help-27

[devout snow]

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[lament prawn]

does anyone know what the red curve represents?

[lament prawn]

what is the eqn to plot it. https://eaglepubs.erau.edu/introductiontoaerospaceflightvehicles/chapter/steady-level-flight-performance/

[devout snow]

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[lament prawn]

.close

[devout snow]

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[devout snow]

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[atomic idol]

How do I prove these statements are true? I’m suppose for use induction but the explanation for that isn’t very clear and idk if I did it right
I just wrote down some vague ideas

[atomic idol]

How do I prove n+1 other than just plugging in a specific value

I feel like saying „it’s true for n=1 and n=2“ isn’t enough to show it’ll always work

[harsh sierra]

Yeah you need to prove n=1, which is the base case, then assuming its true for n=k, you want to show its true for n=k+1

[atomic idol]

What does assuming it for n=k even mean
Wouldn’t that assume that it’ll be true no matter what n is?

[harsh sierra]

Assume the statement is true for a specific n

Whatz important its not any n, its some specific, and we aant to show it implies the next case

For example in 2,
You want to show
1 + 2 + ... + n = n(n+1)/2
implies
1 + 2 + ... + (n+1) = (n+1)(n+2)/2

[atomic idol]

So I need to rewrite it in a way that uses n+1?

[harsh sierra]

Its not exactly rewriting

[atomic idol]

Im not familiar with English terminology too much

[harsh sierra]

Yeah so no problem

[atomic idol]

How would I write the first one using n+1

That big pi symbol just means the same thing as the gaus thing with multiplication instead of addition right

[atomic idol]

What about this tho? I’m still pretty confused

[harsh sierra]

Plug in n+1 for n

Thats the statement for n+1

[atomic idol]

I got that part

Idk how it’d look for this specific function because I don’t know how to deal with the multiplication

[harsh sierra]

Its the same as addition, just multiplciation

[atomic idol]

Would I write it as sth similar to pi(1)*pi(2)…p(n+1)?

[harsh sierra]

Similar yes

You would write it as
(1-1/4)(1-1/16)....(1-1/(4(n+1)^2))

[devout snow]

@atomic idol Has your question been resolved?

[atomic idol]

Since I don’t know the amount of steps (because n isn’t clear)
How do I calculate that value to prove that it’s right

[harsh sierra]

You know the amount

The amount is n

What you need to use is that
a1 * a2 * ...* an * a_(n+1) =
(a1 * a2 * .... * an) * a_n+1

Where here you can use the assumption that the statement holds for n

[atomic idol]

The assumption is what proves it?

[harsh sierra]

The assumption helps

That is usually the point of induction. You can't solve the nth case easily by itself, but you can solve it from the previous case

[atomic idol]

Like factorials

[stiff karma]

I need your pfp @atomic idol

?av 1083137780877643896

[atomic idol]

Im Sorry I don’t have it as a picture on my phone

Found it on google

[stiff karma]

Ok I have it

[atomic idol]

Omfg

What have I started

[devout snow]

@atomic idol Has your question been resolved?

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[crystal dawn]

.close

[devout snow]

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[crystal dawn]

<@&268886789983436800> here too

[devout snow]

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[devout snow]

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[cerulean ruin]

I am

[cerulean ruin]

kind of stuck here

my idea is to show that

$$ \forall n \in \mathbb{Z}^{\geq 3}, f^{(n)}(z_0) = \dfrac{n!}{2\pi i}\displaystyle\int_{C_R} \dfrac{f(z)}{(z-z_0)^{n+1}) dz = 0, \quad C_R = {z: |z - z_0| \leq R}$$

[woven radish]

toast
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cerulean ruin]

$$
[
\forall n \in \mathbb{Z}{\ge 3}, \quad
f^{(n)}(z_0)
= \frac{n!}{2\pi i}
\int{C_R} \frac{f(z)}{(z - z_0)^{n+1}} , dz
= 0,
\quad
C_R = { z : |z - z_0| \le R }.
]
$$

[woven radish]

toast
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cerulean ruin]

essentially i want to show this

but i am not sure how to use f(z) here

im guessing we want to solve this integral from 0 to 2pi

oh do we plug in the circle equation z_0 + Re^{i\theta} into z

oh theres this fmla

[supple knot]

you just need to bound the integral and let R go to infinity

bound the nth derivative in absolute value and then bring the absoluate value bars inside the integral

| int (stuff) dz | <= int |stuff dz |

[devout snow]

@cerulean ruin Has your question been resolved?

[devout snow]

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[devout snow]

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[hazy cypress]

how do i go from
$$\sqrt{\cos2\theta+\frac{\sin^22\theta}{\cos2\theta}}$$ to $$\sqrt{\frac{\cos^22\theta+\sin^22\theta}{\cos2\theta}}$$

[dire forge]

what are you trying to do? simplify it?

[woven radish]

Weathers

[dire forge]

oh okay

you're putting everything under a common denominator

so that you can add the two expressions together

it's like when you do 2 + 1/2 = 4/2 + 1/2 = 5/2

[hazy cypress]

that's almost an apples to oranges comparison

i guess you could say:

[lethal wraith]

replace the numbers with functions

same thing

[hazy cypress]

$\frac{cos^22\theta}{cos2\theta}$ but that doesn't look right

[woven radish]

Weathers

[lethal wraith]

i wanna hear the joke now

[lethal wraith]

that's an intermediate step

[hazy cypress]

i mean

.......

oh.

[lethal wraith]

doesn't have to be reduced

[versed juniper]

sometimes in math you make things more complicated to get a simpler result

[hazy cypress]

$$\sqrt{\cos2\theta+\frac{\sin^22\theta}{\cos2\theta}}$$
$$\sqrt{\frac{\cos^22\theta}{\cos2\theta}+\frac{\sin^22\theta}{\cos2\theta}}$$ $$\sqrt{\frac{\cos^22\theta+\sin^22\theta}{\cos2\theta}}$$

[woven radish]

Weathers

[hazy cypress]

like this, right?

[lethal wraith]

yes

[hazy cypress]

that's so stupid

.closed

.close

[devout snow]

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[lethal wraith]

but then you have sin^2 + cos^2

what does that tell us

[hazy cypress]

it's a part of a bigger problem, it cancels with another sqrtcos2theta

[lethal wraith]

ah

[devout snow]

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[frosty pendant]

need help with proofs
If every room in a house has an even number of doors, then show that the number of door which
open outside must be even as well.
how to show/prove this without graph theory

[full geode]

Why don’t you want graph theory?

[frosty pendant]

we've been asked to figure out a way to prove it without graph theory

[full geode]

Maybe try induction on the number of rooms?

[versed juniper]

doors that connect rooms count as 2 doors, 1 for each room
what about doors that connect outside?

[frosty pendant]

how

[full geode]

Well consider 1 room. Any amount of doors you place will lead to the outside. Then supposing you have k rooms you can look at what happens when you add an additional room

It must be in a spot that was once “outside” right? What does that give you ?

That or it’s disconnected and there is nothing to do

[frosty pendant]

oh so we can add rooms to show that? i thought the number of rooms woulf have yo be fixed

[full geode]

Well to get n rooms, you need n - 1 rooms

[frosty pendant]

right

but then even if i use induction how do i solve it for k in the first place?

[full geode]

What do you mean solve for k?

[full geode]

Try working with this

So remove some room and use your induction hypothesis to make an argument on what changes and how including that room affected things, there’s not really a strong algebraic way of writing this without a model

[frosty pendant]

ah okay

thabk you so much

thank*

[full geode]

No problem!

[frosty pendant]

.close

[devout snow]

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[devout snow]

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[eager stone]

I dont understand what happened at the 5^2= 2(mod 23) step

[supple knot]

25 - 23 = 2

[eager stone]

we can do that..?

[supple knot]

one of the basic properties of modular arithmetic

[eager stone]

ok ill study it again lol ty

[supple knot]

https://brilliant.org/wiki/modular-arithmetic/

[eager stone]

.close

[devout snow]

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[devout snow]

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[brisk quail]

please can somebody explain to me why these two work? i have some trouble understanding the topic

[drifting sierra]

,rccw

[drifting sierra]

Why the sums on the left equal the formulas on the right?

[brisk quail]

yes

[drifting sierra]

,, \sum_{r=1}^n 1 = \underbrace{1+1+...+1}_{n\text{ times}}

[woven radish]

Nel

[drifting sierra]

r goes from 1 to n

[brisk quail]

ahh yes this is straight forward

thank you

what about the other?

[drifting sierra]

,, \sum_{r=1}^n r = 1+2+...+n

[woven radish]

Nel

[drifting sierra]

As to why that equals n(n+1)/2, check out https://en.wikipedia.org/wiki/Triangular_number#Formula, there is a visual proof and a lot more details if you care to dive deeper

[brisk quail]

i'll check it out now, tyvm

[devout snow]

@brisk quail Has your question been resolved?

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[marble surge]

Hello I’m wondering what do I use to add on to step 3 that BEA and BED are supplementary

[marble surge]

<@&286206848099549185>

[thin fern]

what is a linear pair?

is that just two angles that form a line

[marble surge]

No

Actually yes right?

Cause 2 adjacent angles would be needed

[thin fern]

Yeah it is I just googled it

so your answer is correct

[drifting sierra]

I think you misread OP's question

[drifting sierra]

What options do you have?

[thin fern]

Oh they're asking about step 2

I think?

[drifting sierra]

Yes, which presumably is something required for steps 3 and 4

[marble surge]

I thought AEB was congruent to CDB because they were vertical angles

But then I realized how wrong it was

[drifting sierra]

What happens when you click on that box

[drifting sierra]

Scroll down to show the rest?

[drifting sierra]

odd

Oh, maybe "classify a triangle"?

[marble surge]

Oh ok so say Bdc is a triangle?

Isosceles I mean

[drifting sierra]

Not BDC

[marble surge]

Oh nvm I mean BED

[drifting sierra]

Yes

[marble surge]

So pick because it has 2 congruent sides?

[drifting sierra]

Yes

I would've expected two steps though, the other being that BED = BDE (angles)

[marble surge]

Oh shi it was actually this instead😭

[drifting sierra]

Oh

That's kinda stupid

[marble surge]

I agree

Thanks for the help though

[drifting sierra]

Yeah I expected the answer to be either one, but really it should have been both

Anyway...

[marble surge]

Wait do you know what would be the missing one for this?

.close

[devout snow]

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[honest garnet]

bro help im doing the same hw

[honest garnet]

on deltamath

[solar goblet]

what have you tried?

can you show your current work?

[devout snow]

@honest garnet Has your question been resolved?

[devout snow]

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[fickle chasm]

guys do i take precalc in 9th grade and go to tj or should i go to acl and take integrated math 1

[fickle chasm]

i'm in al2 trig in 8th

idk what to take next yr

[upbeat forge]

perhaps look at both subjects and see which topic(s) are more useful/appealing to you.

[thin fern]

tj as in TJHSST

?

[devout snow]

@fickle chasm Has your question been resolved?

[fickle chasm]

ya

[devout snow]

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[young siren]

Hello! I need a lot of help with solving these types of problems. i missed a lot of my classes and am very confused and stressed as I have a test this friday

[kindred urchin]

hey so what exactly are you confused about?

[young siren]

I don't understand trigonometiric equations/identities. I have a really hard time starting these problems because I don't know where to start. My teacher gave us a lot of equations to identify and use with the problems, but i honestly don't know them

[kindred urchin]

Okay, what do you know about the trigonometric identities

Let's start there

[young siren]

Well... lol. Um, i guess i can kinda understand how to solve soome equations. for example: verifying. My teacher taught us to start with the more complex side and break it down into sin's and cos's. But then he started throwing in equations we should identify. Problem number one i understand completely, so I don't really need help with that one

I also get the gist of (cos,sin)

[kindred urchin]

Makes sense, yeah that's generally what you should do if working with tans and sec's are a bit complicated

however what makes them different are the domain restrictions

are you aware of the unit circle?

[young siren]

Yes, I have one on me

we have to use that to get probably most of our answers

[kindred urchin]

yeah that makes sense, will you have that for the test on friday

?

[young siren]

yes

[kindred urchin]

okay great

on problem 2a, how would you start

[young siren]

well, I had asked for some help so I feel most confident with that one. I know to start witht he sin in the parenthesis. on the unit circle sin 7pi/6 is -1/2

[kindred urchin]

Great

[young siren]

I know the answer to this one, but i dont know how to solve it exactly. I dont want to memorize, I need to know how to do it

[kindred urchin]

okay so you understand that the inside of the inverse sine function becomes -1/2 right

[young siren]

yes

[kindred urchin]

do you know how the inverse trig functions work

[young siren]

I would like to say yes, but i dont feel confident to explain

[kindred urchin]

okay

at its most basic level, they are the inverse operations of the trigonometric functions

similar to how ln(e^x) just becomes x, inverse sin of sin(x) just becomes x

[young siren]

okay

[kindred urchin]

the trigonometric functions take in a value theta and give out a ratio

the inverse functions do the opposite by taking in a ratio and giving outa value theta

theta is just the angle

[young siren]

right

[kindred urchin]

so the domains and ranges of the inverse functions are swapped for the original trig functions

so lets take an example

whats sin(45) or sin(pi/4)

[young siren]

like from the unit circ itself?

[kindred urchin]

yeah

the answer is the same regardless of if its from the calculator or unit circle or anything

[young siren]

root 2/2

[kindred urchin]

rigth

[young siren]

for sin (45)

[kindred urchin]

so then

that implies the inverse sine of sqrt(2)/2 is what

[young siren]

2/root 2...?

im sorry i am so bad at math

and not very confident

[kindred urchin]

no its completely fine

the inverse sine function essentially takes in a ratio value and gives out an angle

so you know that sin(45) = root 2 divided by 2

the inverse sine function does the opposite

so the inverse sine of root 2 divided by 2 is 45 degrees

it essentially does the opposite or undoes the sine function by going backwards

[young siren]

ok so like cos(45) inverse would aslo be root2/2

[kindred urchin]

sort of

[young siren]

or does cos not apply in the same way

[kindred urchin]

the inverse trigonometric functions do not take in ANGLE or RADIAN measures

they are not the inputs, they are the outputs

they're just undoing each other

does this sort of make sense?

[young siren]

yeah

so for problem 2 part a), confirmed by my teacher the answer is -pie/6. I can understand how he got there, kinda

i get really confused on which way you have to go on the unit circle

[kindred urchin]

I see

well

how would you evaluate sin(pi/4) using the unit circle

you would look at the angle/radian, pi/4, and then look for sine value that corresponds to that measure right?

[young siren]

right

[kindred urchin]

so working backwards

when you're asked to find the inverse sine of root 2 divided by 2

you would look at where root 2 divided by 2 is on the unit circle first, and then see which angle corresponds to that value

[young siren]

right

[kindred urchin]

does that make sense

[young siren]

yes

[kindred urchin]

so for the problem 2a

[young siren]

I can get up to the point where you have to find the corresponding "coordinate". But i don't understand the direction. Like how do you get to pie/6 but then its negative?

[kindred urchin]

wdym

[young siren]

I literally do not know... I just know the answer is -pie/6. my tecaher explained that the other -1/2 cannot be used because its too close to 2pie? but it works at pie/6. But then you somehow make that a negative

in the fourth quad^ for -1/2

[kindred urchin]

okay

i get it

alright so the inverse sine function have domain restrictions

[young siren]

[-1,1}

[kindred urchin]

yeah

because of that you can only give out values from -pi/2 to pi/2

[young siren]

right

[kindred urchin]

thats why you can only use the first and the fourth quadrant

now -pi/6 is another way to write 11pi/6

which is in the domain of the arcsine function

[young siren]

so why -pi/6 and not 11pi/6

for the answer

[kindred urchin]

it's because 11pi/6 isn't in the range for arcsin(x)

even though they're the same value

it's sort of confusing but bear with me

the outputs of the arcsin function must always be within -pi/2 to pi/2

so we always convert our final answer to fall within that range

[young siren]

i think i am getting a better understanding

also quick question because I do have to go

[kindred urchin]

for sure, no worries

[young siren]

when I look at the unit circle, I envision a graph. so can I imagine 180 deg and 0 (-1,0)(1,0) as the domain and basically the y-axis as range

like was that the concept the whole time lol

so (-1,0)(1,0) = [-1,1]

[kindred urchin]

oh yeah, something like that fs

[young siren]

word

[kindred urchin]

yeah

[young siren]

okay thannk you

I really appreciate the time LOL

[kindred urchin]

for sure man np

have a good one

[young siren]

you too

[kindred urchin]

and gl on the test

[young siren]

thank you

.close

[devout snow]

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[devout snow]

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[keen sundial]

what did i do wrong in q 3

[carmine veldt]

last term is 10x25 + 3

[keen sundial]

oh I see

how did i even get 263

so from 0 to 25

are there 25 terms?

or 26

or is that from 1 to 25

[restive river]

i NEVER NEW KURAPIKA LIKED MATH

sorry caps

[carmine veldt]

there are 26 terms, you're right

but check the last term in the summation

n goes from 0 to 25 right?

[uneven coral]

I'd suggest you to remove the +3 (just add 26*3 later), and do solely the 10n part

That'd be easier imo

[keen sundial]

wdym

could u write it down I think ill understand

[uneven coral]

Like it's basically (0 + 3) + (10 + 3) + (20 + 3)..........(250 + 3)

So add the +3s

Then it's simply (0 + 10 + 20 + 30 +.......+ 250) + (3*26)

[full sand]

Hi does anyone know about boleean algebra?

[carmine veldt]

basics

send your question in diff channel

this one's occupied

[full sand]

I'm having an issue but I dont know where the mistake is

okay thanks

[devout snow]

@keen sundial Has your question been resolved?

[kindred urchin]

🙂

[uneven coral]

@keen sundial you did it?

[devout snow]

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[drowsy parcel]

How and what the teacher did in 2nd method..
i don't get it..
maybe it is derived from the first method..

please help..i am stuck here

[pseudo basin]

entire work is 60 units, A does 6 units per day and B does 5

if A didn't fuck off for the last day she would have done 6 more units of work meaning 66 total

at a combined work rate of 11 units/day that means 6 days

[drowsy parcel]

why add 6..
isn't 60 the whole work

[pseudo basin]

if A didn't fuck off for the last day she would have done 6 more units of work meaning 66 total

in our scenario we are told A didn't work on the last day

so we imagine, well what if she DID work

[pseudo basin]

i also literally said "A would have done 6 more units of work"

[drowsy parcel]

https://tenor.com/view/brain-not-braining-confused-mental-health-brain-needs-oxygen-gif-2319166761334254514

what should i do

[pseudo basin]

idk, i don't speak hindi so i can't translate it for you

[drowsy parcel]

that is the issue...
there is no help further by him

[pseudo basin]

by whom?

do not call me "bro".

[drowsy parcel]

teacher..

i think i made her mad

[pseudo basin]

you didn't

[uneven coral]

Let's respect each other eh? It'd be nice if you deleted that

[dark surge]

Unrelated but are you by chance british

[uneven coral]

What makes you say that?

[dark surge]

Idk its just the way you type

I usually see british people say eh

[atomic hedge]

No, that’s a Canadian thing

[dark surge]

is it?

oh yeah now that I think about it

[pseudo basin]

!redir

[devout snow]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[dark surge]

😔

[uneven coral]

@pseudo basin the hindi part is only the translation of the question

[pseudo basin]

no i mean like

translate the stuff im saying into hindi

[uneven coral]

Alright then

[pseudo basin]

cause wizard is struggling to understand me

[pseudo basin]

from here

[uneven coral]

Alright

[uneven coral]

Haan to bhai poora work 60 units hai

A 6 units per day karta hai aur B 5 units per day

[drowsy parcel]

maybe that helps.ok go

[uneven coral]

Agar A aakhri din na jaati, to vo 6 units zyada kaam kar leti

Yaani total 66

Saath mein work karne ka rate 11 units/day tha na, to 6 din donon saath mein kaam karte to 6×11 total 66 units kar lete

Abhi ke liye ham bas maan rahe hain ki A aakhri din tak thi thik hai, question mein bola nahi thi lekin use solve karne ke liye ham ye maan lete hain ki chalo bhai thi

[drowsy parcel]

to 60 me A ka akhri din add nhi hai?

[drowsy parcel]

.

[uneven coral]

basically haan

60 tab hai jab A ek din pehle chali gayi

[drowsy parcel]

pr kaise.
60 to total work hai na

[uneven coral]

Aur 66 tab hota agar A us aakhri din bhi kaam karti

[uneven coral]

Acha dekho

Agar main roz 5 rupees, aur tum roz 6 rupees kamaate ho

Aur ham dekhte hain ki bhai 6 din mein total kitna kama lein

To mere 6×5, aur tumhare 6×6

Total 66 rupees kamae

[vital edge]

Might be a better idea to think of how much work must be left on the last day

Instead of adding an extra 6

That might be more intuitive no?

[uneven coral]

lekin agar last day tum paise contribute na karo, to phir sirf 6×5 aur tumahre honge 5×6 (6th din nahi gina)

Tab total 60 rupees hi honge

basically aisa hai

[uneven coral]

True, but I'm just following on with what Ann started

[vital edge]

Fair

[drowsy parcel]

to LCM total work nhi btata?

[uneven coral]

Haan batata hai, lekin tab jab donon ne same dinon tak kaam kara ho

Acha nahi ruko dekho

Are yaar, matlab zyada complicate ho gaya ye

A 10 din mein kar legi

B 12 mein kar lega

LCM se total work nikal aaya

joki hai 60

[drowsy parcel]

hn

[uneven coral]

Haan to ab donon ko alag alag 60 se divide kar lo

A ka aa jayega 6, aur B ka 5

Kya hua?

[drowsy parcel]

hn..theek hai...so A 1 din me 6 unit kaam krdega..aur B 5 unit.

[uneven coral]

Haan basically

To ab, pehle x din, donon bhai behen saath mein kaam karenge

Yaani (6 + 5) kaam hoga har din, aur x dinon mein (6 + 5)x kaam hoga

Joki 11x ho gaya

Ye samajh gaye?

[drowsy parcel]

haan

[uneven coral]

Theek hai

Ab aakhri din A to chali gayi, yaani ek din B ne akele kaam kara, to total work aakhri din sirf 5 hua

Ab total work 60 tha, jo pehle nikaal liya tha

To in dinon mein, jo total kaam hua hai, vo hai 11x (pehle x din, jab donon saath mein kaam kar rahe the) + 5

Donon ko baraabar kar lo

11x + 5 = 60

[drowsy parcel]

okay..

fir x nikal aega...joki a and b dono kaam days hoga...
ek din B ne kaam kiya...

[uneven coral]

Haan bhai

To jo x nikalega

usmein vo ek din jod dena jismein B ne akele kaam kiya, total number of days mein to vo din bhi aaega na

[uneven coral]

Ab isko solve kar ke x aa jayega 5

Usmein 1 din B waala jod do

Total din aa gaye 6

[drowsy parcel]

are yr...
fir method 1 pe aa gye..
sorry yr...
method 2 smjna haj muje..
muje time chahiye smjne me..
mai maths me weak hu yr..

[uneven coral]

Koi nahi bhai, method 2 thoda zyada jhilau hai, lekin haan zyada tez hai to usko bhi dekh lete hain

Haan to ismein waheen aa jayega

Maan lo A ne saare din kaam kara hota

[drowsy parcel]

ok

[uneven coral]

To pehle to hamara LCM 60 aa raha tha, jismein A ek din pehle chali gayi

[drowsy parcel]

haa

[uneven coral]

Ab maan lo agar A us din bhi kaam karti, tab total work done 60 + 6, yaani 66 ho jaata

Tumko bhai kaheen koi samasya ho to khul ke bol do, koi dikkat nahi

[drowsy parcel]

ek aur question bheju...to shyd usse jyada clear ho pae..usme bhi yhi method use kiya hai..pr minus kiya hai total work me..shyd us ke saath comparison me smjana easy ho jaega

[uneven coral]

Haan bhai jaisa tumhe sahi lage

[drowsy parcel]

method 2 only..

yr thoda complicated hogya hai

[uneven coral]

Waheen to, lekin shayad agar samajh jao to help mil jaye

Acha bhai isko aise hi samjho

Maan lo main 24 rupees aur tum 32 rupees kama rahe ho din ke

[drowsy parcel]

hn

ok

[uneven coral]

Aur hamne LCM nikaala, to aaya 96, yaaki mujhe 96 kamaane mein 4 din lagg jayenge aur tumhe 3

[drowsy parcel]

hn ok

[uneven coral]

Bhai yaar ye complicated ho raha hai

[drowsy parcel]

Break lena better hoga

[uneven coral]

Kehne ko main normally samjha to loon, lekin usmein kaho samajh aaye na aaye

[drowsy parcel]

thank you for the efforts..
mai isi ko reread krlunga baad me..

[uneven coral]

Are bhai ab saral shabdon mein bolun, to inhone jo makkari us ek aadmi ne kari ek din na aake, use bhi jod ke (ki maan lo vo agar us din aa gaya hota) donon ke total work/day se divide kar diya hai, isse inhe total din mil gaye hain

Kyunki aise mein donon ne same days kaam kara hai, to seedha total work done nikaalo, donon ka work done/day se divide maar do, aur aise mil jayega number of days

Ab bhai isse koi help mile to le lena🙏

[drowsy parcel]

I will reread this later...

.close

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[solar goblet]

quick question, a function's domain need not be {1, ..., n} right?

since for n > 2 there isn't any one-to-one function

thanks

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[midnight echo]

wait

.reopen

[devout snow]

✅ Original question: #help-27 message

[solar goblet]

yes?

[midnight echo]

this is not true fyi

wait nvm I'm wrong, confused it with onto

disregard that

[solar goblet]

yeah

[midnight echo]

Yeah that's all, your initial statement is still true

[solar goblet]

alright thanks

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[arctic field]

i would think that given the wording of the question, they are asking explicitly for functions f : {1, ..., n} -> {0, 1}

[solar goblet]

yeah that's what i think too

[solar goblet]

but considering the chapter is about counting, this interpretation seems more likely

[arctic field]

i mean the answer to a would simply be 0

theres nothing wrong with that

[harsh sierra]

That is only if n > 2

[solar goblet]

this is the same book that gave me a really bad time on the logic chapter because their exercises are pretty poorly worded, so i wouldn't trust that they really meant your interpretation

[arctic field]

[harsh sierra]

I agree with Coriolanus

I don't think there is a mistake

[devout snow]

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[solar goblet]

.i checked the solution, your interpretation is correct, my bad

[prime sphinx]

how do u solve this

[soft umbra]

What have you tried?

[prime sphinx]

idk what the two equations should be

[brisk raptor]

how do you express perimeter in terms of length and width

[prime sphinx]

2(L x W)

[brisk raptor]

the problem gives you the second equation

[hardy trail]

Btw the operation is not multiplication :)

[hollow bolt]

since the length depends on the width, i'd let w be the variable for width

then try to express the length in terms of w or the width

[prime sphinx]

ohh yeaaa

2 ( L+ W)

[hardy trail]

Better 👍

[hollow bolt]

@prime sphinx

then you can plug in the values for l and w into 2l + 2w, or when factored, 2(l + w)

[prime sphinx]

ooohh i got it

thankss

[hollow bolt]

yeah then it's just solving for w

[prime sphinx]

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[hasty slate]

Solve 4cos(5x)=2 for the domain [0,1/2π]
I do all the algebra steps but then I have no idea how to find the actual solution to finish the question off. Please send help!

Photo attached is what I have so far

[fossil locust]

,rccw

[fossil locust]

$\frac{1}{15} + k \frac{2}{5}$ does not become $\frac{7}{15} k$

[woven radish]

south

[hasty slate]

What does it become?

[fossil locust]

$\frac{1}{15} + k \frac{6}{15} = \frac{1 + 6k}{15}$

[woven radish]

south

[hasty slate]

I have looked at the answer in the back of the lesson and I can't figure out how they've come to that conclusion

[fossil locust]

you have the same 3rd line as the solution

so you can plug in (integer) values of k in there directly

1/15 pi + 0 * 2/5 pi = 1/15 pi
1/15 pi + 1 * 2/5 pi = 7/15 pi

then adding 2/15 pi to 7/15 pi makes the answer be over 1/2 pi

ah, you must have written the wrong domain in your book

the solution is using the domain [0, pi)

[hasty slate]

Oh man, I did write the wrong domain, you're right it is [0,π].

So they've in order done k=0, k=1 for the left side. Now I'm staring at the right side, how is that possible?

[fossil locust]

same thing, so you can't choose k = 0 there cause that would be negative

it's k = 1 and k = 2

[hasty slate]

Omg I understand it now, thank you so much. How do you determine what K to use? Or is it really just trial and error?

[fossil locust]

yes

it should be easy to see which k you should start from

then keep adding the 2/5 pi or the coefficient of k, as that's the same as increasing k by 1

stop when you're outside the domain

[hasty slate]

I do not find it easy to see, but I guess that comes with practice

[fossil locust]

mhm!

[hasty slate]

Thankyou thankyou thankyou!!

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[eager stone]

how do I show
5 | (n⁵-n)

[eager stone]

tried 5q+r 0<=r<5

but I dont know how to go further

[dire umbra]

U need help?

[eager stone]

yes

[pallid moth]

what exactly is n

[dire umbra]

There are 7 ways

[eager stone]

i gotta show it for every n>=1

[dire umbra]

if (p) is a prime number, then for any integer (n), we have (n^{p}\equiv n\quad

[woven radish]

Celestial
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dire umbra]

(Mod p)

[eager stone]

n^p equiv n (mod p)

right

[dire umbra]

If (n\equiv 0\quad (\bmod 5)), then (n) is a multiple of 5.

[woven radish]

Celestial

[dire umbra]

In evry possible case for (n), at least one of the factors in the product (n(n-1)(n+1)(n^{2}+1)) is a multiple of 5. Therefore, the entire product (n^{5}-n) is divisible by 5

[woven radish]

Celestial

[eager stone]

how do i show that 😭

[dire umbra]

.

[fossil locust]

great, if you use the binomial theorem on (5q + r)^5 - (5q + r)

you should get that the remainder mod 5 is r^5 - r

so you can just test n = 0, 1, 2, 3, 4

[fossil locust]

are you familiar with the fact that in any 3 consecutive integers, at least one of them is a multiple of 3?

so n(n + 1)(n + 2) is always a multiple of 3, for example

[eager stone]

yes

but mod being 5 is confusing me

[fossil locust]

yeah cause n^2 + 1 is the odd one out

[eager stone]

i dont think we learned the binomial theorem

[fossil locust]

if you use modular arithmetic here, then n being 0, 1, -1 (mod 5) already works

so n has to be -2, 2 (mod 5)

[eager stone]

hmmm alright

i think i get it

[fossil locust]

yeah, so if you take modular arithmetic for granted, it's not bad

you could also directly test n = 0, 1, 2, 3, 4 cause that's every single number mod 5

every integer n is congruent to one of these

[eager stone]

lol true i will do that if I cant think of another solution during the exam

thanks!

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[bitter narwhal]

any tips to catch up on math that you ignored your entire life?
like literally entire life, but now im stuck in a CS major where I have 4 math subjects which are way out of my league and aptitude as well

[upbeat forge]

Khan Academy and Math LibreTexts, and some dedication, should be a great start.

go through the topics in order, and start from whichever topic you feel like you don't fully understand.

of course, feel free to skip whatever you don't actually need, using the math subjects to guide you in this decision.

[bitter narwhal]

Ooh alright, I just started with Algebra 1 in khan academy
Ill check out Math LibreTexts!

[upbeat forge]

good luck in your math journey! do you have any other questions?

[devout snow]

@bitter narwhal Has your question been resolved?

[bitter narwhal]

not rn ty!

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[bitter narwhal]

yess

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[drowsy compass]

i got a vector that is parallel to the one given in part a

is that fine

[drowsy compass]

i got <1,4,-9>

[lunar harbor]

no

You found vector CB, not vector BC

[drowsy compass]

ah, that sucks

does this logic work?

was my approach wrong in the first place

[lunar harbor]

This logic is fine

[drowsy compass]

oh wait

how the heck did i make that mistake 🤦

thats how you know its too late to do math

i literally did the opposite of what i wrote

thanks

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[glad sigil]

can any1 give a hint ?

[carmine veldt]

eliminate t

[harsh sierra]

You have 4 options, plug them in both lines

Maybe that is not the most clear way to put it

But generally think what a point on both lines must satisfy

[glad sigil]

im not sure if they ask for the same t

[harsh sierra]

Its not the same t

[glad sigil]

ok

[glad sigil]

@carmine veldt thats what i came up with but sadly not more than that

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[harsh sierra]

.reopen

[devout snow]

✅ Original question: #help-27 message

[harsh sierra]

Sorry for leaving

So you have two options:
1 - check whether the 4 options are on both lines by solving two linear equations
2 - solve the equation
t1v1 + v0 - v1 = t2v2 + v0 + v2
This gives you a familiy of possible vectors which are in the intersection.

Where option 2 is basically what you already did

[devout snow]

@glad sigil Has your question been resolved?

[glad sigil]

all good i solved it actually by the choices of the question and i understood its for v0 since itll be x1(t1-1) and itll be t1=1 and t2=1 all the time ig

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[dense rampart]

Do my inner (x) limits not specify the region from parabola to slanted plane? How come I need to elaborate the middle limits instead of just knowing their exact values?

[dense rampart]

<@&286206848099549185>

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@dense rampart Has your question been resolved?

[dense rampart]

.close

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[ionic jasper]

yo guys ill write an exam tomorrow and i need help

[ionic jasper]

idk how to aproach task 9 and just did something i felt correct

[faint gorge]

,rccw

[ionic jasper]

can any1 confirm this for a)

[faint gorge]

Sieht gut aus

[ionic jasper]

ok

fand ich nur bisschen komisch da ich nicht wuste ob 1 spalt jetzt eine einheit ist

oder wie das sonst wäre

danke dir

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[real pond]

how in the world did it go from line 2 to line 3

[young spade]

factor out (k+1)!

[real pond]

ah i see

makes sense now thanks

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[orchid wasp]

How u do this

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[woven dagger]

Im struggling on this problem, and idk what i did wrong 😭

[supple knot]

are you supposed to simplify the values

[woven dagger]

No

[supple knot]

pain

[woven dagger]

I dont have to simplify

Desmos says im right bruh

Idk anymore tbh

Am I missing something from the question or am I interpreting the question wrong?

[supple knot]

where did the 8 go?

or the 8/2 = 4 ?

[woven dagger]

I distributed the 1/2 and pulled out another 1/2 from the trig identity

Desmos said I did the integral right, I just dont know if my interpretation of the question is right at this point

[supple knot]

maybe your setup is wrong then

[woven dagger]

Yeah this is what I was thinking tbh

But idk how else I would set it up

[hazy plover]

Pi

[supple knot]

try -pi/3 to pi/3 instead of pi/3 to 5pi/3

[woven dagger]

Because after graphing desmos, I see the appropriate interval for theta is pi/3 to 5pi/3

[woven dagger]

Alr\

Yeah I get -pi this time and it's also wrong 😭

-pi/3 to pi/3 shouldnt work, since it's specifically asking for outside 3cos(theta) but inside 1+cos(theta)

So pi/3 to 5pi/3 should be the only correct interval

[devout snow]

@woven dagger Has your question been resolved?

[woven dagger]

<@&286206848099549185>

[devout snow]

@woven dagger Has your question been resolved?

[woven dagger]

<@&286206848099549185> Sorry to bother, but can someone pls lmk if Im setting up the integral wrong?

[tender cobalt]

trying this out myself rn, but you have to account for that 3 cos(theta) is sometimes negative, which is hidden in its graph

you need to split the integral into three components, two of them use 3 cos (theta) when 3 cos(theta) > 0

the middle one uses 0 as its lower bound when 3 cos(theta) < 0

[woven dagger]

Bruhh

[tender cobalt]

ok Ive confirmed that that works

directly integrating the area I see with a calculator confirms the answer is somewhere close to π/4

this is the same as the answer Im getting with what Ive said

[woven dagger]

Yeah the assignment says pi/4 is the right answer now 😭

Ty for your help

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[tender cobalt]

np

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[burnt drift]

So I'm having problems with d

[burnt drift]

As I understand it, a linear transformation preserves addition and scaling

[lyric hornet]

what is (\mathcal{P})?

[woven radish]

ΠαϳαμαΜαμαΛλαμα

[burnt drift]

polynomial vector space

any polynomial

So I think I need to check if T(f)(t1) + T(f)(t2) = T(f)(t1 + t2) and cT(f)(t1) = T(f)(ct1)

the latter seems obvious to me

wait actually come to think of it f might be the input

yeah, I think f may be the input

[lyric hornet]

you need to get that for two f,g in P we have that T(f+g)(t)=T(f)(t)+T(g)(t)

that is additivity

then you're right, to check scaling its cT(f)(t)=T(f)(ct)

[burnt drift]

gimme a sec

[runic prawn]

the input to T, yes

[burnt drift]

so just to conceptually clear this up, our complete output still does not have t filled in

we just get an integral with t in it as output

[runic prawn]

yeah i guess so

[lyric hornet]

pardon, but random question, do you know HouseBlaster well? He's one of your dc friends

[runic prawn]

the output is a function

which inherently doesn't involve "t"

[burnt drift]

I know him

what a coincidence lol

[burnt drift]

I see

[lyric hornet]

lmao that's funny, knew him from like 1st and 2nd grade

small world

[runic prawn]

it's like how f is a function but not f(x). the latter is a number for any x in the domain

[burnt drift]

I understand

so to find the kernel it makes no sense to find ts for which the integral equals zero, only fs

[runic prawn]

yep

[burnt drift]

so f(s) = 0

[runic prawn]

kerT = {f in P: T(f) = (0 function)}

[burnt drift]

would the image of T be the polynomial space?

[burnt drift]

great. it's very intuitive. do I have to prove it or is it self evident

[runic prawn]

a subpace of the polynomial space

[lyric hornet]

if (\operatorname{dom}(T)\subseteq\mathcal{P}n) then (\operatorname{im}(T)\subseteq\mathcal{P}{n+1}) notably, since integrating gives you a higher power

[runic prawn]

img(T) is not P

[burnt drift]

I guess we can't get a constant

[runic prawn]

yes

[burnt drift]

So going through the answer key the image is apparently this

any idea why g(0) = 0?

[acoustic leaf]

,, \int_0^0 f(s) \dd s = 0

[woven radish]

cloud ☁

[lyric hornet]

woah happy pink name cloud

[burnt drift]

wdym

wouldn't that be 0 for any function

[acoustic leaf]

yes, exactly

[burnt drift]

that wouldn't mean that the function goes through the origin, though

[acoustic leaf]

the input is the function $f$, the output is the function $g$ where
[ g(t) = \int_0^t f(s) \dd s ]

[woven radish]

cloud ☁

[acoustic leaf]

this integral property says that if we plug in t = 0 to g, we must get 0 out

[burnt drift]

how could a function not satisfy this

[acoustic leaf]

it must be true for every f

which is why the image of T, which is the set of all possible outputs, must only include output functions which have the property g(0) = 0

[burnt drift]

sorry, I'm just not following. isn't that a bit verbose and redundant with a fundamental property of integrals?

[acoustic leaf]

you were asking why we must have g(0) = 0, and the integral property is the reason

[burnt drift]

but, like, if it's just always true, then we wouldn't have to specify it right

could we write img(T) = {P}

[acoustic leaf]

no, because that wouldn't be true

the function h(t) = 3t + 4 is in P, but it is not in image(T). so the two sets can't be equal

[burnt drift]

so I'm sort of becoming confused here, why wouldn't that work

[burnt drift]

wait it has a constant, I sort of see

[acoustic leaf]

image(T) is the set of all possible outputs of T, given any input in the domain

P is the set of all polynomials

image(T) and P are not the same set, because P contains functions that image(T) doesn't

[burnt drift]

OHH

I see

g(0) = 0 just means that there are no constants

everything must be dependent on t

[acoustic leaf]

it's equivalent to saying the constant term must be 0, yes

[burnt drift]

thanks

thanks all of you

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[keen sundial]

why is there an absolute value in r

[acoustic leaf]

|r| < 1 is a convenient shorthand for -1 < r < 1

[keen sundial]

so r has to be greater than -1

[upbeat forge]

and less than 1.

[keen sundial]

can it be -0.5?

[upbeat forge]

yes, I suppose.

[keen sundial]

but tthere is an absolute value sign

[upbeat forge]

|-0.5| = 0.5.

[keen sundial]

Yeah so its 0,5

not -0.5

[upbeat forge]

? it doesn't matter though.

if r = -0.5, the absolute value will turn it positive. that's nothing wrong with it.

[keen sundial]

so 0.5 < 1

just means the negative version?

[upbeat forge]

wdym by "negative version"?

[keen sundial]

negative version of 0.5 (-0.5)

[upbeat forge]

and how does that have any relationship with this?

[keen sundial]

nvm, I should prob stop overthinking

[upbeat forge]

excellent strategy.

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[woven dove]

How to solve this convolution problem?

[woven dove]

I'm trying to understand so hard convolution but nothing clicks in my mind.

I watched a few video about it still no idea how to do "calculations"

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[lyric barn]

Can someone help me I did something wrong but I can’t see where

[restive river]

could u send the problem :DDD

[pure stone]

show the original problem

[restive river]

ill give it a try and see how u did

[lyric barn]

Thank you!

[pure stone]

first, do you know why we don't take x = -2^(-1/4) as solution?

[lyric barn]

Ah because you can’t have a negative answer in the log?

[pure stone]

well there's log_2(x) in the first equation

x must be greater than 0

you're doing correct so far

[lyric barn]

Ohh I think I subtracted the powers wrong when calculating for the y value

[pure stone]

y = 2^(-1/2) /2 is basically 2^(-1/2) * 2^(-1)

[lyric barn]

Ah yeah so then u get 2^(-3/2) right?

[pure stone]

yep

[lyric barn]

Thank you!

By the way could u help me with another question it’s trig related plss

[pure stone]

do note that x > 0, y > 0 and 4x√y > 0 next time

[pure stone]

send it here

[pure stone]

what is the derivative of sin x

[lyric barn]

Cosx

[pure stone]

then take a look at your u'

[lyric barn]

Ah but don’t I keep the 2?

[pure stone]

Keep the 2

[lyric barn]

Ohhhh yeah the - sign

Sorry I’m tired I’m making silly mistakes, thank you!

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[autumn sluice]

Dobleron, a chocolate is known for its distinctive shape and its taste. The quality assurance team of the brand undertakes quality test in two steps:
• Step 1: 30 out of every 990 chocolate bars are randomly picked for physical examination, where the shape of the bar is examined.
• Step 2: For every 30 chocolate bars whose shape is examined, 1 chocolate bar is randomly picked for chemical examination, where the composition and taste of the chocolate is examined.
If one chocolate bar is randomly picked out of 990 bars, what is the probability that it was not chemically examined, provided it was physically examined?

I got 29/30 but the answer in my book is showing 29/990

[carmine veldt]

you're right

book answer shows the joint probability I think

[spice token]

I see

You missed step 1

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[hard talon]

I'm stuck on how to start a proof for this question. I've got

$P(n):n=2^k\cdot l, \quad k\ge 0, l$ odd.

But then, I'm not sure. I'm very new to induction proofs also and this type of question is unfamiliar

[woven radish]

charnixe

[hard talon]

like i'm not even sure how to approach the basecase cuz of the k and l stuff 😭

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[hard talon]

ykw I gotta sleep anyway

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[lyric barn]

Can someone help me please? I’m doing part b (I used Theta instead of the other letter because I’m used to it and this is just practice btw)

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[lyric barn]

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[restive river]

I solved 54 as choice D, but it’s supposed to be A, can someone explain why?

[restive river]

since its a square, the two perpendicular forces would be equal in magnitude right? So i moved the force that was going in CD to BE but as 1/2 F while CB remains F and used the Pythagorean theorem and got 4 as F but apparently its wrong

[upbeat forge]

not always. if CE was a diagonal of the square, yes. but CE is not a diagonal and thus that assumption does not work.

[restive river]

so the two forces arent equal in magnitude?

I kinda thought about that cause CE doesnt look like it bisects the angle between the two forces either but i just assumed so

[upbeat forge]

definitely not. you can see in the diagram itself that the length of the vertical component is very much shorter than the horizontal component.

in fact, because E is the midpoint of AB, the vertical component of CE is exactly half the horizontal component.

[restive river]

i did that, i assumed that the vertical component is 1/2 F meanwhile the horizontal is F and tried to solve it using Pythagorean theorem but i got 4

[upbeat forge]

F is 4N, correct.

but are you looking for F?

[restive river]

well it asked for the force in the direction of CD, so i think that’s F? but im not sure

[upbeat forge]

you are asked for the vertical component, no?

is the vertical component F?

reminder that you are asked for the magnitude of the force vector in the direction of CD, not CB.

[restive river]

oh wait so i just have to multiply it by half to get the vertical component

[upbeat forge]

exactly so.

[restive river]

okay but as you’ve mentioned, CD and CB are not equal in magnitude right?

[upbeat forge]

CD and CB are equal in magnitude.

but that's because they are two sides of the same square.

then again, you're not asked for the magnitude of CD.

you are asked for the magnitude of the component of CE in the direction of CD, and that's a bit of a different story.

that would be F/2, as you've found out.

[restive river]

right i understand, thank you alot

[upbeat forge]

glad to have helped!

[restive river]

.close

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[dusky belfry]

I got a question, I have a question asking me to "In simplified form, determine the polinomial $P(x)$, knowing that "quociente" (idk how to pronounce that) and the rest of $P(x)$ by $B(x) = x^2 -3x + 1$ is respectivelly, $Q(x) = 2x - 1$ and $R(x) = x+1$"

[woven radish]

ℕ∈ℝD ALERT: Gonçalo Gonçalves

[pseudo basin]

quotient and remainder are the english words

anyway you should know the formula that links the four polynomials in a division.

@rain summit hold your horses for a moment

[dusky belfry]

yeah

hold it

I'll type it

[pseudo basin]

was about to ask you to state this formula

[dusky belfry]

P(x) = Divisor * Quocient + R(x)

sorry

forgot the name was divisor

[rain summit]

B(x) is the divisor?

[dusky belfry]

let me see

yeah

it is

so $P(x) = B(x) \times Q(x) + R(x)$

[woven radish]

ℕ∈ℝD ALERT: Gonçalo Gonçalves

[rain summit]

apply this exact formula and substitute the values in

[dusky belfry]

did not think of that tbh

I thought

[rain summit]

It's kinda obvious to me

[dusky belfry]

I had to apply ruffini's law in reverse

sorry,

yaknow how self study is

[rain summit]

Oh you're self studying?

[dusky belfry]

yeah

school doesn't really help

[rain summit]

Try putting the values the question gave you into this equation

[dusky belfry]

allr

a sec

[pseudo basin]

yeah you overthought it

[dusky belfry]

trying to make it pretty sorry, a few more minutes

[rain summit]

You can just expand all that out and simplify

[dusky belfry]

ok

iy's done

let me check

it's done

.close

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[dusky belfry]

thanks @pseudo basin and @rain summit for helping

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[sterile pewter]

how do I find lim of n to infinity this equation?

[sterile pewter]

I tried using this, but then if I move y and n^(a-1) to the top, it'll be theta * y^-1 *n^(1-a), which will go to infinity as n go to infinity, which won't work then

[knotty sage]

what do you mean by "integrate n to infinity" ?

[sterile pewter]

wait srry find lim of n to infinity

[cinder bobcat]

there is a way to make this work. b is obviously 1, and c is 0. what can you make d(n) to use this formula?

[sterile pewter]

as I'm saying this, I noticed that I forgot the mention the range of the values

x > theta> 0

but a <1, and can be negative