help-36

[maiden rapids]

.close

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[shell condor]

.I didn't even do much😭🥲 (still thanks man)

[drowsy epoch]

.sufficient a existence

[proper lion]

i was not mentione still thx

[drowsy epoch]

@proper lion u were a ghost

[somber fog]

Have a nice day!

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[spiral olive]

Hello, can anyone help me solve this problem ?

[spiral olive]

It is wrote in french, so here is what does it mean in english :
The figure shows two perpendicular segments inside a square.
Three lengths are indicated.
What is the length labeled h in the figure?

[versed crater]

What have you tried

[spiral olive]

Good question, nothing since i am not having ideas, I am not able to find a relation or something like that, i tried make equations, it didnt worked, then tryed to maybe apply pythagore theorem, but I am not able, I am struggling idk what to do

[versed crater]

Hmm

[versed crater]

If I set some numbers to 0

Do we see any patterns

How does the side length a affect the ? section

[spiral olive]

I dont get it

What do you mean by a, its 7.7

[versed crater]

I’ve shifted that diagonal so the 4.7 is now 0

[spiral olive]

4.7?

You mean 7.7?

Also, when you mean shifting the diagonal, you mean that ?

[spiral olive]

Is it like that ?

Sorry guys if its easy and i dont find it, but I just really dont get it idk why

[versed crater]

I also shifted the vertical looking one

I wanted you to look at the problem in general

[spiral olive]

Oh

Wait

[proper lion]

you can consider one of the vertices as origin then write an expression slope of the lines

for eg if you consider top left point to be origin, let side length of square be L so
F =(4.8, 0)
A = (L,0)
G = (L,-h)
B= (L,-L)
C =([L-7.7], -L)
D =(0,-L)
E = (0, -[L-6.2])

you can fin the slope of line FC let it be m , slope of other line would be -1/m

at the end your L terms cancel out and you find h

lengthy method

[spiral olive]

I dont get it for the moment, but i will get it dont worry

[proper lion]

alr the method which frosst suggests is more suitable ig this is just bruteforce

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[proper lion]

<@&268886789983436800>

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[spiral olive]

Answer D ; anyways thx @versed crater and @proper lion ; I still wonder why didn't I have the idea you had @versed crater, like i dont think i even had the idea to kind of "move" both diagonal, i was just focusing on triangles or equations ..

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[pliant coral]

I need help with completing a square

[pliant coral]

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[earnest shuttle]

bro i know the answer is c but like how did they go from the first step to having x times 4^4x^5 + 1 factored outside

[earnest shuttle]

oh yeah ur asked to find the derivative of it

[drowsy epoch]

i mean each summand has a factor $4$ and $4^{4x^5}$ and $x$

[leaden spire]

what do you have to do here? choose the correct option that shows the correct steps and solution?

[earnest shuttle]

oh wait ur rigfht

[earnest shuttle]

yeah you ahve to choose the one that is the correct derivative

[leaden spire]

then just use the product rule of the differentiation, assuming that you know it, so you can figure out the answer

[earnest shuttle]

ya i did that but idk why the final has 4x^5 + 1 in it

[drowsy epoch]

,, 4^{4x^5} \cdot 4^1 = 4^{4x^5+1}

[leaden spire]

4^(4x^5 + 1) = 4^(4x^5) * 4^1, so maybe they are also factoring out 4 too

[earnest shuttle]

ohh i see

ok thanks man

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[late gazelle]

For this question: Show that $f(x)= x\pi - \sin(x)$ is a increasing function for all values of x. I get $f'(x) = \pi - \cos(x)$ but im not sure how to go from here

[soft zealot]

KB

[somber fog]

Ok, what ideas you have?

[late gazelle]

i mean, if its always increasing than the derivative should always be positive ? or at the worst going to 0 and then back to positive

[somber fog]

Yeah, you're right.

[late gazelle]

so do i solve: f'(x) > 0

[somber fog]

Exactly!

[somber fog]

Do you have more questions?

[late gazelle]

so $x > \arccos(\pi)$ ?

[soft zealot]

KB

[late gazelle]

hmmm, thats not a number

[somber fog]

You cannot calculate $\arccos(\pi)$.
Because the arc cosine function only accepts values between -1 and 1. And pi is greater because it is 3.14..

[soft zealot]

Ga³¹Br³⁵I⁵³9000✞

[late gazelle]

or going back to cosx > pi

x = -1

• 2pik

wait

cos pi = -1

not the otherway around lol, mb

how is this possible?

pi-cos is increasing and decreasing

like, looking at graphs of it it is a oto function. but idk how to show that... Considering i dont know how to solve this inequality

[somber fog]

Think that this inequality cannot be resolved by clearing, it is resolved by dimensioning.

😉

[late gazelle]

what is that

[somber fog]

The largest value that cosine can have is 1. Right?

[late gazelle]

yra

[somber fog]

$$\text{Minimum of } f'(x) = \pi - 1$$

[soft zealot]

Ga³¹Br³⁵I⁵³9000✞

[late gazelle]

oh

i think i get it

[somber fog]

Yeah!

[late gazelle]

wait no nvm

bc this is the rate of change

oh wait

its never 0 or less than 0

yea no this makes sense now

[somber fog]

Yeah, here is.

Do you understand now?

[late gazelle]

yea

its wired tho, we never do this sort of thing

[somber fog]

Do you have more questions?

[somber fog]

Yes, is strange but it's good to know.

[late gazelle]

ill probably be like on and off, some of this looks fine until i get stuck like on this one

[somber fog]

Np, it's normal. If you have a question at a later time, you can open another channel, and I will help you.

🙂

[late gazelle]

ty for the help

.close

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[somber fog]

You're welcome. Have a nice day!

[late gazelle]

.reopen

[final saddle]

✅ Original question: #help-36 message

[late gazelle]

For this question: Using decomposition, show:

$h(x) = (\pi-x) - \cos(x)$ is oto

[soft zealot]

KB

[somber fog]

What have you tried?

[late gazelle]

well i decomposed it

so i got f(x) = pi-x

and g(x) = x+ cosx

then i took the derivative of both

and f'(x) is always decreasing, and g'(x) is always increasing

but idk what to do with that information to show h(x) is oto

[somber fog]

Ok, send what did you get.

[somber fog]

Think about: $g'(x) = 1 - \without(x)$. Like $g'(x) \geq 0$. And $f(x)$: It is a straight line with a slope of $-1$. Being a line, it never repeats values.

[soft zealot]

Ga³¹Br³⁵I⁵³9000✞
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[somber fog]

And h(x) is the composition of two functions that are individually one-to-one, right?

[late gazelle]

yea

so in general, if f and g are oto, then the composition of f and g is also oto?

[somber fog]

Yeah.

Because g goes up and the other always goes down, when combining them there is no way for the function to retrace its steps. Right?

[late gazelle]

i suppose

[somber fog]

Do you understand better?

[late gazelle]

i guess yea

[somber fog]

🙂

[late gazelle]

I got another question, write an increasing oto function, that satisfies f o g = g o f. is it just f = g and f,g = x

i mean, its boring but it does work right ?

[somber fog]

What have you tried?

[late gazelle]

f,g=x

[somber fog]

Yeah, it's boring, but you are right.

[late gazelle]

it doesnt have to be a fancy solution, it just wants an example

[somber fog]

Then, it's correct.

[late gazelle]

hmm ok. ty

[somber fog]

Do you have more questions?

[late gazelle]

nah, ive finished this chapter. unless I want to move onto inverse functions which I might? idk, i got derivatives to revise now (test on tuesday)

[somber fog]

Review derivative and if you later have doubts about inverse functions you can open another channel. Good luck with derivatives.

😉

[late gazelle]

ty

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[gritty flume]

Number b3)

[somber fog]

I see, what's your question?

[gritty flume]

It’s wrong

[final tangle]

how are you getting that last lime

[unkempt rune]

cos(180 - a) => cos(90+90-a) => cos(90-(-90+a)) => sin(a-90)

Substitute a as theta

do th same for rest

[forest veldt]

hi lol

[final tangle]

you have something of the form
pq + rs
But you seem to be treating that as
p + q + r + s

[drowsy garden]

Isn't it supposed to be 0?

[gritty flume]

Yeah I looked at the wrong answer

[unkempt rune]

its 1 dude

i mean for b1

tan(180+ a) == tan(90+90+a) == tan(90-(-90-a)) == cot(-(90+a)) == -cot(90+a)

[final saddle]

@gritty flume Has your question been resolved?

[unkempt rune]

in second part tan(180-a)== tan(90+90-a) == tan(90-(-90+a)) == cot(a-90)

cot(-(90-a))== -cot(90-a)

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[next thorn]

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[bronze grotto]

$\frac{1}{sinx} * \frac{1}{sin2x} * \frac{1}{sin3x} * \frac{1}{sin4x}$ \\
Where x = $\frac{\pi}{9}$ \\
How do I show that this equals $\frac{16}{3}$ without just entering this into a calculator?

[soft zealot]

glaedr_

[keen storm]

calculating it with pen and paper

[bronze grotto]

um well yes but what I meant was can I get fractional outputs from those sin functions by tweaking around with the argument of those sine functions

[keen storm]

u could firstly make it all into 1 fraction

[bronze grotto]

how so

[keen storm]

1/sinx * sin2x * sin3x * sin4x

[bronze grotto]

.....

[keen storm]

?

u could do that

[sand badge]

sin 3x = sin 60deg = root(3)/2

Now

This is an identity for product of sines

[bronze grotto]

wow

thanks so much @sand badge 🙂

[sand badge]

It is a pretty standard question in JEE mathematics

And there they wouldn't ask us to prove it but instead they'd ask us to find the answer

[bronze grotto]

.close

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[foggy ingot]

.reopen

[supple wren]

how does x+1/xe^x+e^x factor down to e

[loud sundial]

$xe^x+e^x=e^x \cdot x+e^x \cdot 1=e^x (x+1)$

[soft zealot]

Civil Service Pigeon

[supple wren]

x = -1

[loud sundial]

$$\implies \frac{x+1}{xe^x+e^x}=\frac{x+1}{e^x (x+1)}=\frac{1}{e^x}=e^{-x}$$

[soft zealot]

Civil Service Pigeon

[loud sundial]

ok plug this in then

and you get e

[supple wren]

but how do we get (x+1) on the bottom too?

I'm just totally confused on how it arrives at that conclusion

[supple wren]

sorry I missed that

okay thank you

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[loud sundial]

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[wicked imp]

Could someone verify if this is correct?

Problem: Use row reduction to solve the linear system whose augmented matrix given
as below.

Give me a moment to show my solution:

[loud sundial]

You can check this yourself in #bots

,w rref{(1,2,1,2,1,2),(0,0,1,-1,-1,4),(2,4,3,3,3,4),(3,6,6,3,6,6)}

[soft zealot]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[wicked imp]

Oh sweet!

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[wicked imp]

.close

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[idle acorn]

i need help in question anybody

Solve for
𝑥
x:

2
𝑥
−
3
𝑥
−
1

3
x−1
2x−3
​

=3

[slow coral]

next time lease format this so it easy to read

i can not tell how is the equation is suppose to look

[idle acorn]

Solve for x:

i have trained to solve it 3 times i gave up

[proper lion]

!occupied

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[slow coral]

multiply both side by $x-1$

[soft zealot]

IdelUser404

[final saddle]

Please don't occupy multiple help channels.

[slow coral]

.close

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[keen storm]

<@&268886789983436800>

[mossy flare]

good lord

[keen storm]

dude these people r sooo cringeeee

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[worthy wedge]

bruh

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[weary meteor]

does a specific one need to be on top of it doesnt matter?

[keen storm]

wdym

[weary meteor]

here ill send it

$x^2+3y=27$

[soft zealot]

Yybepic

[polar spruce]

whar

[weary meteor]

$x^2+y^2=45$

[soft zealot]

Yybepic

[keen storm]

ok

how would u approach this

[weary meteor]

i thhink i would subtract the first oen from the second one

i know how to do it

but i dont know which im susposed to subtract from

[keen storm]

where do you need to subtract?

[weary meteor]

i need to subtraact the equations

[keen storm]

you could do either y²-45 or x²-45

[weary meteor]

does it matter which i subtrracted from?

[keen storm]

ye

cz a-b is not equal to b-a

i would do it diffrently tough

[weary meteor]

because if subtract first one from second one

it gives me 3y-y^2

which isnt in order

[keen storm]

,

[weary meteor]

im told they woyuld cancel out the x^2

[keen storm]

who told u that

[weary meteor]

ixl

ima try it rq see if im right

[proper lion]

you can

[weary meteor]

yeah i was correct

thanks for the advice

.solved

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[final saddle]

@soft pewter Has your question been resolved?

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[sour kindle]

This was the exercise i was given, "Prove that there is a one-to-one map from RP2 to the disk with opposite boundary points identified"

i have seen how the map is constructed by keep the vertices of RP^2 and then bending the edges so they are curved like a circle.
But i feel like it doesnt work, or maybe i misunderstand "disk with opposite boundary points identified"

[sour kindle]

like in RP^2 the figure is upside down compared to the disk

[final saddle]

@sour kindle Has your question been resolved?

[foggy ingot]

the exercise is abt the quotient space not the raw disk

[final saddle]

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[lyric pawn]

not sure what method to use

[drowsy epoch]

use the power rule

[lyric pawn]

ohhh

bro

i was overthinking it

[drowsy epoch]

if that cleared your doubt actually, you may close the channel

[lyric pawn]

yeah sorry i was trying to figure it out

my teacher never taught me how to do that reverse power rule ngl

[drowsy epoch]

looks correct, good job

[lyric pawn]

.close

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[pine sand]

all option seems correct to me

[clear moon]

is that asterisk in option C referring to the derivative?

D is true by invertible matrix theorem or just general understanding of determinants, A is true becuase A * A^-1 = I and determinants are multiplicative so you can take det of both sides. B is true because to undo two applications of the transformation A, you can apply two applications of the inverse

[pine sand]

I'm not sure about asterisk

[clear moon]

ah i looked it up i think the apostrophe means the transpose

i usually have seen it with a T as the notation

in that case you're right, it seems all four are true

[pine sand]

Thanks

It is derivative? If you have read somewhere

[clear moon]

no it's not the derivative, it usually doesn't make sense to talk about the derivatives of a matrix (at least when the entries are all numbers, which seems to be the case here)

[pine sand]

Can I think of B?

A is non zero but it is square can be 0

[clear moon]

it looks like an alternative notation for the matrix transpose

[pine sand]

Then inverse doesn't exist?

[clear moon]

that's not possible if A is invertible

if A is invertible then det(A) != 0

so if you're claiming that A^2 = 0 then it would mean that det(A^2) = det(0) which means that det(A) * det(A) = 0 which would mean that det(A) = 0

which is a contradiction since you assumed A was invertible

in general for invertible matrices A it's true that (A^k)^-1 = (A^-1)^k

[pine sand]

Yeah you are right

Thanks

.close

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[round sleet]

d-c=l

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[crimson ermine]

say that riddhima is a dragon. according to vacuous truths any statement can be made on her and it will always be correct as the statements cant be disproven as dragons dont exist. cant we deduce by a similar reasoning that any statement made on her cannot be proven which means that all of the statements are false?

[somber fog]

Think that things that do not exist cannot be false because for something to be false you need proof against it. Since there are no dragons, there is no proof, and since there is no proof, there is no lie.

[opaque ember]

just apply the same reasoning to the negation of the statement. we get that the negation is true, ie the statement is false

[tranquil pine]

Math is so fascinating, isn't it? Thanks for the help tho

[somber fog]

Yeah, it's fascinating. You are welcome.

Do you have more questions?

[opaque ember]

a false statement implies every other statement is both true and false. this is sometimes called the principle of explosion

[tranquil pine]

Can't believe it used to be the subject I hate

[tranquil pine]

Not at all I'm gonna practice some word problems now, easy work

[somber fog]

Good, have a nice day.

[tranquil pine]

I think it's the language which connects different people together

[tranquil pine]

You too, cya

[opaque ember]

luna is not OP

[tranquil pine]

Yea I'm not lol

[somber fog]

Yeah, excuse me.

[tranquil pine]

No problemo

[tranquil pine]

Calculus is what gets me confused mostly

It isn't in my grade's curriculum but will physics be any harder in the senior classes?

[somber fog]

You can ask in help channels or in https://discord.com/channels/268882317391429632/326138772477575180. If you have questions.

[tranquil pine]

Ah alrr

[crimson ermine]

dont you also need to prove that its false

how can you be sure about it

[somber fog]

A statement only becomes False when you can find a counterexample. For example all dragons are yellow is False, since you must show yourself a dragon that is not green. Since there are no dragons, you cannot present the evidence necessary for the phrase to be false.

[crimson ermine]

dragons dont exist so we cannot prove that there exist non-yellow dragons. by the logic of vacuous truth does this not imply that all dragons are yellow?

[supple jolt]

yes, and at the same time you cannot prove that the statement is true

[somber fog]

Yeah.

[supple jolt]

and unless you first prove that the statement is true, i have no reason to believe it, neither am i obligated to prove that it's false

so the only logical conclusion is to not believe your statement

until you provide evidence

[somber fog]

Do you understand it better now?

[supple jolt]

also

what if you said

[crimson ermine]

but you can use the same logic to show that all dragons arent yellow and to prove statements which are logical inverses of each other

[supple jolt]

"Riddhima exists"

then it would be true becuase it is a statement about a dragon

but at the same time its false because there are no dragons

[somber fog]

Yes.

[crimson ermine]

that breaks logic tho

[supple jolt]

yes so the question allows for all sorts of paradoxes

[somber fog]

Yeah.

[crimson ermine]

so whats the point bro 🥀

[supple jolt]

the point is that your propositions fall under their own weight

[somber fog]

In the world of formal logic, if something doesn't exist, you can say anything about it and be right.

[crimson ermine]

vacuous truths are frequently used in problems related to transitive relations isnt that wrong

[somber fog]

No, it's just technically necessary

[crimson ermine]

doesnt that make formal logic paradoxical

[supple jolt]

nope

[somber fog]

No.

[crimson ermine]

you just admitted that

[somber fog]

In this case, no.

[crimson ermine]

what 😭

am i just dumb

[supple jolt]

say that riddhima is a dragon. according to vacuous truths any statement can be made on her and it will always be correct as the statements cant be disproven as dragons dont exist. cant we deduce by a similar reasoning that any statement made on her cannot be proven which means that all of the statements are false?

[crimson ermine]

yes thats my question

[somber fog]

What ideas you have?

[supple jolt]

so you understand that
not provable ≠ false
not disprovable ≠ true

?

[somber fog]

Apply that in your question.

[crimson ermine]

yes

but by vacuous truths not disprovable = true which means that vacuous truths are lies right?

[supple jolt]

no

[somber fog]

No, because you can-t prove it.

[supple jolt]

vacuous truth does not mean “anything that can’t be disproved is true”

[crimson ermine]

wait what

then what does it mean

[crimson ermine]

In mathematics and logic, a vacuous truth is a conditional or universal statement (specifically a universal statement that can be converted to a conditional statement) that is true because the antecedent cannot be satisfied. -- wiki definition

[supple jolt]

vacuous truths are statements that count as true because there are no cases that could make them false. For example, if dragons do not exist, then “all dragons have wings” is vacuously true, because there is no dragon that proves it wrong

and also

[somber fog]

It's not that everything that is not denied is true. It is that every promise whose initial condition is not fulfilled is true.

[supple jolt]

"riddhima is a dragon" is not a vacuous truth

so " according to vacuous truths any statement can be made on her and it will always be correct as the statements cant be disproven as dragons dont exist." is false

[crimson ermine]

so all dragons dont have wings is simultaneously true which is absurd

so vacuous truths are lies

[supple jolt]

yes

[supple jolt]

no

they are true but empty statements

[somber fog]

Yeah.

[somber fog]

No, because you can't demonstrate it.

[crimson ermine]

like in transitive relation problems some relations just are like {(1,2),(6,7)} so by vacuous truths textbooks say that these relations are transitive but by vacuous truths isnt it simultaneously true that they arent transitive which is paradoxical which is bad and wrong

[supple jolt]

https://tenor.com/view/south-park-67-67-south-park-67-south-park-6-7-gif-13284396363591354590

[supple jolt]

well for it to be non-transitive you would have to disprove the fact that it's transitive

[crimson ermine]

for it to be transitive youll have to disprove that its nontransitive

[supple jolt]

you can prove that it's transitive

and you can also prove that it's not non-transitive

also

"dragons have wings" and "dragons have no wings" arent negations of eachother

[crimson ermine]

you can also prove the opposites of both statements

[supple jolt]

but

a relation is transitive and a relation is nontransitive are negations of eachother

[supple jolt]

so both of these are true by vacious truth

[supple jolt]

but the same doesnt apply here

because vacuous truths dont also make the negation true

it only makes the positive condition true when there are no counterexamples

[crimson ermine]

why

[somber fog]

Because if there aren't connections you can't show anything.

[crimson ermine]

right so doesnt that mean that the answer is uncertain rather that true

[supple jolt]

which answer

[crimson ermine]

whether the relation is transitive or not

[supple jolt]

"the relation is transitive" is true

so it's negation must be false

if P is true, then not P is false

so "the relation is not transitive" is false

so there's no paradox

[crimson ermine]

in {(1,2),(6,7)} a third relation does not exist. why doesnt this mean that the relation is not transitive as for a function to be transitive it must have 3 pairs

[supple jolt]

a relation can still be transitive even if it doesnt have 3 pairs

[somber fog]

Not necessary.

[final saddle]

@crimson ermine Has your question been resolved?

[crimson ermine]

how

a relation is transitive if we have ab and bc and then ac right?

[sturdy cypress]

you didn't make it have ab and bc

[final saddle]

@crimson ermine Has your question been resolved?

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[final tangle]

do you hav e a pic of the original question

dunno why you're using theta
but also why did you ingore the sin and +1

you use whatever your function given to you is

you don't get to pick and choose parts you like / don't like

which last function

was that your work or someone elses

wdym

are you saying what you posted is the corrected version?

oh

i see now

you only did the argument

you want f(7pi) - f(-pi)

you evaluate f(x) at each value and subtract

no

[proper lion]

what does growth mean

[final tangle]

f(-pi) =
f(7pi) =

[proper lion]

isnt that what he did ?

[final tangle]

he left it as f(x)

notation is important

[proper lion]

oh mb

oh

@final tangle this might be different than f(7pi) - f(-pi) ig

[final tangle]

wait, is it asking for the increase over the interval
or when it is increasing in the interval

[proper lion]

the interval where it increases
does it mean this ?

[final tangle]

do you have a full worked example to compare to

where

that doesn't seem like the full work

because there's no relation to growth there

most likely some translation issue that I'm not entirely sure what they want then
going to try get someone more familiar with this

ok so it seems to be asking

or when it is increasing in the interval

in which case do you know when sin is normally increasing?

its determine when f(x) is increasing, not the amount of increase

e.g. these intervals

do you know the general interval for when sin(t) is increasing

not quite

that'd be one interval

do you know how to get the general interval from that?

simply add 2k * pi to each end

(-pi/2 + 2k pi, pi/2 + 2k pi)

then you want to consider in general when the function is increasing
which you could do from solving:

$-\frac{\pi}{2} + 2k\pi < \frac 25 (x - \pi) < \frac{\pi}{2} + 2k\pi$

[soft zealot]

ραμOmeganato5

[proper lion]

try to find intervals which are in x belongs to [-pi,7pi]

you multiply by 5/2 not 2/5 but thats correct

no you did some mistake

how is -5pi/4 + 5kpi + pi = -4pi/4 + 5kpi

and 5pi/4 + 5kpi + p = 6pi/5 + 5kpi

how show what you did

oh you cannot because -5pi/4 is a fraction

you need to take LCM

then add

Lowest common multiple

yes

you have $\frac{-\pi}{4} + 5k\pi <x< \frac{9\pi}{4} + 5k\pi, x\in [-\pi,7\pi]$

[soft zealot]

ghost

[proper lion]

replace by integer and check if the bounds lie in [-pi,7pi]

what is the value of sin(-pi/2)

you can also check the derivative of sinx, cosx you want sinx to be increasing, find the interval where cosx > 0

sin(-pi/2) = -1 as x goes to 0 then pi/2 sinx goes from -1 to 1 then it decreases

they took it from graph of sinx

your function had sin(2/5(x-pi)) we consider (2/5(x-pi)) = t then set -pi/2 +2kpi <t < pi/2+2kpi

it is not it came from sinx,
we put you function to be sint then applied those bounds, so you know that (2/5(x-pi)) should lie between them for the functon to be increasing

you solve for the intervals

lol ic

[final saddle]

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[proper lion]

np but you were mostly done

it was a language issue otherwise wouldnt have taken so long

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[quartz galleon]

am at step 2(sin/cos)/1 + (sin/cos)(sin/cos)

don't know how to continue

[high jolt]

there's a pythagorean identity you can use to simplify the denominator

[quartz galleon]

what do you mean

[high jolt]

are you aware $\tan^2(\theta) + 1 = \sec^2(\theta)$?

[quartz galleon]

no

[high jolt]

oh

[quartz galleon]

we don't have that in our syllabus

about sec

[high jolt]

hm

maybe u can do it without that lemem check

[high jolt]

ok so you have

$$\frac{2 \frac{\sin(\theta)}{\cos(\theta)}}{1 + \frac{\sin^2(\theta)}{\cos^2(\theta)} }$$

[high jolt]

correct?

[quartz galleon]

correct

[high jolt]

make the denominator into one fraction

what fraction do you get?

[quartz galleon]

i honestly have no clue

[high jolt]

just add 1 + sin^2(theta)/cos^2(theta)

by using common denominators

[quartz galleon]

i cant find a common denominator 😭

[fathom meteor]

how can you express 1 in terms of the fraction's denominator

[quartz galleon]

im lost

[fathom meteor]

think of the fact the anything divided by itself is 1

[fathom meteor]

dont need to use these yet

[quartz galleon]

Ok

[quartz galleon]

so there's a way to figure out what sin^2/cos^2 is?

[fathom meteor]

thats not the thing im hinting at

use that fact to represent 1 in terms of the denominator of the small fraction

[quartz galleon]

???

i am so lost what the hell

[fathom meteor]

$\frac{x}{x}=1$

[soft zealot]

holathere

[fathom meteor]

x can be anything

[quartz galleon]

yeah

[fathom meteor]

including the denominator of the fraction, cos^2(theta)

what does that give you

[quartz galleon]

no clue

[fathom meteor]

plug in x=cos^2(theta) into this identity

[quartz galleon]

is it 1

[fathom meteor]

write out the identity in terms of cos^2(theta) by replacing the two x's with that term

[quartz galleon]

replace the two x's with cos^2(theta)?

[fathom meteor]

yes that is a substitution that preserves the identity

what do you get

[quartz galleon]

so cos^2(theta)/cos^2(theta)

[fathom meteor]

=1, exactly

[quartz galleon]

yes

but where does that apply here

cuz we have sin^2(theta)

[fathom meteor]

so in the oriiginal big fraction's denominator you can represent 1 as..?

[quartz galleon]

cos^2/cos^2?

[fathom meteor]

yes

sry i gtg now hope you can get it solved though

[quartz galleon]

yooo i figured it out lets fucking go

.close

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[pseudo cradle]

Where am I going wrong here?

[pseudo cradle]

the answer key shows $26 \sqrt{5} - 15 \sqrt{2}$

[soft zealot]

Vortac

[desert mantle]

7*4

[pseudo cradle]

doh

that was dumb... thank you

.close

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[amber spruce]

I was being a little calculus guy and then I got to this question. Ive never done an induction proof before so I had to look it up and its like you prove that 1 case is true, like n=1, and then make an inductive step that if k is true than k+1 is true? Anyway, I tried my best but I'm kinda stuck

[amber spruce]

Yeah idk what I'm doing

[tender pollen]

ok, there's a famous relation with the binomial coefficients called pascal's identity

[\binom{n-1}{r-1}+\binom{n-1}r=\binom nr]

[soft zealot]

literally anything

[tender pollen]

you can prove this by algebra

is this what you were looking for? @amber spruce

[amber spruce]

Okay so I've seen that before but how would that help me here

Like I'm unsure what I'm even trying to achieve here

[tender pollen]

just differentiate your k case, right?

and then apply the product rule again

[amber spruce]

I did that no?

[tender pollen]

reindex the sum

split the sum into two and reindex the sum, i believe

[tender pollen]

[\sum_{r=0}^k\binom kr\frac{\text d^{r+1}u}{\text dx^{r+1}}\frac{\text d^{k-r}u}{\text dx^{k-r}}=\sum_{r=1}^{k+1}\binom k{r-1}\frac{\text d^ru}{\text dx^r}\frac{\text d^{k-r+1}u}{\text dx^{k-r+1}}]

[amber spruce]

Okay wait

Okay so

Oh

[soft zealot]

literally anything

[amber spruce]

I see

[tender pollen]

sorry had to correct that a few times

add the r=0 case in

[amber spruce]

So I see

Okay that works in the end

But my question is that how can we just re index using r - 1

Or can we reindex with anything in reality but r-1 is the most useful thing to reindex with?

This is my first time doing this like ever

[final saddle]

@amber spruce Has your question been resolved?

[storm haven]

well all youre doing in reality is a substitution

let r = i-1 (equivalently i=r+1)

r goes from 0 to k

so i goes from 1 to k+1

[amber spruce]

O

Hol up

[amber spruce]

Yeah I'm confused as heck

I'll just figure it out tomorrow

[final saddle]

@amber spruce Has your question been resolved?

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[tender heart]

Can someone help me with this inequality? Idk if what i did was right but im quite stuck either way

[severe canyon]

Hint: how would you solve $\frac{3x-1}{x + 1} \leq 5$?

[soft zealot]

Alberto Z.

[severe canyon]

I hope you don't do
3x - 1 ≤ 5
x + 1 ≤ 5
right? 😅

[tender heart]

No!

Since there is x at the denominator

So I would do -5

On both sides

And then bring it to the same denominator

I first thought to take the thing at the top and solve it and then determine the sign and then same thing for the bottom one yet idk if this is alr

[kindred mortar]

There is a problem between lines 2 and 3

[tender heart]

Oops

[kindred mortar]

You cant get rid of the denominator like that

[tender heart]

Oh wait right

Let me try to fix it, I got an idea

Now do I make a system with -5<=fraction and fraction>=5?

And find solutions for each?

[kindred mortar]

it can work yes

[tender heart]

Alright! I will try to see if I can solve it!

Is this alr up untill now?

[candid pulsar]

3x - 5x is not -8x

[tender heart]

Oopsies, its -2x🥀

Tysm for spotting this! I suck at simple stuff

[candid pulsar]

i think the rest is correct yeah

from here normally you make a sign table to solve the inequality

now that you know the roots

[tender heart]

Is this alr?

And the the final solution should be S1 U S2

I think

[kindred mortar]

looks correct

[kindred mortar]

You have to verify both equations

[candid pulsar]

the values at x = -1 and -1/2 alone looks a bit shady

at x = -1 its (-ve number)/0 and at x = -1/2 its 0/(+ve number)

so it should be
undefined at x = -1 and
0 at x = -1/2

other than those two points the rest is right

[tender heart]

Yeah

Oops

So then itll be -1) and [-1/2

At the solution

[candid pulsar]

yeah

[tender heart]

(I didnt write the whole, js like those two)

[tender heart]

Alrighty!

[somber fog]

Do you have more questions?

[pliant elk]

!done

[final saddle]

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

[tender heart]

Im done now, tysm yall!!!

.close

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[somber fog]

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[warm python]

Consider $x^2y^3= a^n$. It follows that a must be a variable of $x$ and $y$. We then have $(x^py^q)^n=x^2y^3
pn=2;qn=3$follows. It then follows that $n$ must be $1$, $q$ must be $3$ and $p$ must be $2$. But now $xy^5$ isn't generated. We thus have a contradiction

[soft zealot]

wai

[final saddle]

@warm python Has your question been resolved?

[warm python]

.close

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[solar lichen]

how come you need to split the limits into 2 parts. from 0 to 2 and then 2 to 5

the area underneath is all positive

so how come you cant do 0 to 5 as the limit?

[final tangle]

you have two different functions from 0 to 2 and from 2 to 5

[solar lichen]

ohhhh so the area under x^2 + 2 first which is 0 to 2

and then 2x+10 qhich is 2 to 5?

[final tangle]

-2x+10

[solar lichen]

yes

alright tysm 🙂

.close

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[pine sand]

a^2-b^2=288...how many integer pair of (a,b)

[pine sand]

(a+b)(a-b)=288

[spare summit]

consider factoring 288 too.

[ebon agate]

You can ask multiple questions on one channel

[pine sand]

Ahha... thanks

288=2.3.3.4.4

Ohh wait

[tiny kraken]

not recommended btw

[ebon agate]

Why?

[spare summit]

if a Helpful isn't present then your current question can be buried

[ebon agate]

Oh right

[pine sand]

2^5.3^2

[ebon agate]

Fair enough

[spare summit]

don't prime factor. do regular factoring

[pine sand]

Owww

1
2
3
4
6
8
9

.....

Many

[ebon agate]

18 tuples exactly

[spare summit]

list them in a x b format

so that you don't duplicate factors later

[ebon agate]

Btw jig

If a+b=even

and a-b=odd

Will a and b be integers?

[pine sand]

1.288
2.144
3.96
4.72
6.48
8.36
9.32
12.24

[ebon agate]

And the reversed list too

[pine sand]

No idea

[spare summit]

reverse is not necessary as long as you remember the reverse of each pair

[ebon agate]

Fair enough

[spare summit]

you don't want to spend all day listing factors

[ebon agate]

Also negatives too

[pine sand]

[ebon agate]

36 tuples dang

[pine sand]

Omg

Painful

[ebon agate]

No we just have to answer how many

Not bad

[ebon agate]

Add these equations

[somber fog]

Do you understand it better now?

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@pine sand Has your question been resolved?

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[gleaming anchor]

<@&268886789983436800>

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[chilly rover]

Guys I am so confused in 25-30

[somber fog]

What have you tried?

[chilly rover]

so for 27

i got z = x^2

i know its a parabola

but I am confused

how to know which axis is it directed along

thats the tough part

[somber fog]

We need the behavior of y, right?

And $y = \frac{1}{1+t^2}$.

[soft zealot]

Ga³¹Br³⁵I⁵³9000✞

[chilly rover]

how did u know that we need the behaviour of y

[somber fog]

Imagine that the y-axis is a hallway or a tunnel. Since y = \frac{1}{1+t^2}, the parabola stretches toward you when t=$0 and moves away toward the wall when t is very large (because y approaches 0).

<@&268886789983436800>

[half geode]

what was that jst rn?

[unborn forge]

Usually it’s either scams or nsfw

[somber fog]

Never join this links.

[unborn forge]

But fortunately this server has very fast moderation

[soft zealot]

PopCat

[keen swallow]

$x+1=x$

[soft zealot]

PopCat

[final saddle]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[keen swallow]

im testing it ok sorry

[unborn forge]

Go to #latex-testing

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[somber fog]

!occupied

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@chilly rover Has your question been resolved?

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[fast onyx]

Im asked to think of a real world example for derivatives of composite functions, but i can't think of anything specific. Any suggestions?

[loud sundial]

The effectiveness of a drug depends on concentration, and concentration depends on time

So if you want to find the rate at which the effectiveness of a drug is wearing off (aka $\dv{E}{t}$), that's the derivative of a composite function

[soft zealot]

Civil Service Pigeon

[loud sundial]

Namely $E(C(t))$

[soft zealot]

Civil Service Pigeon

[loud sundial]

that's just one example ofc

[fast onyx]

that's perfect 🙂

[loud sundial]

you can probably dig in #math-pedagogy if you want to find a more in depth discussion on this

[fast onyx]

i would but this is more than sufficient

[loud sundial]

[fast onyx]

thx! @loud sundial

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[fast onyx]

c.lose

mfml

.close

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[south dirge]

<@&268886789983436800>

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[drowsy epoch]

What have you tried?

[pliant snow]

I just don't know how to start, I understand we are trying to make an equation so it show how many time it show up for every time you go around the unit circle

[drowsy epoch]

Ok maybe let's start like this, what do you know about tanθ generally

[pliant snow]

It’s sin/cos

[drowsy epoch]

now what happens at cos(θ)=0

[pliant snow]

Cos is 0 at pi/2, and 3pi/2

[drowsy epoch]

yes but what does that mean for tan

[pliant snow]

It’s Undefined

At those points

[drowsy epoch]

yes and how does tan behave around such points

[pliant snow]

Vertical Asymptotes

[drowsy epoch]

tan goes to inf/-inf

So yeah

You need therefore to find out when when cos(θ/2)=0

the equation θ=... contains all vertical asymptotes

[pliant snow]

It’s pi/2 and 3pi/2 so how would I make an equation for that.?

[drowsy epoch]

cosine is periodic

these are not all solutions

[pliant snow]

It happens at every time it rotates pi units right?

[drowsy epoch]

how so?

[drowsy epoch]

you'll need some additional param to define

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@pliant snow Has your question been resolved?

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[sick olive]

Hi

[sick olive]

What is the least possible value of $(xy-1)^2 + (x+y)^2$ for real numbers $x$ and $y$?

[soft zealot]

Hate

[sick olive]

How can we quickly solve this problem?

Which of the following is equivalent to [(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?]

[soft zealot]

Hate

[sick olive]

Also this one

I need help

[karmic glen]

what have you tried?

[sick olive]

I don’t really know how to solve it

And I don’t wanna input a guess and check method

I think we could first simplify it

[karmic glen]

yes

[sick olive]

Wait I have to go eat

[karmic glen]

okay

[sick olive]

Could you explain the 2 problems and I can

View it later

[karmic glen]

sure

for 1 its a quadratic in x

[south dirge]

Ill give you a hint! try opening the bracket and then grouping the like terms togehter

[karmic glen]

multiply by (2-3)/(2-3) and use a^2-b^2 formula

[south dirge]

wouldnt this be little lengthy?

[karmic glen]

oh wait lol

just a^2 + b^2 >= 0

[karmic glen]

yeah mb

[sick olive]

Nvm I came back

[sick olive]

Hm can u show me

Oh I get it

[south dirge]

try this!

[karmic glen]

its just a square is >=0 so the smallest is if both brackets are 0

[sick olive]

Ok

Wait

[south dirge]

sorry that doesnt seem right!!

[karmic glen]

what

oh that has no solutions

okay

right

[sick olive]

Xy^2-2xy+1 and x^2+2xy+y^2

[south dirge]

Yea!

Do you see smth that cancels?

[sick olive]

2xy

Adding them gives

[south dirge]

Okay! rewrite then

[sick olive]

Xy^2+x^2+1+y^2

Then we can simplify I think

[south dirge]

Yes!!

This is the crucial step!

[sick olive]

I’m so bad at simplifying

[south dirge]

Thats fine think a little

[south dirge]

(i was too 😭)

[sick olive]

Lemme think…

I’m stupid..

[whole halo]

try writing it as $x^2y^2+x^2+y^2+1$

[soft zealot]

mtt

[south dirge]

should i give a hint?

[sick olive]

No

Almost got it

[sick olive]

X^2+1

Times

Wait

Y^2+1

[south dirge]

Yeah!!

[sick olive]

yess

[south dirge]

So you are saying (x^2 +1) (y^2 +1) right?

[sick olive]

Yes

Least possible value is

Wait..

[south dirge]

hold on lets first see this!

we need to minimise this so we want x^2 and y^2 as small as possible right?

[sick olive]

Yes

Small as possible is 0

[south dirge]

Yeah!!!

[sick olive]

So X=-1

I think

[south dirge]

No no what?

[whole halo]

(-1)^2 is 1, not -1

[sick olive]

Oh

Sorry I’m stupid

X=0?

[south dirge]

Please dont say that!

[south dirge]

yup!

[sick olive]

Ok and y is also?

[south dirge]

You tell me

[sick olive]

Yes?

[whole halo]

go say what number you think y should be

[sick olive]

Y is 0

Maybe

[south dirge]

Correct

[sick olive]

Pretty sure

Yess

Tysm

[south dirge]

So the minimum value of the give expression is ?

[sick olive]

LEMEM go back to the problem rq

1

The minimum value is 1

Yesss

[south dirge]

Thats right!

[sick olive]

For this one

Could you also do multiply by 2-3 and divide by it?

Just a passing thought

[south dirge]

Yeah you can

Why dont you try that?

[sick olive]

Most ppl told me it’s 3-2

Lemme try it

Oh yeah it does work

Thank you so much

How do I close ur

It

[south dirge]

!done

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[sick olive]

.close

[final saddle]

Channel closed

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[sick olive]

Thank you again

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[vital crag]

Domain of arcsine is -1 to 1

This has no solutions in reals

[drowsy epoch]

what the greatest possible value for 2 sin (pi(x-3) + 4

you certainly know that sin(x) is at most 1

then 2sin(x)?

then 2sin(x)+4

yep

notice: the pi(x-3) inside sine, also changes sine horizontally

horizontal translation/stretching

so in a sense we dont need that information to conclude that 2 sin (pi(x-3) + 4 reaches at most 6

well the left side is at most 6

that's the domain

when it comes to the range we know that the upper bound is y=6

i wouldnt view it like that 😭

like imagine it graphically if your function is always at most reaching height 6, how can you expect it to have something for height 11

well you have mathematically a contradiction

yeah left side is at most 6 so equating that to something greater to 6 is nonsense

it's prob for practice

you are not supposed to solve equations on standby mode

like this

sometime you should also be able to interpret the equation to figure out if it's possible at all before you do some computations

saves you time

[drowsy epoch]

then alarm bells should start ringing knowing this

yes

you can apply the same concept for cosine

wdym

well cosine has the same default range as sine