help-33

[sick dawn]

Yes

Then we can find it

Can the bot show it as working

[potent fog]

the bot?

[sick dawn]

The bots that are in the server

[potent fog]

what's the expression you want texed

[sick dawn]

The ratio we just did

But simplified

[potent fog]

this right

so what's the base

and what's the height

[sick dawn]

One of my friends did this is it. Orrect?

[sick dawn]

Nvm dont do that ill have to get off soon

[sick dawn]

He took the sides as 1 and 2

[marsh citrus]

@sick dawn Has your question been resolved?

[marsh citrus]

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[still temple]

hello! i'm hoping to get some help understanding the logic behind solving a trigonometric function - in this case, sin.

i've been given a question that asks me to "find the values of theta where -pi ≤ theta ≤ 4pi given that sin(theta) = −0.05". my approach was to:

1. isolate theta with arcsin, which gives us theta = -0.0500
2. theta is in the fourth quadrant of the cycle from -2pi to 0, so given CAST, theta's related angle is in the third quadrant
3. the third quadrant's range starts at -pi, so we can take -pi - (-0.0500) to find the related angle. this comes out to -3.0916
4. now that we the solutions to the first cycle (-3.0916, -0.05), we can add 2pi to both solutions for the second period, and add 4pi to find the third period.
5. this gives us solutions at -3.092, -0.05, 3.192, 6.233, 9.475, 12.516

is my approach correct? i'm not totally confident in trig yet so i'd love for someone to run their eyes over this question and lmk if i did anything wrong. thanks!

EDIT: ofuckofuck i typed sin in the text above, but i meant cos. replace sin/arcsin with cos/arccos lol!!

[iron meadow]

yes, your reasoning is all correct, by the by this is a good illustration of sin(x) ~ x for small x because sin(x) has a slope of 1, same as x at x = 0 so its values are approximated by y = x nearby

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

hmm ok thanks for the sanity check! this is a really good way to visualize it in my mind too, thanks for sharing. i feel smart for realizing the same applies to y = x+(pi/2) and y = cos(x), where you can kinda approximate it between x~=-2.3 and x~=-0.6 :D

[marsh citrus]

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[still temple]

appreciate the help

[marsh citrus]

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[vast crown]

Sorry

Wrong channel

.close

[marsh citrus]

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[spare peak]

Can someone help me with the question, I've been stuck on it for weeks now:

Use the result that arg(z1/z2) = arg(z1)-arg(z2) to draw the complex locus given by the equation arg (z+1)/(z-1) = pi/2

[spare peak]

can someone pls help

[sand fable]

let z = x + yi

[spare peak]

yeah

I did that

but idk what to do next

[sand fable]

such identity

[spare peak]

uhh

yeah

[spare peak]

so I did x + iy + 1 - (x + iy - 1) = pi/2

[sand fable]

u r missing the arg() functions

[spare peak]

arg ( x +iy + 1 - ( x + iy - 1 ))
= arg (x+iy+1-x-iy+1)

and that equals to arg (2)

idk what to do from here

help pls

[sand fable]

u should write on paper

ur expressions r extremely wrong here

[spare peak]

uhh so how am i supposed to do it

[sand fable]

so

applying arg(z1/z2) = arg(z1)-arg(z2) to arg((z+1)/(z-1)) = pi/2

[spare peak]

yeah

[sand fable]

we have, arg(z + 1) - arg(z - 1) = pi/2

[spare peak]

yes

[sand fable]

right?

ok

[spare peak]

yes

[sand fable]

then, z = x + yi

[spare peak]

yeah

[sand fable]

arg(x + yi + 1) - arg(x + yi - 1) = pi/2

[spare peak]

yes

[sand fable]

do u know how to rewrite arg(a + bi)

[spare peak]

no

do you mean polar form or smth

[sand fable]

arg(a + bi) is normally seen as arctan(b/a)

[spare peak]

oh yeah

[sand fable]

it's not technically always correct since signs but ye

[spare peak]

yeah, but u just at 180 if it isn't correct right?

[spare peak]

but how do you do that right now cz if you put that in it's arg(y/x+1) - arg(y/(x-1)

pls help i've been stuck on this for 2 weeks

[sand fable]

no

it's arctangent not arg

[spare peak]

oh mb

i meant tan sry

arctan*

[sand fable]

nw

arctan(y/(x + 1)) - arctan(y/(x - 1)) = pi/2

[spare peak]

yeah

[sand fable]

then, to relate x and y, we take tan() of both sides

[spare peak]

uhh

yeah

[spare peak]

so y/x+1 - y/x-1 = tan(pi/2)

[sand fable]

no

tan(arctan(a) - arctan(b)) is not (a - b)

[sand fable]

refer here:

[spare peak]

oh ok

[spare peak]

but doesn't the arctan cancel out?

help pls

<@&286206848099549185>

.close

[marsh citrus]

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[marsh citrus]

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[ebon cradle]

Take t = 0 in both equations

[ebon cradle]

The v_p(t) as well

[marsh citrus]

@hushed maple Has your question been resolved?

[vast crown]

I've attempted part a) of the question that looks like this:

Can someone explain to me how to sketch b?

I understand that the real points on the vertical and the imaginary on the horizontal are the same

Does this mean that they are mapped in the same way as in a? (Of equal magnitude and equal angular argument)

I’m particularly interested in what happens to the line y-x=1

Thank you

Sorry I accidentally sent it in ur channel

[hushed maple]

no problem

[marsh citrus]

@hushed maple Has your question been resolved?

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[marsh citrus]

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[wise willow]

Hello ik this is an easy question but i dont know how to do it 😞

[wise willow]

I understand all other parts but not part c

[silk sparrow]

Take the equation of the plane ax + by + cz = d, a b and c are the 3 coefficients of your cross product

Substitute the coordinates of a known point in to get d

Cancelling some common factors should get you to the equation of P1

Give it a go and if you can't arrive at that equation lmk

[wise willow]

Ok i’ll try now ty

Ok i got the answer but like i dont understand why we are doing this sorrry

Like how do yk the coefficients are the cross product of ab and ac specifically

And how does it relate to the normal and stuff

Normal stuff confuses me aswl

[wise willow]

It can be any point??

[silk sparrow]

A unique plane is defined by a point in space and a vector that is perpendicular to that plane

(if you didn't have the normal vector, the plane could take any orientation)

Basically if you have a point P0(x0 y0 z0) that lies on a plane we're interested in, and a normal vector to it n=(a b c), any point P(x y z) will lie on the plane if the vector drawn from P0 to P is perpendicular to our normal vector

Two vectors are perpendicular to each other if the dot product is 0

And if you expand out the dot product and move the constants to one side, you will arrive at the general equation of a plane that I used above

[silk sparrow]

any point on the plane yes

[silk sparrow]

If you want to derive it, you can start at $n \cdot (P_0 - P) = 0$

[elfin berry]

TayBee

[wise willow]

How can a something be perpendicular to a single point ?

[silk sparrow]

Poor wording on my part, the vector is normal to the plane that the point lies on, not the point itself

But you define a plane by a point (think of it as the original point where your plane starts from) and a vector that lies normal to the plane

To work out if any other point lies on the plane

You find the vector from your original point to your other point

And if that's perpendicular to the normal vector, the point lies on the plane

[wise willow]

😭

Is this the same thing that we use the right hand rule to say whether its going into the plane or out of the plane

[silk sparrow]

Here's a pretty picture, your plane is in green, and you can see that if you took any point on that green rectangle then drew a line from the corner of the rectangle where the normal vector comes out of (which I defined as P0) to the point on your plane, it will always be perpendicular to AxB

[silk sparrow]

Effectively yes

The RH rule is the quick way to determine the direction of a cross product

[wise willow]

Oh ok that kind of makes sense but wouldnt a j component make the plane 3d?

Or is it alrdy taking into acc the j from the two lines

Like each of them have their own positon

[silk sparrow]

No because if you take the cross product of 2 3D vectors ijk you will end up with the picture I pasted above

B looks like it is going into the page so it's probably a 3D vector anyway

A plane is by definition a 2D surface

[wise willow]

Ok wait if thats cartesian form then what about like the x-a over d = y -a over d etc

Or is that completely different

Because i understand vectors when we put it into r = a + kd

[silk sparrow]

This is the parametric equation of a line

So if we have

$x = x_0 + ta \ y = y_0 + tb \ z = z_0 + tc$

[elfin berry]

TayBee

[silk sparrow]

You can rearrange them all to make t the subject

Since t is constant, you can then equate them together

And thus you have the symmetric equation

Obviously it all interrelates but for now just try to work with what you need for the question

The question you had is using the plane equation ax+by+cz = d so just take the parameters as you need them to solve the problem

Don't mix up forms unless it is made necessary

[wise willow]

Wait i kinda get the normal thing now coz if u have a plane with a normal and u have a random eqn ax plus blah blah = d then the cross product is gonna give u a line that crosses the plane and also u sub in another line that intesects it so ur essentially just equating two eqns?

[silk sparrow]

yeah pretty much haha

[wise willow]

Ok cool ahhaa

Ty for ur help

[silk sparrow]

No problem :)

[marsh citrus]

@wise willow Has your question been resolved?

[marsh citrus]

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[open kayak]

From $2^k(2n + 1) = 2^{k'}(2n' + 1)$, how do we justify that $k = k'$ and $n = n'$?

[open kayak]

Do we just say "by comparing LHS and RHS"?

[desert socket]

(2n+1) is an always odd number

while 2^k is even

so you can compare in this case

[open kayak]

Ah

Thank you

[next raft]

natural numbers wasnt specified

[open kayak]

Actually they are natural

Thanks

.close

[marsh citrus]

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[marsh citrus]

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[cobalt garnet]

How many possible words of length 6 can be formed if exactly two letters can be repeated?

[cobalt garnet]

26x26x25x24x23x22

Is that right?

[marsh citrus]

@cobalt garnet Has your question been resolved?

[cobalt garnet]

<@&286206848099549185>

[brittle holly]

Is the word made up of only 2 letters or 2 repeated letters + extra

[cobalt garnet]

It's should be a 6 letter word where exactly two letters can be repeated. Like 'wwords'

[brittle holly]

So all should not be unique right?

[cobalt garnet]

Yes

[brittle holly]

K I'll try

[cobalt garnet]

In the text book the answer is 118,404,000

[brittle holly]

Sry nvm

[brittle holly]

Btw how can wwords satisfy this, bcz exactly 2 should be repeated

[cobalt garnet]

There are 2 W's

So two letters are repeated

[brittle holly]

I thought wwoord kind of thing were w and o were repeated

[cobalt garnet]

Maybe that. I don't know. The wording in the textbook is quite ambiguous and it doesn't give any examples

[brittle holly]

K I'll try my luck with what I understood

[cobalt garnet]

This is my way of thinking. First we answer the question of how many words can be formed such that two letters are the same. That would be 26 different words. And now we calculate how many different 4 letter words can be formed from 25 letters since one letter was used to form 2 letter word where two letters are the same

And then multiply that together to get the answer

[brittle holly]

Can be true

[cobalt garnet]

But the answers don't match

If two letters 'must' be the same then the above answer should satisfy the question

But in the textbook it just says 'exactly two letters are repeated'

[brittle holly]

Thanks for your patience but I couldn't

[cobalt garnet]

It's ok. I think the answer in the textbook is incorrect

[cobalt garnet]

.close

[marsh citrus]

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[marsh citrus]

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[wind whale]

If $a_{n-1} = \frac{a_n - n}{n} \forall n \in \mathbb{N} \setminus {1}, a_1 = 1$\
then find, $\lim_{n \to \infty} \prod_{k=1}^{n} \left( 1 + \frac{1}{a_k} \right)$

[elfin berry]

ginny

[wind whale]

from what i did, it's $\frac{a_{n} + 1}{n!}$ but idk what that is

[elfin berry]

ginny

[wind whale]

so i guess for the status factoid it's "i have begun but got stuck"

[marsh citrus]

@wind whale Has your question been resolved?

[wind whale]

<@&286206848099549185> if it's still a thing

[marsh citrus]

@wind whale Has your question been resolved?

[wind whale]

[marsh citrus]

@wind whale Has your question been resolved?

[marsh citrus]

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[late root]

What exactly is a domain and range of a f(x) function?
Is the range the lowest value of f(x) and the highest value of f(x) like 2< f(x) < 5 and the domain all the values of x like 3 < x < 5?

[marsh citrus]

@late root Has your question been resolved?

[quartz plank]

Yeah

Domain is all x values you can input

Range is all f(x) you get

But the domain doesn’t always have to be one straight segment so to speak

It can be a combination of different parts

[marsh citrus]

@late root Has your question been resolved?

[late root]

Ok ty

.close

[marsh citrus]

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[worldly lantern]

.open

.reopen

[marsh citrus]

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[remote fjord]

,rccw

[remote fjord]

Ok so here’s the question and the answer but I’m having a problem seeing how I go from what I was able to get to that

I got the mass for both, the static friction for the table and boxes, the accleration, and I just realized I may have the Fk for 1-2 wrong….

Wait no I don’t

Nvm yes I do

But I don’t think kinetic friction is important in this right?

[shadow warren]

i think the acceleration might be wrong, can you show your work?

[remote fjord]

Ye

I used this

,rccw

[shadow warren]

is W the pull in the diagram

[remote fjord]

Shoot that’s for a different problem

Gimmie a sec

Oh I got it

I did just 10kg instead of 14

[shadow warren]

alright

should be around 4.8 or so

[remote fjord]

What

F=ma?

[shadow warren]

the acceleration

[remote fjord]

Ye

150=14*a

So I got like 10.7

[shadow warren]

so its $F_{net}=ma$ yes?

[elfin berry]

y0shi

[shadow warren]

what is our F net?

[remote fjord]

Yes

150

[shadow warren]

not exactly

aside from that force

what other force is present

along the x axis

[remote fjord]

Friction

[shadow warren]

yep

[remote fjord]

How do I include that tho?

[shadow warren]

so lets say the force being applied is F

$F_{net}$ can be broken up into $F-f_s$

[elfin berry]

y0shi

[shadow warren]

since F net is really just the vector sum of the forces

when we're using newton's second law, we have to make sure we include all the forces in the direction we're analyzing

[remote fjord]

Okay so we have 150N and friction

But how do I include friction into the formula

[shadow warren]

so what does friction equal?

[remote fjord]

I legitimately don’t know how I can do this formula without friction but idk how to include it

Kinetic friction is .600
Static friction is .800

[shadow warren]

okay but what is the formula for the force of friction?

[remote fjord]

Static or kinetic

[shadow warren]

well it would be kinetic in this case

[shadow warren]

if we're looking at the two-block system

[remote fjord]

The last one?

[shadow warren]

yep

[remote fjord]

But do I use the sum of the weight or just one

[shadow warren]

so since we know that $F_{net}=F-f_s$, then by substitution $F-f_s=m_{tot}a$

[elfin berry]

y0shi

[remote fjord]

.6*137.2

[shadow warren]

so what is N equal to

[remote fjord]

Of the two blocks 137.2

[shadow warren]

yeah

thats right

now we have everything and we can solve for a

[remote fjord]

Fk= 82.32

But what do I do with that now

[shadow warren]

^

[remote fjord]

That’s the kinetic friction of the whole thing between the table

[shadow warren]

we used newton's second law to derive that equation

we know F, f_s, and m_total

[remote fjord]

But I don’t have static force

[shadow warren]

oh wait thats my bad

i meant f_k

[remote fjord]

Uh

,rccw

[remote fjord]

That?

[shadow warren]

yep

but thats equal to ma

by newton's second law

[remote fjord]

,rccw

[remote fjord]

So solve for A?

[shadow warren]

yep

[remote fjord]

Ok

What’s next tho

[shadow warren]

so now we have a

we know that the two-block system must be accelerating at this a

and that also means that each of the blocks are accelerating with this a

so if we now just analyze the upper box

we can use newton's second law to find what f_s is

[remote fjord]

109.76

[shadow warren]

that seems a little too big

[remote fjord]

F_s*newtons?

[shadow warren]

so the equation should be $F_{net}=ma$, then $f_s=ma$

[elfin berry]

y0shi

[shadow warren]

where m is the mass of the top block

[remote fjord]

Why just the top block

[shadow warren]

since we're only analyzing the system of only the top block

[remote fjord]

189.34

[shadow warren]

thats still too big

howd you get that answer?

[remote fjord]

Used Newtons

19.32

Oh that’s the actual answer

[shadow warren]

yep thats good

[remote fjord]

Why does it work though?

Since it’s asking for the friction force of the upper box

Dosent this just give the force?

[shadow warren]

doing this gives the net force

and the net force for the upper box is just the force of friction

since theres no other forces exerted on it

[remote fjord]

But there’s friction between the boxes

Like if I were doing this how would I find a formula for Fnet usually

[shadow warren]

this is the friction between the boxes

[shadow warren]

well you would first identify the system youre analyzing

in this case we analyzed only the upper box

then we would write out newton's second law for one of the directions

it would be best to draw out a free body diagram for problems like this

it helps clarify things

[remote fjord]

Kinda

More or less

Like I don’t think physics is like hard in terms of the math but more or less of why am I allowed to do it like this is what I’m having trouble with

[shadow warren]

so what part of what we did was confusing

the equation?

[remote fjord]

What I was suppose to do with it after I got all my numbers

[shadow warren]

like how to connect the acceleration to the force of static friction?

[remote fjord]

Ye I think

[shadow warren]

well on what two objects is the frictional force acting on?

[remote fjord]

The blocks

[shadow warren]

yep so that means we have to analyze either the top block or the bottom block

we can choose whichever one

i just chose the top block since its more simple

[remote fjord]

But why am I going from using newtons for everything else to using kg for the answer?

Like if I gave the newton equivalent of 19.32 would that be an accepted form of the answer?

[shadow warren]

19.32 is in newtons

we just multiplied two different quantities to get newtons

since a newton is kg m/s^2

[remote fjord]

Ye kg*9.8

In this case

[shadow warren]

well 4.8 since that was our acceleration

[remote fjord]

Oh so I’m out case f_s=f_net

And f_net was m*a

So f_s=ma

I think

[shadow warren]

yep

[marsh citrus]

@remote fjord Has your question been resolved?

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[sullen ice]

btoh of these i'm off by a little bit

[sullen ice]

i checked with a z chart and its the same issue

i used mu = np and standard deviation = sqrt(npq)

oh

whats "correction for continuity" ?

how do you know to add or subtract .5?

to x-mu

[marsh citrus]

@sullen ice Has your question been resolved?

[marsh citrus]

@sullen ice Has your question been resolved?

[marsh citrus]

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[sullen ice]

.reopen

[marsh citrus]

✅

[glass silo]

For the P(X > 9), you want the normal approximation, call it Y to be P(Y > 8.5): the idea is that each specific binomial value, you want to go 0.5 up and down

So e.g. if you wanted to approximate the binomial P(X = 9), you'd do that as P(8.5 < Y < 9.5) to cover both 0.5 above and below, if you wanted something like P(X < 9), you'd instead do P(Y < 9.5) (go 0.5 up to cover the "bar" that corresponds to 9), so on and so forth

[marsh citrus]

@sullen ice Has your question been resolved?

[sullen ice]

oh okay

thank you

.close

[marsh citrus]

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[ember flame]

This is false right.?

!status

[marsh citrus]

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

[ember flame]

4

[slim spire]

should be fixed if you put 3x1-5x2 in the 3rd slot instead because it's x3=...

[ember flame]

Oh shit

Sorry

[ember flame]

But still by inspection

There will never be a 4 in x2 for both vectors in the basis right?

So it’s obviously false?

[slim spire]

where's 4 coming from? I see if you plug in the basis vectors into the formula it fails yea

[ember flame]

I was talking about these 4s

For both vectors x2

Both of them will never show up

For lack of a better word

[slim spire]

ah yea makes sense

[marsh citrus]

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[marsh citrus]

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[amber sleet]

can someone explain this process. i get its implicit differentiation but im lost now. (its my professors work from our notes)

[proud ice]

What about it do you not understand?

[amber sleet]

nvm

ive been doing my math hw for like 3 hours so my braincells are gone

i just did a mental dumb

😔

.close

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[marsh citrus]

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[steep storm]

how many functions defined on a set of n points are possible with values 0 or 1? how many of these functions have precisely m'1's in their range

[steep storm]

its combinoatorics

[static quarry]

m1's?

[steep storm]

yeah

wait ill send a pic

22

[static quarry]

ohh, m 1's

what have you tried?

[steep storm]

i dont know where to start

frankly i dont understand the problem

[static quarry]

suppose n is 3

your set could be labeled {1,2,3}

[steep storm]

yeah

[static quarry]

how many functions can you define on this set which have values 0 or 1 at each point?

you can write them all out explicitly

[steep storm]

got it

can you like rephrase the question

if possible

[static quarry]

didn't i just do that

[steep storm]

what about teh second part

wait i got itmb

[static quarry]

oh like in the above example

how many of those functions take on the value 1 exactly m times

where m is either 0, 1 2 or 3

[steep storm]

right okay

[static quarry]

for example f(1) = 0, f(2) = 1, f(3) = 1

that one has value 1 for two points in the domain (namely 2 and 3)

[steep storm]

the answer to the problem is

given as

2^n; n choose m

[static quarry]

sounds right

[steep storm]

$$2^n; \binom{n}{m}$$

[elfin berry]

flurry

[steep storm]

idk why id di that

[static quarry]

you should be able to rephrase it equivalently in terms of subsets of {1,2,3,...,n}

[steep storm]

ok thanks

ill try and understand by repeatedly going over your explanation for the next week

[static quarry]

week??

[steep storm]

yeah

not a fault on your part

week is an exaggeration would be like 3-4 days

[static quarry]

ok well give it some thought and if it's not clear, you can always come back and raise another doubt here

[steep storm]

okok

thanks

.close

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[marsh citrus]

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[azure trench]

If I want to find determinant of 4x4 matrix, is there only 1 method which is making it echelon

[azure trench]

Or is there any quicker method

[verbal ibex]

id say cofactor expansion is quicker depending on the implementation details

[azure trench]

Can u use that formula for any matrix

[marsh citrus]

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[restive anvil]

I have this differential equation :

so i get the homogeneous solution (i don't know if it's the correct term in English sorry if not)

the problem get to the particular solution where in my book it says we notice that y0(x) = x is suitable what does that mean ?

[restive anvil]

I mean I understand what that means but I don't understand how he sees it without doing calculations

[gleaming pecan]

it's just a matter of practice and experience, if you don't have it, use classic methods

[restive anvil]

and could you explain to me what the process is here to see this result

[gleaming pecan]

y = x, y ' = 1 when you put it into your non-ghomogenmous eq, you can see imemdiatley

that L = R

so yoru professor behaved in a such way

when i write book i do the same such tricks

but

student who has no such experience

and time durgn exam

must use classical emthods

so do not waste time for guessing solution

but find it usign avaiaobel tools

is it clear?

[restive anvil]

a yes okay

yes yt

[gleaming pecan]

smiles

[restive anvil]

[gleaming pecan]

)

[restive anvil]

ty

[gleaming pecan]

yw

[restive anvil]

.close

[marsh citrus]

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[marsh citrus]

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[fickle gale]

Help please

[fickle gale]

Determine the vector equal to the sum ST+MK+PL+KP+LS.

Write the name of the vector on latin letters. For example:AB

Please help me

<@&286206848099549185>

[marsh citrus]

@fickle gale Has your question been resolved?

[fickle gale]

<@&286206848099549185>

<@&286206848099549185> Help me please!!!

Please help me, I have work due soon

[still temple]

You just need to find the name of the resultant vector?

Usually the name of the resultant vector after vector addition is given by the First letter of the first vector and the last letter of the last vector. So if a vector addition is of AB + BC +CD, then the resultant vector would be AD

@fickle gale

[fickle gale]

Yes I need a name

AD name is incorrect

I have one try left

[still temple]

I didn't give you the answer. I'm explaining the process to you

[fickle gale]

Ah, ok

[still temple]

In the above vector sum of your question, what is the first letter of the first vector and the last letter of the last vector?

You combine them to get the name of the resultant vector

[fickle gale]

I tried and it doesn't work

Determine the vector equal to the sum ST+MK+PL+KP+LS.

Write the name of the vector on latin letters. For example:AB

If I'm doing it right, am I making the name just M?

[still temple]

No the name is two letters

[fickle gale]

MT or TM?

[fickle gale]

Are any of them correct?

[still temple]

Yes

But Mk continues on with KP but ST doesn't have anything to continue with

So MK should be the first vector here

[fickle gale]

MK's answer?

[still temple]

no

[fickle gale]

Oh no I got it

MT's answer?

[still temple]

Yes it should be

[fickle gale]

Yes, it worked, thank you very much

[still temple]

glad to be of help

[fickle gale]

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[marsh citrus]

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[sturdy lagoon]

this should be really easy but I forget how to get this bottom standard deviation value

[crude basalt]

The rule is $\sigma_{x}=\frac{\sigma}{\sqrt{n}}$

[elfin berry]

The Great D

[sturdy lagoon]

ah ok

[sturdy lagoon]

thank you

so .5

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[mint lily]

Does anyone have steam games\

[devout mauve]

some people do indeed have steam games

tho I fail to see the relevance for this channel

[marsh citrus]

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[marsh citrus]

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[limpid lily]

i am stuck in the last part

where i dont understand how they got y_h

to that

if someone can explain it would be greatful ❤️

[shadow warren]

so lets say we just plugged in -1+2i or -1-2i

we would have $e^{(-1+2i)t}$

im just gonna choose -1+2i since it really doesnt matter in the end

[elfin berry]

y0shi

[limpid lily]

why is there a *t?

[shadow warren]

we can distribute the t and separate the exponential using exponential rules:
$e^{-t+2it}=e^{-t}e^{2it}$

[elfin berry]

y0shi

[shadow warren]

well when solving these differential equations, we assume that the solution is e^rt

and thus we get the auxiliary equation to get r

and we just simply plug it back in

but in this case t would just be x

[limpid lily]

okay so thats just a rule

where u always use it with a t?

[shadow warren]

well it can be any variable

in this case you used x for

ill change it to x so its easier to understand

[limpid lily]

yea i get that part

[shadow warren]

from here, $e^{2ix}$ can be plugged into euler's formula $e^{i\theta}=cos{\theta}+isin{\theta}$

[elfin berry]

y0shi

[shadow warren]

where $\theta=2x$

[elfin berry]

y0shi

[shadow warren]

hence we'll get $e^{-x}(cos(2x)+isin(2x))$

[elfin berry]

y0shi

[shadow warren]

this is the combination of both of the solutions

and the constant of integration goes in front of the cos and the sin

[limpid lily]

uhhh okay i see

[shadow warren]

where i is really just negligible

since it can be regarded as a constant

[limpid lily]

so eulers formula is as i remember for cos and sinx right?

[shadow warren]

you dont really have to prove this every time, you can just remember the general solution

[shadow warren]

well yeah it relates exponential with cos and sin

mainly exponential with imaginary exponents

[limpid lily]

yea i have the test tommorow so yea...

[shadow warren]

yeah so probably good to remember that if $r=\alpha + \beta i$, then the general solution would be $e^{\alpha x}(C_{1}cos(\beta x)+C_{2}sin(\beta x))$

[elfin berry]

y0shi

[limpid lily]

okay i see

thx for the help

but its there

[shadow warren]

alright

ywyw

[limpid lily]

like a certain way i can know when i have to use eulers formula?

[shadow warren]

when you have something like $e^{ix}$

[elfin berry]

y0shi

[shadow warren]

when theres something imaginary in the exponent

but usually you dont need to know how to use this formula for differential equations like this

[limpid lily]

yea becuase the thing is the exam i need 9 poins out of 30

[shadow warren]

since they kinda just give the answer to you without explaining where it comes from

[limpid lily]

and i focus mainly on this kinda questions

[shadow warren]

i see

good luck!

[limpid lily]

like this

[shadow warren]

ah

these are kinda bad too since its not equal to 0

and you have to do even more work

[limpid lily]

the first one was easy

but the second picture was annoying

to do

[shadow warren]

yeah i can tell

[limpid lily]

when its homo and particualr solution

i need to do

but i will go back and practise and practise thx for the help! ❤️

[shadow warren]

alright you're welcome!

[limpid lily]

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[limpid lily]

hi

[limpid lily]

this is wrong right?

we have y=sin x

y'=cos x

and y''=-sinx

and y'''=-cos x

Omg nvm lol

i have done right xD

just tired hahah

or?

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[ionic owl]

Stuck on this problem, not sure how to find what its asking, could someone point me in the right direction?

[ionic owl]

Actually I think I got it?

Using |A| + |B| + |C|−|A ∩ B|−|B ∩ C|−|A ∩ C| + |A ∩ B ∩ C| and plugging in the numbers
40 = 22 + |A ∩ B ∩ C|
18 = |A ∩ B ∩ C|

Is my logic flawed?

[mint lily]

Do you

[marsh citrus]

@ionic owl Has your question been resolved?

[ionic owl]

<@&286206848099549185>

[drifting schooner]

how did you get 22 for |A| + |B| + |C|−|A ∩ B|−|B ∩ C|−|A ∩ C|?

should be 28+17+13-6-7-10 = 35

[marsh citrus]

@ionic owl Has your question been resolved?

[ionic owl]

Bad mental math

But is the process correct

40 = 35 + |A ∩ B ∩ C|
5 = |A ∩ B ∩ C|

[marsh citrus]

@ionic owl Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[frail raptor]

umm so I need some help with an integrals problem, x^2 + y^2 = 9, squares perpendicular to y-axis. So basically I did this problem but my GC is saying the answer aint possible

[sand fable]

what method/answer u get/use ?

[frail raptor]

i just drew a representative triangle

lemme show you what I did

i made a silly mistake

i thought the graph would be from 0 - 8

0 - 9

but its from 0 - 3

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[native tapir]

did he do it wrong?

[native tapir]

shouldnt it be x2?

because what if the man starts

in the first slot

[quaint elm]

then two women would stand next to each other

[steep storm]

It’s correct

[native tapir]

oh ok THX

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[hard gull]

Write an equation of parabola whose axis is parallel to y-axis and points (0,3) , (3,4) and (4,11) lie on the graph

[hard gull]

!status

[marsh citrus]

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

[hard gull]

1

[quiet anvil]

@hard gull the general form of a parabola is f(x) = ax^2 + bx + c

You know f(0) = 3, f(3) = 4, and f(4) = 11 from the points

[maiden trout]

!status

[marsh citrus]

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

[quiet anvil]

@maiden trout status was already answered

[stiff sundial]

Put the 3 values of (x,y) and solve the linear equation in terms of a,b,c

[hard gull]

isnt general form ax^2 + by^2 + 2gx + 2fy + c = 0?

nvm

[stiff sundial]

But it's given its parallel to x axis

[hard gull]

yeah i saw so y = 0

[stiff sundial]

Y*

[quiet anvil]

Parallel to y-axis actually, our bad

This should be a function of y

Wait not

[hard gull]

ok so this is simulatneous equation and i find a b and c

[quiet anvil]

This is fine, the axis of symmetry, not the directrix

Should be a function of x.

[quiet anvil]

Yes exactly

[hard gull]

ok seems easy enough , ty

.close

[marsh citrus]

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[stiff sundial]

Yes, y is a function of x

[hard gull]

.reopen

[marsh citrus]

✅

[hard gull]

how do we know this ?

im new to parabolas and stuff like this

[quiet anvil]

From the points you quoted

[hard gull]

is f(x) = y?

[quiet anvil]

Yes

[hard gull]

hm

ok

.close

[marsh citrus]

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[ivory hull]

How does C(t)=C_0exp(kt) differentiated give dC(t)/dt=kC(t)?

[hard gull]

$\frac{d}{dx}e^{f(x)} = e^{f(x)} \cdot \frac{d}{dx}f(x)$

also this is occupied , next time go to available channels for help

[elfin berry]

JustToPro

[hard gull]

.reopen

[marsh citrus]

✅

[hard gull]

sorry to disturb again , but the answer im getting doesnt match with the answer on the book

the book says
5x^2 - 14x - 3y + 9 =0
im not getting that y term in there

oh wait nvm im dumb , f(x) = y

and we get 3y

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[red coral]

is there any other way to build the graph according to the given function, besides building the table and giving x values?

[flat grail]

yes there is, have you tried to factorise it?

[red coral]

nope

[flat grail]

can you see a factor that you can pull out straight away?

[red coral]

u mean like this?

[flat grail]

exactly

so we know that the graph goes through the origin

[red coral]

yes

[flat grail]

can you factorise x^2-2?

[red coral]

how do i do that😅

[flat grail]

well you know that itll be (x-a)(x-b) as its a quadratic

so you need to pick an a and b such that a*b =2, a-b=0. can you think of any?

this is also called the "difference of 2 squares" if you know it by that term

[red coral]

i cant think of anything😭

[flat grail]

$(x+\sqrt{2})(x-\sqrt{2})$

[elfin berry]

lewis_f04

[flat grail]

take a moment to verify to yourself that this works

and that $(x+\sqrt{2})(x-\sqrt{2}) = x^2-2$

[elfin berry]

lewis_f04

[red coral]

ok lemme see

okay verified

[flat grail]

you see?

[red coral]

yes it does give x²-2

[flat grail]

so now we know that this graph has roots at x=0, x=sqrt2, x= sqrt

do you know how to differentiate?

[red coral]

what do you mean by differentiate?

[flat grail]

does $\frac{dy}{dx} = 3x^2-2$ look at all familiar to you?

[elfin berry]

lewis_f04

[red coral]

does dy/dx has to do with coefficient?

thats what u mean?

[flat grail]

i wouldnt use it in your tests or anything if you havent been taught it, but dy/dx gives the gradient of the function at all points. This is useful for what youre doing because you can use it to find where the curve has a minimums and maximums, so you can see where its "bumps" are

[red coral]

so for example in y=mx + c this dy/dx gives m ?

[flat grail]

exactly

[red coral]

oh yea i get it

[flat grail]

so you need to solve 3x^2 -2 = 0

[red coral]

how did we get 3 tho?

[flat grail]

the rule for differentiation is $\frac{d}{dx} ( x^n) = nx^{n-1}$

[elfin berry]

lewis_f04

[red coral]

is there any other way to find it beacuse i havent been taught this way

[flat grail]

your example with $\frac{d}{dx} (mx+c) = m \cdot 1 * x^0$

[elfin berry]

lewis_f04

[flat grail]

not that i know of, sorry, because you wont be able to tell where the bumps are

[red coral]

oh okay i understand

[flat grail]

those are all the points we know so far, but what they do between those points you cant know, unless you plug numbers in to check, like 1 and -1

[red coral]

we can find the m by this rule right

and after i get the m i solve for c? so i can build the graph

no i think im wrong

c is given right

[marsh citrus]

@red coral Has your question been resolved?

[red coral]

.close

[marsh citrus]

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[midnight estuary]

Let a be a non-zero complex number such that |a| $\neq$ 1. Let P be the point a in the complex plane, and let Q be the point $\frac{1}{\bar a}$. Let $C_1$ be the circle ${z : |z| = 1$} and let $C_2$ be any circle passing through P and Q. Show that $C_1$ and $C_2$ intersect orhtogonally. [Two circles are said to intersect orthogonally if the tangents at a point of intersection are perpendicular to each other.]

[elfin berry]

Normed

[midnight estuary]

.close

[marsh citrus]

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[marsh citrus]

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[midnight estuary]

Let a be a non-zero complex number such that |a| $\neq$ 1. Let P be the point a in the complex plane, and let Q be the point $\frac{1}{\bar a}$. Let $C_1$ be the circle ${z : |z| = 1$} and let $C_2$ be any circle passing through P and Q. Show that $C_1$ and $C_2$ intersect orhtogonally. [Two circles are said to intersect orthogonally if the tangents at a point of intersection are perpendicular to each other.]

[elfin berry]

Normed

[midnight estuary]

I've made the diagram but not able to show that the circles intersect orthogonally, can someone help?

[marsh citrus]

@midnight estuary Has your question been resolved?

[marsh citrus]

@midnight estuary Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[zealous sail]

stuck on the lucky number & patrick NFL

[frosty turret]

can someone helpe with i dont get what technique to use here inclusin exclusin?We aim to create a unique friendship bracelet with a minimum of 3 beads, using only three available bead colors. A bracelet is considered "unique" if no two consecutive beads share the same color. For instance, for a bracelet consisting of 4 beads, there are 18 unique configurations possible.

(a) Let's define $c_n$ as "the number of unique bracelet configurations possible with $n+3$ beads." Given this definition, it is noted that $c_1=18$. What is the value of $c_0$? Additionally, derive a recurrence relation for $c_n$.

[elfin berry]

samenn.

[zealous sail]

<@&286206848099549185>

you are needed.

high demand

[steep storm]

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[zealous sail]

OR u can just answer

[steep storm]

!occupied

[marsh citrus]

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).

[steep storm]

i would if i knew

i have to read the question first

wait

[zealous sail]

TRYYY

12°15'37.5"N, 7_°34'53.2"W heres what i came w but its incomplete + shore some are wrong

[marsh citrus]

@zealous sail Has your question been resolved?

[zealous sail]

NO

IM STILL WAITING FLURRY

[steep storm]

IM STILL SOLVING

actually i stopped and went to sleep

[zealous sail]

UGH DO BETTER.

[steep storm]

bruhv no

[zealous sail]

( its ok i can wait tyt )

[marsh citrus]

@zealous sail Has your question been resolved?

[zealous sail]

<@&286206848099549185>

[pliant siren]

what does mee6 need help for

ah ok youre fake

[severe dawn]

LOL

[zealous sail]

wym im fake im him

[marsh citrus]

@zealous sail Has your question been resolved?

[zealous sail]

<@&286206848099549185>

[marsh citrus]

@zealous sail Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[spring stirrup]

how do i find the domain of 0 < x< 20

[slim spire]

are you asking why 0 < x < 20 is in the solution?

[spring stirrup]

they had to

make up a domain

but idk how they got it

[slim spire]

the squares just get too big

[spring stirrup]

howd u get 20??

[slim spire]

because it's half of the short side 40

it's hard to see without drawing it out to check what the max is

[spring stirrup]

Oh ok i dont have to do any algreba to figure out the domain ?

[slim spire]

uh from the algebra instead you could say (40-2x) would be negative if x>20 and a negative volume doesn't make sense

[spring stirrup]

@wary kite can you explain what he just said

I dont under stand

[wary kite]

the side length is 40 if you split it into 2 squares with max length of 20 there wouldn’t be enough room for the squares

[spring stirrup]

ohh

[marsh citrus]

@spring stirrup Has your question been resolved?

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[open panther]

cann someone help me out with this problem i have no clue on how to set it up

[crystal lintel]

set it up as a discrete time markov chain

you are probably meant to draw a tree diagram or something

[open panther]

i couuld but the next problem asks for 23 times

and who drawing a tree diagram 23 iterations out 💀

[crystal lintel]

ok maybe discrete time markov chains are the idea afterall

[open panther]

whaat is a discrete time markov chain

[sand fable]

what class is this from?

[open panther]

discrete math

[crystal lintel]

never discussed markov chains?

or stepping backa bit do you know matrix multiplication?

[open panther]

yuh i do

i just got no clue abt markov chains

[crystal lintel]

have you had a problem anything like this in the class before

[open panther]

nah

i havnt

thas why im completely los

[sleek lake]

maybe the 23 step problem is different?

[open panther]

same thing

[sleek lake]

it just says "same thing but 23" ok

wait no it's not

[open panther]

wait

you right

💀

i may be blind

but still how do i set it up

[sleek lake]

yeah, no idea

[open panther]

shit

cooked fr 💀

[crystal lintel]

this question seems... weird?

[sleek lake]

it doesn't even make sense, we don't know how many

[crystal lintel]

rose and manuel could just pass to each other

or manuel could pass to carlos

you don't know lol

unless it's supposed to be an expectation problem

[open panther]

yeah idek what to do this problem got me tweakin for hours

[marsh citrus]

@open panther Has your question been resolved?

[marsh citrus]

@open panther Has your question been resolved?

[marsh citrus]

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[marsh citrus]

Available help channel!

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Remember:
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[glossy bobcat]

can sb help me solve this

I dont know what can I do with sin*cos

[static quarry]

what are those exponents? 4?

[sand fable]

is that a 9 or a 4

[glossy bobcat]

4

[static quarry]

well sin^4 - cos^4 can be factored in a useful way

[glossy bobcat]

sin^2-cos^2 maybe

but I dont know what next

[static quarry]

well there's a trig identity for cos^2 - sin^2

and there's also one involving cos(alpha)sin(alpha)

those will probably be useful

[sinful thistle]

wait if im reading this correctly

$\sin x \cos x = \frac {7}{19}$

[elfin berry]

ren

[static quarry]

therefore 2sin(x)cos(x) = ...

(that's a hint)

[sinful thistle]

$|\sin^{4}x-\cos^{4}x|=?$

lisetn to bungo

*listen

[static quarry]

it's a minus isn't it?

sin^4 - cos^4

[glossy bobcat]

yes

[sinful thistle]

oh mb

[glossy bobcat]

2sinxcosx is sin2x

[elfin berry]

ren

[sinful thistle]

correct

[static quarry]

yep

[sinful thistle]

what's cos(2x) in terms of:
a. sin 2x
b. trig identities

[static quarry]

and we've established that sin^4 - cos^4 is the same as sin^2 - cos^2

[sinful thistle]

^^

[static quarry]

and there's a trig identity for the latter

[glossy bobcat]

and then sin2x= 14/19

[sinful thistle]

mhm

[glossy bobcat]

but i dont know if it useful

[sinful thistle]

very good keep going

[static quarry]

it is useful

[sinful thistle]

okay look, dyk difference of squares

$|\sin^{4}x-\cos^{4}x|=|(\sin^2 x + \cos^2 x)(\sin^2 x - \cos^2 x)|$

omfg

[glossy bobcat]

yes

[sinful thistle]

minus

not plus

[elfin berry]

ren

[sinful thistle]

that's better

you understand this?

[glossy bobcat]

I know

[sand fable]

they know

[sinful thistle]

now since sin^2 x + cos^2 x = 1 we can factor that out

[static quarry]

yes, i was recapping, and trying to guide to the next step

[sand fable]

no like

[glossy bobcat]

sin^2-cos^2

[sinful thistle]

mhm

[sand fable]

i was telling ren

[glossy bobcat]

then

[sinful thistle]

what's the trig identity for that

[static quarry]

oh i see

[sinful thistle]

yea i was just explaining again, i came later

[sinful thistle]

what's cos (2x)?

try to express cos (2x) in terms of sin(2x) and in the usual trig identities

@glossy bobcat