help-42

[thick sinew]

whenever you have any two terms

which share a factor

you can exclude it from both using the bracket

e.g. 5*x² + 7*x²

both terms contain x² as a factor

so we can exclude it:

x² * (5 + 7)

or for instance a*b*c + c*d*e

where a,b,c,d,e are some variables

both terms contain c as a factor

so we can exclude it:

c * (a*b + d*e)

likewise with 2^n*16 - 2*2^n

both contain 2^n as a factor

so we can exclude it:

2^n * (16 - 2)

[remote mural]

Oooooo

I finally understand

Wait a Second i'm gonna do It and show you

[spice phoenix]

you did a much better job of explaining it lol

[thick sinew]

we could also exclude the factor 2 as well (although it would be useless)

2^n * 2 * (8 - 1)

you can factor out as you wish

you just need to remove that factor from the terms

we could factor out 17 too (useless again, but just for showcase)

2^n * 2 * 17 * (8/17 - 1/17)

you can even factor out from more than two terms at once:

7a + 8a - 5a + 2a

here we could factor out a

a * (7 + 8 - 5 + 2)

[remote mural]

So

Now, do I cut 2^n in the numerator for the 2^n on the denominator?

[thick sinew]

yes

which leaves you with (16-2)/16

= 14/16

= 7/8

[remote mural]

Aaaaaaaaaa, youre incredible

Thank you so much

[thick sinew]

np 🦇

[remote mural]

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[atomic belfry]

am i allow to ask physics question?

I have done like a lot of force and motion questions but this question questions me

[blazing coyote]

Do you know vector addition ?

[calm coral]

@atomic belfry Has your question been resolved?

[atomic belfry]

ik parralegrom method

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[remote mural]

seems very easy but i have no clue how to do it ngl

[tight trellis]

!occupied

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[proper panther]

Could someone help explain the concept of independent events to me? This question asks which pair is independent

[proper panther]

I think independent means that the probability of getting A will not affect the probability of getting B

[tight trellis]

Yes pretty much, it means that if you do get A, then the probability of getting B as well is exactly the same as it would have been if you did not get A

[proper panther]

So how come B would be a dependent event? Is it cause in order to get the two exact same, we would rely on the first being equal to the second?

[tight trellis]

Expressed, this can be shown as $P(A n B) = P(A) \cdot P(B)$

[potent lotus]

DJJ

[tight trellis]

The n is an and symbol, I don't know the tex for it

[proper panther]

its ok

[proper panther]

Do you know why B would be a dependent event in this problem?

[tight trellis]

Well if you don't get a 5 on the first die, the probability of getting 2 5s drops to 0

[proper panther]

But then how come A is not dependent then?

[tight trellis]

If you roll a 4 on the first die, the chance of not getting a 4 on the second is 5/6. If you roll a 3 on the first die, the chance of not getting a 3 on the second is 5/6

So nothing changes since you're always just "avoiding" 1 number

[proper panther]

Oh so basically, since the probability still exists you can technically say they are independent

but in the case of 5,5 , you would NEED to roll a 5 on the first try otherwise the prob is 0

💀 reading probability problems is so difficult for me sorry

[amber bolt]

you can't call any of A B C independent on its own, it's like asking what makes a number equal and if 83 is an equal number.

[proper panther]

Sorry the question asks which pair is indepedent , it got cut off

[amber bolt]

i see

[proper panther]

I know A and C are independent, but I want to know what makes for example A and B or B and C dependent

[amber bolt]

events are dependent if one ecent "helps" another, or "hurts" it
A B are dependent, B helps A to the point where A is guaranteed

you could also tell that A helps B

[proper panther]

How so?

Is there a formula or something I could use to prove?>

[amber bolt]

the chance of B is 1/36, but if A happened, then it's 1/6

it changed

[proper panther]

So P(B) = 1/36
P(A) = 5/36

right?

[amber bolt]

sorry i misread A

it's the other way around, B "hurts" A

the "formula" is you divide A&B / B

[proper panther]

Oh wait, so we're checking to see if probability of A and B can happen. But since B requires you to have 5, 5, then obviously A cannot happen since it requires the two to be different

[amber bolt]

two dice are different and the outcome is 5,5 has probability 0

0 / (1/36) so P(A|B) = 0

[proper panther]

Wait how did you get a zero?

[amber bolt]

and P(A) isn't 0, so having |B changed it

you also did

you said "cannot happen"

[proper panther]

Bruh

Thank you lol I can see it now @amber bolt

[amber bolt]

so that's exactly how itt works, independent events is when the probability for A stays the same before assuming B and after

or the other way around, it's always symmetrical

only need to check one direction

[proper panther]

I see so in the case of A and C

C requires that you get a 1 on the first roll, but A requires the two dice are different so the probability is there still

[amber bolt]

yes, so the calculation would help prove it

[proper panther]

So it would be P(A|C) right?

[amber bolt]

sure

P(A&C) /P(C)

[proper panther]

1/1296 / 1/36

and I see that equals P(A)

[amber bolt]

it's 5/36 i think for P(A&C)

[proper panther]

O right

5/1296 / 1/36 = 5/36

which is P(A)

[amber bolt]

i don't know what's 1296

[proper panther]

P(A&C) = 5/36 * 1/36

P(C) = 1/36

[amber bolt]

no

[proper panther]

Oh did I do it wrong?

[amber bolt]

multiplying is a shortcut, you can;t use it before you know the events are independent

[proper panther]

Oh so in these problems I cant use those formulas then

[amber bolt]

right

you need to count outcomes

there's 5 ways that C happens and A happens, so it's 5/36 because of that

(5/36)/(6/36) = 5/6

and on its own A is 30/36 = 5/6

[proper panther]

You have to manually count all of them? Theres no way to shortcut it ?

[amber bolt]

no

[proper panther]

T__T

[plush marten]

¿Necesitas ayuda para ser sigma?

[proper panther]

@amber bolt ty for your help good sir!

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[onyx quiver]

what did i do wrong to lose a mark?

[turbid swift]

-36/-18 should be positive 2

[soft ferry]

your teachers highlighted it pretty well

[onyx quiver]

but i didnt rearrange it wrong?

[turbid swift]

you need brackets around it

[onyx quiver]

ohhh ok cuz he circled the -2

[turbid swift]

-(2x-2)

[onyx quiver]

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[remote mural]

Did I do the k-map correctly?

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@remote mural Has your question been resolved?

[remote mural]

Honestly have no idea if i filled out the k-maps correctly

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[remote mural]

.reopen

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✅

[remote mural]

<@&286206848099549185> HELP

[wintry echo]

In the first example, where you put 1, would they be the binary representations of the numbers inside m????

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@remote mural Has your question been resolved?

[remote mural]

I have no idea

yes?

[wintry echo]

How did you put the 1's in the table?

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[umbral lark]

why is there dy/dx after the derivative of sec (y) (sec(y)tan(y) ) and not also with the derivative of tan? i honestly just cant understand why it wouldnt be only when we take the derivative of -y

[inner zealot]

Give me one moment.

This should clear things up.

[inner zealot]

Take a look at that. If you have any questions, feel free to ask.

[terse saddle]

because when deriving trigonometric functions u multiply by the derivative of whats inside. in case of tan x u multiply by 1, and in sec y u multiply by dy/dx

bc d/dx(x)=1 and d/dx(y) = dy/dx

[umbral lark]

ahh right, i guess it makes sense now, thanks!

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[solid steppe]

yo can someone help me with a exam i have soon

[solid steppe]

i wonder which question i should focus on where i have limited time left to focus for the exam

would be rly wonderful if someone helps me

[ornate geyser]

what language is this

[solid steppe]

swedish

[solid steppe]

i need 9 points of 30 points for the exams

and right now i am focusing 1 and 2

the first part are about Complex numbers and algebraic equations, Vector spaces and bases

and the second part is about Eigenvalues, eigenvectors and mapping matrix, Orthogonality , Eigenvalues cont. (orthogonal diagonalization and quadratic shapes), Differential equations

as u can see on the exam pages

/help

!help

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[solid steppe]

anyone plz?

?

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[solid steppe]

.reopen

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✅

[solid steppe]

<@&286206848099549185>

@kind wagon yo bro can u help me?

[kind wagon]

i wish i could help but im not that advanced in math yet

[solid steppe]

uhhh okay thx anyways 🙂

[wraith robin]

which q

[solid steppe]

i have a test bro and i need 9/30 points and i wonder which question i need to focus on to complete the exam

right now i am focusing on question 1 and 2 but i wonder if there is other question i should focus on i have only tomorrow left to practice

[wraith robin]

5

[solid steppe]

on both the exams?

[wraith robin]

yes

5 and 6 are similar and not too difficult

[solid steppe]

okay

thx

here bro i have a question u see the first part

where he starts to do [-3 + i)^4 =...

i dont understand the rule behind it

[solid steppe]

is there a particular rule on it or what is it called?

[wraith robin]

not a particular rule, just solving second order homogeneous/inhomogeneous odes

[wraith robin]

binomial theorem

sorry gtg now

[solid steppe]

okay np thx

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[limber igloo]

I have no idea where to start at 😭
I tried to like make the two big triangles congruent but I'm just more confused now

[calm coral]

@limber igloo Has your question been resolved?

[limber igloo]

<@&286206848099549185>

<@&286206848099549185> pleaseeee 😭

[timid oracle]

What about making the small triangles congruent?

[limber igloo]

Well that was my thought process but then it just stopped

Wouldn't have made any sense

[timid oracle]

why not?

[limber igloo]

I don't know 😭

I just couldn't go further from that

[timid oracle]

hint is that its AAS

[limber igloo]

Well I wouldn't have a problem proving the two triangles congruent since I can just reason that a parallelograms diagonal makes two congruent triangles

I can also do SSS=SSS, since we have two sides congruent by given and AC by reflexive

I don't know how to go further

[timid oracle]

if the two triangles are congruent then EG=GF

so that just automatically completes it

[limber igloo]

How

Would that be a part of the two triangles?

∆ADC and ∆ABC

[timid oracle]

im talking about triangles AEG and CFG

[limber igloo]

Oh that's why

That's gonna be a long process than, cuz I can see how your thinking

Alright I'll do that, thank you ;')

That's all I have, not sure if the reasoning of 4 is correct

[limber igloo]

And I give up

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[quiet island]

Can somebody help me find a ODE for a IVP with only one solution?

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[remote mural]

i need help iwth number #3

i do not know how to get started

ik the concepts and stuff

[flat laurel]

@remote mural

find y=3 in the Y- Axis, then pick the corresponding x values from the x axis

or maybe draw horizontal line y=3 and see what x values it intersect the curve

[remote mural]

Ohhhh

so y=3 corresponds with x=-4

and that is my y,x

[flat laurel]

No

Find which inputs will give you the output 4

[remote mural]

ohhh okay

i got it now

thanks!

[flat laurel]

x=5 is one answer

find the other one

[remote mural]

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[remote mural]

would i do 360 + (-35)

and get 325 to find my coterminal

[swift laurel]

yes

[remote mural]

thanks!

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[steel tide]

Caab we solve this question by options?

[solar spruce]

what do you mean

[steel tide]

By putting values of each option

Is this possible?

Discard method

Hey charbit ❤️

Please guide me

I guess c and d not possible

[upper sparrow]

(while you could try each option, it's probably easier to just generate the basis directly, I'd think: but you could check that each matrix in the options work, and that each set is linearly independent - if you have both, then that's your basis, if you have neither one nor the other, it isn't what you want)

[upper sparrow]

Explain why you think so?

[steel tide]

I trued to add them

Option D first matrix 2-1+2=\0

For c last option is not satisfying

Option a is not lineary independent

B ❤️

[upper sparrow]

it is linearly independent, there's another problem with it

[steel tide]

Sorry a is lineary independent

Determinnant is not 0

B is lineary dependent

So a is true

[upper sparrow]

B is linearly independent too - note that you're considering each matrix as a vector, rather than its columns

[upper sparrow]

What's the form of each element of W? Do you see anything strange with the matrix from (a)?

[steel tide]

I did not understand it

[steel tide]

Please tell me more

What should I apply?

[upper sparrow]

As in, look at how W is defined

[steel tide]

Ops c=0

[upper sparrow]

There ya go

[steel tide]

Lol

The answer was hidden in it

Should I ask another one?

Last one only

(1,1,1) by first option

(1,0,2) by second option

So like this i only imagine option a is correct

Tq

Chaaaaar bit

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[grim bolt]

hi

[grim bolt]

is x+1/x=2 a quadratic equation?

[covert scroll]

Ig it's hyperbole

[grim bolt]

no i am not asking about the graph, i am asking whether it is a quadratic equation or not

[covert scroll]

Quadratic means x²

[grim bolt]

it is reducable to x2-2x+1=0

[patent tundra]

it is not

[grim bolt]

why is it not a quadratic if i can reduce it to x2-2x+1=0?

[civic dirge]

The graph of what?

The graph of (x + 1)/x?

,w graph (x + 1)/x

[grim bolt]

x+(1/x)=2

[potent lotus]

Results provided by WolframAlpha

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Requested by kaynex

[civic dirge]

That's definitely not a picture of a quadratic

[grim bolt]

,w graph x+(1/x)=2

[potent lotus]

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[patent tundra]

Depends on what graph you mean I took the expression you wrote as (x+1)/x = 2. However going by order of operations you are correct

[civic dirge]

You can change it to a similar problem, where you're instead looking at a quadratic!

But the original problem does not feature a quadratic.

[grim bolt]

I was of the same opinion, and argued that since x+(1/x)=2 is not a polynomial, it is thus not a quadratic as a quadratic equation is a polynomial of degree 2

[covert scroll]

Literally half of the hyperbole

[patent tundra]

It is quadratic except it is undefined at x = 0.

[grim bolt]

so will it be a quadratic?

That is what i thought too, but there are only two options in a test true or false

[patent tundra]

I would regard it as a quadratic. Sometimes the meaning of a word will not be clear and can have an ambiguous meaning. I guess you can put down False

[grim bolt]

Thank you! I do believe this is an ambiguous question and should not have come in the test but oh well

Thanks everyone

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[swift sage]

I need help with the cylindrical shells set up here

I know the formula I need is for cylindrical shells about the x-axis which is V = 2pi * integral from c to d of y*f(y) dy

Since y = x^(1/3), x = y^3

so my attempt was

V = 2pi * integral(0, 2) of (y * y^3) dy

but this comes to an incorrect result

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[split meadow]

how do i find a

[split meadow]

yes

can you show me

[crystal forge]

^ law of sines

notice side a is opposite of angle A, side b is opposite angle B etc.

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[low orchid]

can anyone please help me to solve sums of kinematics I am 14 years old but I love maths so I am trying maths of 12th std

[prime skiff]

just send ur question

[low orchid]

ok sir

[prime skiff]

then js wait for help

[exotic falcon]

what is your doubt?

[low orchid]

this is my doubt

[topaz raft]

This doesn’t look like kinematics sir

[winged smelt]

You ask a very good question sir

It interests me so very much to be of help to you

What seems to be your doubt?

I just see the question solely

[low orchid]

this is surface area I have not asked my kinematics doubt yet

[winged smelt]

This seems like a 10th grade question more than a 12th grade

[exotic falcon]

this channel is occupied if you have a doubt about what grade you are in you can ask elsewhere

[winged smelt]

I'm extremely sorry sir

[heady bolt]

what exactly do u want to know?

[low orchid]

which formula will we use and how will we calculate?

[heady bolt]

if u are baffled how to find the solution, its easy, vol of bucket = vol of sand (use the apt equation and find the vol. of sand) and vol of conical heap= vol of sand (from this u can find the radius of the heap). substitute the value of r in slant height equation to get the slant height

[low orchid]

ok I will try this First

[calm coral]

@low orchid Has your question been resolved?

[wide jewel]

Volume will remain same even if the shape has been changed,
Which means that volume of cylindrical bucket = volume of conical heap

Try to find the volume first

[calm coral]

@low orchid Has your question been resolved?

[calm coral]

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[calm coral]

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[kindred turret]

What do i write in a conclusion and reflection on a comparative statistical analysis?

[kindred turret]

<@&286206848099549185> please help meeeeeee

[calm coral]

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[calm coral]

@kindred turret Has your question been resolved?

[calm coral]

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[stark dove]

If |z| = 1, what's the maximum of arg(z-5)?

[stark dove]

Is it just pi or is it something else?

[blazing coyote]

You basically want to find the max of arg (x-5+iy$

[stark dove]

yeah

[blazing coyote]

hmm

so this is analogus to finding the maximum angle of the vector (x-5)i+yj

that would be 180/π, no?

I could be wrong

[stark dove]

yeah that's what i was thinking too

idk i remember doing something similar where i had to find the angle between the tangent from a point and the real axis

I can't remember when we do that :(

[calm coral]

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[remote mural]

What I am doing wrong

[surreal patio]

there should be a $2a\sqrt{a}$ instead of $2\sqrt{a}$

[potent lotus]

JustToPro

[surreal patio]

cuz $(a+b)^2 = a^2 + 2ab + 2b^2$

[potent lotus]

JustToPro

[surreal patio]

also this is wrong

[remote mural]

Can you circulate like you did ?

[surreal patio]

i was pointing out the mistake'

where i circled in the 2nd brackets there is $2\sqrt{a}$ which is incorrect

[potent lotus]

JustToPro

[surreal patio]

also i dont understand the 2nd step

[wide jewel]

The first line is the question right?

And the second being your answer?

[surreal patio]

yes

[wide jewel]

Where did (a+rt(a)) (a-rta) go?

[surreal patio]

hm?

i dont understand this step

cuz if u are multiplying the first 2 brackets , u dont get that

and where did the other terms go

[wide jewel]

From the question, they aren’t in the second step

[wide jewel]

Yeah

[surreal patio]

is this ur work or someone elses?

[calm coral]

@remote mural Has your question been resolved?

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[remote mural]

Hi in what order should go through these chapters?

[blazing coyote]

I'd probably start with theory of equations

then Binomial theorm

logs

induction

[remote mural]

Does that have any pre reqs

[blazing coyote]

other than basic algebra, idts

[celest stream]

is that arihant

[remote mural]

Okay what next?

[remote mural]

Yes

[blazing coyote]

sets, relations and functions

then complex numbers

followed by matrices

and determinants

[celest stream]

are u a new 11th grader?

as in fnished 10th just now

[remote mural]

No

[celest stream]

finished 11th and going to 12th?

[remote mural]

Yes why?

[celest stream]

me too

[blazing coyote]

just a suggestion, try understanding the intution behind matrices and dets

[celest stream]

first of all induction isnt there for jee or cbse

[blazing coyote]

still good to know

very good to know

[celest stream]

igs

[remote mural]

Okay

[remote mural]

Oh I don’t mind, I just want to learn

[blazing coyote]

I'll suggest a playlist for that

[remote mural]

Please do

[blazing coyote]

linear algebra by three blue one brown (3B1B)

[remote mural]

Anything else I should keep in mind?

[blazing coyote]

yes, concepts are important in maths, understand the concepts, they are more important than methods

[remote mural]

Okay noted

[blazing coyote]

and most importantly enjoy your maths journey, all the best !

[remote mural]

Thank you so much!!

[blazing coyote]

ATB for all your entrance exams too

[remote mural]

I appreciate it!

[remote mural]

Thank you, I’m not really planning on appearing in any yet

[blazing coyote]

What do you want to study? There are different exams for different streams

[remote mural]

Math and physics

[blazing coyote]

For engineering :_ JEE and all the state entrance exams

for maths ISI, CMI's entrances

[remote mural]

But I need to work on my chemistry too

[blazing coyote]

for science NEST and IAT

[remote mural]

Because it’s really bad

[blazing coyote]

Don't worry, class 12 chem ain't that bad

[celest stream]

ISI and CMI dont take jee?

[blazing coyote]

Idts

I should hope not

[celest stream]

why

[blazing coyote]

they are for maths

[blazing coyote]

the best in maths in the country could be garbage at chem for instance

[remote mural]

My organic is bad so ehh

I’ll take screenshots of these

[celest stream]

dont try mugging up evrything for organic

[blazing coyote]

don't worry, it gets better, I was terrible too, walked away with 88% in 12th in chem

[remote mural]

Is it okay if I DM you if I have any follow up questions about the same?

[blazing coyote]

which isn't great, but not terrible either

[remote mural]

It’s okay if you refuse

[blazing coyote]

who? me?

[remote mural]

So what do I do?

[remote mural]

Yes

[blazing coyote]

I'd rather not, sorry, I prefer not to DM people I don't know IRL

sorry

but just message here

[remote mural]

Dang that feels good to hear

[celest stream]

try and understand whats happening, coz if u start mugging up everything u wont be able to do shit in exams coz there is way too much to mug up

[remote mural]

Oh it’s okay

[remote mural]

Do you have any recommendations for any resources for the same?

[celest stream]

i think ms chauhan is decent for theory

rlly good for questions tho

[blazing coyote]

NCERT and NCERT examplar is enough for CBSE

[celest stream]

but for organic, there is very less in ncert, atleast in 11th

[blazing coyote]

I'm talking about 12th, but yeah

true

[celest stream]

but for inorganic and physical its like god

u actually understand stuff in that

[blazing coyote]

this is getting off topic(Not maths)

[remote mural]

Alright I understand

Thank you sooo much guys!

I really appreciate it

Have a good night!

I’ll close the channel for now

Thank you again

.close

[calm coral]

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[calm coral]

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[whole dock]

Does multiplication really ALWAYS give you the same result no matter in what order its done and if it does WHY

Im in high school and im still unsure about this

[calm coral]

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[calm coral]

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[dreamy prawn]

How do I do part a.)

radius= 2+2sin(B)

T cos(B) = mg

T sin(B) = m w^2 * r

is that the right equation

[calm coral]

@dreamy prawn Has your question been resolved?

[calm coral]

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[eager token]

Hii how would i do this

[soft ferry]

have you done logs

[eager token]

Yeah

I rewrote it as logbase3 of 80 = 2x

But idk what to do now

The answer is ln80/2ln3

[soft ferry]

i’ve honestly never seen ln in my life but i’d just divide it by 2 and that would be the final answer

[spice phoenix]

take natural log on both sides

[eager token]

Wait but why would one be over the othrr

[eager token]

Wait why

[soft ferry]

u need it in terms of x

n u got 2x

[eager token]

But how would u get ln80/2ln3

[spice phoenix]

any log would be fine

[eager token]

So after putting log in front if both would u divide out 3^2?

[soft ferry]

ohhhh In

log3(80) = In(80)/In(3)

[eager token]

But what about the 2

[soft ferry]

You said 2x

[soft ferry]

so you divide it all by 2

[spice phoenix]

any log is fine

[eager token]

Why would the 3 get squared instead of 80

[spice phoenix]

it's a property of logs

[eager token]

Oh 😭

[spice phoenix]

you can just split $\log_3 80 = \dfrac{\ln 80}{\ln 3}$

[potent lotus]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[spice phoenix]

you'll get the same thing

[eager token]

Okkk tyy

Wait is that the change of base formula

[spice phoenix]

idk what you guys call it 💀

[eager token]

Oh 😭😭

TYY

I think i get it now

[spice phoenix]

you can also use any other base instead of e tho if thats what that formula is

[eager token]

Ohhhh ok

[eager token]

Wait whats the property called

[spice phoenix]

$\log_a x = \dfrac{\log_b x}{\log_b a}$

[potent lotus]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[eager token]

Doesnt the power property say that 80 should be squared

[spice phoenix]

$\dfrac{1}{2}\log_3 80 = \log_3 80^{1/2}$

ok why is it like this

[potent lotus]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[eager token]

So for this

Wouodnt it be logbase 3 of 80^2

[spice phoenix]

80^1/2 yes

you take whatever number is before the log to the power

[eager token]

So why is 3^2

[spice phoenix]

forget it

lol

[eager token]

Oh 😭

[spice phoenix]

do this way

[eager token]

Okkk

Tyyy

.close

[calm coral]

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[calm coral]

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[sweet totem]

I'm trying to solve this system of equations but forgot how it works
[
\frac{d}{dt}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}1&-2\1&-2\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}
]
where $x,y$ are both functions of time $t$.

So I thought I could just split this into
\begin{align}
\frac{dx}{dt}&= x-2y\
\frac{dy}{dt} &= x-2y
\end{align}
Then differentiate (1)?
[
\frac{d^2x}{dt^2}= x-2y-2x+4y = -x+2y?
]
but not sure how to go from here.

[potent lotus]

Faq

[sweet totem]

Am I just going about this the wrong way?

I just realised I could try the eigenvector method

[calm coral]

@sweet totem Has your question been resolved?

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[remote mural]

how do I find the critical points from here??

[languid oak]

A critical point of a function y = f(x) is a point (c, f(c)) on the graph of f(x) at which either the derivative is 0 (or) the derivative is not defined. So ı guess you already found them?

[marsh forge]

You already found the critical points!

Wait nvm

Well

[strange lichen]

find the values of x that satisfy sin(x) = +- sqrt(0.5)

@remote mural

those are ur critical values

[remote mural]

oh okay thank u

[marsh forge]

You can simplify $\sqrt\frac{1}{2}$ to $\frac{\sqrt{2}}{2}$

[potent lotus]

prime

[remote mural]

ohh

thank u very much!

.close

how do I close it 💀

[strange lichen]

do .close again

[remote mural]

ty

.close

[calm coral]

Channel closed

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[marsh forge]

It should become $\frac{\pi}{4}+\frac{k\pi}{2}$ where $k \in \mathbb{Z}$

[potent lotus]

prime

[remote mural]

i know this is silly but what does k e Z mean

I see it a lot and I don't really get it

[marsh forge]

The $\in$ basically means that it's inside of the set thingy. Like the big Z denotes the set of all the integer numbers. The double like R denotes all real numbers and there's more.

[potent lotus]

prime

[remote mural]

oh that makes sense thank you

[marsh forge]

So $k\in \mathbb{Z}$ just means that k holds true for all integers

[potent lotus]

prime

[remote mural]

so it just means that's the equation for all zeros?

[calm coral]

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[fleet arch]

hello, I have no idea of how to get this limit

[fleet arch]

how should I start?

[calm coral]

@fleet arch Has your question been resolved?

[fleet arch]

<@&286206848099549185>

[calm coral]

@fleet arch Has your question been resolved?

[calm coral]

@fleet arch Has your question been resolved?

[calm coral]

@fleet arch Has your question been resolved?

[fleet arch]

thanks!

[calm coral]

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[hazy carbon]

<@&286206848099549185>

[calm coral]

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[dry hill]

hiiiiii

[dry hill]

anyone there?

[turbid swift]

yes

[dry hill]

i have a huge question on my project

its about 3d vector and reprojection

[fathom shuttle]

!filetype

[calm coral]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

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[past cape]

im unsure how sec2x became 1

[sinful pasture]

sec(0)=1/cos(0)=1/1=1

[past cape]

is there a name for this relationship?

[sinful pasture]

sec(x)=1/cos(x) ?

[past cape]

yea, id like to search this up to learn it

[sinful pasture]

reciprocal trigs

[past cape]

also with the ^2 can i write it as (1/cos(0)^2?

[sinful pasture]

yes

[past cape]

awesome!

thanks!

.close

[calm coral]

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[rugged saffron]

Did I mess up here?

I assumed I could use sine rule

however the answer sheet used cos rule and obtained a different value

wondering if I made a mistkae applying the rule somewhere

<@&286206848099549185>

<@&286206848099549185>

[light pasture]

How did you get the 110 40 30 triangle
If it is a 110 40 30 triangle then angle DCB would be 360-90-110=160
But 160+35>180

[calm coral]

@rugged saffron Has your question been resolved?

[rugged saffron]

Given the question, bearing of point c from a is 50

therefore angle a is 90-50 = 40

Bearing of point c from b is 300

300-270 = 30

I don't understand what you are doing with DCB triangle?

The diagram is representing a 3D model, the bottom triangle does not share angles with the vertical one (angle of elevation)

[rugged saffron]

<@&286206848099549185>

[light pasture]

Oh I'm sorry I thought it was a 2d triangle
Sorry

[calm coral]

@rugged saffron Has your question been resolved?

[calm coral]

@rugged saffron Has your question been resolved?

[rugged saffron]

<@&286206848099549185>

[calm coral]

@rugged saffron Has your question been resolved?

[rugged saffron]

<@&286206848099549185>

[worn bronze]

Hey sorry but what exactly is bearing?

I may be able to help if I get know what that means

@rugged saffron

Alright I think I can understand what it means from context

[rugged saffron]

Bearing is basicaly from 000 degrees to 360 degrees

from north

[worn bronze]

I think there's a mistake in the question

If you calculate CB

[potent lotus]

Kappa

[worn bronze]

And then apply cosine rule

[potent lotus]

Kappa

[worn bronze]

This comes out to be around 204.2

Your calculations are right though

And if you apply sine rule considering CB, you get AB = 156.6

The bearings must be wrong I guess

If you fix angle ACB, with the given information (excluding bearings) you can find the other two internal angles for triangle ABC

So ACB = 110 must give you angles which are probably not 30 and 40

$\frac{AB}{\sin\left(110^\circ\right)} = \frac{AC}{\sin\left(\angle CBA\right)} = \frac{BC}{\sin\left(\angle BAC\right)}$

[potent lotus]

Kappa

[worn bronze]

You know AC and BC so you can find angle CBA and angle BAC

[calm coral]

@rugged saffron Has your question been resolved?

[calm coral]

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[upbeat knoll]

hi

[strange lichen]

oh this is interesting

draw a right triangle

and label the side lengths

and label the angles: alpha, beta, theta

@upbeat knoll

[upbeat knoll]

hmm

ok lemme try that

ok

i have done that

i dont see where yout going with this tho

[strange lichen]

now find arctan(x)

well first show me ur triangle

[strange lichen]

well u need the side lengths

it gives u one side length

x

and another is 1

so

label those

[upbeat knoll]

which sides do i label as which

[strange lichen]

think about it

[upbeat knoll]

Is this fine ?

,rotate

[strange lichen]

well we need an angle that satisfies: arctan(x/1) = angle

x cant be the hypotenuse

[upbeat knoll]

Oh ok

Got it

,rotate

[strange lichen]

yea

now label some arbitrary angles

[upbeat knoll]

ok

ah ok i see

both add up to 90

cause of angles in triangle

[strange lichen]

yep

[upbeat knoll]

nice

thanks man

i appreciate it

idk what the ans scheme was doing this actualy makes sense

,close

.close

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[light acorn]

Consider we have two people flipping a fair coin. Given at least one of them flipped a heads, what is the probability that both flipped heads?

[light acorn]

I feel as though I have the answer, 1/3, but I don’t know how to formally get there using conditional probability

oh shit wait nvm

[carmine sorrel]

$P(A|B) = \frac{P(A\cap B)}{P(B)}$
So you need to find the probability that they both flip heads (without any conditions)\
And the probability that at least one of them flipped heads (without any conditions)\
Then divide the two

[pale blade]

try making a tree

[potent lotus]

Zybikron

[carmine sorrel]

or a tree works too

[light acorn]

Pr[At least one heads] = 1 - Pr[No heads] = 1 - 1/4 = 3/4 right

I was just so confused about how \frac{P[A] * P[B]}{P[B]} could evaluate to 1/3 but it’s (1/2 * 1/2)/(3/4)

right?

Wait no

Yea i’m still confused

since the events are independent it intuitively seems to me that P(A \cap B)/P(B) = P(A)

Pr[A] = 1/4 since it’s 1/2 * 1/2

[finite iris]

Ye so (1/4)/(3/4)

[light acorn]

wouldn’t it be ((1/4)*(3/4))/(3/4)?

= 1/4

[finite iris]

You got all the values you needed for Bayes

Don't overthink

[light acorn]

.close

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[solid steppe]

hi

[solid steppe]

i am stuck on a part

i have x^18-2k* (2^k/x^3k)

how can i simplify this?

[solid steppe]

the thing on this part is no needed mate to show the rest but sure 1 sec

here u go mate

[remote mural]

binomial nice

what do u need to show

how many rational terms

or what

u dont really need that

u need

[solid steppe]

x^8 term

[remote mural]

use this

[solid steppe]

i have

[remote mural]

where a = x^2 and b = - 2/x^3

[solid steppe]

yea bro

[remote mural]

ok

calculate it

[solid steppe]

thats exactly what i have done

well

thats the part i am stuck

lol

[remote mural]

not exactly

u have sum

u dont need sum

[solid steppe]

dont understand what u mean bro

[remote mural]

just plug the values in my formula

and that's it

[solid steppe]

so i dont need to simplify it?

[remote mural]

no

just plug values

the values you need

[solid steppe]

can u show how u mean ?

[solid steppe]

bro

k is not equal to 8 xD

[remote mural]

why not

[solid steppe]

its what we need to find

[remote mural]

yeah

so that's how we find it

[solid steppe]

find what?

u already say its equal to 8

[remote mural]

the 8th term

[remote mural]

k is equal to 8

not the 8th term

the 8th term will come from that

[solid steppe]

so what do u get in the end

[remote mural]

magical numers

numbers

[solid steppe]

bro

i will show u

[remote mural]

calculate and find out

[solid steppe]

how its done

[remote mural]

sure

please do

[solid steppe]

we need to simplify it

so we get 18-5k

so we can make it equal to 8 becuase its in the x^8 term

so k=2

and not 8

[remote mural]

can you be very clear what you need to find

[solid steppe]

i need to find the the coefficient for the x^8 term

[remote mural]

so (a+b)^8?

[solid steppe]

huh?

bro the whole point on this question

is to find what k =?

and when u find it

it just place it in the equation ....

and then u have the answer

i found where i did wrong

i forgot the minus

what u said was completely false mate @remote mural

but thx now i understand the question even better 😉

.close

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[coral rivet]

How do I turn -x^2+x+6 to (x-m)^2+n

[coral rivet]

Im trying to use this guide https://www.youtube.com/watch?v=pf9LkX8hpTQ by completing the square

but by halfing x it just doesn't work out when I try to test if it's the right one by turning it back

[calm coral]

@coral rivet Has your question been resolved?

[wanton wave]

It won't quite be in this form, you still need the leading coefficient a

in general the form is a(x-m)^2+n, notice the a in front

I'm not sure if that's what is tripping you up

Anyway, where exactly are you getting stuck?

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@coral rivet Has your question been resolved?

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[keen grove]

Whats the name given to the II symbol?

[untold compass]

it's a capital pi

[calm coral]

@keen grove Has your question been resolved?

[calm coral]

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[remote mural]

there are tricks here im not getting

[calm coral]

@remote mural Has your question been resolved?

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[blazing coyote]

So to start

I was thinking I'll assume the centre to be an arbitrary point (a,b)

and the distance to be an arbitrary value r

so $\sqrt{a^2+b^2}=r$

[potent lotus]

Why am. I here

[blazing coyote]

or a^2+b^2=r^2

which ressembles the eqn of a circle

intutively though

this feels wrong

[graceful dust]

i think it's good to think of this one purely geometrically

[blazing coyote]

I feels its a hyperbola

*it's

[graceful dust]

how can it be

[blazing coyote]

wdym

[graceful dust]

consider the circles of radius 1 that pass through the origin

where are their centers?

[blazing coyote]

yeah, that would be a circle, would it not?

[graceful dust]

ya

[blazing coyote]

oh

of fixed radius

yeah, then this makes sense

[graceful dust]

circles of any radius would give you the whole plane

[blazing coyote]

got it, thanks!

[graceful dust]

[blazing coyote]

which cartoon is that emoji from?

[graceful dust]

her name is my melody

[blazing coyote]

thanks! .close

.close

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[blazing coyote]

.reopen

[calm coral]

✅

[blazing coyote]

so f(1)/f(0)=1/5

or f(10)/f(0)=1/5^10

a rough approximation will be given by teh GP summation

$\frac{5\cdot f\left(0\right)\left(1-\left(\frac{1}{5}\right)^{10}\right)}{4}$

[potent lotus]

Why am. I here

[blazing coyote]

now that I think of it

$0<\int_0^{10}f\left(x\right)<10f\left(10\right)$

right?

[potent lotus]

Why am. I here

[blazing coyote]

but $f(10)=f(0)\frac{1}{5^{10}}$

[potent lotus]

Why am. I here

[blazing coyote]

I get the feeling I'm not using all the information

hmm, let me try again

it's decreasing

so the area is between $0f(0) and 10f(10)$

[potent lotus]

Why am. I here

[blazing coyote]

<@&286206848099549185>

.close

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[quiet island]

if we have y' = f(x,y) = y/x; y(x_0) = y_0, when does the theorem of existence and uniqueness not guarantee uniqueness?

[quiet island]

I think it is in any line that passes by the origin

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@quiet island Has your question been resolved?

[calm coral]

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[weak silo]

Hi I just need to be pointed in the right direction with 13b

[weak silo]

My current working:

$\sqrt{3}+ \sqrt{2} = \frac{p}{q} + \frac{m}{n}$

[potent lotus]

TEDDY

[weak silo]

Wher the HCF of each fraction is 1

I then square both side and equation the surf parts which is