help-26

[sacred cloud]

the second photo is my solution

[gray ridge]

oh lol

[sacred cloud]

HELLO

[gray ridge]

did you check my responce?

back then

[sacred cloud]

yup

i dont really get it thoughhh

[gray ridge]

oh

lemme try to explain

[sacred cloud]

also what is the iterator? we dont really use that term in school

sorryyy

[gray ridge]

summation within summation is indeed nasty but once you get the idea, you'll get it

[gray ridge]

oh, it's the variable that is used to count from the bottom number to the top number

in a summation notation

[sacred cloud]

so the iterator for it is 3 & 4?

cz the summation gets 4 terms then the one inside it gets 3

[gray ridge]

nah, the iterator is n for the interior summation

[sacred cloud]

OH

oops

[gray ridge]

no worries

[sacred cloud]

is my solution wronggg 😞

[gray ridge]

should be wrong... sadly

[sacred cloud]

oh noooo

cld u explain why or is it confusing

[gray ridge]

it's alright

but we need some time

let's take a easier example

[sacred cloud]

i have 2 diff answers because i wasnt sure if i could go over the upper limit for the summation inside

[gray ridge]

ah...

[sacred cloud]

since following the limits u get 3 terms for the summation inside but 4 terms for the outer summation

[gray ridge]

$\sum_{n=2}^{3}\left(\sum_{i=1}^{4}(n+i)\right)$

[thorny flame]

Biscuity

[gray ridge]

therefore there will be a total of 3×4=12 terms

[gray ridge]

like usual calculation, we will tackle the things inside the brackets first

[gray ridge]

$\sum_{i=1}^{4}(n+i)=(n+1)+(n+2)+(n+3)+(n+4)$

[thorny flame]

Biscuity

[gray ridge]

all good till here? @sacred cloud

[sacred cloud]

mhm

[gray ridge]

now

[sacred cloud]

so u expand the one inside first?

[gray ridge]

we plug this into the outer one

[sacred cloud]

u can do this

🙏🙏

[gray ridge]

so, this is =4n+10

now we have simplified it

and we can plug into the outer one

$\sum_{n=2}^{3}\left(\sum_{i=1}^{4}(n+i)\right)=\sum_{n=2}^{3}(4n+10)=(4(2)+10)+(4(3)+10)$

[thorny flame]

Biscuity

[gray ridge]

all good?

[sacred cloud]

waitt im analyzing

[gray ridge]

sure

take your time

[sacred cloud]

OKAY im following

so u use the series from the inner summation to do the outer summation?

[gray ridge]

yep, and that's the way to do it

[sacred cloud]

omygawd

thank u SMMM

[gray ridge]

but note that only the iterator will change

[sacred cloud]

yup!!

iterator is n right

[gray ridge]

and if otherwise stated, we can treat others as constants

[gray ridge]

for the outer one, yea

[sacred cloud]

OKAYYYY

i get it

omg

[gray ridge]

back to the original then?

[sacred cloud]

waitt am i allowed to ask another problem

joke

i dont get how to solve the original one

[gray ridge]

oh yea, sure, since I'm free right now

[sacred cloud]

cz the iterator for the outer one is inside the inner one

[gray ridge]

oh

[gray ridge]

we can treat the outer one as constant inside the inner one

[sacred cloud]

what does that mean

[gray ridge]

lemme show

[sacred cloud]

sorryyy im not really good with terms in math

[gray ridge]

it's sum of n=2 to 4 (mr-np)

so we can just do:
(mr-2p)+(mr-3p)+(mr+4p)

that's the inner part

[sacred cloud]

and then u plug that into the outer one?

[gray ridge]

simplify and plug, yea

[sacred cloud]

OH simplify first yeah

OK I GET IT GUYSS

[gray ridge]

[sacred cloud]

this is the other question but i didnt answer it cz i was confused

[gray ridge]

do tell

[sacred cloud]

wait oops it didnt crop

[gray ridge]

reading

ah

[sacred cloud]

OK this, cz i know how u supply other parts and stuff

but i was confused cz the lower limit didnt change

[gray ridge]

haha

it's the same, just go from 1 to 4 on LHS, and 1 to 3 on RHS separately

[sacred cloud]

buuut how am i supposed to solve for x if i dont know how the iterator increased/decreased

since thats what we usually do, base it off the iterator / lower limit

[gray ridge]

with summation sign, we always go from bottom to top, increasing 1 for each step

[sacred cloud]

do i expand and evaluate first? and then solve for x?

omygawd i didnt think of that earlier

THANK YOUUUUU

i get the concept now

[gray ridge]

yay

[sacred cloud]

i hope u have a nice day/night

[gray ridge]

you too!

[sacred cloud]

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[sacred cloud]

WAAITT

.reopen

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✅

[gray ridge]

?

[sacred cloud]

for #4 a

when it says in terms of r and p

what am i supposed to write

js the equation without the values of the variables?

[gray ridge]

you need so write something like
(number)r + (another number)p

just the expression, yea

[sacred cloud]

OKIEEEE

am done

thank u so much 💗💗

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[pine viper]

differentiate with respect to x, i dont understand the working :') answer is 3(1 - 2xcot2x)cosec2x

[restive inlet]

which part of the working don't you understand

did they show any steps?

[pine viper]

no 😭 thats why i dont understand

i tried using quotient rule but didnt end up with the answer

[restive inlet]

what do you have after applying quotient rule?

[pine viper]

( 3(sin2x) - 3x(2cos2x) )/(sin2x)^2

idk if im wrong for this or im missing a step

[flint stump]

sin2x/sin2x = 1 (you isolate single sin2x from denomintor)

similar for cot

and the second sin2x in denominator becomes cosec

so both answers are the same, just they didnt write it as fractions

theres no multiplication denoted, the x you see is indeed the term x and not the operation

[pine viper]

whered u get cot

[restive inlet]

sry, misread that

[flint stump]

cos2x/sin2x

[pine viper]

u seperate the sin2x^2 into sin2x . sin2x

[flint stump]

yes

[pine viper]

then cross out the 3(sin2x) with sin2x?

making (3 - 3x(2cos2x) ). 1/sinx

whats what i got

[flint stump]

3x(2cos2x) also needs to be divided by sin2x

[pine viper]

so then

[flint stump]

coz you have subtraction

[pine viper]

(3 - 3x(2cot2x)) . 1/sinx ?

[flint stump]

1/sin2x

[pine viper]

oh ya

mistype

[flint stump]

thats csc2x

[pine viper]

thats still different from answer tho right

[flint stump]

you take 3 common out of the numerator term, and you make the csc2x, and you get the answer

[pine viper]

3(1-x(2cot2x)cosecx

aaah

okay

i didnt know u can make x(2cot2x) into 2xcot2x

ok thanks

[flint stump]

np

[pine viper]

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[obsidian plank]

Angle incident and reflected ray is 90⁰ then how is this possible shouldn't they use m.m1=-1 in perpendicular lines

[obsidian plank]

Shouldn't be slope of reflected ray be root3

<@&286206848099549185>

[grand trellis]

Incident and reflected rays aren't perpendicular

[obsidian plank]

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[grand trellis]

Dunno how to start

[pliant yacht]

Plot the points

Draw a line a parallelogram

[grand trellis]

Already did

Doesn't really help much tho

[pliant yacht]

Draw a line thru the origin that divides it into congruent pieces

Solve gradient

[grand trellis]

I am not supposed to find it by accurate plotting

I need to solve using coordinate geometry

[pliant yacht]

Why not lol

[grand trellis]

Because that's too easy lol

[pliant yacht]

Baha

I see

[grand trellis]

• the points are like 153 and 10 💀

[pliant yacht]

Oh i didnt rlly read it

[grand trellis]

Oh alr

[pliant yacht]

Can i see ur graph

[grand trellis]

It's really rough

[pliant yacht]

That’s fine

[grand trellis]

The line which divides is drawn a bit wrong too ;-;

But doesn't really matter tbh

[pliant yacht]

Well

You wanna divide it diagonally from corner to corner

And then u will see your line

[grand trellis]

How

It can also be a little tilted in the middle

[pliant yacht]

Which u will notice that it matches the 2 sides of ur parallelogram that have the same gradient as that line

[grand trellis]

Alr lemme try it then brb

[grand trellis]

Wait

The slopes of AD and BC are undefined ;-;

[pliant yacht]

U can solve it

[grand trellis]

The difference in the x terms is zero

[pliant yacht]

y 2 - y 1
————
x 2 - x 1

[grand trellis]

114-45/ 10-10

Anything divided by zero is undefined ;-;

[pliant yacht]

Wuh oh

Oh i see

[grand trellis]

The options also have 99 , 19 and 9 in the denominator 😭

[pliant yacht]

I think you drew ur parallelogram wrong

😭

[grand trellis]

Do you know some property?

[grand trellis]

What ;-;

[grand trellis]

Like if a line divides a parallelogram into 2 congruent parts is it perpendicular to something maybe

[pliant yacht]

(10, 114) and (10, 45) shld be directly above and below eachother

[grand trellis]

Ik 😭

My graphs are really bad

[pliant yacht]

Same for (28, 153) (28, 84)

[grand trellis]

Yea 😭

[pliant yacht]

Well that changes everything 😭

[grand trellis]

I just made one for the general idea

[grand trellis]

How ;-;

[pliant yacht]

This is a square

[grand trellis]

Huh?

[pliant yacht]

Or rectangle

I wld draw it again

[grand trellis]

Probably isn't 😭

[grand trellis]

Alr wait lemme make a more accurate one

Something like this probably

[pliant yacht]

Okay

So when u draw ur line through the origin to cut it into half

You will notice it goes from corner to corner

(10,45) and (25,153)
Those r ur points

[grand trellis]

How do we know that for sure tho ;-;

[pliant yacht]

Now solve the gradient

[pliant yacht]

Bc that’s how congruency works

[grand trellis]

20/108

Which is uhh

10/54

5/27

[pliant yacht]

Oh btw idk why i said it’s a square/rectangle i just meant u have ur slants on the wrong side (it’s 1am 😵)

[grand trellis]

It's fine lmao

[grand trellis]

This isn't an option btw ;-;

[pliant yacht]

How

y2 - y1

[grand trellis]

y2-y1/x2-x1

[grand trellis]

Yea

Oh wait

[pliant yacht]

Erm

Ur almost there

[grand trellis]

Mb 😭

[pliant yacht]

Lol

[grand trellis]

I'm getting 36/5 😭

There has to be some property

Which we gotta use

[pliant yacht]

Ayo

Now im confused

[grand trellis]

😭😭

[pliant yacht]

Ohh

U did it wrong

[grand trellis]

How ;-;

[pliant yacht]

153 - 45
—————
28 - 10

108 / 18

Simplify

[grand trellis]

36/6

That's just 6 ;-;

[pliant yacht]

Wait wtf

[grand trellis]

😭🙏

[pliant yacht]

Well this makes the most sense on how to solve it

[grand trellis]

Yea ig

I'll try by myself ;-;

Thanks for trying to help tho

[pliant yacht]

Sorry 😭 let me look at the question again

[grand trellis]

It's fine if you don't get it btw

I just realised diagonals do divide a parallelogram into equal pieces so you are technically correct ;-;

Thing is there must be one more line somehow

[pliant yacht]

Okay i got it

It’s ur drawing bro 😭

[grand trellis]

How?

[pliant yacht]

Let me draw it

[grand trellis]

Ignore it it's pretty bad 😭

[grand trellis]

Alr 🙏

[pliant yacht]

So it’s slightly tilted so u cant just use the 2 corners of the parallelogram.
U have to find the mid point of the parallelogram.
Then ur 2 points r now the mid point and (0,0) since it passes thru the origin

Then solve for gradient

[grand trellis]

Oh ;-;

Alr lemme do it rq

WAIT

NO WAY

IT WORKED

[pliant yacht]

Hallelujah

[grand trellis]

99/19

WWWWWW

[pliant yacht]

🙏🙏😮‍💨😮‍💨

[grand trellis]

HOW DID YOU KNOW BUT

THAT IT WOULD PASS THROUGH THE MIDPOINT

[pliant yacht]

I was trying to imagine the parallelogram based of ur drawing

[pliant yacht]

Bc that’s how congruency works

[grand trellis]

OH 😭

TYSMMMMMM

[pliant yacht]

you’re welcome!

[grand trellis]

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[haughty gale]

1+-i/2 was my answer but it was wrong?

How would I do this

[cursive patrol]

can you screenshot how you are inputting your answer?

[haughty gale]

@cursive patrol

[restive inlet]

the values are correct, but not explictly in the specified form

[haughty gale]

What form should it be like

[restive inlet]

a+bi

[haughty gale]

then how would I do it

[restive inlet]

e.g.
1/2 + i/2
where you have two terms with a + sign separating the real and im term

[haughty gale]

So it would be 1/2 + i/2 and 1/2 - i/2

[restive inlet]

yeh

[haughty gale]

Thank you

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[vivid pivot]

How is this not the answer?

[vivid pivot]

I did it with integration by parts

[sudden temple]

missed a negative

[vivid pivot]

Where?

[sudden temple]

from 2nd to the 3rd

a negative from integration f e^{-4x}

[vivid pivot]

Thanks

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[pseudo pond]

Is my work correct so far? Like am I doing this difference quotient thing correct? (3.a.ii)

[quartz birch]

Does no seem wrong to me appart from a h² term missing

Are you running into any problems

[reef fjord]

yes, apart from what jsem pointed out

[quartz birch]

I gotta change my nickname 😄

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[mystic osprey]

aren't neither of these correct?

[mystic osprey]

if he jumps in the pool his distance should be 0 since it measures distance above the pool

however if he returns to the same platform

his distance should be identical

graph a never reaches 0

graph b never returns to the original point

none of these are correct

[dense lily]

It says he climbs back up towards the the platform

Which means he approaches the height of the platform

[mystic osprey]

I picked b anyway

but the wording kind of makes it sound as if

he reaches that point

[dense lily]

b is the most correct

[mystic osprey]

i agree

either way I just wanted to make sure if I was missing something

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[pseudo pond]

Where is it missing I don’t see?

[raven thistle]

this is caluclus 1 if im correct?

[pseudo pond]

Yea

I think

[raven thistle]

uhhh your handwriting is a bit messy what do u exactly need

i seen this a bit late my apolpogies

[pseudo pond]

Uh he said I missed a h^2 idk where I missed it

[raven thistle]

u did

he is right

1-2h+h^2

here

(1-h)^2 =1-2h+h^2

theres no square on the h

@pseudo pond

[pseudo pond]

Alr thank you

I appreciate it

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[raven thistle]

no problem

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[balmy pine]

Solve in whole numbers: a^2 + b^2 + c^2 + d^2 = 2abcd

[balmy pine]

I tried to make a Viet theorem about some variable, but I didn't make any progress

I think no solution without (0,0,0,0)

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[balmy pine]

<@&286206848099549185>

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[earnest pine]

I don’t think there is a non zero solution. I could be wrong though

Took me a sec to find the lemme, but I believe that it is a consequence of

This holds of n>3 too, hence why.

Obviously you can’t easily find the determinant, so you’ll have to use some iff properties of the determinant to show that it is 0. Thus only one solution

That solution must be the trivial solution.

I could be wrong tho, I kinda scribbled down some working with very little rigour, but I think it checks out

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[stuck peak]

hello i’m kinda stuck with question 1c like after where u state that n=k+1

[stuck peak]

this is what i got up to

and i tried to follow my teachers steps but i ended up getting n^2 as my answer??

[golden mesa]

looks like you're substituting n=k+1 but it's not substituted correctly on RHS

[stuck peak]

oh i see

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[stuck peak]

do i bring the rhs to the left after?

[golden mesa]

typically, you'd use the expression corresponding to n=k to prove for n=k+1, going LHS to RHS

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[peak nimbus]

help

[peak nimbus]

how

[pseudo sonnet]

what are you asked to do with this?

[peak nimbus]

factor

[pseudo sonnet]

sure

any ideas?

[peak nimbus]

difference of cube?

[pseudo sonnet]

we'll get there! :p

always start with greatest common factors

you can make your life easier

[peak nimbus]

yes

[pseudo sonnet]

so, what should we do?

(also, brb)

back

[peak nimbus]

idk

[pseudo sonnet]

what's the greatest common factor of the two terms?

[peak nimbus]

2

[pseudo sonnet]

2?

just the number 2?

neither term has a coefficient of 2 though

[peak nimbus]

sorry

1?

[pseudo sonnet]

not quite...

,, a^{7}b^{2} - ab^{2}

[thorny flame]

higher!

[pseudo sonnet]

can you see anything these two have in common?

[peak nimbus]

ab

b^2

[pseudo sonnet]

ding ding ding

well, you could take it one step further

both terms share an a too

so the GCF of the two terms is ab^2

try factoring that out, and see what you get

[peak nimbus]

ok

do i use (a-b)(a^2+ab+b^2)

[pseudo sonnet]

first, can you show me what you got?

the entire expression

[peak nimbus]

im cooked

[pseudo sonnet]

do you not have the expression?

tell me honestly

[peak nimbus]

no :D

[pseudo sonnet]

I see

do you know how to factor a GCF?

[peak nimbus]

like

[pseudo sonnet]

if I gave you $x + xy$, and asked you to factor, what would you do?

[thorny flame]

higher!

[peak nimbus]

x*x+y

idk tbh

[verbal shadow]

cooked fr

[peak nimbus]

yo

chill

math is my weakness

[verbal shadow]

you have 1 x and y x's

u have 1+y of x

(1+y)x

[peak nimbus]

yea

[verbal shadow]

in this case x is the gcd

try factoring your expression

[peak nimbus]

o

ok

so

(a-1)(a+1)

waitno

[verbal shadow]

which expression are you factoring exactly?

[verbal shadow]

yup

try to factor the gcd which is ab^2 out of each term individually

if we want to factor $x^2y^3$ out of $x^4y^7$ we can write it as $(x^2y^3)(x^2y^4)$

[thorny flame]

someone1010

[peak nimbus]

ab^2(a x a x a x a x a x a x a) ( b x b)?

[verbal shadow]

you have created the term $(ab^2)(a^7b^2)$ which is equal to $a^8b^9$ rather than $a^7b^2$

[thorny flame]

someone1010

[verbal shadow]

you want $a^7b^2 = (ab^2)(a^{\text{something}}b^{\text{something else}})$

[thorny flame]

someone1010

[verbal shadow]

remember the rules of exponents

$a^x * a^y = a^{x+y}$

[thorny flame]

someone1010

[verbal shadow]

you can use the converse of this to seperate factors

if u wanted to factor $a^2$ out of $a^8$

[thorny flame]

someone1010

[verbal shadow]

we know that $a^8=a^{2+6}=a^2*a^6$

[thorny flame]

someone1010

[peak nimbus]

a^7 = a^4+3

[verbal shadow]

yup that is a valid way to seperate them

try to seperate $a^1b^2$ out of $a^7b^2$ now

[thorny flame]

someone1010

[peak nimbus]

a^6?

[verbal shadow]

yup

so now you can write it as

$(ab^2)(a^6)$

[thorny flame]

someone1010

[verbal shadow]

$ab^2=(ab^2)(1)$ is trivial

[thorny flame]

someone1010

[verbal shadow]

leaving us with the expression

$(ab^2)(a^6) - (ab^2)(1)$

it's minus im sorry

[thorny flame]

someone1010

[verbal shadow]

can you see the next step?

[peak nimbus]

where did the other ab^2 come from

[verbal shadow]

the original expression was

$a^7b^2-ab^2$

[thorny flame]

someone1010

[verbal shadow]

you found that this is equal to

$ab^2(a^6) - ab^2$

[thorny flame]

someone1010

[verbal shadow]

i thought it was kind of trivial that ab^2 was equal to itself so i just factored 1 out

sorry for the confusion

[peak nimbus]

yeas

ykw

imma just ask my teacher

cuz its late and i cant focus rn

thank u for the help

[verbal shadow]

ok, best of luck

[peak nimbus]

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[solid axle]

need help with this

[loud oasis]

try expanding the square

[solid axle]

ok

then what

[loud oasis]

use the power rule

[sweet shard]

,tex .exp rules

[thorny flame]

riemann

[solid axle]

ok thanks i got it

what about this one

i tried to do u sub

with u = y^3+y

and i got 20 but its 21/4 somehow

[sweet shard]

This is the wrong substitution

[solid axle]

why

[sweet shard]

What's du=?

[solid axle]

du=3y^2+1*dx

[sweet shard]

Nope

Forgot chain rule on the first term

Also there shouldn't be a dx

[solid axle]

wdym forgot chain rule

[sweet shard]

derivative of y^3 is 3y^2 times dy

[somber carbon]

ye ur u isn’t rly gonna help a lot

I first look for stuff inside parentheses or roots to set u to

[topaz sinew]

@solid axle Has your question been resolved?

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[sweet salmon]

how do i do (i)

[neon iron]

excuse me is it a test quiz exam

cause we cant help with those

or is it just a graded assignment

[sweet salmon]

no, not an exam

its a june 2015 exam paper

[arctic briar]

it cant be exam cuz like they have access to technology

[sweet salmon]

yeah

need some help with part (i)

[arctic briar]

*not in the fgorm of calculators

[neon iron]

ah kk

dw

[pearl fog]

i mean, it can very much so be an online exam

[arctic briar]

atp it's open book then

ig u can refrain from giving too many hints then

[topaz sinew]

@sweet salmon Has your question been resolved?

[brave coral]

Start with the definition of E[X] for a discrete random variable

$$ \operatorname{E}[f(X)] = \sum_{x=0}^{\infty} P(X = x) f(x) $$

[thorny flame]

StrangeQuarkAL

[brave coral]

X is a standard distribution but idk if you're allowed to assume its prob. generating function

[topaz sinew]

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[sweet salmon]

.reopen

[topaz sinew]

✅

[sweet salmon]

@brave coral is it possible if you can continue

i got that part

my original plan was using this

[brave coral]

X doesn't follow a binomial distribution btw

[sweet salmon]

how do you know

isn't the probability distriubtion function this?

k = x

[brave coral]

That's definitely not the same thing

[sweet salmon]

how come?

[brave coral]

p^r is a constant for one

[sweet salmon]

yeah ok

[brave coral]

If you try your substitution, it's fairly clear

It also helps that I already know what the probability density function is

You don't need to know what distribution it follows anyway to solve it

Although, if you are meant to know, it would make things easier for you

[sweet salmon]

ok sre

[brave coral]

One sec, I gotta get on my laptop cuz it's a bit much to expain

[sweet salmon]

ok so we have this right

[brave coral]

Yep

[sweet salmon]

what do we do next?

[brave coral]

Lets start by simply moving the constant out of the sum

After which, combine the e^ux and (1-p)^x into one term

[topaz sinew]

@sweet salmon Has your question been resolved?

[sweet salmon]

ok

@brave coral

[brave coral]

Cool

[sweet salmon]

what we do next?

[brave coral]

Do you know the binomial expansion (for negative, rational exponents)?

[sweet salmon]

nope

could you please explain

[brave coral]

Neither of these ring a bell?

[sweet salmon]

think so

think they kinda do

[brave coral]

Well I hope you can follow along

[sweet salmon]

how can we use it

so we swap the x and r with each other?

from the p and (1-p)

[brave coral]

Before we even get there, we need to manipulate the first one I sent

[brave coral]

Notice it's a sum to infinity like this one

the (r k) is defined by

[sweet salmon]

yeah

[brave coral]

Lets replace r with -r

[sweet salmon]

yeah

[brave coral]

$$ \binom{-r}{k} = \frac{-r(-r-1)(-r-2)...(-r-k+1)}{k!} $$

[thorny flame]

StrangeQuarkAL

[sweet salmon]

ok thats really nice

[brave coral]

We have k factors in the numerator, for which we are factor out -1

$$ \binom{-r}{k} = (-1)^k \frac{r(r+1)(r+2)...(r+k-1)}{k!} $$

[thorny flame]

StrangeQuarkAL

[sweet salmon]

ok i see

[brave coral]

great

[brave coral]

now, looking back at this formula, what would it look like if we replaced r with r + k - 1

@sweet salmon

[sweet salmon]

it would be( r+k-1)(...)(r+k-x)/x!

[brave coral]

How did you get the last term (r + k - x)

also for now, keep the k as k, don't replace with x just yet

[sweet salmon]

+1 from last ter

[brave coral]

(r + k - 1 ) - k + 1 = r

[sweet salmon]

so (r+k-1)(r+k-2)...(r+2k-2)

[brave coral]

read this again

[sweet salmon]

oh right (r+k-1)(r+k-2)...(r)

[brave coral]

Yup

[brave coral]

Can you see how that's related to this?

[sweet salmon]

its the other way arround

[brave coral]

Yes!

[sweet salmon]

(-(r+k-1) k)

C(-(r+k-1), k)

i mean C(-r,k)

[brave coral]

Right, C(-r, k) = (-1)^k C(r + k - 1, k)

[sweet salmon]

oh right but we have the (-1)^k

[brave coral]

$$ \binom{-r}{k} = (-1)^k \binom{r + k - 1}{k} $$

[thorny flame]

StrangeQuarkAL

[sweet salmon]

the C(r+x-1,x) = C(-r,x) * (-1)^x

[brave coral]

Yep

[sweet salmon]

with the k

[brave coral]

Wait no

oh yeah nvm

that shouldn't be a k

look at what you've replaced

yup

that's all good

[brave coral]

$$ p^r \sum_{x=0}^{\infty} \binom{-r}{x} (-e^u (1 - p))^x $$

[sweet salmon]

what about the (-1)^x

[brave coral]

exactly right

my bad

[sweet salmon]

and als othe combinatoric

[brave coral]

lol, yeah

[thorny flame]

StrangeQuarkAL

[brave coral]

I combined the (-1)^x and the (e^u (1-p))^x together

[sweet salmon]

ok

[brave coral]

we pretty much have enough to rewrite it as the left hand side

[brave coral]

let x = 1 (in the equation here) so that

$$ (1 + y)^r = \sum_{x=0}^{\infty} \binom{r}{x} y^x $$

[sweet salmon]

ok

so the other way around i see

[brave coral]

mmhmm

[brave coral]

So, what can we replace y and r with to make it equal to the equation here

I'll replace k with x to get you started

[thorny flame]

StrangeQuarkAL

[sweet salmon]

wait i forgot the 1 -exp(u)(1-p)

i got it thanks

[brave coral]

you forgot the 1

[sweet salmon]

yeah just forgot the 1

[brave coral]

yeah

[sweet salmon]

yeah got that

damn, that was a good question

how did you think of using the negative binomial

[brave coral]

yeah it was

My teacher once tasked us with proving the generating functions

[sweet salmon]

lol damn

so you got the same question?

[brave coral]

it's similar but not quite the same

[sweet salmon]

Ok I see

[brave coral]

The way you get it into the fraction form had to be the same though so I just looked back at my notes

[sweet salmon]

I have another moment generating function question, do you think I can DM it to you

My laptop just died

[brave coral]

I'd prefer if you opened a new help channel. Much better to have the active server at your diposal

[sweet salmon]

Ok I see

[brave coral]

Yeah. Anyway, great work. Seems like you really got the explanation

[sweet salmon]

Thanks

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[knotty adder]

I

[topaz sinew]

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[knotty adder]

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[fallow igloo]

Do you have a question?

[knotty adder]

I do

[fallow igloo]

.reopen

[topaz sinew]

✅

[fallow igloo]

Post it here 🙂

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[sharp pagoda]

How does 17x^2 + 4x -12 = 0 get simplified to the answer in the latter screenshot?

[night imp]

4 * 17 * 12

Can you calculate that again

[sharp pagoda]

woops

716

[night imp]

Try again @sharp pagoda

,calc 41712

[thorny flame]

Result:

816

[sharp pagoda]

when I factor out 832 (16+816) I get this

[night imp]

2 * 416

😅

[sharp pagoda]

you're right

my arithmetic is very off today I'm sorry

[night imp]

No worries

It happens

[sharp pagoda]

okay so that factors down to 8sqrt(13)

why does that end up simplifying to the answer that I was given?

[night imp]

If you have something like
(2+2b)/4
We can simplify it by dividing 2 in numerator and denominator
(1+b)/2

That's what your answer paper says

Your answer was (-4±8√13)/34

We can simplify it further by dividing numerator and denominator by 2

[sharp pagoda]

that doesn't affect sqrt(13)?

or I guess the proper way to think about it is sqrt(13) multiplied 8 times

[night imp]

Yesss

8√13 is 8 * √13

[sharp pagoda]

okay, I think I get it now

tysm

[night imp]

No problem

[sharp pagoda]

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[lament jay]

this is a little sad but could anyone explain to me how a purely negative derivative can result in a integral having positive values. aren't both derivative areas negative as well?

[lament jay]

like, i get that the values should be positive in terms of 1/x being 1/x

but i'm missing the connection to the derivative

[smoky sparrow]

if you wanted to find the area of g(x) between x = 1 and 10 for example

you'd integrate g(x) to get f(x) and then sub in x = 10 and x = 1

f(10) - f(1) = 0.1 - 1 = -0.9

which makes sense, our original function was entirely below the x-axis, so the area we get should be negative

I think you've mixed it up: when you integrate g(x) you get f(x)

if you integrate f(x) = 1/x you get ln |x|

[lament jay]

no no i get that g is the derivative of f

wait

[lament jay]

alright i still don't 100% get it i think but it's a little better than before

thanks a lot for the help man

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[smoky sparrow]

npnp

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[golden sky]

how do i do a

[sonic escarp]

identify the given angels in the diagram

[golden sky]

80 and 240 right

[sonic escarp]

show them in the diagram.

[golden sky]

Iike this?

[sonic escarp]

yes.

what is angle x?

and if you know x, can you say something about y?

[golden sky]

x=y?

[sonic escarp]

well. what is x?

[golden sky]

60

ohhh

so thats where you get it

tyty

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[neon iron]

,,\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}

[thorny flame]

!Kiz__

[neon iron]

Okay

It was easy I just over thing it lol

[sudden temple]

!done

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[neon iron]

.close

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[sick brook]

its a logarithm question

is finding x possible?

(2x-1) / x = log 6/ log 2

[sudden temple]

I think from the structure yeah

[sick brook]

idk whats the next step

[sudden temple]

I might just separate the fraction on the left hand side

and solve for x

[sick brook]

howww i dont understand sorry

[sudden temple]

$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

[thorny flame]

77²

[sick brook]

2x-1 / x = log 6 / log 2

2x / x - 1 per x = log 6 / log 2

2x - 1 = xlog 6 / log 2
2xlog 2 - log2 = xlog 6
2xlog 2/log 6 - log2/log 6 = x
xlog2 / log 6 = x
log 2/ log 6 = x/x = 1

wtff did i did wrong somewhere

oh

the 2xlog

ohh theres a rule that makes

plogb/ploga = alogb is that correct

@sudden temple

[sudden temple]

yeah

[sick brook]

2x-1 = x.^2log 6
2x-1/^2log 6 = x
x = 2x-1/^2log 6

thats as far as i can go

[sudden temple]

lol i just lost your channel a few moments ago

and was opening every channel

[sick brook]

https://tenor.com/view/party-nods-american-psycho-donald-kimball-patrick-bateman-gif-3873187078374324019

thanks for helping

i still dont know if im right or not btw

[sudden temple]

wait lemme see

[sudden temple]

after the fourth line from end, you should just 2xlog2 - xlog6 = log2

then just factor out the x

and divide the rest on both sides

this should be it

[sick brook]

apparently i did wrong in early steps

i restarted and im stuck on here?

[sudden temple]

seems fine to me

[sudden temple]

what are you confused about?

just use the property of log

to bring those exponents down

[sick brook]

ohh

[sick brook]

uhh did i make some mistake somewhere

[sudden temple]

dude!

log(ab)=loga+logb

[sick brook]

which part are u pointing out

[sudden temple]

3rd step

[sick brook]

ohh wait lemme try to understand them

[sick brook]

that doesnt make log. log but log +log?

[sudden temple]

obviously bro

that's the property of logarithm

[sick brook]

ik this rule but i dont understand how to bring the real number exponent to log xD

is this supposed to be correct then im sorry

@sudden temple

[sudden temple]

lol

[sudden temple]

yeah

[sick brook]

wat

alrr thanks

[sick brook]

I DONT UNDERSTANDD I ALWAYS STUCK AT SOME PART

[sudden temple]

,rotate

[sick brook]

im so stupid damn

[sudden temple]

did you got it?

[sick brook]

no

[sudden temple]

bro leave the x multiplied on the right hand side

and just expand the left hand side

then combine the x terms and then factor out x

[topaz sinew]

@sick brook Has your question been resolved?

[sick brook]

whats this supposed to mean

[sick brook]

with what

like this i guess??

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[neon iron]

i can put the exponent part as

[timid lintel]

ohh ty ty

[sudden temple]

2^{that thing}

with \ instead of /

[neon iron]

$2^{\frac x{2r}}$

[thorny flame]

kr

[neon iron]

this should work

[mossy pike]

I am form 7th so how can I add fastly and smoothly

[neon iron]

dawg

[neon iron]

thanks yeah

ight so i got it like

that

then i'll have something like

$2^{\frac x{2}} \cdot 2^{\frac x{4}} \cdot 2^{\frac x{8}} \ldots$

[thorny flame]

kr

[neon iron]

what does this converge to

[sudden temple]

$a^{b}{\cdot}a^{c}=a^{b+c}$

[thorny flame]

77²

[neon iron]

of the top of i head i can take out like $2^{\frac x{2}}$

[thorny flame]

kr

[neon iron]

oh

https://tenor.com/view/lil-yachty-drake-oprahs-bank-account-meme-laptop-gif-20803826

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[neon iron]

hi

[neon iron]

i need help in deciding whether i should study what i like or study for tommorows test

[sweet shard]

help channels are for specific math questions. ask #discussion or #serious-discussion

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[halcyon willow]

ik how it gets solved using triangles and expanding the cos term

[halcyon willow]

but in a reference i see they find x and y in quadratic with an imaginary i

and then they take the ratio which i dont understand

can anyone help?

x=cos⁡α±isin⁡α y=cos⁡β±isinbeta

can anyone give the ratio?

[sweet shard]

,tex .sum diff trig

[thorny flame]

riemann

[halcyon willow]

ik thiss

obv

i know how to solve

i am just thinking about this specific method they used

in this question

[sudden temple]

can you show their work?

[sweet shard]

show the entire "specific method" then

[halcyon willow]

they just solve the quadratic and find x = cos⁡α ± i sin⁡α, y = cos⁡β ± i sin β

then they take x/y+y/x = 2 cos (a-b)

how the ratios form that i wanna know

oggg

they use

the conjuagte multiplication thing

[sudden temple]

why is it like the derivation of the formula of cos(a-b)

[halcyon willow]

in complex numbers

right

this was completely out of my mind

lol

i forgot thats how you multiply imaginary numbers lol

yea got it nvm

.close

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[topaz sinew]

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[limber elbow]

i am a student who is new to math
idk the basics too
and i need help with everything (almost)

[limber elbow]

<@&286206848099549185>

[twin heath]

Hello!

Let me try to help you out

What do you know about functions?

[twin heath]

Do you know how the f(t) notation works here?

[limber elbow]

just the basic

about how i can solve for a function

idk how to find domain and range

i mean i do

but i never get it right'

[limber elbow]

no

i don't get math

a lot of times

could u help?

[fallow heart]

I'm a bit busy now, sorry

[limber elbow]

anyone else that can?

[twin heath]

Okay so

F(t) is like a machine

[leaden drum]

The f(t) notation here tells you to input the values that the questions asks for in the equation

[twin heath]

You put in some value of t, and you get something out

[twin heath]

For example, f(0) just means to replace all t with 0

[leaden drum]

(in the "t" in the equation)

[limber elbow]

yea that i got

like in place of x just put, 0, right?

[twin heath]

Yep!

[limber elbow]

how about part b?

[twin heath]

You just replace it with a and math it out

In this case, the a in the parentheses cancels out

[limber elbow]

i really don't get that

[leaden drum]

As for the variables, the "a" ones, think of them as placeholders. For instance,
f(a) equals to a^2
but why?

Well the equation is f(t) = a^2 - (t - a)^2

we put "a" in place of the "t"

This gives us a^2 - (a - a)^2

(a - a) is zero.

0^2 is also zero.

What's left is a^2

Which is the answer for f(a)

[limber elbow]

do u mind solving it or explaining over vc?

[twin heath]

Can't I'm in a doctors office rn lmfao

[limber elbow]

lol okay

[limber elbow]

makes some sense

lemme try solving it

[leaden drum]

As for part b;

the number in front of the "f(a)", 3, tells you to multiply the result of f(a) by 3

[limber elbow]

okay..

[leaden drum]

Part B tells you to sum the outputs (results, basically) of the functions it gave you.

[limber elbow]

alright

imma try it once

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@limber elbow Has your question been resolved?

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[neon iron]

Can somebody explain to me the logic behind part a please I don’t understand the answer

I mainly just don’t understand this

Does this mean find all of the values that cause the rate/gradient to decrease ?

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

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Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.