help-41

[plush latch]

but every number is a coefficent to 3^k

[wheat elm]

I think I'm kind of getting it.

[plush latch]

ye

do you think u can do the second one?

[wheat elm]

I think I can start it

[plush latch]

nice

[wheat elm]

orrr not lol fuck

You spend one week in the hospital and it fucks you

[plush latch]

make sure u write a little like summary that basically just says as the statement is true for n=k+1 it must be true for all n

[plush latch]

oh its chill I can help

[wheat elm]

Okay, so let's see if I translate it right

For this question it's asking that for every positive interger n...

and it seems I think recursive?

[plush latch]

for equation 2 start out by writing a formula in terms of n for the value of the sum of the sequence

[wheat elm]

is this where we use the weird s?

[plush latch]

You dont really need to here tbh

I mean you can if you want to look fancy but there isnt really any need

[wheat elm]

Okay.

[plush latch]

so the sequence is arithmetic

each term has a difference of 2 and the sequence starts at 1

so you can use the sum of series formula

where a is the first term and d if the difference

if you do there you get that the series is equal to

[wheat elm]

ferm term?

[plush latch]

first mb

oh wait mb doing this just proves the whole thing without induction anyway

ye you might need to use the sum notation then

[wheat elm]

Then let's do the sum notation cause this does have to be induction

[plush latch]

I will write down how to do the question on a whiteboard then send a screenshot

[wheat elm]

let's see if I can get it before you lol. (*( I doubt it)

[wheat elm]

I need to install the mathrwriter lol

[plush latch]

whats that

[wheat elm]

a teacher of mine mentioned there's a mathmatical notation document editor

[plush latch]

oh what so you type out maths notation in text

[wheat elm]

Yup

[plush latch]

do u get what I wrote on the white board tho?

[wheat elm]

sort of?

I'm kinda seeing the patterns but the words just... aren't there

[plush latch]

wdym the words?

[wheat elm]

well I can kind of see how we got to the last point, and how it proves the last thing, because essentailly that formula is just a distribute K+1^2, and we manaaged to kind of dissasemble it from the summation notation earlier

[plush latch]

ye

[wheat elm]

so we've proved I beleive, yes?

[plush latch]

ye

all u need to do is write the conclusion but ye thats essentially it

[wheat elm]

the conclusion being that 2n-1 = n^2?

or 1+(2n-1) = n^2?

[plush latch]

the conclusion is just that because the statement is true for n=k+1 it must be true for all values of n

[wheat elm]

Thank you

You made this make a biiit more sense. At least through the pain medications

[plush latch]

np

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[errant flint]

can someone help me with question 15

i understand you gotta use cosine law

but is the total angle the boat turned 120

and not 150

[celest cove]

turned 30
but if youre doing a triangle from the start, turn and finish
the angle at the turn inside the triangle would be 150

exterior angle would be the 30

[errant flint]

could you explain that again

like the second part

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[split sail]

U still need help?

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[meager gale]

i can do the q this way, im just wondering if theres a better way to do it

[weak zinc]

The fact you know that’s a right angle, you can e.g. relate that to the tangent at said point

[meager gale]

if i find tangent at that point, its 2x

idk what to do after that

cause the (1,2) is just a point

so idk what i can do

also if i try perpendicular distance, i get an abs value, which i kinda dont want to deal with

[weak zinc]

You e.g. wanna consider the line [segment] between (1, 2) and (x, x^2), and the same tangent you have

[patent raptor]

https://tenor.com/view/cat-cuddle-gif-27222129

[weak zinc]

See e.g. product of gradients is -1, dot product is zero etc etc

[weak zinc]

[meager gale]

hmmm

gimme a bit

[meager gale]

thinking about these, it feels ike one way or another, i need to make a tangent but.

do i find the normal?

and sub in (1,2) as a pt?

[weak zinc]

I mean, you could find the normal and consider that (1, 2) is on it

[meager gale]

o

ignore the fact that i didnt put =0 at the end

[weak zinc]

(Or that you know the normal gradient is, say, (x^2 - 2)/(x - 1), and that of the tangent is 2x, so [as per the “product is -1” thing, 2x * (x^2 - 2)/(x - 1) = -1, and…)

[weak zinc]

…which gets you the same thing

[meager gale]

so theres no escape from just factorising the polynomial really sadly

and subbing it in

[weak zinc]

I guess so

[meager gale]

:( ic

thanks chartbit

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[sonic musk]

im not quite understanding how this works

[sonic musk]

i think that it has to do with one function being greater than the other but im not quite following

[red sandal]

what happens if a function bigger than another function converges

what can you say about the smaller function

[sonic musk]

then it converges

?

[red sandal]

yeah

if a smaller function converges you can't say anything about the bigger function; it can diverge or converge

[sonic musk]

so since 1/x^3 is the larger function and we know it converges we can state that the smaller one converges

for the first part at least

[red sandal]

yeah

[sonic musk]

but then for the second part 1/x^3 is smaller, so we can't do the same

am i understanding this right?

[red sandal]

yep

[sonic musk]

omg thanks

this was much easier than the textbook

[red sandal]

got you

[sonic musk]

.solved

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[cobalt shard]

(x-a)^2 / z^2 + (y-b)^2 / z^2 + (z-d)^2 / k^2 <= h

[cobalt shard]

Can this be simplified? (Idk how to use wolf bot thing)

Also what I mean by simplified is could it be dumbed down while still keeping the same variables and functions

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[thick seal]

i feel like i don't have to share context for the problem

its just plugging in kidn of thing

w initial is 200^2

w^2 is 50^2

all over

2(314)??

its confusing me

i do that but i don't get the right answer of -59.7

[hard ermine]

It is using the rotational kinematic equation

[thick seal]

yes

Like but

I am aat the part of plugging and chugging the easiy part

but i cant get to -59.7 for some reason in my calc

[hard ermine]

Can you show your work?

[thick seal]

i put in my calc

1 sec

i take pic

like i felt like its just basic plugs

but i end up yeah getting a negative answer but some weird fraction

[hard ermine]

-9375/157 is about -59.71

[thick seal]

OH

why am i getting af raction number tho

;/

i see tho thnxS!!!

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[azure ember]

hopefully somebody knwos phsycis bc I DONT UNDERTSAND WHY THE EMISSITIVTY IN THE ANSWER IS 4. i did everything right but the 4

[azure ember]

omll im going insane the question says e=1??

and this is the equation in the data booklet

P=eoAT^4

[mint nacelle]

nice to see you're doing IB physics

so a sphere has surface area 4pi r^2

but at any one moment, the sun only shines on a disk of radius r at any one time
so the area illuminated is pi r^2

so the power radiated per unit area of the sphere

there's going to be a (pi r^2) / (4 pi r^2) = 1/4 in there

to understand concepts like these in more detail, I highly recommend the Tsokos textbook

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[copper kite]

prove that there no positive real integers such that x^2 - y ^2 = 10
(proof by contradiction algebraically, i cant test values)
what i did was i just put it as (x-y)(x+y) = 10, and after some solving got x = (10-y(x-y))/(x-y) hence we get x in the form of p/q, so its rational and not an integer. But i dont think that this is correct. I also tried another approach where x = root ( 10-y^2) but then idk where to go from there

[copper kite]

would showing it by like taking x and y as a both an even and both as odd number, and do two different proofs, ( becuase for x^2 + y ^2 to be equal to 10, both x and y have to both either odd or even)

so take x as 2a and y as 2b, and then take x as (2a+1) and y as (2b+1) and show it dosent work in both those cases

[mint nacelle]

you're overcomplicating this by a lot

any real number squared cannot be negative

so if x = 4 and x^2 = 16, then there are no solutions

you just have to test x = 1, 2, 3 and you can swap x, y so that is sufficient

[pseudo crescent]

isnt it false btw?

[copper kite]

i know i know thats what i want to do

but my teacher keeps saying in proof by contradiction

[mint nacelle]

lmao yes (x, y) = (3, 1) or (1, 3)

[copper kite]

i have to do algebrically only

and i cant do trial and error

[pseudo crescent]

gl getting your contradiction then lol

[mint nacelle]

that's a strange request

also the thing you want to prove isn't even true

[copper kite]

ya thats true

i think the wquestion is bugged

i think it should be x^2 - y^2 = 10

but the questgion says x^2 +y^2 =10

but even then

[mint nacelle]

the strategy would be (x - y)(x + y) = 1 * 10 = 2 * 5

[copper kite]

ya thats what i did first

then got x as p/q

[mint nacelle]

well then you don't get integer values of x, y

[copper kite]

ya

so then it would contradicted

[mint nacelle]

cause x - y and x + y must have the same parity

yep

[frank zealot]

wait the question is to prove that there exists no positive couple of integers satisfying x² + y² = 10?

[mint nacelle]

yes but it has to be a typo

[frank zealot]

obviously

10 = 3² + 1²

[frank zealot]

it could be this

[copper kite]

Yes, Im solving it using this only

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[spice dagger]

I want to learn about composition of functions when main function is given into piece wise intervals

[spice dagger]

I tried to search over google but i didn't find any source...help

[viral coyote]

Where is t(u)

[amber wasp]

@spice dagger Has your question been resolved?

[spice dagger]

Actually it is x not u

[viral coyote]

Still
Where is t(x)
You can't find (f o t)(x) with only f(x)

[spice dagger]

It is fof(x)

@viral coyote

[viral coyote]

Ah

[spice dagger]

Sry for bad handwriting

[viral coyote]

Put each part in the x of itself
Idk how to explain it right

For example
In the first part
Put (-x) in (x) so it will be -(-x) which means x

In second put (x) in (x) so it will stay x

In third part
Put (2-x) in x so (2-(2-x)) so x

In this question (f o f)(x) will be = x for all x belongs to R

Oh god
Math is beautiful

Please if you didn't understand me ping helpers they will be better at explaining

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[fierce edge]

1/3 the area of square ABCD < area of triangle CDE?

um

[cunning birch]

what is the data

just the figure?

[fierce edge]

yep

like okay if EDC was right then it's easy

[cunning birch]

if E is closer from A than in this picture... like very close....

then area of CDE is kinda area of CDA...

[frank zealot]

u can draw the height coming from the point E to the base CD in triangle EDC

[fierce edge]

hmmm

[frank zealot]

youll notice its equal to AD

and find each area

and compare

[fierce edge]

wait what

drop an altitude from E?

[frank zealot]

from E

yes

[fierce edge]

and why do you know that'd be AD

😭 am i missing something silly

i could be

[frank zealot]

cause

say the altitude from E is the segment [EH]

EHDA would be a rectangle

thats cause any quadrilateral with three right angles is defined as a rectangle

[coral wigeon]

[cunning birch]

anyways if you really wanna solve for area of CDE better to take 1/2 CD * DE * sinCDE

name c the square side

1/3 area of square ABCD is c^2/3

[fierce edge]

i'm lost isn't it just 2 right angles

[fierce edge]

heyy

[fierce edge]

oh yeah wait

can i like bound this?

to account for the sin

lemme try

[frank zealot]

the square gives u two right angles, and the altitude gives u a right angle

[cunning birch]

again if you don't have any specifications on where E is

take E close enough to A

area of CDE is kinda area of CDA so c^2/2

area of ABCD/3 is c^2/3

[fresh ocean]

Why do I have the area of CDE as exactly half the area ABCD?

[fierce edge]

i tried this:

1/2 c^2 sin(CDE) < 1/2 (CD) (DE) sin(CDE) = A

eh 0< sin(CDE) < 1

technically ED > AD = c

[cunning birch]

where'd you get k(k+1)/2

[fresh ocean]

similarity

[cunning birch]

with?

[fresh ocean]

EAF and CDF are similar

so height is k times

base is same for CFD and EFD

so area is k times

[cunning birch]

uh

if lengths are multiplied by k

wait

[fresh ocean]

only height is multiplied

base is common

[cunning birch]

what height are you using for EDF

[fresh ocean]

FEA and FED are same height diff base, so diff area

[fresh ocean]

AE

[cunning birch]

so AE * DF/2

AE is k^2

[fresh ocean]

yea

no not k^2

k*(k+1)

[cunning birch]

why

[fresh ocean]

its k times height of CDF

and that is k+1

[cunning birch]

ok yes

[fresh ocean]

CD is height, DF is base

[cunning birch]

ok

I now agree

[fresh ocean]

All my bases are on the side AD

And heights AE and CD

[cunning birch]

yes

I'm all good

[fresh ocean]

So why is this inequality question, when there is a clear equality relation here 😭

[cunning birch]

IDK

[fierce edge]

😭

[cunning birch]

the computation requires you to do that

for 1/2 > 1/3

and thus false everytime

[fresh ocean]

I am so frigging dumb. We dont even need this shit. The base is CD and height is AD for the triangle

ofc area is half

😭

[cunning birch]

rip

[fierce edge]

i'm still confusedd how the height is AD

did you just move E to a more convenient place

so that you get that the height is AD?

[fresh ocean]

Perpendicular from E on the base CD

that height has same length coz the lines are parallel

[cunning birch]

draw a line perpendicular to (CD)

and the line parallel to (CD) that passes through E

no matter where that line is

[cunning birch]

the (perpendicular) segment between those parallel lines always has the height as its length

[fresh ocean]

frick and I just read thru above messages and youssef said this already. AAAAAAAAAAAAAGH

[frank zealot]

yeah man

[fierce edge]

😭 no i think i'm done

i've been trying to draw the perpendicular

okay i'll just draw it

?

or idk assume this is the perpendicular

depending on what angle D is

[fierce edge]

can u send the picture of this

i'm probably being dumb or smth

[fresh ocean]

this one is right

[frank zealot]

did u seriously draw this when i asked u to draw the altitude from E lmao

[fierce edge]

😭 that was in my mind

tbh in my mind i had multiple caseworks

for when D is right, acute, and obtuse

[fierce edge]

okay okay

[fierce edge]

so i didn't know which altitude

we're using

[frank zealot]

theres one altitude for each point of a triangle

how do u

nvm

[fierce edge]

doesn't it depend on whether EDC is an acute, right, or obtuse triangle

if it's right then the altitude is on ED

[frank zealot]

no

[fierce edge]

if it's obtuse it's outside

if it's acute it's inside

[frank zealot]

oh ur talking abt that

but still theres one altitude

[fierce edge]

yeah so wouldn't this be right if D was acute

[frank zealot]

for each point

[fierce edge]

yeah but 😭

i meant i had to do the casework for which type of triangle?

i didn't know what u meant

are we assuming that

D is obtuse?

[frank zealot]

well hope the view is clearer

D is obviously obtuse

[fierce edge]

but the question maker said u can't rely on the diagrams

which don't have to be drawn to scale

ughhh

[fierce edge]

yeah it's a much easier question i think

EF || AD

and yeah ED > EF = AD

[frank zealot]

u can prove its obtuse

so theres no issue

[fierce edge]

how

[frank zealot]

ok

we agree that AED is a right triangle in A yes?

ADE would be an acute angle

0 =< ADE =< 90

and ADC = 90

so

[fierce edge]

yes

[frank zealot]

90 =< ADE + ADC =< 180

[azure ember]

thank you!! looks like i need ALOT of revision for this topic 😭💀 I'm still a bit confused but ur explanation has helped me move one step further to understanding!.. sorta! thank you so much

[frank zealot]

by euclids axiom of angle addition

90 < EDC < 180

oops it should be strict

[fierce edge]

wait what 😭

one is like "on top"

this drawing is confusing my senses

[frank zealot]

its

[fierce edge]

but yes i agree that CDA is 90

[frank zealot]

literally easy to interpret

ok look

[fierce edge]

and also agree that ADC is acute

[frank zealot]

no it isnt

[fierce edge]

it is

wdym

[frank zealot]

u meant ADE?

[fierce edge]

oh yeah

i meant ADE is acute

ADC is 90

wait

u meant this?

bruh

[frank zealot]

.

[fierce edge]

i thought the square 😭 is above

the triangle

idk this drwaing is weird

sorry

[frank zealot]

either way

[fierce edge]

ok yeah okay that's obvious

[frank zealot]

u answered the question

gj

[fierce edge]

😭 thank you

the drawing looks like 3d to me

cuz of EC

and so i was confused

for the most part

wait i still am

[frank zealot]

are u kidding me

[fierce edge]

LOL

😭

i'm sorry 😭

no i get the math but what is that shape supposed to be

[frank zealot]

its olay im used to this

what shape

[fierce edge]

EDCBA

[frank zealot]

oh my god u made me look at it in 3d

damn

[fierce edge]

😭 i can't see it how you're seeing it

i'm confused

i have the whole 3d shape in my head

just help me with that part idk how

but if u can

[frank zealot]

which part

[fierce edge]

how are u seeing this in 2d

i still think it's 3d 😭

divert my attention LOL

idk

[frank zealot]

i

idk i have better eyes

[fierce edge]

:((

[frank zealot]

i

i see it 3d yes but cant u js control ur mind

and view it 2d

or try drawing the line EC continuous instead of

line by line

[fierce edge]

why tf is it in dashed line anyway

[frank zealot]

or even better, delete that line

no idea

[fierce edge]

dashed line is supposed to represent a line you can't see

in 3d

but it exists

[frank zealot]

yeah this isnt technology

ok so

uh

1/3 area of ABCD < area of DCE

thats all

goodbye

[fierce edge]

NOOOO not the math

😭

I need help seeing this in 2d LOl

okay i think i'm just giving up on this nonsense

but yes i get the math

thanks for that

[frank zealot]

js delete it

we didnt even need it

[fierce edge]

i'm going to stare at it

till i can see the 2d

😭

[frank zealot]

do as u please

[fierce edge]

okay i got it

the EC line does intersect with AD

my head doesn't want it to intersect

cuz they don't in 3d

mmmm

okay thanks guys

it's pretty easy now

.close

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[keen pawn]

false $A ={1}; B={2}$

[grizzled pagoda]

A dense set

[keen pawn]

$A \cup B={1,2}$

[grizzled pagoda]

A dense set

[pseudo crescent]

and X?

[keen pawn]

$X =A \cup B$

[grizzled pagoda]

A dense set

[keen pawn]

Invoking bezout;s lemma, gcd(42,7)=7.
\
So $42a+7b=7$, not$42a+7b=1$

[grizzled pagoda]

A dense set

[pseudo crescent]

hmm

bezout lemma says that you can find a,b with 42a + 7b = 7

[main patrol]

I haven't taken set theory but something doesn't seem right about the first proof

[keen pawn]

I think this could use some work

[pseudo crescent]

but does it say that that there are no a,b, with 42a + 7b = 1?

[keen pawn]

no

hmm

[pseudo crescent]

there is an extended version which implies it, but you dont need it

[keen pawn]

got it

[pseudo crescent]

this proof can be understood by a 6th grader

[keen pawn]

$\frac{42a-1}{7}=b$

[grizzled pagoda]

A dense set

[tulip tapir]

did you finish the other one

[keen pawn]

no

[tulip tapir]

*sighs*

[pseudo crescent]

no fractions please

[keen pawn]

$42a \equiv 1mod(7)$

[grizzled pagoda]

A dense set

[tulip tapir]

the other other one? modular arithmetic?

[keen pawn]

I have to check this

[tulip tapir]

uhm

is "a" an integer?

[keen pawn]

yes

[pseudo crescent]

It's not a proof, its a counterexample

[tulip tapir]

that's not possible then???

[keen pawn]

This is false, so not possible

[tulip tapir]

42(integer) = 7(6 times integer)

ah

[pseudo crescent]

can you write out the proof?

[tulip tapir]

physics?

[pseudo crescent]

Why is this true?

[tulip tapir]

i assume he got it from this

[pseudo crescent]

i dont recall fractions in the defn of modulo though

[tulip tapir]

they aren't

[pseudo crescent]

shh

[tulip tapir]

got it

[pseudo crescent]

i just want wai to formulate the proof in the right way

[keen pawn]

yeah, $42a-1=7b$

[grizzled pagoda]

A dense set

[pseudo crescent]

okay, and?

[keen pawn]

So I have to chck if $42a \equiv 1mod(7)$

[pseudo crescent]

Why?

[pseudo crescent]

does this imply 42 = 1 (mod 7)?

[tulip tapir]

uhmmmmm

wai

check again

[grizzled pagoda]

A dense set

[pseudo crescent]

you're using unnecessarily complicated tools

but okay

[keen pawn]

ok, 42a-1 isn't a multiple of $7$ under any circumstance, thus not possible

[pseudo crescent]

this one is fine

[grizzled pagoda]

A dense set

[pseudo crescent]

Why?

[keen pawn]

Let $a$ be even

[grizzled pagoda]

A dense set

[keen pawn]

Okay, yeah, bad way to prove

The reminder is always 5 by the division algorithm

[pseudo crescent]

it's not

[pseudo crescent]

so 1, 6 or 5?

[keen pawn]

Wiat

I'm rolly confused

*royaly

[pseudo crescent]

try looking back at the original question

look at the left hand side

look at the right hand side

[keen pawn]

This?

[pseudo crescent]

no, 42a + 7b = 1

[keen pawn]

okay

[pseudo crescent]

and no fractions

[keen pawn]

$a=bq+r$

[grizzled pagoda]

A dense set

[keen pawn]

so $42a=-7b+1$

[grizzled pagoda]

A dense set

[pseudo crescent]

can you name some numbers of the form 42a + 7b? What property do they all have? And does 1 have this property?

[keen pawn]

They are all divisible by 7

[pseudo crescent]

yes

is 1 divisible by 7?

[keen pawn]

no

oh

shit

I see

okay

Got. it

thanks!

[pseudo crescent]

I feel like you are sometimes too proof-mindseted

[keen pawn]

yeah

[pseudo crescent]

maybe just try examples sometimes

and get some intuition before starting a proof

[keen pawn]

got it.

[pseudo crescent]

starting a proof is useless if you dont even have idea yet

there is nothing to cry about, you're able to both get an idea and do the proof

you just need to do it in the right order

[keen pawn]

Got it

Got to go for dinner now

.close

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[pseudo crescent]

Enjoy your meal :)

[keen pawn]

thanks

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[surreal badger]

Find the value of

$$\cot 80^{\circ} (\tan 50^{\circ} - \cot 20^{\circ}) + \cot 70^{\circ} (\tan 40^{\circ} - \cot 10^{\circ}) +$$
$$\cot 50^{\circ} (\tan 20^{\circ} + \tan 80^{\circ}) + \cot 40^{\circ} (\tan 10^{\circ} + \tan 70^{\circ}) -$$
$$\cot 20^{\circ} (\cot 80^{\circ} - \tan 50^{\circ}) - \cot 10^{\circ} (\cot 70^{\circ} - \tan 40^{\circ}) $$

[grizzled pagoda]

m13

[arctic oracle]

good luck

[surreal badger]

lol

its so long

[arctic oracle]

i assume its non calculator

[surreal badger]

yea

i think there is a pattern

got it

its 0

lol

.close

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[mellow venture]

I'm following a trigonometry book and it explains radians with this theorem that i don't understand: "The angles at the center of a circumference and the corresponding arcs constitute two classes of directly proportional quantities"

[unique rose]

Yh that's rlly confusing 😭😭

But I think what it's tryna say is

What's the formula yk for circumference of an arc

In terms of radians it becomes theta*r and therefore they're directly proportional quantities

W the constant of proportionality being the radius

But it's super duper weird and confuddling

[mellow venture]

Yeah

Do you know any other way to understand it correctly

[unique rose]

To understand radians?

The way I understand it

Is yk how there's 360 degrees in a full rotation

Similarly there's 2pi radians in a full rotation

That's all radians are

But they're rlly useful

This redefinition

Because the formula for the arc length of a circle in degrees is theta/360 * 2pi*r

But in radians u won't divide by 360 you'll divide by 2pi

And then it'll become theta*r

Which is WAY cleaner

[mellow venture]

What do you mean by yk?

[unique rose]

You know

[mellow venture]

Ok ahah

I understand thanks

.close

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[mellow venture]

.reopen

[amber wasp]

✅

[mellow venture]

Sorry i found another thing in the book

After it introduces the theorem he used l/r = l'/r'

[elder harbor]

,rotate

[mellow venture]

I don't understand why he said that before and then used another proportionality

[elder harbor]

wdym

do u want me to prove this?

or

[mellow venture]

Yes

[elder harbor]

ok

so let's find the arc length of both circles

the smaller one:
${l = \frac{\alpha}{2\pi}r}$

[grizzled pagoda]

k

[elder harbor]

the bigger one:
${l' = \frac{\alpha}{2\pi}r'}$

[grizzled pagoda]

k

[elder harbor]

@mellow venture do u agree with this?

[mellow venture]

Yes

[elder harbor]

ok

so let's rearrange

${l = \frac{\alpha}{2\pi}r \implies \frac{\alpha}{2\pi} = \frac{l}{r}}$ and ${l' = \frac{\alpha}{2\pi}r' \implies \frac{\alpha}{2\pi} = \frac{l'}{r'}}$

[grizzled pagoda]

k

[elder harbor]

since ${\frac{\alpha}{2\pi} = \frac{\alpha}{2\pi} \implies \frac{l}{r} = \frac{l'}{r'}}$

[grizzled pagoda]

k

[mellow venture]

So they are directly proportional

[elder harbor]

yes

${l \propto r}$

[mellow venture]

But why the theorem says that the angles and the arcs are direct proportional

[grizzled pagoda]

k

[elder harbor]

u can actually see the proportionality from either equations

${l = \frac{\alpha}{2\pi}r$. Let ${k = \frac{\alpha}{2\pi}}$, so ${l = kr \implies l \propto r}$

[grizzled pagoda]

k
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[mellow venture]

Oh ok

Thanks k

.close

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[modern acorn]

Hi

[modern acorn]

Can I get a lot of help

By chance can anyone explain a few topics of ninth grade math's to me

[modern acorn]

Alrr it's okay then dw about it, thanks :))

Ik

[solar gust]

Khan academy is the best way to learn about topics, we here for specific question about exercice

Indeed

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[timid vector]

Can I have some help with this? I really don’t know what to do I forgot

[shadow stump]

well we have a quadratic here

[timid vector]

we do?

how do I even get it in the form ax^2 + bx +c

and also how do I find the value of p where it has equal roots

<@&286206848099549185>

[hollow salmon]

Treat (p^2 + 11), (-12p), and p^2 as coefficients.

[timid vector]

and what does that mean

sorry I’m just not the best with math terminology my teacher always says it very simply so 😭

[hollow salmon]

a = p^2 +11
b = -12p
c = p^2

[timid vector]

ahh okay

[hollow salmon]

Then it just becomes a matter of finding the values of p for which the discriminant is greater than 0.

[timid vector]

so trial and error I’m guessing?

[hollow salmon]

Do you know what the discriminant is?

[timid vector]

b^2 - 4ac?

[hollow salmon]

Yes. And when the discriminant is greater than 0, that implies two Real roots.

[timid vector]

I’m concerned

I don’t think I’m doing this right

[hollow salmon]

What did you get?

[timid vector]

just to clarify

I am subbing in a b and c and solving for p?

[hollow salmon]

Yes.

[timid vector]

so tell me why I’m getting 47p^2 = p^3

[hollow salmon]

You're mathing wrong.

[timid vector]

okay what is (-12p)^2

[hollow salmon]

144p^2

[timid vector]

I got that

and 4 x (p^2 + 11) is 4p^2 + 44

right

[hollow salmon]

Yes.

[timid vector]

and I still have to times that by p^2

cos that’s c

[hollow salmon]

Yes.

[timid vector]

nvm

wiat a sec

I’m so dumb

I collected the terms 144p^2 - 4p^3 + 44p^2 to

144p^2 + 44p^2 when it’s takeaway

[hollow salmon]

That should be - 4p^4 btw.

[timid vector]

oh uh

oops

I’m getting 25p^2 = p^4

[hollow salmon]

That's one way to approach the problem. What do you want to do next?

[timid vector]

OH WAIT

can I divide both sides by p^2

???

[hollow salmon]

You could, but doing so eliminates a potential answer.

Be careful when dividing out variables.

[timid vector]

wait no I wrote the question down wrong it’s EQUAL roots not just real

so I just need 1 value

so I’m guessing it is just 5

[hollow salmon]

So you need to find p when the discriminant is equal to zero then.

It's not only just 5.

You would be better served by keeping the terms on one side set equal to zero and find the factors.

$25p^2 - p^4 = 0$

[grizzled pagoda]

Kookiemon

[hollow salmon]

Factor the left-hand side(LHS) and find the roots.

[timid vector]

can I take out a common factor like p^2 to get p^2 (25-p^)

then p = 0 and p=5?

[hollow salmon]

Yes. Well p equals +-5. This is where a condition in the problem that p > 0 comes into play eliminating two of the possible solutions.

[timid vector]

oh ur right

thank u sm

[hollow salmon]

yw

[timid vector]

sorry for any of my silly mistakes

.close

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[topaz agate]

Im struggling with every possible question ive looked at the mark scheme to get the answers but there is no method explained in them

[topaz agate]

<@&286206848099549185>

<@&286206848099549185>

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[topaz agate]

.close

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[mystic harbor]

i got f(2) = 4a-2b+3
lim from left side = 0
from right side = 4a-2b+3
whats the next step?

[misty compass]

What is the question?

[mystic harbor]

Find the values of a and b that make f continuous everywhere.

[quick ridge]

what are the conditions for continuity

[mystic harbor]

if f(2) is = to the limit from both sides

[solar gust]

How did you get 0 from left side ?

[mystic harbor]

wait i messed up

0/0 from left side

[solar gust]

Yeah

But you can get that off

[mystic harbor]

l hopital rule?

[solar gust]

Nah just factor top

a² - b²

X² - 2²

[mystic harbor]

ok sure

[solar gust]

(x-2)(x+2)

[mystic harbor]

ok

[solar gust]

And simplify top and bot

[mystic harbor]

thats just x+2

so left side is 4

so after that what do i do?

[formal dome]

for the function to be continuous everywhere:
a) it has to be continuous in each segment. 1st segment is everywhere but 2 (but it's not defined there), the others are.
b) it has to be continous in the borders, in this case, x=2, and x=3
For it to be continuous in each border, the side limits need to be equal, and equal to the value of the function. So compute the value of the function in x=2, at x=3, and both side limits on 2 and 3

[mystic harbor]

im confused

do i plug 2 and 3 in every function

[amber wasp]

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[meager gale]

i am in desperate need of help with this question

[meager gale]

no idea what to do, no idea where to start

this should be done with using differentiation

<@&286206848099549185>

[torpid cedar]

Reading...

OK

Let's start by making a diagram

Ping me when you have it

[meager gale]

actually

nah

@torpid cedar

[prisma onyx]

okay so how can we relate the areas of the two triangles

[meager gale]

area ast and area abs is in ratio 1:2

doing the sine area rule, 2xy=bc

looking at cosine,

wait lemme write

ive tried doing smth with this before but idk what it would lead to

nvm i got it

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[glad tiger]

I can't figure out how to evaluate this limit. I keep getting [0/0]. I know that it's supposed to be negative infinity from the left and positive infinity from the right, but in order for me to even get infinity I need a non-zero numerator. Does anyone know what I should do? I'm not allowed to use the sin(a)-sin(b) formula.

[pallid canopy]

1-1-1 = ?

[glad tiger]

there should be a two in front of that cos^2

question still applies tho

[pulsar coyote]

Isn’t the denominator factorable

like quadratic

[amber wasp]

@glad tiger Has your question been resolved?

[glad tiger]

yes, but that doesn't give me anything useful. It's just (2cos(t)+1)(cos(t)-1)

[glad tiger]

here's the full thing @pulsar coyote

[pallid canopy]

Where is the original question

[pallid canopy]

I mean original

Not your work

[glad tiger]

yeah, the very top one. It's a parametric equation

[pallid canopy]

Take a picture or screenshot

[glad tiger]

I already did the horizontal and vertical tangents

I'm struggling with the one where both the top and bottom are 0

[pallid canopy]

,tex .double angle

[grizzled pagoda]

riemann

[pallid canopy]

At that step try subbing sqrt(1-cos^2(t)) for sin(t), then let u=cos(t)

Polynomials are generally easier than trig functions

[glad tiger]

wait I have an idea

if I multiply the top and bottom by cos(t)+1, I get

success

limit does not exist because left != right

[pallid canopy]

Domain of t=?

[glad tiger]

oh right

0 <= t <= 2pi

wait but

that would mean it's infinite?

and a vertical asymptote

because there is no negative

I think

[pallid canopy]

It's either + or - infinity yes

Do a sign analysis to see which

[pallid canopy]

You have 4 factors to check the sign of

[glad tiger]

(I know I can't do this)

but it helps

that's 3sin(t)

so then if t < 0 (which is can't be with the restricted domain), then it'd be -2/0^-1, but then if t > 0, then it'd be -2/0^+

but that doesnt make sense... the derivative graph shows the opposte

actually wait no it does because I forgot the negative sign on the bottom

[amber wasp]

@glad tiger Has your question been resolved?

[glad tiger]

okay so upon further inspection, I made a series of errors

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[verbal lark]

One message removed from a suspended account.

[verbal lark]

One message removed from a suspended account.

[outer hull]

what is s?

[verbal lark]

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[outer hull]

does it have any conditions on it?

like s > 0 or some sort

[verbal lark]

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[outer hull]

!xy

[amber wasp]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[verbal lark]

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[verbal lark]

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[modern mauve]

Apply chain rule as well

[verbal lark]

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[verbal lark]

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[little widget]

your 2 looks like an $\alpha$

[grizzled pagoda]

kaue

[verbal lark]

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[modern mauve]

Yeah, my writing is bad lol

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[verbal lark]

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[pallid canopy]

Chain rule

,w diff arccsc(x)

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by rie.mann

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[forest portal]

what stops this from being “D”? (answer is C)

D is 2,3,4

Dont need a long answer just reviewing for my test right now

[full elk]

I think graph 2 is too flat to be a double root

It looks more like a quartic locally

[forest portal]

Ahh ok

[full elk]

Yeah ok I just saw they're all degree 4, and all x-intercepts are shown. So that means that graph 2 has all it's x-intercepts at x=-1

So it has multiplicity 4, not 2

[forest portal]

Oh thats a good way to be 100% sure

Also

Tried this middle question twice, got no where at all

Do you know where / how to start?

missing words “where c and d are intergers”

[blissful swan]

It might help to recall that if p, q, r are all polynomials, pq is a factor of r if and only if both p and q are

[mint nacelle]

yes so factorise and solve x^2 - 3x - 4 = 0

what numbers can you substitute into the cubic polynomial?

[forest portal]

x-4 x+1 is the factors

[forest portal]

How do i know if theres c and d in the cubic tho?

[mint nacelle]

yeah

and so.......

[forest portal]

Is that possible to divide tho?

1 sec

[mint nacelle]

no, you have to approach this with the factor theorem

[forest portal]

Ohh

Remainder therom

[mint nacelle]

I mean polynomial long division would be super messy

yeah

actually there is another method based on comparing coefficients

[forest portal]

Itd be f(-1) or f(1)

i always mix those ip

[blissful swan]

Factor theorem is probably the easiest way but division and comparing coefficients will work too

[mint nacelle]

well write x + 1 as x - (-1)

[forest portal]

Well i just substitute in -1 for x, = 0 right

[mint nacelle]

no need for any division btw, (x^2 - 3x - 4)(x + b)

compare the coefficient of x^2 with -6

[mint nacelle]

yeah

[blissful swan]

yeah thats why i said using factor theorem is easier than division

[mint nacelle]

it is -1 for x + 1

cause (-1) + 1 = 0

[forest portal]

Think i messed up somewhere

cuz then i got 2d = 0

[mint nacelle]

-1 - 6 - c + d = 0
-7 + d = c

[mint nacelle]

you should be able to correct your work now

[forest portal]

Oh

[forest portal]

i substitute this into the original equation?

Or after i substitute the -1

[mint nacelle]

wait you didn't use x - 4 yet?

[forest portal]

Didnt touch it

[mint nacelle]

you also have f(4) = 0 remember

you need that

[forest portal]

so after i get this, can i just substitute back into the original?

.close

[amber wasp]

Channel closed

Closed by @forest portal

Use .reopen if this was a mistake.

[amber wasp]

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[regal flume]

hiii can i send pictures?

[amber wasp]

Please don't occupy multiple help channels.

[mint nacelle]

yes

[mint nacelle]

send pictures not files we have to download

ok

[regal flume]

is that good? :)))

[mint nacelle]

okay, so if we increase each side by the same amount, say x

can you come up with an expression for the volume of the new box?

[regal flume]

volume=height times length times width or something to that degree

[mint nacelle]

yes

so what are the new sides?

[regal flume]

…idk that’s why i’m asking for help sorry

i didn’t exactly learn this

[mint nacelle]

if we increase 4 by x what do we get

[regal flume]

ummmm 4..?

5

what does x equal

if it’s 2 then the answer is 6 prob

[simple nexus]

Can’t you just use algebra to figure it out?

Or do you need a graphical solution

@regal flume

[amber wasp]

@regal flume Has your question been resolved?