help-13

[hollow trail]

yes

really it's the derivative of 2x + 1

if we think of y = f(g(x))
then
f(x) = √x
g(x) = 2x + 1
then by the chain rule:
dy/dx = df/dg • dg/dx = f'(g(x)) • g'(x)

[deep otter]

that makes sense to me

why do we multiply this by dx/dt though

I think I'm just confused on how y is a function of t if t isn't present in y(x)

[hollow trail]

because x is a function of t

so we start by finding dy/dx, which is
dy/dx = 1/2(2x+1)^(1/2) • 2
then we have
y(x(t)), since x is a function of t
so
dy/dt = dy/dx • dx/dt

[cedar kiln]

@deep otter Has your question been resolved?

[deep otter]

I think that makes sense

thank you

.close

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[green meadow]

So I don't know if I need to put a specific equation in or not but I need help learning how to solve quadratics for a test tomorrow and this is where I thought of to ask for help first.

[green meadow]

.close

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[turbid isle]

isn’t the average value of a function tbe same thing as rienmums sums but just in intergal notation?

[scenic chasm]

You mean if rienmums is the same thing as integral?

[hollow minnow]

average of a function as in ( \frac{1}{b-a} \int_a^b f(x) \mathrm d x)

[wraith dagger]

Pure

[scenic chasm]

Oh

[turbid isle]

ya

[cedar kiln]

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[hearty mist]

what test would i use to solve this?

[dire geode]

Try any

[hearty mist]

could i use root test?

[digital cliff]

go for it

even if youre not sure, just try it, see what comes of it

[hearty mist]

alr

does this make sense?

[digital cliff]

sure, i dont think l'hopital was necessary though

[hearty mist]

was there a better way to do it

wihtout that rule

its like my go to rule for limits

[supple flume]

In my opinion if it works for you and you find it easy keep at it. (It’s probably better to find more ways though just in case)

[hearty mist]

oh alright

but is the work correct?

[digital cliff]

yeah i never said it was wrong, youre in the clear

[hearty mist]

oh alright thx

im doing one more problem but ill like it if u could check it

should i keep this help open?

[digital cliff]

sure

[hearty mist]

so

alr

[cedar kiln]

@hearty mist Has your question been resolved?

[hearty mist]

sry send the wrong one

but that does look correct?

[hearty mist]

would u be able to check for me?

!help

[cedar kiln]

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

[digital cliff]

would just be 2<x<4

[hearty mist]

why wouldnt it include 2 and 4

it converges no?

[digital cliff]

have you checked that?

you know it converges for |x-3|<1 ie 2<x<4

you havent checked to see if it converges at 2 or 4

[hearty mist]

its the first one

here

i sent two pics

[digital cliff]

aha

alrighty then

[hearty mist]

so am i correct?

[digital cliff]

what have you done here, youve shown the terms tend to 0, you havent strictly shown the series do, though i can say they do

[hearty mist]

im showing the terms for alternating test

so im using b sub n to prove that the series converges

[digital cliff]

oh right, sorry, its been a while

[hearty mist]

yea

its ok

[digital cliff]

yeah i think youre good to go

[hearty mist]

alright thx

how would i close this?

[cedar kiln]

@hearty mist Has your question been resolved?

[cedar kiln]

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[crimson sedge]

In how many ways can a rook from the bottom left corner of an 8 x 8 chessboard reach the top right corner in exactly 4 moves? (The rook must not be on the top right corner prior to the 4th move.)?
The answer key was 532 how did they arrived with the answer

[cedar kiln]

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[hollow storm]

Can somebody help me with this please? Prove that the R^2 plane cannot be written as a finite union of lines. Prove that it is not even a finite union of circles.

[hollow storm]

it's google translated sorry

[dull oxide]

Circles one is easier than lines as a finite union is bounded

[hollow storm]

both finite union of lines and circles are bounded?

[mental trail]

no, that's the point, only for circles

[dull oxide]

Circles are bounded, lines are not

Don't know how to prove for lines

[hollow storm]

oh so

the circles themselves are bounded

or the union of circles

sorry if i am dumb

[crimson delta]

the only thing I can think of for lines is that their union isnt open. but that seems rather overkill, surely it should be easier?

I suppose we can check that for some other line you can have at most n intersections

[hollow storm]

how do you write this in terms of a function?

[crimson delta]

why a function

[hollow storm]

uh i dont know

but if a line has a dimension of R and the plane has a dimension of R^2

can we prove somehow that R --> R^2 is not surjective and so the plane cannot be written as a union of lines

no wait i am tripping

R and R^2 have the same cardinality,, right?

[crimson delta]

yes

[hollow storm]

dumb question

if R^2 is infinite and the union of lines is finite

[crimson delta]

well the lines are still infinite

[hollow storm]

oh right

[indigo yoke]

Hello how can i Create a frequency distribution, Create cumulative frequency distribution, Calculate mean, median, mode, STD and variance from this data set .
Im so confused (like no clue & i lost my notebook so im even more screwed)
Thanks

[cedar kiln]

@hollow storm Has your question been resolved?

[hollow storm]

sorry, the translation was faulty. it wasn't circles but circumferences. can we still prove that the plane can't be written as a finite union of circumferences?

[crimson delta]

yes

same argument still applies

[hollow storm]

alright thank you

for the union of lines i'll ask the professor

[cedar kiln]

@hollow storm Has your question been resolved?

[hollow storm]

.close

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[golden relic]

May I ask for help in this matter

[olive crescent]

.ccw

,ccw

[fallen moat]

,rccw

[golden relic]

Ccw?

[cedar kiln]

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

[golden relic]

1

[fallen moat]

First, name the center as O
hint: consider angle WOS and angle VOT

how are they related (in measures)

[golden relic]

12x-2=5x+10+something

Right?

[fallen moat]

yes

[golden relic]

Something is VOU I think

[fallen moat]

and with that, you can find VOU

[golden relic]

That's where I got stuck

[fallen moat]

how?

[golden relic]

12x-2-5x+10 right?

[fallen moat]

nope

[golden relic]

How then

[fallen moat]

12x-2=5x+10+VOU
then
12x-2 -5x -10 = 5x+10+VOU -5x -10

[golden relic]

Ohh

Its -10

7x-12?

Or wrong

[fallen moat]

next

[golden relic]

Ok

[fallen moat]

consider measure of WOT (left side)

what information will we get

[golden relic]

180?

[fallen moat]

good

[golden relic]

What's next?

Subtract 12x-2?

[fallen moat]

wait...

i think we missed an important info

[golden relic]

What is it?

[fallen moat]

why TOU and VOU are marked?

with same color

do you know what does that mean?

[golden relic]

Oh that's just an indication that it has the same angle with WOS

[fallen moat]

i am thinking

with the same angle notation

is meaning they are of the same measure

(which may have covered in class, but I don't know if that true)

i.e. TOU=VOU

does this make sense?

[golden relic]

I don't get it

[fallen moat]

oh, e.g.

[golden relic]

I ask my teach she said that the red mark indicates that WOS=VOS+TOU

[fallen moat]

for isosceles triangle, we have the base angles equal

[golden relic]

Yes

[fallen moat]

i thought it's about the same as this one

same marking means angles of same measure

[golden relic]

I think Its different

[fallen moat]

oh ok

[fallen moat]

if that's the case

my help ends here, since I don't know how to proceed.

[golden relic]

Oh

[fallen moat]

like without that info i "assumed"

I don't know how we can find x

[golden relic]

Hmm

[fallen moat]

sorry man

[golden relic]

It's fine

[fallen moat]

but I would personally suggest to ask a bit more about the 2 angle notations with the same color

I really think they are somewhat related to something crucial

[golden relic]

I can't ask her rn since it's already late sorry mate

[fallen moat]

it's okay, in case you wanna continue with that question, treat those two red notations as equal in measure.
if not, you can just wait to ask her tomorrow

[golden relic]

Let us continue

[fallen moat]

after this VOU we find, we know that
VOU=TOU
so we have....?

hint: TOU=5x+10

[golden relic]

2(5x+10)?

[fallen moat]

nah

[golden relic]

They are both 5x+10?

[fallen moat]

you can do that too

but my method would be:

7x-12=5x+10

[golden relic]

Oh

[fallen moat]

hence we can find x

[golden relic]

Ohh

[fallen moat]

and with this, we can find all measures of other angles

[golden relic]

Is X=11?

[fallen moat]

yep

x=11

i think you can handle the rest 😁

do tell if you are stuck

[golden relic]

Ohh ok thanks

Thanks

.close

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[green coral]

what is the total lenght of the wire

[spice torrent]

60

[alpine rune]

the total length is 60

the area is 120

and its a rectangle

now what's the area of a rectangle?

[spice torrent]

Lw

[alpine rune]

Lw?

oh length*width

alright

[green coral]

width im guesing

[alpine rune]

now, what's the perimeter of a rectangle?

[alpine rune]

gotcha

[spice torrent]

2l + 2w

[alpine rune]

cool

so 2l + 2w = 60

l*w = 210

2 equations, 2 variables

[spice torrent]

Oh I see

L=210-w

[green coral]

no?

[alpine rune]

no

[green coral]

its in multiplication

[spice torrent]

Oh ya

[green coral]

so... l = 210 (?) w

[spice torrent]

210/w

[green coral]

e substitute

[spice torrent]

Got it

[green coral]

quadratic in w , solve the thing

[spice torrent]

Thanks

[green coral]

cool

[spice torrent]

.close

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[errant wasp]

i think im sort of confused, could someone explain the definition of the quotion of a set?

[umbral dew]

are you sure that's a correct question?, first time hearing something like "quotion of a set"

[olive crescent]

quotient set?

[errant wasp]

uhh sorry maybe i did this bad

i can show how it was defined in class

[cerulean sail]

Are you happy with equivalence classes in general?

Dopamine

[errant wasp]

yeah, basically all the elements of a set that have the same relation right?

[cerulean sail]

Yep pretty much - all the related elements

The quotient of a set under an equivalence relation is basically the set of all the equivalence classes you can get

So e.g. taking the integers and the equivalence relation to be the modulo 2 one (two elements equivalent if their difference is divisible by 2) then you have only equivalence classes [0] and [1], so the integers “quotiented” out by that equivalence relation is {[0], [1]}

(Which I’m sure is somewhat familiar!)

[errant wasp]

oh i gotcha

[cerulean sail]

Does that at least help a bit?

[errant wasp]

of all the equivalence classes you can get --> i think im confused about this

[cerulean sail]

so basically you know like with the previous example I said

You could consider [2], but that equivalence class is the same as [0]

Similarly [3], [5], [7] etc etc is really just [1]

[errant wasp]

[0] and [1] belong to the same equivalence class, a quotient or whatever is just the set of all elements belonging to the same equivalence class?

[cerulean sail]

[0] and [1] are in the quotient of Z (with respect to the “difference divisible by 2” equivalence relation)

(That’s partly why the notation Z/2Z is used for it)

But of course [0] and [1] are distinct equivalence classes

Basically you have the equivalence classes, which is the set of all elements that are equivalent to some representative

Then go through all possible equivalence classes (there may be finitely many, or infinitely many), and put them in a set

[cerulean sail]

The quotient is that set you get

[errant wasp]

yeah i think the part that there are multiple equivalence classes in a set confused me

[cerulean sail]

Yeah it’s like a set of sets there (but of course those equivalence classes in the set don’t need to be the same!)

[errant wasp]

like it could also include other modulo equivalence classes?

[cerulean sail]

For another example of an equivalence relation, you can define one on the real numbers by saying that a pair of them are equivalent if their difference is an integer, and there are infinitely many equivalence classes you get from that one

[cerulean sail]

It would contain different equivalence classes - but under the same equivalence relation you’re considering

[errant wasp]

because there are infinitely many differences?

[cerulean sail]

Basically yeah, in fact each real number would be equivalent to one within [0,1) and you have infinitely many numbers in there (I’m sure someone will come around and elaborate on what that becomes, better than I can!)

[errant wasp]

like the equivalence class of integers pairs that have a difference of 0 and have a difference of 1

[cerulean sail]

Yep basically

[errant wasp]

okay thanl you that makes a lot of sense

[cerulean sail]

But if e.g. you changed it to be “difference divisible by 5” the equivalence classes you get will change and as I’m sure you know, Z/5Z and Z/2Z are different

[errant wasp]

z/5z --> set of elements congruent to mod 5 right

congruence --> a = bmodn => a - b is divisible by mod n

[cerulean sail]

Yep, basically congruence being one of the typical equivalence relation examples!

[errant wasp]

also this is the integers modulo 4 right

honestly im ngl im still really bad with congruences, and understanding why a = bmodn is so useful

[cerulean sail]

I’m assuming that’s what they mean

[cerulean sail]

Awww they are something and can take time to get used to

[errant wasp]

i only have like 2 weeks im ngl

can i ask another question

[cerulean sail]

Yep sure!

[errant wasp]

can you help me go over this proof

the equivalence relation is juse that there is a bijection between A and B

[cerulean sail]

Are you happy with the part where they basically say that two sets are equivalent iff they have the same number of elements?

[errant wasp]

yes

[cerulean sail]

Cool cool and how they found the equivalence class [{1,2}] as well?

[errant wasp]

no not really

[cerulean sail]

Alright, so from the part above, a subset of X is equivalent to {1,2} if and only if it has two (distinct) elements

So you want to find the subsets of X that have two elements in them

{1,2} is one of them quite clearly

You can get the other by picking two of the three elements from X to make your subset

Does that make it any better?

[errant wasp]

ohh gotcha

okay

[cerulean sail]

As for all the equivalence classes

[cerulean sail]

As the subsets are equivalent iff they have the same number of elements

[errant wasp]

ohh

[cerulean sail]

The possibilities are having a subset with no elements, a subset with 1 element, a subset with 2 elements and a subset with 3 elements (all of those will be in the same respective equivalence classes!)

[errant wasp]

so like {2} is equivalent to 1

[cerulean sail]

{2} is equivalent to the set {1} yep

Both having 1 element

Similarly {3} is equivalent to both!

[errant wasp]

so thats why you do like [{1}] ~= in P(x)/ ~

so you could have included {2} instead of 1 there right

[cerulean sail]

Yep, you can choose whatever you want (in the equivalence class) to represent it, and remember that equivalence classes are identical or disjoint

So [{1}] = [{2}] = [{3}]

[errant wasp]

im also sort of confused about the role of the powerset here

also can i brb in 15 mins, i need to walk to class

[cerulean sail]

Basically you’re defining the equivalence relation on the power set, not X itself

[cerulean sail]

Sure, ping me when you’re back

[errant wasp]

also the bijection is solely to do with the size of the set

ie like p({1, 2}) --> p({3, 4}) maps to eachother because their powersets are the same size

[cerulean sail]

Yeah, the bijection you make basically is based on the size of the set you’re trying to make a bijection from

If you had a set of two elements, you need to pick two elements to map to as per the previous example

[cerulean sail]

are you using p to represent the power set here?
It is a true statement sure…

[cedar kiln]

@errant wasp Has your question been resolved?

[errant wasp]

yeah is there a flaw in this?

because once you take the power set, you want to check that every element in the powerset maps to an element in the other powerset?

because wouldnt it be p(a) --> p(b)

@cerulean sail

[cerulean sail]

Well, erm, you want to consider that in the power set, that any two sets you find will have a bijection between them if they have the same number of elements

You’re more wanting to e.g. think that there is at least one bijection between {1,2} and {3,4}

[errant wasp]

between the set or the elements in the set

[cerulean sail]

Between the elements within the set* sorry

[errant wasp]

then whast the binjection of the elements of {1,2} and {3,4}

[cerulean sail]

You can e.g. define f(1) = 3 and f(2) = 4, which is clearly invertible (you can also do, say, f(1) = 4 and f(2) = 3)

[errant wasp]

gotcha

i also forgot invertability is a way to define injections/surjections/bijections

[errant wasp]

then idk still why is the poweorset important

[cerulean sail]

Because you're defining the subsets of X as equivalent, not the elements of X themselves

[errant wasp]

thanks, can i also ask about partial orders

[cerulean sail]

Erm I may not be the best to answer about those you might be better off starting a new channel and asking there, don't want to explain anything poorly for you

[errant wasp]

thank you for helping!

[cerulean sail]

Always a pleasure to

[errant wasp]

.close

[cedar kiln]

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[cedar kiln]

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[cold briar]

Given point P, (a, b, c) how would I find the distance from the x axis?

[cold briar]

is it just distance to (a, 0, 0)?

[subtle spruce]

yes

[cold briar]

.close

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[small raptor]

8x - 5 = 2x + 6

[rich blaze]

You have to solve for x?

[small raptor]

which number needs to be moved

[primal socket]

Awkward question

[rich blaze]

You can "move" both of them 🤔 it probably wants you to have the x on the left side, and the numbers on the right side

[small raptor]

right

[primal socket]

First of all we still don't know the objective so right now I don't see the point in "moving" anything. Second of all we don't "move" terms. We add or substract on both sides.

[rich blaze]

Are you sure thats the whole exercise?

[cedar kiln]

@small raptor Has your question been resolved?

[cedar kiln]

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[marsh turret]

not sure how to do

[chilly geyser]

You can find the length of BC by using cosine law

Then, apply sine rule to find the measure of angle B.

[marsh turret]

sooo would i use the 11 and 10 or the 34

[silver fable]

Find BC using second equation first

Then u should be able to calculate cos of the angle using first equation

[marsh turret]

soooo it wouldnt be C i assume

[silver fable]

wdym?

[marsh turret]

like the answer cause the number is too small

idk if i did this right but i got a^2=121+100-2ac.cosB

[silver fable]

a² = 121 + 100 -2×11×10×cosÂ

[silver fable]

You mixed 2nd and 3rd equation i think

[marsh turret]

ah yes that what i meant

[silver fable]

There's no need to use the 3rd equation

If we add 1st and 2nd equation we get:
$\ a^2+c^2=a^2+b^2-2ab \cos B +b^2+c^2-2bc \cos A$

[wraith dagger]

Adam Chebil

[marsh turret]

.close

[cedar kiln]

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[hollow storm]

Hello, can somebody please help me with this?

[hollow storm]

Let f : R → R be a function defined by the formula f(x) = ax + b. Prove that if f fixes two points of R then f is the identity on R

So this means that if it fixes two points like f(x1) = x1 and f(x2) = x2 then for all x f(x) = x ?

[cedar kiln]

@hollow storm Has your question been resolved?

[cerulean sail]

Might be worth noting that $\frac{f(p_1) - f(p_2)}{p_1 - p_2} = a$ for any choice of distinct $p_1, p_2$

[wraith dagger]

@cerulean sail

[cedar kiln]

@hollow storm Has your question been resolved?

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[hollow storm]

thank you

[cedar kiln]

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[bright sorrel]

I have these 3 vectors and I'm trying to find the basis

[bright sorrel]

seems to me like a_1, and a_2 are linearly independent

but then a_2 and a_3 are also linearly independent

can I just choose a_2 and a_3

or a_1 and a_2

oh wait they are linearly independent to each other

so

they form the basis

[pastel vault]

a basis for a subspace means that we have vectors v1, v2, v3 etc.

such that av1 + bv2 + cv3 + ... = (x, y, z, ...) where a, b, c... are not all 0

so they're finding the column space of that matrix

there's a typo: the first vector in the basis should be (3, 6, 0)

the first column of the original matrix

[bright sorrel]

yeah, yeah

so all of them are the basis

right?

[pastel vault]

no

only (3, 6, 0) and (2, 1, 3) are the basis vectors

for the subspace

@bright sorrel

[bright sorrel]

right, how did you see this?

taking the first coordinate from each vector?

[pastel vault]

cause the answer states the first and the third columns are pivot columns

so you go back to the original matrix

and read off the pivot columns from there

[bright sorrel]

can you pls point it out?

[pastel vault]

in the RREF

[bright sorrel]

I think I'm confused from the language barrier

[pastel vault]

you look at the RREF to find the pivot columns

[bright sorrel]

can you pls point it out using a software or something? :c

[bright sorrel]

right, those are linearly independent

[pastel vault]

yes

[bright sorrel]

but aren't second and fourth column also?

[pastel vault]

I think they took a shortcut

that's not actually the RREF

[bright sorrel]

😭

[wraith garnet]

Is this chatgpt...

[bright sorrel]

yes

[pastel vault]

shit

[wraith garnet]

well thats ur first issue

[bright sorrel]

right so

I have the vectors

(3, 1, 2, 4) (6, 2, 1, 1) and (0, 0, 3, 4)

using a matrix I deduce it to

[pastel vault]

https://www.wolframalpha.com/input?i=rref+[{3%2C+1%2C+2%2C+4}%3B+{6%2C+6%2C+2%2C+1}%3B+{0%2C+0%2C+3%2C+4}]

PLEASE STOP USING CHATGPT

oml

okay the pivot columns have a 1 in each column in RREF

[bright sorrel]

(3, 1, 0, 0) (0, 0, 1, 0) (0, 0, 0, 1)

[pastel vault]

so the 1st, 2nd, and 3rd columns are actually pivot columns

[bright sorrel]

Right

I see it

[pastel vault]

anyways that's not the point

just read off the rows of the RREF to find the basis

as you've noticed they're all linearly independent

so the basis is just the original 3 vectors

(3, 1, 2, 4); (6, 2, 1, 1); (0, 0, 3, 4)

chatGPT sucks at stuff related to maths

[bright sorrel]

definitely

.close

[cedar kiln]

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[static loom]

can anyone help

[static loom]

( sec a , b , c)

hi

[gritty viper]

What are the common differences for the first sequence in section A

[static loom]

hm

[gritty viper]

Also the "nth term" column, the answer should not be a number

[static loom]

apologies.

so could you help?

[gritty viper]

Do you know the answer to this?

[static loom]

Also can you elaborate what the common diff is

Not really.

[gritty viper]

The difference between one number in the sequence and the next

[static loom]

3

[gritty viper]

I shouldn't have said "common"

[static loom]

2nd right?

[gritty viper]

Eventually

[static loom]

is it 3

[gritty viper]

No

Let's look at the differences for now

3 because 3-0=3, right?

[static loom]

yes

[gritty viper]

so what's the next difference

[static loom]

5

[gritty viper]

right

keep going

[static loom]

7

9

[gritty viper]

brb

[static loom]

alr

same wanna continue this in 7 min?

currently doing a tutor class but tutor is delaying answering the ques so came here.

[gritty viper]

Sure if I'm available

I might not be

[static loom]

@gritty viper

Are you available?

<@&286206848099549185>

[modest lichen]

search up quadratic sequences

[static loom]

Yeah since then I have made some progress but wondering if they are correct

Do you want to confirm with me

[dire thorn]

I will confirm if u send

[static loom]

Ok

So for the second row

Here

Ignore first row.

[modest lichen]

no

[modest lichen]

same

[static loom]

For the second sequence on section A , will the 6th term be 38 and 7th 51?

[dire thorn]

yeah

[static loom]

Alright

Also

Can you assist on the nth term part

[dire thorn]

yeah let me think about it for a sec

[static loom]

alr

[dire thorn]

@static loom have u seen smth like this b4

[static loom]

Yes

A and b , c are constants

[dire thorn]

do you know what those constants are for

[static loom]

No

[dire thorn]

ok so your learning about quadratic sequences. A quadratic is anything of the form an^2+bn+c (n is a variable, a,b,c are constants)

If we can find a,b,c, you plug them into that representation and thats is the formula for the nth term

[static loom]

so for the second sequence , what would the constant be for?

[dire thorn]

Do you know what the 2nd difference would be

[static loom]

No

Could you explain how I get it

[dire thorn]

Yeah

so the difference between the terms if 3,5,7,9...

Agree?

[static loom]

Yes

Ah yes

[dire thorn]

nvm I was trippin but thats the next part

[static loom]

Difference between those terms is 3,5,7,9

Difference between those temrs

are 2

[dire thorn]

Yeah

exactly

[static loom]

Ah ok

[dire thorn]

so 2 is the 2nd difference

solve 2a=2 for a

[static loom]

Alright

a = 1?

[dire thorn]

yeah

lets now do the 2nd part. Whats u2-u1 (2nd term - 1st term)

we need to solve for a,b, and c before we get the formula

[static loom]

2nd term - 1st term

So would those terms be

the 3,5,7,9 or the actual sequence terms?

[dire thorn]

actual sequence

[static loom]

oh ok

2nd is 6

1st is 3

Its 3

[dire thorn]

yeah

3a+b=3 solve for b. Youve found a=1 so solve for b in 3(1)+b=3 aka 3+b=3

[static loom]

so 3+b=3

a = 1

b+3 = 3

3+b is equal to 3

I dont know what to do after the stpe

step

[dire thorn]

What number does b have to be in order for 3+b=3 to be true

[static loom]

B has to be 0

[dire thorn]

yeah so a=1, b= 0

[static loom]

In order for 3+b equal to 3

[dire thorn]

now we find c

use 3rd formula here

[static loom]

Pass me the equation for c , ill see if i can do it. Also , any quicker ways to find it?

[dire thorn]

plugging in the values we already know we get 1+0+c=3

[dire thorn]

I do not think there is any quicker way to find it using a 100% process. If you pay close attention you will probably (not 100% sure) find the solution has the same form as the last entry for each section and could skip a bit but its better to go through the process

[static loom]

Check if my answer is correct ill type it out

Will it be n2 + ( the rest )

[dire thorn]

if n2 means n squared yeah

[dire thorn]

here we find c=2

we have a=1, b=0, c=2, so plugging that into an^2+bn+c we get n^2+0b+2 aka n^2+2

[dire thorn]

I gotta go now but it should be exactly the same process for all the problems. Solve for a using 1st formula in the picture im replying to, solve for b using 2nd, solve for c using 3rd

[static loom]

Yeah

Oh alr

Ill talk later if ur on.

[cedar kiln]

@static loom Has your question been resolved?

[cedar kiln]

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[crimson sedge]

How do I begin

[dire thorn]

$\frac{x^a}{x^b}=x^{a-b}$

[wraith dagger]

Anton Yakunin.

[crimson sedge]

?

[dire thorn]

Apply it to your question

[crimson sedge]

What do u mean

[dire thorn]

Ok lets do this differently

30 = 3 * 5 * 2, x^4=x * x * x * x, y^6 = y * y * y * y * y * y
20 = 2 * 2 * 5, x^3 = x * x * x, y^2 = y * y

[crimson sedge]

Ok

[dire thorn]

How would you simplify $\frac{3\cdot 5 \cdot 2 \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y}{2\cdot 2\cdot 5 \cdot x \cdot x \cdot x \cdot y \cdot y}$

[wraith dagger]

Anton Yakunin.

[crimson sedge]

x^3 and y^2

For the bottom

And x^4,y^6

[dire thorn]

nah

Cancel stuff

[crimson sedge]

[dire thorn]

you know how a number divided by itself is 1? use that to simplify this

If it appears on the top, and on the bottom, then you can just delete them

for example we can cancel a 2 from the numerator and denominator to get

$\frac{3\cdot 5 \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y}{ 2\cdot 5 \cdot x \cdot x \cdot x \cdot y \cdot y}$

[wraith dagger]

Anton Yakunin.

[dire thorn]

Keep doing this until you cant anymore

[crimson sedge]

I don’t get it man

Sorry

@dire thorn

[dire thorn]

dw I will start much simpler then we'll get back to this

[crimson sedge]

Ok

[dire thorn]

You know what $\frac{7}{7}$ is

[wraith dagger]

Anton Yakunin.

[crimson sedge]

Yea

[dire thorn]

What is it

[crimson sedge]

1 whole

[dire thorn]

Yeah how about $\frac{38}{38}$

[wraith dagger]

Anton Yakunin.

[crimson sedge]

1 whole

Aswell

[dire thorn]

yeah so whenever a number is divided by itself, it is equivalent to 1

[crimson sedge]

Ok

[dire thorn]

Lets move on. What is $\frac{10}{2}$

[wraith dagger]

Anton Yakunin.

[crimson sedge]

5

[dire thorn]

yeah

Do you agree with this? $5\cdot 2 = 10$

[wraith dagger]

Anton Yakunin.

[crimson sedge]

Yes

[dire thorn]

ok lets replace the numerator with that

$\frac{5\cdot 2}{2}$

[wraith dagger]

Anton Yakunin.

[dire thorn]

Notice we can get 5 as the result by cancelling the 2s

[crimson sedge]

Yes

[dire thorn]

Ok now we will introduce variables

what is $\frac{x}{x}$

[wraith dagger]

Anton Yakunin.

[crimson sedge]

X?

[dire thorn]

What does x represent

it represents a number we do not know

[crimson sedge]

10?

[dire thorn]

Use this

[crimson sedge]

10

[dire thorn]

Yeah

[crimson sedge]

Sorry

Wait no

[dire thorn]

As long as x is not 0 of course because we cant divide by 0

That doesnt really matter though

[crimson sedge]

Ye

[dire thorn]

Ok whats $\frac{5 \cdot 5 \cdot 5 \cdot 5}{5 \cdot 5}$

[wraith dagger]

Anton Yakunin.

[dire thorn]

Explain the process

Dont just give number

[crimson sedge]

So there’s 5 up top and 2 5s on the bottom so we do 5^5 and 5^2

Or correct me

[dire thorn]

yeah thats the long way of doing it. Although that works for now, thinking that way will make it harder to do it with variables (also its 5^4 but im pretty sure that was typo)

[crimson sedge]

Oop yeah

My b

[dire thorn]

Remember what I did here

I did not do 5*2=10, then do 10/2 = 5. I just immediately cancelled the 2s and found the answer to be 5. I never multiplied anything

We can do the same thing here

[crimson sedge]

Ok

[dire thorn]

I will look for a video on this I think it will be more helpful

https://www.youtube.com/watch?v=PcEwj5_v75g

Watch this first

Then watch this after

https://www.youtube.com/watch?v=WSQh4o3yu4I

[crimson sedge]

Ok

Ok I watched

But I gtg

Sorry

@dire thorn

[sinful field]

do you know how to solve it now ?

[crimson sedge]

Not fully

But I’m getting there

[sinful field]

good for you, just keep trying

[crimson sedge]

This was my question btw

[sinful field]

i would give you the answer but i will wait for your guy that helped above, if he doesn't respond in a while i will help.

[bright sorrel]

x^4/x^3=x^(4-3)

y^6/y^2=y^(6-2)

[crimson sedge]

Hmm?

[bright sorrel]

?

[crimson sedge]

What is that

[bright sorrel]

help

ig

as someone else stated above

x^y/x^z=x^(y-z)

[crimson sedge]

Uhh

[dire thorn]

Along with the videos I sent, you should look up more about variables and unknown constants. The message I am replying to (same thing as the first message I sent), is a formula/rule. If you can understand what variables and unknown constants are, you can apply formulas like these and solve the problem in seconds.

[sinful field]

btw (x^2) is the same as (x to the power of 2), just if its confusing

[crimson sedge]

.close

[cedar kiln]

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[crimson sedge]

1.5x^0.5 = 12x^-2

whats x???

i tried to use log but idk

[dire thorn]

log is usualyl for when something is raised to the power of x. In this case you can divide one of the variables on both sides and use exponent rules

$\frac{x^a}{x^b}=x^{a-b}$

[wraith dagger]

Anton Yakunin.

[scenic chasm]

is that (1.5)x^(1/2) or (1.5x)^(1/2)?

[crimson sedge]

how do i remove the power?

x^-2.5 = 1.5/12

[dire thorn]

how would you remove the power for smth like x^2=1? Do same thing here

[crimson sedge]

umm the first part

1/2 in each side

so do i do -1/2.5 in each side?

[dire thorn]

Yeah

assuming you do the appropriate operation

but that is the number you use yeah

[crimson sedge]

yh ok

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[bold pond]

3x =0 in mod 18

[bold pond]

so i know that x is equal to 0,6,12 but i need to solve it for all possible solutions

im using the euclidean algoritm but having difficulties

[crimson delta]

well I mean that is all solutions

[bold pond]

but i havent proved that thats all possible solutions

its asking to prove it and i dont feel like writing every number out

so ik to use the euclidean algorithm

[crimson delta]

well ok

you are looking for solutions to 3x+18y=0

equivalently, x+6y=0

and from there its easy

you dont need eea

[bold pond]

ok uhh 😅 i just started this topic so im not sure but how would this give the all 3 solutions? so we have x+6y=0 and thus also x=12y

[crimson delta]

x=-6y

and you can pick any value of y you want

[bold pond]

but negatives cant exist in mod no? so -6=12 in mod 18

[crimson delta]

well I started with 3x+18y=0

there I meant integers x,y

this then turns into just x=0,6,12 mod 18, yes

first solve it in the integers

then restrict to mod 18

[bold pond]

ok i think i get it, thank you

.close

[cedar kiln]

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[fossil thorn]

Oops sorry didn't see you typing 😭😭

[fossil thorn]

I don't get what im doing wrong

[cedar kiln]

@fossil thorn Has your question been resolved?

[errant chasm]

Yeah tough one

I think I got it tho

It's 92/47 K

U forgot the K

@fossil thorn

[fossil thorn]

ohhhhh

tyyyy

.close

[cedar kiln]

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[crimson sedge]

Systems of linear equations with 2 variables, i have a few of them that I'd appreciate if someone could explain to me

[crimson sedge]

1. |3x-y=13

|2x-3y=-4

2. |3x+2y=4x-5

|3x+2y=y+1

3. If |x+3y=4

|2x-7=1

Find x-y=?

[crimson sedge]

^

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[lusty ridge]

Yes

You can rearrange 3x - y = 13

By isolating for example -y

so by subtracting 3x on both sides

so you have -y = 13 - 3x

And times it by -1 to remove the minus

so it will be

y = -(13 - 3x)

Now try and insert this known y, into -2x + 3y = -4

Ok wait

I have a simpler method

I just realized

[crimson sedge]

Can u just like write it out for me in some form

Like digital

[lusty ridge]

3x - y = 13
-2x + 3y = -4

Let's multiply the first equation by 3 so that the coefficients of y in both equations are opposites:

That'll be

9x - 3y = 39
-2x + 3y = -4

And now

We add the two equations to get rid of y

(9x−3y) + (−2x + 3y) = 39 + (−4)

Removing paranthesis:

9x − 3y − 2x + 3y = 39 − 4

That is simplified to

7x = 35

Simply divide both sides by 7

to find x

x = 5

And now insert the value of x into one of the original equations

and isolate y

I dont have time you can now, I think you got it now

I know you're smart enough

basically just

3x − y = 13

3(5) − y = 13

15 − y = 13

-y = -2

times both sides by -1

to get rid of minus

y = 2

All in all:

x = 5
y = 2

You can confirm these by inserting both values in original equations to see if they add up

no problem 😊

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[tough dock]

On a recent test for Geometry 1, there was the question "What goes up must come down. The ball went up. It will come down. Decide weather inductive reasoning or deductive reasoning is used to reach the conclusion." I answered deductive reasoning because it does not use specific observations and pattern recognition, it uses existing theory (newtons 2nd law) to formulate a hypothesis. I was marked incorrect on this question, but I would like to argue for my grade because I believe it is deductive reasoning. Am I correct?

[solid juniper]

what a weird question

[tough dock]

Yeah

This teacher adds a lot of non objective questions, it's super annoying

Oh yeah also this 1 question would push me from a B to an A

<@&286206848099549185>

[minor quest]

hi

[tough dock]

Hello, the question is the long message above, 15 minutes thingy

[tough dock]

Can you help me sir

[minor quest]

sure

what grade r u

[tough dock]

9th grade

Geometry

[minor quest]

alr

lets see

im from canada, so the currciculum might be a bit different

my math teacher is absolutely dogwater

he doesnt teach anything

he jsut writes questions and expects us to do it

lol

[tough dock]

Its alr, mine is also terrible

[minor quest]

im in math 9 enriched

so we learn like 1 unit in like3 days

rn we're doing imaginary number and polynomials combines

anyway what u need help with

[tough dock]

This

[minor quest]

first of all, its "whether", not "weather"

[tough dock]

I'ma math kid not an English kid mb

[minor quest]

ok

Given the exact wording of the prompt, the question may be searching for a contrast between inductive and deductive thinking.

and i see ur using newton law

if u r not sure, u can always ask ur teach

also, why do i feel like this is physics

[tough dock]

It's a geometry bruh moment

[minor quest]

mmm

If the question emphasized the idea of observing a specific instance (the ball going up) and making a generalization (it will come down), they might be looking for an inductive reasoning response.

[tough dock]

My teacher puts random shit that isn't related to standards on the tests

[minor quest]

my teacher asked us the janitor's name at the very end of a test...

[tough dock]

That was the entirety of the question

[minor quest]

and that was worth 30%

[minor quest]

ye ye ik

just pointing it out

[tough dock]

Yeah

But in general I feel like I don't agree with the teachers awnser on this

Because it isn't pattern recognition

[minor quest]

i dont agree with the treachers answer on this because it is not math

[tough dock]

And it's presented not in the context of a scenario

[minor quest]

and ur teacher is dogwater

[tough dock]

By that I mean that they don't say "a ball goes up and comes back down"

They say what goes up must come down

[minor quest]

what if its filled with helium😏

it'll float

[tough dock]

Which makes it sound like it's a law

[minor quest]

m

[tough dock]

And not a specific observation

[minor quest]

hang on

gimme 1 min

brb

gotta ask my question too

🙂

alr im back

[tough dock]

Alr

But do you think I should challenge the teacher on this

[minor quest]

yes

and challenge the teacher on giving questions that are not related

[tough dock]

Yeah I'm gonna double check the standards on that for my area

[minor quest]

this is what im learning

lol

pretty simple

[tough dock]

Oh is that precalc?

[minor quest]

...

nah

grade 9

[tough dock]

Oh wait

Nvm

[minor quest]

https://curriculum.gov.bc.ca/curriculum/mathematics/9/core

yeah

[tough dock]

That doesn't look complicated

[minor quest]

thats our curriculum

[minor quest]

its too easy

im built different

[tough dock]

Fr

[minor quest]

he also hands out these CYU problems

check ur understanding

[tough dock]

Wish me luck, my teacher is very stubborn

[minor quest]

he handed one ot today

gl

[tough dock]

She barely ever admits she's wrong

Even though she's wrong a lot

She just blames us

[minor quest]

this our new cyu

simple

as i said

the whiteboard

i feel like american curriculum is so different

like, its not even related

its clearly physics

she needs jesus

[tough dock]

W

T

F

[minor quest]

ye

[cedar kiln]

@tough dock Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[jolly swan]

To approximate the length of a lake, a surveyor starts at one end of the lake and walks 245 yards. He then turns 110 and walks 270 yards until he arrives at the other end of the lake. Approximately how long is the lake?

[jolly swan]

This is what I have so far

[ocean mural]

do u know law of cosines

[cedar kiln]

@jolly swan Has your question been resolved?

[jolly swan]

Yes

Thats why I labeled them

b^2 = a^2 + c^2 - 2(a)(b)cosc

Or something along the lines of that

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cedar kiln]

@faint yoke Has your question been resolved?

[cedar kiln]

Channel closed

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[cedar kiln]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[shrewd forge]

hey guys
is this true or false?

i try to prove X is finite iff dim is finite
and the direct part is easy since there now could be only a finitely many functions
but how to prove the converse part ?

.close

[cedar kiln]

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Available help channel!

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crimson sedge]

Hi, for question (a), I used the definition of Taylor Series

$f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(0)x^n}{n!}$

[wraith dagger]

XO

[crimson sedge]

For (b), am I expected to use this same formula?

And do the ratio test

Setting $a_n = \frac{f^{(n)}(0)x^n}{n!}$

[wraith dagger]

XO

[gritty galleon]

Yeah

[crimson sedge]

Using nth derivative, all that?

[gritty galleon]

Yeah ig

I thinku can find a formula here for the nth derivative

[crimson sedge]

Can you actually?

Well

f(0) = 1

f'(0) = 1

f''(0) = 2

f'''(0) = 6

So I would say that nth derivative at zero is given by n(n - 1)

[gritty galleon]

Check for the 4th too

[crimson sedge]

But that doesn't work for n = 0

,w derivative of 6/(1-x)^4

[wraith dagger]

Results provided by WolframAlpha

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Requested by hidingandseeking

[gritty galleon]

Wdym

[crimson sedge]

Doesn't work

For 4th

Derivative

,w derivative 2/(1-x)^3

[wraith dagger]

Results provided by WolframAlpha

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Requested by hidingandseeking

[gritty galleon]

Oh

[crimson sedge]

Are my derivatives correct

[gritty galleon]

Till second it's correct

[crimson sedge]

f^(3)(x) also correct right?