help-17

[blazing depot]

i dont rlly know there use

[silent burrow]

you can look at it that way

[blazing depot]

so i basically try to get the slope of some complex line

( not linear )

[vocal sleet]

@blazing depot Has your question been resolved?

[blazing depot]

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[vivid kelp]

Can someone explain to me why in the identity we cant use that 1/k = 0 but after we refactor we use lim n- Infinity for 1/n to 0?

[silent burrow]

in the second line?

[vivid kelp]

From the identity compared to the last line

So we are doing all these rearanging instead of (1 - 1/k) being (1 - 0) but in the last line use that

Maybe for context, we are talking about sequences here.

<@&286206848099549185>

[vocal sleet]

@vivid kelp Has your question been resolved?

[vocal sleet]

@vivid kelp Has your question been resolved?

[hard atlas]

where exactly would you want to use that 1/k goes to 0? in the first two lines you have a product. depending on fast/slow each factors goes to 1, the total product could go to pretty much anywhere

[vivid kelp]

Thats pretty much the question

So we cant use it because we are working with a product ?

[vocal sleet]

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[vivid kelp]

.close

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[kind zephyr]

does anyone know how you would do this question?

this is what I have so far, i need to find the v max now

i have found the critical points and did the dessian test

[vocal sleet]

@kind zephyr Has your question been resolved?

[vocal sleet]

@kind zephyr Has your question been resolved?

[ocean iron]

Mb

[kind zephyr]

<@&286206848099549185>

[kind zephyr]

anyone can help?

[vocal sleet]

@kind zephyr Has your question been resolved?

[vocal sleet]

@kind zephyr Has your question been resolved?

[main spade]

The maximum obtained by the conditions V_x = V_y = 0 will give you the maximum of V(x, y) in the interior of the region V is defined on. You need to determine if the maximum lies on the boundary of the region, i.e., find the maximum of V(x, y) on the boundary.

Let (X, Y) be the point of the maximum. If you have determine that V(X, Y) > V(x, y), where these (x, y) are the boundary points, you can use the second derivative test to determine if V(X, Y) is a relative (and absolute) maximum).

[kind zephyr]

oh ok, so to find that you need to use the points that give you the maximum, and with the constant 'h' i have found that to be 1/2 do I input that too

[main spade]

How did you find that h = 1/2?

[kind zephyr]

from the hessian test

[main spade]

It seems to me that h is just a arbitrary positive constant.

[hybrid flicker]

you shouldn't be able to find the value of a constant that could equal anything

[kind zephyr]

oh i see, so you don't need to find anything for it

[main spade]

I don't think you can unless you've been given extra information.

[hybrid flicker]

What have you done to find the critical point? Looking back at this, I'm having a hard time processing what you did

[main spade]

This is a surprisingly loaded question in terms of the theory used.

[main spade]

Same. Try and start from the beginning. Show us how you determined your critical points.

[kind zephyr]

i found the critical points from -2x(2h-y), ignoring the h. Then that comes to -2x = 0 and y =2 I than subbed that into the first derivative of fy and fx

[main spade]

Ignoring the h is a critical error.

You're solving the problem for h = 1 then.

[kind zephyr]

yea

[hybrid flicker]

And we're asking you to solve for h = anything positive

[main spade]

Please do not set h = 1.

[kind zephyr]

right ok

also with finding the critical point are we using the first derivative formula for fx ro fy?

[hybrid flicker]

The critical point verifies $f_x(x,y) = f_y(x,y) = 0$

[twin meteor]

rafilou2003

[hybrid flicker]

I can see that you only used one of them

[kind zephyr]

oh

[main spade]

A critical point is the points (a, b) where V_x(a, b) = 0 and V_y(a, b) = 0.

[kind zephyr]

oh i see

and then you sub x,y critical points into the second derivative with respect to x,y

[main spade]

These are important because a maximum or minimum inside the boundary (not on the boundary) can only occur at critical points.

[kind zephyr]

yep got it

[main spade]

You need to solve the system V_x(x, y) = V_y(x, y) = 0 for all pairs of (x, y).

[kind zephyr]

for the first deravitive?

is the problem with the x^2?

[hybrid flicker]

Ok I take back what I said sorry

[kind zephyr]

how would you find the critical points for fy?

[kind zephyr]

oh nah, no worries

[hybrid flicker]

Solve the equation $f_y(x,y) = 0$

[twin meteor]

rafilou2003

[hybrid flicker]

It will give you information about the critical point(s) of f

[main spade]

Firstly, we require V_x(x, y) = -2x(2h - y) = 0, which implies x(y - 2h) = 0.
You determined correctly that V_y(x, y) = 2h - 2y + x^2 = 0.

You can determine from x(y - 2h) = 0 that either x = 0 or y = 2h. You can assume x = 0 and then y = 2h in the second equation.

[kind zephyr]

ohh yea

so what happend to 2x?

[hybrid flicker]

If -2 × ... = 0, then multiply by -1/2 on both sides...

[kind zephyr]

oh i see

so for V_y(x,y) would x be 0?

[main spade]

Yes. This is a necessary condition we are imposing.

[kind zephyr]

oh ok

so with that are we saying that both V_x adn V_y has both y=2h and x=0?

[main spade]

We consider separately x = 0 and y = 2h in the equation 2h - 2y + x^2 = 0.

[kind zephyr]

and the h is just a constant that stays with 2?

[main spade]

Obviously 2h is 2h. :/

[kind zephyr]

right yep

with the points found on V_x and V_y do you then sub that into the second derivative of buth V_x and V_y

[main spade]

Show your workings for both cases please.

[main spade]

You first need to show us what you got for the critical points.

[main spade]

,rotate

[kind zephyr]

this was found using using the -2x(2h-y)

[main spade]

I don't accept these results.

You have numerical numbers when you should not. (Numerical numbers lol.)

You have numbers instead of the results containing the arbitrary constant h.

You will have to let me know if you're willing to adjust your solution.

[kind zephyr]

oh right yea, so (0,1) should be (0,h)?

because for the Fy it's 2h-2y+x^2

is that still numerical?

[main spade]

You need to try and explain your workings in a much nicer way.

[kind zephyr]

yea i agree

[main spade]

Something like: Set x = 0 in the second equation 2h - 2y + x^2 = 0, from this we obtain 2h - 2y = 0, y = h.

[kind zephyr]

and for the Fx it should be (+or- sqrt(2),2h)

[main spade]

Use this to guide a explanation for the other option.

[kind zephyr]

oh i see ok

[main spade]

Set y = 2h in...

[kind zephyr]

i checked to see if the critical points are correct a calcualtor and it got the same as what i got, other than the h constant

so with that found wouldn't (0,1) be the only critical point to consider as it is within the shaded parabola?

[main spade]

I refuse to conform to the (0, 1) stuff, sorry.

[kind zephyr]

right ok

wdym by the y = 2h above?

is that an example

[main spade]

I wanted you to mimic the explanation for the case of y = 2h.

[kind zephyr]

oh

[main spade]

I wanted to help with your explanation skills.

[kind zephyr]

set y = 2h in the equation 2h-2y+x^2 =0, this would give x = sqrt(2)

like that?

[main spade]

Set y = 2h in the second equation 2h - 2y + x^2 = 0, from this we obtain 2h - 4h + x^2 = 0, x^2 = 2h, x = +- sqrt(2h).

[kind zephyr]

oh yes makes sense

[main spade]

It is nice to refer to it as a second equation as it would be clear you would be referring to the system:
V_x(x, y) = -2x(2h - y) = 0, x(y - 2h) = 0
V_y(x, y) = 2h - 2y + x^2 = 0.

[main spade]

Now as you referred to earlier you can exclude some of these critical points from further consideration because they do not lie inside the region which V(x, y) is defined on.

[kind zephyr]

yeap

so you would only sub in those points

[main spade]

Clearly, you need only test the point (0, h).

[kind zephyr]

yea

to see if it is a local maximum

[main spade]

You will use a theorem such as this to test the point.

Note: f_11 = f_xx and f_12 = f_xy, etc.

You need to find the values of V_xx, V_xy = V_yx and V_yy at (0, h) to use the test.

[kind zephyr]

what are these values used for?

[main spade]

This is the test you have been trying to use the entire time. We use them to show hypothesis 3 is satisfied at (0, h).

[kind zephyr]

so this is like testing if it is the right values to use?

[main spade]

All the work we have done so far has been done to determine the coordinates (X, Y) in the region we are defined that satisfy hypothesis 2.

To use the theorem (also referred to as the second derivative test) we must check the remaining hypotheses are satisfied.

V_xx = -2(2h - y), V(0, h) = -2(2h - h) = -2h < 0 since h > 0.

This shows hypothesis 4 is satisfied.

[kind zephyr]

ohh i get it

[main spade]

I'll translate hypothesis 3 into a more familiar form:
3. (V_xy)^2 - V_xx V_yy < 0 at (X, Y)

Remember this could be written as a determinant.

[kind zephyr]

ohhh is this the hessian test?

[main spade]

There's probably many names for it.

My book uses Theorem 2 lol.

[kind zephyr]

becase that would be (2x)^2 - (-4h+2y)(2h-2)

does that look right

[main spade]

I would not leave it as algebra personally. Especially since V_xx(0, h), V_xy(0, h) amd V_yy(0, h) are the only ones needed.

[kind zephyr]

also this is less than 0 which means it is a local maximum?

[kind zephyr]

oh i see

[main spade]

This shows hypothesis 4 is satisfied only.

We need to calculate (V_xy)^2 - V_xx V_yy < 0 at (0, h).

What do you get?

[kind zephyr]

after subbing in (0,h) into this function here?

[main spade]

Obviously, that's your algebraic result for any point (x, y).

[kind zephyr]

i got 2h

[main spade]

I got V_xx(0, h) = -2h, V_xy(0, h) = 0, V_yy(0, h) = -2. I get -4h.

[kind zephyr]

ohh yea i got the signs mixed up

[main spade]

You have now shown that the point (0, h) is a relative maximum.

[kind zephyr]

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

[kind zephyr]

yes

[main spade]

I'm following how it's done here.

Since the critical point occurs on the boundary of these region I'm not sure if you need to test all the boundary for a maximum.

[kind zephyr]

i think only the one that is in the region

[main spade]

I mean more often that not the critical point you consider would be inside the region and not on the boundary and you would need to perform an analysis of the value of V(x, y) on the boundary but since this is on the boundary I'm not sure you need to bother.

[kind zephyr]

oh ok i see

[main spade]

Actually, no you do.

[kind zephyr]

so with knowing that this is a maximum and that it is 4h? would you then find the V_max?

[main spade]

This critical point you found might possibly be the smallest relative maximum and the other points on the boundary might result in a bigger value of V(x, y).

[kind zephyr]

ah ok, would those points still be in the shaded region?

[main spade]

You could see what V(+-sqrt(2h), 2h) gives.

[main spade]

Yes. The maximum could be on the boundary, we need to check.

If (0, h) is the place of largest relative maximum you might not need to test the boundary.

Well V = 0 at both of them.

[kind zephyr]

did you find v = 0 by plugging in the (0,h) into the first given equation in the question?

[main spade]

V(0, h) = h^2. Clearly there can be no greater value of V(x, y).

[kind zephyr]

ooh

[main spade]

No. I tested the other critical points.

V(+-sqrt(2h), 2h) = (2h - 2h)(2h - 2h) = 0.

[kind zephyr]

where is the h^2 from?

[main spade]

What else could V(0, h) possibly refer to?

[kind zephyr]

yep ok

[kind zephyr]

ohh

ok i get that thanks

so Vmax would then equal h^2

[main spade]

It would appear so.

[kind zephyr]

so that would be the final answer then h^2?

[main spade]

It would appear so.

[kind zephyr]

cheers thank you for all the help, really appreciate it, sorry for the dumb questions👍

.close

[vocal sleet]

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[deft sedge]

to find y''', should you first distribute the 10 first before finding dy/dx or can you do that last?

[blissful agate]

however you like to

you will get the same answer

[deft sedge]

i see

so is the answer 60t or 60 for y'''?

apparently its 60, but not sure why

[blissful agate]

y''' is 60

because y'' is 60t

[deft sedge]

y' = 5+30t^2

which is velocity

[blissful agate]

where do you get the minus from

[deft sedge]

y''= 60t

[blissful agate]

yeah

[deft sedge]

typo

y''=acceleration

[blissful agate]

yes

is P(t) the rate

[deft sedge]

lol

[blissful agate]

to find acc take the derivative twice

[deft sedge]

i did tho

60t

[blissful agate]

yeah

[deft sedge]

lowkey im lost

forgot

[blissful agate]

what is the problem

[deft sedge]

how to find acceleration

[blissful agate]

now

[deft sedge]

so its 60 or 60t

y'' = 60t?

which is also accelerlation ?

[blissful agate]

yes

this is all to the question

i dont know why i am overcomplicating this probelm

[deft sedge]

dont think it goes that deep tho

c right

[blissful agate]

yeah

[deft sedge]

ty p29

u the man

[blissful agate]

is it right

[deft sedge]

yea

[blissful agate]

good

sorry again

for confusing you

[deft sedge]

na i confused u

i said y'''

lol

[carmine leaf]

wait what does

[deft sedge]

in the start

[carmine leaf]

60t people per year mean 💀

[deft sedge]

apparently lotta ppl be fucking in 1996

😄

from 10 - 119760

@carmine leaf

if ur asking what does acceleration of population mean

then i have no clue bro

[carmine leaf]

no, not that just a weird sentence imo

[deft sedge]

like wym

how do you mean

[carmine leaf]

nah is fine

acceleration is 60 per year people per year

makes sense i guess

[deft sedge]

it is a weird sentence tho

you aint wrong

60 x ppl per year squared

looooool

mymathlab be trippin

yo koter

what would it even look like

on a graph

acceleration of 60t

is that the change of speed?

this shit always confuses me

i wanna fully understand it

once and for all . . .

😂

velocity and acceleration

confuses me

[carmine leaf]

you just keep adding a change

rate

change of rate

change of change of rate

so in a car example the rate can be distance

change of rate is the rate at which distance changes (velocity)

and change of change of rate is the rate at which the velocity changes (acceleration)

so for example when you first drive into the motorway. you need increase velocity because the speed limit is higher

which is acceleration. you are going from lets say 60km/h to 90km/h

but if you are now already on the motorway your acceleration is pretty much 0 because you're just keeping a steady velocity

make sense?

[deft sedge]

YES

legit u just solved it for me

ty bro

speed and velocity vary how though?

velocity changes right

[carmine leaf]

velocity is directional and speed has no direction

if something is moving it has speed

[deft sedge]

drug dealers have speed too

i heard

[carmine leaf]

or +- velocity depending on what direction you consider positive

[deft sedge]

hm u kinda lost me tho

is velocity

not like

15x?

or something

idk

[carmine leaf]

yes for 15x right

[deft sedge]

speed is 15

[carmine leaf]

your x can be negative or positive

but for speed

you do |15x|

[deft sedge]

maybe these are poor examples

[carmine leaf]

because even if its moving down

you still consider it +speed

[deft sedge]

slowing down?

you mean?

[carmine leaf]

no slowing down is deacceleration

[deft sedge]

how can speed go down

[carmine leaf]

ok lets use car example again

[deft sedge]

thats deacceleration

like u said

[carmine leaf]

if you instead of total distance

use distance from home

and you record someone driving from home to work and back

the distance would increase until you reach work

and then decrease again when you drive from work to home

[deft sedge]

ok

[carmine leaf]

so the velocity is positive from home to work

and negative from work to home

but the speed is always positive. it only cares about if youre moving or not

doesnt matter if its a decrease or increase

actually a better example.

if you are walking backwards

velocity= -4km/h

speed = 4km/h

[deft sedge]

im pretty sure

if u walk backwards

ppl gonna think u on speed

💀

[carmine leaf]

💀

[deft sedge]

i gues that makes sense

but sort of not

[carmine leaf]

its not important in math i dont think

just physics

cause you decide what positive direction is for example

[deft sedge]

im eating an orange rn

[carmine leaf]

like are you going from 0m to 8000m in the air when a plane takes off

or 0m to -8000m

1st is positive velocity and 2nd is negative velocity

[deft sedge]

i think thats silly

thats a direction thing

[carmine leaf]

yes exactly but it matters in physics

like velocities can cancel

[deft sedge]

not a change of speed thing

rate of change of speed

[carmine leaf]

if something is causing you to go -5km/h and something else 5km/h then you just end up standing still

[deft sedge]

no

i moved back

then forward

back to my original location

i didnt stand still xd

[carmine leaf]

no but like at the same time

like if 2 people push you idk im bad at physics

i cant bring up good examples lol

[deft sedge]

yo

ur good bro

i think i get it now tho

strange how velocity works tho

are the other ones

speed and aceleration

can u have negative

of those

or just accleration

?

[carmine leaf]

just acceleration

[deft sedge]

and velocity

[carmine leaf]

yes

[deft sedge]

got it

ty

[carmine leaf]

but even in math its important information

[deft sedge]

def bro

[carmine leaf]

like if a price is going down

[deft sedge]

ye this is why i asked

😄

[carmine leaf]

-velocity is prob way more useful to know

than someone just saying it has a speed of idk 0.55dlrs/day

just forget speed is a word ngl

never used

[deft sedge]

never used the word speed?

in math?

o.o

or in general

um so lets say we are talking about someone's bank account

lets say its always increasing at a steady rate

thats speed

if it changes like they spend money on discord nitro

then velocity decreases

and the change of that speed is deacceleration

[carmine leaf]

i think you misunderstand. its not that velocity is always negative

[deft sedge]

im aware

it can increase too

[carmine leaf]

but its just weird to even use speed then

just use velocity

[deft sedge]

ik its weird

but

they ask for this sht dont they

in math

well

nvm

if its

[carmine leaf]

idk maybe

[deft sedge]

money

lol

fuck it

do i have the right idea tho

if their money stays in the bank

then thats steady

= speed

[carmine leaf]

wdym stays like nothing happens?

[deft sedge]

like they locked his back account

cant add or subtract

$

[carmine leaf]

thats 0 velocity and 0 acceleration

[deft sedge]

0 speed?

[carmine leaf]

yes

i mean its not changing at all

[deft sedge]

ok this is good

r u sure speed is 0 tho

not x?

[carmine leaf]

yes im sure

[deft sedge]

if ur driving at 50 mph

ur speed is 50

no?

[carmine leaf]

yes

[deft sedge]

ok

so

if he has 100 dollars in bank

and doesnt touch it

his speed is 100

[carmine leaf]

no because in the driving example

you are measuring distance

and distance is not standing still

[deft sedge]

tbh

acceleration, speed, velocity

should stay out of

problems

not relating to physics

xd

[carmine leaf]

sure just think rate of change i guess

can be more intuitive imo

bank you have f(t)=100 so rate of change f'(t)=0

car you have d(t)=50t so rate of change is d'(t)=50

[deft sedge]

whats the t in 50t

t dstance?

time

lol

[carmine leaf]

no d(t) is distance at some t

so in 2 hours (t=2)

[deft sedge]

some point?

[carmine leaf]

you have travlled d(2)=50 * 2 =100 distance (miles)

[deft sedge]

thats distance tho

not speed

o.o

well

if u travel 100 miles

ur speed is not 100

it varies

u cant go 0-100

steady

then 100-0

well unless ur me

i went from 0-100 real quick multiple times

[carmine leaf]

when you drive a car what else are you measuring speed for?

its distance

[deft sedge]

one time at this big festival party, slapped this chicks ass

first second of meeting her

then picked her up

and took her back to my cousin's place

in bolivia

😂

im a mad man

[deft sedge]

so u dont get pulled over

like idk bro

anyway ima drop this

i got so much to do

ill revisit

it

another time

ty for giving me more info tho

koter i got a big problem for u

i need ur help solving this one tho

legit if we can do it step by step that would be awesome

i got absolutely destroyed

trying to find the answer to this one

[carmine leaf]

quotient or product rule pick one

[deft sedge]

ngl

quotient is harder

but

i wanna imrpove

lets do quotient

[carmine leaf]

$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

[twin meteor]

Køter

[deft sedge]

heres what i got [(x^5)/(x(ln(x)) + (5x^4)/(ln(x))] / (x^10)

[carmine leaf]

youre close

your problem the derivative of f(x)=ln(x)

[deft sedge]

1/xlnx

ym

[carmine leaf]

,w d/dx ln(x)

[twin meteor]

Results provided by WolframAlpha

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

Requested by Køter#5504

[deft sedge]

y' of lnx =
1/lnx * 1/x

is 1/xlnx

[carmine leaf]

,w d/dx ln(x)

[twin meteor]

Results provided by WolframAlpha

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

Requested by Køter#5504

[deft sedge]

im not blind lol

well sometimes i am

but yes i saw this

[carmine leaf]

ok so why do you still think its not 1/x

[deft sedge]

log(x) = (1/ln_10) * (1/x) = (1/xln_10)

dy/dx of lnx tho, i have no idea

apparently log and ln are the same

but dont really care to get into that rn

[carmine leaf]

wolfram uses log=ln

so what i just showed you is the derivative of ln(x)

also please be more clear

when using functions

first you typed logx so no parenthesis which is alright

then you type ln_10 * 1/x

which makes no sense at all

and then you typed ln_10 alone in the end

which makes no sense either

functions have inputs

ln_10 makes no sense by itself

ln_10(x) means you are inputing x into the function

[deft sedge]

i dont make up the rules

xd

[carmine leaf]

this is ln(10)

and ln(a)

[deft sedge]

but ye mb about my lack of parenthesis

[carmine leaf]

ln doesnt have a base

i mean it does its log_e

but you dont type it out

so you cant do ln_10 it makes no sense

ln(10) and ln(a)

[deft sedge]

ln is log_e?

[carmine leaf]

yes

[deft sedge]

ok so

what do i do

how do i get 1/x

as dy.dx

of lnx

[carmine leaf]

what happens if a=e?

[deft sedge]

u get lnx

lol

[carmine leaf]

yes on the LHS you get d/dx ln(x)

what about the RHS?

[deft sedge]

https://tenor.com/view/bye-felicia-hiding-humiliated-leave-me-alone-girl-bye-gif-21567771

[carmine leaf]

im asking you to replace one thing

a=e

[deft sedge]

we were asked to find dy/dx lnx

so Log_e x

now i gotta find

dy/dx

of log_e x?

[carmine leaf]

you just said the same thing twice

[deft sedge]

ok i guess it cancels out on rhs

[carmine leaf]

what cancels

[deft sedge]

ik for clarification

purposes

was trying to not confuse

mb

[carmine leaf]

im not necessarily asking for the end result. i just want you to tell me what happens when a=e

and i can help simplify later

[deft sedge]

it cancels out

[carmine leaf]

if you dont see it

[deft sedge]

ln e

[carmine leaf]

yes

im not a mind reader

idk waht 'it' is

before you say it

[deft sedge]

ln and e

=1

ur left with 1/x

[carmine leaf]

ln(e)

yes

good

burn this into your memory

d/dx ln(x)=1/x

[deft sedge]

i tattood it into my mind

[carmine leaf]

good

its an important result you will use all the time

so back to your original problel now that you know

type it out

[carmine leaf]

again close

first off its - not + look at quotient rule i sent

$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

[twin meteor]

Køter

[carmine leaf]

2nd you had $f(x)g'(x)=5x^4ln(x)$

[twin meteor]

Køter

[carmine leaf]

you cancelled the x^4

but where did the 5 go

[deft sedge]

oh the 5

ok

[sudden compass]

why hasnt this closed yet...

[deft sedge]

because we arent done jerking each other off to 1+1 =2

[sudden compass]

....

[deft sedge]

if you dont like it, go cry in the lobby voice channel

ima hold this channel hostage for 72 hours

.close

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[azure horizon]

Can someone help w question 3 how do I do that

[azure horizon]

I can’t rationalise it

[hasty pulsar]

,rotate

[wary mantle]

Is this during an exam?

[azure horizon]

Niet It’s a past paper I printed off myself by MadasMaths

I acc can’t believe that some ppl actually manage to cheat bc here you get watched like a fecking hawk

[wary mantle]

There is the identity $(a-b)(a^2 + ab + b^2) = a^3 - b^3$.
Can you continue with that hint?

[azure horizon]

I will try

Ty

Could that also work w fractional powers?

[wary mantle]

It holds for all $a, b \in \mathbb{R}$

[azure horizon]

Ok so if I change out the powers to fractions is that a win?

[wary mantle]

What do you mean?
You can look at $\sqrt[3]{4}$ as just some $a \in \mathbb{R}$

[azure horizon]

If I make that equal a^3

[wary mantle]

Well, you could of course do that, but there is a way I find easier to rationalize the denominator.

[azure horizon]

Ok

[wary mantle]

$(a-b)(a^2 + ab + b^2) = a^3 - b^3$. \ $a \rightarrow a^3$ \ $b \rightarrow b^3$

[azure horizon]

Ohhh

Ty

[wary mantle]

There is one number that we want to raise to the third power

[azure horizon]

Nope I’m still not getting it

[brisk imp]

Big balls 69

Jkjk

That’s wasn’t funny nvm

[wary mantle]

The idea is to multiply numerator and denominator of the fraction by $(a^2 + ab + b^2)$.

[wary mantle]

In this case, we have $\frac{3}{\sqrt[3]{4} - 1}$.

We want to raise $\sqrt[3]{4}$ to the third power so that the $\sqrt[3]{}$ vanishes.

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[clever swallow]

can someone explain the second step in (1)? I'm entirely clueless, and even unsure as to where the additional eigen value came from??

[clever swallow]

this is a proof for why the sum of the diagonal is equal to the sum of eigen values, and why the product of eigen values is the det(a)

[paper depot]

what additional eigenvalue?

[clever swallow]

there's lambda 1 and lambda 2, but in the first line now we're taking the difference with another lambda

[paper depot]

because we're looking at the charpoly

the unindexed lambda is just the variable in the charpoly

[clever swallow]

charpoly?

I don't know what that is

[formal rock]

Characteristic polynomial

[clever swallow]

ah i see a few pages later they mention the characteristic polynomial of a matrix

[formal rock]

Mhm. We can call it charpoly for short

[clever swallow]

i see that it obviously looks like the factored form of a polynomial, but I still don't see why this is valid..?

and how does this relate to the determinant

[formal rock]

$(\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2 -2(\lambda_1 + \lambda_2)\lambda + \lambda_1 \lambda_2$?

[twin meteor]

VulcanOne

[formal rock]

λ with no subscripts is the variable

Like x

But we use it to find the eigenvalues λ1 and λ2

[clever swallow]

makes sense, but why is that equal to the determinant?

like sure ok, but why?

is that just how its defined and i gotta accept it or is there logic behind it

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[simple beacon]

Solve 2x4 – 128x = 0 using the most efficient method based on learning of this past unit,

[tidal jackal]

this looks perfect @simple beacon !

[simple beacon]

thanks

are you able to assist with one more

@tidal jackal

[tidal jackal]

Sure

please ping me though when you send it

[simple beacon]

When a polynomial p(x) is divided by x + 2 the remainder is 3. Determine the remainder when the following polynomials are also divided by x + 2:

a) p(x) + x + 2 Remainder: __________ b) p(x) + x + 1 Remainder: ____________

Justify your answer for (b) above.

@tidal jackal

[tidal jackal]

Thanks, ill check your work now

Correct.

But for a you could have just factored out the x+2

(x+2)(p(x)+1)

[simple beacon]

is it still good

@tidal jackal do you have time for one more

[tidal jackal]

Sure

[simple beacon]

Solve x^3 + 3x^2 - 4x < 12 using an interval table

@tidal jackal are you still there

[tidal jackal]

didnt see it

i won't check your work, but ill check the answer

so if you get the right answer, your work is most likely correct as well

[simple beacon]

alright man

[tidal jackal]

ok wrong answer

[simple beacon]

dam

[tidal jackal]

why is (-3,-2) negative?

i checked the table and you are correct that 2 factors are negative and one is positive

what is the sign of negative*negative?

[simple beacon]

positive

[tidal jackal]

correct

so then the function is positive in (-3,-2)

[simple beacon]

so if I change that simple error

im good to go

[tidal jackal]

nope

[simple beacon]

dam

[tidal jackal]

look in -2<x<2

one factor is negative, two are positive

what is negative*positive?

[simple beacon]

positive

[tidal jackal]

-1*1=?

[simple beacon]

-1

I meant negative

[tidal jackal]

yup

[vocal sleet]

@simple beacon Has your question been resolved?

[simple beacon]

@tidal jackal so the solution should be written as x equivalent to (3,2) u (-2,2)

[tidal jackal]

no.

(-infinity, -3) U (-2,2)

your answer is x<-3 and -2<x<2

[simple beacon]

ok thanks

[vocal sleet]

@simple beacon Has your question been resolved?

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[vast shale]

Help me prove this one:

Let $A\in m_{n\times n}\left(\mathbb{R}\right)$ If $A$ is invertible from the right then $A$ is invertible from the left

[twin meteor]

CStudent

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

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Channel closed

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[cinder oyster]

.close

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[echo plaza]

you can find the x coordinate of the vertex of a parabola using -b/2a, but how is this derived

[flat whale]

calculus

[echo plaza]

i cant use calculus

so i have to use a different method

[flat whale]

completing the square

[echo plaza]

what will that achieve?

vertex form?

[flat whale]

try it

[echo plaza]

i see how it will work

[echo plaza]

can i multiply both sides by 10?

[flat whale]

sure? i have no idea where your equation is

[echo plaza]

H(t)=-4.9t^2+24.5t+1.5

i actually have to divite

it gets ugly

[echo plaza]

can i do something better?

[flat whale]

that's different from what you typed

Are you solving for when H(t) = 0?

show the full set of instructions

[echo plaza]

do i have to?

if so than this becomes easier

[flat whale]

it's a question for you

[flat whale]

.

[echo plaza]

you mean show the original problem?

[flat whale]

then the answer to this is no

use $h_{M}$ for maximum height. it's just a constant. now you can use completing the square methods

[twin meteor]

riemann

[echo plaza]

i'm not sure how i would use that

[flat whale]

re-write your equation with $h_M$ instead of $H(t)$ and put the entire equation into vertex form to find the vertex

[twin meteor]

riemann

[echo plaza]

so this stays the same (except for replacing H(t))?

[flat whale]

yes. again, h_M is a constant

[echo plaza]

i just feel like i'm doing something wrong

the numbers are so messy

[flat whale]

probably just algebra mistake. show your work

[vocal sleet]

@echo plaza Has your question been resolved?

[echo plaza]

this is absolutely ugly, and i am not allowed to use a calculuator

[lusty fox]

do you need to simplify

[echo plaza]

probably not

but i dont know if this is right

does it look fine?

[lusty fox]

wait what is going on here

why did you do those steps

just take the derivative

[flat whale]

.

[lusty fox]

ah my bad

[flat whale]

keep going

[echo plaza]

i get this.
h_M=-49/10(t+245/-98)^2-49/10(15/-49-(245^2/98^2)
wouldnt the max height be the k value though? -49/10(15/-49-(245^2/98^2)

[flat whale]

what is k value

[echo plaza]

the y value of the vertex

in a(x-h)^2+k

[lusty fox]

only when a is negative

[echo plaza]

a is negative here

[lusty fox]

yes

[echo plaza]

so the answer is -49/10(15/-49-(245^2/98^2)?

well that doesnt seem right

because

it cant be negative

im so confused

?

are you there?

[lusty fox]

@echo plazaozrim bhaim volchim

i think its actually not that bad

245/49 is actually just 5

[lusty fox]

simplify step three

and it should be a lot easier

[echo plaza]

?

[lusty fox]

dont tell me you cant read english-hebrew

[echo plaza]

no im just a begginer at hebrew

[lusty fox]

yet your name is in hebrew

interesting

[echo plaza]

yes

[lusty fox]

I said "helpers come and go"

[echo plaza]

ah

[lusty fox]

now dont do all the +-

keep it simple

and its not h_M

its just h

h_M is k

[echo plaza]

i get this now 49/10((15/49)+(25/4))

much nicer

[echo plaza]

yeah thats what i thought

thanks

toda raba

[lusty fox]

no problem

[echo plaza]

.close

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[peak axle]

If I have a matrix in RREF, how can I use that to determine which vectors represented by it are linearly independent?

[lyric fossil]

the set containing the vectors with pivot rows is linearly independent

if you include any other vector (if there are any) it becomes a linearly dependent set

[peak axle]

This is my RREF for reference

So in that case would it be the first, second, fifth, and sixth columns?

[lyric fossil]

yes

you can also swap any vector for any coming after it, up to the next pivot

[peak axle]

Is there a way for me to determine the value of those vectors given the rest of the information there?

[lyric fossil]

for example you could take the first, third, fifth, and seventh

[lyric fossil]

what do you mean value

[peak axle]

Like would my basis vectors just be {1,0,0,0,0,0,0}, {0,1,0,0,0,0,0}, {0,0,1,0,0,0,0}, and {0,0,0,1,0,0,0}?

[lyric fossil]

yes

or like i said earlier you can swap any of those with the vectors after it, up to the next pivot

[lyric fossil]

this is a basis for the column space

actually

im not sure about that

are you looking for a basis for your column space

[peak axle]

Is there a way to find the basis for my original set of vectors given that? My original vectors had non-zero entries in rows 5-7

[lyric fossil]

what do you mean by basis for the original set of vectors

are you asking for a basis for the space spanned by the columns of the original matrix?

if so take the column vectors where your pivots are right now

that is, the first, second, fifth, sixth vectors in your original matrix

[peak axle]

This is my original matrix which I made from 7 vectors in a set S, I'm trying to find a subset of vectors in S that are linearly independent and form a basis for the subspace spanned by S

[lyric fossil]

then do this

[peak axle]

Oh that's it?

[lyric fossil]

yeah should be

[peak axle]

That's pretty magical

And since there's four vectors I'm guessing the dimension of aforementioned subspace would be 4?

[lyric fossil]

oops

[lyric fossil]

yea

[peak axle]

Cool thanks

.close

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[empty river]

if I have a function $\frac{x-4}{x^2+x-20}$
I would need to simplify the rational expression to find the x-intercept right, since setting numerator of this = 0 andd solving for x finds x-ints?

[twin meteor]

notnick

[rose raft]

4

[empty river]

how would it be 4 when x-4 is one of the restrictions?

[rose raft]

oh nvm

[empty river]

nvm i just solved my own question

.close

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[haughty monolith]

How do I draw a diagram for this to help understand what I’m finding

[outer warren]

the angle in the [] seems to indicate the bearing, have you done stuff with that before?

[haughty monolith]

What?

[placid spear]

do what he said any you will get your answer

[haughty monolith]

That made no sense

This is a kinematics question and I’m solving for the displacement

[placid spear]

multiply v by 8 draw the gex if 2.6

[outer warren]

ignore them

[haughty monolith]

What do I do

[outer warren]

the angle in the [] seems to indicate the bearing, have you done stuff with that before?

[haughty monolith]

No, what me to show you my formula sheet so you know what kinda question this is

Want*

[outer warren]

this question from the looks of it is a geometry question related to bearings

[haughty monolith]

I don’t know what bearings are

[outer warren]

<@&268886789983436800> chatgpt and unhelpful spam

[haughty monolith]

Bro your not funny

@outer warren do you have an idea yet?

[outer warren]

yeh, do a quick search on bearings if you don't know what they are

[haughty monolith]

.close

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[modern radish]

Hey

[modern radish]

I've been having trouble with this problem all day. It uses the attached definition of pi, I know I need to prove somehow that $3(10/71) < a4 < pi < b4 < 3(1/7)$

[twin meteor]

WegenLisbeth

[modern radish]

I have no clue on how to even start though and I've spent about 2 hours staring at it today.

I have a4 and b4 calculated in excel, if that means anything.

<@&286206848099549185>

[vocal sleet]

@modern radish Has your question been resolved?

[modern radish]

Maybe it is sufficient to show that the base case is between the given values and n+1 is as well.

[vocal sleet]

@modern radish Has your question been resolved?

[modern radish]

.close a chad in the discussion channel for proofs helped me

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[tame river]

How do I solve this?
-_- Wow. Images got butchered.

[tame river]

More specifically, how do I solve this part?

[halcyon compass]

seems like linear function transformations you familiar with those?

you can either plugin values and get the line graph

or if the teacher wants a more detailed and proper answer you can explain all those operations how they manipulate the y=x function

[tame river]

. . .

[halcyon compass]

didn't understand what i said?

[tame river]

no.

[halcyon compass]

well you need to learn more of whatever teacher explained in class than, google function transformations

translation, dilation, and reflection are the 3 transformations

[tame river]

...

-_- I recognize those through.

[halcyon compass]

but again shortcut would be just plugin values and draw the points

[tame river]

yes okay so

how to do that?

[halcyon compass]

just put x= something

and solve for y

[tame river]

just start throwing numbers into x and y?

right then so

[halcyon compass]

just x and solve what y is for the x you throw in

[tame river]

y+2=2(3-2)

ok

[halcyon compass]

you can do .close If you have no further questions

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[covert salmon]

Need Help With Linear Algebra

[covert salmon]

How do I do this?

[maiden iron]

do what ?

[covert salmon]

How do I find the answers...

To the bottom

[maiden iron]

algebraic multiplicity of -1 is the degree of the monomial (λ+1)

[covert salmon]

And how would I find that degree?

[maiden iron]

you dont know what the degree of a monomial is ?

[covert salmon]

isn't the answer just 2

[maiden iron]

no

[covert salmon]

sorry

1

[maiden iron]

2 is for 1

[maiden iron]

it is

[covert salmon]

it's just 1,1,2 in that order?

[maiden iron]

yes

[covert salmon]

wow

ty

sir

Have an outstanding day!

[maiden iron]

Thx

[covert salmon]

.close

[vocal sleet]

Channel closed

Closed by @covert salmon

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[maiden iron]

u too

[vocal sleet]

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>.
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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.

[knotty tangle]

Can someone help me with this problem I don't know what I am doing?

[rapid swift]

you have nothing?

[knotty tangle]

No not really threw my paper into the trash.

[rapid swift]

what did you have on the paper

[knotty tangle]

Some random mess

[rapid swift]

you have no idea whatsoever?

[knotty tangle]

No I was not present in class when the teacher was going over notes so I've been trying to figure it out myself before next class.

[rapid swift]

can you not ask your teacher or something? also this doesn't seem to involve just one class

this has a few different concepts

surely you've seen some of them before?

[jagged bone]

WHERE DO I ASK FOR HELP

[knotty tangle]

Well its pre calculus

You go click on a server

where it says its available

and post your question there

[rapid swift]

#❓how-to-get-help

[knotty tangle]

ye what they did

[rapid swift]

I'm not exactly sure how to help you here if you have absolutely nothing to go off

I would suggest looking up some videos on Khan Academy or a similar resource about the words mentioned in the problem

[knotty tangle]

Alright well I have to get going thank you for chatting.

.close

[vocal sleet]

Channel closed

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Use .reopen if this was a mistake.

[vocal sleet]

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.

[vocal sleet]

@gloomy venture Has your question been resolved?

[icy spear]

Not sure if that converges

Shouldn't that be x=3

Nvm

Its the area above the curve, not under it

[fluid obsidian]

Badly phrased question

Anyway just find it

Nothing much to do here

[vocal sleet]

Channel closed

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