help-23

[dense lark]

give me a moment, its a pretty direct question

lemme go thru my notes rq

got it

alr so

[simple galleon]

youre missing an integral by 1/y

anyway

e to both sides

[dense lark]

have u tried this

[simple galleon]

,,\ln y = 2\ln (x+1)-4\ln (x+4)-\frac3{x+1}+c\\implies e^{\ln y} = e^{2\ln (x+1)-4\ln (x+4)-\frac3{x+1}+c}

[flat frigate]

bored amogi

[patent citrus]

find C first for y = 1/256 and x = 0, then make an eqn for y for x = 2 and C and raise to the power of e

[simple galleon]

yes, she did

[dense lark]

this equation

[dense lark]

alr

[barren fjord]

Ah ok

[simple galleon]

from there it should be easy enough to get y

[dense lark]

🥷🏼 ur in 10th

[simple galleon]

remember to write x>0

[patent citrus]

@barren fjord

[simple galleon]

wait

x>-1 mb

[patent citrus]

yes

[barren fjord]

Is this ok

Should it be (x+4)^-4 actually

[patent citrus]

terms shuld be multiplied right

[barren fjord]

What

[patent citrus]

how are they in addition

[barren fjord]

It’s log laws

[simple galleon]

2ln(x+1)-4ln(x+4)=ln((x+1)^2/(x+4)^4)

[patent citrus]

and what about exponent laws

[simple galleon]

,,2\ln(x+1)-4\ln(x+4)=\ln(\frac{(x+1)^2}{(x+4)^4})

[flat frigate]

bored amogi

[patent citrus]

dude am i tweaking or they shuld be multiplied

[simple galleon]

they could also be multiplied

[patent citrus]

also?

[simple galleon]

making it ln((x+1)^2(x+4)^-4)

[barren fjord]

Is the bottom one okay

[patent citrus]

no

[simple galleon]

dividing by (x+4)^4 is the same as multiplying by (x+4)^-4

[patent citrus]

yea

[barren fjord]

Why are you teaching him😭😭😭

[simple galleon]

so you could do either

[patent citrus]

all the 3 are in multiplication

not addition

[barren fjord]

Is my expression okay @simple galleon

[simple galleon]

the bottom line? no you still have to divide

[barren fjord]

Wdym

[patent citrus]

bro

[simple galleon]

(x+1)^2/(x+4)^-4

[patent citrus]

i said the same thing

..

[simple galleon]

i was confused too for a sec

[barren fjord]

Idrk what ur saying

Sorry

[barren fjord]

But it’s e to the power of ln

Which undoes it

[patent citrus]

you do it after you split

because exponent comes first

[simple galleon]

,,e^{\ln A+\ln B}=e^{\ln A}e^{\ln B}=AB

[flat frigate]

bored amogi

[simple galleon]

or:

,,e^{\ln A+\ln B}=e^{\ln(AB)}=AB

[flat frigate]

bored amogi

[patent citrus]

so this is the correct 1

here

[simple galleon]

either way, the + become a *

it does not get 'undone'

[barren fjord]

Ok

Thx

.close

[safe radish]

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[simple galleon]

@barren fjord please add absolute value bars to the ln

or write that x>-1

[runic narwhal]

Hi

[safe radish]

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[median sentinel]

What is the e (Euler’s number ) exactly , like why that certain number and we use it in log it turns to ln

[stoic dune]

e^x is the function that's the derivative of itself. That's a big deal if you care about derivatives

ln is just a shorthand for log base e.

[azure harbor]

https://en.wikipedia.org/wiki/E_(mathematical_constant)

Why the fuck is it capitalized

[stoic dune]

E

[azure harbor]

E

[broken forum]

E

[solar hazel]

X.E

[primal bone]

E

[arctic raven]

[\frac{\mathrm d}{\mathrm dx}\left(a^x\right)=\lim_{h\to0}\frac{a^{x+h}-a^x}h=a^x\lim_{h\to0}\frac{a^h-1}h]and $a=e$ uniquely satisfies [\lim_{h\to0}\frac{e^h-1}h=1]

[flat frigate]

Flip

[primal bone]

I think it's only capitalised due to page indexing reasons

It shouldn't be tbh

(and I'm fairly sure there will be evidence to the contrary)

[tawdry plover]

E

[arctic raven]

isn't that meme from 2016?

[primal bone]

https://klipy.com/gifs/marikiplier-e-meme-lord-farquaad-1--k01KS87Z7A0PNMQKA4QWTZZ7X7Q

And its several variants?

[arctic raven]

yes

lord farquiplier

[primal bone]

"my name is welcome"

Anyways @median sentinel did you have any follow-up questions

-# also holy shit the name didn't have to tie up so well with the aforementioned meme lol

[safe radish]

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[median sentinel]

no thank u

[median sentinel]

yeah\

[tardy mango]

it's a bot so you don't have to reply btw

.close

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[astral glacier]

Damn @primal bone is a bot now?

[primal bone]

@astral glacier It's this response, genius

[astral glacier]

God damn it

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[errant rapids]

do you know the definition of cross/cartesian product of sets?

[reef bear]

Yes

[fair hound]

first, notice that the problem can be reformulated in the following way:

1. how many pairs (a1, a2) can I make, ie how many elements are in A²?
2. for each, how many pairs (b1, b2) are suitable?

each choice of a pair in A² and a suitable pair in B² will give an element of R

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[distant elm]

are these correct

[safe radish]

@distant elm Has your question been resolved?

[mild plaza]

No idea

[distant elm]

??

[plucky elk]

Yes it all looks right

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[runic basin]

I’m really bad at limits and need help

[wind cypress]

do you know how to evaluate an absolute value when approaching from a certain direction?

with respect to limits

you essentially split it into two cases (when you come from the left ("negative side") and the other from the right ("positive side"))

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@runic basin Has your question been resolved?

[wind cypress]

Do u still need help?

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[lean crystal]

Did I solve this problem correctly?

[weak kite]

Yes this is correct

[opaque fern]

Yes

[lean crystal]

great thanks!

[weak kite]

Your welcome

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[echo gazelle]

Can I pls have some help with this

Idk how to start

[split fulcrum]

Hi

do you know how to find the circulation in general over a vector field F

[echo gazelle]

Ya it’s

Field dot product with the

Dr

I’m kind of confused with parametrising

[split fulcrum]

well

what condition do we need to impose since this is a unit sphere and x=y

[echo gazelle]

Am I supposed to find something that satisfies this

[split fulcrum]

you have the right idea with (0,0,1) -> (0,0,-1) but your t isnt quite right

[echo gazelle]

Oh

[split fulcrum]

first

unit sphere means what?

[echo gazelle]

Radius is 1

[split fulcrum]

so x^2 + y^2 + z^2 = 1 right

[echo gazelle]

Yah

[split fulcrum]

but what does the question say

x=y

[echo gazelle]

Ya

[split fulcrum]

soo what do we have now

[echo gazelle]

2x^2 + z^2 = 1

[split fulcrum]

do you recognise this form at all

[echo gazelle]

Circle

Side ways circle..

[split fulcrum]

well

its called an ellipse

but yeah

[echo gazelle]

Oh

Yah

[split fulcrum]

do you know the standard form of an ellipse

[echo gazelle]

Um circle equation but with different coefficients

[split fulcrum]

its (x^2 / a^2) + (z^2 / b^2) = 1

[echo gazelle]

Oh

[split fulcrum]

yes

its just a thing like if youve seen it before then you know it

@echo gazelle we know how to parametrize an ellipse.. hopefully but the goal is to making our equation look like the standard form of an ellipse

[echo gazelle]

Do I need to know that..

[split fulcrum]

im not sure how you could do this in a way that doesnt involve the ellipse

this is probably the easiest way

[echo gazelle]

They did it using spherical coordinates

Pi/4 and r=1

[split fulcrum]

oh yeah

youd still get the same parametrization either way but you can do that

you know sin = cos because of x=y

[echo gazelle]

Wait wat

How do I do it

[split fulcrum]

whats the transformation you use for spherical

[echo gazelle]

Wdym

0,0,1 to 0,0,-1

[split fulcrum]

im referring to x,y,z

x becomes what?

y becomes what?

in polar coordinates you let x=rcos, y=rsin but what does that look like in spherical

your material must have included this somewhere do you have notes on it

[echo gazelle]

cosu sinv

sinv is the radius

[split fulcrum]

and what about y

[safe radish]

@echo gazelle Has your question been resolved?

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[small sandal]

For 10a. Im kind of lost at this part, idk how to continue:

[median sand]

Reduce first row.

Divide the first row by 2 to make the first coefficient be 1.

[small sandal]

mmh kk, but idk about my 4th row

what do I do with it?

[primal ice]

I see that you've kept the first row starting with 2. You should always make the first pivot to be 1

So I'd swap rows

Swap r4 to r1

[small sandal]

you can swap rows?

[primal ice]

And then row reduce as normal

Ofc you can

It's one of the steps of gauss jordan elimination

[small sandal]

but how does that help in this case, I dont have a leading term

[median sand]

The fun fact is that since the solutions of 4 equations are all zeros, you can just write Ax=0 and row reduce A to save time writing.

[crisp hull]

Ye thats one of elementary operation

[small sandal]

uhh, idk what that means... sorry

[median sand]

And you always aim to get the first pivot of the first equation to be 1 first.

[small sandal]

but how does it change anything? how is it even useful

[median sand]

Do you know RREF?

[primal ice]

Because the first pivot needs to be one

In order to pivot to the next row

[small sandal]

uh it can solve the equation on my calc, but idk what i means

[prisma elk]

you dont have to put in RREF to solve it right?

[median sand]

It is Gaussian elimination ,so.

[small sandal]

like rref(A) gives me some mattrix thingy

[primal ice]

I think you do

[small sandal]

yea that one

[primal ice]

If you don't rref then how will you get the solutions

[crisp hull]

To get a easily reduceable form

[prisma elk]

-# from REF, without having the pivots as 1

[primal ice]

Why not just use gauss jordan elimination? It's much more simpler. I believe

[small sandal]

uhh what is that

[crisp hull]

Look at the question

[primal ice]

Oooo augmented matrix, I missed that part completely

[vague phoenix]

You want to do sth like this @small sandal

make column with 1 entry (1) only, other is 0, and make numbers of 1 like a staircase

[primal ice]

But didn't it mention augmented matrix

[small sandal]

this is augmented no?

[vague phoenix]

It is the same for augmented matrix

[small sandal]

or at least this is how we went through it in class

but tbh, they didn't explain anything, they just said a lot of fancy terms that I didnt understand lol

[crisp hull]

So you gotta get the reudced echelon form and then solve by substitution ig thats i remember

[primal ice]

Tbh, this is the rref form and it will give you the same answer anyway

[vague phoenix]

Not until you get to university and do math major 😂

[primal ice]

So it's X1=1 X2=3 x3=-3 x4=0

[small sandal]

they said doing simultaneous with matricies is like "Uni lvl" or something

[vague phoenix]

Kindergarten maths 😂

[prisma elk]

https://www.youtube.com/watch?v=Z1pm64Vhb2Q (if you are confused)

[small sandal]

is this not what im already doing?

[prisma elk]

you were doing the other method

to get row echelon form

[small sandal]

they look identical

ngl

[vague phoenix]

Gauss elimination here is basically just elimination or substitution in simultaneous equations you did in college (or high school), but written in matrices

[small sandal]

mhm

[vague phoenix]

Says 5x +3y = 10 and 2x+4y = 6, when you subtract them to delete y (or x), that is a row elementary transformation in gauss elimination

Or if you switch the places of these 2 equations, it still give the equivalent solution, that is row swapping that one of the helper mentioned before

[small sandal]

but row swapping changes nothing, so what's the point of it?

[vague phoenix]

To make it more convenient to solve, but it is a transformation

[small sandal]

in our case, it makes it harder no? bc row 4 has like no terms (if we swap with row 1)

[crisp hull]

It does to achieve the reduced echelon form

[small sandal]

mhm ok

[small sandal]

ill see if i can make anymore progress on ^ ^

[vague phoenix]

Try divide first row by 2

[small sandal]

im doing it rn

[vague phoenix]

Then move on next column

and make this column be 0 1 0 0

next one is 0 0 1 0 etc

You are lucky that this one have no free variables, next one is funnier

[small sandal]

free variables?

[vague phoenix]

just solve this one first

[small sandal]

so, R_3 - 8R_1

i divided by R_1 by -2 btw, since that gave positive 1 as leading term

[vague phoenix]

Show work btw

[small sandal]

like each individual calculation?

oh wait nvm

that doesnt work

hmm, 4x4 is much harder than the 3x3's

[vague phoenix]

They are the same

First line will be 1 1 2 0 0

[small sandal]

oh im dumb

i read my first row as {-2, 2, 4, 0 | 0}

but its positive 2, 2, 4, 0 | 0

[vague phoenix]

Bro

[small sandal]

so, 31z=0, z=0 is valid right dumb question (ignore)

[vague phoenix]

what did you get for first line

[small sandal]

1 1 2 0 | 0

[vague phoenix]

good

now you want to make that line to be 1 0 sth sth sth

[small sandal]

ill send what I got rq, we should be able to make R_1 be 1 0 0 0 | 0 with this right

[vague phoenix]

not so fast

[small sandal]

R_1 - 2R_3

[vague phoenix]

Honestly a better move can be made here

You can make R1 <- R1+R2

Then take R2 divides by -1

[small sandal]

R_1 <- R_1 + R_2 ?

what does the <- mean

[vague phoenix]

Which is, the new row 1 can be made by adding R2 to R1

if you add them together you will have 1 0 -1 1 0

[small sandal]

oh ic

[vague phoenix]

You want to finish column by column, top to bottom

[small sandal]

can I divide Row 4 by -5?

that makes it 0 0 1 0 | 0

[vague phoenix]

yes

[small sandal]

and i can use the 1 better than the -5 for future steps

ooo, can I do this to R_3 : R_3 + R_4 then, R_3 / -8 then, R_3 + R_2

that gives me the -8 to be a 0

so it should simplify it yea?

[vague phoenix]

you can simplify row 2 first

then take 8R2+R3 to make the second variable of R3 be 0

There are many ways to do gauss elimination, but normally i would solve like that

[small sandal]

-# this is just a guess, but since we want the leading 1's by the end of this, and since we cant change the "output" part of the matrix, is x, y, z, w all just 1?

[small sandal]

mhm ok

[vague phoenix]

leading 0 and each column that have number 1 should have other entries to be 0

[small sandal]

but for my w variable, i have 0 0 1 0 | 0

that means w doesn't have a solution right?

[vague phoenix]

no

it is 0

[small sandal]

0w = 0?

[vague phoenix]

and not w btw

it is z

0 0 1 0 0 means 0x+0y+1z+0w = 0 -> z = 0

[small sandal]

ahh ok, so we can "swap" it with row 3 then

since that puts it in the uhh correct looking form right?

[vague phoenix]

it is still correct if you leave it like that, but mathematicians want it to be beautiful, so swap row there is "recommended"

[small sandal]

mhm kk

uhh

what did I do wrong

i got -4z = 0, and 1z=0

[prisma elk]

I think you can make the fourth row a row of zeros

[small sandal]

i can do that?

like R_4 * 4

then, R_4 + (R_3)

[prisma elk]

yes

which means that you need to introduce a free parameter

[small sandal]

a what

...

[prisma elk]

So you have 4 variables and 4 equations but one of the rows is a row of zeros

so you are down to 3 equation and 4 variables

[small sandal]

uh yea

[prisma elk]

to make it the solution consistent, you introduce a free parameter

where it would have inf solutions

[small sandal]

like w = lambda

[prisma elk]

did you finish putting in rref form

[small sandal]

uh no

idk how to continue from here

ills send a pic of my step

take R_3 and divide by -4

and R_1 + R_3

to eliminate the -1 in the first row?

[prisma elk]

oh no dont do that

[small sandal]

but then there is that pesky second row with its {0, -1, -3, 1 | 0}

that cant be removed

[small sandal]

uh why?

[prisma elk]

let the -1 be for now

remove the -3 in the second row

also make the -4 in the third row to be a 1

[small sandal]

yea ok

so, R_2 : R_2 + 3R_3

to get rid of that -3

[prisma elk]

ye

[small sandal]

well now its just 1's, -1's, and 0's

[prisma elk]

make the -1 in row 2 postive

[small sandal]

so, R_2 * (-1)

[prisma elk]

ye

show me what you got

[prisma elk]

ok now that is in rref form

[small sandal]

uh how

shouldn't it be leading 1's and 0's all around?

[prisma elk]

uh

[small sandal]

or is that not rref?

[prisma elk]

,w rref( (2, 2 ,4,0, 0), (0,-1, - 3,1 ,0 ), (3,1,1,2,0), (1,3,-2,-2,0))

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by interstellar3398

[prisma elk]

it does not have to be leading 1s I think

I mean

surrouding 0s

[small sandal]

oh, so this is where we gotta parameterize it?

[prisma elk]

yea

[small sandal]

so uh, x -z = 0

and, y - w = 0

those are two we look at now yea?

[prisma elk]

also that w = lambda you free parameter

[small sandal]

oh I c a problem

[small sandal]

this isnt what I have

[small sandal]

like answer wise

[prisma elk]

oh ic

[small sandal]

idk how they get their 1 on the right

OH iC

[vague phoenix]

1 line is linear dependent

[small sandal]

i replaced by 1 in the corner with a 0

on an earlier step...

[vague phoenix]

For question b, it also have 1 linear dependent vector, since it can only have at most 3 pivots

[small sandal]

what does pivot mean

[vague phoenix]

Number 1 at the beginning of the row

other ways, it will be infinite solutions

[small sandal]

ok i fixed it

[small sandal]

got this answer now

[small sandal]

eh

wha is this

[prisma elk]

that is what pivot is: the first non_zero number

but continue on your problem

[small sandal]

now its just doing parameterization or something

i think

[prisma elk]

back substitution

[vague phoenix]

I don;t think it will be necessary for OP

at this level

[small sandal]

and idk how to do it...

[small sandal]

yea I haven't come across it

[vague phoenix]

What is your final matrix

[small sandal]

^ ^

[vague phoenix]

Can you write it as simultaneous equations?

[small sandal]

1x + w = 0

y - w = 0

z = 0

0w =0 ?

[vague phoenix]

Last row will not count, since it is a row of zeros

[small sandal]

mhm ok

[vague phoenix]

So you see z = 0

Let $w=t$ $\forall t \in \mathbb{R}$

[flat frigate]

Nefer

[small sandal]

mhm

[vague phoenix]

1x + w = 0
y - w = 0 can rewrite as x = -w = -t and y = w = t

Then, the solutions will be:

x = -t
y = t
z = 0
w = t, for all t in R

[small sandal]

so why does the w come back in the end?

wasn't it all 0's?

[vague phoenix]

you want to find all 4 solutions

x w y z

x y z are all assigned to a variable, but w is not, it is a free variable

[small sandal]

oh, thats what you meant by "free variable" earlier

[vague phoenix]

because you can parametrise it

w can be anything

[small sandal]

so in general, if I end up with something like: {0 0 0 0 | 0} as my final row, or as one of my rows; then that becomes say W = t

and my other "equations" are in terms of W ==> in terms of t

so everything is in terms of one variable t

[vague phoenix]

Not really

a row of 0 indicates that your system of equations have 1 free variable

[small sandal]

but depending on if it ends up being like, x + 3w = -1 for example, then x is in terms of w and by extensions, x is in terms of t

[vague phoenix]

you could set x to be the free variable too and let w in term of x

[small sandal]

so it just depends on what you want it to be, and what the equations have it end up being

[vague phoenix]

Spot on

[small sandal]

ok

this is so much work for 1 question

reminds me of long division of polynomials, just a lot of basic operations chained together

ty for the help

[vague phoenix]

long division for?

Are you doing partial fractions?

[small sandal]

in what context

[vague phoenix]

long division of poly

what are you using for

[small sandal]

uh simplifying integrands (is what we are using it for currently)

so would this be like a similar process to the matrix thingy

or is this different entirely

[vague phoenix]

Looks nice

[small sandal]

ty for the helps

.close

[safe radish]

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[gleaming coral]

Hello so ive got a theoric question

[gleaming coral]

Is it possible that a funcion has both oblique and horizontal asyntotes

[thin bridge]

yes

[gleaming coral]

how

like claude responded the same and said ofc a funcion could have both if for example funcion doesnt have a horizontal asymptote in +inf, but has in -inf

and i asked, ok send me an example of where a funcion has both

and he thought for a few mins and couldnt find one

[thin bridge]

don't have an equation on hand,
but you can construct one to behave something like

[gleaming coral]

but if that is a single funcion

like not in parts

ohhhh

You can only have both by constructing a piecewise-defined function. You evaluate the horizontal asymptote at one end of the graph (e.g., (x \to -\infty)) and the oblique asymptote at the other (e.g., (x \to \infty)

[flat frigate]

lu01ck | Παυλος

[gleaming coral]

so if its not a pieces function

its impossible

[wild copper]

Not necessarily. It's just the easiest way to define a function.

[gleaming coral]

?

[wild copper]

Also, it also raises the question, what counts as piecewise function? Is (x+sqrt(x^2))/2 a piecewise function?

[gleaming coral]

piecewise funcion is a function which changes itself throughout the domain

for example

1 - |x|

[wild copper]

That's still pretty ill-defined.

If you allow squaring and taking a square root, you can encode the branching.

[gleaming coral]

what

[gleaming coral]

this freaks me out

how is in geogebra it shows me a different funcion in comparison to 2x/2

[wild copper]

sqrt(x^2) is just |x| in disguise.

[gleaming coral]

and you cant just say that x=sqrt(x^2)

i understand that when you have smth to the sqare root

you get like +- its value

but

what the fuck

[gleaming coral]

i understand that for the negative values its the same

[thin bridge]

try that

[gleaming coral]

damn

[thin bridge]

look up constructing a hyperbola

with your choice of asymptotes

solve for y, and choose which branch to keep

[gleaming coral]

that just makes a hyperbola

[thin bridge]

1 branch (not the whole thing), making it a function
which satisfies the conditions you mentioned

[gleaming coral]

so you say it has both asymptotes?

[thin bridge]

yes

[gleaming coral]

let me see

crazy

thx

.close

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[hoary seal]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[honest perch]

you cannot conclude that 0 < limg < 3

but you can conclude that 0 <= limg <= 3, if the limit exists

[crystal palm]

example: f=0, h=3, g=sin(x)+1, then g does not have a limit, even though it is bound between f and h

[safe radish]

@hoary seal Has your question been resolved?

[hoary seal]

I don’t get it

[wary cape]

it can be divergent just not to infinity

[hoary seal]

Actually let’s just say step 1 because I’m a little lost

I do understand what limits are

[crystal palm]

As in, this is an example for f g h, such that the assumptions in the question hold true. And in this example g does not have a limit, so in general the limit is not between 0 and 3 as it does not have to exist in the first place

[hoary seal]

Ah

So for part a, it’s the 4th option

[wary cape]

yes

[hoary seal]

And for b, all of them are not true

Correct?

[wary cape]

isnt it the 4th one

[hoary seal]

It can’t be the fourth one because it’s bounded

G is not going to infinity

[wary cape]

yes so that

[hoary seal]

Yes but also the other options

Because we said limit of g doesn’t exist

[wary cape]

top three can be true
if the limit exist

[wary cape]

we cannot determine if the limit of g exist

<@&268886789983436800>

mr beast is not this generous vro

[golden nebula]

<@&268886789983436800>

[hoary seal]

Exactly

So

Those 3 statement must be false

Because they are giving us limits

[wary cape]

no as in
we cannot determine if there is a limit
doesnt mean there isnt a limit

[hoary seal]

Ohhh

Got it ty

What about this question

[wild copper]

How does row operator relate to the determinant?

[hoary seal]

It should be the first one

[wild copper]

?

[vague phoenix]

row operator doesnt affect determinant

[obsidian oracle]

Depends on which row operation

But adding a multiple of a row to another row will not change the determinant

[vague phoenix]

If add a multi of one row to other will not change det

If swap rows or multiply a row by a constant k will change

[hoary seal]

In the form Rn + (alpha)Rn where n can be any row

It doesn’t change the determinant

[vague phoenix]

What is Rn here?

[hoary seal]

It could be any row

[obsidian oracle]

Replacing R_i with R_i + (alpha)R_j (where j is different from i) will not affect determinant

[wild copper]

As long as it's different from the row it applies on.

[hoary seal]

Yeah it’s not the first option

For part a

Nor the third one

[hoary seal]

I see

So how do we tackle this question

[wild copper]

The solution is literally given to you

[hoary seal]

For the first one is it saying

$2(R_6) + R_7$

[flat frigate]

ø

[hoary seal]

But what row is this expression equal to

[obsidian oracle]

"We are adding 2 times row 6 to row 7"

So literally, you take A

And on row 7, you add twice row 6

[hoary seal]

So $R_7 = 2(R_6)$?

[flat frigate]

ø

[obsidian oracle]

not like that

Take Row 7 of matrix A

and add 2R_6 to it, not instead of it

that gives you the new row 7 of matrix B

[wild copper]

It's $R_7 = R_7 + 2R_6$

[flat frigate]

Xwtek

[hoary seal]

Ahh

That makes sense

But the question didn’t say to add r7

[obsidian oracle]

it said to add 2R_6 TO R_7

so you need to add them together

[hoary seal]

Nvm your right

I mean the question didn’t say to equate it to r7

[obsidian oracle]

it's not equated to R_7 per se

we should have written $R_7 \leftarrow R_7 + 2R_6$

[flat frigate]

Rafibirthday2003

[obsidian oracle]

For the matrix B, we replace the 7th row of A by the new calculated row "R_7 + 2R_6"

[hoary seal]

For b, let’s say we originally had det(A) and then we swapped rows. Woudnt that become -det(A)

Which should be the same as -det(H)

[obsidian oracle]

H is the result of swapping the rows of A

so det(H) is -det(A)

[wild copper]

Yeah. I've used programmer's equal sign.

[safe radish]

@hoary seal Has your question been resolved?

[hoary seal]

Thank you

Got it

.close

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[thick nest]

Hey

.close

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[mortal marten]

the rule of the function g is the form g(x) = sqrt -(x-h)+k h > 0 k > 0

Which of the following is true
A) dom g = [h,+infinite]
B) ima g = ]-infinite,k]
C) the initial value of the function g is negative
D) the function g has no zero

[mortal marten]

not A or B

how do i know if initial value is negative

[cunning pasture]

g(x) = sqrt -(x-h)+k
It's not clear what this is

[mortal marten]

usually its g(x) = a sqrtb(x-h)+k

actually its not c cause if the initial value was negative it would be -sqrt

here its just sqrt

[primal ice]

Can You post a screenshot of the question?

[muted sapphire]

well for the function to work, we must have the inside of the sqrt to be >= 0

[mortal marten]

yea

[muted sapphire]

so the domain is all x such that -(x - h) >= 0

which is -x + h >= 0

thus x <= h

[mortal marten]

when its -x + h >= 0

then u flip the sign when u addiiton to h =< x?

[muted sapphire]

No, -x + h>= 0 which is h >= x which is x <= h

[mortal marten]

oh yeah wait

so it cant be a

well i chose d

[safe radish]

@mortal marten Has your question been resolved?

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[stiff nest]

are these the only ways

i differentiated the expression using first principles

then got 8x - 5

and subbed in x = 2

and got 11

why doesnt the mark scheme accept this

[primal bone]

It does, it's essentially Alt 1

Mark schemes aren't exhaustive in methods per se (nor can they be all the time)

[fathom jewel]

[primal bone]

This is literally in the mark scheme

You've found the derivative, via first principles as required

That leads you up to the "and so" part

Then you've done a substitution

[dim wraith]

lol

[primal bone]

That's the rest of it

If you want a mark breakdown (which is overkill for assistance here lol but):
B1 - demonstrate an understanding of what "first principles" means
M1 - acknowledged the need to do something algebraic to get to some "first principles" fractioN
A1 - substituted the right things in
M1 - ended up with some function which you can use to then calculate the right gradient
A1 - correct gradient calculated

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[mortal marten]

i gotta show that this expression = cosec(x)

[mortal marten]

first i did sin(x)/(sqrt(cos^2(x)) X sqrt(tan^2(x)/(sin(x)+1)^2

then sin(x)/cos(x) X tan(x)/(sin(x)+1)^2

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[dense oriole]

I have a weird/stupid basic theoretical question just food for thought. Lets suppose in a concept perfect scenario where we have infinite precision an object of length 1 like [0,1] interval. We then decide to cut the object so we cut it at exactly 0.5 mark, considering this is a perfect hypothesis where there is no loss of the "material" how can we have 2 perfect halves after? I mean if we cut at exactly 0.5 mark where does 0.5 go? I know it is the shared boundary of the 2 parts. One could say that then we have [0,0.5] and [0.5,1] but i think thats in the case of theoretical intervals when we analyze like a function or something but not when we talk about a "physical object" since one part can be at only place at a time. i dont know if you actually understand whats making it difficult for me to understand it.

tl;dr In a perfect scenario where you have an object of length 1 with infinite precision, if you cut it in half where does the halfway mark (0.5 point) go? How can you have 2 perfect halves?

Thanks in advance,

[small sandal]

if I cut an object of length 1 I get two halves of length 0.5 right
If I then recombine them, I get 0.5 + 0.5 which is 1.

[wild copper]

The answer is: you don't care where the halfway mark go, the measures of both halves of the object are still both 0.5

[dense oriole]

but if you think about it physically when you have the whole object of length 1 and you cut at 0.5, well there is something at the 0.5 marl , doesnt that need to go somewhere?

[wild copper]

You decide

[dense oriole]

i mean this makes sense numerically but it kind of confuses me when i think about this example i gave like a more geomtric way

[stoic dune]

Can't be done physically, much like how you can't have an infinitely precise length

In math we don't really worry about these silly things

[dense oriole]

so this is like absurd by definition?

yeah i figured but i dont know i just had it as a weird though and couldn't actually find an answer myself that satisfied me

[wild copper]

If you strikethrough the "physical" part of the question, it's actually a pretty interesting food of thought.

You'll learn about measure theory this way.

[dense oriole]

yeah i guess so

i get these kind of though sometimes

so we have a final one liner to close this?

like a solution or an explanation

[wild copper]

All of the [0, 0.5), [0, 0.5], (0.5, 1], and [0.5, 1] have the measure of 0.5. Whether the boundary is included or not does not change the measure.

If you want to close it now, type .close

[dense oriole]

thanks for the help guys

.close

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[safe radish]

Please don't occupy multiple help channels.

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[dreamy kestrel]

Hi

[dreamy kestrel]

w is a complex number here

why cant we bring all terms to one side and put real parts equal to zero and imaginary parts to 0

and find w

I did that but I got an absurb value, not a complex number w

[wild copper]

Because w is also complex?

[dreamy kestrel]

🤔

[wild copper]

I would collect the terms with w on one side, and the terms without w on another side.

Complex numbers form a field, so we can divide.

[dreamy kestrel]

thats what answer sheet did. But I dont understand, if we just bring all terms to one side and put real parts to zero and imaginary to zero

Cant we find w

by that

[wild copper]

Assume you want to solve 3 - wi = 0. With your procedure, we set 3 = 0, and -w = 0. Already we've encountered a contradiction.

[zealous tundra]

did u try putting w as a +ib?

[dreamy kestrel]

Oh yes! that could work too

Nice dude

[dreamy kestrel]

I see I see

Thanks man

I get it

.close

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[kind adder]

Need help

[nimble wyvern]

!da2a

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[kind adder]

Ok

The question is

[nimble wyvern]

drumroll noises

[kind adder]

1 sec

[austere goblet]

I'd recommend you just send an image of the question if you have it.

[nimble wyvern]

-# just send anything at this point ;-;

[kind adder]

Given the function f: \mathbb{R} \rightarrow

[nimble wyvern]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[kind adder]

I am stuck to b

I think I need to do Bolzano

[earnest nacelle]

<@&268886789983436800>

[nimble wyvern]

<@&268886789983436800>

[kind adder]

Emmmmm

[nimble wyvern]

wow shitty internet didnt send for 4 seconds

[kind adder]

So I do Bolzano ?

[open wedge]

<@&268886789983436800>

wow im late

😭

[kind adder]

Yo what I do I am still stuc

[nimble wyvern]

wow they posted the link in 9 channels and the mods still not here

<@&268886789983436800>

[kind adder]

Just ban him bro

[austere goblet]

we wish we were mods

[nimble wyvern]

wtf 2 at once

[kind adder]

Ok now at the b I do Theorem of Bolzano ?

And monotony

??????

[austere goblet]

if by the Bolzano theorem you mean the intermediate value theorem, then yes

[kind adder]

Οk wait a sec to do it

[austere goblet]

also please be patient.

[kind adder]

Ok mb

[austere goblet]

gotta need some time to read and plan out answers. you don't want a rushed, half-baked answer too, I hope (that is assuming the helper knows at a glance what to do)

[kind adder]

Ok

I did it

Now c

[vague phoenix]

Bolzano-Weierstrass

[kind adder]

I did that for c is good ?

[whole hill]

Well, c was asking you to prove that the function is 1:1

And THEN, find its inverse

[kind adder]

Yea I did it now I do the graph

For d

[whole hill]

It doesn’t seem like you did

[kind adder]

It’s f^-1(x) = -ln(x-2)

[whole hill]

But that’s not what you wrote

[whole hill]

That’s not in this picture, lol. Also, how did you prove the function is 1:1? Finding an inverse to prove its 1:1 is circular reasoning

[kind adder]

I found the derivative to be negative for all of R, so f is strictly decreasing.

So is 1-1

Then I solve f(x) =y

Go see the first picture

[whole hill]

I can’t read your handwriting

[kind adder]

Ik hahahaha

I just solve it and I found it

simple when I derived before I didn't say that f is differentiable

[whole hill]

But Yk bro. That’s a really smart way to do it. The way I typically do it is:

Assume

f(x1) = f(x2)

Where x1 and x2 are arbitrary inputs.

And if I solve, and get

x1 = x2

Then the function MUST be 1:1 since the only way we get the same output is if the x values are the same

Anyways, as for D, what did you get?

[kind adder]

Yeah, exactly! Since I already proved it's strictly decreasing (f'(x) < 0), it's automatically 1-1. Your method is the classic way to do it, but using the derivative saves a ton of algebraic work in this case

[whole hill]

No, Ik. Which is why I commend you. It’s a really clever way to do it

[whole hill]

Anyways, have you done your graphs for ‘d’

[kind adder]

Wait a monomer for d

Moment *

[whole hill]

Torque

[kind adder]

I did it

He tells me to do the graphic representation in advance, so I don't mind

[whole hill]

There’s an issue with your inverse graph. You’ve got the correct orientation, and correct vertical asymptote. But the range of the parent function, y = ln(x) is all real numbers. So, the graph should extend to -♾️

[kind adder]

Now

[whole hill]

It was closer the first time. lol.

You still need the vertical asymptote of x = 2

You still need the graph to be flipped because of the negative out front

And you need the graph to approach -♾️ as x approaches ♾️

[kind adder]

This one ?

[whole hill]

Bingo bro!! That bottom one is correct

[kind adder]

Ok

I have one more exercise

[whole hill]

I can’t see on your screen shot. But for the integral one, is there a -45 at the end?

[kind adder]

I translated to English so you can understand it

[whole hill]

Oh, what language do you originally speak, if you don’t mind my asking

Cool bro!!

[kind adder]

Greek

[kind adder]

Where you from

If you don’t mind of course

[whole hill]

So, for this one, since you know that the DEFINITE integral is gonna be a constant, just replace it with C for now.

So you’d get:

f(x) = (10x³ + 3x) * C - 45

Whatever you get you can substitute it for f(t), then take the definite integral from [0, 2].

Substitute C back into the original equation f(x), and then you’re gonna get that:

f(x) = 20x³ + 6x - 45

[kind adder]

This one ?

[whole hill]

I’m from Colombia, but I live in Canada

[kind adder]

Ok

[whole hill]

,rotate

[whole hill]

Yep, now, do the integral, and solve for I

Remember, I is a constant, so you don’t have to worry about doing any fancy math with it. lol

[whole hill]

Bingo!!!

Now, substitute “I = 2” into our original expression of f(x)

[kind adder]

I did it

Now the b

[whole hill]

Can I see what you did?

[kind adder]

I solve it

It was easy one

[whole hill]

I meant, can I see what you did for this one

[kind adder]

It’s on the top

Of the picture

Ant b

I did it

[whole hill]

Good job bro!!

[kind adder]

Now c

How I start ?

[whole hill]

I don't see a "c"

[kind adder]

I found g’’

Now I think I have to ingestion tow times ?

<@&286206848099549185>

<@&286206848099549185>

[pale dock]

Stop?

[kind adder]

Mb

[pale dock]

Dont ping helpers twice in a row

And why would you even TRY to ping everyone in a server of 330 thousand?

[kind adder]

Mb sorry

[safe radish]

@kind adder Has your question been resolved?

[kind adder]

No

[whole hill]

Just an FYI, helpers are less inclined to help you if you ping a lot

[kind adder]

Mb

[kind adder]

Can you check my work

I think I solved

[kind adder]

.close

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[gilded star]

2^9?

chat gpt solved already

.close

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[hoary seal]

this is wrong

[zinc hornet]

If you look at the notation, you have the u and vs swapped

There is also no way you have a cos(x) without the scalar inside

[hoary seal]

isnt u(x)= e^(15*x)?

[zinc hornet]

If you look at your screenshot, u(x) is the one you differentiate

$uv' = uv - \int vu' $

[hoary seal]

$ is the bot working $

[zinc hornet]

It doesn't really matter, I would hope you can read that either way

[hoary seal]

but im just matching the given statement to the general statement

and it aligns

[burnt notch]

But the website is using the other way round...

[zinc hornet]

I'm not sure what you mean by that. Multiplication is commutative. You can see that you are associating the wrong function based on how they are using the notation.

In their notation, u(x) is the function you differentiate, v(x) is the one you integrate.

It doesn't really matter what you call u or v, but in this case it does because you have to fill in a box

[hoary seal]

so we cant assume too early, we need to look at the derivative of both u(x) and v(x)?

[zinc hornet]

The question is defining the notation at the top

You need to use their definition

[hoary seal]

how do u know that exp(15*x) is not u(x)?

[zinc hornet]

Because you have no way of generating a 1/15 in the denominator

u'(x) = 15e^(15x)

[hoary seal]

ok so u(x)= sin(8*x)

Lhs is first case

RHS is second case

[zinc hornet]

You could technically define u(x) to be e^(15x) but it is more of a technical nit picky reason that is just going to confuse you.

It is definitely not the intended answer

[hoary seal]

So what I’m doing

Here is good so far

[zinc hornet]

well, your u's look exactly the same as v's so I don't know for one, and also the derivative of sin(x) is cos(x).

You have an integrand, e^(15x) sin(8x) and you want to split it into two pieces, one piece you integrate and one piece you differentiate. It looks like you are differentiating both sides there.

[hoary seal]

No I wrote it to test, so my correct one is on the right hand side not lhs

This is wrong too

[zinc hornet]

You really need to just look at the formula at the top and make sure you are matching the symbols correctly

Write down what is the u(x) and what is the v(x) in your p(x) and check that it makes sense for what you are doing

[hoary seal]

u(x)=sin(8*x)

[burnt notch]

Yup

[hoary seal]

to find v(x) i can just integrate e^15*x

[burnt notch]

Well, what is your choice for v**'**(x)?

[hoary seal]

v(x)=e^(15*x^2)/2

[burnt notch]

Huh????

Where the heck is that coming from?

Also, I asked for v**'**, not v

[hoary seal]

Im trying to integrate e^(15*x)

[burnt notch]

.

[burnt notch]

By the way, this is a HUGE and serious mistake if that is how you think you integrate e^(15x)

[hoary seal]

how do we integrate exponentials again?

i know that we dont bring the 15 down

[burnt notch]

Haven't you been taught it? 🤔

[hoary seal]

yeah last year but I think I forgot

[burnt notch]

Well, then go revise it

If you're doing integrals you need to know how to integrate exponentials VERY WELL

[hoary seal]

ah so its

e^(15*x)/15

+C

🙂

.close

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[stiff nest]

hey

[stiff nest]

for this

ur supposed to do it like this

but can you also assume that there are no positive integers a and b with a even such that a + 2b = sqrt(8ab) ?

[tardy mango]

What would that accomplish

Ex. 3a+4b=13 has no integer solutions where a is even (why?) but there's definitely solutions with a odd, such as (a,b)=(3,1)

[stiff nest]

hm

[primal bone]

The question doesn't ask for this in any case

The assumption for a contradiction must be that a and b are positive integers, a is odd, and that that equation holds

(if a is even, who cares)

[safe radish]

@stiff nest Has your question been resolved?

[safe radish]

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[vale onyx]

idk how to deduce lambda

[vale onyx]

is it because NB has no a component and u can write it as like mu b

[safe radish]

@vale onyx Has your question been resolved?

[vale onyx]

veri much no

i think it is time.. <@&286206848099549185> help pls? 🙏

[hard crest]

ON should be expressible solely in terms of b

given N is on the line OB

[vale onyx]

ok

.close

[safe radish]

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[vale onyx]

.reopen

[safe radish]

✅ Original question: #help-23 message

[vale onyx]

ok i just dont get it

[vale onyx]

is it like this

1/2 lambda is mu

[hard crest]

i think you can find lambda

you know that ON = (2 - 3/2 L)a + 1/2 Lb which that a component needs to be 0 right

[vale onyx]

ook

but the scale factor for NB is not Lambda right

[hard crest]

i don't think so but i'm not sure

[small flame]

can someone help with 9th algerba rq?

[vale onyx]

when u have the version with only b

[vale onyx]

use an open help channel

[hard crest]

i haven't run the numbers

[small flame]

can u # it

[hard crest]

use any of these

[vale onyx]

bro its on the side try #help-12

[hard crest]

but you should be able to get an actual answer for what lambda equals

like an actual numbr

[vale onyx]

ye 4/3

[hard crest]

probably 2 or 3

ah cool

[vale onyx]

.close

[safe radish]

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[safe radish]

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