help-0

[alpine sable]

Fr

?

[blazing widget]

Yes

[alpine sable]

I’m 13

[blazing widget]

Bro im in 5th

[alpine sable]

THATS CAP

No like fr

Tell me

[blazing widget]

7th

[alpine sable]

Dawg…

[blazing widget]

.close

[lone heart]

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[alpine sable]

U sure?

U understand?

[blazing widget]

Yes thanks

[alpine sable]

Ok

[chilly orchid]

Hi, spanish chat?

[lone heart]

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[hollow imp]

what does this mean??

[hollow imp]

it told us to calculate the hours or sth??

[lone heart]

@hollow imp Has your question been resolved?

[zinc haven]

do you have a question?

[hollow imp]

yesss

about the previous math question gr5th

here

what does this question request us to do

[zinc haven]

that question has already been resolved. unless you want to challenge the current solution, you should take this to dms or another channel

it is their question, not yours

[zinc haven]

open a new channel

[blazing widget]

for him

[zinc haven]

look, you should open a new channel. this is getting messy and hard to follow for potential helpers

[blazing widget]

my question was answered

im answering his

[hollow imp]

no,i just want to know the request of the math ques

[blazing widget]

this is the request of the math question

they want you to create an inequality

[hollow imp]

tysmm

[zinc haven]

then you should still not open a help channel for that. again, take it to dms

[blazing widget]

do .close

[lone heart]

@hollow imp Has your question been resolved?

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[sage yew]

The answer to this is D right?

sorry A**

[high imp]

<2,5> means moving x by 2 and y by 5, right?

[sage yew]

i mean this is about transformations

so i’m guessing so right?

@high imp

[fickle latch]

Is this a test?

[sage yew]

mock test

why?

[high imp]

If u mean moving x to another place by factor 2

[fickle latch]

Just curious. You seem quite urgent! And we can see the title in the top left corner.

[high imp]

It must be C

[sage yew]

No not urgent, I have until tommorow

[sage yew]

As I said this is a mock test

we have a real test on thursday

it’s just the graphs that are kind of getting me; and I can’t seem to grasp the concepts

[sage yew]

How come? I got A for some odd reason, D also as a second choice

oh my for

i’m so stupid

[high imp]

All A(0;0) would be A'(0+2 ; 0+5)

[sage yew]

i moved them 2 unity’s to the left 🤦‍♂️🤦‍♂️🤦‍♂️

which is negative

i just realize it’s a positive

[fickle latch]

I see. Alright then.

Well, yes, transformations can be quite difficult, if you cannot understand the denotations. <2,5> indicates both a horizontal and vertical translations.

[sage yew]

Yes I know that

my apologies

when it said 2, i did it 2 units to the lest

left

though it’s a positive im supposed to do it to the right

i forgot there’s negatives on that graph for a minute

wait a minute

i got answer choice D

@fickle latch @high imp

[high imp]

Just move point A twice to the right and five times up, and see the figure that matches these coordinates.

[high imp]

Look at B'

[sage yew]

oh

sorry

it’s C 😅

do you guys help on trigonometry by chance?

[fickle latch]

Of course. We help with everything. Forward your question.

[sage yew]

Not really a question but; I’m taking trigonometry next year, I’ve taken college algebra and it’s overall easy and easy to understand/grasp

my only problem is anything with shapes

don’t get me wrong i hate shapes

and i’m unsure what will prepare me for trigonometry

[round zealot]

formula page

its a lot of memorizing

[sage yew]

is it easy overall?

[round zealot]

I don't know what it will be like for you

are you good at math

[lone heart]

@sage yew Has your question been resolved?

[lone heart]

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[wintry wadi]

is it possible to solve for angle CAB here?
the question only gives CBA = 60° and asked for angle CBA + angle CAB
ik angle CAO = 30° but won't angle CAB depend on the location of B?

[high imp]

!original

[lone heart]

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.

[wintry wadi]

here, it doesn't show anything else
it says

if angle ABC = 60° and it is added to angle ACB, it will result in an angle of ... degrees

from the answer choice, it seems like 120° is the correct answer so ACB = 60°, bc the answer choices are
30°, 90°, 120°, 180°, and 200°
but no idea where that comes from

[high imp]

It could be 90

CAB

[wintry wadi]

yeah, if CB is parallel to CO

then ACB = 30

[high imp]

That's OCB

180 = 60° + CAB + ACB

CAB = 180° - 60° - ACB

[wintry wadi]

I meant if it was like this, then ACB = 30

[wintry wadi]

and then what though?

[high imp]

Question didnt give u 120° angle

Maybe it's asking about range?

@wintry wadi

[wintry wadi]

yeah, but that's a basic fact no? the central angle is double the ABC type of angle, idk what it's called

[wintry wadi]

the answer choices are just numbers

also I think this proves that CAB depends on the location of B

right?

[high imp]

So, the shape of triangle ABC doesn't matter.

Then, make ACB = BAC

Or ACB = 2 BAC

[wintry wadi]

I mean, it's not given, so yeah, ACB could be any angle in some range

so it's probably a typo then

[high imp]

The angle is variable, how could you know?

[high imp]

I think so

Or maybe it meant something else

The maximum value of the angle CAB

[wintry wadi]

yeah, it seems like the max is 120°

so 180 could be the answer

anw, I think it's just a mistake in the problem, so I'm gonna close this

thanks!

.close

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[silver tinsel]

I'm supposed to write this as a single logarithm, and I'm definitely doing e wrong. But I also don't know how to start..

[twin nimbus]

1/log_a(b) = log_b(a)

The above identity is true, but can you prove it?

[silver tinsel]

I'm not really sure how to

[twin nimbus]

Hint, log_a(a) is 1.

@silver tinsel Use the above along with the change of base formula

[silver tinsel]

Wait it makes sense now

Did I do it right??

Same question, but part c. The correct answer is log (1000a²/b) but idk how to get there

[stone coyote]

do uk the base of Log here

is it 'e', is it '10' is it '1' .....?

[silver tinsel]

10

[stone coyote]

ye

so 3 = 3 log_10 (10)

do u understand what im tryna tell

[silver tinsel]

I do, yeah!!!!

Thank you!

For d, is that a right way of simplifying it?

[stone coyote]

its wrong

[silver tinsel]

Oh dang

Cause that's what photomath showed, but I wasn't sure if it was right

[stone coyote]

nvm thats correct

mbmb

[lone heart]

@silver tinsel Has your question been resolved?

[silver tinsel]

Oop okay, thanks!

[lone heart]

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[sick vigil]

Hello, I'm working on integrated integrals and I'm having issues seeing the bounds for this one integral

[sick vigil]

How do I write latex here

Anyways, this is my integral, and I want to rewrite it as f(x,y,z) dx dy dz

I can instatly see that z ranged from 0 to 8, but I'm having a hard time justifiying the limits for y

[twin nimbus]

You have the order wrong

It's inside to outside

Not left to right

[sick vigil]

I'm not sure what you mean

I'm not evaluting the integral, just rewriting it, and I want new bounds

[sick vigil]

I see

However, If lets say I did want to write the integral in a differnt order, I know that 0<y<8 and 0 <z<y so 0<z<8, and now I need to argue y's bounds

[twin nimbus]

But that's not valid

Because z's bounds are not constant

They depend on y

If we are considering a region where y = 3 the. The bounds of z are 0 to 3

It makes a triangle in the zy plane

[nocturne veldt]

by putting dollar signs around the expression $\frac 12$

[ocean seal]

00100000

[sick vigil]

I'm not sure what you mean

I think we are able to assume that z is bounded by a function of x and y

so z is bounded by the xy plane and z=y plane

[tough scroll]

y varies from 0to8, x varies from yto8, z varies from 0toy

you want yo change the order to x then y then z

[sick vigil]

correct

[tough scroll]

The best way to do this in my opinion, is that this is a volume intgeral and limits are specified

try to imagine what shape is specified by the limits

And then easily you can find

[sick vigil]

Some sort of tetrahedrion

[twin nimbus]

https://www.desmos.com/3d/zdyvkcmavd

This is your area of integration, if it helps you visualize it

[sick vigil]

Ah, close

I'm not sure what to call this shape... anyways, I can see that z does range from 0 to 8, y ranges from z to 8

and z ranges from y to 8

hmm, I mean, I can figure it out when I have the graph, but my issue is I won't having a graphing calcuator on an exam

[tough scroll]

Algebraically 0<y<8, x<y<8 and 0<z<y

so 0<z<y<x<8

So x varies from 0 to 8, y varies from 0 to x and z varies from 0 to y

[sick vigil]

I see

I'm going to try another order and see if I understand now

So, the algebraic interpretation is alot more intuitive, however this is mostly bc we are dealing with pretty simple bounds, so I imagine this is just an easy case

[tough scroll]

$\int_{0}^{y}\int_{0}^{x}\int_{0}^{8}f(x,y,z)dxdydz$

[ocean seal]

Light Yagami

[tough scroll]

i think geomteric would be good for difficult bounds

[sick vigil]

My issue is that I'm not very good at drawing solid regions

I guess I just need to practice drawing more

[twin nimbus]

You've got three interesting planes, xy, xz, and yz. Try drawing just the intersections in these planes to help get an idea of how the shape is, well, shaped.

You can also project the change of the shape onto your 2d plane by using shading. Ultimately, it's a visualization aid though, so don't be too hung up on making it pretty, it just has to help your brain grasp it

[sick vigil]

I have aphantasia, so I can only visualise concepts, thats why the algebraic apporach is easier for me to see

Basically, if I don't draw it its impossible for me, and my drawing sucks 😄

[twin nimbus]

Well, practice makes perfect then!

[sick vigil]

Let me try drawing it

Oh, I saw the issue I had originally, I drew the wrong intersection

Anyways, thank you guys for the help, sorry for any confusions I might have caused 🙂

.close

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[vale wigeon]

that B is flying off somewhere

do you mean: $A^{\log_A(B)}$?

[ocean seal]

Ann

[vale wigeon]

can't help but notice you have every single pronoun role btw. what are your pronouns

[high imp]

Maybe log^B

[vale wigeon]

and also more crucially can you like, write on paper what you mean

if you somehow don't mean $A^{\log_A(B)}$

[ocean seal]

Ann

[vale wigeon]

that or post a screenshot/picture of whatever it is that you saw

[hushed locust]

,rccw

[dawn spruce]

just rip the paper in half bro what is that 🙏

[tired mirage]

i dont wanna sound dumb, but isnt that just b?

[hushed locust]

i think the only reasonable interpretation is that it is a typo and it should say [ a^{\log_a b} ]

[ocean seal]

cloud

[tired mirage]

ah, i just remember e^ln(c) (c is a constant)

e^ln(c) = c

and ln(c) is just log_e(c)

[high imp]

$a^{(Log_a)}^b$

[vale wigeon]

from where is it known that this is not supposed to be $a^{\log_a(b)}$?

[ocean seal]

Ann

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

[high imp]

B is up to log

[tired mirage]

ah, its the b that looks like a power

[tired mirage]

this

i think its a typo

[stiff trench]

then the log doesn’t have an argument which doesn’t rly make sense

theres needs to be smth the logarithm is logarithming

[tired mirage]

isnt log just log_10 in most cases?

ah, so it should be log_A

yea i think the editor messed up the latex for e)

i think they meant to type [a^{log_a(b)}

[a^{log_a(b)}]

[stiff trench]

u need backslashes before the brackets

[tired mirage]

$a^{\log_a(b)}$?

[ocean seal]

Potus

[stiff trench]

if ure just starting logarithms i think it should be $a^{\log_a(b)}$ but id ask if not

[ocean seal]

blahaquil

[stiff trench]

yea

[tired mirage]

alr nice

[stiff trench]

bc exponents and logs cancel out

[tired mirage]

bc thats one of the log properties

[vale wigeon]

... i think you're like

severely, SEVERELY overthinking

this bad typesetting

and producing more and even worse typesetting as a result

[tired mirage]

yup

that equals b

youre gonna encounter a lot of those if you pick stem (looking at $e^\ln(c)$

[ocean seal]

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

[tired mirage]

yeah, if log_a is the power of a

i can show you the proof if youre still confused

alr

first we have $y = a^\log_a(x)$

[ocean seal]

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

[tired mirage]

we can log both sides such that we obtain $\log_a(y) = \log(a^\log_a(x))$

[ocean seal]

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

[tired mirage]

$\log_a(y) = \log_a(a^{\log_a(x)})$

[ocean seal]

Potus

[tired mirage]

there we go

one of the log properties suggests $\log_a(u^v) = v*\log_a(u)$

[ocean seal]

Potus

[tired mirage]

hence we obtain $\log_a(y) = \log_a(x)\log_a(a)$

[ocean seal]

Potus

[tired mirage]

since $\log_a(a) = 1$ we finally obtain $log_a(y) = log_a(x)$

[ocean seal]

Potus

[tired mirage]

we can now cancel out $\log_a$ from both sides to obtain $y = x$

[ocean seal]

Potus

[tired mirage]

thus we obtained $y = {a^\log_a(x)} = x$

[ocean seal]

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

[tired mirage]

@sudden pelican all good?

we can assume a = 10 and get the same result if we used log_10

oh mb

its my preference, you can also use log_10 or ln and you will gain the same result

i used log_a to stay consistent with your question

@sudden pelican all good?

oh damn 💀 im so dumb

yeah, i used log_a to cancel out the a

wait

hmm

ok i remember now, you need the log_a to cancel out the a, if it was 10^log_10(b) i woulve used log_10 to cancel out the 10

yeah my mistake on my end

it has to be log_a to cancel out the a, and cancel out the log_a to obtain y = b

yup

exactly

youre gonna use that property a lot in stem

a lot of $e^\ln(a)$

[ocean seal]

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

[vale wigeon]

btw @sudden pelican just so you know our TeXit bot can (and does, as you saw) render LaTeX code

it is no doubt easier to type your code directly into discord than use whatever other app/service/etc

[tired mirage]

🤝

[vale wigeon]

also your shit is bolded for apparently no reason

wdym

[tired mirage]

@vale wigeon oh yeah quick question, if i have a $dt = ds$ can i just distribute the $\int$ to both sides to obtain $\int{dt} = \int{ds}$?

[ocean seal]

Potus

[tired mirage]

just need your opinion on this

[vale wigeon]

...why are you asking me this in another person's help channel

[tired mirage]

ah mb, ill make a new help question

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[limber willow]

hey, could someone solve this for me please? i want to see a worked solution as i'm not able to write anything, this is just to kinda revise for my test which is in an hour. ONLY PART C

f(x) is **4x+1/x+4
f^-1(x) is 1-4x/x-4

[vale wigeon]

could someone solve this for me please?
nope, we don't do that here sorry

[lone heart]

@limber willow Has your question been resolved?

[limber willow]

@dawn spruce thank you so so much, i was confused at the beginning steps (equating) as i wasn't sure how cross miltip'ication worked which was a huuuge issue, your calculations are rlly helpful!! tysm for helping

.close

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[fresh solar]

hi guys i'm not sure if this is a question i should ask here because i'm writing a short math discussion paper but i'm having some trouble with LaTeX formatting. There's a giant gap between my first equation f(z) = z = x+iy and the previous paragraph and i'm not sure how to get rid of the empty space there

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[snow merlin]

I am stuck at this question cause when i use L'hospotal rule I get 1/3 as answer but using desmos it says 1/2 and the answer key also say 1/2

[pulsar aspen]

@snow merlin show your entire attempt

[wind cloak]

Directly using L'H seems kind of nasty on this

I think you should simplify the inner integral using the tan(A - B) identity first

[serene scarab]

how is it nasty?

[wind cloak]

Because of the full Leibniz rule

[ocean seal]

neon

[serene scarab]

yea this uses modified

[snow merlin]

This is how I did it where did I do mistake

[wind cloak]

you didn't differentiate the integral correctly

[wind cloak]

refer to this

[snow merlin]

Oh but i get it thanks

.close

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[wind cloak]

Maybe the integral is actually doable I don't know though

[snow merlin]

Yeah I was doing it for 10 mins and I was stupid to not notice the mistake in differentiating

.close

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[tulip lagoon]

1. Consider the function , defined by

The area of the plane figure determined by the graph of the function , the Ox-axis, and the vertical lines and , with , is:

a)
b)
c)
d)
e)
f)

[tulip lagoon]

this is the problem

[lone heart]

@tulip lagoon Has your question been resolved?

[tulip lagoon]

<@&286206848099549185>

[slim kite]

Did you try expanding out the square and taking u = cos x

I think that'll remove the sin t and we'll get it in the form of 1/√(c+u^2)

[lone heart]

@tulip lagoon Has your question been resolved?

[tulip lagoon]

yes i already did that

the problem isnt integrating

i get this

[slim kite]

There's 2 x cost

U forgot the x

[proper tangle]

,rccw

[tulip lagoon]

OHHHH

damn

[proper tangle]

that will rotate your image counterclockwise by 90°, so that it's easier to read on desktop app

[tulip lagoon]

i didnt see it

i still dont know what to do after

[proper tangle]

,w int sqrt(x^2+1), x=0..1

[ocean seal]

Results provided by WolframAlpha

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

Requested by tkl1

[proper tangle]

it should involve exponential function

which is something transcendental

[tulip lagoon]

oh yeah mb

[proper tangle]

also you need "dx" at the end of each integral

[tulip lagoon]

yeah mo my bad i considered x^2 like a simple x

[tulip lagoon]

yeah ik

)))

[proper tangle]

you need a hyperbolic trigonometric substitution to evaluate such integrals

[tulip lagoon]

i think i can just integrate by parts

[proper tangle]

or just trigonometric substitution

[tulip lagoon]

yeah no i dont know how to do that one))

[proper tangle]

https://en.wikipedia.org/wiki/Trigonometric_substitution#See_also

or Euler's substitution
https://en.wikipedia.org/wiki/Euler_substitution

[tulip lagoon]

yeah okay thanks ill try using these

ill ask again for help if anything))

.close

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[alpine sable]

how would I go about solving this?

[alpine sable]

im not sure what that is

oh like (x+1)ln3?

and distribute?

[tight marten]

no

well yeah sure

[alpine sable]

is there an easier way or do i distribute for (x+1)ln3 = (2x+4)ln5

[tight marten]

ah i just meant you should do it to both sides first

but you got it

[alpine sable]

and why wouldnt raising both sides to the e work?

[tight marten]

you could

but this is the easier way

[alpine sable]

ok thanks for the help

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[alpine sable]

.reopen

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✅

[alpine sable]

just came across another problem im a bit stumped on.

so far ive gotten to 8x = 100 + 100sqrt(x)

but im not sure if there is a simpler way other than dividing by 8 and what would I even do after that?

condensed into: log(8x/1+sqrt(x)) = 2

exponentiated by 10 to remove log: 8x/1+sqrt(x) = 100

multiplied both sides by 1+sqrt(x) and ended up with 8x = 100+100sqrt(x)

obviously it looks like the next step is divide by 8 but after that how do i go forward?

what should I do?

yeah it would become something like 25/2 if im not wrong

how though

4? so make it 2x = 25+25sqrt(x)?

what do i do after that? am i forced to do the quadratic formula?

yeah

how exactly does that work with a square root though?

ax^2 + bx + c = 0 no?

there is no square though

let me try that then

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[gaunt mirage]

What is 6 Squared Multiplied by 3 cubed?

[gaunt mirage]

So if cubed is just multiplying 3 times would leave us with 3 X 3 X 3

So if 3 X 3=9 that would give us 9 X 3

Which is 27

Then 6 squared is 36 so 27 X 36=972

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[tulip bone]

can somebody explain uh

[tulip bone]

injective bijective and surjective functions

im having trouble understnding

[sharp skiff]

do you know one one functions

?

[tulip bone]

yes

one x to one y value

[sharp skiff]

Exactly

That's called injection

[lament wyvern]

injective: one to one
surjective: covers the whole range
bijective: both

[tulip bone]

remember kids

make sure ur relationships are like injective functions

[lament wyvern]

slightly longer:
injective: different inputs means different outputs. if you have an output then you know there is only 1 possible input associated with it
surjective: for any possible member of the codomain, there is some input that will produce it
bijective: both, which means for every member of the codomain there is exactly one input that could produce it. i.e. the reverse is also a function

(exercise: explain why the inverse of an injective but not surjective, or surjective but not injective, function is not a function.)

[tulip bone]

wait whats bijective

no im with my girlfriend for a year

[lament wyvern]

bijective === injective + surjective

[sharp skiff]

One one onto

[tulip bone]

more even

[tulip bone]

and surjective?

[lament wyvern]

surjective: covers the whole range

surjective: for any possible member of the codomain, there is some input that will produce it

[tulip bone]

well yes i read that

but its too technical

what does it mean

like lets say diagramatically

[lament wyvern]

if your function is f : A -> B, and f is surjective, then for every element b in B, there exists some a in A such that f(a) = b

you cover the whole range

a nonsurjective function would be something like f: R -> R where f(x) = 5 for all x. there is no a such that f(a) = 6, for instance

[tulip bone]

the whole range but not the codomain?

[lament wyvern]

oh sorry. the whole codomain

range = codomain for a surjective function

[tulip bone]

not necessarily

[tulip bone]

OH

[lament wyvern]

it is now I've edited it 😉

[tulip bone]

now it makes better sense

[lament wyvern]

good catch though, thank you

[tulip bone]

so if b has elements lets say 1, 2 and 3, a has elements x , y, then its not bijective since one of the elements doesn’t have a match in A

[lament wyvern]

f : A -> B where |A| = 2 and |B| = 3 cannot be surjective because there arne't enough. possible as for each b, yes

[tulip bone]

alrighty i think ive a fair idea, ty

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[eternal hare]

hi

[eternal hare]

how would i answer this in the form aπ+b

i got my answer as 61.5π it is probably wrong though

[lament wyvern]

can you write an expression for the perimeter of the two pieces individually

[earnest slate]

did you add the two radii in the quarter part

[eternal hare]

sorry but im not sure how to

shall i send a pic of my working out?

[lament wyvern]

yes

[earnest slate]

2 times 21

[eternal hare]

ah so its 42 my bad

but how do i get +b?

[lament wyvern]

general thoughts - it's better to write out what you are doing, not just what you've calculated

2 pi 21 is great except....2 pi 21 is what exactly? where does the 1.5 come from? (but krishil has this)

[earnest slate]

so what is 42(pi)

ok

[eternal hare]

yeah i think the 1.5 is wrong but i did 1/4 of 6 cause its a quarter circle

[eternal hare]

the circumference of the circle

[earnest slate]

yeah the bigger circle

[eternal hare]

yep

[earnest slate]

now what is the radius of quarter

[eternal hare]

is it just 6mm or would i have to quarter that

[earnest slate]

its 6 mm

[eternal hare]

👍

[earnest slate]

radius is distancw from centre to circumference

[eternal hare]

wait so for the +b

[earnest slate]

you will have to 1/4 the circumference

[eternal hare]

would i just do 2π6

substitute 3 for pi

and add that to 42?

well 42π

[earnest slate]

you also have to add these 2 radii

[eternal hare]

OHHH

thank you so much

[earnest slate]

no

dont substitute that

[eternal hare]

wait could i quickly do that very quickly then send you my answer?

[earnest slate]

sure

[eternal hare]

45π+12?

[earnest slate]

yes

[eternal hare]

thank you for the help

[earnest slate]

nws

[eternal hare]

.close

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[tulip lagoon]

i tried at a) to expand sin2xsin3x

in the end i got sumn like sin^4(pi/12)(cos^4(pi/12)-1)

and there has to be an easier way to do this exercise then this

and at b) i have no idea

maybe just converting cos^2 int 1-sin^2

but still idk

[bleak heath]

in a) try using the formula $sinxsiny= (cos(x-y) - cos(x+y))/2$

[ocean seal]

Aria

[tulip lagoon]

i dont really see where this gets me

[bleak heath]

it'll make the integration easier

[tulip lagoon]

oh okay fair yeah ur right

and then i use sinxcosx form

[bleak heath]

Use this formula to convert sin2xsin3x

[tulip lagoon]

yhyh got it thanks

its a lot easier after i do that

[bleak heath]

Yeah

[tulip lagoon]

i dont have sin(pi/12) i have onli pi/6, pi/3, pi/2

and at b)?

i have 0 clues

[bleak heath]

Uhh wait

Maybe use the formula $cos^2(x)=(1+cos2x)/2$

[ocean seal]

Aria

[lone heart]

@tulip lagoon Has your question been resolved?

[tulip lagoon]

and then cos(2narccos(x))?

[bleak heath]

you can write 2arccosx in the form arccos(2x^2-1)

but I don't know what we can do of narccos(2x^2-1)

[tulip lagoon]

<@&286206848099549185>

b)

<@&286206848099549185>

[lone heart]

@tulip lagoon Has your question been resolved?

[tulip lagoon]

<@&286206848099549185>

[tacit arch]

what do you get after you use this

[tulip lagoon]

cos(2narccos(x))

1+ integral of cos(2narccos(x))dx

[tacit arch]

did you try double angle identity

[next stratus]

double angle would just give the same thing back tho

[tacit arch]

do you know what double angle identity is

[next stratus]

cos2x?

[tacit arch]

that's not an identity

[next stratus]

cos2x=2cos^2x-1

[next stratus]

its this but rearranged

[tacit arch]

oh yes i was thinking of the sin(2x) one

[next stratus]

but uhh he has cos, there isnt any sin unless your asking him to multiply and divide by sin^2(narccosx) but i feel that would just complicate it

[tacit arch]

try usub with $u= \arccos(x)$

[ocean seal]

riemann

[alpine nacelle]

yeah that + double angle identity + other trig identity and you get back to primitives you can calculate

[next stratus]

huh but don't you get integral -cos^2(nu) sin u du, how do you deal with the n

[tulip lagoon]

exactly

[tulip lagoon]

i tred

tried*

[next stratus]

@tacit arch?

[tulip lagoon]

alrdy

[tacit arch]

similar to this one but with different left side

,tex .prod2sum/sincos

[ocean seal]

riemann

[tulip lagoon]

i tried sumn like this alrdy but only get -cos((2n+1)u)/((2n+1)*4) from 0 to pi

and also cos(2n-1)...

[tacit arch]

can you just show your work then

[tulip lagoon]

2n-1 where is 2n+1

[tacit arch]

the entire steps

[tacit arch]

starting here

[tulip lagoon]

yeah one sec its written everywhere

lemme write it again and ill show u

idk what to do from here

[next stratus]

btw there should be - instead of + bw the two terms here

[tulip lagoon]

why?

[next stratus]

its x-y not y-x

2sintcosnt=sin(t+nt)+sin(t-nt)

[tulip lagoon]

oh yh ur right

mb

[tacit arch]

why can't you evaluate this

[tulip lagoon]

looks too hard😓

idk you think this is the only way?

[tacit arch]

why is it hard

[tulip lagoon]

it looks to complicated to only be solveable like this

[tacit arch]

what part of evaluating this is hard exactly

[tulip lagoon]

cause i also need to fit it to In=254/255

[tacit arch]

yea that's later

if you keep jumping around in the problem i can't help you

[tulip lagoon]

fair ill evaluate that

but wdym by keep jumping

[tacit arch]

.

[next stratus]

cosine of an odd multiple of pi is -1 and cosine 0 is one, that should be enough to help you evaluate

[tacit arch]

i wasn't asking you

[next stratus]

sorry wrong reply

[tulip lagoon]

oh yeah i forgot n is natural

yeaaaah

okay

yeah

thats what i was missing

yeah thanks a lot

yeah i got to the end of it

[tulip lagoon]

also here the sin(t) reduced because i wrote + instead of -

so i also had another -1/2cost from 0 to pi

which is another +1

and i got at the end 2n-2/2n-1=254/255

thank you very much @tacit arch and @next stratus 🙏 🙏

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[odd field]

Yo

[odd field]

I’m new to this and I struggle with maths. Is this right place here?

[warm quarry]

But yes this place is really good

[vale wigeon]

do read #❓how-to-get-help and #rules

but otherwise yes

[warm quarry]

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[vale wigeon]

and post your question(s) here

the more specific, the better

[warm quarry]

Just ask your stuff here and there is plenty of people that are willing to help you :)

[odd field]

Thanks everyone when I go home I’ll send a few things I want to look at

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@odd field Has your question been resolved?

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[austere veldt]

Hi guys,can someone please help with this question?

[topaz sage]

What's the question? A question contains a question mark (?).

I have even asked ChatGPT but it too can't find a question mark.

[next stratus]

i think they meant help simplify

[topaz sage]

What's the question?

I will call the moderators if you don't respond in 5 minutes.

[next stratus]

actually convert tan to sin and cos first will help much better

[austere veldt]

Yeah

[austere veldt]

I can't learn like that(.Can you show it?

[next stratus]

did you try converting?

[austere veldt]

...I know the answer,but need to solve

[austere veldt]

Yeah

But it didn't help

[next stratus]

now you see the 1 add it to the fraction obtained from tan

then try using a trig identity on the numerator

[austere veldt]

Hmmm

Let me try

I got this

What I supposed to do?

Oh

I found it

Zero, actually

Thank you brother @next stratus

[next stratus]

ur wlcm

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[upbeat pilot]

Let ABC be a triangle with ∠BAC = 70°, ∠ABC = 50°, and M be the midpoint of side AC. Let P and Q be points on lines BC and BA, respectively, such that the circle passing through P, M, B is tangent to BA and the circle passing through Q, M, B is tangent to BC. What is the measure of the angle ∠PQB?

(A) 60°
(B) 70°
(C) 80°
(D) 90°
(E) 100°

[topaz sage]

draw diag first

[upbeat pilot]

I already know that the arc BP and the arc BQ measures 100°

[upbeat pilot]

What diagonal mate?

[topaz sage]

diagram

[upbeat pilot]

ah

[upbeat pilot]

Yeah, I drew it in my notebook

[topaz sage]

show it to us here by sending it here

[upbeat pilot]

It is kinda bad, but here you go

[topaz sage]

,rotate

[upbeat pilot]

<@&286206848099549185>

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@upbeat pilot Has your question been resolved?

[upbeat pilot]

.close

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[formal dew]

why if 2^n-1 is prime then n is prime without using the binomial summation ? like is there another proof?

[vale wigeon]

contrapositive

if n = pq with p and q both >1 then 2^n - 1 factorizes

via, say, the geometric progression sum formula

[formal dew]

isn't this the summation widely known proof ? I'm maybe making it harder a bit 😅 because I'm searching for a proof mainly using modular arithmetic

[vale wigeon]

2^pq ≡ 1 (mod 2^p - 1) or something

if you want modulo stuff

[vale wigeon]

still with this underlying logic tho

[formal dew]

yes I got this part as a given information but how to deduce the rest using this exactly is what I am struggling with

I am not understanding the reasoning behind clearly, it's a bit foggy

[vale wigeon]

do you know that if a ≡ b (mod m) then a^k ≡ b^k (mod m)

[formal dew]

yes of course

[vale wigeon]

2^p ≡ 1 (mod 2^p - 1) is clear

raise both sides to the q'th power

win

[formal dew]

this is not the part that's foggy, but the summation stuff and how to do it if I wanted to not use it (2^n-1 is prime then n is prime), the whole internet is showing the summation proof while I am aiming to use another proof to solve this according to our curriculum

[pallid scarab]

I don't see what part of Ann's proof is foggy

She didn't use any summation stuff

Proof by contrapositive: if n is composite, then 2^n -1 is composite

say n = pq

then 2^p = 1 (mod 2^p-1)

2^(pq) = 1^q = 1 (mod 2^p-1)

2^n - 1 = 0 (mod 2^p -1)

2^n -1 is divisible by 2^p -1

2^n -1 is composite

where is the "summation stuff"?

[formal dew]

I thought she was talking about (a^p-1)((a^p)^q-1+(a^p)^q-2+...+a^0-1) 😅 because it's a widely known proof

[formal dew]

thankkk you so much for the explanation

both of you ^^

[pallid scarab]

thank Ann, I just regurgitated her proof xd

[formal dew]

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[crimson tendon]

do you know the answers?

@acoustic lance

[acoustic lance]

I believe it is 11

[crimson tendon]

single correct?

or multiple?

[acoustic lance]

Well, the answer sheet just says D

[crimson tendon]

i see

[real minnow]

[lone heart]

@acoustic lance Has your question been resolved?

[acoustic lance]

<@&286206848099549185>

[acoustic lance]

Does anyone know how to do this? 💀

[upbeat hornet]

Yes

You can substitute each k into the polynomial to get four linear equations, which you can then solve

[upbeat hornet]

or you can consider f(x) - 2024x. This polynomial has roots at 1, 2, 3, and 4. So you can tell that f(x) - 2024x is divisible by (x-1)(x-2)(x-3)(x-4). Since f(x) - 2024x is a quintic, that leaves a factor of p(x-c) for some reals p and c.
But since the coefficient of x^5 in f(x) is 1, we must have p = 1, and then this determines f(x) up to one degree of freedom. Then you can get some linear equations and solve

[acoustic lance]

Oooo, thanks for the idea

[upbeat hornet]

Do you know about polynomial interpolation? This uses similar ideas

[acoustic lance]

My current knowledge is only limited to highschool maths

[long basin]

Haha

[ancient carbon]

somebosy please help bru khan academy got me fucked up

[long basin]

Well I did not get it so I went for chat gpt

[long basin]

Its 15

[ancient carbon]

im actually so dumb im using a discord server for math

[acoustic lance]

But the answer sheet says it's 11

[upbeat hornet]

Oh, the idea of polynomial interpolation is relatively straightforward: for any sequence of numbers, say 1, 2, 4, you can make the next number be anything you like, according to a polynomial formula, by just solving a system of linear equations

[long basin]

But gpt bro says 15 with solution

[acoustic lance]

I dunno man, gpt could be wrong sometimes

[long basin]

It wasn't a normal question fr some complicated shit

[upbeat hornet]

in this case you can let p(x) = ax^3 + bx^2 + cx + d, then solve for a, b, c, d given that p(1) = 1, p(2) = 2, p(3) = 4, and p(4) is whatever you want

so you can make a cubic go through any four points

[upbeat hornet]

gpt can be confidently incorrect

[long basin]

Give an exact number does that mean in integer?

[ancient carbon]

idk probably

[upbeat hornet]

please go to another help channel, like #help-38

[ancient carbon]

okay im sorry guys..!

heh.

[long basin]

Ig Its 1.2

[long basin]

1.2

[ancient carbon]

ty king

[acoustic lance]

Cool, I'll go do some more research myself, thanks for the insight, I think this channel have been open for long enough, I'll close it now

.close

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[warped topaz]

Im working with some weird numbers, which makes me think ive made a mistake 🤔

First to find the intersection on the y axis set x = 0

cos(2[0]) + sin(3[0] + pi/3)

cos(0) + sin(pi/3)

(1) + (sqrt(3)/2)

= (2 + sqrt(3))/2

[vale wigeon]

so far so good, that's your g(0)

[warped topaz]

Then to find the slope of tangent line l at point A we take g'((2 + sqrt(3))/2)

[vale wigeon]

nope

g'(0).

[warped topaz]

Ohhhhhh

Thats right

[vale wigeon]

not g'(g(0))

[warped topaz]

Ok I think that was the root of my issues 1 moment I will see if I can solve it myself

Yep got it 😄

Thanks!

❤️

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[tough lintel]

So is it good of me to think I can construct a subset S of the power set, such that if x is in that S there is no function phi : A -> P(A) which will map elements to S?

[upbeat hornet]

Could you elaborate or use logical symbols to say that?

[tough lintel]

Im not very good at this, and im just doing it as a hobby.

Im not good at writing up math either.

However, I think since the cardinality of A is less than the cardinality of |P(A)| then its obviously not surjective IMO.

Therefore there is a subset S of the power set

in which there in no mapping for the x in S

where phi: A -> P(A) exists in which phi(x) in S

i guess i could be more formal and define s as

S = {x in A | x not in phi(A)}

i was kinda naively thinking theres no inverse maping for a set in S which is in the powerset of A for a mapping phi

[tough lintel]

does that make sense?

[upbeat hornet]

regarding the last line of this message, that is the very thing you are required to prove

it would be good to write your Assumptions at the top and the thing you eed to prove at the very bottom, so you don’t accidentally mix them up and get circular reasoning

[upbeat hornet]

Not really. I still don’t know the definition of phi is

[tough lintel]

its just a random function that maps A to the power set of A

[upbeat hornet]

Try carrying out your definition on a concrete example, like A = {1, 2}

[upbeat hornet]

but I think this idea is the idea you need, after looking more closely

you just need to turn that into a rigorous proof

[tough lintel]

is it not rigerous enough saying

like

i mean i think if say theres a random function that is this mapping then is that rigerous enough?

[upbeat hornet]

you have defined S, but now you have to somehow prove that the existence of S means that the cardinality of A is strictly less than the cardinality of P(S)

[tough lintel]

i agree, this is because theres no function thats surjeective from A to P(A), which means there is an set in P(A) which doesnt have a connection through a function with A

and if it did then it would be a contradiction by how i defined S lol

but im prob going circular as you said

[lone heart]

@tough lintel Has your question been resolved?

[upbeat hornet]

this is because theres no surjective function from A to P(A)
How do you justify this statement?

[tough lintel]

because you can make a set s which doesnt have a functional mapping. Therefore this isnt onto, there isnt a mapping from A to P(A) that is onto

And if anything its related to the fact that the cardinatlity of A is less than its powerset.

so this means theres no x that can be mapped to S

unless im off lol. sorry not good at this

[upbeat hornet]

what do you mean by “have a functional mapping”? do you mean “there is no element of A mapping to that set”?

if so, then you need to justify that first statement

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@tough lintel Has your question been resolved?

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[tame fox]

Hi, could someone tell me if the set of all real numbers is a subset of itself? If not, is there any way for it to be?

[vale wigeon]

any set is a subset of itself.

[exotic canopy]

$A\subseteq A$ for any set $A$

[ocean seal]

artemetra

[vale wigeon]

R is not special in this regard.

[tame fox]

Ohhh i see, a friend asked me yesterday and i told her that its not. Ill know better not to say that anymore. Thanks!

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[native lagoon]

If the point P on the curve, 4x^2 + 5y^2 = 20 is farthest from the point Q(0, −4), then (PQ)^2 is equal to
(A) 21 (B) 36 (C) 48 (D) 29

[native lagoon]

i tried it and my answer is 64 which is not given in any options

[molten spire]

Ans 36

[proven depot]

send method

anyways take x = acos(t) and y = bsin(t)

[native lagoon]

I did this

[native lagoon]

How?

[proven depot]

find PQ^2

you will get a quadratic in terms of sin after you convert everything into sin

[molten spire]

@native lagoon
Wait am explaining you in message

[native lagoon]

I will try that too

[native lagoon]

Okay

[slim kite]

See, try drawing the diagram first

And go by logic (if it is allowed)

[native lagoon]

Yh it's mcqs only

[slim kite]

Then draw the diagram first

And locate Q

[native lagoon]

Let me try

[molten spire]

Ya I share you just check out message

[native lagoon]

Got it thanks

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[alpine sable]

help b

i genuinely lost

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[coarse elm]

Optimization gr 12 calculus

The answer is: 14 062.5 m^2 and dont know what to do

oh wait i asked google and it showed me the steps lol

i dont understand why its 750 = 2l + 5w

clearly by my drawing its 8l

[velvet veldt]

you divide up each length into 4 segments, but you don’t need to do that

[coarse elm]

but the original question says whats the area of one box

so i thought I have to divide it up so its length * width

[velvet veldt]

i mean it works, but is just more algebra

[coarse elm]

oh its still suppose to work??

why didnt mine work then

🥲

[lime bobcat]

Everything seems fine but the units are 3,2 m²

If there is a mistake I don't see it

[coarse elm]

but the answerkey says this

oh nvm i think i found my mistake

[lime bobcat]

Where is it?

[coarse elm]

ugh lol still not getting the answer

this part

i just need to multiply -150 by 5 and then divide it by -16

[lime bobcat]

750/16

[coarse elm]

yup

so 46.875

but if i still plug that in i get 3515 m squared bu the answer is 14 062.5

[lime bobcat]

Multiply 3515·4

😉

[coarse elm]

OH!

But isn't that not fair though

The question clearly states, what's the largest area of each box?

"of each"

😒

[lime bobcat]

I know but problem posers make mistakes hahaha

[coarse elm]

Oh well

Thank you so much for noticing that though

[lime bobcat]

I solved many problems like that and most of the times the solution asked is the dimensions of the bigger fencing

[coarse elm]

Thank you!!

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[proud wyvern]

where am I going wrong

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[topaz ridge]

ignore the top half of my work, look at the bottom one in pencil. The second image is the problem - a vectors assignment. I don't know what im doing wrong and I keep getting them wrong.

[violet jetty]

w = <4,-5>, not <4,5>

[topaz ridge]

oh

so then b = 3 and a = 7?

i still got it wrong 🥲

[violet jetty]

show your work

[violet jetty]

how is -5 - 4 = 9?

[topaz ridge]

dont two negatives add together?

[violet jetty]

they multiply together. -5 * -4 = 20. But -5-4 is just -9, it is four less than negative 5

you might be confusing it with -5-(-4) which is equivilant to -5+4

But here you have -5-4

[topaz ridge]

so it would -9 then divided by 3 which would be -3??

[violet jetty]

looks like it

[topaz ridge]

okok

and then a would be positive 1?

[violet jetty]

yeah

[topaz ridge]

kk ty

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[alpine sable]

Hello

[mental python]

hi

[forest marsh]

Hi

[mental python]

do you have a question?

[alpine sable]

Yes,

How many five-digit numbers of the form 𝑎𝑏𝑐14 can be made:
a. 488
b. 388
c. 688
d. 588

[alpine sable]

(This is a question from my entry exam that I can't solve with 5 other people...)

currently in a call, we can't do anything to it

[mental python]

is that the whole question?

[alpine sable]

yes

[mental python]

ohh, i see what theyre getting at

whats your first instinct

of amount of possibilities

[alpine sable]

We somehow got to 578 but never to any of the actual answers

ChatGPT also doesn't help

[mental python]

thats not what i asked

[mental python]

!nogpt

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

[mental python]

!show

[lone heart]

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

[pulsar kernel]

idgi

of the form abc14 ? any more info

[alpine sable]

There isn't

[pulsar kernel]

abc are supposed to be numbers 0-9 ?

[alpine sable]

that's it

yes

[mental python]

||a cant be 0||

[alpine sable]

true

[mental python]

bro thats not for you smh

[pulsar kernel]

Oh

[olive sentinel]

this doesn't look correct

options do not match up

[alpine sable]

Also, abc can't be the same, any of them, it can't be aab14, for example 11314

[olive sentinel]

ahh

[mental python]

.....