help-42

[tacit linden]

whats steps 1 and 2

[calm coral]

@real coral Has your question been resolved?

[dusty torrent]

Basically you need to multiply both sides by the logarithm with the same base as expoent

[magic otter]

What exactly is being substituted back into the expression in step 4.?

[magic otter]

What is this saying? That x=b at the end?

[calm coral]

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[real coral]

this was the problem. So, here, I could easily get 10^b = 2 from the given. then replacing 2 with 10^b. then I can multiply b with the log. Then, I put the b on top of x. there I got stuck

[real coral]

Thank you Piru. You saved a lot of my time 😄 ❤️

[tacit linden]

what were step 1 and 2 in the original image you sent

[lean hemlock]

D?

[tacit linden]

its probably not really a good idea to use chatgpt

usually

[cerulean cradle]

Its d clearly

[real coral]

Yes, I often don’t find any relevant proof or explanation from it. They just somehow get to the answer. And says, this is it😅

[cerulean cradle]

!status

[calm coral]

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

[cerulean cradle]

@real coral what are you up to ?

log_10(2) = b
or 2 = 10^b----(i)
now 2^(log_10(x))
= 10^b(log_10(x)) using 2's value from (i)
= 10^(log_10(x^b))
= x^b

[real coral]

My major is Psychology. But in my high school, I had problems with log and calculus. So got some free time, and brushing them up.

Just for self satisfaction 😅

[cerulean cradle]

i hope this helps

[real coral]

Definitely,, thank you so much.. ❤️❤️

[steel yoke]

can u show why 10^log_10(1) = 1

[brittle isle]

well it sort of follows from what a logarithm actually is

log_10(a) answers the question "what do i have to raise 10 to to get a?"

[tidal grotto]

inverse function moment

:3

[brittle isle]

then 10^log_10(a) is doing exactly that, raising 10 to that thing, whatever it is

so you'd expect it to be a

you could also realize that log_10(1) is 0

so you just have 10^0 = 1

[steel yoke]

thankss

[calm coral]

@real coral Has your question been resolved?

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[cerulean cradle]

Therfore third root is p

[winter elbow]

yeah, that’s the key

[brittle isle]

am i stupid, i'm looking at this and not seeing why it's p

[winter elbow]

Vieta

[cerulean cradle]

ita the vietas formulae

[winter elbow]

$\alpha$ + $\beta$ + $\gamma$

[brittle isle]

i must've learned the formula at some point but we might've called it something else

the roots add up to something?

[cerulean cradle]

Yeah those with sum and product of roots

[potent lotus]

Sufferrrrring rn D:

[brittle isle]

oh damn

i knew this for quadratics but didn't realize it generalized

[cerulean cradle]

Its true for n degree

Just combinations

[smoky bone]

explain me guys

[winter elbow]

$\alpha$ + $\beta$ + $\gamma$

[potent lotus]

Sufferrrrring rn D:

[winter elbow]

Do you know how to find its value?

alpha beta and gamma stands for the three roots of the function

[smoky bone]

i know for a quadratic, alpha plus beta equals -b/a of the equation -> ax^2 +bx +c = 0

[tidal grotto]

$x^{3}-px+qx-r=(x-\alpha)(x-\beta)(x-\gamma)$

[potent lotus]

🫎 Moosey 🫎

[tidal grotto]

expand RHS and compare coefficients

[smoky bone]

correction -> px^2

[smoky bone]

$x^{3} - (\alpha+\beta+\gamma)x^{2} -\alpha\beta\gamma$

[potent lotus]

silentsolo7269

[smoky bone]

q must be equal to 0?

[tidal grotto]

do you see why you get gamma=p from this?

[smoky bone]

no

[tidal grotto]

look at coefficient of x^2

[tidal grotto]

here

[smoky bone]

ahh i see now

coz alpha plus beta equals 0 i gotcha

so condition is that p is the third root and q is equal to zero, right? @tidal grotto

[tidal grotto]

there's definitely some terms and such you forgot in this expansion

but the x^2 coefficients are right

i'm assuming you just simplified down

[smoky bone]

yeah i forgot the (betaXgamma+alphaXgamma)x term

[tidal grotto]

theres' another x term

alphabetax

[smoky bone]

yeah yeah yeah

okay thanks buddy!

imma close and ask another ques now

.close

[calm coral]

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[smoky bone]

.reopen

[calm coral]

✅

[smoky bone]

@tidal grotto buddy where did that rhs came from?

[tidal grotto]

every cubic has 3 roots

@smoky bone

[smoky bone]

ok..

[tidal grotto]

so every cubic can be factored like this (x-alpha)(x-beta)(x-gamma)

the reason theres no constant in front is because it matches the coefficient of the leading power in the cubic, (1)x^3

[smoky bone]

ahh gotcha thanks

.close

[calm coral]

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[smoky bone]

prove that the relation congruence modulo n is an equivalence relation in the set of integers.

[calm coral]

Please don't occupy multiple help channels.

[pseudo wedge]

!status

[calm coral]

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

[naive meadow]

Please don't open multiple help channels

[pseudo wedge]

Prove reflexivity, symmetry and transitivity

[smoky bone]

ok

[smoky bone]

of?

[pseudo wedge]

The relation

Your relation is given by $$aRb\iff a\equiv b\text{(mod n)}$$

[potent lotus]

kheerii

[naive meadow]

Please close this or your other one

[tidal grotto]

what is the definition of an equivalence relation, what properties must it satisfy?

[smoky bone]

must be reflexive, symmetric and transitive

[tidal grotto]

what does reflexive mean? symmetric? transitive?

you must prove these properties hold for congruence modulo n

[tidal grotto]

using this

[smoky bone]

Reflexive
a R a
as a=a

symmetric
a,b R b,a
as f(a,b) = f(b,a)

transitive
a R b and b R c => a R b
as f(a) = f(b) = f(c)

i suppose

@tidal grotto

[tidal grotto]

perhaps it would be more illuminating if you used the other notation for congruence modulo n

$a\equiv b\text{(mod n)} \iff \exists k$ s.t. $a=nk+b$

[potent lotus]

🫎 Moosey 🫎

[smoky bone]

yeah this one seems bit easier to prove. thanks!

.close

[calm coral]

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@remote mural Has your question been resolved?

[grim bolt]

AE is a straight line because

angle EBA is 45, angle BEA is 45 and angle BAE is 90, meaning BAE is a right angles triangle and the sum of all interior angles is 180

i mean you have marked it as 45

hmm, but how else how you make 45 degrees if ae was not a straight line

[distant flare]

[grim bolt]

just give me a moment

@remote mural

let us assume ADE was not a straight line

Ignore the line BD please, i drew it by mistake

doesnt matter

[distant flare]

We aren’t dealing with curved lines here anyway

[grim bolt]

no problem

Actually @remote mural , there is a much simpler proof for this

no need to draw figures also

assume ADE is not straight

so ADEB will be a quardilateral

for a quadrilateral, sum of interior angles is 360

A+D+E+B=360

A=90, B=45, E=45

thus we get, D=180

remember, a straight line is 180 degrees, this contradicts our assumption that ADE is not straight

thus ADE is a straight line

ignore that diagram

i made an error in it

I just extended AD to touch the line BE at point E'

but since both angles E' and E are 45 degrees, this is not possible that they are different points

so E and E' are same points

so ADE is a straight line

[distant flare]

Hey there

If he’s not here I can help

What’s the issue

@remote mural

Hmm

Original question?

Let me see the original diagram

What’s the question here

Find the angles?

I see

Yea ur diagram is rly confusing

Do u have an unlabeled one

Ok better

Ok so

It is a straight line

Simply put it can only be a straight line because sum of angles ends up to be 180

So right angle triangle

Yk sum of angles in a right angle triangle is 180

Yes?

R u familiar

Here

Wait

Agree?

It has to be straight in order to be 45 degree

If it wasn’t straight

Then wouldn’t add up to be 180

U get what I mean

Then it wouldn’t be a right angle triangle

But it’s 90 degrees as shown

Actually no quadrilaterals need it to be straight too

Sum of angles in right angle triangle is 180

That actually says everything

?

Wait

This angle you labelled 45

If ur not assuming you can say

Sum of angles is 180 meaning it’s a right angle

Hm ok

Honestly I need original image

Ok

Yes

One sec

Thinking of a better way to tout it

Put

[blissful field]

You want to know why AE is a straight line?

Well

Its a right triangle

A triangle

A shape drawn with 3 straight lines between 3 points

In euclidian geometry)

We dont question the straightness of lines in euclidian geometry (the geometry you learn) bc all lines that look straight is straight and all lines that doesnt look straight isnt straight

It would still be a straight line

Angles dont change the straightness of straight lines

[blissful field]

1 degree off of what? Angles right?

You’re asking what would happen if we decrease the angles

Its not. It needs no proof. As i said, in the geometry we learn, a straight line is a line that looks straight. Since DE looks straight then its straight. No exam will tell you to prove that DE is straight or not

[distant flare]

[blissful field]

Again, its still a triangle and a triangle is 3 straight lines conncected and thus DE is straight

In fact, for us to apply triangle rules on that triangle DE NEEDS to be straight

[distant flare]

[blissful field]

If you want to find the area of smth thats bent you dont use euclid geometry you use calculus

Which you’ll learn

Euclid geometry only works when all lines are straight anyways.

@remote mural

[blissful field]

(Discluding circles)

[distant flare]

I think he dipped

[blissful field]

Ok another question for you, imagine a circle, done? Ok. Prove that circle is bent not straight

Well you dont need to bc its obvious

Same thing with triangles

The lengths of triangles are always straight

Quadrilateral??? Is that a shape?

Holup

[distant flare]

A quadrilateral is defined as a two-dimensional shape with four sides, four vertices, and four angles.

[blissful field]

Ooooh

I get what you mean

Ok its not a quadrilateral

Why? The line isnt bent. How? Thats the way it is.

Bc theres no trick in straightness of a line

If it was quarersersdtal then it would’ve informed you

Math is solving problems not figuring out if a line is bent or not

Infact

[distant flare]

It did say

[blissful field]

I doubt theres any proof for that at all

[distant flare]

Quadrilateral

[blissful field]

Abde?

Adeb is a triangle….

Uhm

Dice

Show us the rest of the question

[blissful field]

He gave you the name of the quadrilateral

CEDB is a quadrilateral tgo

The way i was taught was “if you look at a triangle and it looks like a triangle then its a triangle

There is no proof in why ADEB is a triangle

Its 3 lines connec

[distant flare]

Don’t understand why u need to prove anything

Is a triangle

No test will ask
This

[blissful field]

If we flip bcd then it’ll still be a triangle

You just flipped it

Name of the shape?

What shape are you talking about

Give the name

[distant flare]

Where did u come up with this question

[blissful field]

Well the shape DBCE is indeee a quadrilateral

Its a triangle. No way its anything other than than Even if you flip BED then ABED would still be a triangle

Bc its 3 straight lines connected

How are they straight? They look straight

If they werent straight then it wouldnt even be a quadrilateral

A quadrilateral is 4 straight lines connected

[blissful field]

The person that drew that made an error (they mentioned it) so their error is How they drew DE… DE is a straight line

And thus

AbED is a triangle

This what ABED would look like

You gotta wait until the pic loads

Bc my wifi is slow

But it’d show u a trianfle

[distant flare]

Thx car 🙏

[blissful field]

@remote mural

[blissful field]

This is ABDE btw

Wait

[distant flare]

Yuh

[blissful field]

Wait a sec

A pic is loading

.

[blissful field]

@remote mural this is what ABED looks like when you flip the triangle

This image is incorrect

[distant flare]

Stop referring to incorrect image

It just make u confused

🙏

[blissful field]

DE is drawn in the same direction as AD

There is no DE goes upward

DE goes where Ad goes

[distant flare]

WHICH IS STRAIGHT

HORIZONTAL

A——D———E

[calm coral]

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[calm coral]

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[stark cloak]

Can someone explain the difference between subpolynomial, o(n) and polylogarithmic O(log(n) ^c) for c a constant?

[stark cloak]

Apparently theyre not the same but I thought they were

[alpine stone]

Well, why did you think so?

[stark cloak]

Because on wikipedia theyre listed seperately

If you go to wikipedia Timecomplexity

[alpine stone]

I don't see a valid reason to believe o(n) = O(log(n)^c) from that

Anyway, consider (n + log(n))/2

[calm coral]

@stark cloak Has your question been resolved?

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[real flax]

hello, i have this one geometry problem that i cant seem to find a solution to

so i have AB and A'B' two segments place wherever, with M the middle point of AB and M' the middle point of A'B'. I am asked to proove that the middle of AA', the middle of BB' and the middle of MM' are collinear

[real flax]

i did manage to solve it using coordinates but it is a problem from a 9th grade book i found, and it actually requires me to do it in another way

ive been trying to find some lines to use the Thales theorem but im just so lost

this is what i got stucked on

(i extended AB and A'B' to meet in a point P)

from there i tried to proove that A'' and M'' and on the same line as M'' and B''

oh yea i noted B'' as the middle point of BB' A'' of AA' and M'' of MM'

[calm coral]

@real flax Has your question been resolved?

[eternal stirrup]

I’m pretty sure you can do this “coordinate” free with linear algebra

If you’re allowed to do it that way

Had a similar exercise once

[real flax]

wait what do you mean linear algebra? doesnt that mean calculus and stuff? or am i being dumb

[eternal stirrup]

Oh well not really in this case. You can think of the points as vectors from some origin O, which we don’t really care about

And this makes some geometry problems a lot easier to solve. But this is assuming you’re allowed to do this

[real flax]

oh vectors never been my strong part xd

[eternal stirrup]

Okay, just ignore my suggestion then. I guess it’s better to do it as intended

[eternal stirrup]

How would you generally show three points are colinear?

[real flax]

well i would proove that all 3 are on the same line... i know this is literally the definition but i dont know how else to say it ::sku

💀

[real flax]

i think the vector method is the way to solve it

i mean the intended method

i have not worked with vectors in like 3 years so thats why im a little gray on the solution

[eternal stirrup]

Well that’s the thing, I’d say your first way might be perfectly legitimate unless you know they must of done it coordinate free.
So if you know general of showing three points are collinear without coordinates you’ll know it’s possible

[real flax]

i do know because i was told that it is not the intended way

[eternal stirrup]

Again, did they specify how? It still might be legitimate to use coordinates

Like e.g. showing the slope is the same

And it helps immensely since we know how simple a mid point is expressed with coordinates too

[real flax]

the problem is in the vector chapter of the book, and like a few pages back there was the Thales theorem ]

[eternal stirrup]

Oh I see so you’re actually allowed to use vectors?

[real flax]

yes

wait wrong reply

yes.

i think im** supposed **to use vectors not just allowed

[eternal stirrup]

Okay right that changes a few things. So in your textbook do they go over how vectors can be parallel and moreover how to adapt these to showing when three points are collinear?

[real flax]

i cant find anything specific about adpating parallel vectors to show that three points are collinear but i remember seing something about pallaleism for vectors

[eternal stirrup]

Okay, how comfortable are you with vectors in general? Do you know how to, say, interpret one from just two points?

[real flax]

you mean vectorial summing?

i did that back when i was in hs shis 😅 , but i do remember some stuff so i guess im relatively comfortable

[eternal stirrup]

Okay, so to get you started. Say we have three given points A, B and C

To represent the segment AB as a vector we can take the end point minus the start point, so B - A would be one such vector. But any scalar multiple would be correct too—and so they’re parallel.

So what you want to show is that say (not the same letters)

AB and AC are parallel.

I.e

B-A = t(C-A) for some constant t

(Draw this!)

[real flax]

gimme 5 💀

[eternal stirrup]

Yes take your time

[real flax]

wait

how can ab and ac be parallel if A its a common point?

[eternal stirrup]

I’m not using the same points as you do if that’s what you’re asking within?

Just in general, if they are parallel then the equality above holds for some constant t

[real flax]

wait youre actually cooking

[eternal stirrup]

It’s a lot clearer if you attempt to draw a picture

So in your case it’s those mid points that you mentioned you want to use instead, but these you you can describe in terms of A, B, A’ and B’.

So for example M = (A+B)/2

[real flax]

Am i dumb?

[eternal stirrup]

,rotate

[real flax]

Im not sure how else to place C so AB and AC are parallel

[eternal stirrup]

The letters don’t matter here, you can switch place between A and B and it would still be true

[real flax]

yes

oh

[eternal stirrup]

As long as C lie on the same line that AB spans you’re fine

[real flax]

how do i proove that theyre parallel tho M13_Think

[real flax]

ye

[eternal stirrup]

Well, that’s what I tried to describe above

Read it carefully

It’s not supposed to be easy, take your time it’s worth digesting it slowly; that means you’re learning

[real flax]

im not sure what you meant by B-A, are we subtracting points or point's coordinates

[eternal stirrup]

And I encourage questions

[real flax]

i mean the coordinates of the points*

[eternal stirrup]

Oh yeah might be unclear, yea so coordinate wise subtraction, but I sense that you don’t even need to bother expressing it explicitly

You’ll probably find that it’s unnecessary

So in some sense it’s actually not coordinate free

[real flax]

yea thats why i was a bit confused

[eternal stirrup]

But in some sense it is, since you don’t even need to know anything about them

So just try to keep them as is, that is B-A, treat it as a point

[eternal stirrup]

And I would use this suggestion

[real flax]

Basically this?

[eternal stirrup]

Not quite, you don’t really need to express them explicitly

Take A, B and C as given already

It’s not necessary, since you’ll come to the same conclusion anyways

The powerful part here is you can express your mid points in terms of the given points, you will eventually see that they must be parallel if you just trust the process

[eternal stirrup]

Try this suggestion!

[real flax]

oh tahts what you mean by algebra

[eternal stirrup]

What you’ll notice is that this is infact almost the same as doing it by coordinates; but in a much more streamline way

[eternal stirrup]

Yeah!

[real flax]

wait

hold up

does this mean

Or did i just write something stupid

[eternal stirrup]

It makes perfect sense!

[real flax]

Did i just solve it 💀

[eternal stirrup]

Well I think you were supposed to show the three midpoints were collinear; unless that’s just a reduced way of doing it? Maybe I’m missing something

Oh wait yeah that must mean they’re collinear lmao

[real flax]

If the middle point of a

Yea

[eternal stirrup]

Oh yeah, you’re done!

[real flax]

Anticlimactic ending 💀

[eternal stirrup]

Lmao

Good work

[real flax]

Gimme 5 to consult with this one 9th grader and see if this is how its meant to be done

[eternal stirrup]

Pretty short and neat right?

[real flax]

If its not i give up

[eternal stirrup]

Oh damn

[real flax]

Suspiciously short

Welp he cba

I think its the intended way thank you so much again!

[eternal stirrup]

Ur welcome!

[real flax]

.close

[calm coral]

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[calm coral]

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[potent lotus]

l'agit

[warm warren]

What have you tried

What’s the definition of the expectation

You just need $ dollar signs around

The text

[potent lotus]

l'agit

[warm warren]

The last 2 are the same thing

Do you know about $\mathbb E[X] = \sum_{n\geq 0} nP{X=n}$

[potent lotus]

Frosst

[warm warren]

If you sum over all the things in the sample space

That’s the same as summing over the all the outcomes of X

Because $X^{-1}(\mathbb N_0) = \Omega \cup {}$

[potent lotus]

Frosst

[warm warren]

If you sum over the support you’ll count everything in the sample space once

Let’s say $\Omega = {A, B, C}$

[potent lotus]

Frosst

[warm warren]

And $X(A) = 2$, $X(B) = 5$, $X(C) = 2$

[potent lotus]

Frosst

[warm warren]

This is saying go through each event

[warm warren]

This is saying go through each integer

And use the measure P on the preimage

Let’s say A B C were all 1/3 chance for simplicity

[warm warren]

Then by the expectation is $1/3 * 2 + 1/3 * 5 + 1/3 * 2$

[potent lotus]

Frosst

[warm warren]

If we look at this instead

This says

What’s the preimage of 0? (Under X)

Well it’s empty, X maps nothing to 0

P({}) = 0 by construction of measures

Ok next

What’s the preimage of 1? (Under X)

Well it’s empty, X maps nothing to 1

Ok next

What’s the preimage of 2? (Under X)

Well…?

What does X map to 2?

[warm warren]

Look here

Yes

So the preimage of 2 under X is the set {A, C}

Then you now find 2 * P({A, C})

Well you know P is additive so this is just P(A) + P(C) = 2/3

So the nP(X = n) for n = 2 term is 4/3

What about n = 3?

What’s 3P(X = 3)

[calm coral]

@red wigeon Has your question been resolved?

[wheat yacht]

.help

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Commands:

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• help: .help
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Type .help <command name> for more info on a command.

[wheat yacht]

Someone help me with this

Hii

Can I send you the image.

Aa

Where is the

Helpers

[calm coral]

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@red wigeon Has your question been resolved?

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[blazing coyote]

To prove:- That the addition of subspaces is commutative for two subspaces, say U, W

[potent igloo]

How is U+W defined?

[potent igloo]

Oh ok. Not too hard to prove then

[blazing coyote]

I know

but I wanted to check if the way I write proofs is fine

Let $s\in U$ and $t \in W$ be arbitrary elements of their respective subspaces

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

then $s+t \in U+W$ by definition

[potent lotus]

ƒ(Why am. I here)=I don't know

[potent igloo]

You can shorten this to "Choose any $s\in U" and "t\in W$"

[potent lotus]

SWR

[blazing coyote]

But we already know addition is commutative

so $t+s=s+t$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

and $t+s \in W+U$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

but $t+s = s+t$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

thus it follows that $s+t \in W+U$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

$\implies W+U=U=W$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

Thus the addition of vector sub-spaces is commutative

QED

[pallid halo]

no, that only implies $U+W \subseteq W+U$

[potent lotus]

Bungo

[blazing coyote]

ah

so I reverse this in the opposite direction

[potent igloo]

Or just say "we can prove the opposite by the same strategy"

[pallid halo]

yep, just reverse the roles of U and W, and of s and t, and the same proof works

[blazing coyote]

oo

[tidal grotto]

you can say WLOG at the beginning or smth along those lines

[blazing coyote]

thanks so much everyone!

.close

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[tame plume]

could someone help me set this question up? I'm not really sure where to start

[rustic osprey]

I’d split the circle into two parts: above the x axis and below the x axis

Can you find the equation of each of these curves (with y isolated)?

[tame plume]

would it be y = 16 - x^2

would it be y = sqrt(16 -x^2)

for one

and then for the other one

y = -sqrt(16 - x^2)

wait

i think i can do it now

do I now just evaluate both of those integrals from -4 to 4

i only really have to do one right

and then multiply it by 2

[rustic osprey]

Yeah

[tame plume]

thanks

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[waxen mica]

need help finding the maximum ive found the minimum but wanna find othe range of values for it

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[calm coral]

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

[uncut island]

dude i answered this question before

whats your answer

there are 2 red 7s

so the first one is 2/52

[gritty grove]

it means that once you get the card

it doesnt go back into the deck

[potent igloo]

@remote mural, please stop spamming helpers ping. You've been told multiples times.

[uncut island]

I didnt correct you for the second one because it was correct

yes

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[median cloak]

Hello

[median cloak]

Bonjour

Given f(x) is a 3 degree of polynomial

I see the problem

Nevermind

.close

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[harsh night]

A square of area 1848 is inscribed in a square of area 2016. Find the side lengths of the triangles.

[harsh night]

hint pls

[strange lichen]

draw a picture

label the side lengths

u can figure out the rest

[harsh night]

?

[dull wagon]

have you drawn a diagram

[calm coral]

@harsh night Has your question been resolved?

[harsh night]

yes

[dull wagon]

show your diagram

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[hearty coyote]

Pls help

[blazing coyote]

[potent igloo]

You need to ask a question to recieve help.

[exotic falcon]

Sometimes just my mere presence is enough

[calm coral]

@hearty coyote Has your question been resolved?

[hearty coyote]

Naur ur too slow

My exam was today

[tacit linden]

What

[calm coral]

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[blazing coyote]

The total number of books in a library is $2n+1$ , one is allowed to borrow minmum one book, and at most n books. If the total number of selctiosn is 63, find

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

so I as thinking

$c(2n+,1)+c(2n+,2).....c(2n+1,n)= 2^{2n}-1$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

is that right

so $2^{2n} =64$

or n=3

[potent lotus]

ƒ(Why am. I here)=I don't know

[pallid halo]

i agree with this answer

there's a nice shortcut that i can share if you are interested

[blazing coyote]

please do

[pallid halo]

ok, i'll illustrate with n=2 and then generalize

if n=2 then 2n+1 = 5, so there are 5 books

before we apply the restrictions from the problem statement, a person could borrow 0,1,2,3,4,5

and because of symmetry, the number of ways to borrow 0,1,2 is equal to the number of ways to borrow 3,4,5

so:

# of ways to borrow 0,1,2 = (# of ways to borrow 0,1,2,3,4,5) / 2

[blazing coyote]

yes

[pallid halo]

sorry formatting

but # of ways to borrow 0,1,2,3,4,5 is 2^5

so dividing by 2 we get 16 ways to borrow 0,1,2

now 0 is not allowed

so we have to subtract 1

leaving us with 15 ways

now generalizing:

there are 2^(2n+1) ways to borrow 0,1,2,...,2n+1 books

so there are half as many ways to borrow 0 through n

i.e. there are 2^(2n) ways

now subtract 1 since 0 is disallowed

leaving 2^(2n) - 1

set that equal to 63

and you get the same answer as yours

[blazing coyote]

yeah, got it

thanks!

[pallid halo]

sure

cheers

[blazing coyote]

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[native raptor]

im not sure how to do this have been thinking abt it for a while

[native raptor]

honestly im not sure what do u get finding the area of a distance time graaph

graph

i did it as 50t=70-80t but im not sure how im supposed to use my graph for that just kinda thought of it aft solving i got 32 mins so it wld be 12.32 but ans is 33

12.33

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[shy iron]

does anyone know for the parts that say M1 if i have to do that

[shy iron]

pile if i wrote p(x greater than or equal to 8) then put it into my cal

would i still get full marks?

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@shy iron Has your question been resolved?

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[blazing coyote]

if $\frac{1+3p}{3};\frac{1-p}{4} ; \frac{1-2p}{2}$ are the probabilities of mutually exclusive events find the range of $p$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

options:-

between 1/3 and 1/2

between 1/3 and 2/3

between 1/4 and 1/2

between 1/3 and 2/5

[hollow totem]

what have you tried

[blazing coyote]

wait a minute

formatting it

I have two ideas

one

the sum of the 3 probabilities will be between zero and 1

and simultaneously the each indiviual proabbility will lie between zero and 1

[hollow totem]

yes

the most efficient way to do this

each probability must be positive, and the sum must be less than 1

note that this implies the other conditions (sum is positive and each individual probability is less than 1)

[hollow totem]

so check the values that satisfy every one of these

[blazing coyote]

so I have

$\ 0\le\frac{\left(1+3p\right)}{3}+\frac{\left(1-p\right)}{4}+\frac{\left(1-2p\right)}{2}\le1$

[potent lotus]

ƒ(Why am. I here)=I don't know

[hollow totem]

yeah

as I mentioned though, you don't need the 0 as you're gonna test the 3 individual terms for positivity anyways

[blazing coyote]

oh boy

so I have to solve each inequality seprately

aand then find the intersection space with this?

[hollow totem]

yeah

[blazing coyote]

I think I'll do it later then

sorry

[hollow totem]

fine by me lol

[blazing coyote]

.close

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[hollow totem]

as long as you know how

[blazing coyote]

I know how

yeah

I've done something very similar before

[calm coral]

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[swift tide]

So first find the roots of the characteristic equation

I got 6 and -3

$y_{h} = Ae^{6t} + Be^{-3t}$

But since $\varphi (t) = 6e^{6t}$ is similar form to $Ae^{6t}$, our particular solution should be $y_{p} = kte^{6t}$?

[potent lotus]

potatology

[swift tide]

yes

[versed silo]

yeah that's right

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[void frost]

\begin{problem} Describe the geometric position of the lines with the equation [g_s: \vec x = \mrm{2 \ 1 \ 3} + t \cdot \mrm{3s \ -4s \ 1}.] \end{problem}

[void frost]

They have the common intersection (2, 1, 3)

Anything else..?

[ancient helm]

is there more context to the question?

[blazing coyote]

find the direction cosines of the line perhaps ?

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[vocal island]

Hi, I'm having problems with this limit (senx = sinx)

[vocal island]

I have already used l'hopital 2 times, getting a huge expression, but I don't know if it's the correct way

I can't use Taylor yet, so I don't know how to proceed

[topaz raft]

just wondering if senx is sinx how come the spelling for the numerator also isnt senx lol

[vocal island]

This is the graph and this is when x->0+ which is 0.166667 = 3/5

[vocal island]

Teachers wrote that, it's the same

This is the first use of lhopital

This is the second use of lhopital

[south thorn]

what is sen

[vocal island]

Sin(x)

[south thorn]

oh mb

[vocal island]

And this simplified becomes this

Which is still a 0/0

[hollow lion]

Maybe try splitting the fraction, then distributing the exponent in denominator and simplifying each fraction to get rid of the x in denom

[vocal island]

At the first use of l'hopital or where?

[hollow lion]

Before l’hop

[vocal island]

Okay, wait me a second

[south thorn]

idk it might help but maybe try changing it? idk:

[vocal island]

Let's try with this

[south thorn]

ok xd

[hollow lion]

Now manipulate them so they are easy to differentiate

i.e. get rid of products of x terms

[vocal island]

Like this?

[hollow lion]

I was thinking
[ \frac{x^{-1/2}}{\sin (x)^{3/2}} \dots ]

[south thorn]

maybe: $\frac{1}{\sqrt{x sin(x)}sin(x)}$-$\frac{1}{\sqrt{x sin(x)}x}$

[potent lotus]

clonesolopros

shsgd

[south thorn]

same thing xd

[vocal island]

How can I get to that expressión?

[south thorn]

$\frac{\sqrt{x sin(x)}}{x sin^2(x)}$-$\frac{\sqrt{x sin(x)}}{x^2 sin(x)}$

[hollow lion]

x^1 / x^3/2 = x^(1-3/2)

[potent lotus]

clonesolopros

[hollow lion]

[ \lim_{x\to 0^+} \frac {x^{-1/2}}{\sin (x)^{3/2}} - \frac{\sin (x)^{-1/2}}{x^{3/2}} ]

[south thorn]

$\implies \frac{\sqrt{xsin(x)}}{xsin(x)}(\frac{1}{sin(x)}-\frac{1}{x})$

[potent lotus]

clonesolopros

shsgd

[hollow lion]

That seems easier to l’hop

[south thorn]

$=\frac{1}{\sqrt{xsin(x)}}(\frac{1}{sin(x)}-\frac{1}{x})$

[potent lotus]

clonesolopros

[south thorn]

and use L'H?

[vocal island]

So which one should I try?

Also, with your permission, I tag the helpers to have more options

<@&286206848099549185>

[south thorn]

yeah np

tbf, mine only has 1 square root

[hollow lion]

Whichever you find is simplest, but I reckon there is a more elegant solution available

[south thorn]

like which one

[hollow lion]

Not sure, just have a feeling l’hop isn’t necessary

[vocal island]

I tried wolfram alpha, could this be useful?

.close

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[lime reef]

in last equation, dt is negative. can I simply take the minus sign out of integration?

[lime reef]

(also I didn't show, but limits of integration become 1 to -1 after substitution)

[hollow lion]

Yes

[south thorn]

what is t?

[lime reef]

💀.

[south thorn]

I cant see it :/

[clear delta]

t = cosø

[south thorn]

isnt this just integrating: $sin^3(\theta)$

[potent lotus]

clonesolopros

[lime reef]

I saw someone change limits to remove negative sign. is changing limits necessary? can't my answer be simply negative

without changing limits

[hollow lion]

[ -\int_a^b \dd x = \int_b^a \dd x ]

[potent lotus]

shsgd

[lime reef]

.close

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[chrome plume]

2.70 how to do it

My book shows the answear is e^3/2

Here is my attempt

[calm coral]

@chrome plume Has your question been resolved?

[naive meadow]

,rotate 90

[calm coral]

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[sharp wasp]

The weight of a component depends on the raw material used for its construction:

The weight of Type A components is uniformly distributed between 2.0 and 3.0 kilograms.
The weight of Type B components is uniformly distributed between 2.3 and 3.5 kilograms.
The weight of Type C components is uniformly distributed between 2.9 and 4.0 kilograms.
In the warehouse of a factory, there are many such components stored. The percentages of the components are:

Percentage of Type A components: 35%
Percentage of Type B components: 20%
Percentage of Type C components: 45%
Let B be the weight of a randomly selected component.

Α.Determine the Probability Density Function (PDF) of the Weight B.
Β.Determine the mean and variance of the weight of a randomly selected component.
C.If 100 such components are placed in a box, what is the probability that the total weight of the box exceeds 305 kilograms?

[sharp wasp]

This is for A, but i dont know how to find the mean and variance

[narrow dome]

The mean is the Integral from -inf to inf of x * f(x) dx

And the variance is M[X^2] - M[X]^2

[sharp wasp]

so the integral from 2 to 4?

or i add all the areas

[narrow dome]

Yeah, it's the same as you did there, but you have a x next to every constant

[sharp wasp]

so the same but i add a x for each?

[narrow dome]

This should be better

And the variance is

[sharp wasp]

i got it thank you

and then i just put instead of x x^2

[narrow dome]

Yes

There is also this formula

m is M[X]

[sharp wasp]

good

it was simplier than i thought thank you

[narrow dome]

This one is much easier to use

[narrow dome]

Np

[sharp wasp]

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[eager beacon]

Would this be convergent or divergent? or conditionally so

[vocal summit]

converge. it obviously converges conditionally because [\lim_{n\to \infty}\frac{n^2}{e^n}=0], but you can use the ratio test to test the absoluteness of its convergence (it probably does).

[potent lotus]

fish

[eager beacon]

ah well I see why

thank you

[vocal summit]

and yeah it does converge absolutely

[eager beacon]

I figured for all values of e^n

[upper sparrow]

"at least conditionally" - at least the definition of conditional convergence is converging, but not absolutely

[eager beacon]

.close

[calm coral]

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[calm coral]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[remote mural]

HALP

[remote mural]

ok anyone online?

i'm thinking its non collinear

what about this

[potent igloo]

We can't help you with tests or quizzes

[remote mural]

i think conditional

oh ok

[keen pike]

Is this for a test?

[remote mural]

yuss:> online course one

[keen pike]

alr gl

[remote mural]

nw then thank u 🙂

.close

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[calm coral]

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[ebon gull]

How do I get it via the other element

I did vertical but horizontal isn't making sense

[calm coral]

@ebon gull Has your question been resolved?

[calm coral]

@ebon gull Has your question been resolved?

[calm coral]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[vital rose]

Hi, the question is:

A rescue ship is working with a rescue helicopter to save some boaters in distress. They use a lighthouse to determine their relationship between each other and the stranded boaters. The angle of elevation between the lighthouse and the helicopter is 15 degrees, and the helicopter is flying at an altitude of 150 metres. The angle of depression between the lighthouse and the rescue boat is 20 degrees. The angle of depression between the lighthouse and the distressed boat is 8 degrees. The lighthouse is 15 metres tall.

How far apart are the vessels?
What are the angles between the helicopter and the rescue boat with the distressed boat?

[vital rose]

I've been working on this for about an hour and I spend some time on it in school; it's a problem that uses trigonometry. However, my answers are a bit different for question 2) compared to my teacher's.

[vital rose]

Here's a rough sketch I drew like 15 min. ago:

https://cdn.discordapp.com/attachments/1239758296538157106/1251000291126542416/IMG_1113.jpg?ex=666da50f&is=666c538f&hm=179ea97d669cc3949c3b7c5efb523fd83ace168b2e0fcd2aaaae1176e23adf96&

[distant flare]

Hey there

[vital rose]

It's not to scale but my answer for question a) matches the answers; however, for question b) it doesn't!

[vital rose]

Hii!

[distant flare]

So question a is correct

[vital rose]

Yes!

I also got 65.52 metres.

Whereas for question b) they got:

Angle at the rescue boat: 126 degrees
Angle at the distressed boat: 48 degrees
Angle at the helicopter: 6 degrees

[vital rose]

See, I really don't see how it's 48 degrees for the distressed boat because the helicopter is like 500 metres right of the lighthouse and the distressed boat is only 100 metres.

[distant flare]

👌 let me check it out

[vital rose]

Therefore, the angles at the distressed boat and the rescue boat should be greater than 90 degrees, shouldn't it? Unless, I'm interpreting it wrong...idk.

[vital rose]

Thanks. For this, I used trigonometric functions and cosine laws.

[distant flare]

Let me see

Ok so

[vital rose]

Yes, finally!

[distant flare]

One sec

Double checking

What’s your value for b?

[vital rose]

Question b?

[distant flare]

Ye

[vital rose]

I have no idea

I dont even properly understand the question tbh!

[vital rose]

In this diagram

For B

We're finding the angles here

[distant flare]

Small tip just label which is which so it’s eaisrr also

If not it gets quite confusing

[vital rose]

Oh I did but I write lightly

And end up erasing it

[distant flare]

Ah okok

[vital rose]

Because I made the diagram too small so I couldnt fit a lot

But I drew like a GIANT one at school and it's still there 😭

[distant flare]

😂

[vital rose]

What did you get for question b?

Im not even sure I'm doing it correctly

[distant flare]

Same answer

[vital rose]

But I feel like my angle values are right

[vital rose]

SAME AS ME?!

Or same as the question's answers??

[distant flare]

Questions answers

[vital rose]

Oh-

What did you do?

[distant flare]

Ur diagram is incorrect

[vital rose]

Do you know where I messed up?

[distant flare]

Ok so to go back first question

You used 8 degree and sin yes?

[vital rose]

8 degrees and 20 degrees

[distant flare]

Yes

[vital rose]

I used toa

To solve for the adjacent angles

i mean sides

[distant flare]

Ah okok

[vital rose]

And then I did pythagorean to find their hypotenuses

Same for the helicopter thing

[vital rose]

Then I drew a triangle to form the triangle going from theh elciopter to the 20 degrees

And I used cosine law to solve for the 486.32 metres

Actually I tried sine law first but it gave me like 38 something degrees which made no sense

[distant flare]

Yea I would need to draw a new diagram for you

[vital rose]

Oh, do you have time to?

Or if you know where I made a mistake

[vital rose]

Like that 142 degrees is the fishy one

[distant flare]

Is this urgent? Not rn I can hit you up later though

Yea it’s just misinterpreted

[vital rose]

cos-1 (43.86^2+486.32^-521.60^2/(2x43.86x486.32)

[vital rose]

This should solve for that 142

[vital rose]

Oh no it isn't urgent dw! Thank you so much though

What time do you think you'll be available? It's around ~10 pm for me

[distant flare]

Am

12am

[distant flare]

Yea I’ll hit you up when im free sorry brother

[vital rose]

It's the weekends so I'll be awake

[vital rose]

It's fine!! Tysm though 🦋

[distant flare]

Cool

[calm coral]

@vital rose Has your question been resolved?

[vital rose]

No

Im waiting Mr. Bot

[calm coral]

@vital rose Has your question been resolved?

[calm coral]

@vital rose Has your question been resolved?

[calm coral]

Channel closed

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Available help channel!

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

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

[hot prairie]

integrate sin^3(x)cos(x)

[hot prairie]

idk what to do here

please help

[lunar trench]

Integrate it

Hope this helped

[blazing coyote]

do you know integration by substitution ?

[hot prairie]

idk

i was taught reverse chain rule

[blazing coyote]

you may have heard of u-sub?

[hot prairie]

no

[lunar trench]

Damn

You need to read on it

Dont think this can be solved without it

[winged pond]

Inverse chain rule should be the same as substitution I think

[jolly jacinth]

it is

so reverse chain rule works

when you have a function

being multiplied by its derivative

[hot prairie]

which one to u substitute

[lunar trench]

Is that the one with I and the other sign

[hot prairie]

like idk what one you derive

[jolly jacinth]

sometimes multiplied by a constant

[lunar trench]

Aaaa

[jolly jacinth]

so mate

lets look at this function

[lunar trench]

Probably derive sin3 and integrate cos no?

[jolly jacinth]

are you serious mate

listen to me

[lunar trench]

No

I am not your mate, pal

[clear delta]

you're thinking of IBP

[lunar trench]

What

Is that

[jolly jacinth]

IBP is like trying to kill a mosquito with a bazooka

[hot prairie]

bruh im getting confused

[jolly jacinth]

for this problem

ok

[hot prairie]

who's the one i should listen to

[jolly jacinth]

me

[hot prairie]

ok ok

[jolly jacinth]

so

reverse chain rule

[hot prairie]

yeah

[jolly jacinth]

works when you have two functions

one is a function

[hot prairie]

yes

[jolly jacinth]

the other is the derivative of said function

does that make sense

[hot prairie]

uhh

yeah

[jolly jacinth]

so we have the integral of sin^3(x)cos(x)

is there a function here

[hot prairie]

yeah

2

[jolly jacinth]

which

[hot prairie]

sin^3(x) and cos(x)?

[jolly jacinth]

so

what is the derivative of sinx