help-0

[shadow sparrow]

$\frac{(0.96)(0.72)}{(0.96)(0.72)+(0.04)(1)}$

[ocean seal]

Huntifer

[solemn juniper]

looks fine

[shadow sparrow]

Alright, that makes more sense, I was just getting confused with the latter part, thank you!

.close

[lone heart]

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[vapid shuttle]

I was hoping someone could check my work on this

[ocean seal]

Austin

[vapid shuttle]

g is differentiable because it is a composition of differentiable functions

anywho

[ocean seal]

Austin

Austin

Austin

Austin

Austin

[vapid shuttle]

and done?

[mortal trellis]

well in the last step you dont know whether C is bigger than 1

[vapid shuttle]

oh right

in the inserting the constant multiple of C

like the very last step?

[mortal trellis]

yes

[vapid shuttle]

hm

feels like there should not be difficulty there

[mortal trellis]

no its just a technicality

[vapid shuttle]

right but I don't even immediately see a way to fix it

[mortal trellis]

do we need the specific C that you get from uniform cont?

[vapid shuttle]

no

just that c infront equals c on top

so we just need to split the C

such that

e^(C/a)=C

no nvm

that wouldn't work

[mortal trellis]

what happens if we use 2C instead of C

[mortal trellis]

btw this is lipschitz continuity directly, not uniform continuity

[vapid shuttle]

oh right

mistake

lipschitz is stronger than uniform

if we use 2C

then

e^(2C|x|)

is our upper bound

[mortal trellis]

lets stay at this step. we know that at the end we will need C>=1. so can we already say at this point whether we can claim that for C?

[vapid shuttle]

yes can't we

the lipschitz constant is 1

since |g'(x)|<=1

so then we can make the step I made at the end knowing C>=1 ?

[mortal trellis]

well you could also have |g'(x)|<=0.7 maybe

[vapid shuttle]

ok

[mortal trellis]

you are slightly overthinking this

[vapid shuttle]

but then C>=.7

and we do have <=1

so C>=1 also

[mortal trellis]

if this difference quotient is always <= 0.5, then it is also always <=2

[vapid shuttle]

oh

[mortal trellis]

so without problems we can just increase C until it is >=1

[vapid shuttle]

yeah okay

yeah it is <= C

lol

so we can just say

if C < 1

then

make it bigger

lol

make C >= 1

and it is still larger

then the last step holds

[mortal trellis]

yes

[vapid shuttle]

I knew it had to be something simple

did it look good other than that?

[mortal trellis]

I didnt spot anything else

[vapid shuttle]

awesome ty for your help then

https://tenor.com/view/hug-love-hi-bye-cat-gif-5848836

.close

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[lone heart]

@ancient cargo Has your question been resolved?

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[ancient cargo]

.reopen

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✅

[ancient cargo]

<@&286206848099549185>

[visual vault]

Bro

(No. Of students that completed it)/(total no of students) x 100

[ancient cargo]

53 percent right

[visual vault]

Idk

Do it urself it’s basic math

[ancient cargo]

i did

and got it wrong

lemme do it again

no need for profanity and agression towards my ethnic grp

[plucky oasis]

How do I do it

[ancient cargo]

bro lemme close it first

.close

[lone heart]

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[wild umbra]

open a new channel

[lone heart]

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[zinc shore]

can someone help with this question

[lone heart]

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[sick smelt]

In the given Isoceles Triangle, the base is QR, find Angle <PRS

[sick smelt]

PM = PN = QR
MQ = MN

[lone heart]

@sick smelt Has your question been resolved?

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@sick smelt Has your question been resolved?

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@sick smelt Has your question been resolved?

[coarse vale]

What do we need to find the angle in terms of.
Is this all the data provided in the question?

[humble widget]

So, if PRQ is an isoceles triangle that means all of his angels are equal, so <pqr=<prq=<qpr which means the 180°/3=<pqr=<prq=<qpr.
Also we have that <PRS = <QRS - <PRQ
If <QRS= 180° and <PRQ=180/30=60° That means <PRS= 180 - 60 = 120°

@sick smelt

[glacial patrol]

How do u know prw is isosceles

Prq

Oh rip im dumb he says it

That's the part I've been missing

I was just going off of the diagram

[humble widget]

It happens to all of us sometimes

[sick smelt]

thx, im gonna try to understand this now

[sick smelt]

u say that

& forget what an isoceles triangle is

xD

[glacial patrol]

Wait ur work work still not right tho, not all 3 angles of prq are equal

[humble widget]

Yeah lol

[sick smelt]

yea lol

[humble widget]

I messed up

[glacial patrol]

No I forget that you called it an isosceles triangle >. >

[sick smelt]

now yk, can u solve now?

[glacial patrol]

Ik what it is

[sick smelt]

@humble widget forgot lol

anyway, can u solve?

[glacial patrol]

oh ye

[sick smelt]

ok I'll wait

[humble widget]

I'll try

[glacial patrol]

OK I got the answer now buuuuuuuuuut

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[glacial patrol]

So instead we're gonna guide u to it lmao

Aight hmmm where to start

Say <PMN = a and <PQR = c, u can try to write <MPN in terms of a and then also in terms of c

Soooo try doing that

idk ping me if uve done that but I might've fallen asleep by then

I'm not implying ur gonna take long im just rly tired

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[mental wadi]

Hello! I'm having a hard time wrapping my head around a probability problem.

What are the chances of drawing 4 tiles from a bag of 26 different letters that form a 3 letter word. There are 769 different 3 letter words

[swift tulip]

lets see

herm

so does it mean that you could make a three letter word using three of the four tiles you picked?

@mental wadi

[mental wadi]

yes

[swift tulip]

if so then its not possible to determine without knowing the specific list of possible words

[mental wadi]

how come?

[swift tulip]

well i think

because see

if you have 4 tiles

then that means that you dont use a tile

however

wait

im not sure

it woud be very hard to prove that you dont need the list of specific words

i think

so i am going to assume you can determine

where did you get the number 769?

is that just a random number/

[leaden condor]

i'm guessing 769 means there's 769 possiblities

[mental wadi]

No, I was curious about this question and created a program that went through a CSV of all the words in the Collins Dictionary.

That's how I found 769. This number does not include the words that would be impossible to make with this cenario like ZZZ, and I also chose to remove anagrams. Without all this, it's 1347 words

[swift tulip]

see there is a chance that

say you pick

AEBC

or i guess

ABCE

then you can probably form more than one word out of it

with a set of 4 lettesr

you can form 24 different words but they might not exist

ig you could complementary count

[mental wadi]

what's that?

[swift tulip]

like you count the bad cases and subtract from mthe total cases

im confused

honestly

wait

ohh lol im being stupid

wait

huh?

see out of the 24 possible formed words from a picked set of 4 letters

you might have only one that is in your 769

oh wait i think i figured out a way

see if i pick a random 3 letter word out of all the distinct letter three letter words

[sour dove]

Just a quick side step question

were you told to find all valid words? Often in these problems (if it's HW) they just want the total number of words you can make, valid or not. Also, it helps to know whether or not your drawings are with or without replacement. If without replacement, is it with 26 total titles or there duplicates? If there are duplicates then how many total tiles are there?

[swift tulip]

or ig out of all the possible combinations of three different letters

there are 26 choose 3 combinations of three different letters

choosing one of those comes with probability 1/2600

wait

no

huh?

mb its 769/2600

and then

there are 22 options for the fourth letter

wait no

im so confused

i have come to the conclusion

[mental wadi]

I was hoping for a solution taht included valid words

only

[swift tulip]

that you need the list of words

ok and i am not looking through a list of 769 words!

do you mind if i go

im sorry

[mental wadi]

And then what would with the words

[swift tulip]

i wanted to help you

but peace

☮️

up, up, and awaayyy

[mental wadi]

What would I need to do with the words

[sour dove]

so same question from above

is this with replacement?

[mental wadi]

replacement?

[sour dove]

like if you draw a tile, you mark it down and you toss the tile back into the bag?

yeah

so like you said ZZZZ is something you can draw

[mental wadi]

No you do not put it back into the bag

[sour dove]

but in order for that to happen, you'd need to either:

1. Have 4 of each letter making the bag of letters total to 4x26 = 104 total tiles
2. You have 26 letters each tile representing 1 letter. As soon as you draw it, you put it back into the bag so you can draw that same tile

[mental wadi]

you have 26, then take one so you have 25, then draw from that so you have 24

[sour dove]

ok so then we'll need more tiles than 26 if we want something like AAAA or AAAB or AABB to be possible

so repeated letters are not possible then

[mental wadi]

They are not included in the 769 words

I've also removed them

[sour dove]

right I know

but probability is (preferred outcome)/(total number of outcomes)

it's important info for the total number of possibilities

[mental wadi]

Ok

[sour dove]

but it sounds like we can't do that. So what you're saying is that you would not have valid 3 letter words like "APP"?

[mental wadi]

I would not

[sour dove]

hmm ok, so you've already calculated that with the 769?

(sorry just getting a feel for the problem)

[mental wadi]

No worries, I appreciate you considering all this

Yes that is already calculated with the 769

[sour dove]

ok cool

the hard part is that extra letter we're drawing

let me think for a second

I want to say it's a simple as 769/(26C4)

but I"m guessing that's incorrect

[mental wadi]

That's what I thought too

[sour dove]

hmm

[mental wadi]

But in my head, it doesnt' really make sense

a 5% chance only?

[sour dove]

yeah that doenn't sound right lol

tbh combos are not my strong suit outside of the basics (I wish I was better...). Do you have access to these groups on the left?

[mental wadi]

No I selected pre-university instead

[sour dove]

damn I like your gusto if youre pre university haha

nice challenging problem

ok so here's what I would do if I was you: I'd switch to early university (you'll get access to both pre and early sections), then ask in #probability-statistics . You're probably more likely to get more refined help with this. There.

Keep this channel open for like 30 more minutes, ping @Helpers and if no one shows up, I'd close and reopen a new one (normally I don't like doing that but if someone hasn't answered in quite a bit of time I think it's valid)

[mental wadi]

Ok thank you

[sour dove]

if you were just choosing 3 tiles instead of 4 this would be much simpler lol

oh wait, huh

maybe

if you include 4C3 in the numerator to indicate the tiles you are choosing

something like this

$\frac{729}{\binom{26}{4}} \cdot \binom{4}{3}$

[ocean seal]

MellowDramaLlama

The following error occured while calculating:
Error: Undefined symbol choose

[sour dove]

,w (729 / (26 choose 4)) * (4 choose 3)

[ocean seal]

Results provided by WolframAlpha

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

Requested by mellowdramallama

[sour dove]

hmm... 19.5%... that does sound more reasonable lol

[mental wadi]

Why did you multiply it by that?

[sour dove]

because of the 4 tiles we chose, we need to choose 3 of those, right?

I could be completely wrong

I would confirm with my reccomendation above and ask but it's getting closre to a more reasonable answer

[mental wadi]

Ok thank you

[glacial patrol]

principle of exclusion inclusion?

And lots of it

I really do think this is gonna require a lot of spamming of principle of exclusion inclusion lmao

[sour dove]

oh thank goodness someone knows what they're talking about lmao

I was starting to panic

I forgot about the exc/incl principla though

[mental wadi]

whats that?

[glacial patrol]

https://brilliant.org/wiki/principle-of-inclusion-and-exclusion-pie/ basically for instance if you wanna count the number of ways to do A or B

Then you start by adding A + B

But you're adding the part A and B twice

So you substract that by the number of ways to do A and B

And that generalizes to like more than 2 sets

But hmm

I dunno if what I thought we could do PIE on was right

I'm starting to think that there's actually no neat solution based on skimming this stackexchange answer https://stats.stackexchange.com/questions/74468/probability-of-drawing-a-given-word-from-a-bag-of-letters-in-scrabble

[mental wadi]

Hmm ok

There are no wildcards in this problem

[glacial patrol]

But that seems more complicated

Yeah

[mental wadi]

Ok so... there's no solution?

[lone heart]

@mental wadi Has your question been resolved?

[mental wadi]

No

[lone heart]

@mental wadi Has your question been resolved?

[lone heart]

@mental wadi Has your question been resolved?

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@vast stump Has your question been resolved?

[vast stump]

<@&286206848099549185>

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<@&286206848099549185>

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[alpine sable]

tjenare

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@vast stump Has your question been resolved?

[gusty ocean]

.

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[upbeat eagle]

heya

[upbeat eagle]

Im having a problem,

Does 2PR = RQ mean PR:RQ ratio= 1:2

.close

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[cunning tulip]

.reopen

.open

how i dot this

[cunning tulip]

if i put everything to one side and the other is 0, can i use b^2 - 4ac

[green knoll]

can you see any way to put everything to one side?

[unkempt robin]

I'm guessing you are referring to the discriminant for solving a quadratic equation

[cunning tulip]

yes

[unkempt robin]

Simplify moving everything to one side doesn't get close to a quadratic equation in the form $ax^2 + bx + c = 0$ for you to use the quadratic formula

[cunning tulip]

i think remove those under things

i dont know in english

[ocean seal]

USS-Enterprise

[cunning tulip]

but if i multiply everything by x(x+1)

[unkempt robin]

Yes, you should first work on getting rid of fractions

[unkempt robin]

Correct.

[cunning tulip]

i mean x+7

i get

420 (x+7) = 420x - 2x^2 - 14

?

[unkempt robin]

Yeah*

[cunning tulip]

then multiply further

[unkempt robin]

Yes

[cunning tulip]

420 x + 2940 = 420x - 2x^2 -14

then to one side

420x + 2940 - 420x +2x^2 + 14

[unkempt robin]

= 0**

[cunning tulip]

2x^2 + 2954 = 0

?

[unkempt robin]

Ah nevermind I see the error

[unkempt robin]

14x

Not 14

[cunning tulip]

oh

where did i make mistake

oh i see

[unkempt robin]

You had $x \cdot (x+7) \cdot 2$ right

[ocean seal]

USS-Enterprise

[cunning tulip]

yea

i see now

[unkempt robin]

$(x^2 + 7x) \cdot 2 = 2x^2 + 14x$

[ocean seal]

USS-Enterprise

[unkempt robin]

14x not 14

[cunning tulip]

i forgot to multiply by x

[unkempt robin]

Yes

[cunning tulip]

and only multiplied by 7

ok

so 2x^2 + 14x + 2940

= 0

then discriminant

[unkempt robin]

Exactly

Hang on!

Do you really wanna work with 2940?

[cunning tulip]

aah

[unkempt robin]

You see that all coefficients share a factor of 2

[cunning tulip]

yes

x^2 + 7x + 1470

[unkempt robin]

= 0

😄

[cunning tulip]

thx

[unkempt robin]

There you go

[upbeat eagle]

nvm

[cunning tulip]

yea

x1 and x2

but only 1 is correct

coz this is from a question asking how many people can fit in a house or sum

so if there is negative x its wrong

196 - 5880

hmm

discriminant is negative

so no answer

[upbeat eagle]

You have to put it into b^2- 4ac no?

[cunning tulip]

yes

i did

[cunning tulip]

here

but discriminant is negative so there is no answer right>

?

[unkempt robin]

Note

You squared 14

Not 7

Should be 49 - 5880

Because b^2 = 7^2 = 49

You used the b we had before we divided everything by 2 😄

[upbeat eagle]

I think there is some error

in the math

[unkempt robin]

No, for the original equation there are no real solutions

[upbeat eagle]

ah

right

[unkempt robin]

Can you send the original question The Broken Sigma?

You talked about how many people can be in the house and we got 2 complex solutions so you probably set the equation up wrong

(we were in the process of solving the equation correctly though)

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[vast arrow]

How can I solve this without geometry and plug them in

[vast arrow]

I know how to do it with geometry but I wanna do it without

[ivory pivot]

what have you tried ?

[jagged cobalt]

split the integral ig

[vast arrow]

no were still learning these and im not sure how to start about it

[ivory pivot]

remember that you can always do that $\int_a^c..=\int_a^b..+\int_b^c..$ with $a<b<c$

[ocean seal]

everg

[vast arrow]

right

nvm I think I got it

[ivory pivot]

🙂

[vast arrow]

I believe I got it let me give it a shot

thanks!

.close

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[alpine sable]

i need help

.reopen

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[lone marlin]

How

[lone heart]

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[lone marlin]

Hellnaw

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[steady basin]

to 4sf

could someone check and see if they get the same as me plz

anyone?

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[surreal estuary]

Hello.
f(x) = (x^3 / 3) + 3x^2 - 7x
f'(x) = (3x^2 / 3) + 3(2x) - 7 = x^2 + 6x - 7 is positive so it's monotonically-increasing

Would this be enough as proof?

[alpine sable]

I don't understand what you mean, can you explain more?

[surreal estuary]

(x^3 / 3) + 3x^2 - 7x

I have that function

And they ask if its monotone

If yes then have to say if it's increasing or decreasing

Domain is R

[limpid turret]

What is f'(0)?

[surreal estuary]

What do you mean

Is not it -7

0^2 + 6*0 - 7

[alpine sable]

F'(0)= -7
F'(1)= 0
F'(2)= 9

Well, we conclude in this way that the function is increasing in an ordinal manner (I am not good enough at mathematics)

[surreal estuary]

That true, but when I looked at graph in Google it did show that it's decreasing too

[alpine sable]

Really?

[surreal estuary]

Yes

Try -4 and -3

[alpine sable]

do you mean this?

[surreal estuary]

Yes

[alpine sable]

Bro it is increasing

[surreal estuary]

How is -inf to -3 increasing

Is not it decreasing?

[alpine sable]

My English is bad. Do you mean -inf ------> Minus infinity?

[surreal estuary]

Yes

[alpine sable]

Well, I don't understand you here. Can you explain?

[surreal estuary]

All I am saying is that on that graph minus infinity to -3 values are decreasing and then they increase again

[alpine sable]

Oh that's right

Did mean now

F'(-4)= -15
F'(-3)= -16
F'(-2)= -15

I feel complicated 😂

[surreal estuary]

Hm that what I meant

[lone heart]

@surreal estuary Has your question been resolved?

[surreal estuary]

I have been researching some and if I understood correctly

If derivative is over zero it's increasing

If understand zero it's decreasing

So if I look for discriminate and find out that there are 2 zeros it means it is not monotone

If there is 1 zero then I will check f'(x-1) f'(x) and f'(x+1) to see what's it like

And if it's 0 then just check any value to see if it's increasing or decreasing

Is this correct?

[lone heart]

@surreal estuary Has your question been resolved?

[lone heart]

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[swift saddle]

how is this wrong

[gray isle]

dy/dt isn't 8

[swift saddle]

its 8t

which i just forgot to type but its still consistent with the next step

[gray isle]

you're conflating x with t in the following steps

at the point (where) t = 2, (not x=2)

[swift saddle]

shoot

!close

/close

i tried

[topaz nacelle]

.

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[solid mural]

What's the meaning of "c be the fractional proportion of a + b"?

[exotic canopy]

context?

[exotic canopy]

i think they mean fractional part

[worn fox]

fractional part presumably

[exotic canopy]

yeah

$\lfloor x \rfloor - x$

[ocean seal]

artemetra

[worn fox]

e.g. fractional part of 3.14159265358 is 0.14159265358

[solid mural]

Is it must be start from the first decimal (0.1415... )? Or it can be 0.0415... ?

[exotic canopy]

wdym

[worn fox]

the whole decimal

[exotic canopy]

yes

[solid mural]

I see, thanks all

.close

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[paper temple]

Hi I need some help with this:

A function f : Z × Z → Z is defined as f (m, n) = 3n − 4m.
(a) Prove or disprove: f is injective.
(b) Prove or disprove: f is surjective.

[lone heart]

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

[paper temple]

Not sure where to begin

[rose sigil]

experiment with some specific values for (a)

[opal jolt]

By definition, if f is injective, an element can only be mapped to an original (m, n) pair
If there's more than one pair that result in the same value of f, then f is not injective

If f is surjective, it means that for every value that the function can take, there's at least one pair (m, n) that gives that value

[hybrid spear]

can someone explain how to solve this logarithmic equation pls?

[opal jolt]

!occupied

[lone heart]

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).

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[elfin kernel]

would this suffice as a proper answer?

Because i see some other answers compare the original series to 1/sqrt(n^2 + 2n + 1)

[worn fox]

you're comparing it above by a series that diverges, this doesn't tell you anything

[elfin kernel]

OH

[tall topaz]

Try limit comparison with the same series

[elfin kernel]

ok

[elfin kernel]

Would this work

Should be by Lim test at the end

Limit exist and Bn diverges so An must diverge?

[tall topaz]

Yes

[elfin kernel]

Cooool

if I used comparison for the same problem

since I have An and Bn in fraction form, then I want the reverse of what part b is saying?

[tall topaz]

You need a series that’s smaller than 1/sqrt(n^2 + 1) that’s also divergent

1/n happens to be bigger so it doesn’t work

[elfin kernel]

ohhhh

so I take the entire thing to be An

ok cooool

thank you!

.close

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[dawn barn]

aside from the corrections I've already made, what else am i doing wrong?

[valid crypt]

it can be simplified more

[dawn barn]

how so?

[valid crypt]

I don't see a problem with the math rn

ln(e) = 1

[dawn barn]

ohh shit

i read the question wrong smh

i saw e^8 and thought it was x

[valid crypt]

well the math should still work out

just replace the x with an 8 in the final equation

and solve for k

[dawn barn]

how can i solve for k if theres2 variables tho

if it was just x wouldnt it be impossible?

idk

[valid crypt]

what are the two variables?

[dawn barn]

x and k

[valid crypt]

ok start from the third line but replace e^x with e^8 for the upper bound

[dawn barn]

yh i get that

[valid crypt]

yay you found k lol

[dawn barn]

thanks for ur help

.close

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[median oar]

I’m struggling to express why choosing z = -R “minimises” the bottom

I can kind of explain it visually but when one of the numbers on the bottom becomes complex (ie. the points aren’t colinear with the origin) I can’t really make sense of what “minimising” means formally

Should I be trying to use calculus?

[keen plinth]

△≤

[median oar]

I thought so too somehow that might help but I couldn’t quite figure out how to phrase it

[keen plinth]

take |z + 3| and bound it below using the reverse triangle inequality

recall that |z| = R

[median oar]

I think they wanted us to use sup length thing (that’s the hint on the answer sheet)

Or is that the same thing

[keen plinth]

this is part of it

[median oar]

So I have $\underset{z\in{|z|=R}}{\text{min}}(|z+3|^3 |z+2|^2)$

[ocean seal]

Frosst

[median oar]

Then I split each thing up?

[keen plinth]

yes

[median oar]

That makes a lot of sense

But I’m not really directly using the sup length thing

But more using it to then further bound it with triangle ineq

[keen plinth]

thats how its usually done

the ML bound is pretty much just the triangle inequality for integrals

then you triangle inequality again the integrand to get an upper bound

[median oar]

Ok but now my signs are wrong

[keen plinth]

if you're trying to find the minimum, an upper bound isn't very useful

[median oar]

That’s very true

Aha what about this

,rotate

[keen plinth]

sure

[median oar]

So I’m here and it’s blaringly obvious that z = -R is the best choice

But I can’t figure out how to say it

[keen plinth]

△≤

[median oar]

But I only have 1 abs value so I need to make something up?

[ocean seal]

snow

[median oar]

Oh..

I think I just dont remember this

[keen plinth]

let's use it

[median oar]

,rotate

[median oar]

Ok so that’s good then

That gives a lower bound which gives an upper bound for the max then I have my inequality

Now this is great and all when the points 3 and 2 are colinear with the origin

But what about when I have say (z + 1)(z - i) on the bottom?

Something like this

Bc I can’t just split the minimum anymore the same way I did before

[keen plinth]

why not

[median oar]

And even the way I split it last time is kinda sketchy because the z chosen was suppose to minimise the product

It had to be the same z for both minimums

[keen plinth]

you're not trying to find a perfect upper bound

you're just trying to find a particular upper bound

[median oar]

So it’s fine?

It’ll not be an equality but inequality?

[keen plinth]

min ab >= min a min b

same with max but other way

[median oar]

This becomes >=?

Ah then I bound the min from below with the splitting

Which makes my max even bigger

Which is fine

Very clever indeed

[lone heart]

@median oar Has your question been resolved?

[median oar]

Aha so like this

Should’ve been a - in the question

[keen plinth]

looks fine

[lone heart]

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[median oar]

Fank yew

[lone heart]

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[summer dirge]

not really understanding this part of the proof

why is this the case?

[vapid shuttle]

if there wasn't, then a wouldn't be the least upper bound

[rose sigil]

draw a picture

[vapid shuttle]

if eps>0 and a-an >= eps (for all n), then a-eps >= an (for all n)

right?

but if eps>0, a-eps would be smaller than a

and yet still an upper bound for an

which contradicts the fact that a is the smallest upper bound for an

right?

[summer dirge]

give me a moment

[rose sigil]

analyst austin

[vapid shuttle]

so true

Let me know if that doesn't make sense ING

[lone heart]

@summer dirge Has your question been resolved?

[vapid shuttle]

@summer dirge do you have another question?

[summer dirge]

not as of yet, i'm just still thinking about this lol

[vapid shuttle]

Okay feel free to tag me if you need clarification

[summer dirge]

its actually the logic thats tripping me up

not the argument

we have a statement in the form (P and Q) implies R, yes?

[vapid shuttle]

if a is the least upper bound of A => for all eps>0 there exists N such that a-aN<eps

so not quite

[onyx compass]

hello can you help me with logarithms ?

[vapid shuttle]

!occupied

[lone heart]

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).

[summer dirge]

why is eps > 0 not an assumption?

[vapid shuttle]

give me any epsilon > 0

and I can give you a natural number

such that a-aN < epsilon

that is all this is saying

it isn't true if epsilon is negative/0

[summer dirge]

then i think the issue for me is the wording

spivak writes "if eps > 0"

[vapid shuttle]

yes that is fine

what's wrong with that

[summer dirge]

does he mean if eps > 0, then ..., or for all eps > 0, ...

[vapid shuttle]

they're equivalent here

if eps > 0 means the statement should work for any eps > 0

since the epsilon is arbitrary

except for the fact that it is > 0

[summer dirge]

i see

it wasnt stated explicitly so i didnt catch on ig

[vapid shuttle]

fair enough

it's not so intuitive in the way he states it

[summer dirge]

.close

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[weak goblet]

i basically need this to be proven, the question is on the top of the page of both

[weak goblet]

like this entire thing to be proven i guess, my professor said i didn’t prove it correctly

and i have to use induction

(ba^n b^-1)(bab^-1)
(ba^n)(b^-1)(bab^-1) [probably allowed]
(ba^n)(b^-1 b a b^-1) [associativity]
(ba^n)((b^-1 b a) b^-1) [maybe necessary step idk]
(ba^n ((a)(b^-1))) [definition]
this would prove it but i have to use induction

[lone heart]

@weak goblet Has your question been resolved?

[weak goblet]

no

[lone heart]

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[vapid steppe]

does $\frac{x+1}{(x+1)^2}$ have a vertical asymptote at $x=-1$

[ocean seal]

Jash

[ocean whale]

Did you try plotting it?

[vapid steppe]

its no calculater problem

wait i meant does it have a removable discontinuity

[vast arrow]

Do you know how to find a VA

[vapid steppe]

ye

[vast arrow]

How do you do it : )

[vapid steppe]

if $\lim_{x\to a}f(x)=\pm\infty$ then $f$ has a VA at $x=a$

[ocean whale]

It has an asymptote at x = -1 so it'll never reach the discontinuity at x = -1

[ocean seal]

Jash

[ocean whale]

You realize he made a mistake and was asking about removeable discontinuity, right?

Not about VAs

[vast arrow]

ohh

not until now lol

thank you

[vapid steppe]

is there something called a limit function

$f(x)=\lim_{t\to x} g(t)$

[ocean seal]

Jash

[long axle]

That would be interesting

Would require the existence of a totally separate function

[vapid steppe]

.close

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[pearl tangle]

Nour drove from the Dead Sea up to Amman, and her altitude increased at a constant rate. When she began driving, her altitude was
[400] meters below sea level. When she arrived in Amman
[2] hours later, her altitude was
[1000] meters above sea level.
Let
[y] represent Nour's altitude (in meters) relative to sea level after
[x] hours.
Complete the equation for the relationship between the altitude and number of hours.

[ocean seal]

conraks'

[median osprey]

Find the constant rate and use it for developing a relationship

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@pearl tangle Has your question been resolved?

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[drifting stump]

Greeting. I want to know if my calculation is correct

Thank you very much

[alpine sable]

It’s correct

[drifting stump]

Thank you so much

I really appreicate it

.close

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[alpine sable]

solution(find x) to this problem? where [ ] is Greatest Integer Fraction
The Teacher and i both found that x belongs to (1,4/3] but i found that 7/5 is also the solution to the equation so whats the actual answer i cant figure it out

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@alpine sable Has your question been resolved?

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[dire terrace]

how wouldve i have done any of this?

Im so confused is there videos where i can learn how to do this

[native cloud]

Number 4 is basically just substitution

[safe pecan]

number 1 is just rearranging for A

root both sides

2*

[dire terrace]

you cant put root

[safe pecan]

number 3 is just simple fractions

[safe pecan]

no like

to get A alone you take the square root of both sides

which means putting both sides to the power of 1/2

$\sqrt{x}=x^{\frac{1}{2}}$

[ocean seal]

clacii

[dire terrace]

then it would be x^-3?

im confused

[safe pecan]

wait i'm mind blanking gimme a sec

[dire terrace]

how would i even do this

[safe pecan]

yeah i think that should be the answer idk

either that or i literally can't do basic maths

[safe pecan]

$A=\text{log} _3 9^{x^2}$

[ocean seal]

clacii

[safe pecan]

that's it

rewriting exponentials to log

[dire terrace]

it has to be a number apparently

[restive yew]

you could just reduce that to 2x^2 ?

[dire terrace]

how would you reduce it?

[restive yew]

using properties of the log function

$A=\text{log} _3 3^{2x^2}$

[ocean seal]

hriday2408

[dire terrace]

it would be 2x^2

?

[restive yew]

final answer? yeah

[dire terrace]

using power property

I would do the same with this?

i dunno what they mean by perfect square

[lone heart]

@dire terrace Has your question been resolved?

[dire terrace]

how would i start to solve this?

[lone heart]

@dire terrace Has your question been resolved?

[floral bronze]

YOU'RE GIVEN A^2 ,
THEY WANT A
THEREFORE, TAKE THE SQUARE ROOT OF EACH SIDE AND SIMPLIFY

ALL UR QUESTIONS ARE ABOUT PRACTICING INDEX LAWS, SO HERE. TAKE SOME LOGS TOO

SQUARE ROOT MEANS N=2

[lone heart]

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[vague bone]

the interquartile range is just q3-q1 right?

so 80.5-68=12.5

but like this is what the answer said

[sullen dome]

You are correct. Seems like they made an error.

[vague bone]

ohh

okok

thanks

[sullen dome]

That's just the full range

[vague bone]

yeah thats what i thought

i wasnt too sure if it was just me or the answer

lol

alr thx

.close

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[karmic hornet]

!help

[lone heart]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[karmic hornet]

Please help me

[sullen dome]

What have you done so far?

[karmic hornet]

ive done like all the labelling

i forgot how to get the answer

[sullen dome]

So the first important thing to notice without replacement

Do you know what that means?

[karmic hornet]

no\

[sullen dome]

So if you have 7 balls. And take away 1 ball. Without replacement. i.e you don't put it back. How many do you have?

[karmic hornet]

6

[sullen dome]

Yeah, so it's the same for the animals

You have 7 animals to begin with. Then you chose one, so now you can only chose from 6

And you have 2 ways to get 1 koala and 1 wallaby

So either you chose and you get 1 wallaby then 1 koala. Or you chose 1 koala and then get a wallaby

[karmic hornet]

i see

[sullen dome]

So you add those 2 up

[lone heart]

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[tough dawn]

I worked out this problem and I just wanted to confirm that I did everything correctly.

[alpine sable]

.reopen

[tough dawn]

🤨

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@tough dawn Has your question been resolved?

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@tough dawn Has your question been resolved?

[devout helm]

u made a mistake, right at the end 2c=3 means c = 3/2 not 2/3

[lone heart]

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[thin sedge]

T

[thin sedge]

If you are looking for inverse sin or arcsin

It restricts it and the answers are only in the shaded PURPLE part ?

[foggy sage]

depends on which branch you are using but yes kinda like that

[thin sedge]

ok I’m in prec calc

ok so

[foggy sage]

basically in this case

you are using the branch [-pi/2, pi/2]

[thin sedge]

if it’s for arccos or cos inverse then it is the top half of the circle ?

[foggy sage]

you can also use the branch [pi/2, 3pi/2] for example

[foggy sage]

not always

thats if you choose to use the branch [0, pi]

if you use [pi, 2pi] it would be the lower half of the circle

this is because sin and cos are periodic so their values repeat after a while

[thin sedge]

Ok

so yes if my teacher uses 0,pi

[foggy sage]

yep

then it is the top half of the circle for arccos

and the right half for arcsin

[thin sedge]

Does the range really matter ? They r both same

I mean

[foggy sage]

well the majority of times you would just use the branch [0, pi] for arccos and [-pi/2, pi/2] for arcsin

[thin sedge]

Domain

or

idk

[thin sedge]

ok yea

[foggy sage]

but it's not correct to say it is always those

[thin sedge]

We will probably go over that later

[foggy sage]

but yes, if you are using these branches then your understanding is correct

btw these are the most commonly used branches so they are often also called the principal branches

[thin sedge]

that’s cool

my teacher just call it

domain restrictions or range restrictions

[foggy sage]

yep essentially the same thing

[thin sedge]

Principal branches is a cooler name

Thx

.close

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[fallen lotus]

.help

[lone heart]

Commands:

• clopen: .close, .reopen, .solved, .unsolved
• consensus: .poll
• factoids: .tag
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[fallen lotus]

Help

[fallen lotus]

Question 4 a)

Spearman's rank correlation coefficient

I have the formula, but don't know how to get the values or rank the data please help

.unsolved

<@&286206848099549185>

[cinder egret]

whats the formula

[fallen lotus]

1 - 6 Summation of d^2 / n(n^2 - 1 )

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[fallen lotus]

I give up

.close

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[desert bone]

hey guys, i am trying to solve this integral by u-substitution, but i can't cancel the x^6 and i can't solve it, can someone give me any hint

[gray isle]

is that:
$$\int x^6\br{3-\frac 2x}^6 \dd{x}$$

[ocean seal]

ℝαμΩℕωⅤ

[desert bone]

yes

[gray isle]

apply exponent laws to combine those powers of 6

[desert bone]

ok so i shouldn't use u-substitution right?

[gray isle]

you can use it later if you feel the need

what do you have after doing what i recommended

[rare gale]

there must be a more elegant way than this

[gray isle]

seems pretty elegant to me

like a 2.5 step solution if you're efficient

[desert bone]

i have integration of (3x-2x)^6

[gray isle]

how are you getting the -2x

[desert bone]

(3x-2)^6 sorry

[gray isle]

and you can use u-sub here if you want
if you're good with you chain rule, it isn't explicitly needed

[desert bone]

i used u-sub, and now i solved it, it is (3x-2)^7 / 2 + c

thank you

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[upbeat eagle]

Hia

[lone heart]

Please don't occupy multiple help channels.

[upbeat eagle]

How do I find the position Vector of B relative to O

OA = 15cos30 - 15 cos30
+

AB = -9cos60 - 9cos60

Wouldnt this be correct?

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[upbeat eagle]

Unloved

[floral bronze]

THE GREEN LINE WOULD BE 9SIN(60) SINCE ITS THE SIDE OPPOSITE THE ANGLE

AND PURPLE WOULD BE 9COS(60) SINCE ITS ADJACENT TO THE ANGLE

I MEANT 60

[upbeat eagle]

yes

So what I wrote is correct

But forwhatever reason When I add up the both sin (J) , and cos (i) answers

I get the wrong answer

--> position Vector of B relative to O

[floral bronze]

OA = 15COS30 I - 15SIN30 J

DO U SEE UR MISTAKE?

[upbeat eagle]

no?

[floral bronze]

THE GREEN LINE IS 15COS30
PURPLE IS 15SIN30

[upbeat eagle]

yes..

[floral bronze]

YOU HAVE 2 COS TERMS HERE

THE 2ND IS SUPPOSED TO BE SIN

AND SAME PRIBLEM HERE, ONE'S SUPPOSED TO BE SIN

[upbeat eagle]

so

OA= 15 cos30 i - 15sin 30 j

AB= -9cos60 -9 sin 60

Am I understanding it correctly?

[floral bronze]

OA IS CORRECT YEAH

BUT AB ISNT

WELL, UNLESS U MEANT
-9COS60 J -9 SIN 60 I

[upbeat eagle]

no ^

-9cos60 i, - 9sin 60 j

Hmmm

[upbeat eagle]

Pruple is - 9 sin 60 no?

Sin goes down

Cos goes across

[floral bronze]

THE VERTICAL J COMPONENT HERE IS COS THIS TIME

[floral bronze]

NOT ALWAYS

[upbeat eagle]

Oh, I thought always

[floral bronze]

IF WE WERE TALKING ABOUT THE ORANGE ANGLE YOUD BE CORRECT

BUT WE'RE DOING THE BLUE ANGLE

[upbeat eagle]

Ok so

[floral bronze]

SIN IS ALWAYS OPPOSITE THE ANGLE
COS IS ALWAYS ADJACENT

[upbeat eagle]

OH I get it

I just always thought sin is up/down, cos is across <-->

Ok

Understand it :>

[floral bronze]

🔥 EPIC DESPACITO

[upbeat eagle]

Epic gamer moment

https://tenor.com/view/dance-anime-cute-girl-kotori-itsuka-gif-17438856

I dont get the right answer stilll

So

OA = 15cos30 i -15 sin30 j

AB= -9sin60 i - 9 cos60 j

Right?

and to find OB, its just add? to find i & j

[upbeat eagle]

Wait I get,

5.1 i, -12 j