help-27

[devout snow]

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[edgy nacelle]

ok so ive been stuck on this question as my answer is weird

[edgy nacelle]

iii)

ive already posted it but stuck again

this is what ive done

can somone point me in theright direction and confirm which parts i have done right

[neat solstice]

do you have to have an algebraic formal way of finding the solution?

this is the type of question i would do by visualising it

[edgy nacelle]

i think so, but i dont know, im open to any solutions as we are not gievn any solutions to work from for past papers at my uni

[cold bone]

I see no error in this btw

[edgy nacelle]

is there anything else i can do with my solution above

[rose rock]

You are basically there

[runic prawn]

you can check your answer on desmos

[rose rock]

-pi/3 is not the range of theta tho

[edgy nacelle]

nice spot , thanks

sorry omg, i meant iv)

not iii)

[rose rock]

Oh

[edgy nacelle]

so sorry about that ahah

[cold bone]

okok

[edgy nacelle]

i meant iv)

im not sure where i have gone wrong but my answer for φ(0)being 1/0=1/8 seems odd

[runic prawn]

i think they're asking for $$ \lim_{\alpha \to 0} \phi(\alpha) $$

[woven radish]

mayer-vietorUs

[runic prawn]

again you can check this on desmos

[iron sun]

Since they mention series expansion, it might be useful in this instance to make use of the maclaurin expansions of sqrt(1-x) and sqrt(1+x) up to some suitable big o error

[rose rock]

Even (powers of a) terms (a^2, a^4, a^6, ...) cancels out in sqrt(1 - a) - sqrt(1 + a)

You shouldn't have that a^2 term there

[edgy nacelle]

ahhhhh

thanks

i literally just didnt put a minus there for some reason

[rose rock]

No problem. mistakes happen

[iron sun]

Didn’t notice that you already did an expansion!

[edgy nacelle]

so all i will have is 1/8?

so as a goes to 0 nothing happens?

[iron sun]

Yeah 1/8

[rose rock]

From what I can see, yeah

[edgy nacelle]

cool

thanks

.close

[devout snow]

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[devout snow]

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[kindred belfry]

Can someone please please please check my work?

[kindred belfry]

Not the result

[kindred belfry]

But rather.. do i use the formula correctly..

[charred scroll]

2sinxcosx is equal to sin(2x)

[kindred belfry]

So..
6sinxcosx is equal to sin6x?

[charred scroll]

you can write 6sinxcosx as 3(2sinxcosx)

[kindred belfry]

Hmm right

What else i did wrong..

[charred scroll]

I'm not quite sure if that's what the question wants you to do but you can also replace (cosx)^2 - (sinx)^2 with cos(2x)

[charred scroll]

wait nvm, while this is true, it'd be easier to factor 3cosx in your question

[kindred belfry]

Oh..

[charred scroll]

use trigonometric identities to replace cos(2x) in the denumerator with something you can cross off with the top

[kindred belfry]

This? How..

[charred scroll]

oh I made a mistake too... sorry about that

[kindred belfry]

Oh, its okay..

I still havent got it tho..

[charred scroll]

I think I got it

you can factor 3cosx from the top, and replace (sin2x)^2 using the sin(2x) identity

and continue from there

[kindred belfry]

Hmm lemme try

[kindred belfry]

Uhh what do i do now..

[charred scroll]

uhhh... do you know the sin(2x) identity?

[kindred belfry]

No

Sorry

[charred scroll]

ahh

sin(2x) = 2(sinx)(cosx)

so (sin(2x))^2 = 4((sinx)^2)((cosx)^2)

[kindred belfry]

Ouhh

[charred scroll]

you can also factor cosx from the top to make it easier by writing 3cosx(1-2sinx)

[kindred belfry]

Oh right

[kindred belfry]

I think i still dont understand what to do with the denumenator

[charred scroll]

u can factor (cos(x))^2 from the denumenator

[kindred belfry]

Hmm how..

[charred scroll]

you got this in the denumenator:
(cosx)^2 - 4(sinx)^2 (cosx)^2
so you can write it as:
(cosx)^2 . (1 - 4(sinx)^2)

[kindred belfry]

Ohhh

[kindred belfry]

Hmmm

[charred scroll]

nice

[kindred belfry]

What can i do next..

Just insert the x?

[charred scroll]

take a look at the numenator and denumenator and see what you can cross off :D

[kindred belfry]

Ohh okie

[kindred belfry]

I think i got it, thankyou so much

[charred scroll]

cheers!

[kindred belfry]

You are a genius

.close

[devout snow]

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[gilded plinth]

Can someone help me solve this?

nvm i was looking at the wrong thing in the answer key lol

.close

[devout snow]

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[native trench]

When should I use implicit differentiation? Sometimes the derivative of y is 0 and sometimes it is (dy)/(dx). When do I choose the one and when do I choose the other?

[torn bane]

if y depends on x, then the derivative of y with respect to x will always be dy/dx

[native trench]

I’ve seen that dy/dx is 0 in some cases, when should I use that?

[torn bane]

if it's 0, then y does not depend on x, or in other words is constant with respect to x

[native trench]

Ah ok, thanks.

.close

[devout snow]

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[devout snow]

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[urban moat]

x^(x+1)=(x+1)^x I know that is something like 2.2 but I dont undesrstand why can anyone help me?

[devout snow]

@urban moat Has your question been resolved?

[lusty sapphire]

I think linear approximation is the only way you can solve this

[urban moat]

can u say something more about that?

[torn bane]

the equation has some relation to the lambert w function i think, so there isn't an analytic solution

[urban moat]

what does a lambert function do here?

2^x=x right?

[lusty sapphire]

It does not, sadly.

[urban moat]

ok so how do we solve it?

[lusty sapphire]

You can't find an exact solution. You can only approximate it

https://en.m.wikipedia.org/wiki/Newton's_method

[urban moat]

I know but maybe there is an algebraic thing or pi or something else

I know its here

[lusty sapphire]

There is not

[urban moat]

OK

thx

[devout snow]

@urban moat Has your question been resolved?

[devout snow]

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[shrewd briar]

my teacher marked this wrong, but Im pretty sure I did it correctly?

[shrewd briar]

I found the area of the larger sector then subtracted from the smaller sector,
They said to subtract the radius first, then find the area of that smaller radius

[spiral hornet]

oh bcuz this is not a circle

[spiral hornet]

one is 7in and one is 10in

so ur not geting the 360 and pi

[shrewd briar]

yeah but im doing 95/360

or I could convert to radians and do rad * radius

[spiral hornet]

ye and ?

[shrewd briar]

that takes a portion of the full area

[spiral hornet]

where did u get pi and r and stuff

it is not a circle

so it is not portional to be taken a portion

[shrewd briar]

its 95 deg of a circle

so i take 95/360 of a full circles area

that works out

[spiral hornet]

then ur only getting this

[shrewd briar]

bro what

[spiral hornet]

the other one is 10 in

ok it is an ugly drawing but ye

u get it

[shrewd briar]

I'd like for someone else to help me please.
this is just wrong

[spiral hornet]

:(

me sad

i go cry

[shrewd briar]

its misinformation.

[spiral hornet]

1 sec for 1 last chance

[shrewd briar]

you're just wrong

not being helpful at all, so please let someone else help

[spiral hornet]

here

no thanks for my new beutiful drawing

i also highlighted the diference

[shrewd briar]

thats legit what I did

95/360 * pi * 10^2 - 95/360 * pi * 7^2

[spiral hornet]

oh i get what ur saying

but if u minus a really big one - a average big one = a small one

but that one is big..

try to draw it

[shrewd briar]

I know where I messed up now

[spiral hornet]

nvm ill make room for others

[spiral hornet]

oh

💀

[shrewd briar]

I should have done the area of
10 - 7 = 3
so then

95/360 * pi * 10^2 - 95/360 * pi * 3^2

instead of subtracting the rad of 7

[spiral hornet]

ye u shoudl take the really big one to minus a small one to get that

[shrewd briar]

alr. Sorry for being agressive earlier, you were just being kind of confusing

@spiral hornet

[spiral hornet]

nah all good

:3

[devout snow]

@shrewd briar Has your question been resolved?

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[ionic prism]

can anyone help me understand why $0\leq \theta \leq \pi$ in the following problem and not $2\pi$?
I have the mass density $\rho\left(x,y\right)=\frac{1}{\sqrt{x^{2}+y^{2}}}$ of a disk that is given by $x^2+(y-1)^2=1$
I converted everything to polar and got the density to be $\rho\left(r,\theta \right)=r^{-1}$ and the circle is described by $r=2\sin \theta$
so then that means the bounds will be $0\leq r \leq 2\sin \theta$, and bounds for $\theta$ which i originally thought should be $0\leq \theta \leq 2\pi$ but as I've said the answer for the bounds of it is $0\leq \theta \leq \pi$.
can anyone explain this?

[woven radish]

horizon2.0

[loud dagger]

It's because the polar curve $r=2\sin(\theta)$ makes two loops over $0\leq \theta \leq 2\pi$, try plugging in $\theta=0$, $\theta=\frac{\pi}{2}$, and $\theta=\pi$ and you'll see that's enough for a full loop already.

[woven radish]

DonDoesMath

[ionic prism]

oh i see, is there some general way to know about this?

like for any curve $r=f(\theta )$ to know the domain of $\theta$?

[woven radish]

horizon2.0

[loud dagger]

Polar functions are usually defined for all $\theta$ (unless there's a discontinuity), the tricky part is that they often loop on themselves and we're interested in finding where a loop starts/stops. In the case of $r=2\sin(\theta)$, it's a bit of a coincidence that the period of sin is $2\pi$ but it loops after just $\pi$ because right as the angles go past $\pi$, $r$ turns negative and happens to form the same circle as previously. The only way I'm aware of to deal with it is plugging in test values and just getting used to polar functions being weird.

[woven radish]

DonDoesMath

[loud dagger]

So that $r=2\sin(\theta)$ example is still defined for all $\theta$, but we only care about one loop of it in this context.

[woven radish]

DonDoesMath

[ionic prism]

i understand, do you know the proper terms for the "checking if looping" we've discussed about? plugging in values sounds wrong, like i feel there has to be some theorem about it

[loud dagger]

I don't know about a theorem, but you can also choose any valid starting $r$ value you want and solve it algebraically. If we take $r=0$, then $0=2\sin(\theta)$ so $\theta=...,-2\pi,-\pi,0,\pi,2\pi,...$

[woven radish]

DonDoesMath

[loud dagger]

Actually, now that I think about it, that's not necessarily true for all polar curves, my apologies

I'm not aware of a better way than trying test points and making a rough sketch

[devout snow]

@ionic prism Has your question been resolved?

[devout snow]

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[lyric plaza]

Genuinely stuck here, what am I supposed to do? Am I supposed to simplify the first term so I also get a radical with the same radicand (2^1/3)?

[rose rock]

Good so far. You can simplify $\sqrt[3]{64}$

[woven radish]

StrangeQuarkAL

[devout snow]

@lyric plaza Has your question been resolved?

[lyric plaza]

Ahh, okay now I see. Thank you!

[devout snow]

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[restive river]

the answer should be x < -1 but in the 2) i have a different answer, i could change the method for solve this but this one looks easier

[thin inlet]

why did you flip the inequality sign in the other cases?

i.e. why is it white?

don't flip it for any of the cases, thats your issue

[restive river]

i got almost the same

[restive river]

i thought that cause i was multiplying by a negative number

[devout snow]

@restive river Has your question been resolved?

[thin inlet]

ok

so you have |x+5|, this is x+5 for x>=-5 and -(x+5) for x<-5

then you have |x-3|, this is x-3 for x>=3 and -(x+3) x<3

the last inequality is invalid, because if x>3, then x is not less than -5

[restive river]

didnt understand this

[thin inlet]

-(x+5) for x<-5

[thin inlet]

x-3 for x>=3

we don't relate these expressions, because they occur at non overlaping x-values

[restive river]

like this?

[thin inlet]

the definition of |x| is x if x=>0 and -x if x<0

[devout snow]

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[restive river]

.reopen

[devout snow]

✅

[restive river]

okay i think i got it now

so i have to analyze case by case

and see if that is true?

[thin inlet]

yes

alternatively you can use |x|=sqrt(x^2)

[restive river]

noe

i dont think it is allowed

[thin inlet]

it'll still work

[restive river]

ik but its for entrance exam for uni

[thin inlet]

oh you could definitely use that too

[restive river]

sure?

[thin inlet]

yeah

it's much faster and clearer that way

[thin inlet]

,rccw

[thin inlet]

ah

if you're only allowed to use those properties

then i would steer clear from using sqrt(x^2)=|x|

[restive river]

what info u are substracting if u do that anyways?

[thin inlet]

?

i suppose you could square both sides and get something similar in beginning

similar to using sqrt(x^2)=|x|

and this would preserve the inequality since |a|>0

[restive river]

Is there any reason why this should not be done?

[thin inlet]

why what should not be done

[restive river]

im asking

like why i cant use sqrt(x^2)=|x|

is there a math reason that is not compatible with that property?

[devout snow]

@restive river Has your question been resolved?

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[hexed cedar]

can someone walk me thru this please

[hexed cedar]

i dont know how to get the parameters of integration

[devout snow]

@hexed cedar Has your question been resolved?

[hexed cedar]

<@&286206848099549185>

[thin fern]

!status

[devout snow]

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

[thin fern]

do you mean for the line integral

[hexed cedar]

yes

[hexed cedar]

1

[thin fern]

So this is Stokes right

[hexed cedar]

Yes

how do I set up the integral?

once i set it up i think i can solve it

[thin fern]

Basically, stokes theorem tells us the double integral is equivalent to the line integral of the boundary of S

[hexed cedar]

right

[thin fern]

so you need to set that up

which looks like a pain to solve considering the F we're given

[hexed cedar]

yeah

[thin fern]

F isn't conservative either

Oh

we need to parametrize the boundary first

then it should simplify nicely

[hexed cedar]

how do we do that?

[thin fern]

i.e. we need to parametrize $z=4-x^2-y^2$

[woven radish]

Ari

[hexed cedar]

what is the boundry?

Ohh

Ok

in terms of t?

or what

[thin fern]

sure

[hexed cedar]

ok wait how would i do that

[thin fern]

Since this is a circular region

I'd say something like $x=\cos{t}$, $y=\sin{t}$, $z=4-t^2$ would be good.

[woven radish]

Ari

[thin fern]

Or even better

nvm I thought I had a better one

[hexed cedar]

wait how are you coming up with these

im so lost

[thin fern]

because we have -(x^2+y^2)

which is -r^2

so this tells me I should be setting x as cos and y as sin since we have circular regions

[hexed cedar]

oh

ok

[thin fern]

wait

No mb the boundary is just the circle with radius 2

So $x=2\cos{t}$, $y=2\sin{t}$, $z=0$

[woven radish]

Ari

[thin fern]

mb

[hexed cedar]

wait i dont get it

why 2cost

and 2sint

why the

2

[thin fern]

So draw the graph

or try to picture it

The boundary of it (since we're only looking at the region above the xy-plane$ is the level-set of the paraboloid at z=0

If you set z=0 we have x^2+y^2=4 which is a circle of radius 2

[hexed cedar]

i have no idea how to picture 3d graphs lol

is that a problem

if im doing vector calc

[thin fern]

Yes

[hexed cedar]

rip

[thin fern]

you don't exactly need to picture it here

you just need to recognize taking the level set at z=0

[hexed cedar]

ok

at z=0 its a circle with radius 2

[thin fern]

right

[thin fern]

how do we parametrize that

[hexed cedar]

ummmm

with r and theta

?

oh frick

i need to translate this entire damn thing?

[thin fern]

Well we know r, it's 2

[thin fern]

yes

[hexed cedar]

ok give me a moment

ill send a picture

ok does this look right

sorry its rotated idk what happened

[thin fern]

r=2

don't use r

[hexed cedar]

oh

ok

i thought

wait

dont i do like dr dtheta

if i replace it all with r wht do i do later on in the integral

[thin fern]

No we're doing a line integral

dr dtheta is like

solids

[devout snow]

@hexed cedar Has your question been resolved?

[wary urchin]

what is the problem?

Could you please tell me?

[devout snow]

@hexed cedar Has your question been resolved?

[devout snow]

@hexed cedar Has your question been resolved?

[devout snow]

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[restive river]

Problem 17

I did it multiple times but kept getting 22

And then the positive m is 19/3

(x was supposed to be m)

[devout snow]

@restive river Has your question been resolved?

[spiral hornet]

this is kinda long

y can do it this way ig

$x^{2} - 2x\sqrt(x^{2} - 5x - 3) + x^{2} - 5x - 3 = 16$

[woven radish]

≅ Semicolons

[spiral hornet]

=> $( x - \sqrt(x^{2} - 5x - 3)) ^{2} = 16$

[woven radish]

≅ Semicolons

[spiral hornet]

i think it is easier if u find x that way

1 sec

[spiral hornet]

i got x = 19/3 but there is no 22 answer

[restive river]

Yea that's what I'm confused about lol

okay then I thought it was just me

ty

I'll ask my teacher

.close

[devout snow]

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[devout snow]

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[worthy cargo]

Woke up early this morning for review, looking for someone to answer questions about my practice quizzes

[lost laurel]

Problem 1 looks alright

[faint gorge]

-sqrt(5) and sqrt(5) should be included

In Problem 2 you forgot the square root

sqrt(5+h) + sqrt(5) instead of sqrt(5+h) + 5 in the denominator

[worthy cargo]

This is actually the answer key the professor provided us haha I am asking questions related to them

[worthy cargo]

I think when he took the conjaguate the sqrt cancelled

[faint gorge]

no

You multiplied numerator and denomimator by sqrt(5+h) + sqrt(5)

[worthy cargo]

I see

This one I do not understand however

[faint gorge]

draw the sine function with the sign respectively

from 0 to 3π/2

[worthy cargo]

Okay so this can be solved accordingly to the unit circle

3pi/2 sin value is -1

Sin (0) value is 1

[faint gorge]

Yea

[devout snow]

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[fallen harness]

why do we use those arrows for x,y?

[fallen harness]

like <x,y>

is it just another form of a bracket

[stone stump]

yes

[fallen harness]

alr

why is there no line in between x and y

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.reopen

[devout snow]

✅

[devout snow]

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[fallen harness]

<@&286206848099549185>

[native trench]

Because it is not a fraction?

[fallen harness]

but why do they write it like that

[native trench]

It is just a notation rule, like the notation (x, y) for coordinates?

[fallen harness]

oh alr

[boreal helm]

@fallen harness

.close

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[cold bone]

What are the different methods to solve this?
Could someone explain

[lean rover]

u can use the vformula

cos(t)cos(s)=1/2[cos(t-s)+cos(t+s)]

[cold bone]

I have found my answer this way

I am just curious if there is any other way

[lean rover]

ah ok idk any other way

[cold bone]

okok
thanks for the help btw

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[jaunty mantle]

can someone translate this into probability

[jaunty mantle]

im finding it very difficult to understand what they are doing in the first paragraph

why is this $\int_0^\infty l(x + t),dt$ important?

[woven radish]

frosst

[jaunty mantle]

as i understand it, $l(x) = \mathbb P(X > x)$ where $X$ denotes the lifespan of someone

[woven radish]

frosst

[jaunty mantle]

either way we somehow want to get to $\overset{\circ}{e}(x) = \mathbb E[X|X>x]$

right?

[frozen aurora]

isn't l the count of people

[woven radish]

frosst

[jaunty mantle]

yes but it'll be the probability of i just set l_alpha = 1

[frozen aurora]

what's alpha?

[jaunty mantle]

the initial value of people

how many people are alive at x=0

[frozen aurora]

ah

[jaunty mantle]

idk why you actuaries use actual numbers, 1 is perfectly fine

and allows for l_x to be interpreted as a probability (or at least i think you can do that)

[frozen aurora]

idk if that's the right thing to do in this case. the first integral finds the total number of YEARS lived after age x of the whole population

[jaunty mantle]

so if i've got 2 people

and they live from age 30 to 35

[frozen aurora]

also i don't do actuarial anything lol

[jaunty mantle]

the integral gives me 10?

[frozen aurora]

i think so

[jaunty mantle]

lol haha ok

[frozen aurora]

i do agree that it's written in a very confusing way

[jaunty mantle]

if i just ignore the top part and whatever nonsense it's saying, can i just go from here

[frozen aurora]

but combining with the next sentence after the integral it makes sense

[jaunty mantle]

it is not clear to me that dividing by some l_x gives the average years lived

[frozen aurora]

the integral gave you 10 and then you divide by the total population which is 2, so indeed 5 years have been lived

[jaunty mantle]

ok but why aren't we formulating it in the language of E[X|X>x]

[frozen aurora]

i actually think that's possible, let me try to rewrite it like that

[frozen aurora]

or maybe not nvm

how i see it is they are actually computing the CDF

so the $e_x$ rather is $e_x = P(X> x)$

[woven radish]

artemetra

[frozen aurora]

yeah

[jaunty mantle]

$\mathbb E[X|X>x] = \int_\mathbb R xf_{X|X>x}(x),dx$

[woven radish]

frosst

[frozen aurora]

that makes much more sense

[jaunty mantle]

is there supposed to be a circle on top

[frozen aurora]

yeah

i forgot the latex for it

[jaunty mantle]

but isn't that just l_x

[frozen aurora]

nope

[jaunty mantle]

hmm

[frozen aurora]

l_x is population

[frozen aurora]

a number

e_x is probability

[jaunty mantle]

but if my initial population is 1

[frozen aurora]

that a person lives past age x

[jaunty mantle]

then e_x = l_x

[frozen aurora]

no?

[jaunty mantle]

dumb actuarial notation:
Let $x$ denote the current age of someone. Let $t$ be some time measurement (in years). Let $X$ denote the lifespan of someone.
\begin{align*}
_tp_x &= p(x, t) = \mathbb P(X \geq x + t | X > x) \
tq_x & = q(x, t) = 1 - p(x, t) = \mathbb P(X < x + t | X > x) \
{t|}q_x &= p(x, t+1) - p(x, t) = \mathbb P( x + t + 1 \leq X \leq x + t | X > x) \
\intertext{Life Tables: }
l_x &= l(x) = \mathbb P(X > x) \
d_x &= d(x) = \mathbb P(x + 1 > X > x) \
\intertext{Lastly, actuaries like to use (l\alpha = 100000) but we are not dumb, so}
l\alpha &= 1
\end{align*}

[woven radish]

frosst

[jaunty mantle]

this is what i've got for what these symbols mean

is l_x not the same as e_x

[frozen aurora]

l_x is not related to probability at all

or like

[jaunty mantle]

ok if i start with 100 people

and by year 5 i have 97 people left

[frozen aurora]

yeah, 97 is your l_5

[jaunty mantle]

how is that different to i start with a probability of 100% of being alive

and by year 5 i have a probability of 97% of being alive

[frozen aurora]

one is measured/sampled and the other one is calculated?

if $l_a$ was 1, then you would get
$e_x = \int_{t=0}^\infty l_{x+t} dt$

[woven radish]

artemetra

[frozen aurora]

i don't see how e_x can be equal to l_x if it's the integral

[jaunty mantle]

this is hte average time you have to live once you are at age x

[frozen aurora]

yea

no

wait wtf

[jaunty mantle]

how much time left

e_x is average number of years lived after age x

[frozen aurora]

yes

got it

[jaunty mantle]

so given that X > x, what is the E[X]

so it's E[X|X>x]

[frozen aurora]

okay i am confused now too ahah

[jaunty mantle]

if i use my definition here, i'd get

$\mathbb E[X|X>a] = \frac{1}{\int_{(a, \infty)}f_X(x),dx}\int_{(a, \infty)}xf_X(x),dx$

[woven radish]

frosst

[frozen aurora]

from https://infinityisreallybig.com/2019/04/04/the-complete-expectation-of-life/

[jaunty mantle]

see i feel like that makes sense

im seeing things i recognise here

ahhh

this is where the UUD comes in

that's this part

this sounds like a terrible model

[devout snow]

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[maiden imp]

Can anyone help me with this?
343. Scrooge McDuck's money bin is surrounded by a 4.0-meter-high fence at a distance of 12.0 meters. The Beagle Boys are trying to climb the fence from the outside using a ladder to reach the wall of the money bin. How long must the ladder be at least to reach the wall of the money bin?
Hint: Let x be the distance from the bottom of the ladder to the fence and y the distance from the top of the ladder to the ground. You can find the relationship between the variables x and y by using similar triangles.

[maiden imp]

Here is the question. I translated it in english because the original is in finnish

It is in a book with logarithms and exponents so I think you maybe need to use them. Also you could try to make a function and of the derivative function you see the smallest possible lenght of the ladder

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[solar osprey]

a) In how many ways can you draw a straight line from A to one of the numbers, and another straight line from B to a different number?

b) How many of the configurations in a) satisfy the condition that the lines do not intersect?

c) In how many ways can you draw straight lines from A, B, and C to three different numbers so that no lines intersect?

I forgot how to do basic math pls help 🫠

[solar osprey]

this is what i got so far, does it look right?

a) for each possible line for from A there are 8 possible lines from B. 9x8=72
b) the lines won’t intersect as long as the line from B goes to a number that’s higher than the number A is connected to. So if A->1 then there’s 8 possibilities and so on. (9-1)+(9-2)+…+(9-9) = 36
c) A<B<C. idk how to calculate the possibilities for that

[devout snow]

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[solar osprey]

<@&286206848099549185>

[devout snow]

@solar osprey Has your question been resolved?

[storm rose]

For b and c you can just compute combination by binomial coefficient

[solar osprey]

how

[storm rose]

In first one if you choose two numbers you can only two ways connect it with A and B and one intersect one not

[solar osprey]

?

[storm rose]

Lets choose two number

For example 1 2

You have two ways to connect it to A and B

A 1 B 2 and A 2 B 1

And one is intersecting one not

You can all possible connection put in diiferent set where you put connection in the same set if they connect to the same numbers for examp A1 B2 and A2 B1 but A3 B1 is in another set

You create factor set

You get it?

@solar osprey

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[restive river]

Is this possible

[twin rose]

Yeah

[olive snow]

1/8 = 8^-1

[twin rose]

True for all x

[restive river]

I will show all my steps and I need advice on where I got it wrong

[olive snow]

But this is true so there is nothing wrong with this

[restive river]

Okay

7/9 not 9/7

[olive snow]

There is a mistake at 2nd line

(1/8)^1-5x = (1/8)*(1/8)^-5x

[restive river]

Okay I see

.flip

.r

.rotate

Who said the person that stops learning stops living

[olive snow]

Idk, such a cool guy if he exist

I always thinked that

[olive snow]

Actually since 1/8 = 8^-1

You will get 2^6x+7 = 8^((-1)*(1-5x))

[restive river]

Then distribute the -1

[olive snow]

So 2^6x+7 = 8^(-1+5x)

How would you proceed then ?

[restive river]

Thank you guys

[olive snow]

Indeed

[restive river]

.close

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[olive snow]

Gj

[restive river]

Ty

[devout snow]

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[whole oyster]

I am stuck

[inland plover]

formula of a slope is y2-y1/x2-x1

[whole oyster]

im confused where i plug in x^2 and x=-1

[inland plover]

x^2 and x=-1 seems irrelevant in this question from what i see the question is formulated weirdly

if you want to find the actual slope tho you can calculate de derivative of x^2 with x=-1

and check which is closer to that answer

[whole oyster]

strange

I am struggling tbh

[inland plover]

find the definition of a derivative

[whole oyster]

so i do 5(x+0.001)^2-4.5(x+0.001)-5x^2-4.5x all over h?

[inland plover]

yes and then replace x = 4

[whole oyster]

i got -35959.995, that doesn't feel correct but I'm unsure

[whole oyster]

do i use this without replacing x with 4 to find the slope first?

[inland plover]

it should be close to 35.5

[whole oyster]

hmmm i am way off

[inland plover]

but h = 0.001 in your case

and you can either replace x=4 at the begining or the end it shouldnt matter

[whole oyster]

i keep getting the same answer, where am i doing it wrong?

[inland plover]

your signs are wrong

[whole oyster]

I'm not sure what you mean

[inland plover]

the -1 is distributed so you should have +4.5(4)

[whole oyster]

ohhhh dangit

i always forget that

[inland plover]

and no need for the lim since you already replaced h

[whole oyster]

it still works with the lim there right?

in a calculator i mean

[inland plover]

its not accurate math

[whole oyster]

okay i got 35.505, so now the part asking for the slope?

[inland plover]

35.5 is the slope

[whole oyster]

and also the y value?

sorry

so how do i use the slope to find f(4)?

[inland plover]

you dont

[whole oyster]

okay, I'm unsure how to find the y-value at x=4

[inland plover]

replace x=4 in your original function

[whole oyster]

okay i got 62

so now for the last part of the question, i have to write mx+b

not sure how to do that, which i feel like i should know how to

[inland plover]

yes you should

[whole oyster]

😅

35.505x+62?

nope

[inland plover]

you have the point (4, 62), m=35.505 and solve for b

[whole oyster]

i feel stupid, i am still confused

62=35.505(4)+b? and then solve for b right

[inland plover]

yes

[whole oyster]

well, i got that wrong so i have to redo it all 😵‍💫

[inland plover]

b=-80.02

but you need to round 35.505 to 35.5 as specified

and then do the calculation

[whole oyster]

the correct answer was something like 35 * x - 35 -80.02, strange format i don't understand

[inland plover]

couldnt tell ya the way to do it is how i told you

[whole oyster]

okay i got it solved, thank you

.close

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[ancient egret]

Guys can someone help me with Laplace transformation of this piecewise function
I have a strong feeling that the answer is incorrect because it doesn’t take into consideration T +4 can somebody please verify?

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[violet forum]

Hi I want to take the derivative of an array with repsect to another array. I am doing this in python. my current approach is to use numpy.gradient

[violet forum]

I also used spline interpolation, and finite difference (first order) but the results are quite different in all of these

I was looking at Chebyshev polynominals, can these be used to take the derivative?

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[violet forum]

<@&286206848099549185>

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[silk arrow]

I want to participate in math olympiad (new) somebbody tells me where from to start?? i really don't know

[silk arrow]

.

[devout snow]

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[stark wigeon]

Hi! Im also joining this year!! Where are you from and what year do you go to?

I am from switzerland so i went to the swiss math olympiad website and over there was all the information on how to join

[silk arrow]

iam in class 10th from bangladesh

[stark wigeon]

your country should have a website in which you can look for all the details

or you can ask your highschool for more information,

[silk arrow]

oh yeah i forget about it let mme check it out

[stark wigeon]

idk about bangladesh but in switzerland the signup period is by the end of september

and you can start doing the first exam on the 1st iirc

[silk arrow]

can u describe ur preparation about math olympiad ?

i mean explain is it really hard or somethign like that

[stark wigeon]

Ive never competed so i know just as much as you hahaha, i have. What i know about the swiss math olympiad (SMO) Is that there are 3 rounds.
The first round is purely logical and if you pass you are officially competing in the SMO. then you get lessons on advanced topics that arent taught in school. The second exam will be on these topics, aswell as the third/final exam. The people who get the 6 highest score in the third exam compete in IMO (which is in australia for 2025)

[silk arrow]

btw how to get medal [;) after competed with worldwide??

[stark wigeon]

yes, IMO is the worldwide competition (internarional math olympiad) and you can win medals there

im not home but on my pc i have a folder with a bunch of different papers that you can practice on! If you send me a dm in 1 or maybe 2 hours idk i can send you the files as well as more information on the exam. Keep in note that i only know how its done in switzerland...

[silk arrow]

sure

oky then i will dm u after 2 hours

[stark wigeon]

:)

goodluck!! hope to see you in australia at IMO ;)

[silk arrow]

hahah yyeah

;)

thanks for helping

[stark wigeon]

yw :)

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[faint zinc]

!show

[devout snow]

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

[heady rock]

i dont get it, i think if you would calculate with every step i would get it

[faint zinc]

Use a common denominator. Eliminate the denominator (don't forget the restriction that x can't be 8), and then solve for x.

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[solar harbor]

I can't even comprehend what's goin on

[devout snow]

Please don't occupy multiple help channels.

[solar harbor]

ok! sorry!

.close

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[frigid tartan]

If somebody could point me in the right direction that be great. I dont know what formula to use.

[neat solstice]

have you set up any equations?

[devout snow]

@frigid tartan Has your question been resolved?

[frigid tartan]

whoops

[frigid tartan]

i dont know how to set an equation up

there must be a formula right

[neat solstice]

we put water into the reservoir at a constant rate

so a constant amount of Liters per Minute

but we also lose a constant amount of Liters per Minute due to the filter

you can think of this rate of filling the reservoir as a velocity:
When you drive 50 km/h, what is the formula for the distance driven after a time t?

you can do the same here

[frigid tartan]

time distance rate?

distance = time*rate

[neat solstice]

yes

[frigid tartan]

so for rate its d/t

[neat solstice]

so here, it is:
Water = Rate * Time

that is how we define a constant rate

here however, we also have a second rate for losing water

so we get:
Water = Rate_1 * Time - Rate_2 * Time

there is a minus in there because that is the rate at which we LOSE Water

[frigid tartan]

that rate is the 5/11 minutes it takes to fill right

[neat solstice]

5/11 minutes is a time, not a rate

[frigid tartan]

rephrase

losing water during the filling

right

[neat solstice]

the rate at which we fill it has to be in the units Liter/minute
the 5/11 is a measure of time or minutes

[frigid tartan]

rate is (2)/(5/11) then right

for the filling

[neat solstice]

what to do here right now is to get another equation for what happens once the reservoir is full
since we don't fill it anymore, the first rate will disappear and we have one constant less to worry about

[neat solstice]

this would be correct if we wouldn't also lose water whilst filling the tank

[frigid tartan]

ok

[neat solstice]

in other words, Rate_1 - Rate_2 is exactly what you mentioned, the 2*11/5

[frigid tartan]

then the first rate would be the (3)(1/3)

[neat solstice]

start by setting up a new equation

that 3+1/3 is a measure of time again

[frigid tartan]

(2)(3*3/1)-(2)(11/5)

ratetime-rate_1time_*

(20/3)(5/11)-(22/5)(4/3)

[neat solstice]

Let me sum up some info:
(x) After 5/11 Minutes, the tank is full (2 L)
(x) We now lose Water at a constant rate (let's call it B)
(x) After 3+1/3=10/3 more Minutes the tank is empty (0 L)

The general formula we wanna use is
V(t) = C*t + V0
, where C is the rate of change and V0 the starting Volume/Amount of Water

In our case, we start with 2 Liters and our rate of change is B

Note: We can approach this in two ways
(x) set a new zero-point for time and we are ready to go: V(t) = 2 - Bt
(x) offset the function like this: V(t) = 2 - B(t-5/11)

The difference is that in the first approach, we will get V(3+1/3)=0, this is what I mean with a new "zero-point" for time

In the second approach, we would not write V(3+1/3) = 0, but V(5/11 + 3+1/3) = 0

The first approach is faster
The second approach is nicer since it can also be plotted easily in a graphing tool

[frigid tartan]

lets try the second approach

i need to understand it fully before i move on

[neat solstice]

for some intuition what this means: the graph shows the current amount of water depending on the time
reference: we start at V=0, then after some time we reach the max, then we reach V=0 again

the green part is what we get with the second approach, here we continue time as usual
the purple part is where we set the new 0 to make calculation faster but as we can see, it starts at 0 and makes less sense graphically

We know generally: V(t) = C*t + V0

for us: V0 = 2, C = -B

so we get:

V(t) = 2 - B*t

now, this V0 is not at time t=0 but at t=5/11, so we shift the graph by that

we get:
V(t) = 2 - B*(t-5/11)

that is exactly what we see as the green graph

[frigid tartan]

B is the rate of change right?

[neat solstice]

it is the rate at which we lose water through the filter

[frigid tartan]

I just want to clarify that the rate is the rate to empty minus the rate to fill

correct?

[neat solstice]

after we fill the reservoir, we stop filling
so after that point in time, the rate is simply minus the rate at which we empty

but yes, at the start the rate is the rate at which we fill minus the rate at which we empty

[frigid tartan]

ookey dokey

i have the graph on desmos

B should be that rate that we solved for

[neat solstice]

yes exactly

[frigid tartan]

i tried solving and got 156/55

for B

[neat solstice]

2 - B(T-5/11) = 0, for T=5/11 + 3+1/3

since 5/11 + 3+1/3 is the time at the end when it is empty again

[frigid tartan]

yea

[neat solstice]

B(T-5/11) = 2
B = 2 / (T-5/11)

T - 5/11 = 5/11 + 3+1/3 - 5/11 = 3+1/3 = 10/3

B = 2 / (10/3) = 2*3/10 = 3/5

[frigid tartan]

how did you get 10/3?

[neat solstice]

the 3 1/3 is 3 + 1/3 right?

3 + 1/3 = 9/3 + 1/3 = 10/3

[frigid tartan]

oh

for some reason my brain said it was 4/3 lololol

[neat solstice]

i feel that lmao

[frigid tartan]

so the B we just solved for

3/5

is not the combined rate yet

no wait it is

[neat solstice]

exactly

[frigid tartan]

oh

[neat solstice]

now, we can use the other equation again

remember:
V(t) = At - Bt
was the formula for the filling process

but now we know what B is

[frigid tartan]

A is the other rate

[neat solstice]

yes

we also know:
V(5/11) = 2
thus:
At -Bt = 2, for t = 5/11

[frigid tartan]

oh so we dont even need to solve for A

we just plug in and its there

[neat solstice]

well now we have an equation with only one variable (A) so we can now solve for A

thus we get the solution we need

[frigid tartan]

the rate in which the water is added

5 l/min?

[neat solstice]

yep should be right

[frigid tartan]

wao

you are my hero

😭

thank you so much

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[tawdry venture]

i thought the formula i highlighted was supposed to be e(x^2)-(e(x))^2

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[hasty flame]

with this info, i did these problems, can someone verify if they are correct?

these need to be correct for the rest of the problems, i will add those after

[devout snow]

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[hasty flame]

<@&286206848099549185>

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@hasty flame Has your question been resolved?

[hasty flame]

<@&286206848099549185>

[lone mason]

yes

sry cant help im more of a alegebra guy not a geometrical one

[restive river]

u seem to have the solution

what do u need help with?

@hasty flame

[hasty flame]

checking my work

because the next part is confusing

[hasty flame]

i'm not sure how to do the rest of these

i think that 3 is correct but idk at all based on 4 5 and 6

[hasty flame]

<@&286206848099549185>

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[frail sun]

what does that last part mean? it makes no sense.

[frail sun]

says set z=0 then it magically finds the point???

[faint gorge]

shit is getting more advanced

[frail sun]

it just doesn't make sense at all

they magically get those numbers and call it a day when it's supposed to tell me how they did it

[magic wind]

IN the very top, you have the equations x+6y-z=2 and 2x-y+z=3. Plug in z=0 and solve the system of linear equations.

If you multiply the first equation with 2 and subtract the second equation, you get 13y = 1 => y = 1/13.

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[faint gorge]

I would write it into an augmented coefficient matrix

[dense lynx]

do you know when a system has infinitely many solutions?

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[umbral vortex]

I need help with this notation.

[umbral vortex]

I can figure out how to go from y(cx+d)=ax+b to x=(b-dy)/(cy-a) (edited)

But this notation is confunsing me, can someone help me understand?

[sly kraken]

not y= ❌ it should be x = (b-dy)/(cy-a)

[umbral vortex]

Yeah, yeah, sorry

[sly kraken]

can you elaborate about the confusing part?😅

[umbral vortex]

What does the arrow mean?

Sometimes he manipulate the left side and sometimes he manipulates right side of the equation

From context I can discern which is which, but for me it's unintuitive.

[lunar harbor]

this is just bad notation tbh

I have no idea what it means

[sly kraken]

yea

[umbral vortex]

Instead of doing the whole equation, it's written like it's one side at a time.

The equal sign at the start makes me think of y(cx+d) = (b-dy)/(cy-a) which is wrong

[sly kraken]

yup that is wrong... it wants to say x=... but notation is like that🤷‍♂️

[umbral vortex]

damn, I hope there's not more things like this in the book.

Thanks, guys

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[warm nymph]

Pls help

What’s the next step

[glossy dew]

hmm, you kind of messed up one of the terms, try rechecking

[warm nymph]

Ty

[feral agate]

Check the last term again

[warm nymph]

I got it

It was my dumbass friend

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[toxic estuary]

HI guys, i cant solve these two roots of polynomials questions.Please help,the last three pictures are my calculations for the questions.

[devout snow]

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@toxic estuary Has your question been resolved?

[toxic estuary]

No

[topaz umbra]

um

have u tried solving this by checking the discriminant and all?

[devout snow]

@toxic estuary Has your question been resolved?

[toxic estuary]

Not yet,i dont really understand how to check the discriminant

[devout snow]

@toxic estuary Has your question been resolved?

[cedar berry]

@toxic estuary bro the equation can be divided to (x^2-x-1)(x^2+2x+2)

in 5. (iv)

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[fair mantle]

equation of the plane that contains these two lines

[fair mantle]

I got that the lines are multiples of each other, what does that imply? Don't know how to proceed from there

Are they parallel?

<@&286206848099549185>

[proven sun]

depends

if the y-intercepts/constant terms are the same, and one slope/gradient is a multiple of another, then they are not parallel

[proven sun]

lines will be parallel if they have the same slope/gradient but different y-intercepts

oh

wait

I just realised

are we working with vector descriptions of lines?

[fair mantle]

yeah

[proven sun]

ah

my mistake

so if you have a vector which is a multiple of another vector, then yes, they are parallel

[fair mantle]

so how would i go about finding the equation of the plane?

[proven sun]

for two given vectors in a plane, I forgot how to find the equation of the plane

I will try to figure it out, but maybe ask for other helpers too

[devout snow]

@fair mantle Has your question been resolved?

[fair mantle]

<@&286206848099549185>

[stark wigeon]

nvm im stupid

[fair mantle]

that would give 0

[stark wigeon]

yea lol

you can treat the vectors not as arrows but as coordinates in the 3d space, and then find the plain that passes through those coordinates? but i havnt done linear algebra yet so im just guessing based on what i know

[fair mantle]

it's analytic geometry not linear algebra :p

[stark wigeon]

ah

well i havent done that either so ahaahaha

idk why actually cause we where supposed to do it this year but idk

by parallel vectors do you mean colinear or?

[fair mantle]

no, they don't have any points in common

they aren't colinear

wait

actually they might be colinear

hold on

[stark wigeon]

if one is a multiple of an another then it would be collinear and thus there are an infinite amount of distinct plains that passes through the vector

[fair mantle]

that's true

i think that might be it

[stark wigeon]

could be, if they are parallel and not colinear im not sure i can help with much more

but in the case that they are parallel and have no points in common that means that the 2 vectors are defined both by its head and tail so im not sure how you would work with that

[fair mantle]

yeah i think that's it. if it's not i will know soon anyways lol

thanks

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[stark wigeon]

yw:) have a good day

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[lone stirrup]

I'm trying to understand how the whole integer method works

[lone stirrup]

My book says that

is it just an expression in algebra or am i wrong?

[regal storm]

Basically when you multiply in rhs you'll get lhs

[lone stirrup]

I don't follow

[regal storm]

Wym?

[lone stirrup]

The book states the image above, but where do they get d and e from?

[regal storm]

d and e are constants

[lone stirrup]

but what is d and e?

[regal storm]

d x e = c

[vernal monolith]

Its not guaranteed that it can be factorized like that though for every real b and c

x²+x+1 cant be factored

[lone stirrup]

I'm just more curious on where d and e come from if we have b and c starting variables

[vernal monolith]

And has discriminant lesser than zero

[vernal monolith]

d and e are factors of c

Not any factor though, they are factors of c such that d+e = b
@lone stirrup

[lone stirrup]

I see so it's basically an expression

Some sort of rule?