulean sail]

Honorary member

[cerulean sail]

Real analysis

[worthy ridge]

Ya think it's possible to learn real analysis in 30 days

from rudins

10 hrs a day training or someth

[cerulean sail]

Principles of mathematical analysis?

[worthy ridge]

yeah

Maybe with MIT OCW

[cerulean sail]

Hmm what have you covered so far for analysis?
If you've covered like "normal" real analysis then it should be doable (so e.g. Cauchy/convergent sequences and stuff like that, in the context of real numbers and that) but if not then

[autumn fox]

I'm having my exam in Analysis in 2 weeks

Just have to remember 12 proofs. It's not that bad

[worthy ridge]

I'm doing multivariable calculus in the same semester

implications are that I haven't done it yet

[cerulean sail]

Daaaammnnnn best of luck, sure you'll be alright of course if there aren't many proofs to remember

[cerulean sail]

oof hmmm well I mean, anything is doable with enough effort

[cedar kiln]

@worthy ridge Has your question been resolved?

[cedar kiln]

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[spare carbon]

does anyone know what it means by upper limit

?

i am aware on the normal formulas i am just unsure which to use

[cedar kiln]

@spare carbon Has your question been resolved?

[spare carbon]

<@&286206848099549185>

[royal loom]

If you can construct the proper interval, the upper limit will just be the very top value of said interval

for example, if 0<x<99

99 is the upper limit

[spare carbon]

so just the larger number?

[royal loom]

The larger number that sets the roof of the interval

[spare carbon]

so in this example

6.0 would be the upper?

[royal loom]

yeah, or 5.98 if you don't want to round

[spare carbon]

(not sure why it was rounded but whatev)

ok that helps

[spare carbon]

got any input on this?

unsure on what it means by "true mean"

[royal loom]

<@&268886789983436800> User keeps going to various other channels to ask for help and ignores multiple replies instructing to read #❓how-to-get-help and #rules

[spare carbon]

so uh

any idea?

[spare carbon]

@royal loom any idea on what the question is asking for in "true mean"?

i know its a thing but i havent seen it until now

[cedar kiln]

@spare carbon Has your question been resolved?

[cedar kiln]

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[crimson imp]

Hi

[crimson imp]

How do I do this

[lone dune]

do you know how similar triangles work?

[crimson imp]

I did 2x-6/20=39/x+8

[lone dune]

yeah thats the first step

then you can solve for x from that equation

[crimson imp]

Then I got 2x^2+10x-48=780

[lone dune]

yeah you can solve the quadratic equation

from there

to get 2 values of x

and only one of them would work

[crimson imp]

I did factoring and I was confused

[naive totem]

Problem 9

If the cardinality is 2, why are there three elements listed out?

Can someone explain this to me?

[crimson imp]

?

[lone dune]

#❓how-to-get-help

this channel is already taken

[lone dune]

what did you get?

[crimson imp]

2x^2+10x-828=0

I have never do big number factor before

[lone dune]

well

you can factor out 2

and get 2(x^2 + 5x - 414) = 0

do you know the quadratic formula?

[crimson imp]

Ohh I didn’t know you can simply it

[crimson imp]

Yes but wouldn’t the number get even bigger because it would square the number?

[lone dune]

well the quadratic formula would be easier than factoring in this case

you just need to do a bit of calculations

[crimson imp]

Okay let me try that

18 and -23

The positive one is the answer right?

@lone dune

[lone dune]

yeah

because you cant have a negative side length

[crimson imp]

Thank youu

.close

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[patent moss]

can someone walk me through this

[foggy merlin]

what have u tried ?

[patent moss]

i have base case of n=1 and thats true

\

im not sure if this is right

[foggy merlin]

add $\frac{1}{\sqrt{n+1}}$ on both sides

[wraith dagger]

Ｈｅｒｅｌｓ

[foggy merlin]

you'll have $\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} \geq \sqrt{n} + \frac{1}{\sqrt{n+1}}$

[wraith dagger]

Ｈｅｒｅｌｓ

[foggy merlin]

$\sqrt{n} + \frac{1}{\sqrt{n+1}} = \sqrt{n} + \frac{\sqrt{n+1}}{n+1}$

[wraith dagger]

Ｈｅｒｅｌｓ

[foggy merlin]

and of course this thing is greater than sqrt{n+1}

[cedar kiln]

@patent moss Has your question been resolved?

[patent moss]

.close

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[spring tide]

Hi I just need someone to check my answer

[spring tide]

$$\int \frac{5}{4x^2 + 4x - 1}dx = \int 5(4x^2 + 4x - 1)^{-1}dx$$

$$ = \frac{5\ln(4x^2 + 4x - 1)}{8x + 4}$$

I made up a random question to practice but when I try and put it into Mathway it won’t give me an answer

So I need someone to check for me

[wraith dagger]

Nomzz1

[jade sonnet]

That’s not how these things work, you can’t divide through by the derivative like that, if you differentiate it it won’t give you back the original function you integrated. I’d try partial fractions

[spring tide]

Why not though, is that not the reverse of the chain rule?

[high spruce]

yeah I see what you did, you can only do that with a constant

[spring tide]

Ah

[high spruce]

you can't do the same thing with variable

[spring tide]

I’ll close the channel for now then come back when I’ve worked out an answer

.close

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[latent fox]

is there a multiply sign missing here

[slow thicket]

its implied but yes

[cedar kiln]

@latent fox Has your question been resolved?

[hot iron]

what do you need help with exactly

[slow thicket]

they already said

[hot iron]

the question regarding if there's a multiply sign?

@slow thicket

[slow thicket]

yes

[cedar kiln]

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[limber basin]

Hi! Does anyone know how to simplify this?

[alpine bone]

do you know what is the conjugate of sqrt(x) - 1

[limber basin]

sqrt(x) + 1?

[alpine bone]

yes of course

expand the fraction by sqrt(x) + 1

[limber basin]

I did that

on the bottom i got x-1

[alpine bone]

great

x - 1 on the top and x - 1 on the bottom cancels out

so you get sqrt(x) + 1 left

[limber basin]

oh so i dont need to expand the top?

[alpine bone]

yes

if you can cancel out them

[limber basin]

ohhh

[alpine bone]

why would you need to expend them?

[limber basin]

I see so its sqrtx + 1

[alpine bone]

yes

[limber basin]

thank you

[alpine bone]

you're welcome

[limber basin]

For this one

since the bottom is like difference of squares

can i do (x-1)(x+1) in a sqrt?

like this

.clos

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[thin cape]

my kindergarten teacher told me look at my shoes and add them but im still stuck i said 8 but its wrong someone please help

[granite knoll]

<@&268886789983436800>

[fair geyser]

why is it important that you answer correctly

is your teacher threatening you?

[thin cape]

yes

[thin cape]

r u snitching for asking a question??

[fair geyser]

you should pee on their belongings

when they leave thier desk, if they leave something behind, pee on it

if they literally never leave anything, worst case you can pee on the desk

[thin cape]

ah ok

thanks man

what say when they call mama

[fair geyser]

tell on them

they threatened you

[thin cape]

ok

ty

.close

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[dull oxide]

Test?

[flat birch]

no just a review sheet

[violet rapids]

who got they phone on a test

[flat birch]

ong 😭

[dull oxide]

for the third one, you can use power of a point theorem

[flat birch]

uhh whats that

[dull oxide]

https://en.wikipedia.org/wiki/Power_of_a_point

[flat birch]

could u walk me through it

plz

[dull oxide]

For second problem, mark O as the center of the circle, and A at the x angle. Then draw the segments OW, OU, and OA

[cedar kiln]

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[sweet pasture]

wait nvm i figured out how to do it

wait

it's use the quotient rule right?

then sub in two

I think

[granite knoll]

try it, and see what you get

[sweet pasture]

9/4

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[upbeat loom]

So every derivative in this must be in respect to time im guessing, is that correct? this is not explained well in the book or in the lecture

[upbeat loom]

this is the question

[cedar kiln]

@upbeat loom Has your question been resolved?

[tame wraith]

chain rule

df(x)/dt = df(x)/dx * dx/dt

[upbeat loom]

thanks, so basically, every derivative taken, needs to end up in respect to time for these types of problems?

[tame wraith]

yh

[upbeat loom]

thanks so much

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[slim gorge]

how to turn sin(pi/3-a) into a form that can be evaluated with cosa/sina??

[fading hazel]

compound angle formulae

[slim gorge]

.close

[cedar kiln]

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[slim gorge]

thanks

forgot for a sec

[cedar kiln]

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[lost whale]

Can anybody help me with this, been on it like for 30 mins

[cedar kiln]

@lost whale Has your question been resolved?

[hallow quartz]

@lost whale

hey

[lost whale]

hi

could you help me pls

its been 30 mins bruh im so stuck

[hallow quartz]

yes

ha all good

[lost whale]

thank you so much man

i figured out that the d value is -20

[hallow quartz]

i’m guessing u have done simultaneous equations before

[lost whale]

bc of the y intercept

kind of

[hallow quartz]

ye makes sense

is this a calculator question

[lost whale]

yeah

[hallow quartz]

do you have a simultaneous equations solver on it?

if not it’s fine u should probably know how to do it without

[lost whale]

no i dont think so

[hallow quartz]

all good

do you know the remainder theroum

[lost whale]

yes i do

[hallow quartz]

ok good

if you know the factors and it’s remainder u can plug in the root into the equation and it will equal the remainder

[lost whale]

i am lost lol

[hallow quartz]

all good

if x+1 is equal to -5

then if we plug x=-1 in all x’s in the equation the equation will equal -5

[lost whale]

wait can i show u waht I have now

[hallow quartz]

yes

[hallow quartz]

yes correct

now you have two equations with two variables

mb

[lost whale]

What now?

[hallow quartz]

use simultaneous equations

[lost whale]

how would i do that?

she never taught this bruh 😭

[hallow quartz]

b-c=16

9b-3c=48

[lost whale]

oh, I try substituting?

[hallow quartz]

yes

[lost whale]

okok

[hallow quartz]

elimination or sub

def sub tho

[hallow quartz]

i probably have a different name for it

[lost whale]

ohhh okok

i kind of got it now

[hallow quartz]

what’s ur answer

[cedar kiln]

@lost whale Has your question been resolved?

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[nocturne walrus]

how do i do the rest?

[short blade]

did you substitute it in?

[nocturne walrus]

sorry what do you mean?

[short blade]

did you substitute the u and du in?

[nocturne walrus]

how should i do that?

[short blade]

do you know how u substitution works?

[nocturne walrus]

no

[cedar kiln]

@nocturne walrus Has your question been resolved?

[nocturne walrus]

<@&286206848099549185>

[cedar kiln]

@nocturne walrus Has your question been resolved?

[nocturne walrus]

.close

[cedar kiln]

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[mossy heron]

Does the square root include the power of 5 or not?

[slow thicket]

wdym

@mossy heron

[mossy heron]

5/2 is the power of what's inside the parenthesis

[mossy heron]

I don't know if I put the power inside the square root, it would still allow me to use the trig sub or not.

if it's inside the square root, it would be 2tan(theta), but if not then it would be 2tan(theta) to the power of 5

[slow thicket]

well you can do it either way

[mossy heron]

What do you mean?

[mossy heron]

Yeah, but I meant if I can replace the square root even if I have the power inside

the thing is that I don't know if I can apply that if the power is inside the square root

I suppose it doesn't because (x^2-a^2)^5 shouldn't be the same as x^2-a^2 but I'm not sure

<@&286206848099549185> I had another channel with the same question and nobody answered. Please help.

[cedar kiln]

@mossy heron Has your question been resolved?

[mossy heron]

Yeah, makes sense. Thank you!

.close

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[wild linden]

hi

[wild linden]

its not correct and idk why plz help

[cedar kiln]

@wild linden Has your question been resolved?

[wild linden]

make it make sense please

[wild linden]

anyone know this

[wild linden]

i put in 5 and it was wrong

idk

[cedar kiln]

@wild linden Has your question been resolved?

[cedar kiln]

@wild linden Has your question been resolved?

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[timid anchor]

what does "min" do? picks out the minimum of the two inside "{}" ??

[worthy ridge]

yep

minimum of the two expressions when evaluated

[timid anchor]

ok thank you

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[crimson olive]

is it correct to get a negative value for GP ?

[crimson olive]

i feel like i did this wrong as it doesnt make sense tht the GP would be negative

[radiant topaz]

your first operation is wrong

[crimson olive]

I see it now, thanks

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[crimson sedge]

what do I do from here?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[wild spear]

My only thought on how to calculate that integral is by using partial fractions, however tedious it is.

[hidden osprey]

complex numbers

or partial fraction decomposition

[cedar kiln]

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[crimson sedge]

why can't u do trig sub

[crimson sedge]

how

[hidden osprey]

good luck

[crimson sedge]

why doesn't trig sub work

[hidden osprey]

be the first one to make it work

i believe in you

[tulip stream]

type x in terms of theta

[crimson sedge]

?

oh yeah x=sqrt(tan(theta))

[tulip stream]

you let x²=tanθ

go on from there

see if it solves

[crimson sedge]

how do i integrate sqrt(tan(theta))?

[tulip stream]

its 1/sqrt(tanθ)

not straight up that

[crimson sedge]

oh yeah

that doesn't make it much better...

[tulip stream]

yeah it only gets more complicated

[crimson sedge]

how do u recognize when to do u sub or trig sub or partial fractions with problems like these

[tulip stream]

partial fractions is most likely when the denominator can be factored

[lyric leaf]

as a rule of thumb if you've got an expression in a square root and not much else, a u-sub'll probably work

[crimson sedge]

this problem works with trig sub

[tropic oxide]

it's probably less a question of whether you can do trig sub and more now annoying it's gonna be

[lyric leaf]

oh yeah, the u-sub on 1/sqrt(tanx) looks gross lol

[tulip stream]

addition to this

[crimson sedge]

ok so when should I use trig sub

[crimson sedge]

in cases other than this

[tulip stream]

not sure how that one is done but i would let x^3 = u then try to do it like that

[crimson sedge]

let x^3 = tan(theta)

and it works

[tulip stream]

oh yeah i meant tan but typed x instead

oh nvm i meant u, then i would do another sub from u to tanθ

u=tanθ

[crimson sedge]

why would the same concept not apply to this

[tulip stream]

well its entirely different than format of 1/(1+x²)

[tulip stream]

i guess on this guess chain rule helps eliminate some stuff and convert it clean

[crimson sedge]

but it reaches that format by subbing in for x^2

[tulip stream]

x² brings 2x by chain rule

and it caused the major problem we have seen which is sqrt(tanθ)

[crimson sedge]

in this problem d/dx of x^3 = tan(theta) results in 3x^2 dx = sec^2(theta) dtheta

aren't these similar then

[tulip stream]

not too similiar

[tulip stream]

main difference is you can have clear sub which simplifies everything nicely (which is the 2nd one you sent) while on first one, no matter what you sub its still complex to solve

i have seen a solution where you can use this fact, then use partial fractions to solve

on bottom i typed (x+1 ±...) on both factors but meant (x² + 1 ±...)

still takes so long despite that

[crimson sedge]

ok thanks

.close

[cedar kiln]

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[kindred orbit]

it's supposed to be $A_{30}$

[wraith dagger]

JA.

[kindred orbit]

the only thing i can make out is that $n(A_1 \cup A_2 \cup A_3....A_{30})$ = n(B_1 \cup B_2 \cup B_3....B_n)$

[wraith dagger]

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

[kindred orbit]

and i don't know where to go on from here

[tropic oxide]

$\sum_{i=1}^{30} n(A_i) = 10 \cdot n \paren{\bigcup_{i=1}^{30} A_i}$

[wraith dagger]

Ann

[kindred orbit]

i don't know what you mean

[tropic oxide]

which part of my notation is escaping you?

[kindred orbit]

the sigma notation'

i haven't learnt that yet

[tropic oxide]

but you have learned indexed union notation?

[kindred orbit]

yes but

not sigma notation

[tropic oxide]

sigma notation is the same but for numbers and with addition

$n(A_1) + n(A_2) + \dots + n(A_{30})$

[wraith dagger]

Ann

[kindred orbit]

hmmm so

that's what the sigma notation is notating?

also i don't understand the part where it said that "assume that each elements of S belongs to exactly ten of the A_i's and to exactly nine of the B_j's"

i have understood everything else except that part

[tropic oxide]

yes it is

[tropic oxide]

an element of S belongs to at least one of the thirty sets called A_i, by definition of the indexed union.

a priori it may belong to two or more of those sets, or even to all of them.

you are told that for each element of S, the number of sets A_i to which it belongs is exactly 10.

[kindred orbit]

why at least one?

soooo wait

[tropic oxide]

thats the definition of union...

$x \in A_1 \cup A_2 \cup \dots \cup A_{30}$ if and only if $x \in A_1$ or $x \in A_2$ etc. up to $x \in A_{30}$

[wraith dagger]

Ann

[kindred orbit]

soooo

n(

oops

some of the sets in $A_1, A_2,... A_{30}$ have the same elements?

[wraith dagger]

JA.

[tropic oxide]

yes

of course they do

(note however this does NOT mean that some of the sets in the A-family are equal. they don't need to be)

[kindred orbit]

yeah yeah i know

so since S belongs to exactly 10 sets of Ai's that means the number of elements in $A_1 \cup A_2 \cup A_3... A_{30}$ can be represented as $\fract{5 x 30}{10}

[wraith dagger]

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

[tropic oxide]

so since S belongs to exactly 10 sets of Ai's
no. bad.

[kindred orbit]

uh i'm new to latex sooo uhh

5 x 30 /10

[tropic oxide]

i wasn't criticizing your latex, i was criticizing your phrasing

[kindred orbit]

??

[tropic oxide]

S itself is a set, and it cannot be an element of any A_i

$\frac{5 \cdot 30}{10}$ is what you wanted to type, i think.

[wraith dagger]

Ann

[tropic oxide]

it's \frac and not \fract.

[kindred orbit]

yes that's how

oops lol

so that goes the same for the B elements

i mean the B sets

[tropic oxide]

yes. same reasoning, different numbers

[kindred orbit]

$\frac{5 \cdot 30}{10} = \frac{n \cdot 3}{9}$

[wraith dagger]

JA.

[kindred orbit]

n = 45

is that correct

[kindred orbit]

also i meant each element of S i probably forgot to type that

but not the S as a whole

[tropic oxide]

appears so.

[kindred orbit]

alright thank you very much for the help

.close

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[uncut carbon]

Is it okay to ask cooking questions

[south tundra]

As long as that is connected to maths somehow

[fair mortar]

||calculate no of atoms of carbon present in 200m^3 of cake||

[uncut carbon]

It's about bread

[south tundra]

So is math relevant or not?

[cedar kiln]

@uncut carbon Has your question been resolved?

[uncut carbon]

No but do you have a good recipe

Like the measurements

[south tundra]

No

Ugh

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[winter torrent]

Since your trinomial is a quadratic equation (an equation of degree 2), there are two possible identities I immediately think of...

$\left( a + b \right)^2 = a^2+b^2+2ab$;
$\left( a - b \right)^2 = a^2+b^2-2ab$

[wraith dagger]

Void Walker

[winter torrent]

Therefore your factor(s) look something like $\left( z + \right)$

[stone flame]

why do you have to do \left(

doesnt ( work just fine

[wraith dagger]

Void Walker

[winter torrent]

Never did for me

$($

[wraith dagger]

Void Walker

[stone flame]

$z^{2} +16z$ can be rewritten into $z^{2} + 2(1)(8)z$ or $z^{2} + 2(-1)(-8)z$

notice how it is similar to the form of $ax^{2} + 2abx + bx^{2}$, where b is the thing you want to put in

[wraith dagger]

Tangerine

[torpid wren]

Thank you both very much

[wraith dagger]

Tangerine

[torpid wren]

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[near hearth]

I need help, this is about Marketing Equilibrium. I have the solution but idk if this is right, my teacher only shows the example where the y has no number (example: 3x - y - 20 = 0) something like that.

[near hearth]

also the (i) question is about

(i) Determine which is the demand and which is the supply curve
(ii) Graph the functions and estimate the Market Equilibrium Price and Quantity

<@&286206848099549185>

[cedar kiln]

@near hearth Has your question been resolved?

[near hearth]

Please I need help

[near hearth]

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[near hearth]

Here is my solution and the question, idk how to get it because of the y has a number on it? Does the number in the y is canceled or it would be minus or what to the number idk

[near hearth]

Market Equilibrium Point

<@&286206848099549185>

[cedar kiln]

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[runic garnet]

Do u know how to solve a system of linear equations?

[gritty galleon]

what are you exactly asking here? to solve the system of equations or the entire question

[runic garnet]

O wait that’s ur second image

[gritty galleon]

and you dont need to 'guess' values

[near hearth]

I just need to know about the supply and demand

but I need to like put the y like on my solution

My teacher's example doesn't have number on the y like on the given 3y and 4y

[runic garnet]

Do u know about the law of supply and the law of demand?

[near hearth]

no

[runic garnet]

Google it - it will help u. In terms of the system of equations, do u know how to solve a system algebraically

[near hearth]

But I'm here also to know if my solutions of finding supply and demand is right or wrong

the 2nd image

[runic garnet]

How come there’s 4 different solutions?

There should only be 1

Which brings me to my thrice asked question - do u know how to solve a system of linear equations algebraically?

[near hearth]

I'm not unsure of my solutions because the y has number

[runic garnet]

How are u getting ur solutions?

Are u guessing

[near hearth]

Here's what my teacher did

[near hearth]

the one on the given has number on it

the y exactly

[runic garnet]

Yes ur teacher rewrote the first equation into slope intercept form in order to ascertain the slope and y intercept (presumably to graph the line)

But did ur teacher show how to solve it algebraically?

[near hearth]

Like the graph?

[runic garnet]

See that in red

Ur teacher solved for the equilibrium point

Using the elimination method

Are u familiar with the elimination and substitution methods?

[near hearth]

elimination is what my teacher did on y, substitution on that part idk

[near hearth]

So does that mean the 3y and 4y in the given is the letter y will be eliminated or with the number itself?

[runic garnet]

Show me what u mean

I want to make sure ur thinking of it properly

[near hearth]

Oh and the demand

[runic garnet]

Yes. That’s just this

[near hearth]

What happen to the y with number like in the 1st pic?

Oh ok

Wait

[runic garnet]

No, that’s not how u get slope and y intercept

U have to manipulate such that y has a coefficient of 1

[near hearth]

How?

[runic garnet]

https://youtu.be/-Eze0i92WMY

[near hearth]

but should I divided them just to get like from the graph? Like 2/3 is 0.67 then m = 0.67 and -8/3 is -2.67 then b = -2.67?

[runic garnet]

Yes but it’s tougher to graph using decimals

Use fractions, it’s easier to do rise/run

[runic garnet]

This is correct

[near hearth]

another one

[near hearth]

I tried using Geobra idk how

[cedar kiln]

@near hearth Has your question been resolved?

[cedar kiln]

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[smoky terrace]

@fair mortar I have a question about your solution for the polygons. If I try this with a square, with radius 1 (=half the diagonal if you know what i mean), then i get 1 for some reason

[smoky terrace]

but i think it should be two actually

[fair mortar]

yes

wait

is r is 1

then side is

[smoky terrace]

$2sin(45)cos(45)$

[wraith dagger]

Emiel

[smoky terrace]

this is what i did

[fair mortar]

wait

[smoky terrace]

radius is one so that disappears

[fair mortar]

d=aroot2

so d is 2

2 = a root 2

a is root 2

[smoky terrace]

wait what is d?

[fair mortar]

so area is 4 cm sq?

[fair mortar]

diagonal

actually the point is to find the side length using r

i think i would have messed up somewhere

the formula shouldnt be taht smal

[smoky terrace]

no it is 2cm^2

when the diagonal is two

[fair mortar]

ah es my bad

[smoky terrace]

bc the side is sqrt(2)

[fair mortar]

or maybe it is r + istead of te r *

i forgor lol

the basic idea is

to find the side length using radus

then use the reqular apothem formula

[smoky terrace]

i think your formula is wrong

because it is a linear graph

it shouldn't be

oh wait no

and this is the formula for the area

@fair mortar pls help

[fair mortar]

so uhm

take a square for example

let radius be 10

[smoky terrace]

x is number of sides btw

[fair mortar]

theta is the interior angle so it is 90 for a square

so radius bisects it and now theta would be 45

draw the apothem

so

sin 45 = apothem/radius

wait no

use cos

cos 45 = base/ radius

so now you have base

side is 2 times the base

now you have base and apothem

find permeter and use the formula

[smoky terrace]

can you please recheck your formula

[fair mortar]

taking this square,