help-13

[sick ruin]

Overall, yeah

[pale kiln]

Question a gives me two options tho

[cedar kiln]

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[pale kiln]

Oh

[pale kiln]

Welp

[sick ruin]

Yeah they’re asking “where is y increasing” (first box) and “where is y decreasing” (second box)

[pale kiln]

Ohhh I see

I see

So

The range of this in interval notation is

Oh dang I was gonna figure the answer then ask

Its aight

The next question are similar

And I can use your answer as reference

Thank you

Oh

Aight one sec

Give me one sec

[2,4] U [4, infinity)

Got it

You there?

Oh yo

Ok

Is my answer wrong?

Ohh

I see

It says it was wrong tho

Hmm

One sec

Oh

Interesting

Can I screen shot all of this?

For future reference?

Okay

Im confused on another thing

Its either 6

Or -1

But they say its neither

Unless the answer is -6???

Oh its 1

I got it

[cedar kiln]

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[crimson sedge]

hey

[crimson sedge]

https://prnt.sc/_pQ9ZqonSiNG

how do i check the domain of this function

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

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[shadow scarab]

how does, ln x = 5 (e^5 = x) become e^ln x = e^5???

[dense hornet]

by raising e to the power of "both sides"

[shadow scarab]

wdym

[dense hornet]

it's just an informal way of expressing the step out in words

basically, what they did here is they took one side of the equation, say, ln(x) and then raise e to the power of ln(x), you get e^(lnx), however recall that ln(x)=5, hence e^(lnx) is also e^5

[shadow scarab]

ik about the right side. But i dont get how the left side become e^ln x

[dense hornet]

hence we get the equation e^(lnx)=e^5

[shadow scarab]

i see.

but dont we have to do the thinbg on both sidces?

sides*

[dense hornet]

yeah, you can think of it as such

hence like i have said

[dense hornet]

.

[shadow scarab]

i see. Thx

.close

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[tepid chasm]

I am answering a definite integral. I am left with 1/2(-ln3 - ln5). Up to how many decimal places should I use for ln3 and ln5? Or does it not matter?

[jaunty mural]

You leave it in exact form

Why would you use decimal places at this stage...

[tepid chasm]

Aight, leaving it exactly as it is

.close

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[solid quarry]

thats a cool questionnn

i think whichever goes last has an advantage

that would be the odd numbered turns since 19^2 = 361

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.reopen

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[solid quarry]

hi

[solid quarry]

this question

is it ideal for the second player to mimic the first?

in a way reflected about the diagonal

then even if the first player plays a perfect game, each advance will be mimicked

up until the final turn where the first player could get one more point

[solid quarry]

oh but thats not possible

is the first player bound to win if they just fill every cell they can with a 1?

no, theyd have to put at least 19 0s to keep from tying

@celest ledge halp

[celest ledge]

What makes you think I am able to solve competition questions…😂

I just happened to solve two , I generally can’t solve them at all

[solid quarry]

u solved the chess one

this one is also on a grid

[celest ledge]

😂the chess one is definitely not a competition question

[cedar kiln]

@solid quarry Has your question been resolved?

[misty crescent]

just was

in vc

[celest ledge]

Well, at least I can make Alice not lose now

She puts all 1 for her places at first row, that gives her at least 10

Then starting from second row.she places 1 when the above placed by bob is 0, she places 0 when the above is 1

So sum of any column is at most 10

I am substituting it to make it a Winning strategy

Well seems like that’s how far I can go. I can only make Alice win or draw.

If bob is smart then this way it’s definitely a draw

[solid quarry]

im not sure i get it

why must sum of every column be at most 10?

[celest ledge]

It’s not necessarily that order they place numbers but the thought is still valid. Alice place 1 at (1,1). If bob place a number at first row then Alice places a 1 at first row. If bob places a number 0 at (x,y) Alice places a 1 at (x+1,y) when x is even, Alice places a 1 at (x-1,y) when x is odd. Similarly If bob places a number 1 at (x,y) Alice places a 0 at (x+1,y) when x is even, Alice places a 0 at (x-1,y) when x is odd.

You see, after it’s done, (2k,y)+(2k+1,y) must be 1

So the sum of each column is either 9 or 10

And Alice put 10 1 at the first row, so A>=10, and B<=10

[solid quarry]

does it matter if bob places in those spots before? even though that would lead allice to get more points in the first row

[celest ledge]

? Then they just switch identity

Bob becomes alice, it doesn’t matter

[celest ledge]

Wait I was so dumb…

Why don’t we just let Alice put 0 all the cases😂

First Alice place 1 at (1,1). If bob place a number at first row then Alice places a 1 at first row. If bob places a number at (x,y) Alice places a 0 at (x+1,y) when x is even, Alice places a 0 at (x-1,y) when x is odd.

[solid quarry]

oo

[celest ledge]

It sounds like it gives Alice more winning chance but the result seems unchanged.we still have A>=10, B<=10

Anyway I still use first way, at least bob has to calculate harder in his head to make it a draw😂

[solid quarry]

i keep second guessing your logic

[solid quarry]

if bob places at (1,2) allice places at (2,2), (1,3) allice places at (2,3) and so on until bob has 19 in a column?

placing vertically makes more sense

im gunna sleep on it

[celest ledge]

No bob place at (4,8) then Alice place at (5,8). Bob place at (11,2) then Alice place at (10,2)

If bob place at (1,x) then Alice place at (1,y) for any x ,y

And 9 or 10, not 19

[cedar kiln]

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[crimson sedge]

Hi can anyone tell me how to find this
So, what I got as my solution is C(r+b-k+1, r) <- events where you obtain the first red ball
C(a, b) is a over b (binomial coefficient)
But in the solution, they seem to have C(r+b-k, r-1) somehow

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[honest vortex]

@crimson sedge

is that the question

number 2

[crimson sedge]

Yeah

2.a

[honest vortex]

the probability of picking red is

r/(r+b)

right

and it says without replacement

so wouldn't it be

r

/

((r+b) * (K * -1) )

I may be very wrong

which I probably am

[crude pine]

Why "*"

[honest vortex]

that's a multiply

bro please help

It's not even my question

yet I'd like to know

[crude pine]

I know but why would you multiply it

[honest vortex]

I put multiply because every time

you take one out

you don't replace

right

so then if k is the number of trials

then I multiplied -1 by K

so it would mean for K amount of trials there was 1 ball take away each time

wait would the numerator be R-1

because there is no replacement?

[crude pine]

Yes i still dont understand why would you multiply

[honest vortex]

would it be

[crude pine]

I think its -(k-1)

[honest vortex]

oh

my friend says it would be this

then equate it to K

[crude pine]

In first step -0
In second step -1

[crude pine]

Okay but it says THE FIRST red ball

[honest vortex]

yeah exactly

so you can't have r - 1

[crude pine]

(r-1) doesnt make sense for me

Yeah yeah

[honest vortex]

yeah we agree on that

[crude pine]

r/((r+b)-(k-1))

Am i wrong

[honest vortex]

can you explain the -(k-1))

I'm not too sure why

I just want to know

[crude pine]

On the first step there shouldnt be any missing ball-> 1-1=0

On second step there should be (r+b)-1 balls

2-1=1

[honest vortex]

Ohhh okay

[crude pine]

Maybe i am wrong but idk

I think its true

[honest vortex]

maybe I should try writing on pape

r

wait

what am I doing

bruh

[crude pine]

Huh

What

1.st (r+b)
2nd (r+b)-1
3rd (r+b)-2
4th (r+b)-3

[honest vortex]

yes

[crude pine]

Thats why k-1

[honest vortex]

so in theory it should be

k - 1

YES

[crude pine]

Yeah

We solve it

Solved

[honest vortex]

thank you bus

[crude pine]

No problem also thank you too

[crimson sedge]

Oh

xD

I now understand what the problem actually meant

You were basically emptying the whole box full of balls

By taking balls one by one

[crude pine]

Yeah

[cedar kiln]

@crimson sedge Has your question been resolved?

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[sand ether]

Need help on this last part

[dire geode]

what kind of help

[sand ether]

how to show the last part

[dire geode]

did you do i, ii, and the first half of iii?

[sand ether]

yup

nvm thx

.close

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[dense berry]

what does a tetrahedron even look like

[dense wing]

4 triangles together in 3D space

[dense berry]

this is for yr 11 trig and we've never heard of a tetrahedron

so one triangular base and three triangles for sides?

[dense wing]

pyramid with triangle base

[dense berry]

ohh

[dense wing]

(Also... just goggle what a tetrahedron is)

[faint dirge]

I prefer bong

it's a much better search engine

[dense berry]

so for 8a, do I need to find the centre of the base

i'm so lost how the heck do i even start this

all i know is that all triangles have side length x and angles of 60 deg

[faint dirge]

ok so it's not actually that bad

8b is easier than 8a I can't visualise shapes

[dense berry]

i am stupid too it's okay

i need like some 3d shape thingy

[faint dirge]

just a sec

I got it

a is to the left, b is to the right

so a is actually easier

it's a right triangle

[dense berry]

right

[dense berry]

yeah mines is much worse ignore it

[faint dirge]

so you know the hypotenuse is s, you can find the base side

[dense berry]

how?

[faint dirge]

it's the radius of an equilateral triangle

[dense berry]

it's the what

[faint dirge]

it will be a constant proportion based off the side length

[dense berry]

i've never heard of this

[faint dirge]

radius, distance from center to corner

it's very obscure, used for annoying geometric stuff like this

[dense berry]

i see

[faint dirge]

usually you only have to use side length, but the other stuff has a name too

flashbacks to the orthocenter, circumcenter and incenter of a triangle

[dense berry]

i don't think i've heard of it (yr 11 remember) but i'm already dreading it

i always did hate geometry

can i make a wild guess that the angle would 45 degree?

[faint dirge]

I have no idea, I have not tried any of these numbers

but I doubt it

[dense berry]

wait

[faint dirge]

some basic trig and logic (cutting 60º in half gives you 30º) tells you the radius is s/sqrt(3)

[dense berry]

i think i have an idea

what the hell

okay wait hear me out

the green triangle here is isosceles right?

[faint dirge]

yup

[dense berry]

therefore the two nonright angles are 45 right?

[faint dirge]

nope

because the center angle is not right

[dense berry]

it's not?

[faint dirge]

nope

[dense berry]

dang

[faint dirge]

it's 120º iirc

anyways, we already got it

[faint dirge]

as per this, S/2/radius = cos(30), radius = S/sqrt(3)

[dense berry]

it's just that i want to apply stuff i already know

[faint dirge]

so now we have two sides of this triangle, the hypotenuse and the adjacent (S and S/sqrt(3) respectively)

[dense berry]

so anything involving sin rule, cos rule or the such

[faint dirge]

this is all done using trig

[dense berry]

but i don't know how to use them

[faint dirge]

trig is the idea that for any angle, triangles must follow certain ratios

[dense berry]

ohh

[faint dirge]

if you take two right triangles, both with angle 38.38º, the opposite side divided by the hypotenuse is the same as in any other (right) triangle's opposite side divided by the hypotenuse

[dense berry]

okay yeah i get that

same ratios and such

[faint dirge]

you can expand trig to nonright triangles, through the law of sines or cosines, but it's a pain P:

[dense berry]

yeah i agree

one second i'll brb

[faint dirge]

anyways, here we say that the ratio of S/2, our adjacent, divided by a, our radius/hypotenuse, is the same as the cos(30º), which is sqrt(3)/2

if you do the math, you get a=S/sqrt(3)

[dense berry]

wait wait can i try a different approach

[faint dirge]

you can try

[dense berry]

uh what the fuck

i got (in my CAS) that x=b

oh wait i'm a dumbass

[faint dirge]

anyways, given our base side here is S/sqrt(3), and the hypotenuse is S, then cos(angle) will equal S/sqrt(3)/S = 1/sqrt(3)

[dense berry]

the angle is 30 not 60

i got it

i think

[faint dirge]

tell me when you find 8a then lol

[dense berry]

is it 54.74 degrees?

[faint dirge]

yup

[dense berry]

it issss

yeaaaaaaaasssss

[faint dirge]

:D

[dense berry]

wooohoo

god that was a pain

so this is what i did

[faint dirge]

the other one's process is basically identical, dw (or maybe do worry?)

[dense berry]

i probably should

i used the sine rule

so x/sin(120 = b(the base of the blue triangle)/sin(30)

[faint dirge]

well.... yes

you could have simply used trig the way it was originally created to be used, but this works perfectly fine too lol

[dense berry]

i just found the other method

...hard to digest?

[faint dirge]

wdym?

[dense berry]

like it didn't sit right with me

i don't really know

this makes more sense to me in a weird way

but anyway

[faint dirge]

oh yeah dw about it

[dense berry]

now how do i go about 8b?

[faint dirge]

if your method works, and you understand it, that's fine

it's just an atraditional method

https://media.discordapp.net/attachments/903430334681608232/951993778200264705/Screenshot_2022-03-11_at_19.03.16.png

so now we need the center angle

we are going to do the exact same process

we will find the base side, and the hypotenuse

[dense berry]

wait which angle

hang on hang on

[faint dirge]

the angle between the front facing face and the base face

the dotted line is meant to be vertical

[dense berry]

oh

[faint dirge]

but as you can see, I am not an artist

[dense berry]

wait that's it?

dang i really lost myself

[faint dirge]

lol

[dense berry]

i thought it was the angle at the top of the two sloping sides

[faint dirge]

anyways we come back to here

https://media.discordapp.net/attachments/903430334681608232/951995135351193660/Screenshot_2022-03-11_at_19.08.28.png

instead of using the radius (a)

we use the "h" value as our base

you can find that however, there are three easy methods

law of sines, normal trig, or special right triangle memorisation

[dense berry]

wait wait

[faint dirge]

in order of how hard they are, starting with the hardest

[dense berry]

just a second please

[faint dirge]

no worries

[dense berry]

is it the yellow or the red angle i'm trying to find

oh wait

[faint dirge]

red angle

the yellow angle is just 60º

[dense berry]

yeah it's red i'm just dumb

[faint dirge]

this is a pretty good image

so, we want the hypotenuse, which is the distance from the center of the base of the triangle to the opposite point/angle

[dense berry]

right

[faint dirge]

and we want the base, which is the "h" value here, the distance from the center of the side to the center

then we find the ratio and take the arccos of that ratio, giving us our angle

easy peasy

[dense berry]

hang on

i'm slowwwlly processing this

3d shapes is so much pain

so

[faint dirge]

mhm

which is why we reduce it to 2d projections and shapes

[faint dirge]

this is the 2d projection we use to understand what we need to find

and to find those sides we need, we just use 2d triangles

[dense berry]

right i think i got it

and i need to find the height of the green tringle

[faint dirge]

yup

[dense berry]

*30 degrees for the angle not 60 🤦‍♂️

[faint dirge]

mhm

[dense berry]

soo

73.22 degrees?

dang it no it's not

WHY THE HECK IS IT NOT

i'm going to die here

okay okay

finally

oh my god FINALLY

i got it i hate trig i hate geometry i hate maths

thanks so much sycthe for ur help if u see this

.close

[cedar kiln]

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[unborn hound]

kinda stuck here

not sure where to go next

[cedar kiln]

@unborn hound Has your question been resolved?

[unborn hound]

.close

[cedar kiln]

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[hybrid orbit]

how can i find the standard deviation and the mean when im only given the z score

[hybrid orbit]

i tried doing the let x thingie but it ended up just cancelling each other out

this formula

[dense wing]

you can't find both with only 1 z-score

[hybrid orbit]

this is where im confused with lol

idk if the SD should be given but usually it is lol

[dense wing]

$Z\sim N(0,1)$

[wraith dagger]

Mosh

[dense wing]

if $X\sim N(\mu,\sigma^2)$, then $\frac{X-\mu}{\sigma}\sim N(0,1)$

[wraith dagger]

Mosh

[hybrid orbit]

what i did was
-2.37 = (.0089 - u ) / o
o = (.0089 - u) / -2.37

then plug that in the original equation but yeah it just cancelled out lol

[dense wing]

where the hell is .0089 coming from?

[hybrid orbit]

the z table

[dense wing]

Post your N(0,1) table

[hybrid orbit]

N(0,1) ?

[dense wing]

The standard normal.

[hybrid orbit]

https://www.ztable.net/

i just use this lol but our teacher makes us round it off to 4 decimals

[dense wing]

so yeah, 1/2-0.00889

[hybrid orbit]

oh 1/2 cuz its the mean?

[dense wing]

cause you want the area b/w 0 and -2.37

cause $P(Z\leq 0)=\frac{1}{2}$

[wraith dagger]

Mosh

[dense wing]

again, where $Z\sim N(0,1)$

[wraith dagger]

Mosh

[hybrid orbit]

wait wheres the zero from

[dense wing]

the mean of Z is 0

[hybrid orbit]

the problem is between mean and -2.37

[dense wing]

the variance is 1

yeah

it's a N(0,1) rv

[hybrid orbit]

is this by default?

[dense wing]

Z is standard normal....

so yes

the standard normal has the standard normal mean and variance

[hybrid orbit]

wait so if the mean isnt given we just assume it is 1 in normal distribution?

i mean 0

[dense wing]

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

[dense wing]

NO

You're told the curve is the standard normal

Thus you know it's centered at 0

[hybrid orbit]

ohhh

i see

its only zero if it is in standard

i get it now lol

[dense wing]

no

converse

if it's standard normal, the mean is 0

[hybrid orbit]

The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1

[hybrid orbit]

this?

[dense wing]

you can easily have $N(0,2)$, which has mean 0 but not be the standard normal

yes

[wraith dagger]

Mosh

[dense wing]

$N(\mu,\sigma^2)$ has mean $\mu$ and variance $\sigma^2$

[wraith dagger]

Mosh

[hybrid orbit]

so if im looking for the area between
i use
1/2 - z score

if its to the right or left
1 - z score

[dense wing]

Don't try and make this into formulas.

Just use properties of the normal density function.

It's symmetric about the mean, area under it over all of R is 1

You just use logic about what the region you're looking for is like, and solve based on your table

[hybrid orbit]

Thanks!!

ill look into the properties lol

[dense wing]

Only proper rule you should remember is 68-95-99.7 rule imo

[hybrid orbit]

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[warm vector]

We’re finding x degrees here

[livid hound]

what have you tried so far

[warm vector]

I tried using the polygon formula

Ok I got it

Thanks

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[bold hinge]

How much does a circular sector deviate from a triangle

[bold hinge]

In terms of area

[cedar kiln]

@bold hinge Has your question been resolved?

[cedar kiln]

@bold hinge Has your question been resolved?

[astral crown]

can't you calculate the difference

in terms of the angle

[bold hinge]

i could but that's not satisfying

i want to do it in terms of differentials

[astral crown]

so you're looking at something like the proof involving polar integration

[bold hinge]

proof of what? that a triangle deviates from a sector?

lol

[bold hinge]

show me this it sounds interesting actually

[astral crown]

its something along the lines of area bound by a polar curve can be approximated using triangles or sectors

triangles will have like 1/2 r^2 sin(theta) and sectors will use 1/2 r^2 theta

[bold hinge]

really?

[astral crown]

but the approximation is better ofc for smaller theta

[bold hinge]

so this just becomes sin(theta) - theta

[bold hinge]

the triangle only really exists for theta less than 90 degrees doesn't it?

[astral crown]

yea, its the sum of many smaller triangles/sectors

[bold hinge]

or i guess 180 degrees

[cedar kiln]

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[scenic pasture]

how to do part b

[flint plinth]

if you plug in $e^{i\theta}$ to the formula for the series from part (a), what do you get?

[wraith dagger]

OurBelovedBungo

[scenic pasture]

i get (e^itheta - e^(n+1)itheta)/1-e^itheta

[flint plinth]

looks right

ok now notice that the sine series is the imaginary part of the exponential series

so you need the imaginary part of your formula

[scenic pasture]

oh ok

ok thanks will see what i get

how do i get just the imaginary part since the denominator has e^itheta

so wont all of it be imaginary

[flint plinth]

one general approach is that if you have a complex number $w$, then its imaginary part is $\frac{w - \overline{w}}{2i}$, you could apply that to $w = \frac{e^{i\theta} - e^{i(n+1)\theta}}{1 - e^{i\theta}}$ and simplify

[wraith dagger]

OurBelovedBungo

[cedar kiln]

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[scenic pasture]

Thanks for the help

[cedar kiln]

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[errant palm]

pls help me

[errant palm]

nvm

i got it

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[slender flower]

@livid hound

[slender flower]

whats the solution of the semi

[livid hound]

apply formula for area of a semicircle

[slender flower]

where>?

so y is 0?

[livid hound]

wdym

[slender flower]

y=sqrt(16-x^2)

[livid hound]

that's the equation of a semi-circle (arc)

[slender flower]

then?

whats next?

I thought x is 4

[livid hound]

wdym

[slender flower]

whats next?

[livid hound]

do you understand that you're being asked for the area of the shaded region

[slender flower]

IDK

can you show me what the solution >?

[livid hound]

we're getting there...

you first need to understand the problem

[slender flower]

then

so 8 is the x ?>

@livid hound

[livid hound]

whut

wdym by "is the x"

[slender flower]

can you show me how ?

[livid hound]

forget about the graph

for a sec

do you know the formula for the area of a circle?

[slender flower]

A=πr2

[livid hound]

use ^ to denote exponentiation

and hence what would be the area of a semi-circle

[worthy snow]

||A/2||

[slender flower]

1/2

1/2 πr^2

[livid hound]

$A_{\text{semi-circle}} = \frac12 \times \pi r^2$

[wraith dagger]

ℝamonov

[livid hound]

can you identify the radius of your semi circle?

[slender flower]

4?

[livid hound]

yes

[slender flower]

then?

[livid hound]

well you have the formula for the area of a semi-circle

and you have the radius of your semi-circle...

[slender flower]

whats next ?

[livid hound]

use your radius and the formula

[slender flower]

what formula?

[worthy snow]

sub in the radius to the formula to find the area of the semi circle

[livid hound]

the formula for the area of a semi-circle

that you just gave me

because that's what you have in your question

[worthy snow]

ramonov u mind helping me after this (#help-12)

[slender flower]

y=sqrt(16-x^2)

this one ?

[livid hound]

no

[slender flower]

oh I see

Wait

[livid hound]

that's what was supposed to be the equation of the curve

[slender flower]

pie r^2 ?

[livid hound]

no e in pi

pi r^2 gives the area of a full circle

1/2 * pi * r^2 gives the area of a semi-circle

1/2 πr^2
the formula for the area of a semi-circle
that you just gave me

[slender flower]

6.28

[livid hound]

how are you getting 6.28

[slender flower]

1/2 x 3.14 x 4

[livid hound]

you're not applying the formula properly

specifically you ignored the part where radius is squared

and you shouldn't be using an approximation for pi

[slender flower]

How?

wait

[livid hound]

$A_{\text{semi-circle}} = \frac12 \times \pi r^{\red{2}}\neq \frac12 \times \pi r$

[wraith dagger]

ℝamonov

[livid hound]

$\pi \neq 3.14$

[wraith dagger]

ℝamonov

[slender flower]

25.12

[livid hound]

leave pi as pi unless otherwise stated

[slender flower]

8 pi

i see

@livid hound

thats only the solution ?

[tribal drum]

1/2(pi r^2)

r is 4

1/2(pi 4^2)

[livid hound]

dunno what else you want

[tribal drum]

1/2 (pi 16)

[livid hound]

the question asks for the area of the region,
and you found it

[slender flower]

ok thanks

[worthy snow]

dont write it like that, preferably if you keep the pi in the equation of the area

[cedar kiln]

@slender flower Has your question been resolved?

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[golden geode]

what wrong did i do

[cedar kiln]

@golden geode Has your question been resolved?

[golden geode]

<@&286206848099549185>

[inland tartan]

gonna try help on this one because i was listening to https://www.youtube.com/watch?v=3YDOZjrVMAY when i saw this

your handwriting's a little weird, no offense. can you tell me the actual question?

i can explain the question and you can double check your working as we go along

[golden geode]

lol none taken stress taking over

hold one let me get it

question a

[inland tartan]

here's me solving the turning point, which you incorrectly calculated as (2,3)

what i'm doing here is finding the turning point of the original quadratic (x^2-4x+3), and then finding the minimum through completing the square (hope you know that), and taking the absolute value of the Y coordinate. i do this because the absolute value just makes the negative points positive (see the pic i posted of the 2 graphs)

you've factorised and found the "roots" (points where y=0) correctly though - just factorise the quadratic, and find when it's equal to 0

does the turning point/minimum solution make sense @golden geode ?

[golden geode]

hold up i am still kinda new to this topic

hw you got this

ohhh i got it thanks

[inland tartan]

basically, you can complete the square to find the minimum of a turning point - it's a little long so i can link you a video

[inland tartan]

you sure?

[golden geode]

yep

[inland tartan]

aight nice. tell me if you need anything else

if you don't you can close the channel with .close

[golden geode]

thanks alot

[inland tartan]

np!

[golden geode]

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[lament flint]

i dont understand the solution

[bold hinge]

You should know this polynomial to be special

[lament flint]

how is it special?

[bold hinge]

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

The solutions to this are the 3rd roots of 1

[lament flint]

yea

but why does 6 divide k

[bold hinge]

,w ((1+isqrt(3))/2)^3

[lament flint]

its sqrt(3)

[bold hinge]

Huh I can't read sorry

[wraith dagger]

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Requested by EndTimes#8053

[lament flint]

should it be 3, since theyre cube roots

oh

theyre cube roots of -1

[bold hinge]

My bad

Yeah so that hopefully clears things up

3×2=6

[lament flint]

also why does 1-e = 1/e?

is that just by trying 1/e with the roots

[bold hinge]

1-x = 1/x
x-x^2 = 1
x^2-x+1=0

[lament flint]

oh

[bold hinge]

Yep.

[lament flint]

ok ty

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[bold hinge]

👍🏻

[cedar kiln]

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[karmic totem]

Hello! I am currently following a paper on physics simulation for a hobby project and I have come across a problem:
Given two vectors $v$ and $x = r \cross v$, I need a 3x3 matrix with the following property: $mv = x = r \cross v$.
If I remember correctly, one cannot simply divide two vectors and expect to get the right matrix (I wouldn't know how to anyways). So which strategy would one use to calculate the needed 3x3 matrix?

[wraith dagger]

Cube

[cedar kiln]

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[karmic totem]

<@&286206848099549185>

[karmic totem]

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[scenic crest]

Why did definite integral of second function(1) become "x" and not "e-1" in 2 line of solution?

And also why did {(lnx)^n.x|^e 1} become "e-0"?

[cedar kiln]

@scenic crest Has your question been resolved?

[cedar kiln]

@scenic crest Has your question been resolved?

[cedar kiln]

@scenic crest Has your question been resolved?

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[toxic flame]

please can someone suggest which part needs simplifying

[cedar kiln]

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[celest ledge]

I am about to use pigeonhole

We probably need to show that {Σ{x: x from A} mod (1+2+…+n)} contains 1+2+…+n different elements

[terse roost]

you would need them to be consecutive in addition to that

and you might be able to prove by recurence that there is a subset A of {0,...,n-1} such that every number between 0 and n(n-1)/2 is the sum of the elements of A

and i'm pretty sure you can do the same for A being a subset of {m,...,m+n-1} and dealing with numbers between m and n(n+m-1)/2

[celest ledge]

You mean n(n+2m-1)/2?

[terse roost]

yes sry

By the way you would need m to be >= 1, m=0 doesnt work

between n(n+2m-1)/2 and m there are (n-1)(n+2m)/2 + 1 numbers if i'm not wrong. This is above 1 + ... + n if m>=1

then it's done if the recursive reasoning works

but it doesnt

my bad

[celest ledge]

Okay I got it. So we can prove his statement when m>=1

m=0 we know it’s false

[terse roost]

the recursive thing doesnt work

[celest ledge]

You mean this?

[terse roost]

yep

[celest ledge]

I proved it

[terse roost]

what if m = 5 and n = 2 ? you have 5 and 6 as numbers and want every numbers between 5 and 11 as sums. How do you build 7 ?

[celest ledge]

Shit

I thought I proved it

[terse roost]

what you can build are km + p(k) where p(k) can be decomposed as a sum of k elements between 0 and n - 1

[celest ledge]

We can only prove this when m=0 or 1 I think

[terse roost]

yep

hum

maybe i have an idea

if we can prove that the km + p(k) all take different values mod n(n+1)/2 it would work right ?

then suppose they don't : km + p(k) = k'm + p(k') mod n(n+1)/2

there might be some way to find k'' and p(k'') that makes this work

[celest ledge]

That’s impossible I think

The former are 2^n -1 numbers

But the latter is (n+1)n/2

[terse roost]

what if m < (n+1)n/2 ?

plus its "mod"

[cedar kiln]

@cinder tide Has your question been resolved?

[terse roost]

let's say km = m' mod n(n+1)/2 with m' between 0 and n(n+1)/2 - 1. The recursive thing works for m = 1 so there is q and p(q) such that p(q) = m' = km mod n(n+1)/2. Then there is p(n-q) such that p(q) = n(n+1)/2 - p(n-q) (p(n-q) contains the terms not use to build p(q)). There fore, p(n-q) = -m' mod n(n+1)/2 and km + p(n-q) = 0 mod n(n+1)/2. This means we just need to prove that it is possible to have n - q = k

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[celest ledge]

Got it, We need to have a q-subset whose sum =(n-q)m mod n(n+1)/2. But it seems transform back to our original question 😂

[cedar kiln]

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[crimson sedge]

please help me

[celest ledge]

Infinity many

Or you want them to be what? Positive? Non-negative?

[crimson sedge]

yess

[celest ledge]

C(19,3) all positive solutions, C(23,3) all non-negative solutions

[crimson sedge]

but i need the even number of it

[celest ledge]

All them need to be even?

Positive or non-negative?

[crimson sedge]

yess

non negative and even

[celest ledge]

C(13,3)

[crimson sedge]

how?

[celest ledge]

It’s the same as 4 non-negative integers whose sum is 10

It’s equivalent to 4 positive integers whose sum is 14

[crimson sedge]

wait....

[celest ledge]

Put fourteen 1 on a line, use three objects to separate those 1 to create four blocks, it’s the same as choose 3 places to put those objects from 13 places

[crimson sedge]

how about there are both negative and positive?

[celest ledge]

So C(13,3)=286

[crimson sedge]

ohh

[celest ledge]

What?

A number can not be both positive and negative

[crimson sedge]

Ohhh....

Ok

.close

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[real bone]

Do I not have to use ln to calculate the derivative of 11^x -10^x?

[kindred storm]

@real bone Yes, but it's 11ˣ = eˣ ˡᵒᵍ⁽¹¹⁾.

[real bone]

Or did I forget the product rule

[kindred storm]

So, you'll get log(11) eˣ ˡᵒᵍ⁽¹¹⁾ as the derivative, which equals log(11) 11ˣ.

[real bone]

idk how you found that derivative

[kindred storm]

I converted 11ˣ to eˣ ˡᵒᵍ⁽¹¹⁾ using the logarithm rules from algebra.

Then I took the derivative of that.

[real bone]

I actually don't remember that log rule

Ok so

[real bone]

Ty

[kindred storm]

It's like rule 7 combined with an exponent rule.

11ˣ = (11)ˣ = (eˡᵒᵍ⁽¹¹⁾)ˣ

[real bone]

Rule 7 seems confusing

[kindred storm]

Well, basically exponents and logarithms are inverses.

[real bone]

Oh true

B^k =logb of k

Right?

[kindred storm]

So, logₓ(xᵃ) = a and xˡᵒᵍˣ⁽ᵃ⁾ = a.

No, log(bᵏ) = k log(b).

[real bone]

Wait like if I wanted to make the inverse of b^k

[kindred storm]

You have 11ˣ

But that doesn't have a nice derivative rule.

eˣ does, though.

So, you want to change the base to e instead of 11.

The way you do that is to move the 11 into the exponent.

The way you do that is to do eˡᵒᵍ⁽¹¹⁾ ˣ.

[wraith dagger]

Chai T. Rex

[kindred storm]

You take the base whatever the new base is logarithm of the old base and put that in the exponent.

So, if the new base is c, you take the base c logarithm of the old base.

[wraith dagger]

Chai T. Rex

[real bone]

Is that cuz x^logx is 1?

Log base of x I mean

[wraith dagger]

Chai T. Rex

[kindred storm]

Never mind.

Let's look at rule 7 again.

[wraith dagger]

Chai T. Rex

[kindred storm]

So, you can change the base from k to b using rule 7.

Here's how you do it.

[wraith dagger]

Chai T. Rex

[kindred storm]

First, we parenthesize the 11.

Then, we can use rule 7.

[wraith dagger]

Chai T. Rex

[kindred storm]

We've used rule 7 inside the parentheses.

Does that make sense so far?

[real bone]

Is log 11 the same as loge 11

[kindred storm]

log by itself usually means logₑ or ln.

In calculus, I mean.

[real bone]

Oh

[kindred storm]

In other subjects, it can have a different base.

Like in computers, it can have base 2 instead of e.

But we're doing calculus, so it's usually base e.

[real bone]

Ok

[wraith dagger]

Chai T. Rex

[real bone]

I forgot the rules of what the derivative of a logarithm is

[kindred storm]

Oh, it's a constant.

log(constant) is a constant.

Like log(5) is a constant.

[real bone]

Which is zero

[kindred storm]

So log(11) is also a constant.

Well, let's be careful.

For the derivative here, we use the chain rule.

[real bone]

Oh

[kindred storm]

Because it's a function, eˣ, but with a different argument than x.

Like if you have sin(10 x), that's like sin(x), but it's different than x.

So, you use the chain rule.

So, we do the chain rule.

[wraith dagger]

Chai T. Rex

[real bone]

e^log(11)x * 1

[kindred storm]

How did you get the 1?

[real bone]

Derivative of the exponent

[kindred storm]

OK, what's the derivative of log(11) x?

[real bone]

Well I thought you said log of anything is a constant

Oh

[kindred storm]

Right, so it's a constant times x.

[real bone]

Then it would be log(11)

[kindred storm]

Right.

[wraith dagger]

Chai T. Rex

[kindred storm]

And then we can work backwards on the e part.

[wraith dagger]

Chai T. Rex

[kindred storm]

So, we have log(11) 11ˣ.

Does it make sense how we got that?

[real bone]

Yes

[kindred storm]

So, bˣ's derivative is log(b) bˣ.

Which works when the base is e.

log(e) eˣ = 1 eˣ = eˣ.

[real bone]

But it's the same as taking ln of the function doing the derivative and multiply by the original function

[kindred storm]

Well, if you have bˣ, it's the same as ln(b) bˣ.

The derivative is.

You can either remember to do the chain rule and the logarithm rules or you can remember the derivative of bˣ as log(b) bˣ directly.

It's not the ln of the function doing the derivative, though.

It's the ln of the base.

So, bˣ means log(b).

Does that make sense?

[real bone]

Yea but log(b) is ln(b)

[kindred storm]

Yes, in calculus it usually is.

[real bone]

Well my original approach to the problem was to take the ln of function and find the derivative and multiply the result with the original function

So if I did that why did I get the wrong answer

[kindred storm]

With L'Hospital, you just take the derivative.

You can't use the regular limit laws with things that are in indeterminate forms, like 0/0.

They have restrictions on using them when certain things like that happen.

So, you need to use L'Hospital or some other method.

[real bone]

Yea but another way of finding a derivative of a function is the way I used

[kindred storm]

Do you mean logarithmic differentiation?

[real bone]

Yes

So I took that approach to finding the derivative and got the wrong answer some how

[kindred storm]

OK, let's try that.

[real bone]

And thank you for showing me another way btw

[kindred storm]

(11ˣ)' = 11ˣ (ln(11ˣ))'

= 11ˣ (x ln(11))'

= 11ˣ (ln(11) + 0)

[real bone]

Yep

[kindred storm]

= ln(11) 11ˣ.

[real bone]

Oh

I think I forgot product rule

[kindred storm]

Oh, OK.

[wraith dagger]

Chai T. Rex

[real bone]

I will multiply by the original function after of course

[kindred storm]

No, the first part is right, but the second and third parts aren't.

You shouldn't get x ln(11) - x ln(10) in the top.

[real bone]

xln(11/10)?

[kindred storm]

Well, the first expression's limit is 0/0.

Which is indeterminate.

So, you can use L'Hospital's.

But when we used L'Hospital's we got ln(11) 11ˣ - ln(10) 10ˣ on top, not those.

[real bone]

I thought lhopital's just said to take a derivative of the numerator and of the denominator separately

[kindred storm]

Yes, and then the new numerator is the derivative of the old and the new denominator is the derivative of the old.

[real bone]

Did I mess up in applying the ln?

[kindred storm]

No, you didn't write the result of the derivative.

The result was 1 on bottom and ln(11) 11ˣ - ln(10) 10ˣ on top.

[real bone]

Ok so applying ln was correct

[kindred storm]

You have x ln(11) - x ln(10), which is not the same as ln(11) 11ˣ - ln(10) 10ˣ.

[real bone]

xln11 - xln10 all over lnx

[kindred storm]

Why are you applying logarithm?

The derivative is done.

[real bone]

Logarithmic differentiation

[kindred storm]

Yes, and we got ln(11) 11ˣ - ln(10) 10ˣ on top, right?

[real bone]

I dodnt

[kindred storm]

Well, we have (11ˣ)' = 11ˣ (ln(11ˣ))'.

Where did you go from there?

[real bone]

Cuz I was gunna do 1. Take ln
2. Take derivative
3. Multiply by original function

[kindred storm]

No, that's not how logarithmic differentiation works.

It's not take the logarithm and then do the derivative.

It's the original expression times the derivative of the logarithm of the original expression.

f'(x) = f(x) (ln(f(x)))'

Don't forget the f(x) in front, multiplied with the derivative of the logarithm.

[real bone]

Yea I was gunna multiply by f(x) after

[kindred storm]

OK, but then you can't write it all as one expression.

You can't have x ln(11) - x ln(10).

[real bone]

I'm still wanting to make sure I did ln(f(x)) correctly

[kindred storm]

You need to separate it out.

It's better to write exactly what you're doing.

[real bone]

Ok like doing each step for each term separately?

[wraith dagger]

Chai T. Rex

[kindred storm]

Yes, because you're taking the derivative of each term in the numerator.

[real bone]

Oh

True ok

[kindred storm]

Not of the whole numerator.

[real bone]

So if we just deal with 11^x

We do xln11 then find it's derivative

[kindred storm]

Yes, but write out what you're doing so that it can be followed along with.

[real bone]

Ok

So ln(11^x)

Is xln11

Then I do product rule and chain rule I think

Erm not chain rule

[kindred storm]

Yes, the product rule is correct.

[wraith dagger]

Chai T. Rex

[kindred storm]

See, here the left side tells what the right side, your work, means.

[real bone]

Product rule gave me ln11 + x/11

[kindred storm]

How did you get x/11?

[real bone]

Derivative of ln11

Times x

[kindred storm]

What's the derivative of ln(11)?

[real bone]

1/11

[kindred storm]

No.

ln(11) is a constant.

,calc log(11)

[wraith dagger]

Result:

2.3978952727984

[kindred storm]

It's 2.40 and so on.

[real bone]

Oh its it only ln/x that has the derivative of 1/x?

[kindred storm]

Right.

It has to have a nonconstant variable.

[real bone]

Ok my bad

So ln11 + 0

[kindred storm]

Right.

[real bone]

And the same for 10^x

[kindred storm]

Yes, but first finish (11ˣ)'

Multiply the derivative of the logarithm with the original function.

[real bone]

11^x * ln11

[kindred storm]

If you forget to do that, there's a way to remember.

You just write what you're figuring out on the left.

[wraith dagger]

Chai T. Rex

[kindred storm]

That way, you'll know you're not done yet when you get ln(11), because you don't have (11ˣ)' yet.

[real bone]

Oh ok

[wraith dagger]

Chai T. Rex

[kindred storm]

And then do the part in parentheses on the right.

[wraith dagger]

Chai T. Rex

[real bone]

The bottom is just x/x

[kindred storm]

That way, you use the ' to keep track of what you differentiate.

Sorry, keep track of what you still need to differentiate.

[real bone]

So 11^x * (ln11) - 10^x * (ln10)