help-38

[carmine spade]

you have used arcsin

[marsh forum]

right

[carmine spade]

why

[solid kiln]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[carmine spade]

and why do you want to know about e^2u?

[marsh forum]

It doesn't really matter IMO

just wrote it

[marsh forum]

oh, right

I don't need to care about it

[carmine spade]

i dont think this approach is fruitful

i think it will work ... eventually ... with a lot of work

[marsh forum]

yeah, Induction is probbly the way

[carmine spade]

yeah its quite quick

when it says 'prove by X or otherwise' generally using X is the easiest way

but not always

its good to trust your gut and go with something if you think it could be good

[marsh forum]

so using induction

$tan(n\theta + \theta) = \frac{ tan(n\theta)+ 2}{-tan(n \theta)+3}$

[carmine spade]

numerator is right

denom is not

[solid kiln]

ƒ(Why am. I here)=I don't Know

[carmine spade]

can you write out the general additional formula for tan(a + b)

[marsh forum]

tan(A+B)= tan(A)+tan(B)/(1-tan(A)tan(B))

ooo

right

[carmine spade]

yeah thats right

[marsh forum]

$\tan\left(n\theta+\theta\right)=\frac{\left(2+\tan\left(n\theta\right)\right)}{1-2\tan\left(n\theta\right)}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[carmine spade]

yep

what is our inductive statement going to be

[marsh forum]

now we assume $tan(n \theta)$ is ratioanl, right

[solid kiln]

ƒ(Why am. I here)=I don't Know

[carmine spade]

i.e. what can we say tan(n theta) =

[marsh forum]

let $tan(n\theta)= a_n$ where it's a rational number, which can be written with an odd denominator

[carmine spade]

very nearly

how can be be more explicit

[solid kiln]

ƒ(Why am. I here)=I don't Know

[carmine spade]

even more

what is the form of a rational number

[marsh forum]

p/q

[carmine spade]

yep

so tan(n theta) = p_n / q_n where q_n is odd

[marsh forum]

yup, I think I got it from here

thanks!

[carmine spade]

nice

[marsh forum]

so that would be

$\tan\left(n\theta+\theta\right)=\frac{\left(2+\frac{p}{q}\right)}{1-\frac{2p}{q}}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

$\tan\left(n\theta+\theta\right)=\frac{\left(2q+p\right)}{q-2p}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

now q is odd and 2p is always even

q-2p is always odd

and 2q+p is always rational

[carmine spade]

better yet 2q + p is always an integer

[marsh forum]

right

thanks!

I have one question

though this will probably only be covered in RA

why is induction a valid method

[carmine spade]

not done yet, last step (really should be first step for sake of exam technique, but its not so important) is to check the base case

[carmine spade]

what is RA?

[marsh forum]

Real analysis

[carmine spade]

what do you mean why is induction a valid method ?

[marsh forum]

no, like why does it work

we assume somethinfg is true

and base everything off that

[carmine spade]

that is the importance of this ^^^

the base case

all we have done so far is say "if it is true for n, then it is true for n+1"

[lyric nymph]

maybe this is what you are looking for

[carmine spade]

so currently we dont know that anything is true or not

we need to "tip the first domino"

in this case tan(1 * theta) = 2 = 2/1

[lyric nymph]

and this is the well ordering principle

[marsh forum]

thanks!

[marsh forum]

that's tan(2\theta)?

[carmine spade]

how so?

[marsh forum]

that's the base case

right?

[carmine spade]

tan(theta) = 2

in the question

[marsh forum]

yes

[carmine spade]

so this is the base case n = 1

[marsh forum]

ah

ok

[carmine spade]

and our proof says that whenever we have n true n+1 is true

so here we have n=1 true, so n=2 is also true

[marsh forum]

ah

[carmine spade]

then since n=2 is true, n=3 is true

and so on

[marsh forum]

I see

[carmine spade]

that is how induction works and why the base case is so important

without it we havent really proved anything

[marsh forum]

can I close the channel now?

[carmine spade]

if youre content then by all means

[marsh forum]

it's 1:30 am here, kind of have to sleep

thanks

[carmine spade]

its your channel haha, feel free to do as you please

anyway, gn

[marsh forum]

thanks

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[trim jolt]

Please don't occupy multiple help channels.

[delicate lance]

how does this become 2x + 18 - 32 ?

[simple haven]

Let u = (2x + 18)^(1/5)

And notice you get (u - 2)(u^4 + 2 u^3 + 2^2 u^2 + 2^3 u + 2^4)

This form should tickle the back of your mind

[delicate lance]

it doesnt tickle

please

[simple haven]

Hint

Geometric series

Second hint can you verify (x^n - 1) = (x - 1)(x^(n-1) + x^(n-2) + ... + x + 1) ?

What happens if you have x^n - y^n ?

[delicate lance]

hmm

[simple haven]

Exactly

Can you use this?

[delicate lance]

Yes

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[subtle ether]

Im so lost as how to do c

[dusky flint]

did you already show b

[subtle ether]

i mean its just simultaneous equations

I didnt rly bother

idrk why that would help with part c either

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[placid grail]

i have to find cdf of Y = X^2 where X is standard normal

[dull island]

hello

let me look through it

[placid grail]

hello

sorry it got cut off but i made a = sqrt(y)

[dull island]

check the third line

[placid grail]

ohhhhhh

[dull island]

you count the middle twice

[placid grail]

that makes sense as to why the number is bigger than 1

gotcha thanks

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[graceful zenith]

How do I do c

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@graceful zenith Has your question been resolved?

[graceful zenith]

<@&286206848099549185>

[last rose]

if you've proven a and b

then you can just add the two right?

[graceful zenith]

Ohk

Ok

[graceful zenith]

Ok I did it

How do I do d tho

Oh I already got c

The qu after that tho

Np

Yea

Oh

Ok

Thank you

Ok

Tysm

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[wooden rampart]

you're looking at (c)?

[vast valley]

yeah

idk even how to start

[wooden rampart]

are you familiar with derivatives yet

[vast valley]

yeah

[wooden rampart]

cuz this seems like competition math

lol okay

[vast valley]

i can do calc 1

[wooden rampart]

yeah that's sufficient

[vast valley]

ohh

i find derivative

substitute

gradient

for that point

[wooden rampart]

yep

[vast valley]

point gradient formula

for the line

then i just simultaneous

to find second intercept

is that right

oh wait i dont rlly even need that

ight lemme try solve rq

[wooden rampart]

yeah try it first, sounds like you're in the right direction

[vast valley]

wait

how do i find point A

to find the gradient of L

[wooden rampart]

that's the tricky part

first you find the derivative of the function

[vast valley]

3x^2-1

[wooden rampart]

yes that's the general slope of the function

wait sorry give me a second

[vast valley]

ok

im guessing i need to use part b somehow

[wooden rampart]

ngl i lost where i was

but the direction is definitely correct

[vast valley]

ngl im lost

its some simultanous equation

[split chasm]

you can combine parts a,b

consider how you'd normally approach finding the intersection points of two curves

[vast valley]

a simultaneous equation?

[split chasm]

yeh,

[fossil folio]

y = x^3 - x + 3
y' = 3x^2 - 1

||slope at point A is (y - (-3)) / (x - (-2)) = (y + 3)/(x + 2)

these slopes must be equal so

3x^2 - 1 = (y + 3)/(x + 2)

since y = x^3 - x + 3

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

(3x^2 - 1)(x + 2) = x^3 - x + 6
3x^3 - x + 6x^2 - 2 = x^3 - x + 6
2x^3 + 6x^2 - 8 = 0
x^3 + 3x^2 - 4 = 0

(x - 1)(x^2 + 4x + 4) = 0
(x - 1)(x + 2)^2 = 0

therefore x = 1
and y = 1^3 - 1 + 3 = 3

y = 2x + 1||

[wooden rampart]

oh wow i actually had the answer all along, just miscalculated. yes you are right

[split chasm]

!nosols

[trim jolt]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[split chasm]

also don't need to approach it that way from work that's already been done

[wooden rampart]

you know that the line must pass through (-2, -3), which you addressed in part b. you also know the slope m

[vast valley]

how do u find the slope

[wooden rampart]

it's the gradient that you found using the derivative

[split chasm]

go back to what I said

[vast valley]

why would i combine part a)

[split chasm]

consider how you'd normally approach finding the intersection points of two curves

if you were just asked to find the intersection points between
y =x^3 -x +3 and y+3 = m(x+2)
what would be your first step

[vast valley]

and how do i equate those

thats all i got too so far

i cant find m by subbing in (-2,-3)

[split chasm]

have you ever solved stuff like simultaneous equations before?

[vast valley]

oh wait

[wooden rampart]

@vast valley the equation that you got in part b is essentially this: https://www.desmos.com/calculator/xkvr4obhqz

the problem is that you want to find the value of m that makes the line tangent to the curve

[split chasm]

sub the y from one equation into the other

[wooden rampart]

use the slider of m to see what i mean

[vast valley]

ohh ok

[wooden rampart]

so you want that line to intersect the curve right?

that's where simultaneous equations come in

if you solve for y of the line you get y=m(x+2)-3, and you want that to intersect y=x^3-x+3, so you substitute the y from one equation to another as ℝαμΩℕωⅤ said

[vast valley]

yeah thats what im tryna do rn

[wooden rampart]

do yk what you're solving for

[vast valley]

idk im so cooked 😭

[split chasm]

can you do just the substitution of y from the first equation into the second

[vast valley]

i tried unless im doing something wrong

[split chasm]

show what you have

[wooden rampart]

that second line is sufficient

you just solved for m in terms of x

The thing is that we also know m=the derivative of the curve

by definition of derivative

[split chasm]

um...

[vast valley]

so m = 3x^2-1 when X=A

honestly idk where im going

[wooden rampart]

i'll let ℝαμΩℕωⅤ take this bc he's going in a diff direction

[split chasm]

where are those first two equations at the top coming from

[vast valley]

substiting for y

[vast valley]

using these 2

[split chasm]

won't give you what you wrote

[vast valley]

wait wdym

[split chasm]

where's that red
$$x^3 - x \red{- 2} = m(x+2)-3$$
coming from \
and why are you setting $m$ equal to $x^3 - x +2$

[solid kiln]

ℝαμΩℕωⅤ

[vast valley]

oh i wrote 1\

and basically i moved three over

so it is x^3-x+2

then i divide both sides

by (x+2)

[split chasm]

still wrong

[wooden rampart]

there's a denominator

[vast valley]

its a fraction

[split chasm]

first equation is still wrong

[vast valley]

oh wait its plus 3 im on something

so its plus 6

[split chasm]

and there's no need to do the division

[vast valley]

oh ok

[split chasm]

after fixing that, rearrange the equation you have to general form

[vast valley]

wait wdym

[split chasm]

rearrange that equation
$$x^3 - x \blue{+3}= m(x+2)-3$$
to the form
$$ax^3 + bx^2 + cx + d = 0$$

[solid kiln]

ℝαμΩℕωⅤ

[vast valley]

what do i do with the m(x+2)

expand it out?

[wooden rampart]

yes

and get everything to one side

[vast valley]

ok

so x^3+0x^2-(m+1)x+(6-2m)=0

[split chasm]

messed up a sign

-mx - x = -(m + 1)x

yes

[vast valley]

ok so what do i do now

[split chasm]

now this is the same equation you have from part a)

[vast valley]

oh wait yh

ohh

and i can find m=2

ok i got it now

but why does a=1

[split chasm]

wdym a=1

[vast valley]

i mean

point A

x value is 1

[split chasm]

that'd be the intersection point which you can get from various methods

[vast valley]

ohok

[split chasm]

but since the question doesn't ask for that, you don't have find it
(this method bypasses that)

[vast valley]

yeah ok

ty

[wooden rampart]

idk if it's okay for me to send this but

this was the method i used in case you wanted a different way

@vast valley

then you'd find the equation of the line

[split chasm]

^ is a common approach to use if you didn't already have the first two parts done

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[native radish]

Help

[native radish]

Help

Anybody pls help

Help

Yes

Sin2x = 2sinxcosx

How show me

@torpid mountain

Cos^2 2x ) / sin^2 2x ?

Is it

Yeah

Yeah I don't

Now next

What ?

@torpid mountain

How ?

@torpid mountain

@torpid mountain

Pls hurry my sister is on the door to snatch away the phone

[solid kiln]

..,,

[native radish]

Solve it on the paper pls quickly and send it

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[native radish]

Explain the fourth line

Pls

Sorry for bothering you

@torpid mountain

Pls man

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[tardy hemlock]

If an exercise asks you to sketch $\frac{4x - 5}{(x+1)(x-2)}$ with its asymptotes, zeros and {\bf monotonic behaviour}, is there a quick way to determine the monotonicity (without having to take derivatives) or do you just sketch it and the monotonicity will "automatically be in there"?

[nova spire]

Yeah i would just take the derivative

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[tardy hemlock]

So no quick way

Alright

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[fresh pendant]

if we know that r(a1,a1+a2,a2+a3)=r(a4,a1+a4,a2+a4,a3+a4), For the ranks to be equal, a4 might need to be expressible in terms of a1 a2 and a3? Why?

[fresh pendant]

Currently I know for the second set, the presence of a4 in all vectors suggests that a4 plays a crucial role in the dependencies

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[wet ember]

So we're given this function $p(y)$

[solid kiln]

Decoway

[wet ember]

and basically this is a part of an algorithm (simon's algorithm) meant to find some bitstring s

y is also a bitstring, and $p(y)$ calculates the probability that we choose y out of $y \in \Sigma^{n}$

[solid kiln]

Decoway

[wet ember]

nvm .close

.close

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[spring epoch]

Is there a possible way to find the square root of a number through calculations only and not memorization or a calculator?

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[digital cloak]

Babylonian method?

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[tulip glade]

If anyone knows how to do any of these with showing work please 😭🙏

[twilit meteor]

okay let#s take this one q at a time

so how far ave u made it with this one

[tulip glade]

Solved it j unsure if it’s correct

[twilit meteor]

ok well let's see ur answers

[tulip glade]

Amplitude: 2
Max: 9
Min: 5
eom: y=7
Period: “2pi”/3

[twilit meteor]

ye that all seems right

nice work

[tulip glade]

That’s all correct?

[twilit meteor]

ye

[tulip glade]

bet

Next problem?

[twilit meteor]

that or we're wrong in the same way

[tulip glade]

lmaooo

[twilit meteor]

i cant remember what a terminal point is so let's do this one now

what do u have so far?

[tulip glade]

I feel very wrong abt this one

[twilit meteor]

well, maybe u are and maybe u aren't. i won't know until i see your working

[tulip glade]

This what I got 😭🙏

[twilit meteor]

ok lemme see

this might be a resolution thing but this should say $$\cot(-\theta) = \frac{\cos(-\theta)}{\sin(-\theta)}$$, not $\frac{\cos(\theta)}{\sin(\theta)}$

[solid kiln]

Out Of Nosh

[twilit meteor]

although u get to the right conlcusion of cot(-x) = -cot(x) so it's not a huge thing

[tulip glade]

So it should be fine without it too right?

[twilit meteor]

well it's still a mistake, just one that gets corrected later, since what i assume ur doing is writing out cot(-x) in terms of its deffinition

anyway im not sure what's happening here but there's an easier way of doing this

[tulip glade]

Oh ok

[twilit meteor]

namely $\cot(-\theta) \cos(-\theta) + sin(-\theta) = \frac{-\cos^{2}(\theta)}{\sin(\theta)} - \sin(\theta)$.

[solid kiln]

Out Of Nosh

[twilit meteor]

brb gotta eat

[tulip glade]

Kay

I should be good with the rest

[twilit meteor]

anyway turn that into one fraction and you'll get there

[tulip glade]

I think they’re correct

Thanks for your help

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[bold jasper]

in the following problem is there a smarter and faster way to compute the determinant than by doing the cofactor expansion?

[bold jasper]

the matrix is symmetric and only has ones so i feel almost like they made it to be solved by using some clever trick using symmetric matricies or something

[digital hull]

Reduce it to an upper triangular form, it's the fastest way

[bold jasper]

isnt reducing going to omit some eigenvalues

from what i know row matrix A doesnt have the same eigenvalues as its reduced form?

or maybe its more correct to say "its not guaranteed to have the same eigenvalues as some of its row-reduced forms"

[marble wharf]

well ok sure but who cares about eigenvalues. you want the det

and row reducing doesnt destroy that

(as long as you are careful with switching rows and multiplying them by some number)

[hallow kite]

just keep track of negatives from row swaps, and that your determinant scales when you scale

otherwise row reducing preserves determinant

[marble wharf]

you can actually guess two eigenvalues

and then use the trace to find the third

oh wait nvm the two you can guess are the same

hmm

definitely feels like some clever trick could be possible. but you'd definitely be faster if you just row reduce

[trim jolt]

@bold jasper Has your question been resolved?

Channel closed

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[wheat tide]

i have started by integrating f'(x)

[unkempt smelt]

your x^(3/2) term is incorrect

it should have a 2a/3 factor

$\frac{2a}{3}\cdot \frac32=a$

[solid kiln]

Flappie

[wheat tide]

dont i add one to the power and divide the coefficient by the new power

is a / 3/2 = 2a/3

[unkempt smelt]

this is correct

y=2x^2+2a/3x^(3/2)+bx+C

do you know where to go from here?

[wheat tide]

not sure

my thought process is to put the stationary point into it

[unkempt smelt]

stationary point into where?

[wheat tide]

here

as the x and y values

[unkempt smelt]

that wont do much yet

a stationary point means that the derivative is 0

[wheat tide]

so should i find the put the y value to 0

to get the x intercept

[unkempt smelt]

yes

[wheat tide]

ok

[unkempt smelt]

actually, also this, mb yes

but its because you know the point is on the line

[wheat tide]

which should i do first

or does it not matter

[unkempt smelt]

doesnt mater

matter

you need all of them

you have 3 unknowns (a,b,C)

so you need at least 3 equations

[wheat tide]

not really sure how to solve this

[unkempt smelt]

$f(x)=2x^2+\frac{2a}{3}x^{\frac32}+bx+c\f'(x)=4a+a\sqrt{x}+b$

[solid kiln]

Flappie

[unkempt smelt]

we have one point, which is a stationary point

the point must lie on the line

so f(4)=3

its a stationary point, so f'(4)=0

and the y-intercept is at -5, so f(0)=-5

does this make sense?

[wheat tide]

uh

i put the stationary point in

[unkempt smelt]

yeah, thats the first equation you get

now do the 2nd and 3rd

[wheat tide]

how

do i put -5 as y and solve for x?

[unkempt smelt]

no, the last statement says that it crosses the y-axis at -5

if it crosses the y-axis, that means x=0

[wheat tide]

ok..

so how do i get the other equations

set y to 0?

[unkempt smelt]

these are your 3 equations

f(4)=3
f'(4)=0
f(0)=-5

[wheat tide]

ok

so i have gotten c as -5

[unkempt smelt]

thats correct

[wheat tide]

im definitely getting this wrong

i got b as -1.618253365

[unkempt smelt]

is that -(sqrt5+1)/2?

[unkempt smelt]

keep it in exact form unless necessary

[wheat tide]

oh woa it is

my calculator didnt give it in surd form

i was just rounding off and got the right answer

[unkempt smelt]

its the golden ratio

,w golden ratio

[solid kiln]

Results provided by WolframAlpha

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

Requested by flappie.

[wheat tide]

so my final value for a is -33+3root5/8

or -4.96

b is -1.62

and c is -5

.close

[trim jolt]

Channel closed

Closed by @wheat tide

Use .reopen if this was a mistake.

[trim jolt]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[elder wing]

Can anyone help me spot the mistake?

[rotund grove]

,w golden ratio

[solid kiln]

Results provided by WolframAlpha

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

Requested by anshulyolo121

[elder wing]

sorry, what does it have to do with the problem, I can't see.

[rotund grove]

Same.

[rotund grove]

think out of the box

[trim jolt]

@elder wing Has your question been resolved?

[trim jolt]

@elder wing Has your question been resolved?

[trim jolt]

@elder wing Has your question been resolved?

[bright lion]

blud what

[inland orbit]

This is correct. It might differ from the answer given in your book by a constant

Your solution differs by one given by WolframAlpha by pi/2, which is fine

[elder wing]

is just that the graphs look a bit different, that throws me off

but I'll go with it

[inland orbit]

what is the solution given?

[elder wing]

something around the lines of too large

is just that I usually forget some rules like adding an absolute value and end up with a integral that has an opposite sign on some interval, I don't know a lot about complex numbers, but that yellow line going the other way just made me think something may be wrong

[inland orbit]

If you graph your solution alongside that one in desmos, they have the same shape. And if you shift yours down by pi/2 they are in fact identical

Yours is much simpler, but I think wolfram's works for complex numbers where yours may not

but I'm not sure about that last part

[elder wing]

ok then, I'll think I'll leave it at that, thank you.

[inland orbit]

Sure thing 👍

[elder wing]

.close

[trim jolt]

Channel closed

Closed by @elder wing

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• 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.

[mental pike]

Hello, I need help with solving an Augmented Matrices. It’s my first time and I’m confused on what Elementary Row Operations to use.

[patent crow]

send the matrix

[mental pike]

Hello

[patent crow]

hello again

[mental pike]

Here it is

[patent crow]

,rotate

[patent crow]

oh dear it's like seeing my own handwriting

do you know the gauss-jordan row eliminition/row reduction algorithm?

[mental pike]

Sorry. I wrote it while in a car 😢

[mental pike]

No I am just learning Matrices

[patent crow]

do you know how row operations are written?

ie R1 + R2 -> R2

or 3R2 -> R2

[mental pike]

The textbook said to use the 3 Elementary Row Operations: Interchange Rows: Swap the position of two rows in the matrix.
Multiply a Row by a Non-Zero Constant: Multiply all elements of a row by the same non-zero value.
Add a Multiple of One Row to Another Row: Replace one row with the sum of that row and a multiple of another row.

[patent crow]

notation varies, but are you familiar?

yes, those are the three

[mental pike]

Row 1 plus row 2 to row2?

[patent crow]

yup, replace row2 with 1 times row1 added to row2

you also have R1<->R2 but that almost never happens

[mental pike]

Ok so basically just add 1 to 2?

[patent crow]

so, you have
[1 2 6]
[2 1 9]

[mental pike]

Yes

[patent crow]

the goal is to a matrix of reduced row echelon form, if you want to solve for each value

that means if you have n rows, your first n columns are all zeros except one 1, and each row starts with a 1 with only zeroes above it

[mental pike]

You want
[1 0]
[0 1] right?

[patent crow]

in this case,
[1 0 X]
[0 1 Y]

which becomes x + 0y = X and 0x + y = Y

in other words, each row gives you the solution for each variable

[mental pike]

Yes

I’m wondering what operation I should do from the original Augmented Matrices, that’s what I’m confused about.

[patent crow]

typically you go column by column

so you try to make the first row start with a 1, then make every row beneath that have 0s under that 1

then you go to the second row, make it start with a 1, then make every row beneath that have 0s under that 1

etc...

[mental pike]

Ok

[patent crow]

then, once you have reached your last row, you go upwards, and subtract so that each slot above the 1 becomes a 0 (and since every element before the 1 is 0, it doesn't affect the rest of the matrix)

so in this case

https://media.discordapp.net/attachments/908078008609431674/1254609771990876190/444537905919164428.png?ex=667a1de7&is=6678cc67&hm=0d1fbe6848d43014a54062a8aa3fd9cfe972ae954b52b8324ddaf96d696a025c&=&format=webp&quality=lossless&width=1442&height=1082

row 1 starts with a 1 already

how would you make row 2 start with a 0?

note: never multiply a row by 0

[mental pike]

Subtract by 2

[patent crow]

elaborate, what row operation is that exactly

[mental pike]

That is 3

[patent crow]

Add a Multiple of One Row to Another Row: Replace one row with the sum of that row and a multiple of another row.

[mental pike]

Add a Multiple of One Row to Another Row: Replace one row with the sum of that row and a multiple of another row.

[patent crow]

ok, what exactly are you doing?

like, write it in 6R7+R8->R8 notation (add 6 times R7 to R8 and make it R8)

[mental pike]

I’m confused about the notation.

[patent crow]

for instance, to write adding three times the second row to the first row

you write 3R2 + R1 -> R1

to multiply a row by 7, you do 7R3 -> R3

to swap two rows, you do R18 <-> R1

[mental pike]

Ok

So then you could do -R1+R2

-2+2 equals zero

[patent crow]

sure

no

-R1+R2 -> R2
gives
[1 2 6] (row 1 untouched)
[2-1 1-2 9-6] = [1 -1 3] (-R1 + R2 -> R2)

[mental pike]

Ok.

[patent crow]

hint: you should multiply R1 by -2 not -1 before adding

[mental pike]

-2R1 + R2

?

[patent crow]

what does that give you?

[mental pike]

0 0/ -3

Oh wait

0 3/-3

[patent crow]

I don't see how?

your column 2 element is wrong

[mental pike]

-2x2=-4 then that +1 = -3

[patent crow]

you wrote +3

0 -3 -3 is correct

so now you have

[1 2 6]
[0 -3 -3]

now, we move on to row 2

how do you make it start with a 1?

[mental pike]

Oh sorry typing mistake.

We have to check which of the three operations work correct?

We likely have to do this: Multiply a Row by a Non-Zero Constant: Multiply all elements of a row by the same non-zero value.

Because if we added to the second row it would distrust the 0

[patent crow]

yup

[patent crow]

not sure what this means! we normalise every row as starting with 1 because otherwise it makes it more annoying to get a useful answer out of our solution, and it normalises things

[mental pike]

Because if we had to add a number wouldn’t it disrupt the zero by changing it?

OHHHH

we multiply the second row by -1/3

It makes the -3 a 1

[patent crow]

Yes

you NEVER add constants to a row

Only other rows and multiples thereof

[mental pike]

Oh ok, I meant adding a row would change the 0

So now from here we only have one more element that’s needed to be changed.

The 2 in row 1 column 2

[patent crow]

Yup

[mental pike]

So could we do

R1 + (-2)R2

(4,1)

[patent crow]

yup

you can check by plugging it in too

1(4) + 2(1) = 6 ? yes
2(4) + 1(1) = 9 ? yes

[mental pike]

It took a while but it makes sense now. I was looking at it more closely and the answer clicked.

[patent crow]

that's wonderful!!

[mental pike]

Thanks so much for the help.

.close

[trim jolt]

Channel closed

Closed by @mental pike

Use .reopen if this was a mistake.

[trim jolt]

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.

[delicate lance]

Is displacement basically position?

and what is a maximum posivtive velocity

[quick hound]

yes

highest max speed in positive direction

[quick hound]

wait before you do more question

do you have a formula booklet

[delicate lance]

no ;-;

[quick hound]

if your taking calc 2 i gurantee you have one

this for final right?

[delicate lance]

yes

[quick hound]

no formula booklet?

[delicate lance]

no formula

booklet

[quick hound]

shit

[delicate lance]

he doesnt give

[quick hound]

i gotchu

[delicate lance]

tyyyyyy

[quick hound]

you took physics before?

[delicate lance]

nope

😦

[quick hound]

thats tough

this question would be free with physics

[delicate lance]

damn

i wouldve had a 98 in calc if i didnt have to do vectors

everyone else took physics so it was so easy for them but it was so hard

for me

[quick hound]

displacement is perpendicular distance from original point

[delicate lance]

oh ok

[quick hound]

velocity is speed with a direction

so negative velocity means their going backwards

[delicate lance]

ohok

[quick hound]

you cant have negative speed that doesnt exist

[delicate lance]

true

so maximum posible velocity jus means

plugging the critical point and domain into the original equation to find the maximum velocity?

[quick hound]

yes

[delicate lance]

so why do they find acceleration

[quick hound]

for critical points

when acceleration is 0

its either max or min velocity

[delicate lance]

ohoh

[quick hound]

you know how the 3 things work right?

[delicate lance]

Kind of

[quick hound]

derive displacement = velocity

derive velocity = acceleration

and reverse for integrals

[delicate lance]

wait yo udid your calc exam ? grade 12?

[quick hound]

nah

uni

[delicate lance]

ohhhhh

[quick hound]

year 1

[delicate lance]

so i have hope

[quick hound]

ye

[delicate lance]

are you canadian like me

[quick hound]

nah

some of my friends went there though

similar stuff

[quick hound]

just apply this

same concept of how you find critical points of a graph

when gradient=0 its the critical point

how do you get the gradient?

thats the acceleration

[delicate lance]

oh okay

do you know hwo to do this?

last question and ill stop btohering

[quick hound]

alg

im bored out of my mind rn

[delicate lance]

Im sad as hell cause i purposefully did not study and now im facing the consequences

[quick hound]

me when there are consequences for my actions

[delicate lance]

😭 😭

so what is this differentiation thing called

[quick hound]

no idea

[delicate lance]

I couldnt care less ab this question

lets mov on

[quick hound]

i dont know i learnt this

think*

[delicate lance]

Im ginna cry

I SHOULDVE STUDIED

I dont know what i did all day

do you remember optimization questions

[quick hound]

kind of

they were hard for me though

prob hardest topic

[delicate lance]

Tell me why i got perfect on the optimization test and now i dont remember ANYTHING.

well ur in uni im in hs so

obv its harder for u

[quick hound]

ye

gl man

[delicate lance]

how do oyu differentiate lnx

to the powr of

2

[quick hound]

chain rule

[delicate lance]

how did they get that 1/x is the derivative of lnx

[quick hound]

uhh

memorization

[delicate lance]

oh ok

[quick hound]

derivative of ln(u) is (u'/u)

derivative of x is 1

u is x

so 1/x

[delicate lance]

ohhok

[quasi saffron]

ln x is 1/x right

?

[quick hound]

yes

[delicate lance]

ye

[quasi saffron]

ah see

[quick hound]

the derivative if ur asking that

[delicate lance]

so guys how da hell do i do this

[quasi saffron]

yes

[quick hound]

i havnet learnt derivatives of x in power

[delicate lance]

ohok

[quasi saffron]

wait

[quasi saffron]

i think u need to simplify first i guess

[delicate lance]

he multiplied both sides by ln but why?

[quasi saffron]

wait

i think to eliminate

[quick hound]

oh

thats how you do it

[delicate lance]

tf how does the ln become 1/y

[quick hound]

thats y

not 4

[quasi saffron]

ln y

[delicate lance]

OHH

[quick hound]

remember derivative of lnx

[delicate lance]

SO if its lnp it would be 1/p

[quick hound]

yes

[delicate lance]

4000 neurons just activated

[quick hound]

before we continue

whats ln(2x)

[quasi saffron]

lmao

[delicate lance]

1/2x

[quick hound]

u sure?

[quasi saffron]

2/x?

[delicate lance]

im gonna cry

[quick hound]

this

u' is first derivative

[delicate lance]

u' u ????????

[quasi saffron]

ah

[delicate lance]

1/x

[quick hound]

ok start

u =2x

whats first derivative of 2x

[delicate lance]

2/2x

[quasi saffron]

oh it doesnt matter

[quick hound]

there

okay good

[quasi saffron]

no u did right

2/2x

[quick hound]

yeah 2/2x is right

[delicate lance]

1/x

[quasi saffron]

yes

nice

[quick hound]

okay now ln(2x+5)

[delicate lance]

1/x+5

[quasi saffron]

so ln is an expression right?

[quick hound]

we need this pattern in ur head before looking at that or it will be hard

[quick hound]

ln is short for log e (u)

log e

basically

[quasi saffron]

ahh got it

[delicate lance]

bros learning with me and helping at the same time

wtfff 😭

[quick hound]

perfect

[delicate lance]

does bro also have an exam in 2 days

[quick hound]

take it slowly

whats derivative of 2x+5

[delicate lance]

2

[quick hound]

ok

now first derivative/ original

[delicate lance]

2/2x+5

[quick hound]

good

now back to this

f(x) he substituted with y

took ln of both sides

[delicate lance]

so he brought ln so he could bring the exponent down

[quick hound]

do you know how log rules work?

hopefullyy es

yes

[delicate lance]

i forgot

all about them

^_^

[quick hound]

we are cooked

[delicate lance]

^_^!

Still havent gone over vectors unit bt

btw

im FR cooked

[quick hound]

i skipped vectors in calc

[quasi saffron]

come in handy

[quick hound]

yeah that works

[delicate lance]

ohhhhhh

a white spider just flew on my keyboard

I got the heeby jeebies

[quick hound]

is ur ass in australia

[delicate lance]

no MF 😭

I mean ur not a mf because your helping me , Sorry.

[quasi saffron]

lmao

[delicate lance]

b^k = b^k holy shit

[quick hound]

yep

[delicate lance]

I dont understand rule 7

[quick hound]

okay

what exactly does a logarithm give u

[delicate lance]

tbh

no clue

[quick hound]

amazing

[delicate lance]

^_^

[quick hound]

fiddle with ur calc more

[delicate lance]

it gave me 0

[quick hound]

okay basically lets look at log 2 (5)

the value it gives

is the number that 2 has to be raised to = 5

[delicate lance]

2^(The vlaue you search for on the calcuator) Gives you 5

[quick hound]

yes

[solid kiln]

lILi

[delicate lance]

wow

I learn from

tactile

learning

Ngl i think im on the spectrum :/

anyways continue

[quick hound]

didnt u get it

[delicate lance]

I did get it

[quick hound]

for the working up there its just long derivation

[delicate lance]

WOw

You're helping me so much

you're such a nice person

[quick hound]

im just bored

[quasi saffron]

bored could be in a million definition

lol

[delicate lance]

chess player

[quick hound]

nothin tod oo

[delicate lance]

Who is tihs person that keeps typing

@dreamy sleet

[quick hound]

are you allowed calc?

[dreamy sleet]

Hi :>

[delicate lance]

yes

[delicate lance]

Hi

[quasi saffron]

u use the scientific calc right?

[delicate lance]

casio fx-300es plus natural vpam

[quasi saffron]

i think that will be helpful

okay nice

[delicate lance]

literally

everyone in my class has this calculatr

its embarrassing because i want to be different

[quick hound]

theres this neat derivative button

if you dont know about it

use it in a crutch

[delicate lance]

i dont have it

[quick hound]

h fuck

[quasi saffron]

lol

[delicate lance]

All this robux and im worried about a silly exam........

[quick hound]

what the hell

?????

thats alot of robux

[delicate lance]

I could feed 5 families with this

[quasi saffron]

bro

[quick hound]

bro i could buy so much deepwoken with that

aint no way

[quasi saffron]

🙂

[delicate lance]

deepwoken

ik the owner of that game

[quick hound]

damn

[quasi saffron]

bro u can buy casio fx-991 plus with that, so u can have derivatives button

[delicate lance]

MY TEACHER DOESNT LET ME USE DERIVATIVE BUTTONS

😦

[quasi saffron]

LOL

[quick hound]

how he gonna know?

real question

[delicate lance]

he checks everyones calculators in every test

especially the exam, he says hes gonna pick up peoples erasers out of no where

[quick hound]

just use it to double check ur answers

since you need working anwy

[delicate lance]

omg

u helped me so much

i see the correlation here

[quick hound]

yeah u needed that

[delicate lance]

help

whats teh secnod step say

[quick hound]

huh?

same thing but with f(3)

theres 1/y on left side rmember

[delicate lance]

did they move the y to the other eqn?

and then changed it to f(3)?

[quick hound]

yes

[delicate lance]

wait why would it be f(3)

cant i just leave it as

y

[quick hound]

yeah but

what u gonna do with y

[delicate lance]

idk

kill it

[quick hound]

you need the value of y'

right

[delicate lance]

yes

[quick hound]

so you move y to right

and you get the value of y which is f(3)

[delicate lance]

oh ok

last question

how you do dis

[quick hound]

log rules

[quick hound]

think about this

[delicate lance]

ok i’ll

wtf

where did the yln4 come from

[quick hound]

you got the first step right?

yay or nay

and also log rules

[dreamy sleet]

,tex $\frac{\mathrm{d}}{\mathrm{d}x}[\log_a(u)]=\frac{1}{u\cdot\ln(a)}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}$

[delicate lance]

i understood first step

[solid kiln]

lILi

[quick hound]

not that

[dreamy sleet]

fixed variables, sorry

[delicate lance]

can u elaborate

[quick hound]

he dont get how the 4^y moves y down

[delicate lance]

i’m a bit slow

WIAT

I do get it

[quick hound]

ok

[dreamy sleet]

oh i didn't read the full thing

[delicate lance]

YO UHAVE TO PUT LN DOWN TO UNLOCK THE EXPONENT AND BRING IT TO THE BACK

[quick hound]

yes i guess

[delicate lance]

u guess?

i just had the most genius discovery ever

[dreamy sleet]

oh shit, my bad, i realise what's happening

[quick hound]

how do you wrie that stuff

[delicate lance]

[dreamy sleet]

LaTeX, I can send you a message explaining some of the basics if you'd like

later though

[quick hound]

sure

[quick hound]

understand the rest of it?

[dreamy sleet]

ok now let me write out what actually applies here

[delicate lance]

nope

[dreamy sleet]

,tex $\ln(b^k)=k\ln(b)$