help-10

[wary badger]

you don’t have good notes do you?

like the note taking app

[foggy merlin]

so was my original fbd wrong or do u just have a diff preference

[wary badger]

it was wrong

[foggy merlin]

fr

thanks i appreicate u ima ask a friend later

[wary badger]

i can draw it for you now

i just pulled out my ipad

😭

[foggy merlin]

sure

@wary badger i get it now i apperic ate u fr

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[twin marlin]

Hello, id like to know why its 5/2 not 5

[wary badger]

u = x^2 + 5, du = 2x dx

[twin marlin]

Ahhh darn

Yeah

Thank you

[wary badger]

you’re welcome

[twin marlin]

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[violet coral]

hey, i need help with this problem, it is intoductory linear algebra:
Use the dot product to find a vector vVector = (v1, v2) that is perpendicular to the vector uVector = (8, -3). Thanks

[patent osprey]

my fav perpendicular vector

is (-b, a)

[violet coral]

alr

i'll try it out

[violet coral]

Yup it worked. Thanks,

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[delicate trout]

can someone please help me with this? idk how to do it

[tough reef]

Do you know which of the graphs the function is, for starters?

[delicate trout]

No i dont

im not sure how to figure that out

[patent osprey]

you can use a graphing calculator

[delicate trout]

oh ok

[patent osprey]

ill bless you today

[delicate trout]

ty

would you happen to know how to find x

[patent osprey]

mhm

where is the graph discontinuous

in this pic

[delicate trout]

um

at the bottom?

[patent osprey]

wait first what does it mean for a graph to be continuous

[delicate trout]

it means theres no gaps in the graph

[patent osprey]

kinda

try (2n+1)π/2

[delicate trout]

i honestly dont really know

[patent osprey]

and nπ

KY

[delicate trout]

what

im confused

[patent osprey]

try (2n+1)π/2

and nπ

[delicate trout]

on the graph?

[patent osprey]

mhm

[tranquil geyser]

When does tan²(x) have gaps

Also when does ln(x) have gaps

[patent osprey]

you don't even need to do that

they want you to use a graphing calc

for this problem

so you can spot the disc. immediately

[tranquil geyser]

Doesn't the OP have to understand how it works?

We can skip this

If she does

[delicate trout]

i put this in desmos and nothing shows

[patent osprey]

I don't think so

[tranquil geyser]

.

[delicate trout]

I dont know

[tranquil geyser]

Okay when does tanx have gaps

[delicate trout]

at x=π/2?

[tranquil geyser]

At x = π/2 + nπ

,w cos x = 0

[warm shale]

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[tranquil geyser]

Okay so when tanx is undefined then tan²x is also undefined because tan²x = (tanx)²

So now when is lnx undefined

@delicate trout

[delicate trout]

is it also undefined?

[tranquil geyser]

Ln x is undefined for x≤0

So (tan²x) is never less than 0

[delicate trout]

oh ok

[tranquil geyser]

So we have our gaps at 0, π/2

So the first option is out

Now to find which of the other graphs is most correct

[delicate trout]

how do you know where pi/2 is?

on the graph

[tranquil geyser]

Pi/2 ≈ 3.142÷2

[delicate trout]

ohh ok

[tranquil geyser]

You can select a random y value where the function is defined like 2 and
Solve for x in 2=ln(tan²x)

,w solve 2=ln(tan²x)

[warm shale]

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[tranquil geyser]

,w tan^-1 e

[warm shale]

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[tranquil geyser]

@delicate trout

Then at x≈1.22

y=2

[delicate trout]

oh ok

[tranquil geyser]

Then just look for the graph that has y = 2 somewhere between 1 and 1.5

[delicate trout]

okay

thank you for the help

[tranquil geyser]

Which one is it

[delicate trout]

the last one?

[tranquil geyser]

Yes should be

,w plot ln(tan²x)

[warm shale]

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[tranquil geyser]

Now you can use the graphing calculator to verify

[delicate trout]

ok

[tranquil geyser]

That's all

[tranquil geyser]

!done

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[timid silo]

what have you tried

[odd mural]

Answer is b

[timid silo]

u = log(x) would've saved u a lot time

[odd mural]

Oh

Shall I try that way?

[timid silo]

yea

[odd mural]

Oki

Anything wrong?

@timid silo

;-;

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[quiet drum]

how did they go from the 1st line to the 2nd line

[worn yoke]

factor our t. then you get
0 = t(15 - 4.9t)

[obtuse pebble]

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[timid silo]

i am having trouble with this problem

i will send my attempt at a solution

i am making a silly mistake

i dont know where it is

i should have the degree of the numerator = deg(denominator)

<@&286206848099549185>

can someone check my work plz

ive been stuck on this problem for 3 days and its a very annoying problem

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[low trail]

Is (5x²-17x-15)÷(x-4)=5²+3x-3?

[quiet drum]

Where does the ‘t’ outside the bracket go ?

[deft magnet]

first term should be 5x?

[obtuse pebble]

@low trail Has your question been resolved?

[low trail]

Yea I just forgot

5x

[low trail]

Thanks

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[full vault]

guys for this question the solution found the annual effective rate of 2.5%. Why dont cant we just find the interest rate such that there is less than 25$ of interest for 7 days.
Anyone can help?

[obtuse pebble]

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[safe fjord]

help

[timid silo]

what's the question

[safe fjord]

oh let me pull it back up

sorry i didnt hear the notification

The region R is bounded by y = x2 − 4 and y = −3x is divided into two pieces by the line y = a,
where −4 ≤ a ≤ 12. R1 is the part of R that lies below y = a and R2 is the part of R that lies above y = a. Suppose it
is known that the area of R1 and the area of R2 are equal

Graph the regions R, R1, and R2 and label the boundary curves. Then, without performing any
calculations, estimate the value of a from your graph.

help

<@&286206848099549185>

[wintry swift]

!status

[obtuse pebble]

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

[safe fjord]

1

[wintry swift]

you know how to graph functions?

[safe fjord]

yeah

[wintry swift]

then do it.

[safe fjord]

i know what bounded means aswell

yeah i have that part

[wintry swift]

well then you have already started and your answer 1 was wrong.

anyway. show your sketch.

[safe fjord]

i did it in desmos

[wintry swift]

make a screenshot

[safe fjord]

i dont get the a part

okay

[wintry swift]

assume a = 5 for example.

[wintry swift]

so you nearly did: Graph the regions R, R1, and R2 and label the boundary curves.

[safe fjord]

so is r everything betweeen -3x and x^2-4

[wintry swift]

yes

[safe fjord]

the r1 is everything above green line

and r2 everything under

[wintry swift]

the other way round.

[safe fjord]

but how would i get r1 and r2 to be ==

[wintry swift]

r1 is below, r2 above

[safe fjord]

oh r1 everything under

but how would i estimate to have the regions equal to each other

it would definatley have to be lower then 5

[wintry swift]

you have gridlines, at the moment 5 lines for a step 2, right?

[safe fjord]

wait wha?

im cconfused can you reword the question

every 4 blocks is 2

[wintry swift]

sorry, 4 lines for a step 2.

[safe fjord]

yeah

[wintry swift]

make them smaller, for example 10 lines for a step 2.

ok, 10 may be to much

[safe fjord]

is this okay

[wintry swift]

well, yes. its a kind of compromise, more lines would give better results, but needs more work. you decide,

count the squares inside the regions givs you a size of the ares.

[safe fjord]

okay how many do i have to count like how high up

[wintry swift]

count for R, the half the resulted value and then count from the bottom to top to get the postion of the line.

[safe fjord]

and some are not full squares

[wintry swift]

well thats what i said. number of lines determine the quality of the result.

you can count the squares fully inside -> lower bound for the area
count the squares needed to cover the region fully -> upper bound for the area

[safe fjord]

can i just average out the limits?

or will that not work

[wintry swift]

well ouf coures, you are asked to estimate.

its your decision how you estimate.

[safe fjord]

oh that will probaly be more efficent

[obtuse pebble]

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[verbal oak]

What would be the correct way of showing this?

[verbal oak]

I have tried using the integral method to show that $int_{1}^{\infty}\frac{1}{n^2+1} = \frac{\pi}{4}$ so $S \leq \frac{\pi}{4} + S_1 = \frac{\pi}{4} + \frac{1}{2}$

[warm shale]

louis from brazil

[verbal oak]

Forgot the \ before int but it doesn't really matter

But I'm not really sure how to show that S is greater than π/4

S being the infinite sum of the series

[tardy epoch]

There's more than one correct way

Did you learn the arctan integral yet

[verbal oak]

yep that's what I did

Or do you mean integral of arctan literally

[tardy epoch]

Sorry I meant arctan = integral 1/(1+x^2)

Use Riemann sum on the integral to bound. The series in the middle should be one of the left or right endpoints

[obtuse pebble]

@verbal oak Has your question been resolved?

[verbal oak]

You mean something like $\int_{1}^{n}f(x)dx \leq$ Infinite Sum $\leq \int_{1}^{n}f(x)dx + S_1$?

[warm shale]

louis from brazil

[verbal oak]

I meant infinity instead of n

Kinda sleepy rn

[tardy epoch]

Yea

[verbal oak]

Okay then

Is there any textbook that shows why this works?

The one my professor is using doesn't really say explicitly that

[tardy epoch]

Google Riemann sum

[verbal oak]

okay then

Thank you!

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[upper shuttle]

does a finite integral domain necessarily have an identity

[dusk widget]

hey esthesia

I don't know any ring theory, but perhaps this MSE post may be useful?
https://math.stackexchange.com/questions/118266/prove-that-every-finite-domain-contains-an-identity-element

oh, and congrats on Helpful!

[upper shuttle]

i didn't even notice

thanks

[dusk widget]

is the stack exchange post useful

[upper shuttle]

this is amazing yes it helped

[dusk widget]

[upper shuttle]

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[tawdry vortex]

sry for tagging

[tawdry vortex]

@crude coral

i went afk sry

SO it basically disprove this experiment right

Numbers r involved so pls its math 😭

[crude coral]

why u say that

[tawdry vortex]

cuz it say particle have wave propert at the end

[crude coral]

the slits are really really small , almost equal to the wavelenth of electrons

[tawdry vortex]

oh

so basically they have them

but invisible basically

[crude coral]

invisible?

[tawdry vortex]

like

hard to see

i guess

the waves of particles

[crude coral]

yeah ofc

[tawdry vortex]

but isnt that for any waves

like wave of light do we rly se e dat

off topic i guess

[crude coral]

the experiment we are using basically uses refraction

and refraction is a property of wave

so yeah , waves can do that

[tawdry vortex]

o

[crude coral]

thats why we conclude that particle of electron is like a wave , cuz they show this refraction property

[tawdry vortex]

AHhh

I see

Ok

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[wheat pebble]

Need some help here not sure what I am doing wrong

[latent walrus]

-int(Sn)>=-I(f)>=-int(tn)

[wheat pebble]

Yeah

Then when you add them, it doesnt cancel out to the result

[latent walrus]

you can only really 'add' them as
int(s)-int(t)<=I~(f)-I(f)<=int(t)-int(s) i believe

[wheat pebble]

Ah wait, yours is a little different to mine.

Yes I agree with you. But their answer is different

[wheat pebble]

You added the negative but also switched the sign. I thought it was one or the other.

Ah i see now its not. okay

Another attemp based on your answer

[obtuse pebble]

@wheat pebble Has your question been resolved?

[wheat pebble]

<@&286206848099549185>

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@wheat pebble Has your question been resolved?

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[thin cave]

Hi I did the ans which is 343/6 .

[timid silo]

so do u have a question ?

[ionic isle]

congrats bro

[thin cave]

But I wanna to understand that why integrate -4 to 3 for the curve does the area below x-axis included .... ???

no absolute thing ?

[timid silo]

if u just integrate the function x^2 +2x-4 , u dont get the area u are looking for

i dont rly get ur question actually

[thin cave]

alright so my steps is integrate -4 to 3 (x+8) - (x^2+2x-4) dx = 343/6 right ?

why the region below x-axis included ?

[timid silo]

sorry i dont know

[wheat pebble]

If you do integrate $x^2+2x-4$ it is negative, so it does include that area under the curve.

[warm shale]

Sands

[wheat pebble]

So you might just need to get a bit creative.

You can obviously work out the area under x+8 within those limits.

So you basically just need to work out what those 2 small triangles are, you can do that by finding out where $x^2 +2x-4$ is equal to zero, and integrate between -4 and the first zero point and the second point and 3.

[warm shale]

Sands

[wheat pebble]

Hope that makes sense

[obtuse pebble]

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[nocturne junco]

What did I do wrong? (The answer is supposed to be x <= -5/4)

[nocturne junco]

.close

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[open mountain]

What does that mean that the flux of a vector field is coming out of the surface S?

[open mountain]

Can it be both incoming and outgoing?

[obtuse pebble]

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[unreal musk]

Bear in mind that you're trying to solve $c_1 p_1 + c_2 p_2 + c_3 p_3 = q$, from which you could e.g. compare coefficients for both

[warm shale]

@unreal musk

[unreal musk]

That'll give you linear equations which should hopefully be alright to solve

Well, with respect to the B basis, or the "standard" basis (say {1, t, t^2})?

If you want to do that, write out what c1p1 + c2p2 + c3p3 is in terms of 1, t and t^2

[obtuse pebble]

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[obtuse pebble]

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[young crescent]

How do I solve this bro I think I have autism

[junior granite]

signs of 8y and y^2 aren't quite right

[young crescent]

Ignore everything after the second line

[junior granite]

your way is alright

just the sign of those two were off the track

[young crescent]

So what now

[timid silo]

wdym

x= 2+sqrt(y)

subsitute in second equation

[tawdry sail]

$y = x^2 + 4x + 4 \ \sqrt{x} + x^2 + 4x +4 = 4$

[timid silo]

u messed it

[warm shale]

<rajel />

[tawdry sail]

this is how i would go

[tawdry sail]

why?

[timid silo]

i was just refering as to your TEX was not good

[obtuse pebble]

@young crescent Has your question been resolved?

[young crescent]

I don't even know myself

[obtuse pebble]

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[heady turtle]

how to move forward

[tender tusk]

+-1

wait maybe not?

[heady turtle]

yea I tried but failed

[tender tusk]

trig sub maybe?

[heady turtle]

failed

[zenith raft]

is that + infinity in the bound?

mellin transforms

[obtuse pebble]

@heady turtle Has your question been resolved?

[heady turtle]

omg what's that ? I searched but how can I use that in this q?

[zenith raft]

the integral is (almost, up to a shift by 1 which is no problem) the mellin transform of $x\mapsto \frac{1}{(1+x^2)(1+x^a)}$

[warm shale]

layla is not harper

[zenith raft]

but even though mellin transforms are very well studied i'm not sure how much that helps unfortunately

[heady turtle]

ok thank you for providing a new pespective, I'll take a look

.close

[obtuse pebble]

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[snow falcon]

Can someone help me with part b

I dont understand it

I’d appreciate the help

[last pilot]

imma be very honest, the text is tiny

we aren't gonna be squinting at our screens just to decipher what's on there

maybe write it down here?

[ruby path]

Did you find the value of k?

Oh wait nvm okay that's (a)

Do you understand what (b) is asking for

[snow falcon]

No I don’t

That’s what I need help with

[ruby path]

You can do it classically

[snow falcon]

I dont get the x1+x2 bit

[ruby path]

See the number of pairs of observations (it's ordered) whose sum is 4

[ruby path]

It just means you've "drawn" twice

With a specific order

[snow falcon]

What else were you gonna say bro ?

[ruby path]

Nothing else?

[snow falcon]

Ohhh Thank uuu

I got that

Appreciate you

[ruby path]

(and then divide by the total number of pairs possible)

That should net the right answer

[snow falcon]

I get that now thank you

Bro idk why I was stuck on that 😭

[ruby path]

It's alright probability notation is weird anyways

[obtuse pebble]

@snow falcon Has your question been resolved?

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[thorny hawk]

can someone pls help me solve this with steps????

[azure wraith]

first

is this a sin or a cos graph

[thorny hawk]

this is cos right?

[azure wraith]

..

its a sin graph

[old lily]

it could be modelled by both cos and sin but yes maybe sin is easier

[gentle knot]

Yes it s a sinusoidal

[azure wraith]

yeah

lets do the easiest method

[gentle knot]

We calculate the amplitude , midline, period and then we do the phase shift and then function , no?

[azure wraith]

yes

[thorny hawk]

oh ok

[azure wraith]

if we do sin

we dont have to phase shift

[gentle knot]

oh, yes

[old lily]

can u say what's the amplitude of the wave

[old lily]

open a new channel

[dreamy trellis]

K

[thorny hawk]

to calc amplitude i learnt that its max value - min value divided by 2?

[gentle knot]

I think we need to estimate

[old lily]

yeah u can do that

[thorny hawk]

so in this case it would be 2.5 - 1.5 / 2 right

[old lily]

yes

0.5

[gentle knot]

0.5

meters

[old lily]

you also need the wavelength

[gentle knot]

and the midline is the same but whit +

[old lily]

ie distance between peaks

[thorny hawk]

so the standard equation is y = a sin (bx + c) + d

so amplitude is 0.5 in this case

how do i calculate (bx+c) + d

<@&286206848099549185>

anyone pls

[obtuse pebble]

@thorny hawk Has your question been resolved?

[woeful torrent]

start with y = sin(x).

how can you change this function so it models the period of the wave?

how can you change this function so it's at the same height as the wave?

[thorny hawk]

ok

what do i do after

[woeful torrent]

what happens if I do y = A sin(x)?
what happens if I do y = sin(B x)?
what happens if I do y = sin(x) + C?
what happens if I do y = sin(x + D)?

play around with them. can you see why changing them does what it does?
https://www.desmos.com/calculator/ao1cxj2ahl

[thorny hawk]

got the answer, tysm

how do i calculate b from this?

because its in degrees what should i do

<@&286206848099549185>

<@&286206848099549185>

[opaque smelt]

Which function is equal to 1 when x=0?

[thorny hawk]

i dont get it

[obtuse pebble]

@thorny hawk Has your question been resolved?

[obtuse pebble]

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[timid silo]

Prove these

[red ice]

start from sin^2 theta + cos^2 theta = 1

first divide everything by sin^2 theta

then go back to the original equation, and now divide everything by cos^2

[obtuse pebble]

@timid silo Has your question been resolved?

[last pilot]

@timid silo close this channel, you have another one

[obtuse pebble]

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[wooden tree]

hi

[wooden tree]

is there anything more that im missing?

can someone help me please im so lost

.close

[obtuse pebble]

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[obtuse pebble]

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[dapper briar]

ive been struggling with this question for a long time ...

A space ship uses up 1/1,000,000 of its weight to speed up by 1 m/s.

We want to build a space ship that weighs 100 metric tons without the fuel, and we want to speed it up to 1/10 of the speed of light.

Relativistic effects aside, that is, according to the much simpler Newton mechanics, how much fuel will that space ship need when it starts?

[opaque mesa]

hmmm

can you show the original question

[dapper briar]

that was the question i copypasted it

i tried making each speed up different and calculating it but it got really specific and weird

for each push

[opaque mesa]

uses up 1/1000000 of the fuel to speed up by 1 m/s or the weight

[dapper briar]

yes

[opaque mesa]

it was a question with a choice

[dapper briar]

i think the weight of the ship

[opaque mesa]

it wouldnt make sense then

[dapper briar]

im not sure

i guess weight cant be used only fuel

then its fuel

[opaque mesa]

can you show the original question

@dapper briar

[dapper briar]

just this

[dapper briar]

thats all it had

[opaque mesa]

did they say the answer?

[dapper briar]

no

but i want to try solving it anyway but its very weird

before that we did debt

debt income things like that

[opaque mesa]

mhm

when velocity increases by 1 the mass of the fuel decreases by 1/1000000 of it

so from it you can write an equation

$\frac{dm}{m} = -\frac{1}{1000000} dv$

[warm shale]

General_Jacob

[dapper briar]

dm means change in mass right

[opaque mesa]

yes

you can integrate both sides

$\int_{m_0}^{m_f} \frac{dm}{m} = \int_{0}^{v} -\frac{1}{1000000}dv$

[warm shale]

General_Jacob

[opaque mesa]

m0 is starting mass

mf is final mass

[dapper briar]

oh

[opaque mesa]

v is final velocity

[dapper briar]

idk calculus

[opaque mesa]

i guess you cant solve it then

[dapper briar]

does a logarithm do the same thing because we did that before this question

[opaque mesa]

the left side would be equal to the natural log of mf / m0

$\ln(\frac{m_f+ 100000}{m_0+ 100000}) = -\frac{1}{1000000}v$

[dapper briar]

ohh

so 1 is every speed up

so the second speed up would be 2/100000

[opaque mesa]

the final mass of the fuel would be 0

[dapper briar]

hmm

[opaque mesa]

hmm

maybe we should add 100 metric tons to the denominator and numerator

[dapper briar]

so it would be 1

[warm shale]

General_Jacob

[opaque mesa]

idk

$\frac{100000}{m_0 + 100000} = e^{-\frac{1}{1000000}v}$

[warm shale]

General_Jacob

[opaque mesa]

i dont know how to solve it

[dapper briar]

oh

its weird

[dapper briar]

what happened when you used calculus before

did it work

then

[opaque mesa]

it didnt

[opaque mesa]

i dont think its correct

[dapper briar]

oh

..

[opaque mesa]

m_0 would be equal to 10^18 kg

i think i solved it

[obtuse pebble]

@dapper briar Has your question been resolved?

[obtuse pebble]

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[heady oxide]

Can someone please explain how you find the area of a sector? Its a circle but there's a highlighted bit of 60° and you have the find the area of that the entire area of the circle is 48m squared

[restive acorn]

Basically

Sectors and arcs of circles

Are proportionally related to the total area and circumference of the circle

Imagine a 180 degree arc (which would be made with a diameter)

That 180 degree arc is cutting the circle in half

So the area of the sector it corresponds to is half the total area

Cut a circle into quarters

The central.angle is 90

[heady oxide]

How do you work it out tho?

[restive acorn]

90/360 = 1/4

You have a 60 deg

[heady oxide]

Wait do u do 360 bc a circle is 360 degrees

[restive acorn]

Yes

You basically do

[heady oxide]

Did u simplify it

To 1/4?

[restive acorn]

Measure of central.angle/360

[restive acorn]

Yes

[heady oxide]

And that gives u the area ?

[restive acorn]

This calculation gives you what % of the total

So if the whole area is 600

And you have central.angle 60

60/360 = 1/6 of total area

So you do

1/6 × 600 = 100 as area of that sector

[heady oxide]

So 600 is the area??

[restive acorn]

I made up a number in this case

I made up total area of the entire circle = 600

[heady oxide]

2ait so u just divide the degree by 360 and that gives u the percentage

[restive acorn]

Yesss

[heady oxide]

But then what abt the area

[restive acorn]

Well did they tell you like

Radius or something

[heady oxide]

Should I send u the question

[restive acorn]

Yea

[restive acorn]

Normally they'll give you enough information to find:

a. The percentage

b. The total area

In this case they tell you total area

The entire circle has what area?

[heady oxide]

48

[restive acorn]

Perfect

So then your central angle is what portion of the circle

[heady oxide]

Soo how do u get the area of the 60 degrees

[restive acorn]

Answer this next

[heady oxide]

60/360

[restive acorn]

Perfect

So that reduces to 1/6

[heady oxide]

How

[restive acorn]

60/360

Both divisible by 60

[heady oxide]

Ooo ok yhx then whag do u do

What

Thx

[restive acorn]

Ok so you know 2 things:

Total Area = 48

Area of Sector is 1/6 of Total

[heady oxide]

Yh

[restive acorn]

Link those two together

What is 1/6 of 48 then?

[heady oxide]

Wdym 😭

Do u do 6 x 48?????

[restive acorn]

Do

1/6 × 48

If i tell you i have 100 dollars and I'm gonna give you half how do you figure that out?

You do

100 × 1/2

= 50

[heady oxide]

÷ 2

[restive acorn]

Yea

[heady oxide]

Y would u do 1/2

Nvm
I get it

[restive acorn]

Divide by 2 is same as multiply by 1/2

[heady oxide]

Yhyh mb

[restive acorn]

So similarly

To get 1/6 of a number

You do

1/6 × number

Basically you do

(Deg/360) × [Total Area of Circle] = Area of Sector

[heady oxide]

How do u tim3s a number by a fraction

[restive acorn]

Well

48 is

48/1

So call it

1/6 × 48/1

Then multiply "straight across"

[heady oxide]

So it's 48/6

[restive acorn]

Yesss

Perfect

[heady oxide]

But then what's it as a an area

[restive acorn]

That number but

meters squared

[heady oxide]

How do u turn it into that

[restive acorn]

Just say

8

m^2

[heady oxide]

How'd u get 8

[restive acorn]

48/6

You can divide

[heady oxide]

How

Wdym

[restive acorn]

Fractions mean divide

48÷6

[heady oxide]

Oooooo

[restive acorn]

That's the real meaning of 48/6

[heady oxide]

Oh I get it

Thankyouuu

[restive acorn]

Yw!

[obtuse pebble]

@heady oxide Has your question been resolved?

[heady oxide]

Yes

[obtuse pebble]

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[obtuse pebble]

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[sinful falcon]

Okok I’m entirely sorry for not responding to the help earlier I’m trying to find the derivative of (problem 41)

[sinful falcon]

This was the help I was given

I’m kinda confused on what to multiply over the terms something though

[viral blade]

this sentence makes no sense

[sinful falcon]

Okay

One second

[viral blade]

are you confused on why $a + \f bc = \f{ac+b}c$

[warm shale]

mommymorphism aficionado

[sinful falcon]

Sorry

.close

[obtuse pebble]

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[restive gorge]

You can also do this and then differentiate
[ x^{\frac{4}{5}}(x-4)^2 = \left ( x^{\frac{2}{5}} \right )^2(x-4)^2 = \left ( x^{\frac{7}{5}}-4x^{\frac{2}{5}} \right )^2 ]

[warm shale]

bacc the sigma😔🤞

[restive gorge]

to keep it as simplified as possible

[obtuse pebble]

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@late flower Has your question been resolved?

[obtuse pebble]

@late flower Has your question been resolved?

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[rancid rose]

Can someone please check

[teal turret]

homelessness

thats what happens

[rancid rose]

right these answers are lying

I think I got it right tho can u check

<@&286206848099549185>

[timid silo]

Which one

[rancid rose]

The second one

Pls

did I get it right @timid silo

[timid silo]

Wait I'm in school

[rancid rose]

Oh okay Sorry

[timid silo]

I didnt take a good look yet

But I did notice smth

[timid silo]

I think so too

It would likely make sense if you pick b for the first one then C for the second

[rancid rose]

do u think it could be black market

[timid silo]

Possible

[rancid rose]

I know it can’t be a,c for a

So it leaves b,d

I think d is also a possibility vut it’s tricky

Cuz b also makes sense

I used chat gpt it’s saying b for a

Also here’s the hint it might help

I think it’s B,B

could u double check pls

<@&286206848099549185>

[timid silo]

I think thats correct

but dont take my word for it I suck at econs

[rancid rose]

Oh okay thank u I’m hoping this is right 😭

It’s either B,D for a

I keep rereading

B is wrong

It was black market 😭

[timid silo]

I got my econs teacher had a look

[timid silo]

yeah 😅

[rancid rose]

Honestly I struggle in micro too much

[timid silo]

because he said that the wording of the question is ambigious in the sense that it doesnt say whether the second question is independent or not

[timid silo]

I dont like econs in general lol

[rancid rose]

It’s the wording that’s just trying to trick u I always struggle in this

[timid silo]

Is this for GCSE

[timid silo]

We have 1 paper of only MCQ Econs questions and they always try to trick you with the scenarios

[rancid rose]

This is for micro on achieve

It’s an online ebook

[rancid rose]

yeah I don’t like that and im worried since my midterm is this week

[timid silo]

GL 🙏

[rancid rose]

Thank you!!!

[timid silo]

Yeah i think so too

it intersects at $7 so 6 would be suitable

wait

[rancid rose]

are u sure cuz I do get 5 points off that’s why I’m worried

[timid silo]

Uh again dont take my word for it I suck at econs

but I think since the graph intersects at $6

waitttt

[rancid rose]

I think a is right

It’s just b

I don’t think it’s 7

So it has to be 5-6

[timid silo]

$5?

because its below the equilib

or else it wouldnt be binding

[rancid rose]

yep and it intersects at 5

would that make sense

[timid silo]

idk ask more helpers, econs aint my strong suit 😝

[rancid rose]

oh okay lemme ask

<@&286206848099549185>

[tardy epoch]

Don't use chatgpt for math

[rancid rose]

I’m kinda desperate tbh

Is it right at least

[tardy epoch]

.

[rancid rose]

Free chatgpt be giving me wrong answers too

So I have to double check

Could u just lmk

The answer was 5

😭

Can I get help in my next question

<@&286206848099549185>

I think it’s shortage and 300 pls help

[obtuse pebble]

@rancid rose Has your question been resolved?

[rancid rose]

<@&286206848099549185>

Close

.close

[obtuse pebble]

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[obtuse pebble]

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[stark moon]

I'd like to know how to do questions like these and how do I even do this question

[stark moon]

I will brb please help me out

<@&286206848099549185>

nvm I got it

.close

[obtuse pebble]

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[echo knoll]

How do I solve this question?

[obtuse pebble]

@echo knoll Has your question been resolved?

[obtuse pebble]

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[valid geyser]

let $G$ be a cyclic group of order $n$, generated by the element $a$.
The Euler $\phi$ function is defined by
[
\phi (n) = | { 1\le a \le n, \ \gcd(a,n) =1 }
]

Show that $Aut (G) \simeq (\mathbb{Z}/n\mathbb{Z})^\times$, where the latter is the multiplicative group of classes mod $n$ that are coprime to $n$. In particular, $|Aut (G)| = \phi (n)$.

[warm shale]

kaxi

[valid geyser]

Intution is $Aut(G) \simeq C_{n-1}$

[warm shale]

kaxi

[valid geyser]

So suffices to show $C_{n-1} \simeq (\mathbb{Z}/n\mathbb{Z})^\times$

[warm shale]

kaxi

[obtuse pebble]

@valid geyser Has your question been resolved?

[valid geyser]

Sadge

@upper shuttle

[upper shuttle]

sadge

[valid geyser]

i feel

lonely and alone

[upper shuttle]

LOL

[valid geyser]

😔

im fr though

like feels like everyone is ignoring me

and idk whwat to do

not like this server, although kinda

[upper shuttle]

:(

i got it

for n=1 its trivial

n=2 its trivial too as theres only 1 group of order 2

for n>2, we can pick the k=n-1 and it would be coprime to n.

i wanted to say k generates the group but im not sure

[valid geyser]

😔

i dont think so

its some number theory

[upper shuttle]

nvm n-1 has order 2

i give up i shall google

https://math.stackexchange.com/questions/2597389/proving-that-eulers-totient-function-is-multiplicative-by-abstract-algebra

@valid geyser

[obtuse pebble]

@valid geyser Has your question been resolved?

[valid geyser]

😔

[upper shuttle]

😔

are u ok

[obtuse pebble]

@valid geyser Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[zenith tide]

but the answer's supposed to be

[obtuse pebble]

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[dark orchid]

aaaaa how do i find the hours?

[dark orchid]

got temp from plugging in 24 in t

perhaps is it like this to get the hours?

[dark orchid]

one problem, forgot how to do this to solve

[obtuse pebble]

@dark orchid Has your question been resolved?

Channel closed

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[dark orchid]

.reopen

[obtuse pebble]

✅

[dapper briar]

what does that mean did you solve it

did it work

.reopen

$\frac{100000}{m_0 + 100000} = e^{-\frac{1}{1000000}v}$

[warm shale]

𝔂𝓮𝓼

[dark orchid]

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[dark orchid]

.reopen

[obtuse pebble]

✅

[dark orchid]

<@&286206848099549185>

[zenith tide]

find local max

differentiate once and equate it to zero

and find values for T

[zenith tide]

differentiate the equation you differnetated

substitute T in the second order derivative

the value for which you get a negative second order derivative is

the maximum

[dark orchid]

https://tenor.com/view/wtf-confused-look-gif-24835232

[dark orchid]

how?

[zenith tide]

do you know calculus

[dark orchid]

taking pre cal

rn

[zenith tide]

im not sure about how to solve it without calculus

[dark orchid]

wait there was video there the whole time?! tf

bruh

[zenith tide]

what

do you know what the answer is?

[zenith tide]

try to learn the basics of differentiation and it'll be easy

[zenith tide]

-?

the question asks you the time taken right?

[dark orchid]

yeah

[zenith tide]

man im not sure

now im confused

this is what im getting from what ive learnt

[zenith tide]

<@&286206848099549185>

not sure what's wrong with my calculation

[plush pollen]

0.036 isnt 0.36

for part a

you are right about differentiating and finding the stationary point here

but 2(-0.018)=-0.036

not -0.36

[zenith tide]

that still isn't his answer

tho

[plush pollen]

because hes wrong

[zenith tide]

no

the video says

10.5

[plush pollen]

and hes wrong

its 11.2

[zenith tide]

ah i see

no it's 12.6 i did the calculation wrong too

💀

[plush pollen]

? it is 11.2 i can promise you that

[zenith tide]

oh yeah

im dumb nvm

[dark orchid]

how did u guys solsolve it?

[plush pollen]

its a concave down quadratic, so you can differentiate it (or complete the square to find the vertex) and set it to 0 to find the local maximum

[zenith tide]

ayodele, he doesn't know calc

[plush pollen]

ah

complete the square

you will have the vertex

[dark orchid]

ah ok

[plush pollen]

factor out -0.018

[zenith tide]

these gotta be the answers then

[plush pollen]

yup

[zenith tide]

epic

it's like i know most of the concepts

[plush pollen]

tell me what you get from completing the square

[zenith tide]

but i keep making errors

in my calculation

that's what's ruining my score in my math tests

@plush pollen do you think ill be able to give sat?

[plush pollen]

give sat?

[zenith tide]

the exam

[plush pollen]

oh, im british i dont know much about the sat

[zenith tide]

the one after school

[zenith tide]

so well i wanna leave my country once im done with my scjoolingt

[zenith tide]

what'll i have to do to get into a decent college there?

[plush pollen]

most universities dont require an admissions test, merely your exam grades to align with their requirements

i personally did the MAT and TMUA, admissions tests for Oxford, Imperial, and Warwick

[zenith tide]

can i dm you later?

[plush pollen]

sure

[zenith tide]

im honestly worried about what ill be doing after my school

[plush pollen]

try not to worry about it too much, everyone is stressed but i promise its much scarier before than after

[zenith tide]

ill dm u

this isn't the right place

[plush pollen]

nope, close

you factored -0.019

not -0.018

the question has -0.018

[dark orchid]

yeah mb

forgot to send

-_-

[obtuse pebble]

@dark orchid Has your question been resolved?

[obtuse pebble]

Channel closed

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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.

[left mesa]

Help me convert somthn into logarithmic functions

[left mesa]

This one

[plush pollen]

you have 3^(y-1) = 10x

log_3(10x)=y-1

so y=log_3(10x)+1

[left mesa]

Wait lemme try

[left mesa]

Give me a sec

[left mesa]

What happened to the denominator 5?

OHHH

Wait

[limber ocean]

switched to the other side

2x times 5

[left mesa]

Thanks

[left mesa]

Can you check if I got this one right?

[limber ocean]

sure go ahead

[left mesa]

This one

<@&286206848099549185>

[left mesa]

@limber ocean

[limber ocean]

sorry mate I was making breakfast

this is way off

[left mesa]

Oh mb

???

[limber ocean]

I can see the errors but it's much simpler than you think

[left mesa]

Then how?

[limber ocean]

what's your current uni/high school level?

[left mesa]

Grade 11

[limber ocean]

alright so familiar with logs?

[left mesa]

Yeah our lesson today

[limber ocean]

alright so

first step is to multiply each side by 5