help-10

[unreal musk]

Do you know the area that the lawns (see the grass) make up?

[pine sapphire]

the lawn?

so 4(10 x 10)?

sorry am very laggy rn

oh is the lawn the giant semi circles?

OH SO

we will find their area!

pi x r^2

the radius is 5

and since there are two "full" circles

its 2(pi x 5^2)

or

2(25pi)?

YAYAY

ok!

now we must make it equal to the value of the flowerbeds

so we got 4pi(5 - x/2)^2

because

we got four little circles

ok!

now we can cancel out pi?

[unreal musk]

Yep yep

[pine sapphire]

yayayay

oh and we can divide 50 by 4 to isole the exponent??

and then root it?

oh

ok no rooting!

[unreal musk]

One thing though, while you can, you may not want to

[pine sapphire]

should i expand the brackets?

okok!!

[unreal musk]

Yep, remember they ask you to show the thing they said

[pine sapphire]

oh yes!

i should not solve for it

ok so should i put everything over the common denominator of 4?

ok now i will combine like terms!

pretend i wrote - 0

= 0

OKOK I DID IT TYTY

but how did u know they were searching for area?

[unreal musk]

(one thing, be careful about putting the equal signs here, as you could be taken to be implying the lines above are the same as the ones below, which they aren't )

[pine sapphire]

oh yes!

the amount of times i lost communication marks from that

crit c is the death of me

wait but how did u know the question was looking at area?

[unreal musk]

And, well, the fact they gave you a statement about the area (the whole "4 times as much area" statement) kinda gives you a constraint, and of course, knowing the areas of both the flowerbeds and the lawns kinda hints that you have the equation there

[pine sapphire]

OHH OK THAT MAKES SENSE

so its kinda based off the clues the question gives!!

tyty so much!!

🥛 🐟 🐡

❤️

.close

[obtuse pebble]

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[quiet quartz]

Can someone explain part a

[quiet quartz]

I got to the stage of 3+- 2i

.

.

Pls someone help

……

.close()

[quick totem]

Just use Vieta's theorem. In this case the product of the roots equals minus the last coefficient. So a+13/a+46=52. Solving this quadratic gives you the values of the roots

[quiet quartz]

Yes but i only got 2 roots

For solving that

I got alpha = 3+2i and 3-2i

[quick totem]

you get 2 sets of possible roots, with 3 roots each

[quiet quartz]

So how would 3rd root be found

[quick totem]

you have their expressions in the statement

a, 1/a and a+13/a+46

[obtuse pebble]

@quiet quartz Has your question been resolved?

[quiet quartz]

No

[obtuse pebble]

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[open spoke]

Which gradient descent is better: L infinity, L 1, or normal gradient descent?

[brazen viper]

Most gradient descent problems use L2. But the best option depends on the underlying function. It's hard to define "better".

[obtuse pebble]

@open spoke Has your question been resolved?

[open spoke]

What about the function makes Ln the best option?

[brazen viper]

What I mean to say, is that you can construct functions such that moving along a cardinal direction is the best choice because there are deep global minimums in the corners of the domain but the middle of the domain is a small local minimum.

These are probably not something you're going to often see outside of pathological cases though.

[obtuse pebble]

@open spoke Has your question been resolved?

[open spoke]

So L1 and Linf are like more straight jagged movements but L2 is more curvy so its usually better to use l2?

[brazen viper]

@open spoke L1 distance is min(x_i) and Linf is max(x_i), so the gradient is only affected by a single direction in the general case.

L2 is the quadrature sum, so it has components in every direction, and this is usually what you want.

Sorry, my bad, not min, but just plain sum

Manhattan distance

So it is affected by all of the components, but the gradients in a local area tend to be all aimed along a single direction.

[obtuse pebble]

@open spoke Has your question been resolved?

[obtuse pebble]

@open spoke Has your question been resolved?

[obtuse pebble]

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[cloud bone]

hello i have this question what do i need to do in other to find vertical Asymptote in ratzional premitive?

[cloud bone]

got this question stack at it

<@&286206848099549185>

[tardy epoch]

https://www.cuemath.com/calculus/rational-function/

[obtuse pebble]

@cloud bone Has your question been resolved?

[cloud bone]

thanks

[obtuse pebble]

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[golden mango]

For geogebra users how i chose the angle of A or whatever and the size of triagle sides ?

[golden mango]

sorry for bad english

ping when reply

[obtuse pebble]

@golden mango Has your question been resolved?

[golden mango]

<@&286206848099549185>

hello ?

[timid silo]

hi

[golden mango]

hi

also how i change my roles

[tardy epoch]

id:customize

[upbeat island]

for angle A i would make a measured angle and then extend it and find the intersection with BC

[golden mango]

bro it show this

[tardy epoch]

https://discord.com/channels/268882317391429632/customize-community

[golden mango]

same

mb becausei selected every role

[tardy epoch]

https://support.discord.com/hc/en-us

[golden mango]

.closed

[upbeat island]

example
https://www.geogebra.org/classic/fagzh2p4

oh

[golden mango]

thanks anyways

.close

[obtuse pebble]

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✅

[golden mango]

.close

[obtuse pebble]

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[slim ibex]

you see the numerator

in bottom left

what if it was all squared

(a+sqrt(b))^2

do i just do the conjugate of it and using it multiply top and bottom as usual

[civic zealot]

,w expand (a + sqrt(b))^2

[warm shale]

Results provided by WolframAlpha

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

Requested by zybikron

[civic zealot]

squaring it doesn't get rid of the radical

or are you asking if the whole denominator was squared to start?

[obtuse pebble]

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[bitter phoenix]

oh i figured it out

.close

[obtuse pebble]

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[surreal nexus]

I want the area of the little box in the rectangle

[surreal nexus]

i already did it and got 111.5x - 100.5 + 19.5(x**2)

i js want a confirmation

[primal vortex]

Can you show your work

Howd you get to that conclusion

[obtuse pebble]

@surreal nexus Has your question been resolved?

[surreal nexus]

i did it at school but basically i calculated the two triangles 's remaining parts and when i got the area of the 4 triangles i removed it from the rectangle's area

[primal vortex]

Sounds right, but id like to see your process

[obtuse pebble]

@surreal nexus Has your question been resolved?

[surreal nexus]

<@&286206848099549185>

[surreal nexus]

i understand but i only need the confirmation from another person so i know waht to expect

[primal vortex]

sorry just got him gimme a sec

[primal vortex]

I'm getting a different answer

[surreal nexus]

can i maybe see how u prooceeded ?

[primal vortex]

first thought is finding the missing lengths by setting (3x+4)+(4x-6)=x+p(x) and finding it that way, the missing length on the bottom is 2x+28 and on the right is 6x-2

[surreal nexus]

positive

[primal vortex]

just for the x^2 term I get 14x^2

wait

15.5x^2 forgot a negative

f\left(x\right)=\frac{-\left(x-20\right)\left(3x+4\right)}{2}+\frac{\left(5x+9\right)x}{2}+\frac{\left(4x-6\right)\left(2x+28\right)}{2}+\frac{\left(\ 6x-2\right)\left(2x+1\right)}{2}

$f\left(x\right)=\frac{-\left(x-20\right)\left(3x+4\right)}{2}+\frac{\left(5x+9\right)x}{2}+\frac{\left(4x-6\right)\left(2x+28\right)}{2}+\frac{\left(\ 6x-2\right)\left(2x+1\right)}{2}$

[warm shale]

SayMyName

[primal vortex]

for the little triangles does this look right

[surreal nexus]

dont be a smartass

[primal vortex]

what

[quartz scaffold]

ignore him

[dawn vale]

i belive its just pythagorouse

(might be wrong)

[timid silo]

most likely

[obtuse pebble]

@surreal nexus Has your question been resolved?

[obtuse pebble]

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[edgy mango]

.reopen

#help-10

[edgy mango]

.reopen

Hi

[obtuse pebble]

@edgy mango Has your question been resolved?

[edgy mango]

No

[shut lagoon]

What is your question?

[edgy mango]

For system of equations substitution I kinda understand it but not really I just need my answers checked for 2 of them

[obtuse pebble]

@edgy mango Has your question been resolved?

[obtuse pebble]

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[timid silo]

i did the hint but how is that helpful😭

[timid silo]

for x is integer and rational
f(x) = xf(1) so f(x) = cx

next im thinking would be to show for x is irrational but...

maybe thats not the right thing to do hahahah

[gilded needle]

well yea that is the next thing to do..

do you have a plan for how to do it?

[zenith raft]

bungo sir

[gilded needle]

slayla ma'am

[timid silo]

no bc i cant write an irrational # in terms of jsut integers and rationals to start doing algebra

[gilded needle]

btw, the statement is false if you drop the continuity assumption, which is rather cool

[gilded needle]

you can approximate an irrational using rationals, right?

3, 3.1, 3.14, 3.141, ...

[timid silo]

ig

[gilded needle]

maybe you can find a more mathematical way of saying that..

think about what properties Q has, as a subset of R

[timid silo]

can we say

we can find a sequence (p_n) \in Q subset R st it converges to an irrational number x
so f(x) = lim n-> inf f(p_n)

[gilded needle]

those statements are true, can you justify why?

[zenith raft]

oh yea? write down a counterexample for me

[timid silo]

2nd part bc of continuity via sequences

[gilded needle]

i choose not to

[timid silo]

1st part rational is dense?

[gilded needle]

yep!

[gilded needle]

yep!

[timid silo]

so f(x) = xf(1) and we r done

i see

[gilded needle]

indeed

[timid silo]

thanks

[gilded needle]

yw

[timid silo]

doesnt feel right i didnt get stuck for 30 mins

[zenith raft]

that’s the power of bungo

[timid silo]

ill ask another💀

[gilded needle]

next time let me know and i'll delay giving a hint for 30 min haha

[timid silo]

i started with listing the definitions (been helpful ngl)

for all M \in R, there exists N_1 \in R st for all x \in R (x > N_1 then f(x) > M)

similarly there exists N_2\in R st for all x\in R (x < N_2 then f(x) > M)

assume N_1 > N_2

for epsilon > 0 consider the interval [N_2 - e, N_1 + e] = [a,b]
the restrict f to g:[a,b]->R

by boundedness inf f(x) in that interval = f(x_0) <= f(x) for x in that that interval

how do i show for all x

[zenith raft]

in about 21 mins bungo will be back

[timid silo]

hahahahaha

in the meantime i can latek that last question

[unreal musk]

Why did you choose the interval [N_2 - e, N_1 + e] rather than just [N_2, N_1]?

[timid silo]

because i wrote x > N_1 above😭

oooo

maybe i do it for [N_2, N_1] and for x>N_1 and x<N_2 M is already less than every f(x). M can also be written as f(x_1) for some x_1 bc f is continuous
and in the interval [N_2, N_1] inf f(x_2) is smaller
so for some x_o we have f(x_o) = min(f(x_1), f(x_2)) <= f(x) for all x???

[unreal musk]

Doing it for x > N_1, x inside [N_2, N_1] and x < N_2 may be a nicer idea you know within [N_2, N_1] f will attain its inf of course...

M can also be written as f(x_1) for some x_1 bc f is continuous
are you sure about that? What did you choose your M to be?

[timid silo]

any positive number

i see the problem

kind of

if i do f(x) = x^2 +2
surely M=1 is working but we dont have an x

[unreal musk]

Yep, situations like that

You could, nonetheless, choose an M of your liking, get the N_1 and N_2 such that you have f(x) > M for x > N_1 and x < N_2, and (assuming N_1 > N_2) consider the interval [N_2, N_1], for which you know there's some minimum in that interval that f attains, let's call it M_0 and choose x_0 in [N_2, N_1] such that f(x_0) = M_0

Then how does that M_0 compare to M?

[timid silo]

probably M_0 < M

<=

[unreal musk]

That may or may not be the case

If M_0 <= M, then you're happy and don't need to do anything more from that point, but what if M_0 > M instead?

[timid silo]

some f(x) for x<N_2 and x>N_1 might be less than M_0?

can i choose M thats already in f and then say x for min(M,M_0) later?

[unreal musk]

How do you mean "choose M thats already in f"?

[timid silo]

we said for all M .......................f(x) > M, so surely true for M = f(x_1) for some x_1 as well

[unreal musk]

You could choose that, though whether that helps much or not...

I would think to instead take that M0 we found, and as we know that it's a real number, we can also apply the divergence to infinity to it again

[timid silo]

what does that mean😭

like f goes to inf for x-> inf and -inf so the minimum cant occur there, so the minimum has to be in a bounded region?

[unreal musk]

As in, you know there's some other variable, say N3 or something, such that f(x) > M0 whenever x > N3 (or whatever)

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

yess

[timid silo]

i was thinking because f(x) > M = f(x_1) for x>N_1 and x<N_2
so the only possible values of x for which f(x) <= M would be in the interval [N_2, N_1]
so x_1 \in [N_2, N_1]

and f(x_0) <= f(x_1) as its the minimum

[unreal musk]

Ahh, that’s fair enough I think yea I think that’s fine then

[timid silo]

😭 oh god bless

[timid silo]

wouldnt i get just another interval from here like [N4,N3]🥹

[unreal musk]

Well, there would be another interval, but from there you know that in the second interval everything would be above some minimum, and outside you’re forced to be above M0 anyway (is the sketch of the idea I had)

But yours works out nicer I think

[timid silo]

its that 12 am powerspike

[unreal musk]

Awwww

[timid silo]

thank you

and i think bungo went to bed hahahahahaha

.solved

[obtuse pebble]

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[dusk widget]

LMFAO

[obtuse pebble]

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[timid silo]

would anyone explain how we got the recursive relation a_n + a_(n-1) = 4^(n-1), i dont get the logic for this

[timid silo]

the problem statement: "In a basketball match, A,B,C,D,E pass ball among them. A gives the 1st throw and receives the ball after nth pass. Find the number of ways A can get the ball after nth pass"

i dont understand how total sequences for first (n-1) passes = 4^(n-1)

i do kinda get it cuz one can throw to other 4 ppl but the recursive relation relating to this feels weird

[zenith raft]

there are 4^(n-1) strings of characters from the dictionary {A,B,C,D,E} with no consecutive same elements, and each one corresponds to an n-1 length passing sequence. a_{n-1} of them end in A, and in that case the next throw can’t be to A

so that’s why a_{n-1} is subtracted

[timid silo]

Ok yea that makes sense, tq

.solved

[obtuse pebble]

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[spice pawn]

From Khan academy:

y = Asin(Bx-C) + D
y = Acos(Bx-C) + D

From the web:

y = Asin[B(x-C)] + D
y = Acos[B(x-C)] + D

Which of these are correct?

[sinful bear]

if possible can someone help with number 1? thanks a lot

[spice pawn]

oh, sorry bro

[sinful bear]

im sorry

[spice pawn]

no it's okay

[spice pawn]

you reopen it

[sinful bear]

how?

[spice pawn]

.close

[obtuse pebble]

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[sinful bear]

.reopen

.open

[mint vector]

just ask a question i think

[sinful bear]

thanks

[spice pawn]

.close

.reopen

[obtuse pebble]

✅

[spice pawn]

.close

[obtuse pebble]

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[obtuse pebble]

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[night plume]

did i do this wrong, my teacher has a different way of solving it but i got the same answer

[frigid burrow]

You have done it correctly

But I think your teacher wants you to use summation properties for this

Like sun of first n natural numbers

@night plume

[night plume]

im going to ask them if theyre fine with doing it that way later

[night plume]

but im trying to do it this way on a similar problem and it doesnt work anymore?

[frigid burrow]

Oh

Can you share that problem?

[frigid burrow]

Summation(a+b) = summation(a) + summation(b)

Use that

[night plume]

how do i do this

i got 21

[frigid burrow]

Can you write down all the properties you know about summation?

[night plume]

like the formulas?

[frigid burrow]

Yup

[night plume]

u mean these?

[frigid burrow]

Yess

[night plume]

so i try to solve this first right

[frigid burrow]

Use this and then send me an ss of what you get

[night plume]

but this doesnt have the i thing on the right side

[frigid burrow]

I'm gonna recommend putting brackets

This way you won't get confused

?

[night plume]

im confused

this dude did it that way idk why he set it up like that

[frigid burrow]

Well let's look at a different examples then

Give me a minute

$\sum_{i=1}^{4} (i+1)$

[warm shale]

Wumpus Man

[frigid burrow]

How would you do this?

[night plume]

14

id swap it to i 1 -4 then add them all up

[frigid burrow]

I think you mean something like 1+2+3+4 and then +4

So 14

[night plume]

yep

[frigid burrow]

Ok now

$\sum_{j=1}^{10}\sum_{i=1}^{4} i+1$

[warm shale]

Wumpus Man

[night plume]

ok

so i know the right summation is 14

now what do i do with the one on the left?

[frigid burrow]

Just 14 + 14 + 14 + .....

I'll add the brackets for you

$\sum_{j=1}^{10}[\sum_{i=1}^{4} (i+1)]$

[warm shale]

Wumpus Man

[night plume]

wait

so is the sum of the right summation

become the right side of this summation?

[frigid burrow]

Umm what?

The right side of the summation becomes the right side of this summation

What that makes sense

Yip

Yup

[night plume]

so since this is 14

[frigid burrow]

We call that thing argument

[night plume]

i put 14 on here

[frigid burrow]

$\sum (x)$

So x is argument

[warm shale]

Wumpus Man

[frigid burrow]

I think you got it

[night plume]

$

[frigid burrow]

But to make sure I'll write it

[night plume]

$\sum (14)$

[warm shale]

Railey

[frigid burrow]

$\sum_{j=1}^{10}[\sum_{i=1}^{4} (i+1)] = \sum_{j=1}^{10}[14]$

[warm shale]

Wumpus Man

[frigid burrow]

Yup

[night plume]

ok so i just solve that

but how do i do that?

[frigid burrow]

^

[night plume]

i just multiply 14 with 10 right

[frigid burrow]

Yup

[night plume]

so its 140

bruh

ok back to this

[frigid burrow]

Yup do you see what I was tryna explain?

[night plume]

yes

u essentially made it into 1 summation by solving the summation on the right?

[night plume]

how will it apply to this

[frigid burrow]

Yup

Let's first solve the Sigma on right then

[night plume]

ok

6?

[frigid burrow]

$\sum_{j=1}^{3} (ij)$

It will be in terms of i

[night plume]

its not clicking for me

[warm shale]

Wumpus Man

[night plume]

so i put in 1 , 2 and 3 for j

[frigid burrow]

$\sum_{j=1}^{3} (ij) = i \cdot \sum_{j=1}^{3} (j)$

[warm shale]

Wumpus Man

[night plume]

so 6i?

[frigid burrow]

Get it?

Yup

[night plume]

and then
4
sigma
i=1

[frigid burrow]

Yup

What will that reduce too?

[night plume]

6 + 12 + 18 +24

it works

LETS GOOOOOO

[frigid burrow]

Nice

[night plume]

so howd u write all that latex

thats what its called right

[frigid burrow]

Yes

Honestly idk myself

I just ask someone else for the symbols

I have experience with programming so that helps

[night plume]

ok cool

i appreciate the help

im studying for my exam 2 and i can get to the other topics lol (matrices)

[frigid burrow]

Ohh gl

!done

[obtuse pebble]

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[night plume]

.close

[obtuse pebble]

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[echo forum]

I'm a bit unsure of how to set this up for part a

[obtuse pebble]

@echo forum Has your question been resolved?

[echo forum]

would I just use dv=2 and dT=2?

[restive gorge]

I am not a physcist but it seems in a) you are trying to set up a flux integral?

[restive gorge]

How did you parameterize C_1?

[echo forum]

I did C_1=<1+2t,2+2t>

[restive gorge]

dV = 2 dt
dT = 2 dt make sense then

[echo forum]

ok

so id be integrating 2(1+2t)/(2+2t)+2k from t=0 to t=1?

[restive gorge]

Now that I think about it, does it have to do with Green's theorem?

[echo forum]

uh

maybe?

[warm shale]

𝔸dωn𝓲²s

[restive gorge]

so yes

[echo forum]

ok i can do that, that shouldnt be too hard

I end up with 3-ln(2)+2k

[restive gorge]

,w Integrate[2(1+2t)/(2+2t)+2k, {t,0,1}]

[warm shale]

Results provided by WolframAlpha

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

Requested by adoniszgks

[echo forum]

hm

oh wait i think i see what i did

hold on

yeah ok i see now

i got 2-ln(2)+2k

[echo forum]

which seems right based on this

so that leaves the 2nd part of a

[restive gorge]

Yup

[echo forum]

I'd assume I'll have to do 2 integrals here.

My first one would be 2k from t=0 to t=1. where I parametrized the curve as <1+2t,1>.

my second would be 1/(1+t) from t=1 to t=3. I parametrized the curve as <1,1+t>

[restive gorge]

I think you need a third

[echo forum]

would I?

what for?

[restive gorge]

(1,2) to (3,2) and (3,2) to (3,4)

It's basically like this

_|

[echo forum]

right

[echo forum]

and my first integral here was for (1,2) to (3,2). My second was for (3,2) to (3,4)

[restive gorge]

Your first is over C_1 (1,2) to (3,4)
Your second first is over (1,2) to (3,2) and your second second is over (3,2) to (3,4)

[echo forum]

ok then i guess i should have said i need 2 integrals for the second part of a. and that these are the integrals i came up with

[obtuse pebble]

@echo forum Has your question been resolved?

[obtuse pebble]

@echo forum Has your question been resolved?

[obtuse pebble]

@echo forum Has your question been resolved?

[obtuse pebble]

@echo forum Has your question been resolved?

[royal mulch]

Hey @echo forum I took a peek at part (a) to see if I could reason my way through it. I have attached some work if you wish to discuss it

Since the two paths give different final outputs, we can see that the field is not conservative (ie. we don't have independence of path)

[echo forum]

right, that makes sense

[royal mulch]

I was unsure how to help with part (b) and (c), but (a) felt OK

[echo forum]

Yeah im not sure how to do b. I have a guess for c though

For b, I would probably say no because it depends on the path taken?

but im not certain that thats actually good reasoning

[obtuse pebble]

@echo forum Has your question been resolved?

[echo forum]

How would I do part C?

[echo forum]

.close

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[spiral lintel]

Is there a way to calculate a term in the Fibonacci sequence without having to just add the preceding terms until I reach the term I desire to find?

Sequence: 0, 1, 1, 2, 3, 5, 8, 13, ...

What If I want to find T27, is there a more practical way? Or is the Fibonacci sequence just one of the annoying things In Math that takes a long route?

[worn yoke]

there is a direct formula in terms of powers of the golden ratio, but since that involves powers of expressions with square roots it's not really easier to do by hand

[spiral lintel]

So do you suggest I just stick to adding preceding values until I reach my desired term? What does this formula you mentioned look like?

[fossil crag]

you can also use faster rec formula F(2k+1) = F(k)^2 + F(k+1)^2

[fossil crag]

to get F27 for example, you'll need F13 and F14 with this formula

[worn yoke]

,, F_n = \cfrac{1}{\sqrt{5}}\left(\cfrac{1+\sqrt{5}}{2}\right)^{!n} - , \cfrac{1}{\sqrt{5}}\left(\cfrac{1-\sqrt{5}}{2}\right)^{!n}

[fossil crag]

and then from the ground up to F14

[warm shale]

cloud

[worn yoke]

copied from the wikipedia page

[spiral lintel]

Which is the issue I see, I know this formula: Tn = (Tn-1) + (Tn-2)
But In order to follow that formula, I would need to know the 2 previous terms, meaning I would need to calculate them too, so In the end my only logical solution is just writing down the sequence up until the thing I need

n = Term number im desiring to find

T = term

[fossil crag]

well yeah compute up to F14 to get F13 and F14

[timid silo]

I flipping love this server ❤️

[fossil crag]

and then shortcut to F27 with this

[fossil crag]

otherwise this is easy with a calculator

[spiral lintel]

That formula Is especially advanced for where I'm standing on sequences, especially in my country's education methods, so I think I'll be safe just adding, I doubt the examiners would do anything crazy, since they never have before

But thanks for helping! Just opened this to make sure I'm not missing anything else, I'm going to stick to adding precedings, since that formula is not anything we've learnt in class

[kind hawk]

well you could apply the same formula for F13. and surely there is a variant which works for F14

[spiral lintel]

I feel like this is something that may have stumped many students at some point

But yeah

My issue is solved

.close

[obtuse pebble]

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[fossil crag]

$\displaystyle \forall (p,q)\in \mathbb {Z} ^{2},F_{p}F_{q+1}+F_{p-1}F_{q}=F_{p+q}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

so F(2k) = Fk(F(k-1) + F(k+1))

or alternatively F(2k) = F(k)(2F(k-1) + F(k))

so yeah with F(k) and F(k-1) you can have both F(2k) and F(2k-1)

[obtuse pebble]

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[remote gazelle]

how do i do this problem

[fossil crag]

you have a volume V = {(x,y,z), 0<= x <= 1, etc... }

[fossil crag]

here they've chosen to represent V in this order: bounds for x, bounds for y when x is fixed, bounds for z when x and y are fixed

so what is one of the other possibilities for example?

[remote gazelle]

dy dx dz

[fossil crag]

ok so that would involve finding the bounds in which order

[remote gazelle]

bounds for z bounds for x bounds for y?

[fossil crag]

yes

so, accounting for all the points in the volume V

what's the least possible value z can take

and the biggest possible value z can take

[remote gazelle]

biggest possible value z can take is 1

biggest possible value x can take is also 1

[fossil crag]

yeah but that doesn't matter

we're only interested about the first bounds for now, z

once you have that

we FIX a value of z

and now we have to find the bounds for x, given that z is fixed

[fossil crag]

like in this order example

we started by saying that "choosing x first, x can be anywhere between 0 and 1"

and then "once x is fixed, y is between sqrt(x) and 1"

so the bounds for y, since we had to choose it afterwards, depends on whatever we fixed beforehand

[remote gazelle]

wait i think i got the idea

for dy dx dz

z boudns should be 0 to 1 right

[fossil crag]

yep

[remote gazelle]

then x bounds should be 0 to (1-z)^2

then y bounds should be sqrt(x) to 1-z

[fossil crag]

should be that, lemme check

0 <= z <= 1-y <= 1 - sqrt(x)

so sqrt(x) <= 1-z

x <= (1-z)^2

yep

[remote gazelle]

ok cool tysm

imma try same thing for dx dy dz now

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[candid flare]

dumb question, but what is this trying to prove?

[candid flare]

is it trying to prove the uniqueness of f, that it only maps one value to each aN?

[alpine bison]

it seems like it is proving that is well defined such a sequence... obviously for every n, the map f_n is unique because they are given in the assumptions.You could have more sequences a_n, b_n ec. that satisfy the given condition. The proposition proves that actually there is only such a sequence

[obtuse pebble]

@candid flare Has your question been resolved?

[obtuse pebble]

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[candid flare]

i am sorry, i dont think i understand. what do you mean by "only such a sequence"?

.reopen

[obtuse pebble]

✅

[candid flare]

like every bit of text makes me think the proof is about assuring that any aN is never redefined, and defined for each natural number N.

[obtuse pebble]

@candid flare Has your question been resolved?

[alpine bison]

with such a sequence I meant a sequence such that

[candid flare]

That's what i meant, from what i understood, what the thing is trying to do is:

(i) Take a function f that transforms natural numbers
(ii) define a base value, ie: a0 = c.
(iii) set an++ = f(an)

and then prove that this sequence will

(i) never overwrite the values (ie: if aN = 5, then aN is always 5) (this is done via peano axiom 4, which is that every natural number has a unique successor and that if m++ = n++, then m = n)

(ii) be defined for each natural number N (ie: the sequence expands indefinitely) (this is clear because of induction)

am i understanding this correctly?

[alpine bison]

yes I think so ... I would just point out that in (i) you are taking an infinite number of functions, one for every natural number n

[candid flare]

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[topaz fjord]

Hello

[topaz fjord]

Need help with this equation: KE = 1/2mv^2

How do I solve for the variable v?

What's my first step?

@me if your here

[modest lichen]

Isolate the v^2 and then you know that if x^2 = a then x = sqrt(a) or -sqrt(a)

[obtuse pebble]

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[neat glacier]

anyone can help me with this question? sorry i totally forgot high school maths after a long time away from school

[red ice]

the options don't make sense

[dark vector]

i guess d?

but yno south is right

[red ice]

yeah are you sure you have the right question for those options

[neat glacier]

yep the question and the options are as such.

[primal vortex]

Where did you get this

[serene elk]

this question looks like a concussed man wrote it

[neat glacier]

maybe its an error question.. hmm...

[primal vortex]

It looks like someone photoshopped it

[neat glacier]

its from a model test paper

[serene elk]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[serene elk]

is it the original?

[primal vortex]

Which part confuses you more, scientific notation or simplfying variable fractions?

[primal vortex]

Oh, yeah makes no sense

[neat glacier]

error question then? hmm?

[primal vortex]

But what kinda help are you looking for?

[neat glacier]

i tried to simplify it but cant seem to reach an answer for it

[red ice]

do you see how this is equal to $\frac{7.92 \cdot 10^{-2}}{16 \cdot 10^{-6}}$?

[warm shale]

south, just south

[primal vortex]

These might help, treat 10 as a

[red ice]

we used the product rule

[obtuse pebble]

@neat glacier Has your question been resolved?

[obtuse pebble]

@neat glacier Has your question been resolved?

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[pliant zodiac]

Guys, i don’t understand why that “4” is just ‘4’ and not ‘4!’

[fossil crag]

you choose where to place the 5

4 options

then you choose and place the 3 other digits among the 6 left

6P3

alternatively, ONLY choose the other 3 digits in the 4 digit sequence

6C3

and then place the 4 digits you now have

4!

[pliant zodiac]

oh wait that's tricky

ooohhh

permutation is about the position

[fossil crag]

yes

nPk is about choosing k numbers out of n and ordering them

so when you do 4 * 6P3

all you have to do is place the digit you already have (5) on one of the 4 tiles

then there are three tiles left to fill

so fill them, left to right, with 3 digits out of the 6 possible

[pliant zodiac]

sooo 6P3 is like talking about the 6 choices left x 5 choices left x 4 choices left to fill in the 3 leftover spaces?

wait that sounds wrong

[fossil crag]

it is kinda that actually

6 possibilities for the leftmost tile that isn't filled

5 for the next

4 for the last

[pliant zodiac]

wait if there is 4 possibilities for the number '5', and we put it as 4

when there is 3 possibilities for the other numbers (example '6'), why do we put it 6 not 3?

[fossil crag]

?

we're not choosing which tile to put a number on

we're choosing which NUMBER to put on the tile

since we haven't selected them yet

we already know 5 is one of the selected digits, so we place it on one of the 4 available tiles

but there are still the left, middle and right tiles to fill

you can put one of 6 digits on the left tile

after choosing what to put on the left tile

you choose one of 5 remaining digits on the middle tile

and finally one of 4 remaining digits for the right tile

[pliant zodiac]

why is it not 3 x 2 x 1 since there are 3 available tiles left, 2 available tiles left, and 1 available tile left?

[fossil crag]

again

we are not choosing which tile gets which number

we are choosing with NUMBER gets which tile

the tiles are already placed

the numbers aren't even selected yet

[pliant zodiac]

ohhhh i think i get itt

[pliant zodiac]

andd for the number '5', it is 4 tiles bcs the number is already selected but the tile haven't been selected rightt?

[fossil crag]

we are placing the number 5

for the other digits, we need to choose and place them

[pliant zodiac]

i seee

[fossil crag]

so, if you already had chosen which digits to place with 5

say 1,2,3

but not placed them yet

then you would have to choose which tile to place them on

[pliant zodiac]

ohhhhhh

that makes senseeee

[fossil crag]

so number "1" gets 3 spots to choose from

number "2" gets 2 spots

and number "3" gets the remaining spot

but that only makes sense if you chose the digits before

so 6C3

[fossil crag]

in total, this way to choose gives us:
4 * 6C3 * 3 * 2 * 1

[fossil crag]

it's this alternative way I proposed, 6C3 * 4!

[fossil crag]

either that, or "place the 5 and immediately fill in the other 3 spots with 6 digits"

4 * 6P3

[pliant zodiac]

thank you so much for your super detailed explanation

i understandddd

yaaayyyy

[pliant zodiac]

also i wanted to thank you and other helpers for helping me so much particularly in these probability, permutation, combination questions, i got a really good score in my exam (which i never expected bcs i've always struggled with questions of probability)

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[proud forge]

How do i solve the equivalent resistance at terminals a-b of the network?

[proud forge]

Im on this process but not sure if im doing it right and dont know to continue

[obtuse pebble]

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@proud forge Has your question been resolved?

[obtuse pebble]

@proud forge Has your question been resolved?

[obtuse pebble]

@proud forge Has your question been resolved?

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[grand dune]

can someone pls help

[tardy epoch]

its the sum of two geometric series

[grand dune]

but how can i split it in to 2?

[opaque dome]

The fraction ?

[restive gorge]

Guten Abend

[opaque dome]

Just split

And you can "distribute" the sum

Idk what the word is

Like a(x+b) = ax + ab

[restive gorge]

you can foil the sum

[opaque dome]

Right foil

[grand dune]

but doesnt it need to be in a form of q^k so that i can put it in a geometric series

[restive gorge]

dann bring es in die form

potenzgesetze

[tardy epoch]

it will after you distribute

[opaque dome]

You know its serious when adonis speak german

[restive gorge]

i just said he can bring it into the form using power rules

[opaque dome]

Yup dw i understand german xd

The switch was just funny

[restive gorge]

i am gekocht

[grand dune]

aber minus -1 stört dabei

[restive gorge]

4^(k-1) = 4^k/4

[grand dune]

omg ich hab die aufgabe falsch gelesen

[restive gorge]

ich hab sie rückwärts gelesen

[grand dune]

danke

[obtuse pebble]

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[grand dune]

.close

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[timber yarrow]

hey i am not really working on a specific problem but i remember this weird question i got on a test once and got wrong. It was to find the derivative of -2e^5t and 8e^-2t and for my answer i placed -2e^5t * 5 * ln(-2e) and 8e^-2t * -2 * ln(8e) but got it wrong

[timber yarrow]

the correct answers were the same thing but not with the ln's so im just confused lol

[tardy epoch]

show how you got your answer

[timber yarrow]

through my notes?

idk in my notes its says like when i encounter a problem like that its like a^n = a^n * n' * ln(a)

[obtuse pebble]

@timber yarrow Has your question been resolved?

[worldly hull]

-2 and 8 are just real values right?

When you have to differentiate something of the form "k . f(x)" with respect to x, what happens to the coefficient k?

[worldly hull]

This formula is correct by the way (well not the equal sign but that's just notation)

[timber yarrow]

mm im just confused because its a parametric question so im just confused when they get dy/dx they dont seem to use the ln(x)

you know what i mean? its a lil confusing sorry :3

[worldly hull]

I'm confused, I don't understand what you are trying to say

[worldly hull]

You have not answered this though, do you know the answer?

[obtuse pebble]

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[opaque dome]

Seems fine

[wild swallow]

you just did a direct proof and wrapped it in a contradiction proof...

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[cobalt anvil]

I tried doing this question and got it wrong, how should i approach these types of questions?

[fickle lantern]

This is just Fundamental Theorem of Calculus

$\frac{d}{dx}\int_a^x f(t) dt = f(x)$

[warm shale]

LTHMath

[fickle lantern]

You're overcomplicating it

[cobalt anvil]

oh do i just substitute u with 7x and 5x and find the difference?

[fickle lantern]

Yep

[cobalt anvil]

but do i still need to multiply each integral with 7 and 5

[high lily]

yes,

[cobalt anvil]

okayy thanks !

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[spring stag]

Can anyone help me with this question 8. c)

[spring stag]

for 8. c)

[acoustic sentinel]

I would first of all assume that the hill's slope doesn't change much and that she goes down at a roughly constant speed. So while she's on the hill, you could probably model her height as a straight line downwards against time

while she's in the chairlift line, she's staying put. So her height would just be a flat (horizontal) line. It's not changing over time.

While she's in the chairlift she's going up at a roughly constant rate, but half as slow, so line with lower slope

I would just say that you could model her height with lines, "piecewise" depending on whether she is going down the hill, waiting, or going up

"piecewise linear" is probably the correct terminology to use

[spring stag]

thank you so much, I understand it now

one more question

is it okay if I sketch out what the graphical model would look like or would I have to create a combined function to model it on desmos

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[wet bear]

~tex Our equation is $4a+16b+48c+\cdots = S$. There can be at max 16 variables (S is at max $4\times 10^6$) and the coefficients are of this form $i\cdot 2^{i+1}$. We want the number of total non negative solutions.

[warm shale]

yashashwi

[wet bear]

What methods are there to compute this

[upbeat island]

a, b, c are natural numbers ?

i'll rename to a_1, a_2, a_3 for convenience. a good start might be to notice that a_16 is either 0 or 1 given the bounds of S.

if all other a_i = 0, then S = a_16 * 16 * 2^17
a_16 = 1  =>  S = 2097152
a_16 = 2  =>  S = 4194304 > 4e6

[wet bear]

Yes, but we need total number of ordered tuples satisfying the equation

[upbeat island]

ok.... we've just bounded a_16 for all the permutations

maybe try a_15 and see if a pattern emerges (idk but this seems like a good try)

you say "compute this"-- are you doing this by hand or programming?

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[gritty rose]

Hi so I have an optimization problem I'm doing as practice and I got the answer, but the website suddenly subtracted 0.99 from the maximized area and I'm not sure why. I was hoping people here could help me understand if it's just weird or if there is a reason for that.

[gritty rose]

I'm confused why it went from 6158.40 to 6157.41

[obtuse pebble]

@gritty rose Has your question been resolved?

[languid sierra]

lemme check that for you

The difference between 6158.40 and 6157.41 maybe due to rounding errors in intermediate steps....

[gritty rose]

It says only to round on the final step so I'm not sure

And 6158.40 was the correct answer

If it's just a rounding error then I won't worry too much about it

Was just kinda confused where that number came from

Alright thank you! 💜

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[flat roost]

can someone pls explain the how tf has he integrated this 💀 ?

[lyric ember]

you can solve $\int e^x \sin(x) dx$ using double partial Integration

[warm shale]

tobi

[flat roost]

now whats that 💀
can you send an article/vid on it ?

[lyric ember]

you mean Integration by parts?

[flat roost]

nope , double partial integration

this thing

[lyric ember]

If you take the derivitve of sin(x) twice you get -sin(x) so if you do Integration by parts once you get an Integral with e^xcos(x) if you do Integration by parts on this Integral, then you get e^xsin(x) which is the same Integral as on the left side. so you can solve for your Integral using basic algebra

https://www.youtube.com/watch?v=2JKPTCBaI4w&ab_channel=TheOrganicChemistryTutor

[flat roost]

oh ok mb mb , i got confused because of the y in the ques .

thanks 🫡

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[split isle]

i have an outline for a proof but it feels weird since it doesnt really use the stuff in the chapter leading up to this question

suppose d and d' are gcd(a,b), then d = ax + by, and since d' | a and d' | b, a = d'k, b = d'l for some k, l in D

then subbing in gets d = d'kx + d'ly = d'(kx + ly)

and d is in <d>

and since d = d'z for some z in D, d is in <d'>

then since the intersection of <d> and <d'> is nonempty, <d>=<d'> (since <-> partitions D)

and therefore d and d' are associates (by a thm proved like 2 chapters ago)

[dark stirrup]

what is <d>?

[split isle]

the ideal generated by d

[dark stirrup]

I'm supposing you got this by doing the same argument with d and d' switched?

[split isle]

<d'> = {d'x : x \in D}

[dark stirrup]

Oh duh

[split isle]

i guess my only issue is like

i feel like im not using the info given

like this doesnt use the fact that d' is a gcd of a and b

and i feel like questions dont usually give unnecessary information

also i dont think it uses anything from the chapter

oh wait

it does use 1 theorem

[dark stirrup]

What's the other theorem?

[split isle]

that if d = gcd(a,b) then there exist x,y in D such that d = ax + by lol

[dark stirrup]

but what about d'? Where do you assume its "greatest" condition?

[split isle]

i dont

well i use the fact that its a common divisor of a and b

but not that its a greatest one

idk i think ill just write it out properly to make sure i havent assumed anything i shouldnt have

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[tame roost]

Hello Ladies and Gentlemen

[tame roost]

I'm just confused with a slight issue

I have to make a function in c where a range of integer numbers 0...255 returns a range of values to 0...65535

I thought about multiplying the function with 256 but this doesnt work, since 1 is subtracted from iz

I just dont know how to start it off

[polar fossil]

i would go simpler than that

what's the goal exactly?

[tame roost]

Input: 0 ... 255
Output: 0 ... 65535
Linear course

[polar fossil]

okay, what does linear mean?

[tame roost]

That output is directly proportional to input

[polar fossil]

mmk, so we can write that as f(x) = ax

and you want f(255) = 65535 right

[tame roost]

Yep indeed

[polar fossil]

ok so what's a?

[tame roost]

257

[coral swan]

nice Picture

[tame roost]

Wait it's easier than I thought

I thought I have to do subtractions and stuff

Sinde I start from 0 instead of 1

Ye sorry guys didn't mean to waste time

Thanks though

[marsh geyser]

!done

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[fathom kindle]

wait how do i do this

without looking at a graph

or making one

[azure wraith]

Sec is 1/cos

[fathom kindle]

ya

[azure wraith]

Look where cos is 0

[sudden aspen]

Since it makes the denominator equal to 0

[azure wraith]

Cuz 1/0 is asymptote form

[sudden aspen]

making it undefined

(you know your unit circle, right?)

[fathom kindle]

ya

so

its at 90 degrees

and

270

[sudden aspen]

the question is asking for radians

[fathom kindle]

ya

so pi /2

[sudden aspen]

so what are those in radians?

and

[fathom kindle]

3pi/2

[sudden aspen]

nicely done

[fathom kindle]

yay

tysm

[sudden aspen]

So, how would you write the answer?

[fathom kindle]

oh

is it like when feta = pi/2 or 3pi/2, there is a vertical asymp

im not sure

[sudden aspen]

You can say:

"y = sec theta is undefined at the values theta = pi/2 and 3pi/2 in the interval 0 to 2pi"

[fathom kindle]

ya thats good

thank uu

[sudden aspen]

ofc

[fathom kindle]

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[remote pivot]

So for this problem, my textbook tells me to use integration by parts to solve this, but I was wondering, if this popped up on an exam, how would I know to use integration by parts?

[high lily]

at a glance other approaches like sub don't seem to work due to the type of composition you have

[remote pivot]

so i guess if u sub doesnt work, then i should go through and try whatever other methods i know?

[high lily]

yeh

[remote pivot]

alright thanks

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[dawn thunder]

I just want some clarity, I dont need to find the original matrix right? That seems too complicated for what theyre asking here, V would just be the colums morphed into a 3x3 matrix, and then D would just be a diagonal matrix of the eigen values, and then the fundamental matrix would just be e^(x * eigen value) * the corresponding eigen vecotor in a 3x3 grid?

[dark stirrup]

I dont need to find the original matrix right?
No.

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[dark stirrup]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[dark stirrup]

V would just be the colums morphed into a 3x3 matrix, and then D would just be a diagonal matrix of the eigen values, and then the fundamental matrix would just be e^(x * eigen value) * the corresponding eigen vecotor in a 3x3 grid?
yes

[dawn thunder]

sickkk

thanks

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[true widget]

how do i reflect points across lines like (-1/5)x-9.8?

[true widget]

like for example (9, -8)

[deft fable]

find the perpendicular line

then the intersection

then the reflection point is equidistant from the intersection point along the perpendicular line

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[potent hound]

f'(x)=secxtanx+sinx

[potent hound]

when set to 0

how to solve for x?

[ember frost]

hint: ||multiply both side by cosx||

[potent hound]

don't do this to me 😮

what makes u think to multiply both sides?

<@&286206848099549185>

[grand arrow]

whats the question

oh

[potent hound]

i tried writing it all in cos sin

but got stuck

[grand arrow]

-sin x = secxtanx

did u get there

[grand arrow]

wait i dont think theres a solution

yeah

[potent hound]

u sure?

[grand arrow]

nvm

i misread

[potent hound]

help :

me 😦

[grand arrow]

wait no

theres no soln

cos^2x = -1

cos^2x = sqrt(-1)

[potent hound]

Always positive 1 I know

[grand arrow]

so u have no real solutions

[grand arrow]

wait no

u do

[potent hound]

so this is one of the calculus problems where they are asking u find the extrema values

but after finding the derivative

im not sure how to solve when its set to zero

[grand arrow]

hm

yeah i think u still have no real solution

[potent hound]

im starting to really hate these problems

teacher trolls u with "trig style questions"

g(x)=sin^2(x)+x ask u to find local min or max

when u look up the graph

[grand arrow]

actually no

the solution set is any n*pi

try to solve it without moving sin to the right

[grand arrow]

$ \frac{1}{\cos \theta}*\frac{\sin \theta}{\cos \theta} + \sin\theta $

[potent hound]

so when the sign doesnt change from + to -