help-0

[ocean seal]

Kepe

[thick lynx]

@tall oar

[tall oar]

Thank youuuu

[lone heart]

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[lean plover]

my question

is the answer

what is that answer saying? i dont understand it?

this is the question that that answer is the answer to btw

[solid cargo]

It’s like u can’t measure the probability when the data acquisition is biased right? Cz ur using that probability to generalise over the relation of fungi and bacteria over all buildings so u can’t do it if u don’t collect random samples

[inner vector]

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

[lone heart]

@lean plover Has your question been resolved?

[lean plover]

ah alright

so its equivalent to saying 'in order to reduce sampling bias'

.close

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[chrome salmon]

That's a bit trickey but maybe you can make cases

What's your work on it?

So far what have you tried?

[ripe rain]

Ya I think we must make cases

Like 1< i j k <10

[white jolt]

@ripe rain I've a doubt is it from a jee advance exercise just asking

[ripe rain]

Ya

[pale kestrel]

I feel like there's something you can draw here

[little drum]

I feel like there's a formula for these sums

which I forgot 🤦‍♂️

[ripe rain]

In this case sum will be $^{11} c_3$

[ocean seal]

Deep Hariya

[ripe rain]

?

[little drum]

oof

[ripe rain]

This will be for 1<i j k<10

[little drum]

$a_1(a_2(a_2 + a_3 + \ldots + a_{20}) + a_3(a_3 + \ldots + a_{20}) + \ldots + a_{19}(a_{19} + a_{20}) + a_{20}(a_{20}))$

[ocean seal]

Ansh

[little drum]

For i =1

[ripe rain]

Ya and i guess we must make cases?

[little drum]

$= a_1 (a_2+a_3 + \ldots + a_{20})^2$ ?

[ocean seal]

Ansh

[little drum]

nope some terms missing

[little drum]

I'm a 100 percent sure this bs has a formula to it

And a real nasty trick as well 🤦‍♂️ lemme try n figure

[ripe rain]

Hm

I am unable to figure out anything

[little drum]

$a_2(a_2 + a_3 + \ldots + a_{20}) + a_3(a_3 + \ldots + a_{20}) + \ldots + a_{19}(a_{19} + a_{20}) + a_{20}(a_{20})$

[ripe rain]

If it helps, this is problem of combinatorics

[ocean seal]

Ansh

[little drum]

Nope it doesn't help lol. I can figure at least that much

[ripe rain]

Ya

[little drum]

$= a_{20}(a_2 + a_3 + \ldots + a_{20}) + a_{19}(a_2 + \ldots + a_{19}) + \ldots$

[ocean seal]

Ansh

[little drum]

Add them up

[chrome salmon]

There's a generating function method

[little drum]

$=(a_2 + \ldots + a_{20})^2 + \sum_{i = 2}^{20} a_i^2$

[ocean seal]

Ansh

[little drum]

hence

[little drum]

$= \frac{a_1}{2} \qty((a_2 + \ldots + a_{20})^2 + \sum_{i = 2}^{20} a_i^2)$

[ocean seal]

Ansh

[little drum]

hmm

so your sum becomes $\frac{1}{2}\qty((a_2 + \ldots + a_{20})^2 + \sum_{i = 2}^{20} a_i^2)\qty(\sum_{i=1}^{19} a_i)$

?

[ripe rain]

Nope

Didn’t get it

[ocean seal]

Ansh

[little drum]

what's the answer btw (@_@;) if you have it

[ripe rain]

4585

[little drum]

oof

Lemme see

Question = $\frac{1}{2} \sum_{i=1}^{19} a_i \qty(\sum_{k=i+1}^{20} a_k^2 + \qty(\sum_{k=i+1}^{20} a_k)^2)$

[ocean seal]

Ansh

[little drum]

@ripe rain try this maybe?

can't make it more concise (@_@;)

or wait, maybe I can Lemme think

hmm?

[ripe rain]

Never mind

[little drum]

what +_+

[ripe rain]

Lemme digest it

[little drum]

Ahaa I got it!

[ripe rain]

I think there’s something wrong

[little drum]

I think I got what you meant when you said this from combi

[pale kestrel]

reuse indices, what

[little drum]

hmm?

[pale kestrel]

uhhhh

can u put a bracket around that middle summation smh

confused me

[little drum]

$\frac{1}{2} \sum_{i=1}^{19} a_i \qty(\qty(\sum_{k=i+1}^{20} a_k^2) + \qty(\sum_{k=i+1}^{20} a_k)^2)$

[ocean seal]

Ansh

[little drum]

hmm

ouch tho.. this still pretty stupid

[pale kestrel]

well you can simplify some of this to closed form

yh?

oh wait

what are the a_k

[little drum]

mhm

[pale kestrel]

oh ok, so closed form is possible

[little drum]

but still pretty stupid ;-;

aaaaa

[pale kestrel]

huh why?

what?

[little drum]

this is a combi question shuri

[pale kestrel]

This can surely simplify to 1 sum easily

[little drum]

obviously there's a possible translation

that gets you a concise form in one line

(@_@;)

[pale kestrel]

$\frac{1}{2} \sum_{i=1}^{19} a_i \qty(\qty(\sum_{k=i+1}^{20} a_k^2) + \qty(\sum_{k=i+1}^{20} a_k)^2)$

Or just use algebra . ..

[little drum]

coughs

\frac{1}{2} \sum_{i=1}^{19} a_i \qty(\qty(\sum_{k=i+1}^{20} a_k^2) + \qty(\sum_{k=i+1}^{20} a_k)^2)

[ocean seal]

Shuri2060

[pale kestrel]

$\frac{1}{2} \sum_{i=1}^{19} a_i \qty(50 + 900)$

[little drum]

wut

nuuuu

[pale kestrel]

oh i didnt readddddddddd

the bottom

[little drum]

k from (i+1) to 20

[pale kestrel]

hue hue i knew that

split into 2 case ig

its not that painful is it?

$\frac{1}{2} \sum_{i=1}^{10} a_i \qty(\qty(\sum_{k=i+1}^{10} a_k^2) + \qty(20+\sum_{k=i+1}^{10} a_k)^2 + 40)$
$+\frac{1}{2} \sum_{i=11}^{19} a_i \qty(\qty(\sum_{k=i+1}^{20} 4) + \qty(\sum_{k=i+1}^{20} 2)^2)$

@little drum better?

oh wait i fudged the top right

[ocean seal]

Shuri2060

[little drum]

Lmao

[pale kestrel]

yess?

[little drum]

I got 3520 though

And I'm wondering coz I'm sure I didn't mistake the summation part at least

For any i, the inside sum is $\sum_{i+1 \leq j \leq k \leq 20} a_j a_k$

[ocean seal]

Ansh

[little drum]

$= a_{i+1} (a_{i+1} + \ldots + a_{20}) + a_{i+2}(a_{i+2} + \ldots + a_{20}) + \ldots + a_{20} a_{20}$

[ocean seal]

Ansh

[little drum]

which is also the same as writing

$= a_{20}(a_{i+1} + a_{i+2} + \ldots + a_{20}) + \ldots + a_{i+1} a_{i+1}$

[ocean seal]

Ansh

[little drum]

so you add them up and take half and get what I wrote

but this actually gives a wrong answer after calculating?

[ripe rain]

Did you calculate?

I haven’t

[little drum]

-,- I got 3520 ... kinda

okay nvm

I miscalculated

the sum might be right 🤦‍♂️

[ripe rain]

Well then lemme check

[little drum]

,w sum from 1 to 10 (n^2 - 61n + 950)

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Ansh#3502

[little drum]

,w sum from 11 to 20 (20-n)(21-n)

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ansh#3502

[little drum]

,calc 6530/2 + 330*4

[ocean seal]

Result:

4585

[little drum]

@ripe rain let's gooooooo

Lmao

Got final expression as $\frac{1}{2} \sum_{i=1}^{10} (i^2 - 61i + 950) + 4\sum_{i=11}^{20} (20-i)(21-i)$

[ocean seal]

Ansh

[little drum]

hmm

[ripe rain]

I didn’t get the last part though

smh

[little drum]

which part?

[ripe rain]

$4\sum_{i=11}^{20} (20-i)(21-i)$

[little drum]

lmao

that's only half of it tho

[ocean seal]

Deep Hariya

[little drum]

ah

XD need to retry it then

Also

U sure you're supposed to do it this way? and not .. maybe make some combinatorial argument?

[ripe rain]

Well this is allowed but it maybe a bit too long way

Time consuming tbh

It is question of jee

[little drum]

took 3 mins wdym -,-

oops

not 3 really

[ripe rain]

Xd

[little drum]

Yeah there's supposed to be an easier way obv

[ripe rain]

More like 30 min

But yeah

[little drum]

wait tho

$\sum_{i<j\leq k \leq 20} a_ja_k$

[ocean seal]

Ansh

[little drum]

j and k can be written interchangeably

[ripe rain]

Idts

[little drum]

I was expecting it to break to $\qty(\sum_{i<j\leq 20} a_j )^2$

[ripe rain]

j=<k

Ahhh

[ocean seal]

Ansh

[ripe rain]

But this includes case when j>k?

[little drum]

hence, our bigger expression lol

Wouldn't it be much easier to take cases and simply evaluate the sum?

1 : 1(1 + ... + 1 + 2 + 2 + ... + 2 ) <- for a couple of times and done -,- takes 2 mins max

hmpffff

I'll ask someone abt this one though

( altho I believe I've been given this one before as a class question )

[ripe rain]

Ya well that was my approach during the test but it ended up taking 10 mins

[little drum]

Lol

You prepping for JEE?

[ripe rain]

Maybe cause i am weak in calculation

[ripe rain]

Ya

[little drum]

XD

Generalizing for a simple i < 11 and i > 10 case would've been much simpler

in which case you can directly use the sum thingy I shared

and rack some brains get the ans.

[ripe rain]

I made like 3-4 different cases

[little drum]

I think that much wasn't really needed

[ripe rain]

But i got the ans correct tho

[little drum]

Like you saw, I only made cases for i, and the j,k summation was cleaned through Gauss' Summation technique

[ripe rain]

Lol

[ripe rain]

Ya

[little drum]

(fyi there's no such technique. it's just the childhood way how Gauss got the sum of n natural no.s)

[little drum]

mhmm

btw, this question from PnC? or summation n series?

[ripe rain]

Will notify you tomorrow with solution of my teacher tho if interested?

[little drum]

sure :o

[ripe rain]

Its cumulative test so can’t really tell

[little drum]

hmmmm >_< It's totally gonna be a half-assed 3 line solution lmao

JEE be expecting newtons at the exam

[ripe rain]

Lets see tho

Well thanks then

.close

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[alpine sable]

Help on qs 11 please

[covert agate]

denote G = (1, 4)

consider the ratio of PG to GB to find P

and consider slope of GB to find Q

[lone heart]

@alpine sable Has your question been resolved?

[barren glade]

or he can just form an equation and do y=0

he has an advanced role

[covert agate]

probably just joined

i had every role when i joined lol

analysis number theory etc

i knew none and i still know none

[barren glade]

no you're pretty smart ive seen u around the help channels

[covert agate]

doesn’t mean i’ve attained enough knowledge on ‘advanced math’ lol

i’m struggling to learn sequences

[barren glade]

so you were a god back then

[covert agate]

huh no

[barren glade]

it's a joke

because you had every role

[covert agate]

probably shouldn’t hang around discord at 3

[barren glade]

oh are u indian by any chance it's 2am for me

1am**

[covert agate]

hong konger

[woeful heart]

same

[covert agate]

!?

[barren glade]

lol jinx

[covert agate]

jeng

[woeful heart]

how old r u

[covert agate]

form 3

[woeful heart]

damn

[covert agate]

you?

[woeful heart]

ur young

[barren glade]

old enough

[woeful heart]

im uni student

i thought u might be uni student also

cause u knew the double angle formulas for sin/cos

form 3 shouldn't know that

[covert agate]

coincidence zeh

[barren glade]

how old are form 3 ppl?

[lone heart]

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[vapid crypt]

how do I solve this?

[undone hinge]

you know the pq formular?

[quiet patio]

What do you need to do?

[vapid crypt]

what is that?

[undone hinge]

solve for a probably

[vapid crypt]

solve for a and b

[vapid crypt]

for a and b

I only know the quadratic formula for one variable

[undone hinge]

i guess you have to first solve it for a and then for b

[vapid crypt]

I set b to any value? then solve for a?

like b=1

[undone hinge]

nah you have to do it for every number

so let b be b

[vapid crypt]

what

b=b gives same formula

I don't understand

[undone hinge]

you dont get 1 or 2 values for a here

you have to do it dependent on b

[undone hinge]

or you can use the abc formular

in this situaton your "x" is a

[vapid crypt]

ok I got

[undone hinge]

where did you get that 42b^2 from?

[undone hinge]

and yea you have to write a instead of x because in the equation above is also a instead of x

[vapid crypt]

ok lets just assume a=x in this case

[vapid crypt]

b^2 - 4ac = (4b)^2 - 4(1)(b^2) = 16b^2 - 4b^2 = 12b^2

whoops I got a different answer

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@vapid crypt Has your question been resolved?

[vapid crypt]

.close

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[candid minnow]

can I get some help understanding why this is true

this is the solution but I don't understand where (-1)^3 came from and how det(A) = -det(A) implies Ax=0 has non trivial solutions

[placid zinc]

For a matrix of size n:
det(kA) = kⁿdet(A)

Which is a fancy way to say "Since multiplying a row by k will multiply the determinant by k, multiplying all n rows by k will multiply the determinant by kⁿ"

det(A) = -det(A) implies det(A) = 0 implies non-trivial solutions.

[candid minnow]

yea but why does det(A) = -det(A) imply det(A)=0 ? @placid zinc

[placid zinc]

Solve for x:
x = -x

[candid minnow]

ok mb i thought it worked dif for determinants

thanks

.close

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[ionic bison]

ok, real talk

[ionic bison]

why does equating coefficients work

[tacit arch]

• Ask your math question in a clear, concise manner.

[ionic bison]

yeah sorry

[bleak ridge]

Can you give example

[ionic bison]

so is it true that if given x^2+x+3 = Ax^2+Bx+C
the only solutions for A B and C are 1, 1, and 3 respectively?

[blissful whale]

yes, if you want to have it true for all x.

Here is a short proof

if you subtract one side from the other

you get

(A - 1)x² + (B - 1)x + (C - 3) = 0

assume one coefficient was not zero (i.e. for example A != 1 and A - 1 != 0)

then you could put in a very large x dominating the equation making this definitely not 0 in total.

but we want to find A, B and C so that it holds for all x

so A = 1 and B = 1 and C = 3 is necessary.

[ionic bison]

you know what, that makes sense. Thanks!

.close

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[eternal prairie]

Are my answer and the solution not equivalent? (I have been given 0 points on this)

[frosty eagle]

bruh no that's totally not equivalent

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[undone agate]

Totally stuck on this question. Is my diagram right that this is square pyramid? And slant height is 4?

I dont know what to do with the perimeter being 56 though

I thought we could do 56/4 = 14 for each square side but that doesnt make if we try to make a right triangle with 7 as the base and 4 as slant height

Plus perimeter would be all the sides added together

[ionic bison]

that picture is too small @undone agate Ican't read it

[undone agate]

Joey cuts a square out of a piece of construction paper and then tapes an equilateral triangle to each side of that square. He then traces the composite figure onto a piece of paper. If the perimeter of the composite figure Joey draws is 56cm and the height of each triangle is 4cm, what is the area of the composite figure?

I know we have to use something with the fact it is equilateral

but we cant assume that the sides are equal to the square sides, i think

[ionic bison]

have you drawn the composite figure?

[ionic bison]

what do you mean it doesn't work?

[undone agate]

Yes i drew it in a picture

[undone agate]

Well if we draw a right triangle inside, then the base would be 7 with slant height (hypoentuse 4)

So that cant be a triangle because hypotenuse is smaller

Plus i think perimeter means you add the square sides plus the 4 sides of the triangle that dont overlap with the square

[ionic bison]

draw the composite picture, so laid out flat on a table

it should look something like a 4 pointed star

[undone agate]

Seriously??

The way i saw it is that i cut out a square

Then glued a triangle to each side

Which makes a square pyramid

[ionic bison]

yeah but the composite picture

one sec

Never said he connected the sides

does that make sense? (sorry my drawing is terrible)

[undone agate]

Oh.......

So the base would be 14 of each triangle, and square side

Then we just find area of each one and add them up?

So 14 x 14 + 4(1/2 * 14 * 4)?

[undone agate]

Yeah it does. I waa trying to imagine having a paper toy square pyramid and tracing it

And that didnt make sense

[ionic bison]

well the perimeter is 56 right

of the composite picture

[undone agate]

Oh right so it is 56/8

[ionic bison]

exactly

[undone agate]

Thanks!

[ionic bison]

👍

[lone heart]

@undone agate Has your question been resolved?

[eternal prairie]

why not

[lone heart]

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[eternal prairie]

can someone explain why they might be different

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@eternal prairie Has your question been resolved?

[lone heart]

@eternal prairie Has your question been resolved?

[void birch]

what equation

oh nvm

you should ask that in the logic channel since not many logic post secondary's roam in the help servers

[lone heart]

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[radiant topaz]

i dont know how this last pic was achieved

[royal lark]

Bottom became sin^2 (x) and since you had sinx / sinx you could make that 1, leaving just 1 - cosx on top and the remaining sinx on the bottom

[radiant topaz]

thanks thats all i needed

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[warped sluice]

I'm confused on whether to factor this or not. Can i get some help

[bleak ridge]

Is this just for practice

@warped sluice

[warped sluice]

yes

these are the im kendall hunt practice problems

[bleak ridge]

I mean usually youd just distribute the negative and subtract each part

The real part and the imaginary part individually

[warped sluice]

ok

got it

ty

[bleak ridge]

[warped sluice]

pretty sure its B

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[final badge]

hello

[final badge]

Did I do it right for the second one?

for Point B

gravity number is 9.8m/s

;-;

I feel like im probably forgetting a step

eh

[lone heart]

@final badge Has your question been resolved?

[final badge]

hm.

.close

[lone heart]

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Please don't occupy multiple help channels.

[buoyant kayak]

1.) don't occupy multiple channels
2.) don't ping helpers right away
3.) don't ping random people

#❓how-to-get-help #rules i'd suggest some light reading

<@&268886789983436800>

[sly mantle]

👢

.close

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[hard wing]

trying to simplify for question l

shouldI multiply negative into square root of 7 + x

[pine kettle]

question l (L)

[hard wing]

yes!

[pine kettle]

Well

If we first try to plug in 0

We get divide by 0

Which means we have to keep working and have to transform the equation

[hard wing]

yea ik

[pine kettle]

So

[hard wing]

i need to rationalize the numerator

but not sure how to proceed with it

[pine kettle]

You will want to multiply by the conjugate, have you heard of this before?

[hard wing]

yes

multiply by what though

√7 -x + √7 +x ?

[pine kettle]

Make sure its a "giant one"

[hard wing]

?

wdym

√7-x - √7 + x multiplied by √7-x + √7 + x

am i correct?

[wary stream]

That's supposed to $\sqrt{7-x}+\sqrt{7+x}$, right?

[ocean seal]

dldh06

[hard wing]

yeye

[wary stream]

That's what sills meant by giant one

[pine kettle]

so $\cdot \frac{\sqrt{7-x}+\sqrt{7+x}}{\sqrt{7-x}+\sqrt{7+x}}$

[wary stream]

Giant sqrt

[hard wing]

?

oh ok

[wary stream]

Because yours only was over the 7

[hard wing]

yeah

so my end result

oh

[pine kettle]

I HATE LATEX!!!

[ocean seal]

sills

[hard wing]

yes

yea thats what i meant

[pine kettle]

there we go

[wary stream]

This looked like $\sqrt{7} -x+\sqrt{7} +x$

[ocean seal]

dldh06

[hard wing]

oh no

lol

how do i do this

[hard wing]

to better illustrate lol

how about this??

[misty dagger]

distribution

[hard wing]

is this correct?

[wary stream]

Where did 2 come from?

[misty dagger]

Lol

[hard wing]

its a different question***

[misty dagger]

I was about to say

[hard wing]

this is the starting equation

[misty dagger]

Ye look right

It's +

2+squareroot4+x

Not -

[hard wing]

?

[misty dagger]

You start 2-squareroot 4+x. I assumed you start new problem

[hard wing]

yea

so this, yea?

[misty dagger]

Ye

Also that's not -2

Lmao

[hard wing]

oh why?

just 2?

[misty dagger]

Unless you put -2 in beginning

[hard wing]

no

[misty dagger]

Then no

[hard wing]

the starting equation is just 2

oh ok

would this be correct?

[misty dagger]

No that is funny

[hard wing]

?

would both 4 and x be negative?

[misty dagger]

That x should be cancel out

[hard wing]

yes i know

but after conjugating

this is what im left with for now

its correct right?

ok isee

i got it correct thanks!

[hard wing]

sorry

for this

[misty dagger]

You should simplify down more since I didn't do the math with you lmao

[hard wing]

ye ik

[misty dagger]

I just know bottom x is just gone

[hard wing]

i am left with this right?

(before simplifying)

original equation is

@misty dagger

[misty dagger]

Is this a new problem

[hard wing]

yea

[hard wing]

original equation is as follows

[misty dagger]

Didn't the helper explains?

This

[hard wing]

yeah

but i just wanted to make sure

[misty dagger]

Ye it's right

lol

[hard wing]

ok but like

[hard wing]

the top cancels each other out right?

7 - 7

-x + x

?

but the final answer is

-1/square root (7)

how -1?

[misty dagger]

-2x

lol

[hard wing]

?

[misty dagger]

Something is wrong with your distribution

[hard wing]

-x + x

is it this?

[misty dagger]

(squareroot(7-x)-(squareroot(7+x))(squareroot(7-x)+squareroot(7+x))

Right

[hard wing]

yea

[misty dagger]

If you would think if this as (7x+1)(2x+1)

[misty dagger]

How would you done it?

[hard wing]

mutliply 7x and 2x

7x and 1

1 and 2x

1 and 1

[misty dagger]

So?

(squareroot(7-x)-(squareroot(7+x))(squareroot(7-x)+squareroot(7+x))

[hard wing]

oh

so its

7-x -7-x

?

after multiplying

[misty dagger]

Ye

So

[hard wing]

ah okay

[misty dagger]

You know what to do the rest

[hard wing]

wait

but im getting wrong answer

[misty dagger]

??

[hard wing]

numerator is -2x right?

then igot rid of the X

[misty dagger]

Ye

It's right

[hard wing]

im lef with -2/ √7-x - √7+x

but wouldnt this be -2/0?

[misty dagger]

???

[hard wing]

??

is this right?

square root 7 - square root 7

[misty dagger]

it's adding

lmao

[hard wing]

o

so how did they get

-1/root 7

what is cancelled out

[misty dagger]

Wut u mean?

??

what is square root 7 + square root 7

[hard wing]

square root 14

oh

-1/square root of 7

oh

fuck me

[misty dagger]

lol

[hard wing]

thank you lmao

[misty dagger]

yup

[hard wing]

fuck fuck me

thank you

[lone heart]

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[alpine sable]

Could someone please help me explain this?

[wary stream]

Find the roots

[alpine sable]

Hwo do I do that

[pine kettle]

The roots of a function are when:

[alpine sable]

When?

[pine kettle]

When the graph intercepts with the x axis

So 0 when y = 0

A ball is thrown up, it goes up, then comes back down, so when y (height) = 0, is when the ball is back on the ground

[alpine sable]

Would there be a formula I could follow?

[pine kettle]

Uh not really

This is understanding properties of quadratic functions

[alpine sable]

Oh ok thank you

.close

[lone heart]

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[teal glen]

Can anyone check if my part A solution is correct for 1.61?

[teal glen]

$1-\frac{\text{52 x 48 x 44 x 40 x 36 x 32 x 28}}{\text{52 x 51 x 50 x 49 x 48 x 47 x 46}}$

[ocean seal]

fake tree

[teal glen]

This is what I have

please ping me if you reply, am working on part b of the problem right now

[tacit arch]

Show how you arrived at this

[teal glen]

sure. Let me take a photo, 1 second

Not much work but that's just what I have written down

52, 48 (1 denomination eliminated), 44 (2 eliminated), ...

[lone heart]

@teal glen Has your question been resolved?

[teal glen]

<@&286206848099549185>

[tacit arch]

I don't understand why 48

Oh it must be different from the first card drawn so 52-4

[teal glen]

Yea, does that look correct? or way off

[naive valley]

Looks right to me.

That page has some fairly cool questions, what book is this?

[teal glen]

It's called knowing the odds by john walsh @naive valley

[naive valley]

Nice, I didn't know about this book.

[teal glen]

The book and class is kinda tough for me at least 😵‍💫

Kinda confusing how we go over the borel sigma algebra stuff and never touch it again (at least so far in the problems)

[naive valley]

Yeah, I just took a look at the first few pages on Amazon and it's an immediate crash course in sigma algebras and monotone classes without much chatter. Usually books take a bit more time to motivate this stuff!

[lone heart]

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[neat sierra]

is that all the question gives

[lone heart]

@grizzled garden Has your question been resolved?

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[lone heart]

@cunning trout Has your question been resolved?

[cunning trout]

.close

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[dapper sparrow]

How does it become 20√x19?

when simplifying

how is the first sqrt(x becomeing 20sqrt(x10)?

[little drum]

,rotate

[little drum]

would be much easier for you if you revise the law of indices

$\sqrt[n]{x^m} = x^{\frac{m}{n}}$

[dapper sparrow]

n being what?

[ocean seal]

Ansh

[dapper sparrow]

n and m being what?

[little drum]

Of course I'm talking about m and n being complex numbers with no real parts ?

Is that even relevant to the context?

...

[dapper sparrow]

i was given notes on this but i need more in depth guide to it all

[little drum]

For the moment though, let's deal with your question first?

[dapper sparrow]

so in this particular situation what would n and m be?

[little drum]

Here, you're dealing with $\sqrt[4]{x}, \sqrt[5]{x}$ and $\sqrt[2]{x}$

[ocean seal]

Ansh

[little drum]

so your n and m are clearly 2, 4 and 5 with m being 1 for each radical no?

[dapper sparrow]

yes

first one being x1/2?

x^1/2

[little drum]

So, we can rewrite them as: $x^{\frac{1}{4}}, x^{\frac{1}{5}}$ and $x^{\frac{1}{2}}$

[ocean seal]

Ansh

[dapper sparrow]

an then you just multiply them left to right?

[little drum]

multiply?

Yeah. in case they're parent and daughter radicals

but if the radicals were multiplied initially, you'd add the indices no?

In your question, the first part shows nested radicals so you'll multiply the indices, and the right second part shows a product of radicals in which case you'd add the indices

[dapper sparrow]

im lost

[little drum]

Recall / Revise properties of indices

$\sqrt[4]{\sqrt[5]{\sqrt{x}}} = \qty(\qty(x^{\frac{1}{2}})^{\frac{1}{5}})^{\frac{1}{4}}$

[ocean seal]

Ansh

[dapper sparrow]

that helps a lot

[little drum]

$= x^{\frac{1}{2}\cdot \frac{1}{5}\cdot \frac{1}{4}}$

[ocean seal]

Ansh

[little drum]

using properties of indices

[dapper sparrow]

wait no im lost again

[little drum]

However, $\sqrt[4]{x} \cdot \sqrt[5]{x}\cdot \sqrt{x} = x^{\frac{1}{2}} \cdot x^{\frac{1}{4}} \cdot x^{\frac{1}{5}}$

[ocean seal]

Ansh

[dapper sparrow]

rn i have 40sqrtx+40sqrtx

[little drum]

of course you'll have that :/

When multiplying common bases with uncommon indices, you .... don't multiply the indices

That's all I got to say. Check google for properties of indices. :D

[dapper sparrow]

alright ill come back here if i dont get it...

.close

[lone heart]

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[rapid lava]

does anybody know how to do these 😭

answers with steps would be greatly appreciated!

i understand the formula but idk how to apply it

[buoyant kayak]

are you familiar with the unit circle?

[rapid lava]

somewhat

[buoyant kayak]

can you tell me what values of x sin(x)=-1/2?

[rapid lava]

this paper is due in 10 minutes 💀

[buoyant kayak]

that's quite unfortunate

[rapid lava]

it truly is

would x be 1

or

am i really far off

<@&286206848099549185>

so so lost

[tacit arch]

Pick one problem and show what you've tried and someone can help

[rapid lava]

am i close

[tacit arch]

On the right track yea

Find cos(theta)

[rapid lava]

for num 2?

was 2 even right 😭

i will literally venmo somebody rn who can do this

im down bad

like literally

[tacit arch]

Yea 2. You just need to use sohcahtoa to find cosine of the angle. Then you'll have enough to calculate sin(2theta)

[lone heart]

@rapid lava Has your question been resolved?

[rapid lava]

can somebody show the questions solved with steps 😭

ill send $20 frr

[warm patio]

20$?

[gray isle]

Do not offer money for doing homework assignments, and vice versa.

#rules

[warm patio]

@rapid lava

True

[lone heart]

@rapid lava Has your question been resolved?

[lone heart]

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[frank rapids]

can someone help me with this question? (permutations)
i’m having trouble with part b).

[frank rapids]

my work: (parts a & c are correct i suppose; i just dk how to solve part b)

[hasty elk]

unless i'm very mistaken, it's like selecting a team with the four main members, and two people as well to act as reserves

[frank rapids]

i think there can only be 4 members in total out of the original 10 no?

[hasty elk]

right, but unless i'm mistaken, in part b the "team" is four members, and two people who can fill in when someone gets sick

[frank rapids]

if that’s the case, then how would i solve it?

i know that the final answer is supposed to be 151 200 (from the back of the book)

151,200*

[hasty elk]

the final answer is a little weird with the current question phrasing, but i think with that i know what the question means

it means that you have four members

and then you have Reserve 1 and Reserve 2

each of these positions are distinct

[frank rapids]

ahhh ur right! so i solved it the way u explained —if theres two Alternate members (basically just two extra members) & i got 151,200.

thx!

[hasty elk]

yay!

[lone heart]

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[sweet vessel]

is this equation impossible to get a closed form solution for? tan(x)+x=y, solve for x

[rapid mist]

Wait, is that just the normal tangent function shifted up by x?

[sweet vessel]

yeah

its like a stepping tangent function

https://www.desmos.com/calculator/hfoysceonb

[rapid mist]

Geometrically it's just the tangent function that goes inclined.

[sweet vessel]

Well, nevermind my actual question is how can I find the polar equation for the involute function

[rapid mist]

What's the involute function?

[sweet vessel]

inv(α)=tan(α)−α

basically what I said earlier

[rapid mist]

is α just the input?

@sweet vessel

[sweet vessel]

yeah

but basically I want θ=inv(α)

so I want to find α

to plot it in polar equation

[rapid mist]

Just making sure, is inv() just 1/()?

[sweet vessel]

no its involute function

[sweet vessel]

this

α is the input

[rapid mist]

A function in itself?

[sweet vessel]

yes

[rapid mist]

Wait nevermind, there is an = sign there.

@sweet vessel

[sweet vessel]

hmmm nevermind

I don't think its possible without newton's method

or other approximations

.close

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[plain torrent]

I’m supposed to solve for f(t) = 2.5t^2 + 6t

[plain torrent]

i think i messed up somewhere but idk where

im been trying for a while

[gray isle]

t or x

[plain torrent]

i replaced t for x

[gray isle]

your f(x+h) is incorrect in the first line

[plain torrent]

oh

i see

(x + h)^2 yeah

x^2 2xh h2

[gray isle]

that's not quite what i was referring to

but that expansion was also a mistake

[plain torrent]

but how though

should it be

(2.5x + h)^2

[gray isle]

no

you replaced the t in 2.5t^2 with (x+h)
but only replaced the t in 6t with x instead of (x+h)

[plain torrent]

are you saying it should be 6(x+h)?

[gray isle]

yes

[plain torrent]

oh ok

thank you

.close

[lone heart]

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[teal glen]

Hi, second question (not sure if this is allowed though). Johnny has not studied for tomorrow's test, so his probability of passing is 1/4. However, if he cheats, he will raise his probability of passing to 3/4. He has not decided whether or not to cheat, but he is leaning that way: the probability that he cheats is 3/4. Suppose he takes his test and passes it. What is the probability that he cheated?

[lone heart]

Please don't occupy multiple help channels.

[teal glen]

Oh sorry, I thought 2 were allowed, my bad.

[lone heart]

@teal glen Has your question been resolved?

[teal glen]

<@&286206848099549185>

Is this just conditional prob.

P(Johnny cheated given that he passed the exam)

oh wait no is it more nested?

is it P(cheat)P(some conditional) + P(not cheating)P(some conditional)? seems like it

[ocean dirge]

can someone help

[vale wigeon]

@ocean dirge get your own channel. #❓how-to-get-help

[rapid mist]

3/4. It's in the question, if that's not what you mean? WORD YOUR QUESTIONS BETTER!

[ocean dirge]

oh

[teal glen]

that's not right? @rapid mist

you might be missing the nuance in the question....

[vale wigeon]

the4spaceconstants is confusing P(cheat) with P(cheat | pass)

[teal glen]

Yea I think so as well...

[vale wigeon]

you have here:
P(pass | cheat) = 3/4
P(pass | no cheat) = 1/4
P(cheat) = 3/4

P(pass) = P(pass|cheat)P(cheat) + P(pass|no cheat)P(no cheat) = 3/4 * 3/4 + 1/4 * 1/4 = 10/16

= 5/8

P(cheat|pass) = P(cheat)P(pass|cheat)/P(pass)

bayes' theorem

[rapid mist]

Pass probability: 1/16 + (3/4)^2

[teal glen]

oh ok got it, thanks forgot about bayes theorem

[rapid mist]

5/8, we thought the same way.

[rapid mist]

9/10?

[teal glen]

Don't think it is 9/10 at least based on the conditioning given

sorry, you got it right

I got the conditioning the other way around

I was looking at the alternative for bayes rule

[rapid mist]

Things got squared.

(3/4)^2, (1/4)^2

[teal glen]

Sure I'm just trying to understand how they come to be usually

the computation is irrelevant most of the time

[lone heart]

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[hazy stirrup]

i need help with c

i got d=absolute value(2q-q^2+3)/sqrt(5)

i did some research but i found out that you're supposed to use derivatives

[vocal edge]

umm do you know derivatives?

[hazy stirrup]

no

[vocal edge]

oh

[hazy stirrup]

its meant to be doable without them

[vocal edge]

ok lemme think

[pine kettle]

Oh mbmb

@dull oak check out #❓how-to-get-help

[hazy stirrup]

all g

[pine kettle]

@dull oak #❓how-to-get-help

[dull oak]

can u talk in dms @pine kettle

ill delete my messages

[severe sluice]

maybe subtract the equations and find vertex

[pine kettle]

@dull oak #❓how-to-get-help

[hazy stirrup]

yeah i tried that but the vertex isn't the closest point

<@&286206848099549185>

[severe sluice]

maybe you got the vertex of y=x^2

[hazy stirrup]

yeah its 0,0

[severe sluice]

yep

so that's incorrect

[hazy stirrup]

mhm

[severe sluice]

you need to find the vertex of the distance formula

[severe sluice]

btw this is also incorrect.

[hazy stirrup]

oops its -3

typo

[severe sluice]

still incorrect

there shouldnt be a /sqrt(5) term

[hazy stirrup]

what

isn't that the perpendicular distance formula

[severe sluice]

what?

oh nvm

just find the vertex

[hazy stirrup]

vertex of what

[severe sluice]

the distance equation! it's a quadratic

[hazy stirrup]

ill try that

[vocal edge]

https://www.desmos.com/calculator/zy7xw89q0e
it's the vertex of distance equation

[severe sluice]

maybe they meant perpendicular distance

[vocal edge]

ohhh

😐

[hazy stirrup]

wait no thats fine

you got the point

the one closest to the line

how

@vocal edge

[vocal edge]

turn on this equation.
Minimum distance is at the vertex of this parabola (equation for distance).

[hazy stirrup]

wait why

[vocal edge]

cuz it's the minimum value of distance

I think they probably mean perpendicular distance in your question

meglo is right

😐

[hazy stirrup]

yep sry forgot to mention this was from a perpendicular distance worksheet

guess i give up

ill just wait for tmr

its midnight

.close

[lone heart]

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[earnest lodge]

anyone can help me check this question?

it's partial fractions

[vague coral]

$(x-1)^3 = (x-1)(x-1)^2 = (x-1)(x^2 -2x+1)$
$$\frac{x^3}{(x-1)^3} = \frac{A}{x-1} + \frac{Bx+C}{x^2-2x+1}$$

[ocean seal]

Herels

[earnest lodge]

is doing it my way not allowed?

[vague coral]

you were wrong tho

[earnest lodge]

oh

[vague coral]

wait a sec

[earnest lodge]

I was looking at an example on YouTube

and followed it

[vague coral]

you were right I think

[earnest lodge]

but when I searched on Wolfram

they gave this

so I'm very confused

[vague coral]

I think I got it now

since the two polynomials have the same degree, just do a long division

[earnest lodge]

I dont get it

[vague coral]

I got :
$$1 + \frac{3}{x-1} + \frac{3}{(x-1)^2} + \frac{1}{(x-1)^3}$$

[ocean seal]

Herels

[vague coral]

just divide x^3 by (x-1)^3

[earnest lodge]

is that what the qn wants? tho

my teacher told me to do like this

like that formula

[vague coral]

Well yeah but its kinda impossible to use this because the two polynomials have the same degree

So its better to do a long division

[earnest lodge]

I saw this on a web

[vague coral]

Your exercise is x^3/(x-1)^3

[earnest lodge]

yea

but isn't the method the same since the denominator is cubed?

[vague coral]

In your case, the numerator and the denominator have the same degree

so, no

[earnest lodge]

ohhhhhhhhhhhhhhhhhhhhhhhhhh

okay makes sense

[vague coral]

a quick long division and its done

[earnest lodge]

uh how do I do long division for this

[vague coral]

x divides by x-1 and then you cube it

or if you have good eyes,
$\frac{x}{x-1} = \frac{x-1+1}{x-1}$

[ocean seal]

Herels

[vague coral]

so $\frac{x}{x-1} = 1+\frac{1}{x-1}$

[ocean seal]

Herels

[earnest lodge]

ahhhh yess yes

I did it

it looks like the solution from the website thank you

[lone heart]

@earnest lodge Has your question been resolved?

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• 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.

[brittle sun]

(This is a math question in the context of Induction, but my question relates to Algebra)
The following is a solution to said math question