help-10

[median dome]

$|z-1|=1\implies z=1+e^{i\theta}, \theta\in (-\pi, \pi]$

[warm shale]

kheerii

[limber nebula]

Sorry i'm not familiar with e^i theta notation

only cis

[median dome]

$z-1=\cos\theta+i\sin\theta$

[warm shale]

kheerii

[median dome]

how about this

[limber nebula]

oh ok I see

how do I use this to answer the question>?

[median dome]

you can get z down to a more manageable form

$z=1+\cos\theta+i\sin\theta$

[warm shale]

kheerii

[median dome]

try using trigonometric identities here

[limber nebula]

so the question becomes possible values of
$\arctan \left(\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)+1}\right)$

?

[warm shale]

cherry

[median dome]

uhh, not exactly

this is the argument of z

[limber nebula]

so wouldn't possible values of that be the same as possible values of Arg(z)?

[obtuse pebble]

@limber nebula Has your question been resolved?

[obtuse pebble]

@limber nebula Has your question been resolved?

[limber nebula]

<@&286206848099549185>

[obtuse pebble]

@limber nebula Has your question been resolved?

[limber nebula]

<@&286206848099549185>

[tender trench]

Stop spamimg bruv

[gloomy tusk]

x + 6 lies on AE and 3x -8 lies on EB?
i guess u can just equate x + 6 = 3x - 8
solve for x, then substitute x back in the first eq

[obtuse pebble]

@limber nebula Has your question been resolved?

[tulip prawn]

how is that spamming?

[obtuse pebble]

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[cosmic sonnet]

you cant use sin rule for the first one

cause for sin rule you need two sides and one angle

you have no angle

so you need to use cos rule

now think a little about if he has made any mistakes

also for the second one

you can not use sin rule

because you dont have the opposite angle for any of the sides given

so you would use the cos rule

I think you need to revise your sin and cos rule

[cerulean summit]

my teacher does not teach

[cosmic sonnet]

real

[obtuse pebble]

@cerulean summit Has your question been resolved?

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[frigid cliff]

elp im confused

[frigid cliff]

this has got something to do with centroids

<@&286206848099549185>

[molten stirrup]

every point is random ?

no angle nothing ?

[frigid cliff]

um there are side lengths

for example b is the centroid

and 2x-5 is the length of YD

[molten stirrup]

oh

[molten stirrup]

that helps

[frigid cliff]

yeahh its that kind

i need to find x but I dont know how

[molten stirrup]

so b is the intersection of 3 bisector ye ?

centroid and stuff

[frigid cliff]

idk i dont think so tho

[frigid cliff]

ohh yes

for this

but idk how to identify since it looks pretty equal

[molten stirrup]

ye

note that

in bisectors there is a ratio of 2/3

pretty hard to prove

[frigid cliff]

wait so

[molten stirrup]

here for eg

[frigid cliff]

can u tell which side is longer

wait nvm

let me try to figure this

[molten stirrup]

ye use the ratio

[frigid cliff]

DY = DB + BY

[molten stirrup]

i d often "figure out" . i jsut often look at the diagram and guess what is 1/3 and 2/3 lol

[frigid cliff]

BY = DY - DB

ok

wait

2x-5-3x-18

-x-23

[opaque dome]

Becareful to sign

U have to put parentheses

[frigid cliff]

huh

[opaque dome]

Its 2x-5 - (3x-18)

[frigid cliff]

okok tysmm

2x-5 -(3x-18)

2x-5 -3x+18

-x+13

tysmm

BY = -x+13

-x+13 = 2(3x-18)

-x+13 = 6x-36

-7x = -49

x = 7

2(7)-5

14-5

9

pls check work s

[molten stirrup]

🎊 🎉

[opaque dome]

Gg wp

[frigid cliff]

is this correct

[molten stirrup]

x = 7 is corect

[frigid cliff]

ok tu

ty

.close

[obtuse pebble]

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[frigid cliff]

.reopen

[obtuse pebble]

✅

[frigid cliff]

i did this and got 13

im not sure if im right tho

[molten stirrup]

13 👍

[opaque dome]

Ye

[frigid cliff]

this be 12 right?

i dont trust my brain after multiple late nights

[molten stirrup]

...

i should make mistakes in math

thats common

its ok to not see how u did wrong

but did wrong and not do it again..

[pastel gyro]

Do you know what the ratio EL:LY is equal to? If you can find out/know the ratio, the rest is easy.

[lusty willow]

excuse me yall

[opaque dome]

True, you can remember that EL = 2/3 EY and LY = 1/3 EY

[opaque dome]

?

[lusty willow]

see this:

[molten stirrup]

set a new one

[lusty willow]

how in gods grace is this related

[opaque dome]

Set a New channel

[lusty willow]

mm?

sorry im just ne

new8

[molten stirrup]

in the math help w green tick

[lusty willow]

can you send the link?

[molten stirrup]

select one of those help -

[lusty willow]

wait

[molten stirrup]

not its literally

[opaque dome]

Just abovz

[molten stirrup]

the math help-avalable category

[lusty willow]

oh ok

[molten stirrup]

just type there ull get diverted here

[obtuse pebble]

@frigid cliff Has your question been resolved?

[obtuse pebble]

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[rough stump]

if cot(θ) + cos(θ) = p and cot(θ) - cos(θ) = q , then (p^2 - q^2)^2 in terms of p and q is

[rough stump]

!status

[obtuse pebble]

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

[rough stump]

2

I try to put the value in the equation

and get [ 4cotθcosθ ]^2

idk what to do next

[latent walrus]

(p^2-q^2)^2 in terms of p and q?

[rough stump]

ya

[latent walrus]

(it already is?)

do you mean in terms of theta?

[rough stump]

ya but they say

give answer in p and q

[latent walrus]

hm

im not sure how one would interpret that

[latent walrus]

this is fine though

[rough stump]

ya

but

when i am solving this i could not get the answer

[latent walrus]

do you have a picture of the question as given?

[rough stump]

nope

sorry

[latent walrus]

im afraid im not too sure what it wants then, sorry

[rough stump]

np

thanks btw

[obtuse pebble]

@rough stump Has your question been resolved?

[rough stump]

<@&286206848099549185>

[slim cove]

I agree with AZ, the question doesn't really make sense

Are you sure this is the original wording of the question

Also, does it give you answer choices? Like multiple choice?

[rough stump]

ya

i am sure

cuz answer is 16pq

[rough stump]

<@&286206848099549185>

[pastel gyro]

We have $\left(p^2 - q^2 \right)^2 = \left(p + q \right)^2 \left(p - q \right)^2 = \left(2 \cot(\theta) \right)^2 \left(2 \cos(\theta) \right)^2 = 16 \cot^2 (\theta) \cos^2 (\theta) = 16 \cot^2 (\theta) \left(1 - \sin^2 (\theta) \right) = 16 \left(\cot^2 (\theta) - \cos^2 (\theta) \right)$ and now factorize and replace the factors by $p$ and $q$ accordingly, and you're done.

[warm shale]

PowerUp

[rough stump]

wait

let me solve this way

can you explain me 2nd last step

after we take common

[pastel gyro]

You mean what to do after we get to $16 \left(\cot^2 (\theta) - \cos^2 (\theta) \right)$?

[warm shale]

PowerUp

[rough stump]

no

[pastel gyro]

This step: $16 \cot^2 (\theta) \left(1 - \sin^2 (\theta) \right)$?

[warm shale]

PowerUp

[rough stump]

16cot^2(theta)cos^2(theta)

ya this one

how you do it

[pastel gyro]

Are you familiar with the trigonometric identity $\cos^2 (\theta) = 1 - \sin^2 (\theta)$?

[warm shale]

PowerUp

[rough stump]

yup

got it tahnks

[pastel gyro]

I just replaced the cosine squared with that identity

[rough stump]

Thank you

[pastel gyro]

Npp

[rough stump]

bro

Thanks again

bye have a good day

[pastel gyro]

Np 👍

[rough stump]

.cllose

.close

[obtuse pebble]

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[obtuse pebble]

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[barren shadow]

gpt gave me 2 diff answeres , isn't ls x>-1 and ld x<-1

[barren shadow]

it's not continuous because ls is undefined

[smoky vigil]

!nogpt

[obtuse pebble]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[barren shadow]

yes i get that that's why i asked here

do i take for left side ln(x+1) and right side x+3?

and for k(x)? do i use -2?

[smoky vigil]

you have them mixed up

ln(x+1) for x>-1

x^2+3 for x <-1

[barren shadow]

and what do i do with -2?

this is what confuses me

[smoky vigil]

both limits must be equal to that value in order for the function to be continuous

[barren shadow]

Yes

[smoky vigil]

The function $k(x)$ is continuous at $x=-1$ if $\lim_{x \to -1^{+}}k(x)=\lim_{x \to -1^{-}}k(x)=k(-1)$

[warm shale]

🫎 Moosey 🫎

[smoky vigil]

if one of these is not true, then the function is not continous at k=-1. There are several types of discontinuities

[barren shadow]

well one is undefined

but my question was , do i not use -2 there?\

or?

[smoky vigil]

if one of the limits is undefined, then the function is not continuous

at x=-1

[barren shadow]

yes i have concluded it's not continuous

ls=ld=k(x) right?

[smoky vigil]

you only need to use -2 if both limits were equal to the same value. if they were both equal to -2, then it would be continuous function, if they were both equal to same value, but not -2 (say, 3), then it would be a discontinuous function. it would have a removable discontinuity/jump discontinuity

[barren shadow]

Ook , i get it now so they would've had to be -2

[smoky vigil]

if you wanted to use -2 in anything, yes. but if the limits are not equal to each other, of if one is undefined, just halt there and say the function is not continuous at x=-1, maybe discuss the type of discontinuity, but that's all

[barren shadow]

ok so for a different exercise for example

[smoky vigil]

not continuous

[barren shadow]

do i not use ls=ld here?

[smoky vigil]

you can

[barren shadow]

how so

[smoky vigil]

it will be the same arguement

well, what is $\lim_{x \to 0^{+}}\frac{1}{x}$

[warm shale]

🫎 Moosey 🫎

[barren shadow]

so left side is 1/x and right side 0?

[smoky vigil]

right side is 0^+ and left side is 0^- yes

[barren shadow]

but the x > 0 and x < 0 don't apply

[smoky vigil]

???????

[barren shadow]

pardon

this is the formula i've known , x->xo , x>xo ; x->xo x<xo

[smoky vigil]

$x \ne 0 \equiv$ $x > 0$ or $x < 0$

[warm shale]

🫎 Moosey 🫎

[smoky vigil]

this piecewise function is basically a compact way of saying m(x) is defined as 1/x if x>0 and 1/x if x<0 and m(0)=0

[barren shadow]

I see

Thank you Moosey

[smoky vigil]

if you're not equal to 0, your greater than 0 or less than 0

does that make sense

[barren shadow]

Yes

I got that now , it was just logic

i will close this thread now thanks for the help

.close

[obtuse pebble]

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[obtuse pebble]

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[karmic hedge]

It says my answer is wrong because the second derivative is negative, am I crazy or is it not positive?

3sin(x)

3sin(5) = 0.261...

[haughty coyote]

sin(5) is, in fact, negative

[karmic hedge]

Is it wrong cause of the degrees?

I just put 3sin(5) into wra

[haughty coyote]

Yeah try radians now

[karmic hedge]

ok I see...

Got it now hahaha

Thank you!

❤️

.close

[obtuse pebble]

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[worn ermine]

brain not working

to find the width

of smaller rectangular prism

width of bigger one is 10

[reef tendon]

hello again percy

[worn ermine]

yo

[opaque dome]

Thales ?

[reef tendon]

what does 4 represent

[worn ermine]

height of smaller prism

since

both together make a height of 9

[reef tendon]

oh

[worn ermine]

the height of the bigger prism is 5

[reef tendon]

didn't see that

[worn ermine]

so

ok

[reef tendon]

cuz you outlined the smaller one out

[worn ermine]

but I can't find width

[opaque dome]

Draw the diagonal of the Lil prism and keep drawing to hit the line under

[worn ermine]

ye

[worn ermine]

hm

[reef tendon]

maybe it's not drawn to scale

[worn ermine]

it isn't

drawn to scale

[opaque dome]

Oh

[reef tendon]

idk

[timid silo]

dont matter, you’ll just be using similar triangles

[opaque dome]

Anyway with values he can do thales

[reef tendon]

pretend the height=width of smaller prism

[worn ermine]

\wbat isthales

[reef tendon]

oh yeah

but you don't know where it hits and on which length

[opaque dome]

Thales theorem with similar triangle

[worn ermine]

teach

[reef tendon]

ooooh

mb

im stupid

[worn ermine]

can you teach me thales

with this problem

[opaque dome]

Wait im trying to see if it works first

[worn ermine]

you can't possibly need to use thales theorem for this

we didn't learn that

[opaque dome]

Oh ok

[worn ermine]

there has to be a way that is 10 million times simpler

a way that we just have to look at the image to figure it out

[opaque dome]

Isint it just a square on the sides ?

Like 4*4 ?

So the width is 4 too

[worn ermine]

o

wait

bruh I'm dumb

[opaque dome]

So the volume of Lil one is 192 and big one is 600 so added its 792 cm^3

[worn ermine]

I covered up the congruent lines

with

[opaque dome]

Lmao

[worn ermine]

can you check this one for me

to find the triangle

we know base and height

so 12*7/2

42

*h

17

714

is the volume

[opaque dome]

It is right

[obtuse pebble]

@worn ermine Has your question been resolved?

[worn ermine]

👍

[obtuse pebble]

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[heady turtle]

any examples for this? I mean the n indep invariants, I'm not sure if i can think of any

[bronze mica]

I suspect they're the coefficients of $\det(A-\lambda I)$ as a polynomial in $\lambda$

[warm shale]

Edward II

[heady turtle]

why?

[bronze mica]

Because this polynomial is invariant (relatively easy proof), and any polynomial will show up so the coefficient should (?) be independent

except for the leading term btw

[heady turtle]

is there any geometric meaning of the polynomials?

[bronze mica]

so for a 2x2 matrix this is the polynomial $\lambda^2-\lambda\operatorname{tr}A+\det A$

[warm shale]

Edward II

[bronze mica]

I don't believe there's anything geometric

I mean

[bronze mica]

(up to sign)

[heady turtle]

Mmm then how can we prove it's invariant? (I know it's a relatively easy proof but i have no idea currently

[bronze mica]

$M^{-1}AM-\lambda I= M^{-1}(A-\lambda I)M$

[warm shale]

Edward II

[heady turtle]

oops

ok i see it now, so there's also no intuitive way to understand? tbh i just use poly to calculte but I have never figured out what is poly, what it represents

[bronze mica]

also eigenvalues of A are roots of the polynomial

so by Viete's formulae you can see the other invariants can also be written as like

in the case $n=3$ if a matrix has (complex) eigenvalues $\lambda_1,\lambda_2,\lambda_3$ (with multiplicity as roots of the polynomial)

the third invariant would be $\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1$.

[warm shale]

Edward II

Edward II

[heady turtle]

Viete's formulae ok a new word, I'll look it up

[heady turtle]

so do you have any ideas about this?

[bronze mica]

also known as Vieta, and I don't know why I said formulae because I normally use theorem

[bronze mica]

I think it's just a useful thing to know

like there's more facts about the polynomial you could find, but I don't think any of them are particularly intuitive?

[heady turtle]

like just a tool for calculating eigenvalues?

[bronze mica]

I can't think of anything right now

[heady turtle]

Ok, i see, thank you anyway!!

.close

[obtuse pebble]

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[umbral lagoon]

Ian walks from one end of a straight track to the other, then turns around and walks back. At
the same time, Mike leaves from the same place as Ian and rides an e-bike back and forth from
one end to the other until Ian returns to their starting point. Assuming both travel at constant
speeds with Mike’s speed exactly 9 times that of Ian’s, how many times will they meet on the
track after they start? [Note: The meeting at the very end should be counted.]

[uneven palm]

wait a minute..

#help-0 has the same question

[umbral lagoon]

rlly?

oh it is

[uneven palm]

yeah I haven't really looked into it but their convo may help lol

[umbral lagoon]

i still have no idea D: lmao

.close

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[dire jackal]

I don't think this is correct

b^2 - 4ac to determine the amount of solutions for the x-intercepts

[tender tusk]

which part?

[tender tusk]

correct

[dire jackal]

Yeah, but I might be typing it in wrong

because

it gives domain error

Which means no solution

However, this was written in my review packets and I dont quite know/rem

[tender tusk]

what did u type in?

[dire jackal]

square root (-4)^2 - 4(1)(3)

[tender tusk]

yeah that looks correct

[dire jackal]

So I assume this actually has no soltutions

rather than 2 like my old notes

[tender tusk]

no it does have 2 solutions

[dire jackal]

wait then

[tender tusk]

if the discriminant>0 then there are 2 real solutions

[dire jackal]

well it saids domain error

so I don't quite know whats with that

[tender tusk]

what are u typing that into?

[dire jackal]

am I supposed to solve the whole quadratic equation

[dire jackal]

like the calc Im using?

or?

[tender tusk]

yeah is that a graphing calc or something? idk how they work

[dire jackal]

Im not sure, its a texas intrument TI-30XS

[tender tusk]

whats the goal of this question? to find the no. of solutions or the solutions itself

[dire jackal]

you state amount of real solutions and graph it

well sketch a possible graph

[tender tusk]

right

well i dont think you'd need a calculator anyways

[dire jackal]

oh

well I used one but yeah

[tender tusk]

the discriminant is b^2-4ac btw, not sure if that will solve ur calculator issue

[dire jackal]

okay, well, uhm do you know how to uh get a positive discriminant so it is 2 solutions?

or am I screwing up bc -4^2 is actually 16 not -16 which would make sense somewhat but also not at the same time

[tender tusk]

u should have typed (-4)^2 which would be 16

[dire jackal]

oh

wait

hold up

I am supposed to put parentheses when I do b^2

over the b

[tender tusk]

yes, (-4)^2 is different to -4^2

[dire jackal]

okay, I see, but my calculator is like cracked or something dumb like that

if I do that

it gives me some interesting answers

so next time I'll just solve (b)^2 seperate

thank you!

[tender tusk]

all good

[dire jackal]

.close

[obtuse pebble]

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[patent anchor]

how do i do this

[brazen viper]

So not reflexive and not irreflexive means there's going to be at least some relations that are self-self but not all. Not transitive means there are at least some self-other relationships, because we need an example of (a,b), (b,c) but not (a,c)

Solution spoiler: so we can have something like the set is {a, b, c}

||R = {(a, a), (a, b), (a, c)}||

||This relationship is not reflexive because (b, b) and (c, c) are not members. It is not irreflexive because (a, a) is. It is not symmetric because (c,b) is not a member, not transitive because (a, c) isn't, and is antisymmetric because (a, b) and (b, c) both don't have their symmetric counterparts.||

@patent anchor

[obtuse pebble]

@patent anchor Has your question been resolved?

[patent anchor]

ooh okayy thank u <3

[obtuse pebble]

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[stoic elm]

How do i use herons formula?

[stoic elm]

with the angles 31.5, 79.9 and 68.6

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[scenic geode]

Does anyone knows how to do this

[tender tusk]

hmm i think u can inscribe another square inside this square

if u look at each region

you'll notice that there are 2 outer sides of the region that exist in each region

u can construct a triangle using those sides in each region

not sure if i coan describe this by word

i'll leave the rest up to you

[scenic geode]

ohh

I think I get the concept

Tysm!

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[broken shard]

I’m a little confused here. If every subgroup of H has a preimage, doesn’t that imply that the homomorphism has to be surjective?

[kind hawk]

well the image of that preimage doesnt have to equal E, it can be only a subset of E

[broken shard]

ahhh okay that makes sense

thank you

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[timid silo]

Hi guys I'm a dumb 13 year old 7th grader, I need help with basic math for you pls DM me 😭😭

[gilded needle]

if you have a question, ask it here

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[ebon sand]

does it matter if I keep the 2 in that bracket

or if I bring out the 2^-2

I get the same answer with it cancelling out with 4

so I'm guessing it's right

[safe edge]

u gotta “bring it out”, or it’s different

[ebon sand]

so (4)(2^-2)(1+3/2x)^-2 is fine?

[safe edge]

mhm

[ebon sand]

ok ty

[safe edge]

it’s equivalent

[warm canopy]

The 2 isn't being raised to the power -2

[safe edge]

yeah he fixed that

[ebon sand]

wait what

[safe edge]

nah u fixed it

it aight

[safe edge]

he’s talking abt this one

which was wrong

[ebon sand]

ok

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[wraith barn]

can i have help with this?

[wraith barn]

Q2

[iron edge]

Show the above information

[wraith barn]

the above info isnt necessary

its just that

thats all

<@&286206848099549185>

[obtuse pebble]

@wraith barn Has your question been resolved?

[opaque dome]

ima try

no drawing ?

[iron edge]

There must be something up there 👆

[opaque dome]

try to draw something always when it's not given, it will gives u ideas

[obtuse pebble]

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[iron edge]

.reopen

[obtuse pebble]

✅

[wraith barn]

Can i ask

Ah

Wait

Are we allowed to just join them like that?

[opaque dome]

it's called Chasle's relation

[wraith barn]

btw top of the page

[opaque dome]

you can join them since any vectors like AC can be expressed as AB+BC

[wraith barn]

Sorry i should be more clear

The vector ab and pq

Like your diagram

What if they were going in different directions

[rigid plaza]

I recommend doing 2d plots for points instead of a number line when testing examples and trying to get a better intuition

[rigid plaza]

The algebra expressed below the number line doesn't use the fact that diagram assumes all points are collinear so it is all good

[wraith barn]

Hmmm

basically

where im confused is

if a,b,c,p,q,r are points, then why couldnt it be like this

and then we would be unable to say that about is pq

am i just still thinking scalar

[obtuse pebble]

@wraith barn Has your question been resolved?

[wraith barn]

<@&286206848099549185>

!noans

[obtuse pebble]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[obtuse pebble]

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[wraith barn]

.close

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[final lily]

where does the -1 come from? i have watch linear regression derivative too but I still don't understand it

[final lily]

why it taking the -1 from the intercept, what rule is this?

[thin pawn]

You‘re taking the derivative wrt the intercept, and if you look closely, there‘s a minus sign before intercept. That means its derivative will be negative 1

[wooden cipher]

chain rule

[thin pawn]

Also, yeah, maybe I misunderstood the question here 🫠

[thin pawn]

Oh i‘m really blind, I didn‘t see this, sorry :\

[final lily]

the chain rule that I found is something like this

I don't find anything that talk about being able to take out the number beside the x on d/dx

[tender tusk]

there are like 50 chain rules i swear

the derivative of [f(x)]^n = nf'(x)[f(x)]^(n-1)

[final lily]

oh is that how the rule go, thx

wait no

when should I use
3x^2 to become 6x

and when should I use
3x^2 and become 6x × 3??

like when do you take out that number beside the x on d/dx

[tender tusk]

ur doing it a bit wrong

the function is 3(x)^2

[thin pawn]

If you use the chain rule for 3x^2, all you‘ll get is 3 * 2 * x * x'

And x' is 1

[final lily]

oh! I see thx guys

[obtuse pebble]

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[dusty spear]

claim

[dusty spear]

need to be guided through this

[cunning burrow]

So you are told to use bayes theorem

[dusty spear]

wait I'm going to rewrite what I already have

and show you where I'm stuck

this is correct so far yes?

[glacial obsidian]

Yes

[cunning burrow]

We do not know the exactly probability a selection will occur only the ratio. But let make the assumption the probabilities are as above

[dusty spear]

this is still correct yes

[cunning burrow]

Yes. You can alway transition from the definition of conditional probability to bayes theorem

[dusty spear]

okay so the part I'm stuck on is how do I find the probability of not S

[cunning burrow]

So S is if a change occurs in profit

[dusty spear]

yes

[cunning burrow]

Perhaps it will be easier to think about if we take the complement of the denominator

So we have 1 - P(S)

Now we need to find P(S)

[dusty spear]

yeah that works too

I still can't figure out how

[cunning burrow]

Let S_A denote if A make a change, similarly S_B if B makes a change etc

So P(S) = P(S_A or P_B or P_C)

[dusty spear]

interesting

I'm trying to recall if I've seen this before

probably not though

this is new to me

[timid silo]

is that gcse??

[dusty spear]

it's undergraduate discrete maths

[cunning burrow]

P(A)P(S|A) + P(B)P(S|B) + P(C)P(S|C)

This is from using total formula of probability.

[timid silo]

ohhh u r uni studuent

[cunning burrow]

I wish had less intuitive way to explain it

[dusty spear]

what does P(A)P(S|A) mean?

multiply them?

[cunning burrow]

yes multiply

[dusty spear]

hold on I'm going to process this

[cunning burrow]

I get the feeling I made a mistake

[dusty spear]

it looks legit no

does b over a1 mean B | A1

[cunning burrow]

I think the assumption I am making is that P(A and S) and P(B and S) are completely separate that is mutually exclusive

[cunning burrow]

The notation you are using does not seem to be standard. It mostly likely meant to mean B|A but I would not use this notation

[dusty spear]

okay so

the answer I got from your solution lines up with everyone elses

so it's either you're right or everyone is coincidentally wrong

[cunning burrow]

Maybe try the principle of inclusion-exclusion as a double check. Are we assuming only one person is hired or there multiple people hired?

It could also be no people are hired

[dusty spear]

I would hope we're dealing with the case that only one guy is hired

because of the initial statement of the odds of them being picked

[cunning burrow]

I see. I will leave the rest to you

[dusty spear]

I'll reply here if I run into anything

thanks for everything so far

.close

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[tawdry swan]

Hello, I want to start relearning maths again, and I have covers some topic, like solving equation, linear equation and some basic operation. I am bit confused about which topic are related to prepare for Calculus.

Is there book Or something like a list of topic that are necessary for calculus, @everyone

[gray kiln]

that field is called precalculus

and you can search for precalculus books or precalculus courses online. many are decent enough

[tawdry swan]

Will it help me with algebra?

[gray kiln]

It can but you probably need to practice a bit on some basic algebra

if you don't really know how to like says factoring polynomial

[tawdry swan]

It's this topic from algebra 1 or 2?

[gray kiln]

I don't know what you are talking about algebra 1 or algebra 2 given that I don't know what curriculum your school is using

Now, I will say that the bare minimum is to be able to factor polynomials and solve inequalities that involve polynomials. You sometimes need to learn how to dividing polynomial (long division) and partial fractions. You need trigonometry and stuff but that is not in traditional algebra class.

[tawdry swan]

Okay, got it

Thank you.

[obtuse pebble]

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[crisp ember]

would rlly appreciate any help on what to do next for part b im a bot confused

[crisp ember]

please

[weary reef]

Well, Phi approaches to Phi_0, what happens to the denominator?

[crisp ember]

cos this is what i know: well for +φo we get the numerator approaching 2φ and the denominator approches 0
but this is undefined
inside a Ln
so im not too sure what to say about it, and for -φo we get x(φ) approaching ln(0/-2φ)
and ln(0)is undefined

[weary reef]

We can see the numerator is gonna be 2 phi

[crisp ember]

yeah

[weary reef]

Yes but what happens to the fraction? Its value approaches...

[crisp ember]

infinity?

but why?

[weary reef]

Consider $\frac{1}{x}$, as x tends to zero, what happens to the value?

[warm shale]

StrangeQuarkAL

[crisp ember]

it gets divided by a smaller and smaller value

so it gets closer to 1

[weary reef]

1 / 0.02

[crisp ember]

ah

i see

thanks

[weary reef]

no problem

[crisp ember]

.close

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[weary reef]

It won't be the same for -phi_0 btw

but similar

[crisp ember]

what do i do for ln(-0/2phi)

@weary reef

is this negative infinity as ln(0) is undefined

.reopen

[obtuse pebble]

✅

[weary reef]

Approaching -phi_0 from the right, yes, it's negative infinity seeing as the fraction is < 1 as it approaches

[crisp ember]

ok

could u explain further

and approaching from left does it approach infinity

@weary reef

[weary reef]

Just to be clear, do you mean approaches phi_0 or -phi_0

[crisp ember]

ln(0/-2phi)

is what i get when phi approaches phi_0

sorry yes

ln(0/-2phi_0)

[weary reef]

don't you mean -phi_0?

[crisp ember]

ah yes sorry my typibg, been doing this all day aha

yes thats what i mean

as it approaches -phi_0

[weary reef]

Like we did before, the numerator is approaching zero (from a positive value), while the denominator approaches 2phi

The thing to notice is that (phi_0 + phi) / (phi_0 -phi) is a fraction less than 1 "during" this approach

and the ln of a number less than 1 (but greater than 0) is always negative

[crisp ember]

but would it not be negative anyway

cos it ln(0/-2phi_0)

also its not greater then zero its less than zero

this does not makse sense

.close

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[crisp ember]

.reopen

[obtuse pebble]

✅

[crisp ember]

@weary reef

[obtuse pebble]

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[vital moat]

For number three I need someone to check my work to see if it’s right

[dark stirrup]

.close

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[west tulip]

( - 5 , - 7 ) , ( 13,2 ) ) and ( ( - 5,6 ) \

[west tulip]

find the circumcenter orthocenter centroid and nine points for the triangle whose vertices are ( -5 ,-7) , ( 13,2 ) and ( - 5,6 )

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[vast echo]

im struggling with this, trying to do something with similar triangles or the pythagorean theorem but im not getting anywhere

[tranquil sonnet]

hint: replace pentagon with any regular polygon and it will be true

[vast echo]

thats not a proof of it working for pentagons though

it obviously works for squares and i could probably prove it for triangles with some basic algebra but a pentagon is much more complicated

[tranquil sonnet]

no

[wooden cipher]

consider scaling factors

[prime pilot]

what does [] mean in relation to polygons?

[tranquil sonnet]

area

[wooden cipher]

area

[prime pilot]

]oh ok

[vast echo]

im trying something with that but i cant seem to get it to work

[wooden cipher]

what do you have so far

[vast echo]

let me write it down

i have that the scaling factor of BC to AB is sqrt(BC^2 + AC^2)/BC but its such a gnarly expression

maybe im just stupid and am missing something obvious but im really lost here

[wooden cipher]

sqrt(BC^2+AC^2)

look familiar?

[vast echo]

its the pythagorean theorem, thats how i know thats the scaling factor in the first place

[wooden cipher]

ok, are familiar with the fact if you scale a side of a shape, the area increases with the square the scaling factor

for example if you have a square of side length 1 and scale it by the 3, the area gets scaled by 9

[vast echo]

i know that for triangles and squares but i havent proven it for pentagons

or a regular n-gon

[wooden cipher]

if youve proven it for triangles, triangulate your polygon

then boom free proof for any n-gon

[vast echo]

true

[wooden cipher]

so anyway, i would startby letting P be the area of a pentagon with side length 1

[vast echo]

i dont know that area

[wooden cipher]

dont need to know it

try to find P_A in terms of P

[vast echo]

itd be P*BC^2 wouldnt it?

i might see a path

[wooden cipher]

yeah

[vast echo]

i think ive got it

P(BC)^2 + P(AC^)^2 = P(BC^2 + AC^2), so we get the scaling factor for a new pentagon

the square root of which is sqrt(BC^2 + AC^2), which we know is the side length of P_C

i think thats proof

[wooden cipher]

sounds good

[vast echo]

alr thanks

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[karmic hedge]

I think I know how to solve this but im stuck on something

sin(x)=cos(x)
sin(x)/cos(x) = 1
tan(x) = 1
x = tan^-1(1)

The domain of tan^-1 means it should be pi/4 and 7pi/4

But as you can see from the graph its pi/4 and 5pi/4

I dont really understand...

[tender tusk]

do u know integration?

[karmic hedge]

Sort of

[stoic yacht]

are you sure tan(7pi/4) is 1?

[karmic hedge]

No thats what im confused about

[stoic yacht]

first of all arctan only ever spits out one value since it's a function, so arctan(1) is pi/4

but tan has period pi

so all solutions to tanx = 1 are pi/4 + k * pi where k is an integer

this is where 5pi/4 comes from

tan(7pi/4) = -1 btw, not 1

[karmic hedge]

I get this

[karmic hedge]

But I dont get how you got 5pi/4

[stoic yacht]

you agree tan(pi/4) = 1?

[karmic hedge]

Yes

[stoic yacht]

arctan has range (-pi/2, pi/2) so arctan(1) = pi/4

tanx is periodic with period pi

which means adding or subtracting multiples of pi to the angle won't change the tan of that angle

[karmic hedge]

Sry this means the blue line in the image right

[karmic hedge]

^^

[stoic yacht]

yes

[karmic hedge]

Ok im with you so far

[karmic hedge]

Yes makes sense

[stoic yacht]

great, then you only care about angles x satisfying tanx = 1 on the interval [0, 2pi]

since that's where your integration problem is happening

pi/4 is one such x satisfying tanx = 1 on that interval, but if you add pi, 5pi/4 is another valid angle in that interval satisfying tanx = 1

[karmic hedge]

Wait I dont understand this sry

[stoic yacht]

you're trying to find areas between sinx and cosx for 0 <= x <= 2pi

[karmic hedge]

Thats anywhere on the unit circle right?

[stoic yacht]

which means you care where sinx = cosx for 0 <= x <= 2pi, or where tanx = 1 for 0 <= x <= 2pi

so you know that x = pi/4 is one such angle satisfying tanx = 1 and it's in your interval

any other valid angles would be shifted by factors of pi

so see if you can add or subtract pi from pi/4 to get another angle x in 0 <= x <= 2pi

adding pi gives 5pi/4 which is in that interval

[karmic hedge]

Sry give me a moment I am a bit lost I need to reread your messages a few times 😅

[karmic hedge]

Why between sinx and cosx?

[prime pilot]

thats wht theyre asking in the q: to find the area between the graphs y=sinx and y=cosx

[karmic hedge]

Nice name haha we are brothers

But ok

[prime pilot]

🤣

[karmic hedge]

Ok wait I need to reread these messages a few times again ahhaha

So sorry about this

Your profile picture is my brain right now Odina

[prime pilot]

🤣 which part is losing you

[karmic hedge]

I was struggling to understand the tan bit but now were bringing sin and cos into it

I want to ask how sin and cos are related to tan but thats a whole other thing I am already struggling to grasp this other stuff haha one thing at a time

[prime pilot]

well we want to find the area between sinx and cosx on the graph, right?

[karmic hedge]

Yes

[prime pilot]

red is cosx and blue is sinx

[karmic hedge]

Yeye I get you

I was trying to find the x intercepts

I know you can see it on the graph but I wanted to work it out mathematically

But thats what I was struggling with

[prime pilot]

yeah

[karmic hedge]

I understand that tan inverse of 1 is pi/4 + k

[prime pilot]

so then we do sinx = cosx

tanx = 1

[karmic hedge]

And you add numbers to k to get the other values

[prime pilot]

so x = π/4, using the arctan function

[prime pilot]

exactly

but we only want the first 2 positive intersections

[karmic hedge]

But I thought that the domain of tan inverse was between pi/2 and -pi/2

[prime pilot]

range*

[karmic hedge]

5pi/4 isnt in that domain

range yeah

sry

[prime pilot]

yeah so π/4 is wht arctan gets

reason why it cant be multiple is because functions must allocate an input to a single unique output

so it wouldnt make sense to say arctan(1) = π/4, 5π/4, 9π/4, ...etc

but in reality there are multiple angles that give 1 when put into the tan function

and to get from one to the other u jus need to add/subtract π

[karmic hedge]

Why pi

[prime pilot]

thats why to get our next point of intersection, we add π to π/4, to get 5π/4 and we can use those 2 values for our integration

[dreamy fox]

Is my method correct?

[prime pilot]

you could jus say thats a property of tanx

like it's always like that for tan

you can prove it using addition formulae asw

[karmic hedge]

What do you mean due to symmetry sry

[karmic hedge]

Ok I can remember that

[karmic hedge]

So this is something else? It doesnt apply to this question?

[dreamy fox]

If you look carefully if you flip portion C and portion D in a particular manner, it'll be of the same shape of B. We often use such visual observations to simplify the solution.

[karmic hedge]

Ahhhh I see

[dreamy fox]

C and D are obvious mirror images

[karmic hedge]

thats a neat trick 😅

I will NEVER notice this in an exam tho hahahaha

[dreamy fox]

We use such tricks for competitive exams of our country since time is a very important factor in it.

[karmic hedge]

Hahaha im so bad at spotting things like that tho

[dreamy fox]

And it's all mcq, no one's gonna look at your method

[dreamy fox]

It takes practice, not that hard

A very famous professor of oxford once said

[dreamy fox]

That a good mathematician is lazy. He/she will always try to think of a method before trying to solve it using pen and paper directly.

Such that it requires lesser amount of physical work (writing)

[karmic hedge]

Haha I will learn the material first and then learn fancy tricks like this 😂 but thank you for showing me

I think I need a break now tho

My brain is fried LOL

Thank you everyone for your help, you are all very kind!

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[slim violet]

Hi! Im trying to do this HW problem and im stuck on the final part of this problem. I already solved most of it, I just don't understand how to convert K_2 to X_2 in the solution

[obtuse pebble]

@slim violet Has your question been resolved?

[slim violet]

<@&286206848099549185>

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[brisk yacht]

i need insurance that i am correct

[empty maple]

what did you find

[brisk yacht]

x

[empty maple]

x?

[brisk yacht]

yes

[empty maple]

so x=x?

[brisk yacht]

yea but i need to solve for.

x

[empty maple]

yes and you should find a value for x?

[brisk yacht]

yes

[empty maple]

so what value did you find for x

[brisk yacht]

4

but

thats an npv

[empty maple]

so 4 is not a solutions since you can't divide by 0 right?

[brisk yacht]

yes

[empty maple]

and there is no other value for x that can make this equaiton possible

which means there is no solutions

[brisk yacht]

yes

ty

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[haughty coral]

~help

[haughty coral]

lemme send screenshot of work

this is my work

would i write it as f'(x)=13?

is my work right?

[dark stirrup]

Where's the -22 coming from?

[haughty coral]

i used (f(x)-f(a))/x-a

[dark stirrup]

oh

[haughty coral]

f(a)= 22

[stone verge]

ok

[dark stirrup]

Then yes, looks good

[haughty coral]

so how would i write the actual answer

is it f'(x)=13 or do i just box 13

[stone verge]

I think f'(x) = 13 is a more complete picture

but

I'd write it as the limit definition of derivative

so, lim x->2 f(x) = 13

prolly

[haughty coral]

ok ty

[dark stirrup]

You would start with $f'(2)=\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}$ then arrive at $13$.

[stone verge]

or check what other examples in text book give

[warm shale]

SWR

[haughty coral]

.solved

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[haughty coral]

ty

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[fluid briar]

Is this A or D???

[wraith rivet]

lol

bro came from gohars server

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[dusky arrow]

how did we get that quadratic

[cunning burrow]

$If we know log_a (z) = log_a^2 (1/2z-1/16). We can take the sqrt of the argument and base of the argument. Doing this on RHS we get log_a(sqrt(1/2z -1/16)$

[warm shale]

team132

[cunning burrow]

From there getting the quadratic equation is easy

[obtuse pebble]

@dusky arrow Has your question been resolved?

[dusky arrow]

can u still continue

[cunning burrow]

So you have log_a (z) = log_a(sqrt(1/2*z -1/16))

Equate both of the arguments in the logarithms

And you will end up with quadratic

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[tepid yoke]

Yeah send your question

Do you know first of all how to find the ratio between fruit and biscuit mass

To find the ratio of fruit mass : biscuit mass (fruit mass to biscuit mass) you divide the number of units of fruit by the number of units of biscuit

To find the number of units of each, you count the colored squares shown for each

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[near pasture]

Question 4

[past solar]

note that a polygon has an exterior angle of n degrees given by 360/x

[obtuse pebble]

@near pasture Has your question been resolved?

[near pasture]

Perfect

So
n=360/x

Right?

[past solar]

yup where x is the sides

[near pasture]

Alr

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[rough stump]

The absolute least value of √2sinx - √2cosx

[rough stump]

A) -2 B) 0 C) 2 D) None of these

[tepid yoke]

Do you know what the absolute least value of sin x or cos x by themselves is?

[rough stump]

Not a clear idea

but i think minimum value that function give

[tepid yoke]

Yeah it is the smallest possible value the function can give as an output

Sine and cosine fluctuate between -1 and 1 when they are not multiplied by anything

[rough stump]

ya

so what is correct option

[tepid yoke]

So you can start with an inequality -1 <= sin x

[rough stump]

-√2 < = sin < = √2

right cuz √2 are Multiplying

[tepid yoke]

But try taking the sqrt 2 out as a factor

[rough stump]

• 1 < = sin < = + 1 and -1 < = cos < = 1

[rough stump]

??

√2 ( sinx - cosx )

like this

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[rough stump]

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