help-23

[polar pulsar]

ye but for dis. distributions integration's unecessary

discrete*

[dull sequoia]

well continuous and discretes will both work like this im just generalising

[long shore]

true thats way better

[polar pulsar]

im not saying you cant use integration, i just mean that its kinda useless when you already know the values

[dull sequoia]

i was trying to show LOTUS, where you can just apply the g to the x in the definition

[long shore]

whats the notation of $f_X(x)$

[flat frigate]

suds

[dull sequoia]

that's the pdf of X

[polar pulsar]

LOTUS?

[dull sequoia]

if X is discrete you have p_X(x) or P(X = x)

[long shore]

okay

ok so its just different notation

[dull sequoia]

law of the unconscious statistician

which says this (for measurable g)

[polar pulsar]

that's not real is it 😭 who came up with that

[dull sequoia]

no it's real lmao

[polar pulsar]

wild name

makes sense why he's unconscious tho

imagine doing more work for no reason

😭

[dull sequoia]

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

i promise it's real man

[polar pulsar]

😭

in the unconscious guy's defence, the definition is the only thing against that thinking, theres no reason why you couldnt define it the other way round 😭

[long shore]

ok no worries thanks guys

helped a lot

bc otherwise i wouldve just learnt thats how u calculate it and not really understand

.close

[safe radish]

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[hazy sundial]

So we have a function defined from [0 ; +inf) with values in R ,with the property that
Abs( f(x)- f(y) ) <= Abs (sin x -sin y ) , x y being postive real numbers , and we need to prove this function is periodic and limited

[solar hazel]

does limited mean bounded?

[hazy sundial]

Yes excuse my poor wording

I firstly replaced y with x+2k pi k being a integer and proved that its periodic

But for bounded I got to the relation -2 <= f(x) -f (y) <= 2

And idk how to continue

[mortal sandal]

for all x,y?

[hazy sundial]

yes

[mortal sandal]

fix x

say x=0

[solar hazel]

amen dreyuk

[mortal sandal]

🙏

[hazy sundial]

then vary y and because we got a periodic function we can subtract or sumn to get only f (y) bounded?

[mortal sandal]

I don't think we need periodic

[mortal sandal]

Only this

[light shoal]

when in doubt, use the triangle inequality

[hazy sundial]

so we get f(0) and f(y) and how do we continue?

[mortal sandal]

isolate f(y)

[solar hazel]

isn't it very simple

[hazy sundial]

Im not seeing your vision yet

[solar hazel]

for all x we have -2 + f(0) <= f(x) <= 2 + f(0)

[hazy sundial]

Ohhh

Thanks everyone

.close

[safe radish]

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[mortal sandal]

No problem

[solar hazel]

[safe radish]

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[harsh sapphire]

Hi guys, would someone be able to help me know where to go next for this question?

[harsh sapphire]

This is what I have done so far

I asked my teacher and he literally said his brain wasnt working when he looked at it lol

[solar hazel]

beautiful handwriting

[harsh sapphire]

aw thanks haha

[brave wolf]

shouldnt it be $\sin\left(2kA\right)+2\sin\left(A\right)\cos\left(\left(2k+1\right)A\right)=\sin\left(\left(2k+2\right)A\right)$

[flat frigate]

MathIsAlwaysRight

[harsh sapphire]

hm

noo i dont think so

i substituted the LHS (excluding cos(2k+1)A to where i assumed true for n=k

[brave wolf]

$2\sin\left(A\right)\left(\cos\left(A\right)+...+\cos\left(\left(2k-1\right)A\right)+\cos\left(\left(2k+1\right)A\right)\right)=\sin\left(\left(2k+2\right)A\right)$

[flat frigate]

MathIsAlwaysRight

[brave wolf]

yeah, but it should look like this, no?

so you have to distribute the 2sin(A) to the first part and to cos((2k+1)A) and then substitute

[harsh sapphire]

noo

im substituting the n=k for sin(2kA)

substituting what was stated on the LHS where it equaled the RHS for n=k

[brave wolf]

yeah, i understand that

You're substituting the first equation into the second one, right?

[harsh sapphire]

ohh

yea i assumed that it would be in the brackets but my teacher told me it wouldnt be

he might have been wrong though

[brave wolf]

It is in the brackets

[harsh sapphire]

sorry i get what youre saying

well thats annoying haha

[brave wolf]

$\sin\left(2kA\right)+2\sin\left(A\right)\cos\left(\left(2k+1\right)A\right)=\sin\left(\left(2k+2\right)A\right)$

[flat frigate]

MathIsAlwaysRight

[harsh sapphire]

let me rewrite

[brave wolf]

anyways, when you're here it's probably just a bunch of trig identities

[harsh sapphire]

would you be able to steer me in the right direction? just not sure where to go

do i just let that be the 2k+1 and move to working on one side only?

[brave wolf]

you could use the sin(A) - sin(B) formula if you know that

[harsh sapphire]

ohh yea

[brave wolf]

that's probably the simplest way to do it

[harsh sapphire]

so sin((2k+2)A)-sin(2kA)?

and i expand it to sin(2kA+2A)?

[brave wolf]

sure, if you want to you can

but you could also just apply the identity

[harsh sapphire]

oh wait thats not in my formula sheet

uuugghhrrghhg

[brave wolf]

hmm you can prolly avoid it

[brave wolf]

I'd try doing sin(A + B) and cos(A + B) then

until there is no 2k+1 and 2k+2

just sin(2k) and cos(2k)

[harsh sapphire]

okay

do you think id still need to expan

[brave wolf]

yeah, probably

[harsh sapphire]

okaay let me do that rq

[brave wolf]

I'd expand these two

[harsh sapphire]

ohh right i forgot that i could do that

okay will do

soryr i got a snack

[safe radish]

@harsh sapphire Has your question been resolved?

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[echo gazelle]

how would i approach this q?

[main mural]

compare the limits when y=x and y=-x

[normal moss]

approach 😂

sorry

Or just use polar coordinates

[echo gazelle]

wait why?

[frozen marlin]

just do it :>

[echo gazelle]

o

is that the process everytime tho

[frozen marlin]

not necessarily

sometimes they agree, sometimes they don't

[echo gazelle]

ok lemme do9 it

[frozen marlin]

alright

[normal moss]

That's not the point of math, or teaching math.

[frozen marlin]

my apologies

but here it's kind of obvious why we pick these two paths

i'd rather have the helpee work it out before i explain to them

considering this specific method doesn't work for every limit

[normal moss]

Familiar as breathing for some, foreign as Mars for others.

[frozen marlin]

okay take over for me then

i don't mind

[frozen marlin]

did that sound passive aggressive

mb didn't mean it that way

[frozen marlin]

yep. what does that tell you

[echo gazelle]

theyre not the same so theres no limit

[frozen marlin]

correct

i'll let USSE explain

[normal moss]

No, I didn't mean it that way at all. Just trying to say that everyone is at different stages. What might be obvious to you might not be to someone else

And we should strive to help them understand

[echo gazelle]

is it because it passes through 0,0

[normal moss]

Not just say to just do it

[frozen marlin]

again -- what could i have done there?

i'd like to know so i don't do it again, i'm not a helpful anymore, so bit rusty

[normal moss]

Well first

I'd ask the OP

[echo gazelle]

guys i think thats me

[normal moss]

If they understand why we are even testing

[echo gazelle]

no

[normal moss]

Right

Do you understand the definition of a limit in multiple variables?

[echo gazelle]

it can be appraoched from a lot of places

[normal moss]

Infinitely many

But the key idea is

[echo gazelle]

ohh cute nice

[normal moss]

That we must approach the same value along every path for the limit to exist

And so, from logic

If we find two paths which give different values, then we already know the limit does not exist

[echo gazelle]

yas

[normal moss]

As it is in contradiction with this

Correct?

[echo gazelle]

yeah

[frozen marlin]

oh. i assumed you meant explain why we go thru the paths y = x and and y = -x, USSE

[normal moss]

Well I am getting to that

But first I wanted to see if the general concept was clear

[echo gazelle]

thx for explaining cuz i dont really udnerstand my teacher

[main mural]

and for that reason i'd say it's typically easier to prove that a limit doesn't exist than if it does lol, because for non-existence you only 1 counterexample, but for proper limits you'd need to typically involve the usual epsilon delta definition

[normal moss]

And then we test with lines like y = x, y = -x or the axes x = 0, y = 0, and broadly y = kx. Because they turn the two-variable function into a single variable-function

Making the limits easy to compute

[normal moss]

Yes, of course 😂

But a very important point is

[echo gazelle]

ohh

[normal moss]

Testing many lines can either disprove the existence of the limit (if we find two different values)

Or only suggest the limit exists

If we test say 15 lines and we get the same value

That does not mean the limit exists

[echo gazelle]

so for this i can use x=0 and y = 0 too?

[normal moss]

Yes, of course

Or y = 4x, y = ex

[main mural]

sure, you'll get that the limit along those paths will be 0, which is still distinct from 1/2

[normal moss]

Because we haven't checked every path

[echo gazelle]

how can we check every path

[normal moss]

We can't

[normal moss]

That's why this

[echo gazelle]

oh ok

doesnt y = kx

[normal moss]

That's why we use polar coordinates then usually

[echo gazelle]

test everyu path

[normal moss]

No

Curved paths matter as well

You can also test y = x^2

[echo gazelle]

oh

[main mural]

nope, there are limits which exist for every y=kx but fail for example for y=x^2

damn USSE we are in sync lol

[normal moss]

But polar coordinates cover all paths

[echo gazelle]

ohh im learning that next

i think

[main mural]

wait, do they?

[normal moss]

Yes

[echo gazelle]

do u guys like calculus

[normal moss]

You change the function to depend on r and theta

r approaches 0

And theta represents all possible directions from which you can approach the origin

And so if we figure out the limit depends on theta

The limit doesn't exist

As that implies the function approaches different values at different paths

[normal moss]

I do

[main mural]

riight ok yes i remember that

that was the question i failed 1 year ago on my multivariable exam

[echo gazelle]

what is it approaches something so its not undefined so like (1,1)

do u just substitute

[main mural]

yup

no problems there

[echo gazelle]

ohh ok

[normal moss]

And just a sidenote

If we used polar coordinates for your example

[main mural]

you only get problems when you have 0/0 or inf/inf

[normal moss]

We get $f(r, \theta) = \cos{\theta} \cdot \sin{\theta}$

[flat frigate]

USS-Enterprise

[normal moss]

after maybe 2 lines so very easy*

and we clearly see it depends on theta

And hence the limit does not exist

[echo gazelle]

ok tysm guys

i will be back later

[normal moss]

I wish we had more calculus

We have only Analysis

[echo gazelle]

omg..

[normal moss]

So we barely do any actual examples

[echo gazelle]

this stuff is damn hard

[normal moss]

I found discrete math harder

😭

[echo gazelle]

whats discrete math

...

[main mural]

fr?

[normal moss]

Whatever isn't continuous

[echo gazelle]

is that like proofs

[normal moss]

numbers, graphs, logic, combinatorics

[echo gazelle]

I HATE

COMBINATORICS

[normal moss]

Yes

[echo gazelle]

everyone was so good at it

but im so bad at it

😭

[main mural]

real asf dude

[normal moss]

But I guess that's just me

I love analysis the most probably

[echo gazelle]

im ok at proofs

like induction and stuff

converse was it

idk i forgot i did it last year

[normal moss]

I am always amazed by combinatorics because it requires such a different mindset I guess

[slate current]

I love combinatorics

[normal moss]

I still remember schoolmates who got 2 and 3 (C and Ds) at math got As and Bs at combinatorics specifically

[echo gazelle]

i think im trash at it casue i cant envison it

[normal moss]

Oh me neither

I have combinatorics next year

wish me luck

[echo gazelle]

good luck

[normal moss]

🙏

Well look

I hope you get what we were doing now

And if you have more questions feel free to come back 🙂

[echo gazelle]

ok thx so much

.close

[safe radish]

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[surreal kayak]

sm1 vc i need help rq

[brave wolf]

we dont have vc help in here, can it be done in help channels?

[warm warren]

Just post the question

[safe radish]

@surreal kayak Has your question been resolved?

[surreal kayak]

k

[crimson python]

calculate the acceleration through the equations of motion and the find force

time can also be calculated through equations of motion

[lean otter]

change the units too

[safe radish]

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[lean otter]

speed is in m/s but distance is in cm

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

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[safe radish]

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[manic patrol]

<@&268886789983436800>

[safe radish]

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[pliant citrus]

when burnt from flammable end (first pic) the velocity of burning is 3v
when burnt from the other side the v is 2v

[pliant citrus]

ler me show u my solving

hold on

[open wedge]

what's this translates to

[primal bone]

See the second image

[primal bone]

The top part's already been translated in the message itself

[open wedge]

i meant this

oh alr

[pliant citrus]

i keep finding 13,8

but the answer says 12,8

i genuinely think the answers wrong

[gritty tartan]

wat does 4,8 mean

[pliant citrus]

seconds

[gritty tartan]

Huh

[pliant citrus]

it takes 4,8 seconds for the flame to reach there

[gritty tartan]

4.8 seconds?

[pliant citrus]

yea

[gritty tartan]

oh okay

i was confused at first

[pliant citrus]

yeah punctiation changes from language to language ig😭

[gritty tartan]

nothing on how long the matchstick is?

[pliant citrus]

nope

but you can give any number

[gritty tartan]

so just the velocity of burning?

[pliant citrus]

cus you js need the time

[pliant citrus]

yes

lets say the stick is 12x long

the time it takes to burn from flammable end wd be 12x=3v* 4t

so 4t

[gritty tartan]

wait what

i thought itd be 3t

Oh wait r u putting in the units

[pliant citrus]

why?

[pliant citrus]

yeah

we dont know the exact time and v

[gritty tartan]

okay

[pliant citrus]

the slower you are the more time it takes to burn

[manic patrol]

after some bashing independent of yours, I agree with your answer.
most likely a typo on the answer key?

[pliant citrus]

yeahhh i think so too

[manic patrol]

probably raise it to your teacher in this case.

or if the answer key has the working as well, maybe show it and we can run through it to see what they're assuming and/or doing.

[pliant citrus]

ya

.close

[safe radish]

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[gritty tartan]

oh wait hold on a minute

bruh

[safe radish]

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[craggy wyvern]

How do I calculate Perimeter and Area of this circle like thing. 2 boxes are 1 c/m

[craggy wyvern]

I know the formula for a normal circle

[earnest nacelle]

if I tell you that you need to apply the formula for the area of a semicircle four times, can you do it?

[craggy wyvern]

Yes ig, semi circle is r^2•pi:2 right?

[earnest nacelle]

(1/2) pi r^2, yes.

but there will be three different radii at play here

[craggy wyvern]

Like this?

[earnest nacelle]

there's one more circle at play

[craggy wyvern]

Like this

[earnest nacelle]

that's more like it

but of course, remember you only have semicircles here

don't actually use the formula for the area of a regular circle

[craggy wyvern]

So I calculate all semi circles and subtract them from the biggest one?

[earnest nacelle]

the three smaller ones from the biggest one, yes

[craggy wyvern]

Ok great, thank you so much 😊

And have a nice rest day!

.close

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[earnest nacelle]

likewise!

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[faint glade]

A mistake at the foundry has caused a 48.9 litre alloy of bronze to contain only 8% of tin instead of the necessary 12%.  
How much tin needs to be added to the bronze to fix the mistake?

How do I approach this problem?

[faint glade]

I found the original amount of tin which is 3.912L

[graceful quail]

Say you add x litres of tin, what would the new amount of alloy and new amount of tin be

[faint glade]

but I'm confused about how I should think about setting up the equation

$48.9 + x$?

[flat frigate]

Vortac

[lean otter]

for amount of alliy correct

[faint glade]

for tin, would it be t = 48.9x?

[graceful quail]

This was your original amount of tin

So what do you get once you've added x

[faint glade]

$ 48.9 + x = 3.912y$?

Its some kind of ratio, I'm just unsure what

I'm having a lot of issues determining the order of mixing problem steps

Some are SLEs, some arent

So right now, I have:

48.9L @ 8 %tin
3.912L tin

xL @12% tin
y Tin

[graceful quail]

Your new amount of tin would be 3.912 + x as you added x tin

[faint glade]

so

xL @ 12% tin
3.912 + x tin
```?

[graceful quail]

You have a total volume of 48.9 + x and a volume of tin of 3.912 + x, so take their ratio to find the % of tin

[faint glade]

like, $\frac{3.912}{48.9} = \frac{3.912 +x}{48.9 + x}$?

[flat frigate]

Vortac

[graceful quail]

As $12% = 0.12$ we need $\frac{3.912+x}{48.9+x}=0.12$

brub

As $$12% = 0.12$$ we need $$\frac{3.912+x}{48.9+x}=0.12$$

As $$12% = 0.12 we need \frac{3.912+x}{48.9+x}=0.12$$

[flat frigate]

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

[graceful quail]

What is going on man

[lean otter]

$12 % = 0.12$ we need $\frac{3.912+x}{48.9+x}=0.12$

[flat frigate]

hail

[graceful quail]

Ty

[lean otter]

\% for it

[faint glade]

That's the part that confuses me

how to recognize how to setup the equation

I guess I just do a lot of problems until I have it memorized?

Coffee worth $1.05 per pound is mixed with coffee worth 85¢ per pound to obtain 20 pounds of a mixture worth 90¢ per pound. How many pounds of each type are used?

I find this much more straightforward, since you can represent it as

x + y = 20

and

1.05x + 0.85y = 20(.90)

[graceful quail]

Correct

[faint glade]

Its just an SLE

[frozen marlin]

good job yay

[faint glade]

but problem one, its not clear to me

[graceful quail]

You need to think about it intuitively

If you want to fix this problem you need to add some tin (since there's not enough)

So what happens after you add more tin?

You get more tin, but also you get more alloy

[normal moss]

backslash before %, \%

Oh nevermind

I didn't read the chat

[vernal portal]

Initially the alloy needed 12% of the total volume as tin but due to prblm in factory it ended up with only 8% tin so we are tryna add some tin to make it 12% tin

[normal moss]

🤦‍♂️

[faint glade]

this makes sense

We need to preserve the ratio

[vernal portal]

So it had 8% tin meaning 8% of 48.9L which is 3.912L now lets say we added xLitres of tin now the amount of tin is 3.912+xL BUT amount of total alloy is 48.9+xL

So we need 3.912+x/48.9+x to be 12%

[vernal portal]

.

[faint glade]

So solving this, I get x = 0.04

which I don't think is right

[graceful quail]

Show us your working

[faint glade]

$\frac{3.912 + x}{48.9 + x} = 0.12$

$0.08 + x = 0.12$

[flat frigate]

Vortac

[lean otter]

no space after $

[faint glade]

$x = 0.04$

[flat frigate]

Vortac

[graceful quail]

Where did that come from?

[flat frigate]

Vortac

[faint glade]

3.912/48.9

[graceful quail]

You can't do that division because theres a +x in the way

[faint glade]

I guess the + x would cancel out id I did this

[graceful quail]

It doesn't cancel because it's added

So you need to multiply both sides by 48.9+x

[faint glade]

doh

I should've known that...

[lean otter]

,, \frac{a+b}{c+b} \neq \frac{a}{c} + b

[flat frigate]

hail

[graceful quail]

The undergrad's dream

[broken forum]

https://tenor.com/view/confident-pigeon-pigeon-doctor-tuff-meme-gif-2300550870562275327

[faint glade]

so x = 2.2227?

[graceful quail]

Check if it works

[broken forum]

,w (3.912 + 2.2227)/(48.9 + 2.2227)

[flat frigate]

Results provided by WolframAlpha

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

Requested by nyxzore

[graceful quail]

Close enough

[graceful quail]

What do your teachers want you to round to?

[faint glade]

2 decimal places

[graceful quail]

What happened to x on the RHS on line 3

[faint glade]

I forgot to write it, whoops

[graceful quail]

So what's your final answer?

[broken forum]

woah, your notation is strange

[faint glade]

They need to add 2.22L of tin

[broken forum]

on line 2

be sure to write $(48.9+x)$ in brackets on both sides

[flat frigate]

Nyxzore

[broken forum]

otherwise it looks like $x\frac{3.92+x}{x+48.9}+48.9$

[flat frigate]

Nyxzore

[broken forum]

which is very different and u don't get that nice cancellation

which u used

u wiff me?

[graceful quail]

https://giphy.com/gifs/1yMvhR4M47Okw4n8tt

@faint glade You here?

[faint glade]

Whoops, sorry

yeah

Thanks a bunch for the help, much appreciated

I took notes

[graceful quail]

You can use .close if you're done

[safe radish]

@faint glade Has your question been resolved?

[faint glade]

.close

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[spare dragon]

Okay for a slightly more confusing one - I am doing a lil research on the mean value theorem. How would you go about proving that the MVT for differentials, that being f'(c) = s(x) within an interval, doesn't work for a discontinuous function where there does exist a tangent that fulfils this simple goal even with the discontinuity? In the image, there would be a tangent on the low slope of the x^2 graph that has the same value as the secant from a given point on f(x<0) to a point on the parabola, given it is on the right of the maxima ofc

[peak estuary]

pick a different counterexample

[spare dragon]

Okay so when the formula demands continuity, it is not because it will always require it, but because it sometimes doesn't work?

[peak estuary]

if it is continuous then it will always work. if not then it sometimes works and sometimes doesnt

[spare dragon]

I see, thank you so much!

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[craggy thunder]

Hi there, I am looking at 3 planes that i know intersect, is there an intuitive way to determine if they intersect on one line, or 3 lines? I can get by by finding a point on each and testing if it is shared

[covert yoke]

@craggy thunder their normal vectors are coplanar if they intersect in a single line. That being said three general planes can intersect at three lines even if there are coplanar normal vectors

[craggy thunder]

right, im trying to differentiate between them sharing 1 intersection line or 3

i know they intersect

[covert yoke]

What I'm saying is that coplanar normal vectors is a necessary but not sufficient condition

If you have both coplanar normal vectors and some point that all three planes share this would be both necessary and sufficient. @craggy thunder

[craggy thunder]

i see, so i always need to find a point

[covert yoke]

Dunno

[craggy thunder]

well if i know all the normal vectors are coplanar

say if i wanted to find the line they intersect on

could one or three lines?

what im trying to ask is how do i know if they intersect at one or three lines, aside from testing points

[covert yoke]

You could find any point on the line that two planes share and if that point is on the third plane it's one line otherwise three

[craggy thunder]

oki doki i'll just do that

thanks

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[kindred slate]

for part b, i found the edge connectivity lamda to be 2

what does it mean then to find a cutset of 3 vertices

my def of cutset involves edges, not vertices. so im confused

[safe radish]

@kindred slate Has your question been resolved?

[harsh condor]

do you need help with edges

[kindred slate]

i got {ce, ef, ed} and {cd, ed, dh, dg}

for the cutsets for part b

[harsh condor]

yeah

[kindred slate]

nice

thx

[harsh condor]

np

[kindred slate]

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[hearty fox]

we have : a+b+c= a^2 + b^2 + c^2=1 and x/a=y/b=z/c ( a,b,c are not equal to 0)
prove: (x+y+z)^2 = x^2+y^2+z^2

[hasty wagon]

what have you tired?

[hearty fox]

the last step

[hasty wagon]

which is?

[hearty fox]

K= x/a=y/b=z/c
=>a= xK; b=yK; c=zK
=> xK+yK+zK=1
K(x+y+z)=1
x+y+z= 1/K
=>(xK)^2+(yK)^2+(zK)^2=1
x^2.K^2+y^2.K^2+z^2.K^2=1
K^2(x^2+y^2+z^2)=1
x^2+y^2+z^2=1/K^2

[hasty wagon]

note that for the first step
K=x/a....
x=aK

[hearty fox]

hmm

[hearty fox]

oh i know it

K= x/a=y/b=z/c
=>a= xK; b=yK; c=zK
=> xK+yK+zK=1
K(x+y+z)=1
x+y+z= 1/K
=>(x+y+z)^2=(1/K)^2
=>(xK)^2+(yK)^2+(zK)^2=1
x^2.K^2+y^2.K^2+z^2.K^2=1
K^2(x^2+y^2+z^2)=1
x^2+y^2+z^2=1/K^2
=> x^2+y^2+z^2=(1/K)^2

=>...

[hasty wagon]

then, use the 2nd line and put it into a+b+c=1

oh

you did that

x+y+z=1/K
(x+y+z)²=(1/K)²=x²+y²+z²

[hearty fox]

yay

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[echo gazelle]

casn i pls have some help for dis

[echo gazelle]

so i do x = 0 and y = 0

bruh

ok

🙁

ok thanks a lot

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[median scaffold]

How did log3 9 become 2?

[vague phoenix]

Apply rule 6

Log 3 of 9 = log 3 of 3^2

[solar hazel]

or just recognize 3^2 = 9

log_3(9) is the number 3 needs to be raised to to return 9

which is 2

[vague phoenix]

excellent explanation above! Then just chuck the equation into the rule to find out @median scaffold

[solar hazel]

well my point was you don’t need to use rule 6 at all

[vague phoenix]

Basically the same log 3 of 3 is 1 and pull 2 to behind the log

[safe radish]

@median scaffold Has your question been resolved?

[safe radish]

@median scaffold Has your question been resolved?

[median scaffold]

So you turn the 3 to 9 first then turn it to 3² and take the 2?

[median scaffold]

Tjank youyy

So if its log8(64) the answer is 2?

[vague phoenix]

Correct

[median scaffold]

How about log2(8)?
3?

[vague phoenix]

9 to 3^2

Correct

[median scaffold]

Ohh okayy thanks so muchh ^^

[vague phoenix]

You understood it perfectly, well done

[median scaffold]

Logarithms giving me a headache

[median scaffold]

Tysmm

[vague phoenix]

Hahaha same

!done

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[median scaffold]

Thanks so muchh!!

.close

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[vague phoenix]

If u have anything just ask me🙏👌

[median scaffold]

Okayy ^^ thank youy

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[pseudo tree]

log is the inverse function of the exponential

[pseudo tree]

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[worldly lantern]

find k so that line (d): y=2x-k+3 and parabol (P): y=x^2 intersect each other and 2 different point A(x1;y1) and B(x2;y2) satisfy the following equation: x1x2(y1+y2)=-6

[worldly lantern]

this was someone else’s doubt

and it can be solved by subbing

i was interested in something else tho

can we do this?

in first step i made it variable and said that (x2,y2) must be solutions

in second step i reversed the process of T=0 in conics

and got the eqn

is this possible and what do i do now

[vague phoenix]

Why dont you use viète formula?

The original question is from a vietnamese student and at this grade we learn about viète

[worldly lantern]

whats that

[vague phoenix]

Viièta if you dont know

He does know about it

https://en.wikipedia.org/wiki/Viète's_formula

[worldly lantern]

how does this solve it

cant relate tbh

[lean otter]

in simple words this one

[worldly lantern]

thats vieta’s

right

that does it

thanks

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[lean otter]

arent they the same people

[worldly lantern]

idk

wikipedia says no

[lean otter]

oh

[vague phoenix]

Viète

French sounds way more luxurious than english🙌

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[bronze monolith]

Why does solving (2x-4)/x >= 7 like a linear inequality lead to an inconsistent result

[lean otter]

when multiplying both sides by x you consider x not equal to zero as well as x >0 . if x is <0 inequality flips

w drawing

[bronze monolith]

but you dont know the value of x, so how are you supposed to know whether to flip the inequality?

:cries:

[tough anchor]

You take into account all cases of x. x > 0, x = 0, and x <0.

[bronze monolith]

but you cant flip the inequality and also not flip the inequality

so do you create multiple inequalities?

[lean otter]

yes take multiple cases

[bronze monolith]

interesting

oh wait

the solution itself implies that the solutions of x are negative right?

how did i not see that

dumb dumb idiot haha

stupid dumb dumb gilbert

i understand it 2% better now, thanks

❤️

ill close the channel for now

.close

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[bronze monolith]

.reopen

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✅ Original question: #help-23 message

[bronze monolith]

btw is that inequality linear

[coarse birch]

the initial problem?

[bronze monolith]

yea

[coarse birch]

(2x-4)/x >= 7 is nonlinear

[bronze monolith]

ah ok

[coarse birch]

whats definition of linear?

[bronze monolith]

does it mean the expression has to be a polynomial of degree 1?

[coarse birch]

yeah

so

[bronze monolith]

and this one has a negative power

[coarse birch]

you have x^-1 as well as a linear numerator

so while the 2x-4 is linear, the / x makes it non-linear overall

[bronze monolith]

i see

also its univariate right

[coarse birch]

yeah

you need to trust your fundamentals

[bronze monolith]

but dont people call equations like this linear

(5-3x)/2x = 8

[coarse birch]

the good thing about math, is while we constantly add new ideas, it cannot contradict prvious information (as long as you stay within the same algebraic system)

uni = one

variate = variable

[bronze monolith]

you might find an equation like this in a middle school algebra problem

[coarse birch]

so yes, its univariate as there is only one x value

that equation is also nonlinear yes

now when you multiply across by 2x it becomes linear, but as given it is non-linear

[bronze monolith]

under the topic of "linear equations"

ah ok

[coarse birch]

5-3x=16x is a linear equation

[bronze monolith]

do univariate equations produce lines tho

[coarse birch]

(5-3x)/2x=8 is nonlinear

univariate just means involving a single variable

it doesnt have anything to do with lienar

[lean otter]

you multiplied by x but did not change the sign we can do that by assuming x>0 however doing so results in contradicting solution where x is negative hence you can say there exists no solution for x > 0

[bronze monolith]

sorry i meant univariate linear equations

[bronze monolith]

thats wonderful

[coarse birch]

something is univariate if it contains one variable

you don't need to ask anyone about that anymore

something is linear if the expression contains degree 1 as the highest power

you dont need to ask anyone about that anymore

as long as you check those conditions --> you will be correct

[bronze monolith]

i gotchu man

thanks

[coarse birch]

np at all

what was your solution earlier btw?

for the original inequality?

[bronze monolith]

i havent solved it yet

[coarse birch]

ahh i see

[bronze monolith]

i still dont fully grasp how this can be solved

[coarse birch]

okay so

these problems are very tedious

you need to be very careful about the positiveness/negativeness of the numerator and denominator

[bronze monolith]

i get that you can check for a contradicting solution tho

OH WAIT

[coarse birch]

there are a couple of different strategies

[bronze monolith]

there are no solutions for x > 0

because if you assume x > 0 you end up with a contradiction

but if you assume x < 0 youre chilling

you are a genius @lean otter

[coarse birch]

I'm not sure that is true gilbert

[bronze monolith]

NOOOO

[coarse birch]

have you done inequalities that don't include division by x, something like simpler linear inequalities?

[bronze monolith]

yeah i can do basic inequalities without variables in denominators

[coarse birch]

Okay, so what is your strategy when solving something like 2x-1=5, for example?

[lean otter]

there is another way to do this rather than solving all cases you can shift 7 to LHS, you will get a single expression >= 0 you can check the sign of this expression for different values of x

[bronze monolith]

that'd be the same as solving the linear equation

[coarse birch]

sorry 2x - 1 > 5

[bronze monolith]

the sign isnt flipped in that case

[coarse birch]

mistyped

[bronze monolith]

i get why multiplying/dividing by -1 flips the sign

[lean otter]

nah i am dumb

[coarse birch]

@bronze monolith not so fussed about the final answer, but what would your strategy be to solve something like 2x - 1 > 5 ?

[bronze monolith]

ah ok

add 1 to both sides, divide both sides by 2

i.e. reverse the operations done to x

[coarse birch]

yes i understand that, but what are you trying to find by doing that?

Which x value makes it = 5*, correct?

[bronze monolith]

yes

[coarse birch]

right, and then how do you identify the answers for x

so we understand that x = 3 implies 2x - 1 = 5

how do you solve 2x - 1 > 5 then?

[bronze monolith]

same thing but we find the range of x values that make 2x -1 > 5

right?

is that what youre asking?

the solutions to the inequality

[coarse birch]

yes, so if you do that with that specific equation you will end at x > 3

[bronze monolith]

yea

[coarse birch]

however, another way of doing it is to simply solve = 5 first

then test values either side of the identified solution

so test any value > 3, and any value < 3, to check if those ranges are true or false

Now this is NOT necessary for a simple linear equation, because you can re-arrange, but for some more complex inequality, particularly those where there are multiple zeroes or mixes of positive and negatives, it is often a good strategy to do this.

So after identifying each value that makes the original "equation" true, then you test values either side of those x values to check each part of the range

[bronze monolith]

but how are you supposed to know there arent some wonky turning points on either side where suddenly solutions appear?

is it because you would end up with multiple solutions = 5?

[coarse birch]

yes

exactly

[bronze monolith]

if it was some weird non-linear equation

[coarse birch]

GOOD logic

[bronze monolith]

so they'd ahve to cross = 5 again

what if they decide to teleport past 5?

[coarse birch]

so if it is x^2 + 4x + 3 > 1, for example

^ that is a good question

it isn't often where that happens

but one of those times is where you have a fraction with variable top and bottom --> it can swap from positive to negative

most of the time you dont have that issue --> if a function is continuous then you don't need to worry about that

so for something like a quadratic inequality, you will typically get two answers for the "equals"

then you will test on the left of the first value, between the two values, and on the right, and you will know

but for these division ones, you have to be super careful about the positive / negativeness of the expression

try this for a moment:
(2x-1) / x > 0

What about the numerator/denominator would make this positive, and what would make it negative? Don't need specifics for this problem, think about positive/negative combinations

[bronze monolith]

the numerator and denominator would have to both have the same sign to make it positive

or have opposite signs for it to be negative

[coarse birch]

right okay, so then to solve the inequality, you find what intervals for x would make those conditions true

so what would have to be true for x, so that both numerator and denominator are positive?

[bronze monolith]

hmmmmmmmmmmmmmmmmmmmm

x > 0 ?

since 2x - 1 is always greater than x for positive x values

right?

[coarse birch]

well thats for the denominator, but is that good enough for the numerator

you tell me

maybe think about solving 2x - 1 > 0?

[bronze monolith]

oh yeah

so i'd have to find the values of x which satisfy both 2x - 1 > 0 and x > 0

[coarse birch]

yes, so what is that interval?

[bronze monolith]

x > 1/2

[coarse birch]

perfect

[bronze monolith]

so that would be the solution to the original?

[coarse birch]

do the same for both negative now

also both negative right

[bronze monolith]

x < 1/2

[coarse birch]

x > 1/2 implies both positive therefore overall positive

x<1/2 wouldn't make the denominator negative though would it?

[bronze monolith]

ohh

x < 0 then

[coarse birch]

are you sure?

[bronze monolith]

2x-1 < 0 when x < 1/2
x < 0 when x < 0

so x < 0 would be the range in which both parts are always negative

[coarse birch]

good

[bronze monolith]

then what about 0 < x < 1/2?

[coarse birch]

well thats negative

[bronze monolith]

oh sorry

yes of course

ahhh

[coarse birch]

because logically, the only way it could be positive is if both are the same (num and denom)

and you already exhausted those conditions

so what remains is not positive

[bronze monolith]

genius man

you area genius

[coarse birch]

okay

so now

your problem

is a combination of the two concepts

for it to be > 7

you need to think BOTH about postiveness/negativeness of the overall expression

but its not simply that, you also need to solve it for = 7 to know where the critical value is

whereas with this one, the critical value was 0 so simply positive or negative

you need this one to be BOTH positive AND greater than 7

so try to come up with conditions that make both of those true

and the reason this is necessary, is because if you simply multiply across by x to clear it from the numerator --> this introduces extraneous (incorrect) solutions, because it "forgets" that originally we were dividing by x.

[bronze monolith]

im gonna need to read through the chat for a sec

ill ping you

[coarse birch]

ok

[safe radish]

@bronze monolith Has your question been resolved?

[bronze monolith]

@coarse birch may i ask what you meant by "mixes of positive and negatives" here?
"but for some more complex inequality, particularly those where there are multiple zeroes or mixes of positive and negatives, it is often a good strategy to do this."

[coarse birch]

same as the ineqality for > 0 we did before

both numerator and denom > 0 or < 0

[bronze monolith]

ah so like mixes in a fraction

ok cool

[safe radish]

@bronze monolith Has your question been resolved?

[bronze monolith]

@coarse birch

is this accurate

i tried to solve it using those ideas where the red arrow starts

it feels like im oscillating in and out of grasping how it works

one minute i understand, the next i forget what it was that i even understood

[coarse birch]

Hmm there are some good ideas but it’s not quite accurate yet.

We didn’t quite do a full example for this type of problem. You have used some of the ideas correctly but missed some of combining the ideas together

Yes you need x>2 for it to be positive, BUT this doesn’t mean it’s greater than 7. Only when x>2 it might be possible to be greater than 7. You still need to find the interval

You are correct that for 0<x<2 it will be negative and therefore can’t be >7

You should also consider what values of x would make the overall fraction positive by making both the numerator and denominator negative, as well. This is the other case where it will be positive and may have some results for =7

And to find when it is >7, try solving the equation for equals 7 first, then testing values.

[lean otter]

alternatively what you can do is shifting everything to LHS , it will result in an expression >= 0 now you have less work as you just need to check when the expression is positive

[coarse birch]

This is true ^ however requires basically the same logic / check after

[bronze monolith]

ohhh

so you check if its ever equal to 7 for values of x which make it positive, and then test around those values?

so you check within each interval

[coarse birch]

It’s the combination of both ideas 1. Solving for x for the equation to find the value/s that are equal to 7 (these are called critical points) , and 2 double checking / ensuring it is ALSO in intervals that make the fraction positive

[bronze monolith]

wait what am i saying rn

[coarse birch]

2. Is necessary because when you solve for x one of the steps is multiplying the equation by x, and when you do that you create extra answers compared to the original inequality that are incorrect due to the behaviour of the fraction

[bronze monolith]

so you find the critical points, and then the inequality solutions will be on either side of those critical points within the positive/negative intervals

[coarse birch]

Yes basically

[bronze monolith]

i dont really get this

how does it create "extra answers"

[coarse birch]

Look, I prefer teaching this way - focusing on logic / reasoning and building strong understanding, rather than just giving students efficient ways to solve specific problems. I do this because it’s more compatible to other problems and developing a stronger understanding of math as a whole.

However, without a whiteboard / the ability to talk rather than just message it’s maybe easier to give you a simpler method that works for this problem (but may not work in all).

Try squaring both sides. Then solve the inequality like you normally would.

This means that the issues with the denominator being negative go away, because it’s always positive.

[coarse birch]

Due to the way inequalities work - when you multiple or divide by a negative you need to flip the sign

Simple example, if I just do the algebra blindly.

(x+1)/(x-3)> 2
Multiply each side by x-3:
x+1 > 2(x-3)
Expand
(x+1) > 2x-6
Subtract x from each side
1 > x - 6
Add 6 to each side
7 > x
Conclusion
x < 7

So supposedly, for all x < 7, the expression is greater than 2.

But try testing x=-1/2
it is false.
Also try x=2.

These are created because when x = these values, the numerator is negative.

[bronze monolith]

are you a teacher irl?

[coarse birch]

So there are a few strategies you can use address this problem:

1. What we have been doing - you can solve the equation for critical points, then test values either side of the critical points but only for intervals that both make the fraction positive or negative (to match RHS, e.g. positive for >7),

OR

2. You can square both sides, multiply by the denominator and solve the resulting quadratic inequality (will still require solving for critical points and testing either side, but removes the need to check for the positive/negativeness of the fraction)

OR
3. You can simply think about what makes the denominator positive, multiply it across and solve the equality normally (algebraically) and then separately find the interval that makes the denominator negative, multiply across and FLIP the inequality sign (because you have multiplied by a negative), then solve that inequality normally (algebraically).

Yes I am a qualified teacher / quite active tutor

[bronze monolith]

thats good to know!

[coarse birch]

You in AUS?

[bronze monolith]

nope, South Africa

[coarse birch]

Hah fair

What’s the time for you?

[bronze monolith]

how old are you?

4:43pm

[coarse birch]

I have hired a couple tutors who are from South Africa haha

I’m 31, you?

[bronze monolith]

im 21

[coarse birch]

Uni study or?

[bronze monolith]

not yet

so you even manage other tutors?

why does your role say undergrad maths?

[coarse birch]

I own a company

[bronze monolith]

i see i see

[coarse birch]

I have about 25 tutors

[bronze monolith]

whats the company called?

[coarse birch]

Well my degree is Bachelor of Education (Secondary - Mathematics & Physics Major)

[bronze monolith]

do they tutor A Level maths?

[bronze monolith]

dang

[coarse birch]

We don’t have A levels in Australia so the curriculum isn’t exactly the same, but I definitely have tutors who have knowledge of all/majority of A level content, myself included

[bronze monolith]

right

and do your tutors follow a similar logical method of teaching as you?

[coarse birch]

Different tutors have different styles, but we focus on 1-on-1 and tailoring approaches to the individual students.

[bronze monolith]

thats amazing

[coarse birch]

My preferred way is good for understanding / building logical ideas and foundation of math.

But it’s not the most time efficient short term, especially if a student doesn’t practice enough.

I can also teach through shortcuts and tricks for specific problems but I don’t think it’s as good for long term success

[bronze monolith]

wow

thats exactly how i want to learn maths

[coarse birch]

I can tell.

[bronze monolith]

could you suggest any resources for that kind of learning?

or should i just get a list of topics in the right chronological order and then work through each using a variety of resources

and only move on once i understand a topic fully

[coarse birch]

When we are talking back and forth, you are roughly following and maybe some bits don’t make sense. But instead of rushing to the answer you are taking time to review and think over it and check your own understanding

A lot of students just want this answer now

But that’s not the best way to learn math imo

[bronze monolith]

if i lived in Australia i would pay an exorbitant sum of money for your services

[coarse birch]

What is your goal?

[coarse birch]

Haha

If you wanted to do something regularly we could possible schedule an online session, but we’d need to discuss outside of this channel. Swap to dms

[bronze monolith]

1. you cant properly adapt to the variety of challenges you might face in real problems
2. theres no genuine feeling of satisfaction

yeah that'd be sick man

[bronze monolith]

well, my goal is to finish A Level maths this year, yet not by brute forcing a bunch of past papers, but understanding the content as though it were a delightful hobby

im willing to literally spend 10 hours a day if need be

since otherwise i may end up homeless (not really)

[stoic dune]

<@&268886789983436800>

[coarse birch]

Hey just curious, what’s the @mod tag for?

[stoic dune]

Someone spammed. They're gone now

[primal bone]

I'm gonna still argue that learning the curriculum solely from shortcuts, then, is not as productive - you're jeopardising your understanding of how/why those shortcuts would work in the first place, i.e. you're building a castle on sand

[safe radish]

@bronze monolith Has your question been resolved?

[safe radish]

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[safe radish]

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[gentle summit]

Hello, I need help with question b). Where did the one I lined in red come from? (This is the answer sheet) (The second picture is my answer for a) )

[primal bone]

The volume of the cone that gets removed; it's generated by spinning the line y = x + 2 about the x-axis

You can just plug that into the integral too; but you're missing something here

Actually you're missing two things - something where the orange arrow is pointing, and something to the left too (the brown arrow)

[gentle summit]

Ahhhhh I just gotta use the one for volume

I think I understand,, this'll generate the same answer right?

[primal bone]

ah wait

You're talking about b?

The answer you've written seems to be for a

(ah shet you DID say that mb lol)

[primal bone]

But yh — you're calculating a volume for part b, so you need the pi times y^2-looking-integral

[gentle summit]

NAHHH dw it happens

Alright I understand now, thank you!

.close

[safe radish]

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[gentle summit]

Oh hell nah

[primal bone]

Server invite?

yh the mods took care of it already, they speedy

[safe radish]

Available help channel!

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[lean otter]

Hello! I would like help on this question if that is alright - I'd appreciate it!

I'm struggling with part(b) only as it confuses me

[graceful quail]

Find x s.t. 2 + 3x/4 = 1.925

[lean otter]

What do you mean?

[graceful quail]

You have $(2+\frac{3x}{4})^{6}$ written in a different way

[flat frigate]

WeAreIngram

[graceful quail]

Simpler way

And need to compute 1.925 ^ 6

[lean otter]

Oh, and how do I do that?

[graceful quail]

So if we have 2 + 3x/4 = 1.925

[lean otter]

Yes

[graceful quail]

You can use what you have as an approximation

$1.925^{6} = (2-0.075)^{6} = (2 - \frac{0.3}{4})^{6} = (2 + \frac{3(-0.1)}{4})^{6}$

[flat frigate]

WeAreIngram

[lean otter]

I don't understand it

Can I show you what the mark scheme says?

[graceful quail]

Yeah

[lean otter]

How on earth is 3/4 = 0.075

[graceful quail]

Did you read my equation?

[lean otter]

Yes but I don't understand it

[graceful quail]

What part?

[lean otter]

Everything after the equals symbol

[graceful quail]

1.925 = 2 - 0.075

[lean otter]

Why would I isolate the 2 to figure this out? I think I don't understand this part

[median vigil]

you are trying to use an approximation you found for (2 + 3x/4)^6 to approximate 1.925^6. it might be a good start to find a value of x so that (2 + 3x/4)^6 = 1.925^6

[graceful quail]

Because you need to write it in the same format as your equation in (a) in order to use it

[lean otter]

I think I don't understand the topic on approximations itself so I'll just review it instead

Thank you for helping me, I appreciate it

[graceful quail]

No probs, you can use .close if you're done

[lean otter]

.close

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[void path]

<@&268886789983436800>

[safe radish]

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[safe radish]

@spiral saddle Has your question been resolved?

[delicate shore]

!status

[safe radish]

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

[vague phoenix]

Can you translate

[safe radish]

@spiral saddle Has your question been resolved?

[hard crest]

que as hecho ya?

[safe radish]

@spiral saddle Has your question been resolved?

[safe radish]

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

[fickle mantle]

Can I get help

[surreal plover]

ok

number 15 right?

[fickle mantle]

Yes pls

[surreal plover]

well i gotta tell you something

theres no real solution to that

unless you use imaginary numbers

[fickle mantle]

Ig

But I wanna solve it

[surreal plover]

because you can square root a negative number

ok

ill help you

so do you have an idea of what to do first?

[fickle mantle]

No

[surreal plover]

aight ill guide you

[fickle mantle]

I thought if square rooting -16

But it doesn’t work

[surreal plover]

thats why its impossible without imaginary numbers

lets pretend we can

[limpid lodge]

Have u studied complex nums yet?

[fickle mantle]

WiFi wht Tht is

[surreal plover]

he hasnt

​

[fickle mantle]

Idk

[limpid lodge]

Oh then r u sure its negative 16?