help-0

[near hollow]

and for 1 by 1 you get 1/2

[near hollow]

Nope

Too many combinations for 3 by 3

And 4 by 4 I dread to think

well for 2 by 2 the set is only size 3

but for 3 by 3 the size is going to be larger than 10

and 4 by 4 much bigger again

ext.

[near hollow]

there are 4 vectors in the span (1 ,0), (0,1), (0,0), (1,1)

so 2^n different possible vectors

that we can choose from

Ah ok I miss read it, this is an unsolved problem that we could have a look at not that we need to solve

sounds promising what we have got to tho

thought about maybe using inclusion exclusion and that'll then open up an inductive step or turn it into a recurrence relation

so we creating the condition Ci = vi and vi+1 are linearly independent and we are looking for |C1 n C2 n C3 n C4...Cn-1|

Which can be rearranged using inclusion exclusion to get |C1 n C2 n C3 n C4...Cn-1| = (|C1 u C2 u C3 u C4...Cn-1|- SUM(|Ci|)+SUM(|Ci n Cj|) ... (-1)^(n-1)SUM(Ci n Cj n Ck...Cz))/(-1)^n where S ={i,j,k....z} and |S| = n-2

We can use induction or a recurrence relation for the intersection of the n-2 conditions and we can work out the union part as it's satisfied if any of the vectors are different to each other

hmm not sure I follow

yes

but im not sure how you plan on counting the different linearly independent sets that you can form through the knowledge of that

[lone heart]

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[olive oar]

If I want to answer the following question: If an ant were to walk across the curve $f(x) = x^4 - 3x^3 + 2x + 1$ for 3 units. What is the largest change in $x$ that is possible to accomplish?

[ocean seal]

ohNoiAmHere

[olive oar]

Could I just say that this happens when the average value of the slopes of the tangents is minimal

so when $$\frac{1}{b-a}\int_{a}^{b} f'\ dx$$

[ocean seal]

ohNoiAmHere

[olive oar]

is minimzed

(i asked this question before but couldn't come up with this method, we ended up with something using simpsons rule)

the issue is how do i get it over 3 units?

like $$3 = \int_{a}^{b} \sqrt{1 + (f')^2}\ dx$$

[ocean seal]

ohNoiAmHere

[olive oar]

so i have two equations but what do i do from here

(i think before i had 10 units)

[near hollow]

Have you done Lagrange multipliers?

[olive oar]

no

im looking at them now and they definitely look like the way to go

[near hollow]

Yeah basically with lagrange multipliers you need a restraint (in this case the restraint is the length along the curve) we'll call g(x,y) and then the function you're maximising (x-x0)

We can then set a new function $\phi(x,y) = x - x_0 - \lambda \cdot g(x,y)$

[ocean seal]

Max..

[near hollow]

then partially differentiate with respect to x = it to 0 and then usually the same but with y but I feel that will cancel

[olive oar]

wait why can i call the average of tangents x-x_0?

or is that b-a?

[near hollow]

we're trying to maximise the start and finish distance

lemme try do this one myself first then ill help otherwise i might confuse lmao

[olive oar]

k

[near hollow]

Idk lmao it seems just as hard, might be worth looking at tackling the integral

[olive oar]

howd would i go about that though

the top one inst an integral cause of ftc

but the bottom one seems painful

and f' should be an absolute value in the first right?

[lone heart]

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[viscid moat]

wtf happened here

okay sure he multiplied by (x/2)/(x/2)

but where did x^2/sin^2(x)

come from

and 1-cosx=2sin^2(x/2)

[sudden hinge]

they just multiplied and divided by x^2 / (x/2)^2 to get a term of the form sin(x)/x for which we know what the limit is going to be

[lone heart]

@viscid moat Has your question been resolved?

[viscid moat]

can you please explain again ?

i didnt understand

[viscid moat]

OH

i forgot

about that

didnt pay attetion

ur right he just multiplies by x^2/x^2 and (x/2)/(x/2)

okay thank you

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[alpine sable]

Please help

[alpine sable]

<@&286206848099549185> please help me

[near hollow]

can you send a better image of it?

[alpine sable]

It's actually from my module. I have the entire picture of the module

Question 31

[near hollow]

like a closer picture tho lol just so I can actually read it properly

[alpine sable]

Uhh it's actually send to me by my friend so I can't physically click a closer pic

Sorry for the inconvenience

😅

[near hollow]

what does it ask you to find

does it say n/1000 or sqrt(1000)

[alpine sable]

n/1000

[near hollow]

ok ok

[alpine sable]

Basically they gave this weird expression

Root 1+1/1²+1/2²+ root 1+1/2²+1/3².....root 1+1/1999²+1/2000²=n-1/n

whats the value of n/1000

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[tacit arch]

.close

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[zenith compass]

Let $n\in\mathbb{N}\setminus{0}$.

Suppose that there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$.

Since $n\in\mathbb{N}\setminus{0}$, we have $n\in\mathbb{R}$.

Since $n\in\mathbb{R} \setminus{0}$, $n^2>0$.

Hence it is impossible that $n^2<-1$.

Hence the statement (D) is false.

[ocean seal]

Trenton

[zenith compass]

I am proving (b)(IV)

But it seems to be so trivial

Dk whether I made a mistake or not

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@zenith compass Has your question been resolved?

[lone heart]

@zenith compass Has your question been resolved?

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[sacred yacht]

My little sister needs help with this and I don't even know how to solve it-

[buoyant kayak]

does she know how to read a graph?

and/or do you...?

[last ether]

You reflect it across the x axis

That basically means to multiply each y value of each coordinate by -1

[sacred yacht]

Ah.

Alright.

[last ether]

And then the moving of the units

That should be self explanatory

[sacred yacht]

Mhm, thanks.

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[lone heart]

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[urban scroll]

What am I doing wrong? 😕

[elfin snow]

a and b are the legs while c is the hypotenuse

and idk why you put a minus sign there

[urban scroll]

oh the reason I put the minus is because I thought on the problems where the hypotenuse was already found you have to subtract instead of add

[elfin snow]

it seems like you are just like 'memorizing' an algorithm without actually understand what's going on

using the picture, what are the lengths of the legs of this triangle?

[urban scroll]

13 and 15ft

[elfin snow]

no

[urban scroll]

oh

[elfin snow]

15 is the hypotenuse

the hypotenuse is the longest side in every triangle

the legs are the two sides adjacent to the right angle

(in a right triangle of course)

does this make sense?

[urban scroll]

yes

[elfin snow]

so what are the lengths of the legs?

[urban scroll]

well from the answer i got it would be 7.5 and 13ft

[elfin snow]

but it's not 7.5

you don't know what it is yet

all you know is that it is x

makes sense?

[urban scroll]

yes

but your supposed to solve what x is right

and i got the wrong answer

[elfin snow]

yes

we will get to that

[last ether]

you need to find x, so lets assign each number as a letter in the Pythag theorem

[elfin snow]

I got this umbra

so icy, in the pythagorean theorem, a and b represent the legs while c represents the hypotenuse

[urban scroll]

yes

[elfin snow]

so what will c be equal to?

[urban scroll]

15ft

[elfin snow]

so a and b are 13 and x

it doesn't matter which is which

[urban scroll]

ohhh

[elfin snow]

but for us let's call a=13 and b=x

so let's substitute what we know into our formula

what does it look like now?

[urban scroll]

this???

[elfin snow]

no

we said b=x and c=13

[urban scroll]

what about a

[elfin snow]

a=13

[urban scroll]

so a and c equal 13

[elfin snow]

sorry sorry

c=15

[urban scroll]

oh ok

so a=13 b=x and c=15

[elfin snow]

yes

plug that into your formula now

[urban scroll]

I don't think I wwrote that correctly

[elfin snow]

no you didn't

there won't be a b in your equation anymore

it is replaced by x

[urban scroll]

Oh

Ah ok

so now do I just solve it

[elfin snow]

well first what do you have rn?

[urban scroll]

that's all rn

[elfin snow]

okay so let me show you what it's supposed to be

$13^2+x^2=15^2$

[ocean seal]

guh node

[elfin snow]

there's no b here

13 and x are the legs

15 is the hypotenuse

does this make sense?

[urban scroll]

yes

[elfin snow]

alright so how would you start to solve this

[lone heart]

@urban scroll Has your question been resolved?

[urban scroll]

SORRY

I was talking to my dad

[urban scroll]

I'm confused

[elfin snow]

ur good lol

what are you confused about

[opaque inlet]

15^2 not 15 btw

[lone heart]

@urban scroll Has your question been resolved?

[urban scroll]

did I do it right now?

[elfin snow]

yeah

but you need the label

and no x=

so your answer would be 7.5ft

which I realize is what you got initially

but you were lucky since you made a lot of errors in your working

I hope this helped you to understand the entire process without remembering a certain algorithm

[urban scroll]

yeah thanks

.close

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[lunar girder]

Hello

[lunar girder]

.close

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[dense nova]

His can someone help me

[dense nova]

I need to do the derivative of arccos of a/h with respect to t

Idk how to do it

[lone heart]

@dense nova Has your question been resolved?

[dense nova]

<@&286206848099549185>

[calm tusk]

a,h and t related in some way?

[lone heart]

@dense nova Has your question been resolved?

[dense nova]

Yes

A and h are functions of time

[calm tusk]

okz

use this

[dense nova]

Ok thankyou

[glad badger]

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[vestal aspen]

how to find the length of tangent when equations of the circle and radical centre is given?

[vestal aspen]

any hints?

[gray isle]

you seem to be omitting some info

can you post a specific example

[vestal aspen]

yes

[vestal aspen]

(in this question, I got the point(which is the radical centre) but not getting the length(which is the length of tangent))

[lone heart]

@vestal aspen Has your question been resolved?

[lone heart]

@vestal aspen Has your question been resolved?

[lone heart]

@vestal aspen Has your question been resolved?

[calm tusk]

you know the radius and the distance btw radical center and the center of the circle .. a right angle triangle ..

[calm tusk]

the first one is it even a real circle?

[lone heart]

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[vague pumice]

How do you find the scalar?

Do i just multiply it by zero?

Oh its not possible right?

[upbeat gorge]

Do you know what’s the product of a scalar and a matrix?

[vague pumice]

Im sorry, no the professor just gave us the activity with no lectures

[upbeat gorge]

ah

So what happens is that you just multiply every term in the matrix by the scalar

[vague pumice]

The scalar is the zeroes?

Ah no

[upbeat gorge]

For example

Do the other two and then we can proceed to the next step

[vague pumice]

Okay wait

[upbeat gorge]

Good

Now, the next step has to do with adding the matrices

[vague pumice]

Ah

[upbeat gorge]

Do you know how to add matrices?

[vague pumice]

2c1

6c2

6c3

Ah no no

Wait wait

[blazing saffron]

They're just vectors

ie; top row is just adding across c1 + c2 + 3c3

Continue for the other two

[vague pumice]

Ah okay

[blazing saffron]

It's really a system of equations

Represented differently

3 equations 3 unknowns

[vague pumice]

Can you check if i got it

Wait

[blazing saffron]

No

They're 3 equations

[vague pumice]

Ah the one that you use the elimination method?

Or the subtitution

[blazing saffron]

Add the top row for the first equation, the middle row for the second, the bottom row for the last,
\[c_1 + c_2 + 3c_3=0\]
\[ 2c_1 + 3c_2 + 7c_3=0\]
\[-1c_1  -2c_2 -4c_3=0 \]

Just adding

Across

[ocean seal]

Skid

[blazing saffron]

It's a system of equations

[vague pumice]

Wait

I got it

[blazing saffron]

The c_i notation is irrelevant it is equivalently

\[x + y + 3z=0\]
\[ 2x + 3y + 7z=0\]
\[-x  -2y -4z =0 \]

[ocean seal]

Skid

[blazing saffron]

So

[vague pumice]

2c1 +2c2-6c3 =0

[blazing saffron]

What's that

You just solve the system of equations

Or

If you've learned reduced row echelon form that can be used as well

[vague pumice]

System of equations?

[blazing saffron]

Yes

[vague pumice]

Is this the final answer?

Ah wait

[blazing saffron]

No

https://www.mathsisfun.com/algebra/systems-linear-equations.html

[vague pumice]

X =0
Y=0
Z=0

[blazing saffron]

You can't do that

[vague pumice]

I solved using elimination method

[blazing saffron]

The problem says they can't all be 0

It's a valid solution

But not for the problem

[vague pumice]

Ah

[blazing saffron]

You won't get real number values by solving the equation you will get like x=2y, y=-z

In this instance, you can let z equal any value

Then compute x and y

z=1 would be the easiest

[vague pumice]

Is there really an answer here

It say if possible?

.close

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[blazing saffron]

Yes, consider c_1= -2, c_2 = -1, c_3 = 1

There's infinite answers

[vague pumice]

.reopen

[lone heart]

✅

[vague pumice]

Ah okay ill try to study what link you have sent thank you again sir

.close

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[alpine sable]

For inequalities with one unknown, is there a preferred order to write it?
I've evaluated 15 < 1-7x to -2 > x, but is x < -2 equally valid? Is there a preferred convention?

[alpine sable]

do you write 2=x or x=2? same with inequalities, x is preffered to be on the left but both are valid

Thanks, I will put x on the left then.

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[vast bobcat]

pretty sure my working for first problem is wrong and I don't know how to approach second problem

[lone heart]

@vast bobcat Has your question been resolved?

[gilded vessel]

you approach second problem the same way you approach the first

[vast bobcat]

but is my first approach correct?

[normal wing]

Personally, I’d do
2^2 = 4 = -1 (mod 5)

Same for the 3

For the second question

[rotund plank]

No

You’ve factored stuff out wrong

Also what one are you starting with

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[spring jackal]

Hey

[jagged imp]

Hey

[surreal jewel]

hey

[limpid spade]

het

[spring jackal]

I’m not too sure where to start with this

[limpid spade]

gey

hey

[spring jackal]

Lol

[limpid spade]

common denumerator

[surreal jewel]

set x can’t be equal to 0 first

[limpid spade]

$\frac{3x^2 - 2}{x}$

[ocean seal]

Ugu'yaränikeri'u

[surreal jewel]

then multiply by x both sides

[jagged imp]

I'd argue that getting a common denominator is a little inefficient when you can just multiply both sides by x in the first place but proceed

[surreal jewel]

exactly

[limpid spade]

betas

smh

[surreal jewel]

just set $x ≠ 0$
and then it becomes $3x^2-4x+2=0$

[ocean seal]

Trollo

[limpid spade]

yay

[spring jackal]

2/3?

[surreal jewel]

what

[spring jackal]

I’m being told the answer is 2/3

But i have no clue how

but the thing i cant explain is the changing from /9 to /3

[gray isle]

how are you getting 9

[spring jackal]

Oh my days I’m dumb

That should have been /6

Alright well thanks anyways that’s all I wanted to know ig I just wrote the wrong number

.close

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[fossil mason]

Hi, could someone help me?

[echo socket]

Trying to find OBC and ACB angles?

[lone heart]

@fossil mason Has your question been resolved?

[fossil mason]

OBC

sorry didnt see your message :C

[echo socket]

I think you can set up a system of linear equations for OBC and ACB

[fossil mason]

are there any other simpler way

I thought of doing that too

[echo socket]

OBC + ABC + 55 = 180
50 + 130 + (25 + ACB) + (50 + OCB) = 360

Can't think of anything else for now

[fossil mason]

how did u get the first eqn

[echo socket]

Sum of angles in a triangle should be 180

[fossil mason]

huh

[echo socket]

Look at the triangle on the right

[fossil mason]

wait do u mean obc

[echo socket]

Well you could consider that triangle too

[fossil mason]

sorry i cant seem to solve it

[echo socket]

Hmm, though it looks like such system of equations wouldn't have a solution

Let me try considering the OCB triangle then

Yeah same here

Oh there's a contradiction in your diagram

OBA and BOC angles should be equal

[fossil mason]

oh i see

assuming its 30 too

i still cant solve it

WAIT

I FOUND A WAY TO SOLVE IT TYVM

.close

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[ionic bison]

is a "unit eigenvector" different? because I'm fairly certain these are correct

[ionic bison]

I've no clue what I'm doing wrong here

[zenith lodge]

recall the definition of an eigenvector

if [a b] is an eigenvector, is [2a 2b] also one?

[lone heart]

@ionic bison Has your question been resolved?

[lone heart]

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[cinder sundial]

it asks for the area of triangle ACD

and its 8

[rotund plank]

Show what you were given in the question and all your workings

[cinder sundial]

then i started think about what is it length

[cinder sundial]

i do, its all in the picture

[zenith lodge]

try to find a formula for the area from the lengths of AC, CD and angle ACD

[cinder sundial]

i wonder what is it length

so (1/2) * 4 * 5 * sin(90+theta) = the area

[rotund plank]

What lengths are you actually given in the question

[cinder sundial]

cuz its bigger than 90 degree, it hard for me to reconize the height of triangleACD

[zenith lodge]

*its length

but still, i don't know which length you're referring to

[cinder sundial]

the height, im sorry

[zenith lodge]

ohh

[cinder sundial]

is it sin(90+theta) * 5 or sin(90+theta) * 4

and i wonder how it works?

[zenith lodge]

both are valid

remember that for each triangle there are 3 "height"s

[cinder sundial]

could you draw it out on the picture

[zenith lodge]

if you want the height with base AC, it's sin(90+theta) * 4

[cinder sundial]

cuz i can only find one

[rotund plank]

Do you know $\frac{1}{2}ab\sin{C}$?

[ocean seal]

dk.dkn

[cinder sundial]

i do

but i wonder how it works on this case

[rotund plank]

Then work out theta

C = 90 + theta

[cinder sundial]

this case is kinda special

[rotund plank]

We have the other two lengths

[cinder sundial]

theta is bigger than 90

lets say i just want the height of ACD triangle with AD as its base, then whats the height?

[rotund plank]

I’m pretty sure that doesn’t matter

[cinder sundial]

with AD as its base

then whats the height?

[rotund plank]

Whether the angle is 90° less than 90° or above it’s irrelevant

[cinder sundial]

it matters

i want to know how it works

not just put the formula in and get the answer

[rotund plank]

It doesn’t matter

[cinder sundial]

can someone answer it

[upbeat gorge]

So if you know the 1/2 bh formula, 1/2 absinC comes from that
if a is your base, bsinC gives your height; similarly, if b is your base, asinC gives your height

[rotund plank]

Exactly

[upbeat gorge]

Now in this case finding the height would not be the wisest move in the world

But that’s where it comes from

[rotund plank]

I have no idea why you’re so fixated on the fact C is larger than 90°, it’s a general formula

Working at the height is like working backwards from the formula

[cinder sundial]

is there a video that proves the formula when theta is bigger than 90

[cinder sundial]

i got it

it seems legal to me

no matter how big or small the theta is

the formula will alway works

[rotund plank]

Yes

[cinder sundial]

i think the question is solved

thank you

[rotund plank]

Yea np

[lone heart]

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[dark umbra]

Hello guys, for a population of N, if you randomly extract 50% of N for a random sample to try to find the true mean value of N, can we claim that we are 95% confident that the true mean value lies somewhere between the confidence interval, or is there some fault in this logic

[dark umbra]

for e.g., you wanna find the mean time of how fast teenagers eat, so you take about 50% of students from a population of students to find the actual mean. Since I took about 50% for sampling, do I need to increase my confidence level to 99% or is 95% suffice to assume?

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@dark umbra Has your question been resolved?

[lone heart]

@dark umbra Has your question been resolved?

[noble sinew]

depends

[lone heart]

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[tight locust]

question

[tight locust]

why does 100/9801 yield:
0.0102030405060708091011121314151617181920...

all the way up to 97

[noble sinew]

https://math.stackexchange.com/questions/102682/what-is-special-about-the-numbers-9801-998001-99980001

[wary stream]

9 is a unique number, like that post scapeprof posted, 1/9^2, 1/99^2, 1/999^2, etc, you'll get a pattern

,w 1/9^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by dldh06#2659

[wary stream]

Zero through 9 repeats for 1/9^2

,w 1/99^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by dldh06#2659

[wary stream]

Then it's 00 to 09 repeats

Then 000 to 009 or something like that

[tight locust]

yeah that's pretty cool

[rotund plank]

I like the power series explanation

[wary stream]

And also, the other weird thing about 9 is, $0. \bar{9} = 1$ too

[tight locust]

yes ofc

[ocean seal]

dldh06

[tight locust]

it's weird how 98 is missing though

what are the rational convergents for champernownes constant? are they also in this form

[wary stream]

That's right, it skips 8

Forgot about that

https://www.youtube.com/watch?v=daro6K6mym8&ab_channel=Numberphile

[wary stream]

I was mistaken, it's all the two digits except for 98

[wary stream]

@tight locust This answers your original question, fyi

[lone heart]

@tight locust Has your question been resolved?

[lone heart]

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[silk crystal]

hi guys, a rather simple question: solve for x, f(x) = e^x + e^(-x)?

[near saffron]

Solve for x in terms of f(x)?

Multiply both sides by e^x to get (e^x)f(x)=e^(2x)+1

[silk crystal]

sorry

[near saffron]

Then it becomes a quadratic equation

[gray isle]

do you have the exact wording of the question

[silk crystal]

well, i am trying to solve an integral actually (entry-level calculus), (e^x + e^-x) / (e^x - e^-x), so i substituted u = e^x + e^-x, and i need to get the derivative of g(u), and to get g(u) i need to isolate x, s.t. x = g(u)

if i am understanding this correctly:)

so i found myself not being able to isolate x

[near saffron]

Try substituting u = e^x - e^-x and you get du = (e^x + e^-x)dx so the integral becomes du/u

So in that way you don't need to isolate x to find dx

(e^x + e^-x)dx is in the numerator

[silk crystal]

just a moment, i am processing this:)

ah right

great

but, out of curiosity, if i wanted to isolate x, what would be the right way?

[near saffron]

so f(x) = e^x + e^-x and you multiply both sides by e^x
now you have e^x f(x) = e^(2x)+1, and then you subtract e^x f(x) to get it all on one side. Recall e^(2x) = (e^x)^2
(e^x)^2 - f(x) e^x + 1 = 0, and you can use the quadratic formula to solve for e^x. You can imagine it as a variable.
e^x = ( f(x) + sqrt( (f(x))^2 - 4 ) ) / 2. Then you take the natural log of both sides to isolate x
x = ln( (f(x) + sqrt( (f(x))^2 - 4 ) ) / 2 )

Since ln(a/b) = ln(a) - ln(b), you can put x = ln( f(x) + sqrt( (f(x))^2 - 4 ) ) - ln(2), which makes the derivative simpler because derivative of a constant is 0

[gray isle]

recognise hyperbolic trig definitions

[silk crystal]

i see

alright, great, thanks for the help!

[lone heart]

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[swift igloo]

cant understand how do they calculate A:B:C

A:B and B:C i get but is there some formula to get A:B:C?

[lone heart]

@swift igloo Has your question been resolved?

[swift igloo]

<@&286206848099549185>

[alpine sable]

,rotate

[alpine sable]

you have to make B equal in both

so they multiply first ratio by 10, second by 9

then you get equal B in both

and can compare them

[simple spire]

is this channel available for help?

[swift igloo]

i see thankss!!

[swift igloo]

can u also tell me how did they find the ratio of all three numbers here

,rotate

[swift igloo]

in question 6

how did they get 4:6:3

[alpine sable]

they divided original ratio by 2.75

[swift igloo]

why 2.75

[alpine sable]

cause it gives pretty numbers

[swift igloo]

uh

[alpine sable]

i mean it gives natural numbers

it is not required to simplify like that to solve it

you can do it with original ratio

they just simplified to make the ratio look better

[swift igloo]

ohh so if i dont simplify i will add 11, 16.5, and 8.25

[alpine sable]

but ratios are often written with just natural numbers so they maybe did it for that reason too

[swift igloo]

and then divide by 8,25 and multiply by 19.5

this is what they did in our school

[alpine sable]

19.5*8.25/(11+16.5+8.25) = 4.5 yeah

yeah idk what lakh is it like thousands or what

[lone heart]

@swift igloo Has your question been resolved?

[swift igloo]

yeah its a currency

1 lakh = 10 ten thousand.

[lone heart]

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[ebon bridge]

hello

[ebon bridge]

i got this as the answer

is this correct?

[solemn grove]

/rotate

how do you rotate?

[buoyant kayak]

,rotate

[solemn grove]

thanks man

[solemn grove]

it’s r in the bottom

not r^2

right?

@ebon bridge

line where you wrote C’(x)

[ebon bridge]

its a r^2

[solemn grove]

r x 1/r^2 = 1/r

oh

AFTER DIFFERENTIATING

right

apologies

[ebon bridge]

no worries

[solemn grove]

yeah, looks alright ig, didn’t do any of the arithmetic

but the method looks right

you just need to actually find the cost now

plug the x into C(x)

and you’re good to go

@ebon bridge

[ebon bridge]

kk tysm

but @solemn grove one thing

just remembered

the radius

doesnt satisfy the height restriction

which is 10 cm

[solemn grove]

wait

I FORGOT THERE WERE RESTRICTIONS

and no, no

what’s the height

14 is the longest length

[ebon bridge]

this

[solemn grove]

no, like what’s the height of the cylinder

@ebon bridge

[ebon bridge]

its 10 cm

@solemn grove

[solemn grove]

that’s the dimension of the rectangular prism my guy

I’m asking about the cylinder

[ebon bridge]

u use the height of the rectangle

[solemn grove]

no?

you have an eqn for h in terms of r

[ebon bridge]

i acc dk

im so stuck

[solemn grove]

we have r

sub it in

[solemn grove]

line2

@ebon bridge

[ebon bridge]

ye when i do that

i get height which is past the restriction

[solemn grove]

which is?

[ebon bridge]

i got 15. 62 which is over 10 cm(the restriction)

[solemn grove]

no

[ebon bridge]

wdym?

[solemn grove]

you can flip the rectangle on its side

to get this

new height 14

actually, use h =14 and try find x from that

[ebon bridge]

ok

[solemn grove]

I still think you might be able to fit whatever cylinder you found in here

[peak zephyr]

the questions defines the height as 10 though

[solemn grove]

no

it says height of the prism

[peak zephyr]

it says

[solemn grove]

it’s just another dimension

rectangular prisms don’t have radius

but cylinders do

does that mean you can’t fit it in

?

bro

you can literally flip the rectangular prism however you want

it’s still the same shape

[peak zephyr]

bro i get that but when subbing in 4.4 into the height equation at the beginngin it exceeds the height of the prism

[solemn grove]

and you’ve done nothing different

[solemn grove]

that’s the height of cylinder

[peak zephyr]

yeah
the height of the cylinder exceeds the height of the prism

[solemn grove]

tilt it

[solemn grove]

1. height of prism is just another dimension, the names for length, width and height of rectangular prism are interchangeable
2. simple pythagoras will tell you the longest rod you can fit in here

is about 20cm

so there is more room there than you think

@peak zephyr

it doesn’t show much yet

but there is potential room for 15cm

[peak zephyr]

how do we know that the radius of 4.4 will fit inside the prism when tilted

[solemn grove]

radius of 4 is diameter of 8

8.8 even

[peak zephyr]

yeah

[peak zephyr]

wdym

[solemn grove]

4.4 x2

[peak zephyr]

is it possible to solve this without tilting the cylinder

[peak zephyr]

right

[solemn grove]

bro this looks more like a calc question than an optimisation question

only one independent var

you could prolly formulate what the max radius could be

[solemn grove]

but I think what he has done is probs enough

[peak zephyr]

max radius would be 5.5 i belive

[solemn grove]

you only need 1 more cm for height

[solemn grove]

haven’t worked that out yet sorry, but I think you’re excluding cases where h<r

[peak zephyr]

tru tru
how would i include that

[solemn grove]

in theory you could have a 7cm or probs even bigger one

dw too much about it

and as for fitting this in

you need less than 2 cm height

but you have 1+cm in length and 2+ cm in width

so you know, there is defo room for the said cylinder

when you tilt it just slightly so that it occupies more width and length instead of height

voila

@ebon bridge

your answer should still be ok

the original radius and height I mean

[ebon bridge]

oh ok

[lone heart]

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[lone heart]

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[somber torrent]

Sat math, no calc
How can I solve this quickly? the officail explenation took me like 10 minutes to write everything out and the entire section you get 25 minutes for 20 questions

[lone heart]

@somber torrent Has your question been resolved?

[somber torrent]

<@&286206848099549185>

[storm grotto]

putting it into matrix form and solving it with gaussian elimination will prob let you write less stuff down

[dense sleet]

Ok so i think you can do it in few mins

$$2\left(x - \frac{1}{3}\right) - \frac{3}{2} \left(y - \frac{1}{6}\right) = 0$$
$$3\left(y - \frac{1}{2}\right) - \frac{8}{3} \left(x - \frac{1}{6}\right) = 0$$

[ocean seal]

Pluton

[dense sleet]

If you write it as

$$2\left(x - \frac{1}{3}\right) - \frac{3}{2} \left(y - \frac{1}{2} + \frac{1}{3}\right) = 0$$
$$3\left(y - \frac{1}{2}\right) - \frac{8}{3} \left(x - \frac{1}{2} + \frac{1}{3}\right) = 0$$

[ocean seal]

Pluton

[dense sleet]

So y can be substituted

But x. Hmm..

$$2\left(x - \frac{1}{3}\right) - \frac{3}{2} \left(y - \frac{1}{2} + \frac{1}{3}\right) = 0$$
$$3\left(y - \frac{1}{2}\right) - \frac{8}{3} \left(x - \frac{1}{3} + \frac{1}{6}\right) = 0$$

[ocean seal]

Pluton

[dense sleet]

Then substitute

$$2\left(a\right) - \frac{3}{2} \left(b + \frac{1}{3}\right) = 0$$
$$3\left(b\right) - \frac{8}{3} \left(a + \frac{1}{6}\right) = 0$$

[ocean seal]

Pluton

[dense sleet]

It is a little simpler than before but idk for how much

[lone heart]

@somber torrent Has your question been resolved?

[lone heart]

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[nocturne gazelle]

Passengers arrive at a bus stop as a Poisson process with rate λ. Each passenger is headed
towards one of two destinations A and B with probabilities p and 1 −p, respectively. Buses
to A arrive as a Poisson process with rate μA, while buses to B arrive as a Poisson process
with rate μB . The arrivals of the buses and passengers and the destination of each passenger
are all independent.
How many people on average get on the first bus to A?

[lone heart]

Please don't occupy multiple help channels.

[wary stream]

Don't use multiple channels

[lone heart]

@nocturne gazelle Has your question been resolved?

[nocturne gazelle]

no

[lone heart]

@nocturne gazelle Has your question been resolved?

[nocturne gazelle]

no

[lone heart]

@nocturne gazelle Has your question been resolved?

[nocturne gazelle]

no

[lone heart]

@nocturne gazelle Has your question been resolved?

[nocturne gazelle]

no

[still onyx]

Still need help?

[lone heart]

@nocturne gazelle Has your question been resolved?

[lone heart]

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[exotic imp]

how do i find region R with using integration

heres my step but i got it wrong

[lone heart]

@exotic imp Has your question been resolved?

[exotic imp]

no

<@&286206848099549185>

[carmine oar]

@exotic imp would you mind take it once again?

it is a bit unclear

[exotic imp]

my answer?

[carmine oar]

yes

thanks

[carmine oar]

thanks

As region r2 is below the x axis

it's area is -ve

so you will have to remove the -ve sign as the area is positive

and for r1, the result is incorrect

[lone heart]

@exotic imp Has your question been resolved?

[lone heart]

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[next cliff]

Hello there

[next cliff]

.close

[lone heart]

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[royal obsidian]

can someone help me with this olympiad problem

[royal obsidian]

Create a polynomial function, 𝑞(𝑥), defined by being a difference of cubes with a root at 𝑥=2

<@&286206848099549185>

.close

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[barren olive]

hey guys i been solving this question and im unsure if i've done something wrong, because i can't find the solution but i introduced a u-sub for e^x -1 , du = e^x dx , and e^x = u + 1 and ln u = x

by using algebraic manipulation i multiplied numerator and denominator of the original integral so that i can get it in terms of u, which led to integral of ln u / (u)(u+1)

[lone heart]

@barren olive Has your question been resolved?

[near hollow]

$\pi^2/6$ ?

[ocean seal]

Max..

[near hollow]

@barren olive

[barren olive]

idk the answer, but what was your method?

@near hollow

[near hollow]

$\frac{x}{e^{x}-1} = \frac{xe^{-x}}{1-e^{-x}}$ now recognise if we pull out the x this looks like the geometric series infinite sum.$x\frac{e^{-x}}{1-e^{-x}}$

[ocean seal]

Max..

[near hollow]

With first term $e^{-x}$ and common ratio $e^{-x}$

[ocean seal]

Max..

[near hollow]

So our integral is now $\int_{0}^{\infty} x \sum_{i\ge 1} e^{-ix}$

[ocean seal]

Max..

[near hollow]

Which you can then integrate term wise

(Integrate $xe^{-ix}$ or use the laplace formula )

[ocean seal]

Max..

[near hollow]

We find this is just $\sum_{n\ge 1} 1/i^2 = \frac{6}{\pi^2}$

[ocean seal]

Max..

[barren olive]

@near hollow interesting stuff, i could understand everything from your 2nd step onwards however i don't agree with how you could equate x/1-e^-x = e^-x/1-e^-x

[barren olive]

i.e, this step

[near hollow]

Ok so all I've done is multiplied top and bottom of the fraction by $e^{-x}$

[ocean seal]

Max..

[barren olive]

well shouldnt the denominator become into ${e^-x -e^-2x}$ then ?

[ocean seal]

royal

[near hollow]

No because it's $e^x$ originally

[ocean seal]

Max..

[barren olive]

OHHH

my bad my bad i completely overlooked that

ty this solution works out neatly

[near hollow]

It's a hard thing to spot at first but with practice you'll start to recognise this stuff

I usually look for a reflection property

[barren olive]

I haven't learnt laplace yet

[near hollow]

But in this case I couldn't find one

[barren olive]

ive only learnt calc 1 in my school so i was pretty confused on how to solve this

its a putnam integral so i guess i bit off more than i could chew xD

[near hollow]

It's $\ell(f(t)) = \int_{0}^{\infty} f(t)e^{-st}dt$

[ocean seal]

Max..

[near hollow]

And there's a table for the different f(t)'s you can have

For instance f(t) = t has formula $1/s^2$

[barren olive]

and for every laplace type function, you can rewrite it as a geometric series?

[ocean seal]

Max..

[near hollow]

No no no

The original integral is not a laplace integral or laplace function

[barren olive]

this integral is right?

[near hollow]

When looking at functions sometimes you might spot a power series that it can be written as, if it can be written as a power series then usually it's game over

[near hollow]

Yes

[barren olive]

ohh i see how this comes into play then

since the solution would be 1/s^2 and in this case

s^2 is nothing but i^2

also, how did our initial x disappear then

[near hollow]

Exactly

[near hollow]

It didn't, when we had that sum multiplied by x if we expanded the sum we are just left with a sum of laplace integrals

$\int \sum xe^{-ix}$

[barren olive]

oh yeah because its infinite it doesnt really matter

[ocean seal]

Max..

[near hollow]

It does matter we can't just discard it

It has been used

$\int \sum xe^{-ix}= \sum \int xe^{-ix}$

[ocean seal]

Max..

[barren olive]

alright so the way i understand it f(t) = x and the e^st = e^-ix

[near hollow]

$\sum \int_{0}^{\infty} xe^{-ix} = \sum \frac{1}{i^2}$

[ocean seal]

Max..

[barren olive]

ok that makes sense to me now, ty

[near hollow]

👍

[barren olive]

really appreciate the help, couldn't have solved this without you @near hollow

[near hollow]

Keep up the grind bro starting young is the best way to success

[barren olive]

haha im trying about to finish hs soon

preparing for finals and another olympiad style test afterwards

[near hollow]

Good luck in that

Best prep is doing practice and learning method of solutions to problems you couldnt answer

[barren olive]

yep i feel the same way, ive just been knocking out past paper questions

i think i should close the channel so someone else can use it for help, we can talk more in DMS tho

.close

[lone heart]

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[fossil mason]

I need help with a and b

[barren olive]

make use of the chain rule for the first question

the question tells you dx/dt = 2

and gives you an eqn for y wrt x

so you can find dy/dx

using dy/dx * dx/dt you can find dy/dt

[fossil mason]

I think I got it

[barren olive]

did you get the final answer as 24units s^-1 ?

[fossil mason]

I sent an img

But it’s still loading

But yes HAHAHAH

[barren olive]

hahah nice

lemme take a look @ q2

eh its basically the same thing, make use of the chain rule again

pretty straight forward

[fossil mason]

But dx/dy

Do I rearrange ?

[barren olive]

yes or you could just invert it

cus 1/dy/dx = dx/dy

[fossil mason]

My img is still loading

Again

But I get it, lemme try

[barren olive]

sure

[fossil mason]

I think I got it

[barren olive]

show me your working

if you'd like me to check

[fossil mason]

Hang on it’s loading

Sorry my house wifi is very unstable

[barren olive]

np but i think the 4th step might be wrong im gonna check again by doing it tho

oh nope its right my bad

[fossil mason]

Ok ty

[barren olive]

i couldnt really see the 3 infront of the x^2 which is why i thought it might've been done wrong

[fossil mason]

Ohh okies

That’s all tysm!!

.close

[lone heart]

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[limpid spade]

Jesus

[oblique sparrow]

je||sus||

[quasi scarab]

omg

[marsh sonnet]

At least they try it

"show your work"
the work :

[quasi scarab]

sry, i cant verify ~20 pages

but put them in wolfram alpha

and it shows the graph

and it can derivate too

[limpid spade]

literally

all of these can be checked on online

[marsh sonnet]

Yeah

[quasi scarab]

also

its a little full of tautology

[lone heart]

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[dense compass]

what happens if you combine the areas AB and ECD?

[dense compass]

what do you notice about the shape?

[oblique sparrow]

oh i see a half circle

[dense compass]

not exactly, as in if you were to move the shaded area AB over to the other side and combining it with the shaded area CDE what do you get?

[oblique sparrow]

is this stupid

ba and cde side is same?

[dense compass]

well you don't know for sure that D is exactly half way between EC so not sure if you can do that

[oblique sparrow]

its isosceles

[dense compass]

oh yeah sorry

didn't spot that

you could definitely do it like that

[dense compass]

But I think this will be simpler

does that make sense?

[oblique sparrow]

yea

[dense compass]

so you can easily find the area of that triangle

[quasi scarab]

try to divide it into known-shaped ones

[undone rivet]

well I did it like this

Not sure if I am correct though

I haven't touched this part in some time

wait @quasi scarab
what you sent was way easier than my method

why did you delete it?

[quasi scarab]

it was just an idea

but it was wrong

[undone rivet]

wait it was wrong?

that seemed right to me

[quasi scarab]

maybe like this

or did u do like this?

start with B

you know the sides and we need the angles

@undone rivet

[undone rivet]

uhh no I didn't use that method
what you have done is that you found the parallelogram area and minus the A+B+A areas right?

that method should also be correct

also wait

It can't be equal

on those 2 sides

A and A on either sides can't be equal

as on left side it is a triangle

and on right side it is a sector

[quasi scarab]

yeah, its another triangle, sry

[undone rivet]

If so that method should be right too

[undone rivet]

What I did here was basically find the parallelogram area of OECD and sector area of OED and take the difference which would be the right hand side shaded area
now do the same thing for the left hand side ; sector area - triangle area

and then I took the total

which adds up to 40

[lone heart]

@oblique sparrow Has your question been resolved?

[lone heart]

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[alpine sable]

is the primitive and the intergral the same thing?

[tacit arch]

Primitive and the antiderivative are the same thing

To calculate the area under the curve of a function, called the integral, over a range [a,b], you evaluate the function's antiderivative at the endpoints and take the difference

[alpine sable]

ah i see

[lone heart]

@alpine sable Has your question been resolved?

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[pale mauve]

Can anyone help with the first problem ?

[lone heart]

@pale mauve Has your question been resolved?

[tacit arch]

what have you tried?

[pale mauve]

So far I used this graph, b/c I’m completely lose in the problem and thought that could be the answers for the first one. But I don’t really know honestly

[tacit arch]

do you understand what the numbers in this table represents

[pale mauve]

Nope I watch a yt video but I still feel lost

[tacit arch]

show the title/description of the table

[tacit arch]

you're reading it correctly. 0.0968 is correct

[pale mauve]

For the second problem do I need to do it different ?

[tacit arch]

slightly