help-0

[proven haven]

where do i get a graph

[grave quartz]

well on some paper lol

[proven haven]

im graphing it rn

[grave quartz]

ye do and u will see that again there is only one answer lol

[proven haven]

yea so this what i got

whats that

[grave quartz]

ok now dot to dot it

[proven haven]

there

what that

[grave quartz]

ok I think u should get that lol

[proven haven]

quadratic?

[grave quartz]

U need to revise graphs of functions

yes

[proven haven]

ok

[grave quartz]

linear is straight

[proven haven]

I need help on 1 more question

well a lot more actually

[grave quartz]

ok

[proven haven]

if u dont mind

I got like 10 in total but they simple

[grave quartz]

Ill have a look at one more then im going because im tired and its late for me

[proven haven]

alright

[grave quartz]

so first thing is first... Simplify B and C

I know the answer but do this anyway

[proven haven]

how do i simplify those

[grave quartz]

well expand the brackets and get a equation y =ax+b like A and D

[proven haven]

y = -3x-1 for B???

[grave quartz]

no

try again

[proven haven]

sec

I think i got it

is it C

[grave quartz]

correct

im off so Ill love u and leave u !

[proven haven]

alr thank you nye

bye

.close

[lone heart]

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[wheat isle]

I’m not sure what to values to put in into the formula

[lone heart]

@wheat isle Has your question been resolved?

[merry depot]

The notation P(A | B) means "Probability of A given B"
So you have written P(F | NR) which i assume means Probability of Fire given No Rain, which isn't one of your questions.

And you have also written P(A n B) / P(B) which is equal to P(A | B)

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[blissful hound]

i factorized this problem but when i used geogebra it gave me a weird difference between the answer in the video and mine

[blissful hound]

should i have not factorized it and why?

or my answer is wrong?

[merry depot]

factoring it doesn't do anything. so you're answer is probably wrong.

[blissful hound]

oh thanks,let me do it again real quick

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[frail spear]

Hello

[frail spear]

I need help

[mental flame]

send your question

also for future reference, send your question with your initial message

[frail spear]

I don't know where to start to solve this problem

[lone heart]

@frail spear Has your question been resolved?

[frail spear]

.close

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[hazy fulcrum]

I need help solving

The shape of f(x) = |x|, but shifted 12 units up and then reflected in the x-axis.

[hazy fulcrum]

The answer given in the book is f(x) = -|x| - 12

But shouldn’t it be + 12 instead of - 12?

[fierce narwhal]

it's reflected in the x axis

so before it's reflected, it's +12

[hazy fulcrum]

Wait let me put it on desmos again to see it clearly

[fierce narwhal]

before reflecting vs after

[hazy fulcrum]

So it turns into -12?

[fierce narwhal]

yup

[hazy fulcrum]

Can I see it as - (|x| + 12)?

[fierce narwhal]

yes, which is the same as -1(|x| + 12)

the -1 is distributed to both terms as -|x| - 12

[hazy fulcrum]

OHHH

OKOK I UNDERSTAND IT

thank you

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[rotund crater]

How do I calculate the area of ​​a rectangle using integrals?

[vapid shuttle]

Why do you want to do this?

whatever the height of the rectangle is, say 5, integrate that along the length of the rectangle (if the length is 10, do integral from 0 to 10)

you should see very quickly that this is just the same as doing base * height

[rotund crater]

why cant be -1?

[vapid shuttle]

it is an area

areas aren't negative

but why are you asking this?

[rotund crater]

because after I want programming with java and I want demostrate this.

[vapid shuttle]

to demonstate what?

after what?

[last ether]

Unless you're talking about net integrals

[vapid shuttle]

OP refuses to give context here

so I am assuming that he is referencing the area of an actual rectangle

[rotund crater]

To show what is the area of ​​the rectangle

[last ether]

Do you know how integrals work in the first place? Start with that

[rotund crater]

yes

[vapid shuttle]

why do you need to do an integral for this? Area equals base * height

[last ether]

Alright so you're aware that if the width is just the distance between the bounds (b-a) as long as b>a

And then your height is just gonna be the integrand

Because you're gonna add all the skinny rectangles between y = height and y = 0 from x = a to x = b

And that'll give you the area of a rectangle of a certain height with a width of b-a

[rotund crater]

It's that I want to use integrals, I don't want to neglect myself and I forget how to integrate a function.

[last ether]

I mean I just told you what to do

[vapid shuttle]

( so did I earlier )

[last ether]

I'm surprised Austin didn't hit me with the !nosols

But I'm too cool for that

[vapid shuttle]

@worn fox spotted

[last ether]

Reported tbh

@worn fox jail time

[rotund crater]

what happen aaa

[last ether]

Read from here

[rotund crater]

i see thx for the help

[lone heart]

@rotund crater Has your question been resolved?

[lone heart]

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[wheat isle]

https://media.discordapp.net/attachments/490557019623915520/1141133340896927875/image.png

https://media.discordapp.net/attachments/490557019623915520/1141133340636885002/IMG_9582.png

Btw the F|NR is supposed to be NF|NR

[wheat isle]

I don’t know what values to use here

[lone heart]

@wheat isle Has your question been resolved?

[ancient saddle]

Hello, are you trying to find P(NF|NR) ? Just asking because the problem doesn't ask for that

[wheat isle]

Oh I thought it was

[ancient saddle]

Mmm not really

[wheat isle]

Find the probability there is no bushfire given that it has rained

Oh so P(NF|R) then

[ancient saddle]

Yes

You take for granted that it has rained, so the probability that rains doesn't matter

Just look at the R branch

R → NF

[wheat isle]

0.46 times 0.79?

[ancient saddle]

No, just 0.79

It's given that it has rained, we don't consider its probability

[wheat isle]

So what should I do

[paper mango]

as mentioned for the a) you only take the value 0.79

since you start from the node for Rain as it's given

@wheat isle are you stuck on the others

[wheat isle]

Is part b the same thing

wait no given there is a bushfire so do we go backwards or

[paper mango]

it's the other way around

yeah

[wheat isle]

So 0.21?

[paper mango]

I didn't calc the values, but no

if you looked at the upper branch

You want to know P(R|F)

@wheat isle and by Bayes Rule: P(R|F) = P(F|R) * P(R)/P(F)

so if you know P(B|R) and P(R) and P(F) then you can calculate that percentage

P(R) is 0.46

given by the branches

P(F|R) is 0.21

and P(F) is the sum of all Fire paths

so P(F) = 0.46*0.21 + 0.54*0.85

sry for the occasional absences, must be annoying given that you have the channel open for 1.2h

[wheat isle]

All good

[paper mango]

if you're still on it I can dedicate here now iyw

[wheat isle]

Yeah I’m looking at the questions again

So I’m trying to find P(R|F)

[paper mango]

yes

are you familiar with Bayes' Rule

[wheat isle]

I got introduced to it but I haven’t used it much

[paper mango]

bc it'll come in handy here

[wheat isle]

Okay

[paper mango]

generalised P(A|B) = P(B|A) * P(A)/P(B)

therefore if you want to calculate P(R|F):

P(R|F) = P(F|R) * P(R)/P(F)

@wheat isle so which of these probabilities can you directly read from the graph?

[wheat isle]

P(R) and P(F) I think we don’t have P(F|R)

[paper mango]

not yet yes

ah sry

the other way around

you already know P(F|R) from the graph

but not yet P(F)

[wheat isle]

Oh we do?

[alpine sable]

let P(B) be the probability of a bushfire, P® be the probability of rain, and P(B|R) be the conditional probability of a bushfire given that it has rained. similarly, let P(B') and P(R') be the complements of P(B) and P® respectively.

according to the image, we have:
P(B|R') = 0.85 P(B|R) = 0.21 P® = 0.46

using the formula of conditional probability, we can find the following probabilities:

A: P(B'|R) = 1 - P(B|R) = 1 - 0.21 = 0.79

B: P(R|B) = P(B|R) * P® / P(B)

• to find P(B), we use the law of total probability:

P(B) = P(B|R) * P® + P(B|R') * P(R') = 0.21 * 0.46 + 0.85 * (1 - 0.46) = 0.5164
therefore,
P(R|B) = 0.21 * 0.46 / 0.5164 = 0.1873

C: P(R'|B) = 1 - P(R|B) = 1 - 0.1873 = 0.8127

D: P(R|B') = P(B'|R) * P® / P(B')

to find P(B'), we using the fact that P(B') = 1 - P(B)

P(B') = 1 - 0.5164 = 0.4836
therefore,
P(R|B') = 0.79 * 0.46 / 0.4836 = 0.7525

[paper mango]

we'll keep it here for checking after

[paper mango]

you have P(R) as 0.46

[wheat isle]

Wow thank you I’ll make sure to check later

[paper mango]

then P(F|R) is 0.21

[wheat isle]

Yeah

[paper mango]

since it's the upper branch starting from R

and you don't know P(F) yet

[wheat isle]

Right

[paper mango]

if you don't yet know the probability of a single event yet, you'd use the principle of summation of paths

just add all paths together that end at F

so 0.46*0.21 + 0.54*0.85

[wheat isle]

Is there a rule that shows this?

[paper mango]

well you can deduce it via logic as well

[alpine sable]

choice d has a 75.25% chance
choice b has a 18.73% chance

[paper mango]

the sum of probabilities of all events that lead to A

is the probablity of A

[wheat isle]

Haven’t heard of that before

[paper mango]

I could also try to make some odd example if it's not clear

[wheat isle]

Yeah sure that would b great

[paper mango]

hm k, imagine asking the question what's the probablity of you dying tomorrow, so you could look at all the possible paths of choices you could take, but you only add up the probabilities of the paths that actually lead to death

dk why I came up with that

[wheat isle]

Lmao

[paper mango]

if you add all of these probabilities together

you get the probability you seek

because all of these paths lead to the result whose probability you search

ok so you have all probablities to calculate P(R|F)

which means b) is done

c) is P(nR|F)

[wheat isle]

Wait why do we multiply

[paper mango]

you add each path

and for each path you multiply within

the probability of a single full path is the product of its probabilities

[wheat isle]

Why don’t you do 0.46 + 0.21 to add a path

Oh

[paper mango]

well after 0.46 you don't end up at F

but you want the probability of F

[wheat isle]

Wait so I don’t understand what paths did you add here and why

[paper mango]

you start at the beginning of the graph

and end up at F with a path

there are only two paths which end up at F

so P(F) is the sum of these two paths

[wheat isle]

Because we want P(R|F) right

[paper mango]

in words: "The probability of a Fire = the probability of Fire if it's Raining + the probablity of Fire it's not Raining"

[paper mango]

regarding the paths P(F)

which we'll use to calculate P(R|F)

since P(R|F) = P(F|R) * P(R) / P(F)

and we didn't have P(F)

[wheat isle]

So to get P(F) we added all the probabilities that give us F

[paper mango]

yes

[wheat isle]

Ok think I get that now

[paper mango]

kk, best to write them down

because with this kind of task

[wheat isle]

Yeah

[paper mango]

you can just write all relevant probabilities down

and then answering the questions takes a minute

[wheat isle]

Alright

[paper mango]

k so b) is done

for c)

it's asking for P(nR|F)

[wheat isle]

yep

[paper mango]

which we can't read from the graph, as we can't start at F and end up at nR

therefore, again Bayes' Rule:

P(nR|F) = P(F|nR) * P(nR)/P(F)

[wheat isle]

Didn’t we do that for P(R|F)

Oh wait no

We didn’t nvm

I thought we could go backwards from the graph

[paper mango]

partially, I wanted to get to that

now, we could again calculate the values etc.

but as you just mentioned P(R|F)

how can we only use P(R|F) to calculate P(nR|F)

[wheat isle]

Hm

Let me see hold on

[paper mango]

well it's the same as if you have P(A) and want to calculate P(nA)

[wheat isle]

I’m not entirely sure

[paper mango]

well if I tell you the probability of an event A is 60%

what's the probablitiy of event A not happening?

[wheat isle]

40%

[paper mango]

yes

likewise for P(R|F) and P(nR|F)

[desert cairn]

Solve for Fa in terms of m, g, and theta. I cannot use T in my response. T sin(angle) =Fa and T cos(angle)=mg. Can I get help on this one?

[paper mango]

if the chance of R happening under the assumption of F is 60%

then the chance of R not happening under the assumption of F is 40%

[wheat isle]

#❓how-to-get-help

Wait what did we calculate P(R|F) to be

[paper mango]

idk I just told you how you'd calculate it

you'll have to write them down :D

[wheat isle]

This is P(R|F)? Or P(F)

[paper mango]

this is P(F)

[wheat isle]

Okay

[alpine sable]

yall still not done?

[paper mango]

crrntly fairly stepwise

[wheat isle]

Oh wait do we have P(F|R)

[paper mango]

ofc

from the graph

0.21

[wheat isle]

I thought we weren’t allowed to go backwards?

[paper mango]

not backwards?

we don't go backwards

P(F|R) means you start at R

and end up at F

so you go from left to right

[wheat isle]

Oh yeah

[paper mango]

sry gtg but you have the solutions above

just remember Bayes Rule and the path summation

which gets you through

[wheat isle]

Yep tysm for all the help

[lone heart]

@wheat isle Has your question been resolved?

[lone heart]

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[lone heart]

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[void crypt]

How do i find c?

[nimble fern]

f(x-13) is not f(x)-13

[void crypt]

so f(x) would be transformed to the right then?

[nimble fern]

yep, it will be transformed to the right by 13 units

[lone heart]

@void crypt Has your question been resolved?

[void crypt]

<@&286206848099549185>

[buoyant summit]

try yourself it's easy

[void crypt]

.close

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[paper sage]

@tardy stag

[sour verge]

Lagrange multipliers, no?

[paper sage]

yeh

[sour verge]

So region is bounded, function must attain max and min on the disk. Now this can happen either when the gradient is null (inside the disk) or somewhere on the boundary.

What troubles you?

[paper sage]

I don't know how to proceed after i found -1/4 which I got from plugging inside the function x=0 and y=1/2

[sour verge]

So that's one possible value, but just on the inside of the disk. You need to check the boundary using Lagrange multipliers

[paper sage]

May you help me with that

[sour verge]

Yes. So these possible extremas on the boundary occur when the gradient of your function is a multiple of the gradient of the restriction.

I trust you've already found the gradient of the function

And it shouldn't be too hard to compute the gradient of the boundary equation

[paper sage]

but in our case our boundary is a inequality so how can we do ?

[sour verge]

The boundary is just the equality at this point.

You've already dealt with whatever is inside the disk with your first test.

You have x^2 + y^2 = 1

[paper sage]

so I have already dealt with x^2+y^2<1

[sour verge]

Yep!

[paper sage]

oh okay

[paper sage]

so now i have to deal with this

[sour verge]

Exactly, this is where Lagrange multipliers come in handy. You need the gradient of g(x,y) = x^2 + y^2

[paper sage]

yes ofc so we will have g'x=2x and g'y=2y

times lambda

[sour verge]

Yep, now with lagrange multipliers, you need points that satisfy 4x = L * 2x and 2y-1 = L * 2 y

[paper sage]

so in the end we will have 4x=lamda*2x and 2y-1=lamda(2y)?

[sour verge]

Yessss

[paper sage]

and now i want to solve for x or lamda?

[sour verge]

At the end of the day, you want x and y, but you can always find some lambda that works and work with that.

[paper sage]

then in our first case I will have 0=lamda2x-4x which is same as 0=2x(lamda-2)

[sour verge]

Yep!

[paper sage]

which gives us x=0 and lamda =2

what happens next?

[sour verge]

either one or the other

Case 1 : x=0

[paper sage]

okay

where do I plug this x=0?

[sour verge]

In that case, what does your boundary say?

[paper sage]

so If I plug inside our boundary we will have 0^2+y^2=1

[sour verge]

Yes

[paper sage]

which give us y=+-1

[sour verge]

Spot on.

Now for case 2 : L = 2

[paper sage]

where do I put this?

inside our bounadry?

we don't have any L in our bounadry

[sour verge]

There's no L in the boundary, but there is one in the second equation..

Sooooo

[paper sage]

what do you mean by second equation?

[sour verge]

2y - 1 = L * 2y

[paper sage]

yes ofc sorry

so if we plug in that

we will get 2y-1=4y

solve for y we will have -1=4y-2y

which is -1=2y which is y=-1/2

[sour verge]

Yes, and then the boundary says that ...

[paper sage]

so you mean I should plug in y=-1/2 inside our boundary

[sour verge]

Yes

[paper sage]

okay then I will x^2+1/4=1

so i will get x^2=1-1/4

x^2=3/4

so we will get x=+-square root of (3/4)?

[sour verge]

Yess

[paper sage]

okay what happens next

[sour verge]

Now you've got a bunch of possible points : (0,1/2), (0,1), (0,-1) ...

You can just plug them in the function, and determine from those your extrema

[paper sage]

but I still didn't get the right answer

the fmin is -1/4 which i got from the (0,1/2)

but fmax=9/4

and i didn't got that

the points (0,1) and (0,-1) are giving me 0

so how do I find 9/4?

[sour verge]

What about the other 2?

[paper sage]

2(0)^2+1^2-1 gives me 0

[sour verge]

Those ones

[paper sage]

ohhh

yeah ofc

sorry

i didn't try them

let me see

yes i got right

[alpine sable]

Hi can u help me?

[paper sage]

thank you som much for the help

[sour verge]

No worries, hope you learned a bit through it all 🙂

[paper sage]

Your help was so smooth

[paper sage]

yeah i did

[alpine sable]

hello can yall help me pls?

[paper sage]

So one question

[sour verge]

#❓how-to-get-help

[paper sage]

Every time i'm checking directly the fucntion it means i have already dealt with inequality?

Then after I just deal with equal on my bounary

[alpine sable]

idk where that is

its algebra

[sour verge]

Yeah, essentially, what happens is that the first test you do is a test for an extrema over the whole of R^2. In that specific case, it happened to fall into your domain, so you must take it into account. However, by "cutting" out your domain, you might create some sharp edges that make some "artificial" extremas, which is why you must check the boundary separately.

[alpine sable]

pls help me

[paper sage]

bro stop

[sour verge]

read the #❓how-to-get-help channel

click on it

it will explain

this channel is occupied.

[paper sage]

oh okay so it happened to fall into your domain you mean that -1/4 is under my domian since it's lower than 1?

[sour verge]

Not really. It's because the point you found, (0,1/2) is within the domain x^2 + y^2 <= 1.

[paper sage]

oh okay you mean that point

[sour verge]

You could have a function that has an extrema outside of your restricted domain, in which case you don't need to take it into account.

Yeah

[paper sage]

okay I understand now

[sour verge]

Great!

[paper sage]

thank you so much

have a good day! 🙂

[sour verge]

Have a good day/night then

[paper sage]

.close

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[hollow lark]

Can anyone help? I'm really not sure how to narrow it down

[hollow lark]

I'll send my working but I've made no progress

[upbeat gorge]

Hmm I think I’ve seen this question stem here already
But anyway

[hollow lark]

sorry if it has I haven’t seen it before

[upbeat gorge]

Caveman idea: if you want log base 3, try taking log base 3 of all the 5 numbers

[ocean seal]

CST

[upbeat gorge]

*two compound inequalities

[hollow lark]

thank you I think I understand that

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[marsh solar]

In an undirected random graph of 8 vertices, the probability of an edge being present between a pair of vertices in 1/2. What is the expected number of unordered cycles of length 4?

Actually there was a similar question for cycle length of 3.
It's answer was 8C3 × (1/2)^3.

Based on that, for 4 length cycle I thought it would be 8C4 × (1/2)^4. But I think this is wrong because there are also structures with 4 vertices and 4 edges which doesn't forms a cycle.

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@marsh solar Has your question been resolved?

[marsh solar]

<@&286206848099549185>

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@marsh solar Has your question been resolved?

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@marsh solar Has your question been resolved?

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[modern jungle]

Is it event possible to solve? Is it enough data? I feel like it is, but not sure. I know how to find the area of the segment of the small circle
Segment area = r² × arccos((r-h)/r) - (r-h) × √(2 × r × h - h²). But cant figure out how to do it for the big circle.

[lone heart]

@modern jungle Has your question been resolved?

[modern jungle]

<@&286206848099549185>

[mossy siren]

.close

What do u need helo wiht

[modern jungle]

to express R in terms of r and h

[lone heart]

@modern jungle Has your question been resolved?

[radiant siren]

Where did you get this "area = r² × arccos((r-h)/r) - (r-h) × √(2 × r × h - h²)" formula from?

[modern jungle]

this website https://www.omnicalculator.com/math/segment-area

[radiant siren]

OK

Do I feel correctly that you are trying to just "fit a formula" to solve it? That you are not sure?

If my feeling is right, then let me suggest to jsut try to SOLVE it by ... well... solve it yourself, with not help of complex formulae, just using "simple" formulae!

Unfortunately I have to go now, but I promise you I shall solve it myself and will send you my solution if you wish

How urgent is it?

[modern jungle]

yes I want to find a formula for R . Not urgent. If you feel like you can solve it do it please and send to me

[radiant siren]

I am new to discord

I shallstart solving it for you and send to you and we'll continue a little later today or tomorrow. We'll get there! How can I find you here to tell you when I am ready?

Shall I just DM you?

[modern jungle]

yes

[radiant siren]

(I must go now)

OK I shall DM

And if you still need help by then we'll concentratertogether and will solve - rather YOU will solve, I'll only help you to see how to THINK - deal?

[modern jungle]

ok

[radiant siren]

Will DM you later today

bye for now!

[modern jungle]

bye

[lone heart]

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[modern jungle]

.close

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[modern jungle]

hard|not possible to solve

[lone heart]

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[jade cobalt]

Can anyone walk me through this?

[wind cloak]

x = -1 is a root

So divide by (x + 1)

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[vestal yarrow]

please help me with my method to solve this

i dont really know what to do, somebody pls help guide me

[kind cosmos]

lemme try

[vestal yarrow]

ok thank you

i found a similar question in a past paper but they use a differnent equation

why do they use this equation?

oh they are using the envelope equation in that question?

however the answer on my question says use trajectory equation twice

[vestal yarrow]

are you still working on it?

i have to move on to prevent wasting time, please ping me if you can help

<@&286206848099549185>

@kind cosmos

<@&286206848099549185>

pls

[kind cosmos]

by putting eqn of trajectory twice

the answer isnt comin

the things which i thought

[vestal yarrow]

yeah i dont know what to do i showed when i tried to use it twice

[kind cosmos]

like

if the 2h height is just touched

then it must be the height of projectile

if we look at it that way

[vestal yarrow]

i mean im using the points (H,H) and then (3H, 2H)

[kind cosmos]

yes i got it

[vestal yarrow]

but in the question it says we can vary alpha, i dont know what that means really

[kind cosmos]

the angle and velocity can be varied

it is also a confusing part of ques

[vestal yarrow]

but we are trying to find the velocity?

why should we vary it

and by vary it does it mean i just plug in random values of alpha?

[kind cosmos]

we have to find the velocity in which it just touches both the tops

[vestal yarrow]

yes

[kind cosmos]

minimum value

[vestal yarrow]

so my thinking was set both equations equal to eachother then solve for V

yeah thats when you use = in the equation right? rather than less than or more than signs

[kind cosmos]

but shouldnt it also be equal to height of projectile

[vestal yarrow]

can you show me your workings?

[kind cosmos]

its messed up lol

[vestal yarrow]

still i need to see im very lost

[kind cosmos]

can i dm you?

[vestal yarrow]

yes sure

why do u need to dm tho

[lone heart]

@vestal yarrow Has your question been resolved?

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@vestal yarrow Has your question been resolved?

[alpine sable]

@vestal yarrow

there is an alternate form of the equation of trajectory of the path of a projectile

y=xtan(angle of projection)*[1-(x/range of the projectile)]

so we have to clear two walls, one which is H units tall and H units away, and another which is 2H units tall and 3H units away

$y=xtan\theta(1-x/R)$

[ocean seal]

dimpledoink

[alpine sable]

put y=H and x=H here, let this be the first equation
put y=2H and x=3H, let this be the second equation
divide these equations, find R in terms of H, let this be the third equation
plug R back into the first equation and you'll get the value of tan alpha
from there you can find sin alpha, and cos alpha
use the formula for the range of a projectile and plug in sin alpha and cos alpha, it goes (2*v^2(initial)*(sin(alpha)*(cos(alpha))/g
equate this R to the third equation, and you'll have the answer

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[vestal yarrow]

thank you sm

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[alpine sable]

If your cut is parrallel to base, I think so

[hushed locust]

you can think of a cone as a triangle that’s been rotated around the center. so if the cut is parallel to the base, the new smaller cone’s triangle cross-section will be similar to the original cone’s triangle cross section

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[marsh kelp]

the question is
function: f(x,y) = cos(xy^2)
for what angle theta with the positive x-as is the Dthetaf in point (1,2) it's maximum size?

[floral current]

Could I get some help on a and c?

[marsh kelp]

this one is occupied

could u delete it and use a free one

[floral current]

Oh lol xd

[tepid fulcrum]

Dutch?

[marsh kelp]

ja

[tepid fulcrum]

Kan je je vraag in nederlands typen

[marsh kelp]

ik moet voor die functie de hoek theta vinden met de positieve x-as waarvoor de richtingsafgeleide Df in punt 1,2 maximaal is

functie: f(x,y) = cos(xy^2)

[tepid fulcrum]

Waar is theta

[marsh kelp]

vgs mij is theta xy^2

maar da word ook in de opgave ni vermeld

[tepid fulcrum]

Ik snap je vraag niet want je vult al een punt in

[marsh kelp]

da moet met vectoren

want de richtingsafgeleide heeft ook zijn eigen formule die je moet toepassen denk ik

en daarvoor moet je partieel afleiden

[tepid fulcrum]

Ja maar vaak hoe het zit met extremum vragen vul je geen punt in

Je zoekt die punt

[marsh kelp]

ja daarom dat ik het ook niet snap

maar het is een examenvraag geweest

[tepid fulcrum]

Heb je foto

Van de exacte vraag

[marsh kelp]

ja

secondje

@tepid fulcrum kun je ff dms kijken

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@marsh kelp Has your question been resolved?

[marsh kelp]

wanthan doesn't believe in himself and thats why i need a different helper

[tacit arch]

.close

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[quaint birch]

How do I do this

do I have to use squeece theorem?

[azure swan]

Yes

What property does cos(x) have?

[quaint birch]

wdym property?

range?

[azure swan]

Yes

The range is here important

[quaint birch]

-1 <= cos(x) <= 1

[azure swan]

Now you have everything you need to apply the squeeze theorem

[quaint birch]

but idk how to deal with 5/x^2

inside of cos(x)

[azure swan]

You don't have to

[quaint birch]

or is it just -x^2 <= x^2cos(x) <= x^2

and then i find the lim of -x^2 and x^2 when x -> 0

which would be 0

so the answer would be 0?

[azure swan]

-1 <= cos(5/x^2) <= 1

[azure swan]

Yes

[quaint birch]

ah ty

.close

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[sharp bay]

I am so confused in those question, i only need the answer pls

[vapid steppe]

for the x^2+3x+9 one cant you just divide both sides by 9

[safe tartan]

Seems sus

[lone heart]

@sharp bay Has your question been resolved?

[tepid fulcrum]

Find the roots

Then do a(x-root1)(x-root2)

There should be a number with ‘a’ and no x

That would be equal to 1

With ‘a’ found you have A en B as well

Huh there is no root

Complex numbers?

[winter light]

I would use the formulas of sum and product of the roots of a 2nd degree equation

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[peak pecan]

Why do AE/AC = DE/BC ?

[proven leaf]

Isolate △ACB and △AED, knowing ∠C=90º and ∠E=90º. See if you can figure out why :)

[peak pecan]

wooo

I don't understand

lol

[uncut torrent]

triangle abc and ade are similar

[peak pecan]

yea ?

[uncut torrent]

AE/DE = AC/BC

rearrange

[peak pecan]

but I don't see link between those two triangle

[uncut torrent]

consider the angle BAC

same as angle DAE

[peak pecan]

Yea ?

[uncut torrent]

so tan(BAC) = tan(DAE)

[uncut torrent]

hence this relation

[peak pecan]

em

don't understand

[uncut torrent]

well u see why this is true right

[peak pecan]

no ?

That's the reason that I don't understand

[uncut torrent]

angle BAC = angle DAE

[peak pecan]

yea ?

[uncut torrent]

oh do you know what tan is

[peak pecan]

yea

but not much

[uncut torrent]

tan(angle) = opposite/adjacent yeah

[peak pecan]

yea

[uncut torrent]

so tan(BAC) = BC/AC

u can see in the diagram

and tan(DAE) = DE/AE

[peak pecan]

eh

sure

[uncut torrent]

since BAC = DAE, tan(BAC) = tan(DAE)

[peak pecan]

eh

[uncut torrent]

therefore BC/AC = DE/AE

[peak pecan]

I see some relation now

[uncut torrent]

rearrange this to get to your original relation

[peak pecan]

sure

[uncut torrent]

AE/AC = DE/BC

[peak pecan]

thanks

[uncut torrent]

nw

[peak pecan]

.close

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[rigid crater]

CAN SOMEONE HELP ME

[rigid crater]

I HAVE A TEST IN 1 HOUR

[fossil pecan]

,rotate

[fossil pecan]

what's ur question

[rigid crater]

How do u find the X slope in this

@fossil pecan

[fossil pecan]

in which problem

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@rigid crater Has your question been resolved?

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[native oasis]

i need help

so i found that the first statement is true

well no i didnt

im confused as to whether it can be true

since x>3 does not include 3

idk im rlly confused with the entire thing

[near apex]

Other definition of function does that. Compute f(3) using that.

Btw, you don't need that even.

It just asks you limit of those expression, regardless of function.

[wary stream]

You can technically pick a number greater than 3 but not equal to 3, like 3.00000000001

[native oasis]

Yeah I see now

Ok, on to the second statement

i found that to be true

but

i dont know whether statement 3 is

since in order for their to be a limit it has to be the same from negative and the positive side i believe

so would all 3 statements be true?

[near apex]

What you believe is indeed correct.

[near apex]

Think of what you said just now, before this.

In order for limit to exist, it must be same from left and right side.

[native oasis]

yes

and from the left side its

[near apex]

Is that the case here?

[native oasis]

yes i think so

coming form left side is 1

and

wait no

[near apex]

No. It's not.

[native oasis]

wait

ok from the RIGHT side

its 1

and left side is 0

[near apex]

Yes.

So, now, does the limit exist?

[native oasis]

No

[near apex]

Good.

[native oasis]

thank you very much

have a good day

.close

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[prisma lintel]

$\int sin\omega t + cos\omega t . dt$

[ocean seal]

Bugatti W16 Mistral

[prisma lintel]

Omega here is a constant

I got the part where the integration would be
-coswt + sinwt

but there is something missing in the answer

[umbral star]

are you sure that the antiderivative of $\sin(\omega t)$ is just $-\cos(\omega t)$?

[ocean seal]

qianqian07

[prisma lintel]

nopw

not sure

there must be some other multiplier

[umbral star]

try using $u=\omega t$

[ocean seal]

qianqian07

[umbral star]

then $du=\omega dt$

[prisma lintel]

then w must be multiplied

[ocean seal]

qianqian07

[umbral star]

$dt=\frac{du}{\omega}$

[ocean seal]

qianqian07

[umbral star]

can you finish from here?

[prisma lintel]

m trying

OHH

nvm

I just replaced dt with du without multiplying omega

ahh ty @umbral star

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[umbral star]

👍

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[queen merlin]

<@&286206848099549185> I have to factor -20xy+4y^2+25x^2, been stuck for a while and would really appreciate any help you can give!

[mental flame]

!15mins

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

[pseudo cosmos]

What do the ^ mean

[queen merlin]

to the power of, so 4y^2 is 47 squared

[pseudo cosmos]

Idk if I’m stupid but isn’t 4 times 4 16 then 16 times two is 36???

[mental flame]

Y is being raised to the power of 2, not 4

[queen merlin]

thats true but i dont see the relevence

[vapid shuttle]

4y^2 is 4 times y to the power of two

[pseudo cosmos]

Yeah then it’s be 4y squared

[vapid shuttle]

no

that is (4y)^2

this is 4*(y)^2

[pseudo cosmos]

Yeah

That’s what I’m saying

[vapid shuttle]

(4y)^2=16y^2 not equal to 4y^2

[pseudo cosmos]

Yes

Ohhhitd be 4 times y squared then???

Right?

[queen merlin]

yes

[pseudo cosmos]

YES I GOT THIS

GIVE ME A MINUTE

[queen merlin]

i have the answer if it helps i just need to know how to get there

its (5x-2y)^2

[pseudo cosmos]

Are u just simplifying

Nvm

Sorry j thought u we’re simplifying

I can try and figure out factor in though

[mental flame]

Try getting the expression in the form of
a^2-2ab+b^2
So we can use the formula
(a-b)^2=a^2-2ab+b^2

[pseudo cosmos]

Damn ok sorry I do not know anymore I thought u were simplifying

[queen merlin]

whats the name of this formula?

[mental flame]

There isn't a name that I know of.

[lone heart]

@queen merlin Has your question been resolved?

[vapid shuttle]

binomial squared?

the name doesn't matter

[lone heart]

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[proven leaf]

is there anyway to approach: $\frac{y''}{y'^2}=\frac{y+2}{e^{2x}(y+1)}$?

[proven leaf]

[ocean seal]

MrFancy

[placid zinc]

Christ you've got to warn somebody before posting gore here

Such an ugly DE my eyes

[proven leaf]

so, no?

[placid zinc]

Where did you get it from lol. Are you expected to be able to do it?

[proven leaf]

I made this bad boy myself, I got bored

[placid zinc]

That's brave lol

[proven leaf]

y=W(e^x) where W(x) is the product log

[placid zinc]

oh ew

[proven leaf]

I was wondering can I solve the DE? or is just like, stay the fuck away?

[placid zinc]

I'm afraid to say "definitely never touch this" because weird methods exist, but DEs much simpler than this can't be done.

[proven leaf]

Think a stack exchange guy would be able to do this? I don't wanna touch that website and become corrupted but im really curious if this can be done and another question arises, what is the simplest DE that cannot be solved?

[proven leaf]

huh I just found it, well thanks man :)

.close

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[placid zinc]

Oh you did find a method? Share lol

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• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mint estuary]

Wut is the answer to this

I got an answer, but idk if its right

[tacit arch]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[mint estuary]

I jus wana know if wut i got is right

(-infinity, -1) U (1, infinity)

I think^

[royal socket]

Not quite

Where did -1 and 1 come from

[tacit arch]

No

[mint estuary]

From graph

[mint estuary]

^

Wait sorry

Wrong question

on what open intervals is g(x) concave up?

^^

[mint estuary]

For this question i got. (-inf,-3) U (0,2)

there

<@&286206848099549185>

[mint estuary]

Uh

Hello?

[tacit arch]

Oh

Yes that's a different question

[mint estuary]

?

[mint estuary]

I need both

[mint estuary]

..

help?

[tacit arch]

2a is still wrong

Decreasing means for two values, the slope is negative

Pick two points on (0,2). What's the sign of the slope

[mint estuary]

?

slope???

isnt it when Y is decreasing

can we do a voice call?

Pls reply quick if u can, its late for me

[tacit arch]

No thanks

[mint estuary]

ok

explain better

[tacit arch]

I was just explaining to you what decreasing means

[mint estuary]

..

[tacit arch]

You have to understand this

[mint estuary]

Dont get it

r u saying

that the questions asks when the slope is decreasing?

[tacit arch]

What don't you get

No

[mint estuary]

then?

[tacit arch]

Slope < 0 is decreasing

Slope > 0 is increasing

[mint estuary]

yea

[mint estuary]

Negative?

[tacit arch]

What's the slope on (0,2)

Yes

That means it's decreasing

[mint estuary]

yea?

i wrote that?

[mint estuary]

I know

[tacit arch]

Oh that's a 0

Alright that's right then. Nevermind what I said

[mint estuary]

U tryina say the slope is from y/x

2 values*?

[tacit arch]

Change in y and change in x

Not y and x

There's a difference

[mint estuary]

yea i kno

slope formula thingi

y-y1 x-x1

ok what second question

2b wut

[tacit arch]

Yes looks right

[mint estuary]

Many put 0

Few of my friends

but idk why tho

[tacit arch]

Good for them?

[mint estuary]

R U sure its correct

[tacit arch]

I can't read their minds

[mint estuary]

confident?

Tis question 2

What open intervals is f(x) concave down

[mint estuary]

If u say so

[tacit arch]

?

[mint estuary]

..

no i mean if u say it right ig

I gota trust

[mint estuary]

Need for this

[mint estuary]

^

Reply quick so i can stop bothering u

[tacit arch]

No thanks

[mint estuary]

Bruh

Can u c the graph?

Its like L quality

<@&286206848099549185>

[lone heart]

@mint estuary Has your question been resolved?

[mint estuary]

.close

[lone heart]

Channel closed

Closed by @mint estuary

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[quartz rapids]

I'm struggling with a taylor series manipulation problem:
The goal is to find the analytic sum of the series: $\sum_{n=1}^\infty \dfrac{(-1)^n5^{2n-1}}{(n-1)!}$

[ocean seal]

Sean

[quartz rapids]

So far, I've been able to manipulate the maclaurin series of sin(x) to get:
$-\sin(5)=\sum_{n=1}^\infty \dfrac{(-1)^n5^{2n-1}}{(2n-1)!}$
(5 being in radians)

[ocean seal]

Sean

[quartz rapids]

my roadblock is attempting to manipulate the left side to somehow get rid of the 2 coefficient in the denominator's factorial

no amount of eulers reflection or factorial manipulation seems to be getting anywhere for me

[tacit arch]

Sine is not the correct function

,tex .maclaurin

[ocean seal]

riemann

[tacit arch]

First one with y = -x

[quartz rapids]

ohhhhh I gotcha, Ill give that a shot real quick

I've gotten to: $-e^{-x}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{\left(n-1\right)}}{\left(n-1\right)!}$

[ocean seal]

Sean

[quartz rapids]

I think I got it but desmos is saying otherwise

Does this look right @tacit arch ?

Nvm wolfram alpha says im right so I’m calling that a W, thanks for the help!

.close

[lone heart]

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chilly idol]

i need help on 25

question 25

i got the intervals

(1, +♾️ ) (1,2) (2, +♾️ )

i also got x>1 and x>2

i dont know what to do next

.close

[lone heart]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[cinder roost]

So I have found an equation for the mobility (degrees of freedom for a scissor lift mechanism).

Basically using the equation: M = 3(n-j-1) + 1j = 3(n-j-1) + j

I found that with a n value of 8, and a j value of 9, the mobility M is equal to 1.
As I increased the number of links (by 2), the j value became 13, and the M still equaled to 1.
So basically, increasing the number of links to increase the layer for the scissor lift (2 is added for every level you want to go up), the M will always equal 1.

How can I used the equation 3(n-j-1)+j=1

With 𝑛 = 𝑛_0, (𝑛_1 + 2), (𝑛_2 + 4), …[n_0=8]
And 𝑗 = 𝑗_0, (𝑗_1 + 3), (𝑗_2 + 6), … [𝑗_0=9]
to have an equation that applies to all cases m, and m+1

[lone heart]

@cinder roost Has your question been resolved?

[cinder roost]

<@&286206848099549185>

[lone heart]

@cinder roost Has your question been resolved?

[lone heart]

@cinder roost Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

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[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[upper flame]

for #98 I understand how to write the area as a function of x but how do I find the domain of the function?

(^^this is the answer but I dont understand how to get to it)

[echo socket]

What condition on $a$ makes the expression $\sqrt{a}$ defined in real numbers?

[ocean seal]

A Lonely Bean

[upper flame]

I dont understand the question ;-;

squaring it?

[echo socket]

Okay, can you tell me which of these numbers exist?
[ \sqrt{2}, \sqrt{42}, \sqrt{-3}, \sqrt{0} ]

[ocean seal]

A Lonely Bean

[upper flame]

1 2 and 4

i think

[echo socket]

Yeah, because the numbers under the square root symbols there are nonnegative, right?

[upper flame]

yess

[echo socket]

So, $\sqrt{a}$ is defined whenever $a \ge 0$

[ocean seal]

A Lonely Bean

[upper flame]

yes

[echo socket]

Given this, when would $\sqrt{36 - x^2}$ be defined?

[ocean seal]

A Lonely Bean

[upper flame]

when it is greater than or equal to 0

so we would set that to greater than or equal to 0!

[echo socket]

Yup, can you solve for the values of $x$ such that $36 - x^2 \ge 0$?

[ocean seal]

A Lonely Bean

[keen plinth]

specify "it"

[upper flame]

i get plus and minus 6

x is less than or equal to plus and minus 6

[upper flame]

when the expression is greater than or equal to 0(?)

[echo socket]

Hopefully what you mean is that $-6 \le x \le 6$

[ocean seal]

A Lonely Bean

[upper flame]

ermm

[keen plinth]

you should make clear what you mean by "it is greater than 0" since we don't know what you mean by "it"

[echo socket]

Okay let's walk through solving $36 - x^2 \ge 0$ together, shall we?

[upper flame]

did i do this wrong 😭

[ocean seal]

A Lonely Bean

[upper flame]

yes plss

[upper flame]

got itt

[echo socket]

Yeah there's no such valid step as what you did in the final line

[echo socket]

This channel is taken

So, generally $a^2 \le b^2$ implies $\abs{a} \le \abs{b}$

[ocean seal]

A Lonely Bean

[upper flame]

yes

[echo socket]

So, $x^2 \le 6^2$ would mean that $\abs{x} \le \abs{6}$ or $\abs{x} \le 6$

[ocean seal]

A Lonely Bean

[upper flame]

yess

[echo socket]

Which happens if and only if $-6 \le x \le 6$