help-0

[candid dune]

alright ill do that

[charred pilot]

sqrt(10816)

=104

[candid dune]

yeah 104

alright thank u so much

[alpine sable]

Can someone please help, I think it’s x=1

[charred pilot]

x=1 is one of the answers

But there are more

[charred pilot]

Did you understand how I used the slope to find the other points?

[alpine sable]

X=1,3,5?

[charred pilot]

Yes, assuming the endpoints are excluded

[alpine sable]

Thanks a lot

[candid dune]

i will read again what u said

[lone heart]

@candid dune Has your question been resolved?

[lone heart]

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[nimble jasper]

Hey! Is this correct? I first did 1/3R1, then R2+R1, then R3+3R1

[nimble jasper]

<@&286206848099549185> 😓

[lone heart]

@nimble jasper Has your question been resolved?

[nimble jasper]

ahhh can anyone help me out

[tight pier]

yes but the rank is off

[nimble jasper]

Oh really? Can u please explain why

Because I thought rank was based off of rows that were linearly independent and

None of these rows are the same as each other

[tight pier]

yes

now tell me are they linearly independent based on your form?

R2 and R3 are multiples of each other

[nimble jasper]

sorry I thought linearly dependent means they’re all different rows

But you’re right

[tight pier]

no

[nimble jasper]

So since r2 and r3 can become each other

[tight pier]

So you basically have a zero row

[nimble jasper]

would the rank be 2 then

[tight pier]

ye

[nimble jasper]

Ahh I see, well thank you!!

hmm it says I got it wrong

[tight pier]

yea

you haven't reduced entierely

but i thought that was irrelevant

[nimble jasper]

I probably should’ve had it fully reduced then

[tight pier]

yea

mb

[nimble jasper]

But it’s weird bc the question only asked REF

but no worries

[tight pier]

no worries

[tight pier]

you dont wanna skipp basically entries

like the red path does

[nimble jasper]

Ahhh okay

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@nimble jasper Has your question been resolved?

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[proud saddle]

x=1 since its a rule that where you see a pointy structure you cannot differentiate those , in this case at X=1

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[peak pelican]

I have solved this question 3 times, getting the same answer each time, i've asked chatGPT and everything for help...

I need a math wizzard to tell me what I am doing wrong

[hushed locust]

can you show your work?

[peak pelican]

yeah

Let me try to flip

[hushed locust]

,rccw

[peak pelican]

oh

im sorry for having caveman writing 🙏

basically multiply by 2 the second equation, so the -y cancels the 2y on top

then 10x = 9

div 9/10

x = 9/10

plug into original equation 2x + 2y = 1

19/10 simplifies to 9/5 + 2y = 1

2y = 1 - 9/5

[void swift]

wheres the equal sign ^

yea thats right

[peak pelican]

2y = -4/5

-4/5 div 2 = -2/5

y = -2/5

so is the website wrong?

I am so lost

[void swift]

u mean y=-2/5

[hushed locust]

you got the same answer this time

[peak pelican]

right

[peak pelican]

it ask for the y value

[void swift]

yes and u said x=-2/5 before so i corrected u

[peak pelican]

im so annoyed because I get a free CLEP voucher if i score good but this stupid website keeps telling me im wrong

[hushed locust]

the site says "answer: -2/5" which is the same as what you got

[peak pelican]

no

thats mine

it doesnt show the "right" answer

it displays mine

forgot to lead with that

[void swift]

maybe try -0.4 ?

dunno what maths software if any they r using

[peak pelican]

I cant retry the same set, but I wanted to confirm that I am not like messing up some super small detail in my solving

[hushed locust]

your answer is correct, it may be a formatting issue or other site problem

[peak pelican]

so much for a goverment funded site

https://learn.modernstates.org/

[peak pelican]

thank you

[void swift]

could be d2l shenanigans or an incorrect config for the problem

who knows

[peak pelican]

they cant even get the most basic shit right mindblowing

.close

[lone heart]

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[lone heart]

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[tribal axle]

Hi! I wanted to ask some help in clarifying what the supremum does in the definition of the lesbegue integral.
From what i understand the supremum of a set takes its least upper bound
Example: $sup{1, 2, 3} = 3$
But here i have: $\int_X f d\mu := sup{I(h) : h \in S^+, h \leq f} \ $
Where $h$ are simple functions: $h = \sum_{i=1}^{n} c_i \chi_{A_i} \$
And $S^+ := {g: X \to \mathbb{R} : g \text{ simple function}, g \geq 0}$

[ocean seal]

woomy

woomy

[lone heart]

@tribal axle Has your question been resolved?

[tribal axle]

<@&286206848099549185>

[nimble fern]

I really wish i can help, but I didn't learn much about lesbegue integral.
But one thing, supremum of a set may not always be the least upper bound of the elements of the set.

e.g. sup[0,1) = 1

[tribal axle]

huh, in this case taken the supremum meant "taking thinner and thinner step functions"

basically you approximate the function you are integrating with step functions

[nimble fern]

ah i see

[tribal axle]

like a riemann integral

[nimble fern]

sorry for the confusion

[tribal axle]

and the supremum meant taking the thinnest step functions, kinda like the limit

[tribal axle]

all good, im also confused as to how the supremum is working like a limit

[tacit arch]

Supremum of a limit of bounded simple functions is well defined yes

[tribal axle]

Could you provide a definition that i can look more into

[tacit arch]

Starts around page 50
https://www.stat.rice.edu/~dobelman/courses/texts/qualify/Measure.Theory.Tao.pdf

[tribal axle]

Thanks!

.close

[lone heart]

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[lone heart]

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[fervent ferry]

hello, i'm trying to get a geometric meaning of the row space of a matrix.
The column space is the linear combination o the column vectors of the matrix. To have a visual intuition of the transformation i imagine what happens to the orthonormal basis vector when i apply the matrix to them. So in a sense the matrix is transforming the space and if i just plot the columns vectors of the matrix it would tell me where the basis would go, and i can see the effect of the transformation.

I don't what am i seeing if i plot the row vectors. can anyone help me out?

[lone heart]

@fervent ferry Has your question been resolved?

[mortal trellis]

you can see the row space as being orthogonal to the null space

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[lone heart]

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[void carbon]

Given an acute triangle ( \triangle ABC ) with ( AB \neq AC ). Let ( I ), ( O ), and ( H ) denote the incenter, circumcenter, and orthocenter of ( \triangle ABC ), respectively. Point ( M ) is the midpoint of ( BC ). The line ( MI ) intersects ( AH ) at ( N ), and the circle with diameter ( MN ) intersects the circumcircle of ( \triangle BHC ) at ( P ) and ( Q ). Points ( X ) and ( Y ) lie on lines ( MP ) and ( MQ ), respectively, such that ( X, A, Y ) lie on the circumcircle of ( \triangle ABC ) in that order. The line ( XY ) intersects ( AB ) and ( AC ) at ( D ) and ( E ), respectively. Given that ( IO \parallel BC ), prove that ( MD = ME ).

[ocean seal]

6EQUJ5

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[alpine sable]

this is the question i’m stuck on. i have been stuck on for hours. i don’t rlly know how to sketch it since it doesn’t gives me a right angled triangle and when it does the bearings are off

[lone heart]

@alpine sable Has your question been resolved?

[finite flax]

You didn't draw a picture?

c is the distance from E to G.

Bearings are measured clockwise from north

Draw cardinal axes at the points and approximate the direction and distance from the point

Use basic Geometry theorems to fill in one unknown angle

My notation is all shit

Using A B C when we started with E F G

Anyway

Law of cosine finds c = EG

Law of sines finds angle B, which fills in the complete bearing angle from E to G

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[smoky crane]

Hello. I have got as question in my math book I have to proove that:
\newline
\newline
if $a_1 \equiv b_1 \mod m$ and if $a_2 \equiv b_2 \mod m$ then $a_1 \pm b1 \equiv \mod m a_2 \pm b_2$ and then $a_1 \mul b_1 \equiv a_2 \cdot b_2 \mod m$ 
\newline
so I wrote:
theorem:
\[if $a_1 \equiv b_1 \mod m$ and $a_2 \equiv b_2 \mod m$ then $a_1 \pm b_1 \equiv a_2 \pm b_2 \mod m$ and $a_1 \mul b_1 \equiv a_2 \cdot b_2 \mod m$
\]
proof:
\[
as $a_1 - b_1 \mod m \implies (a_1-b_1) | m$
as $a2 - b2 \mod m \implies (a2-b2) | m$
\]
so as:
$(a1-b1) | m$
and as
$(a2-b2) | m$
then:

$(a1-b1) - (a2-b2) \mid m$

it follows that:
$(a1-b1) + (a2-b2) \mid m$

and as:
$(a1-b1) \mid m$
and:
$(a2-b2) \mid m$
then:
$(a1-b1) \cdot (a2-b2) \mid m \newline$
\newline
Is the custom proof that I wrote correct please?```

[ocean seal]

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

[sour mica]

a1 - b1 is 0 mod m, and a2 - b2 is also 0 mod m. So they are certainly equivalent. But a1+b1 is 2*b1 mod m, and a2 + b2 is 2*b2 mod m. So theres no reason for the sums to be equivalent

The theorem would be true if it were a1 ± a2 is equivalent to b1 ± b2 (mod m)

[smoky crane]

hello. I am very confused is it a - b | m or m | a - b

[sour mica]

m | a - b is true in general since we can expect a-b to be larger than m or equal to m

[smoky crane]

then is a - b | m true?

[sour mica]

not necessarily since a-b can be 2m, and 2m is not a factor of m

[smoky crane]

ah ok thanks. then my proof is wrong?

[sour mica]

your proof only tackles the a1-b1 part and not a1+b1 part...

so since the "theorem" is itself invalid, I suppose you should also include a disproof for the a1+b1 part of the problem

[tiny sky]

i think that's just a typo tho

like surely it's to prove a1 + a2 = b1 + b2 mod m

[sour mica]

yea thats what I said earlier

[smoky crane]

I mean should I do m | a - b in my proof instaead of a - b | m to proove it?

[sour mica]

You are technically doing that, but you seem to have messed up the notation

like its true if you write 2|6, but if you write 6|2 then its false

[lone heart]

@smoky crane Has your question been resolved?

[smoky crane]

ok I see perfectly the mistake then

[lone heart]

@smoky crane Has your question been resolved?

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[smoky crane]

can I ask again when I will have finished latter please?

[lone heart]

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[lunar plank]

what is the coeffecient of x^3 in the expansion of (1+3x-x^2-x^3)^3

[lunar plank]

<@&286206848099549185>

[alpine sable]

!15min

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[alpine sable]

Just expand it using foil

[lunar plank]

hell nah

[rose sigil]

no need for that

[lunar plank]

why use foil, there's prob an easier solution

how tf do you use the binomial theorem here

or maybe i shouldn't use the binomial theorem

[rose sigil]

think about all the ways you can choose 3 summands in 1+3x-x^2-x^3 and multiply them to get an ax^3 term

[lunar plank]

wdym by that? like (-x^3), (-x^2)(+3x)?

[rose sigil]

not sure what you mean

[lunar plank]

cuz i tried adding the coefficients of these.
since (-x^3) + (-x^2)(3x)
(-x^3) + (-3x^3)
-4x^3
so the answer is 4?

i tried that and it was the wrong answer

how to solve hahaha

i tried ai, it gave me different answers

[rose sigil]

no surprises there

[rose sigil]

i don’t really have the energy to explain what i mean by this, sorry

[lunar plank]

oh damn

[cosmic basin]

Hello I need help w something this is a lil bit stupid but I can’t figure it out :(

[lunar plank]

bro why are you here T_T

i also need help lmao

[cosmic basin]

huh

[lunar plank]

bro.

[cosmic basin]

SORRY THIS IS MY FIRST TIME USING THIS DISCORD SERVER

[lunar plank]

it's my help channel

ohhh

[cosmic basin]

I DIDNT KNOW MY BAD

[lunar plank]

its okay

do you see the math help (available) chats?

go there

[cosmic basin]

sorry again

[lunar plank]

its oke, i was like that the first time i was here

and anyways

<@&286206848099549185>

[lone heart]

@lunar plank Has your question been resolved?

[deep condor]

$$ (1+3x-x^2-x^3)(1+3x-x^2-x^3)(1+3x-x^2-x^3) $$

[ocean seal]

RadMeerkat62445

[deep condor]

To find the coefficient of x^3, multiply products of terms that can give x^3 term

[lone heart]

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[sonic sedge]

yo

how do i solve thisss

[twin nimbus]

Factor the numerator and denominator

[sonic sedge]

yea i did stuff like that before

see this one? its easy

but this one is much more complicated and i dont understand it

[twin nimbus]

Hint, if there is a point wise discontinuity, then (x+2) will be a factor

[sonic sedge]

am supposed to multiply the numerator and denominator with 4+x power 2 right?

[twin nimbus]

(because x+2 = 0 when x = -2)

[vast tapir]

try to use the difference of squares property

factoring will work just fine

[sonic sedge]

if i do the regular factoring that i sent in the other pic

it wouldnt give me the right answer

[vast tapir]

no dont try to multiply or divide, just factor something out

ill give you a hint, the bottom factors into x^2(2+x)

[vast tapir]

no not like this

[twin nimbus]

You can't solve this one this way

[vast tapir]

look at the top

[twin nimbus]

You have to find the common factor between the numerator and denominator

[vast tapir]

hint: ||(a^2 - b^2) = (a+b)(a-b)||

use this if you dont get anywhere after a bit

[sonic sedge]

as in something to divide both of them to remove the power?

like 2

to make the 4x power 2 a 2x

[vast tapir]

no see here, the top and bottom both share a factor

you dont have to multiply and divide by anything, but just have to make that observation

[sonic sedge]

am not good with factorization rules

if theres any old rule that am supposed to know from a younger grade

show it to me

[vast tapir]

you can see the top is a difference of two squares

[vast tapir]

.

[twin nimbus]

Also this

[sonic sedge]

so i just use this on the bottom numbers

[winter light]

@sonic sedge

[sonic sedge]

what does that even mean ok i made the observation now what

[twin nimbus]

Here is a similar problem solved as an example

[winter light]

You use this thing you're talking about when you have roots (mainly)

[sonic sedge]

and your rule wouldnt reven work on the bottom numbers cause its not (a^2 - b^2) its (a^2 - b^3)

[winter light]

The key is factoring both numerator and denominator. The rest is just simplifying and plugging in

[sonic sedge]

well yea thats obviously what i need to do

[winter light]

And what's troubling you?

[sonic sedge]

is that its not factoring

[twin nimbus]

\begin{align*}
\lim_{x \rightarrow 3} \frac{x^2 - 9}{x^2 - 5x + 6} &= \frac{(x+3)(x-3)}{(x-2)(x-3)} \
&= \frac{x+3}{x-2} \
&= \frac{6}{1} = 6
\end{align*}

[ocean seal]

OmnipotentEntity

[twin nimbus]

This is a similar problem to yours

We factor the numerator and denominator, find the common factor that vanishes at the limit point, cancel it, and evaluate at the limit point

In the case, because the limit is as x goes to 3, the term (x-3) vanishes

So this is the term we must eliminate (because 0/0 is not defined)

In your case the limit is x goes to -2

So the term you need to eliminate is (x+2)

Which implies that if this limit exists and is finite then it must be the case that the numerator and denominator share factors of (x+2)

@sonic sedge does the above make sense, if not do you have questions?

[sonic sedge]

yea one sec am just wondering how did you reach this conclusion

[twin nimbus]

Difference of squares

[sonic sedge]

you just rooted them right

you rooted them

[twin nimbus]

First mentioned here

[sonic sedge]

the 9 doesnt have a power

so this rule doesnt apply

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

[twin nimbus]

9 = 3^2

[sonic sedge]

ahhh ok

and for the bottom part

how did we get this

[twin nimbus]

You factor the quadratic. I observe (-3)(-2) = 6 and (-3)+(-2)=5

[sonic sedge]

ohhh you just did the normal quadratic factorization ok

what numbers multiply = 6 and add = -5 right?

[twin nimbus]

Exactly

[sonic sedge]

alright i understand but one more question

would cancelling the x here be wrong?

because you cancel stuff in division

[twin nimbus]

Cancelling the x meaning you get what exactly?

[sonic sedge]

-9 / -5x+6

[twin nimbus]

-9/(-5x + 6) ?

[sonic sedge]

yes

[twin nimbus]

Yes, that would be wrong

[sonic sedge]

damn ok

any reason why?

[twin nimbus]

You can only cancel things that multiply

Not things that add

Notice that I do cancel the (x-3) later on

Because they multiply

[sonic sedge]

you cant cancel minus and addition?

[twin nimbus]

Not in a division

[sonic sedge]

alright thank you sm

.close

[lone heart]

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[vestal crown]

Can someone explain c and d

[lone heart]

@vestal crown Has your question been resolved?

[mossy jasper]

Show the working you did for parts a and b and it should be a lot easier to demonstrate

But basically only 2 means you need EXACTLY 2 that DON'T have defects, and at most 1 means there must be either 0 or 1 that DO have a defect

[vestal crown]

Why i cant just write(0.950.950.05)

@mossy jasper

[mossy jasper]

Oh you mean you wanted to explain how the answer book did it

[vestal crown]

Yes

[mossy jasper]

Let's talk about c to start.

You have 3 chips
only 2 do not have defects
The probability that they DO have one is 0.05, so the probability that they DON'T is 0.95

There are multiple combinations of chips for the next 2 parts, unlike the first 2.

Either chip 1 and 2 don't have defects, chips 2 and 3 don't have defects, or chips 1 and 3 don't have defects.

I hope this helps a bit for the start at least

That's why they have so many probability listed, because the different placements are the different chips since you assume each of them have a unique option

[vestal crown]

Ohhhhhhh

[lone heart]

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[shell junco]

hm

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[cunning flower]

hi

[cunning flower]

for this probability question

where it ask for probability of exactly a pair of cards from 3 cards picked in a deck of 52

I got a ridiculous which doesnt make sense

[prime badge]

there are 52×51×50 outcomes

no that wouldn't help

[cunning flower]

o right thats why

[prime badge]

you would divide the numerator by 6 instead

should work after that 🤔

[cunning flower]

wdym by divide the numerator by 6

[prime badge]

7488 × 3 / 6

[cunning flower]

oh so 52 x 51 x 50 = 22100 x 6

[prime badge]

but that's the same as doing 52×51×50

why did i think it wouldn;t work

yeah ok either one

[cunning flower]

are u suppose to just do 52 x 51 x 50 or do 52C3

in a question like this

[prime badge]

you can do 52c3 if you remove the order from the numerator as well

i don't know what you're supposed to do

it's the same answer, so it doesn;t matter what you;re supposed to do

[cunning flower]

wait but 52C3 is 52!/49! x 3! which is 52 x 51 x 50 / 6

[prime badge]

7488 × 3 gives you ordered hands, each unordered hand appears 6 times

so you divide by 6

and then 52c3 would be right

[mossy jasper]

What does the c mean? Never learnt that notation

[cunning flower]

the choose in n choose r function

[mossy jasper]

ohhh that one

Then yeah there's 52C3 total combinations of 3 cards

[cunning flower]

alr

is there a scenario when i should use 52C3 and when i should just use 52 x 51 x 50

[mossy jasper]

52C3 is not the same as 52 * 51 * 50

52C3 = 22100, which is the total number of possible 3 card combinations. 52 * 51 * 50 = 132600

[cunning flower]

ah alright

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[stiff falcon]

Can someone help me with Q8?

I completely forgot how to lol

[cinder compass]

h is 90

[random dirge]

why

did you just look at it or is there a trick to it

i'm also confused

[stiff falcon]

How?

[cinder compass]

it's always true

[stiff falcon]

It might now be though?

[random dirge]

oh yeah that's true

[cinder compass]

wdym

[stiff falcon]

Its not marked in as a right angle though?

[winter light]

Yeah because they want you to figure it out

[cinder compass]

the ange subtended by a diameter is 90 degrees

[winter light]

Note that the largest chord is marked with a circle (which is meant to be the centre of the circle). They marked it so that it should ring you a bell

[cinder compass]

given the other questions, i feel like that's too much to ask

[cinder compass]

u can prove it if you want

draw a radius the point

and use isosceles triangles have equal angles

[stiff falcon]

But how is it 90?

[cinder compass]

that's what i'm saying

it will take a while to explain

[cinder compass]

but u do it like this

[stiff falcon]

Are you 100%sue its 90

[cinder compass]

yes

[stiff falcon]

It won't work

[winter light]

There's a theorem stating that

[stiff falcon]

Because you would have 22+90+45

[winter light]

Why 45°?

[stiff falcon]

Which = 157

[stiff falcon]

That's half of 90

[winter light]

Mmh I'm not following you

Prolly @cinder compass does, since he's helped you from the beginning

[cinder compass]

i is not half of h

[stiff falcon]

H is 68

[winter light]

Yes

[stiff falcon]

Is it though

[proud obsidian]

h is 90

[stiff falcon]

But how?

[pseudo ice]

[pseudo ice]

Did you mean h, or i?

[stiff falcon]

Idk

[stiff falcon]

I

I is 68

[proud obsidian]

inscribed right angle theorem

[pseudo ice]

i is 68, sure

[stiff falcon]

Thanks

. Close

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[manic flame]

Hey

[manic flame]

So the question says the probability function is:

$$P(\xi=x)=\binom6x0.1^x0.9^{6-x},\quad x=0,1,2,3,4,5,6$$

[ocean seal]

Totalani

[manic flame]

It then asks me to get these but honestly not sure hwo to being with these.

actually I think I get it

for a I ahve to add x= 1 and 2 and 3 together

will try it to confirm

I must be doing something wrong cuz im not getting the correct asnwer

$$\begin{aligned}&P(\xi=0)&&=\quad\binom600.1^00.9^6=0.531,\&P(\xi=1)&&=\quad\binom610.1^10.9^5=0.354,\&P(\xi=2)&&=\quad\binom620.1^20.9^4=0.098.\end{aligned}$$

[ocean seal]

Totalani

[manic flame]

I get 0,983, the book says 0,468 ?

Wait don't think I'm suppose to add 0

That explains it

We guchi boys

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[hollow trench]

Hello, can someone help me with equations with 2 absolute values? |x-2| + |x-3| = 3

[jagged cobalt]

sure

[hollow trench]

till now I've done just equations which had just one absolute value, for example,
|x-2| = 3 => { x - 2 = 3
{ x - 2 = -3 and so on

but I'm quite confused when we have 2 absolute values

[storm marten]

4 cases arrive

treat them separately

[wheat abyss]

hello

[gritty bramble]

3 u mean

[wheat abyss]

what do you call two guys that love math? algebros!

[jagged cobalt]

youll have to split it into a few more cases
x>2, x<2
x>3, x<3

so theres 2<x
2<x<3
3<x

[wheat abyss]

no, incorrect

[gritty bramble]

this is a help channel

go to #discussion

[wheat abyss]

x<3

okey

[hollow trench]

so

x>2, x>3

[gritty bramble]

its x<2, and also 1 more case: 2<x<3

[hollow trench]

what do I need to write exactly

[gritty bramble]

u have to first write the cases and then check for each case and depending on what the expression inside modulus is (negative or positive) open the modulus according to the definition of modulus function

[hollow trench]

can you explain me please how we ve got to

x<2, x>3 and 2<x<3

[proven depot]

another method:- square both sides, solve for x

you will get two values of x, both of which are correct

you just need to know how to factor quadratic equations for the end result

[hollow trench]

|x-2| + |x-3| = 3

so

you mean (|x-2| + |x-3|)^2 = 9

?

[proven depot]

yes

be careful when simplifying

[hollow trench]

2x^2-10x+13 + 2|x-2||x-3| = 9

I don't think I made any mistakes

[proven depot]

how did you get that

[hollow trench]

|x-2|^2 + 2|x-2||x-3| + |x-3|^2 = 9

|x-2|^2 isnt just (x-2)^2?

[proven depot]

oh yes mb

[hollow trench]

and you just add the quadratics

[proven depot]

remove the abs value and multiply

(x-2) *(x-3)

[hollow trench]

2x^2-10x+13 + 2(x-2)(x-3) = 9

?

[proven depot]

ye

[hollow trench]

4x^2-20x+25 =9

so 4x^2-20x+16=0

[proven depot]

yea subtract 9 from both sides

divide by 4

I mean you can simplify it

to x^2 -5x +4

[hollow trench]

oh

x1 = 1

x2 = 4

[proven depot]

now just plug it in the original equation

both values

check if it satisfies or not

[hollow trench]

4-2+4-3 = 3

and

|-1| + |-2| is 1 + 2

so it's 3

yea

it satisfies

thank you so much

[proven depot]

yea and I think those are the only two values

[hollow trench]

hmm

someone said earlier something about 3 or 4 cases

[proven depot]

I will see on wolfram wait

[hollow trench]

actually, if you've got time, can you help me with one more thing?

[proven depot]

yep those are the only integer solutions

[proven depot]

ok

[hollow trench]

how can I do

|x+|x-3|| = 5

[proven depot]

1 min

hmm

@hollow trench so you need to solve two equations

first x+|x-3| = 5

x+|x-3| = -5

and in those two equations you get two more cases for each equation

so 4 equations

you get few complex solutions

[hollow trench]

complex?

haven't studied this yet

[proven depot]

yea

[hollow trench]

It's my first year of highschool

we should study this next year

[proven depot]

if you get them you leave it

[hollow trench]

ohh

[proven depot]

bc this is Real numbers

you need real solutions

[hollow trench]

oh, the exercies mentions to do this equations in R

[proven depot]

yea

[gritty bramble]

basically you find the point where the absolute function turns zero, so |x-2| is zero at x=2, now x can be >2 or <2, similarily for |x-3| x=3 and x can be >3, <3
if you combine x>2, x<2, x>3, x<3 you get three conditions: x<2, x>3 and 2<x<3

[proven depot]

in each equation isolate variables and constants and then solve for x

[hollow trench]

hmm

x+|x-3| = -5

x-3 = -5-x

2x = -2

x = -1

and x-3 = 5-x, 2x = 8, x = 4

[proven depot]

plug those values and verify

[hollow trench]

1 + | -2 | = 5

that's definetly not right

[proven depot]

yep

[hollow trench]

I am wrong somewhere?

[proven depot]

I think while making equations

[hollow trench]

Or I'm supposed to get values that won't satisfy

4 + 1 is 5

well u said

x+|x-3| = 5, -5

x-3 = 5-x

[proven depot]

x = 4 is the answer

only 1 solution

[hollow trench]

just 4?

hmm

but shouldn't there be 2 solutions

[proven depot]

why

[hollow trench]

I mean

we've done 2 equations

the one with +5 and the one with -5

[proven depot]

its the only integer solution satisfying the original question

[hollow trench]

hmm

alright

thank you very much for making me understand

have a nice day!

[proven depot]

no problem

there are complex solutions

but this is the only integer solutions and also present in Real number system

[hollow trench]

I can't wait to study complex numbers :D

but unfortunately you don't really learn much about them in highschool

in my country

[proven depot]

ah

you can watch online lectures

on youtube

[hollow trench]

maybe just that i^2 = -1 and the conjugate, if I wrote it right

I'll go for it

[proven depot]

yea

[hollow trench]

But I won't use them very much at school, and without practice I will probably forgot what I'm learning from youtube

[proven depot]

I think khan academy has a course

for beginners

[proven depot]

oh alr

[hollow trench]

I'll go and watch it, thank you for telling me

is complex numbers in algebra 1 or algebra 2?

or in none of them

[proven depot]

I don't really know

[hollow trench]

oh, I'll just search

have a nice day

[proven depot]

you too

close this channel

[hollow trench]

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[viral falcon]

I have already shown the last part, but how do show that f(x-a) has a Fourier transform in the first place?

[tacit arch]

plug f(x-a) into the definition and do a substitution

[viral falcon]

Yeah I know

But I'm wondering how to show: "if f(x) has a Fourier transform, so does f(x-a) for any a in real numbers"

[viral falcon]

is it enough to just show that?

[tacit arch]

if this is a functional analysis class then you just need to show f(x-a) is in L^1

[viral falcon]

It's not functional analysis, but alright. Don't know what L^1 is either .-. Haven't learned anything about it

In the class i'm taking right now we're learning Laplace- and Fouriertransform, PDE's and Complex Analysis. Doubt that there's a umbrella term for those topics

[tacit arch]

then yes

[viral falcon]

Okay, nice

Thanks

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[elfin pulsar]

Hey !
Can i write this to say that i take a couple a,b,c in R^3 but with a different of 0 ?

[tacit arch]

why

or are you asking how to write it in latex?

what does "different of 0" mean?

[elfin pulsar]

a =/= 0

i think i do not need to translate this one but if needed tell me

i want to know if i can write :

or if it's nonsense

[tacit arch]

oh yes $R^*$ is normal to exclude 0 because 0 doesn't have an inverse in the ring R

[ocean seal]

riemann

[elfin pulsar]

So i wrote it right ?

[tacit arch]

no haha

[elfin pulsar]

explain pls

[tacit arch]

$a \in \R^$ and $(b,c)\in\R^2$ is more clear. or also $(a, b, c) \in \R^ \times \R^2$ is also okay

[ocean seal]

riemann

[elfin pulsar]

Okay so i just forgot the X thx

that was my thought

THX

!!!!

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[fast oar]

Working on this problem for a calc III class

[fast oar]

No clue how to do it, I’ve tried some partial stuff, some algebraic stuff, making y a function of x, etc.

I know the answer is for x>0 but have no clue how to prove it

[tacit arch]

,rccw

[marsh echo]

is basically asking u to find the domain for whic hthat is a one-to-one function

is graphing it possible?

and then try teh horizontal line test?

[ruby valley]

Yea

[fast oar]

That’s what I did but I’m looking for a proof

[ruby valley]

I also would graph it out

[void swift]

well thats rather nontrivial to graph

[ruby valley]

But u don’t have to

[void swift]

im thinking maybe smth to do with the monotonicity of the function?

if u can differentiate it maybe u will discover smth interesting /shrugs

[ruby valley]

But u can use the Formel

[fast oar]

Partials with respect to x and y don’t offer anything very interesting

[ruby valley]

Like the steps

[fast oar]

Formel?

[ruby valley]

Like steps to solve a function without graph

[marsh echo]

implicit diff

[ruby valley]

U don’t now?

[fast oar]

I did implicit diff already, doesn’t give anything

[void swift]

maybe spam implicit function theorem

[ruby valley]

Yea

Like smth like that

U didn’t got many information to graph it out

So u have to calculate

[fast oar]

Oh I think that would work

Let me read up on the theorem

Wait but doesn’t implicit just tell you that a function can model points in the neighborhood?

How would I use it to show bijectivity

[ruby valley]

U have to do it like 2-3 times

[fast oar]

Wdym

Like I just show that there are functions on different y values and where they end?

So like, I show that there’s a top curve function with one point, a bottom curve function with another, then show where the bottom curve function ends?

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[formal oracle]

I dont get it why burgers equation isnt linear?
As i understand i cant do u*∂u/∂x which would equal to u ∂u^2/∂ux?

[formal oracle]

uu = u^2 and ux = ux and in that case it would fit that it's isnt linear since it's not to the first power

[tacit arch]

u^2 is not linear in u

[formal oracle]

How would product of u and ux break the rule of them not being to the 1st power

Well yes u^2 is not linear in u but how do i know it's u^2? As i understand i can't just do u * u = u^2

[tacit arch]

u * du / dx is also nonlinear

[formal oracle]

In this picture red x is multiplication.

My tought process was that if i multiply u by those values i get the equation on the right and that would break the linearity.

Can i think of it like that or that's wrong way to go about it?

[tacit arch]

wot

u * du/dx is multiplication

between u

and du/dx

du^2 / (dux) is nonsense

[formal oracle]

Okay. homogeneous linear PDE is if sum of terms, each of which involves dependant variable (u in this case) or one of it's derivative to the 1st power.

How does u du/dx break this rule?

[tacit arch]

u = u^1

and du/dx = (du/dx)^1

so u * du/dx has a power of 2

[formal oracle]

Thank you, now i get it

I need to practice writing the right questions straight away lol

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[hushed crown]

<@&286206848099549185>

[manic eagle]

What's up

[alpine sable]

you are supposed to ask the question and wait

pinf the helpers if no one answers it

after 15 min

[azure needle]

Where's the question

[tacit arch]

!15m

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[brittle lake]

Number 69 a, would I find derivative of f first of g? Like it would be 4(2) then I find the inner derivative of g to multiply it by which is 6, so it would be 4(2) times 6 right?

Or am I dumb or stupid

[lone heart]

@brittle lake Has your question been resolved?

[brittle lake]

<@&286206848099549185>

@mighty cosmos

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[brittle lake]

@mighty cosmos help me next time

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Uhh

.osr

.close

What the

.reopen

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[brittle lake]

Guidance with number 72 please?

I know that the derivative of f(x) is about -1

Since part a is asking me to find h’(2)

I’d find f’f(x) then f’(x)

Well, I’m stuck on that part

So if someone could aid me

🙏

[indigo saddle]

Apply the chain rule to h first

[brittle lake]

Didn’t I just do that

[brittle lake]

With this

[indigo saddle]

Show me

Write it out explicitly

That’s often the first step in getting unstuck in a problem

[brittle lake]

I mean it’s just this but replace x with 1

f’(f(1)) * f’(1)

[indigo saddle]

Where did you get 1 from?

[brittle lake]

I mean 2

I’m tripping

[indigo saddle]

Right

So you have h(2) = f’(f(2)) * f’(2)

Now, can you figure out what f’(2) is?

[brittle lake]

-1

[indigo saddle]

Ok

Can you plug that into the equation?

Also for f(2)

[brittle lake]

I figured it out.

.close

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[brittle lake]

.reopen

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✅

[brittle lake]

Problem 86. a

Would velocity be derivative of the equation of motion

Motion is still position right

[tight pier]

yes 1st derivative

[brittle lake]

This is actually so lame (the problem)

Okok

So

[tight pier]

ok (so true)

[brittle lake]

Acos would just be -Asin

And wt + 8

Just w?

[tight pier]

i see what you saying (yes)

[brittle lake]

Oh ok thank you (thank you)

[tight pier]

no problem (no problem)

[brittle lake]

Given that v = -Asin(w) is that my answer for a (?)

[tight pier]

wrong derivative

[brittle lake]

die

How did I get it wrong

[ocean seal]

bacc

bacc

[brittle lake]

Oh

Yeah

https://tenor.com/view/pou-gif-27568079

[tight pier]

damn what happened to poo

[brittle lake]

Uh

Alcohol

Meth

Something like that

But it’s ok you worry about him later

[brittle lake]

so given this derivative how would I find the velocity at time t?

Would

Actually

Isn’t it just the derivative

[tight pier]

yes

[brittle lake]

Is velocity 0 when the harmonic motion is at the end or beginning

I mean

End and beginning

I’d assume so

Because

It’s stopping

Due to negative acceleration

But

How would I find this value with

My derivative

Oh

I see

Why am I thinking through texts

LOL

I’m writing my thoughts

Acoustic moment(?)

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[brittle lake]

I got it

(Is it 8)

[tight pier]

s'(t) = 0 is b)

Find all solutions of that

[tight pier]

wtf

[brittle lake]

Me

[tight pier]

divided by sin

that's a sin

[brittle lake]

You can’t do that

Oh

Skull

[tight pier]

You can instead divide by -A too

[ocean seal]

bacc

[tight pier]

Now you can apply arcsine

on both sides

[brittle lake]

Whoa whoa

[tight pier]

whoa whoa

[brittle lake]

How do I apply arcsin I’ve never used it before

[tight pier]

You apply the inverse function

[brittle lake]

this is the last problem so it’s suppose to be like a trick challenge thing I think

[tight pier]

If you apply the inverse function on your function, the they cancel each other out

[tight pier]

you are the trick challenge

[brittle lake]

Noooo

[ocean seal]

bacc

[tight pier]

You can also deduce that sine is 0 for multiples of pi

[ocean seal]

bacc

[tight pier]

So you can solve this for t

[brittle lake]

@tight pier you are a life saver thanks I did it

https://tenor.com/view/sml-jeffy-life-savers-mints-life-savers-mints-gif-7839346603373039805

[tight pier]

Did I save your life?

[brittle lake]

Yea

Unlike @mighty cosmos

[tight pier]

haha

[brittle lake]

Can I ask

Are u in college

[tight pier]

yes

[brittle lake]

Or are u one of those high school genius

[tight pier]

i am one of those college idiots

[brittle lake]

.close

[tight pier]

.close (didn't close what the heck)

[brittle lake]

That’s lie if that was true you would drop out of math and be homeless addict

[lone heart]

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.

[tight pier]

i am homeless addict with math

[brittle lake]

You opened up a help channel

How can I help

[tight pier]

spread your light across the world

[brittle lake]

You’re violating the rules, ask help for math designated questions only

[tight pier]

aight bet

[brittle lake]

(I sometimes ask chemistry questions here)

[tight pier]

you should moderate the server

[brittle lake]

No I don’t wanna be on discord too much I only use it mainly for help with subjects I need help with 😢

[tight pier]

cool

.close

[lone heart]

Channel closed

Closed by @tight pier

Use .reopen if this was a mistake.

[brittle lake]

Do you study often

[tight pier]

no

[brittle lake]

You seem like you do

Oh

nice.

[tight pier]

nah

i believe in my natural intelligence

[brittle lake]

Alpha response

https://tenor.com/view/omega-gif-21656744

[tight pier]

Lady, that's an omega.

[brittle lake]

Oh

Ok I got to go thank you for your help man 👨 🙏

I’m sure you are a very fun person to hang out with

Tell your friends I said hi

Or something

Ok bye

[tight pier]

😂

[tight pier]

I will tell my 0 friends hi

[brittle lake]

(I’m gonna assume this is a joke)

But I’m sure many people will like your humor

I think you’re a fun person

[tight pier]

no

[brittle lake]

Yes

[tight pier]

my humor is not for the weak

nor for the majority

only for truly chosen ones

[brittle lake]

I feel like you would be in a club with lots of people to talk like this around

I’m sure you’re fun you just haven’t found the people yet then

:)

Ok bye for real now

[buoyant saddle]

bro what is this

💀

[brittle lake]

He asked for help with

A derivative

It was

Find derivative f(x) = x^2

[buoyant saddle]

this some edating

[brittle lake]

LOL

[buoyant saddle]

💀

[tight pier]

that's alpha dating

go back to the kitchen knife

[buoyant saddle]

discord romance

[brittle lake]

I already got someone 🤐

[buoyant saddle]

tell him i’m sorry

[brittle lake]

oh ok

He says sorry

[tight pier]

idc

[buoyant saddle]

😬

[brittle lake]

🥶

[buoyant saddle]

yikes

😭😭

[brittle lake]

You made him mad

[buoyant saddle]

2024

[brittle lake]

Better say sorry

Before it’s too late

[tight pier]

lol i am not mad

[brittle lake]

(we know)

[buoyant saddle]

how many egirls do you have bro

😭

[tight pier]

maybe you should stop calling girls e-girls and you'd get people to like you

[buoyant saddle]

i’ll let you discord lovers get back to it

[buoyant saddle]

you’re hilarious

[lone heart]

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.

[lone heart]

@floral bone Has your question been resolved?

[alpine sable]

What have you tried?

You can use the Exponential Growth formula?

So you can you $x_t = x_o(1+gr)^{t}$.

[ocean seal]

Sukiyaki

[alpine sable]

For your first question, 22 years will elapse since 1950. Your initial volume of oil will be 530,000,000.

$x_t = 530000000(1+0.075)^{22}$.

[ocean seal]

Sukiyaki

[alpine sable]

So it would roughly be 2.6 Billion Tonnes.

@floral bone

I guess you can perform the same for the following 3 questions and you should be fine..:)

[lone heart]

@floral bone Has your question been resolved?

[lone heart]

@floral bone Has your question been resolved?

[lone heart]

@floral bone Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.

[burnt reef]

school never taught me this property so I'd really appreciate it if it can be done without this

thanks regardless, more knowledge is always better

[livid sage]

to prove, u = -x

[burnt reef]

I'll try it and get back to you thanks

[burnt reef]

can you show me the proof please

@livid sage <@&286206848099549185>

[wanton wigeon]

Yo

[burnt reef]

hey.

[wanton wigeon]

Yo

Let there be a function f(x)

Simple proof

See

Let there be a integral of f(x) from a to b

[burnt reef]

okay

[wanton wigeon]

Now what I do is

I sub

x=-t

[burnt reef]

yeah

[wanton wigeon]

Ok

So now it would become integral of -f(-t)dt from -a to -b

as dx=-dt

[burnt reef]

yeah

[wanton wigeon]

Now you know that

if function is even

Then f(x)=f(-x)

[burnt reef]

yep

[wanton wigeon]

Ok

So now

I say function is even

[burnt reef]

alr

[wanton wigeon]

so we can say f(-t)=f(t)

[burnt reef]

yes

[wanton wigeon]

So we get -f(t)dt from -a to -b

[burnt reef]

yes

[wanton wigeon]

So we will have

Wait

[wanton wigeon]

You want this property right

💀☠️

Why I am telling you the other

[burnt reef]

hey can you please go a little faster

i don't want to be stuck on this for any longer

[wanton wigeon]

Okok

See we begin fast

We have integral f(x)dx from -a to a

Now I say f(x) is even

So

f(-x)=f(x)

So now we sub x=-t

so dx=-dt

So we get

f(-t)-dt from a to -a

[burnt reef]

go on

[wanton wigeon]

Now see

we know f(-t)=f(t)