help-41

[cursive jay]

at some point

[paper ibex]

My exact doubt is, for instance if the question was instead "find the vertex of the transformed function y=1/4(x+1/3)" , here 1/4 will become 4 and so 4*5 gives us 20...am I not supposed to work with this logic, so working backwards I should get 1/4 too

[cursive jay]

Wait why does 1/4 become 4?

Let's work through that one actually

y = 1/4 f(x + 1/3)

You already calculated that x = 13/3

[paper ibex]

OH

Wait

I'm so sorry man

LMAOOOO😭😭

Bro there ain't no way it just clicked

I confused vertical and horizontal

[cursive jay]

Oh haha

[cursive jay]

Well done

[paper ibex]

Look cuz here

For the last one, this is where I'm supposed to take the reciprocal

Oh damn bro I need to head to sleep seriously

[cursive jay]

Did you work out what was going on with the first pic you sent?

[cursive jay]

Don't we all 🥲

[paper ibex]

No 😭😭 ill try what u told me earlier once I wake up cuz I dint think anythings getting into my head atm

This right?

[cursive jay]

Yes that

Get an equation for y, solve for y, and then from that you can work out x

[paper ibex]

Alr I'll take a screenshot cuz I'll lose this channel

[paper ibex]

Okay tysmm bro

[cursive jay]

Good luck 🙂

[paper ibex]

Tyyy

.close

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[split sail]

hello

[amber wasp]

Please don't occupy multiple help channels.

[honest dagger]

.close

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[lime jasper]

How do u find the asymptotes of 1/x^2 - 1

[pallid canopy]

$\frac{1}{x^2 -1}$?

[grizzled pagoda]

riemann

[lime jasper]

Yes

[pallid canopy]

bottom bullet point

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[rain imp]

can somebody help me create a function to model this graph?

[rain imp]

I understand va's are +- 3

so in the denominator I will have (something)/(x+3)(x-3)

and the ha is y = 1, so the numerator has to be degree 2 to match the degree 2 denominator

my attempt was something like y = x^2/(x+3)(x-3)

but when i graph this on desmos, it doesn't match the graph

[shrewd elm]

you need something with odd symmetry near x=0

if your numerator was x instead of x^2, that would be true

example:

,w plot -x/((x-3)(x+3))

[grizzled pagoda]

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[rain imp]

yea if i set it to -x

but the HA

is off ;[

[shrewd elm]

if you like the general shape but you want the HA to be 1 instead of 0, just add 1

[rain imp]

OHHH

so I can just vertically translate

[shrewd elm]

yep

[rain imp]

tyty

[shrewd elm]

yw

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[split sail]

Im stuck on this equation

The delta is supposed to be 25 and ive got 95

What Did i do wrong?

[crude kite]

What exactly is delta denoting?

[outer hull]

discriminant i assume

[split sail]

its b squared - 4ac

[split sail]

Im doing quadratic equations

[crude kite]

So what's your a, b, and c?

[split sail]

a =1 b = -2 c=-24

[crude kite]

That equation says x^2 - 2x - 24, right?

[split sail]

Oh -2

[split sail]

Yeah

[crude kite]

Plugging it into the formula, what do you have?

[split sail]

-2^2-4•1•(-24)

[crude kite]

Parentheses

For the -2

But what does that result in?

[split sail]

1/4?

[crude kite]

How

What's (-2)^2?

[split sail]

-4

[crude kite]

No

$-2^2 \neq (-2)^2$

[grizzled pagoda]

CaptainNova22

[split sail]

Ohhh

4

[crude kite]

What's -4 * 1 * -24?

[split sail]

96?

[crude kite]

Yeah

So now add them

[split sail]

-4-96 is 100

[crude kite]

No it's 4 + 96

Because (-2)^2 is 4

4 - 4 * 1 * -24

[split sail]

Okay so 4+96=100

[umbral osprey]

96

[crude kite]

Yes

[split sail]

I need to do something else

Now

Bc in the answers its x=6 or -4

[crude kite]

You found delta

It looks like the problem is wanting you to find the roots

[umbral osprey]

delta math?

[crude kite]

Discriminant
@umbral osprey

[umbral osprey]

oh

[split sail]

Ohhh okay i need to put it in the formula

(-b+delta sqrt)/ 2a

[crude kite]

It's sqrt

Not squared

[split sail]

So (-(-2)+sqrt100)/2 • 1

Is (2+ sqrt 100 )/ 2

[crude kite]

It's also plus and minus

[split sail]

Yeah thats why there are 2 answers

[split sail]

do i multiple it by 2 ?

[crude kite]

Why?

Just simplify it

[split sail]

Can u help me i have no idea how because of the sqrt

[crude kite]

Do you know what a sqrt is?

[split sail]

Oh wait nvm i right infront of me

Sorry

i tried to complicate it when the sqrt is 10

thank u very much much i think thats it tbf

im not the sharpest knife in the drawer

.

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[native thunder]

I need help. I have fallen, and I can't get back up

[viral root]

tried partial fractions?

[pallid canopy]

.close

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[outer hull]

more like spinal fractures

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[nocturne plaza]

I don't know what they're asking me to do? :')

[vestal hatch]

do you know what a riemann sum is?

[nocturne plaza]

I think that's what we're doing neeext?

Idek anymore x-x

[vestal hatch]

well that's basically what you're doing here

[nocturne plaza]

Ah ;-;

[vestal hatch]

"assume" the speed from 0-12 is 30, assume speed from 12-24 is 28, etc

[nocturne plaza]

ahhh ok, I sorta see what you mean? 🤔

ok yeah I think I get it

hopefully 😅

[vestal hatch]

this is a left-hand sum if we're using the points on the left to guess at the intervals

if you have the speed at 60 you could do a right-hand sum too

where 0-12 is 28, 12-24 is 25, etc

or you could even average the two but that's not necessary

[nocturne plaza]

ahh oki, I'll try it. Thank you :D

I shall close the channel for now :)

[vestal hatch]

np

[nocturne plaza]

.close

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[mossy eagle]

What is the triple angle formula for cos6x

[vestal hatch]

wym triple angle formula

you want to write it in terms of cos(2x)?

[honest dagger]

$\cos6x=\cos(2x+4x)$

[grizzled pagoda]

SWR

[mossy eagle]

Is it 4cos³2x-3cos2x
Or 4cos³3x-3cos3x

[vestal hatch]

6x is triple what angle

[mossy eagle]

Using the idea of cos3x=4cos³x - 3cosx

[vestal hatch]

right ok

what is 6x the triple of

2x?

so that's what should be the 'x' s

cos(3y) = 4cos^3(y) - 3cos(y)

y = 2x

[mossy eagle]

Ohh

So it can't be written directly u need another variable?

[vestal hatch]

like

using the formula "cos(3x) = 4cos^3(x) - 3cos(x)" is just confusing when you're trying to find cos(6x)

you have to set x = 2x and that doesn't make any sense

[honest dagger]

!xy

[amber wasp]

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

[vestal hatch]

so i just changed the formula to use y instead

[winged panther]

!nosols

[amber wasp]

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.

[winged panther]

Can't give you soln. But the way to derive it is as Dreyuk suggested

cos 6x = cos 3(2x) = 4cos³2x-3cos2x

and then you use, cos 2x = 2cos² x - 1

[mossy eagle]

Oh I see

[mossy eagle]

I was doing a problem that involved me to use that identity of cos3x then I was just wondering what if we had cos6x

[vestal hatch]

,w expand 4(2cos^2(x)-1)^3 - 3(2cos^2(x)-1)

$32\cos^6x-48\cos^4x+18\cos^2x-1$

[mossy eagle]

How did he come up with this

[vestal hatch]

[vestal hatch]

6x = 2x + 4x?

[mossy eagle]

It is harder than I expected lol

[mossy eagle]

Nvm it's just 4+2

[vestal hatch]

oops messed it up

[grizzled pagoda]

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[vestal hatch]

ehre it is

[grizzled pagoda]

Dreyuk

[mossy eagle]

Thank you

[vestal hatch]

np

[mossy eagle]

can we be friends

my math is quite bad idk if u can help

[vestal hatch]

sorry i dont take friend requests from mathcord helpees

i end up with too many people DMing asking for help

i'd rather just come here and help when i feel like it

[mossy eagle]

Ok

.close

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[keen pawn]

Find a simpler form of $\neg(\neg P \lor Q) \lor ( P \land \neg R)$
\
$\neg(\neg P \lor Q) \lor ( P \land \neg R) \equiv (P \land \neg Q) \lor (P \land \neg R) \equiv P \land ( \neg Q \lor \neg R) \equiv P \land \neg( Q \land R)$

[grizzled pagoda]

What a wonderful world!

[keen pawn]

Is this fine?

[winged panther]

Tis okay

[keen pawn]

Thanks

.close

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[nocturne plaza]

this is super quick, but is #5 the right one to use if I have a 3 in front of the i? Could I just multiply the whole thing by 3? (Included a badly drawn image of the problem I'm working on, haha)

[vestal hatch]

yes you can factor the 3 out of the sum

that's rule number 3 on your list

[nocturne plaza]

oh, I assumed the a_i disqualified it, but I maybe misinterpreted what they were saying 😅

woops

aight, ty :)

.close

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[thick canyon]

how to find ${d/dr} \fracc{r )1-r)²}$

[grizzled pagoda]

bagelguy3
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thick canyon]

how to find d/dr r/(1-r)^2 $$

[grizzled pagoda]

bagelguy3
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thick canyon]

d/dr r / (1-r)

[split sail]

$\frac{d}{dr} (\frac{r}{1-r})$

[thick canyon]

yes

[split sail]

Yuck

[split sail]

Quotient rule?

[thick canyon]

wait no

mb

I messed up the question

theres no square in the denominator

[split sail]

.

[night snow]

lmao

[thick canyon]

how

look

r (1-r)^-1

use power rule

we ge t

(1-r)^-1 + (1-r)^-2

lemme calc g(r) real quick

[split sail]

Yeah you could do that too

[thick canyon]

see

so u geat

get

wait what

do u agree?

oh wait

I GOTTA MULTIPLY BY R

i forgot

so thats where i went wrong

[night snow]

there shud be a -ve sign too

[thick canyon]

i see

what -ve?

theres no -ve

i kept forgetting to multiply the second term by r

cuz f(r) is r

g(r) = (1-r)^-1

[night snow]

when u differentiate (1-r)^-1
u get -1.(1-r)^2

[thick canyon]

yes

and then u multiply by the derivative of the inner function

[night snow]

oh ye mb

[thick canyon]

which is -1

[night snow]

mbmbmbmbmb

[thick canyon]

so the two negatives cancel each other out

thx for ur help

.close

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[grizzled crystal]

hey

[grizzled crystal]

i saw a problem someone here had and tried to solve it

[grizzled pagoda]

Martin

[grizzled crystal]

Now I tried 2 different options and got 2 different results

[grizzled pagoda]

Martin

[grizzled crystal]

Option 2 (discs):
$$V=\int dV, dV=\pi r^2 dz$$
$$r^2=-2\ln{\frac{\pi z}{5}}$$

[grizzled pagoda]

Martin

[grizzled crystal]

nvm

got it

.close

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[keen pawn]

Simplify $\neg(\neg P \land Q) \lor (P \land \neg R)$\$\neg(\neg P \land Q) \lor (P \land \neg R) \equiv (P \lor \neg Q) \lor (P \land \neg R) \equiv P \land ( \neg Q \lor \neg R) \equiv P \land \neg ( Q \land R)$

[grizzled pagoda]

What a wonderful world!

[sinful sapphire]

i don't get it

[keen pawn]

hmm?

[sinful sapphire]

is this 4 answers to one question?

[keen pawn]

Thse are my steps

[sinful sapphire]

oh it's a chain

yeah ok

this is wrong

[keen pawn]

The connectors should be switched?

[sinful sapphire]

you can tell because ~Q is sufficient on the left

but the right requires P

[keen pawn]

So it should be $P \lor ( \neg Q \land \neg R)$

[grizzled pagoda]

What a wonderful world!

[sinful sapphire]

same deal, ~Q was sufficient but not anymore

[keen pawn]

hmm?

What am I doing wrong

[sinful sapphire]

i dont know

[keen pawn]

I'm sure it's fine until $(P \lor \neg Q) \lor ( P \land \neg R)$

[grizzled pagoda]

What a wonderful world!

[keen pawn]

then P is common, so using the distributive law

So it should be $P \lor ( \neg Q \land \neg R)$

[grizzled pagoda]

What a wonderful world!

[sinful sapphire]

same problem

[grizzled pagoda]

frownyfrog

[sinful sapphire]

i don't know what steps

[keen pawn]

huh, what about the R

is this a tautology

[sinful sapphire]

~R matters if P

P is sufficient on its own

[keen pawn]

oh

Got it

thanks

.close

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[torpid cloak]

is this correct?

[torpid cloak]

because when I ask chatGPT it gives different asnwer

[sterile swan]

!nogpt

[amber wasp]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[torpid cloak]

yeha but there is no other help

[amber wasp]

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

[torpid cloak]

ain't nobody helps here maybe once in a year so I've gotta ask chat for help

[sterile swan]

!volunteers

[amber wasp]

Helpers are just people volunteering their time to help you. Be polite and patient.

[torpid cloak]

Thats what I am sayin there is harly helpers to help I am not saying there are no help but its hard to get helper... so I've gotta ask chat. I post here just in case someone replys

[winged panther]

Yes this is correct.

[torpid cloak]

what about this path

[winged panther]

If you select the other neighbour at the first step, F -> G -> H -> E -> K -> F is next

But yeah, next we have F G H K E F which indeed is shorter by 10m

[torpid cloak]

but how am I finding that as I start from F, E is in fact shortest path

[winged panther]

the "double nearest neighbour" method, apparently requires you to check for "2 nearest (cheapest) neighbors" at each step - so:
F -> E | G, following G, H -> E | K,

there's a simpler way to do this though, let me check my notes for DAA

[winged panther]

Use Prim's algorithm

You'll get a minimum cost spanning tree that looks like:

K <- H <- G <- F -> E
And since you require the "shortest trip", connect E -> K and you have a minimum cost trip as required

[amber wasp]

@torpid cloak Has your question been resolved?

[winged panther]

So you get the path F -> G -> H -> K -> E -> F

Could also use Kruskal's but Prim's is convenient here

@torpid cloak.

[amber wasp]

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[split sail]

In the second equation in part c the one at the right how would I know whether | 2(5)-b| is positive or negative

[split sail]

a > b and a and b are positive constants

[cunning birch]

try both options, see which one makes sense

[split sail]

🥲

[cunning birch]

one of them should give you a-b = some value that contradicts what you already know

[amber wasp]

@split sail Has your question been resolved?

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[lusty pine]

find the sum of all positive integers less than 100 such that they are the diffrence of 2 squares

[lusty pine]

i mean obv (a-b)(a+b) but i cant think of any ways to progress

maybe there are many terms that cancel out?

[gloomy tide]

do you mean squares

[lusty pine]

oops

[gloomy tide]

tbh it's low enough that you can just kinda do it by hand

the number needs to break down into two factors that differ by an even number

so it needs to be not prime and either divisible by 4 or odd

[lusty pine]

2^2-1^2=3 which is prime

[gloomy tide]

blehhh ok yes i forgot about that one

[lusty pine]

ok i see

[fresh ocean]

All odd numbers are differences of cosecutive squares

[gloomy tide]

ok maybe not the prime bit

[lusty pine]

so ammount of odds less than 100 is 50 and the number divisible by 4 is 25, so 75?

wait no i forgot what wr were counting

[gloomy tide]

i believe you wanted the sum

[lusty pine]

yeah whoopie

[fresh ocean]

Also, I dont think multiples of 4 would work. Like difference of which squares is 4?

[gloomy tide]

2^2 and 0^2

[fresh ocean]

Ohh im dumb

[lusty pine]

total is 100×99/2=4950

[exotic trout]

any 2 numbers with both their sum and difference even or at least either their sum or difference is divisible by 4 work right

[lusty pine]

subtract by the 2 mod 4, its 2(25)+(0+24)(25)/2×4=50+1200=1250

so 4950-1250=3700

but ans key is 3695?

[winged panther]

._>

[lusty pine]

wait 1 and 4 cant be written as diffrence of 2 squares

[fresh ocean]

1=1-0, 4=4-0

[lusty pine]

0 is not a positive integer

ah wait

[winged panther]

Maybe constraint is, r, s in r² - s² should be natural numbers

[lusty pine]

ok fuck im silly

ok ty guys!

.close

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[gloomy tide]

you didn't say that

[lusty pine]

yeah mb

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[fading cliff]

Suppose we have a rectangle ABCD with BC=2cm and AC=4cm (diagonal) where would the point E lie if vector(BE)+vector(BC)=vector(AC)? (they're limiting us to using basic vector addition and difference)

[fading cliff]

<@&286206848099549185>

[short minnow]

Imagine this:

what has BE to be then?

Sry, made a mistake in the graphic

[fading cliff]

The thing is i got vector BA but apparently it may not the answer

So E would've been overlapping with A

[woeful sable]

not exactly vector BA

I found it to be overlapping with C.

[short minnow]

remember those are vectors. they don't have an exact position

[woeful sable]

BE + BC = AC

[short minnow]

you can moce vector BC to the position of AD and it's still the same vector

[woeful sable]

BC = AD

AD + BE = AC

[fading cliff]

So BE = DC?

[woeful sable]

-EB = DC

yes

oh right

it still overlaps with A

you're right

[fading cliff]

They are asking us later what's the nature of triangle ACE and what's its area so it can't be the same point

[short minnow]

it is neither the point C nor the Point A, but the vector DC is equal to BE

that gives you that E has to be 2 right from B

[short minnow]

that is what moving the vector BC to AD leads to

[fading cliff]

How did we get this btw?

oh

[woeful sable]

not exactly though, isn't DC equal to EB?

cause the vector is going from D to C which is left to right

and in BE the vector is going from B to E which is right to left

[fading cliff]

Honestly I'm still confused

[woeful sable]

gosh i hate vectors

[short minnow]

no,
for graphical vector addition you need to make the vector that is the sum and one of those being added to start at the same point.<
Thats why you move BC to AD
the other vector that you add is the one leading from the tip of the fist vector to the end of the result vector.

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[zenith heron]

Just to be clear, what does the diagonal lines between the parabola mean?

Does it mean where the inequality is true?

If you want to know the answer, the fastest way is to test a value between those lines to the inequality

For example, x should be between -3 and 3

So pick a number between it

And also, its a hollow circle so it could only be < or >

Yep

Np

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[weary shoal]

How

[weary shoal]

To calculate sqrt5(250)

With accuarnacy 1/1000

Using taylor series

[winged panther]

$\sqrt[5]{250} ?$

[grizzled pagoda]

Arya

[weary shoal]

Yes

The hint is to use a formuła $\sqrt[n]{x+1} ?$

[grizzled pagoda]

FIFI27

[weary shoal]

So n=5

X=250?

[winged panther]

start with making it in the form x + 1, x << 1

$250 = l^5 + x$

[weary shoal]

oh

[grizzled pagoda]

Arya

[weary shoal]

so 2^5+7

[winged panther]

so that, when you take out l from under the root, you're left with (1 + x/l^5) where x/l^5 << 1

[weary shoal]

then 2*(1+7/5^2)*)1/5)

[winged panther]

???? 250 = 32 + 7?

[weary shoal]

my bad

3

not 2

[winged panther]

yes, so $\sqrt[5]{250} = 3\sqrt[5]{1 + \frac{7}{3^5}}$

[grizzled pagoda]

Arya

[winged panther]

and now you're good to use (1 + x)^{1/n} series

[weary shoal]

okey

so my n is 2?

the level of derivative i need

[winged panther]

(@_@;) You need accuracy upto 3 decimal places

[weary shoal]

ithink i do sth wrong

no

w8

i checked

won't the accurancy be good enough?

[winged panther]

,calc 3*(7/243)*(7/243)/25

[grizzled pagoda]

Result:

9.957831631357e-5

[winged panther]

,calc 7/243

[grizzled pagoda]

Result:

0.02880658436214

[winged panther]

since this number is already 0.02880... you won't even need x² term from the taylor series

[weary shoal]

yeh

[winged panther]

just go (1 + x)^n = 1 + nx with this for three decimal places

[weary shoal]

i used x0=0

then

f(0)=0

[winged panther]

$3\sqrt[5]{1 + \frac{7}{3^5}} = 3 \cdot \left(1 + \frac{1}{5} \cdot \frac{7}{243}\right)$

[weary shoal]

f'(0)

and stoped

[grizzled pagoda]

Arya

[weary shoal]

exactly

as you wrote it

[winged panther]

that's good enough then

,calc 3*(7/(243 *5) + 1)

[grizzled pagoda]

Result:

3.0172839506173

[winged panther]

3.017

[weary shoal]

okey i got it now i see the trick

thank you

.close

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[winged panther]

:D

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[silver barn]

can someone explain this to me. I was so confident it was 4

[silver barn]

nevermind i got it

.close

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[sharp dock]

yo

[sharp dock]

I have a test on this tomorrow and I have no idea how to do this

[unborn plaza]

$\log(ab)=\log(a)+\log(b)$

[grizzled pagoda]

Axe

[unborn plaza]

you can use this property to solve the first one

[sharp dock]

How?

What's a and b?

[unborn plaza]

a and b is anything that's multiplied inside a logarithm

[sharp dock]

Huh

[unborn plaza]

ab is a product

this property says that you can rewrite it when you have a log of a product

m^4 n^2 is a product

it's a product of m^4 and n^2

[sharp dock]

yea?

So a is 4 & b is 2?

[unborn plaza]

a is m^4 and b is n^2

[sharp dock]

alr

What rule do we use to solve it?

[unborn plaza]

we can use this rule to simplify it

then we can use another rule to simplify it more

but let's take the first step first

[sharp dock]

Alr so

log(ab) = log(m⁴) + log(n²)

?

That's what u said right

[unborn plaza]

log(m⁴n²) = log(m⁴) + log(n²)

we can always separate like this

[sharp dock]

So.. what now?

[unborn plaza]

no matter what the first thing & second thing are

now you can take exponents to the front

[sharp dock]

4logm+2logn

[unborn plaza]

yes exactly

oh and don't forget the base 5

[sharp dock]

Base 5?

Oh yea

mb

4logbase5 + 2logbase5

Hmm

[unborn plaza]

just remember the little 5 when you write it with a pencil

on the homework

[sharp dock]

Yea ik

[unborn plaza]

then it's done

[sharp dock]

Oh that's it?

Thanks!

[unborn plaza]

yes

but

the other problems are trickier

[sharp dock]

Do u mind helping me with the other ones? I'd really appreciate it

[unborn plaza]

sure

[sharp dock]

Ty

Here

[unborn plaza]

(b) right?

[sharp dock]

Yep

[unborn plaza]

first, the cube root is the same as a (1/3) exponent

[unborn plaza]

so you can use this rule

[sharp dock]

Why are they 1/3?

[unborn plaza]

$\sqrt{x}=x^{\frac{1}{2}}$\
$\sqrt[3]{x}=x^{\frac{1}{3}}$\
$\sqrt[4]{x}=x^{\frac{1}{4}}$

[grizzled pagoda]

Axe

[unborn plaza]

and so on

and in our case,

$\sqrt[3]{27x^3y^2}=(27x^3y^2)^{\frac{1}{3}}$

[grizzled pagoda]

Axe

[unborn plaza]

it's the same idea

[sharp dock]

Oh so whatever number is outside of the sqaure root we just put 1 over it?

Like if it's 6

We put 1/6?

[unborn plaza]

yes

$\sqrt[6]{x}=x^{\frac{1}{6}}$

[grizzled pagoda]

Axe

[sharp dock]

Ah I see

Makes more sense now lmao

And sqaure root of 27 is 3 right?

[unborn plaza]

we call it cube root

[sharp dock]

So it becomes 3xy⅓?

Or 3xy (x³)⅓ (y²)⅓

I'm so confused

[unborn plaza]

well we can distribute the 1/3 exponent if we want

but we can also bring it to the front since it's a log

[sharp dock]

What's easier?

[unborn plaza]

i think it's about the same

here

$(27x^3y^2)^{\frac{1}{3}}$

[grizzled pagoda]

Axe

[unborn plaza]

this was close

only the y is wrong

[sharp dock]

Y is wrong?

[unborn plaza]

it's not y⅓

it's y^(2/3)

because you combine the 2 that was already there with the 1/3

here let's take it step by step

[sharp dock]

so 3xy⅔

[unborn plaza]

yes

[sharp dock]

Ah alr

[unborn plaza]

ok you got it

[sharp dock]

So do the 1s just add up up to 2?

[unborn plaza]

no

[sharp dock]

Or do we use the 2 and 3 from the x and y

[unborn plaza]

i'll show

[sharp dock]

Alr

[unborn plaza]

$(27x^3y^2)^{\frac{1}{3}}=27^{\frac{1}{3}}(x^3)^{\frac{1}{3}}(y^2)^{\frac{1}{3}}=3x^{3\cdot\frac{1}{3}}y^{2\cdot\frac{1}{3}}$

[grizzled pagoda]

Axe

[unborn plaza]

we multiply the exponents

3 * (1/3) and 2 * (1/3)

[sharp dock]

Huh

That's alot

[unborn plaza]

yeah

[sharp dock]

Honestly on the test

Ima just add the 1s

Or the 2s

And use that as my power

Like x³ and y²

So

Just make it

3xy⅔

And the base is 3

So

Logbase3(3xy⅔)

Right?

And than after that I have no idea what to do

[unborn plaza]

after that we can use the first rule i showed you

on the inside you have 3 and x and y^(2/3) all multiplied together

we only had 2 things before

but now we have 3

that's okay, we can still use the rule

so it becomes log + log + log

[sharp dock]

So it becomes: logbase3³+logbase3^x + logbase3^y⅔

[unborn plaza]

it becomes:

$\log_3(3)+\log_3(x)+\log_3(y^{\frac{2}{3}})$

[grizzled pagoda]

Axe

[sharp dock]

yea that's what I meant

That's it?

Or do we move the 2/3

[unborn plaza]

you can simplify more

[sharp dock]

To the front

[unborn plaza]

yes you can move 2/3

exactly

and you can also simplify log_3(3)

_3 means base 3

[sharp dock]

Wym?

Can u show me

[unborn plaza]

$\log_3(3)=1$

[grizzled pagoda]

Axe

[sharp dock]

Oh cuz they cross out?

[unborn plaza]

yes

[sharp dock]

Oh ok

Ty

I'll just not do the final question on the test tmrw

[unborn plaza]

you're welcome

[sharp dock]

Cuz we have 3 options

And I'll choose this one

[unborn plaza]

all right good luck

[sharp dock]

Take care

.close

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[tiny tree]

hi! i wanna know if the following is correct
k,p,x<n; gcd (x,y)=1. WLOG assume k>p
(k-p)x≡0modn
(k-p)y≡0modn
=>n divides gcd((k-p)x,(k-p)y)
=> n divides k-p
=>(k-p)>=n
a contradiction

[honest dagger]

How do you figure k-p>n?

[tiny tree]

k and p are smaller than n

also

they're all positive

[honest dagger]

2 divides 0, but that doesn't mean 0>2

[tiny tree]

right, my bad but i forgot to mention k != p

[honest dagger]

then $(k-p)x\equiv 0$ mod $n\to (k-p)x|n\to x|\frac{n}{k-p}$. But then $y|\frac{n}{k-p}$ too, which contradicts $\gcd(x, y)=1$ unless $n=k-p$, which is not possible if $0<k,p<n$

[grizzled pagoda]

SWR

[tiny tree]

(k-p)x=0modn x divides n/(k-p)
but what if n=12, (k-p)=3 x=8 ?
(k-p)x is divisible by n but i don't get how x must divide (n)/(k-p) I think we just get x must divide some integral multiple of n/(k-p)

[tiny tree]

is my reasoning incorrect?

[honest dagger]

oh oops I think I had the mod logic backwards

sec

[tiny tree]

ook

can u tell me if my logic is incorrect?

<@&286206848099549185>

<@&286206848099549185>

[honest dagger]

@tiny tree is y<n also, or can it be anything?

[tiny tree]

yes y<n

[honest dagger]

So, 0<k,p,x,y<n

[tiny tree]

yesyes

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@tiny tree Has your question been resolved?

[tiny tree]

.ckose

.close

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[twilit cove]

can I ask anyone a question regarding ordinary least squares

[twilit cove]

linear regression, but in code

heres a snippet of the data im working with:

MPG,cylinders,displacement,horsepower,weight,acceleration,name
18,8,307,130,3504,12,chevrolet chevelle malibu
15,8,350,165,3693,11.5,buick skylark 320
18,8,318,150,3436,11,plymouth satellite
16,8,304,150,3433,12,amc rebel sst
17,8,302,140,3449,10.5,ford torino
15,8,429,198,4341,10,ford galaxie 500
14,8,454,220,4354,9,chevrolet impala
14,8,440,215,4312,8.5,plymouth fury iii
14,8,455,225,4425,10,pontiac catalina
15,8,390,190,3850,8.5,amc ambassador dpl
15,8,383,170,3563,10,dodge challenger se
14,8,340,160,3609,8,plymouth 'cuda 340```

Im using linear regression to predict the MPG based on weight

    # if len(b.shape) != 1:
    #     raise ValueError("X must be 1d-array")
    # if len(X.shape) !=2:
    #     raise ValueError("X must be 2d-array")

    yp = X@b
    return yp

X= np.array([[1,0],[1,-1],[1,2]])
b= np.array([0.1,0.3])
linearModelPredict(b, X)
print(f"dimension of the array:", linearModelPredict(b, X).shape)```

and im using mean squared error as my loss function, I also computed gradient

    #error checking
    if len(beta.shape) != 1 and len(y.shape) != 1:
        beta = beta.reshape(-1)
        y = y.reshape(-1)
    if len(X.shape) != 2:
        raise ValueError("X must be 2D array")
    
    #predictions
    predictions = linearModelPredict(beta,X)

    mse = np.mean((y-predictions)**2)
    # print((y-predictions).shape)
    # print(X.shape)
    gradient = -2*(X.T @ (y-predictions))

    return mse, gradient

X= np.array([[1,0],[1,-1],[1,2]])
b= np.array([0.1,0.3])
y = np.array([0,0.4,2])

mse, gradient= linearModelMSE(b,X,y)
print("MSE is:", mse)
print("gradient is:", gradient)```

this is the first question, Use the above functions to fit your model to the car data. Use the MPG as the target (y) variable and only the weight variable as the independent (x). Fit the model with a constant. Then use your model and the fitted parameters to make predictions along a grid of equally spaced weights within the original range of the weight variable. I think my code is right for this

y = df['MPG']

#call the fitting function
r2, beta_optimal = linearModelFit(X, y, linearModelMSE)


plt.scatter(df['weight'], df['MPG'], alpha=0.5, label="given data points")

weight_grid = np.linspace(df.loc[:,'weight'].min(), df.loc[:,'weight'].max(), 100)
new_X = np.column_stack((np.ones_like(weight_grid), weight_grid))
mpg_pred = linearModelPredict(beta_optimal, new_X)


plt.plot(weight_grid, mpg_pred, 'r-', label=(f"R^2 (goodness of fit):{r2:.3f}"))
plt.xlabel('Weight')
plt.ylabel('MPG')
plt.title('Relationship Between Cars\' Weight and MPG')

plt.legend()

print(f"R^2 is: {r2}")```

Second question is: Now use sklearn's linear_model to fit the model with all the available data. Plot the data and add a line for the predicted values as you did in the previous question. Also report the $R^2$ value for the fit. Im not sure what it's asking

[grizzled pagoda]

Aff07
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit cove]

Im not sure how using sklearn's linear model differs from mine, because if Im just looking at weight for x variable, using sklearn's linear model seems to graph the same

unless im understanding the question wwrong, and we're supposed to have an X design matrix with data of cylinders, displacement, horsepower, weight, acceleration

I forgot to mention, I also made my own model fitting function

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@twilit cove Has your question been resolved?

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@twilit cove Has your question been resolved?

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@twilit cove Has your question been resolved?

[twilit cove]

.close

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[odd vault]

Hellow guys I am a little bit confused by the phrasing of the of equations 2.31 and 2.32. Reversing the roles of t and tau the equations remain the same no?

[shrewd elm]

not unless u*(tau - t) = u*(t - tau)

[odd vault]

If I am not mistaken tau is the dummy integration variable in the first equation and so is t in the second.

[shrewd elm]

yea

but if u* is not an even function, then you don't necessarily have u*(tau - t) = u*(t - tau)

[odd vault]

hmmm I see

kind of hard to visualize

[shrewd elm]

yea i'm not sure physically what's going on here but the two integrals won't be the same in general even though they're obviously related

[odd vault]

fair enough. Thank you fo rthe advice!

[shrewd elm]

yw

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[split sail]

1.For every $\epsilon >0$,there are infinity many n such that the distance between $a_n$ and 0 is less than $\epsilon$.

[split sail]

Give an example for a_n

Will a_n =2^-n satisfy the condition?

[unborn plaza]

it should, yeah

[split sail]

Nice

One mroe

2.For every $\epsilon >0$ for all but finitely many n the distance between $a_n$ and 4 is less than $\epsilon$

Banged my head over this problem

An example

So

[unborn plaza]

the wording is a little messed up

but i think i get it

[split sail]

I have to take an a_n such that the limit at infinity is 4

[split sail]

Yeah 😔

[unborn plaza]

i think you just need to remove "such that"

[unborn plaza]

thanks

[split sail]

Oh yeah

It's more clear now

So I'm guessing I have to take a certain sequence such that it increases for a range of n and then after crossing that range it starts decreasing

Which Is wrong?

[unborn plaza]

you can make it as simple as the first one

in fact a_n=4 works

[split sail]

[unborn plaza]

[split sail]

But

[split sail]

Wouldn't there be infinitely many n

[unborn plaza]

0 a_n where it fails

infinitely many where the distance is < epsilon

0 is finite

[split sail]

Exactly

So infinite n ?

[shadow stump]

"for all but finitely many n"

[unborn plaza]

it means for all n with only a few exceptions

[split sail]

Yeah the "all n" is also a pain

I didn't get it

[unborn plaza]

can you think of a sequence with 4 as a limit?

[split sail]

the one you mentioned i.e a_n=4

Or

[unborn plaza]

a more interesting one?

[split sail]

a_n = 4(1 - [1/n])

[] is just brackets no step function

Lol step means it would just be 4 anyways

[unborn plaza]

ok i'll choose epsilon = 1

how many exceptions?

[split sail]

4

1,2,3,4

[unborn plaza]

yeah

so it's true for all but four n

four is finite

[split sail]

BRUH

MAN

[shadow stump]

if we chose a smaller epsilon then we would have more exceptions but still finitely many

[split sail]

I OVERCOMPLICATED THE QUESTION ITSELF

BRUHHH

Thanks

Oh one more thing

For negation statements

[split sail]

For this one I wrote

1.For atleast one $\epsilon >0$,there are infinity many n such that the distance between $a_n$ and 0 is greater than $\epsilon$.

[split sail]

Correct?🙏

@shadow stump @unborn plaza ?

[unborn plaza]

it should be greater than or equal to

[split sail]

https://tenor.com/view/facepalm-crowd-gif-11749272

[unborn plaza]

i think this is right if a_n is defined for all natural numbers

wait

[split sail]

Isn't a_n always defined for all natural numbers

?

[unborn plaza]

yeah it's a sequence so yeah

wait

no

it's wrong

the negation is this:

For at least one $\epsilon > 0$, there are finitely many $n$ such that the distance between $a_n$ and 0 is less that $\epsilon$

[grizzled pagoda]

Axe

[split sail]

I gtg for class

Can I ping you later?

[unborn plaza]

yeah you can but i might not answer if i'm away

i'll be going to sleep in a few hours

but i don't mind if you ping

[unborn plaza]

when i said "finitely" here i meant "only finitely"
or i should have just said "not infinitely many"

[amber wasp]

@split sail Has your question been resolved?

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[grizzled pagoda]

lin.fei

[fresh ocean]

Isnt that a standard identity from school syllabus?
$a^3 + b^3 + c^3 - 3abc = (a^2+b^2+c^2-ab-bc-ca) (a+b+c)$

[grizzled pagoda]

Facter10Br4g

[fresh ocean]

So you dont need to prove it. Just use the result

But, if you feel its appropriate, you can prove it.

But, imo, simply mentioning the identity is enough

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[wide jackal]

!status

[amber wasp]

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

[wide jackal]

4

im getting

3 times 1/2 and 1 time 5/8 so that becomes , 12 + 5/32 = 17/32

cos three quadrants have half area

and the circle one 4 pieces together are quadrant of a quadrant, i.e 1/16, and then there are 10 of those so

[foggy kayak]

not that easy

[wide jackal]

sorry for ghosting u btw

[foggy kayak]

the circle one's a bit diffreent

[foggy kayak]

np

[wide jackal]

circle one is ez imo

bcs

[foggy kayak]

what about left top quadrant

[wide jackal]

this

is a quadrant

[wide jackal]

thats half

[foggy kayak]

ok

[foggy kayak]

ok now consider the other parts of the top right

[wide jackal]

yeah

[wide jackal]

so u got those exact math one more time just not together

tthat makes half of the square

then u get 2 of the piieces instead of 4

so that's half of a quadrant

ie. 1/8

now multiply all by 1/4 bcs quadrant

and 17/32

im not sure tho

cos no ak

[foggy kayak]

if you only consider the top right quadrant for a second, rearrangling we get 1/2 + 1/8

so 5/8 only 2nd quadrant

[wide jackal]

yea

i said that

[foggy kayak]

i gotta catch up

[wide jackal]

lol

[foggy kayak]

ok so thers are just 1/2s

[wide jackal]

ye

[foggy kayak]

and we divide by 4 for quadrants

[wide jackal]

yea

[foggy kayak]

1/8 * 3 (for 3 quads)

• 5/8 * 1/4

[wide jackal]

.close

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Channel closed

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[rough delta]

Find all natural solutions for (a,b) such that a!=b^2+44

I have found (6,26) and I guess it is the only solution but how do we prove that?

[indigo cloud]

mod n for a convenient n

at least I think that should work

[rough delta]

Well with mod ten we know that b ends with 6

[winged panther]

For a ≥ 7, b² + 44 = 0 (mod 7) does not have a solution

[indigo cloud]

!nosols

[amber wasp]

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.

[winged panther]

and 4

[winged panther]

From here, can you check for the rest?

since b² + 44 ≥ 44, you have a restriction for "a" as well

[rough delta]

Yeah, I think I did somth to delete that

[winged panther]

yes, 4! = 24 < 44 ≤ 44 + b². So yeah 4 would not be a soln.

so what values of "a" are you left with

[rough delta]

Uhh so b^2=0,1,2,4 but none makes 0(mod7)

[winged panther]

You should be checking b² + 44 = 0 (mod 7)

[rough delta]

Ok, so none of them makes 0

How do you come up with mod 7?

[winged panther]

you need to check mod for a couple numbers before you hit a home run. I checked mod 10 first as well, before switching to mod 7

[indigo cloud]

mod 7 is a good idea after you found your solution with 6

[rough delta]

Because we know that we are evaluating a>7, a<7 I have tried

[indigo cloud]

because you know that 7! and every number after that will be divisible by 7, so then it has a different behavior mod 7 compared to before

modulo arguments can only show that there are no further solutions if something "changes" in the modulo setting

[winged panther]

So now?

[rough delta]

I have tried mod 11 because of 44 but didn’t work

[rough delta]

Yeah so only sol 6!! Thank you so muchh!!

[winged panther]

Did you check a = 5?

,calc sqrt(120 - 44)

[grizzled pagoda]

Result:

8.7177978870813

[rough delta]

Yeah

Didn’t work

[winged panther]

welp, 6 it is then

✅

[rough delta]

Thank so much Aya for your time and patience❤️

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[civic spindle]

https://cdn.discordapp.com/attachments/908076667224854568/1329342859949903894/foil-method-formula.png?ex=678aa743&is=678955c3&hm=c70232249cf1e850345f6f5e091d41c09a5ebb5b43351c22c3e4ace08760a0aa&

[amber wasp]

Please don't occupy multiple help channels.

[civic spindle]

man i had to make a second channel cuz i couldnt find the one i just made

legit dissapeared and wouldnt show up on logs

well then

[civic spindle]

ok so if i had this

[foggy kayak]

then close the other one

[civic spindle]

no

cuz its gone

its literally dissapeared

i have no idea where it is

FINALLY

alr someone pinged me in it

wtfd

is going on

it keeps closing my channels

i wouldve lost this one too

if it didnt show up on searchbar

F

[foggy kayak]

i think you clocked the portion above the channels

[civic spindle]

ok but how do i do this?

[civic spindle]

tyy

[foggy kayak]

what's there to do about?

[civic spindle]

this

https://cdn.discordapp.com/attachments/908076667224854568/1329340243228229674/343767005373136896.png?ex=678aa4d3&is=67895353&hm=7b6db41f340b11f786f34238a729757e14faa41a37d875a3b9e8035005290f65&

bc i do not know how to simplify it

if it has

a fraction

[civic spindle]

what do i do if b was a fraction like b/e

[foggy kayak]

square the numerator and denominator

first use (a+b)² formula

[civic spindle]

afk for now

.close

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Channel closed

Closed by @civic spindle

Use .reopen if this was a mistake.

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sacred magnet]

Guys help how to find length ad i already found length ac which is 867

[hollow wolf]

(Our server's bot @amber wasp is currently down, consider using the #1021175428326633542 instead)

[untold torrent]

Help

Indont know what 1 is asking for

.close

[civic spindle]

https://media.discordapp.net/attachments/908076667224854568/1329342859949903894/foil-method-formula.png?ex=678aa743&is=678955c3&hm=c70232249cf1e850345f6f5e091d41c09a5ebb5b43351c22c3e4ace08760a0aa&=&format=webp&quality=lossless&width=1081&height=608

uhhh

i guess this is mine? cuz he closed it so

so uhm

[civic spindle]

what do you do if B was a fraction like B/I e.g., 5/2

[spring kayak]

I have a big problem. Can i solve this differential equation :
y'' = - a * sin(y)
Please ?

[bronze locust]

How to get the length of any of these lines

?

i thought i get chat box :c

i need to get the lenght of a line and the same line for the image

[spring kayak]

Simply use the Pythagorrean theorem

[bronze locust]

i guess

[spring kayak]

Do you know vectors ?

[bronze locust]

uhh

[bronze locust]

i dont have a , b , or c

[spring kayak]

Ok

Lets say, to go from U to T, you go 4 units to the right, then 6 units upwards.

Your 4 is the same thing as your a, and the 6 is the same thing as your b

[bronze locust]

ohhh

[spring kayak]

With each of these lines, you can draw a right-angled triangle

[bronze locust]

i got 7.21