help-17

[vast shale]

so like

the sum of an arithmetic series is given by the formula[
S_n = \4n2\9{2a + (n-1)d}
]

[vast shale]

n is the amount of terms there are, so the series

4 + 7 + 10

has 3 terms so n = 3

and a is the first term in that series, so a = 4, and d is the difference between terms. This would be d = 3 because you add 3 each successive term

is this fine for an explanation of this? @onyx swallow

[onyx swallow]

yes

[vast shale]

oki so in ur question it's said that "the first 48 terms of an arithmetic series..."

among n, d, and a what does 48 represent here ya think?

[onyx swallow]

n probably?

[vast shale]

yeah

cool

so

,align
S_n &= \4n2\9{2a + (n-1)d}\
S_{48} &= \4{48}2\9{2a+47\2d}

[vast shale]

is this fine by ya ?

[onyx swallow]

yes

[vast shale]

ok

[vast shale]

so next on in the question

it says that

"the series we just wrote is 4 times the sum of the first 36 terms of the same series"

so like

so first of all how do we write out the series for the first 36 terms?

[onyx swallow]

do I simplify this then multiply everything by 4?

[vast shale]

no

[vast shale]

answer this as if that was the only thing i told you

[vast shale]

like, ignore this for now

[onyx swallow]

is it

S_36= 36/2 [2a+35d]

[vast shale]

yeah!

so now

the question says S_48 is 4 times bigger than S_36

how do u write that out in an equation?

[onyx swallow]

S_48 = S_36*4

[vast shale]

yea

aight so time to do some ALGEBRUH

[vast shale]

replace S_48 and S_36 with their expressions and simplify

[onyx swallow]

wait I think I got this wrong

u said S_48 is 4 times bigger?

[vast shale]

nah u didnt

yeah

students make that mistake a lot but like

"x is d times bigger than y" is x = d*y not the other way around

[onyx swallow]

oh wait nvm

[vast shale]

ye

[onyx swallow]

the wording is so weird-

it's making me doubt myself

[vast shale]

u r correct dw

[vast shale]

do this

try and solve for a

[onyx swallow]

do I keep the S_48 and S_36 in the equation?

[vast shale]

no

replace them with their expressions

so like

,, \4{48}2\8{2a+47d} = 4\8{\4{36}2\8{2a+35d}}

[vast shale]

solve for a

[onyx swallow]

which bracket do I expand first on right side?-

the inner one?

[vast shale]

both lmao

order doesnt matter

u can do the first one inside if u want to

i am expecting u to get me an expression of the form

a = ???

from the above

[valid isle]

you can multiply both sides by 2 first that will make things easier

[onyx swallow]

ok so I did

48/2[2a+47d] = 4( 36/2 [2a+35d] )
24[2a+47d]= 4( 18[2a+35d] )
48a+1128d= 4( 36a+630d )
48a+1128d= 144a+2520d
48a-144a=2520d-1128d
-96a=1392d
a= 1392d/-96
a=-14.5d

[vast shale]

aight!

great

so thats correct

now lets get to the final bit of the question

"Find the sum of the first 30 terms of this series"

So we construct another series of the form [
S_n = \4n2\9{2a+(n-1)d}
]
but $n= 30$ and $a= -14.5d$

[vast shale]

can you substitute in the values for n and a, and simplify ur expression? @onyx swallow

[onyx swallow]

S_n=n/2[2a+(n-1)d]
S_30=30/2[(2•-14.5d)+(30-1)d]
S_30=30/2[-29+29d]
S_30=15[-29+29d]
S_30=-435•+435d

[vast shale]

uh

like

in ur second line

the 2 disappeared

[onyx swallow]

it did?

[vast shale]

well no discord just erased the multiplication sign 💀

but the d disappeared yes

it should be
S_n=n/2[2a+(n-1)d]
S_30=30/2[2(-14.5d)+(30-1)d]

what is 2(-14.5d)

[onyx swallow]

-29d

[vast shale]

yea

but like

u made it -29

so like

[onyx swallow]

ohh

[vast shale]

S_30=15[-29d+29d]

what is that

[onyx swallow]

S_30=0

[vast shale]

yep!

thats it

g'job

[onyx swallow]

tysm for the help today!!

[vast shale]

nw

hope you feel better soon

i gotta go now so u can open a new channel if needs be

gluck

[onyx swallow]

trust me, I won't when it comes to math 💀

[vast shale]

[onyx swallow]

welp, bye

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[vast shale]

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[echo fiber]

dude i dont get part 3 wth am i supposed to do?

[echo fiber]

these are my answers for the other 2 parts but part 3 i dont get

oh wait do we just 30l - l^2 find the vertex

[drifting terrace]

yes

[echo fiber]

and after we find the vertex just subsitute it into the equation?

[drifting terrace]

yep

[echo fiber]

then wth was chat gpt talking about

with all that complicated talk

[drifting terrace]

lol

!nogpt

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Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[echo fiber]

oh

good to know

[drifting terrace]

also in case this ever shows up again

[echo fiber]

wait but i have another question

[drifting terrace]

the optimal rectangular area for some fixed perimeter is always a square

[echo fiber]

oh thats awesonme

but uh

what does this mean

heres the graph brw

[drifting terrace]

well do the results match?

looks to me it does

[echo fiber]

yeah they do

15(x axis) is equal to 225

but i mean what am i supposed o write

hes gone

rip

[drifting terrace]

idk just say they match or something

[echo fiber]

alr thanks

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[desert glen]

How do I do this?

[desert glen]

This is what I've done so far:

[tawny nacelle]

starting with cot seems like a good idea

[desert glen]

im not really sure how this helps though

[tawny nacelle]

look at your 1's

try to find common denominators

[tawny nacelle]

wait how did you go from 2nd step to 3rd step?

[desert glen]

i just took the reciprocal and multiply it

[desert glen]

so do you want the one to equal $\frac{\sin}{\sin}$

[twin meteor]

guy

[tawny nacelle]

what do you mean?

[desert glen]

i just flipped sin/cos so i was able to multiply them

[tawny nacelle]

okay, this seems good

[tawny nacelle]

but this doesn't work when you have a sum in the numerator and denominator...

[desert glen]

oh

i didnt know that

ok lol mb

[tawny nacelle]

this is good though

can you go from here?

what should the next step be?

[desert glen]

im not really sure what the next step is after this though

its not squared so i cant use sin^2+cos^2=1

[tawny nacelle]

lemme see

[steel snow]

This should work now right

[tawny nacelle]

yes, you can do that step now

I'm just trying to see what can be done after that...

[steel snow]

Sin+cos will be on top and bottom

[loud walrus]

This doesnt look the appropriate way tho

[desert glen]

when are you not able to use this step though? im a bit confused as to why it didnt work before

[steel snow]

So it is simplified to one thing

[desert glen]

[tawny nacelle]

yes, it will

hm

that works

[steel snow]

You need one denominator so don’t use it for multiple fractions

[steel snow]

So like this doesn’t work (imagine the left stuff is 1) because they aren’t combined

kind of

Idk how to explain

[desert glen]

ah i see

i think i get it

[steel snow]

It just needs to be exactly one fraction divided by one fraction

[desert glen]

so u basically need one fractino and not multiple

[steel snow]

Yes

[steel snow]

do you see anything that cancels out

[desert glen]

yes lemme draw it out so i can visualize it tho

one sec

which is just cot

i think right?

[steel snow]

Yes

Congrats

[desert glen]

yep that was the right answer

thanks guys!

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[abstract palm]

how to do this limit

[silk osprey]

try simplifying the inside first

[abstract palm]

Like this?

[silk osprey]

that isn’t legible

are those N

[abstract palm]

sorry

[abstract palm]

yes

[silk osprey]

so how’d you get the N+2

for the exponent

[abstract palm]

multipliing by N

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[rain spire]

hey, does anyone know why the kronecker-delta just disappears here?

[lyric fossil]

is this using einstein summation convention?

[rain spire]

yep

why can't i and j be equal?

[lyric fossil]

do you have more context?

oh

if i = j then you just get xlpl - xkpk = 0

since those are implicitly summing over all values of l and k

and if i=/=j then the terms go away as well

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[rain spire]

Ah thanks!

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[gaunt willow]

what part are you stuck on

[hexed dagger]

Idk where to start

[gaunt willow]

so how do you get the solutions for dy/dx = -2/x^3 + 2e^(2x-2)

its separable right

[hexed dagger]

Yeahh

[gaunt willow]

ok now you can integrate both sides

[gaunt willow]

careful its e^(2x-2) not x^(2x-2)

[hexed dagger]

Ohh thank you

@gaunt willow

[gaunt willow]

yeah

so for what x is that defined

[hexed dagger]

Do I need to graph it?

@gaunt willow or?

[gaunt willow]

no

just like

look at the function

when is 1/x^2 defined

when is e^(2x-2) defined

[hexed dagger]

1/x^2 is undefined at 0

But e^(2x-2) is defined everywhere, right?

[gaunt willow]

yes

[hexed dagger]

So D?

[gaunt willow]

yes

[hexed dagger]

Ok, thank you so much

[gaunt willow]

np

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[glad socket]

are ellipses generally in the equation (i)x^2 + (j)y^2 = k

[glad socket]

where i, j, and k are any real number

besides 0

[spice vortex]

that’s a circle

[glad socket]

2x^2 + 3y^2 = 1 is an ellipse no?

5x^2 + 7y^2 = 3

???

[spice vortex]

yes that one is

[glad socket]

so is it a circle or ellipse

[gaunt willow]

ellipse

[glad socket]

ok thank you

[spice vortex]

it depends if you can do x^2/a^2 +y^2/b^2 = 1

in this case you can

[gaunt willow]

but they can also be

(x-h)^2/a^2 + (y-k)^2/b^2 = 1

(off centered)

[glad socket]

without /a^2 would be a circle right

and /b^2

[gaunt willow]

if a = b its a circle

[glad socket]

oh ok

thanks

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[quartz forge]

How do u solve the differential equation ?

[flint idol]

separation of variable

[quartz forge]

Yh but

But we do t know x(t) in terms of t

So how does the integral on the right work

[flint idol]

you didnt exactly separate the variables

lets say x(t) = y

would it be easier to see it that way?

such that $\frac{dy}{dt}=-ky$

[twin meteor]

y0shi

[flint idol]

you have to get all the y's on one side

and the t's on the other side

[quartz forge]

O right

Yes

[quartz forge]

Where did x(0) come from

[flint idol]

constant of integration

when integrating you should have $ln{\abs{y}}=-kt+C$

[twin meteor]

y0shi

[flint idol]

raise e to the both sides we'll get $y=e^{-kt+C}$

[twin meteor]

y0shi

[flint idol]

using exponential rules:
$y=e^{-kt}e^C$

[twin meteor]

y0shi

[flint idol]

but e^C is just another constant

which really is the initial population

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[empty tusk]

help

[empty tusk]

mostly on

just

how to set up the proportion equation

thats really the only part i need help on

someone help pls

this is like

1/12 ixls i need to do

in 2 days

[rough dew]

you can use similar triangles

triangle YWZ and triangle WXY are similar because they share 2 angles (90° and angle W) and a side (XW)

first, find WY

[empty tusk]

38

[rough dew]

nevermind, you have WY

so now, WX / WY = WZ / WX

[empty tusk]

thank u so much

may i ask how u set that equation up tho

[rough dew]

so in similar triangles, ratio of corresponding sides will all be equal

in the diagram, hypotenuse of YWZ is WY while hypotenuse of WXY is WX

wait

other way around

in the diagram, hypotenuse of WXZ is WX while hypotenuse of WXY is WY

thats the first ratio, WX / WY

[empty tusk]

mhm

[rough dew]

for the other ratio, we need WZ in it

WZ corresponds to WX

so second ratio is WZ / WX

[empty tusk]

sorry if this is a dumb question but how do they correspond?

im getting it tho so far

[rough dew]

so the two triangles share an angle, which is angle W

so the corresponding sides of angle W will be corresponding

in the smaller triangle, corresponding side of W is XZ, and in the larger triangle, corresponding side is XY

since XZ and XY correspond, and WX and WY correspond, that leaves the last pair of corresponding sides, which is WZ / WX, the ratio we need

[empty tusk]

mk

i will now attempt to solve rq

[rough dew]

👍

[empty tusk]

k i got 15.157

is this correct?

[rough dew]

one sec

yep got the same answer

[empty tusk]

alr ty

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[rich rock]

what does the beginning part mean, outside of the { }

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[rich rock]

<@&286206848099549185>

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[cloud horizon]

Hey, @rich rock. I'm so sorry for the late response! The beginning part of the question outside of the curly braces {} is asking whether the union of the sets defined within the braces is finite, countably infinite, or uncountably infinite. You are required to justify your answer.

Here's a breakdown of the question:

• The union of sets from ( n = 1 ) to ( n = 7 ):
  $ \bigcup_{n=1}^{7} $

• The set of all points ( (x,y) ) in the 2-dimensional real number plane that satisfy the equation ( x^2 + y^2 = n^2 ):
  $ {(x,y) \in \mathbb{R}^2 | x^2 + y^2 = n^2} $

• The intersection with the set of all points ( (x,y) ) that satisfy the equation ( y^2 - 4x^4 = 0 ):
  $ {(x,y) \in \mathbb{R}^2 | y^2 - 4x^4 = 0} $

• The question asks whether this set is finite, countably infinite, or uncountably infinite.

To answer the question, you would need to determine the nature of the points that satisfy both equations simultaneously and whether these points form a set that is finite, countably infinite, or uncountably infinite. Remember to provide a justification for your answer as required by the exam question. If you need further assistance with solving this problem, feel free to ask!

[twin meteor]

ItzJezze

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[warped tangle]

How do we find the limit $\lim_{x\rightarrow\infty}\ln{(1 - e^x)}$?

[twin meteor]

Bennxy

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[frank iris]

Is it not 1/2bh?

base is 2.2

and height will be tan(39)*2.2

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[frank iris]

any <@&286206848099549185>

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[languid dune]

how do I do this

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[wispy finch]

,rotate

[wispy finch]

,rotate

[wispy finch]

can someone help me to pursue this with polar coordinates

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[winged violet]

how did they get the - (6!/4!)(7C4)? i dont really get it

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[winged violet]

<@&286206848099549185>

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[brisk moss]

i have no clue how what those expressions are counting and i got a different answer

5*(6 choose 2)*(8 choose 4) is smaller than the expression on the paper but i can't see what's wrong with it
5 permutations of MIIII, 6 choose 2 ways to embed the P's into each permutation of MIIII, 8 choose 4 ways to embed the S's into each permutation of MIIIIPP

what would this not count?

7!*(8 choose 4)/(4!*2!)- 6!*(7 choose 4)/(4!) = 6300 and 5*(6 choose 2)*(8 choose 4) = 5250

[winged violet]

i think the idea for the first term was you take the S's out first so you are left with MIIIPPI

and then theres (7!/4!2!) to permute that
then theres 8 choose 4 spots to put the S's such that they are not adjacent to each other

[brisk moss]

yea that makes sense

[winged violet]

i think its missing like the cases where 2S's can be adjacent to each other but P's are separate? like the question is asking no 2 S's are not adjacent AND P's are not adjacent

so like

if 2 S's are adjacent and P's are not, then that still does not fulfill the criteria

[winged violet]

or vice versa

[brisk moss]

oh huh... it doesn't read to me like that but prolly what they meant if my answer is wrong oops misread what you said

[winged violet]

i think the wording is kind of ambiguous

LOL

actually

you might be right

hmmm

[brisk moss]

i'll just try to do it with the other interpretation and see if i get the same answer as them

i think i get something much bigger then

i'm done with this bs ask whoever wrote the solution

[brisk moss]

actually i misread something you said: yes, that was how i interpreted it, and that's what my solution counts?

[brisk moss]

any permutation generated this way has no adjacent S's or P's

i can't think of any adjusted problem where the answer is what they give

[winged violet]

maybe yeah

thanks for the help 🙏

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[brisk moss]

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[brisk bison]

is there a specific property that needs to be fulfilled for an iterated integral equal to product of integrals? Like if i have a triple integral of a function, is there a property that the function needs to meet to allow that equation to equal a product of three integrals of different functions?

[heavy yoke]

if the function can be split into a product of functions of one variable each, e.g. f(x,y,z) = g(x) h(y) l(z)

[brisk bison]

right that makes sense

[brisk bison]

ty for the help

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[lyric fossil]

i think that is sufficient but not necessary

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[zealous ledge]

Can we do this at the end ?

[zealous ledge]

,rotate

[zealous ledge]

,rotate

[zealous ledge]

,rotate

[zealous ledge]

<@&286206848099549185>

[torpid olive]

Yes this is legal

e^C is just another constant

[zealous ledge]

k thx

[torpid olive]

and in a sense you could say that C1=e^c

[zealous ledge]

a little question

[torpid olive]

go ahead

[zealous ledge]

why we can remove C from the exponent ?

like can we remove another term in the exponent of e ?

or it is an exception ?

[torpid olive]

I wouldnt say necesarrily

The C that is in the final answer is just representitive of e^C

and we know that e^c is a constant

And you know by exponenet laws. if they have the same base then you can seperate them such that (e^-1/x^3)(e^c) right?

[zealous ledge]

yea ik this

[torpid olive]

C is technically not removed from the exponent but is another way to express that the e^C as a constant

does that make sense or nah

[zealous ledge]

oh okay i got it thx again

have a nice day/evening !

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[vast shale]

Hey I got a problem in this differentiation question

[vast shale]

5.E

[tawny nacelle]

right, what’ve you tried?

[vast shale]

First I used the quotient rule

[tawny nacelle]

why the quotient rule?

[vast shale]

Felt you could only use the quotient rule in this tbh

Idk what I did

[kind light]

isnt the quotient rule for fractions?

[vast shale]

Wait sorry

[tawny nacelle]

quotient rule is used when you have a quotient of functions..

you have a product here

[vast shale]

Not the quotient rule

My bad

I didn't use product either

I think i used chain

Let me re-check.

I did like

Sinx d/dx (1+cosx)^2 + (1+cosx)^2d/dx sinx

I am fairly new to this chapter so I don't really know what I am doing

[kind light]

im not that skilled either, but i think for the left one you use the chain rule?

[vast shale]

After that I am confused because. It was something like

(sinx) ^2 (1+cosx)d/dx(1+cosx)+(1+cosx) ^2(cosx)

[vast shale]

Like why d/dx a second time? That's my question

[vast shale]

Yes i thinkso

[kind light]

is this like the key?

[vast shale]

Key?

What's that?

[kind light]

like the answer

[vast shale]

Oh

No no

That's the second step

The answer is.

2sinx(1+cosx)(-sinx)+(1+cosx)^2(cosx)

[vast shale]

That's the answer

[kind light]

ok uh i dont think the (sinx)^2 on the left is correct, did you mean like sinx*2(1+cosx)d/dx(1+cosx)

also im pretty sure you just continue with this

[vast shale]

Well, i am confused about that line. I just copied the answer because I couldn't understand why it happened

[vast shale]

In the answer it was written that way so I am not sure.

[kind light]

$$sin(x)\frac{d}{dx}(1+cos(x))^2$$
using chain rule(d/dx(f(g(x))=f'(g(x))×g'(x)
$$sin(x)2(1+cos(x)) \frac{d}{dx}(1+cos(x))$$
$$sin(x)2(1+cos(x))(-sin(x))$$

[twin meteor]

Skill_Issue

[vast shale]

Why is x separately in a bracket?

I don't know, I am so confused how am I gonna do integration 🥲

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

But I geuss close it

Guess,*

Close

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[vast shale]

all i did so far is this:
$\sum_{k=1}^n \dfrac{k}{(k+1)!}$

[twin meteor]

Derivative

[vast shale]

and then i simplified to this

$\sum_{k=1}^n \dfrac{1}{(k+1)(k-1)!}$

[twin meteor]

Derivative

[vast shale]

but now i am lost

[hidden kelp]

You made a mistake, (k+1)! == (k+1)k!

Wait unless you just made a couple steps in one let me see

[vast shale]

well actually, i did make a mistake i went one step too far

[hidden kelp]

Oh no you‘re correct my mistake

[vast shale]

i should keep it in your form

maybe i can get some time of choosing formula

$\binom{n}{k}$

[twin meteor]

Derivative

[hidden kelp]

Oh maybe yeah

[vast shale]

but ye this i dont know what to do hahah

combinatorics are very hard

[twin meteor]

ℑμΤ𝛄𝛗θ

[vast shale]

yes

[twin meteor]

ℑμΤ𝛄𝛗θ

[vast shale]

yes

[hidden kelp]

Surely there‘s a way to link them up

[vast shale]

the problem is the k+1

and we have k at the top not k!

[hidden kelp]

To get k! At the top we could multiply by (k-1)!

[vast shale]

ahh

[hidden kelp]

I don‘t have paper near me so you‘ll have to lmk if it leads to something please xD

[vast shale]

ok i will i am doing it

now we have

$\dfrac{k!}{k!(k+1)(k-1)!}$

[vast shale]

this is what we have now

[twin meteor]

Derivative

[vast shale]

so, we cancel out the k!'s

[hidden kelp]

well if we want to go the binomial way we want the k! on top

[vast shale]

ok so we need to manipulate more

[hidden kelp]

Unless you recognise this form 🤷

[vast shale]

i dont recognise this haha

we need something else at the bottom we need n!

or whatever n equals

could be 1, 2

etc

[hidden kelp]

Yeah if we get the k‘s sorted out we can find n easily

I‘ll go fetch a piece of paper, i really want to see where this goes haha

[vast shale]

ok

thanks for your help

i will try to see if i can do something in the meantime

[hidden kelp]

Ifk if this is the route to take, i can‘t seem to get to anywhere useful

[vast shale]

ye same

its really tough lol

[hidden kelp]

Wolfram answer doesn‘t seem to have binomial so yeah, maybe not

[vast shale]

i see

[vast shale]

how does it solve it?

[hidden kelp]

It doesn‘t give work :(

[vast shale]

hmm

welp

ill move onto another question

[hidden kelp]

Sure. I‘ll work a bit more and if I find anything I‘ll lyk

[vast shale]

ok thanks!

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185> there is a problem listed above, Its very difficult and if someone can give me a hint that would be greatly appreciated .thanks

[vast shale]

here

[stone flower]

This?

Looks like a taylor series problem

N/(n+1)!

Split n into n+1 and -1 and try solving

Looks some combination of e

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

ok i will see cuz its in a chapter about combinatorics

[vocal sleet]

@vast shale Has your question been resolved?

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[oblique rivet]

how to do b

[edgy sapphire]

the first bit is just 1-P(no 1 is rolled), but since you're conditioned on no 5 or 6's, you have to do that with the P(A|B)=P(A and B)/P(B) equation

[vocal sleet]

@oblique rivet Has your question been resolved?

[oblique rivet]

yes

so P(B) will be just (4/6)^n yes?

[edgy sapphire]

right

[oblique rivet]

now A intersection B

is P(A intersect B)= P(5>X>=1)

Bascially X can be 1,2,3,4

??

[edgy sapphire]

yea so P(A intersect B) is a little weird right, so you have to find it with the complement instead

instead of 1-P(no 1 is rolled), the thing you want is
P(no 5,6) - P(no 1 is rolled | no 5,6)

so do P(intersect B) where A is no 1 is rolled, since that's easier and just X can be 2,3,4

[oblique rivet]

ok one min

[edgy sapphire]

actually I'm wrong wouldn't it just be P(no 5,6) - (P no 5,6 AND no 1)

idk why you'd use a conditional

[oblique rivet]

thank you very mcuh, i think it works

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[oblique rivet]

.reopen

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✅

[oblique rivet]

.close

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[fiery jungle]

Can anyone help me with this?approach

[fiery jungle]

ups, image is up now

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[tiny bramble]

Could someone help me either by going through this with me or providing me a source that can help me learn it?

[vocal sleet]

@tiny bramble Has your question been resolved?

[vast shale]

hi

are u stuck on part a

[tiny bramble]

yes

[vast shale]

ok

so for a) i)

we know that s(x) = 1, and s(x) = (kx+1)/(x-3)

so we can say that (kx+1)/(x-3) = 1

and then multiplying x-3 up

kx+1 = x-3

[tiny bramble]

yeah im following

[vast shale]

oh i misread the question

this should still work

lets rearrange for x

x(k-1) = -4

so x = -4/(k-1)

which is undefined if k = 1

so i think maybe k = 1 for the first part

[tiny bramble]

how did you get to this im confused

[vast shale]

ohh

[vast shale]

going back to this

lets minus x from both sides

and minus 1 from both sides

so

kx - x = -4

we can then factorise the x out of the left side

so

x (k-1) = -4

and then divide through by (k-1) to find x

does that make sense?

[tiny bramble]

yes gotcha

[vast shale]

would you like me to walk you through a) ii) as well

i think its a bit more obvious

[tiny bramble]

i'll try that myself and then confirm my answer with you if that's okay

[vast shale]

sure

[tiny bramble]

how did you get this bit

[vast shale]

you cant divide by 0

[tiny bramble]

ohhh yes okay

[vast shale]

so if k did equal 1, then k-1 = 0

which is undefined

although its different approach for part ii

think about using the discriminant

[tiny bramble]

is the answer -x?

[vast shale]

i dont think so

but i havent done the question

i can do it rq if you want

[tiny bramble]

if this is right then -x + x is 0

so would be -x

but idrk

[vast shale]

its a completely different process in this case

oh i made a mistake 1 moment

(kx + 1)/ (x - 3) = -x

kx + 1 = -x^2 +3x

x^2 + (k-3)x +1 = 0

FOR NO SOLUTIONS B^2 - 4AC < 0

try this

ill let you try go from there

[tiny bramble]

why = 0?

oh nvm misread

dw

i dont get how u got from the 2nd to the 3rd

to x^2 + (k-3)x + 1 = 0

[vast shale]

okay i just added x^2 to both sides

and then minused 3x from both sides

and that gave me x^2 + kx -3x +1 = 0

then i factorised the x terms

so x^2 + (k-3)x + 1= 0

[tiny bramble]

is the next step -1?

yeah is that the next step im not too sure

[vast shale]

no

[vast shale]

do you know what the discriminant is

for a quadratic, there are no real solutions if b^2 - 4ac < 0

[tiny bramble]

so b is 1, a is (k-3) and c is 1 ?

wait idk

im gonna watch a quick video on it

[vast shale]

yes!

[tiny bramble]

in the video im watchin it says b is (k+3), a is 1 and c is 1

which one is correct?

[vast shale]

for this exact question?

im fairly sure that its k-3 for this question but i might be wrong

[tiny bramble]

someone told me k = 3

but that doesnt make sense to me

[tiny bramble]

got it!

1 < k < 5

[tiny bramble]

i put this into the discriminant formula

and then got k^2 - 6k + 5 = 0

factorised it, so (k-1)(k-5)

and that means if k is less than 1, it'll equal a positive, and if k is more than 5, it'll equal a positive also

@vast shale lmk if u agree

[tiny bramble]

now imma try iii by myself

[tiny bramble]

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[slim spoke]

(context: discrete maths, chromatic number, vertex coloring) In the exam we had to prove a vertex coloring with max 4 colors concerning a (1,1,1,2,3) Graph but when I put it into the term I got 3,37 which is smaller than 4 and therefore false. Surprise, surprise, it was wrong so im wondering if I have misunderstood the concept? If anyone could maybe give their insight that would be great

[vocal sleet]

@slim spoke Has your question been resolved?

[edgy sapphire]

I'm just guessing but wouldn't (1,1,1,2,3) be 8 total so you get sqrt(16+1/4) which is more than 4?

[slim spoke]

2|E| basically means you take into consideration that you visit a cortex by the edge two times (sry I have this course in german so Im not that good with the english jargon) so it is 8

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[blissful gale]

if 8cm two long chords subtend a central angle of 60 then radius of thr citclr is

[blissful gale]

i used the formula l=r theta

but it didnt seem to work out

cus theres no options

sorry mt inteenate is slow

[silent tusk]

Btw I think the radius of the circle is

8

think

[blissful gale]

didnt u just google that 😂😂😂

its not…cus two long chords

idk

[silent tusk]

lol

i did

thats why i said think

[blissful gale]

🥲

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[heady stag]

why am i gettin 11/12 if i integrate with y first and 23/24 if i integrate with x first

[crude arrow]

can you show ur work

[heady stag]

yes sure

@crude arrow

[crude arrow]

give me a second to read this over

[obsidian stream]

I don't know why they teach you to find the P(2x+y>1) by taking 1 - P(2x+y<=1) when it's straight forward calculus either way, but your bounds are wrong. x goes from 0 to 1/2.

[crude arrow]

^ ah yes

when X = 1, we have 2 * 1 > 1

[crude arrow]

eh I mean it's an extra step

[heady stag]

oh yeaa it makes sense now

[crude arrow]

aight chef

[heady stag]

yes youre right

[crude arrow]

looks like ur cooking

have a good night

[heady stag]

these bounds are hard man dang

[crude arrow]

ah just think through it chef

don't worry

[heady stag]

aight thanks man

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[vast shale]

you got it

[crude arrow]

remember as long as you have the stove on, you keep cooking

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[wicked dune]

Could I please get help on finding resources on conditional distribution tables and how to make them in R? I've been trying to find some to no avail.

[woeful stream]

are you trying to generate a random variable

hmm maybe not

it's just a table isn't it

so the question is, what form is your data currently in

[wicked dune]

It's a dataset grabbed from a database, I can pull up a website for it if ya want

I unfortunately don't have the data itself handy. It's a set of numbers that I transformed into categorical data though.

https://www.icpsr.umich.edu/web/NAHDAP/series/35

[vocal sleet]

@wicked dune Has your question been resolved?

[woeful stream]

there are probably methods in R to make tables

so to make the conditional distribution table is probably just putting in the right numbers

[vocal sleet]

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[finite hatch]

hi

[finite hatch]

am trying to prove pascals rule by doing

lhs = rhs

but

it doesnt work

wat i do wrong?

oh i see.

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[turbid jungle]

The following things I have done is make A:B = 25:5 and make C:D = 12:18

But I am stuck as to how to do the rest

[copper crypt]

First of all do you understand why it’s even theoretically possible to solve this?

You have 4 variables and 3 equations so you don’t have enough information to find the values for each variable

But you can find their ratios

[turbid jungle]

I know

I am trying to work out the ratio

[copper crypt]

I’d suggest making two more variables tbh, like x and y

And say that A is 5x, B is x, C is 2y, D is 3y

And then solve for an equation with just x and y

So that way you can get all four variables in terms of x, or all in terms of y

[turbid jungle]

I’ll try

So x = 10y/6

So if I rearrange I will get 25y:5y:6y:9y

So the ratio is 25:5:6:9 ?

@copper crypt I am doing gcse level higher in the exam this question is worth 3 mark should I still make an x and y variable for it

[copper crypt]

I have no idea what that is

But yeah I think you should

Adding two variables sounds like it’s more work but i think it’s faster

[turbid jungle]

GCSE level is equivalent to what a 15/16 year old is learning

[turbid jungle]

Ok

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[rapid swift]

I have this geometry problem inspired by some dude's pool shot technique I saw on Instagram. Are angles AcC annd BcC necessarily congruent, and how would I prove it?

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.reopen

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[fading willow]

Cc and Bb are parallel

and u have Bc going throw them

it looks like a Z shape

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[tulip spear]

how do I find the critical points of this implicit derivative? I'm not sure R^2/r^2 is correct either.

[brazen scroll]

Use that R=(r×s)/(r+s)

Then find derivative to r

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[tulip spear]

thx

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[vast shale]

If S(sexagesimal) and C(centesimal) are what is known for a non-zero angle, reduce E. E=S^2 + C^2 + SC/SC should output 271/90

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[nova fulcrum]

$\frac{\tan(x)^2 + \cot(x)^2 + 2}{(\tan(x)^2 + 1) \cdot (\cot(x)^2 + 1)} = \cos(x)$

[twin meteor]

Alex

[cedar cape]

multiply the denominator see what happens

[nova fulcrum]

i am a bit confused because i get no solutions for this

[cedar cape]

remember tan^2 * cot^2 = 1

[nova fulcrum]

but the answer i get doing this is x=0 but that cannot be a solution

[cedar cape]

it’s not 0

what is the whole fraction equal to?

[nova fulcrum]

to 1

[cedar cape]

so cos(x) = 1

[nova fulcrum]

and cos(x) is 1 when x =0

[cedar cape]

and now you use the formula

[cedar cape]

that is partially true

[nova fulcrum]

any solution i get for cos is undefined in the lhs

[cedar cape]

what is cos(2pi)

[nova fulcrum]

1 but that is also undefined for cot

[cedar cape]

yeah cot is indeed undefined for x=0

[nova fulcrum]

so the equation has no solutions?

[cedar cape]

seems like it

wait a second

no it really has no solutions

[nova fulcrum]

gr8 so i was right

sure 100% right? Xd

[cedar cape]

because you get x=2kpi

[nova fulcrum]

yes

[cedar cape]

and sin(2kpi) always equals 0

no solutions

[nova fulcrum]

perfect, thx

.close

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[zinc silo]

f(x) = x^2+x-6/x (its a fraction) find the degree of numerator and denomerator and find the table of values

[twin meteor]

Adam Chebil

[zinc silo]

the first 1

[urban laurel]

do you how to find the degree of polynomial ?

[zinc silo]

no

[urban laurel]

it's the highest exponent

[zinc silo]

ok

[urban laurel]

it's the highest of the degrees of the monomials
x², x and -6x^0 are the monomials in this case

[urban laurel]

do u know how to find the degree of the numerator now ?

[zinc silo]

yes

would there be a horizontal asymptote?

[urban laurel]

no, but what does that have to do with the question ??

[zinc silo]

this is the full question

[twin meteor]

Adam Chebil

[zinc silo]

so what is the degree oif numerator and denominator

[urban laurel]

u said that u know that

[zinc silo]

i need you to check if i got it right

i got 2 for denominator and 1 for numerator

[urban laurel]

2 for the numerator and 1 for the denominator

[zinc silo]

dang i got it mixed up

what is the table of values

<@&286206848099549185>

[lapis maple]

I'm pretty sure the table is for important values
So y intrrseps
x interseps
Asyantops
And other numbers
The things that would help you graph the function

[zinc silo]

i just need the x and y values

5 times

its algebra 2

[lapis maple]

For the x y values just put in when x=-5,2,0,-1000(or another incedble small number),10000 (or another really big number)

To show your limits and intercepts/asentopse

[zinc silo]

.close

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[cursive shadow]

How do they get from f'(x) to f''(x)

[cursive shadow]

it starts from f(x) = x*4^x

[tight sonnet]

they took derivative of f'(x) 🤨

[cursive shadow]

Yes, but i will show where i get stuck.

Because you have to use the product for the second time, right?

[tight sonnet]

not really

or yeah you do

mb

regardless where are you stuck?

[cursive shadow]

I will write it down and send a photo

So i figured out f'

[tight sonnet]

but ln(4) is a constant

[cursive shadow]

Now i want to figure out f'', however the answer model does not show the second derivative of ln(4)

Huh?

[wicked glen]

green fn

[tight sonnet]

ln(4) is a constant (xln(4))'=1*ln(4)=ln(4)

[cursive shadow]

But what about the rule ln(x) = 1/x?

[tight sonnet]

that derivative of ln(x) = 1/x ?

but ln(x) is a function

and ln(4) is a constant

like derivative of x = 1

but derivative of 4 = 0

or derivative of x^2= 2x and derivative of 2^2=0

[cursive shadow]

Oh

Let me try

[tight sonnet]

maybe to see its easier

[twin meteor]

The following error occured while calculating:
Error: Undefined function ln

[tight sonnet]

ln(4)=1.38....

its just a constant nummber

so you have 1.38x

[cursive shadow]

Okay, i will write it down and see where i go

[tight sonnet]

very important note: derivative of a function ln(x) at the point x=4 is actually 1/4

but here ln(4) doesnt mean that in this context

cause no one would write ln(4) meaning this

if you need to find derivative of a function at the point it will be very clear you need to do this

here ln(4) is just a constant like 2,3, 10

and derivative of a constant is always 0

[cursive shadow]

Is this it?

[tight sonnet]

1+xln(4) need to be under bracket

so 4^xln(4)(1+xln(4))

but yeah thats correct otherwise

ln(4) lmao

typo

and see that you can factor 4^xln(4)

and you get your answer

[cursive shadow]

I believe i got it

I will try it a few times and see where i go

Thanks

[tight sonnet]

ok, np youre welcome 👍

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[fluid pumice]

hence means i HAVE TO use something that i somehow derived from part (i)

part (ii) is the question

[lyric fossil]

consider using a trig identity on the denominator

[fluid pumice]

perhaps something to do with sin^2 + cos^2 = 1?

[lyric fossil]

yes

[fluid pumice]

but how will that get me closer to

that

[lyric fossil]

well

i promise you it will

[sinful nacelle]

i assume this means use the derivate somehow, which to me implies integration by parts^^