help-13

[errant vector]

alr

lets solve for y'

[ember lichen]

a;thpigh

i think u made a mistake

• y

not + y

[errant vector]

ye

sorry lol

its late

[ember lichen]

all good

[errant vector]

anyways im gonna move y and 2x to the other side ig

-xy' + 6yy' = y - 2x

and factor the y'

y'(-x + 6y) = y - 2x

if i make any silly mistakes point them out ig

[ember lichen]

makes sense

[errant vector]

so ig you were right

or symb was right

y' = (y-2x)/(-x + 6y)

i was confused cuz it was missing the y'

y' = (y-2x)/(-x + 6y) = 1

alright well this is a problem

[ember lichen]

why are we making y' = 0?

[errant vector]

1

[ember lichen]

should've we make it 1?

[errant vector]

sorry

yeah

lol

[ember lichen]

then i assume we just multiply the denom. out, yeah?

[errant vector]

im not sure if we're meant to solve this or not

there are 2 variables

[ember lichen]

yeah

that's what im confused about

[errant vector]

it does say "any" tangent

so like

it might just want us to leave the point it touches untouched

wait but that wouldnt make any sense cuz the problem would be too easy 😭

[ember lichen]

ans:

[errant vector]

ok wait

alr so the slope is 1

and the points are (-12,f(-12)) and (12,f(12))

im confused on how we were supposed to figure out 12,-12

i look stupid

[ember lichen]

no ur not

i don't get it either

[errant vector]

hmm

[ember lichen]

[errant vector]

ima ask alpha

wolfram alpha doesnt know

[ember lichen]

https://tenor.com/view/skull-funny-aughhh-emoji-twiggle-gif-25776741

[undone star]

im not too caught up with all the convo

[ember lichen]

you know it's fucked up when alpha doesn't know

[undone star]

but from a glance

it should be implicit?

[ember lichen]

yes

[errant vector]

yeah

[ember lichen]

we alr did that

[undone star]

so ur ifnal answer should be in terms of y' and y

[ember lichen]

this question is fucked up and i have like 10 more like these

[errant vector]

basically, what values of x make y' = 1

[ember lichen]

answer here

[errant vector]

except its implicit

like so

[ember lichen]

wait

[errant vector]

f(x,y) = 0

[ember lichen]

dy/dx = 1
dy/dx = that stupid equation

sub in 1

solve for y

then substitute it in

into the original equation

no?

[errant vector]

stupid equation = 1

[ember lichen]

no the stupid f'(x)

stupid equation

[errant vector]

(y-2x)/(-x + 6y) = 1

[ember lichen]

yes

[errant vector]

ye

[ember lichen]

i got y = -x/5

[undone star]

thats straonge

i got negative 3x/5

i might have done something wrong

give me a sec, i do it on paper

[ember lichen]

nah i checked it on desmos

it's -x/5

[errant vector]

alr

[ember lichen]

that is like our derivative

wait no

it's the uh

what the fuck is it

tangent?

[errant vector]

?

thats another elliptical

ellipsis

whatever they're called

[ember lichen]

yeah

i got it mapped in desmos

and uh

it's an equation

OH

WE HAVE THE TANGENT

[errant vector]

defined in terms of x and y

f(x,y)

[ember lichen]

SO WE SOLVE FOR INTERSECTIONS OF THE TANGENT

ON THE ECLIPSE

[errant vector]

OH

I SEE IT

BRUH

[ember lichen]

this is fucked up

[errant vector]

How was anyone supposed to know that

what is this

precalc?

[ember lichen]

IB math calc + vectors

[errant vector]

ib math?

[ember lichen]

AP but for europeans, but i live in canada so they colonized us

[errant vector]

ah

vectors 💀

that is a damn good problem but i hope i never see it on a test

[ember lichen]

LMAOOO

my teacher

[errant vector]

😭 why was that so hard

[ember lichen]

has like 10 more lined up

nah it's not that

it's cuz he is makes math contests 💀

[errant vector]

i still dont understand how we were supposed to figure it out w/o a graphing calculator

[ember lichen]

so bro will throw us the most diabolical problem known to man

ok so uh

-x/5

[errant vector]

why would it have been the intersections tho

[ember lichen]

idfk either

i gotten so used to this

i don't even wanna know anymore

we got the normal?

yeah the normal i think

[errant vector]

nah

the angle is 90deg

isnt*

[ember lichen]

true

idfk then

[errant vector]

😭

[ember lichen]

LMAOOOO (i cant take this any longer with math. why did we invent this shit bro. im going back to the babylonian times not to warn them of god's warth, but to sabatoge their progress in math and return us back to cave men)

[errant vector]

lmfao

alr ig .done

[ember lichen]

ok funny oart

i dont know what to actually do

so like

so we take -x/5

and sub it in?

or like

[errant vector]

nah

[ember lichen]

do we make it equal?

[errant vector]

we just graph -x/5

and then the we get the (x,y) of the intercepts

and then we plug it in for yk

a

dy/dx = 1

is basically

[ember lichen]

😭

[errant vector]

f_2(x,y) = 1

because 2 vars:x,y

and then we take the intersections from -x/5 and plug em into f_2(x,y)

wait this is getting so confusing one sec ima latex it

[ember lichen]

what?

f_2?

what is f_2?

-x/5?

[errant vector]

$$\frac{y-2x}{6y-x} = 1 \to y = -x/5$$

wait

[wraith dagger]

amukh1

[ember lichen]

si

[errant vector]

f(x) is the original function x^2 - xy..

$$f(x) = -x/5 \to (a,b),(c,d)$$

[wraith dagger]

amukh1

[errant vector]

ab,cd are the two coordinates we get from the intersection

so we make it equal instead of substituting

so therefore,

$$f(x) = -x/5 \to (a,b),(c,d) \implies \frac{(b,d) - 2(a,c)}{6(b,d) - (a,c)} = 1$$

alright there

[wraith dagger]

amukh1

[errant vector]

see

we get ab,cd

[ember lichen]

where did uh

2 and 6

come from?

[errant vector]

and they satisfy the equation

from dy/dx

[ember lichen]

ohhh

alr

[errant vector]

y - 2x / 6y - x

im saying (b,d)

cuz there are 2 y values, b and d

cuz 2 intersections

so now we solved that equation

we can say

f(k) + (x-k)

where k is

(a,c)

wait sh-

alright there

thats my solution

still confused on why its the intersection but whatever

get it? kinda? lol

[ember lichen]

but how would you be a,b and c,d?

see i did make it equal

i just an uglier

x^3 equation

[errant vector]

¯_(ツ)_/¯

idk

i cant scroll that far back rn

to see

but ill assume your right

[ember lichen]

wait

im stupid

gahh

it still doesn't solve shit

i think we gotta subtitute in

so we don't have to deal with a y

[errant vector]

its solved

no look

if we set them equal

we should be able to produce two ordered pairs

one sec gotta get the question again

$$

[ember lichen]

i do that right

[errant vector]

ohh

fuck

[ember lichen]

$$
5x^2 - 5xy + 15y^2 - 5 \cdot 132 - x = 0
$$

[wraith dagger]

dannylewastaken

[errant vector]

its not an explicit equation

[ember lichen]

i think we sub in

then solve

[errant vector]

wait im slowly getting more confused 😭

we're solving for the intersections rn right

[ember lichen]

yes

[errant vector]

alr yeah however you prefer it should work

mine doesnt work cuz thats not explicit

i forgot ig

[ember lichen]

BOOM

SOLVED IT

HOLY FUCK

yeah

you sub in y as -x/5

then make those 2 equations equal

then solve

then it's just quadratic

[errant vector]

wow

thats amazing

[errant vector]

which 2

[ember lichen]

-x/5

then the original

[errant vector]

-x/5 = the original subbed y -> -x/5?

[ember lichen]

yes

oh wait

[errant vector]

and it gives the intersections?

[ember lichen]

i am dumb

LMAOOO

[errant vector]

nice notes

you take those online? or on paper

it looks online but the handwriting is too good for online

[ember lichen]

it's online

i have an ipad

[errant vector]

ah

what app are the notes on?

[ember lichen]

goodnotes

OH WAIT

I FORGOR

THE MULTIPLY

5*132

LMAOOO

[errant vector]

oh lol

do you have the apple pencil or smthn else

i used to use goodnotes but the notes didnt come out very well

[ember lichen]

veyg

it's wrong

im gonna

die

[errant vector]

😭

atleast we somewhat figured it out

[ember lichen]

did i sub in wrong?

[errant vector]

we just cant find the roots lol

[ember lichen]

we spent

over 1 hour

on this shit

[errant vector]

bruhh

[ember lichen]

the next problem

is worse 💀

[errant vector]

😭

can i see it

[ember lichen]

WAITY

IT EASIER

WHAT

HOW

THIS IS SO MUCH EASIER

WHY THE FUCK

OH MY GOD

IM GONNA KMS

[errant vector]

lmfao

[ember lichen]

THIS IS LITERALLY DERIVE

[errant vector]

and plug

thats it

lmfao

[ember lichen]

THEN PUT IT INTO THAT STUPID Y_1 - Y_2 = SLOPE(X - X_1)

BRI

BRO

WHY

[errant vector]

6(x-3) -2

[ember lichen]

life is unfair

[errant vector]

there

that should be it unless im tripping

[ember lichen]

yeah

ur right

[errant vector]

ig u coulda done that too lol

[ember lichen]

nah i have to my teacher is persistent on that

but gahh

how to find roots

bruh

did i sub it in wrong?

[errant vector]

im not sure

:google:

[ember lichen]

i-

line 2

LMAOO

i moved it over

and din't change the sign

[errant vector]

TF

THEY DIDNT TELL ANYONE ACTUALLY HOW TO SOLVE

NOO

oh wait

maybe you set both equations to 0

and then set them equal to eachother??

ohhh i think thats it

[ember lichen]

what 😭

setting it to zero

is just x = 0

[errant vector]

lemme just do it in my notebook and send a pic

[ember lichen]

well you should get

x = +- 10

[errant vector]

oh wait no

im so f- confused

how does one find the intersections between an implicit function and an explicit function

so y = -x/5

-y - (x/5) = 0

right

and x^2 - xy + 3y^2 - 132 = 0

so -y - (x/5) = x^2 - xy + 3y^2 - 132

😭😭 i hope this is right

[ember lichen]

wait

-y - (x/5) ?

OHH

you moved it

[errant vector]

ye

[ember lichen]

oh

that makes more sense...?

wait no

[errant vector]

i set them = 0

[ember lichen]

how do you even

solve it

[errant vector]

so they are equal to eachother

yeah thats the part idk..

lol

WHYY

[ember lichen]

bro you gotta sub in y

[errant vector]

i think i might go into spiraling depression

[ember lichen]

dawg im getting so close to x

its making me go insane

[errant vector]

lol

[ember lichen]

it's x = 9.7 and -10.1

💀

[errant vector]

wow thats like .3 and .1 off

[ember lichen]

yeah

clearly i am going insane

the more i stare at this proble,

[errant vector]

its been way more than an hour at this point 😭 — Today at 12:45 AM

[ember lichen]

the more i want to

[errant vector]

like 15 mins

ig thats not alot more

[ember lichen]

question if wanting a cs degree is worth it

i could move to the netherlands

and get a game dev degree and actually have fun

but no i choose to stay and suffer

[errant vector]

oh im thinking of doing cs when i go to college too! :D

[ember lichen]

bro also wants to suffer too

[errant vector]

yeah lol

[ember lichen]

bro this is messed up

i can't solve for x

bro what??

[errant vector]

this problem is too much

ask your teacher abt it im actually curious as hell how thats supposed to even fing be solved

like tf

[ember lichen]

amukh

you are a genius

albert if you will

[errant vector]

:0

[ember lichen]

no wait

you were almost albert

yeah you set both to zero

[errant vector]

LOL takes it back

[ember lichen]

but never had to

it was literally

i shit you not

sub in

and solve

[errant vector]

for what

x?

[ember lichen]

y

[errant vector]

im dead inside

[ember lichen]

take the orignial equation

replace y wiith -x/5

im gonna

die

[errant vector]

😭

my mind is too young for this pain

[ember lichen]

maybe the hood isn't that bad

[errant vector]

lol

[ember lichen]

nowait

we arent done yet

we need to uh

equation

that is easier though

[errant vector]

are u using an apple pencil?

[ember lichen]

yeah

[errant vector]

damn

wait i was gonna get one and i stopped after i saw it was 129 bucks 😭

[ember lichen]

LMAOO

it's worth it imo

like papers no more

and like its cool as shit since funny questions like these

ill put them into my note taking app

and make a note for myself to remember

that this is how i solve this question

oh my god

i finished solving it

oh my god

i don't feel happy

i just am

zamn so this is what the buddha meant when he said attachment meant suffering. thank you @errant vector for helping me through this experience

.close

[cedar kiln]

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[silent plover]

hi can someone help me for case 2

[flint plinth]

if y = 7/(3x), can you solve for x in terms of y?

[silent plover]

ah

i got it

ty

[cedar kiln]

@silent plover Has your question been resolved?

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[quartz dome]

how do i find the domain and range

[south tundra]

Generally a square root exists as long as the number under the root is nonnegative, right?

So we must have 3 - 2x - x^2 >= 0

But, also,

A fraction exists as long as its denominator is nonzero

Meaning we can't have 3 - 2x - x^2 = 0

[quartz dome]

i waited so long that i figured it out nearly

i got up to having (x-1)(x-3) root under the 1

so the domain should be x can not be equal to 3 and 1

but the answer says its −3 < x < 1

can u explain how they got that

typo oops

[south tundra]

You had (x - 1)(x - 3) < 0, right?

[quartz dome]

yes

[south tundra]

Hm, wait though

Could you show your work?

[quartz dome]

sure

which part?

[south tundra]

The thing is that you should have started 3 - 2x - x^2 > 0

Which is equivalent to (x - 1)(x + 3) < 0 instead

[quartz dome]

wait then lemme check i think i messed up

This is my working

[quartz dome]

wait why does it go from > to <

[south tundra]

From there you factor the last inequality into (-3 - x)(x - 1) > 0, right?

Here you can multiply both sides by -1, which flips the sign

Hence the (x + 3)(x - 1) < 0

[quartz dome]

ohh alr

why is it −3 < x instead of x < -3

[south tundra]

Let's say x < -3

Then x + 3 is negative, right?

[quartz dome]

oh yeah

[south tundra]

And so is x - 1

And the product of two negative numbers gives a positive number

[south tundra]

But we need < 0 here

There is an intuitive and a formal way to solve this equation

You can imagine the graph of the parabola (x + 3)(x - 1)

(Or sketch it)

It should look like this

,w graph y = (x + 3)(x - 1)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by A Lonely Bean#7998

[south tundra]

Doesn't matter how accurately you draw/imagine it, the important thing to remember is that it passes the x-axis at -3 and 1

And that the parabola is directed upwards

[south tundra]

Now, we need the solutions to (x + 3)(x - 1) < 0, right?

So, we need to look at the part where parabola goes below the x-axis

[quartz dome]

jyes

[south tundra]

As you can see, that happens when x is between -3 and 1

So the solution is -3 < x < 1

[quartz dome]

oh lol i get it

[quartz dome]

except this part

nvm got it

ty

and for range do i just sub in some stuff?

[south tundra]

For the range you first need to know what values does the quadratic expression take

In order to do that, we must know the maximum/minimum (in this case, minimum) value of the parabola

That can be done in two ways:

1. Complete the square
2. Plug in x = -b/2a into the expression (a being the coefficient of x^2 and b being the coefficient of x)

Here it is simpler to complete the square

Cause x^2 + 2x - 3 is clearly (x + 1)^2 - 4

Can you guess the minimum value of this?

[quartz dome]

did u get x^2 + 2x - 3 from multiplying by -1

and is the minimum (-1, and something)

[south tundra]

I just expanded (x + 3)(x - 1)

Good note though, we needed to know the maximum of 3 - 2x - x^2 instead whoops

But its maximum is equal to the -minimum of (x + 3)(x - 1)

[quartz dome]

oh ok

[south tundra]

Okay, let me redo this part if you are confused

[quartz dome]

thanks

[south tundra]

We need to know the maximum value of 3 - 2x - x^2

We can complete the square in order to do that

[quartz dome]

yes

[south tundra]

3 - 2x - x^2 = 3 + 1 - 1 - 2x - x^2 = 4 - (x + 1)^2

[quartz dome]

do we need to know the minimum

oh was that what u were doing earlier

[south tundra]

Yes

[quartz dome]

mb

[south tundra]

The maximum of this is 4

So we have shown that 0 < 3 - 2x - x^2 <= 4, right?

(0 < because we are picking the values for which the quadratic expression is positive)

[south tundra]

So sqrt(3 - 2x - x^2) <= 2

[quartz dome]

where did the <= come from

<= 4

[south tundra]

<= 4 is the same as saying 4 is the maximum value of that

And we have shown that 4 is exactly that

So 3 - 2x - x^2 <= 4

[quartz dome]

oh ok

[south tundra]

3 - 2x - x^2 <= 4 and therefore sqrt(3 - 2x - x^2) <= 2, right?

[quartz dome]

yes

[south tundra]

And 1/sqrt(3 - 2x - x^2) >= 1/2

So, the range is [1/2, +inf)

[quartz dome]

what happened in between that

[south tundra]

How did we get here?

[quartz dome]

u go their reciprocals by flipping the fraction?

[south tundra]

Yeah you can do that generally

[quartz dome]

i was asking what happened between this and the message after it

[south tundra]

Flipping the fraction changes the inequality sign

Oh

You mean that

We wanted to know the range of 1/sqrt(3 - 2x - x^2), right?

Aka the set of values that it could take

And we have shown that 1/sqrt(3 - 2x - x^2) is always bigger than or equal to 1/2

So it takes every value from 1/2 and further

[quartz dome]

yes i understand

[south tundra]

So the set of its values is simply [1/2, +inf)

[quartz dome]

the answer page says the answer was y > 12

is that the same thing?

[south tundra]

I have look at the graph of our function a while ago and confirmed that the range is actually [1/2, +inf)

So the answer page is wrong

,w graph 1/sqrt(3 - 2x - x^2)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by A Lonely Bean#7998

[south tundra]

See, the graph goes from 1/2 upwards

[quartz dome]

i dont get it but ill trust u on that ig

maths is hard imma read over this

ty for helping

.close

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[proven silo]

hi

[proven silo]

how to prove if the angle suspened by the arc is equal to the double of it in any part of circle

[livid hound]

prove if the angle suspened by the arc is
you mean central angle?

you can look up proofs of it

proof of inscribed angle theorem

[dusk finch]

https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-inscribed-angles/a/inscribed-and-central-angles-proof

[proven silo]

.close

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[signal flame]

i am lost on the one with 0,5 😇

[tropic oxide]

do you mean to say, "I know how to solve direct proportion problems like this, but the presence of the number 0.5 paralyzes me"?

[signal flame]

well i’m just lost on how to find out B below the 0,5th

[vagrant elbow]

Have you found the value of the constant of proportionality

[signal flame]

nope

[vagrant elbow]

Do you know how these problems work

[signal flame]

not sure

[vagrant elbow]

Alright

What does it mean for a and b to be directly proportional?

[cedar kiln]

@signal flame Has your question been resolved?

[cedar kiln]

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[silver sphinx]

help plz

[south tundra]

Have you tried anything?

[strange wind]

determining the nature is basically if is a max or minimum point, which can be shown by d2y/dx2 being negative or positive
Sketch should be fine

[silver sphinx]

idk what do i do for x^3(x-4)

[south tundra]

It is asking you to find the stationary points of the curve y = x^3(x - 4) and determine their nature

[silver sphinx]

like i did 3x(x-4)

[livid hound]

3x(x-4)
where's that coming from

[south tundra]

Meaning you need to find the points where the derivative of x^3(x - 4) is 0

[silver sphinx]

yeah

[silver sphinx]

dy/dx= x^3(x-4）

[livid hound]

no

[silver sphinx]

i think?

[livid hound]

y = x^3(x-4)

[crimson sedge]

Can you differentiate this expression?

[south tundra]

To find the derivative you can use the product and power rules

Or expand the brackets and just use the power rule

Anyways, you should get that dy/dx = 4x^3 - 12x^2

Finding the zeroes of this should be simple

[silver sphinx]

okay so

it should be

3x^2+(x-4)3x^2

or no

[silver sphinx]

wait how

[south tundra]

y = x^3(x - 4) = x^4 - 4x^3

dy/dx = 4x^3 - 12x^2

[cerulean sail]

Assuming you tried the product rule here?

[silver sphinx]

yeah

[south tundra]

Applying the product rule right away should have yielded 3x^2(x - 4) + x^3

[silver sphinx]

oh okay

[cerulean sail]

[you differentiated both x^3 and (x-4) for that first term, it looks like]

[south tundra]

But simplifying this still directs you to dy/dx = 4x^3 - 12x^2

[silver sphinx]

yeah i got this

[south tundra]

So, can you find the zeroes of dy/dx?

[silver sphinx]

wdym zeroes

[south tundra]

The values of x for which dy/dx = 0

[silver sphinx]

so x=0

[south tundra]

No

4x^3 - 12x^2 = 0

Generally a number c is a zero of a function f if and only if f(c) = 0

[silver sphinx]

okay

[south tundra]

So by the zeroes of dy/dx I mean the solutions to 4x^3 - 12x^2 = 0

(Since dy/dx and 4x^3 - 12x^2 are the same as we have shown)

[silver sphinx]

so x=3

[silver sphinx]

alright

[south tundra]

There is another solution

[silver sphinx]

so 4x^2(x-3)=0

x=0 and x=3

right?

[south tundra]

Yeah

So x = 0 and x = 3 are the stationary points

Now, it is also asking you to "determine their nature"

That means "Determine whether they are the maximum/minimum/saddle points"

We can do that by looking at the sign of y'' at those points

Can you find the expression for y''?

[silver sphinx]

y=0 and

y=-27

so (0,0) and (3,-27)

[south tundra]

Right, those are the stationary points

[silver sphinx]

but if i do d^2y/dx^2

[south tundra]

Now we need to know the signs of y'' at x = 0 and x = 3

[silver sphinx]

when x=0 d^2y/dx^2=0

[silver sphinx]

wdym by signs of y

[south tundra]

y''*

[south tundra]

Whether it's positive/negative/zero

[south tundra]

Yeah, so x = 0 is a saddle point

[silver sphinx]

so 0 and positive

[south tundra]

Now check d^2y/dx^2 for x = 3

[silver sphinx]

36

so it's a min pt (3,-27)

[south tundra]

Yeah

[silver sphinx]

okay

graph

like this right

[south tundra]

Yeah

[cedar kiln]

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[cedar kiln]

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[wild kindle]

The rectangle HDEG has the same area as the square ABFG.
In addition, the lengths DE = 12 cm and GE = 27 cm are given. Calculate the
Size of the hatched area ABFH.

[wild kindle]

The rectangle HDEG has the same area as the square ABFG.
In addition, the lengths DE = 12 cm and GE = 27 cm are given. Calculate the
Size of the hatched area ABFH.

[cedar kiln]

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[graceful gyro]

Why can't I find the voltage through capacitors when s is closed

They are in series so

[scarlet garnet]

,rccw

[graceful gyro]

2× 9 = 18

Cause total charge could be find by eff C and V right ???

<@&286206848099549185>

[clear berry]

Can't help you with physics problems

[cedar kiln]

@graceful gyro Has your question been resolved?

[cedar kiln]

@graceful gyro Has your question been resolved?

[cedar kiln]

@graceful gyro Has your question been resolved?

[gritty galleon]

I feel I should be ablle to do this but i cant help, sorry 😦

[distant cedar]

Hi @graceful gyro

Where are you facing issues? This is a physics problem btw not maths

[cedar kiln]

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[pale badger]

Sorry 1 sec

[crimson sedge]

Send a picture. People won’t download files

[pale badger]

Can someone explain to me question 6 (c) (iii)

[pale badger]

Idk why it sent as a file

[cerulean sail]

Did you manage part ii, by any chance?

[pale badger]

Yes

[pale badger]

I got up to the third line of (iii) in this image

But i dont understand how they turned (cos2x+1/2)^2 into zero?

[cerulean sail]

Any real number squared is nonnegative, in particular, $\pqty{\cos(2x) + \frac{1}{2}}^{2} \geq 0$

[wraith dagger]

@cerulean sail

[pale badger]

Oh shit yeah

Alri thanks

Feel stupid now lol

[cerulean sail]

It happens sometimes, don't worry

[pale badger]

.close

[cedar kiln]

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[ebon badge]

can someone help me?

i translated it so and i dont really know math words in english

[cedar kiln]

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[ebon badge]

a is a parameter it has 1 value

i translated the question and i dont know any math terms in english

[cedar kiln]

@ebon badge Has your question been resolved?

[cedar kiln]

@ebon badge Has your question been resolved?

[cedar kiln]

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[eternal dagger]

could someone help me with this practice problem please?
i think its a combination question but im unsure how to solve it

[cedar kiln]

@eternal dagger Has your question been resolved?

[ebon badge]

.reopen

wow man why did you have do that

thats just an annoying thing to do

[eternal dagger]

my bad?

[ebon badge]

wtf is that?

what grade are you in?

[eternal dagger]

statistics

[eternal dagger]

undergrad

[ebon badge]

thats in collage?

what?

how?

did you already learn the stuff i asked for help for?

[eternal dagger]

probably but i forgot it

ill leave the channel

i thought ur question had been resolved

[ebon badge]

btw i solved your problem

the percentage of left handed kids is 11.1%

the percentage of boys is 55.5%

and the rest just subtract from 100

[eternal dagger]

ok thanks

.close

[cedar kiln]

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[cedar kiln]

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[strange wind]

i dont know how to approach this problem

the far i have gotten is
For x0 = 0, we have x1 = 1, x2 = 2, x3 = 5, x4 = 26, x5 = 677, x6 = 458330, x7 = 210066388901, x8 = 441875919508520019

[cedar kiln]

@strange wind Has your question been resolved?

[crimson delta]

do you know what modulo is?

[strange wind]

not really, is it the same as modulus?

|this|

[crimson delta]

no

[strange wind]

uhm could you explain it? or connect it tho this question?

wait @ebon badge was this your channel?

[ebon badge]

im just interested what grade do you think should learn this?

yes

[ebon badge]

yes

[crimson delta]

well it was but now its not anymore

[strange wind]

it showed me it was available

sorry man

[ebon badge]

oh ok

[strange wind]

ill leave from here share your question again

[ebon badge]

can you help me with my question

[strange wind]

lol lemme check it out

[crimson delta]

barney get a new channel

[strange wind]

@crimson delta ill ping you to my new channel its fine

[crimson delta]

this is your channel

dont leave

[strange wind]

ohh

okay

For x0 = 0, we have x1 = 1, x2 = 2, x3 = 5, x4 = 26, x5 = 677, x6 = 458330, x7 = 210066388901, x8 = 441875919508520019

thats where we were

[crimson delta]

it will take a bit longer to explain modulo so I'm currently thinking about how to do this without it

[strange wind]

okay sure, ill be waiting its fine

some of the ideas that i had was proof by induction

[crimson delta]

hmm I can't think of anything

so the idea about modulo that the only thing that matters is remainders

we want to show that eventually x_k - 2 is divisible by 30, so phrased differently, eventually it has remainder 0 after dividing by 30

which happens when x_k has remainder 2 when divided by 30

now the nice thing about remainders is that they play nice with addition and multiplication

namely, if a has remainder r and b has remainder q, then ab has the same remainder as rq and a+b has the same remainder as r+q

this makes the numbers you have to handle much smaller

for example, x5=677 in your example has remainder 17 and x6=458330 has remainder 20

and now notice that if you instead just work with the remainders, then 17^2+1=290 also has remainder 20

[strange wind]

wait but working with remainders wount prove that its divisible by 30

like x5=677 we know is not a multiple of 30

[crimson delta]

yes it has remainder 17

but if the remainder of x_k is 2, then x_k-2 is divisible by 30

and that's what you want to show

[strange wind]

ohh yeah, basically is the remainder is any prime factor of 30 we have our answer right?

[crimson delta]

no

btw you made a mistake when calculating x8

,calc 210066388901^2+1

[wraith dagger]

Result:

4.4127887745906e+22

[crimson delta]

..

44127887745906175987802

[strange wind]

ohh lemme do it again

[strange wind]

yeah

thats the one

[crimson delta]

which you might notice is in fact 2 above a multiple of 30

but the nice thing here is that we could have noticed that easier by just working with the remainders

20^2+1=401 has remainder 11

and then 11^2+1=122 has remainder 2

(and that's how I noticed you made the mistake in the first place)

[strange wind]

tho the only problem here is that we need to prove for all possible values of x0

[crimson delta]

or do we only have to prove it for all possible remainders that x0 could have

[strange wind]

i mean yes, thats one way of putting it

but with the remainder approach what proof do we use?

like how can i prove this?

[crimson delta]

bruteforce

I can't think of anything nicer right now

[strange wind]

i am doing this on paper, how do i plan to brute force on this

[crimson delta]

well its only 30 possible remaindeers

like with your earlier sequence you started with the remainder 0 and then got 1, 2 (at here you would be done btw), 5, 26, 17, 20, 11, 2

which shows that if x0 has any of those then eventually you reach a 2

[strange wind]

dont you think there could be another way, coz i mean 30 possibilities is quite a number to do by hand

if i was using a program/python to do it, thats fine

[crimson delta]

well you will notice that because you are working with squaring then a lot of values fall together

(essentially for the reason that x^2=(-x)^2)

[strange wind]

well before i start with the whole bruteforce, do you think there is a easier way

something like using sum of a series or something in those lines?

[crimson delta]

no I can't think of anything

[pastel path]

So say like there was an equation 4x+5y<=180
how do i calculate x and y so that their sum is the greatest combination (of whole numbers)

i dont know if this channel is empty but it doesnt say a name haha

[strange wind]

@pastel path

its not empty sorry

[pastel path]

mk my bad.

thanks for letting me know!

[strange wind]

np mate

[strange wind]

hmm alright thanks

[strange wind]

.close

[cedar kiln]

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[livid crescent]

does this math look correct

[cedar kiln]

@livid crescent Has your question been resolved?

[dire geode]

What even is the problem

[cedar kiln]

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[scenic fable]

I dont understand

[scenic fable]

sad moment

[high spruce]

:(

[scenic fable]

I tried to understand

but I cant

I want to learn math

and english

sad

very sad

pls tell me advice

[high spruce]

what's your question

[scenic fable]

my book is in english

and is math

In a meeting there are 201 people of 5 different nationalities. It is known that, in every group of 6, at least two are the same age. Prove that there are at least 5 people from the same country, of the same age and of the same sex.

[high spruce]

ok this is pigeon hole principle

are you familiar with it?

[scenic fable]

noyt

I can help with math exercise

or not

my dear acquaintace

tance

[high spruce]

basically, the pigeon hole principle means that if there are n+1 holes and n pigeons then one of the holes will have at least 2 pigeons

so by this logic, at least 41 people will have the same nationality, and at least 101 people will have the same sex

I'm trying to figure out the age condition though, give me a minute

[scenic fable]

I dont understand

because I dont speak english

[high spruce]

what language do you speak

[scenic fable]

spanish

but normally

[high spruce]

hablo

[scenic fable]

can you speak in englih

i learn english and math

bruh

[high spruce]

bruh

[scenic fable]

whay

what is bruh

[high spruce]

brah are you trolling

ok anyway

[scenic fable]

not

[high spruce]

so let me try and clarify the quetsion

there are 201 people
that means at least 101 are of one sex

is that clear?

[high spruce]

porque tu no hablo espanol con mi

[scenic fable]

use trasladertor google

is very bad

[high spruce]

?

Entiendo espanol mi amigo

[high spruce]

do you understand or not

[scenic fable]

I want to learn english

math not

when i have a ask of math

i go to her

here

[high spruce]

I'm trying to teach you maths

this isn't an english learning discord

[cedar kiln]

@scenic fable Has your question been resolved?

[cedar kiln]

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[dusty ibex]

am I trippin or isnt all of these wrong?

[gritty viper]

you're tripping

[dusty ibex]

can you show me how to work these out?

i substitube 1 first into the formula on the right side then i do 2 then 3

[gritty viper]

well I mean one thing you could do is try to find a counterexample to one

[dusty ibex]

wym?

[gritty viper]

well 3 of them should be true, one isn't right

So if you find an n value that doesn't work you know that's the false one

[dusty ibex]

yea but I cant find neither that is true

take number 1 for example

i say 2(2+1) that is 6

I dont think that is what i should get eithere

[gritty viper]

?

[high spruce]

the first one can be easily derived using the formula for the sum of the first n integers

this guy

add the first and last, second and second to last, third and third to last and so on terms together

and from there you can derive formula

[dusty ibex]

ok so i add 2 and 6 = 8 then 4 and 4= 8 then 6 and 2 = 8

like that?

[autumn fox]

Generally the sum of an arithmetic sequence can be found by

$S_{n} = \frac{n \cdot (a_{1} + a_{n})}{2}$

Where $a_{1}$ is the 1st term of the sequence and $a_{n}$ the n'th term

[wraith dagger]

Mikkel

[dusty ibex]

I dont understand this

💀

[high spruce]

@dusty ibex do you know the derivation for this?

it's the same as what Mikkel is describing

just the 'simplest case' I guess

[cedar kiln]

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[cedar kiln]

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[terse locust]

Hello, I don’t know where to begin?

[cedar kiln]

@terse locust Has your question been resolved?

[zenith sail]

@terse locust Don't worry about Sierra's simulation. Only one of your choices gets the true probability right anyway.

[terse locust]

Alright! How would I go about finding the probablilty

[zenith sail]

consider all the possibilities when you roll a pair of dice

how many sum up to 7 or 11?

[terse locust]

so 36

[zenith sail]

yep

[terse locust]

would that be 8

[zenith sail]

yep

[terse locust]

Would it have be to be E?

[zenith sail]

yep 👍

[terse locust]

Thanks for the help!

[zenith sail]

np

[terse locust]

.close

[cedar kiln]

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[finite moth]

Hey, I'm having trouble with the idea of the inequality here in the induction step.

Obv we're assuming $3^n \geq 2^(n+1)$

[wraith dagger]

Солдат удачи

[finite moth]

But with the inequality, couldn't I literally just multiply it by 3^n and $2^n) on either side

as 3^n preserves the inequality

and 2^n is established to be smaller than 3^n

[dire geode]

Write the math what you mean

[finite moth]

So, given it's an inequality $3^n * 3^n = 3^(n+1)$

[wraith dagger]

Солдат удачи

[finite moth]

because we're trying to prove $3^(n+1) \geq 2^(n+2)$

[wraith dagger]

Солдат удачи

[finite moth]

how do I make these non cancer exponents, whatever

[dire geode]

No

[finite moth]

How would I approach this problem then?

The inequality aspect confuses me

[dire geode]

3^n times 3^n equals 3^(2n)

[finite moth]

How would you then get n+1, by adding them together?

[dire geode]

Use 3>=2

3>2 works as well

,tex .exp rules

[wraith dagger]

riemann

[dire geode]

To get 3^(n+1), you use first row above with x=n

[finite moth]

oh, so I would multiply by 3?

$3^n * 3^1 (3) = 3^(n+1)$

[wraith dagger]

Солдат удачи

[finite moth]

Okay, that makes sense.

Can I do the same on the other side since it is an inequality and 3>2?

but with 2

[dire geode]

You'd chain the inequalities

Something like 2^(n+1) * 2 < 3^n * 2 < ... < 3^(n+1)

Maybe <= instead of <

[finite moth]

I haven't seen it done like that before for this kind of inequality proof, I thought since it's strictly still less than the LHS adding something would work.

[cedar kiln]

@finite moth Has your question been resolved?

[cedar kiln]

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