help-23

[safe radish]

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[real turtle]

Does there exist a natural number n such that 7^n = 2024..... when written out in its decimal expansion

[real turtle]

I've tried the first few numbers and it doesn't seem like you can create something like that and i've messed with it 10^n and all but nothing seems to actually work

i can't prove this won't work

<@&286206848099549185>

[steel stag]

the question seems pretty weird, usually there's not much you can do with leading digits. like it's all basically random if you keep going https://artofproblemsolving.com/community/c2532359h3210961_the_first_digits_of_powers_of_2_and_benfordamp039s_law

[real turtle]

i was thinking along the lines that no power of 7 can have the form number then 0 then number yk

but i struggle with proving it

[steel stag]

does the question say something about the ending digits? idk seems like a part is missing

[real turtle]

nope that is the entire question: Does there exist a natural number n such that 7^n = 2024... when written out in its decimal expansion?

Character for character

[edgy isle]

,w 7^9

[flat frigate]

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Requested by rbit.sr

[real turtle]

yh i figured it out after :(((

still have no idea how to solve it

[edgy isle]

ill think about it but yeah its tough

[steel stag]

still no idea why but

[real turtle]

wow

-3

oop

[zinc pumice]

can I ask the context of this question? What are you doing, number theory?

[real turtle]

Yh it's a problem sheet my friend made w some problem solving esque questions

[edgy isle]

ok notice how when taking 7^((m+0.306318)*ln(10)/ln(7)) we are always getting 2024 as the leading digits for any natural number m

[real turtle]

Where's the 0.306318 come from?

[edgy isle]

so I think one just needs to get this exponent close to an integer

[edgy isle]

fractional part of ln(20245)/ln(10)

added a 5 for more safety

[real turtle]

this is where i got so far w the problem

[zinc pumice]

Anyway I think this is a calculation mistake by wolfram. I am not sure but: 1)I think that the answer to the question should be no (because of how it is written, and the fact that it is not humanly possible to come out with that example). 2)If you ask wolfram alpha for "a power of 7 that starts with 2024" it says that there are none (https://www.wolframalpha.com/input?i=can+a+power+of+7+start+with+2024, it could be wrong though)

[real turtle]

u can change that to then say the fractional part of nlog107 needs to be between yap and yap

[edgy isle]

I just need to find an m that makes this term approximate an integer

[real turtle]

not much to go from here

[safe radish]

@real turtle Has your question been resolved?

[edgy isle]

,w 7^545046

[flat frigate]

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Requested by rbit.sr

[real turtle]

nice

aight this has gone on too longg imma close it

[edgy isle]

had to write a little script, just needed m*ln(10)/ln(7) + ln(20245)/ln(7) to be as close as possible to an integer. Pretty sure wolfram should be correct

[real turtle]

aight that's good i'll close

thx

.close

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[safe radish]

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[cold jasper]

Hello. Im kinda confused with this (or maybe overthinking. im too fried today)

I need to write a property that demonstrates this is truth, however im kinda confused wether this is associative or commutative because the agrupation doesn't change to be considered associative, it just switches places with the four.
So would it be commutative or associative?

[marsh walrus]

youre using the commutativity of addition

[devout shale]

Indeed

It might be easier to see if you just say something like, Call (2+3):=u

then its u+4=4+u

which hopefully is more clearly commutativity

[cold jasper]

Ohh alright, so it would be considered associative if the number inside of the parenthesis switch/change, right?
for example:
(2+3)+4=(4+3)+2

[devout shale]

You wrote that wrong

(2+3)+4=2+(3+4)

would be associativity

[cold jasper]

Ohh alright, thank you so much!

.close

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[modern bloom]

Prove: For the surface (\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{a} ) (where ( a > 0 )),
the sum of the intercepts on the three coordinate axes of the tangent plane at any point is a constant.

[flat frigate]

miyo

[rocky quarry]

Could you use implicit differentiation to find the partial derivatives?

[safe radish]

@modern bloom Has your question been resolved?

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[past crystal]

Why is this incorrect? Two of the points are on one plane, and the other two are on a different plane. So it's not coplaner... right? Coplaner means all the points are one one plane?

[past crystal]

<@&286206848099549185>

[magic junco]

nope, there is a slight mistake

three points form a plane, not two

[past crystal]

But isn't it still not coplanar though?

Since coplanar means all on the same plane

[magic junco]

the four points are not on the same plane, yes

your second statement is incorrect, three points can determine a plane

the rest are all good

[past crystal]

ohhh I see, thank you so much I was so confused to why I got the question wrong 🙏

[magic junco]

np, have a good one

!done

[safe radish]

If you are done with this channel, please mark your problem as solved by typing .close

[past crystal]

.close

[safe radish]

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[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[fathom adder]

What is your question ?

[median vault]

i got no idea what they did here

im assuming they used their graphics calculator to do something but i dont understand the working out

[fathom adder]

Not quite

[median vault]

what formula is that that they are substituting into

[fathom adder]

Is it a binomial distribution ?

[median vault]

wait ill snd the quesiton

sry i was patting my cat

[fathom adder]

Yeah its a binomial, but sigma = sqrt(np(1-p)), i dont see how they get it down

Sigma = Sqrt(64*(0.36)) = 8sqrt(0.36) = 8*6/10 = 2.4

[median vault]

do u know what formula theyre using for sigma

[fathom adder]

No, idk what formula they're using

[fathom adder]

But the yellow here is obtain with gdc

[median vault]

do you know what to input to the gdc

[safe radish]

@median vault Has your question been resolved?

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[willow owl]

This is not cool T-T

Can someone help me with it, the final answer I got was just the constant c, everything was subtracted off

As I use the power reducing formula, then subtract them, only the +c remains

What is wrong with my solution

[cedar void]

my nightmare from now

[willow owl]

If you'd ask me where did it go wrong, I have no idea TT

[cedar void]

but wait

everything doesn't cancels

[willow owl]

Yeah it does

[cedar void]

you sure

[willow owl]

I think so..

[cedar void]

missed a negative

[cedar void]

see here

[willow owl]

😳

Omg wait

I didn't notice that until now

[cedar void]

lol

[willow owl]

I got the answer, thank u very much, it was a beginner mistake 🥹

.close

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[cunning pasture]

I have a question of myself

[cunning pasture]

made by me

I have a doubt

If AEB and DEC are similar, by CPCT the corresponding angles are equal.

So angle BAE = angle EDC

And so on.

WIll the other pair of triangles be also similar?

hi, wheir the helpers gone to?

at

[safe radish]

@cunning pasture Has your question been resolved?

[cunning pasture]

<@&286206848099549185>

[safe radish]

Please do not ping individual helpers unprompted.

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[cunning pasture]

what is this bot even saying?

[solemn mantle]

the bot is yappibng

[cunning pasture]

Please type here your helpful texts while I go for cycling for 20-30 minutes.

[red delta]

No

Come back

[cunning pasture]

yes i have

What's the matter of laugh?

<@&286206848099549185>

another 15 mins have passed therefore I can ping helpers

[karmic hedge]

Yes

The quadrilateral is then cyclic

[cunning pasture]

prove it

<@&286206848099549185>

[karmic hedge]

It’s just angle chases

Converse of inscribed Angie theorem

[cunning pasture]

prove it lies on a circle

[cedar void]

how in the world did you even used CPCT before proving the congruency?

[cunning pasture]

I mean CAST

Corresponding angles of similar triangles

[cedar void]

okok

[cedar void]

yeah your statement holds

other pair of triangles will be similar

[safe radish]

@cunning pasture Has your question been resolved?

[safe radish]

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[safe radish]

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[lime dust]

,rotate

[glossy cypress]

Beautiful

[cosmic cloak]

eeee

[lime dust]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[cosmic cloak]

i hate triangles

[cosmic cloak]

i completely dont know how to solve this.. i just started this

[lime dust]

You can start by finding the angles of the triangle for example.

[glossy cypress]

Okay so let's start with this angle

Damn it is NOT loading

Okay uh 💀

Let's start with <ABD

What do you think that's equal to?

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[foggy echo]

determine the length of the sloping edges of the roof, i.e. the distance from "A" to "B"
(the roof pitch is 45 degrees, all measurements are in metres)

[foggy echo]

I don't know how to approach this to be fair

<@&286206848099549185>

[safe radish]

@foggy echo Has your question been resolved?

[safe radish]

@foggy echo Has your question been resolved?

[frozen veldt]

have you tried picturing a right triangle going from A to the right angle and from there to B and then back to A?

you basically want to figure out the hypotenuse of that triangle

[foggy echo]

right

[frozen veldt]

and there's one side you're missing

but maybe another pythagorean theorem / sneaky right triangle could help you out

btw you have a funny bio

[foggy echo]

tihi

@frozen veldt how do you get helpful rank btw?, cause I'm a math-obsessed alien, and I'm good at helping people

[foggy echo]

yeah that's what I thought aswell

honestly just made this to get a second opinion

lol

[frozen veldt]

well now you can feel safe in knowing that your approach won't lead to a dead-end

[foggy echo]

haha yeah

.close

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[safe radish]

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[burnt spire]

not sure what im doing wrong here

[cerulean coyote]

can you show your work?

[burnt spire]

yes

uhh

plug in 5.9 into the curve

so it'd be 3/(5-5.9)

so 3/-.9 = -3.3(repeat)

plug that into the secant equation

(-3.3(repeat)-(-3))/(5.9-5)

wait

I may be missing a step i did

nvm i didnt

[cerulean coyote]

i see your mistake. it should be $\frac{-3-(-3.3\dots)}{6-5.9}$

[flat frigate]

fish

[cerulean coyote]

instead of -3.3-(-3)

since -3 is at point 6, and -3.333 is at point 5.9

[burnt spire]

ohh

i was basing it off of the where they were in the mpq

i goofed 😭

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[jolly pasture]

none of the answers here match up; i just selected the closest one. Shouldn't the answer be...

[jolly pasture]

$f'(x)=3x^2 + \frac{10}{x^6}$

[flat frigate]

Matz

[jolly pasture]

like the exponent is wrong no? or am i doing something wrong

[rigid inlet]

the exponent is wrong

[jolly pasture]

okay awesome

thanks!

.close

[safe radish]

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[rigid inlet]

.reopen

[safe radish]

✅

[rigid inlet]

hold

they actually do have the right answer on there

[jolly pasture]

oh hello

thanks for pointing it out

ohh the -6 exponent!!!

okay

that's so weird though

our teacher told us to never leave them as such

[rigid inlet]

i imagine your teacher didn't write these

[jolly pasture]

I sure hope not

thanks either way!!

:))

I'll change my answer now

.close

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[safe radish]

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[prime locust]

To start, maybe first try finding a basis for $\mathbb{H}_1^{\perp}$ and a basis for $\mathbb{H}_2$

[flat frigate]

Ryan

[prime locust]

The first basis will have a k in it

[safe radish]

@spiral saddle Has your question been resolved?

[prime locust]

@spiral saddle did u try this yet?

[prime locust]

fantastic job.

next is H2

[spiral saddle]

So?

[prime locust]

yet again, great job.

next question for you: what must the dimension of T be?

hint: now that you found bases, what are the dimensions of H1^(⟂) and H2?

[spiral saddle]

The corollary

Do you speak French?

[prime locust]

i dont speak french sorry

but yes you are correct

dim(T) = 2

well, we're trying to create T

and we want it to have dimension 2

next step is doing something called "extending a basis"

have you heard of something like that before?

[spiral saddle]

I have heard but dunno what it means

[prime locust]

gimme a min

okay before i begin let's be clear what we're doing here

H1^(⟂) has a dimension of 1
H2 has a dimension of 3

to find T, we need to do what's called extend a basis of H1^(⟂) to a basis of H2

[spiral saddle]

how do I do that shit}

[prime locust]

sorry im taking my time here. trying to fact check myself before i say something wrong lol

[spiral saddle]

do we need to use S for that shenanigans

[prime locust]

i think we can deal with S later

but basically, since H1^(⟂) has dimension 1,

we need to add 2 more vectors to the basis of H1^(⟂) in order to get a basis of H2

[spiral saddle]

ye dim 1 vs dim 3 so what?

[prime locust]

so like, you found that H1^(⟂) = <(1, 0, k ,2)>

now you need to find 2 more vectors, let's call them v1 and v2, such that H2 = <(1, 0, k ,2), v1, v2>

does that make sense so far?

[spiral saddle]

yeah

but how

[prime locust]

if you can do that, then T = <v1, v2>

so now that we're clear what we're doing

i will show you how to extend a basis in general. gimme a minute to write this all out

If $V=\langle v_1, v_2, ..., v_k\rangle$ is a subspace of $W=\langle w_1, w_2, ..., w_n\rangle$, then you can \textbf{extend the basis} for $V$ to a basis of $W$ by first putting all of those vectors together in a set like this: $\beta={v_1, v_2, ..., v_k, w_1, w_2, ..., w_n}$. Now the thing is that this set might not be \textit{linearly independent}, so to fix this issue, we \textit{remove} as many $W_i$ from $\beta$ "as possible", as to not change span$(\beta)$, until the set becomes linearly independent.

[flat frigate]

Ryan

[spiral saddle]

what

[prime locust]

There is an algorithm for doing this.

which i will show you

[spiral saddle]

can you show a drawing

[prime locust]

erm not really

[spiral saddle]

that would be fine

[prime locust]

"extend a basis of V to a basis of W" means you add vectors to the set {v1, v2, ..., vk} until it spans W

[spiral saddle]

ye I had that figured until you started rambling about linear independence

[prime locust]

correction, until it spans W and is linearly independent

[spiral saddle]

without explaining with a drowing

[prime locust]

thats what it means to be a basis

[prime locust]

what i should have said here is:

"extend a basis of V to a basis of W" means you add vectors to the set {v1, v2, ..., vk} until it becomes a basis of W.

[snow robin]

a basis spans and is linearly independent

you don't have to state it is linearly independent again

so you were right the first time

[prime locust]

what are you referring to here?

[snow robin]

this

[spiral saddle]

can someone hand hold me throughout this exercise

[prime locust]

i was not right the first time

[prime locust]

yes i will show you how

but first you must understand the concept

[spiral saddle]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[snow robin]

i was looking at this sorry "extend a basis of V to a basis of W"

[prime locust]

np

[spiral saddle]

can someone help me with my algebra homework

[prime locust]

anyway, @spiral saddle
i will show you how to do this specific problem once you understand this:

"extend a basis of V to a basis of W" means you add vectors to the set {v1, v2, ..., vk} until it becomes a basis of W.

[spiral saddle]

ok

[prime locust]

in this case, we need to add 2 vectors to {(1, 0, k ,2)} until it becomes a basis of H2

[spiral saddle]

ye

v1 and v2

[prime locust]

now i will show you how to find those vectors

you first create a matrix, whose column vectors are all the vectors from the basis of H1^(⟂) and the vectors from the basis of H2

so that would be a 4x4 matrix whose columns are (1, 0, k ,2), (2, 1, 0, 0), (1, 0, 1, 0), and (-1, 0, 0, 1)

can you draw that matrix?

[spiral saddle]

,w rref {{1,0,k,2},{2,1,0,0},{1,0,1,0},{-1,0,0,1}}

[prime locust]

wtf

where did k go

[prime locust]

thats a matrix whose ROWS are those vectors

i think the vectors have to be the COLUMNS of the matrix

[prime locust]

great work, you've done this step

NOW, row reduce

note that you're gonna end up with k's in different spots in the matrix

but that's alright. row-reduce it like you would any other matrix

[prime locust]

uhm

im not exactly sure what happened here

the first two steps for row-reduction here are
R3 - k * R1 --> R3
R4 - 2 * R1 --> R4

do you see why?

[prime locust]

i'm quite lost at what you're doing here

[spiral saddle]

you show me how to row reduce}

[prime locust]

i am not your servant lol

[prime locust]

also, i just did in this message here

[prime locust]

start from here again but do this as your first 2 steps:
R3 - k * R1 --> R3
R4 - 2 * R1 --> R4

[prime locust]

that's how to row reduce

this is what you will get after doing so

the reason i did it this way:
you want to make the first column have a pivot (which is where the "1" is) and 0's below that pivot

does that make sense?

@spiral saddle unfortunately it's getting late and i have work tomorrow. if you still need help you can ping helpers again. best of luck 💜

[spiral saddle]

@

<@&286206848099549185>

I messed up again

[safe radish]

@spiral saddle Has your question been resolved?

[safe radish]

@spiral saddle Has your question been resolved?

[safe radish]

@spiral saddle Has your question been resolved?

[spiral saddle]

SOMEONE HELPPPPPPP

<@&286206848099549185>

[spiral saddle]

<@&286206848099549185>

[icy bison]

which software is this

how do u write like that

[spiral saddle]

is my phone

[icy bison]

which app?

[spiral saddle]

the app samsung notes

[icy bison]

u write with hand or stylus?

[spiral saddle]

stylus

is the stylus that comes with the phone

[icy bison]

ohh ic

thanks

[spiral saddle]

no worries

[wicked sail]

This looks confusing...

[spiral saddle]

what is the confusing part

[spiral saddle]

<@&286206848099549185>

<@&286206848099549185>

help please

[snow robin]

Have you tried doing what he explained?

[grim kraken]

Oh still this problem. So last time we found the orthogonal complement and now we would need to find an idea of what T is right ?

[safe radish]

@spiral saddle Has your question been resolved?

[flat frigate]

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[safe radish]

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[upbeat ridge]

Hello

[upbeat ridge]

I'm not sure what to do next

Also, how do I show that the two vectors are not parallel?

[manic bane]

Two vectors are parallel, if their cross product is 0.

[median vigil]

two vectors are parallel if one is a scalar multiple of the other

[upbeat ridge]

When two vectors are parallel or perp their cross product is 0, right?

[median vigil]

parellel - cross product 0, dot product maximum
perpendicular - dot product 0, cross product maximum

[upbeat ridge]

Right

Uhh so

How do I proceed from here?

.close

[safe radish]

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[steep magnet]

So its a sin function

I was thinking the equilibrium must be in the middle of 65 and 4
And the amplitude must be half of the distance between 65 and 4

So (65 - 4)/2 = 61/2
61/2 + 4 = 69/2

So it must be 61sin(x)/2 + 69/2

And it happens every 8 hours so the period must be 24/8 = 3

So like 61sin(3x)/2 + 69/2? Feels like this is wrong but this is my best guess

[safe radish]

@steep magnet Has your question been resolved?

[steep magnet]

.close

[safe radish]

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[fresh forum]

Could someone teach me how to resolve trigonometric equations such as

[fresh forum]

the forth/sixth ones

i'm lost when a root is included, also when something else than x is in the parenthesis

[glacial meadow]

Well, you can do two things

[fresh forum]

How so

[glacial meadow]

First of all, for the fourth one, the most common trigonometric functions(sine and cosine) are separated by a factor of π/2

So sin(x+π/2)=cos(x) and cos(x-π/2)=sin(x)

That could help you in the fourth one

[fresh forum]

now how didn't i know that

so i can just

replace .

?*

[glacial meadow]

Yeah

[fresh forum]

oh

then just do arccos of the root / 2 ?

[glacial meadow]

After that, you can just solve for everything that's inside the function and then solve for x

[glacial meadow]

In that case, yeah

[fresh forum]

what about the 6th

[fresh forum]

Could it not be that easy ? i can't see any other case like that

[glacial meadow]

In general, if you've got f(g(x)) with f a trigonometric function, you can consider g(x) the variable, solve the equation for g(x) and afterwards solve for x

[fresh forum]

what would the g(x) be in that case

[glacial meadow]

Assume you've got cos(x+2) for example, you can't replace that with a simple sine function

[glacial meadow]

Any generic function, whatever it may be

For example, if you've got cos(ln x), think of ln x as the variable, solve for ln x

[fresh forum]

Yes exactly these ones

i can't resolve them

[glacial meadow]

And then, once you get the results solve for x

[fresh forum]

understood

[glacial meadow]

And for the example I gave earlier, just do the same but for g=x+2

[fresh forum]

x + 2??

oh yeah

but then like

how to resolve that

[glacial meadow]

Trigonometric equations can get a little more complicated but for now that should suffice

[glacial meadow]

Cos(x+2)=k

u=x+2

Cos(u)=k, u=arccos(k)

[fresh forum]

Well i don't want to alarm you but

[glacial meadow]

X+2=arccos(k), x=arccos(k)-2+2πn, n an integer

[fresh forum]

i may or may not be passing my exam to enter last year of school before uni tommorow... morning

[glacial meadow]

Doubt they'll be asking you for much more then

Maybe equations involving angle sums and double angles, but those are just formulas

[fresh forum]

wait

if i send you my exam

are you willign to help a bit longer ..? pretty please

[glacial meadow]

Well, sure, but if you want me to actually solve the exercises I'll need ya to wait ~10 minutes, I'm not home right now

So I ain't got a notebook and a pen on me

[fresh forum]

just to give you an idea

that's the type of exercises i'll have tmmr

[glacial meadow]

I can't read some of your handwriting

Could you type some of the examples here?

[fresh forum]

yeah ofc

there's the sheet iyw

i can zoom in

that's the first exercise

2nd

3rd

last

[dull river]

hi i am in my senior year and I have just attended few of my math classes of my junior year and now i am screwed real bad so can someone help me learn inverse trigonometry

[fresh forum]

bro this is my channel

We're gonna have beef

[dull river]

yeah okay

what

what did i do

[fresh forum]

nah jk dw

[dull river]

bro

u scared me

[fresh forum]

bahahah no don't worry bro

[dull river]

like seriously

yeah'

JUST HELP ME

[fresh forum]

have a good one

[dull river]

PLEASE

[glacial meadow]

🗿

[dull river]

WHAT

[fresh forum]

CLAIM YOUR OWN CHANNEL

[dull river]

okay

[glacial meadow]

And yeah, all you'll need is some intelligent substitutions, that's it

[dull river]

i jus loving talking in caps

[fresh forum]

SAME

[dull river]

W

[fresh forum]

So all i need is to perfectly know all my formulas ?

[dull river]

I LOVE YELLING

[glacial meadow]

My trig exam was way worse, I remember we had double angle identities and allat

[glacial meadow]

Not really, what you need is to see the convenient substitution in each problem

[fresh forum]

Oh yeah same but i passed derives

[glacial meadow]

For example, if you got a(cos(x))^2+b(cos(x))+c, solve first for cos(x) and then for x

[fresh forum]

Nah this is chinese

so

[glacial meadow]

Then the thing I told you before, If you've got a function inside the trig one solve for the function inside first then for x

[glacial meadow]

That was just me trying to write a standard case, lemme rephrase it

[fresh forum]

(cos(x))² + b.(cos(x)) + c ?

[glacial meadow]

If you've got something that seems like a quadratic equation but with trig functions instead of x, solve for the trig function first and then for x

[dull river]

can someone tell me how this help rooms work i am new on discord

[glacial meadow]

Go to one of the channels under math help(available) and type your problem

[fresh forum]

click on the channel my mouse is on

[dull river]

okay thanks guys

[fresh forum]

so as you were saying

[fresh forum]

yw dw

[dull river]

sorry for bothering yall

[fresh forum]

no worries !

[dull river]

a quick question

um i need to do my trigo from scartch and my derivatives too so from where should i start ??

um guys ??

[fresh forum]

derivatives are hardest for most of ppl but i passed 90%, while i did 10 in trigonometry so start with the hardest one

[burnt notch]

Derivatives, and when you encounter some trig functions go and revise them

[dull river]

okay

[fresh forum]

@glacial meadow

please savior

[glacial meadow]

Yeah, sorry, just got back

Lemme see

[glacial meadow]

You want me to solve some or you have a specific question?

[fresh forum]

Specifiv

[glacial meadow]

Alright, what's the question then?

[fresh forum]

Can you make me just 1 exercise both for trig and limits (and asymptots but not derivatives i passed them)

[glacial meadow]

Sure, which ones do you want me to do

[fresh forum]

mmh

this one for example

[glacial meadow]

The first or the second one?

[fresh forum]

also can you do it with me ? like point to me what to do but i try to do it by ms the most possible

[fresh forum]

second one isn't a problem

[glacial meadow]

Alright, do you want me to solve it myself first,show you and you ask me about my method or to go along with ya?

[fresh forum]

with me plsss

[glacial meadow]

Alright, then I'll give you some hints to see if you can get to it

it can be solved, as well, with a substitution

Which substitution would you use here?

[fresh forum]

mmh

the denominators ?

put everythign in the parenthesis on 24 ?

[glacial meadow]

Alright, lemme rephrase that

What change of variable would you do here?, as in u=g(x)

[fresh forum]

oh

i'm lost

[glacial meadow]

Let's try to think about it in a different way, then, I'll leave the variable changes aside for now

@fresh forum do you know what an arc function is?

as in, what it does

[fresh forum]

surely but i'm not english native so i can't visualize it

would it be like

the calculus inside the circle ?

[glacial meadow]

Well, I'm not a native speaker either, I just want you to think about the relationship between the trig function and its arc function

[glacial meadow]

Not really, but I'm not asking about geometry rn, think about them as functions

[fresh forum]

I don't think so then

[glacial meadow]

What happens to a cosine(x) when you apply arccos to it?

[fresh forum]

oh

you get x

[glacial meadow]

Alright

What would happen if we applied arccos to cos(x+1) then?

[fresh forum]

mmh

the answer would be = to x + 1 ?

[glacial meadow]

Yup

[fresh forum]

you cannot be real

so then

[glacial meadow]

Leaping a bit, if we had cos(g(x)) , with g(x) being any function, what would we get applying arccos?

[fresh forum]

this one would be equal to

x divided by 6

to wich we added 5pi/4 ??

[fresh forum]

f(x) ?

g(x) *

[glacial meadow]

Yup

[glacial meadow]

Basically, arccos undoes the cosine function, arcsin undoes the sine function and arctan undoes the tan function

[glacial meadow]

Yeah, though you'd have to apply arccos to the other side as well

[fresh forum]

the root ?

[glacial meadow]

So what's inside the cosine will be equal to arccos,(-sqrt(2)/2)

[glacial meadow]

Yup

With that, you can solve practically any of the equations you've got there

[fresh forum]

ok ok

now about the limits

[glacial meadow]

Alright, which one?

[fresh forum]

(also its not my nail)

in this one

[glacial meadow]

What's it asking me to do

[fresh forum]

i need to calculate a and b

[glacial meadow]

Alright, what conditions does the exercise give you?

Continuity? Differentiability? Limit equal to something?

[fresh forum]

nothing ig

[glacial meadow]

Show me the full exercise, I might be able to figure out what it says even if it's in french

[fresh forum]

okay

[glacial meadow]

Saw it already, determine the values of a and b such that the function has x=-1 as a horizontal asymptote and y=-4 as a vertical asymptote

[fresh forum]

now

how did you know that

its right

i was struggling to take a screenshot btw sorry

[glacial meadow]

Scrolled up to your image and figured it out from the french text

To the full one

[fresh forum]

nice ig...

so

oh

[glacial meadow]

Do you know how to do it?

[fresh forum]

i don't remember

[burnt notch]

You could revise the theory, then 😅

[fresh forum]

I lost my sheets...

[burnt notch]

There's Internet, or textbooks

[glacial meadow]

I can give you a quick summary but if what you're lacking is study time I really can't help you buddy

[fresh forum]

The summary would help alot

i know almost every formula with sin cos tan cotan

but the others idk

[glacial meadow]

Well then:
Horizontal asymptotes are “impossible values of x“ for the function, in other words(and in most cases), zeroes of the denominator
Vertical asymptotes are the value of f the function tends to, it's the result of lim(x->∞)f(x), but only if the limit is finite

Oblicuous asymptotes are a little harder, since they're full fledged straight lines

Their slope m=lim(x->∞)f(x)/x, n=lim(x->∞) (f(x)-mx)

The asymptote itself is of the form y=mx+n

That's asymptotes in a nutshell, limits at that level are bunch of systematic procedures and I already told ya how to deal with trig equations

[fresh forum]

sorry i was helpign my mom

Well Prim sorry i'm gonna give up for today

if you can just give me the summary

You'll have done much more for me than my teacher

[glacial meadow]

Well, that's fair

[glacial meadow]

What else do you want the summary on?

[fresh forum]

I think only trig and obliques asymptots

[glacial meadow]

Oblique asymptotes are here

[glacial meadow]

As for trig, when you see something that's not just x inside the trig function, apply the arc function on both sides first and then solve for x

[fresh forum]

okay that'll help alot

[glacial meadow]

Remember the formulas for them, and when faced with the exam, remember to think, and not just repeat what you've done before

[fresh forum]

Okay sir than kyou alot

have a very good day and i hope you'll be happy forever

Bbye !

.close

[safe radish]

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[glacial meadow]

Well, see ya later

[safe radish]

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[fringe dock]

can someone explain what log law this is

[fringe dock]

,rotate

[fringe dock]

ignore the top right

[shell field]

That is not really a law

It is more like doing the opposite

Something like if you have x/2

And than you multiply by 2

And you are left with x

[fringe dock]

ah okay

[fringe dock]

if you don’t mind

may you give me a similar question?

[shell field]

Maybe something like this

[fringe dock]

thanks

@shell field

is this right

[shell field]

Well the log part is, but u have to subtract 8

[fringe dock]

oops

my mistake

my brain stopped

there we go

@shell field

[shell field]

That looks better

[fringe dock]

thanks

[shell field]

No problem

[fringe dock]

.solved

[safe radish]

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[safe radish]

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[proud briar]

I need help with converting fractions to decimals with long division, I've been stuck in the long division part for a while and have even asked youtube and google, I need to convert 5/3 and 1/8 into decimals, I know the answer I just don't know how to do long division

[proud briar]

For converting 1/8 so far I have

8/1.0

Above I have 0.

[safe radish]

@proud briar Has your question been resolved?

[proud briar]

<@&286206848099549185>

.close

[safe radish]

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[safe radish]

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[shrewd topaz]

so i used product rule

f(x)=6x
f'(x)=6
g(x)=arccos(x-1)
g'(x)=-1/(sqrt1-(x-1)^2)

(-6x/sqrt1-(x-1)^2) + 6(arccos(x-1))

can i plug in 1 now

or do i still have some factoring left

[junior vale]

What happens if you plug in 1 ?

[shrewd topaz]

0

so keep factoring

how do i do that

common denominator?

[junior vale]

I would recommend rationalization

[shrewd topaz]

wdym

i mean i hope you dont mean what i think you mean

so (6arccos(x-1)) * (sqrt1-(x-1)^2)?

[lime dust]

Why -6 in the first part?

[junior vale]

g'

[shrewd topaz]

6x/1 * -1/sqrt(1-(x-1)^2

[lime dust]

Not negative

Let me recheck, I might be misreading

[shrewd topaz]

derivative of (x-1) is 1

formula for arccos u is -u'/(sqrt1-u^2)

[lime dust]

No

[shrewd topaz]

oh

[lime dust]

Recheck it pls

I think u are missreading

It is positive

[junior vale]

Derivative of y = arccos(x) is dy/dx = -1/sin(y), which is -1/(sqrt(1-x^2))

[lime dust]

Oj

It is arccos

Not arcsin

Apologies i read arcsin

[shrewd topaz]

i was looking at the wrong problem too

[lime dust]

I am the one misreading lol

[shrewd topaz]

i feel like this problem is common with the arc derivatives

[shrewd topaz]

anyways

[lime dust]

Forget what I said, i just read arcsin

[shrewd topaz]

u good

[junior vale]

I always turn them into implicit differentiation

[shrewd topaz]

leibniz notation for the product rule?

[junior vale]

I meant the arc-functions. Turn y = arccos(x) into cos(y) = x. Easier to derive.

[shrewd topaz]

idk

can we move forward now

[junior vale]

Right, so your (x-1)^2 doesn't cause problems.

Don't need to rationalize

I got tripped up on parentheses

What's the slope at x = 1 ?

[shrewd topaz]

okay wait

[shrewd topaz]

do you not get 0

[junior vale]

No, it's not a horizontal tangent.

[shrewd topaz]

right

or did i just plug in x = 1 wrong

cause i keep getting 0

-6+6

[junior vale]

What is arccos(0) ?

[shrewd topaz]

1

so 6 on the right term

[junior vale]

No, that's cos(0)

[shrewd topaz]

1/cos(0) is still 1

[junior vale]

Arccosine asks "what angle is going to get you zero in a cosine function?"

[shrewd topaz]

arccos isnt 1/cos?

[junior vale]

No

That's secant

arccosine is the inverse function. Reverses inputs & outputs

[shrewd topaz]

oh so i plug 0 into u

0/1

is 0

6*0 is 0

so we just have -6

[junior vale]

No, the arccosine part is still not right.

[junior vale]

Consider this question again.

[shrewd topaz]

90

[junior vale]

Now again, but in radians

[shrewd topaz]

pi/2

so i plug in pi/2??

[mossy fractal]

no, you plugged in 0 and you got pi/2

[junior vale]

The arccosine of zero gets you pi/2.

[shrewd topaz]

oh

so thats how i approach arc identities?

[junior vale]

Yes. Alternatively, switch the variables and look at the functional inverse.

[shrewd topaz]

yea inverses are still super confusing for me

much less trig functions

[junior vale]

arccos(x-1) = N means cos(N) = x-1

[shrewd topaz]

i see

[junior vale]

So N is an angle and x-1 is the cosine ratio,

which is really (x-1)/1

So what are you getting for slope?

I like the -6 part. What's the arccos part getting you / what are you getting altogether?

[shrewd topaz]

-6+pi/2?

[junior vale]

almost

[shrewd topaz]

k so we have arccos(x-1)

im plugging in x =1

i get arccos(0)

now i want to find a value that gives me 0 in the cosine function

which is pi/2

but

what do i do with that

[junior vale]

What's attached and how?

[shrewd topaz]

• 6?

so 3pi

[junior vale]

And all together?

[shrewd topaz]

-6+3pi

[junior vale]

Now what?

[shrewd topaz]

thats my slope

okay but i dont understand

because i didnt apply this process to other problems

and i still got the right answer

[junior vale]

Not sure what you mean, but you're looking for the equation of the line that is tangent to the function above at the given point.
You found the slope of a tangent line to the curve at the given point (3pi - 6), so what does "equation of a line" mean you need?

[shrewd topaz]

okay take this problem i did earlier as an example

i just plugged my x value into the y'

ok yk im gonna accept that i have no idea what im going on about

[shrewd topaz]

plug it into our point slope form

[junior vale]

Yep.

[shrewd topaz]

my point is

[junior vale]

We're doing the same problem as from Algebra I, when we were asked to find the equation of a line that had a given slope & went through a specific point.
This one is just more sophisticated.

[shrewd topaz]

why cant i plug arccos u into my derivative formula

[shrewd topaz]

no i understand my issue isnt with that

i just want to understand this now so i dont have to worry about it later

[shrewd topaz]

so this will apply every time

i ask this question every time im plugging a value into an arc function

?

[junior vale]

Yes, essentially. Your arccosine is looking for what angle gets you the stuff on the inside, when cosine is applied to it.

[shrewd topaz]

what about 3pi/2

[junior vale]

Great question. We could also ask what about 17pi/2

[shrewd topaz]

right which value do we pick since they all work

[junior vale]

The issue is that the arccosine function needs to be a function.

So there can only be one answer

[shrewd topaz]

idk what you mean by that

[junior vale]

A function needs to map each input to at most one output.

[shrewd topaz]

which is what the (1,3pi) means

[junior vale]

When you input 1 for x, you get out 3pi for y, and no others.

[shrewd topaz]

is that why we chose pi/2

because thats the only one that could lead to 3pi

[junior vale]

Sort of. The natural domains & ranges for the arccosine function are reverse of the domains & ranges for the cosine function.
What are we allowed to plug INTO a cosine function?

Or what things are we plugging in?

[shrewd topaz]

and 3pi is in that domain

[junior vale]

So if we're looking just at the cosine function, we are plugging in angles. Is there any restriction for a standard cosine function?

[shrewd topaz]

0 and pi

between those two values

[junior vale]

Not really. Your standard cosine function is allowed to have any angular values. I see where you're coming from, though.

We restrict the domain artificially because we need to make the cosine function one-to-one.

[shrewd topaz]

okay then its between 0 and 2pi

if we're looking at a unit circle

[junior vale]

Sure, but you are allowed to plug in 50pi or -50,000,000 in for the angle.
Not really the overall point.

It's periodic, so it comes back, eventually.

Every time it self-overlaps, it is not one-to-one, and can't have a functional inverse (arccosine).

[shrewd topaz]

so what are you getting at

[junior vale]

So when we artificially restrict the domain, it is x∈[0,pi], with range y∈[-1,1]

[shrewd topaz]

for the cosine function

[junior vale]

Yes.

[shrewd topaz]

and the arccos function is the inverse of the domain and range

[junior vale]

Yes

[shrewd topaz]

so x = [-1,1] and y =[0,pi]

[junior vale]

Yep!

[shrewd topaz]

so thats why we chose pi/2

[junior vale]

Yes, it's the only output for arccosine which matches the cosine-ratio value of 0.

[shrewd topaz]

okay then i think i understand

ty

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[junior vale]

No prob 🙂

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[ripe tinsel]

Where did I go wrong here?

[median vigil]

your dot diagram starts to look a bit off towards the right side. i would recommend using a calculator to find where each dot should be

[safe radish]

@ripe tinsel Has your question been resolved?

[quaint relic]

You drew your graph incorrectly.

If the second point is slightly past -7 going from left to right, it is not at -7.2, but some other value.

i would ask "what two values must this value be between"

and look at the drawing you made

[ripe tinsel]

you're right

it should be going backwards

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[ripe tinsel]

.reopen

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✅

[ripe tinsel]

I chose -6.8 and then calculated -1.2 as the slope

Cuz one point is definitely -8 and other one is -6.8

[quaint relic]

You need to be more diligent about negative signs.

🤔

i think

[ripe tinsel]

yeah

common problem with this level of maths

[quaint relic]

it should be (-8 - - 6.8) / (0 -1)

If i remember rise over run correctly

[ripe tinsel]

Okay so then X=10 I can just input this equation and hopefully it works, stay tuned

[quaint relic]

but you wrote something different if you got -1.2 as the answer

[ripe tinsel]

oh its 1.2

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[gentle gorge]

can someone quickly help me understand the AND distributive property? where A+BC = (A+B)(A+C)

[gentle gorge]

when i distribute the right hand side i get A (idempotency) + AC + BA + BC

i don't see how it simplifies to A + BC due to the middle terms

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[royal folio]

Determine the area of ​​the left side wall of the house! (√3 = 1.73)

i don’t know what to do next

[safe radish]

@royal folio Has your question been resolved?

[safe radish]

@royal folio Has your question been resolved?

[eager drift]

Still need help?

Treat the wall as two different shapes, a triangle and rectangle

Calculate their individual areas using the formulas for their respective areas

Then the total area is the sum of the two

[royal folio]

but how to find the area of the triangle

it’ll be 1/2 x 6 x ³√6

[quaint relic]

the base of the triangle is 12, the side is root 3, so we can find the 3rd length with triangle rules, such as law of sines to find the top angle, then 180- both other angles to find the other angle, then law of sines again to find the missing length.

Now draw an imaginary new right triangle that splits the top angle**. We know what the left angle and right angle is, 60 and 90, and we know the length of the upper left

now calculate the new imaginary triangle side length. using the law of sines again.

This gives us the triangles height.

[royal folio]

ouu

i got it

31,14+48

thank you

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[willow owl]

Why is this illegal

Is there anyway I can integrate it without the power reducing formula

[cedar void]

how did you got that 3rd line?

[willow owl]

sinudu= -cosu+c

[median vigil]

the product rule for derivatives says that the the derivative of f(x) * g(x) is not f'(x) * g'(x), so we can't do it in reverse either

[willow owl]

Sorry, I didn't kind of get that 😅

Oh

Ohhhhh

[loud osprey]

i think ur only option here is the power reduction formula

[willow owl]

Ohh okay, no more other options ig. Thank you

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[loud osprey]

i mean u can do integration by parts but

thatll involve more trig functions

and its overall just more complicated

[willow owl]

Yeah no way I'm doing that lmao. But thank you!

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[kindred scroll]

i didnt undrstood the first step of the solution

[rich whale]

They used the property x^(a+b)=x^a*x^b

like x^(1+a)=x*x^a

[fast mason]

They did (x^1)*(x^a)=x^(1+a)

[kindred scroll]

oh yes

i didnt thought it

now i got it thnx

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[west hedge]

help me describe the bar chart

How should I describe it.

The horizontal line shows the sales in billion pound?

fk

[mossy lotus]

This is for ielts. You have to describe it like which sector has the most sales, which has second most and so on. Bigger words and correct grammar is more important

No need for any maths description

Write stuff like ROW has been better in this, while UK in that and so on

[west hedge]

I want to note that the sales is indicated with billion pound on the chart

[mossy lotus]

Yeah, then just write that

[west hedge]

How should I describe the chart

[mossy lotus]

like "The sales are plotted on the chart, with their values in billions of GBP"

[west hedge]

ohhh

[west hedge]

ohhhh

"GBP", my man dropped the professional term.

I'm astounished.

[mossy lotus]

I mean its IELTS is bri'ish

you gotta use words like bottah o watah and such

(for all your study prep purposes, this was a joke)

[west hedge]

I see, you are a veteran.

thanks for your advice.

[mossy lotus]

I also had to write that exam a few years back

I understand your stuggle. GL with it tho

[west hedge]

I'm going to drop an overall 8, for sure.

[mossy lotus]

8 is easy, no worries. You got it

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[leaden harness]

What's wrong here

[leaden harness]

I solved this twice and I still got the same values

I don't get it

[little dust]

buddy none of us can read it

[leaden harness]

woops mb