help-33

[lucid zenith]

This gives you sinx+cosx=0

[valid hinge]

Ohh wait

Yeah

True

Sinx+cosx=0

[lucid zenith]

This implies that tanx=-1

[valid hinge]

Abosulety

[lucid zenith]

Which doesn’t have any solutions

In our interval

[valid hinge]

How you checked?

[lucid zenith]

tanx=-1 at x=-pi/4, 3pi/4…

I just checked that 3pi/4>2

[valid hinge]

0 to 2π

[lucid zenith]

Oh yeah okay then you will have solutions

Be careful when you solve the systems though

$\sin x\ge 0\land \cos x=0 \lor\sin x<0\land \tan x=-1$

[elfin berry]

kheerii

[valid hinge]

sinx<0 and tanx=-1 what do you mean be careful here?

[lucid zenith]

tanx=-1 will have two solutions in [0, 2pi], one where sinx>0 and one where sinx<0

You take only the solution with sinx<0

[marsh citrus]

@valid hinge Has your question been resolved?

[valid hinge]

Ohh i see

So in previous case cosx is 0 at π/2,3π/2

We will discard 3π/2

@lucid zenith

[late geode]

why discard 3pi/2

[marsh citrus]

@valid hinge Has your question been resolved?

[marsh citrus]

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[stone dirge]

What rule would I use for this one?

[late geode]

first step is always to identify common factors and factor those out

[stone dirge]

Yeah I know how to do it

but I used the rules gcf and difference of squares

is there a rule I forgot to name?

Sorry, I didn't make that clear

[latent coral]

oh if you're just naming rules, yeah you factored out the gcf and factored a difference of squares

i don't see anything else

[late geode]

if those are directly linked to your steps then that's it

were there any steps you performed with no justification?

[marsh flower]

Yeah i suppose it’s A

[stone dirge]

Yeah, I was just wonering if there was a rule I forgot to name, I am just trying to practice knowing the rules, thanks guys

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[bitter terrace]

how am I to find D?

[bitter terrace]

i know the distance BD is 2sqrt26

distance between AD and CD is 2sqrt13

[mild perch]

Draw a picture

It should help

[bitter terrace]

algebraically?

i know roughly where the point is

its below all of them

[mild perch]

But do you see how you can get the point using the vectors involved?

Let's do a simpler version of a problem

If (0,0), (1,0), (0,1) make a right triangle

What point can we add to make it a square

[bitter terrace]

(1,1)

[mild perch]

What about (1,0), (2,0), (1,1)

[bitter terrace]

(2,1)

[mild perch]

Right

Do you see a pattern

Think of the sides that make the right angle as vectors

[bitter terrace]

uhhhhh

then should I wait

can I

say vector AB = DC, BC = AD, put that in xi + yj thing and then do simultaneous or is that clunky

[mild perch]

Do you understand how to solve the problem if the right angle was at the origin?

[bitter terrace]

idk how to solve with right angle vectors

AC = AO + OC?

idk man

[mild perch]

Like if we have points (0,0), (-1,1) and (1,1)

Can you solve this

[bitter terrace]

im using common knowledge for this though

its obvious its at (0,2)

[mild perch]

How do you get (0,2) by applying simple operations to the set of vectors I gave

This is why I said a drawing might be helpful to understand what is happening

[bitter terrace]

im not saying I dont know low to do what youre saying, but can you saying without math lingo cause im terminologically dumb

[mild perch]

What part

[bitter terrace]

sorry just gimme a sec

[bitter terrace]

do you mean this?

[mild perch]

Right

[bitter terrace]

cause theyre equal? or because you can add vectors to give your result?

[mild perch]

How do you get the top point using the bottom vectors you drew

[bitter terrace]

equal magnitude, weird direction idk

[mild perch]

You are just adding them

[bitter terrace]

ah true

[mild perch]

The displacement of the opposite point is the sum of the displacement of the two adjacent points from the point

[bitter terrace]

like this?

[mild perch]

The stuff with the same color are the same vector

[bitter terrace]

sorry gimme a sec i’m trying to relate this back to the q

so I have this

im stuck on two parts here now

if i were to find it, I would be finding BD

how would I separate that in order to find D? do I separate into BO + OD = BD?

and with the 2sqrt13, how would I find the exact placement if all i have is that

[mild perch]

The kind of goofy way if you can't figure out how to use the vectors here

Would be to shift the problem to the origin and shift it back

You have AB and BC as vectors right

Use those to get from B to D

Same thing we practiced

Here point be takes the role of the origin in our simpler case

[bitter terrace]

uhh is this better, considering I found the column vector thing from part a

[bitter terrace]

ill try this for the ractice

[mild perch]

Should give you the same answer

[mild perch]

Don't see anyone wrong with it

[bitter terrace]

is this what you were thinking or am I doing a different way again

[mild perch]

This is it yeah

But the way you did is good too

This was just using the most geometrically intuitive way for me

[bitter terrace]

fair

thanks a lot man i appreciate the help

[mild perch]

You're welcome

[bitter terrace]

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[marsh citrus]

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[hybrid prism]

<@&286206848099549185>

[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.

[hybrid prism]

I am not able to under stand the solution of this question

Kindly help

[marsh citrus]

@hybrid prism Has your question been resolved?

[hybrid prism]

Mo

No

<@&286206848099549185>

[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.

[hybrid prism]

<@&286206848099549185>

[vocal leaf]

??

[hybrid prism]

Somebody help me

[vocal leaf]

how can i help

[hybrid prism]

.

I am not able to understand the solution

[vocal leaf]

what is your gender

[hybrid prism]

What??

[hybrid prism]

Male why??

[vocal leaf]

just a minute and i will explain

[sinful thistle]

how does that matter lol

[hybrid prism]

Yheaa

[vocal leaf]

it dosent i'm just asking

[hybrid prism]

See I am indian i don't understand what pronoun are

I am just simple male ... And please solve my problem

[sinful thistle]

okay

ykw

i'll tell u

simply substitute sin x = t

then use the quadratic formula

@hybrid prism

[hybrid prism]

I did .. and not got the answer

[hybrid prism]

I think it's a wrong approach

[sinful thistle]

it is the correct approach.

[hybrid prism]

Please show me the solution

[sinful thistle]

what have you tried

[hybrid prism]

I am not able to understand the solution

I have tried everything

I have asked this same question in another server.. they are stuck for 3 hours..

[cloud lark]

alr leme crack at it

[sinful thistle]

okay first of all

@hybrid prism show me "everything" that you have tried

[cloud lark]

alr just to clarify, when they say the equation has roots (0, 180), they mean x = 0 and x = 180 are the roots?

[hybrid prism]

Ok

Solved using application of derivatives

But it is long method

[hybrid prism]

I want to understand this solution

@sinful thistle have tried it

[cloud lark]

i might get it hang on

this one is some trouble

aw man

i will admit

this one is pretty mind boggling

i couldnt get it

[marsh citrus]

@hybrid prism Has your question been resolved?

[hybrid prism]

Anyone able to understand solution

I am giving up

<@&286206848099549185>

[marsh citrus]

@hybrid prism Has your question been resolved?

[valid hinge]

Because at 3π/2 sinx is negative but we are looking sinx>=0

[spark siren]

with the substitution t = sin x its just straightforward, but lets start at the beginning.

(sin x)^2-k sin x -3 = 0, and we are asked for which k there are two dinstinct real roots in [0,pi].

Do you understand the question?

[marsh citrus]

@hybrid prism Has your question been resolved?

[hybrid prism]

Yes

Then

[spark siren]

ist a quadratic equation in sin x , lets write t instead of sin x, t = sin x., so we get
t^2-kt-3=0
OK?

[hybrid prism]

Yes

[spark siren]

now solve for t.

[hybrid prism]

Can you explain me your solution by writing somewhere

[spark siren]

!nosols

[marsh citrus]

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.

[hybrid prism]

What

It's a illustration not homeworks

[spark siren]

its a simple quadratic equation. you can do this by yourself.

[hybrid prism]

I am stuck .

[hybrid prism]

I have solved this .. but my method is different.. I want to understand illustration method

[spark siren]

write down the soultion for the quadratic equation t^2-kt-3 = 0 and i will do the next step with you.

[spark siren]

well, waited long enough.

[marsh citrus]

@hybrid prism Has your question been resolved?

[marsh citrus]

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[robust silo]

An arithmetic sequence with 2024 terms has first term 27/2and common difference -5/4 How
many of the 2024 terms are integers?
(A) 3 (B) 506 (C) 507 (D) 505 (E) 1011

[unreal geode]

1. 54/4
2. 49/4
   3) 44/4
   7) 20/4
   So every (4n+3)th step is integer

[robust silo]

rlly?

is the answer D?

[unreal geode]

how did you find it

[robust silo]

4n+3=2024
n=505.25
and i just rounded to 505

i dont think i did it right

[unreal geode]

I guess that is a way too but I found it using number of terms formula:
Let's say we have another sequence that goes like: [3, 7, 10, ..., 2023] I have to find number of terms using: (an-a1)/r + 1 => (2023-3)/4 + 1 so 506

[robust silo]

ooo ok

thanks

imma close this channel now 😄

.close

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[wind mauve]

How do I get started on this question?

[marsh citrus]

@wind mauve Has your question been resolved?

[marsh citrus]

@wind mauve Has your question been resolved?

[thorn ridge]

1. To find the limit lim∞ P(Sn ≤ 7n), we need to consider the total waiting time Sn during n mornings. Since the waiting time for bus 372 is exponential with a mean of 10 minutes and for bus 75 is exponential with a mean of 20 minutes, the waiting times can be modeled as exponential random variables.

Let X be the waiting time for bus 372 and Y be the waiting time for bus 75. Then, the total waiting time Sn can be represented as Sn = X1 + X2 + ... + Xn, where Xi is the waiting time on the ith morning.

The sum of exponential random variables follows a gamma distribution. Therefore, Sn follows a gamma distribution with parameters n and λ, where λ is the rate parameter.

Since the sum of exponential random variables is gamma-distributed, we can calculate the probability P(Sn ≤ 7n) using the cumulative distribution function (CDF) of the gamma distribution.

2. To find the limit limn→∞ P(Tn ≥ 0.6n), we need to consider the number of times the bus 372 is taken during n mornings, denoted by Tn.

Given that Tn follows a binomial distribution with parameters n and p, where p is the probability of taking bus 372, we can calculate P(Tn ≥ 0.6n) using the cumulative distribution function (CDF) of the binomial distribution.

By using the given hint and the properties of exponential and binomial distributions, we can compute the required probabilities and limits based on the provided information.

This took me 5 min so please read it

[marsh citrus]

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[dawn sage]

hello! I'm working on a problem and I cannot for the life of me how to take this integral. It starts with
$$\int\limits^{\infty}_{0} \mathrm{e}^{-st} \left(\sin\left(at\right) - at \cos\left(at\right)\right) , \mathrm{d}t$$
which becomes
$$-\frac{\mathrm{e}^{-st} \left(\left(\left(a^{2} s^{2} + a^{4}\right) t + s^{3} + 3a^{2} s\right) \sin\left(at\right) + \left(\left(-as^{3} - a^{3} s\right) t + 2a^{3}\right) \cos\left(at\right)\right)}{\left(s^{2} + a^{2}\right)^{2}} + C$$
how do I then use the bounds to find the definite part?

[elfin berry]

PolloTundra | Aidan B

[hazy lion]

can you just plug in the values?

[dawn sage]

yes, how do I plug in the infinity though? it has been a while haha

[hazy lion]

you argue that $e^{-s \infty}$ is exponentially small

[elfin berry]

jan Niku

[hazy lion]

and that no other term is able to grow fast enough at this limit to counteract the decay

[dawn sage]

makes sense, thank you!

[hazy lion]

since its all polynomial and oscillating

[dawn sage]

so you just defer to the 0 term bc it's infinitely small, gotcha thank you!

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[naive stump]

Hi all,

For context, the highschool exam isn't heavy in linear algebra, its kinda stuck to just 2d and 3d vectors, mostly dealing with projectile, volumes and motion

[naive stump]

im aware of linear independence as a concept but i cannot use it in an exam that im prepping for

is there another way to solve this question without referring to independence

the textbook is a cambridge print but doesnt have worked solutions unfortuantely

[final skiff]

Combine like terms and assume lambda != l and show that they will have to be parallel

[naive stump]

ah okay via contradiction seems easy enough

that works, thx

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[sinful egret]

is this solutions

OK?

[marsh citrus]

@sinful egret Has your question been resolved?

[civic moss]

What does “algebraic sum” mean here?

[sinful egret]

i have no idea dude i saw the solution on the google

if its mean algebraic sum and the

[civic moss]

Where is this problem coming from

[sinful egret]

lenght

it shouldn't be zero

[sinful egret]

its from a book

[civic moss]

Length? These are lines, which are infinitely long

[sinful egret]

a st line is such that the algebric sum of the lengths of the perpendicular drawn on it from any number of fixed points is zero show that the line alwyas pass through a fix point

[civic moss]

This is a garbage question

[sinful egret]

this was the

orignal ques

[sinful egret]

ok but how to solve that

[civic moss]

Passes through a fixed point? What does that even mean?

[sinful egret]

bro if my pea sized mind knew that i wouldn't ask it here

🤣

[civic moss]

LOL

[civic moss]

wtf does algebraic sum of the lengths of the perpendiculars mean

Perpendiculars are usually infinitely long lines with infinite length

[sinful egret]

nah

they mean

that

a point form which a perpendicular line is drawn

to a line

[civic moss]

Any number of fixed points? wtf does that means

Horrible wording

[sinful egret]

something like this

[civic moss]

Fix points p_1, …, p_n in space. Let L be a line. Suppose L has the property that the perpendicular line segments drawn from each p_i to L have total sum equal to 0. Show that L passes through all of p_1, …, p_n.

Is this what they meant?

[sinful egret]

kinda

yes

[civic moss]

Well that’s easy: lengths are always nonnegative. so if the sum of the lengths is 0, then each of the lengths are 0. Thus each p_i must have distance 0 to the line, so L passes through each p_i

[sinful egret]

makes sence

[civic moss]

Garbagely worded question

[sinful egret]

lmao

ok thanks man

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[languid obsidian]

I just want to know if my answer is correct

[latent coral]

that line is sloping down, so you'd expect the slope to be negative, not positive

[languid obsidian]

So -4/-5

[latent coral]

-4/-5 is the same as 4/5

you didn't change the answer by making both negative

[languid obsidian]

😭 right so -4/5

??

[latent coral]

yeah

[languid obsidian]

Okay thank you Neil🫡

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[little bough]

would this be allowed to do?

cancelling out -3x+5y

[dry prawn]

it would not

[elfin berry]

y0shi

[shadow warren]

cant cancel things like that

[little bough]

huh

why

cause it’s negative?

[cunning chasm]

(2+4)/2

is equal to 3

not 4

[little bough]

i mean yeah

but for this

[cunning chasm]

anytime there is an addition/subtraction sign u cant cancel like that unless it's a factor

[little bough]

i cancelled it out and got the right Wnser

oh

is that the difference between the work?

[cunning chasm]

not really sure what im looking at tbh

[little bough]

just the cancelling part

Like in the recent example I sent you I was able to cancel it and I got the right answer but from the example I sent before this I wasn’t able to cancel it

I don’t understand the difference

[cunning chasm]

wrong method, right answer

[little bough]

oh

[cunning chasm]

prob a coincidence

it's like doing 16/64 and cancelling the 6's

[little bough]

damn

[cunning chasm]

it's right but for the wrong reason

[little bough]

okok

tysm

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[still temple]

can anyone help me with this?

[still temple]

At point Q angle is -30°
So sin(-30°) = -1/2
Cos(-30°) = cos(30°) = √3/2
Tan(-30°) = sin(-30°)/cos(-30°) = -1/√3

thank you so so much 💕

👍

[lucid zenith]

!nosols

[marsh citrus]

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.

[still temple]

👍 understandable i didn't know that it's homework

Sorry

that helped me with the other questions, it linked to one another but with different numbers :)

(i learn by getting the answer and working backwards w it lol

[lucid zenith]

Not just if it’s homework, in general you shouldn’t be answering people’s questions for them, lead them to the answer so they can do it themselves

[still temple]

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[steady mural]

could someone

[steady mural]

help with this

I tried multiple ways to get this equation

didnt get it once

[rustic zinc]

do you have work to post

[rustic zinc]

well v_n = 20 / 2^n cuz velocity decreases in half after every bounce

time for bounce cycle (up and down) is gonna be 2t = v_n / g

g is the acceleration due to the gravity

so t = (20/2^n)/(10)

then t = 2/2^n

then 2t is 4/2^n

then total time for all bounces would be like

that would be like T = 2 + 4/2 + 4/4 + 4/8 ...

same as 2 + 2(1 + 1/2 + 1/4 + ...)

parentheses is 2 so total would be

T = 2 + 2*2

T = 2+4 = 6 seconds

now for the equation part

we could use the relationship for n bounces

v_n could also be represented as v_n -1 / 2

then turn it into v_n = v_0 (1/2)^n

which is 20(1/2)^n

then v_n is also 20 * 2^-n

but 2^-n is also 6 - t_n / 4

so v_n = 20 * (6- t_n) / 4

5(6-t_n)

30 - 5t_n

v_n = 15 - 2.5t_n

[marsh citrus]

@steady mural Has your question been resolved?

[marsh citrus]

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[soft canyon]

why the P ⇒ Q (P implies Q) can be written as "P only if Q." whats the meaning of "only if ". What is the key point of difference between the "P only if Q." and "P if Q."?that is more likely a question of language to use,but that is my question.

[teal arch]

If and only if means biimplication

I think

[knotty trellis]

I think that this article explains it quite well
https://www.khanacademy.org/test-prep/lsat/lsat-lessons/logic-toolbox-new/a/logic-toolbox--if-and-only-if

"I wear a hat only if it's sunny" implies "if it's not sunny, I don't wear a hat" which is equivalent to "if I wear a hat, then it's sunny"

[marsh citrus]

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[marsh citrus]

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[echo parcel]

I'm unsure how to approach this question, the double modulus makes it seem as if casework would be really tedious. Any adivicee would be greatly appreciated

[marsh citrus]

@echo parcel Has your question been resolved?

[echo parcel]

<@&286206848099549185>

[winter sedge]

Try taking cases

Suitable cases

And plot the respective lines for those cases

[queen quartz]

So this being a polygon you could look for the vertices

[winter sedge]

For example, one of the cases would be x>2 and y>2

[echo parcel]

like just trying cases in the first quadrant assuming its just a reflection for the others?

[winter sedge]

That won't necessarily work for all questions.

I'm not sure if that'll work here

[queen quartz]

It will be because you consider absolute value of x and y

As such the inequality doesnt vary whatdver sign you put on x and y

Also assuming you get good symmetries is kinda dangerous in general imo

[winter sedge]

Ahh yes true.

[winter sedge]

Real

[echo parcel]

yea i was thinking it was a bit of a shot in the dark

oops

[fossil osprey]

||y'all could've just checked desmos||

[queen quartz]

But here its tfrue

So its a good idea

[queen quartz]

Wym

[echo parcel]

what’s the fun in that

[queen quartz]

Ye possible but depending on what exam you take

You cant always use that

[echo parcel]

yea this is a non-calculator exam

[queen quartz]

Ok so what i think works well is writing y as f(x)

First of all we only look at the edge so the inequality becomes an equality

And say we only look at the positive y part

Then you get

Hold on ive made a mistake

Notice the function isnt perfectly defined as sometimes two outputs are possible

[echo parcel]

hmm
$x-2+y-2\leq4 \$
$y\leq8+x \$?

oops that didn’t put line breaks in

[queen quartz]

Just write y as f(x)

Personnally it helps me

[elfin berry]

noah

[queen quartz]

Because ill be looking for what y is possible given x

[echo parcel]

oh that’s true

[queen quartz]

You need to keep track of every hypothesis youve used tho

Because when concluding youll need to make sure you studied every case

So again

[echo parcel]

could it be more useful to consider a simpler graph and transformation

so as not to need casework

[queen quartz]

Unless youve seen how to get an area by integration (which i have not) we will be looking at the edge to try and identify known polygons

The edge is almost always (here it is the case) the points for which this inequality is an equality

[queen quartz]

If you find one sure, but if we try and answer this question im not sure how that will help us specifically

[queen quartz]

This already gives us quite some info

But we can simplify further

We see the inequality doesnt vary whatever the sign of x and y is

So we can study the case x>=0 y>=0 and the rest will be symmetric

[echo parcel]

yea that’s true

so if we make it an equality and just see what happens

and then perhaps change some parts that don’t fit exactly

[queen quartz]

Making it an equality is not random, it is specifically because it mathematically gives us the edge of that polygon

One thing i cant stress enough

We have to be precise

You cant add or leave out any area

So notice how what we ve only simplified one thing because there is a symmetry and the other is only looking at the edge

Which still allows us to identify the entire shape

So

We now have

[echo parcel]

so if x and y are both positive

y=-x+8

[queen quartz]

Are you sure

If x=1

[echo parcel]

yea that’s true

[queen quartz]

So y =7

?

[queen quartz]

Please show me how you get there

[echo parcel]

well if we take the positive of both modulus

x-2+y-2=4

[queen quartz]

Why are we allowed to do that

Or are we making an additional hypothesis ?

[echo parcel]

oh are we assuming x is greater than 2

and the same for y

[queen quartz]

Exactly

So we have to distinguish cases

Now we are making for

When its true for both x and y, only for x, only for y, for neither of them

So

We could write |x-2| as say a

Then lets see what values y can take based on a

Without any hypothesis besides x and y positive

[echo parcel]

|y-2|<=4-a

[queen quartz]

Yes

So what values can y take given an a

Now we have to make different cases

[echo parcel]

so are the different cases being y<2 y>2

[queen quartz]

You should rather do it based on a

You can see that y-2 cannot be lower than -2

Because y>=0

[echo parcel]

as we are only looking in the positive for both x and y?

[queen quartz]

Yes

But a is positive regardless because its a modulus

[echo parcel]

okay so if its -2 then a<=6

[queen quartz]

Youll see that right now we dont care about the exact expression of a

[queen quartz]

What is -2

[echo parcel]

y-2

so when y is 0

[queen quartz]

What i meant is

Express y based on a

Imagine you know the value of a

Then what values can y take

If a<2

Then 4-a >2

[echo parcel]

anywhere from 6 to -2?

[queen quartz]

We only look at y>=0 and y satisfies the equality above

[echo parcel]

sorry im a bit lost where did this come from

[queen quartz]

No problem lets backtrack a bit

So the idea would be

Given x let s express y

So we are trying, for a given x, to give all of the values y could take, and only values y can take

[echo parcel]

yep

[queen quartz]

So y has to :
Be positive
Satisfy the equality

Now we yave

[echo parcel]

given x is positive right?

[queen quartz]

|y-2| = 4-|x-2|

[echo parcel]

yep

[queen quartz]

We write |x-2| as a

[echo parcel]

|y-2| = 4 - a

hmm so do we want to say what the max and min values of a would be in this situation?

well if x is 0

a is 2

so |y-2| = 2

as y is positive it must be 4

if a is 4, meaning x is 6. then y is 2

still not too sure how i should expand from this like i could just start picking numbers but im not too sure which ones i would pick

[marsh citrus]

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[marsh citrus]

@echo parcel Has your question been resolved?

[marsh citrus]

@echo parcel Has your question been resolved?

[marsh citrus]

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[sharp burrow]

.close

[marsh citrus]

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[uncut beacon]

$\left|6-5x\right|>x^2$

[elfin berry]

DapperJaguar197

[uncut beacon]

Can someone help me with this inequality?

[brave marsh]

Two cases.
If $6-5x \ge 0$, then you have $6-5x > x^2$, which you can solve by factoring and checking when the factors have the same sign (remember $xy > 0 \Leftrightarrow x,y > 0 \text{ OR } x,y < 0.$).

Then the other case is when $6-5x < 0$ and you have $5x - 6 > x^2$, solved in the same way.

[elfin berry]

Azyrashacorki

DapperJaguar197

[brave marsh]

If $6-5x \ge 0$, then $|6-5x| = 6-5x$. \
If $6-5x < 0$, then $|6-5x| = -(6-5x) = 5x-6$.

[elfin berry]

Azyrashacorki

[brave marsh]

That's the definition of the absolute value.

[uncut beacon]

$$I got (x+6)(x-1)<0. I'm not sure what to do next.$$

[elfin berry]

DapperJaguar197

[brave marsh]

$xy < 0$ if, and only if, $x > 0, y<0$ or if $x < 0, y > 0$, i.e. the signs of the factors $x$ and $y$ differ.

[elfin berry]

Azyrashacorki

[brave marsh]

You can apply this to your inequality,.

[uncut beacon]

Idk how 😢

[brave marsh]

You need to consider (x+6) < 0, (x-1) > 0 and then (x+6) > 0, (x-1)<0.

[uncut beacon]

I have no idea on how to do that. Is there like a method layout I can follow?

[brave marsh]

Surely you can solve x+6 < 0 and x-1 > 0?

[uncut beacon]

I can but Idk how to apply this to the main inequality. x<-6 and x>1

[brave marsh]

That means that is x<-6 and x>1, the inequality holds. Does that correspond to any interval?

[uncut beacon]

Not sure. I would guess no

[brave marsh]

Indeed. No x's satisfy both of those.

[brave marsh]

There was another possibility here. x+6 > 0 and x-1 < 0

Can you see if this corresponds to any interval?

[uncut beacon]

-6<x<1

[brave marsh]

Good!

So that's one part of the solution.

And since $6-5x \ge 0$ means that $x \le \frac{6}{5}$, we don't have to restrict this interval. (Remember this was the initial assumption we made to simplify the absolute value).

[elfin berry]

Azyrashacorki

[brave marsh]

Now you need to do the same thing but for $6-5x < 0$.

[elfin berry]

Azyrashacorki

[brave marsh]

As per this.

[uncut beacon]

I get x>6/5

[uncut beacon]

Is there meant to be a case where I consider xy<0?

[brave marsh]

This will be useful. Now you should do the same thing we did but with 5x - 6 > x^2.

[uncut beacon]

I get (x-3)(x-2)<0

[brave marsh]

Good. Now you need to check when the factors have different signs like the last time.

So x-3 > 0 and x-2 < 0.

See if that gives an interval.

And then x - 3 < 0 and x -2 > 0

[uncut beacon]

x>3 and x<2 which wouldn't work. And x<3 and x>2 which is just 2<x<3

[brave marsh]

Ok. So you get the interval 2<x<3.

Do you need to change anything to ensure that x > 6/5?

[uncut beacon]

2>6/5 so I guess no

[brave marsh]

You're right

So you have both intervals then

(-6,1) and (2,3)

[uncut beacon]

Would that be the final solution to the inequality then?

[brave marsh]

Yes.

[uncut beacon]

Great. Thanks a lot for your help

.close

[marsh citrus]

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[spiral plank]

quick question about ftc: is this true

[spiral plank]

like if a smaller number is the upper bound do you just do the negative of the antiderivatives when doing the subtracting part

[sweet pawn]

no

it should be F(a) - F(b)

[spiral plank]

whoops that is what i meant

[sweet pawn]

it's always just upper bound - lower bound

[spiral plank]

what do you do when the lower number is the upper bound though

[sweet pawn]

doesn't matter

[spiral plank]

do you just add a negative to the integral and then flip the bounds

[sweet pawn]

yeah

[spiral plank]

ah ok that’s what i was trying to get at

.close

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[fading furnace]

does anyone know why the value between the functions is equal to half the area bounded by r=theta (from 0 to pi)

[fading furnace]

would the parabola be measured from both sides? like in the thing on the right would it be the same on the other side of the y axis

i got x^2/2 from integrating r = theta by itself and im trying to find something that resembles the value of a polar form equation integrated without using r^2/2

<@&286206848099549185>

[marsh citrus]

@fading furnace Has your question been resolved?

[marsh citrus]

@fading furnace Has your question been resolved?

[marsh citrus]

@fading furnace Has your question been resolved?

[idle ridge]

I'm not sure what you're asking. Are you asking why the area under a polar curve is given by
[\int_a^b \frac12 r(\theta)^2,\dd\theta?]
It's because the formula for a circular arc of angle (\Delta\theta) and radius (r) is (\frac12 r^2,\Delta\theta).

[elfin berry]

cramERIC–rao_bound

[idle ridge]

You can get this from the normal area of a circle formula pretty quickly: the area of the whole circle is (\pi r^2) and then if you only want an angle of (\Delta\theta) out of that, the angle of the entire circle is (2\pi), so the proportion of the circle the arc takes up is (\Delta\theta/(2\pi)), and
[\frac{\Delta\theta}{2\pi}(\pi r^2) = \frac12 r^2\Delta\theta,] like you want.

[elfin berry]

cramERIC–rao_bound

[idle ridge]

@fading furnace does that answer your question?

[marsh citrus]

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[robust silo]

6. A cylinder has radius 10 and height 20. A spiral with constant slope starts at the bottom of
   the cylinder, wraps exactly twice around the cylinder, and finishes at the top of the cylinder.
   What is the length of the spiral?
   (A) √(400π^2 + 400) (B) 2√(400π^2 + 400) (C) 2√(1600π^2 + 25)
   (D) √(100π^2 + 100) (E) 2√(400π^2 + 100)

[robust silo]

<@&286206848099549185>

[leaden monolith]

What have you tried

[marsh citrus]

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[robust silo]

o i solved it lol

but thanks

[marsh citrus]

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[heavy willow]

i got the period and frequency wrong. pls tell me what i did wrong.

[half kettle]

I believe that since the period is the length of one complete cycle of the function, the period can be found by identifying the distance between repeating points

Given this what are our given points?

Reading it u get the concept but u just basically mixed a bit of stuff up

[marsh citrus]

@heavy willow Has your question been resolved?

[heavy willow]

the distance would be 3pi in that case. but the answer the software gave me was 2pi

i am not sure how 2pi was caluclated

[half kettle]

Huh that's weird is it possible the software could be wrong or?

[heavy willow]

idk, i would have to ask my teacher

i really don't get how 2pi was calculated. i'll just ask in office hrs

[half kettle]

Okay then srry if I couldn't help much

[heavy willow]

its ok

[marsh citrus]

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[spice frost]

hey is anyone available to help

[marsh citrus]

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[wary bluff]

How would you find the closed form expression of something like this

[marsh citrus]

@wary bluff Has your question been resolved?

[nocturne hinge]

Generating functions are a beauty, and a mess.

[wary bluff]

what do generating functions have to do with this tho

oh mb i thought u were talking about mgfs 🤦

do i just keep guessing for these type of questions? im not sure how to come up with them

[nocturne hinge]

Uhhh shoot generating function are pretty hard to understand, if you just want to use them here’s an example vid that does a problem like yours https://youtu.be/Pp4PWCPzeQs?si=b9U034dFQ1LjWR4X

Basis of them is creating an infinite polynomial based on the sequence and canceling the infinite part using the recurrence relation given

[wary bluff]

wth thats crazy

alr lemme try

[nocturne hinge]

That gives you the generating function which then can be expand using Taylor series, which then you can take the terms from there (absolutely tedious but it works for I think any recurrence)

[wary bluff]

calculating the first few terms gives us

3, 7, 15, 31, 63, 127,

fn - 3f(n-1) - 2f(n-2) = 0

        f = 3 + 7x + 15x^2 + 31x^3 + 127x^4 ...

-3xf = 9x + 21x^2 + 45x^3 + 93x^4 ..
+2x^2f = 6x^2 + 14x^3 + 30x^4 ..

= 3 -2x + 64x^4 ....

shit this is kinda aids how many values do i gotta callculate?

[nocturne hinge]

Y’know, you could’ve just looked at the numbers and figured out something (which is what I’m fearing your supposed to do)

[wary bluff]

I gotta keep calculating till I get 0?

[wary bluff]

😭 oh

how would I do it

[nocturne hinge]

*+2x^2 double negative

[wary bluff]

?

wym

oh shit

coz fn - 3f(n-1) + 2f(n-2) = 0

[nocturne hinge]

(Wait what class is this

[wary bluff]

computer science

"Theory of Computation"

[nocturne hinge]

Damn

Wrong way of solving then

The numbers generated are 3,7,15,31,63,127,keep calculating and you get like 1023,4047

These numbers ring a bell(almost)?

[wary bluff]

nah

[wary bluff]

how did you know that at one point those numbers will appear?

[nocturne hinge]

8191, 16383

[wary bluff]

bro how are you coming up with these numbers

[nocturne hinge]

By hand/I may have gotten it the hard way already

[wary bluff]

howww

put me on the method

🙏

[nocturne hinge]

It’s The one above 💀

[wary bluff]

3,7,15,31,63,127, 255, 511, 1023, 2047, 4095

damn you were wrong about 4047

but even after calculating sm terms I dont see a pattern 😭

2^i - 1

[nocturne hinge]

Wth

Oh wait almost

[wary bluff]

yuhhh thats it

lets fucking goooo

https://tenor.com/view/black-guy-whtie-and-black-cry-praying-looking-up-gif-553441503240360994

[nocturne hinge]

WHY USE “I” tho???

[wary bluff]

i have another question though I wasted like 30 mins on it

[wary bluff]

lol mb meant n

its like 6 am im tripping rn

but help me with this too pls one sec ill attach ss

0 = 1
1 = 0 = 1
2 = 00, 11 = 2
3 = 000, 110, 011 = 3
4 = 0000, 1100, 0011, 0110, 1111 = 5
5 = 00000, 11000, 00011, 00110, 01100, 11110, 01111 = 7
6 = 000000, 111111, 111100, 001111, 011110, 110000, 011000, 001100, 000110, 000011 = 10
7 = 0000000, 1111110, 0111111, 1111000, 0111100, 0011110, 0001111, 1100000, 0110000, 0011000, 0001100, 0000110, 0000011 = 13

I calculated like uptil H(7) and i cant notice a pattern 😭

[nocturne hinge]

AH

[wary bluff]

u alrdy realized

been like 10 secs

[nocturne hinge]

Also you should check, I can say def it’s almost

[wary bluff]

ye i checked for the first 12 terms i calculated its correct

[wary bluff]

help me with this tho pls 🙏 😭

This one got me going crazy ngl

[nocturne hinge]

Man its supposed to be one forum one question😭
My best guess is it’s more recurrences

[wary bluff]

allow me

🙏

so we have

1, 2, 3, 5, 7, 10, 13, ...

wait isnt this fib seq but shifted by 1?

eh acc idk

[nocturne hinge]

So, basically, you wanna take the terms from the sequence before, and add a 0 to the front/back
Same thing with 1s, but be careful, needs to stay even
Dunno how to account for edge cases, also it’s rly late for me cuz I didn’t realize I’m gonna be typing this long 😐

[wary bluff]

shit alr ill try

ty

.close

[marsh citrus]

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[nocturne hinge]

🫡

[wary bluff]

brah im an idiot i literally calculated the values wrong im p sure its straight up a regulae fib seq

🤦‍♂️

[nocturne hinge]

Yea sorry for typing when this is closed just wanted to add an extra caveat

Make sure to shift it so it gives 3 when n=0 for the first problem.

Currently it’s n=2 that gives 3.

[wary bluff]

[marsh citrus]

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[naive topaz]

I have problem understanding the solution of (V), how did they transform A(k) into a polynomial of (1-k)?

A(k) is the area shaded in the graph*

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[naive topaz]

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[limpid bison]

@jagged moth

[marsh citrus]

Please don't occupy multiple help channels.

[jagged oracle]

Yo

Yo

Me there

@limpid bison

Yo

[limpid bison]

wait its under my name now tho

just look at the math help channel

(available

and ask ur question there

.close

[marsh citrus]

Channel closed

Closed by @limpid bison

Use .reopen if this was a mistake.

[jagged oracle]

Bro I came for helping you bro

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fickle shell]

integration of 1/a^2+x^2 is (1/a)*arctan(x/a)+c, right?

Well a is 1/sqrt(2), no?

[teal arch]

Correct

[quaint elm]

is it? I don't think those are equal

[teal arch]

In these situations, always make coefficient of x^2 1

Then apply formula

It makes sure you are not making an error

[fickle shell]

I multiply by 1/2 to do that right

[teal arch]

Yup

[fickle shell]

And then its 1/sqrt(2) because its a^2 in the formula, no?

[quaint elm]

you multiply what by 1/2

you can't just decide to multiply things by 1/2 for fun

[teal arch]

Multiply and divide

[fickle shell]

The denominator

1/(2x^2 + 1)

1/(x^2 + 1/2)

So a = 1/sqrt(2)

[quaint elm]

those are not the same fraction

[fickle shell]

Ah yeah sry the 1/2 is outside the integral

[quaint elm]

oh well now we are getting somewhere

[fickle shell]

Ah I think I see the issue

Yep I got it

Thanks!

.close

[marsh citrus]

Channel closed

Closed by @fickle shell

Use .reopen if this was a mistake.

[fickle shell]

.reopen

[marsh citrus]

✅

[fickle shell]

Is there a handy formula like *integration of 1/a^2+x^2 is (1/a)arctan(x/a)+c for arcsin questions like this?

I know 1/sqrt(1-x^2) = arcsin(x) but something with a's would be nice hehe

[fierce jewel]

substituting x in terms of sin/cos theta so that u remove the root term in denominator and , replacing dx through the relation of x in sin/cos theta terms

[fickle shell]

Wait, tan?

[teal arch]

Do the same thing I told you

[fierce jewel]

my bad

[fickle shell]

npnp haha

[teal arch]

Remove the coefficient of x^2

[fickle shell]

Ugh its really messy 😝

Is there no handy formula :3

[teal arch]

You will have to u substitute otherwise

[fickle shell]

Oh god

[teal arch]

Not fun

[fickle shell]

Ok

[fickle shell]

Yeah no 😂

[teal arch]

Haha

[fickle shell]

So multiplying the denominator by sqrt(1/4) ?

1/sqrt(1-4x^2)

(1/sqrt(1/4)) 1/sqrt(1/4-x^2)

[fierce jewel]

don't do this, you are complicating ur working

your main goal is removing the root

and the function u have inside the root is 1-4x^2
since in these types we use the knowledge of trigonometric identities
what comes to your mind when u see this
probably 1 - sin^2 theta or 1-cos^2 theta
and since u know that they equal to cos^theta and sin^2 theta , the root term will get removed as well

so just try to substitute 4x^2 part

[fickle shell]

Which root sry?

[fierce jewel]

the root in denominator

[fickle shell]

But we want it no?

for 1/sqrt(1-x^2) = arcsin(x)

[fierce jewel]

do you know the derivation behind this formula?

[fickle shell]

No

[fierce jewel]

yea u need that knowledge

[fossil osprey]

You could create it yourself, you just need to figure out how to solve the integral in general

[fierce jewel]

yeah

its generally recommended to create it yourself rather than remembering formulas like these for integration

[fickle shell]

But why do we want to get rid of the square root in the denominator

[fierce jewel]

i can show u how this formula comes

then u will understand

[fickle shell]

Ok

Yes please

[fierce jewel]

give me a min

[fickle shell]

No rush

[tame dagger]

We want to get rid of the sqrt because there is no adequate direct substitution

We do this by using one of the 3 main trig sub formulas we can derive via the identity $\sin^{2}{(x)} + \cos^{2}{x} = 1$

[fickle shell]

Wait what

I'm even more confused

[fierce jewel]

let me explain

u got the reason cat collective gave right

we don't have a direct formula for integration of sqr root

so we want to get rid of it

we have 3 basic trigonometric identities

sin^2 x + cos^ x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = cosec^2 x

[fickle shell]

Ahh

[elfin berry]

The Cat Collective

[fickle shell]

I remember those I think

[fierce jewel]

usually in these types of integration of questions

[fierce jewel]

inside the sqr root, we usually get a similar looking function to these but in normal variable terms

to tackle these problems, we simply assume them to be equal to a trignometric function in theta terms

[fierce jewel]

for example here

i assumed x to be sin theta

[tame dagger]

$$
b^{2}x^{2} + a^{2}
\
\
a^{2} - b^{2}x^{2}
\
\
b^{2}x^{2} - a^{2}
$$

What

A

Goddangit

[elfin berry]

The Cat Collective

[fierce jewel]

since we have x = sin theta -----1
differentiating it
we get dx = cos theta dtheta------2

now if u substitute both of these

[tame dagger]

OH CMON

[fierce jewel]

look what we get

inside the root , we are getting cos^2 theta, so that removes the root , and
finally we get (cos theta / cos theta ) dtheta

which is just dtheta

and you know integration of dtheta

which will be theta + C

[tame dagger]

And to go back from theta to x you can either draw a right triangle and find the sides or you can use inverse trig functions

[fierce jewel]

now theta can be found using 1
x = sin theta
arcsin(x) = theta

hence ur answer is arcsin(x) + C

[fickle shell]

This is very confusing, I need to reread everything like 10 times, bare with me

[fierce jewel]

take your time

[tame dagger]

Trig sub is pretty weird the first time you try it

[fierce jewel]

very true

it takes a lot of time to get hang of it lol

[tame dagger]

Yea we still struggle with back substituting at the end a lot

[fierce jewel]

and i would recommend not to remember formulas for these types of trig subs

because a simple change gives a different answer

so formulas never work

[fickle shell]

It just keeps getting worse and worse

[tame dagger]

One except we'd say it to memorize the integrals of arcsec(x), arctan(x) and arcsin(x)

[fierce jewel]

yes

[tame dagger]

Deriving these every time is a mild pain in the butt imo

[fickle shell]

When you say you 'assume it to be sin theta' you mean you just sub it in? Or does assuming mean a different thing

I have never really worked with theta before, like I know what it is but never used it like this...

[fierce jewel]

basically

trigo functions give us ratios

and all trigo functions have a range

so when i say assume

i am basically saying that x is equal to some angle theta , whose sin theta is equal to x

eg.

we have 1/2

i can say that 1/2 = sin theta for some value of theta

and we know that sin pi/6 = 1/2

so we are using this idea to our help

[tame dagger]

Normally in maths, assume means that you're substituting something in or having implied bounds or such

Actuslly my example as kinda crap, so deleted it

[fickle shell]

Do you have to use theta? Could you just as easily use 'a' or something? Sry if this is a dumb question

[fierce jewel]

yes yes

we generally use theta for angles that why i said theta

[fickle shell]

Yes you have to use theta? Or yes you can use a? Or yes its a dumb question 😂

[fierce jewel]

you can use anything u want lol

[fickle shell]

Ah cool

[fierce jewel]

sin a or sin b anything lol

[fickle shell]

Okok, I was overthinking that it was a whole new thing

Thats nice

sry im still reading through this its taking a while hahaha

[fierce jewel]

sure take your time

[fickle shell]

Ok...

I think I maybe get it, possibly

[fickle shell]

But I dont see why we need to remove the square root still sry

[fierce jewel]

i mean you could try to do it without trig
using general substitution and etc , and you would realise that there is no such exact way to tackle the problem
we are always stuck somehow because of the square root
using trigo substitution helps us remove the square root function making it easier to solve

its just an algebraic method to solve it

[fickle shell]

Mb after I will do it without trig but one thing at a time haha

[fickle shell]

Whats the first step to solve this

[fierce jewel]

substituting x in such a way that you have a function that looks similar to these identities

so here

we have 1-4x^2 , we want it to look like 1-sin^2 a

ie you can say that 4x^2 = sin^2 a
which means 2x = sin a

now follow the steps like the previous one and you should get to the answer

[fickle shell]

Damn ok it makes sense when you explain it but idk how I will think of this in an exam

Like I wouldnt even notice I could change 4x^2 = sin^2(a) into 2x=sin(a) lol

[fierce jewel]

yeah

you are doing it for the first time

its hard

just practice some solved problems from some book, first try solving it on your own , then look at the solved solution

you would get hang of it then

[fickle shell]

So I sub in sin(a)/2 for x into 1/sqrt(1-4x^2)?

Or am I doing it all wrong

[fierce jewel]

yes

[fierce jewel]

correct

[fickle shell]

1/sqrt(1-4x^2)

1/sqrt(1-4(sin(a)/2)^2)

1/sqrt(1-4(sin^2(a)/4))

1/sqrt(1-sin^2(a))

Ummm

[fierce jewel]

use sin^2 (a) + cos^2 (a) = 1

[fickle shell]

Ah yeah

1/sqrt(cos^2(a))

1/cos(a)

then just integrate?

[fierce jewel]

correct, but remember you have dx, you gotta change that as well

dx/cos(a)
you can't integrate this

you need dx in terms of a

[fickle shell]

Ah yeah forgot about a

Hmmm

[fierce jewel]

remember what you assumed x to be

[fickle shell]

All the formulas that have a have a sin or cos in them which is annoying for integration, no?

[fierce jewel]

yea

trigo sub is what makes integration harder

[fickle shell]

So do I need U-sub after all?

To finish this off

[fierce jewel]

yea , you need to change dx

we have 2x = sin(a)

differentiate this

[fickle shell]

1/cos(arcsin(2x))

[fickle shell]

Differentiate this?

[fickle shell]

Or this

Sry if dumb question

[fierce jewel]

logically answer, which one looks easier

[fickle shell]

This but I dont see why were differentating it?

[fierce jewel]

just differentiate and tell me what u get

[fickle shell]

sin(a)-2x

-cos(a)-(x^2)

[fierce jewel]

you are integrating

[fickle shell]

Oh lol

[fierce jewel]

lol

[fickle shell]

Wait it wasnt correct either

Its just -2 right?

[fierce jewel]

the integration?

[fickle shell]

Differentiation

Cause its with respect to x no?

[fierce jewel]

nope

you have 2x =sin(a)

differentiate it with respect to x

so on lhs, you have 2

[fickle shell]

rhs = 0?

[fierce jewel]

on rhs you have
dsin(a)/dx
multiply and divide by da

[fickle shell]

1*

[fierce jewel]

so u get (dsin(a)/da)(da/dx)

[fickle shell]

wait no it is 0

[fierce jewel]

nope

[fickle shell]

💀

[fierce jewel]

sin(a) isn't a constant with respect to x

[fickle shell]

Im confused

its really not 0?

[fierce jewel]

nope its not 0

its actually which is cos(a) da/ dx = 2
so cos(a) da = 2 dx

this is what u get

[fickle shell]

Why da? It was with respect to x?

[fierce jewel]

yes it was with respect to x

but u can differentiate sin(a) with respect to dx

so we multiply and divide it with da

and you can say , its kind of chain rule,
sin(a) differentiation would be cos(a) , but we have 'a' inside, whose differentiation would be da/dx
so 2 = cos(a) da/dx
2dx = cos(a) da

[fickle shell]

Ok I think I get it

My brain is on fire from so many new concepts 😂

I need a break after this question lol

But ok