help-10

[lusty plinth]

how do i find ....

oh i get it

it becomes half the factorial of the original number

had to use example of 2 4 6 8 10

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[limpid pollen]

how do you come up with (a-1)(b-1) when the question didnt give a hint

[limpid pollen]

the consider part

this is my teacher's handwriting

is there any alternative ways to do this question

[wooden cipher]

its more your teacher skipping steps

[limpid pollen]

what

[wooden cipher]

oh you want to figure out alternate methods

[limpid pollen]

yes maybe if possible

[wooden cipher]

there arent really good alternate methods

this is a common trick in problems especially involving roots of quadratics

looks up SFFT or simons favorite factoring trick

it becomes second nature when you see this setup

[limpid pollen]

mmh ok imma search it

mmh it seems like completing the square

[wooden cipher]

very similar, yes

[limpid pollen]

but i still dont understand where u get (a-1)(b-1)

do u js keep in mind there is a (a-1)(b-1) as a trick

or you can use SFFT to deal with it

[wooden cipher]

you want the p to cancel out so you are only working with a and b

so you subtract the second equation from the first to get ab+a+b=2002 which you can tell you might be on the right path because SFFT is useful

[solar stump]

It's a common trick

If you're dealing with Diophantine equations (i.e. equations for which you want to find integer solutions), say in x and y, and you're told what their sum and product are, a common thing to try is some variation of
(x + 1) (y + 1)

This is because you can expand this and regroup to obtain
[xy] + [x + y] + 1

[limpid pollen]

ongg

omg

ok

so cool

tho this isnt mentioned in textbooks

[solar stump]

It's something that's picked up from when you learn to expand brackets

It's not explicitly taught, because it's really difficult to state

And in practice this is only clear to you now because you know about how expanding and factorising works; if you had this on the outset you'd just be confused why this should even work in the first place

Which... fairs

Now, your teacher notices that both the product and the sum have a "p" in there

[solar stump]

Ideally we could cancel them out in this line, if we subtracted the sum from the product

And in fact we can, if, instead of "+1" in each bracket, we used "-1"

which leads us into this line

Now you've boiled this down to "two integers whose product is 2003"

I don't think you're told whether 2003 is prime, and it takes a while to check tbh, but then the rest of this question makes use of 2003 being in fact prime

[limpid pollen]

sooo complex

[solar stump]

Proving that 2003 is prime?

Or the rest of the work there?

[limpid pollen]

no

umm

[solar stump]

Or the algebra part before this?

[limpid pollen]

just to think of there is a trick (a-1)(b-1) can solve this

in exam i probably couldnt think of (a-1)(b-1)

[solar stump]

Yeah, not complex, just hard to think of on the spot if you've never seen similar examples before

[limpid pollen]

o

[solar stump]

(those are two slightly different things btw)

[limpid pollen]

does this type of question have a name

[solar stump]

^^^

[limpid pollen]

oo

[solar stump]

They're more common in Olympiad papers than in school exam papers tbh

[limpid pollen]

are there any other trick aside from a+1 b+1 and a-1 b-1

[solar stump]

Well, they vary, and they get quite technical in descriptions, ...

There is this though - https://www.youtube.com/watch?v=3U3PbYnkkAM

[limpid pollen]

ahhh math is so scary

but hmm

anyways tysm for you both for helping me out

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[last ore]

Hi is this the most efficient method for this?

[jaunty ocean]

Well, you may know that a number is a unit mod n iff it is coprime to n. So, If n is prime, the only number not coprime to n is a multiple of n itself (i.e., 0).

[last ore]

Ohhh I see

So that also means

That any prime n

Would form a cyclic group asw?

Idk of that’s the best way to phrase it

Like

[compact shadow]

U(Z/p^a Z) is cyclic for odd prime p, U(Z/2^a Z) is cyclic for a =1,2, and is Z/2Z direct sum another cyclic group for a>=3. It’s a classic result, can be found many places, for example basic algebra i by Jacobson, chap 4

(By CRT U(Z/nZ) can be factored as subgroups listed above)

[last ore]

Hold on, I’ve only recently started algebra 1 😭

Basic algebra i, is that a good read?

[compact shadow]

I read it, I think it’s great but I didn’t read others to make a comparison

Some others might say Rotman, Lang, …

[last ore]

Hm I mean tbf, I’m not too interested in algebra

Do yh any good combi/prob books?

[compact shadow]

Enumerative combinatorics by Stanley is great, very beautiful

Probability no idea

[last ore]

Is it basic, like do I need prerequisite?

[compact shadow]

No

The only subject in math that requires background knowledge is geometry

[last ore]

Alright, I shall save it on reading list

[compact shadow]

Like algebraic geometry/ differential geometry. Almost any other field doesn’t require background.

[last ore]

Oh alright

[compact shadow]

(Probability I heard some soviets wrote some good textbooks, not I am any familiar with. They might be old.

[last ore]

I’ll check some out, I’ve started an intro or probability course and I think it would be nice to see examples elsewhere

Alr tysm

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[iron edge]

,rcw

,rccw

[rancid owl]

<@&286206848099549185>

[versed stratus]

!15m

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[iron edge]

Have you tried anything yet?

[rancid owl]

yes

like i am confused

on simplyfying f(x) you get sec(2x)

so according to that both should be continous

[stable halo]

then you check the continuity of cos(2x) or smth I think

because 1/sec(2x) is cos(2x)

[rancid owl]

yes

by that i get both of them continous

but the answer key says discontnuous for i)

[obtuse pebble]

@rancid owl Has your question been resolved?

[valid hearth]

check at pi/2

[rancid owl]

the original function give undefined. but if i simplify the function first and then put the value i get defined value

idk why its showing discontinuous in the ans key

[coral yew]

It seems pretty continuous to me

[rancid owl]

hmm

ig the ans key may be wrong

it shouldn't be but..

if anyone has some idea pls tell

[coral yew]

Well f(x) is sec(2x) right

Then 1/f(x) is cos(2x)

[valid hearth]

hm.. it seems continuous

[coral yew]

And clearly cosine is everywhere continuous?

[rancid owl]

yeah

you are right ig.

thank you

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[solid yarrow]

Exercise 12)
Do I understand correctly that the extra + we are left with at some point turns into +1?

[valid hearth]

,rcw

[valid hearth]

why

[west nexus]

9² = 81 why did you write 82

[winter stump]

Where is the "extra +"?

[inner sierra]

that the hell is this

that does not look like a 9

im confused if it's a 5 or a 3

[calm wedge]

Looks more like an eight to me.

Who am I to judge. My 2 sometimes confuses me while looking like an 8.

[calm wedge]

Oh no no I see what happened.

Since - 9² = -81.

And you wrote -82.

You added a +1 to balance it out.

Because -82 + 1 = -81

[valid hearth]

why tho

[calm wedge]

Otherwise it should be written as -81 -27 + 64.
Which yields the same result as -44.

[calm wedge]

She must've taken it from someone who did it luike that.

Like*

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@solid yarrow Has your question been resolved?

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[fallen flame]

Hi ok so I was given f(x) = ln(9 - x^2)
And asked how to find:
domain of the function
min and max points
increase and decrease

how do i do that
i got e^y = 9 - x^2? but idk what else to do

[safe mauve]

lets keep it as f(x)

remember how ln(x) cant have negative entries?

[fallen flame]

it has to be bigger than 0 right?

[safe mauve]

yes

so when is 9-x^2 greater than zero?

[fallen flame]

always?

[safe mauve]

always?

what about x = 10?

that'd be 9 -100 right

[fallen flame]

Oh.

[safe mauve]

which is less then 0

[fallen flame]

right

[safe mauve]

so solve for x: 9-x^2 > 0

[fallen flame]

x = 3
x = -3?

[safe mauve]

Those are the endpoints yes

we have -3 < x < 3 right?

because at x = -3 or x = 3, we'd have ln(0), which isn't possible

so the domain is (-3,3)

first question done

[fallen flame]

ohhhh

[fossil pendant]

for min / max points and the intervals of which the function decreases and increases ---> find the critical points of the function.

do you know how to find that?

[fallen flame]

ummm

derivative?

[safe mauve]

you could use derivatives

but I think it's easier just looking at the function and thinking logically

[fossil pendant]

assuming you cant use the graph.

[fossil pendant]

this too

but do you know what a critical point IS?

[fallen flame]

when slope is equal to 0?

[fossil pendant]

that's 1 of the 3 types of critical points

[safe mauve]

The goal is max and min right

when is ln(9-x^2) the greatest value

[fallen flame]

im not sure

[safe mauve]

try some values

and see whats happening

[fallen flame]

f(2) = 5?

[safe mauve]

well ln(5)

[fallen flame]

oh

yeah

[safe mauve]

and how about f(1)

[fallen flame]

ln(8)

[safe mauve]

and f(-1)?

[fallen flame]

same thing?

ln(8)

[safe mauve]

Yes!

notice a trend?

[fallen flame]

the positive numbers and their negative versions equal the same?

[safe mauve]

Yes, true

[fossil pendant]

notice that the value of ln(smth) grows if that smth grows. like ln(3) > ln(2) and ln(x) <ln(x+1) (with respect to the domain)

so the "max" of ln(9-x^2) is where 9-x^2 is the biggest it can ever be!

[fallen flame]

x = 2.999999?

[fossil pendant]

you can try to think of that! what value of x, with respect to the domain, which makes (9-x^2) as big as possible

[fallen flame]

wait no

im dumb

umm

x = 0?

[fossil pendant]

AMAZING

true

you see

[rapid fossil]

hi

[fossil pendant]

do i have to explain or

do you get what you did 😭

[silk ibex]

hello

[fossil pendant]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[rapid fossil]

i need someone who knows trigonometry its too hard for mo

[fallen flame]

its equal to 0 when x = 3 and x = -3
so i had to find the midpoint?

[fossil pendant]

ah- what a weird way to do it?

I'm not sure if that always works

but the way i thought of it is

9 and x^2 are both positive

if you want the biggest possible value out of 9-x^2 that means you need the smallest value for x^2

[fallen flame]

ohhh

i see

[fossil pendant]

mhm

okay so back to hydle's method

[rapid fossil]

x should be 0 i think

[fossil pendant]

now you recognize that the function reaches a maximum when x=0

when x=0, ln(9-x^2) =?

[rapid fossil]

8

?

[fossil pendant]

please do not interrupt

[rapid fossil]

srry

[fallen flame]

no its ln(9) right

[fossil pendant]

if you want help with your own question, #❓how-to-get-help

[fossil pendant]

indeed

so that makes the point of which the max occurs be...?

[rapid fossil]

idk trigo can u explain it a lil bit

[fallen flame]

(0 , ln(9) ) ?

[fossil pendant]

indeed!!!

yaaay

[fallen flame]

yippee

[fossil pendant]

now for the min... uhm-

im not sure how you can

"logically" do this

but yeah tis better for you to do the critical points test now

[fallen flame]

well

actually

yeah idk

the answer sheet says theres no min point

[fossil pendant]

critical points are 3, a point of which the derivative ismulti-valued (which usually only happens with piecewise functions so you dont have to worry about it) or the derivative = 0 (what we're gonna look for) or the derivative at a point is undefined (parallel to the y axis)

[fossil pendant]

yes. the critical point tests would've let you know that yourself

[fallen flame]

oh

[fossil pendant]

so now

first of all

this is not a piecewise function

so we dont have to test for that type of critical points

so now get the derivative

to see if it could ever be undefined

[fallen flame]

wdym

[fossil pendant]

as I said earlier, there are two other types of critical points, ones where the derivative = 0 and ones where the derivative is undefined

a value is usually undefined if its divided by zero

so our goal here, to check for both critical points, find the derivative

and ask two questions

"can this =0, and if the answer is yes then solve for the point if the answer is no then neglect this test

and the second question is

"can this be undefined, if the answer is yes then solve for that point. if the answer is no neglect this test"

[safe mauve]

uhh did we do both max and min ?

[fossil pendant]

we did max

[safe mauve]

ah

[fossil pendant]

with your method

[safe mauve]

does he know why there is no min

[fossil pendant]

now im trying to explain how to find critical points to get the "decreasing/increasing" part and how that there is no min point

[fossil pendant]

im trying to make him find that out himself with the critical points

I might not be doing a great job tho...

if you're willing to take over and propose a simpler apprach then sure!

[fallen flame]

umm the derivative of ln(X) is (1/X) right

[safe mauve]

true

[fossil pendant]

i mean- yeah

[safe mauve]

Do you know why ln(9-x^2) does not have a minimum???

??

[fallen flame]

umm logically kind of
i dont understand the math behind it though

[safe mauve]

well lets try 2.999999 as a value

as it lies within our bounds -3< x < 3 right

[fallen flame]

yeah

[safe mauve]

which is -12.0238....

now try 2.999999999999999999999999999999

which is still within -3 < x < 3

[fallen flame]

my calculator says 0

[safe mauve]

which is of course -60

[safe mauve]

wow really?

pretty bad calculator then

[fossil pendant]

how did you get -60

[fallen flame]

ill try with google then

[safe mauve]

The point is, we can keep adding 9's behind the 2.9, which gives a lower and lower value

[fallen flame]

?

[safe mauve]

you forgot the ln infront of it

so the minimum is -infinity, which really isn't a number

so we can just say there is no minimum

[fallen flame]

calculator now says math error

huh.

[fossil pendant]

yes because your calculator keeps approximating 2.999... as 3

thats what calculators usually do

[safe mauve]

yeah the calculator rounds off

[fallen flame]

i see

[safe mauve]

but yeah we can have the minimum be lower and lower

so there is no minimum

[fallen flame]

ohhhhh

[safe mauve]

next, "increase and decrease"

Do you know how to find this?

[fallen flame]

ummm

something about the min/max points

[fossil pendant]

Nothing specific versus this problem but you'r sure gonna get stupid'er functions where logic may be hard to think of

https://youtu.be/Bp9EbV3COVA?si=lTKhR0TpxHoTjmJ8

learn what a critical point is

learn how to find them

and then you can solve everything!

even the most disgusting functions!

[fallen flame]

ooo

[fossil pendant]

yeah infact, ciritcal points forces you to find "increase / decrease" before min/max

[safe mauve]

yeah for these cases that works really well

[fossil pendant]

also teaches you how to find absolute minima nad maxima

local too

[safe mauve]

not sure if that's the point of the exercise but yea

[fossil pendant]

you'll learn all of that

[fallen flame]

I see

[fossil pendant]

they do recognize f'(x)=0 so im assuming they are just learning about the behavior of functions

so yeah!

if you have anything to ask about the video itself

then you may

[fallen flame]

I bookmarked it, Ill watch it in a bit :)

[fossil pendant]

https://youtu.be/V239IIoGVT8?si=Vehl8c7m3IAvA_gi

this too

they're both the same videos but- both are important

the most important thing is that you understand what a critcal point is

[fallen flame]

I think i just missed a good chunk of material in class a few years ago and I'm still paying for it by missing some basic knowledge

[fossil pendant]

because thats what we rely on to find EVERYTHING abouyt functions

well- not everything

[safe mauve]

it's pretty straightforward

[fossil pendant]

but the your-level and my-level everything

[fallen flame]

alright

[fossil pendant]

yeah, great luck

!done

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

[safe mauve]

I think you'll be fine kaz

[fallen flame]

alright

thank you both :)

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[serene otter]

how in the world do we deduce 4)c

[royal basin]

did you do 4b?

[serene otter]

yes

[royal basin]

note that when you put $z=1$ in the left-hand side in 4a), the general term inside the product becomes $$2 - 2 \cos\paren{\frac{\alpha}{n} + \frac{2k\pi}{n}}$$ and you can simplify it via $2 - 2\cos(t) = 4 \sin^2(t/2)$

[serene otter]

it's 2-2cos(a)

isn't it just alpha ?

[warm shale]

Ann

[royal basin]

@serene otter

[serene otter]

if you put one in the first equation it gives 2-2cos(a) tho

oh wait

imma try to do smt I think I got it

yup I got it thanks so much @royal basin !!!

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[radiant scarab]

How to solve 2nd part where Y is to be found

[royal basin]

it's easier to count the ways to select 3 points when there are two or more adjacent. this count will be X-Y (and so, as a bonus, you get to see immediately if D is correct or not)

for this i recommend separately counting the cases of 3 consecutive points vs. 2 adjacent points and the third not

[radiant scarab]

How to write 2 adjacent points and 3rd not adjacent

In mathematical terms

I don't understand what you mean by 3rd not

3 consecutive is 18c3, 2 adjacent is 17c3

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[radiant scarab]

I got your help but I couldn't solve it as you told, I'll reopen the help channel

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[karmic hedge]

Last question for tonight, then I can sleeeeeeeeeeeeeeeep finally

First of all idrk what this is

Is it a 3d shape or what

[hardy widget]

It's the plane that "touches" a surface at a given point.

[karmic hedge]

Ahhh ok I see, so what do we get by getting the derivative with respect to x and then with respect to yand subbing in the values

Here

What does that give us?

Wait I just got it

Thanks!

❤️

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[leaden seal]

Is this the correct way to factor this monomial

[hardy widget]

You're missing the $x^3$ ones. What you have so far is good though!

[warm shale]

Civil Service Pigeon

[leaden seal]

Oh okay got it, thank you

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[leaden seal]

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[obtuse pebble]

✅ Original question: #help-10 message

[leaden seal]

One more question, when I am factoring a negative monomial are all the factors going to be positive, negative, or both?

[slate shoal]

Yeah, factors of negative numbers do have negatives in them

[leaden seal]

So all the factors are going to be negative?

[slate shoal]

some, not all

-8 has factor of 4 and -4

etc.

[leaden seal]

So would -4 be 2 and -2, etc?

[slate shoal]

yes

[leaden seal]

👍 ok

Sorry misclick

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[brazen wedge]

i understand how to graph the parts within the parenthesis but the exponenets are confusing me. i attempted to graph a pint at -8 and it says i need 2 points. im struggling to find where the other pint would be.

[obtuse pebble]

@brazen wedge Has your question been resolved?

[polar fossil]

the roots are where $\blue{(3x-6)^4}\green{(x+8)^3} = 0$

[warm shale]

clumsy

[polar fossil]

which means that either $\blue{(3x-6)^4} = 0$ or $\green{(x+8)^3} = 0$

[warm shale]

clumsy

[brazen wedge]

thank you, i know what to do from here

.close

[obtuse pebble]

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[obtuse pebble]

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[pulsar shard]

Hello! Good evening.

[pulsar shard]

As you can see in the question. This is 101 Statistics. The example laid out is the way to find the probability which is found by first finding the z score and then using a table from the book to match and find the exact probability. This is made easier by using a formula in the calculator called NormalCDF. But when I use it, and input the exact same numbers, my answer is not a 100% accurate. The answer in the book is 0.7698
For me, using the calculator, it comes up as 0.7717.
Yes there almost similar but not quite. My professor has been no help. This is a winter course so I dont have alot of time to figure stuff like this with one on one help. Honestly regretting this course. I have shared my answer from the calculator.

My request is if any one can help me where this inconsistency is occurring? Or why is it. I am positive this formula is correct. Its literally in the previous page. Funnily enough that question had exactly matching answers.

[worn yoke]

i think you included the wrong image from your book?

[pulsar shard]

Yes

Youre right. Sorry. One second

[crisp venture]

Its what happens when you round off too early

Because the z values are in 3sf, there is a slight change in the final answer

[pulsar shard]

Should I stick to the table then?

[crisp venture]

Yes

Stick to what they want

[pulsar shard]

But it gets tedious you know. 2 days ago I did the midterm online and it was 75 minutes with 50 questions man. I barely got it done.

Check this out. The previous page is telling me to use the formula since it makes my life easier. And the answers matched as well on the calculator. Then this example comes and its not matching up again.

[latent bison]

i think they are just showing you that you can either use table or use calculator as 2 methods

[pulsar shard]

The benefit if this formula is skipping the process of finding the z score first.

[pulsar shard]

Yes. But when I use the calculator which is quicker to do so, there is a disparity.

[zenith raft]

what a name

[pulsar shard]

Appreciate it

[latent bison]

the disparity is because the table doesnt have enough space to account for more s.f.

[crisp venture]

What you could do to eliminate the disparity is round the z score to 5sf

[pulsar shard]

14 year old me is to thank hehe

[pulsar shard]

Right right. That makes sense now.

What i am worried about is my professor not recognizing that.

He is telling me not to fret.

[crisp venture]

I would say to use 5sf answers for working, 3sf for final

I dont think they will penalise you for being more accurate

[pulsar shard]

What is sf? Significant factor?

[crisp venture]

Significant figures

[latent bison]

π = 3.1415926535...
π to 3s.f. is 3.14
π to 5s.f. is 3.1416

[pulsar shard]

What i am worried about is during the final, on pearson، I only have to input the answer in. So if it asks a specific decimal point like say round of to 4 decimal place then there is a difference between .67014 and .67019

[crisp venture]

A general rule is to put your working with 2 places more than what the answers want

So for example, if answers in 4dp, working should be in 6dp

[pulsar shard]

I cant do that in Pearson. It has to be exact. Have you used Pearson before? Its like um like a blank that i have to fill in and in brackets on the side, it will mention it has to be this many decimal places.

And I am positive there is no set range to account for close answers.

Its either right or wrong.

[crisp venture]

No I have not

[pulsar shard]

Oh man

Did you ever use a calculator?

During your time?

[crisp venture]

Dog, I did my A Levels last year

[pulsar shard]

In my head you're 40+ now. A

Oh

Hahaha

[latent bison]

same here 🙂

[pulsar shard]

Oh your in the uk

[crisp venture]

No, somewhere else

[latent bison]

and we both know each other 🙂

[pulsar shard]

Ah right. British curriculum

[crisp venture]

Its not British

[pulsar shard]

What is it called then? I used to study in uae and there it was a/o levels as well. I thought it was called British curriculum

[latent bison]

dont you dare tell him where we are from

[crisp venture]

@latent bison , do you want to say?

[pulsar shard]

LoL

Im from the States. Can't be worse bruh

Good ol 'Merica. Especially considering what's happening now.

Anyways. Appreciate the help guys. I'll just pray to god.

[latent bison]

i think we can roll off from here

[pulsar shard]

.close

[obtuse pebble]

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[obtuse pebble]

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[round island]

Where does the 36 inches fit into play here and how do I find the angle needed to solve the problem

[formal island]

you are asked about the angle of elevation from the child POV to the tip of the turret.

if the kid happened to measure 380 feet too then the angle would be 0 anyways

[round island]

…..huh

I’m asked about the angle of elevation from the child’s pov to the top of the turret

[formal island]

and youre given info about the position of the child itself

[round island]

So how do I find the angle from the child’s pov to the tip of the turret and would the 3 feet not cause this to not be a triangle anymore

[hollow mulch]

Anyona have ap precalc answers?

[formal island]

you have to define a new triangle yourself using the position of the child eye's

[round island]

Fuckkkkk ok

[formal island]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hollow mulch]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hollow mulch]

huh what does that mean?

[formal island]

read

[round island]

It means go get your own channel

[hollow mulch]

sorry

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[round island]

Ok so I adjusted the height- how do I find the angle

[formal island]

Just create your own help thread or consult math discussion, if you dont know how just go to how-to-get-help.

[formal island]

you know all three sides formed by this new triangle, and more importantly, one is 90º since this is still a right triangle.

you could use law of sines here.

[round island]

I don’t know all 3 sides-

[formal island]

id advice you draw the problem as you carefully read

[round island]

…I did

[formal island]

well, you wouldnt know they hypotenuse instantly, but you can just do pythagoras.

[round island]

Oh true

Let me do that rq

Could I not just use the law of sins to find an equation for x-

Wait no not sins

[formal island]

using what pair of angle-side?

[round island]

Cosines

[formal island]

oh

[round island]

Using the adjacent and hypotenuse

[formal island]

yeah, you could too.

[round island]

I think that’s what’s being asked of me

[formal island]

But law of sines is just easier i guess, far shorter formula.

[round island]

Not really

Would you approach 377 or 160 as the adjacent angle from 90

[formal island]

Do you remember how law of sines works?

[round island]

I don’t know what I’m doing tbh

This problem is formatted differently from the rest on the paper

And I’m confused

[round island]

I don think that’s what this assignment is about

[formal island]

If law of sines is off limits, i would assume law of cosines also is.

You can do sohcahtoa too

[round island]

Idk what I’m talking about

[formal island]

in fact, soh is just a really specific case of law of sines for right triangles.

[round island]

I’m looking back at my notes and I don’t have anything that would suggest having to find the angle-

I’m very confused

[formal island]

what theorems / formulas you have noted down?

I like to believe pythagoras is there to start with

[round island]

This is just supposed to be trig ratio word problems with right triangles

[round island]

You could be right but nothing else on the page is like that-

[formal island]

.

[formal island]

Well, just sohcahtoa

we can work with it anyways

Remember how to use it?

[round island]

Not…exactly

Should i just skip this problem for now and work on the rest?

[formal island]

not really ig.

,tex .sohcahtoa

[formal island]

SOH CAH TOA is just 3 identities to find the angle based on pairs of sides of a right-triangle.

[round island]

Yeah

[formal island]

Here you have all 3 sides, so you can use any as long as you use the correct pair for your function

[round island]

But my question is why that’s on here if all of the other problems aren’t anything like that-

[formal island]

idk

[round island]

Idk it feels wrong to me

[formal island]

for the sake of convenience, we will use "TOA", would you be able to setup the equation?

[round island]

Tan(theta)= 377/160?

[formal island]

yep.

[round island]

And just divide that to find the angle?

[formal island]

You cant divide by tan.

[round island]

Oh yeah

[formal island]

You use its inverse trigonometric function

[round island]

Ugghhhh fuck

[formal island]

arctan

[round island]

Huh

[formal island]

or $\tan^{-1}(x)$

[round island]

So…….what does that imply

[formal island]

i wont bore you with the details

but, you remember how square root "undoes" x^2

[round island]

Yes

[formal island]

Same deal.

[round island]

Dumb it down- what do I do to make this happen

[formal island]

arctan undoes tan.

so arctan(tan(x)) = x

just apply arctan to both sides

[round island]

So fun fact I can’t do that

I get a bullshit calculator and I’m not allowed to use anything except plus, minus, multiply, divide, and square root

[formal island]

you can just write theta = arctan(377/160)

[round island]

What in the ever loving fuck is arc tan- all I’ve ever heard is cotangent- is that the same thing?????

Ok so update after reading the assignment description I can’t use inverses on this

[formal island]

If this is a common ratio i have 0 clue

It clearly doesnt look like it.

Cotangent is the Multiplicative Inverse (1/x)

Arctangent is the Function Inverse

[round island]

I’m just going to ask her tomorrow I guess

[obtuse pebble]

@round island Has your question been resolved?

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[manic yarrow]

I mean I can use matrix notation and show equality it'll just take forever is there an easier way

[obtuse pebble]

@manic yarrow Has your question been resolved?

[obtuse pebble]

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[brittle oracle]

calculator allowed

[leaden lion]

You can eliminate options by putting values

Take roots of this eqn as -3 and -2 then c=6

And in another case take roots as -1 and 0

[brittle oracle]

is there an algebraic way to do it

or is it just sub and eliminate

i tried to break down b = uv and c = u + v

and then u = v + 1

idk it got very confusing

[red ice]

$r_1 + r_2 = -b$ and $r_1 r_2 = c$

with $r_2 - r_1 = 1$ that becomes $1 + 2r_1 = -b$ and $r_1 (1 + r_1) = c$

[warm shale]

south

[red ice]

yeah...

[brittle oracle]

hm i was on the right track

its b in terms of c tho

thats what threw me off

[red ice]

yeah so you just have to quadratic formula it

[brittle oracle]

frickk

[red ice]

$x^2 + x - c = 0 \implies x = \frac{-1 \pm \sqrt{1 + 4c}}{2}$

[warm shale]

south

[red ice]

multiply that by 2 and add 1

[brittle oracle]

whered b go?

[red ice]

I just let $r_1 = x$ in $r_1 (1 + r_1) = c$

[warm shale]

south

[brittle oracle]

🫠

maybe ill try have another look at it causei m super lost

thanks for ur help

[red ice]

no worries!

[vestal bridge]

quadratic formula

the good news is a few terms fall off

[obtuse pebble]

@brittle oracle Has your question been resolved?

[obtuse pebble]

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[soft scroll]

is this question right

[soft scroll]

im putting it in a calculator and either its wrong or this is wrong or im doing it wrong

[elder vale]

What have you done ?

[soft scroll]

cause i got 5.49 grams/mol

[elder vale]

Do you know moles = Volume in litres/22.4

Or just use PV=nRT

Write n as given mass/molar mass

And substitute everything

[soft scroll]

ya let me retry doing it

[elder vale]

Did you use PV=nRT ?

Or something else

[soft scroll]

ya i used the formula

[elder vale]

Alr let me cross check the calculation

I'm getting the answer as one of the options

So I'm assuming you made a silly mistake somewhere

[soft scroll]

i dont know they barely taught the fundamentals for this

not a great course here

[elder vale]

It's alright if you need help just ask

[soft scroll]

alright so how do you input the constant

r

i have a very loose understanding of what it is

[soft scroll]

.

and are you allowed to multiply liters by atmospheres here

[silent flame]

Yes

R is in L atm

[soft scroll]

so the resulting unit is l/atm

what is

[silent flame]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[soft scroll]

sorry

doing work on a screenshot

[silent flame]

Ok lemme check

[soft scroll]

i dont know what to input for r

because the definition it has there has 4 different units

little confusing

[silent flame]

Put it as it is, you can see the units match, L atm/ mol K

[latent bison]

dimensional analysis

[soft scroll]

oh wait youre right

[soft scroll]

what

[silent flame]

This is n

[soft scroll]

i thought n would be 1

since we're solving for grams per mol

you could just convert that couldnt you

[silent flame]

Nah it is not 1 every time, it changes wrt to others pressure etc..

[silent flame]

Yes

Specifically when we are solving for mass, n is not to be assumed it will depend on the volume etc.

[soft scroll]

does r change depending on the rest of the equation

[latent bison]

to get grams per mol

you can divide total mass by total amount

total amount n doesnt always have to be 1 mol

[soft scroll]

do you enter the values into r

for the answer

[silent flame]

Yes it can change for eg if they are in different units

[silent flame]

We know no one takes that chapter seriously

[soft scroll]

.

[latent bison]

||
its self-explanatory, really
||

[silent flame]

Yes n = PV/RT

[latent bison]

before you enter values in physics you must always ensure the units match

if units dont match, then convert first

otherwise, just plug the values in

[silent flame]

Indeed, has applications

[soft scroll]

yes of course im just asking if i can plug the units ive gotten into the formula it gave

[latent bison]

do the units match?

[soft scroll]

yes

[latent bison]

so can you plug the values in?

[soft scroll]

its just that PV=nRt, r is in the formula

so im hesitant

[silent flame]

So plug the value of r?

[soft scroll]

hm

one sec

[latent bison]

yes you can plug the given R

[soft scroll]

what is the value

im not good with working in physics units

like for say atm in the formula do i plug in 0.95 or something else

in the formula for R

[latent bison]

0.95

[soft scroll]

so the units where it doesnt have a stated value i plug in from the rest ok

[hardy widget]

btw this doesn't rlly belong here

discord.gg/chemistry

discord.gg/physics

one of these would be a better fit

[soft scroll]

mb i was only in here

i'll join those for future ones

but for mol in the equation is it calling for molar mass

or i guess the 5.49g of gas

because theres no stated number of moles

oh wait you dont multiply volume by the 0.0821 L

mass/amount so

27.0044

.close

[obtuse pebble]

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[obtuse pebble]

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[plain badge]

how to find primitive function in this example?

[kind hawk]

do you know how to find a primitive for e.g. 2^x ?

[red ice]

if so, then f(x) = 3^(-x) so then there's an extra - due to the *reverse chain rule

[kind hawk]

or (1/3)^x

[red ice]

oh wait I pinpointed the issue before you sent that

wild

[plain badge]

this seems to be the final answer

[kind hawk]

yes

my hope was more that you might realize that while 2^x might seem easier, its actually the same thing

if you write the original function as (1/3)^x

[red ice]

when you differentiate 3^(-x), you need to multiply by -1 due to the chain rule

so when you integrate it, you need to divide by -1 to undo the differentiation (so that you get 3^(-x) back)

I feel it's more important to realise the reverse chain rule (a shortcut for u-sub)

just cause it's so widely used, even though you don't necessarily need it here

[plain badge]

Thanks for the explanation, I'll go learn the rules.

.close

[obtuse pebble]

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[kind hawk]

for all these more basic functions it is a good idea to differentiate the result you get again

[obtuse pebble]

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[timid silo]

pls help me with my homework

[hardy widget]

Consider divisors

gtg but this is basically it

[timid silo]

okay it didnt work so easy

ill do it by myself

sorry i have one more

thats too easy ig

i have so many time to solve this 2

homeworks

like 80 years

.close

[obtuse pebble]

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[short crescent]

i just have no clue how to set up the bounds of the integral

[short crescent]

i know one of them have to be 0-2 and one have to be 0-(4-x)?

[daring ravine]

can you draw it

[short crescent]

uhh i can try but im not sure how to graph z=4-x

is it just slanted with a slope of -1 and y-int 4?

or z-int i guess

[west stratus]

what is integration i know it is the inverse of differentiation

[short crescent]

area under a curve ig

[daring ravine]

yeah its a plane

[short crescent]

ill just graph it on desmos

[daring ravine]

at least, draw the xy, yz, xz planes

[west stratus]

jee?

[daring ravine]

i highly recommend being able to draw it or sketch it yourself

desmos probably wont be on your test

if sketching 3d is hard (understandable) then the 3 planes is good enough too

[short crescent]

i probably would be able to its just hard for me to keep track of it because im not that good at 3d graphing

[short crescent]

wdym the 3 planes?

x-y, x-z, and y-z?

[daring ravine]

cross section at xy yz xz

[short crescent]

how do i graph that i think it just made it more confusing for me hold on let me try to graph it in 3d

this is my best interpretation of it

the parabola is along the x-y plane it just looks a bit tilted i think

so the z bounds are 0 to 4-x

y is 0 to 4-y^2

and x is 0-4?

@daring ravine

[daring ravine]

the final part includes 0 <= y <= 2 and 0 <= x <= 2

[short crescent]

oh right so it doesnt go all the way to the vertex of x=4-y^2 only halfway right?

[short crescent]

yeah thats what i was thinking, i have a visual 3d image of it in my head i just cant put it to paper haha

but wait

[daring ravine]

now what order are you integrating though

[short crescent]

is the y bound 0 to 2 or 0 to 4-y^2?

[daring ravine]

it depends on the order of integration

[short crescent]

well whatever happens my outermost integral will be dx most likely

because the bounds are constant

so i could do

$\int_0^2 \int_0^{4-x} \int_0^{4-y^2} ydydzdx$ ?

[warm shale]

Krish

[daring ravine]

Try the first inner integral and see what happens

[short crescent]

oh it would just give me terms of y at the end

so that doesnt work

[daring ravine]

seeing as z = 4 - x

x = 4 - y^2

[short crescent]

can i just switch the two inner?

$\int_0^2 \int_0^{4-y^2} \int_0^{4-x} y dzdydx$

[warm shale]

Krish

[daring ravine]

well you have the same problem

generally if the bounds of a variable A is in terms of variable A, you got a problem

let's try to find a single order of integration and work out the bounds for it first

[short crescent]

oh wait i could just have y be 0-2 and x be 0-4-y^2

so same order of integrals, but change the variables to dzdxdy?

that would get rid of the y's at the end

[daring ravine]

sounds like a plan

[short crescent]

ill try that rn then thank you

i got 68/3

<@&286206848099549185>

[tardy epoch]

what is the question now?

[short crescent]

i just wanna check if this is right

well moreso where i went wrong

[tardy epoch]

why do you think it's wrong

[short crescent]

WA gave me a different answer when i went to check

[tardy epoch]

show what wolfram gave

[short crescent]

im off by 12 somehow

[tardy epoch]

you didn't distribute the 4 correctly

[short crescent]

oh my god

that explains why i had an extra 12 because they cancel out

appreciate the help both of you

.close

[obtuse pebble]

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[obtuse pebble]

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[karmic hedge]

Oh wait I just noticed my mistake 😅

.close

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[karmic hedge]

I did everything the same as what they did except this one step

Turning this

Into this using l'hopital

I just subbed the infinity directly in, which gave me -1 instead

Obviously this leads to a different answer so what was wrong with my method?

[hardy widget]

How (in general, trying to do "arithmetic" with infinity is a bad idea)

also I'd just do
$$\frac{b-2}{b+2}=\frac{1-\frac{2}{b}}{1+\frac{2}{b}}$$ tbh

[warm shale]

Civil Service Pigeon

[karmic hedge]

Ah ok

I see my mistake now

Ok thanks!

❤️

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[hardy widget]

[obtuse pebble]

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[worldly sky]

hey, im having some trouble with memorizing or just understanding (bruh) the base changes of a coordinate matrix of a linear application. i do know there are some diagrams to help but i keep messing them up or switching the order of the bases, is there any way to "deduce" it if i forget about the general ""formula""?
(for example, let A be the matrix coordinates of B in C, and M the one from K to T, so its M=PAQ with P,Q inversable, but which bases im changing?)

[kind hawk]

the diagram is good

you start with basic B. you have to translate to basis K. then apply M. then translate from T to C

[worldly sky]

im still not getting it 🥀

sorry for the messy drawing my handwriting aint rhe best either LOL but is this it? its wrong right?

then itd be.... C_BK • A • C_CT?

or TC?

[kind hawk]

I dont know the notation your course uses

C_BK takes vectors written in basis K and outputs vectors in basis B

yes?

[worldly sky]

M_BC is coordinates in base C of f(b_i)

[worldly sky]

yeah

[kind hawk]

so then M_TK = C_TC A_CB C_BK

you read from the right

and all the input and output bases have to much

[worldly sky]

its in opposite order for coord matrixes from base changes ones

[kind hawk]

this is just the kinda thing where notation is always awful, no matter what you do

[kind hawk]

I personally always have to think like this

[worldly sky]

so, this? (i can explain the notation further if u want me to, it messes me up a lot bc it changes sides)

[kind hawk]

you start in some basis, you have to translate to another basis, then you do your matrix in that basis, and then you translate back to yet another basis

[worldly sky]

ill try to remember this 😵‍💫

[kind hawk]

are you sure that is true in your notation

[kind hawk]

and this

[worldly sky]

yeah, for coord matrix u read it as "coordinates in base C of f(B)" and in base changers "coordinates of base B of vectors of K"

its like mixed

[worldly sky]

left C inputs C and turns to T coords, middle M inputs f(b) in C coords, and right C inputs K and turns into B coordinates

idk i always mix them up

i probably have to smash my head against this wall until i understand the notstion dont i

[kind hawk]

I would try to stay away a bit from the notation. and just think about what P and Q are supposed to do etc

[upbeat plinth]

maybe some arrows in subscripts would help?

$C_{B\leftarrow K}$ for change of basis $K$ to $B$ etc

[warm shale]

ロケット・ジャンプ

[upbeat plinth]

arrows might help u keep the correct order of bases

[worldly sky]

im honestly so confused rn 💀

[upbeat plinth]

$M^f_{T\leftarrow K}=C_{T\leftarrow C}M^f_{C\leftarrow B}C_{B\leftarrow K}$

[warm shale]

ロケット・ジャンプ

[worldly sky]

this is what my prof said in class, but some of the arrows are in the wrong way?

[worldly sky]

first arrow from K to B should be from B to K...?

okay i think i got it with some other help from classmates

damn this is messy 😵‍💫

thanks for helping anyways!

.close

[obtuse pebble]

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[frozen coral]

what did I do wrong?

[neon sand]

@frozen coral its better if i tell how i did it

[frozen coral]

go on

[neon sand]

what is the answer @frozen coral

[frozen coral]

i have no idea

they didn't give the answer

[neon sand]

@frozen coral (pi -1)**2/pi+1

im getting this

[frozen coral]

lemme try

[neon sand]

as you wish i can give you the sol. if u want

[frozen coral]

That was correct

[frozen coral]

yeah please share the solution

[neon sand]

@frozen coral wait a sec then

i havent done the computation as its pure bs to do it again and i hadnt done it previously back then so yeah here's the sol.

@frozen coral

[silent flame]

Bhai

!nosols

[obtuse pebble]

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.

[neon sand]

he asked for it

[frozen coral]

im not copying don't worry

i think my answer was correct i just had to simplify it a bit more

i got it thanks

.close

[obtuse pebble]

Channel closed

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[silent flame]

Ok,na bol rha hu kyuki mods allow nahi karte

Mute wagera kr sakts he

Baki no worries

[neon sand]

oh sorry

[obtuse pebble]

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[unreal tangle]

Hello guys a basic question please:
I have f(x) = 1/x - 1+ ln(x)
how can I prove that I(2;f(2)) is an inflection point and what does it mean?

[trim portal]

Do you know how to differentiate stuff?

[unreal tangle]

yes

[trim portal]

and do you know what convexity / concavity is?

[unreal tangle]

no

[trim portal]

concave is when it's kinda bent upwards

convex is when its bent downwards

inflection point is where concave part changes to convex part

[unreal tangle]

oh okay i got it and how do i prove it

that a point is an inflection point

[trim portal]

a function is concave on some interval when f''(x) < 0 for x on that interval and convex if f''(x) > 0

inflection point happens when f''(x) changes sign, which happens when f''(x) = 0 (assuming that the second derivative is continuous at the desired point)

so we first find f''(x), then find points which are suspicious of being inflection points

those are the ones with f''(x) = 0, or where the function is discontinuous

now we need to check only 2, so we focus our attention on 2

[unreal tangle]

okay wait let me try

[trim portal]

we should expect to find that f''(2) indeed is 0

[trim portal]

after that, we should still check whether the concavity / convexity really changed. Because sometimes the function can go convex -> f''(x) = 0 -> still convex

we do that by picking a point slightly below 2 and another point slightly above 2 and checking their f''

[unreal tangle]

f'(x) = -1/x² + 1/x
So f''(x) = 2/x^3 - 1/x²

this is the step one right ?

[trim portal]

yep

now check when f''(x) = 0

[unreal tangle]

in 2

[trim portal]

yep, thats the only root

and there are no relevant discontinuities in f''(x), so we should expect that the sign of f''(x) can change only at 2

so now we have to check whether it really changed

so try plugging in e.g. 1 and 3

find their second derivatives

[unreal tangle]

when x = 3 f''(x) <0
when x = 1 f''(x) >0