help-38

[swift drum]

yes

[unique minnow]

So that looks like

This

Now we need the left bit

And we know that when x is negative,

|x| = -x

(Just like how |-5| = - (-5) )

And now we have f(x) = 3/4 * |x|.

[unique minnow]

So if you replace |x| in the equation with that, what do you get?

[swift drum]

3/4

[unique minnow]

Yeah, f(x) = -3/4*x when x < 0

So that's a line of slope -3/4 going through the origin.

[swift drum]

oh ok

[unique minnow]

So you get the usual cone shape for the absolute value, but it's a bit more open because of the coefficient.

[swift drum]

ohhhhh tysm

That makes sense

Idk why I was overthinking it

[unique minnow]

That's ok

Happens

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[sudden patio]

I'm confused about how to do part c

[sudden patio]

how would I show it for any Qd

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[deft pebble]

don't get it

<@&286206848099549185>

[grim sparrow]

call the length of the rectangle L

then the width is 9/16 * L

the area of a rectange is length * width

so L * (9/16)L = 1 (which is the area of the rectangle)

you can now solve for L to get a number

with that information, you can find the width of the rectangle too, and then the perimeter

@deft pebble

[deft pebble]

k

ill try

@grim sparrow

how i do that

yo

im stuck

i got x^2 = 16/9?

@grim sparrow

yoo

[grim sparrow]

take the square root

[deft pebble]

how

[grim sparrow]

and note that you must exclude the negative solution

because lengths arent negative

[grim sparrow]

wdym how

[deft pebble]

how do i root it

[grim sparrow]

what

[deft pebble]

is it 4/3?

[grim sparrow]

yeah

[deft pebble]

i like did square root on both side

is that how

ur supposed to?

[grim sparrow]

well yeah?

[deft pebble]

k

[grim sparrow]

how else would one do it lol

[deft pebble]

25/6

got it

thanks

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[frozen tangle]

help me plz

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[wraith hinge]

In one school, 135 students traveled with a or b bus companies to go to and from their homes during the holiday. 75 of the students went with company a, and 90 of them returned with company b. 86 students traveled with different companies on their way back and forth. Accordingly, what is the total number of students who went with company b and returned with company a?

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Please don't occupy multiple help channels.

@wraith hinge Has your question been resolved?

[wraith hinge]

@proper kernel

[proper kernel]

did you draw a venn diagram

[wraith hinge]

wait a sec

it's like 3 circles right?

one is a and other is b

another one 86 students

with some intersection with a and b?

@proper kernel

[proper kernel]

it looks like you can make 4 circles based on:

1. enter with company A, exit with company A
2. enter with company A, exit with company B
3. enter with company B, exit with company A
4. enter with company B, exit with company B

these circles are all completely separate from each other

[wraith hinge]

okie

[proper kernel]

wait is there a simpler way?

[wraith hinge]

49 students go back using the same bus right?

[dull temple]

i think you could use 2 circles, one for A on the way there and one for A on the way back

but tbh your way is easier in this case i think

[proper kernel]

so instead of circles, we can consider that a student does either 1, 2, 3, 4
we'll call them A, B, C, D

[wraith hinge]

okie

[proper kernel]

not exactly a good name but:

A. enter with company a, exit with company a
B. enter with company a, exit with company b
C. enter with company b, exit with company a
D. enter with company b, exit with company b

so you know that A + B + C + D = 135

[wraith hinge]

okie

[proper kernel]

and that since B + C = 86, A + D = 49

now try listing the other facts in terms of this A, B, C, D

[wraith hinge]

right

[proper kernel]

youll end up with a system of equations to solve for C

[wraith hinge]

system of equations

[proper kernel]

not exactly the simplest way

but the setup is simple

[wraith hinge]

how do i form equations?

[proper kernel]

when it says 86 students travelled with different companies,

that means students in B and students in C, right

[wraith hinge]

yeah of course

[proper kernel]

that also means no students in A and D, right

[wraith hinge]

right

[proper kernel]

so only B and C can add to 86

therefore B + C = 86

[wraith hinge]

yeah

but i gotta figure out only c

[proper kernel]

you didnt even tell me what equations you got

the two I said arent the only ones

[wraith hinge]

yeah

hold on

is it like the other equations are formed using the initial conditions?

[proper kernel]

ofc just like with B + C = 86

[wraith hinge]

75 go with bus a

and 60 with bus b

[proper kernel]

going with bus a here means entering with bus a

[wraith hinge]

yeah

[proper kernel]

so what equation can you make out of "75 enter with company a"

[wraith hinge]

but

we don't know how many are gonna come back using bus a

or bus b

[proper kernel]

thats a shame isnt it

[proper kernel]

even though Im asking you about this question

and I do expect an answer in the form ? + ? = 75

[wraith hinge]

no no wait

[proper kernel]

at least state the equation before we move on

[wraith hinge]

A+B=75 right?

[proper kernel]

yea thats the one

now they also say "90 of them returned with company b"

[wraith hinge]

yes

[proper kernel]

this means "90 students exited with company B"

[wraith hinge]

B+D=90

[proper kernel]

yep

[wraith hinge]

thank you so much!!

[proper kernel]

did you figure it out?

[wraith hinge]

yeah i will tell the answer wait

[proper kernel]

alr I was going to post this if you were stuck:
so knowing that A + B + C + D = 135, you can get equations like so:
A + B = 75 "75 went with company a"
C + D = 60 so 60 went with company b
B + D = 90 "90 returned with company b"
A + C = 45 so 45 returned with company a
B + C = 86 "86 traveled with different companies"
A + D = 49 so 49 traveled with the same company

you could use just these three to find the value of C
A + B = 75
B + C = 86
A + C = 45

[wraith hinge]

14?

[proper kernel]

,calc (45 + 86 - 75)/2

[solid kiln]

Result:

28

[proper kernel]

Im getting C = 28 instead

[wraith hinge]

wait

lemme tell you how i did

c+d=60 right?

d=49-A

c+49-a=60

-a=c-45

isn't it correct?

[proper kernel]

thats correct so far but it doesnt look like you know where youre going with that

[wraith hinge]

where did i go wrong?

[proper kernel]

try focusing on a smaller set of equations

or try focusing on a specific goal

[proper kernel]

you can use these three to solve for C

[wraith hinge]

yeah but the way i solved is it wrong? or did i do any mistake?

[proper kernel]

thats correct so far but it doesnt look like you know where youre going with that

[wraith hinge]

c+49+c-45=60

[proper kernel]

,calc (60+45-49)/2

[solid kiln]

Result:

28

[wraith hinge]

yeah

okie thank you so much

[proper kernel]

np

[wraith hinge]

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[chilly spear]

hello

[chilly spear]

how sqrt can be lipschitzienne

on [1,+oo[

[grizzled vine]

whats lipschitzienne

?

[chilly spear]

https://cdn.discordapp.com/attachments/1017520785981702185/1201838750137856041/663183019280171018.png?ex=65cb468f&is=65b8d18f&hm=911c298b23663b9f73addf6b74e9adc775a11074bca0ecf1851913b24633e1fe&

lipschitz continous function

sqrtx is not bounded

[broken solstice]

Recall the definition of Lipschitz continuity

here

Sqrt x is not bounded above but its derivative is

on that interval

@chilly spear

[subtle lava]

Alternatively:

$\lvert \sqrt{x}-\sqrt{y}\rvert =\lvert \frac{x-y}{\sqrt{x}+\sqrt{y}}\rvert=\lvert \frac{1}{\sqrt{x}+\sqrt{y}}\rvert\lvert x-y\rvert$

[solid kiln]

Max

[subtle lava]

Then show the function in front of $\lvert x-y\rvert$ attains a maximum and that will be your $C$

[solid kiln]

Max

[chilly spear]

okay thanks

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[granite heart]

how to simplify this

[granite heart]

it should become x + sqrt(x^2 + 1)

[marble wharf]

write the thing inside the sqrt as a product. then use sqrt(ab)=sqrt(a)sqrt(b)

[granite heart]

oh so 4(x^2 + 1)

then sqrt(4) ?

[dull temple]

yes

[granite heart]

ok thanks

how to close

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[dusky tinsel]

is the answer correct

[dusky tinsel]

i got -1

[dusky tinsel]

instead of +ve
this is differnetitiating

since $(1* ln(-x) + (x * -1/x))$

[solid kiln]

™Vlad The Lad

[grim sparrow]

i think you didnt use the chain rule on ln(-x)

d/dx [ln(-x)] = 1/(-x) * -1

so the two minus signs cancel

and you will get the + that's in the answer

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[dusky tinsel]

ty

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[delicate pagoda]

"prove the following error-bound for the rectange-rule of numerical integration"

[delicate pagoda]

How would I go about that? I'd assume there's a more or less "standard" way to do that for all newton-cotes formulas?

I(f) being the integral of a function f with

Î(f) is the approximation by rectangle rule

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[floral inlet]

This looks like proof that as you get into higher maths less and less people are able to help. Good luck

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@delicate pagoda Has your question been resolved?

[verbal gulch]

is this error bound the supremum over all tau_1?

[delicate pagoda]

from what I understand it's "for some tau1 in [a,b]"

[zinc ginkgo]

You have to give the definition of I hat too

[delicate pagoda]

approximation of integral of f by rectangle rule

[zinc ginkgo]

Give the formula

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@delicate pagoda Has your question been resolved?

[delicate pagoda]

you approximate the integral from a to b over f by the rectangle of width (b-a) and height f(a)

we're looking at the error boundary of this approximation

[tepid cobalt]

You can use the mean-value theorem.

I imagine something like

$\int_a^b f(x)dx - (b-a)f(a) = \int_a^b f(x)-f(a) dx = \int_a^b f'(\tau_1)(x-a)dx$

[solid kiln]

OneTrackPony

[tepid cobalt]

Hm, I am hiding an x-dependence in the tau1

Perhaps wait with the tau1 then. Do

$\int_a^b f(x)dx - (b-a)f(a) = \int_a^b f(x)-f(a) dx = \int_a^b f'(\xi)(x-a)dx$

[solid kiln]

OneTrackPony

[tepid cobalt]

Since f is C^2, f' is C^1, and is thus bounded on the closed interval. It has a max and a min.

So you can use some bounds $m \leq f'(\xi)\leq M$ to get

$m\int_a^b(x-a)dx\leq \int_a^b f(x)dx - (b-a)f(a)\leq M\int_a^b(x-a) dx$

[solid kiln]

OneTrackPony

[tepid cobalt]

Notice how $\int_a^b (x-a) dx = \frac{1}{2}(b-a)^2$.

[solid kiln]

OneTrackPony

[tepid cobalt]

Then use the intermediate value theorem to get your tau_1 in the end.

[delicate pagoda]

how are we getting this equality?

[delicate pagoda]

$\int_a^b f(x)-f(a) dx = \int_a^b f'(\xi)(x-a)dx$

[solid kiln]

Bob Goldham

[delicate pagoda]

oh wait that's because we're in C²?

we can assume that there must be some ξ in [a;b] for which the derivative is exactly the slope between f(x) and f(a)

alright got it

that makes sense

thank you for helping out!

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[wraith hinge]

thought this was easy until I got every one wrong

I thought it was sign flips

is it not?

like flipping signs for all terms

[indigo ferry]

For all terms?

What did you get for a)

[wraith hinge]

like

$-2\sqrt{3}+4$

[solid kiln]

mj

[left oriole]

you only flip the sign on the term that contains the radical

or if both do, then on one of them

[wraith hinge]

so then for c

it could either be 2 root 3 - root 2 or -2root3 + root2?

[left oriole]

yea

[wraith hinge]

idk not really comfortable with this flexibility lol

it is what it is I guess

wait but how come @left oriole

[left oriole]

i don't make the rules

it doesn't really matter even for those of the form a + sqrt(b)

you can flip the sign of either term and it'll work

[wraith hinge]

no i meant

how come

you do that

what is the end goal

of doign that

[left oriole]

usually the point is that if you multiply a radical expression by its conjugate, the square roots go away:
$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$

[solid kiln]

Bungo

[wraith hinge]

mmm ok

[left oriole]

which is often useful when doing algebraic manipulations involving radicals

[wraith hinge]

right

ok that makes sense then

thanks

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[shadow dew]

Hi. I need to derive the following function using the product rule. What is the first step?

[shadow dew]

I think I figured it out. You need to use a combination of chain rule and product rule

.close

[proper kernel]

editing the message doesnt work
type .close

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[main otter]

These two have been confusing me

cause for the looped limacon, they set the equation equal to 0

but for the cardioid, they set it equal to 4

why didn't they also set the cardioid equal to 0?

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[modern cargo]

Since irl nobody would tell you which angle is the right angle, how to identify the hypotenuse, opposite and adjacent without knowing the right angle? What if the theta is placed on right angle and the hypotenuse is oppositing the theta?

[split chasm]

wdym by

without knowing the right angle

[blazing geode]

I suppose you have little understanding of a right angle

That you cannot identify it when seeing one

[dull temple]

if you try to compute a trig function of the right angle then a lot of things break

which is why eg tan90º is undefined

[split chasm]

if you aren't told anything about right angles
you could first determine if you have a right triangle by seeing if it satisfies pythagoras

and if it does, you'll have a right triangle
in which the longest side (hypotenuse) will be facing the right angle

[modern cargo]

So is knowing the right angle obligatory before doing trig?

[split chasm]

no

the presence of a right triangle is needed for right triangle trig

there are laws that are applicable to non-right triangle, but those are based from constructions of right triangles

[modern cargo]

How do I find the right triangle without using protractor ?

[split chasm]

as described

wdym by find the right triangle

[modern cargo]

Yeah I just learned it. Cosine & sine law. The problem is that I skipped everything on highschool 😭 (bcs everyone else in my class did too)

[modern cargo]

The angle which is 90°

[split chasm]

you could first determine if you have a right triangle by seeing if it satisfies pythagoras
and if it does, you'll have a right triangle
in which the longest side (hypotenuse) will be facing the right angle

normally if you are given a right triangle, the right angle will be indicated by a small square shape at the angle

[modern cargo]

So hypotenuse is always the longest side?

[split chasm]

yes, of a right triangle

[modern cargo]

So without using protractor, a right angle is an angle between two of the shortest side?

[split chasm]

depends if you know the triangle is a right triangle

you can't assume there will be a right angle between the two shortest sides of any triangle

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[tall needle]

need help... I got y=(x^2+2x-2)/2 for the a.) (domain is done as well)
But b is messing me up-- I got 0=x^2 + (2-sqrt2)x-2(sqrt3)

[tall needle]

This is tech-free as well, so there should be a way. I'm just really confused cos it's a rlly nasty quadratic

nvm i just tried to square root both sides of the equation, retrying again

[left oriole]

this seems kinda gnarly, i'm getting a 4th order equation and not a nice one

i'll let you know if i have an ah-ha moment

[tall needle]

tysm, i'm trying to find the factor for long division but it's not going well

[split chasm]

consider that inverses are relfected over y=x

(don't try to solve the equation you get from f(x) = f^-1(x))

instead you can solve
f(x) = x or f^-1(x) = x

[tall needle]

oh cos they intersect

that's pretty clever

[left oriole]

nice observation

that makes this easy!

[tall needle]

so i got x=sqrt2

uh +-

[split chasm]

and then apply the domain restriction from part a)

[tall needle]

aight

tysm @split chasm @left oriole

how to close?

.close

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[spice fulcrum]

I need help with B

It should be in the interval 4-5 right? so 8% of the population? I get 27*0.08=2.16 but idk to round up or down or if I did a step wrong

Because the answer says its 2

[dapper swift]

Yeah that's the right interval

[dull temple]

i guess round to nearest

[dapper swift]

Yeah this is a bit iffy, I would just round to the nearest integer

[spice fulcrum]

The worked solution just directly says its roughly 2

So im just gonna go along with it and round to nearest

But I still dont get it

A percentage of a population should round up right

Oh well I want to graduate, so I guess its right

Thanks for the help :)

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[tight hornet]

Could I please have an explanation of part a

[narrow silo]

can you figure out how 11/32 was calculated?

ex: ||left riemann sum, right riemann sum, midpoint riemann sum||

[tight hornet]

which area formula do u use?

[narrow silo]

thats what you have to figure out

[tight hornet]

ah okay thanks, im think trapezium?

or rectangle

[narrow silo]

is there more context to the problem?

[tight hornet]

no its just a textbook question sorry

but your meant to add up all the areas and then they're meant to equal 11/32 but i did and it and got 49/32

[narrow silo]

you might have forgotten to divide by 4

because each interval is 1/4 units wide

[tight hornet]

okay i did it again and got 15/32

im using the rectangle area formula btw (length x breadth)

[narrow silo]

yeah this is annoying

cause idk how they got 11/32

im sry but i dont think i can help u

[tight hornet]

thank you for your help anyways you helped me get closer to the answer, so I'll keep trying 🙂

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[latent girder]

Please help I ’don’t understand anything

[knotty locust]

do you know SOHCAHTOA

[latent girder]

nope

[knotty locust]

well then you can't do it

check your notes from class

[latent girder]

the teacher never taught us

[knotty locust]

https://www.khanacademy.org/math/geometry-home/right-triangles-topic/intro-to-the-trig-ratios-geo/v/basic-trigonometry

Then learn it here

[latent girder]

alright thank u

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[primal wraith]

hi im trying to prove this inequality here

$(xi)_{1}^{n}$ is defined to be a sequcne of positive terms and m is defined as their mean

we cant use am-gm inequality to prove it

[solid kiln]

abraxas

[primal wraith]

oops that should be an x_i my bad

oh we are also given the fact that for t>0
$log(t) \leq t-1$

[solid kiln]

abraxas

[marsh forum]

AM GM inequality I would think

[alpine grove]

They cant use that one

[marsh forum]

why not?

[primal wraith]

the next part wants us to use that to prove AM-GM

[marsh forum]

oh

sorry , didn't read that

maybe induction?

[primal wraith]

i tried induction an di got stuck

[marsh forum]

!show

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[primal wraith]

uhh let me rewrite out my induction since i got rid of it

[alpine grove]

log in this case means base 10 btw?

[primal ingot]

Perhaps they want you to use jensens inequality?

[primal wraith]

log e

[alpine grove]

ln ?

[primal ingot]

It doesn't matter

[primal wraith]

uhm we havent been taught that in lectures so we cant use that

[primal ingot]

Okay, use induction.

It is used in the standard proofs of both AM GM and Jensens I believe.

[digital bolt]

probably it's because log(x) < x

[primal wraith]

this is what i managed to do

i was thinking to use the inequality at the top to

turn the

log(k+1) term into something with just k but that terms engative so idk

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[primal wraith]

<@&286206848099549185>

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[zealous canopy]

s(t) = position as a function of time
s'(t) = velocity as a function of time
s''(t) = acceleration as a function of time
does s'''(t) equal to something?

[unique pine]

yes, jerk

though its not much in use

[stark bison]

"Equal" is kind of a wrong word for this

It's rather "What is s'''(t) named"

Because velocity and acceleration are pretty much defined with those derivatives

[zealous canopy]

ouch :c

[unique pine]

not calling you a jerk 😭 its just called jerk

[stark bison]

They didn't call you a jerk, "jerk" is the answer to your question

[zealous canopy]

like jerk as in a sudden movement?

[zealous canopy]

I see I see XD

[unique pine]

you can think of it that way, sure

[zealous canopy]

okay

what would the unit be?

m^(2)/s?

[stark bison]

m/s^3

[zealous canopy]

oh yeah

[stark bison]

The dimension of s increments for each time derivative taken

[zealous canopy]

that makes sense

[zealous canopy]

yeah yeah, v = m/s; a = m/s^(2); so m/s^(3) would naturaly be the next step

what symbol does it have?

[stark bison]

Should be j

Yeah wiki says j

[zealous canopy]

okay, that makes sense

what scenario would it make sense to use j(t)?

[stark bison]

When acceleration isn't linear

I.e., when j is not a constant

Actually j(t) would make sense in any scenario

You can let functions be constant anyway

[zealous canopy]

okay so when you drive a car on a road with twists and turns and varying speed limits, then it would make sense to use j(t)

right?

[stark bison]

Yeah

[zealous canopy]

okay cool

thank you

both

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[wraith hinge]

If the sine is opp/hyp, you can get cosecant by flipping the sine which is hyp/opp right?
Example: sine=5/10
So the cosecant will be 10/5

[zinc ginkgo]

No

cosecant = 10/5

Use Pythagorean theorem to find cosine

[wraith hinge]

It's a typi

Typo

[sly pewter]

Yeah so you are coorect

[wraith hinge]

Thanks!

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[acoustic mantle]

What would be the neccesary initial conditions? How would you figure that out?

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[silver gust]

I don’t know how to calculate x1

Nothing

[buoyant depot]

can someone pls help me out with this: A binary string is a sequence of digits where each digit is either 0 or 1. A switch in
a string is a point where a digit is followed by a different digit, in other words, it’s an
occurrence of 01 or 10 in the string. For example, the string 001101 has 3 switches.
How many binary strings of length 14 have exactly 5 switches?

[silver gust]

.close

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[heavy elbow]

Hi- I missed this bit in my notes, and none of my friends in my class are online. Can someone help me fill them out?

[heavy elbow]

,rotate

[heavy elbow]

nvm got it

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[wraith hinge]

Do hyperbolas get restricted by the asymptotes all the time?

[wraith hinge]

I swear I saw hyperbola which passed right through a horizontal ashmptote

[vagrant prism]

wdym restricted?

[vagrant prism]

you might be wrong

if it passed through the asymptote, there was no asymptote in the first place

[wraith hinge]

You wanna ask

• Is it necessary there are asymptomes to all hyperbolas?

Then yes, all hyperbolas have asymptomes

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[worldly tartan]

.reopen

we have to find total no of meaningful arrangements. An arrangement will not be meaningful if any closing bracket comes before a opening bracket. Means for every closing bracket an opening bracket must already be there

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Please don't occupy multiple help channels.

[wraith hinge]

Okay so

Can you arrange 6 red balls and 6 blue balls for me, if they are all identical except for their color?

[worldly tartan]

yes but here the problem is there have to be a opening bracket for every closing bracket

[wraith hinge]

Means
((()))()()() Is valid?

[worldly tartan]

yes it is valid

but )()()()()()( is not

[wraith hinge]

Okay

[shrewd ridge]

this is like the caterpillars from a few days ago

[worldly tartan]

so there have to be a opening bracket before a closing bracket

[worldly tartan]

i dont know what is caterpillars from a few days ago??

How to find no of meaningful arrangements of brackets?

[shrewd ridge]

way too hard

[worldly tartan]

yes 😩

<@&286206848099549185>

anyone help!!

[shrewd ridge]

https://en.wikipedia.org/wiki/Catalan_number#Second_proof

this seems almost understandable

[worldly tartan]

thank you. i will check

[shrewd ridge]

when you're below the diagonal, (closing brackets) <= (opening brackets)

and if you pass the diagonal you broke the rule

and every path that does that corresponds to a path that brings you to a certain point near the end

so you can subtract them

this bijection is hard to see, sounds true tho

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[shrewd ridge]

@worldly tartan you don't need the closed form though

i guess it's much easier to do the recurrence

ok you left, superslow thinking on my part

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[viscid flower]

Curious about this,

Show that $|\det A| = \sigma_1 \sigma_2$, where $\sigma_i$ is the ith singular value in the decomposition $A=U\Sigma V^T$.

[solid kiln]

jan Niku

[viscid flower]

oh, wait im dumb and mixed up my problems

nvm

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[frosty heron]

Someone help

[hexed shard]

if $\frac{A}{L}=\lambda$

[solid kiln]

The Great D

[hexed shard]

then $\frac{B}{L}$ which can also be written as $\frac{L-A}{L}$ will equal what?

[solid kiln]

The Great D

[frosty heron]

L - Lambda

L - A/L

Right?

@hexed shard

[hexed shard]

almost but not quite

$\frac{L-A}{L} = \frac{L}{L}-\frac{A}{L}$

[solid kiln]

The Great D

[frosty heron]

Yes

[hexed shard]

and L/L is not equal to L but rather...

1

[frosty heron]

So it’s 1 - A/L

[hexed shard]

yes

and A/L we said was lambda

so your final answer is $1-\lambda$

[solid kiln]

The Great D

[frosty heron]

So mu is equal to 1 - lambda

@hexed shard but in terms of A and B mu is A/B

A/B = 1 - A/L

Bro I think you missed mu here

Is this correct

@hexed shard ??

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[raw magnet]

need some help here

[solid kiln]

nosqldb

[raw magnet]

this is what I've got so far

btw c(n,m) denotes the amount of permutations of [n] with m cycles

<@&286206848099549185>

just a hint is sufficient

$\sum {m=k}^{n}\left(\frac{c\left(n,m\right)}{n!}\sum {n=0}^m\left(c\left(m,n\right)\cdot \left(-1\right)^{n-m}x^n\right)\right)$

[raw magnet]

it should be c(m,n)

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[solid kiln]

Mycobacterium

Mycobacterium

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[full axle]

why are the underlined parts true

[full axle]

Arent V and W by definition open and disjoint, so how can they be closed in U?

[left oriole]

that's what it means to be (topologically) disconnected

[full axle]

?

what abt the bottom parts

how is W and V closed in U

[left oriole]

The complement of W in U is V, and V is open, so W is closed

whenever a set is topologically disconnected, it can be partitioned into two nonempty sets, each of which is both closed and open

aka "clopen"

[full axle]

wait but it defined V and W to be open

[left oriole]

yes

but it's also true that V and W are each other's complements

and the complement of an open set is closed

thus they are both also closed

[full axle]

so if i undersand correctly, V, W are open in the complex plane but closed in U?

[left oriole]

they are both open and closed in U

and yes they are both open in C

[full axle]

ok

thx

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[wraith hinge]

question asks to rationalize the numerator

[blazing geode]

Multiply

[wraith hinge]

I've tried multiplying by sqrt(x+h)+x

divided

by sqrt(x+h)+x

[blazing geode]

Multiply the conjugate

[wraith hinge]

mhm

done that

then i have

$

$\frac{h-x}{\sqrt{x+h}+x}$

[solid kiln]

mj

[wraith hinge]

after factoring out an x

from top and bottom

problem is

this isn't the answer

it simplifies more apparently

or

it doesn't and im wrong

[blazing geode]

Someone help him

He’s in urgent need

He cannot breathe being overwhelmed by the question

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[lethal ivy]

Why is there no restriction for y?

[rancid fulcrum]

please help

[hallow spruce]

!occupied

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[lethal ivy]

Im confused why there is no restriction for y?

[hallow spruce]

you restrict one or the other

x =/= +/- 3y is the same thing as y =/= +/- x/3

the two are equivalent

[lethal ivy]

But for others its not like that like ill show u

How come both restricted here?

[hallow spruce]

because you have denominators that involve only x, or only y

[lethal ivy]

No both have x and one has both

[hallow spruce]

the final answer has only y

but also, you can just solve 30xy^4 = 0

[rancid fulcrum]

where to find unoccupied

[hallow spruce]

read the bot message again

[rancid fulcrum]

oh yes

[lethal ivy]

Why does that matter? My teacher says to just look for factored form and do restrictions from there

So she says dont mind anything else

[hallow spruce]

your restrictions are wherever you have a denominator of 0

if you solve 30xy^4 = 0, you will find x=0 or y=0

whereas in your first question, solving x^2 - 9y^2 = 0 gives two possible forms of the desired answer

[lethal ivy]

She says to just look at the factored line

But not reduced part

When looking fir restrictions

[hallow spruce]

what's a "factored line"

[lethal ivy]

Like this:

I circled in black

Thas the line that is fully factored but not reduced yet

This is it for other equation

[hallow spruce]

so in the circled part, you're supposed to solve for when the denominator is 0

[lethal ivy]

Ya like whatever makes denominator 0 thas restriction

[hallow spruce]

though the second question has the added step that you have to solve for both the numerator and denominator being 0

[lethal ivy]

? Wdym

[hallow spruce]

if you have a / (b/c), if b=0 then you have a/0; if c=0 then you have b/0

so you need both b and c to not be 0

[lethal ivy]

O yes ic what u mean

Ya so if y is 0 then that would ruin equation but for first one if y and x both are 0 then that would ruin equation

So y dont y have a restriction ?

[hallow spruce]

if y is 0 and x is anything other than 0 it's fine

which is why your restriction involves x and y at the same time

[lethal ivy]

Alr but how abt this one

If x is 0 then equation ruined

And if y is 0 same thing

Why do we have to write both(

?*

[hallow spruce]

in questions 2 and 3, if x=0, things are undefined regardless of the value of y

in question 1, if x=0, things are undefined only if y=0

[lethal ivy]

So why does y have a restriction

[hallow spruce]

because the reverse is also true

if y=0, the first term is undefined regardless of the value of x

[lethal ivy]

Ic okay. U have to think of the possibilities

It looks like

[hallow spruce]

the first term is this in your 3rd question for clarification

[lethal ivy]

Thx so much

I got it

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[verbal void]

for an arbitrary complex $\lambda$ and $a_1,\dots,a_n,b_1,\dots,b_n\in\bC$, how do we get that $$\sum_{i=1}^n |a_i|^2 + |\lambda|^2\sum_{i=1}^n |b_i|^2 - 2Re(\bar{\lambda}\sum_{i=1}^n a_ib_i) \geq 0$$?

[solid kiln]

CoolShot

[verbal void]

(its a step in proving cauchy schwartz)

prof assumed this to be true with no justification

but i dont understand it

[zinc ginkgo]

Classic proof left to reader

[verbal void]

lol yeah

inductive proof is fine

i remember some other proof from LA too

just never seen this complex one

[zinc ginkgo]

Something like |a - lambda * b|^2>=0

since |z|^2 >= 0 for all complex z.

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[carmine narwhal]

This may be slightly unrelated since its physics, but I am struggling to get the correct answer

[carmine narwhal]

https://cdn.discordapp.com/attachments/1182100918137786421/1203939326124367922/image.png?ex=65d2eadf&is=65c075df&hm=3ecf42235aabdbd9632a35fe10af1cb9671d51c611665b55fa047a81ca62416f&

my work so far

(from a different problem with similar numbers, but smae process

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[verbal void]

looks like an exam man

[verbal void]

uh

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[carmine narwhal]

its pearson

hw

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[graceful zenith]

does this have solutions between 0 and 360

[tepid rock]

What do you think?

[graceful zenith]

no

[tepid rock]

Why?

[graceful zenith]

arc sin -1/2 is -30

so it will be negative for 2nd quad

and positive for 3rd and 4th quad

but since multiplying any degree in the 3rd and 4th quad by 2 (excpet 180) would be greater than 360, there would be no solutions

[tepid rock]

You're right, it has to be greater than 360.

[graceful zenith]

so im right?

[tepid rock]

You are.

[graceful zenith]

ohk thx for clearing that up

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[final cypress]

\

[final cypress]

i need help

[runic dew]

We don't help with exams

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@final cypress Has your question been resolved?

[final cypress]

how is that an exam question

[final cypress]

do you know what homework is

[runic dew]

!status

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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

[final cypress]

4

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[reef patrol]

I don’t get 3rd step. We do with + or -, but he did with multiplication

[reef patrol]

Sin(x+x)= 2sinxcosx

sin(x*x)=?

[jaunty silo]

So you are asking about how $\cos{\left(\frac{x}{2}\times 2\right)}=\cos{\frac{x}{2}}^2-\sin{\frac{x}{2}}^2$?

[solid kiln]

Dri111

[reef patrol]

Yes

I don’t get it

[jaunty silo]

do you know cos(a+b)=?

[reef patrol]

cosacosb-sinasinb

[jaunty silo]

ok then what about cos(a+a), or in other word if substitute b as also a

[reef patrol]

cos^2a-sin^2a

[jaunty silo]

yes

[reef patrol]

But he used multiplication

[jaunty silo]

so $\cos{2a}=(\cos{a})^2-(\sin{a})^2$ right?

[reef patrol]

Uhm

(cos^2a)^2?

[jaunty silo]

ah sorry

[solid kiln]

Dri111

[reef patrol]

Yes

[jaunty silo]

then if $a=\frac{x}{2}$

[solid kiln]

Dri111

[jaunty silo]

would it be same as what you are asking?

[reef patrol]

It is clear to me where we use plus and minus, he used multiplication, how did he use multiplication to do that?

[jaunty silo]

or you can change your question into $\cos{(\frac{x}{2}+\frac{x}{2})}$

[solid kiln]

Dri111

[reef patrol]

And yep that would be true

I just don’t know why he skipped that step

[jaunty silo]

I think he might thought that viewers would already knew it well

[reef patrol]

Okok

Thanks tho

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[tranquil widget]

hi

[tranquil widget]

what does this represent

anyone

[eager sedge]

i think i just THINK idk it might be line with angle 2pi/3

from the positive x axis

@tranquil widget

[tranquil widget]

how do you get that

like

its a point

between -1 and 1

but like

what

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[jaunty silo]

wdym?

[tranquil widget]

i have no idea

💀

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[tranquil widget]

.reopen

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✅

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@tranquil widget Has your question been resolved?

[tranquil widget]

i give up

.close

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[little egret]

excuse me

[little egret]

is this right

[little egret]

for part E

[grim sparrow]

I think you slightly overcomplicated your argument

you could just argue that 4m^2 = 2(2m^2) and is thus even no matter what m is

[little egret]

did i do it wrong then

[grim sparrow]

eh, not necessarily

but it seems a bit roundabout

[little egret]

haha idk i was just like hey i can sqrt this

and i just did it

anyways thanks :P

[grim sparrow]

actually hold on

[little egret]

oh.

[grim sparrow]

by square rooting the right, you also have to sqrt the left

[little egret]

,,,,,,,,,,,,,,,,,,,,,,,

im gonna cry

[grim sparrow]

that just gets you back to n = 2m

[little egret]

ok thanks :P

that wouldve been the stupidest mistake

in a test

hahahaha

[grim sparrow]

Yeah always look for a simpler argument lol

[little egret]

.close

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[fossil cobalt]

i do not know how they reached the answr

[blazing geode]

so you have f'

[fossil cobalt]

yes

[blazing geode]

and f is a two-degree function

[fossil cobalt]

what does this mean

[blazing geode]

Have you learned about calc yet

some high school calc is what i was mentioning

[fossil cobalt]

i am learning antiderivatives now

[blazing geode]

I see

Do you agree that 3x+4 is f'(x) aka the antiderivative of f(x)

as for we learnt that the slope of a tangent line of a function at a certain points could be expressed as f'(x)

[fossil cobalt]

yes

i found the anti derivative of x3x+4

[blazing geode]

yes

tell me what it is

[fossil cobalt]

i got 3/2 ( x^2) +4x + c

[blazing geode]

smart people

intelligent human

how to get the unknown c?

there's something left unused in the description.

[fossil cobalt]

we plugin 2 and 6

[blazing geode]

intellegent people

[fossil cobalt]

3 ( 2) + 4 (2) + c = 6

c= -4

[blazing geode]

what

what

c= -4?

[fossil cobalt]

i plugged in

[blazing geode]

I got c=-8

[fossil cobalt]

opps

yes

[blazing geode]

reckless people

then what should you do

[fossil cobalt]

i plugin -2?

[blazing geode]

what the question was asking you for

[fossil cobalt]

in 3/2 ( x^2) +4x - 8

[blazing geode]

terrific

smart people

[fossil cobalt]

k thanks

[blazing geode]

no problem, elon musk

[fossil cobalt]

this

can you help me

i found x(t)

antiderivative of v(t)

right?

[blazing geode]

man have to clean his anatomy

[fossil cobalt]

nvm i got it

[blazing geode]

smart people

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[astral tangle]

#help-47 message

[astral tangle]

What approach should I follow next?

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[bleak maple]

Can someone help with this?

[trim jolt]

Please don't occupy multiple help channels.

[wraith hinge]

what's your problem with it currently

and what have you tried

[bleak maple]

I don't know how to solve it

[wraith hinge]

what's the objective. Simplification?

[bleak maple]

Yes I did, but I always get wrong anwser

[bleak maple]

yeah

[wraith hinge]

alright

[wraith hinge]

show what you did

[bleak maple]

Gimme sec

[jaunty scarab]

Okay. Start by factoring the insides of every sqrt

[bleak maple]

okay.

[wraith hinge]

just wait a bit until they post their work

[jaunty scarab]

fair enough

[wraith hinge]

we dont even know what made them get stuck

[jaunty scarab]

yeah, better he posts what he already did, mb

[heady mica]

a)6 * sqrt(3) + 4 * sqrt(3) + (-6 * sqrt(3)) + 42 * sqrt(3) = 46 * square root of 3

[bleak maple]

im sorry for waiting so long, I wasn't been in school for 4 weeks, and idk how to solve this

[heady mica]

I can teach you

[bleak maple]

okay thank you

[heady mica]

First

Square root is an operator

Which have some properties

[bleak maple]

Okay

[heady mica]

If everything has the same number as it’s multiple, you can change basis