help-39

[tiny wadi]

for b.)

$$\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}$$
$$S = \pi r^2 + \pi rl \text{ so } \frac{dS}{dr} = 2\pi r + \pi l $$

[jolly parrot]

hoax

[tiny wadi]

where the fuck are these extra terms coming from?

nvm l is also a function of r

.close

[pearl pond]

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[pearl pond]

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[tawdry solar]

!status 1

[pearl pond]

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

[proper nova]

What does the cosine rule says?

[tawdry solar]

,,c^2 = a^2 + b^2 -2ab\cos\left(C\right)

[jolly parrot]

calvin

[north talon]

let the height dropped from B to AC be D. then you can use pythagorus

[tawdry solar]

B to AC?

isn't that just b

[north talon]

<ADB = 90 so in the image BD = h and AD = x

[tawdry solar]

dude

you didn't asnwer my question

AC is a line ???

[north talon]

yeah?

[tawdry solar]

so what do you mean the height dropped from B to AC

thats just h

[north talon]

D is on line AC so that BD is perpendicular to AC

im just naming points

[tawdry solar]

Bd is already perpendicular to ac

[toxic lichen]

that bottom-left point, unnamed in your diagram, Copter gave it the name D

it's where the right angle is

[tawdry solar]

ohh

yes

[toxic lichen]

DAC is one straight line

[tawdry solar]

yes

[toxic lichen]

btw do you have access to the fact that cos(180°-x) = -cos(x)

youre kinda gonna need that

[tawdry solar]

yeah

well for me they laid it differently

-cos(180-x) = cosx

[toxic lichen]

same shit.

[tawdry solar]

okay anywhere

[toxic lichen]

in any case

[tawdry solar]

anyways**

[toxic lichen]

... hold on

their notation for the sides is wack

b is not opposite B and c is not opposite C

what the hell

anyway like let $\angle BAC = \theta$ and then $\angle BAD = 180\dg - \theta$.

and the cosine rule that you wish to prove will look like $$c^2 \overset?= b^2 + (b-x)^2 - 2b(b-x) \cos(theta)?$$ ... wait why is $b$ even appearing twice.

[jolly parrot]

Ann

[toxic lichen]

this diagram is so fucked up.

and you are forced to use it in your proof?

[north talon]

i just realized wtf😭

[toxic lichen]

yikes.

[tawdry solar]

🤷‍♂️

[toxic lichen]

is there any possibility of emailing your professor

[tawdry solar]

im doing practice

its fine

[fiery chasm]

Chill😭🙏🏿

[tawdry solar]

we can re-label the sides if necessary

[toxic lichen]

yeah let's do a relabel

hold on, let me cook one up that isn't messed up

the goal now, with the notations in this diagram, is to prove $$a^2 \overset?= b^2 + c^2 - 2bc \cos(\theta)$$

[jolly parrot]

Ann

[tawdry solar]

yes

[toxic lichen]

what did you mean by this

[toxic lichen]

and now you can, with a clearer conscience, apply Pythagoras on the two right triangles in this picture.

and then also do some soh-cah-toa stuff in BDA.

[tawdry solar]

,,h^2 + x^2 =c^2
\ h^2 + (b+x)^2 = a^2
\ \sin(180 -\th) = \frac hc

[jolly parrot]

calvin

[toxic lichen]

i mean you probably will want to involve the cosine somehow.

[tawdry solar]

woops

,,\cos(180-\th) = \frac xc

[toxic lichen]

given the whole thing we're doing is proving the cosine rule

[jolly parrot]

calvin

[toxic lichen]

that's more like it.

[tawdry solar]

sorry i got carried away

[jolly parrot]

calvin

[tawdry solar]

up here:

[toxic lichen]

what you can do here is consider the expression
b^2 + c^2 - a^2 - 2bc cos(θ)

and try to simplify it down to zero

[tawdry solar]

,,h^2 + (b+x)^2 = a^2
\ h^2 + b^2 + 2bx + b^2 = a^2
\ c^2 + 2bx + b^2=a^2

wait

[jolly parrot]

calvin

[tawdry solar]

wdym

in proofs shouldn't i have to derive the formuila

formula

[toxic lichen]

the formula will happen at the very end

what i am suggesting here is: construct a chain of equalities

which starts with
b^2 + c^2 - a^2 - 2bc cos(θ)
and ends with
0

in so doing, you will have proven b^2 + c^2 - a^2 - 2bc cos(θ) = 0
and the final step will be to add a^2 to both sides of that

[tawdry solar]

so is that all

[toxic lichen]

i mean...

no, it's not all? i gave you a direction to work in, but you haven't actually yet constructed the equality chain.

[jolly parrot]

calvin

[tawdry solar]

and over her

here

i can sub it in

[jolly parrot]

calvin

calvin

[tawdry solar]

is that it?

[toxic lichen]

well, you decided not to follow my directions, so unfortunately i'd have to detangle/decipher your chain of reasoning, which you kind of obscured.

[tawdry solar]

but

but but

i defined cos theta

and then i transposed it

and substituted it in

to therefore derive the cosine ruile

rule

if i present neatly will i get full marks 🥺

[toxic lichen]

you sure didn't present anything neatly to me here and now

[tawdry solar]

ill latex it up

[toxic lichen]

i also dont like your use of the words "define" (which i think you misused) and "transpose" (which probably shouldnt even be appearing)

[bronze heath]

-# after going through it what was done seems good

nah

[jolly parrot]

calvin

[toxic lichen]

ok yeah this works

you still didnt follow my idea but this works

[tawdry solar]

yess full marks

sorry

i didn't understand your idea

but you helped guide me to the answer

which is the most important thing

thank you for your help

.close

[pearl pond]

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[toxic lichen]

my idea was "instead of proving A = ... = ... = ... = B let's prove A-B = ... = ... = ... = 0"

[pearl pond]

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[shadow knoll]

How could i draw the Set {(x,y)| x²-xy+y²<=1} ? It certainly is an ellipse but i dont know how to bring this into the right form to notice that.

[autumn fossil]

Might be a silly way to approach it, but it's symmetric in x,y

that means it must have an axis of symmetry at x = y

so you know it must be tilted at 45°

so major axis is x = y, minor axis is x=-y (perpendicular, and centered at origin)

you could get their lengths by plugging that in

[shadow knoll]

Im not sure i can fully follow 🫠

[autumn fossil]

well if we plug in e.g. x = 3 and y = 2, we get 3^2 - 2*3 + 2^2
if we instead plugged x = 2 and y=3, we get 2^2 - 3*2 + 3^2
which is the exact same thing

therefore if (x,y) is on the ellipse, (y,x) must be there as well

so it must have axis of symmetry at x = y

[shadow knoll]

Okay thats enlightening thanks first of all

How do we know that this has to be an ellipse tho? The symmetry part makes sense

[autumn fossil]

you could use discriminant on it

or we could check whether x = -y is also an axis of symmetry

if it is, then we are left with hyperbola or ellipse

[shadow knoll]

Whats discriminant? Like doing sqrt(..)?

[autumn fossil]

if we then compute intersections of the axes of symmetry with the shape, we will know whether its ellipse of parabola

[autumn fossil]

if we have a quadratic equation (with x,y) which starts like
ax^2 + bxy + cy^2 + .... = 0
then if we check D = b^2 - 4ac (the discriminant, note that its same as with normal quadratic equation), we can determine whether its ellipse, parabola or hyperbola

if D > 0, its hyperbola

if D = 0, its parabola

if D < 0, its ellipse

there is a relatively simple way to explain the intuition behind why that works, or even simpler way in this case would be computing the intersections with the axes of symmetry

we can do whatever u wish

[shadow knoll]

Wow ive never heard of that property thats very useful

So we get D = -5 in our case making it obvious that its an ellipse

[autumn fossil]

yes

[shadow knoll]

Or -4 i think sth like that

Then we look for intersections on symmetry axes and we're done?

[autumn fossil]

it would be (-1)^2 - 4*1*1 = -3

but close enough

[autumn fossil]

yep

those will help you determine where the ellipse "ends"

[shadow knoll]

Well thanks a lot!🙏

[autumn fossil]

you could use it to compute the major and minor axes lengths

[shadow knoll]

.close

[pearl pond]

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[feral sedge]

How much linear algebra do you know

[shadow knoll]

quite a bit

[feral sedge]

You could turn that function into a bilinear form and then diagonalize it

[shadow knoll]

thats buried a bit too deep in my memory i fear

[autumn fossil]

https://en.wikipedia.org/wiki/Quadratic_form

[feral sedge]

the eigenvectors will give you the directions of the minor/major axes and the eigenvalues will be half the axis lengths

or something like that

[shadow knoll]

Ill look into that:)

[feral sedge]

or yeah quadratic form sorry

[autumn fossil]

It's useful especially for "random" quadratic expressions which dont have any symmetry

yours was relatively nice, so i was more inclined for the symmetry argument

usually the multiples of 45° are doable without it

[feral sedge]

yeah that's fair

[pearl pond]

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[bitter herald]

how to solve for $A_1$ in [
2Q\8{\4{\32,A_1\6\sin{\5\pi8}}{\3{N_0}}} = Q\8{\4{\32,A_2}{5\3{N_0}}}
]
where $Q(\2)$ is the Q-function

[pearl pond]

@bitter herald Has your question been resolved?

[tropic saddle]

no clue what the Q function is

but just divide by 2

apply inverse

and then its just a linear equation in A_1

assuming Q has an inverse but if not then good luck anyway

[pearl pond]

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[elder maple]

Im having some trouble here:

[elder maple]

I have gotten, A = 1/2 and B = 1/2

The solution which satisfies the values given, (take note of the answer) would indicate A = 0, and B = 1 , right?

But then if you quickly substitute y =-1/3 when x = 0 in the general solution, you'll find that A + B = 1

So i am somewhat confused here

[lament dawn]

yes but i dont think it satisfies y'=-5/3

[wraith hatch]

what seems contradictory to you?

[elder maple]

oh i reckon i might've worked it out wrong

nothing is contradictory there, i think

A + B = 1, so B = 1 and A = 0

though I got A = 1/2 and B = 1/2

[wraith hatch]

probably a calculation mistake

use the 2nd constraint

[elder maple]

So I'd assume my derivative of the solution is incorrect

[lament dawn]

mhm

[elder maple]

ok perfect got it

i forgot to use the product rule on sin(x)e^sin(x)

so i had an extra + 1, which gave me A-B = -1 instead of A-B = 0

great thank you all!

.close

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[elder maple]

going to reuse this channel here

[elder maple]

can anyone please explain why we do the intermediary step of using z?

I can understand it, but if anyone can explicitly mention why we state for example (in the previous question above):

[wraith hatch]

can u share the whole question

[elder maple]

yes sure

Sorry for the late reply

am also trying to understand

I have solved this correctly, but, is it to be assumed that the substitution of z will always be given?

How will I know to use it in these instances?

[pearl pond]

@elder maple Has your question been resolved?

[pearl pond]

@elder maple Has your question been resolved?

[pearl pond]

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[prime bramble]

<@&268886789983436800>

[unborn abyss]

Mr beast and its consequences have been an unmitigated disaster for the sleepy moderators

[pearl pond]

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[bitter herald]

Dont delete i am about to check in my 2.7k usd

Oh man

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[pearl pond]

Please don't occupy multiple help channels.

[thorn agate]

<@&268886789983436800>

mr beast

[pearl pond]

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[royal harness]

how do I write these in their trigonometric form

[royal harness]

<@&286206848099549185>

[fleet dove]

Ye just one sec

[royal harness]

alr thanksss !

[fleet dove]

idk if it is correct

@royal harness

[royal harness]

what is cis

wait imma send you the whole exercise for you to see

it's question 1

[plucky oriole]

cos + isin => CiS

[royal harness]

how do we write this in its trigonometric form tho 💔

[plucky oriole]

in $re^{i\theta}$ this form?

[jolly parrot]

Shikhar

[royal harness]

that's the exponential form it's not what I want

I need it to be written as cos(theta) + isin(theta)

[plucky oriole]

um..
for $a+ib=r(\cos{\theta}+i\sin{\theta})$
you can prove this in detail..

[jolly parrot]

Shikhar

[plucky oriole]

$\theta=\arctan{\frac{b}{a}}$

[jolly parrot]

Shikhar

[royal harness]

using this would get me nowhere though 💔

[plucky oriole]

ah.. i see.. just accepted defeat.. gud gud

👏

[plucky oriole]

first of all.. this isnt correct

[royal harness]

wait really????

where did I make a mistake

hopefully you're right cuz this is hopeless now

[plucky oriole]

$$\Delta = (-2 \cos \alpha)^2 - 4(1)(1)$$
$$\Delta = 4(\cos^2 \alpha - 1)$$

[jolly parrot]

Shikhar

[plucky oriole]

missed ^2

[royal harness]

yesssss

THANK YOUUUUU

and that that's just equal to sin is that right?

how do I deduce the solutions for this equation

<@&286206848099549185>

[light apex]

max value of n is 0

if im not wrong

[royal harness]

no that's the min value

[light apex]

no wait i mean minimum

yeyeye

[royal harness]

no wait

[timid cipher]

factor out z^n?

[light apex]

u get cos(alpha) = 1/2 + 1/(2*z^n)

minimum value of z^n is 1 so min value of n =0

max is inf

[royal harness]

I know it's in french but look at question 4)a

[royal harness]

ugh I'm confused

[mystic geode]

cant you use quadratic formula?

[light apex]

whats the max value of cos(alpha)?

[royal harness]

1

[royal harness]

we have power of n

[light apex]

seperate cos(alpha) and z^n

[royal harness]

and we have to deduce it form the previous question

[royal harness]

we factor out with z^n?

[light apex]

@mystic geode can you handle this?

i got sm work

[light apex]

assume z^n = a and then do

u get quadratic

[royal harness]

oh alright imma try thanks !

[light apex]

yw

[royal harness]

. close

.close

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[royal galleon]

for number 10 how can we approach this? The two conditions we need are that it is continuous which it cannot be since we have a discontinuity at 0 and the derivative is different when c is positive or negative so how could I make the derivative of a+bx^2 match two derivatives?

[cobalt hinge]

naw, there’s no discontinuity since c is automatically positive

you can’t have |x|<=-1, right?

(that’s my two cents I gotta go)

[royal galleon]

Ye completely missed that thx

[cobalt hinge]

alright I have acquired time to attempt to help you ❤️‍🩹

So how can we make 1/c=a+bc^2?

by plugging in the “split point” or whatever you call it

[jolly parrot]

;(

[royal galleon]

I don't see how we can deal with two derivatives

with one I can find a and b

[cobalt hinge]

well c>0 so we only need to consider that portion with x>0

(i believe)

everything else is continous

[royal galleon]

ok right so a+bx^2 can only be defined when c>0 so we apply the same condition to the top. b = -1/2c^3 and a = (1/c) - (1/2c^2)

[cobalt hinge]

Lemme check

[cobalt hinge]

same b but different a

check when you’re plugging into a=1/c-bc^2

[royal galleon]

for a isn't all we are doing is a - x/2c^3 = 1/c

[cobalt hinge]

i got this:
||-1/c^2=2bc -> b=-1/2c^3
1/c=a+bc^2 -> a=1/c-(-1/2c^3)c^2=1/c+1/2c=3/2c|| check after

[royal galleon]

oh ok a = 1/c + 1/2c^2

[cobalt hinge]

but a+bc^2=1/c? Confused

[royal galleon]

b = -1/2c^3

oh I forgot it is c^2

ok so then a = 1/c + 1/2c

[cobalt hinge]

ye

that’s what I got

[royal galleon]

thx for the help

[cobalt hinge]

yw

🫡

[royal galleon]

.solved

[pearl pond]

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[shrewd rune]

hekko

[shrewd rune]

so how is the first number -1

shouldnt it be

2

[sharp vigil]

for part a)?

[shrewd rune]

ye

[sharp vigil]

how did you get 2?

[shrewd rune]

-1*0 + -1(2) + 2(0) + 2(2)

isnt it row * column

[sharp vigil]

in part a they ask you to add the matrices

when adding matrices you add corresponding entries

[west sapphire]

the answer for AB is wrong, btw

[shrewd rune]

ohh

wait i was multipliying

is -1 + 0

ohhhh

mb mb

[shrewd rune]

thx

[shrewd rune]

thx

.close

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[harsh zealot]

can somebody explain this? Is there something I'm missing because I genuinely don't understand why it's +3 and -5 instead of the other way around.

[prisma kernel]

The order doesn't matter here

Do you want to know how they came up with these values?

[harsh zealot]

yes

[west sapphire]

well it matters in the sense that it can't be (2x-5)(x+3)

if that's what OP was asking

[fierce sky]

Yes, and hence the order DOES matter

[harsh zealot]

i thought the two values have to add up and multiply up to numbers in the equation

[prisma kernel]

🙏

[fierce sky]

You can check by expanding (2x + 3)(x - 5) and (2x - 5)(x + 3)

One will give a -7x term, the other will give +7x

[harsh zealot]

waitttt

[glass meadow]

That works better when the polynomial is monic, that is when the coefficient of the x^2 is 1

[harsh zealot]

explain how i'd get 7x/-7x

i might be slow

[fierce sky]

Surely you know how to expand (2x + 3)(x - 5)?

[harsh zealot]

wdym by monic

[glass meadow]

You can make it monic by just factoring out the 2

...

[harsh zealot]

i probably do but the wording is confusing me so assume i don't

[glass meadow]

that is when the coefficient of the x^2 is 1

[harsh zealot]

alright

did the other dude leave ;-;

[glass meadow]

!vol

[pearl pond]

Helpers are just people volunteering their time to help you. Be polite and patient.

[harsh zealot]

oh mb

[hallow current]

wsp

[frosty sentinel]

if you are too lazy to expand the entire thing and check, you can just focus on the coefficient of x.

example:

(2x + 3) (x - 5)

to get x term in the product, we need to choose a x-term and a constant to multiply together from the two brackets

we can get 2x (-5) = -10x
and 3 (1x) = 3x
add up,
-10x - 3x = -7x, coefficient -7, which is correct

try focusing on the coefficient of x for
(2x - 3) (x + 5), you should get a coefficient of +7

otherwise, you've got to know how to expand (2x - 3) (x + 5)

[pearl pond]

@harsh zealot Has your question been resolved?

[harsh zealot]

thanks bro

[pearl pond]

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[somber meadow]

Can't a trapezoid's area be: ((((b1 - b2) / 2) * h) + (b2 x h)

[glass meadow]

What's b1, b2, h, and why are you using both * and x ? Also there's one too many parentheses in there

[feral sedge]

yes, if you simplify this you'll find it's equal to the standard area formula

[somber meadow]

ok

.close

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[somber meadow]

.close

[timid lodge]

You closed it already just so you know

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[earnest warren]

There is a positive integer $n$ in which all the digits in $n^2$ and $n^3$ are from the set $A={1;2;3;4;5;6;7;8}$ ; there are no repeated digits and if we take all the digits in $n^2$ and $n^3$ we get the set A.

[jolly parrot]

Alexis_Fx

[earnest warren]

Find n

Lucky guess n=24

,calc 24^2

[jolly parrot]

Result:

576

[earnest warren]

<@&268886789983436800>

,calc 24^3

[jolly parrot]

Result:

13824

[earnest warren]

so the digits in 24^2 are 5,6,7 and in 24^3 are 1,3,8,2,4 that would give the set A={1,2,3,4,5,6,7,8}

That said, I'm not sure how to actually do the problem

-# please ping when reply

[pearl pond]

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[earnest warren]

Noo lmao

Misclick

.reopen

[median basalt]

when am i send the math quection in sinhala who can solved it

its like a OL quection

[earnest warren]

What's an OL question

.reopen

[median basalt]

Odrianry Level Examination in SriLANKA

[earnest warren]

Well it's not

But I can try to find similar materials/problems from that

[median basalt]

ill send the question

[earnest warren]

No don't

[median basalt]

?

[earnest warren]

Imma forward my problems so other helpers can see it

Helpers will be confused what's the problem I'm working if you do

[tawdry solar]

send it in an open help channel

#help-2

[earnest warren]

This's the problem, we have to find n

[wraith stag]

i think you can get a band of candidate n and test from there

n^2 and n^3 have to have 8 digits combined

[earnest warren]

I can see that, but that's such a loose bound

[wraith stag]

we can see that n > 9 (not the best bound but its a bound)

[earnest warren]

Yeah

[wraith stag]

n < 99 for sure

too many digits

[tardy reef]

you can even go ahead and say n<33 right?

[wraith stag]

also n < 32

[earnest warren]

That's a good bound yah

[tardy reef]

yea 32 oops

and also, if you look at the last digits of squares and cubes, you are left with numbers ending in 2, 4 and 8

[wraith stag]

n > 19 too

[earnest warren]

Damn why all the bounds are from the fact that the total digits are 8

[wraith stag]

is testing numbers from 20 to 31 good enough

20 and 30 obviously fail

[tardy reef]

only viable ones are 22, 24 and 28

[earnest warren]

It is tbh

[tardy reef]

and 22 goes away as 22*22=484

[wraith stag]

21 and 22 both fail quickly

[wraith stag]

441 too

[tardy reef]

well, 21 goes away already coz square and cube both end in 1

[tardy reef]

.

[wraith stag]

the square has two 4s

[earnest warren]

Hmm okay I see

It turns out to be less interesting than I thought

But thanks you both

[wraith stag]

the point seems to be to confine n

likely extends to other problems you're seeing

[earnest warren]

I thought there were some tricks to deal with the other restrictions

But we only use that at the very end

[wraith stag]

the other restrictions are really hard to check

[earnest warren]

Kinda disappointing

[wraith stag]

start with easiest first

[earnest warren]

That's why I thought it would be interesting if there's a way

I was way too focused on those

Well, good enough, ty yall

Maybe I'm being a little dumb lately

.close

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[pearl pond]

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[bitter herald]

I need help with this bcuz i dont know where to begin

Optimum decision threshold means if z > c then pick 1 if z < c pick -1

c to be determined

[pearl pond]

@bitter herald Has your question been resolved?

[pearl pond]

@bitter herald Has your question been resolved?

[waxen agate]

hex

hm my wifi isn't good...

equal priors equal costs

compare the PDFs... what's the overlap region?

[pearl pond]

@bitter herald Has your question been resolved?

[bitter herald]

what does this mean pure

[bitter herald]

-0.2 < z < 0.2?

[waxen agate]

yeah

i'm on plane so my wifi isn't great

[waxen agate]

P(s_1) = P(s_2) = 1/2 and C_{1|2} = C_{2|1}

equally likely signalling by means equal priors no?

and probability of a bit error P_b... then when we measure bit error rate, we count any wrong decision as 1 error, regardless of which way it goes... so C_{1|2} = C_{2|1}

[waxen agate]

so the optimal threshold can be anywhere in the overlap

[waxen agate]

is @bitter herald ignoring me now

[bitter herald]

Qhats c

C

[bitter herald]

Sry I'm in a spa rn

[waxen agate]

C_{1|2} is the cost of deciding s_1 when the truth was s_2 and vice versa

[waxen agate]

alright blud enjoy

[bitter herald]

Wot

Why is there a coat

Cost

There is no cost mentioned

[waxen agate]

it's just the default assumption whenever the metric is P_b

[bitter herald]

We've solved many problems involving P_e but never once was cost mentioned

[waxen agate]

oh well

costs are only needed if the problem is asking you to minimise Bayes risk with unequal penalties

[waxen agate]

anyways, just pick a convenient c

[bitter herald]

i mean

z = 0

seems logical

[waxen agate]

yeah

c

[bitter herald]

whats c

C?

[waxen agate]

blud

[bitter herald]

oh

ok

so

[bitter herald]

so now we do total probabilioty eright

so what are the probabilities

P_b = P(z < 0, a_1 = 1) + P(z > 0, a_1 = -1)

@waxen agate

is this right

[waxen agate]

it's

P_b = P(decide -1 | a=+1) P(a=+1) P(decide +1 | a=-1)P(a=-1).

which is

P_b = P(z<0 | a=+1) P(a=+1) + P(z>0 | a=-1) P(a=-1).

[cinder flower]

[waxen agate]

with equal priors you have

P(a=+1)=P(a=-1)=1/2

[bitter herald]

blud

thats the same thing as i wrote

[waxen agate]

blud

i don't see your 1/2

[cinder flower]

blud is being disrespectful

[bitter herald]

i mean u didnt write it there either

[waxen agate]

blud might be blind

[bitter herald]

not in that message

.....

so like

,, \lm \int_{-0.2}^0\412\dd z

[bitter herald]

p(z < 0 | a = 1) = 0.2*0.5

[waxen agate]

indeed

[bitter herald]

p(z > 0 | a = -1) = 0.2*0.5

then algebra it

ok

ty pure!!

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[surreal marlin]

guys can someone help me with understanding deravatives

[zenith bramble]

That's awkward

[surreal marlin]

indeed

[zenith bramble]

*71

[plush bramble]

it's your channel, just ask

[zenith bramble]

In the answer key they didn't consider 0 as a point of non differentialbility but they took 2 tho .why so?

[plush bramble]

did you try using the limit definition of derivative

[pulsar flax]

the function is zero in [0,1/2]

and the constant function 0 is differentiable

[thorn agate]

check lim x-> 0+ for 0

and lim x-> 2- for 2

[pearl pond]

@zenith bramble Has your question been resolved?

[pulsar flax]

<@&268886789983436800>

[pearl pond]

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[exotic pasture]

if the length of the focal chord to the parabola y^2=4ax at a distance of 45 units from the vertex of the parabola is 20/81 units, then find the value of a

[exotic pasture]

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[fiery lance]

for c

how is it not:

arc CD = 2r * theta
arc BA = r(0.5 * pi - 0.5 * theta)

arc BA = arc EF

perim = arc CD + arc BA + arc EF + 4r
perim = 2r * theta + pi - theta + 4r
perim = 4r + (2r - 1) * theta + pi

[pearl pond]

@fiery lance Has your question been resolved?

[thorn agate]

You forgot to multiplh pi-theta by r

arc BA = r( pi - theta)/2

arc BA = arc EF

arc BA + arc EF
= 2 arc BA = 2r( pi - theta)/2

=r( pi - theta)

[fiery lance]

i see

.close

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[daring bay]

can someone tell me what i did wrong a)

,rotate

[pearl pond]

@daring bay Has your question been resolved?

[fresh olive]

$$p^2+1-2p+p^2=0,68$$
$$2p^2-2p+0,32=0$$

[jolly parrot]

BBMaths

[fresh olive]

Not -0,32

[daring bay]

oh wtf

wait bro

so here i tried doing these but im unsure

[fresh olive]

,rccw

[daring bay]

im so stressed

[fresh olive]

Why

[daring bay]

i got an exam Wednesday

[fresh olive]

Part a should be around 10^-4

Did you multiply by nCr(75,24)=75 choose 24

[daring bay]

uhh

i just put everything into

binomial distribution

n = 75
k= 24
p= 0,84

[fresh olive]

Does that say 1,000...*10^-23

[daring bay]

25

yeah

1.0000*10^-25

[fresh olive]

Try putting it in again

[daring bay]

yeah

yeah it says 1.00… * 10^-23

[fresh olive]

I think you are putting it in wrong maybe

Or wrong mode more likely

Do you have photos of what you're using to do the calculation

Oh nvm

Sorry

It was me lol

Yeah 10^-23

[daring bay]

phew

tbh im just unsure on the c) and d)

[fresh olive]

b is 0,718...

I think 0,719 is good enough but idk how many decimal places

[pearl pond]

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).

[fresh olive]

Why did you multiply by 40 in c

[daring bay]

uhh

i just thought the 40 would be relevant in some way because

i felt like i had to include them

but the exponents make that up right

[fresh olive]

Yeah the exponents sum to 40

So it has that information

So it only fails for 5th and 20th, there's only one way that can happen

With "It fails for 2 people" you multiply by how many ways you can choose 2 people

but they already chose the people

so no multiplying here

Part d translated badly

[daring bay]

wait so for c) just let the 40 out right

[fresh olive]

$0,84^{38}*0,16^2$

[daring bay]

no multiplying beteeen them?

or yes

[jolly parrot]

BBMaths

[fresh olive]

Yeah sorry

[daring bay]

no tysm

so for d )

the

ill translate

[fresh olive]

n=200

[daring bay]

"out of 200 people, gogo brain works for 157 people back to back and for the rest it doesnt"

[fresh olive]

Oh wait it did translate fine I can't read lol

Yeah $3,4076881372\times10^{-45}$

[jolly parrot]

BBMaths

[fresh olive]

200-(157-1)=200-157+1 because the streak cannot start in the last 157-1 people

[daring bay]

yesh

yes

lets go

ty

kinda satisfied

[fresh olive]

So a,b,d was fine c just remove the 40 times

[daring bay]

yeah ty

free points this section,

also

here, i translated it once for the words but the term it bugged out so

yeah idk what they want

ill open a new channel for this

.close

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[swift spindle]

lets say 'a' is the set {1,2,3} and 'b' is the set {3,4,5}, if i do A U B, is this {1,2,3,4,5} or {1,2,3,3,4,5}?

[toxic lichen]

those are the same set

but typically you would write that as {1,2,3,4,5}

sets don't have duplicates

also math is case sensitive so you should call these A and B throughout for consistency

[swift spindle]

ohh what

[swift spindle]

oh okay thanks

okay tysm

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[swift spindle]

wjahts the diff between belogns to symbol and subset

[swift spindle]

i feel like it does the same thing?

[sharp vigil]

quite different

Belongs to means a single member of the set: $1 \in {1, 2, 3, 4, 5}$

Subset means a set consisting of members of the set: ${1,2,3} \subseteq {1,2,3,4,5}$

[jolly parrot]

cloud ☁

[spare lark]

\in is kinda the singular form and \subset the plural form

[versed mica]

idk that i like this analogy

because a subset can be a singleton still

[swift spindle]

wait

[swift spindle]

so

so {1,2,3} cant belongs to R?

grammar there a bit iffy sorry

[sharp vigil]

{1,2,3} is not a member of the set of real numbers

it is however a subset of the real nymbers

[swift spindle]

ohh i see okok tysm

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[thorny radish]

what do you think

[fading ledge]

nth term test fails

cauchy test

[thorny radish]

yes

[pearl pond]

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[unkempt yacht]

why is leftside skew positive, whereas rightside skew is negative?

[sharp vigil]

the skew is in the direction of the "tail"

[unkempt yacht]

hm, any particular reason to it?

i expected the skew is in the direction of the highest column

[sharp vigil]

i think because the mean is farther in the direction of skew

compared to e.g. the median, which is closer to the visual center (which is the mode in these examples)

[bitter herald]

Think of like an exam where almost everyone got a 100%

But a few students forgot to study ans got 20%

Those few students drag/skew the average down

[unkempt yacht]

that interpretation could work

[unkempt yacht]

what statistical significance do the mean, median, and mode have?

[sharp vigil]

well the mode is the peak value of a distribution, which might be considered an intuitive notion of the center

mean and median both are measures of center, and in simple distributions like this, both are in the direction of skew away from the mode, the mean moreso than the median

(note: this is a rule of thumb, not a strict mathematical rule)

[unkempt yacht]

i see

[pearl pond]

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[quick venture]

can someone help me differentiate this

are we differentiating in terms of alpha?

[toxic lichen]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@quick venture Has your question been resolved?

[pearl pond]

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[fierce aspen]

i dont understand the diagram as well as the q

diagram is confusing cuz how will the mass slide on the vertical wall

[thin sentinel]

uh it is a bead on a hoop
one should consider conservation of energy:
here in this question, the relevant energies are:
gravitational potential energy, elastic potential energy, and the kinetic energy

[rough stream]

Is it supposed to say "when will the length of the spring become R?"

[thin sentinel]

the question is actually
"...velocity of the bead, when the length of the spring becomes R, would be (blank)"
apparently the blank to be filled in isn't shown

[thorn agate]

make equation relating kinetic energy

[pearl pond]

@fierce aspen Has your question been resolved?

[fierce aspen]

ik that

but if i leave it wont it just fall down

or shd i consider it travelling along the wall

[thin sentinel]

give a very small push so that it will slide down along the circle

forgive my handdrawing

[fierce aspen]

ok fine

got it

idk how to do this sum bro

pls help

@thin sentinel

[azure kite]

conserve energy at the initial postion and at a position where the spring wil have max compression

[thin sentinel]

-# nah there is friction at the bottom though, take that into consideration

[azure kite]

oh yeah, mb

you can remove the work done by friction from the initial energy and then equate it to the final block being on spring type

[fierce aspen]

wht abt the work doing while sliding

[azure kite]

take that into consideration
2+xmax

[fierce aspen]

(kx^2)/2 = MgH?

H= 5

[thin sentinel]

anyways still consider energy
in this case the energy at the beginning is only gravity potential
while at the end, the energy becomes potential in the spring + energy wasted in friction

[azure kite]

if friction wasnt there yes, i initially forgot to open the img

mb

[pearl pond]

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[fierce aspen]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[fierce aspen]

huh

but why do we consider friction thing here

[fierce aspen]

like final and initial only shd be sen ight

[azure kite]

energy is only conserved in a system when we have all conservative forces

friction is a...

non conservative force

[fierce aspen]

non conservative

[azure kite]

yeah

imagine this

[fierce aspen]

tbh idk much abt these conservative ad non conservative can u explain a bit brief

[fierce aspen]

thx, pls continue

[azure kite]

the block comes to the floor with velocity v, is it still gonna have velocity v

[fierce aspen]

ye

[azure kite]

drop the physics lets get some logic here! its gonna make sense

[azure kite]

like throughout the floor is it gonna travel in the same velocity?

[thorn agate]

https://tenor.com/view/psych-psychics-gif-18603165

[azure kite]

what is friction

[thorn agate]

basically non- conservative force draws energy from your system

[split void]

ill give a quick hint: divide the problem into two parts: one for the block sliding down the inclined plane, and another for the motion with friction

[fierce aspen]

no

[azure kite]

why is that

[fierce aspen]

ye that ik

[fierce aspen]

friction

[azure kite]

valid aproach!

[fierce aspen]

one more doubt

i have to find the velocity when it just touches the spring right

howd do i do that

[azure kite]

so you get why we cant just conserve energy of initial position and eng

[azure kite]

you remember shm formulaes?

[fierce aspen]

work done by friction is subracted

[fierce aspen]

nop

forgot

[azure kite]

you can either conserve energy at that point while subtracting the work done by friction (suggested as you would learn about how to conserve better as you lack that concept)

or use the Annie method and divide the question into two parts, first calculating the velocity of block initially when it touches the floor

and then use basic kinematics to solve for vewlocity after travelling 2m on the floor

[thorn agate]

https://www.feynmanlectures.caltech.edu/I_14.html
read if you want

[azure kite]

eh nah not required

imo

[thorn agate]

just for non- conservative force, i agree not required

[fierce aspen]

how do i calculate the velocity when it touches the floor
5Mg = Mv^2/2
10g = v^2
v= 10
oh im dumb

thaks

then v^2-u^2 = 2as
but idk accelaration

or shd i do Fcos60/5?

idts

[azure kite]

its gonna be retardation

[fierce aspen]

oh fuck mu N

[azure kite]

what

[fierce aspen]

ye

[fierce aspen]

mb my dum ahh

[azure kite]

take your time

its okay

[fierce aspen]

-.5 * 5g = a

bruh i meant -

[azure kite]

dimention wise wrong

[fierce aspen]

?

[azure kite]

not quite

ma = mu mg

[fierce aspen]

ye

bro im getting 10root 2 wth

[azure kite]

show your work

[thorn agate]

a or velocity?

[fierce aspen]

final veloicty

[fierce aspen]

dh phone

im getting 0 now 💀

v^2 - u^2 = 2as

-2*2*2.5g+ 100 = v^2
v= 100

[azure kite]

^
a=?

[fierce aspen]

-2.5g is what i got

bruh

-.5g

mb

[azure kite]

-2.5g is the force by friction

yeah

[fierce aspen]

u indian btw?

[azure kite]

yeah

[fierce aspen]

jee?

[azure kite]

nah

[dense axle]

jee is fun

btw whats the question

[fierce aspen]

if you went to coaching centre other than fiitjee.

broo

@azure kite the eqn is
1/mv^2 + 1/2kx^2 = -frictional area + MgH right

[pearl pond]

@fierce aspen Has your question been resolved?

[azure kite]

oh nah it was one of the best times i had in my life yet lmfao, very subjective veiw

[azure kite]

for finding the velocity just moment before it touches the spring?

what is this equation?

[fierce aspen]

S= (v^2-u^2)/2as

i got v^2 = 80

[azure kite]

?

for what

[tepid breach]

Hi

[pearl pond]

@fierce aspen Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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[whole knot]

just need help with a quick thing which i cant understand and ai cant help figure it out

[whole knot]

which one of the 4 is correct

[carmine schooner]

what exactly is it you'd like to know?

[whole knot]

and how do i know it incase of changes of the shape

[carmine schooner]

the area of the trapezoid?

[whole knot]

yes

[carmine schooner]

do you know the formula for the area?

of a general trapezoid

notice d is the perpendicular height

[whole knot]

i cant use any helping thing thats the issue

[carmine schooner]

ah, i see

[whole knot]

its just 4 equations and i have to cross which is right

[carmine schooner]

then instead notice you can cooy this shape and make a rectangle with the two copies

[bright apex]

@whole knot are u allowed to use the grids for help

[whole knot]

and my next exam will include a very similar task

[whole knot]

no

[weary ledge]

what are you allowed to use?

[whole knot]

my brain and a pen

[weary ledge]

so you said you know the formula for the area of a trapezoid yeah?

[whole knot]

no, theres gonna be 4 equations for the formula of the area and i have to cross which one is correct and have an explanation

[feral sedge]

No, like the area formula for a general trapezoid

[bright apex]

@whole knot cut the trapezium in halp diagonally use area of triangles?

help*

[carmine schooner]

just use that, it's pretty immediate from there

[crystal dew]

I also find it concerning that you're not allowed to refer to the grid... which is given to you

[quaint ledge]

try plugging in the side lengths into the formula

if you know it

[weary ledge]

this as well

[whole knot]

i can show u the whole thing one sec

[carmine schooner]

!original

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[crystal dew]

I get it if you can't refer to the grid that you drew over the trapezoid yourself or something

but that grid is given by the question, so not being able to use it is a bit

[whole knot]

Which term can be used to calculate the area of the figure?

Tick and give your explanation.

[carmine schooner]

then just by area of a trapezoid it's pretty fast

[random ermine]

or rectangle + triangle(s)

[weary ledge]

so i think the question is asking for proof of the formula? which i think you can get by counting the squares inside the trapezoid

not a proof, just an "explanation"

[whole knot]

and which one do i tick

[carmine schooner]

not allowed to give you solutions, but try eliminating a few

[bright apex]

is it vectors?

[crystal dew]

!noans please.

[pearl pond]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[quaint ledge]

try using the formula for the area of a trapezoid

with the given info

[carmine schooner]

^

[weary ledge]

i dont think youre supposed to do this op also doesnt know the formula

[bright apex]

what is 1,5 a b* c

[carmine schooner]

ah nvm holy ambiguity

[quaint ledge]

im pretty sure i got the ans from that, sorry i didnt know op didnt know the formula

[bright apex]

@carmine schooner

[carmine schooner]

hey i didnt write the question

[weary ledge]

it is trivial if u solve via that

[crystal dew]

if OP is willing to share some extra context about the worksheet/problem, what are the other questions on the same worksheet like?
that'd probably help understand what the question wants

for example, if the other questions are all proofs of areas of shapes from first principles/by picture, then we can assume this is too

[whole knot]

the other tasks are simple trigonometry, this one is the only different

[carmine schooner]

trig?? surely you can use the formula

[crystal dew]

hm, hm. this sounds like you should have the area formula in hand.
but in case not, it sounds like you're dealing with triangles a lot, and you should probably then consider slicing the trapezoid as sheaf mentioned

[whole knot]

yea thats when we will be given a book with formulas and a calculator (once the teacher sees we’re done with the small task and collects it)

[quaint ledge]

i dont rlly see another way (other than maybe bifurcation)

[bright apex]

what grade r u in

[weary ledge]

counting the squares

[crystal dew]

then consider doing this

[quaint ledge]

you cant assume the units tho

[whole knot]

11th

[crystal dew]

surely you have access to the area formulas for a quadrilateral and a triangle?

[weary ledge]

you can surely make it proportional

[bright apex]

it's 3rd option

u can tell there is a square

[crystal dew]

!nosols

[pearl pond]

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.

[whole knot]

i already got it before

but what do i write in the explanation part

[random ermine]

wow thanks for telling us

[whole knot]

having difficulty

[whole knot]

no i got it just know

the answer

[random ermine]

how

[whole knot]

before the guy sent the message

[crystal dew]

then how did you get it if not from that guy?

use that as your explanation

[random ermine]

someone told him

[bright apex]

...

[whole knot]

its what i used

i cut it into 2

square

triangle

[bright apex]

then write that

[whole knot]

alright

thanks for the help🙏🙏

[random ermine]

bruh

what

[crystal dew]

!done then

[pearl pond]

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

[random ermine]

so u suddenly got the solution

[bright apex]

if. ur teacher is not bad and knows u know how u got it then good

[random ermine]

sus af

[bright apex]

and mb for answering

[whole knot]

.close

[pearl pond]

Channel closed

Closed by @whole knot

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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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.

[unreal isle]

I'm looking for a study partner for probabilities any one in interested

[plush bramble]

#probability-statistics . help channels are for math questions

[unreal isle]

ok thank u

[dense jasper]

.close

[pearl pond]

Channel closed

Closed by @dense jasper

Use .reopen if this was a mistake.

[pearl pond]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[fringe fjord]

Hey

[fringe fjord]

Could use some help, Im not too sure what time frame to use for the amortisation formula

Is it the full 120 or 119?

[dense jasper]

,rotate

[dense jasper]

In the future, please use the ,rotate command and send the question as your first message since that's what the bot pins. But you should use the full 120.

[fringe fjord]

Alright, thanks man

[dense jasper]

[pearl pond]

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