help-10

[faint stag]

alright, and N is acting parallel to CA?

[deft hazel]

W_y = W cosθ
W_x = W sinθ

theta is angle between wire and vertical

[faint stag]

yes got that

but the pegs force is still very much confusing

[deft hazel]

The normal force, represented by F_N, is the force that the horizontal peg exerts on the wire to support it. This force is perpendicular to the surface of the peg and points upwards, perpendicular to the wire. Since the wire is not accelerating in the vertical direction, the normal force must balance the weight component in the y-direction. That is:
F_N = W_y
Substituting for W_y, we get
F_N = W cosθ

So, the normal force is equal to the weight component perpendicular to the wire, which is W cosθ.

[faint stag]

why is it not acting on the vertical

[deft hazel]

wdym

it is

[faint stag]

wait let me draw

[deft hazel]

pythagorean theorem

but gtfo this channel

[faint stag]

if N is pointing up, then N = mg right?

[deft hazel]

Newton's Third Law of Motion

It is equal in magnitude but opposite in direction to the perpendicular component of the weight of the triangle, which is mgcosθ.

[faint stag]

N is acting parallel to the vertical

ok wait

this would be easier if u drew a diagram

[lofty tree]

64 btw

[faint stag]

but in my diagram, there are no forces acting opposite to mgcostheta

[deft hazel]

hol up

[faint stag]

i asked chat gpt to draw a diagram

its got our triangle but upside down

x is where the normal force acts

is this how the question wants us to draw it?

[deft hazel]

well

||W|| is typically drawn from balance point (centre of mass)

Not always though depends on the orientation and shape of the object.

In the case of the isosceles triangle hanging on the peg, the weight of the triangle is acting downwards, towards the center of the Earth, and its point of application is at the center of mass of the triangle. The center of mass is the point where the weight of the triangle can be considered to act as if it were concentrated at a single point. In general, the center of mass of a geometric object depends on its shape and mass distribution, and it can be located at any point within the object.

i'm actually goin braindead rn

how would you draw the vector components of a vector orthogonal to the x axis

[faint stag]

to be fair, this is cambridge entrance exams so dont expect it to be easy 😭 😭

[faint stag]

honestly, my main concern right now is how to model the question

once ive modelled it, i can probably do it

[deft hazel]

it's not hard it's just

[faint stag]

does the centre of mass of the wire even act from the centre?

[deft hazel]

tbh i never considered that

but it's an isosceles triangle

[faint stag]

okay, i had a small look at the exemplar but only the beginning because i didnt wanna ruin the answer

i think they found where it acts from

but im honestly not sure what theyre doing

[deft hazel]

let's see

there is no way 2hrs just passed like that

[faint stag]

this is the first part

if u want i can just send the whole thing

[deft hazel]

Nah lets js look at this first

Whats that point on in the triangle on the line

Centre of mass?

[faint stag]

i think

so

[deft hazel]

I see

So they found the centre of mass

[faint stag]

now that i think about it, it wouldnt act on the centre considering its not a solid object

[deft hazel]

Well according to chatgpt let a, b, c be points of the midline

Some maths

then C is (a+b+c)/3

[faint stag]

chatgpt draws it like this

i made it draw the diagram using python turtle cus i wasnt able to understand before

even then this is still confusing

[deft hazel]

Damn

Wait what

[faint stag]

yea its wrong

[deft hazel]

What is a peg

Like

[faint stag]

exactly ??

this question

[deft hazel]

Ohh

Okay well i thought it was a bar

[deft hazel]

Ye

Thats what they meant by horizontal

[faint stag]

well its a horizontal peg?

[deft hazel]

Why such a long ass peg

[faint stag]

wait

what

they mean horizontally

like

wtf????

thats vertically right?

[deft hazel]

Horizontally colinear?

Collinear

[faint stag]

i thought that was an arrow

not a long peg

[deft hazel]

Hol up

[deft hazel]

Lemme just print this out

fuck

[faint stag]

right

[deft hazel]

so what was our issue again

Why F_n=Wcos theta

[faint stag]

yes

[deft hazel]

then the norm of W's resolution (decomposition)

@burnt wedge help

[burnt wedge]

[deft hazel]

sup smegzy

[faint stag]

its the fact that the normal force is acting equal and opposite to the cosine component

[deft hazel]

we've been here for 3hrs

[faint stag]

is whats confusing me

[deft hazel]

the thing is

how did you draw

your components

[faint stag]

thats exactly the thing

i dont know how to

when i drew my diagram, my weight is acting opposite to the normal

so N = W

[deft hazel]

i'm getting so confused now

[faint stag]

its alright, i have a better diagram now

i assumed that the peg was directly in the middle of BC

[deft hazel]

i want to know what a peg is

[faint stag]

now i have to resolve these forces

[deft hazel]

is it a bar is like fabric pegs

[faint stag]

and it should work

nah in maths

its just a dot

or a circle

the way i drew it there

[deft hazel]

why is your

diagram like that

isn't it how we initially thought it was

[faint stag]

nope

the peg cant be in the middle of BC

[deft hazel]

i never assumed it was in the middle

i just read it rests on segment BC

[faint stag]

i think the way u drew it was fine as well

but then weight wouldnt down but a little to the left

im imagining an actual wire trying to fall down the peg

and friction is stopping it

i think this is the only way it makes sense

[deft hazel]

Img load please

the most

physicaly representation of our diagram i can think of

Idk if that's what a peg is

cause when i search up peg i get a trillion things

let the coat hanger be the wire

clearly isosceles

[faint stag]

in maths though we model a peg as just a circle

[deft hazel]

well yes

but a horizontal peg

does that mean

it's a line

like

[faint stag]

good point

maybe it goes horizontally like

into the page?

i honestly have no clue

P is the peg

its just a dot

[deft hazel]

the issue with natural language is it's ambigious when it comes to logic and mathematics

why is your

it's not in equilibirum

[faint stag]

thats examplar

[deft hazel]

oh

[faint stag]

it would be in equilibrium

because F is equal to some component of weight

and N is equal to some component of weight

[deft hazel]

wait what's F

and what's N

N=F_n?

Normal force?

[faint stag]

F is friction

N is normal force

[deft hazel]

oh

[faint stag]

i see why they caclulated the centre of mass

once they get the distance, they can calculate the angle

i guess

im not sure what alpha is

[deft hazel]

let's just

do you have solutions to this q?

or not

[faint stag]

yeah

the examplar

and the official ms i guess

wait ill look at the official ms

[deft hazel]

send me it too

[faint stag]

this is a hint i think

this is the actual mark scheme

[deft hazel]

let's do some maths here let's find the vector projection

of W

[faint stag]

alright

[deft hazel]

that's what we're prolly looking for

idk when i look at wikipedia for a Euclidean vector i go down to the "Decomposition or Resolution" section (the main article links me to "Basis" which is some Linear Algebra shit, but as far as my highschool knowledge goes and the inattention I've given to my class last year we're looking for the vector projection of a vector a on (or onto) a nonzero vector b(also known as the vector component or vector resolution of a in the direction of b)

[faint stag]

yeah i think once we resolve W properly its easy

[deft hazel]

Denoted as

[faint stag]

so are we gonna calculate the centre of mass?

[deft hazel]

yes

well

why do we need the centre of mass

[faint stag]

well the distance from BC to the centre of mass

[deft hazel]

lemme relearn how to find a vector projection

because it gives us the orthorgonal projection of a onto a straight line parallel to b

[deft hazel]

So we could

well

[faint stag]

right

[deft hazel]

i feel like this is a way

but we just don't know how to do it this way

mane this is getting so bloody confusing

hol up

well in order for us to find a_1

we need to determine

what our vector b is

and it's direction

cause

really there are many components that can define a southernly vertical vector

or am i being stupid

oh wait hold up

[faint stag]

how come we are using vectors?

[deft hazel]

er just to visualise

[faint stag]

alright cool

[deft hazel]

So our understanding or well not understanding is to why

[faint stag]

yeah i think so

[deft hazel]

The vector of weight of the wire can be decomposed, one parallel to the contact surface, the scalar rejection ||W||sinθ and the other perpendicular to it, the vector projection ||W||cosθ. Since the wire is not moving vertically, ||F||_n=||W||cosθ.

The Euclidean norm of ||W||cosθ is simply the magnitude of this vector, which represents the force acting perpendicular to the contact surface.

[faint stag]

makes sense

[deft hazel]

So we can move forward from there

which solutions we gonna go with

chatgpt or the one provided by cambridge

i'm acc dead dude

[faint stag]

cambridge would be pereferable

[deft hazel]

aight

lemme print it out

So we essentially resolved

Resolve in the context of physics (and maths) just means to analyse (a force or velocity) into components acting in particular directions

and we justified that

I'm unsure what F, and R is

and is alpha the same angle as ours?

[faint stag]

no alpha is another agnel they lballed

laballed

between BC and the horizontal

[deft hazel]

aight

do you know how to find the centre of mass?

i'll write down how to derive

(2a+2b) x bar=2b(1/2bcos theta)

[obtuse pebble]

@faint stag Has your question been resolved?

[deft hazel]

messaging to keep the channel up

Heres half of it

[obtuse pebble]

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[deft hazel]

.reopen

Fuark

[obtuse pebble]

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[exotic creek]

,w integrate \frac {1}{(2-x)^{2}}

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yomiko#1028

[exotic creek]

shouldn't there be a (-) infront of the answer?

$(2-x)^{-1}$ \ $-(2-x)^{-1}$ = $\frac {-1}{2-x}$

[warm shale]

yomiko

[ancient jacinth]

what is the derivative of 2-x

[exotic creek]

-1

[ancient jacinth]

ye

so when you do u sub, the negatives cancel out

[exotic creek]

oh yeah

yh

thanks

.close

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[obtuse pebble]

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[toxic basin]

yo can someone help me out in understanding something about acceleration and distances
if a=v/t, how can s be equal to at^2/2
woudlnt we substitute v in a=v/t with s/t and get s=at^2

[heavy frost]

yo volex

ill write that down and send ya wait a min

[toxic basin]

tnx gang

[heavy frost]

Lemme get something straight first

Do you agree with me that area of ABCD is equal to S

Because vt=s

[toxic basin]

yeah, i understand that by looking at the graph, v0t+at2/2 would be the correct formula

[heavy frost]

You already understand?

[toxic basin]

but i dont understand it by looking at the a=v/t

if v can be supstituted with s/t

[heavy frost]

a isn't v/t

it's a=deltaV/t

[toxic basin]

im looking at v0=0

[heavy frost]

Oh than yes

Acceleration means how much speed increases per second

and v/t will equal to that (if V0=0 as u said)

[gloomy valve]

and v = delta s / delta t
so substitution doesnt work

[toxic basin]

can i give yall a straight up example

[timid silo]

a -{integrate}-> at -{integrate}-> a/2 t²

[heavy frost]

Do u agree that abhd area+ bch area is entire S?

[toxic basin]

yeah i understand it when looking at the graph but not looking at the formulas only

Take a look at this example: if s1=2cm, s2=8cm, s3=18cm, s4=32cm, t1=0.2s, t2=0.4s, t3=0.6s, t4=0.8s and t5=1s, whats the s5

[gloomy valve]

you need integration if you dont want to do it graphically

[toxic basin]

i calculated v1 is s1/t1 and v2 is s2/t2

then i calculated a=(v2-v1)/(t2-t1) and got 50

then i plugged it into s=at2/2 and got 25 which is obviously wrong

[heavy frost]

I don't think in this case acceleration stays the same

[toxic basin]

it does

[heavy frost]

wait I'll check

I mean graphically is just easier to understand

[gloomy valve]

no v1 is (s2 - s1) / (t2 - t1), v2 is (s3 - s2) / (t3 - t2) and so on

[heavy frost]

Reason why at² gets decided by 2 is that we take average speed

at=v

at²=vt

and v you should take is average one

Because it gets increased

I don't know if u understand my English lol

[toxic basin]

how

[gloomy valve]

you calculate velocity using the differences of distance & time, in this case the neighboring points

the avg velocity would be (s4 - s1) / (t4 - t1) from t = t1 to t = t4

[toxic basin]

omg i think i got it

i went under the assumption that there was no acceleration

[gloomy valve]

but now we dont know if v changes continuously or if it changes abruptly. so you have to make an assumption there or the problem tells you. physically correctly it would continuously change ofc

[toxic basin]

like, for the first 0.2s the speed is constantly v1, then at the point 0.2 it magically becomes v2 and keeps being v2 untill 0.4 etc

then the v=s/t would be correct

but the speed is constantly changing so i should use s=v0t+at2/2 to get the a

then after i get the a, i can calculate s5 simply by plugging it back in the same formula using 1s

am i correct?

[gloomy valve]

well no, v doesnt magically jump between values. the formula s = (1/2)at^2 assumes acceleration is constant and velocity changes continuously.

[toxic basin]

yeah so if it took 0.2s for somehthing to move 2cm, his velocity is 2/0.2 only if the speed is constant, but if its accelerating its a*t

[gloomy valve]

yes if you apply a constant force to the object -> constant acceleration

[toxic basin]

so to calculate the exact speed at any poing i just use a*t and t being that exact second no matter the distance it crossed (which is used only to find the a)

ofc all of this is about the object with v0=0

[gloomy valve]

yes

in general v = at + v_0 as you know

[toxic basin]

alr tnx man i appreciate it

.close

[obtuse pebble]

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[sand laurel]

I have the following equation. And I want to solve for dy/dx.

[sand laurel]

If I use implicit differentiation and use the quotient rule I get

But If I first multiply the original equation by y on both sides and distribute and then use the product rule and solve for dy/dx I get

So why am I get different answers?

[timid silo]

its not correct

[timid silo]

this one is right. by direct implicit differentiation

[sand laurel]

Ok let me redo my work and see what I get

Where did I go wrong

[timid silo]

1st step is wrong

its applicable when it is explicit

[sand laurel]

what do you mean?

[timid silo]

it means it is applicable when the are variably seperable

[sand laurel]

What is applicable

so to be clear I'm not allowed to use the quotient rule?

[obtuse pebble]

@sand laurel Has your question been resolved?

[sand laurel]

<@&286206848099549185>

I dont understand how you went from teh "simplifying this expression, we get: " to "grouping the dy/dx terms on one side..."

I cant manipulate hte algebra to get to the equation you have

[tame narwhal]

is this chatgpt or something? don't just dump all the steps, either

[tame narwhal]

this is correct according to wolfram alpha

[sand laurel]

well so is the second picture if you first multiply both sides by y. So how can 2 equations, that are algebraicly identical (except for at y = 0) give different derivitives?

Is it "illegal" to manipulate the algebra when implicitly differntiating?

[wild swallow]

#changelog message

it's also wrong

[wild swallow]

have you ever considered the possibility that, in fact, the two expressions are equivalent

funny joke

[sand laurel]

Well I get that the expressions are equivalent but their derivitives aren't

which is why I'm really confused

[devout sable]

<@&268886789983436800> chatgpt

[obtuse pebble]

@sand laurel Has your question been resolved?

[timid silo]

See the solution you have uploaded..the first step is wrong because you're treating two different functions as independent and directly you're applying quotient rule..which is not valid because they are one function

If you want I can upload solution by solving two different methods

[obtuse pebble]

@sand laurel Has your question been resolved?

[obtuse pebble]

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[karmic mural]

Why is friction not acting on the lift

[karmic mural]

And where is friction even coming from

[brisk matrix]

i’m confused by your question

what do you mean friction is not acting on the lift

[obtuse pebble]

@karmic mural Has your question been resolved?

[karmic mural]

It says friction is not doing work on the lift

Why

[brisk matrix]

oh sorry i was looking at part a

[obtuse pebble]

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[warm lintel]

Yo i got a very dumb q but it would help loads if yall could answer it

[warm lintel]

basically proving this identity

is easy and all

but

i reach a point

in the last step

idk how they got the 1-

its prob sum am not seeing

or lack of sleep

i have no idea

this is my work

[obtuse pebble]

@warm lintel Has your question been resolved?

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[polar plume]

Could someone explain where the highlighted values are gotten from

[brisk matrix]

from gamma(t)

the second equation in the image

[polar plume]

I'm a bit dumb could you clarify a bit more please lmao

[brisk matrix]

look at the picture you sent

and the definition of gamma

$\gamma(t)$

[warm shale]

maximo

[brisk matrix]

it’s the second equation you have

under the f(x,y,z)

[polar plume]

Ohhh i see it now

Thanks for the help

.close

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[azure anchor]

solve for t, $0=2\cos2t-2\sin t$

[warm shale]

b0ngl0rd

[brisk matrix]

what have you tried

[azure anchor]

i dont even know how to start. i tried simplifying by identities but i ended up getting the wrong answer

[brisk matrix]

use the double angle identity for cos(2t) and then turn the cos^2(t) into 1-sin^2(t)

[azure anchor]

i tried taking the factor 2 out, then using the double angle and did exactly what you said, but somehow it changes the function? i ended up getting a critical value of pi/2, but thats wrong, it should be pi/6

[brisk matrix]

can you show your work

[azure anchor]

wait

1 - sin^2?

omg

[brisk matrix]

i don’t see how you got from the third line to the fourth

[azure anchor]

i didnt realize it was different. i assumed cos(2x) should be similar to cos(x + x)

[brisk matrix]

it is

[azure anchor]

ohhh i got the sign wrong on the formula

[marsh geyser]

cos(2t) - sin(t) = 0
1 - 2sin^2(t) - sin(t) = 0

use this

and replace sen(t) with x, then solve the equation

[brisk matrix]

they could figure this out on their own, and it’s not super helpful if they don’t see what the steps are

[marsh geyser]

i thought u said it before

[brisk matrix]

i didn’t, i explained the steps hoping they’d reach the equation on their own

[marsh geyser]

sry i'm not understanding, u said to him about the double angle identity, what did I say more than that?

[azure anchor]

woops forgot a 2

[brisk matrix]

you didn’t say how you got the second equation is the point

and it’s best if you don’t just drop the equation on them like that

[azure anchor]

could i use the quadratic here by subbing sin(t)=x

[brisk matrix]

and explain the steps so they can reproduce them

[brisk matrix]

that’s the idea yeah

[marsh geyser]

oh sry, i know what u mean now, my bad

[marsh geyser]

you can take advantage of the 0 in the equation to make the right side of the equation simpler

[azure anchor]

maybe dont use quadratic? doesn't seem like i'll get pi/6 this way

oh wait

[marsh geyser]

you found x, now you must find t

[azure anchor]

right, lemme write it out

got it

thanks guys

.close

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[hushed gust]

Hi all! Does anyone know what's wrong with this matlab code?

[brisk matrix]

could it be that you’re reusing f?

[hushed gust]

just changed these 2

[steady aurora]

you use n before you've defined it

your compiler doesn't know what it is when it reads the top line

[hushed gust]

ty bluewolf and maximo :0

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[brisk matrix]

the only restriction comes from the division

what is the only number we can’t divide by?

[high fiber]

zero

or 2 i mean

[brisk matrix]

no zero is correct

[high fiber]

2 in place of x i meant

[brisk matrix]

so the only values not allowed are those that make x^2 - 4 = 0

what are the values of 2 that make that equation true

[high fiber]

x = 2 or x = -2

[brisk matrix]

yes

so those are the only values not in our domain

you could write the domain then as $x \ne 2,-2$

[warm shale]

maximo

[high fiber]

so the domain of f(x) is all real numbers except x = 2 and x = -2?

[brisk matrix]

yes

[high fiber]

do I need to write this in interval notation?

like (-∞, -2) U (-2, 2) U (2, ∞)

[brisk matrix]

i don’t know

i don’t know what the question is

if you understand the domain from just $x\ne 2,-2$ then you don’t really need the interval notation

[warm shale]

maximo

[high fiber]

ah

[high fiber]

now I need to find the x and y intercepts of this function

[brisk matrix]

do you know how you find those

[high fiber]

I think so, lemme show you my work in a few

[tame narwhal]

,rccw

[high fiber]

is there also an y intercept at 0?

[tame narwhal]

,w plot 4/(x^2 - 4)

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by cwatson#5438

[tame narwhal]

you already had the answer

[high fiber]

wait what

[tame narwhal]

there is no x-intercept and the y-intercept is -1

[high fiber]

how do I calculate the functions horizontal asymtotes

[tame narwhal]

I suppose see where its derivative equals 0?

[high fiber]

i mean, I can clearly see there are no horizontal asymtotes, but how do I calculate its derivative

[tame narwhal]

the same way you calculate the derivative of any function

[obtuse pebble]

@high fiber Has your question been resolved?

[obtuse pebble]

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[jovial stag]

Hello I have to find intersection of 2 Affine subspaces. They are represented by those equalities

A1 :
3x_1 + 4x_2 + 6x_3 = 3
4x_1 + x_4 + x_5 = 2
x_2 - 4x_3 = 8

A2 :
x_2 + x_3 - x_4 - x_5 + 2x_6 = 0
-x_3 + x_4 + x_5 - 2x_6 = 0
x_2 - 7x_3 = 0

I tried Top-Down method but it does not have a solution so what it means?

[obtuse pebble]

@jovial stag Has your question been resolved?

[obtuse pebble]

@jovial stag Has your question been resolved?

[obtuse pebble]

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[distant rampart]

4. For what values of $h$ and $k$ is the following system consistent?
   $$
   \begin{array}{r}
   2 x_1-x_2=h \
   -6 x_1+3 x_2=k
   \end{array}
   $$

[warm shale]

dgh

[distant rampart]

Is this correct?

[unkempt vault]

yes

the lhs of the second equation is -3 times the lhs of the first, so the same must hold for the rhss

[warm shale]

dgh

[unkempt vault]

or wait

can you show your row-reduction work?

I think you meant 3 h + k, not h + 3 k

[distant rampart]

yes you're right, mb

.close

[obtuse pebble]

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[distant rampart]

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

24. $\left[\begin{array}{rrr}1 & h & -3 \ -2 & 4 & 6\end{array}\right]$

[warm shale]

dgh

dgh

[distant rampart]

Now, what does this tell us?

24

[obtuse pebble]

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[obtuse pebble]

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[torn tapir]

anyone know how to do similar triangles for geometry

[obtuse pebble]

Please don't occupy multiple help channels.

[unreal musk]

#help-20

.close

[obtuse pebble]

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[fast shadow]

how can I solve this inequality?

[obtuse pebble]

Please don't occupy multiple help channels.

[fierce lagoon]

Stick to one channel

[fast shadow]

.close

[obtuse pebble]

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[high fiber]

how do you find the horizontal asymtotes of a function?

[fierce lagoon]

Consider the limit when x-> infinity and x-> -infinity

Yeah so take the limit as x -> infinity and when x -> -infinity

[high fiber]

how should i format that on paper

[fierce lagoon]

Do you know limits?

[high fiber]

yeah

[fierce lagoon]

So you'll just evaluate $\limit{\frac 4{x^2 -4}}{x}{\infty}$ and $\limit{\frac 4{x^2 -4}}{x}{-\infty}$

[warm shale]

Umbraleviathan

[high fiber]

im getting zero when evaluating both

[fierce lagoon]

Yeah

So your only horizontal asymptote is y = 0

[high fiber]

do you know of any easier way to evaluate these

[fierce lagoon]

,w plot 4/(x^2 - 4)

[fierce lagoon]

Typically if the order of the denominator is larger than the order of the numerator then the asymptote is y = 0

But I'm pretty sure that only applies if the numerator and denominator are continuous for all real values of x

Might not

[fierce lagoon]

But this typically holds true

[high fiber]

oh, so will it work for this one

[fierce lagoon]

If the order of the numerator is larger than the denominator, then there's gonna be an oblique asymptote

And if they're the same, it's just the numerator's leading coefficient divided by the denominator's

[high fiber]

okay, how do i find the vertical asymtotes

i know they are at x=-2 and 2

[fierce lagoon]

Simplify/reduce the function and then set the denominator equal to 0, solve for x

[high fiber]

do i need to do any calculations for this

[fierce lagoon]

Eh

Minor

Just simplifications

[obtuse pebble]

@high fiber Has your question been resolved?

[obtuse pebble]

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[tired sierra]

find the anti derivative of this

a special way of saying intergrate it

I somewhat forgot how to intergrate something like this but to make sure I am doing this right

would this be right?

the red 4 is from me intergrating 2x^2+3

or should it just be 4 x 2

[obsidian isle]

let's back up

we want

$\int 8x(2x^2 + 3)^3 \ dx$

[warm shale]

EndTimes

[tired sierra]

yes

[obsidian isle]

are you familiar with the concept of u sub?

[tired sierra]

yes... but I think that would be a bit excessive here

[obsidian isle]

no, it's actually the easiest way

anything else would involve much more effort

[tired sierra]

give me a quick second if you don't mind?

[obsidian isle]

try letting u = 2x^2 + 3

[tired sierra]

uhh

that'd make 4x

to make the 8 a 4x

we just put a 2 outside the intergration symbol

ok I'll asume so

Oh wow I got the right answer

thanks for the help endtimes

.close

[obtuse pebble]

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[peak ginkgo]

I need help on data analysis and probability

[grizzled shore]

Part a or b

[peak ginkgo]

both actually

[grizzled shore]

Do you know how to find expected value

[peak ginkgo]

no...

[grizzled shore]

This is a discrete probability problem

So we have a few things

$E[X] = \sum_{\text{all }i} x_ip_i$

[warm shale]

Frosst

[grizzled shore]

And $p_i=P(X=i)$

[warm shale]

Frosst

[grizzled shore]

Have you seen these before?

[peak ginkgo]

the first one yes

but i dont really understand it

[grizzled shore]

Ok

Let’s understand that second one first

[peak ginkgo]

okay

[grizzled shore]

p_i means the probability of the ith event

Now we say this is equal to, the probability our outcome = i

That’s the P(X=i)

So an example would be, let X denote the upwards facing side of a 6 sided fair dice roll

Then we can say P(X=1) = 1/6

Or, the probability the face up side is 1, is 1 in 6

We can also say P(X=0) = 0

The probability the face up side is 0, is 0

There is no face that’s 0 so it’s impossible to get it

Following?

[peak ginkgo]

yes

[grizzled shore]

So in this case, we can say p_1 = P(X=1) or slice 1

We’ll call the slice the dart lands on i

So you can land in slice 1 through 8

And since the question says they are equally spaced and the landing is random, p_1 = 1/8, p_2 = 1/8, … p_8 = 1/8

Yeah?

[peak ginkgo]

yes

[grizzled shore]

Now let’s look at this

We know what p_i means

Let’s see what x_i means

It pretty much just means the value of an outcome

https://cdn.discordapp.com/attachments/903430332324405288/1091970137755750521/image.png

So it says if you land in slice 1, the value is $1

If you land in slice 2, the value is $3

Etc etc for each slice

[peak ginkgo]

okay

[grizzled shore]

Now, let’s ask another question

What is the expected value of a dice roll? (6 sided and fair)

[peak ginkgo]

hold on

1/6

[grizzled shore]

No no

If I roll a 3 that’s a value of 3

No rolls are less than 1 so the expected value must be at least 1

Ok how about an easier concept

If you flip heads you win 1 dollar

If you flip tails you lose 1 dollar

What is the expected value of this game

(And how did you get to your conclusion)

(Coin is fair)

[peak ginkgo]

can u explain it again? sorry

[grizzled shore]

Where

Reply to the message where you got lost

[peak ginkgo]

alr wait

[peak ginkgo]

this one

[grizzled shore]

Ok what part of that confuses you

[peak ginkgo]

i just dont understand it in that type of text layout

(the formula)

[grizzled shore]

Do you know what p₁ means

p_1 is the same as p₁

[peak ginkgo]

uhuh

[grizzled shore]

p₁ means it’s a probability with a subscript 1

So it’s got something to do with 1, like it’s dependent on it but not really

You’d see it in things like a₀ + a₁x + a₂x² + a₃x³ + … is a polynomial

Now p_1 = we are just defining what this p_1 means

So we write P(X=1)

Do you understand this notation?

[peak ginkgo]

p times x=1?

[grizzled shore]

No, P is a function

It takes in X=1 as its input

Like f(x)

[peak ginkgo]

so its like p is x=1?

[grizzled shore]

It reads as

P(X=1): what is the probability of our outcome being 1?

[peak ginkgo]

oh

[grizzled shore]

Our outcome = the random variable X

Being 1 = “=1”

Our outcome being 1: X=1

So P(X=1) is a function that takes in “when our outcome is 1” as an input

And gives you a probability as an output

Is that clearer?

[peak ginkgo]

yes

[grizzled shore]

So do you understand the notation p_1 = P(X=1)?

[peak ginkgo]

so the coin flip is 0.5

thx i figured out the problem. its -$0.75 in every toss

imma close it now

thx for the help

.close

[obtuse pebble]

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[bright geyser]

im trying to solve these simulteanously but my results are not working..

using the second eq

im taking conj of both sides

and simplifying i get

z - w = 1 + 15i rigfht

z = 1 + 15i + w

subbing into first

3(1+15i+w) - 2w = 8 + 39i

3 + 45i + 3w - 2w = 8 + 39i

w = 5 - 6i

then z would be

6+9i

wait im correct

i was subbing back in wrong

bruh

.close

[obtuse pebble]

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[obtuse pebble]

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[lethal cape]

what are the steps to simplify expressions?

[mellow glacier]

get the like terms on LHS and RHS

apply the operators

get the variable on LHS and solve RHS to get the answer

[lethal cape]

alright

i get it

thank you ❤️

[mellow glacier]

@lethal cape do .close

[lethal cape]

.close

[obtuse pebble]

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[timid silo]

How do I do b?

[median dome]

try writing the expression in the form given in the first part

then find what a should be to satisfy that

thus you have the value of t

[timid silo]

I don’t get it

Like rewrite it using t formula?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[timber island]

well you're given a cos x - sin x = 1

since a cos x = 1 + sin x

subt. sin x from both sides

[timber island]

you get this

so ur goal is to find an appropriate value of x

when a=2

[timid silo]

How do you know when a=2

.close

[obtuse pebble]

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[tiny isle]

hi

can someone help me to break this down

somehow i can't get 50/3

nevermind

got it

[obtuse pebble]

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[rough bough]

How do I prove that $ x^\frac{1}{7}$ is differentiable using first principle?

[rough bough]

$ x^\frac{1}{7}$

x^(1/7)

[kind hawk]

the tex bot doesnt like spaces

[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

<@&286206848099549185>

[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

<@&286206848099549185>

[rough bough]

<@&286206848099549185>

[stable wyvern]

@rough bough tell me

[rough bough]

Using first properties

f(x) - f(x+h) / h

but (x^(1/7 - x^(1/7+h) )/ h as h-> 0

my concern is what do i do next?

[stable wyvern]

hmm

lemme check the question

is the whole question?

[rough bough]

Yeah

[rough bough]

that's the entire question

[ancient jacinth]

$x^\frac{1}{7}$

[warm shale]

Blighter

[ancient jacinth]

it should be

$\frac{(x+h)^\frac{1}{7} -x^\frac{1}{7}}{h}$

[warm shale]

Blighter

[ancient jacinth]

as h approaches 0

[rough bough]

okay

what do you think we can do here?

[ancient jacinth]

do what newton did

[oblique turtle]

factor the x out

[ancient jacinth]

how?

[oblique turtle]

I don't know latex but

a^1/7 * (1+ a/h)^1/7 - a^1/7

do you know binomial expansion ?

[rough bough]

yes

[oblique turtle]

(1 + x)^a

[ancient jacinth]

what are you trying to write?

[oblique turtle]

$(x)^\frac{1}{7} *(1+h/x)^\frac{1}{7}-x^\frac{1}{7}$

[ancient jacinth]

https://youtu.be/gMlf1ELvRzc
@rough bough check this vid out

where did a come from?

[rough bough]

watched that before

[ancient jacinth]

it helps to understand the binomial theorem

you can apply that here

[warm shale]

Mathu_lmn

[rough bough]

really confused

[obtuse pebble]

@rough bough Has your question been resolved?

[obtuse pebble]

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[tawny void]

Need help with how to do this

[obtuse pebble]

@tawny void Has your question been resolved?

[obtuse pebble]

@tawny void Has your question been resolved?

[tawny void]

<@&286206848099549185>

[obtuse pebble]

@tawny void Has your question been resolved?

[tawny void]

<@&286206848099549185>

[vale fjord]

Get the inverse of matrix M

[vale fjord]

Inverse of matrix M

3 4 1
-2 -3 2
1 1 4

then those numbers below "Decode this message”, make them a 3 by 1 matrix

110
-79
-7

-195
140
15

-120
86
9

157
-117
-12

-151
108
11

-107
76
7

376
-273
-28

Please double check the numbers

Then...

Multiply the inverse of matrix M by each of the 3 by 1 matrix

repeat multiplication until each 3by1 matrix is multiplied

Then

You will get answer. Decode it using the numerical assignment chart

I am not sure if this is the message. But this is what I've decoded:

MEET IN CRYPT AT FOUR

I don't know how to type matrix here in discord

[tawny void]

.close

[obtuse pebble]

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[tame narwhal]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[clear adder]

1

[clear condor]

how much perimeter does the shape have?

[clear adder]

both area and perimeter we have to take it out ourselves

[tame narwhal]

take it out?

[obtuse pebble]

@clear adder Has your question been resolved?

[obtuse pebble]

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[median dome]

can you translate

[glossy haven]

The triangle ABC has sides a, b, c and angles alpha, beta, gamma
The area of the triangle is "A"
Prove/show that a^sin2beta + b^2sin2alpha = 4 times the Area

[glossy haven]

yes yes

I know basic trigonometry, or maybe not so basic, I am at the first semester of university

things like cosine of a sum, sine squared, etc

I tend to brute force this problems, and that almost always works, but I have no time on a test to brute force something and spend one hour

[median dome]

try writing things in terms of the side CD

since that creates the right triangle

remember that $$A = \frac1{2}CD\cdot c$$

[warm shale]

kheerii

[glossy haven]

wait..........

A = 1/2 CDc...........

oh

ah yes

I understand

base * height............

[median dome]

1/2 base * height

correct

[glossy haven]

ahm...

I was thinking about how to do it, I havent seen any methods of obtaining the height yet at my class but...

I learnt this at highschool, but I still cant see how thats useful

so first thing I did was this (just re-write the "sin2a" and "sin2b" at the beginning)

and I've been trapped here 20 minutes

[median dome]

try writing sin(b) cos(b) etc in terms of sohcahtoa

[glossy haven]

ooooooooh makes sense

[median dome]

maybe write DB as x and AD as c-x

[glossy haven]

I will try, ty

okay, arrived to 2 * c * h = 4A, and yeah, thats it

tysm

it was really simple, tysm again!!

[median dome]

no worries

[glossy haven]

<3

[median dome]

yep, try to use sohcahtoa whenever you have a right triangle

[glossy haven]

yes, tysm again again

[median dome]

and label thingsa

no worries

[glossy haven]

.close

[obtuse pebble]

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[sleek mural]

Question

[sleek mural]

By this theorem, are the two angles congruent as well?

Or is it referring to just angles 1 and 2?

,rotate

[obtuse pebble]

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[sleek mural]

<@&286206848099549185>

[cold star]

the theorum refers to angles 1 and 2 only

what the theorum is saying is like if two angles have the same endpoints then they are congruent

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[spice path]

why i wrong

[obtuse pebble]

@spice path Has your question been resolved?

[sleek mural]

But don’t they also have the same endpoints there?

[cold star]

if youre talking about angle b and 1 being congruent, they do not share the same endpoints

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[manic sparrow]

iam a little unable to understand even and odd functions

[manic sparrow]

f(-x) = f(x)

is even

i mean thats the formula i guess

lets say x is 4

what does the output have to be in order for it to be even function

[solar trellis]

There's no output that will guarantee f being even

[vagrant gale]

f(-x) = f(x) really just denotes equality

f(x) can be any function, but if f(-x) is equal to f(x) then the function is even

take for example f(x) = x^2. if you plug in 2 for x, you get f(2) = 2^2 = 4. likewise, if you plug in -2 for x, you get f(-2) = (-2)^2 = 4.

but do note that this property must be satisfied for all values of x

[manic sparrow]

so the output has to be the original value without the negative? or just the original value?

[solar trellis]

No

The output of x is f(x)

[vagrant gale]

No, the output can be literally anything. If the function is even, though, the output for x will always be equal to the output for -x

[solar trellis]

The output of x should be the same as the output of -x

[vagrant gale]

another way to think about an even function is that it is reflected across the y axis

[manic sparrow]

sorry made a mistake let me rephrase, if you input negative value it has to be equals to positive or just the original?

[vagrant gale]

it has to be equal to the positive value

[sleek mural]

What do u mena by endpoint?

[manic sparrow]

ah, for an even function? i see

[sleek mural]

Isn’t it the same arc?

[cold star]

i dont think it has anything to do with arcs for this theorum

its just the two angles have the same endpoints

[sleek mural]

They intercept congruent arcs

[cold star]

arc cd and arc ab are congruent to each other

[sleek mural]

Right

But they’re also the same arcs

Wait, how do u know they’re congruent tho?

[cold star]

there was a theorum that stated it

but i dont have my notebook with me so i dont have it right now

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[sage dagger]

What topics or area of math should I be learning before properly getting into calculus

[sage dagger]

I know there’s a lot of trig stuff

what else?

[robust sleet]

Algebra

[sage dagger]

I’m fine with the algebra right now

[fathom flicker]

This is probably better suited in #discussion, but I would recommend being familiar with lots of algebra and trig for the most part. I mean "pre-calculus" classes typically just go over different types of functions, function notation, algebra review and trigonometry.

[sage dagger]

Okay

How about logarithms?

[robust sleet]

Yes

[fathom flicker]

you should be familiar with logarithims

[sage dagger]

alright

[drifting badger]

Adding to this: you should in general be familiar with most of the well known functions, to the extend you can without calculus

[manic sparrow]

i thought logarithm was simple concept, or maybe what i learned is just basic log?

[bold bane]

A pre-calculus course will give you a general understanding of what you should know. If you want a more deep dive, a course on trigonometry will go into a lot more detail.

[manic sparrow]

log = power to which a number must be raised to to get x

thats it right?

[sage dagger]

I see

By the way what areas of trig should I be familiar with?

Everything?

[manic sparrow]

Pythagorean Theorem i think is one

[fathom flicker]

trig functions, basic trig identities, periodic functions and their behavior, range and domain of trig functions, range and domain of inverse trig functions

[sage dagger]

that’s not just a^2 + b^2 = c^2 ?

[sage dagger]

Alright thanks I’ll keep that in mind

[bold bane]

A course on trigonometry will not only teach you about SOH-CAH-TOA. You will learn about complex numbers, exponential and logarithmic functions, parabolas, ellipses, polar notation, and rotating functions.

[manic sparrow]

thats the one, but i think theres more stuff, working with triangles and finding length etc

cos sin tan all that

[sage dagger]

Ohh, I didn’t know you had to learn complex numbers for calculus

[fathom flicker]

not really, but being familiar is nice

[bold bane]

You don't really need it but it's interesting to know.

[sage dagger]

Alright

I’ll def look into those things

[fathom flicker]

are you planning on self learning calculus?

[sage dagger]

Pretty much yeah. I know it’s not going to be easy but given my situation in highschool right now I’m gonna have to self study it

[fathom flicker]

It's definitely doable! Do you already have a plan of what resources you might use?

I'd strongly recommend looking into this if you don't already have a plan