help-49

[cedar vortex]

Bro got cooked

It’s alr tbo

I gotchu

[radiant gorge]

i did

[cedar vortex]

Lhs you get |10-2y|

RHS you get e^{x+ln(6)}

Which I trust you can simplify

[radiant gorge]

soo you get

-3e^x + 5 or 3e^x+5?

did i pull that out of nowhere

[cedar vortex]

Lowkey

You sorta cooked

[radiant gorge]

am i COOKED or did i cook

cook

damn my grammar 😭

[cedar vortex]

Bro I’m going brain dead

[radiant gorge]

me too

i have like 2 more frqs after this

all for homework

[cedar vortex]

I remember I did like 13 ap frqs in one day

Cuz I waited till the last day

[radiant gorge]

ew oh m ygod

[radiant gorge]

i was super sick so i decided to leave everything

and im left with this 😭😭😭

[cedar vortex]

All good

Just don’t sell on the ap exam

[radiant gorge]

i'll try

but ddi i write the right answer

hopefully

[cedar vortex]

Nah

All good tho

[radiant gorge]

NOOOOOOO

MY ALGEBRA SKILLS 😭😭😭

[cedar vortex]

We move cuz if you don’t know absolute value

Im@not bouta teach you

Sorry

[radiant gorge]

you good

i think i just boofed

im still sick so it's clouding up my brain

[cedar vortex]

Bro stop 😭

[radiant gorge]

😭😭😭😭

[cedar vortex]

I’m still tryna find hamidur on Reddit

[radiant gorge]

im sorry if this sounds stalkerish but we have a mutual omg

[cedar vortex]

How do you know gabe wtf

[radiant gorge]

we both got into yale

yale gc

how do yk him

[cedar vortex]

Gabe saved my life

[radiant gorge]

omg 😭😭

[cedar vortex]

Basically we were both in Columbia and this guy was tryna mug me

And then gabe gave the 2-3 mayweather

Guy ran away

[radiant gorge]

OH he asked me if i wanted to join you guys like a while ago

[radiant gorge]

DAMNNNN

go gabe

🥳🥳🥳

im not super close with him though i just talked to him for like a little bit

[cedar vortex]

You should get close

And do his hw

[radiant gorge]

yale and cant do algebra 2 😭😭😭 cooked

[radiant gorge]

isn't he doing linear algebra

hellllllllll no im struggling with AP CALC 😭😭😭

[cedar vortex]

Anyway we focus on the last q

[radiant gorge]

alralr

[cedar vortex]

Do NOT sell

[radiant gorge]

i will TRY

[cedar vortex]

If you sell

You’re going to jail instead of yale

Fr

[radiant gorge]

nah i had to decline

i didn't get a scholarship

decling yale. insane.

[cedar vortex]

Insane where you going now

[radiant gorge]

thinking of baruch

since it's like way cheaper and a degree is a degree anyways

[cedar vortex]

Nah go to Yale

[radiant gorge]

im not getting any financial help so i can't afford it

[cedar vortex]

Financial aid office go and beg

[radiant gorge]

im gonna be eating sugar soup for breakfast lunch and dinner

[radiant gorge]

I DID AND THEY SAID NO 😭😭😭😭

OK ANYWAYS

[cedar vortex]

Give me 30% of ur income for 30 years and I’ll pay for 2 years of Yale

Income after taxes

[radiant gorge]

for the particular solution y = f(x) found in part c, find lim x-> inf f(x)! 🤓🤓

[radiant gorge]

ok sharktankkk

idk yale sstill on my list but 😭😭 ill see eventually

[cedar vortex]

Nah im not helping you if ur throwing away ur future

[radiant gorge]

BUDDDYYYYYY

😭😭😭

[cedar vortex]

You could go to uni with Gabe!

[radiant gorge]

he mgiht not even go to yale

bro is thinking of doing 6 majors

crazy

and i only got in because of my ECs and essay so like i wouldn't be able to compete with other people

[cedar vortex]

Aight let me help you with ur problem

And I’ll have a lil chat

[radiant gorge]

ok thank you 🙏

[cedar vortex]

Looking back

You didn’t integrate properly

[radiant gorge]

Y = - (e^{x+ln(6)} - 10)/2 and the positive one???

[radiant gorge]

oh no

😭

[cedar vortex]

Should be like 5 or -5

[cedar vortex]

Mhm

[radiant gorge]

oh so dont put the /2

wait

[cedar vortex]

Nah

You fucked up a negative

[radiant gorge]

oops

[radiant gorge]

where did it go wrong

[cedar vortex]

2y-10 = e^{-x+ln(6)}

[cedar vortex]

Yeah it’s y = 5 p sure

Aight we good

[radiant gorge]

oh thank you

[cedar vortex]

If you went to yale, you would have got a tutor who would catch that mistake

[radiant gorge]

STAWP

[cedar vortex]

Anyway you go to yale

[radiant gorge]

you're right but i just cant afford anything 😭😭😭

[cedar vortex]

Make a lotta money and donate it to the math server

[radiant gorge]

smart

[cedar vortex]

I’ll pay the tuition for %

[radiant gorge]

👎

bad risk for you

i feel like i wouldn't have enough motivation at yale either so i wouldn't do very well

[cedar vortex]

Aight

Say good things about me to gabe and Neil

I’ll cya

[radiant gorge]

Alright cyaa

[midnight plank]

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[kind sundial]

From this drawing what can x be

[kind sundial]

A)10 B)20. C)40. D)50. E)110

[lapis citrus]

b

[dim lava]

Actually it’s b

[kind sundial]

Why?

Explain pls

@dim lava

[lapis citrus]

has to equal more than 90 but less than 180

[kind sundial]

What?

Why does it have to be more then 90

?

[dim lava]

Because 180-5x is an angle in that other triangle

[kind sundial]

Ik that

But thats not all pls explain you resening to why it is b)

[lapis citrus]

90 is a square

n that angles bigger

180 is a line

n that angle is smaller

[midnight plank]

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[midnight plank]

@cerulean linden Has your question been resolved?

[midnight plank]

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[dim lava]

The first condition tells you that Adrian's favorite number contains 5 digits, all odd, and that none of the digits are repeated. Meaning that the number is just the digits 1, 3, 5, 7, 9 arranged in some order. That should be enough to get you started, then you can use this information in combination with the next two clues to figure out which order the numbers have to be in.

[magic orchid]

okay thank you

.close

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[ivory stag]

can someone help me do 7 fully

[ivory stag]

idk how to start or how to do any steps

[rain wasp]

a/b=c/d thereforr ac=bd

[ivory stag]

whats a b c d

[cloud wadi]

representing the equations

[ivory stag]

?

[cloud wadi]

the linear and quadratic equations in your question

Also you should notice how x^2 - 4 can be factored into (x-2)(x+2)

[ivory stag]

so like i said earlier idk how to do any steps and idk what a b c d

[cloud wadi]

how did you solve the questions before this?

[ivory stag]

<@&286206848099549185>

[wide marsh]

you have to do something called cross multiplying

[ivory stag]

how would 1 do that

[wide marsh]

any time you have two fractions equal to each other you multiply the denominators to the numerators of the opposite side of the equation

someone has already labelled a picture with a b c and d for you

[ivory stag]

who?

[wide marsh]

.

[wide marsh]

to cross multiply you shuold have a x d on one side and b x c on the other side of the equal sign

or you just can two diagonal arrows if it helps you better visualize without thinking of letters

[ivory stag]

ok i did axd and bxc wat would i do next

[wide marsh]

now that you dont have any fractions you can bring everything to one side of the equals sign and make the other side 0

so rearrange your equation to get a quadratic

and then factor to find roots of x

[ivory stag]

u lost me

[wide marsh]

what do you have

[ivory stag]

how do i this cause i have axd and bxc

cause there 2 diffent things

[wide marsh]

multiply the bottom left with the top right

and the bottom right with the top left

thats called cross multiplying

[ivory stag]

ohhh they equal each other ok

[wide marsh]

yes

[balmy egret]

what do you wanna do in this equestion? find value of x?

[ivory stag]

thats where i was lost

[wide marsh]

this isnt my question but i assume yes find x

[ivory stag]

just say slove rational equations

[balmy egret]

cross multiply the equation first

then on the right hand side you'll find (x2 - 4)(x + 1)

2^2 = 4

(x2 - 2^2)(x + 1)

[ivory stag]

im caught up to here

[wide marsh]

ok have you learned factoring yet

factor the trinomial to find roots of x

[balmy egret]

or 3(x+2)(x-2)=(x^2 - 2^2)(x+1)

[ivory stag]

yea it makes sense

[balmy egret]

on left hand side use identity x2 - y2 = (x + y)(x - y)

[ivory stag]

how would i do that since it has a x^3

[balmy egret]

where?

[wide marsh]

https://www.youtube.com/watch?v=-4jANGlJRSY&pp=ygURZmFjb3RyIHRyaW5vbWlhbHM%3D

[balmy egret]

wait lemee send you my solution

[ivory stag]

these are all x^2

[wide marsh]

oh youre right

well do you know how to factor by grouping

[balmy egret]

see this

[ivory stag]

i forgot wat it called but ik how to get it down to x^2 just kinda a pain

[balmy egret]

is it okay?

[sweet skiff]

I think it is

[wide marsh]

pretty sure you hve a mistake in second line

[balmy egret]

it's not a mistake i just forgot to write the whole equstion

my bad

[balmy egret]

.

[ivory stag]

no

[sweet skiff]

Should be like this right?

[wide marsh]

tbh its not very good to solve these kinds of questions like that

you really should be bringing everything to one side, making the other side 0 and then factoring to solve

you are crossing out the factor (x+2) and pretending its not relevant but it is

hence youre missing one of the roots x=-2

[balmy egret]

well if the equation is easy to solve then it's stupid to make it hard

[ivory stag]

the way hes been teaching me has been super easy

[balmy egret]

(disclamer:- dont use this advise on your maths test)

[balmy egret]

that's nice

[ivory stag]

just need to know what to do next

@wide marsh

[wide marsh]

bro

just factor by grouping and solve for x

if you havent learned that in class then maybe dont worry about it and just ask your teacher

but with these questions just cross multiply then simplify and sovle

[ivory stag]

ight would the answer just be 2

[wide marsh]

no it shuld be -2 and 2

[ivory stag]

right right i forgot x^2-4 could be 2 or -2

[wide marsh]

here is the full solution

[ivory stag]

up mine look the same

thank u so much

[wide marsh]

good job

np

[ivory stag]

.close

[midnight plank]

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[last slate]

hi help

[last slate]

No wrong out

Working

[pearl hull]

what have you done so far?

@last slate

[last slate]

absolutely nothing

completely lost

teacher said to figure it out somehow

[pearl hull]

do you know the line that is perpendicular to x-y=0?

name one of it

[last slate]

what no

[pearl hull]

I see

what's the gradient of x-y=0?

[last slate]

I’m nor sure

what topic is this

I feel like I need to go relearn it

[pearl hull]

same chapter of yours

[last slate]

yeah what chapter is this

the question came on a mixed paper question

[pearl hull]

line? perhaps

we call it circle and line here

[last slate]

Okay

what abt rhis

I know the answer is 1.4

but how?

[pearl hull]

do you know the calculations of log?

[last slate]

do you get 4 —> 2 and increase the power by 1

[last slate]

not allowed to use log

pure algebra

IGCSE

[pearl hull]

sure

4^2 = 2^x

what is x?

[last slate]

3

[pearl hull]

you sure?

[last slate]

wait no what

oh I read that wrong

4 to the power of 2

[pearl hull]

yes

[last slate]

so 2 to the power of 4

my bad

but how does that help

[pearl hull]

chill

be patient XD

[last slate]

Okay

[pearl hull]

what's 4^2x? do you know how to transform it into the form of 2^a?

[last slate]

No I don’t

perhaps 2^3x

or

16^a

[pearl hull]

well

a^2b =(a^2)^b

this is a required skill before this question

for example
16=2^4=(2^2)^2

[last slate]

Okay that makes sense

[pearl hull]

try it

[last slate]

One second

[pearl hull]

ping me when you're done

[last slate]

Do I combine the two numbers

4 and 2

[pearl hull]

yeah, you have to try to transform 4^2x into the form of 2^a

so that you can combine the seperate terms

[last slate]

so..

(2^3)^a?

[pearl hull]

nope

try again 🙂

[last slate]

I don’t know how else to do this

[pearl hull]

4 is 2^2

[last slate]

yes

[pearl hull]

also, you missed the coefficient before x

[last slate]

which is

[pearl hull]

uhh, dont you wanna try it out?

[last slate]

No I’ve been trying these two questions since the morning

[pearl hull]

...

I see

[last slate]

it’s 7PM now

[pearl hull]

4^2x=(2^2)^2x=2^4x

[last slate]

that’s the combination of both those terms

??

Where did you get those from

[last slate]

that doesn’t make that much sense tho

[pearl hull]

it does

[last slate]

the powers inside multiplied?

Oh right power laws

[pearl hull]

keep this in mind

[last slate]

It took me a second right

next step

2^4x multiples by 2^x

[pearl hull]

use the first formula

[last slate]

What toenails

Formalize

formulae

[pearl hull]

here

[last slate]

so 2^5x??

[pearl hull]

👍

[last slate]

2^5 is 32?

[pearl hull]

yeah

[last slate]

Next?

[pearl hull]

find the value of x

[last slate]

how 💀

[pearl hull]

you can start plugging from x=2

as soon as you get 128, you stop

[last slate]

so.. I start guessing??

[pearl hull]

that's what you do without log

[pearl hull]

yeah, start from 2

[last slate]

There’s gotta be a more efficient method than guessing 💀

Also how is the examiner gonna accept that 💀

[pearl hull]

yes there are

[last slate]

and

[pearl hull]

transform 128 into 2^a

[last slate]

💀

Give me a sec

2^7

I don’t know how I got that tho

[pearl hull]

forget this, I just realized x isn't an integer XD

[pearl hull]

yes

[last slate]

yeah 💀

[pearl hull]

solve x

[last slate]

How

[pearl hull]

you have everything you need

bro, do you write down what we have done so far?

it appears that you have written down nothing

[last slate]

yes

[pearl hull]

yeah, you have 2^5x = 2^7

[last slate]

PHHHH

OHH

then I can do

5:7

5/7

[pearl hull]

👍

7/5 precisely

[last slate]

oh yeah that’s what i meant 💀

Thanks chief

[pearl hull]

np

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[last slate]

.close

[midnight plank]

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[last slate]

.repopen

.reopen

[midnight plank]

✅

[last slate]

new question sorry

how do I solve this

[midnight plank]

@last slate Has your question been resolved?

[last slate]

hello

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[frozen tree]

How to make it C = f(C) so I can solve it by the iteration method ?

[fresh sparrow]

multiply both sides by C and divide by v(t)

[frozen tree]

Or could I just

sub V

[fresh sparrow]

yes

same thing

[frozen tree]

and mult. by c

I'll try this first

[fresh sparrow]

yes, it's the same thing as i was saying

you get $C=40\cdot(1-e^{-(C/68.1)\cdot 10})$

[grand pond]

artemetra

[midnight plank]

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[half granite]

What is the degree of this equation?

[half granite]

Can we transpose the sin x then take the arccos of both sides

<@&286206848099549185>

can someone please assist me with this question

[jovial crest]

Not defined

Because it canr be expanded as a polynomial

[half granite]

why can't i take the arccos?

like degree of ln(dy/dx) = x is 1

[jovial crest]

arccos(cosx)=x only when x lies inbetween 0,π

So that equatiom wouldn't be completely accurate

For all x

[half granite]

if i define it say piece-wise? is that allowed?

i mean if i add or subtract pi or pi/2's multiples? is that something we can do?

i'm not sure tbh

[jovial crest]

Even if you do it'd be outside of the range

[half granite]

no like if i say dy/dx = arccos(everything else) + npi

[jovial crest]

Best of luck for boards btw heard you guys have only 3 days for maths

[half granite]

yeah thanks

[jovial crest]

Yeah that's right, but again in the equation they've given us, how they've written it matters

If it were in this way Id say pick 1

[half granite]

okay

im just confused how this is different from ln(dy/dx) = x+ y

Another book said arcsin(dy/dx) = x + y has a degree of 1

[midnight plank]

@half granite Has your question been resolved?

[half granite]

<@&286206848099549185>

@jovial crest can u plz explain

[jovial crest]

so they take sin of both sides

[half granite]

what do u mean by that?

U mean a Maclaurin series? That does exist for arcsin

[jovial crest]

oh yeah i forgot about that

[half granite]

how is degree defined then??

[jovial crest]

power to which the highest order derivaive is raised

[half granite]

in what form?

i was basically asking for a more specific definition

[midnight plank]

@half granite Has your question been resolved?

[midnight plank]

@half granite Has your question been resolved?

[midnight plank]

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[final sky]

how do i properly and mathematically divide the 13 by the 6/5 ratio?

[last slate]

hey so when you wanna divide 13 by 6/5 you're basically flipping the fraction and multiplying so it's like 13 times 5/6 do that and you get 65/6 which rounds to about 10.83 if you wanna keep it super simple just remember dividing by a fraction is like multiplying by its opposite

[final sky]

Ah ok

[last slate]

is it good?

if it's good you can close it with .close

[final sky]

Yep, looks good

thank you

.close

[midnight plank]

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[past aurora]

Let ( A = \begin{pmatrix}
a_{11} & a_{12} & a_{13} \
a_{21} & a_{22} & a_{23} \
a_{31} & a_{32} & a_{33}
\end{pmatrix} ), such that ( \det(A) = 7 ).

\bigskip

Calculate the determinants of the following matrices.

\bigskip

i) ( \begin{pmatrix}
a_{11} & 2a_{12} & -a_{13} \
a_{21} & 2a_{22} & -a_{23} \
a_{31} & 2a_{32} & -a_{33}
\end{pmatrix} )

\bigskip

ii) ( \begin{pmatrix}
a_{11} & a_{12} & a_{13} \
a_{21} + 3a_{11} & a_{22} + 3a_{12} & a_{23} + 3a_{13} \
ka_{31} & ka_{32} & ka_{33}
\end{pmatrix} )

[grand pond]

milanesa a caballo

[frail burrow]

@past aurora

What have you tried?

Do you know about some properties of the determinant?

[past aurora]

I have tried nothing, I have 0 idea.

[past aurora]

does one property comes to your mind?

[frail burrow]

Do you know what happens to the determinant, if you perform row/column operations on A?

For example, what happens if I multiply the first row by a scalar

How does that affect the determinant? Do you know that?

[past aurora]

yes

det(a*k) = k*det(a)

[frail burrow]

No

I'm assuming a is the matrix

I suggest capitalizing the letter when it's a matrix

But also that's not true

k * A where A is a matrix is multiplying EVERY entry by k.

[past aurora]

det(A*k) = k*det(A)

[frail burrow]

I'm just asking what happens if I multiply the first row

As an example

[frail burrow]

Again, that's false

Are you learning using a book maybe?

If you do have a book, it should clearly states the ways the determinant is affected

[frail burrow]

Alright read what A' is

I don't know that language but I can guess what it is

[past aurora]

A' is the resultant matrix after multiplicating one row of A by a scalar K

being $A \in \mathbb{R}^{n \times n}$

[grand pond]

milanesa a caballo

[frail burrow]

Right

[past aurora]

what now?

[frail burrow]

So you understand this?

[past aurora]

spanish . . .

[frail burrow]

Yeah I assumed so

So you understand that multiplying one row by a scslar, also multiplies the determinant by that scalar

[past aurora]

yes

. . .

[frail burrow]

Meaning, if B is the matrix you get after multiplying a row in A by a scalar c, then: det(B) = c * det(A)

Okay,

Now in the first exercise we have something similar

[past aurora]

yes . . .

[frail burrow]

Notice the differences, we multiplied the second column by 2 and the third column by -1

But those are columns and not rows

So what can we do to turn them into rows?

[past aurora]

?

[frail burrow]

Try to think

[past aurora]

not sure

[frail burrow]

It's the transpose

Do you see why?

A^T

[frail burrow]

By the way does this talk about columns aswell?

Or only rows?

[past aurora]

I am not following.

[frail burrow]

Well just answer this...

I really don't think I'm asking for much

Does it mention anything about columns?

[past aurora]

2 talks about swapping rows
3 is just summing a multiple of one of the elements of A so called A' then same determinant

im confused.

[frail burrow]

[
\begin{vmatrix}
a_{11} & 2a_{12} & -a_{13}\
a_{21} & 2a_{22} & -a_{23}\
a_{31} & 2a_{32} & -a_{33}
\end{vmatrix}

\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
2a_{12} & 2a_{22} & 2a_{32}\
-a_{13} & -a_{23} & -a_{33}
\end{vmatrix}
]

[grand pond]

RedstonePlayz09

[frail burrow]

Do you see why this is true? I just transposed the matrix

So the determinant stays the same

[past aurora]

?

I dont see why is this true. but I can accept that $det(A) = det(A^t)$ and live with it.

[grand pond]

milanesa a caballo

[frail burrow]

Okay, that's enough for me

Later you can go over the proof for this if you feel like you want to

Okay, so I'm gonna continue:

[past aurora]

. . . please continue . . .

[grand pond]

RedstonePlayz09

[frail burrow]

So again, you see what I did?

[past aurora]

its wrong doe

I think

let me analyze it deeply

[frail burrow]

I just took the 2 out

Because the whole second row was multiplied by it

[past aurora]

the thing is that only second row multiplies by 2

the other rows doesnt.

. . .

[frail burrow]

The theorem in your book says, that if you get A' by multiplying a row by a scalar, then det(A') = c * det(A) where c is that scalar

It doesn't say "every" row

Just a single row

[past aurora]

shiiiiiit

so what now?

. . .

[frail burrow]

Soooo we continue:

[past aurora]

. . .

[frail burrow]

[
\begin{vmatrix}
a_{11} & 2a_{12} & -a_{13}\
a_{21} & 2a_{22} & -a_{23}\
a_{31} & 2a_{32} & -a_{33}
\end{vmatrix}

\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
2a_{12} & 2a_{22} & 2a_{32}\
-a_{13} & -a_{23} & -a_{33}
\end{vmatrix}
=2
\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
a_{12} & a_{22} & a_{32}\
-a_{13} & -a_{23} & -a_{33}
\end{vmatrix}
=-2
\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
a_{12} & a_{22} & a_{32}\
a_{13} & a_{23} & a_{33}
\end{vmatrix}

\begin{vmatrix}
a_{11} & a_{12} & a_{13}\
a_{21} & a_{22} & a_{23}\
a_{31} & a_{32} & a_{33}
\end{vmatrix}
]

[past aurora]

your shit cut off

@frail burrow

its too long homie

. . .

[frail burrow]

oh its because its too long

[past aurora]

. . .

[grand pond]

RedstonePlayz09

[frail burrow]

ok this is enough for now

yo latex

update it?

omg

[past aurora]

can you elaborate

[frail burrow]

$\begin{vmatrix}
a_{11} & 2a_{12} & -a_{13}\
a_{21} & 2a_{22} & -a_{23}\
a_{31} & 2a_{32} & -a_{33}
\end{vmatrix}

\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
2a_{12} & 2a_{22} & 2a_{32}\
-a_{13} & -a_{23} & -a_{33}
\end{vmatrix}
=2
\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
a_{12} & a_{22} & a_{32}\
-a_{13} & -a_{23} & -a_{33}
\end{vmatrix}$

$
=-2
\begin{vmatrix}
a_{11} & a_{21} & a_{31}\
a_{12} & a_{22} & a_{32}\
a_{13} & a_{23} & a_{33}
\end{vmatrix}

\begin{vmatrix}
a_{11} & a_{12} & a_{13}\
a_{21} & a_{22} & a_{23}\
a_{31} & a_{32} & a_{33}
\end{vmatrix}
$

[past aurora]

. . .

[frail burrow]

WHY IS THERE A MINUS

[past aurora]

just do a line break . . .

. . .

[grand pond]

RedstonePlayz09

[past aurora]

that property about the minus sign, I am confused.

[frail burrow]

good enough

[past aurora]

only one row is multiplied by -1 . . .

[frail burrow]

I just took the -1 out aswell

Yeah

only one row..

[past aurora]

?

[frail burrow]

that's why..

When one row is multiplied by a scalar

[past aurora]

[frail burrow]

the determinant gets multiplied by it aswell

[past aurora]

thanks.

[frail burrow]

oh

I forgot the -2 at the end

[past aurora]

what about ii) now. . .

[frail burrow]

anyways you gotta transpose it back

transpose, take out 2 and -1, transpose back

and u get -2det(A)

so -14

Ok?

[grand pond]

milanesa a caballo

[past aurora]

elaborate.

[frail burrow]

Where?

[past aurora]

oh, first of all.

take a look at second row, this means we can whipe out the "+ x"

because the determinant doesnt change at all in the second row scenario.

[frail burrow]

yeah

You can subtract -3 * the first row

and it reverts whatever they did

and determinant doesn't change

[past aurora]

what now.

[frail burrow]

Write the new determinant

Continue simplifying it

[past aurora]

how do I find the determinant of ii)

?

[frail burrow]

[
\begin{vmatrix}
a_{11} & a_{12} & a_{13}\
a_{21} + 3a_{11} & a_{22} + 3a_{12} & a_{23} + 3a_{13}\
ka_{31} & ka_{32} & ka_{33}
\end{vmatrix}

\begin{vmatrix}
a_{11} & a_{12} & a_{13}\
a_{21} & a_{22} & a_{23}\
ka_{31} & ka_{32} & ka_{33}
\end{vmatrix}
]

[grand pond]

RedstonePlayz09

[frail burrow]

ok?

[past aurora]

yes.

[frail burrow]

Can you try to continue from here?

[past aurora]

k*9 = det of that

but how do we get k.

[frail burrow]

k * 9?

[frail burrow]

Look again

And I'm pretty sure you aren't supposed to find k, there is no way to find it. Just take it as a parameter.

[past aurora]

,,k \cdot \begin{vmatrix}
a{11} & a{12} & a{13}\
a{21} & a{22} & a{23}\
a{31} & a{32} & a_{33}
\end{vmatrix}

[grand pond]

milanesa a caballo

[past aurora]

@frail burrow

what now?

[frail burrow]

What you have now is just k * det(A)

[past aurora]

yes.

[frail burrow]

and det(A) = 7

not 9

[past aurora]

so det of our shit is 7k

. . .

[frail burrow]

yes

[past aurora]

thanks.

.close

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[frail burrow]

Np

[past aurora]

cya.

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[prisma swift]

How should I approach c?

[prisma swift]

Nevermind, I figured it out

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[echo grove]

how do i work out 19.a

[echo grove]

nvm i solved it myself

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[craggy cosmos]

If I have proven that when the geometric mean is maximized, it it is less than or equal to the arithmetic mean, does it follow that the geometric mean is always less than or equal to the arithmetic mean? I can provide my proof of the first part if needed.

[stable dawn]

Would like to see your proof

[craggy cosmos]

I think I made a mistake in the proof now that I am looking back at it but here it is.

The problem was to prove that the geometric mean is leq to the arithmetic mean for postive real numbers x_1, ... x_n with Lagrange multipliers

I actually just switched the less than or equal at the end to equal, so my question is now if I have proven that the geometric mean is equal to the arithmetic mean at the geometric mean's maximum, is it always less than or equal to the arithmetic mean?

Should I also prove that the minimum of the arithmetic mean is geq the geometric mean? (unsure if this would validate the proof or if its necessary)

@stable dawn

[stable dawn]

Try pinging helpers,I'm not that familiar with this proof.

[craggy cosmos]

<@&286206848099549185>

[stable dawn]

Sorry

[craggy cosmos]

All good!

[stable dawn]

[midnight plank]

@craggy cosmos Has your question been resolved?

[craggy cosmos]

Didn't realize this section was for pre-university math questions lol, don't want to occupy a spot so i'll close it

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[waxen needle]

is possible for anyone to solve this?

[midnight plank]

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[old lark]

I am looking for an explanation/proof(?) for something regarding square roots.

when it comes to division of square roots there is a rule stating that √16 / √4 = √16/4
why does this work? why is it true? I've tried rewriting these expressions and breaking it down into its "simplest form" but i still can't wrap my head around it. Also is there a name for this property of square roots?

[dull yoke]

$\frac {\sqrt a}{\sqrt b} = \sqrt {\frac ab}$

[grand pond]

Stephen

[dull yoke]

this right

[violet storm]

well you can think of roots as fractional exponents, then the rules just follow the rules for addition and subtraction of exponents

[old lark]

yeah

[violet storm]

$(a^n)(b^n) = (ab)^n$

[grand pond]

Soosh

[violet storm]

right? and (a/b)^n = a^n / b^n you can distribute exponents over multiplication or division, are you comfortable with that idea already?

[old lark]

i do understand what you are saying about exponents, however i dont quite get how you could think of a square root as a fractional exponent? what would say the square root of 9 look like if it had a fractional exponent?

[violet storm]

$\sqrt{9}=9^{\frac{1}{2}}$

[grand pond]

Soosh

[violet storm]

ok just take this and like square both sides, that can show you why they are equivalent because on the left when you square, the ^2 exponent can be used on both the numerator and denominator individually and you get a/b, while on the right you square and you get a/b

[midnight plank]

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[twilit field]

how would I prove $x^3+x^2+4x$ and -sin(2x) only have one sol

[grand pond]

Why am. I here

[twilit field]

it's obvious that 0 is one sol

but how would I prove that's the only sol?

by proving their difference is never 0(so by taking the derivative of their difference and proving it's monotonic?

[cedar coral]

that question is not properly constructed.
A polynomial has no solutions. It has roots.
Do you mean x^3 + x^2 + 4x = 0?

[twilit field]

yeah

sorry

[cedar coral]

start by factoring. What do you get?

[twilit field]

x(x^2+x+4)=0

oo

oh, it's the only sol the polynomial has too

[runic hamlet]

do you mean x^3+x^2+4x=0 or =-sin(2x)

[twilit field]

=sin(2x)

should have been clearer, sorry

[runic hamlet]

you could compare the derivatives

[twilit field]

how would that help?

[quiet hinge]

If it's strictly increasing or decreasing, only one solution

[twilit field]

oh yeah, shit , how could i miss that. Thanks

so $3x^2+2x+4-2cos(2x)$

[grand pond]

Why am. I here

[quiet hinge]

Yes

[twilit field]

now the minmum value of this derivative is 2

so it's an increasing function

thanks

[quiet hinge]

Also make sure that at some point the function is below 0

Don't forget that

[cedar coral]

it's a cubic. It covers all reals

[quiet hinge]

I know

[twilit field]

yeah, thanks!

[quiet hinge]

Its still important during other problems

[twilit field]

..clsoe

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[half granite]

How do i approach this?

[half granite]

<@&286206848099549185>

[feral sedge]

consider logarithm

[half granite]

oh alright didn't think of that

i'll try that thanks!

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[next sluice]

where does x^(i-1) come from?

[next sluice]

<@&286206848099549185>

some extra info if needed

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[modern marsh]

can i just show an example, k = 1, so n = 1 which is not divisible by 3 or 7?

thanks

[last slate]

yes

[cinder saddle]

please i need help fast

i would not ask unless i need it

please

[modern marsh]

thanks

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[hot fox]

been stuck for a while on what looks like a basic geometry problem : OIA is equilateral with side 1, OB = CD = 1. i want to show AD = cbrt(2).

[hot fox]

i've shown AID is a right triangle, this implies BI = sqrt(3)

there must be some similar triangles around but i can't figure out where

[midnight plank]

@hot fox Has your question been resolved?

[hot fox]

<@&286206848099549185> if anyone has a clue

[feral sedge]

hint: ||menelaus' theorem||

[hot fox]

alright thanks, gonna try

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[uneven sandal]

hello can someone please recommend me good videos or books for Clifford algebra?

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[hot fox]

uh well

[hot fox]

still can't figure it out

i've tried applying the theorem to OAC, which gives

$\frac {BO} {BA} \frac {IC} {IO} \frac {DA} {DC} = 1$

[grand pond]

Gio'

[hot fox]

(i think so)

i.e. AD = 2 / IC

[feral sedge]

good

[hot fox]

well i know $ID = \sqrt{1 - AD^2}$

[grand pond]

Gio'

[hot fox]

but not much about IC

i would need angle IDC

it looks like i'm going to have the same problem since i need to show IC = 2^(2/3)

[feral sedge]

there are still more equation you can make

[hot fox]

hmm

[twilit jetty]

equivalent to menelaus is drawing two altitudes from A to OC and D to OC

[hot fox]

uh i don't know the length of the second one though

wait

[hot fox]

my bad it's AD^2 - 1

so the altitude from D to OC has length $\frac {\sqrt 3} 2 \sqrt {AD^2 - 1}$

[grand pond]

Gio'

[hot fox]

i really don't see how i can show this

[hot fox]

which ones were you thinking about ?

[feral sedge]

you have two option here

at least imo, that i think are fairly nice

1. ||menelaus theorem again||
2. ||law of cosine||

[hot fox]

well i've tried 1. in AIC but i would need an extra point on (AI)

about 2. there's an angle i'm missing

i know nothing about IAC

which triangles would you apply these in ?

@feral sedge

[feral sedge]

for menelaus approach u have exactly 1 remaining option

for loc approach u apply it in a triangle where u already have the central angle

[hot fox]

hmm

[hot fox]

triangle ABD ?

i get AC = sqrt(3) / ID

[feral sedge]

is correct equation

[hot fox]

wait i think i've got it

AD is solution of a polynomial equation of degree 4

gotta factor that out

and i'll be good

it works !!

thank you so much @feral sedge you saved my day

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[wide flame]

hi!! I need some help using the solver function on my calculator, I am trying to solve this equation (I will send a screenshot with my teacher's notes) but when I input it into solver I get -3.33

[hearty rune]

is that 35 or 3s

[wide flame]

3s

[hearty rune]

,w solve 4*10^(-38)=s(3s+10^(-7))^3

[grand pond]

Results provided by WolframAlpha

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

Requested by aldrnari_

[wide flame]

that is what my teacher got

may I show you a picture of what I inputed

[hearty rune]

sure

[wide flame]

let me get my ipad rq 1 sec

I did extra parenthases when normal parenthases didn't work either

it's negative btw the black square covers that

[hearty rune]

thats a solution

there are multiple

[hearty rune]

see its at the top

[wide flame]

how do I get the solution that I need

[hearty rune]

what parameters can you input

[wide flame]

yeah let me get a picture

and you can't change the bounds I don't think

[hearty rune]

curious what happens if you choose x=1

[wide flame]

same thing

-3.33

[hearty rune]

hm, im honestly not sure

[wide flame]

thank you so much btw

[midnight plank]

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[midnight plank]

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[silk grove]

hi

[silk grove]

can i get help on limit questions plz

[hearty rune]

multiply by (a+b)/(a+b)

you have a-b for reference

should be able to work some magic after

[silk grove]

ohhh

can i send pics here

i wana send u it so ik if i got it right

its alot of writing lol

thats good right

i got stuck on this one now

i hate when the limits have trig stuff in it

[hearty rune]

thats seems right except for the last bit

where did 17/10 come from

[silk grove]

the limit

[hearty rune]

oh im high

[silk grove]

sqrt10 + sqrt10

and ya

[hearty rune]

nvm me, youre good

[silk grove]

sorry its not very good handwriting

can uhelp me on the trig one too?

[hearty rune]

not your fault, i just magically forgot sqrt25=5

sure

[silk grove]

what do i start out with

[midnight plank]

@silk grove Has your question been resolved?

[midnight plank]

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[strange barn]

is this correct?

[copper ginkgo]

Yes

[strange barn]

i get the first part

but the second -3e^-t

why isnt it -3te^-2t?

(second part)

[copper ginkgo]

For e it's different

f'(x)e^(f(x))

[strange barn]

what rule is this?

[main current]

Walk us through your thought process. Why are you getting this?

[copper ginkgo]

U differentiated the exponent , and put infront

[strange barn]

okay well

[copper ginkgo]

He's using the x² method prob lol

[strange barn]

well the +3 should be multiplied by -t

making it -3e

and the -t should be reduced 1

to -2e

[main current]

So this is how you'd differentiate a polynomial. But this isn't a polynomial

[strange barn]

oh

[main current]

Know the derivative of y = e^x?

[strange barn]

e^x

right?

[main current]

Bingo! We can use that, and the chain rule, to get the derivative of e^(-x)

[copper ginkgo]

Yea

[strange barn]

-xe^-2x-2e^-2x

im lost man

[main current]

We're differentiating e^(-x) with chain rule.

As always when you do chain rule, you want to identify:

• The outside function
• The inside function

What are they here?

[strange barn]

e is the inside function

and -x is the out side

[midnight plank]

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[midnight plank]

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[gilded spire]

having major trouble w calc!

im not sure how to take the integral with two boundries

because i need to put a boundary on the y axis at y=1 and i also need to put a boundary on the x axis at x=e but I can only use one set of variables in the integral

fs

for sure

.close

[midnight plank]

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[fair tangle]

help yall i got -14/3u+1/3v…answer must be in decimal or integer form so i answered in decimal form but its still incorrect

[midnight plank]

@fair tangle Has your question been resolved?

[midnight plank]

@fair tangle Has your question been resolved?

[midnight plank]

@fair tangle Has your question been resolved?

[hard ore]

Well just from looking I think your answer is wrong

-14/3 u would point way off to the left

Just redo the calculations

[fair tangle]

these are my calculations

[hard ore]

a = b+5

Not b=a+5

[fair tangle]

bruh

that was a stupid mistake

do you notice anything else wrong with my math

okay i just redid the question it’s correct thank you

[midnight plank]

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[dim totem]

How do you find the limit at inf here 🧍‍♂️

[uneven sandal]

for $f(x) = x^x$ ?

[grand pond]

tobi

[zealous schooner]

They want the limit of f’(x)

As x goes to infinity

Which clearly goes to infinity

[uneven sandal]

yeah

[dim totem]

.close

[midnight plank]

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[twilit field]

find the number of triplets $(a_1,a_2,a_3$ that exist such that $a_1+a_2\cos\left(2x\right)+a_3\sin^2\left(x\right)=0$

[grand pond]

Why am. I here

[twilit field]

I solved it by using the fact that $-1+2sin^2(x)=-cos(2x)$

[grand pond]

Why am. I here

[twilit field]

and then saying then using that fact that this is still true if the whole equation is multiplied by a non-zero scalar multiple

however, my book has established relationshipse between $a_1,a_2,a_3$

[grand pond]

Why am. I here

[twilit field]

why is that needed

[runic hamlet]

what kind of relationships

[twilit field]

like $a_1=-a_3/2=a_2$

[grand pond]

Why am. I here

[runic hamlet]

that seems to be roughly in line with the identity you wrote down

[twilit field]

oh

didn't notice

[runic hamlet]

can you post your whole solution and the solution from the book?

[violet storm]

well for the whole equation to = 0, you need the coefficients of the sin^2 x terms to equal 0 and also the coefficients of the other random floating terms to = 0, seems like what you used is a good start

[twilit field]

I may get into legal trouble for that, so I'd rather not sorry

[twilit field]

that's pretty much it

there are some additional statments, like the conclusion

but that's all

[runic hamlet]

for posting a picture?

[violet storm]

so you can rewrite using what you said as:
a1 + a2(1-2sin^2x) + a3 sin^2 x = 0
now you have
a1 + a2 - 2a2sin^2x + a^3sin^2x = 0
for this to be 0 you need
a1 + a2 = 0
-2a2 = a3 i guess, this seems to be what the solution is

[twilit field]

yeah, I'm very paranoid, sorry

[twilit field]

got it. Thanks!

.close

[midnight plank]

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[twilit field]

sorry again

[feral sedge]

nothing wrong with “being paranoid” but it should be noted that realistically nobody is going to get you in trouble for posting a page out of a resource you don’t own (people do this all the time in this server)

[midnight plank]

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[willow lance]

Can someone verify whether my approach for showing that this is an equivalence relation is transitive is correct. The equivalence relation is defined on the group (G,*) Two elements a and b in the group G are considered equivalent if there exists an element h in G such that h * a * h’ = b. Suppose we have h * a * h' = b and h' * b * h'' = c. Can we equate these expressions as follows:
h * a * h' = h' * b * h''
Then, by left-multiplying with h' and right-multiplying with h, we get:
h' * h * a * h' * h = h * h' * b * h' * h
Since h' * h is the identity element e, we obtain:
e * a * e = b * e * e
As the neutral element e does not change the expression, we have:
a = b
Given that b is equivalent to c and a equals b, it follows that a is also equivalent to c.

[placid spoke]

you need to show that if a is equiv to b and b is equiv to c, a is equiv to c. i.e., show there exists g in G such that gag' = c

also, you cannot assume that the h such that hah' = b is the same as the one such that hbh' = c. those values may be different