help-17

[vast shale]

that doesnt make sense

8^2 - 5^2 does not give u the same thing as say 3^2 -1^2 depending on how you modify your reference

am i tripping

[cobalt crypt]

yes you're tripping

pick yourself up from the ground

[vast shale]

okay lets make up an amazing example

well wait

question

x_2 and x_1 are measured from our chosen origin point, yes?

[cobalt crypt]

like

theyre just numbers

[
\int_{x_1}^{x_2} kx \dd x = \int_{x_1 + x_0}^{x_2 + x_0} k(x - x_0) \dd x
]

[vast shale]

okay i think i like get this now

maybe

Pls help me convert this

[dull kayak]

!help

[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[dull kayak]

!status

[vocal sleet]

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

@vast shale Has your question been resolved?

[vocal sleet]

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[pseudo whale]

Curve : f(x)= 1/x on ]0;+infinite[

Starting from a point N on C, construct the rectangle OMNA with M on the x-axis and A on the y-axis. Determine the dimensions of the OMNA rectangle with minimal perimeter.

[pseudo whale]

Im stuck here

[magic wasp]

Can you write the perimeter in terms of only one of the sides?

[pseudo whale]

Like can we determine it with only one side or you are asking for a better instruction traduction ?

[magic wasp]

I'm asking if you can express it

as a formula if you will

[pseudo whale]

With only one side we can’t

[magic wasp]

Ok, express it with whatever you can

[pseudo whale]

Length + width times 2

[magic wasp]

Alright, now can you relate length and width?

[pseudo whale]

What do you mean

[magic wasp]

There are two main things to your problem: OMNA is a rectangle, and N lies on the curve y=1/x

[magic wasp]

Why did you edit that btw?

[pseudo whale]

I wrote times instead of plus

[pseudo whale]

Derivative is useless for the moment right ?

[magic wasp]

Depends what you use it on

If you know the value of OM, do you also know the value of MN?

[pseudo whale]

No

MN isn’t essentially OM divided by 2 right ?

[magic wasp]

I'm not sure you understand what it means for N to lie on the curve y=1/x

[pseudo whale]

No I didn’t

Currently learning derivative in hs

[magic wasp]

If you start at O and go x units to the right, and then go up until you hit the curve, then by how many units did you go up?

[pseudo whale]

Y ?

[magic wasp]

Right... and in terms of x?

[pseudo whale]

1/x

[magic wasp]

Ok, so if x = OM, what's MN?

Man if you take 5 minutes for every reply we'll still be there in 2 hours

[vocal sleet]

@pseudo whale Has your question been resolved?

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[severe willow]

im not sure if i wrote this formula right or not- the given sequence is 5, 8, 11, 14, 17, 20 ;-;

[paper depot]

instead of $5a_1$ you should have just $5$.

[twin meteor]

Ann

[severe willow]

so it should look like (an = 5 + (n-1) 3) ?

[paper depot]

no

$a_n = 5 + (n-1) \cdot 3$

[twin meteor]

Ann

[paper depot]

yes now it's correct

only i would not write $(n-1)3$ because it looks bad stylistically

[twin meteor]

Ann

[severe willow]

what dyou think would be a better way to write it(?)

also thank you

[paper depot]

what i showed

or put the three in front

for a_n = 5 + 3(n-1)

[severe willow]

ohhhh gotcha

alright

thank you for checking this ^^

.close

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[quasi belfry]

Hi

[quasi belfry]

It’s the a

I really don’t know how I’m supposed to do that

I’m apparently supposed to use that

If you don’t know how this works, just explain me how you do it.

I’m literally failing all my math exams and that pisses me off so I’m tryna get better, I need help.

[vocal sleet]

@quasi belfry Has your question been resolved?

[quasi belfry]

<@&286206848099549185>

I wanna die

heeeeeeelp

<@&286206848099549185>

It's been 40 minutes that I need help and still no answers...

I tried to understand, I don't wanna give up but it's probably gonna happen

imma wait 10 more minutes

I still have hope, 10 mores minutes

so it's one hour

<@&286206848099549185>

I give up

nice

.close

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[rigid remnant]

this is wrong, no?

[flat whale]

What exactly do you think is wrong

There's like 6 equations

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[indigo mauve]

Looking for help with this question

[indigo mauve]

This is my current working out

What I am confused by is question d translated polynomial by (-1,2)

[vocal sleet]

@indigo mauve Has your question been resolved?

[indigo mauve]

<@&286206848099549185>

[gaunt glade]

i'm almost certain that in this context it means the graph was moved

i was worried it could have been scaling, but a) this is not directly indicated by the wording, b) you'd probably be seeing a matrix and not a vector

[indigo mauve]

That was my impression but in which direction is it being used I believe I can do it by f(x-1)+2 or f(x+1)-2

It could also be f(x-2)+1 or f(x+2)-1

[gaunt glade]

iirc the awkward x-axis sign swap doesn't occur here, i'm pretty sure (0,0) is supposed to be transformed like this

[gaunt glade]

i think the y-axis part is unambiguous, positive shift is up, negative goes down

of course this only applies to point D, in E you will have to consider these things

[indigo mauve]

So red is the original and blue would be (-1,2) transformation which is f(x+1)+2

Which would be a correct movement based on your explanation here

[gaunt glade]

yeah that looks correct

[indigo mauve]

Brilliant thank you very much

[gaunt glade]

np, i trust E won't be any issue for you then

glad i could help

[indigo mauve]

I haven't looked at it properly yet I am guessing it is the reverse of b

If you can assist that would be great

[gaunt glade]

oh, sure thing

[gaunt glade]

you're right, it's much better to plug x+1 into the factored form of the polynomial

[indigo mauve]

Sorry I am not quite sure what you mean

[gaunt glade]

nothing important really

you want to find the formula of f(x+1)+2, so you could plug the (x+1) into your neatly factored (x-1)(x+1)(x-3) form and work from there or plug it into the originally given cubic polynomial and spend 10 hours doing calculations

mb for unclear wording in the original message

[indigo mauve]

So am I right in saying it would now be

X+1-1=0 x=0
X+1+1=0 x=-2
X+1-3=0 x=2

[gaunt glade]

hmm, not really because the y-shift messes up the roots

$g(x)=f(x+1)+2\f(x)=(x-1)(x+1)(x-3)\
g(x)=2+f(x+1)=2+(x+1-1)(x+1+1)(x+1-3)=\
=2+x(x+2)(x-2)$

[twin meteor]

lifefuel

[gaunt glade]

then all you need to do is multiply it out (and maybe factor it, though the problem doesn't demand it in this case)

[indigo mauve]

Sorry just trying to wrap my head around this been 16 years since I have been in education and just got back into it

[gaunt glade]

sure, ping me if you need further assistance

[indigo mauve]

Thank you

[indigo mauve]

Thanka again I believe I have it now

[gaunt glade]

good job

[indigo mauve]

x^3-4x+2

[gaunt glade]

yep

[indigo mauve]

Finally hahahaha been working on this for ages

I will close this channel and free it up

[gaunt glade]

sure thing, bye-bye

[indigo mauve]

.close

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[hidden kelp]

Can someone help me with this problem? I'm not sure what to do

[lucid bane]

for what angles does sin()=1/2

do you see why i'm looking for where sin()=1/2?

[hidden kelp]

Oh yeah

Because 4x+pi/18 is equal to theta

So I just need to look at the unit circle?

[mild flower]

that will help yeah

[hidden kelp]

So it would be pi/6?

Because its the answer closest to 0?

[mild flower]

,calc 1/2 - sin(4*pi/6 + pi/18)

[twin meteor]

Result:

-0.26604444311898

[mild flower]

doesn't seem to be

[hidden kelp]

Oh wait is the 1/2 supposed to be the y?

And I need to look for the x where y is 1/2?

[mild flower]

you need to find the smallest positive value of x that makes that equation true

you found that sin(ø) = 1/2 if ø = pi/6, which is true, but there are other possibilities

you will likely need to consider all of them

[hidden kelp]

The only other one I see is 5pi/6

But that one looks further away than the pi/6

When I look at the unit circle

[mild flower]

well... ø is farther away yes

but you care where x is

[hidden kelp]

But doesn't the y have to be 1/2?

[mild flower]

yes

[hidden kelp]

I'm so confused 😭

[mild flower]

well no it has to be 0

you know how when you're solving a quadratic you end up with two solutions?

it's like that

[hidden kelp]

Ohhhhh

[mild flower]

but there will be infinitely many (because of multiples of 2π)

[hidden kelp]

So the answer is pi/6 and 5pi/6 because you dont which is closer?

And then you would do the +2pik thing?

[mild flower]

we need to consider both of those, yes

so ø = π/6 +2πk or ø = 5π/6 + 2πk

but what was ø?

[hidden kelp]

Theta is 4x+pi/18

[mild flower]

remember we're trying to solve for x

yeah

so now you can substitute that in

[hidden kelp]

ooooh

kk

[mild flower]

and solve both of those for x

[hidden kelp]

That makes sense

Ill go do that

[mild flower]

(in this case, the π/6 branch will actually be the correct one, but that's not true in general and hard to predict)

[hidden kelp]

Kk thanks a lot

I got 2.27 and 10.64

Oh wait

My calculator is in radians

[mild flower]

leave it in terms of π

and remember the 2πk term don't leave that behind

[hidden kelp]

Oh ok

I did the decimal just so I could easily tell

Thanks for helping me

That helped a lot

.close

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[feral harness]

Claiming

[feral harness]

So I’m trying to graph f(x) given f’(x)

Just need someone to double check if this is good I think I may have messed it up a little

[raven owl]

f(x) should look smoother

[feral harness]

can you elaborate a bit on what you mean smoother?

[raven owl]

currently your f(x) is made of lines. If you take the derivative of a line, you get a constant. This means ur f'(x) would have to jump from constant to constant (and would be flat on intervals), but if you look at the real f'(x) its not doing that.
As |f'(x)| gets bigger, f(x) should get steeper

[feral harness]

ah i see that

thank you for the help man

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[raven owl]

Did u graph them

[minor plank]

Yeah

But it's very big

[raven owl]

u using a graphing calc or an online calc

[minor plank]

Graphing calc

Online calc.not allowed

Nvm got the x into beuy I have another question. Doni.put it here or new Rea

[raven owl]

Either one is fine

making new channel puts ur question back at the top but otherwise it doesnt matter

[minor plank]

.close

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[grand moat]

Encil Zaidi buys a flat worth RM210 000 on credit. He pays a down paymennt and the balance is paid in instalments over 15 years. The flat interest rate charged by the bank is 3.5% per annum. If Encik Zaidi has to pay RM1 000 instalment per month, calculate the down payment that he paid.

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[dawn sandal]

I need help graphing the line

[hoary blaze]

this is a linear function, so you can just find two points that are far apart enough so that you can link them with a line

i'd pick x values that are multiples of 2 for simplicity

for example, x = -8 and x = 8

[dawn sandal]

Like this?

[hoary blaze]

mmh your slope is negative so the line should be going downward as x increases

start by finding out what's y when x = -8

and then what's y when x = 8

(just plug it in your equation)

[dawn sandal]

3

[hoary blaze]

-1 times half of -8 is 4

then you gotta add the + 3

so y = 7 when x = -8

[dawn sandal]

y=7+-1/-2x+3

[hoary blaze]

i just read the b) part my bad

ok so let's try this way

you agree that at x = 0, y is 3?

[dawn sandal]

yes

[hoary blaze]

okay

do you know what the slope formula is

[dawn sandal]

1/2

[hoary blaze]

yes

but how do we get that

if the line was already drawn

or what does it means

[dawn sandal]

i missed the - 1/2 part

[hoary blaze]

okay so the slope is given by delta y/delta x

or change in y/change in x

so if your slope is lets say 1

what does that mean for y when x increases by 1?

[dawn sandal]

1/3?

[hoary blaze]

if change in y is 1 when change in x is 1

that means when x increases by one, y increases by one too

agreed?

[dawn sandal]

1/1

yea

[hoary blaze]

okay so

when you apply the same logic to -1/2

when x increases by one, how does y changes?

[dawn sandal]

by one

-1/-3

[hoary blaze]

okay

when x increases by two, what happens to y if the slope is 1 (the change in y)/2 (the change in x)

[dawn sandal]

4/3

[hoary blaze]

how do you get this

can you find the slope for this line?

[dawn sandal]

2/1

oh

[hoary blaze]

right

[dawn sandal]

i keep thinking y,x instead of x,y

[hoary blaze]

what about this one

[dawn sandal]

1,2

[hoary blaze]

mmmm not quite

so it's the same as last picture

except it's negative

because the line is going downward

[dawn sandal]

so negetive -1/-2

[hoary blaze]

it's still 2/1

[dawn sandal]

-2/-1

nvm

[hoary blaze]

that would be positive

[dawn sandal]

bruh

[hoary blaze]

so it's just -2/1

since the y decrease but the x increases

[dawn sandal]

if a negative is also another negative then it will also be postitive

[hoary blaze]

a negative number divided by another negative number will give a positive number

[dawn sandal]

oooooooooooooh

[hoary blaze]

notice how x is increasing while y is decreasing

[dawn sandal]

yes

[hoary blaze]

that means the change in x will be positive

while the change in y will be negative

[dawn sandal]

2/-1

[hoary blaze]

well that would be -2/1

[dawn sandal]

mis typed it

[hoary blaze]

but that's the same thing in the end

okay so

going back to your question

[hoary blaze]

you can start by drawing a point at x = 0 and y = 3

since it's the initial value

and then you can apply what i showed you to graph the line

[dawn sandal]

ok

did that

[hoary blaze]

keep in mind the slope is -1/2 and not -2/1 like i showed you

[hoary blaze]

can i see?

[hoary blaze]

see

that's -2/1

what we want is -1/2

[dawn sandal]

so the other way

[hoary blaze]

we want the y to decrease by 1 every time x increases by 2

[hoary blaze]

you can say that

[dawn sandal]

/ the going this way

[hoary blaze]

nah

it's still going downward

but you just did -2/1 for the slope

we want -1/2

[hoary blaze]

basically this but going downward instead

[dawn sandal]

how do i rotate that one\

[hoary blaze]

like this

[dawn sandal]

alr

[hoary blaze]

(it's not accurate so forgive me)

but you get the idea

[dawn sandal]

its fine

[dawn sandal]

can you rotate thee image pls

[hoary blaze]

,rotate 90

[dawn sandal]

ty

alr

what about the equation

[hoary blaze]

well you already have it

so idk why it's asking for it again

[dawn sandal]

oh

ok

what about these

[hoary blaze]

ok so for 3

you are given two points

or two coordinates

draw a point on the graph for each one

and draw a line that goes through both of them

[dawn sandal]

im on 2 but we can do 3

[hoary blaze]

well for 2 b)

you just need to draw a triangle under the line

like so

[dawn sandal]

Like that

[hoary blaze]

yea

[dawn sandal]

ok

what do i do for 3

@hoary blaze

[hoary blaze]

see here

[vocal sleet]

@dawn sandal Has your question been resolved?

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[static wing]

Help, I don’t know how to set this problem up. According to the study guide the answer is c!

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[quick rain]

<@&286206848099549185>

[vocal sleet]

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[main tide]

question, would it be frowned upon if i asked for answers and not help?

[frozen bobcat]

yes

[ornate ember]

yeah feel free. This is a resource to help out people, there's not a requirement to help others unless you feel like it

[main tide]

i’m getting two different answers

[ornate ember]

wut? they're the same answer.

Yes you can ask and not help. No it's not frowned upon

[main tide]

oh okay thank you :)

[vast shale]

just send your question and we can address it appropriately?

[brisk moss]

i think you misunderstood or i did

[dull bear]

!noans

[vocal sleet]

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.

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[green crystal]

is this a valid matrix?

[cyan shadow]

yes?

[vast shale]

what do you mean by valid? this is a matrix

[green crystal]

I know that for the third row if it was [ 0 0 0 0 0 1] it would not be valid right?

[thin vale]

It is indeed a matrix

Can confirm this is a matrix

[vast shale]

why wouldnt it be

[cyan shadow]

you are actually asking if this matrix represents a set of equations that have a solution

of which this one does

[green crystal]

yea sorry for the wording

im not sure how to word this but if it was [ 0 0 0 0 0 1] for the third row it woudnt be valid

like it would have no solution

[strange crater]

depends if it's an augmented matrix

[green crystal]

like when i try to get the RREF for this, there would be no solution right

[strange crater]

yes

[vast shale]

$0x_1 + 0x_2 = 1$ has no solution yes

[twin meteor]

Heptagons

[green crystal]

do you mind explalining why?

[strange crater]

because 0 times anything can only equal 0

[vast shale]

and $0+0=0\ne 1$

[twin meteor]

Heptagons

[green crystal]

so this would be consistent

[vast shale]

yes, the only set of solutions here is {2, 0}

[green crystal]

this woudnt work with 0 + 1 = 0 tho

[vast shale]

this represents the equations
$1x_1 + 1x_2 = 2$
and
$1x_2 = 0$
right?

[twin meteor]

Heptagons

[vast shale]

so clearly, x_2 = 0 which we can use to get x_1 = 2

[green crystal]

how does that prove 0 + 1 = 0

[vast shale]

who said it does

[green crystal]

then its irrelevant

[vast shale]

here the equations are
$1x_1 + 1x_2 = 2$
and
$0x_1 + 0x_2 = 1$

there exists no pair of solutions ($x_1, x_2$) such that $0x_1 + 0x_2 = 1$

[twin meteor]

Heptagons

[green crystal]

right but could you apply the same to this

[vast shale]

no, because this has a pair of solutions
x_1 = 2 and x_2 = 0

[green crystal]

ok i think i got it

[green crystal]

so when i said this. I would rephrase and say if it was [ 0 0 0 0 0 1] the matrix would be inconsistent

[vast shale]

yes

in general, any row reduced matrix which has at least 1 row of
$[0, 0, 0, \dots, 0, b] b \ne 0$
has no solutions

[twin meteor]

Heptagons

[green crystal]

and if a given matrix has a zero row that is not at the bottom, it also has no solutions?

[vast shale]

by zero row you mean everything is zero? even the last value?

[green crystal]

yes

[vast shale]

well, it would have infinite solutions

[green crystal]

but its not at the bottom

[vast shale]

cuz there arent enough equations to uniquely determine a solution

[green crystal]

like say the question was: without getting the RREF for this matrix how many solutions would it have

and the zero row is in the middle

[vast shale]

eh i only know RREF

how do you get the RREF of a given matrix?

[green crystal]

gussain elimination

[vast shale]

wait what

im sorry

everything i said was augmented not RREF

[green crystal]

yea yea i know

[vast shale]

i should get out of here

[green crystal]

me too lol

thx for the help

[vast shale]

well in that case i only know augmented form not RREF

im bad at matrices i just googled that bit and thought it was RREF lol

[green crystal]

welp, ty anyway

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[haughty monolith]

does anyone have pictures with lots of math formulas for like what 16 year olds learn?

[haughty monolith]

like

algebra

[grave knoll]

barely any formulas

[haughty monolith]

expression

s

and equations

algebra wise etc

[haughty monolith]

where can i find the few

[grave knoll]

heard of google?

[haughty monolith]

im not getting what im searching for

[haughty monolith]

.close

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[vast shale]

suppose a point is given Q (0,33)

and a parabola is given y^2 =x/4
,
now you have to find a point on the parabola P(h,k)
such that the distance PQ is minimum.

[lucid bane]

calculus optimization question!

[vast shale]

yeah i thought of it

[half imp]

does the cubic work out nicely

Ah ok

[vast shale]

i would get 3 such lines

and

would get lengthy

[lucid bane]

cubic??

[half imp]

What progress have you made

[vast shale]

yeah

[half imp]

Do you have it down to solving a cubic?

[lucid bane]

well, you can focus on top half of parabola, and make it a function of x. It's basically the same question as asking 'how close do the curve, y=sqrt(x/4) come to the point (0,33)

you can think of square distance its easier

[vast shale]

i have thought of using the distance formula and taking its derivative and equating it to 0

[lucid bane]

and minimizing square distance is easier than regular distance

[half imp]

if you try that you should probably use the square of the distance

[vast shale]

yeah

yeah

would that work ?

[half imp]

In the end you'd end up having to solve a cubic equation

[vast shale]

lol

and get 3 points

and the check distances from three points 💀
then tell which is min

😦

[half imp]

Some cubics can be solved without too much trouble using specific methods but if not then it would need the cubic formula which is notoriously complicated

[half imp]

Two of them would probably be imaginary

[vast shale]

how do ya know

please tell

[half imp]

Because moving along the parabola, there is only one point at which distance from (0,33) is not changing

[vast shale]

what?

distance is changing at every point

[half imp]

when the derivative is 0 it is temporarily staying still

Like how a ball stays still in the air for an instant at the peak of the throw

[vast shale]

yup

wait wait

from an external point

[half imp]

So essentially just visualizing there's only one local extrema

[vast shale]

3 normals are possible

and when we are equating the derivative of distance to be 0

are we getting those 3 normals

[half imp]

Are they?

[vast shale]

or the one which is closest ?

[half imp]

I feel like you only get one real solution

But if you do get three then you would choose the one which was closest yeah

[vast shale]

and what would we do in a condition when we get all 3 real ?

[vast shale]

how ?

[half imp]

Calculating all the distances and choosing the smallest one

[vast shale]

alr

got it

thanks

[half imp]

np

[vast shale]

have a great day

.close

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[half imp]

ty u2

[vocal sleet]

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[river parcel]

Im so confused

[vast shale]

evaluate [
\m fx = 2x^{\ff13}+ x^{\ff14} + x^0 \tss{at} x = 16
]

[paper depot]

are we sure the first exponent is 1/3 there

it's not very easy to read from the screen photo

[river parcel]

Omgg i just zoomed in and realized its 1/2

Tyy

[paper depot]

btw what exactly confuses you?

[river parcel]

Well i did the right process but my answer didnt match up with any of the choices cuz i thought the exponent was 1/3 not 1/2

[vast shale]

show your work

[paper depot]

ah.

yeah no that would do it

[vast shale]

also did you correct for the fact that it is x^(1/2) and not x^(1/3) in what you did?

[paper depot]

16^(1/3) is irrational so no wonder you would not get a match

[vast shale]

oh nvm I just read

[river parcel]

It was 1/2

I got 11

[paper depot]

yeah that's correct now

[river parcel]

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[oblique gazelle]

Any ideas?

[vocal sleet]

@oblique gazelle Has your question been resolved?

[oblique gazelle]

No lovely bot, I need more time to get response and argue for my final grade. Thanks

[vocal sleet]

@oblique gazelle Has your question been resolved?

[oblique gazelle]

<@&286206848099549185>

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@oblique gazelle Has your question been resolved?

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[oblique gazelle]

Any ideas?

[vast shale]

sorry what is K11?

[oblique gazelle]

Complete graph

[vast shale]

what does k11 mean

i know what a complete graph is

you mean 12 nodes?

oh ok

[oblique gazelle]

no its not 12 its 11

[vast shale]

ok lol

[oblique gazelle]

No hold on

Apologies

Lmao

Its a path graph on 11 vertices

Yeah much better

[vast shale]

a path graph is a chain right?

[oblique gazelle]

Yeah like a straight line

[vast shale]

yeah alright

[oblique gazelle]

Both these statements look correct to me

[high maple]

Graph theory, huh

[vast shale]

for the second statement, can a path use the same edge twice?

no i guess?

[oblique gazelle]

It can

[high maple]

That's rare for help channels

[vast shale]

huh, how can there be less than 10 such paths then

[oblique gazelle]

Path is when no vertices are repeated

[vast shale]

oh okay

[oblique gazelle]

But in this graph you cant even have such case

Its a stright line, cant go over the same edge twice

[vast shale]

so lets say our graph is
1->2->3....->11
a path will always connect n and n+2, there are 9 such paths i agree

[oblique gazelle]

For one path ofc

[oblique gazelle]

Yep

[vast shale]

wait

is it directed?

[oblique gazelle]

Whats that

No

[vast shale]

are the edges directed or not

[oblique gazelle]

No never heard of such things

[vast shale]

then 18 paths lol not 9

[oblique gazelle]

No no they arent

Its simple graphs

[vast shale]

wait lemme show u

[oblique gazelle]

So yep its 9, I even checked it manually to make sure

[vast shale]

its 18 if its undirected

[high maple]

That's not K11

[oblique gazelle]

Wait how

[oblique gazelle]

There is no K11, ignore that. Made a mistake

[vast shale]

he says its a path graph, my question is if its directed or not

[oblique gazelle]

No its not

[vast shale]

so the answer is 18

[oblique gazelle]

So can you explain then how is it 18?

[vast shale]

not 9

[high maple]

Huh, so your complete graph is not a clique?

[oblique gazelle]

Do you like consider 1~3 and 3~1 paths different?

[high maple]

What

[vast shale]

Yes

[high maple]

Also what

[oblique gazelle]

Its not a complete graph at all, I explained under the first message

[oblique gazelle]

Doesnt it turn into a directed path then?

[high maple]

Ah, path graph on 11 vertices

[vast shale]

yes

[oblique gazelle]

If its undirected then there is no direction then 1~3 and 3~1 are the same

[vast shale]

alright sure

my bad

[oblique gazelle]

Yep

[high maple]

Yea now it makes sense

[oblique gazelle]

This is like the most confusing help channel fr

[vast shale]

so yeah second statement is indeed true

[oblique gazelle]

Yep

[high maple]

But why would you "gotta choose one"

[oblique gazelle]

I even have a proof why the first one is true

[vast shale]

tree graph with sum of degrees 30? you claim that it has 16 nodes

[oblique gazelle]

Thats just what the problem states

[oblique gazelle]

Yeah

[vast shale]

each edge adds 2 to the sum of degrees

[high maple]

Maybe you misread the question?

[vast shale]

so there are 15 edges

hence 16 vertices

[oblique gazelle]

Are you sure tho both are correct?

[vast shale]

unless he means directed path yes lol

[high maple]

Well, assuming the second graph is a path graph

[oblique gazelle]

Exactly

[oblique gazelle]

It is

[vast shale]

also im sorry for confusion, im a competitive programmer and usually in a graph problem i get definitions in the statement

[high maple]

Doubt that

[oblique gazelle]

Nah I am sorry for confusion, I had to correct the question like 5 times xd

So in conclusion

Either im blind as hell and misread one of the statements

Or they made a mistake in the final exam

I chose the first one as my answer

[high maple]

I would go with misread

[oblique gazelle]

And I am like 99.5% sure that it said its a tree graph with sum of degrees 30

[vast shale]

well, we are 100% sure what a tree is and we are 100% sure the first statement is correct

[oblique gazelle]

So we hope I misread the second statement

[high maple]

Yea likely

[oblique gazelle]

Thanks a lot guys

All the best :>

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[void mountain]

Guys is 1200km/90min, 221.7777..m/s?

[void mountain]

km to min

the average speed is 221.7 rounded up to nearest tenths

km/min to m/s

amI correct?

[visual oracle]

seems right

[outer warren]

seems a little off

can you show your calculations

[vocal sleet]

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[wise kiln]

(2x^3-3x)^5 coefficient of x^9

[wise kiln]

Can we solve it without binom expansion?

[sweet birch]

you could but why would you

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[wise kiln]

It would be more efficient?

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[solemn gulch]

difference between separable and linear differential equations?

[solemn gulch]

is this one linear or separable

[paper depot]

linear is y' + p(x)*y = q(x) or anything reducible to that form

an equation is separable if you can separate the variables so that x appears only on one side and y only on the other

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[vast shale]

just want some clarification on this

I chose the origin to be where the spring is neither stretched or compressed

I ended up with like [
mg\m\cos\theta\p{\m\sin\theta D - \m\sin\theta \p{x-D}} - \f12 kx^2 - \mu_k mg\m\cos\theta = 0
]
Where the positive direction of $x$ is downwards along the incline

but this screams wdong

[vast shale]

am I meant to solve for x here?;

is this even correct

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[proven light]

Can someone help me on this physics question? I believe I have the correct answers

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[topaz oar]

I need help solving the equation in the top left, my book says to simplify and identify but I don't know how to simplify further

[tribal moss]

equate the real part and the imaginary part to zero

[tribal moss]

also change minus sign in the denominator into plus, since (a + bi)(a - bi) = a^2 + b^2

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[deft sedge]

Could someone confirm that showing k = θ(n) with respect to n^(2k + 2) = (n/e)^n is proof that a lower bound for the diameter of the cayley graph of permutation transpositions is θ(n)

[deft sedge]

The idea is that n^(2k + 2) is an upper bound on the number of words with word lengths ranging from 0 to k where the characters represent the permutation transpositions

And

(n/e)^n is a lower bound on the number of permutations of n objects (n!)

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<@&286206848099549185>

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[pallid ruin]

For problem 12 calculate the coordinates of the point of intersection of the lines

[pallid ruin]

Do I just find the x and y intercept?

I hate aops wordings

I'm like bad at understanding instructions

[dull bear]

only for (b) and (c) respectively - not for the other two

[pallid ruin]

Wut

So

Whahwjwhwhhwhat

[dull bear]

the lines in a and in d do not need to match up with the x or y intercepts (and in fact, the y intercepts aren't even the same for them at least!)

[pallid ruin]

Oh wait

Ur trying to find the intersection for the uhh

Systems right?

[pallid ruin]

So would the first point be 4,7?

[dull bear]

Yep

[dull bear]

And yep, that's it

[pallid ruin]

So what is b asking then

Ye :D

[dull bear]

"Where does the line y = x - 3 meet the x axis?" (i.e. what is the coordinates of the x-intercept of the line y = x - 3?)

[pallid ruin]

When x is 0?

[pallid ruin]

So then the coordinate would be uhh

(0,-3)

Right?

[dull bear]

be careful! x axis is where y=0

[pallid ruin]

Ohh

[pallid ruin]

How do I find x intercept again?

[dull bear]

Well, you know that y has to be zero, sooo maybe put that in?

[pallid ruin]

Ohhhh

Ok

[dull bear]

Then find what x has to be

[pallid ruin]

Okk

So then c is asking for y then?

[dull bear]

Yep - and you know that x has to be zero

That one should be much simpler(!)

[pallid ruin]

Oh okk

[pallid ruin]

(0,4) ?

Right?

Okk

And then uhh

I do another system?

Waitrr

How do I do this one

[dull bear]

They tell you basically that x=2

[pallid ruin]

Ohhh

[pallid ruin]

Plug in?

Y is 5?

Okk

Tyy

Uhmm

I still need help on something tho

:D

Sorry lol

Should I open another chabnrl?

[dull bear]

You can ask here if you want

[pallid ruin]

Okk

Sp uhh

8 mean 35

Srry

34

[dull bear]

34?

[pallid ruin]

No 45

35

😭😭

35

Srry

[dull bear]

got there in the end

[pallid ruin]

I can uhh

Show u

What I hace

So far

[dull bear]

Yes please

[dull bear]

Be a bit careful note that if A gives B money, then A loses $3 but B gains $3

[pallid ruin]

Ohh

So its (b+3)2

[dull bear]

Yep, doubled

[pallid ruin]

So then the second jd this?

[dull bear]

Yep

[pallid ruin]

Okk

I'll solve it

And tell u

[dull bear]

perfect

[pallid ruin]

B has 37 right?

[dull bear]

I haven't done any work for it hehe

[pallid ruin]

Nooo

It's probably wrong

[dull bear]

Too lazy

[pallid ruin]

😭😭😭😭

[dull bear]

It might be right let me check for you

[pallid ruin]

My number is too biggg

[dull bear]

,w solve a - 3 = 2(b + 3), b - 7 = (a + 7)/3

[twin meteor]

Results provided by WolframAlpha

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

Requested by chartbit

[pallid ruin]

But whatrt

That doesn't make sense se

[dull bear]

Why does it have to be too big and you're right!

[pallid ruin]

If a gives b 3 dollars then b has 4p and a has 80

But b has to have double

😭😭

[dull bear]

4p?

[pallid ruin]

80 I mean

I mean no

40

B has 4p 0

40

[dull bear]

wait the equations might be the wrong way around

[pallid ruin]

😭😭😭😭😭

Let's redo it pls

[dull bear]

Yep, sure!

(just remember that giving money means that one loses and the other gains an equal amount!)

[pallid ruin]

So the first one would be uhm

[pallid ruin]

Idk what to do

😭

[dull bear]

Let A have $a and B have $b to start off with

[pallid ruin]

wut

ill just call the variable x and y

ig

[dull bear]

Well if you want

[pallid ruin]

okk

[dull bear]

Easier for me to keep track if you call them a and b, cause they match up

[pallid ruin]

okk

so a is a

and b is b

there

[dull bear]

Yep, if A gives B $3, then A then has (a - 3) dollars and B then gets (b + 3) dollars, as you decuded correctly

[dull bear]

But B needs to have twice as much as A from there

[pallid ruin]

yeah

a-3=2(b+3)

[pallid ruin]

do you multiply the thing by 3 then on the second equation

where a+7=3(b-7)??

[dull bear]

You could if you wanted - but...

[pallid ruin]

idk what to do

bro

[dull bear]

this isn't B having twice as much as A, this is the other way around

[pallid ruin]

ohh

so its 1/2 b right

a-3=1/2(b+3)??

[pallid ruin]

im assuming thsi is what i got wrong?

[dull bear]

Yea, I think it is

[dull bear]

That works of course - or equivalently, b + 3 = 2(a - 3)

B, with their (b + 3) dollars, has twice as much as A, with their (a - 3) dollars

[pallid ruin]

oh

ok

so let me just solve it

[dull bear]

Yep, don't forget to do the second equation as well

[pallid ruin]

13 dollars for b

and 11 dollars for a?

[dull bear]

,w solve b + 3 = 2(a - 3), a + 7 = 3(b - 7)

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[pallid ruin]

yeyyy

tysm

🫂

[dull bear]

Yep to both

[pallid ruin]

wooo

[dull bear]

A pleasure

[pallid ruin]

okk i can close the channel now

byebye

see u when u help me again lol

or if you help me again

:D

.close

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[plain geyser]

Is this correct?

[twin meteor]

Chai T. Rex

Chai T. Rex

[gritty sage]

Oh, sorry, I just did it and you're right.

The derivative is correct.

[plain geyser]

I know the dy/dx is right
But not sure about dy=

I don't have to make it 1/(2x+9)^1/2

[gritty sage]

OK, so what is delta x in that? Are they doing the limit of (f(x + h) - f(x))/h where h is 0.25?

[plain geyser]

I believe you just plug in x and multiply the formula by delta x

[gritty sage]

Is that what you've done in previous questions like this one with delta x?

[plain geyser]

yes

[gritty sage]

The question seems incoherent to me. delta x and delta y can be used together. dx and dy can be used together, but they never ask for delta y, so there should be no reason to use delta x.

Can you show the part of the book where they explain this?

[plain geyser]

wait I have an example problem

Here is one that

is correct

[gritty sage]

,w 4/sqrt(8(1) + 9) 0.22

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[gritty sage]

OK, so it looks like you're right.

You do (0.25 \frac1{\sqrt{2(3) + 9}}).

,w (0.25 \frac1{\sqrt{2(3) + 9}})

Sorry.

[twin meteor]

Chai T. Rex

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[gritty sage]

How did you get 0.213?

[plain geyser]

(0.22)(4)/(\sqrt(8(1)+9))

[gritty sage]

No, I mean on this ^

Oh, never mind.

I confused the two.

It looks like you're right with 0.065.

I guess they're showing the change from delta y/delta x to dy/dx with this step being between those.

[plain geyser]

ok thx

[gritty sage]

No problem.

[plain geyser]

Is this correct?

[regal parrot]

Was there a mod ping in here

[hollow bobcat]

Ghost ping?

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[loud mango]

Does f(x+h) become the red or blue one

Or is it the same

[livid horizon]

,rotate

[livid horizon]

Red one I think

But

Blue one would be correct as well if there was a +6xh term

[loud mango]

Ohh okay

Thank you

.close

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[vast shale]

hi

[vast shale]

is kirchoff equation only valid at constant pressure ?

[cedar kernel]

are you talking about the chem thing

[vast shale]

yes

[cedar kernel]

not the correct server to ask this tbh

[vast shale]

why

lol

its just math

literally

[cedar kernel]

#old-network

[vast shale]

theyre dead

lol

[cedar kernel]

¯_(ツ)_/¯

i got my answers frequently

one second i would have to reffer with notes , i just studied this thing , have not revised yet

[vast shale]

refer*

ok

[cedar kernel]

oh yeah ofc , we took Cp

[vast shale]

dh/dt = cp only at constant pressure

[cedar kernel]

refering that heat capacity at constant pressure is the change in enthalapy by

Cp

the heat capacity is measured at constant pressure

[vast shale]

dq/dt

yeah

[cedar kernel]

yeah

dH / dT

[vast shale]

how to find gibbs energy with entropy and equilibrium constant data

[cedar kernel]

delta g?

[vast shale]

yes

[cedar kernel]

it was something like delta H = -2.303 nRT log K

then you can use delta g = delta h - t delta s

i think

[vast shale]

how