help-23

[quaint spear]

yes

not

area

OOPS

oh my god

😭

[cedar rover]

don't worry lol

[quaint spear]

THE AMOUNT**

[cedar rover]

yeah, those are the functions we are working with

one of them is the composition of the other two

can you tell which?

[quaint spear]

the discount and the tax

?

[cedar rover]

Yeah those are what we use to compose A(x)

so A(x) will just be our result, let's leave it as it is for the moment

[quaint spear]

ohok

ok

[cedar rover]

let's take it logically

1. I "choose" how much I want to spend
2. I get a discount
3. the discount gets taxed
   and we get our A(x)

Let's give some names to the discount and tax functions, shall we?

[quaint spear]

ok

D and T

[cedar rover]

great names

[quaint spear]

LOL

[cedar rover]

i'm not joking lol

Let's start with D(x), it should return x - (20% of x)

[quaint spear]

ok 😭

ok

[cedar rover]

how can we write this?

[quaint spear]

D(x) = (0.2)x

[cedar rover]

That's super close good job

[quaint spear]

aw what was it

[cedar rover]

0.2x returns 20 % of x

but we need to take that from x

since it's a discount

[quaint spear]

oh so x - 0.2(x)

[cedar rover]

Yup!

[quaint spear]

ohhhhhhokok

[cedar rover]

So
D(x) = 0.8(x)

[quaint spear]

oh

you're right

[cedar rover]

you can also think of it as: I have to pay x money, but i get 20% discounted off, so I just have to pay 80% of what i was meant to (which is x)

[quaint spear]

i shouldve thought of that

yeah

ur so smart okok i get that

so now we do

tax?

[cedar rover]

go ahead ^^

[quaint spear]

UHMMMM

T(d)= D + 0.0625D

[cedar rover]

what do you mean by D?

are you referring to a variable or to the discount function?

[quaint spear]

the

output from the d function we just made

[cedar rover]

putting together the functions is the last thing, that's when we find A(x) (in fact that's what you just did, good job on that)

But let's find just how much we will have to pay with the tax

so
T(x), where x is the money we want to tax

[quaint spear]

oh ok

t(x) = x+ 0.0625x

[cedar rover]

Yes exactly what I meant

Let's just write it as
T(x) = 1.0625(x)

but that's the same

[quaint spear]

ohhh ur right

thats so smart

[cedar rover]

let's think about how we are supposed to get A(x), and then compose it

[quaint spear]

ok

A(d(t(x))) ????????/

[cedar rover]

Oh I'm sorry I choose a poor wording lol

I mean, let's use D(x) and T(x) to find A(x)

so we will have
A(x) = ...

another little thing: remember, inner first

[cedar rover]

if this was A(x) = d(t(x)) this meant:

1. Take x
2. Tax it
3. Discount it

[quaint spear]

ok

[cedar rover]

2 and 3 are swapped

[quaint spear]

oh ok

d(x) + t(x)

sorry im replying slow im just looking thru notes and stuff trying to process everything 😭

[cedar rover]

this means that you

1. find how much money we will have after the discount
2. find how much money will be left after the tax
3. sum them

[cedar rover]

Don't worry

let's work it out together

first of all we have to have some money

x

then let's get the amount of money we have to pay with the discount

D(x)

then let's tax it

T(D(x))

and we are done

[quaint spear]

ohh t(d(x))

[cedar rover]

A(x) = T o D = T(D(x))

let's find the value of A(x)

[quaint spear]

ok

[cedar rover]

literally plug d(x) in t(x)

as it is

[quaint spear]

t(0.8x)

?

wait

no

[cedar rover]

YES!

[quaint spear]

1.0625(0.8x)

?

wai

t

[cedar rover]

Exactly!

[quaint spear]

REALLY

????????????????????

[cedar rover]

Yeah u got it

[quaint spear]

SO

0.85!!!!!!!!

[cedar rover]

A(x) = 1.0625(0.8x)

[quaint spear]

yayyy

ur actually so good at explaining stuff

you made it a lot more processable

do u have to go

[cedar rover]

not really

Do you have more questions?

(As a rule of thumb: divide problems like this in smaller parts, it will become way easier)

[quaint spear]

okok

yeah

theres this one

[cedar rover]

ok

Can you show your work?

[quaint spear]

i have none

😭

[cedar rover]

Ahah don't worry

What would you try in this case?

[quaint spear]

uhmmmmmmmmmmmm

well it says to find the lcd

so i suppose find the lcd

[cedar rover]

yes

[quaint spear]

okok

ITS

x(x-2

0

oops

x(x-2)

[cedar rover]

yes

[quaint spear]

ACTUALLY

OK

what now

multiply by 4??

[cedar rover]

Wait

[quaint spear]

what

[cedar rover]

what equation do you have?

[quaint spear]

well

x(x-2)=1/4

[cedar rover]

you forgot the nominator of your fraction

[quaint spear]

1/x(x-2) = 1/4

[cedar rover]

$\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$

[flat frigate]

Nonna

[quaint spear]

woah..

[cedar rover]

$\frac{1}{x}+\frac{1}{x-2}=\frac{1}{4}$

[flat frigate]

Nonna

[quaint spear]

yeah

so what do i do now

[cedar rover]

$\frac{(1)(x-2) + (1)(x)}{x(x-2)}=\frac{1}{4}$

[flat frigate]

Nonna

[cedar rover]

$\frac{2x-2}{x(x-2)}=\frac{1}{4}$

[flat frigate]

Nonna

[quaint spear]

okay

[cedar rover]

Do you agree?

[quaint spear]

i get it so far

yep

do i multiply by 4 now

[cedar rover]

If you want to

[quaint spear]

what do u mean

whats the right thing to do

[cedar rover]

there isn't a right thing to do lol

What equation do you have now?

[quaint spear]

umm

4(2x-2) / x(x-2) = 1

8x - 8 = x(x-2)

8x - 8 = x^2 -2x

x^2 - 10x +8

?

????????

???????????????//

[cedar rover]

yes that's good

[quaint spear]

is that the right answer

[cedar rover]

.

[quaint spear]

x^2 - 10x + 8

= 0

[cedar rover]

did you type x² - bx + c = 0 or was is there?

[quaint spear]

i typed it

after deleting it

but i get why its there

[cedar rover]

I'm not sure what's meant by "replace the values of b and c to create the equation"

[quaint spear]

v

it wanted me to write it in standard form

so it gave me a formula for standarm form

and i just had to replace the values

[cedar rover]

was it x² - bx + c?

i'm not sure at all about the -b

[quaint spear]

wdym

yeah

it was what u typed

what u saw in the screenshot

was what it was

[cedar rover]

we found a = 1, b = -10, c = 8

so x² - bx + c = 0

should be
x² +10x + 8 = 0

I'm probably wrong lol

[quaint spear]

okok its fine

can u tell me if thats right

[cedar rover]

yes that's right

[quaint spear]

is this right? if not which one did i get wrong

[cedar rover]

Yes it's right

[quaint spear]

which one is it i wanna figure this one out by myself

[cedar rover]

What's an inverse function?

[quaint spear]

like the inverse of its values idk how to explain

ts ok

hold on

[cedar rover]

It's fine, explain the concept

I don't care about the rigorous definition

[quaint spear]

its ok i figured it out im better w inverses

did i get it right

if not which one i wanna figure it out LOL i wanna have a thought process like u

break it down

[cedar rover]

Wait

Which one are we doing

[quaint spear]

the one i just sent

[cedar rover]

ok

f(x) is just the sqrt function

are you familiar with its shape, right

?

[quaint spear]

which one is it tho

i am

the parent

which one is it and ill try to talk it thru

[cedar rover]

let's do it properly

[quaint spear]

i gotta get off disc soon tho 😭

[cedar rover]

ok

Which transformations were used?

@quaint spear ?

[quaint spear]

hmm

-3

so

down 3

[cedar rover]

Yeah

that leaves us with A and C

-f(x) flips a function upside down

f(-x) flips it horizontally

So in this case?

[quaint spear]

-f(x)

[cedar rover]

Yes

[quaint spear]

ohh

its C

[cedar rover]

Yup

[quaint spear]

okok i get it

u should be a teacher

i gtg eat but i feel sm better

[cedar rover]

Bye

[cedar rover]

Lol i'm bad

[quaint spear]

no i like u u explain stuff nice and break it down

okok

[cedar rover]

ok bye

[quaint spear]

ill prob need help tmr LOL

.close

[safe radish]

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[lean otter]

Need help finding value of X

[lean otter]

#help-23

What have you try

Did you try some identities of trigonometry? Like soh cah toa

No I haven’t i need X in order to figure out sine a

I think you can use tangent to figure out x

Am i right?

Tangent is opposite/adjacent right?

Yes

[shadow glade]

$\tan 24= \frac{x}{19}$

[flat frigate]

Sooshon

[shadow glade]

$x=19\tan24$

[flat frigate]

Sooshon

[shadow glade]

and make sure your calculator is in degrees mode not radians

[lean otter]

Ohh will the answer be 14.8 I rounded it up

[timber nova]

8.46

[lean otter]

Bowed u get that?

[timber nova]

tan(24) is 0.44522868530853616392

multiply that with 19

ypu get 8.46 rounded

[lean otter]

Ohhh Alr thanks

[safe radish]

@lean otter Has your question been resolved?

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[jaunty oasis]

hi i need some help with this question so i got 0.8bt=v but it aint on the answer choices xd

[safe radish]

@jaunty oasis Has your question been resolved?

[jaunty oasis]

<@&286206848099549185>

thanks in advance xd

my work:

1 container = 200mL/15mins
1 container = 0.8L/hr
B containers = 0.8B L/hr
B containers = 0.8Bt L/t hrs

Therefore 0.8*Bt L = v L
0.8Bt = v
B=1.25v/t

forgot to send this sry xd

[stray socket]

Seems linear

You have a consistent slope of $\frac{0.2L}{0.25hr}$

Time is in hours

So if we do a bit of analysis

$$B = \frac{0.2v}{0.25hr}thr$$
$$\= 1.25vt$$

@jaunty oasis

Lmk if you have any issues

[jaunty oasis]

should be 0.2L?

[stray socket]

Also since $v = L$ I just replaced the L with v

[flat frigate]

Umbraleviathan

[stray socket]

Ope

Yeah my bad lol

Idk why I put down 0.8

[flat frigate]

Umbraleviathan

[jaunty oasis]

oh

[stray socket]

I did a moment

And

And then that should be 0.8 not 1.25

Which doesn't make sense hold on

[jaunty oasis]

may i know why is it $B=\frac{0.2L}{0.25hr}thr$ instead of $v=\frac{0.2L}{0.25hr}thr$?

[flat frigate]

andrewkong972

[stray socket]

That was an error on my side

So ignore that

[jaunty oasis]

oh ok xd

[stray socket]

Yeah this is atcually kinda weird lol

Is $v$ the number of liters

[flat frigate]

Umbraleviathan

[jaunty oasis]

actually can we assume B=v coz each container = 1L xd

[jaunty oasis]

yea

[stray socket]

I mean no because v = 1, which is a constant

Okay so idk why they don't just use L in the formula

[jaunty oasis]

yes lol

or maybe left side is B x 1L and it cancels out with RHS?

[stray socket]

Well you can say that after 1.25 hours you'll have one B

1 = 1.25v

Cool

B = 1.25t

[jaunty oasis]

B = t/1.25?

[stray socket]

That part is solved

But then

Well

Okay Gimmie a sec

[jaunty oasis]

okie :>

[stray socket]

You can safely say that one L is made per 1.25 hours

[jaunty oasis]

v=1.25t

[stray socket]

Yeah

But then these formulas

Are gonna have v^2 in them then

[jaunty oasis]

yea ;-;

[stray socket]

Which is like

No

Because I got B = 0.8vt

[jaunty oasis]

hmm

t/1.25 = B

could we say this?

(1) v=1.25t
(2) t/1.25 = B
sub (1) into (2)
v=B xddd

my brain is dying halp xd

[jaunty oasis]

how did you get this again?

[stray socket]

Sorry I had to do something

Well

I did the thing

Oh wait

Oh wait it clicked

[jaunty oasis]

nvm xd

[stray socket]

$$B = \frac{1.25hr}{1 L}vLthr$$

Wait no

[flat frigate]

Umbraleviathan

[stray socket]

That

That wouldn't be very nice

Now you get time squared

[jaunty oasis]

1L/1.25hr?

[stray socket]

v is the number of water in T hours correct

[stray socket]

Ratio manipulation

Okay this is dumb

[jaunty oasis]

never heard of that before xd

[stray socket]

If you think logically

vt, in units, is Lhr

Which makes no sense

[jaunty oasis]

yes xd

[stray socket]

Hmm

v L of water in t hours

so v = slope if I read that like 29 million times

Who wrote this question

Well no, v isn't slope

It's like

v/t

[jaunty oasis]

v/t=0.2L/.25 hrs

like thsi?

[jaunty oasis]

i have no idea xd

[stray socket]

The slope is $\frac{v}{1.25hr}$

[flat frigate]

Umbraleviathan

[stray socket]

If you said that v = number if liters

[jaunty oasis]

yeah

[stray socket]

Oh I get it (as I said many times and failed)

Oh this is just poorly written

[jaunty oasis]

XD that was what i was thinking too

[stray socket]

v = number of liters when given time against the rate

[jaunty oasis]

relatable!!!

[jaunty oasis]

yea

[stray socket]

$$v = \frac{L}{1.25hr}thr$$
$$\= 0.8Lt$$

[jaunty oasis]

makes sense

[stray socket]

Well T is hours

[jaunty oasis]

yea

or maybe $vL = \frac{L}{1.25hr}t$

[flat frigate]

andrewkong972

[stray socket]

I mean but then v is the number of liters

[jaunty oasis]

4/5Lt?

[stray socket]

I get this

Oh I am like

Not here physically

Hdiehfjrjd

My brain

[jaunty oasis]

XD

[flat frigate]

Umbraleviathan

[jaunty oasis]

i think $vL = 0.8Lt$?

[flat frigate]

andrewkong972

[jaunty oasis]

coz we need L on the LHS too

[stray socket]

But then that would mean that v = 0.8t

And v is just time

Not the liters

We need to make v = 1

Essentially v = B = 1

[jaunty oasis]

but the equation is basically
no. of litres = rate of water x time?

[stray socket]

Yeah

[jaunty oasis]

oooo

[stray socket]

But then like

That gets me $B = 1.25Lt$

[flat frigate]

Umbraleviathan

[stray socket]

Oh but that's when L, as a changing variable, = 1, = v (which is changing)

So if you like

Magic and shit

1.25vt

But honestly that's as far as I can get without having to take an educated guess

Hm

Maybe there's an easier way

I'm gonna go sleep

Maybe someone can make it easier to solve if they can somehow decode whatever the question's saying

[jaunty oasis]

i have no idea whats the question tryna say too tbh ;-;

anyways thanks man!!

have a good day

<@&286206848099549185> need some backup here xd

[safe radish]

@jaunty oasis Has your question been resolved?

[safe radish]

@jaunty oasis Has your question been resolved?

[safe radish]

@jaunty oasis Has your question been resolved?

[jaunty oasis]

.close

[safe radish]

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[empty meteor]

linear operator has been given as a matrix

find the basis and defect

the answer already has been given but i need explanation

[safe radish]

@empty meteor Has your question been resolved?

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@empty meteor Has your question been resolved?

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@empty meteor Has your question been resolved?

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[glacial lion]

how to get derivative of big number like this

[stray socket]

Chain rule lol

[sweet mica]

Oh, you again

Yes, chain rule

[stray socket]

Yoy

[glacial lion]

ahh okay thanks homies

[stray socket]

Oh also I found out what you meant by obtuse lol

[sweet mica]

Length of Mars daylight

[glacial lion]

that is me

[sweet mica]

Oh, are you the one of trigonometric circumference?

[lean otter]

nice chain rule

[sweet mica]

I meant Collect's length of Mars' daylight

[stray socket]

The thing about coterminacy

[lean otter]

also ig that $\frac{2 \pi}{365}$ refers to the speed of a planet or something?

[flat frigate]

0/0=:Hmmcat:

[sweet mica]

That formula is about the length of the day in Mars

[stray socket]

Why do you just know that on the top of your head lol

[stray socket]

I would expand the inside of the cosine and then just use chain rule

[safe radish]

@glacial lion Has your question been resolved?

[glacial lion]

how do u do chain rule with no exponents??

[safe radish]

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[lone arch]

66 is divided into smaller numbers. One number is 3 greater than twice another number.
How big is the bigger number of the two?

[flat frigate]

Kepe

Kepe

[lone arch]

And gone through them all

Then I noticed 6 - 3 = 2 * 1.5

so the pair is (6, 1.5) and the greater number is 6.

Is there another approach to this?

[sacred wren]

ab=66
a=2b+3

solve the system for a and b

[safe radish]

@lone arch Has your question been resolved?

[lone arch]

(2b + 3)(b) = 66
2b² + 3b = 66
2b² + 3b - 66 = 0

Uh, something is wrong

[lone arch]

They don't exactly have to equal 66 I think

It's just 66 divided into many smaller numbers

[sacred wren]

we can't assume that with the information given

I have no reason to believe it isn't some pair a and b s.t. ab=66

[lone arch]

Yeah, I hate these kinds of statements because of this

thx

.close

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[lean otter]

Pretty sure there is a formula for finding this but I’m not sure what it is

[fossil shell]

what’s the problem here

[lean otter]

I’m not sure how to start

forgot how to do this

[fossil shell]

how would you integrate normally

wrong channel

[lean otter]

(x^2/2) - 3x

or is it u’/ln[u]

something like that?

[safe radish]

@lean otter Has your question been resolved?

[nova creek]

There's indeed a formula for the integral of an absolute value, but it may be easier to just split the integral into two

[safe radish]

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[onyx turret]

Hey

[onyx turret]

Can somebody tell me where I'm making mistakes

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@onyx turret Has your question been resolved?

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[alpine ravine]

Quick question? Is isomorphism commutative?
If A isom to B
then B isom to A?

[alpine ravine]

i though this might be the case because of bijectivity

[wild cape]

whats ur def of "A iso to B"?

[alpine ravine]

there exists a function f:A->B that is a homomorphism and also bijective
and A and B are fields for example

[analog flare]

Pretty sure it is

[wild cape]

ok so maybe this is smth u can try showing urself

say there exists an isomorphism f:A->B

if u can show there exists an isomorphism B->A (say what u think it is before showing its an isomorphism) then ur claim is proved

[alpine ravine]

there exists an isomorphism f:A->B
(A, + *) (B, + *) fields
let g be the inverse of f (also bijective)
g(f(x)) = x
=>
g(f(x)+f(y)) = g(f(x+y)) = x+y = g(f(x)) + g(f(y))
g(f(x)*f(y)) = g(f(x*y)) = x*y = g(f(x)) * g(f(y))

is this ok?

[wild cape]

i expected it to go like g(x+y)=g(x)+g(y) & g(xy)=g(x)g(y) [x,y in B], directly using the def of homomorphism

but the proof is essentially done, it can be quickly fixed to fit expectation

[alpine ravine]

what do you mean

i tried to show that the inverse of f is also an isomorphism

so then there exists a g:B->A which is also an isomorphism

[wild cape]

we already know g is bijective so it suffices to show g is a homomorphism

[alpine ravine]

i did not show that g is bijective

i just wrote this between ()

[wild cape]

im aware hence "already know"

[alpine ravine]

let g be the inverse of f (also bijective)

[wild cape]

its a fact we dont bother to show

[alpine ravine]

g(f(x)+f(y)) = g(f(x+y)) = x+y = g(f(x)) + g(f(y))
g(f(x)*f(y)) = g(f(x*y)) = x*y = g(f(x)) * g(f(y))

this is to show that g is a homomorphism

[wild cape]

yeah thats what i was talking about the whole time

we have g:B->A

[alpine ravine]

oh

[wild cape]

directly going thru the def of homomorphism, we must show g(x+y)=g(x)+g(y) & g(xy)=g(x)g(y) for all x,y in B

[alpine ravine]

you said g(x+y) not g(f(x)+f(y)) that's what confused me
i thought you wanted me to prove this using B directly, not B as the image of f

got it

[wild cape]

the proof relies on writing stuff in terms of f (which u did fine) but ive provided the format each side of the equalities must match

so heres that fix

let x',y' in B, we show g(x'+y')=g(x')+g(y')

f is surjective so x'=f(x), y'=f(y) for some x,y in A

now ill copy ur work

g(x'+y') = g(f(x)+f(y)) = g(f(x+y)) = x+y = g(f(x)) + g(f(y)) = g(x')+g(y')

g(x'y')=g(x')g(y') is shown similarly

[alpine ravine]

yes, I understand

thanks

.close

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[wild cape]

so weve shown theres "symmetry" to iso stuff

.reopen

[safe radish]

✅

[wild cape]

heres smth else before i go

theres also "transitivity"

ie if A is iso to B and B is iso to C then A is iso to C

try showing this @alpine ravine

if ur stuck u can open a channel later and ping me

.close

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[severe crystal]

How do I get 256

[final halo]

Complete the square

[neat kiln]

Vertex?

[severe crystal]

-128 ± 128/3

[lean otter]

128t is the positive term and -16t^2 is the negative term. you are looking at when the object stops in the air, so when 16t^2 == 128t

[severe crystal]

Do I combine the 128 and -128?

[lean otter]

no

did what i just said make sense to you

[severe crystal]

Uh sort of

[lean otter]

so 16t^2 is the force of gravity pulling down, and 128t is your speed.

[severe crystal]

16^2 = 256

[lean otter]

no no

t is the variable here

[severe crystal]

Ok

[lean otter]

do you know how functions work

[severe crystal]

Ye

[lean otter]

good. this is a function.

we are trying to find when the object stops in the air because it: goes up --> stops --> goes down. the highest point is when the object stops

if you can see that

so we look for the point where 128t - 16t² == 0

[severe crystal]

Ok

[lean otter]

now the easiest way to find that point is using the quadratic formula

or if you are willing to do more work, you can factor the expression and find the solution that way

[severe crystal]

Alright

[safe radish]

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[safe radish]

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[severe crystal]

Thanks

[upbeat swan]

.close

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[meager junco]

I instantly went for answer choice A but the correct answer was E

I don't know how to approach this question “for all values x is greater than x”

I made the presumption of making x=5 and didnt go anywhere with it

[tropic steeple]

i guess you can just plug in any x that is greater than 3 (eg 5) and just calculate each of the given choices and the expression given

you get 7/8 which is equivalent to the answer E

[meager junco]

Oh

[upbeat swan]

hi

[meager junco]

Oml why didnt itgink of tha

Hi

Thank you / i have another question

[tropic steeple]

sure, ask

[meager junco]

I chose C, but the correct answer was A

I made an uneducated guess

I tried making the fractions add together

And it didnt amount to A

OH

THERE WAS A MULTIPLICATION SIGN

Oh my god

[tropic steeple]

is that + or / in the end XD

[meager junco]

Thank you

[tropic steeple]

i cant see

[meager junco]

It’s a division sign

[tropic steeple]

ah ok

makes sense then

@meager junco so you understand the answer?

[meager junco]

Wait I got 49

How do you do 5/6 times 1/4

[tropic steeple]

are you calculating it manually?

[meager junco]

Is 5/24?

Yes

[tropic steeple]

yes

[meager junco]

Ive done something wrong

[meager junco]

Complete left to right no?

[tropic steeple]

so, try calculating the left side first

the fractions

[meager junco]

25/24

5/6 when 1/2 +1/3

Then 5/6 times 1/4

5/24

[tropic steeple]

wait

[tropic steeple]

this is what you got?

[meager junco]

As my final

Yes

[tropic steeple]

okay, thats not good lol

lets go from the beggining

what are you calculating first, tell me

[meager junco]

Im adding 1/2 and 1/3 together

[tropic steeple]

okay thats you first mistake

[meager junco]

Oh,

[tropic steeple]

you cant add before multiplying

[meager junco]

Wait do i haveto do pemdas

[tropic steeple]

exactly

[meager junco]

[tropic steeple]

so, what are you calculating first

[meager junco]

So 1/2+1/12

Im gojn to do 1/3*1/4

1/12

Ih gos i gotta divide after the

Which is 5/12

Then i add 1/2 to 5/12

11/12

[tropic steeple]

exactlx

[meager junco]

HOW DO I FORGRT PEMDAS

[tropic steeple]

do you know what to do after?

[meager junco]

Yes add

11+12=23

[tropic steeple]

you understand how you got 11 and 12 to be x and y respectively?

[meager junco]

Yes

With PEMDAS

[tropic steeple]

no, i mean when u got 11/12=x/y

i mean it doesnt matter actually

forget about it

[meager junco]

I have another question

5 more

I chose answer F

And the correct answer was H

I know lattice points are at two lines

Ithink

[tropic steeple]

yeah they are

integer numbers aswell (cause it says its that in the question)

and it cant be on the rectangle

[meager junco]

What does the question mean by “A lattice poin inside but NOT on thr rectangle will be chosen at random.”

[tropic steeple]

it means you intersaction must be inside these rectangle given

but on its edge

[meager junco]

Oh ok

[tropic steeple]

so for example point 0,0 isnt part of your points that you are calculating

and every point on rectangle respectfully

you doing fine there 😄

[meager junco]

So would only 3 pairs

Or 3 points be used in answering the question?

And not (6,4)

[tropic steeple]

there is 15 points in total that are inside your rectangle

[meager junco]

Oh lord

How did you find out that there was 15 points?

[tropic steeple]

you know what integers are?

[meager junco]

Ye

Cant be decimals

[tropic steeple]

yes

maybe its easier if you draw the rectangle

and the x y plane

i will show you

[meager junco]

Ok

[tropic steeple]

so all the intersections inside the rectangle count up to 15

and now you must determine how many of them add up to odd number

so for example point with coordinates (1,1) adds to and even number (1+1=2) and isnt part of your solution

when you determine how many of your points give odd number as a soultion, you divide that number with 15 (total amount of points) and you will get your probability

😄

[meager junco]

OHH

Thank you so much

[tropic steeple]

np

im going to sleep. hopefully i helped you enough

bye!

[meager junco]

Thank yoyy

Night

[safe radish]

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[whole oasis]

I'm getting 1 + DNE, what do I do?

[whole oasis]

Does the limit not exist or is it 1?

[lean otter]

notice that $\lim_{x \to 1} f(x) + g(x) = \lim_{x \to 1} f(x) + \lim_{x \to 1} g(x)$ by the property of limits, you can evaluate them simply. also notice that for $\lim_{x \to 1} g(x)$ to exist, we must satisfy $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^-} g(x)$.

[flat frigate]

Renegade

[whole oasis]

the limit of g(x) as x approaches 1 doesn't exist

[lean otter]

yep, exactly

[whole oasis]

That's the problem

So I'm getting 1 + DNE

[lean otter]

yes, because the one-sided limits are not equal to eachother

as we approach x = 1 from the right, we have 1, but 0 as we approach from the left

the limit for f(x) is negligible because it is a removable discontinuity

so, the answer to your question is just DNE

[whole oasis]

$\lim{x \to 1} f(x) + g(x) = 0?$

[flat frigate]

Zyleaf

[lean otter]

not quite

also you must use _

[whole oasis]

so the limit of the operation doesn't exist

limit as x -> 1 of f(x) + g(x)

[lean otter]

yes, this is the situation we have

the one-sided limits do exist for both, however we cannot evaluate their sum because one limit does not exist

this is often the trouble for piecewise functions

[whole oasis]

Ahh

[lean otter]

so your thinking is completely correct

[safe radish]

@whole oasis Has your question been resolved?

[safe radish]

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[mint flax]

$\bC\cong\bR(i)$ right?

[flat frigate]

gmod

[mint flax]

or is it equal?

since $\bR(i)=\bR\cup{a+bi:a, b\in\bR}$

[flat frigate]

gmod

[mint flax]

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[drowsy charm]

how do i solve this?

[drowsy charm]

amount of squares is 64

[nova creek]

What have you tried?

[drowsy charm]

wait

.close

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[drowsy charm]

how do I solve this? I have no clue what to do.

[drowsy charm]

radius = 13

<@&286206848099549185>

[worthy hemlock]

You have to wait at least 15 minutes before pinging helpers

[drowsy charm]

well, I had to delete the thing, but then i decided to put it back up

I have no absolute idea what to do

[worthy hemlock]

Okay, and you opened a new channel, you still wait

[grave ferry]

@drowsy charm Join point O and D

that forms a radius

and you know radius

and put pythagoras theorem and find ans

[drowsy charm]

ok thx

[grave ferry]

np

[drowsy charm]

lemmejust try it out

[grave ferry]

ok

[drowsy charm]

.close

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[lean otter]

can someone help me write 2 sentences each on how to do roman number operations on addition and subtraction

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[stray socket]

1. Convert to Arabic numerals, do operation
2. Convert back to Roman lol

Idk why anyone would wanna do VIII + IX

[lean otter]

.close

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[safe radish]

@jade vessel Has your question been resolved?

[jade vessel]

<@&286206848099549185>

[frank niche]

What do you need help with?

What are you struggling with this question?

Have you attempted it?

[tame smelt]

I think its the equation

The first part is clear from the graph

[tame smelt]

Ok, to find the equation, it looks similar to the graph of 1/e^x. The asymptote of this expression is y=0 and y intercept is 1. To make the y intercept 11 (as clear from the graph), we multiply our expression with 9 to get 9/e^x, so the y intercept becomes 1*9 = 9 (multiplying with 9 instead of 11 because we are also going to shift the graph up by 2 units to match the asymptote y=2. Then the intercept will become 11). Now match the asymptote by adding 2. We get 9/e^x + 2. Which is decently close! The only thing left is the “stretching” along the x axis. I am not sure how one could find that out, as enough info is not given. Maybe someone else can help on this little bit!

[safe radish]

@jade vessel Has your question been resolved?

[jade vessel]

thanks for the help

[safe radish]

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[vast summit]

formulate and solve accurately real-life problems involving areas of plane regions
Can someone give me another example of this: https://www.desmos.com/calculator/gnfqi7zhbg
Using this image provider below or you can use your own image as an example

[safe radish]

@vast summit Has your question been resolved?

[safe radish]

@vast summit Has your question been resolved?

[safe radish]

@vast summit Has your question been resolved?

[cobalt parrot]

of a desmos thing that does that?

[vast summit]

yep

[cobalt parrot]

lol gl

[safe radish]

@vast summit Has your question been resolved?

[tropic steeple]

XDDD

[stray socket]

Yup

Although the function it's doing is simply just doing what I call "sine wrapping"

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[kind dove]

#help-30 message
@quasi bison I have tried to come up with solution, but unsuccessfully. I dont get the fact why there is a problem with number of edges

[safe radish]

@kind dove Has your question been resolved?

[safe radish]

@kind dove Has your question been resolved?

[quasi bison]

@kind dove have you proved that we cannot have all 4 edges from the removed vertex all going into the same component?

[kind dove]

no

[quasi bison]

try to picture it

[kind dove]

i have tried

[quasi bison]

something impossible will happen if we try to add this vertex back in with 4 edges going into the same component

remember that we assumed that cutting out the removed vertex turned our connected graph into a disconnected one

[kind dove]

hmm

[quasi bison]

i've said all i can without giving it away

[kind dove]

we are giving vertex back to one component or between two components ?

[quasi bison]

what do you mean

[kind dove]

our graph is disconnected so we have two components

[quasi bison]

yes

the removed vertex does not by itself belong to either component

[kind dove]

we are adding vertex back to same position ?

[quasi bison]

of course we are

[kind dove]

hmm, then i dont get what impossible can happen

[kind dove]

some vertex will have more than degree 4

but thats not a problem i think

[quasi bison]

no that is not the issue

why have i been talking about connectedness all this time???

do you not see that the entire graph as pictured in my sketch is disconnected?

[kind dove]

i see

[quasi bison]

right, so this is where the contradiction comes from.

because we started with a connected graph.

[kind dove]

what is the contradiction ?

[quasi bison]

assume that removing a vertex caused G to go from connected to disconnected.

this directly contradicts our assumption!

if the removed vertex had all edges going into the same component, then the graph was disconnected to begin with, which it was not!!!

[kind dove]

hmm

its weird

you mentioned yesterday, that there is problem with an edge count

[quasi bison]

yes

but to actually explain what the problem is, we need to know that the edges from the removed vertex cannot split 4-0 between components

[kind dove]

hmm

tbh, i dont get it much

[quasi bison]

consider the first component. by the argument i laid out above, there are either 1, 2 or 3 edges going to it from the removed vertex.

do you understand this? Y/N

[kind dove]

y

[quasi bison]

since the removed vertex is red (in my coloring), all the vertices in the first component adjacent to the removed vertex are blue.

[kind dove]

y

[quasi bison]

thus they are not adjacent to one another, and upon removal of the removed vertex their degree becomes exactly 3.

[kind dove]

y

[quasi bison]

all other vertices in the first component, and in particular all red vertices in the first component, still have degree 4.

[kind dove]

y

[quasi bison]

in a bipartite graph, the number of edges equals the total degree of all vertices from either part.

thus it equals the total degree of all red vertices and it also equals the total degree of all blue vertices.

[kind dove]

y

[quasi bison]

let me now attempt to get you to finish this up:

by considering both total degrees mod 4, show that this cannot happen.

[kind dove]

i cant finish it

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[lean otter]

please explain this formuka

formula

i used to know itr

but i forgot

.close

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[hybrid flare]

can anyone help

please

[cedar rover]

What have you tried?

[hybrid flare]

3.45 and 31.62

[cedar rover]

what do you mean

[hybrid flare]

like i seclet them

[cedar rover]

turn everything into numerical values

[hybrid flare]

do u know the anser

[blissful gate]

we're not going to give you answers. if you want help, show what work you've done other than guesses

[cedar rover]

Remember that to transform a % into a number, you have to divide it by 100, so 5% is 0.05

[upbeat ridge]

Turn those fractions to x/1000 form

Then come back

[hybrid flare]

okl i didid it ty tyguys

.clse0o

.close

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[lean otter]

What’s the difference between the two? And are both right?

[lean otter]

I don’t see why one of them are wrong

[worthy hemlock]

Which two?

[lean otter]

The ones that aren’t crossed out

I see that the Focus is 1,2
So the graph has to move right which means a negative number in the parenthesis

[worthy hemlock]

One is a vertical stretch, the other is a vertical shrink

[lean otter]

Uh huh, but I don’t get why one of them are wrong…

[worthy hemlock]

Because one has a = 8, the other a = 1/8

[lean otter]

Yeah, but why is it wrong??? I know there’s a difference between them

[worthy hemlock]

.

[lean otter]

Okay let me rephrase

I don’t know which to use in a focus and directrix equation.

[lean otter]

• you gave me the answer, partially without explanation

[worthy hemlock]

So the equation you are would be applying is $(x - h)^2 = 4p(y - k)$. You can find the vertex, because you know the focus is at (1, 2), so the x coordinate is 1, the y is the midpoint between the directrix and focus, so (-2 + 2)/ 2 = 0, y coordinate is 0

[flat frigate]

dldh06

[lean otter]

I know the y = 0 part

[worthy hemlock]

The variable p, is the distance from the vertex to the directrix or vertex to focus

[lean otter]

But uhhh-

2

So that would mean 3 is correct then

[worthy hemlock]

Nope

[lean otter]

What

[worthy hemlock]

Plug everything in $(x - h)^2 = 4p(y - k)$

[flat frigate]

dldh06

[worthy hemlock]

Set it equal to y

[lean otter]

???

ok

[worthy hemlock]

You know h, p, and k

Plug those values in that equation

Manipulate so it becomes y = ...

[lean otter]

What’s H again

[worthy hemlock]

Coordinates of the vertex

[lean otter]

I can’t plug cords of a vertex into an equation though

[worthy hemlock]

Yes you can

[lean otter]

(1-1,0)???

[worthy hemlock]

h is the x coordinate of the vertex, k is the y coordinate

[lean otter]

You could’ve said that from the beginning

That comes out to 0

[worthy hemlock]

No

[lean otter]

1-1

= 0

[worthy hemlock]

Where is 1 - 1 coming from?

$$(x - h)^2 = 4p(y - k)$$
You know h, p, and k,
Plug those values in that equation,
Manipulate so it becomes y = ...

[lean otter]

Actually wait one sec

[flat frigate]

dldh06

[lean otter]

It’s answer 4 when I plug it in

[worthy hemlock]

Then that's the answer

[lean otter]

okay……..

thanks I guess???

.close

[safe radish]

Channel closed

Closed by @sour orbit

Use .reopen if this was a mistake.

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.

[solemn epoch]

To have a basis, vectors must be linearly independent and span the subspace

I have proved the vectors are linearly independent but don't know how to prove that they span the subspace

[obtuse jackal]

It's a family of 3 elements in a 3D vector space

So one property (free or spanning) is enough to show it's a base because it has the right number of elements. Assuming you've seen that in class, which you should have

[solemn epoch]

Hmm I don't recall seeing that in class

[obtuse jackal]

What theory do you have on bases then ? And subsequently on free and spanning families

But anyways if you just want to show it spans S you can just show it spans something that spans S. If you can construct a basis for something, then you can construct any element in the set by expressing the basis in that other coordinate system

So showing you span u, v, w is enough

[solemn epoch]

I have that 1. Bases are not unique, 2. Given a basis for a subspace, every vector in the subspace can be written as a unique linear combination of vectors in that basis, and 3. Any two basis for. The same subspace have the same number of elements

In terms of the theory presented on basis

[obtuse jackal]

How did you prove 3 ?

[solemn epoch]

Oh wait would 3 qualify then for my case

[solemn epoch]

It's given in our textbook

[obtuse jackal]

You see this but don't do proofs ?

That's sad

[solemn epoch]

The proof isn't presented for this