help-19

[bleak shard]

In this case its a bit strange, particularly for the 1 flip scenario

But if half your flips are heads and half are tails

then for 1 flip you'll have 0.5 heads and 0.5 tails

for 10 you'll have 5 heads 5 tails

and so on

Then for each head you gain 2 and for each tails you lose 1

@lime moat

[lime moat]

Ik that

I am struggling with this part:

What are the odds you turn a profit after
1 flip (50%)
10 flips
100 flips
1,000 flips
10,000 flips
100,000 coin flips?

Profit being you get more money than you started with

[lime moat]

<@&286206848099549185>

[bleak shard]

Well you need to have less than twice as many tails

for 10 flips

youll have a profit as long as you get 6 or less tails

Or 4 or more heads if youd rather think of it that way

But its a little more complicated than you might think because it will be the chance you flip 4 head out of 10 flips + the chance you get 5 heads out of 10 flips + 6 heads out of 10 flips + 7 and so on up to 10

[lime moat]

I am just stuck on how to set up a formula, if it even applies to this case

I know that I can have half of the heads as tails + 1, since having 1/3 heads and 2/3 tails = $0

My teacher gave a hint and said Pascal's triangles

[bleak shard]

Does the term Bernoulli trial ring a bell

in this case each coin flip is a bernoulli trial

and when we have a bunch of bernoulli trials it classifies as a Binomial Distribution

[lime moat]

Yes.

[bleak shard]

Which uses a formula like this

Where n will be the number of flips

and k will be how many you need to be heads

But you have to do this for each number of k heads where you'd turn a profit

[lime moat]

So

1 flip (50%)
10 flips

• 4 Heads needed
  100 flips
• 34 Heads needed
  1,000 flips
• 334 Heads needed
  10,000 flips
• 3,334 Heads needed
  100,000 coin flips
• 33,334 Heads needed

[bleak shard]

Yes

Are you allowed to use excel? 😭

[lime moat]

I can use anything

[bleak shard]

Great because calculating all that by hand would be a monster

This is the most accurate approach though I believe.

Oh and I should say in the formula above that p is the probability of whichever event we consider a "success"

in this case flipping a heads

but for this we'll see that p^k*(1-p)^(n-k) will always just be (0.5)^n

since p and 1-p are the same

[lime moat]

Alright, I plugged it into Google Sheets

10 flips

• 4 Heads needed
• =BINOMDIST(6, 10, 0.5, TRUE)
• 82.81%
  100 flips
• 34 Heads needed
• =BINOMDIST(66, 100, 0.5, TRUE)
• 99.956%

Beyond 100 flips, google sheets rounds to 1

[bleak shard]

This seems accurate to me though I haven't verified myself

It makes sense that with more flips it should become increasingly difficult to lose money

[lime moat]

Yeah

[bleak shard]

Since for every 2 flips you should be gaining a dollar essentailly

[lime moat]

Because you need 2 tails to cancel out 1 head

Yep

I just want to make sure that the percentages are correct

Because they look correct

[odd edge]

@lime moat Has your question been resolved?

[frozen valve]

Guys

what 9+16

<@&286206848099549185>

[hot dome]

🤦‍♂️

,w 9+16

[clever fjord]

Results provided by WolframAlpha

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Requested by miyagi4

[frozen valve]

,w 4*5

[clever fjord]

Results provided by WolframAlpha

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Requested by Man of culture#0022

[frozen valve]

HOLY SHIT

,w 1+0+19+19+19+19+888+70183+2938372+90+102+987+13+75

[clever fjord]

Results provided by WolframAlpha

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Requested by Man of culture#0022

[frozen valve]

NO WAYS

,w sin^2+cos^2

[clever fjord]

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Requested by Man of culture#0022

[frozen valve]

no

,w sin^2+cos^2

Its simple bruh

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[frozen valve]

<@&286206848099549185> Whats 67829+21

[odd edge]

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[civic oxide]

67850

[civic oxide]

I think

[marble crane]

3010787

[royal kiln]

What the

[hot dome]

I think this is against the rules bro

[royal kiln]

Yea..

[mystic saffron]

<@&268886789983436800> this person invaded another person's help channel (check above) and is now pretty much trolling

[hot dome]

Addition is booooringgg

[mystic saffron]

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[viscid flint]

that's a risky question to ask here

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[pastel orbit]

?

[hot dome]

Its a reply to troll

[pastel orbit]

oh we're answering something from above?

[hot dome]

Kindie math

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@sturdy juniper Has your question been resolved?

[warped grove]

.close

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[burnt briar]

Can someone explain whats the different between symmetric and cyclic sums

[low locust]

context?

[burnt briar]

4 variable are given (a,b,c,d) whats the symmetric and cyclic sums?

[low locust]

if you have some formula f(a,b,c,d), then the cyclic sum is f(a,b,c,d)+f(b,c,d,a)+f(c,d,a,b)+f(d,a,b,c)

while the symmetric sum goes over all arrangements of the letters a,b,c,d

so it would for example also include f(a,c,b,d) and f(d,c,b,a)

[burnt briar]

If the sum is cyclic, can I write that WLOG max{a,b,c,d} = a

[low locust]

yes

[burnt briar]

Can i also write it as min{a,b,c,d} ?

[low locust]

well not at the same time

can you give more context about what you are trying to do?

[burnt briar]

im trying to understand this

says that p+q+r is symmetric

[low locust]

can you give the entire context and not just three short lines?

[burnt briar]

1. Suppose M and m are the maximum and minimum value of a+b+c with natural numbers (a,b,c)
   that satisfy
   ab + bc + 2ca = abc + a + c + 2

Find M+m

(a-1)(b-2)(c-1) = (a-1) + (b-2) + (c-1)
And (a-1), (b-2), (c-1) = (p,q,r)

so the expression become
pqr = p+q+r

[odd edge]

@burnt briar Has your question been resolved?

[low locust]

p+q+r is symmetric in p,q,r because its the same as q+p+r etc. so if you were to rename the letters p,q,r the value of p+q+r wouldnt change because either way you are adding all three numbers

same for pqr

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[wise bear]

Need help with this

[wise bear]

My answers are apparently worng

Wrong

This is my working

This is all wrong except for the patient H model

I want to know what I did wrong

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[remote prairie]

can someone help me understand derivatives

[remote prairie]

help pls

like what are derivatives

[cinder elk]

Derivatives calculate change in or the slope of a function

Like with linear it's simple since the slope is the same across the entire function, but with non- linear functions not so much

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[glad yacht]

I tried to solve this diff eq:
$y = e^{y' + y'' + y''' \dots}$

[clever fjord]

7aman

[glad yacht]

And reached $\text{li}(y) + \ln \ln y = t + c$

[clever fjord]

7aman

[glad yacht]

Not sure at all on how to solve for y here

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[glad yacht]

.reopen

[odd edge]

✅

[glad yacht]

Is there any possible way of doing it?

$y e^{e^{\text{li} y}} = e^{Ae^t}$

[clever fjord]

7aman

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[small sail]

I’m having trouble starting this one

[small sail]

Like do I actually just give it the area of the circle

[mystic saffron]

can i know what is ave ?

[viscid flint]

average

[small sail]

Oh that’s what it wants

The average between integrals 0-2?

[viscid flint]

look up the average value of a function

your lecture was likely over this exact concept

[small sail]

Funny thing my lecture isn’t until Thursday

[mystic saffron]

||fun fact; do it on friday||

[small sail]

Working Friday :(

[mystic saffron]

:(

[hot briar]

then do it after lecture thursday

or find time

or look up average value of a function

[small sail]

Doing physics work that day

[small sail]

Yup

[hot briar]

you can do physics and math in one day

it is possible

[small sail]

I ain’t that guy 💀

Organic chem tutor to the rescue?

.close

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[queen moat]

In year $N$, the $300^\text{th}$ day of the year is a Tuesday. In year $N+1$, the $200^\text{th}$ day is also a Tuesday. On what day of the week did the $100^\text{th}$ of year $N-1$ occur?

[clever fjord]

Musab

[queen moat]

<@&286206848099549185>

[quasi sparrow]

!15m

[odd edge]

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

[mystic saffron]

I wanna say tuesday but im not sure

not gonna consider leap years in between ? if any

@queen moat

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[vivid ether]

Helps pls

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[magic violet]

hi, we started talking about TVM in my finance class today but we didn't get all the way through it, my BAii calculator also hasnt arrived in the mail yet so I'm struggling to do this since I dont have one

[magic violet]

I thought it was 672 but it wouldnt let me submit that as an answer

[viscid flint]

like if you click on b nothing happens?

[magic violet]

is B right?

[viscid flint]

idk but you said that's what you got right?

[magic violet]

I was randomly putting the values into an online calculator because idk how to do it

[cinder elk]

The equation for compound interest is $A = P(1 + \frac{r}{n})^{nt}$

A = final amount
P = initial principal balance (initial amount put in)
r = interest rate
n = number of times interest applied per time period
t = number of time periods

[clever fjord]

dragonbreath

[magic violet]

lol my professor just realized he didnt explain it and told us the answer

it is infact b

thanks for the help yall

!close

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[rare jolt]

ABCD is a Trapezoid where AB is parallel to DC. angle CAB is a and angle CAD is b.

prove that the ratio between area of triangle CMD to triangle AMB is

$\frac{sin^2b}{sin^2\left(2a+b\right)}$

[clever fjord]

Mathematician

[rare jolt]

firstly i thought i need to calculate The traingle areas ration by its sides

since they are parallel, we can say MDC is similar to MBA

ratio between traingle areas is the ratio between sides ^2.

hence the area ratio is (MD/MB)^2 or (MC/MA)^2 or (DC/BA)^2

lets use the a and b to find whatever ew can express with sides

[rare jolt]

this is where i need help with

or maybe even that way of solving isnt correct

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

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@rare jolt Has your question been resolved?

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<@&286206848099549185>

.close

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[rare jolt]

in this trapezoid AB is parallel to CD and. AB=BC=AD. AE is a diameter in the circle. ADK is and isoceles triangle.

DK=KL.

prove that DM = AE

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<@&286206848099549185>

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[simple timber]

Just 1a

How do I do that?

.close

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[subtle depot]

Using synthetic division how do I solve if k is a root and k = 5sqrt(3) and the equation is -x^5+2x+3

[subtle depot]

Im completely stuck, I can easily do one with only non sqrt roots

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<@&286206848099549185>

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[subtle depot]

.close

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[mystic saffron]

Hey guys. Was trying to do some note copying and review for my test on Friday. I noticed this in the lecture notes?

Should they not be equal to 0? And not 2 halves = 1??

I'll post the profs second page I'm the coming messages. I would love some clarification on this! Thank you!!

[mystic saffron]

Also, please @ me so I get notified

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[onyx burrow]

can someone check my answer?

[subtle depot]

uh yeah, dont you just plug in 1.1

[onyx burrow]

wdym

[subtle depot]

put 1.1 into 5(1.1)^2 + 2

[onyx burrow]

idk

[tropic flare]

how did u get the asnwer

did you just plug it in

[onyx burrow]

yea

[tropic flare]

do you know what the tangent line is

[onyx burrow]

=10x-3

[tropic flare]

if thats the tangent line

i feel like you should plug into that

[onyx burrow]

then I got 7

which is not a option

[tropic flare]

?

try again

sub in 1.1

not 1

[onyx burrow]

7.1

[tropic flare]

uhhh

try again

[onyx burrow]

wdym

[tropic flare]

10*1.1-3

[onyx burrow]

8

[tropic flare]

yes

[onyx burrow]

so thats the answer?

[tropic flare]

i think so

not sure what else it would mean

by use the tangent line

[onyx burrow]

can someone else check if your not 100%

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[compact temple]

hello
I understand how to find the amount of solutions however, im unsure how to do it without calculating the values of x and y

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[tight agate]

hi I'm doing geometry proofs

[tight agate]

I need help on 2 of them

on this one idk what I did wrong or how to make the s not negative

if that's too blurry I can retake

[lament lance]

Thats very blurry

[tight agate]

on this I'm just confused, I'm not sure where to go from the given

[tight agate]

gotcha

is that better?

[lament lance]

I think it is

[tight agate]

cool :D

[lament lance]

Your doing this one?

[tight agate]

on #5 I don't know what I did wrong or how to make the s not negative

on #7 I'm unsure where to go from the given 😭 is it supposed to be a system?

[tight agate]

yeah

[lament lance]

How did u get 2q - r?

[tight agate]

from the first given, 2q-r=p

I plugged it into p=q+2r

[lament lance]

2q-r-q would leave 2q on one side of the equation

[lament lance]

on the 5th row ur 2 seemed to have disapeared

either that or its just really blurry

[tight agate]

the 2 is still there

q-r=2r

where I added -r to 2r and got q=3r

[lament lance]

r = 3r-5
subract 3r on both sides gives
-2r=-5

right?

[tight agate]

ooh yes I see it now

thank you 😭😭 I noticed it when you said it

[lament lance]

ofc (:

[tight agate]

can you help me on the other one as well, I have no clue what to do on it

[lament lance]

lme see

[tight agate]

is that one blurry too

I can't tell

it's not extremely bad for me but I know things can be harder to see for others

[lament lance]

are u aloud to substitute -3 in

[lament lance]

I think its alg

[tight agate]

I don't think so

I think we have to go from the givens

we weren't doing it earlier so I'm assuming we can't now

[lament lance]

are you supposed to do elimination?

[tight agate]

I was thinking that but I didn't know how I could do that and use proofs to do it

I was thinking maybe to rearrange one of the equations to where I could plug it in but I dunno 😭

[lament lance]

I think u goda rearrange them

[tight agate]

I'm gonna try it

[lament lance]

y=-3x-10 for example

do I even make any sense?

I mean like this

[tight agate]

yes

I'm working it out rn and I think that's what I have to do

[lament lance]

y=-3x-10

8x-3(-3x-10)=-21

I dunno man

¯_(ツ)_/¯

[tight agate]

I think that's right

I just got x=9

aggh 😭 what

[lament lance]

what?

[tight agate]

sighh I love geometry

[lament lance]

ok here

[tight agate]

yeah please idk what I did wrong

I have the 1st things right

maybe I distributed wrong

yeah I did

[lament lance]

Distribute

[tight agate]

forgot negative times a negative

is a positive

[lament lance]

8x+9x+30=-21

17x+30=-21

17x=-51

devide both sides by 17

x=-3

[tight agate]

wooo got it

thank u sm

it's late and my brain isn't working so I forgot half the negatives I had

[lament lance]

ofc (:

No worries man lmao

[lament lance]

happens to the best of us

[tight agate]

🙏🙏 tysm again

[lament lance]

yeah have a good night man

:D

[tight agate]

u too!

,close

[lament lance]

almost XD

[lament lance]

U almost have it

[tight agate]

ahh 😭😭

.close

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[soft citrus]

Hello! I need someone to find out what is wrong.

I am new at partial integral

Bruh... My handwriting...

[odd crypt]

Use LIATE

[hollow meadow]

wait what is the question

[odd crypt]

To choose u

Or f(x(

[hollow meadow]

disgusting i use LIPET

[odd crypt]

f(x)*

Same idea

[soft citrus]

Hold on still uploading

[soft citrus]

What is LIPET?...

[hollow meadow]

Log, inverse trig, polynomial, exponential, and trig

its the order in which you choose u in order to make it easier to integrate

[soft citrus]

Alright thank you

But still need find what is wrong

I am not sure what i do

Traditional solving 😁

Disclaimer (im grade 11)

[hollow meadow]

its fine

so first you need to identify u

since 2x is a polynomial it comes before trig so that is your u

[soft citrus]

Is the 2x U?

[hollow meadow]

yeah

[soft citrus]

Alright thanks for your feedback. I just need improve my brain. Any more feedback?

[hollow meadow]

well wait are you in ap calc

or is this a different class

[soft citrus]

Regular

[zinc wing]

whoever wrote that integral needs to be fired

legit looks like abs val 🤣

[hollow meadow]

check out ap daily videos on youtube

you can try to see if there are videos of integration by parts because they are helpful

[hollow meadow]

fax

[soft citrus]

Can i close?

[hollow meadow]

yeah sure have a good day or night

[soft citrus]

.close

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[plain sun]

can someone please help me with question b?

[odd edge]

@plain sun Has your question been resolved?

[odd edge]

@plain sun Has your question been resolved?

[plain sun]

<@&286206848099549185>

[toxic monolith]

but i warn, it is little bit time-consuming, so be patient while using it

[plain sun]

damn I kinda feared having to do that

alrighty ill try it

[toxic monolith]

keeping thumbs up )

[plain sun]

oh wait its actually pretty easy to relate it

since everythings factorised and its easier to visualise

alright i get it now

thanks for the help

[toxic monolith]

yvw)

[plain sun]

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[somber birch]

The answer is not 90.69% what did I do wrong

[undone mauve]

doesn't seem tough

[somber birch]

I didn’t think so but I can’t figure it out

[undone mauve]

what did you try?

[somber birch]

I tried just making the side length of the triangle as 2 then the radius of the circle is 1 so the are of half that circle is pi/2 then the area of the triangle is root3 so put pi/2 over root3

[undone mauve]

area of equilateral triangle is root 3 / 4 * side square

so here it should be root 3, correct

[somber birch]

Yes

[undone mauve]

now the 3 arcs inside it, all have a 60 degree angle, right?

[somber birch]

Yes

I would assume

[undone mauve]

no need to assume. it is an equilateral triangle

[somber birch]

Then yes

[undone mauve]

so the area of one of those, if their radius (of the circle they are from), is pi*r^2 * (60/360)

[somber birch]

Yes

[undone mauve]

now there are 3 of those, so that multiplied by 3

[somber birch]

Multiplied by 3 is 180/360 so half the circle

[undone mauve]

so you should get pi * 1/2 = pi/2, which you got. correct

correct

[somber birch]

So pi/2 correct?

[undone mauve]

yes

[somber birch]

Yeah

[undone mauve]

pi/2 is 1.57079~

[somber birch]

Yes

[undone mauve]

root 3 is 1.732

[somber birch]

Yes

[undone mauve]

so 90.69% I guess

so its correct

which means your answer is right

[somber birch]

Ok then the code to the safe must be wrong because that’s what I put and it was wrong

Thanks so much

[undone mauve]

no worries

[somber birch]

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[surreal bluff]

help

[surreal bluff]

nvm

.close

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[jaunty plank]

how

[wide jackal]

180 degrees = pi radians

[jaunty plank]

thats the answer?

[wide jackal]

no...

[jaunty plank]

oh

[wide jackal]

thats the conversion

[jaunty plank]

lol

so how do i convert

[wide jackal]

degree to radian

[jaunty plank]

uhm

[wide jackal]

uh

[jaunty plank]

wait ill find a way

[wide jackal]

mb

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[pseudo hearth]

How will math help me after the final exams?

[plain ruin]

To next finals

[languid horizon]

I mean if you are goona work in industry that requires knowing math. It will help

If not it really wont

[plain ruin]

It improves logic and etc. 🙂

[languid horizon]

And then finals after that

[languid horizon]

Some people dont need it

[plain ruin]

What work you'll have in future

[sharp oak]

I mean I personally do it, because I find it interesting. Makes programming more fun too, which is another hobby of mine.

[plain ruin]

Lucky people 😦

[sharp oak]

But yeah, I can't imagine even 0.1% of people touch math higher than grade 10, ever. Even some people in specialties like engineers or programmers.

[plain ruin]

What...

Literally a lot of people go to 11 grade

So they touch it

Isn't it

[sharp oak]

I mean for professional careers

[plain ruin]

Ig 20% use it

[sharp oak]

I'm not including myself in this list, as I'm not paid to do math.

[plain ruin]

Businessmans, programmers, engineers, teachers and etc.

A lot

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[sharp bay]

Show that for any number non-zero n the equality holds

[sharp bay]

I tried induction and

[sharp bay]

I got stuck at one point

[halcyon parrot]

i think u'll get 2 HPs

[sharp bay]

HPs?

[halcyon parrot]

harmonic progression

if u combine each of the 2 adjacent tersm with alternate sign on LHS

[sharp bay]

so like 1 - 1/2, 1/3 - 1/4 and so on?

[halcyon parrot]

yea

im not very sure, but u already have a HP on RHS if u get another on the LHS u m8 be able to do smtg

[modern sundial]

This is just plain induction. You should keep trying down that path

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[sharp bay]

very helpful to point out something I mentioned already xD, a tip would've been more useful

(clueless)

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[wide jackal]

the entire question would help

uh

[spring knot]

🤨

[wide jackal]

so youre just supposed to find trigo ratios when theta is 30 degrees?

there is no other information?

[cursive field]

...

i think they mean sine and cosine?

like just type sin 30 = x and so on maybe?

@ornate juniper yk sin and cos right

so what's the issue

...

sin 30 = ?

[real trellis]

.close

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[cursive field]

well that was abrupt

but okay

[real trellis]

pls dont help people with ongoing tests

[cursive field]

oh i'm sorry, i didn't spot that

[wide jackal]

what is the command for that?

!test

[cursive field]

idt there is one

i just checked

[spring knot]

it's @ Mods

[wide jackal]

theres no ! command?

[cursive field]

...
not what he meant

nope

[wide jackal]

hows there a !noans

[cursive field]

lmao

[wide jackal]

but no test

[cursive field]

idk

[spring knot]

idk

actually i think it's !cheating

!cheating

!cheat

[cursive field]

...

[spring knot]

ig not

[cursive field]

no...

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[proven cape]

I didn't understand how he used $a^n-b^n$

[clever fjord]

sofficino9188

[wide jackal]

its an algebraic identity

(x^n - y^n) / (x-y) = (x^ (n-1) + x^ (n-2) * y..... y^(n-1) )

[proven cape]

but how he get the last thing

[wide jackal]

easily understood using binomial

[wide jackal]

.

this

applied there

[proven cape]

i don't understand😭

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[marble ether]

$\lim_1\dfrac{2/x-1}\cdot\dfrac{\dfrac{ln(x)/x-1} - 1}{x-1}$

[clever fjord]

ᑎᗩᑕᖇᗴᝪᑌᔑᗞᗩᗯᑎ596

[marble ether]

eh?

sorry

.close

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[stark elbow]

Just try latexing it again

No need to close the channel, unless you changed your mind

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[elfin warren]

A sailor is with his boat at the point A = (1, 3, 5) and sails in
the direction v = (3, 1, 1)T
. He intends to make a right-angled course change before
to be able to pass through the harbor entrance at the point P = (2, 4, 7).
(a) In what plane is the water surface located?
(b) Determine the point P′ where the course change should be made. i have done a) already and i got the answer 2x-10y+4z=-8. i am not sure how to solve b entierly

[elfin warren]

we get a picture as well

[stark elbow]

What can you say about PP' and v?

[elfin warren]

hmm that they are perpendicular?

[stark elbow]

Yes

Build an equation out of that

First express P' using some variable

[elfin warren]

like P´=(x,y,z)?

[stark elbow]

There is a better way

You know P is on the line he was sailing in

[elfin warren]

yes

[stark elbow]

Sorry, meant P'

Then how can you express P'?

[elfin warren]

has it something to do with t?

P´ = A + t * v?

@stark elbow

[stark elbow]

Yes

n?

[elfin warren]

okay

i take it back with n

[stark elbow]

No need to delete your messages

Don't be embarrased to say something wrong

I make mistakes all the time aswell

[elfin warren]

so just P´ = (1,3,5) + t(3,1,1)

and then write it as (1+3t), (3+t), (5+t)?

[stark elbow]

Yes

(1 + 3t, 3 + t, 5 + t)

[elfin warren]

yes okay

[stark elbow]

Now build an equation

[elfin warren]

you mean like (1+3t) + (3+t) + (5+t)

and solve out t?

t = -6/5

(-13/6, 9/5, 19/5)?

i think that is wrong becuse it is not perpendicular to v

oh i calculated t wrong

t=-9/5

but its still wrong

[stark elbow]

You want v to be perpendicular to PP'

[elfin warren]

oh right

[stark elbow]

Express PP' in terms of t, and build an equation out of this

[elfin warren]

okay (2,4,7) + t(-22/5, 6/5, 16/5)

and then

(2-22/5t, 4+6/5t, 7+16/5t)

[stark elbow]

What is that?

What are you doing

Where are the fractions coming from?

[elfin warren]

I calculated P’ to be (-22/5, 6/5, 16,5)

And put that with p in an equation

i calculated t = -9/5 and then i thought i was supposed to put that back into (1+3t, 3+t, 5+t)

[odd edge]

@elfin warren Has your question been resolved?

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@elfin warren Has your question been resolved?

[elfin warren]

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[pliant barn]

part b help 🙏🏻

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[clever fjord]

kisnar

kisnar

[faint knot]

the absolute value is for the x-2, not the -2x^2

its there because x-2 can be negative

this is in R not Z

[mystic saffron]

yes you can do (x²-4x+4)^x²

probably

[clever fjord]

kisnar

[mystic saffron]

yes

[clever fjord]

kisnar

kisnar

kisnar

[mystic saffron]

no

log can be negative

when is log negative

so 0<x²-4x+4<1

[clever fjord]

kisnar

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[nimble tide]

I had a question if i have a set say, {{A,B},C} = {A,B,C} are these equivalent?

[frigid canopy]

no, two sets are said to be equivalent if they have the same elements if I remember right

[topaz dragon]

No

[nimble tide]

what if i have {{A},{B},{C}}

[topaz dragon]

Assuming A B are sets you are comparing a set of sets with elements that are sets

That is a set of sets

Or a family of sets

[nimble tide]

yeah

[topaz dragon]

Where each element of the set is a set of a set

[zinc glacier]

they have to have exactly the same elements

so the three sets you gave are all different

[nimble tide]

oh right k

[zinc glacier]

{{A, B}, C} = {{B, A}, C} = {C, {B, A}} for example

[nimble tide]

.close

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[clever fjord]

𖤍

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[bleak shard]

My first thought is v=d/t assumes a constant velocity acceleration. But there's an acceleration due to the force of gravity between the planets

And the force due to gravity will change as a function of the position too though.

Which means the acceleration isn't constant.

Unless we are to assume the force of gravity between the planets is negligible. But then the velocity wouldn't change.

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[graceful shore]

I need to find the equation of the tangent to the graph

Can someone tell me what I did wrong?

[mystic saffron]

your derivative seems wrong firstly

[mystic saffron]

sorry misread

but anyways why not substitute x = 3 for your f(x_0)

[graceful shore]

we cannot do that

But let's tey

try*

How to do it?

[mystic saffron]

why not?

thats what linearisation is

[graceful shore]

Oo

[mystic saffron]

,,\m Lx = \m fa + \m{f'}a(x-a)

[graceful shore]

it gives me the result as at the end

[mystic saffron]

yeah

[graceful shore]

Yes, I knew that there was no need to replace it, as it were..

[mystic saffron]

out x = 3 here and simplify

[mystic saffron]

hm?

anyways just substitute in x = 3 for your derivative and function and that should be it from what im seeing

[graceful shore]

thank you

.close

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[worn bramble]

I will translate this just taking the channel

[worn bramble]

Function 𝑔 is given by the formula 𝑔(𝑥) = |- 1 4 𝑥2 + 3𝑥 - 5| for each 𝑥 ∈ ℝ. A fragment of the graph of the function 𝑔 in the Cartesian coordinate system (𝑥, 𝑦) is shown in the figure (units omitted)

Determine the set of all values that the function 𝒈 takes in the interval [𝟗, 𝟏𝟏]. Write down the calculations

I have honestly no clue where to begin

I know too little about functions

Do I calculate g(9) and g(11) and see how it moves?

[ocean bramble]

start by making the equivalent function without the absolute value. You got it graphed, so you clearly see that you will have 3 distinct intervals defined

for each of them, obtain the derivative. First and third interval will show that they are monotonic

on the second, it will show that there's a maximum or minimum at some point (look at the graph)

since you're on a closed interval, there HAS to be a maximum and minimum value within the interval. Either the places where derivative is 0, or the endpoints

since the function is continuous, it will take every value between that max and min values, in each of the three intervals

you can see from the graph that it will take the value 0 as minimum, so you really only care about the maximum in each of the three intervals, and it will take every value from 0 to the max of those 3 maxes

[worn bramble]

I am insanely amazed

[ocean bramble]

how so?

[worn bramble]

Thank you, I will try to make it work 😅

You explained it very neatly

[ocean bramble]

well, this is a pretty standard exercise

now you have to actually calculate each of the steps

[worn bramble]

I have missed a lot of school in the past and I have so many gaps

I am trying to fill all of them before finals and I think right now is the best time to do it

[ocean bramble]

have you studied derivatives yet?

[worn bramble]

Yeah

That I know

[ocean bramble]

then you should have no problem following the procedure

[worn bramble]

I have problems with earlier topics like functions, etc.

trying to catch up

[ocean bramble]

i would recomend consulting the definitions first then

[worn bramble]

Do you have any good resources on that as well?

I only know polish words for mathematical expressions

At least for the more advanced ones

(Btw this problem is from earlier mock final exam)

[odd edge]

@worn bramble Has your question been resolved?

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[zinc wing]

idk where i went wrong when solving this

,tex
$$\sqrt[2]{x}+ \sqrt[2]{y} y\prime = 0, y(1) = 4$$
$$\sqrt[2]{y} y \prime = -\sqrt[2]{x}$$
$$\sqrt[2]{y} \frac{\dd y}{\dd x} = - \sqrt[2]{x}$$
$$\sqrt[2]{y} \dd y = -\sqrt[2]{x} \dd x$$
$$\int \sqrt[2]{y} \dd y = - \int \sqrt[2]{x} \dd x$$
$$\int y^{\frac{1}{2}} \dd y = - \int x^{\frac{1}{2}} \dd x$$
$$\frac{2}{3} y^\frac{3}{2} = - \frac{2}{3} x^\frac{3}{2} + C$$
$$y = -x + C$$
$$ $$

$$4 = -1 + C$$
$$-C = -1 - 4$$
$$-C = -5$$
$$C = 5$$
$$ $$

$$y = 5 - x$$

[clever fjord]

@zinc wing

[zinc wing]

idk where i went wrong when solving this

[sullen ferry]

its a bit more complicated

[zinc wing]

i cant do that?

when i do that, im just changing the value of C

but it doesn't matter cause i simply have to solve for a different value of C, right?

[blazing shuttle]

powers dont distribute over sums

[sullen ferry]

(x+a)^b not equal to x^b+a^b

yes^

[zinc wing]

oh its cause of the power

[zinc wing]

(right?)

,tex
$$\sqrt[2]{x}+ \sqrt[2]{y} y\prime = 0, y(1) = 4$$
$$\sqrt[2]{y} y \prime = -\sqrt[2]{x}$$
$$\sqrt[2]{y} \frac{\dd y}{\dd x} = - \sqrt[2]{x}$$
$$\sqrt[2]{y} \dd y = -\sqrt[2]{x} \dd x$$
$$\int \sqrt[2]{y} \dd y = - \int \sqrt[2]{x} \dd x$$
$$\int y^{\frac{1}{2}} \dd y = - \int x^{\frac{1}{2}} \dd x$$
$$\frac{2}{3} y^\frac{3}{2} = - \frac{2}{3} x^\frac{3}{2} + C$$
$$ y^\frac{3}{2} = -x^\frac{3}{2} + C$$
$$y = \left(-x^\frac{3}{2} + C\right)^{\frac{2}{3}}$$

is this value?

[blazing shuttle]

you dropped the -

[zinc wing]

*is this valid?

[zinc wing]

fixed. (i think)

[blazing shuttle]

its right but a bit weird to say that way

[zinc wing]

yes ik the sqrt[3/2] is bad but the value is correct, right?

[sullen ferry]

3/2 root is a bit weird

you could just write as power 2/3

[zinc wing]

would this be a valid calculator-ready expression

[clever fjord]

@zinc wing

[blazing shuttle]

yep

[zinc wing]

,tex 50)
$$\sqrt[2]{x}+ \sqrt[2]{y} y\prime = 0, y(1) = 4$$
$$\sqrt[2]{y} y \prime = -\sqrt[2]{x}$$
$$\sqrt[2]{y} \frac{\dd y}{\dd x} = - \sqrt[2]{x}$$
$$\sqrt[2]{y} \dd y = -\sqrt[2]{x} \dd x$$
$$\int \sqrt[2]{y} \dd y = - \int \sqrt[2]{x} \dd x$$
$$\int y^{\frac{1}{2}} \dd y = - \int x^{\frac{1}{2}} \dd x$$
$$\frac{2}{3} y^\frac{3}{2} = - \frac{2}{3} x^\frac{3}{2} + C$$
$$ y^\frac{3}{2} = -x^\frac{3}{2} + C$$
$$y = \left(-x^\frac{3}{2} + C\right)^{\frac{2}{3}}$$
$$ $$

$$4 = \left( -1^\frac{3}{2} + C \right)^\frac{2}{3}$$
$$4 = \left( -1 + C\right)^\frac{2}{3}$$
$$4^\frac{3}{2} = C - 1$$
$$8 = C - 1$$
$$C = 9$$
$$ $$

$$y = \left(-x^\frac{3}{2} + 9\right)^{\frac{2}{3}}$$

[blazing shuttle]

x=1

[zinc wing]

correct?

[blazing shuttle]

ya

[clever fjord]

@zinc wing

[zinc wing]

alr thanks so much for your help roketsune!

[blazing shuttle]

np!

[zinc wing]

.close

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[clever fjord]

kisnar

[faint knot]

if P only has real coefficients, it means that P(1 - i) is also = 0

[clever fjord]

kisnar

[faint knot]

thats correct, thats an example of a P(x) that has both 1+i and 1-i as roots

np

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✅

[clever fjord]

kisnar

[faint knot]

yes

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[opal nymph]

Can someone help me with this

[opal nymph]

I tried just rearranging the equation and solving / plugging in A^3 but im not really sure if thats how to approach it

[green elm]

try simply multiplying I - A by the proposed inverse and see if it works

[opal nymph]

Yeah i got this but what exactly do you prove with that

you can replace A^3 with 0 so its just the first 3 terms

[green elm]

well the middle two terms cancel, right?

[opal nymph]

how would it cancel arent they all different

[green elm]

A^2 is the same as IA^2

I is the identity

[opal nymph]

Oh

ok thx let me see then

ok got it thank you

.close

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[opal nymph]

.reopen

[odd edge]

✅

[opal nymph]

sorry one more question

[green elm]

i noticed something important

your name is "waffies" not "waffles"

[opal nymph]

haha yeah

[green elm]

thought i was seeing an optical illusion at first haha

[opal nymph]

gamer tag like that so i made discord same

[green elm]

go ahead with your question 😁

[opal nymph]

For this one i found solutions of A
x1 = -2 -2r -3s-t
x2 = r
x3 = 5-2s -4t
x4 = s
x5=t

for part b im not really sure how to approach it

do you just replace first and third columns

[green elm]

no, they want you to "undo" the row reduction ig

[opal nymph]

oh so you are trying to get a matrix that has those two columns with those numbers

[green elm]

yea

and since you have the augmented matrix, when you "undo" the row manipulations to get those particular columns, you'll get the corresponding b in the last column

[opal nymph]

ok makes sense let me try it

sorry still stuck, what step should i take first

is it better to get row 3 to match first

[green elm]

should be, since you do that last in the "forward" row reduction, you should do it first in the "reverse" version

actually i think the order shouldn't matter

[opal nymph]

yeah i really dont know what im doing im trying different manipulations but im not getting the columns they want

[odd edge]

@opal nymph Has your question been resolved?

[odd edge]

@opal nymph Has your question been resolved?

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[desert marlin]

Prove that for $m\neq n$ that $m\mathbb{Z}$ is not isomorphic to $n\mathbb{Z}$

[clever fjord]

Austin

[desert marlin]

I'm trying to follow along this proof, which seems fine up until the black part

I don't know what generators are, they haven't been covered in my class

Is there a way to explain this part without using them?

@drowsy stump

[drowsy stump]

Basically

Every element in f(mZ) can be written in the form of k*f(m)

Every element in the input is of the form n = n*1, so every element in the output is of the form f(n) = nf(1)

[desert marlin]

what do you mean "in f(mZ)"?

[drowsy stump]

So f(1) "generates" nZ

Meaning that every element in nZ is a multiple of f(1)

[desert marlin]

and why is this a bad thing?

[drowsy stump]

This would mean that f(1) divides n*j for each j in Z

But if f(1) = nk and k > 1, then n*k cannot divide n

Since like n/nk = 1/k which is not an integer

[desert marlin]

so if m!=n this can't be true that f(1) generates Zn ?

[drowsy stump]

Exactly

[desert marlin]

but it must be true that f(1) generates Zn if there is an isomorphism?

[drowsy stump]

Yep

Thats what he meant by mapping generators to generators

[desert marlin]

I don't get what you mean every element in the input is n=n*1

[drowsy stump]

Oh shit i meant f(m) everywhere

Every element in the input is k*m

[desert marlin]

yes

[drowsy stump]

So every element in the output is f(k*m) = kf(m)

So f(m) generates nZ

[desert marlin]

if there's an isomorphism

that is?

[drowsy stump]

Yes

[desert marlin]

So

since

f(m^2)=f(m+....+m)=mf(m)=mnk

m=nk

so

f(m)=f(nk)=kf(n)

for all m

so far so good?

f(m)=kf(n) is where we want to derive our contradiction from

[drowsy stump]

f(m) = kn you mean

[desert marlin]

why

[drowsy stump]

Is n even in mZ?

[desert marlin]

no

I was just going off of this

f(m)=nk

Oh I see

Okay let me restart

yah no I'm lost again @drowsy stump

I found one that I almost understand more I think but still not quite getting the part that I'm supposed to fill in

[faint knot]

theres no way thats a "less than but not equal to" operator
maybe its "less than but not a factor of"??

[desert marlin]

Yah I don't understand it but I'm willing to move past it if you know what they mean about filling in the details about non-surjectivity 😭

.close

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[sand sable]

in what real life situations are quadratic functions applicable

[sand sable]

is it possible to be used in area?

[quartz elk]

throwing a ball

[sand sable]

may i ask for a simple example

[chilly epoch]

What's simpler than throwing a ball?

[sand sable]

no like

an equation

[spiral basalt]

falling objects, assuming no friction, follow a parabola

[quartz elk]

you want an equation

[spiral basalt]

(assuming they have horizontal speed ofc)

[quartz elk]

wait lemme try something rq

[sand sable]

yeah like on how do i form it into an equation

aight tyt

[quartz elk]

$y = x^2$

[clever fjord]

CoderCrafted

[quartz elk]

nah

oh

wow

dang it actually works

cool

[sand sable]

damn sick bot

is that it?

[quartz elk]

no

[odd edge]

@sand sable Has your question been resolved?

[sand sable]

.close

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[steel notch]

can someone tell me how we find that this siereis converges in [2,2)

i thought of using square root test to find the radius

but i ma not sure how to get x^2 alone

so that i have a series of type an*(x^2)

[crisp wadi]

A bit of rewriting and you will notice that this is a geometric series

[steel notch]

using ln?

[crisp wadi]

Nope, just laws of exponents

[steel notch]

writing e^nx^2 as e^n + e^x^2

?

[crisp wadi]

There is no law of exponents implying that

[steel notch]

right sorry i got confused

[crisp wadi]

What I am talking about is
[ \frac{e^{nx^2}}{e^{4n}} = e^{nx^2 - 4n} = ]
[ = e^{(x^2 - 4)n} = \left( e^{x^2 - 4} \right)^n ]

[clever fjord]

孤独な豆