help-0

[pale lance]

Step 1: find dy/dx formula
Step 2: find dy/dx at (0,0)

Like Stephen said, you'll need to implicitly differentiate here

[steel dagger]

oh

why do u do it at 0,0

oh wait

origin

answers just 0

thanks

[long axle]

U solved it?

[steel dagger]

yeah

.close

[lone heart]

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[dim matrix]

How the highlighted part got evaluated?

[dim matrix]

I didn't understand how 4^20 - 3^20/4^20 is evaluated from 1-(-3/4)^20

[naive valley]

1 = 4^20 / 4^20

and also note that (-3/4)^20 = (3/4)^20

because the power is even

[dim matrix]

Okay got it thanks @naive valley @undone holly 👍

.close

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[alpine sable]

.open

[red vault]

a

[red vault]

how do i turn this into slope intercept form

[vale wigeon]

@red vault was 4y = 56 - 9x written by you?

[red vault]

yes

[vale wigeon]

well you're on the right track

how do you think we could now isolate y?

[red vault]

do i divide -9 and 56

for the next step

[vale wigeon]

what do you mean?

[red vault]

am I supposed to divide -9x and 56

[vale wigeon]

... ???

you are supposed to do something to both sides of your equation.

[red vault]

i dont know what to do

[vale wigeon]

if you had the equation 4z = 39 and needed to solve for z, what would you do?

[red vault]

39/4z

so i do 56/4y

is this right

[vale wigeon]

bad phrasing.

you would divide both sides by 4.

and you did not do it properly in your problem.

you should have gone from 4y = -9x + 56 to y = (-9x+56)/4, which would have simplified to y = **-(9/4)**x + 14

[red vault]

is this right

[vale wigeon]

yes

[red vault]

yay

.close

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[thick grail]

So the question wants me to find the contrapositive of this statement: If a and b are both positive and a > b, then 1/a < 1/b

[thick grail]

I was wondering if the negation also applied to the positive

[twilit dune]

Do you know what a contrapositive is?

[thick grail]

yes

you take the converse and

then take the negation of both sides

this is as far as I've got got

If 1/a greater than or equal to 1/b, then a or b

wait it should be negative

[twilit dune]

So you're confused how to negate the first condition?

[south mountain]

Yep, if the original statement is p -> q then it's contrapositive will be ~q -> ~p

[thick grail]

i see

[thick grail]

i was confused about how to negate the second condition

oops

sorry i meant

first

so basically the contrapositive would like something like this:

If 1/a greater than or equal 1/b, then a or b are both negative

[twilit dune]

there are usually many ways to write the negation of a statement

for example you could just attach "it is false that" at the beginning

[thick grail]

i see

but would my statement make logical sense

oh crap

[twilit dune]

you didn't correctly negate the first condition

[thick grail]

i made a mistake

[thick grail]

yeah

If 1/a greater than or equal to 1/b, then (a or b is negative) or (a smaller than or equal to b)

is this correct

[twilit dune]

oh right i didn't see that

well we could first make things easier by writing both the hypothesis and conclusion

hypothesis: a and b are both positive and a > b
conclusion: 1/a < 1/b

[thick grail]

yup

there are multiple ands

makes it kinda weird

[twilit dune]

you negated the conclusion correctly

but you didn't negate "a and b are both positive" correctly

numbers can be neither positive nor negative

[thick grail]

then what can they be

?

[twilit dune]

think about the real numbers, given some value of x how would you classify it as "positive" or "negative"

[thick grail]

we can classify it based on its position on the number line

[twilit dune]

sure, what about defining it using an inequality?

[thick grail]

x > 0 or x < 0

[twilit dune]

indeed, so for what value of x is it neither x>0 nor x<0

[thick grail]

i see

0

[twilit dune]

nonpositive would be the accurate term here

[thick grail]

so a and b can only be 0?

[twilit dune]

im not sure i understand your question

the negation of "a and b are positive" is "a or b are nonpositive"

[thick grail]

ah i see

i kinda get it

so basically it can negative or 0

so you would say nonpositive

[twilit dune]

the rest of your contrapositive statement was correct

[thick grail]

nice

[twilit dune]

indeed

[thick grail]

well thanks so much for your help

really appreciate it

[twilit dune]

you can close the channel using .close if you are done

[thick grail]

.close

[lone heart]

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[thorny moon]

how

[thorny moon]

what does the exponent do to the whole equation?

[pale lance]

Oh my

[thorny moon]

yeah

i have an exam tmr help

[pale lance]

Flips the entire fraction on its head

[wanton nova]

It flips the fraction

Yeah

[thorny moon]

yeahh eyah i know

but

does it affect the other exponents

ajkwhdksa

[pale lance]

All of them yeah

Since the -1 is outside the bracket

[thorny moon]

the possitives become negative?

[pale lance]

Correct

[thorny moon]

ahhhh i see

[pale lance]

Either

[thorny moon]

ok lemme try and solve

[thorny moon]

yesyes\

[pale lance]

You choose to flip the top and bottom OR you make them positive to negative

And vice versa

not both

[thorny moon]

WHAT

i thought

boith

ok

oh omg wait i think i got it

[pale lance]

Let's isolate just x^2 / x^-2 all to the power of -1

[pale lance]

Nice

[thorny moon]

im gonna solve and flip arounf the inside then flip it all at once when im done

[pale lance]

hm

The approach I'd do is just simplify the inside then yes

Then just put it all to the power of -1 at the end

That way you would avoid confusion

[thorny moon]

ya thats what i did

i got the answr 😏

thanks bbg

[pale lance]

Uh

[thorny moon]

how do u close this

[pale lance]

Nice

[thorny moon]

thing

[pale lance]

Here

.close

[thorny moon]

.close

[pale lance]

Like that

[lone heart]

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[thorny moon]

ez

[lone heart]

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[rocky tide]

How do we show ce - df + cf + de is nonzero if cd = 0, c + d is nonzero, ef = 0, and e + f is nonzero?

[tawny condor]

try to factor your expression

oh wait

hmm

[rocky tide]

ce - df + cf + de can't be factored.

[tawny condor]

yeah

didnt see that minus

also wont work anyways

[rocky tide]

So, I'm trying to figure out this problem

And trying to show that it is a binary operation

I did (c + di)(e + fi)

I got ce - df + (cf + de)i.

[tawny condor]

i see

[rocky tide]

But, I want to show that this element is in G.

[tawny condor]

ok let me try to see if I can solve this

[rocky tide]

kk

been stuck on this for an hour lol

(c + di) implies cd = 0 and c + d is nonzero right? and (e + fi) implies ef = 0 and e + f is nonzero.

so, how do we show ce - df + cf + de is nonzero under those assumptions?

you can't

[tawny condor]

isn't this just

cd = 0 means either c or d are zero

or both

but

c + d != 0 means

they are not both zero

so u can probably do this by case work

you have c + di in G, and e + fi in G

and u have 4 cases

c = 0, d != 0, e = 0, f != 0

c != 0, d = 0, e = 0, f != 0

c = 0, d != 0, e != 0, f = 0

c != 0, d = 0, e != 0, f = 0

[rocky tide]

I tried that too

Oh, wait nvm

I just did c = 0

ok lemme try that

wait how does c + d! = 0 mean they are both nonzero? How do you see that algebraically?

[tawny condor]

so first

for c + di to be in G, you need cd = 0

that means at least one of c, d are 0

[rocky tide]

Agreed

[tawny condor]

but if both are 0, then c + d is also 0

which isnt possible

[rocky tide]

Oh

So, contradiction?

[tawny condor]

well this is kinda a proof by contradiction ig

[rocky tide]

Yeah

[tawny condor]

and its really easy to see logically

both c and d cannot be equal to 0

this is true for any element in G

[rocky tide]

right, because that's the only criterion that fails is if c and d are both equal to 0

quick question, if c = 0 then d must not be zero, but could d be zero since d is a real number?

[tawny condor]

no d can't be 0 if you want the element to belong to G

[rocky tide]

failing the equation?

[tawny condor]

everything on the right side (after the |) must be satisfied

so if you assumed 2 elements in G,

then

c, d are real numbers

AND exactly one of c, d are 0

[rocky tide]

alright, because i was just looking at how c, d are in the reals, cd = 0, but if let c = 0, then d is not zero, but d is a real number, so I thought that would be a contradiction

d is defined as a real number

[tawny condor]

0 is a real number

[rocky tide]

yes, i know and d can be 0 but we said d can't be 0

[tawny condor]

d can be 0 if you only say that it is a real number

but if you constraint it further by saying cd = 0, c + d != 0

AND c = 0

then it can only be 0

whatever is after the | just tells you all the conditions that need to hold for an element to be in G

[rocky tide]

so, i should interpret it like if c = 0, then d must not be zero for c + di to be an element of G?

[tawny condor]

yes

exactly

[rocky tide]

but c and d can't be zero because 0 is not an element of G since 0 not equaling 0 is not a true statement . we are only concerned when we can get elements of G, correct?

[tawny condor]

I don't understand what you're asking

[rocky tide]

what I'm trying to say is that 0 is not an element of G because 0 + 0 = 0 but 0 not equaling 0 is not a true statement. Like you said, it has to pass all the conditions for it to be an element of G

[tawny condor]

0 is not an element of G, yes

0 = 0 + 0*i

0 * 0 = 0, and 0 + 0 = 0

[rocky tide]

yes, so we are only concerned with the numbers c and d that give us elements of G

and that's why we break it into 4 cases

[tawny condor]

yes

[rocky tide]

ok ty

.close

[lone heart]

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[tawny condor]

no problem!

[lone heart]

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[ornate condor]

i dont get what they are doing and how its relavent to saying every translation group in R^2 is isomoephic to Z x Z

or if its unrelated could someone explain why R^2 is isomorphic to Z x Z

[worn fox]

Can you show more of the page?

I'm pretty sure it is unrelated

[ornate condor]

http://abstract.ups.edu/aata/matrix-section-symmetry.html

[worn fox]

Yeah they don't prove the theorem

But the translation groups are things defined on a lattice in R²

And you can just pick generators of "up" and "right"

So if the translation sends you idk 5 up and 7 right, you encode that as (5,7) in ZxZ

Hope you see the isomorphism emerging

[ornate condor]

wait cant we go a decimal amount? like .5 up and 0 right

[worn fox]

Scroll up, they're defined on a lattice

[ornate condor]

i get that previously it said it has to be integer amounts of translation for lattices or wallpaper groups? but its just R2

hm so translation groups on R^2 is taken to be on a lattice?

[worn fox]

That's what they have done yes

[ornate condor]

oh

mayb just unclear

idk read a few proofs but they dont rly show the connection between stuff too

[worn fox]

Of course in general the space of all translation on R² is isomorphic to R² itself

[ornate condor]

ill just speed through :c

is ZxZ isomorphic to Z^2?

hm yea it shld b

[worn fox]

Yes

Two notations for same thing

[ornate condor]

oki

thanks

💕

.close

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[charred sequoia]

i only need help for number 3, i need help on how im supposed to get the values of the limits said with the use of the graph given. thank you for reading and i hope you guys can help me <33

[bleak pine]

For a limit of a function to to exist at point a, limit when x approaches a from the left must be same as the limit as x is approaching a from the right side, which must be equal to the value of the function at point a

You can see that is not the case on your graph for points -1 and 3

[charred sequoia]

so are they DNE?

[lone heart]

@charred sequoia Has your question been resolved?

[bleak pine]

Yes

[lone heart]

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[high ether]

if we have a complex eigen vector, is it's conjugate also an eigenvector?

[high ether]

and what's the proof

and are they always distinct, not a linear combination of each other, when it's complex and not real and not purely imaginary?

[worn fox]

Does your matrix have real entries?

[high ether]

yes

[worn fox]

Then consider conjugating both sides of the eigenvector equation

[high ether]

can it occur that we have a purely imaginary eigen vector if out matrix is real?

and diagonalizable

eigen values can be complex

[high ether]

you mean (A-lambdaI)V=0?

[worn fox]

The more simple one

Av=λv

[high ether]

and are they distinct always? Can't they be linear combination of each other?

[grave spoke]

im struggling with a (3x raised to 2 y raised to 3 ) (-5xy) =?

[worn fox]

!help

[lone heart]

Please read #❓how-to-get-help

[high ether]

Also this

[worn fox]

Don't literally come to channel 0

Youre asking a lot of questions before finishing one

[high ether]

You're right sorry

[worn fox]

They are not always distinct you might have a real eigenvector

[high ether]

yeah not interested in that case

complex ones

[worn fox]

Easier question: is the conjugate of a complex number some complex multiple of the number

[high ether]

is it?

[worn fox]

You tell me

[high ether]

i dont think so

except if it's purely complex

with no real parts

[high ether]

right?

no... 1+i and 1-i are conjugates and is a multiple of -i

so actually they all are

complex multiples of each other

[worn fox]

Yeah so think about a vector whose entries are all the same complex number

Eh wait is your scalar multiplication coming from R

[high ether]

No idts

[worn fox]

No its not dw

[worn fox]

So yeah this

[high ether]

Ah

so any vector whose entries are the same complex number

is a multiple of it's conjugate

[worn fox]

There are probably more

But yes they are

[high ether]

and can this occur in real matrices

diagonalizable ones

to have 2 eigenvectors that are a multiple of each other

and why not?

[worn fox]

Well try and find a real matrix that has eigenvector (i,i) or something

[high ether]

so that can occur ???

cause my prof just said it cant

if it's real and diagonalizable

[worn fox]

Idk, mess around and find out

[high ether]

he said for it to be real there always needs to be a complex eigen vector and its distinct conjugate

not sure why that's the case

.close

[lone heart]

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[ivory carbon]

how do you calculate the volume of b?

[ivory carbon]

oops

[ruby current]

calculate the volume of the prism outside and subtract the volume of the prism that is removed

[ivory carbon]

like this?

[ruby current]

why are you subtracting 1260 and then adding it back?

[ivory carbon]

oh

oops

so its just 15,120

[ruby current]

yes

[ivory carbon]

ok

[ruby current]

assuming the height is 18 (cannot see from the image)

[ivory carbon]

yes its 18

can you help me with a word question?

[ruby current]

ask anyway

[ruby current]

can you tell me the volume of a cone given its radius and height?

[ivory carbon]

uh

the volume is 400? 😅

[ruby current]

yes

i'm asking for a general formula for the volume given an arbitrary radius and height

[ivory carbon]

V = 3.14x^2h / 3

[lone heart]

@ivory carbon Has your question been resolved?

[analog falcon]

where x is the radius, and h is the height

so you know 2 variables here. just put in their values and find the third

[ivory carbon]

yes so would it be like this?

how would I get rid of the square root?

I don’t quite understand

[analog falcon]

there are methods of finding squareroots

google

[analog falcon]

but otherwise this looks perfect

[ivory carbon]

but how do I get rid of the x^2 in this equation?

do I divide the 400 and 26.69 by 3 first?

<@&286206848099549185>

[raven haven]

you're darn close

get x^2 by itself

then solve

[lone heart]

@ivory carbon Has your question been resolved?

[ivory carbon]

Ok

[lone heart]

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[final mural]

Hey

[final mural]

An=(7n)!/((7^n)*n!) mod 7

Simplest form of An that is true for every natural number n

I got stuck with even numbers n7+1 =An and uneven n7-1

[carmine reef]

,w (7n)!/((7^n)*n!) mod 7

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by monikanicity#0205

[final mural]

Yes, this thank you

Now what is the simplest form An

[carmine reef]

Do you know how to write a function that oscillates

[final mural]

Is it something like

A sin (wt something something, no 🥲

[carmine reef]

usually it's easier to use something like (-1)^n

[final mural]

Shit that might be it

n7+(-1^(n-1))

This looks right?

[carmine reef]

n7?

[final mural]

7*n

[carmine reef]

nah

that would keep increasing

[final mural]

Hmm

[carmine reef]

also for n=1 it's 8

[final mural]

What about

7n+(-1^(2n-1)) ?

[carmine reef]

6 is the same as -1 mod 7, does that help?

[final mural]

Oh damn yeah

[carmine reef]

Same problem as before, and now for odd n it gives imaginary values

or wait no

it doesn't

but that's just 7n-1

[final mural]

I tried this in maple and it works?

[final mural]

Should prob ping

[carmine reef]

are you taking mod 7 at the end?

[final mural]

Yeah

[carmine reef]

then just get rid of the 7n

[final mural]

Oh wow

That was it thank you!

.close

[lone heart]

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[carmine reef]

this is (-1)^n tho

not (-1)^(n-1)

[final mural]

Oh i meant n-1 my fault

[carmine reef]

?

[lone heart]

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[alpine sable]

Hello, I'm making a game and need to implement a ranked system, I'm planning on using TrueSkill2 but I'm having trouble understanding the math aspect.

[alpine sable]

Here's the link to the paper I'm reading: https://www.microsoft.com/en-us/research/uploads/prod/2018/03/trueskill2.pdf

[daring zenith]

this isnt rly a question 😭

[alpine sable]

Yes it is, let me elaborate!

[crystal mason]

it sure is, but very vast/expansive one

[alpine sable]

As a developer, I'm quite confused about this math.

Here's another paper I'm reading: https://www.researchgate.net/publication/306510969_S1_File/data/57bece2508ae2f5eb32e2b27/pone0149151s001.pdf

1 sec

[crystal mason]

game dev
windows

[alpine sable]

operating system preference isn't something i need to discuss with you.

please help me

[crystal mason]

it was just banter, windows gives great tools nowadays, especially for C# oriented stuff its a must

[alpine sable]

all good 🙂

Do you know somewhere that can help me?

[crystal mason]

probably the paper or googling more about it, do you want to know how to implement it or why this specific algorithm seems to work?

as in, all this explained

[alpine sable]

I know it's a tall order, but yes.

[crystal mason]

have you seen the stackoverflow example at page 3?
The graphs give a good hint how the deviation and variance were chosen

ah bruh it gets even worse

sorry but I don't feel confident walking you through that paper, perhaps you can try to find other resources on it first and then ask for specific subsets you don't understand?

[alpine sable]

Quite possible, the thing is that these papers are basically everything I've been able to find.

[crystal mason]

have you worked out trueskill1 yet?
in all honesty, trueskill 2 seems a bit... too hard for an algorithm based on inference from so many factors

[alpine sable]

I have not worked out trueskill1, do you think I should start there?

I'll make a new help request if I need help with a specific subset.

.close

[lone heart]

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[misty fable]

HHallo i need hellhelphelphellhelphelp 💗

[misty fable]

Ong

[daring zenith]

post q

[misty fable]

I don't get this :((#(_+£+£

[lone heart]

@misty fable Has your question been resolved?

[swift sky]

do you still need help?

[misty fable]

Yes

[near dove]

find their slope

of 2 given point

third point has to be on that line

[misty fable]

Noted, thankuuuu

[near dove]

answer should be -1

[misty fable]

Why is it not -3

[near dove]

when y dropped 2 , x dropped 4

oh

its 3

sorry

i thought its

(k,2)

[misty fable]

HAHWJWAJJAHAHA ITS OK

[near dove]

lmao

[misty fable]

Hamk u I figured this out

[near dove]

always the funny parts

[misty fable]

Can I ask again

NO MISCLICK

[near dove]

sure

[misty fable]

How to solve this pls

[near dove]

insert the values of points to equation

like for example, C

3.(6)-(--6) = 5

18-6 isnt equal to 5

[misty fable]

But

[near dove]

they are x and y values

on analytic board

[misty fable]

Letter a and b has the same uhh

Answer

[near dove]

cant really see them

[misty fable]

Oh hold on

Is it supposed to be -5 = 5

[near dove]

nope

[misty fable]

[near dove]

2nd one is true one

[misty fable]

Ohh okayokay

Thank u 💗

[near dove]

hope you enjoy math

[misty fable]

I want to learn more and I do enjoy it but my teacher is not that good at teaching ☹️

Wait no I don't mean that that's mean

[near dove]

its okay

i have the same problem

sometimes you dont match with correct people

also i think we should close this thread

[misty fable]

Oh okayokayyy

[near dove]

.close

[lone heart]

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[fickle grail]

How do I solve alpha and betha?

[gray isle]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[fickle grail]

I don't know where to begin

[gray isle]

h can be determine from applying
geometric mean of a right triangle
and/or properties similar triangles

after obtaining h, you can determine the angles using trig

[fickle grail]

I don't understand...

<@&286206848099549185>

My first try😬

[modern sedge]

2 equations 3 unknowns....

One more equation would fix it ;)

Maybe sum of angles in triangle?

[modern sedge]

Now maybe try removing h without computing it

I got here, now you have 2 equations and 2 unknowns

[modern sedge]

Okay, now try to substitute 90-beta to alfa, and expand tans

[modern sedge]

Try expanding tans

[fickle grail]

How do I expand tans?

[modern sedge]

sin(x)/cos(x)

that's definition of tan

If you are familiar with the geometry of tan, here is explanation of why you can do that

[fickle grail]

I don't understand it, can u do it with my exercise..? My English isn't good enough

[modern sedge]

Do you know this?

[fickle grail]

Yes

[modern sedge]

Okay, so when you expand it using that identity, you get this

[modern sedge]

Correct, do you know what sin(90-alfa) is?

[fickle grail]

Betha

?

[modern sedge]

Well that's also true, but it's also cos(alfa). sin(90-alfa)=cos(alfa)

Here is proof of what I just said

Do you understand the proof that sin(90-x)=cos(x)?

[fickle grail]

Yes

[modern sedge]

Ok, it can be also proved that cos(90-x)=sin(x)

When you add these identities to equation what do you get?

[modern sedge]

Correct, now try to get rid of the fractions, by multiplying both sides by sin(a) and cos(a)

[modern sedge]

Correct. Do you know this?

[fickle grail]

Yes

[modern sedge]

Okay so try replacing (cosx)^2 with

[fickle grail]

Can I do ÷2 before?

[modern sedge]

Yes, that's good

[modern sedge]

Great, can you think of next step?

[fickle grail]

• sin a ^2 maybe?😅

[modern sedge]

Yep

You get this, what would be next step?

x=alpha

[modern sedge]

Correct, what about next step?

[fickle grail]

÷5

[modern sedge]

Correct, try to solve it by yourself, and tell me when you are stuck

from now it should be pretty easy

[modern sedge]

Are you done?

[lone heart]

@fickle grail Has your question been resolved?

[fickle grail]

Sry, I had something elso to di

[modern sedge]

okay, great. It seems correct

Do you know arcsin()?

Just apply arcsin to both sides, and you get alpha=arcsin(1/sqrt(5)) plug into your calculator, and then remember beta=90-alfa

Alpha should be around 30 and betha around 60

I gtg so I cant help you with rest

[fickle grail]

Thank u so much:)

You are my hero of today:)

[fickle grail]

.close

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[remote bay]

The one in the middle

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@remote bay Has your question been resolved?

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[wind cloak]

Looks like a whole lot of AM-GM-HM to me

[lone heart]

@summer trail Has your question been resolved?

[rocky hatch]

<@&286206848099549185>

Can any1 help with i-Ready

[lone heart]

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[rocky hatch]

.reopen

[tacit arch]

!help

[lone heart]

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[alpine sable]

S1 : The points are given A(1, 0) , B(-2 , 4) , C(-1 , 4) , D(3, 5)

a) Represent the points in plan and write the equation of the lines AB, BC, CA , CD
b) Determine the distance from the points B and D to the right AC (in romanian we say Dreapta AC)
c) Compare the surfaces of the areas ABD , BCD and COD
d) If the point M (m , m+2) is collinear with B and C , calculate the surface area MAD

S3
The points are given A(sin^2a , cos^2a) , same with B and C but it is sin and cos of b and c

b) show that for each a,b,c belongs (it's an E symbol) to R , points A B C is on a right

S4
The points are given A(2,m) , B(m + 1, m) C(1 , 2)
b) Determine m ∈ ℝ so that the surface area ABC is 1

S5
The points A(m, 2m -1) , B(m+1, -m+2) are considered. For what values of m the equality Area of OAB = 23/2 occurs

S7
Determine m ∈ ℝ so that the points A(1, 1) to be at the distance 3 from right BC where B ( 0, 2 - 6m/1-m), C(1, 7m - 1 /m-1)

T1 ex 2
the matrix a is given as
( 2 -1 3)
(-1 4 -5)
(4 -2 6)

e) evolve from the first column after two zeros have been obtained

T1 ex 4 :
The points A(2m + 1 , 3) , B(1, m) and C(-4, 2) are collinear if m = ....

[tacit arch]

wtf you want someone to do your entire homework for you

pick one problem, attempt it, show your work here if you get stuck

[modern topaz]

[alpine sable]

that's like one quarter of the entier homework that i have to do

[tacit arch]

yea that doesn't change anything

[tacit arch]

.

[alpine sable]

solve the S5 if you want

[tacit arch]

did you attempt it?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

yes most of the exercises i knew how to

but some of them i don't know how to solve

i currently written about 17 pages

can you help me with only S5?

[tacit arch]

can you show your attempt

a drawing should be your first step

[alpine sable]

yes one sec

idk where to put m 2m-1 or so in a xoy axis

i just suggested that o is 0 0 but that is incorrect

i know that's why i didn't provided a attempt

as i don't know how to

[tacit arch]

you should not assume either

just use this formula

your z coordinates on A and B can be 0

looks like you forgot a factor of 1/2

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[tacit sequoia]

Can someone help me solve this problem

[lone heart]

@tacit sequoia Has your question been resolved?

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[urban pendant]

help

[urban pendant]

im not sure how to do this, the only thing ive done so far is found the cross product of those two vectors which is <-64, -24, -56>

but im not sure if i can do anything with that

o wait i figured it out

[royal plank]

thats half the work for transforming vector form to normal equation

ah, ok

[urban pendant]

it gave me 8x so i just scaled vector down and plugged in 8(-4) + 3(-11) + 7(7) = -16

!close

!solved

idk how to close

!done

[royal plank]

its .close

.close

[lone heart]

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[molten pivot]

take the cross product of direction vectors and normalize such that first component is 8

[lone heart]

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[verbal owl]

would like to start with an explanation of the solution to 4

[rare gale]

/status

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[verbal owl]

1. on some but just got an idea so gonna try that on a few

[rare gale]

are you familiar with implicit differentiation?

[verbal owl]

i think 4. is A

yeah

[rare gale]

A should be correct

[verbal owl]

okay, lemme try and solve another one

i think 1. is D

@rare gale

3. is B

[molten pivot]

6

[lone heart]

@verbal owl Has your question been resolved?

[verbal owl]

am i doing this implicit differentiation right?

@rare gale

okay, i got it

[lone heart]

@verbal owl Has your question been resolved?

[rustic coral]

.close

[lone heart]

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[rustic coral]

✅

Typo?

✅

Yup 31 and 28

[lone heart]

@alpine sable Has your question been resolved?

[flint pecan]

Yes

Also yes

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[toxic radish]

How do I isolate the shapes

theres like this irregular shape and I don't know how to split it up into one I can solve area

[pseudo ice]

Have you managed to do either parts so far?

for part b, note that the unshaded area[s] are relatively easy to work out the areas of

[toxic radish]

i don't know like the strategy

we're learning circles right now so I assume it has something to do with a circle somewhere in here

a circle that fits in the box would have an d of 10

i dunno if that has to do with the problem though

[pseudo ice]

diameter?

[toxic radish]

yep

[pseudo ice]

Hmmm, well that's technically correct, but...

Notice how the "petals" look curved like a circle...

[toxic radish]

oh i see

would you have to stretch it out a bit to become like a circle

[pseudo ice]

Well you could, or just notice that it's "a bit" of a circle

[toxic radish]

if it's a bit of a circle what would I do next

[pseudo ice]

...I was hoping you'd tell me how much "a bit" of a circle it is

[if you can answer that, you'll have a better idea of how to work stuff out ]

[toxic radish]

i notice like the middle of it you can fit a circle into

[pseudo ice]

Hmmm, well again, technically correct, though not what I was looking for

Think along the lines of semicircles...

[toxic radish]

if you split it along the middle of the petal it creates 2 semicircles?

[pseudo ice]

Hmmm, not quite [I think I'm not explaining myself well tbf, that's on me]

Notice how this blue is a quarter of a circle?

[poor shading, but that's supposed to include the bottom unshaded bit too]

[toxic radish]

so the 1/4 of a circle is the circle that fits into the 10cmx10cm box ?

[pseudo ice]

Basically that's the idea to use, you effectively make 1/4 of a circle from those petal curves

From which things become much easier to find

[toxic radish]

how can I determine that it's 1/4 of the circle

like what information would give me that the petal is 1/4 of a circle

[pseudo ice]

Well notice how e.g. the radius of it would be 10, and there are right angles, implying 90/360 of a whole circle

[right angles from being in a square]

[toxic radish]

ohhh okay so]

the square

creates a right angle

so 90/360

4 squares create 360

ok so

now i know the square = 1/4 of a circle with a radius of 10

riht

[pseudo ice]

Erm, well

[pseudo ice]

You know you have a square, and inside of that you basically have a quarter circle

Of which you can work out the perimeter of a quarter circle quite easily

[toxic radish]

ohh i see the quarter circle now]

its the blue + the petal

right

[pseudo ice]

Yea, my drawing skills were bad but basically those combined yep!

[toxic radish]

I see, so I'll find circumference and then divide by 4 to find the length of the arc

[pseudo ice]

Yep, and that will give you one side of the arc, but remember that there are two of them

[toxic radish]

for radius I get 14.14

sounds reasonable

i think

ok ill plug into formula

[pseudo ice]

Wait how?

[toxic radish]

OH WAIT

i did it wrong 😭

i t hought i had to do pythagorean

i forgot it was a quarter circle

so 10 would be radius

i get 10 pi

for the distance of the shaded part

[pseudo ice]

The total perimeter, yea?

[toxic radish]

yep

[pseudo ice]

Perfect

[toxic radish]

so for the area

what would the strategy be

[pseudo ice]

One good way, in my opinion, is to work out the unshaded areas, as those are easier to find

[pseudo ice]

So if you look at my poor drawing here

You can work out the blue area, and the total area of the square

So you can find the unshaded area easily

[toxic radish]

i see

i will do that noow

i get 57.1cm^2

for the shaded area

[pseudo ice]

Yep I agree with you

[toxic radish]

omg thank you for helping me

🙏

[pseudo ice]

It's a pleasure have a good one!

[toxic radish]

thank you!

.close

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[elder juniper]

hi 😄

[elder juniper]

i need help

so i think its (x, 1/2X)

[wary stream]

Plug in the values and test it

[elder juniper]

it seems so

thats what i did

x is half of the value of Y

but i dont know

is it x*2

or X1/2

[thorny root]

nope

[elder juniper]

what do u think then???

[thorny root]

your solution is correct

[elder juniper]

W

thanks so much

yall a big W

ong

[thorny root]

ya too

[elder juniper]

is this also

[wary stream]

Tbh I was trying to get them to check their work just plugging in the values. It's a good test strategy

[elder juniper]

i did thats how i got the answer!!!!

[elder juniper]

ur W bro thanks 😄

[elder juniper]

is this correct?

[thorny root]

yeah I felt like it was okay to answer once they stated they actually worked on the problem

[elder juniper]

yall such a W

[wary stream]

What's the pattern for the x values in the table?

[elder juniper]

W community

-2

so when i go backwards

its +2

[wary stream]

Good

[elder juniper]

so i got the ans like that

[wary stream]

Exactly

[elder juniper]

😄

W

the question is

if person worked 30 hours

whats pay

every 2 hours i see increase of 50 bucks

so 30/2=15

then i did

15*50 to get ans

is it correct

?

[wary stream]

Sounds good

[elder juniper]

😄

[wary stream]

Overall you are doing good, I suggest just to go with your first inclination

[elder juniper]

😄

i have 2-3 more

and im out

https://tenor.com/view/peace-out-gif-22295199

alr

[carmine reef]

what about the squares made up of smaller squares

[lone heart]

@elder juniper Has your question been resolved?

[lone heart]

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[dapper echo]

Do you guys have any proper study techniques for college? I'm in precalc and I really am struggling with mastering the factoring?

[dapper echo]

.close

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[proud patrol]

<@&286206848099549185> ive been waiting 15min in a another chat but got closed due to no repsonsie

[pliant cedar]

its a regular hexagon

all the internal angles are equal

do u know how to find the sum of interior angles of an n-sided polygon?

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@proud patrol Has your question been resolved?

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[little nova]

Q - v by substitution method

[pliant cedar]

,rotate

[pliant cedar]

u want to solve 1. (v) by substitution?

[little nova]

Yup

[pliant cedar]

right, can u find x using the first equation?

[little nova]

Yeah but Im getting wrong x

[pliant cedar]

what do u get for x

[little nova]

It is supposed to be 0

Wait

[pliant cedar]

im not saying find a value for x, i mean make x the subject of the first equation

[pliant cedar]

like move everything other than x to one side

[pliant cedar]

right

now put this for x in the second one

oops wait

u put the square root in the wrong place

[little nova]

Yeah its fine

[pliant cedar]

it must be y*sqrt(3)/sqrt(2)

[little nova]

I noticed

[pliant cedar]

ok

u do get x = 0

and y = 0

[little nova]

🫠

@pliant cedar

Ohh no

[pliant cedar]

u cant cancel the sqrt2 like that

[little nova]

√3x√3 is 9

[pliant cedar]

no

neither is it sqrt(6)

[little nova]

√9

Thats 3

[pliant cedar]

yes

[proud patrol]

no

[pliant cedar]

well this is no longer ur channel

[proud patrol]

sox

soz

[little nova]

Y = √2

[pliant cedar]

,rotate

[pliant cedar]

yeah -y/sqrt(2) = 0

multiply both sides by sqrt(2)

[little nova]

Bro why I keep forgetting small things like that today

Btw thanks

[little nova]

Thank you

.close

[lone heart]

Channel closed

Closed by @little nova

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[ebon obsidian]

hey can anyone help me wioth a quick absolute ineqaulity question

[naive valley]

one way to solve, |x| < a is true if and only if x^2 < a^2

[ebon obsidian]

i just approached it as like 2 cases

one postive and one negative

but the negative case gave me the wrong answer

[naive valley]

what was that case, can you show it?

[ebon obsidian]

yea let me screen shot it

Here

not sure if this is a valid approach

but the answer was suppose to have the x < 17/7

x>19/5

[naive valley]

you have to be careful when multiplying both sides by 2x - 6, the inequality will change from > to < if 2x - 6 is negative

[ebon obsidian]

but in this case wasnt the 2x-6 postive

like after i first moved the - to the right i changed signs

and then i moved the postive (2x-6) to the right

[naive valley]

how do you know if it was positive or not

there are actually four cases to consider

[ebon obsidian]

ohh

[naive valley]

x + 1 is positive or negative

and 2x - 6 is positive or negative

four possible combinations

some combinations will turn out to be impossible

[ebon obsidian]

so should i just

[naive valley]

but you have to check which ones

[ebon obsidian]

not the the ❤️

< 3

but onlyu change the signs

for the first equation

[naive valley]

might be easier if you do the cross multiplication first:

|x+1| < 3|2x - 6|

that's definitely ok since |2x - 6| is positive (because of the abs val)

now if x+1 is positive and 2x -6 is positive then this becomes x+1 < 3(2x-6)

and there are three other cases

try all four cases

[ebon obsidian]

okok

[naive valley]

each one will have no abs vals

and you can just solve each one

[ebon obsidian]

let me give it a shot

tyty

.close

[lone heart]

Channel closed

Closed by @ebon obsidian

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[high phoenix]

hi

[high phoenix]

anyone able to help me rn

[south mountain]

Sure

[high phoenix]

ok

can you do geomtry/ trig?

[south mountain]

Yep

[high phoenix]

ok

1 sec

ill send a pic of the problem im working on

[south mountain]

Alright take your time

[high phoenix]

any idea ?

[south mountain]

Give me a minute

[high phoenix]

Ok

[south mountain]

,rotate

[wanton pebble]

20 is wrong

[south mountain]

Yep it's supposed to be root of 441

Which is 21

[wanton pebble]

i think your calculation are wrong

approach is right

[south mountain]

35² is 1225

And 28² is 784

We subtract them and get 441

So instead of 20 it should be 21

[high phoenix]

Got it

[south mountain]

I'm getting 114

Can you verify the answer

[high phoenix]

What

Very confused

[south mountain]

I got the perimeter aa 114

As 114*

[high phoenix]

Can you tell me how you got that

[south mountain]

I have a figure