help-13

[dusk goblet]

which is vo^2 /(2a)

so

,calc (1.9)(150)/2.95

[wraith dagger]

Result:

96.610169491525

[dusk goblet]

then

x = vot + 1/2at^2

and you’re solving the quadratic

1/2at^2 + vot - x = 0

solve for t

[odd sierra]

perfect ok

thnx

[dusk goblet]

you’re welcome

[cedar kiln]

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[merry basalt]

How do I do this problem for precalc

[odd verge]

Do you know what causes a "hole "

[merry basalt]

yes

0s on both sides

[odd verge]

No
Just at the bottom

[merry basalt]

oh okay

[odd verge]

It must not necessarily be on top too

[merry basalt]

No it’s just the both yea

[odd verge]

To be clear by hole you mean where it is not continuous?

[merry basalt]

no removable discontiny

[odd verge]

Oh ok

Then factor the bottom

Then use the remainder theorem to find which factor of the bottom is a factor on the top

@merry basalt

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[chilly portal]

Why is this so hard

[chilly portal]

i got sin(x)cos(x) - 2sin^2(x) +1

i swear this is wrong

[cedar kiln]

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[naive hill]

Teacher did this in class, I'm trying to understand why he pulled those numbers specifically to multiply on each side

[whole spear]

I don't understand what that symbol is

is it 2r?

or is it 2 multiplied

oh okay I understand

He didn't really pull any numbers

[naive hill]

2b multiplied to the left side

and a squared times 5 to the right side

[whole spear]

So, to be able to add these two terms together, they have to have the same values in the denominator

So the trick he uses is that he multiplies the values with 1

If you look, he multiplied with 2b both on the numerator and denominator

and 2b/2b is 1

but that way he can rearrange so that on the bottom he gets 20(a^2)(b^2)

[naive hill]

oh i just noticed yeah tysm

i understand it now

[whole spear]

And then on the other one he does the same, but with 5 instead

[naive hill]

He wanted the denominator to be the same like you said

[whole spear]

Ya

[naive hill]

yuh i get it now tysm

[whole spear]

And you can always utilise this trick

[naive hill]

😭

[naive hill]

yeah

[whole spear]

haha happens, it was weird

he seperated the b and 2

instead of writing 2b

and put them each on seperate side

[naive hill]

I know how to mulitply and divide Rational Expressions easily

addition was a little weird but i get it now

Methods are different

[whole spear]

Ya, addition the denominator always has to be the same

Hence why you need to play around with the equations a bit

[naive hill]

i might aswell ask, wb substracting?

[whole spear]

Same concept there

Has to be the same denominator

[naive hill]

Ah ok

[whole spear]

Then you just substract the numerators

[naive hill]

thanks a lot, i appreciate the help

[whole spear]

No worries

Good luck

[naive hill]

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[drowsy rock]

How do you do this? I even asked a Math AI and it said none of the answers are correct

[silver oxide]

use the fact that sec is 1/cos

[drowsy rock]

Ill try

it's inverse secant though

[silver oxide]

yeah same thing

[glad peak]

arcsec ≠ 1/sec

[silver oxide]

when did i say that?

[drowsy rock]

I tried to do implicit but I don't think I did it right

I think the question is just absolutely horrible

I tried using the derivative of arcsec I memorized but that didn't help much either

[glad peak]

Use that d/dx sec = secx tanx

Then you can sub in y, for the secy term. This won’t work well for tany though so you need to use another identity

[cedar kiln]

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[mighty summit]

I'm not sure whether this matrix represents the two matrix the question presented

[mighty summit]

also the entries for the fist matrix seems to be unclear, because $T(v_k) \neq v_k ?$

[wraith dagger]

eurgene114514

[cedar kiln]

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[trail agate]

Hi, I’m working on calc hw and could use a sanity check because I just fat-fingered my way onto the “right” answer and now I’m confused. I thought the answer should be -e^22+4, not 44? Can someone provide a second pair of eyes? 😅

[trail agate]

Like…I’ll take the point on my homework, but…it’s hurting my brain lol

[remote trail]

22=ln(4-x)

e^22=4-x

x=4-e^22

Was what I was thinking

[trail agate]

Ok, so my math was good. Maybe the person who programmed the homework made the exact same keyboard fumble I did.

Thanks for confirming I’m not crazy lol

.close

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@twin dune Has your question been resolved?

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@twin dune Has your question been resolved?

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[wet warren]

how would i evaluate this to change base?
∛3x × 81^(x-4) = ∛9⁶

[wet warren]

whay I tried so far but its not correct

😭

[quiet ledge]

4.6 is not an exact value

[wet warren]

oh

so I shouldn't round it

or is it asking as a fraction

[quiet ledge]

fraction

[wet warren]

tysm !

okayz

2more questions

Q2

I'm struggling to understand how it's a

tbh I didn't understand how I got b either

I tried making it into log form
b = Log 2a C

but still confused

[quiet ledge]

i think ur right tbh

[wet warren]

with b?

wait I think I read it wrong

d is correct answer

😭😭 still dk how

okayz I got that one

nvm

ahaha

last one

[wet warren]

<@&286206848099549185> sorry to ping but I'd like guidance

[cedar kiln]

@wet warren Has your question been resolved?

[rotund yacht]

Maybe this would be helpful? I've never had to really deal with base 2 logarithms in mathematics but this came up

And then these

[wet warren]

Well I tried to change it to exponential form before asking here
ik how to change it but in that example there's only 3 variables
in my example there's 4 if it makes sense

[rotund yacht]

So you have an extra equation?

[wet warren]

extra exponent that I've not had experience with

[rotund yacht]

An extra variable means you are just going to rearrange the same expression but in a way to find the extra variable

[wet warren]

usually as an example it'd be like 5³ = c or so which is ez to transform
but I have no experience w 5³ = c²
yk

idk where to make room for it

[rotund yacht]

Also, btw that example is already in exponential form. You should try changing it to logarithm form.

[wet warren]

yesyes it'd be log 5 n = 3

but idk how to convert with the extra ^2

[rotund yacht]

So is this the question you are trying to solve rn?

Or is this a different question?

[wet warren]

yes

that one

I just tried using a simpler example to help with my confusion

[rotund yacht]

So you got:

log2(5^m) = 3m-1?

[wet warren]

I actually don't remember what my attempts were
its packed away but I thibk that seems correct

[rotund yacht]

Okay, so after log2(5^m) = 3m-1, what do you think you should do next?

[wet warren]

hhh somehow isolate the m

idk how I'd go about this tho

[rotund yacht]

Wait, are you sure this is what you got? I might've made a mistake.

Do you have the equation to transform exponent to logarithm?

[wet warren]

not sure as it's away rn
I haven't made a full attempt at the question anyways I struggled at the first step

[wet warren]

the only equation we have is the first image u sent

[rotund yacht]

Okay, let's just go with it being correct because I did some further research and found this:

[rotund yacht]

So, what do you need to first do to isolate the m?

log2(5^m) = 3m-1?

[rotund yacht]

don't forget that u can use this to ur advantage

[wet warren]

yesyes js give me a sec I'll try pitting it into log form as well

I got

(3m - 1 ) Log 5 2 = m

[rotund yacht]

log5?

[wet warren]

which is like Log 5 2^3m-1 before movements

ye

hold on

[rotund yacht]

Where did you get log5?

You mean log2(5)?

[rotund yacht]

Can you show me your steps?

[wet warren]

yeah let me write it out

hopefully it's legible

blue is the formula u gave

[rotund yacht]

So it seems like we maybe got the wrong form from the get go.

The formula above is right. But is it 5^m or is it 2^m?

This is the problem we are doing, right?

2^3m-1 = 5^m

We are trying to write this as m = 1/(a-log2(c)), now do you think we should do the conversion into logarithm from 5^m side or 2^3m-1 side?

[wet warren]

hmm

not sure which side the conversion would go to

tbh

[rotund yacht]

If we convert into log from 5^m side, what do we get as a result? Write it out.

[wet warren]

I see where we got confused

okay

what result do u want?
I've gotten (3m -1 ) Log 5 2 so far

[rotund yacht]

Is that 3m-1 = log5(2)?

[wet warren]

it's basically Log 5 2^3m-1 = m

[rotund yacht]

Mhmm, that is correct conversion. But, in the form m = 1/(a-log2(c)), do we see a log5?

[wet warren]

ohh

omg no that makes sense

I'll convert with the 2

[rotund yacht]

mhmm, looks good. Now let's simplify further to solve for a and c

[wet warren]

yesyes

hold on let me see

hm

[rotund yacht]

For ur reference

[wet warren]

ik those rules but I don't think it'll be useful in this

[rotund yacht]

Why do you think that?

[wet warren]

I've used it before and it's for expanding

but there's nothing to expand here

example

[rotund yacht]

Yes, but if u expand, it can help you simplify at times.

[wet warren]

If we had Log 2 (3)(4) then we can expand it to Log 2 3 + Log 2 4

but there's nothing multiplying or dividng

[rotund yacht]

mhmm

So, let's look at our example

Can you repost where we are at so far?

[wet warren]

wym repost?

[rotund yacht]

Can you reiterate at what point our equation is

[wet warren]

okay

[rotund yacht]

Like, we just converted from exponent to logarithm, can you show that

The logarithm form

[wet warren]

like explain how we got this ?

[rotund yacht]

I just meant like repost it

[wet warren]

ah

[rotund yacht]

Repost where the equation is

In simple terms

[wet warren]

all we did was convert the beginning equation using the log form

and that's it

[rotund yacht]

Okay, so we got log2(5^m) = 3m-1.

Now, how can this help us get closer to m = 1/(a-log2(c))?

[wet warren]

but now that I'm looking more at it

I think we converted the log form incorrectly

[rotund yacht]

Why do you think that?

[wet warren]

3m -1 isn't in log form
when we did it both shpuld end up having a log

like what u do to one side of the = u do to the other

[rotund yacht]

Let's relook at this

[wet warren]

yes but in that case it doesn't have the exponents on both

hol on lemme draw it

[rotund yacht]

Okay.

[wet warren]

think my brain is clicking

cuz now we shpuld distribute in the 3m -1 to the log 2

[rotund yacht]

Hmmm, okay, but this is incorrect logarithm form

Exponent form is:

2^(3m-1) = 5^m

Now, if we look at exponent form, what is K and what is N?

[rotund yacht]

.

[wet warren]

this log form is used when exponential form only had one exponent not including the one over n

wait just lemme cook a bit

[rotund yacht]

This isn't a matter of algebra but more so following the equation/rule we have

[rotund yacht]

Where does it say that the log form is only used when there is one exponent on one side and not the other? That's not what the rule is saying. It just says if you have 2^K = N, then you can rewrite that as log2(N) = K.

It doesn't matter what K and N are, those are variables. It's just relevant to your equation.

[wet warren]

I'm just saying as past experience working with these logs
I'm super close to solving it
idk how to explain it but it's a process I've gone over before

the issue was the logs in the beginning but when I distributed it on both sides it started working out

I just have to change the 2/5

ah thank you so much for the help !

I got it solved

[rotund yacht]

Well, this is in the incorrect form to begin with. You are solving the wrong question from the beginning. This is what you are solving right now:

2^(Log(5^m)) = 3m-1

Because, you've set K = Log(5^m) and N = 3m-1

[wet warren]

🙏🏽

[rotund yacht]

altho u did get the right answer somehow 🤔 That's so weird lol

[wet warren]

ahaha if it ain't broken don't fix it ig 😭

[rotund yacht]

Here, I think this is how u solve it

move m to the outside of log2(5^m) because of the logarithm rules. The rest is just distribution and factoring and rearranging

[wet warren]

😋😋🙏🏽 ty for the help I'll try figuring it out ur way too

[rotund yacht]

Just giving u this ahead of time in-case ur teacher gives u shit for that other work lol

I'm glad it worked out still, idk how it worked out but the start looked shaky ngl

[wet warren]

😭 real tho but at least we got the answer

.close

[cedar kiln]

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[rotund yacht]

I just realized your algebra is wrong

[wet warren]

how so

[rotund yacht]

Log2() / 3(Log2() - Log2(5))

The Log2s can't cancel each other out

[rotund yacht]

.

This part

[wet warren]

why not

[rotund yacht]

That is 3(Log2) - 3Log2(5), you can't cancel one out with the numerator because it's not like the entire denominator is being multiplied by Log2.

[wet warren]

😔 yikes

[rotund yacht]

Like:
a / (b+c), you can't just say if let's say b = a, you can't do 1 / c. That's not right lol.

[wet warren]

I'll try ur way in the morning then
After I sleep maybe a fresh set of eyes will help

😭 I think it's too late for me to think properly

[rotund yacht]

Sounds good, I think it was the incorrect form from the get-go, hope u feel better tomorrow lol

[rotund yacht]

I feel that 😭

[cedar kiln]

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[hot crag]

can someone explain quotient groups to me? i don't quite get how they work.

[hot crag]

(i know more ring and field theory than group theory, frankly)

please ping

[potent fractal]

@hot crag

1. Consider a subgroup H in a group G. If we multiply H by some elements of g on the left, we will either get H or a set gH that doesn't intersect with H.

Similarly, we can multiply by g on the right

[crimson delta]

then surely you know quotient rings?

[hot crag]

mhm

[crimson delta]

take just the additive part of that

[hot crag]

okay

[crimson delta]

thats a quotient group

[hot crag]

as in

[potent fractal]

If we now consider the set of those non-overlapping subsets of G, we may want to endow it with a group structure

The most natural way would be to inherit the group operation of G somehow

In each of H, g1 H, g2 H, ... we can choose one element and say that (g1 H)*(g2 H) = (g1 g2 H)
We now want to prove that this operation doesn't depend on the choice of g1 and g2.

In order for this idea to work we have to ask for gH=Hg

[hot crag]

yes i do get the bit about the quotient ring having the cosets of H as elements

[crimson delta]

as in forget that the multiplication exists

[hot crag]

uhmmmmmmm

okay

what i don't get is how H is the identity

in what sense is it the identity?

[crimson delta]

in the quotient ring the coset 0+H is the identity of addition

[hot crag]

denascite btw can you explain further? the way i'm used to thinking of quotient rings is essentially remainders

and i don't quite get how those'd work in groups

[crimson delta]

well it doesnt quite work

[hot crag]

oh.

[crimson delta]

I mean remainders also only work in like integers or polynomials

[hot crag]

hm

[potent fractal]

Think of it as of cutting the group into many pieces

[hot crag]

hm okay

[potent fractal]

Each piece is similar to the subgroup H

And the operation on those pieces is naturally inherited from G

[potent fractal]

.

[hot crag]

yeah i got that bit

[crimson delta]

if we quotient by an ideal H, then two elements x,y are equivalent if x-y is in H. in practice for you thats often that y is the remainder when x is divided by whatever generated H

[hot crag]

sorry if i'm not being descriptive

i'm trying to figure out what exactly my problem is too

[hot crag]

mhm

[crimson delta]

if you write that condition multiplicatively then you would have xy^-1 in H. which you should recognise

[hot crag]

yeah

oh i think i might be getting it

so essentially when i quotient by a group G, all elements of G are essentially made identities?

that makes sense
sort of

[crimson delta]

what

[hot crag]

idk either lmao

i'm trying to connect it with quotient rings

[crimson delta]

when you quotient by an ideal such as <p>, you say that elements are the same "up to multiples of p"

[hot crag]

yeah...

[crimson delta]

in other words, up to addition of an element in <p>

or in multiplicative group speak, up to multiplication of an element in H

[hot crag]

*head spins*

hang on lemme try to make sense of this

so for xy in G/H, xy = e if xy is in H?

essentially

[crimson delta]

you are messing up what type of things the elements of G/H are

that certainly doesnt help

[hot crag]

sorry

:(

i think i'll go watch a youtube video instead

i think what's mainly confusing me is that the elements of a quotient ring are going to be the same as the parent ring; not the case w/ quotient groups

[crimson delta]

no no

the elements of a quotient ring are also cosets

often also called residue classes or something

written as [f] or \overline f or f+H or something if you want to be precise with your notation

[hot crag]

yeah no i feel like i have no idea what i'm confused about. just that i am

lemme go watch a video and come back

.close

[cedar kiln]

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[hot crag]

thanks for the help though

OH okay i think i get it

[crimson delta]

you need to do exercises

you won't understand it otherwise

[cedar kiln]

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[oblique flare]

How can I prove it
My process was naming the radii
OB and OD and OP all =1 then I just couldn’t think of a way for it

[loud gyro]

If you want a hint ||inscribed angle theorem||

try to represent the tan (theta/2) in terms of the sides in this diagram

In fact you can try writing all the sin, cos, tan mentioned in the formula in terms of the lengths of relevant sides in this diagram

[oblique flare]

Oh I saw it now
OA=sinθ
AP=cosθ but couldn’t write tan

BOA=1+sinθ

[loud gyro]

can you identify the angle theta/2?

which two segments make that angle?

[oblique flare]

OP and OA

?

[loud gyro]

No, OP and OA make theta

as mentioned in the diagram itself

[oblique flare]

OB and BP ?

[loud gyro]

Yes

can you prove it?

[oblique flare]

let me see

It gives me
tan(θ/2)=cosθ/1+sinθ

[loud gyro]

thats because you have this down wrong

Can you write the basic definition of sin here?

as in ratio of what sides?

[oblique flare]

Oh I got now

Should be
AP=sinθ
And OA=cosθ

Messed up my trig ratios

[loud gyro]

[oblique flare]

Cuz I thought y=sinθ so used that without thinking

[loud gyro]

no worries, happens to everyone

always remember to doublecheck everything

[oblique flare]

Ok

Thx

.close

[cedar kiln]

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[calm depot]

so far with 7a) thingking about the left most side if i place a left slanted tile i get Fn-1 but when it comes to the other two oreinetations i do not know what to do

[cedar kiln]

@calm depot Has your question been resolved?

[crimson sedge]

This is an arithmetic sequence right?

(Sorry if I’m wrong)

[calm depot]

i have been told that it is a recursion which gives the answer as the sum from 1 to n

[mental trail]

uh do they mean F1 = 5

[calm depot]

F1 = 3 as in the number of certain tiling combinations possible

[mental trail]

oh ok yes I see it

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[cedar kiln]

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[fossil berry]

I stuck at b

[flint plinth]

do you have any thoughts?

[cedar kiln]

@fossil berry Has your question been resolved?

[fossil berry]

I have zero thoughts

[flint plinth]

see if you can construct even one injective function

let's say you start with f(1) = A

now there are two possibilities for f(2) because it can't be A

either f(2) = B or f(2) = C

say you chose f(2) = B

then f(3) has to be what

[fossil berry]

Can you please first explain what injective and surjective means. I read the definition but still didnt get it

[flint plinth]

injective is the same as one to one

the function can't assign the same "output" to two different "inputs"

so you can't have say, f(1) = A and f(2) = A

A can only appear once

[fossil berry]

Okay

What is the other one

[flint plinth]

surjective means that all three of A,B,C are used as outputs

so for example:

f(1) = A and f(2) = A and f(3) = B and f(4) = B and f(5) = B is not surjective because it doesn't send any input to C

whereas:
f(1) = A and f(2) = A and f(3) = B and f(4) = C and f(5) = B
is surjective

because A, B and C all show up as output values

[fossil berry]

So surjective means every element in Y have have something

[flint plinth]

every element in Y is equal to f(x) for some x in X

in words, every element in Y is the output for some input

[fossil berry]

I see

Can you give me example in injective?

[flint plinth]

in fact i cannot

it's impossible when there are more input values than output values

[fossil berry]

Why

[flint plinth]

you have five inputs

[fossil berry]

Why it need to be equal

[flint plinth]

and they all need to be assigned to different outputs

so you need at least five output values available

[fossil berry]

Can you get another example apart from the current problem

[flint plinth]

sure

if X = {1,2,3,4,5} and Y = {A,B,C,D,E,F,G}

then there are multiple solutions

for example,

f(1) = A, f(2) = B, f(3) = D, f(4) = F, f(5) = G

A,B,D,F,G are all different

so f is injective

[fossil berry]

How do you calculate that

[flint plinth]

wdym by calculate

[fossil berry]

I mean count the injective functions

[flint plinth]

to specify a function i just need to tell you what the output value is for each input

oh to count

well you can argue as follows

f(1) can be anything so 7 possibilities for f(1) (any of A,B,C,D,E,F,G)

then f(2) has 6 possibilities, it can't reuse the value you assigned to f(1)

f(3) has 5 possibilities

f(4) has 4 possibilities

f(5) has 3

so in total:

7x6x5x4x3

[fossil berry]

I see

Now i started to understand these

[flint plinth]

nice

[fossil berry]

Lets get back to the problem

What about c

[flint plinth]

just gonna recopy it here for reference

ok first of all, can you give an example of a surjective function

for this X and Y

just to make sure you understand it

[fossil berry]

Sorry. We havent had a lecture about this topic but our homework from seminar was this

f(1)->A , f(2) -> B, f(3) -> C

I just have to use all A B C right?

f(4)-> A , f(5) -> A

@flint plinth

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[trail dune]

Why does my method not match da quadratic formula?

[digital cliff]

it does
sqrt(16)=4
(-2 pm 4)/6=(-1 pm 2)/3

your answer is
(-1 pm sqrt(4))/3=(-1 pm 2)/3

same thing

[trail dune]

Sqrt(16) is 4, but da sqrt(4) is not da sqrt(16) doe

[digital cliff]

?

i never claimed it to be

[trail dune]

Calculate da numerator, mine gives you 1, the other gives you 2

[digital cliff]

i just showed you how theyre the same thing

[trail dune]

Oh my bad, ok I understand but I don't understand the sqrt(16) how it balances out w/ da sqrt(4)

[digital cliff]

youre just looking at the numerator, not the whole fraction

the whole reason theyre the same is because they simplify

[trail dune]

I put in da lowest form, n that one is just a step above lowest form yeah?

[digital cliff]

you have
(-1+2)/3

the other is (-2+4)/6

those are the same thing

for the + case

[trail dune]

Ohhhhh okay okay

I get it

Thank you bro

One more question tho, why doesn't the quadratic formula give da lowest form?

[digital cliff]

it would, you just didnt put it into its simplest form

[trail dune]

I se

[digital cliff]

the quadratic formula gives an answer, but any form fitting has to be done manually

[trail dune]

I see, that makes sense

ty 4 da help bro

.solved

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[pliant fulcrum]

i dont even understand what that third step wants

never seen it written like that

work^

.close

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[austere brook]

need help understanding

[austere brook]

how this simplifies

[raven shard]

expand out the entire expression at the start and you can rewrite to get what you see on the bottom

@cerulean sail

[austere brook]

sigma^2 + Npsigma^2 - psigma^2 / N

Which is like

I'm not sure how the middle n drops out

Unless it just does since N/N=1

[raven shard]

yes

[austere brook]

It's just weird that the ^1 dies and then the ^2 also dies when reduced

So the left right n cancel but then the middle n also cancels

Whatever

[cedar kiln]

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[marsh pond]

I need help with this question

[marsh pond]

State whether it is True or False. If True then give a proof. If False then give a counterexample.

I feel like this is false based on contrapositive: "If sqrt(x) is rational, then x is rational"

[dire geode]

Wot

[marsh pond]

https://tenor.com/view/u-wot-m8-gif-20998179

[violet jay]

that's not a counterexample to the contrapositive

[dire geode]

What are you providing with the counter example

[marsh pond]

oh

I messed up then

it is false tho, right?

pretty sure

[nimble mountain]

what is sqrt(pi)

[marsh pond]

I can still give counter example with this? If sqrt(x) is rational, then x is rational

p -> q has equality with ~q -> ~p

[violet jay]

if sqrt(x) = a/b with a,b integers, then can you express x as a ratio of integers?

[marsh pond]

1.7724538509

[nimble mountain]

in terms of an integer over another integer

[marsh pond]

( 1.7724538509 ... ) / 1

its irrational

[nimble mountain]

do u know what is an integer

[violet jay]

1.7724538509..., my favorite integer

[marsh pond]

😄

in between 1 and 2

why do you bring up sqrt(pi) tho

so the answer is true?

[nimble mountain]

what is your train of thought with what u are doing anyway

[marsh pond]

oh, so i just need to show proof with pi, or e

that's it

[violet jay]

i'd recommend considering this question

[nimble mountain]

this is the contrapositive yes and

it says for all

[marsh pond]

i gotta show all irrational numbers?

[silver oxide]

you have to prove it for all irrational numbers, u cant pick a specific example

[marsh pond]

dang

i dont know how to prove it without picking one

[nimble mountain]

u were saying about the contrapositive

[marsh pond]

i can prove with this?
If sqrt(x) is rational, then x is rational

[nimble mountain]

its equivalent

what do you know about the contrapositive

[marsh pond]

am I on the right track here?

[crimson sedge]

What is the question 💀

[marsh pond]

https://tenor.com/view/scroll-you-must-scroll-yoda-star-wars-gif-10500226

[nimble mountain]

u have to show its true for all x

[crimson sedge]

Discord buggin

I can't see it lol

[marsh pond]

like this?

[marsh pond]

only Nitro subscribers can see it

[nimble mountain]

what is that symbol

[marsh pond]

exists

oh it's backwards

[nimble mountain]

u shouldnt be using exists here..

[crimson sedge]

You could assume √x is rational, then we know rational times rational is rational and so x is rational. But it's not so hence proved

[marsh pond]

proof is annoying

[nimble mountain]

i dont get how u got that sqrtR=R nor how it is relevant

[marsh pond]

i'm just grasping at straws here, i really have no clue where to start with the proof

without using an example, that's not good enough

[nimble mountain]

what is a rational number

[marsh pond]

1,2,3,4

[crimson sedge]

Did you read my answer?

[marsh pond]

ya but you said "not so hence proved"

[crimson sedge]

Hence proved that √x is irrational

That's what I meant

[nimble mountain]

what is the definition

[nimble mountain]

definition

[crimson sedge]

See to prove that √x is irrational, we start by letting √x be a rational number and we prove THAT wrong. And hence come to conclusion that it is irrational

[marsh pond]

a rational number is a number that can be expressed as the quotient or fraction ⁠p/q of two integers, a numerator p and a non-zero denominator q.

[nimble mountain]

thank you

so now suppose that sqrt x is rational

what equation can u write from this

[marsh pond]

getting closer?

or should i use R, not Z

[nimble mountain]

i said suppose that sqrt x is rational

[marsh pond]

my definition says integers but the question says real

[nimble mountain]

so u should get

[ \sqrt x = \frac{a}{b} ]

[wraith dagger]

POKÉDANCE

[nimble mountain]

yes x can be any real number

and we are trying to show the contrapositive statement which states

if sqrt x is irrational then x is rational

[marsh pond]

like this?

[nimble mountain]

yea

so what is x

[marsh pond]

(a/b)^2

[nimble mountain]

and is that rational?

[marsh pond]

yes

[nimble mountain]

so the contrapositive is true

and what does that say about the original statement

[marsh pond]

is this fine? should say *contrapositive for last word, Goodnotes still catching up to syncing at screenshot

[nimble mountain]

its fine except that statement where u said the contrapositive statement

maybe say the contrapositive is...

because it seems as though you are just stating the contrapositive as true

other than that its fine

[marsh pond]

this a bit better?

[nimble mountain]

i mean at your third line

add something like

contrapositive:

this is equivalent to...

or something that denotes it is something you are trying to show rather than it being already true

we want to show..

[cedar kiln]

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[late rapids]

Can someone help me with this have no idea where to start

I think I could make a variable a = x/t

[dire geode]

Multiply top and bottom by 1/t^2

And factor out a t^2 from the sqrt

[late rapids]

Don’t know what to do about the square root divided by t

I think if I could get rid of the x the rest of the problem will be straight forward

The x is in the radical being divided by t

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[polar marsh]

where exactly do i go from here?

[cedar kiln]

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[polar marsh]

<@&286206848099549185>

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[vernal shell]

which

[void sand]

where?

[vernal shell]

Do you want to know how the M is positive instead of negative?

if so it's just the distributive property

[void sand]

that would just be the distribution of the -1, if that's what you're asking

[vernal shell]

but if not I still don't know what you're asking

lol

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[vernal shell]

np

[vernal shell]

happy cat :]

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[tepid delta]

aaa

[tepid delta]

<@&286206848099549185> idk how to start or do this its been 1 day

[turbid oracle]

JUST ONLY

That it

[high bridge]

Ok I think I have an idea

Let UP=x and RQ=y

Total fencing used is 4y-2x=30

Total area=y^2-x^2

so it is maximizing y^2-x^2

Hmmmm

Actually no let me take a different approach

Let QP=x

2y+2x=30

And it is maximizing 2xy-x^2

So from equation 1

There is y+x=15

Thus y=15-x

Substitute that into equation 2

[tepid delta]

u lost me here

wait ill let u finish

[high bridge]

You get

2x(15-x)-x^2

30x-3x^2

Now it is completing the square

-3(x^2-10x)

-3(x^2-10x+25) +75

-3(x-5)^2+75

So when x=5, the area is maxed

The dimension will be PQ=5, QR=10,RS=10 ST=5,TU=5,UP=5

There solved

[tepid delta]

i got lost waht happened here

lol

[high bridge]

so the area of the thing is

You can see the area of the thing

two rectangles of lengths x and y with an extra square of length x

This it is xy +xy-xx

2xy-x^2

I need to go to sleep in 5 minutes so if you have question quick

[cedar kiln]

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[maiden fractal]

i got ||13|| as the answer because the only time the roots r integers is when its factorable(?)

so i assumed p and q would be 3 and 2 respectively

but is there any other way to get that

[worldly veldt]

yeah, prime number q would have 2 factors q and 1

[loud gyro]

p and q are prime numbers means that the product of roots is a prime number. That means that the roots have to be +/- 1 and +/- q

[maiden fractal]

OHH

and the only time 1 + prime is a prime is 1 + 2

[loud gyro]

1 is not a prime

[maiden fractal]

mistype

[loud gyro]

yea

[maiden fractal]

yay

thank you

[loud gyro]

p and q are 2 and 3

[maiden fractal]

.close

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[wraith dagger]

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

BhokTjiang

[cedar kiln]

@calm yew Has your question been resolved?

[calm yew]

<@&286206848099549185>

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@calm yew Has your question been resolved?

[distant hearth]

@calm yew

How can i help u today

.close

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[deep burrow]

Hi, I am learning math from scratch because I’ve never been good at it. I failed and barely managed to pass using grace marks. I am watching Organic Chemistry Tutor videos for pre-algebra, and in one of the videos, he talks about PEMDAS. I began to understand the priority of order of operations, but then I got confused. He mentioned that multiplication and division have the same priority, and you should solve them from left to right to get the correct answer. However, in PEMDAS, it seems like multiplication (M) has more priority than division (D). Why is that? I’m very confused. Can anyone help me?

[haughty osprey]

PEMDAS is just a general guide for beginners

In reality multiplication and division have the same priority

[haughty osprey]

BODMAS is another one used

You should solve
Brackets
Multiplication and division
Addition and subtraction

[nimble mountain]

no

[haughty osprey]

Yes?

24/4x3 can be done by dividing OR multiplying first

[nimble mountain]

if u do everything from left to right

but it isn't associative

[hollow trail]

PEMDAS should really be PE(MD)(AS) where the parts in parentheses have equal priority

[deep burrow]

Okay, so you mean multiplication and division (MD) have the same priority, while multiplication with addition or subtraction has a different priority? The same rule applies for division, right?

[hollow trail]

basically the priortity is:

1. parentheses
2. exponentiation
3. multiplication and division
4. addition and subtraction
   going left-to-right on operations with equal priority

[deep burrow]

Okay understood 😊👌🏻

.close

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[candid finch]

I'm unsure how to factor this polynomial

[cedar kiln]

@candid finch Has your question been resolved?

[candid finch]

Will someone help me?

[lean vapor]

Wait

[lean vapor]

Im really sorry for the late reply 😔😔

@candid finch

[candid finch]

That's okay

[lean vapor]

So is it correct ?

[candid finch]

I think so, yeah

Thanks

.close

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[drifting turtle]

Hello, I need help proving the surjective part of this. I can't for the life of me figure out how to.

[flint plinth]

do you know the definition of surjective?

[drifting turtle]

my understanding of it is that

every element of the codomain has a preimage in the domain

or, the range is equal to the codomain

[flint plinth]

yea good

so take an arbitrary element in the codomain

say (a,b)

can you always find an n such that (2n, n+3) = (a,b)?

[drifting turtle]

oh

so

how do i write that in proof

[flint plinth]

depends

[drifting turtle]

do i declare an integer n?

[flint plinth]

are you proving that it is surjective or that it is not?

[drifting turtle]

well

is it

i legit dont know

[flint plinth]

that's where i'm trying to lead you

worry about how to prove it after you work out whether it is or not

you need to find n such that (a,b) = (2n, n+1)

that's equivalent to two equations:

a = 2n
b = n+1

and you need to find an n that solves both

[drifting turtle]

oh but i think it fails in a = 2n

since n = a/2

which isnt guaranteed integer

am i right

[flint plinth]

yep

[drifting turtle]

oh ok so its not surjective

[flint plinth]

so good

now you know it's not surjective

the way to show it formally is to just show one counterexample

like a particular (a,b) that doesn't work

[drifting turtle]

okok, i got it now

thank you very much

[flint plinth]

yw

[drifting turtle]

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[blazing carbon]

I've been doing this problem and I got stuck, I need help

[blazing carbon]

,rotate

[cedar kiln]

@blazing carbon Has your question been resolved?

[blazing carbon]

<@&286206848099549185>

[blazing carbon]

btw all A's in the picture are the same thing

[blazing carbon]

Can i just say that ${x} \subseteq \mathbb{R}$ ?

[wraith dagger]

ω

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@blazing carbon Has your question been resolved?

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.close

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[storm marsh]

@odd verge

[dire geode]

Take it to DMs

[odd verge]

Bro just do what he said

[dire geode]

<@&268886789983436800>

[storm marsh]

k

[unkempt egret]

Seems the situation is resolved?

[dire geode]

yes

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[livid hinge]

Aw dang it

[cedar kiln]

@livid hinge Has your question been resolved?

[crimson sedge]

Can you close this channel?

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[crimson sedge]

hi

[crimson sedge]

b) from A3

that 1/n ln

i have to calculate lim n -> infinity

ive taught of just applying Cezar Stolz but doesnt simplify

clustering criterion doesnt really work either

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[twilit plume]

ok nvm chat gpt said bullshit

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.close

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[twilit plume]

Hello, i'm trying to understand, how we go from left to right? what does it mean to integrate d(sqrt(2+y)) ? and where did the 2 on the denominator go? Thanks

[dire geode]

$d(f(y))$ is shorthand notation for $f'(y) \cdot \frac{dy}{dx}$

[wraith dagger]

riemann

[dire geode]

where f is explicitly a function of y

but y is a dummy variable anyway so it doesn't matter to integrals

$d(f) = f' dy$

[wraith dagger]

riemann

[dire geode]

,w diff sqrt(2+x)

[wraith dagger]

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Requested by rie.mann

[twilit plume]

ok i can solve the integral but the notation is still weird

if i let u = 2 + y

[dire geode]

wut

use FTC

,tex .FTC1

[wraith dagger]

riemann

[bleak orbit]

Guys can I have any sources to take notes from and explains Algebra at an in-depth level along with practice questions. I know the basics I just want to get a refresher on Algebra 1 and master Algebra 2 and jump straight into solving polynomials, functions, precalculus, and all the bigger stuff.

[dire geode]

!occupied

[cedar kiln]

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).

[twilit plume]

i should just solve it normally instead of focusing on this weird notation

[dire geode]

sure sounds reasonable

the point of the weird notation is to "undo" the integral

$\int d(f) = f$

[wraith dagger]

riemann

[dire geode]

physicists eat that shit up

[twilit plume]

ye the guy just wanted to flex

thank you f

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[crimson sedge]

I need help

[oblique flare]

In your head try rotating the triangle XYZ

[crimson sedge]

wdym

is it 2

the scale factor

[oblique flare]

Thought you want to solve but yea it is 2

[crimson sedge]

but my teacher says its 1/2

idk if she made an error or not

[oblique flare]

The reason is what polygon is scale

[oblique flare]

Polygon 1 is scaled by 1/2 to become like polygon 2

[crimson sedge]

is the scale factor 2 or 1/2 then

[oblique flare]

It is 1/2 from polygon 1 to polygon 2

[crimson sedge]

?

you said

its 2 before

[oblique flare]

I just read the question wrong

[crimson sedge]

can you explain how it is 1/2

[oblique flare]

It states what is the scale factor of polygon 1 to polygon 2
So kp1=p2

[fiery plaza]

You can see that in polygon 2, one of the sides is 15.

That 15 has to have a s.f with one of the sides

One of your sides is 30, 15 is half of 30

Now, you technically can't be 100% sure whether 15 was scaled from 30, so you try imagine what it would be like if it were half.

If it were half, then y would be 8, because half of 16 is 8, and x would be 17, because half of 34 is 17

Put it in your pythag formula a^2 + b^2 = c^2, and see if it works, if it does, then there you go, your s.f is 1/2

[crimson sedge]

ok

This was very helpful

Thank you

[fiery plaza]

np

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[sly wyvern]

could anyone please explain the steps to solve this problem I've been trying for 2 hours now and nothing seems to solve correctly

[blazing dune]

Liner algebra and gaussian elimination

[sly wyvern]

how do I identify what to eliminate because I can't get the correct answer

[blazing dune]

You know gaussian elimination method right?

[sly wyvern]

I know what it is but my professor didn't really explain it

so I know I can use it I just struggle to do so correctly

[blazing dune]

It is kinda complicated to explain it so what about a video on YouTube

https://youtu.be/eDb6iugi6Uk?si=zpiNS1EmkNcIyY7e

This might be good
It explains it for 3 variables but the concept can be applied to bigger or smaller grids

If you didn't like it you can check other videos too

[sly wyvern]

ok thank you I'll look into it

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[cedar kiln]

@exotic oracle Has your question been resolved?

[dire geode]

show the entire question

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[celest otter]

Prove:

$\forall k,l\in \mathbb{Z}^+$ the $kl$-th Fibonacci number $f_{kl}$ is
divisible by both $f_k$ and $f_l$.

[wraith dagger]

hamr

[celest otter]

I am thinking of a proof by induction for this one

[gritty viper]

what I like to do is think of the values mod f_k

f_0 is 0

f_k is 0

Try to show that f_k being 0 (mod f_k) basically "resets" the sequence

and so if you write k more terms you'll get another 0 (mod f_k)

[rapid halo]

you can also use binet's formula

fix a k and induct on l

[celest otter]

Ok i will try this

[celest otter]

Yes doing this rn

Thanks guys

[solid juniper]

i thought you were a random person

[rapid halo]

i am

[solid juniper]

[rapid halo]

all of us are random people

motes, dust

[cedar kiln]

@celest otter Has your question been resolved?

[celest otter]

Ok this is kinda what I have so far,

Prove:

$\forall k,l\in \mathbb{Z}^+$ the $kl$-th Fibonacci number $f_{kl}$ is
divisible by both $f_k$ and $f_l$.
[\text{or}]
[
P(k) = \left(f_k | f_{kl}\right)
]
And we w.t.s that $P(k)$ holds true $\forall k\in \mathbb{Z}^+$. From
it is symmetrically true that $\forall l\in \mathbb{Z}^+, : f_l | f_{kl}$

\textbf{Base case($k=1$):}

[\begin{aligned}
f_1 &| f_{l}\Leftrightarrow 1 | f_{l}\
f_l &= \frac{f_l}{1}, : f_l \in \mathbb{Z}\
\end{aligned}]
Thus, $ | f_{l}$ is true.

\textbf{Inductive step($P(n)\Rightarrow P(n+1)$):} Now, assume that
$n$ is a positive integer such that $P(n)$ is true. (In the base case,
we have just seen that $n = 1$ is such an integer.) We will now show
that $P(n + 1)$ is true.

$Proof$: by hypothesis $P(n)$:

[
\forall n,l \in \mathbb{Z}^+, :\left(f_n | f_{ln}\right)
]

Using the assumption, for $P(n+1)$:

[
\forall n,l \in \mathbb{Z}^+, :\left(f_{(n+1)} | f_{l(n+1)}\right)
]
Rewriting $f_{l(n+1)}$:
[\begin{aligned}
f_{l(n+1)} &= f_{(ln+l)}\
&= f_{ln-1}f_{l} + f_{ln}f_{l+1}\
&=
\end{aligned}]
Rewriting $f_{(n+1)}$:
[\begin{aligned}
f_{(n+1)}&= f_{(n-1)}f_{1} + f_{n}f_{1+1}\
&= f_{(n-1)} + f_{n}
\end{aligned}]

[wraith dagger]

hamr

[cedar kiln]

@celest otter Has your question been resolved?

[cedar kiln]

@celest otter Has your question been resolved?

[celest otter]

<@&286206848099549185>

[celest otter]

<@&286206848099549185>

[celest otter]

<@&286206848099549185>

[cedar kiln]

@celest otter Has your question been resolved?

[celest otter]

<@&286206848099549185> pls