help-0

[rough crown]

so Gn would be the sum of (1+2+3+…+n) integers

[native nebula]

like 1 is an integer

then 2+3 is 5 which is another integer

likewise n terms will give n integers right?

[rough crown]

well yeah but to keep it simple, we don’t simplify and keep it as 1+2+3+4+5+… to make use of sum of integers

[native nebula]

yeah

[native nebula]

but the idea is this right?

[rough crown]

nono

we keep them separate

[native nebula]

then how n integers ?

[rough crown]

because adding 1+2+3+… is much easier than 1+5+20+…

[native nebula]

yeah

I got this

[native nebula]

but when you saying n integers you mean this one

right?

[rough crown]

Gn does have n terms, But it is the sum of n(n+1)/2 integers

take a look at each term

first term is just 1

second term is 2+3

third term is 4+5+6

[native nebula]

I got the terms one

[rough crown]

so in the 4th term, it is the next 4 integers and so on

[native nebula]

but i am asking you about the integers

[rough crown]

yes okay

now we are asked to find the sum

[native nebula]

first tell me what is the first integer of Gn

[rough crown]

its just 1

[native nebula]

then second

[rough crown]

it would be 2

then 3 and so on

[native nebula]

then how come it will have n integers ?

[rough crown]

it doesnt

there will be n(n+1)/2 integers

[ancient bridge]

hello

anybody can help me

[rough crown]

#❓how-to-get-help

[native nebula]

it's not a free channel

[rough crown]

this is occupied

okay so the first term has 1 integer, 2nd term has 2 int, 3rd term has 3 int

do u get that?

[native nebula]

yupp

[rough crown]

okay and so on the nth term will aslo have n int yes?

[native nebula]

how you get here?

[rough crown]

thats the total number of integers from All of the terms

num of int in first n terms = num of int in first term+ num of int in second term +… num of int in nth term= 1+2+…+n=n(n+1)/2

[native nebula]

do you think this question is complicated?

[rough crown]

its a bit tricky id say but not too much

[rough crown]

our answer would be the sum of the first n terms, which would be the sum of the first n(n+1)/2 integers

[native nebula]

hmm

understood

[rough crown]

u sure?

[native nebula]

yeah I think 😅

[rough crown]

you don sound sure lol

[native nebula]

actually I should scribe this in my notebook

then it will be into my head permanently

[rough crown]

maybe

try writing out a few of the first few terms

see that they have an increasing amount of integers

and try to sum them up

[native nebula]

yeah okh

but thanx for the help

i appreciate it

[rough crown]

well okay yeh gd luck w ur maths

[native nebula]

haha thanx buddy

.close

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[vale lichen]

It would become cos(infinity) which dne right?

[tawny condor]

Yeah, cos(-inf) to be precise

[vale lichen]

Ok great thanks

.close

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[echo socket]

Try completing the squares in x and y

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@fiery chasm Has your question been resolved?

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[alpine sable]

@eager parcel ask for help with !help

[alpine sable]

.close

[lone heart]

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[eager parcel]

But it’s just bots man

.close

[lone heart]

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[tulip gale]

Someone help me with this?

[tulip gale]

Sorry if its in swedish

[high rapids]

Vad betyder grundpotensform?

[tulip gale]

u swedish?

well obv

[high rapids]

Jag

Ja

[tulip gale]

vet du inte grundpotensform?

potenser

[high rapids]

Aha

"scientific notations"

[tulip gale]

ja fattar hur det blev 1,4

men inte 3

[high rapids]

Oki aså 14 * 10^2

Och sen 1,4 * 10 * 10^2

Därför 1,4 * 10^3

[tulip gale]

aha

okej tack ja fattar

.close

[lone heart]

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[prime dragon]

If something is compounded continuously what number would you put in the place of N

[mortal trellis]

what number? I mean that depends on what I wanna calculate

usually I would choose x for a continuous variable if that is what you mean

[lone heart]

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[alpine sable]

A

[alpine sable]

I need help

[high rapids]

Ask

[ivory pivot]

too small

try to type something

[alpine sable]

Okay

I'll type it down

Jeff is choosing a shrub and a rose tree for his garden. At the garden centre, there are 17 different types of shrubs and some rose trees
Jeff says, "There's 215 different ways to choose one shrub and one rose tree"
Could Jeff be correct?
You must show how you got your answer.

So like, total ways = 215
Let's say every shrub and tree has 5 ways of finding each other's characteristics and their looks.
So, 5x17x2 = 170?

[ivory pivot]

what a story

[alpine sable]

5 = Diff types of characteristics
17 = Total amount of a single object
2 = Both the shrubs and tree = 34

So 5x34

170?

<@&286206848099549185> (Sorry for the ping)

@fallen verge

Can you help?

[lone heart]

@alpine sable Has your question been resolved?

[fallen verge]

why did you ping me?

[alpine sable]

Geoffrey?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[valid island]

@alpine sable

I think u divide it by 17 and show its not a whole number

[lone heart]

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[alpine sable]

please may i have help with this equation

<@&286206848099549185>

im sorry for disturbing you

[pallid scarab]

to get the perimeter of the pond, you basically have to get every side of the right triangle

[alpine sable]

i have done that

and is apparently incorrect

so im doing something wrong apparently

do you know what the answer is?

[pallid scarab]

did you use trigonometry ?

[alpine sable]

yes

sohcahtoa

[pallid scarab]

so the total perimeter is (1 + cos(78°) + sin(78°)) * 1.95m

[alpine sable]

is that multiply by 1.95

[pallid scarab]

1.95m is the length of the hypothenuse, sin(78°)*1.95m that of the adjacent side, cos(78°) x 1.95m that of the opposite side.
So adding those 3 lengths you get the total perimeter

(also verify that your calculator is set in degrees and not radians)

[alpine sable]

gotcha

i was adding 1.95

[lone heart]

@proper mason Has your question been resolved?

[lone heart]

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[alpine sable]

Need help with the circled one

I add 4 plus 24 and I get 28 squared where do I go from there

[slate jolt]

28 =4*7

so sqrt(28)=?

[alpine sable]

No square root

Idk

[slate jolt]

sqrt(a*b) = sqrt(a)*sqrt(b)

[alpine sable]

I guess you can split it up?

7

?

Oh sh

I think I realized somethin

I'm very confused

[alpine sable]

Does it have to be 4 * 7? Can it be 14 * 2

Like this

[slate jolt]

yeah it can be 14*2

but sqrt(14) you dont know the value

its not 7

7²=49

[alpine sable]

Oh I'm dumb

Sorry

Ok ok

So factors of 28 that also have perfect squares

Is this correct?

[slate jolt]

yeah

but you can simplify

all these 2s

[alpine sable]

Yes I have to find the x's

Hmmm

How

[slate jolt]

factorize the numerator by 2

then you have 2*(.....)/2

which is just equal to whats inside the parenthesis

[alpine sable]

Is this correct

[slate jolt]

it is

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[trail meadow]

Two metrics d1 and d2 on X are equivalent if there exists positive constants a, b such that

a* d_1(x,y)<d_2(x,y)<b*d_1(x,y) for all x,y in X

[trail meadow]

Now, I know that this forms an equivalence relation but im asked to show that these define the same open sets in X. (That a set S is open with respect do d_1 iff it is open with respect to d_2)

So suppose S is an open set with respect to d_1 then for all x in S we have B_r(x) is a subset of S

with d_1(x,y) in B_r(x)

and I want to show that S is open with respect to d_2

So I assumed x is in S and I want to let r'=? such that d_2(x,y)<r'

[carmine reef]

ok so

S is an open set in d_1

x is in S

we have an open d_1 r-ball around x that's in S

right

[trail meadow]

yes

[carmine reef]

so d_1(x,y) < r

[trail meadow]

yes

[carmine reef]

now how can we bound d_2(x,y)

using that

[trail meadow]

I thought that we could bound it by d_2(x,y)<b*d_1(x,y)

[carmine reef]

yeah

[trail meadow]

so let r=b*d_1(x,y)?

[carmine reef]

hmm let me think

so we just want to show that there's an open d_2 r' ball around x for any d_2

so we need to find an r' such that the d_2 r' ball is entirely contained in the d_1 r ball

[trail meadow]

yes

oh

i have an idea

choose r'=b*r

then by definition d_1(x,y)<b*d_2(x,y)<b * r=r'

[carmine reef]

well i think the problem here

suppose d_1 and d_2 are the euclidean metric both

but suppose i'm really nefarious and don't tell you that

and worse, i give you a b value of 10000

now you choose r' = 10000

and now you're telling be a 10000-ball is a subset of a 1-ball

[trail meadow]

oh damn

yeah that would be nefarious

[carmine reef]

missing is i give you a ball of radius 1

so i think we need to go the other way

[trail meadow]

other way meaning...?

1/b?

or the other inequality

[carmine reef]

choose r' to be the smaller number

i think a

1/b is a decent candidate as well if a doesn't work but it should probably be a

if b works, a works, since a is always smaller than b

and we want to choose a small enough r'

[trail meadow]

but

a*d_1(x,y)<d_2(x,y)

[carmine reef]

oh

ha

maybe we do want 1/b

ok

d_2(x,y) < bd_1(x,y)

nah this calls for latex

$d_2(x,y) < b*d_1(x,y)$

[ocean seal]

monikanicity

[trail meadow]

we know that d_1(x,y)<r

[carmine reef]

right

$d_2(x,y) < b*d_1(x,y) < br$

[ocean seal]

monikanicity

[trail meadow]

but this may capture too many things

like in the example you gave above

[carmine reef]

ok let's think

[trail meadow]

if b=1000 then we are saying that a 1000r ball fits inside a ball of 1000

[carmine reef]

we want to find an open ball in d2 that fits in an r-ball of d1

we want to find a collection of points r' d2-units away from x that's a subset of a given collection of points r d1-units away from x

we want to make the condition of being r' d2-units away from x strictly stronger than the condition of being r d1-units from x

we want to show for any x,y, if d2(x,y) < r', then d1(x,y) < r

[trail meadow]

we need r' to be strictly smaller than or equal to r in order to be inside of it

[carmine reef]

d2(x,y) < b*d1(x,y)

[trail meadow]

damn this question is weird

[carmine reef]

not a necessary condition actually

let y be an element of a d1 r-ball around x

d1(x,y) < r

d2(x,y) < b*d1(x,y)

i keep losing it lol

d2(x,y) < r' needs to imply d1(x,y) < r

AHA MAYBE

d2(x,y) < r' and a*d1(x,y) < d2(x,y)

[trail meadow]

hmm

[carmine reef]

so d1(x,y) < (1/a)d2(x,y) < r/a

?

i give up

[trail meadow]

hmm

if r'=r/a

nah

thats a bound on d1(x,y)

damn this hard asf 😭

well hold on

we know that it forms an equivalence relation

so why can't we just say that there exists constants c,f such that

cd_2(x,y)<d_1(x,y)<f * d_2(x,y)

[carmine reef]

those constants would be 1/b and 1/a respectively

[trail meadow]

oh right

yeah

[carmine reef]

lol

mood

maybe r' is r/a

[trail meadow]

but how

[carmine reef]

screw it

1/b

ur right r' needs be small

it probably directly follows from

$\frac1bd_2(x,y)<d_1(x,y)$

[ocean seal]

monikanicity

[carmine reef]

then

let r' = r/b

if y is in our d2 ball

then $d_2(x,y) < r'$

[ocean seal]

monikanicity

[carmine reef]

$d_2(x,y) < \frac rb$

[ocean seal]

monikanicity

[carmine reef]

$bd_2(x,y) < r$

[trail meadow]

what

[carmine reef]

lmao

[ocean seal]

monikanicity

[carmine reef]

right

[trail meadow]

yes

okay wait

yeah

1/bd_2(x,y)<d_1(x,y)<r

so 1/bd_2(x,y)<r

[carmine reef]

we don't know d1(x,y) < r sadly

so

$d_2(x,y) < r'$

[ocean seal]

monikanicity

[carmine reef]

$kd_2(x,y) < kr'$

[ocean seal]

monikanicity

[carmine reef]

ok

i don't know why i'm doing this but yesss it's a at last my intuition squares with the problem

let r' = ra

let y be in a d2 r' ball

of x

$d_2(x,y) < r' \
d_2(x,y) < ra \
ad_1(x,y) < d_2(x,y) < ra\
ad_1(x,y) < ra\
d_1(x,y) < r$

[ocean seal]

monikanicity

[carmine reef]

thus choosing r' = ra works, as it ensues any point in a d2 r'-ball will also be in a d1 r-ball and thus in S

[lone heart]

@trail meadow Has your question been resolved?

[trail meadow]

wait

so you're going from d_2 to d_1 right?

like showing that if S is open with respect to d_2 then its open wrt to d_1

[carmine reef]

nah other way

I'm taking a d1-open set

Taking an arbitrary point in the set

claiming the existence of a d1 r-ball around it

finding a smaller d2 r'-ball

[trail meadow]

Oh so you're showing that B_r'(x) is a subset of B_r(x)?

[carmine reef]

since x is general this shows there's a d2-ball (in S) around x for every x

so S is d2-open

[carmine reef]

yeah

[trail meadow]

okay

and now the other direction 😭

[carmine reef]

nah

equivalence classes

we showed for any 2 elements in the same equivalence class, if S is open in one then it's open in the other

no other direction

[trail meadow]

wait but

Oh

they are related to eachother

but how do we know that the other direction will follow like that

i don't understand

[carmine reef]

well

[carmine reef]

you agree with this right

[trail meadow]

yes

[carmine reef]

so suppose A and B are equivalent metrics and S is open in A

Then S is open in B

Now

Suppose B and A are equivalent metrics and S is open in B

Then S is open in A

see how I just switched A and B

If (A,B) is a pair of elements in the same equivalence class

Then so is (B,A)

[trail meadow]

is it because of symmetry?

acc nah its just the equivalence class

[placid bramble]

help me

[carmine reef]

yeah

[carmine reef]

#❓how-to-get-help

[trail meadow]

if b is in the equivalence class of 'a' then 'a' is in the equivalence class of 'b'

[carmine reef]

yeah

every equivalence relation is symmetric

[carmine reef]

So in this sense yes

[trail meadow]

okay

thank you so much

[carmine reef]

no problem!

[trail meadow]

.close

[lone heart]

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[alpine sable]

am i correct

rate is my worst enemy

[carmine reef]

wow

you skipped the entire fraction step

somehow it's mostly good

only complaint is 70+72 isn't 144

[alpine sable]

OH SHIT

AHAHHAHHA

mybad

[alpine sable]

what is that

[carmine reef]

i have no idea what 4(18x) represents even

and the 800

were you copying off something else when you were working?

normally what u do is u say like

[slate jolt]

they just multiply both sides by 5 and 4 to get rid of the denominators

its valid

[carmine reef]

i know but i'm not sure they know

[slate jolt]

why wouldnt they

[carmine reef]

if they were copying off something else

[slate jolt]

yeah but you cant verify this

[carmine reef]

also cause they think rate is their worst enemy

bruh chill

i'm not running the inquisition here

[slate jolt]

yeah dw

[alpine sable]

i multiplied by the lcd

[slate jolt]

but i was wondering why do you find this suspicious

it just didnt occur to me so i wanted to know

[carmine reef]

mainly cause they said they were bad at rates lol

[alpine sable]

no?

.close

[lone heart]

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[carmine reef]

nice job

[carmine reef]

didn't mean to flame u or anything

i believe you

[lone heart]

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[crude jungle]

Someone pls verify if ok

2nd half

[somber escarp]

Correct

Now first half

[crude jungle]

i see i forgot to take the ln of e

i fixed it now

[somber escarp]

?

I thought you were working out what c should be

That's what i meant with 1st half (left part of the equasion)

[crude jungle]

both parts have to do with one another

i messed up the first pic

shouldvd of taken ln

so other pic needs to be done again

i understand it now tho

👍

[somber escarp]

I mean op not following you thoughts but im glad you figured it out haha

[crude jungle]

.close

[lone heart]

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[static breach]

Where wouldn't this be continuous?

[vale wigeon]

nowhere

[lone heart]

@static breach Has your question been resolved?

[lone heart]

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[night dome]

can anyone help me with this

[lone heart]

Please don't occupy multiple help channels.

[languid bolt]

just use one channel bro

#help-9

@night dome

[night dome]

sorry

[carmine reef]

can .close this one

:o other one gone nvm

wait no it's back lol

oh I'm dumb

[lone heart]

@night dome Has your question been resolved?

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[idle steppe]

hi

[idle steppe]

so i know the criteria for a subspace is:

zero vector has to belong
a+b is in the vector
scalar multiplication

my question is, since mu is any real number, that means that the zero vector doesnt exist right?

[slate jolt]

well µ isnt any real number

its the one you chose

but yeah for most choices

this set is not a vector space

[idle steppe]

since 0 is the only value of mu, i can just provide a counterexample that if mu is anything else 0 vector does not exist

and also, may i ask how u typed mu lol

[lone heart]

@idle steppe Has your question been resolved?

[lone heart]

@idle steppe Has your question been resolved?

[carmine reef]

μ

[lone heart]

@idle steppe Has your question been resolved?

[idle steppe]

<@&286206848099549185>

[pallid scarab]

All you have to do is prove mu = 0 is the only possible value of mu such that the set is a subspace.

To do that :
"Suppose lambda, mu are real numbers such that the set {(x,y)|y= lambdax + mu} is a subspace of R²"

"Then using [insert a certain property verified by subspaces], we get mu = 0"

Finally you do it the other way around to show that the values you got for lambda and mu work :
"Take any real number lambda.
Let us prove that {(x,y)| y = lambdax} is a subspace..."

If you need more help ping me @idle steppe

[idle steppe]

The inverse of addition property

[pallid scarab]

What is your question regarding that ?

"Inverse addition property" is for groups, not real subspaces. Because when you're working with a real subspace you've got -1 as a scalar... so "inverse of addition property" falls into scalar multiplication property

[mortal trellis]

ehh

the additive part of a vector space is defined as an abelian group. and then the vector space axioms tells us that scalar multiplication is supposed to be compatible with that

that -1*v = -v follows from that

[lone heart]

@idle steppe Has your question been resolved?

[lone heart]

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[wind silo]

How do you start with direct proof?

[wind silo]

Assume n to be an odd integer 2k+1?

But after that the first part works out to be k^2 + k + 1, and second to be 2k^3 + 3k^2 + 1.5k + 1

How is it possible to match them being equal

[harsh girder]

I think there is a printing error. It should be
$$\lceil \frac{n^2}{4} \rceil=\frac{n^2+3}{4}$$

[wind silo]

woah

if its printing error then no wonder

because i have no idea how cube = square

could i ask if my general direction is correct? for direct proof i assume the first part to be true hence let n = 2k+1 where k is any integers

afterwards i assume q to be true and sub 2k+1 into n in the second part

[golden canyon]

yes, just plug n = 2k+1 and simplify both sides

[wind silo]

thanks a lot! ill check to ensure its actually a printing error

[ocean seal]

秋水

[golden canyon]

I checked and the message above holds

with the square

[harsh girder]

let $n=2k+1$, it's not difficult to prove

[ocean seal]

秋水

[wind silo]

got it! thanks for the help, it should work with n^2 + 3 😃

was stuck trying to make cube = square for past 2 hours

.close

[lone heart]

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[formal ocean]

I’m confused with this

[formal ocean]

I know that the derivative of 12 will be 0

[golden canyon]

do you the derivative of ln(x)?

[runic trench]

1. sum rule
2. evaluate
3. differentiate using constant rule

[formal ocean]

1/x?

[golden canyon]

yup

[formal ocean]

So just multiply it?

2(1/x)

[golden canyon]

yes, the derivative of a constant times a function is the constant times the derivative of the function

d/dx(c * f(x)) = c * d/dx(f(x))

[formal ocean]

So it would be d

[golden canyon]

that's right

[formal ocean]

Okay thx

.close

[lone heart]

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[mighty musk]

how do you multiple the top part

I know the sqrt will cancel out

but will -4 * 4 make it -16?

[stable pecan]

Yes

[mighty musk]

ok

[rough crown]

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

[stable pecan]

To be faster you can always apply this identity

[mighty musk]

thx

[stable pecan]

.close

[lone heart]

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[patent swallow]

how can I make the intersection of a plane and a line 0

[patent swallow]

like not the point 0 but none

there is no intersection

I know nothing about the plane, I just have a line

[arctic mango]

Take any line on a plane parallel to and distinct from the original plane

[patent swallow]

I need the normal of the plane to intersect the line?

[patent swallow]

Normal of the plane . Director vector of the line = 0

right?

that means the line and the plane dont intersect?

well wait no... the line could be in the plane

[arctic mango]

Same technique. Take a plane P' parallel to the original plane P. Find the intersection point p of the line normal to P and the plane P'
Take a line in P' passing through p

[patent swallow]

uhh no idea tbh

[patent swallow]

I dont know what the plane is

I need to create it

[arctic mango]

Take the easiest case

Do you know of any plane?

[patent swallow]

x+b+y=1 for example?

I can create one that I like?

[arctic mango]

Is z = 0 a plane?

[arctic mango]

Think geometrically

[patent swallow]

why not

[arctic mango]

Yes, you can consider any plane whatsoever. First you think geometrically, feel free to draw it on paper

And then you'll realise that the answer you seek will appear right in front of you

All that will remain is to figure out the equations

[patent swallow]

nah I dont get it mate...

[arctic mango]

The normal to z = 0 is the line?

[patent swallow]

nvm

but I need to find a plane

[arctic mango]

What is a plane parallel to the plane z = 0?

[patent swallow]

I have a line and I need to find a plane that doesnt intersect with the line

[arctic mango]

Oh, I see. So you are given a line?

[patent swallow]

yeah 🙂

[arctic mango]

I apologise

[patent swallow]

Maybe I didnt explain myself correctly

[patent swallow]

nah bro chill it's me that I dont get a single thing of this lmao

[arctic mango]

It's just as simple if you think about it geometrically

Draw a line first

[patent swallow]

I got it

the line

[arctic mango]

Find a plane that the line is on

[patent swallow]

something like this

[arctic mango]

And repeat!

The "original plane" will simply be any plane which contains the given line

[patent swallow]

if the director of that line is (-2, -1, 1) then a plane could be
-2x - y + z = 0 ?

[arctic mango]

Does the director satisfy the equation?

First of all, what are you given?

[patent swallow]

the line!

[arctic mango]

What is the line exactly?

[patent swallow]

u(-2, -1, 1)

[arctic mango]

In what form is it given?

[arctic mango]

Passing through the origin?

[patent swallow]

parametric

[patent swallow]

yeah

[arctic mango]

Fair enough. That makes it easier

Okay, so you have this line. What is a plane that contains this line?

[patent swallow]

idk

tbh

gimme a sec

I'll think about this

-2x - y + z = 6 could it be this?

[arctic mango]

The origin doesn't satisfy this equation

[patent swallow]

so = 0

that way it satisfies the origin

[arctic mango]

Then the direction vector doesn't satisfy it

[patent swallow]

yerah

right

I'm just more confused now tbh

well

but

I need a plane that DOESNT intersect with the line

so if the line doesnt satisfy the plane

then we good

[arctic mango]

How do you know it doesn't intersect with the line still?

It can intersect at some other point than the direction vector and the origin

[patent swallow]

I would need to do
param(director) E/E Plane

ok so wait

[arctic mango]

You are basically given two points which describe a line.

[patent swallow]

yeah

[arctic mango]

So find a line parallel to this line
The plane containing this line can't intersect the original line, right?

Well, it actually can. The original way is what works

If I write P(x, y) = x(-2, -1, 1) + y(1, 0, 0) say?
Is this the parametric equation of a plane?
Does our given line lie on this plane?

[patent swallow]

wait

lets go to the begging

[arctic mango]

The way I would think about a plane is like av + bw, where v and w are vectors and a and b are scalars

[patent swallow]

I have a line, that doesn't need to be in a plane named X.

If I find the plane that contains the line, then I need to find a plane (X) that never intersects the plane where the line is contained

right?

[arctic mango]

Very good

[patent swallow]

ok so

[arctic mango]

That's the first approach

[patent swallow]

first step find the plane that contains the line

the line in this case is parameter(-2, -1, 1)

[arctic mango]

And there will be infinitely many of them. We just need to find one

[patent swallow]

a plane has the following form ax + by + cz = d

where a b and c are the normal of the plane

and d is the dot product of the normal times (I forgot what)

it shouldnt be difficult

😭

[arctic mango]

I also forgot

I just know the linear algebra way 😦

[patent swallow]

how is that way

[arctic mango]

That's actually all to do with vectors

[patent swallow]

I'm doing linalg too

[arctic mango]

Let me think for a bit

[patent swallow]

just read in the theory

for an intersection to be null

normal . director = 0

[arctic mango]

Okay. I have an idea

Let's use the vector way to go about it

The first thing we need to determine the plane passing through our line is to find a normal for it. Right?

[patent swallow]

for an intersection to be null

[arctic mango]

The normal of the line and the plane should be the same. So we need to find any unit vector normal to the line

What's a vector normal to the original line?

[patent swallow]

the whole problem here

is that idk what the plane is

I know how to do it with a plane

but idk without it

idk I just wont do it

cant spend 30 mins in the test figuring it out

[arctic mango]

It's fine. Even I'm not able to do it. I've become rusty on my "practical" skills

[lone heart]

@patent swallow Has your question been resolved?

[lone heart]

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[alpine sable]

This is what its asking for

[alpine sable]

43 is the problem I'm having trouble on

This is what I got so far, so I did rational zero theorem and descartes rule of signs but I dont know where to go from here

Should I do synthetic division?

[pallid scarab]

Are you trying to factor this ?

[alpine sable]

I dont know

If I should

[pallid scarab]

What's the goal of the exercice ?

Find x such that f(x) = 0 ?

[alpine sable]

I think so maybe

[pallid scarab]

So... whenever you have this for a polynomial, you're gonna want to factor by roots to simplify the problem.
Because A × B = 0 if and only if one of A or B is 0

[alpine sable]

This is what its asking for

[pallid scarab]

So, recall what the Rational Zero Theorem is

[alpine sable]

I'm sorry my phone is going to die I will brb

Ok I'm back

[pallid scarab]

It says : "If p/q is a rational root of a polynomial with integer coefficients, then..."

[alpine sable]

I could show you the answer if that could help

This is the answer for 43

[pallid scarab]

I know

[alpine sable]

Sorry

[pallid scarab]

I got them already

[pallid scarab]

Dw

[alpine sable]

I'm so confused an I'm scared

[pallid scarab]

Can you complete this statement of the Rational Zero Theorem ?

[alpine sable]

I dont know

[pallid scarab]

(You can note "a0" the non-zero constant coefficient, and "an" the coefficient of highest degree)

Like you write P(x) = a0 + a1•x +...+ an•x^n

With a0...an all integers and a0 is non-zero

Let me get this clear : do you know what the "Rational Zero Theorem" is ? If not, you might want to check your lessons

[alpine sable]

No I guess not

I just know its p/q with p being the number in the back and q being the number in the front

And then you find the factors of those numbers

Atleast I think

[pallid scarab]

p divides the number in the back
q divides the number in the front

In our example :
f(x) = x⁴ + 2x³ + x² - 12x + 8

So if p/q is a root, then p divides 8 and q divides 1

So the only rational roots possible are {1,-1,2,-2,4,-4,8,-8}

[alpine sable]

Yep makes sense

[pallid scarab]

So now, which roots work ?

[alpine sable]

So do I just plug those numbers into the polynomial

[pallid scarab]

And see if it gives 0

[alpine sable]

Ok

Is there a point to doing descartes rule of signs

On this one

[pallid scarab]

When you do find a root, immediately factor f(x) by (x-root) using polynomial division

Another very important thing to check is that, once you've found a root, check "how many times" you can factor by (x-root)

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Are you still here

I found that -1 works

[lone heart]

@alpine sable Has your question been resolved?

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Channel closed

Closed due to the original message being deleted

[raw latch]

dont occupy multiple channels please be patient

[weak iron]

im not

i closed them

[lone heart]

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[ionic dove]

2 numbers sum to a 100 and their product is 200this is what i got, i should factorize it into (y-something)(y-something) how do i get that something?

[vale wigeon]

...why did you write Zero as a word and not as a number?

[ionic dove]

☺

I like to do that

[vale wigeon]

you might be misunderstood if you keep doing that

anyway

you have the equation y^2 - 100y + 200 = 0

do you have access to completing the square or maybe even the quadratic formula

[ionic dove]

Idk what quadric formula is

Explain what i can do as if i was 13 because i am

[vale wigeon]

do you know how to complete the square

if not then now's the time to look that up

[ionic dove]

Idk y value

Do you have any sources

Wait

Do you mean solve for y and substitue

[gray isle]

solving for y will get you y yes

[ionic dove]

So i solve for y and substitute

[wary stream]

#❓how-to-get-help

[gray isle]

if your quadratic doesn't seem factorable and/or you can't be bothered,
quadratic formula won't fail

[ionic dove]

I dont know quadratic

[wary stream]

Because you're interrupting someone's channel

Hence why I linked #❓how-to-get-help

[ionic dove]

How have to temperarly occupy a chennel

So do i solve for y and substitute it

[gray isle]

Idk what quadric formula is

solving for y will get you y yes

[ionic dove]

Ok i will do it real quick

[lone heart]

@ionic dove Has your question been resolved?

[ionic dove]

Can you quadritic formula it for me

[gray isle]

try applying it yourself

identify your a,b,c values and plug

don't let the variable that isn't what you'd be normally used to intimate you

[lone heart]

@ionic dove Has your question been resolved?

[ionic dove]

I have learned quadratics in 20 minutes but i still cant apply it

I tryed solving for y and tryed quadratics but i dont know how, root 9200 is 20 root 23

How do i work with this

[carmine reef]

i think something goes wrong between what you wrote second to last and your circled answer

or no nvm

you just found 1 particular solution

[carmine reef]

you just want to solve for y?

[ionic dove]

.close

[lone heart]

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[static spoke]

answer is B

now close the channel kid

[fluid moat]

Who are you calling kid

Lol

[static spoke]

yogurt

lol

[alpine sable]

but he wants to know how

if you take miles walked over days, you'll get the rate of change

[static spoke]

just multiply ur answer of 1.2 times by 5 and u get 6 miles for every 5 days

fr

[alpine sable]

the fraction simplifies to 6/5 for each row, which implies that Mary walks 6 miles for every 5 days

and you did it right since 6/5 evaluated as a decimal is 1.2

[static spoke]

https://tenor.com/view/kittens-cute-cat-pet-cheeks-gif-16382546

[lone heart]

@alpine sable Has your question been resolved?

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[bleak galleon]

@honest pasture

Your question

[echo socket]

Usually it's helpful to solve for first couple of derivatives and noticing a pattern

f(x) = (x + 9)^(-1)
f'(x) = -1 * (x + y)^(-2)
f''(x) = 1 * 2 * (x + y)^(-3)
f'''(x) = -1 * 2 * 3 * (x + y)^(-4)
...

Do you see the pattern?

[bleak galleon]

yes

[bleak galleon]

@honest pasture

@alpine sable

[tawny condor]

$n! = n \cdot (n - 1) \cdot (n - 2) \cdot \text{ ... } \cdot 2 \cdot 1$

[ocean seal]

RedstonePlayz09

[tawny condor]

Do you know about factorials?

[bleak galleon]

Yes

[tawny condor]

You could use this to express the multiplication.

[bleak galleon]

3! = 321

fuck

[tawny condor]

yeah I get what u meant

XD

space it out

[bleak galleon]

XD

[tawny condor]

3 * 2 * 1

Ok, so what's the general formula you get?

[bleak galleon]

that's the problem for us

@tawny condor is this right

[tawny condor]

you forgot something there

and the exponent is incorrect

[bleak galleon]

Could you work it out please @tawny condor

So that we can have an idea

[tawny condor]

Yes you do... so why did you add n?

[tawny condor]

Try again.

Look

[tawny condor]

He already wrote down the pattern for you guys

I can't just give you the answer

[ancient meteor]

Yes

[bleak galleon]

Put that in de website and check if it works @honest pasture

[ancient meteor]

But still the answer is incomplete

[bleak galleon]

Help him please @ancient meteor

[ancient meteor]

Try to look at this and find what you didnt include in your formula

How can n be negative here .-.

n is the mumber of times you differentiate
How can you you differentiate negative number of times

Yea

Scrim has the right idea

[bleak galleon]

Wow @alpine sable

[ancient meteor]

But a bit incorrect

Ok ill give a hint

• appears when n is odd and doesnt appear when n is even

Now try to include it in the formula

How would that work ?

No

Ok another hint
Any negative number with an even exponent is positive and any negative number with an odd exponent is negative

[bleak galleon]

That's a fact

[distant pagoda]

Hi

just wanted to know

[empty mica]

I got this bit earlier

[distant pagoda]

@honest pasture have u done the part i dmed

like till that part?

[ancient meteor]

Not really wrong but just incomplete

[bleak galleon]

Send ur part here @distant pagoda

[distant pagoda]

ok

[bleak galleon]

R u maths professor @ancient meteor

[ancient meteor]

No

[distant pagoda]

just consider the numerators now
1,-1,2,-6
it's a gp

every term is being multiplied by
-1,-2,-3,-4
which is the ratio itself
and itself is in ap
-1,-2,-3,-4 is the ap

idk more

[bleak galleon]

It's not in gp

[distant pagoda]

i tried my best considering am in 10th grade 😭

bru

😭

[bleak galleon]

Arithmetic progression

Geometric progression

[distant pagoda]

s

[ancient meteor]

(!n)(x+9)^(-n-1)
So this is your formula so far
You are just missing how to include the negative signs for alternate differentiations

[bleak galleon]

Plus first one is not even in GP

[distant pagoda]

yeah same shit

[ancient meteor]

Yes

[bleak galleon]

Bye

Gn

[distant pagoda]

o

@ancient meteor can u give my solution a read

like it's a part of the solution

but yeah

[bleak galleon]

It's not even in gp tho

-1, 1,2,-6

[ancient meteor]

Yes

[bleak galleon]

put that in and check if it works

[distant pagoda]

hmm

idk

[ancient meteor]

Np

[distant pagoda]

10th class knowledge it is 💀

[bleak galleon]

GP means the common product should be same

Like 1,3,9,27

Etc

[ancient meteor]

I used to play
Stopped playing this year to focus more on studies

[distant pagoda]

yeah yeah

got it

just realised

[bleak galleon]

L

Play it @ancient meteor

[distant pagoda]

lol

[bleak galleon]

Lol

[ancient meteor]

F

also i understand hindi

[distant pagoda]

aur vaii

ki hal chal

it is agp my wrong

[stable pecan]

<@&268886789983436800>

<@&268886789983436800>

[sly mantle]

ty

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[ashen scaffold]

hello, can someone help me with: Calculate at what distance the gravitational force on a body with a mass of 5 kg is only half as great as on the earth's surface?

[teal lagoon]

hi

[ashen scaffold]

hello

can you see my question?

[teal lagoon]

no sorry

[ashen scaffold]

okay: hello, can someone help me with: Calculate at what distance the gravitational force on a body with a mass of 5 kg is only half as great as on the earth's surface?

u here?

[abstract fractal]

What have you tried

[ashen scaffold]

bro i have 0 clue, i thought of doing 5kg x 1/2 9,81m/s² but thats so dumb i guess.

[abstract fractal]

What's the force on a body with a mass of 5kg on earth's surface?

[ashen scaffold]

49,05 N?

[abstract fractal]

,calc 5 * 9.81

[ocean seal]

Result:

49.05

[abstract fractal]

Yes

What's half that?

[ashen scaffold]

24,52

24,53*

[abstract fractal]

So we need to find the distance that gives a force of 24.53 N

What's the equation for gravitational force?

[ashen scaffold]

one sec

is it the one with m1 x m2 on top

[abstract fractal]

Yes

[ashen scaffold]

F=G x m1 x m2 : r²

[abstract fractal]

Yep

[ashen scaffold]

ah okay.

[abstract fractal]

You know F, G, m1, and m2

So you just need to solve for r

[ashen scaffold]

so :r or x r

[abstract fractal]

Wdym

[ashen scaffold]

like switch i guess is the word

[abstract fractal]

Wdym

[ashen scaffold]

so its like r= m1 x m2...

[abstract fractal]

$F = \frac{GmM}{r^2}$

[ocean seal]

Gamer Dio

[abstract fractal]

Whats the first step to solve for r?

[ashen scaffold]

i am puzzled, either its :r or x r or root r

[abstract fractal]

There's and r² in the denominator

How do we get rid of it

[ashen scaffold]

:r²?

or just -r²?

ahhh wait multiply by r²

no

[abstract fractal]

Yes

[ashen scaffold]

oh

[abstract fractal]

Multiply both sides by r²

[ashen scaffold]

so F x r² = m1 x m2 x r²

?

x r²

[abstract fractal]

Where'd G go?

[ashen scaffold]

small g?

[abstract fractal]

G

Newtons gravitational constant

[ashen scaffold]

ah

F=G x r² = m1 x m2 x r²

like that?

[abstract fractal]

No

[abstract fractal]

This is the equation

Just multiply both sides by r²

[ashen scaffold]

F x r² = G x m1 x m2 x r²

?

[abstract fractal]

What happened to the r² in the denominator?

On the RHS

[ashen scaffold]

doesn't it just dissapear

[abstract fractal]

Yes, but then why are you still multiplying by r² on the right?

[ashen scaffold]

i don't know actually, english and physics at the same time is difficult

F x r² = G x m1 x m2 ?

[abstract fractal]

This is just math rn

And yes

So how do you isolate the r²

[ashen scaffold]

what does that mean

[abstract fractal]

We want to solve for r

Which means we need r by itself

Which means we need r² by itself

[ashen scaffold]

you want r to be alone`?

ah

[abstract fractal]

But right now we have F * r²

[ashen scaffold]

F x r² = G x m1 x m2 | : F?

[abstract fractal]

Yes

[ashen scaffold]

okay its r² = G x m1 x m2 : F?

then?