help-13

[eager wraith]

i think i can understand better

visually

[gleaming imp]

Yk what that arrow indicates?

It's literally in the following question what do u see on top of MN and NX

an arrow

[eager wraith]

oh

so that’s how it’s drawn

i just drew it

the wrong way

[gleaming imp]

?

[eager wraith]

vector

[gleaming imp]

Yeah!

[eager wraith]

NO

DIRRECTION

DIRECTION

NOT VECYOR

[gleaming imp]

I mean true but

[eager wraith]

DIRECTION

[gleaming imp]

IT goes infinity

[eager wraith]

ok

what about for

22

[gleaming imp]

I must know how you would solve it first

[eager wraith]

which one

[gleaming imp]

22

it's alr if it gives u a dodgy answer

[eager wraith]

uh

so we draw a plane

labeled m

and a line NB

intersecting line B

right

[gleaming imp]

Alright

WHAT

NO

[eager wraith]

WHAT

[gleaming imp]

In which scenarios, a point can be formed?

[eager wraith]

when

[eager wraith]

hm

[gleaming imp]

lets say, The Vector line AB, the point B is already there doesnt matter either tangent or intersected

If u look at line CD and AB, whats wrong with the two lines?

[eager wraith]

E is in the middle

point E

[gleaming imp]

Ok, correct in some way but what im tryna say is

They are on top of each other, AB and CD

[crimson sedge]

နစ်ဂါ

[eager wraith]

Oh ok

[gleaming imp]

The point where two lines meet each other is "new point"

[eager wraith]

Wait so

would the new point in this case be A?

[gleaming imp]

My picture?

[eager wraith]

No in the question

[gleaming imp]

Exactly

[eager wraith]

Aight

[gleaming imp]

So what must the Vector NB do to from a point A?

[eager wraith]

intersect with plane M

[gleaming imp]

YEah

That's it

[eager wraith]

so a plane is just a square right

[gleaming imp]

Yup

[eager wraith]

ok im not sure how to really draw this

[gleaming imp]

draw and show me

[eager wraith]

ok

[gleaming imp]

NB

which comes first

The alphabet

[eager wraith]

yeah idek what i did

[gleaming imp]

?

[eager wraith]

B

[gleaming imp]

bruh Not in alphabet order

the sentence

[eager wraith]

oh

[eager wraith]

u just said

in alphabetical order

[eager wraith]

N

[gleaming imp]

No I meant the alphabet not as in the whole

[eager wraith]

Oh

[gleaming imp]

Single one 😭

[gleaming imp]

Alright, so which do you think would be a staring point

[eager wraith]

N

[gleaming imp]

Yup

[eager wraith]

ok

[gleaming imp]

LOL

[eager wraith]

now what

[gleaming imp]

You put 3 dots there

Look at the question whats on top on NB

[eager wraith]

N sorrow

[gleaming imp]

A vector arrow

[eager wraith]

an atrocious

arrow

[gleaming imp]

So

The line should have an arrow

Also, if you plot the points don't make it extra big black lead voids Lol

just small dot would do the job

[eager wraith]

ok

[gleaming imp]

There you go fam

[eager wraith]

thank you

there’s other questions i did too, could u just tell me if their right

they’re

[gleaming imp]

My Pleasure

Sure I guess

[eager wraith]

i’m confident they’re so it shouldn’t take long

is 10 w

point w

for 10

[gleaming imp]

Yeah

Correct

[eager wraith]

bet

and also

is this J

got

for

28

[gleaming imp]

Isn't it I

It doesn't say not collinear

[eager wraith]

WAIT

I DID

27 ON ACCIDENT

MB

[gleaming imp]

Bruv

[eager wraith]

yeah i for I

got I

wait

isn’t it

C

[gleaming imp]

for number?

[eager wraith]

28

[gleaming imp]

WEhat is you evenn saying lmaoo

[eager wraith]

for 28 is it C

the point that is collinear

with

B and I

[gleaming imp]

Oh right I read wrong

as B and C

yeah it's C

[eager wraith]

aight

[gleaming imp]

mb

[eager wraith]

for 32 is it I

u food

good

wait

32 is B or I

[gleaming imp]

either B or I

[eager wraith]

oh ok so it’s both

[gleaming imp]

"a point"

a

[eager wraith]

what

but then

so it’s only one

[gleaming imp]

yeah

[eager wraith]

which one then

[gleaming imp]

You can choose one I believe

[eager wraith]

oh ok

for this i got R, k, l, s

for 40

wait no

nvm

i did it wrong

i think

wait

i think they are all co-planar

right

[gleaming imp]

Hold on

Do we have to concider the parallel lines?

if they all lies within the same plane

[eager wraith]

I don’t even know

[gleaming imp]

Bruv

[eager wraith]

but i’m pretty sure

they are all coplanar

cuz like

yeah

idk tho

[gleaming imp]

If that rule doesnt apply, I believe its

K,L,R,S

[eager wraith]

but isn’t the point N in the same plane as K and L

and same for P and Q

wait

[gleaming imp]

Ok now that im perplexed

[eager wraith]

i think it’s just R, K, L, and S

ima say that

cuz idek

[gleaming imp]

Yeah same

This ones kinda confusing even for me

[eager wraith]

46 is yes right

[eager wraith]

ye

[gleaming imp]

absolutly

[eager wraith]

and then 56 is never and 60 is always

[gleaming imp]

Yes

[eager wraith]

aight bet

that’s all i have bro

thanks dude

[gleaming imp]

WAIT

WAIT

[eager wraith]

what

happened

[gleaming imp]

A line, if it's a line segment

it can has an end pouint

else if an infinite line

it doesnt

[eager wraith]

on ok

ih

so it’s sometimes

[gleaming imp]

Yeah

[eager wraith]

aight bet

thanks

.close

oh wait

this isn’t my channel

[gleaming imp]

HAve a day you deserve pal

[eager wraith]

you too bro

[gleaming imp]

cya!

[eager wraith]

cya

[cedar kiln]

@autumn nebula Has your question been resolved?

[cedar kiln]

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[hot python]

this right?

[hot python]

the -21 thing kind of caught me up

but i think i figured it out

[cedar kiln]

@hot python Has your question been resolved?

[hot python]

<@&286206848099549185>

[spice phoenix]

Should be correct

[hot python]

eh ok

thanks

.close

[cedar kiln]

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[solemn lintel]

What does it mean when it says “evaluate”? (Ignore the work for 22, my dad was messing around it’s nonsensical)

[late wing]

it's a determinant

first word

[solemn lintel]

so we just have to find the detriment?

(And the other stuff it tells us to do)

[late wing]

yea

[solemn lintel]

oh

sick

thanks man

.close

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[cursive steppe]

Is it x/x+7-11?

[cedar kiln]

@cursive steppe Has your question been resolved?

[cursive steppe]

.close

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[uneven crest]

how would i do this without a calculator?

[uneven crest]

our teacher said use calculator but im curious how we would do it mathematically

[peak minnow]

thats a known limit

[jagged marlin]

thats e

[peak minnow]

if u really wanna prove it or solve it

[jagged marlin]

pick one of the available channels #❓how-to-get-help

[peak minnow]

u could use binomial expansion or log properties

both work and u can prove that its equal to e

[jagged marlin]

Binomial expansion is a good trick, yeah
this limit comes up when you're trying to solve the equation
$\frac{df}{dx} = f(x)$

[wraith dagger]

Dockson

[jagged marlin]

if you use the limit definition of the derivative, you can manipulate it algebraically until you get to that definition of e

[uneven crest]

i dont see where to rly start how do i expand it or use log properties

[jagged marlin]

do you know the binomial theorem

[uneven crest]

no

what is it maybe im jsut unfamiliar with the name

[earnest geyser]

(a+b)^n

Expansion

[uneven crest]

this?

[peak minnow]

$$(1+x)^n = 1 + nx + \frac{(n)(n-1)}{2!}x^2 + \frac{(n)(n-1)(n-2)}{3!}x^3 ..........$$

[wraith dagger]

JustToPro

[peak minnow]

u can use this one here

modified binomial expansion , only works if x = 1

[uneven crest]

i am only in pre calculus so some formulas i may be unfamiliar

[jagged marlin]

limits like these aren't easy to solve analytically, you'll learn to do it in calc 1 and 2

[uneven crest]

ok makes sense why he told us to use calculator then

[earnest geyser]

Or there is a standard method to solve for 1^infnty questions

[uneven crest]

what is that

[earnest geyser]

How do you calculate limits in calculator

[jagged marlin]

plugging in really large values lol

[uneven crest]

just keep plugging in rly big numbers

very cheesy method

and it gets clsoe to 2.72 which is e

[jagged marlin]

isn't there a subtlety around x in that limit being a real number rather than integer

[peak minnow]

i didnt know

i thought it works for all

[earnest geyser]

This one, use this

[jagged marlin]

binomial theorem is probably well defined for non-integer values but it requires the gamma function

[uneven crest]

how do i prove this = 1??

also is sin^2 x = (sin x)^2?

stupid quesiton ik

[jagged marlin]

yes

[uneven crest]

or is sin^2 x = sin (x^2) and not equal to sin x sin x

[jagged marlin]

its the other one

[uneven crest]

sin^2 x = (sinx)^2 right

ok ty

[earnest geyser]

You can memorize it since it is a standard limit

Or use L-Hospital

[uneven crest]

ik but i want to prove

[uneven crest]

im precalc idk what that is

[earnest geyser]

Or use sinx expansion

[earnest geyser]

Then use this

Best way

[jagged marlin]

l'hopital isn't required

isn't squeeze theorem usable here

[whole yarrow]

squeeze 🗣️🗣️🗣️🔥🔥🔥

[earnest geyser]

$$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}....$$

[wraith dagger]

Monarch of Eternal Night

[earnest geyser]

@uneven crest

Now substitute this as sin x

And divide by x

[earnest geyser]

Well it's a method to solve

[whole yarrow]

the dudes in precalc

[jagged marlin]

he hasn't taken calculus yet lmao

[earnest geyser]

I didn't know that

[austere ember]

you can also think about how as sinx takes on inputs very close to 0, sinx looks a lot like x (just graphically, no calculus)

[uneven crest]

lol

[whole yarrow]

fr

[uneven crest]

i have the equation (1-cos^2x)/cosx * cotx this doesnt work right?

[austere ember]

doesn't work for what?

[earnest geyser]

Wdym it doesn't work

[austere ember]

that also just simplifies to sinx

[uneven crest]

yea

[austere ember]

what are you trying to do with that?

[uneven crest]

cause wen i plug in 0 i get 0/0 again

[austere ember]

yeah you're not going to be able to write it in terms of other trig functions in a way that you get something other than 0/0

[uneven crest]

idk im lost

[jagged marlin]

you're not gonna be able to use trig identities

the only way to prove it is through squeeze theorem

[uneven crest]

ig its for calculus

[whole yarrow]

squeeze 🗣️🗣️🗣️🔥🔥🔥

[uneven crest]

idk what squeeze it

[jagged marlin]

or l'hopitals rule

[uneven crest]

but ima learn it

[earnest geyser]

You should use 1-cos^2(x)=2sin^2(x/2)

[jagged marlin]

there's no trig identity you can use that'll make it evaluate to something other than 0/0

neil is right

[earnest geyser]

There is

Sin x= 2sinx/2.cosx/2

So sin x/2 in numerator

[austere ember]

not only is this identity not correct, that doesn't help solve the limit

[earnest geyser]

Isn't the limit 0?

[austere ember]

it's 1

[earnest geyser]

Waot

Wtf

[uneven crest]

ik its 1 cause i memorized that cause its super common but i want to know why

[jagged marlin]

bro this is a standard calc 1 limit

[uneven crest]

ig not possible withonly precalc knowledge

[earnest geyser]

I took 1-cos2x

Mb

[austere ember]

and you're onyl just starting calculus? @uneven crest

[uneven crest]

i am in precalc

[austere ember]

so you haven't learned stuff like l'hopitals or taylor series

[whole yarrow]

you could look up a video

[austere ember]

oh preclac

okay here's a good way to show it

[whole yarrow]

you should be a able to use only precalc to understand squeeze j think

fr

[jagged marlin]

the only way I can intuitively show it is that sin(x) oscillates between -1 and 1

so its absolute value never gets > 1

meanwhile 1/x will approach 0 as x gets larger

[austere ember]

@uneven crest

[austere ember]

we're evaluating the limit at 0, not at infinity

[uneven crest]

makes sense

[jagged marlin]

isn't this just the squeeze theorem

[whole yarrow]

squeeze 🗣️🗣️🗣️🔥🔥🔥

[austere ember]

it is, they're not in calc yet

so i can't show a l'hopital or series solution

[austere ember]

here's that also graphed if you want to see it yourself

like the inequality isn't always true

what matters is that the inequality is true is some neighborhood around 0

[jagged marlin]

there's actually a geometric solution
https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1

[austere ember]

which is where we care about the behaviour of sinx/x

so @uneven crest you can see that sinx/x is squeezed between cosx and 1/cosx

can also see it graphically

near 0

right?

[uneven crest]

yea

i see that

[austere ember]

if sinx/x is upper and lower bounded by those two functions near 0

and both those functions evlauate to the same limit

then sinx/x has its limit simultaneously >= that value and <= that value

therefore the limit must equal that value

notice that cosx tends to 1 as x tends to 0 (you can plug it in directly)

and that 1/cosx also tends to 1

giving you 1 <= your limit <= 1

[uneven crest]

hmm i see based off looking at the graph

pretty interseting

[austere ember]

also here's sinx and x graphed

near 0, sinx is basically indistinguishable from x

so as x tends to 0, sinx tends to x

which means sinx / x tends to x / x which is 1

[uneven crest]

ohhhhh

that makes so much sense

[austere ember]

🙂

[uneven crest]

also since i am learning limits whenever discussing end behavior the limit is always just the horizontal asymptote right

[austere ember]

end behaviour of things like rational functions?

[uneven crest]

ig yea

like when the limit is -inf or inf

[austere ember]

yeah that sounds right

[uneven crest]

ok cool ty

[austere ember]

happy to help

[uneven crest]

👍

.close

[cedar kiln]

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[uneven crest]

any way to solve this algebraically?

[limber marsh]

you can't solve functions!

what are you trying to do with this function?

[uneven crest]

oh mb find the limit as x -> inf and as x -> -inf

limits

[limber marsh]

oh okay, do you know about ln(x) (without absolute values)?

[uneven crest]

yea

[limber marsh]

okay, so |x| = x when x > 0

[uneven crest]

ok

[limber marsh]

so when we take x -> inf, then we can consider only positive values of x, so lim x -> inf of ln|x| = lim x -> inf of ln(x)

[uneven crest]

ok

i see

[limber marsh]

okay, so you already know how to do this one becuase you've seen a graph of ln(x) before

or if not that's okay too

[austere ember]

and if you're not convinced that ln(infinity) doesn't go to infinity since it looks like it tapers off, lnx is just the power you have to raise e to to get x, so what does x have to be for e^x to be infinity?

[uneven crest]

inf

[limber marsh]

okay great

[uneven crest]

ohh i see

this makes more sense

[limber marsh]

so now we can do x -> - inf

but the point is that if x < 0 then |x| = -x

[uneven crest]

yea

[limber marsh]

so we again just have that the limit as x -> - inf of ln|x| = lim x -> -inf of ln(-x)

this is how you do all of these sorts of absolute value questions

[uneven crest]

basically the same then

so it goes to inf cause abs is just positive

[limber marsh]

you have to break it up into cases, where the sign inside the absolute value is positive or negative

[uneven crest]

thanks for the tip

[limber marsh]

yep

[uneven crest]

.close

[cedar kiln]

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[uneven crest]

why is it that this is true when the "outer function" is continuous at the limit of the "inner" function

[austere ember]

doesn't this also require that the inner function is continuous at c?

for g(c) to equal the limit of g at c?

oh no it doesn't

oh wait it does?

[rigid lava]

where'd the ) go

[uneven crest]

idk

thats what my teacher said but he didnt really explain it

[austere ember]

an example i'm thinking of rn is g(x) = sinx/x, f(x) = x, and c = 0

[uneven crest]

what i assume is that they are equal if f(c) is real even if g(c) is not real

kind of like a removeable discontinuity or jump

[austere ember]

then f(g(x)) = sinx/x, so the limit as x appraoches 0 of this is 1

but f(g(c)) = sin0 / 0 which is undefined

which means this equation doesn't hold here

even thouhg the outer function f(x) = x is continuous everywhere

[uneven crest]

hmm

[austere ember]

this is only true when both f and g are continuous at c

then f(g(x)) is continuous at c

which means its limit equals its value

[uneven crest]

ig its something just weird

i wrote this this makes sense write

sperate problem

[austere ember]

that makes sense to me

[uneven crest]

this one is confusng to me

i get to here

[austere ember]

you mean the limit as u approaches infinity of g(u)?

[uneven crest]

u = cscx

i used substuituion

[austere ember]

yes, so then when you let x tend to 0+, u tends to infinity

which means when you turn g(cscx) into g(u)

you also change the limit variable?

[uneven crest]

why 1/sinx goes to inf

because sinx as x -> inf is undefined?

[austere ember]

wait how'd you get x tends to infinity here then?

you had the limit as x tends to 0+

why'd you change x's limit

when you make a substitution, you also change the variable in the limit

[austere ember]

great yeah so you have that it becomes the limit as u approaches infinity of g(u)

[cedar kiln]

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[austere ember]

and notice that's exactly the same as the first limit they gave you but with a variable switched

[austere ember]

this is a lot like example 5 that you just did

[uneven crest]

this is the part that is confusing me can u elaborate a little

because i thought that if u = cscx and if x-> inf its undefined

[austere ember]

no, didn't you already write that the limit as x approaches 0+ of cscx is +inf?

[austere ember]

this is undefined, yes

okay look let me explain

i have a limit i'm evaluating, the limit as x approaches a of f(g(x))

[buoyant latch]

you can prove this via epsilon delta

[austere ember]

it's wrong, i stated a counterexample

both have to be continuous

[buoyant latch]

yeah some version of that statement is correct

oh that one

i can't remember i did it a while ago

[hidden frigate]

if this is the same as example 6, x does not approach infinity, but rather u does

[uneven crest]

OMG

im so stupid

i wrote it incorrectly

[hidden frigate]

not stupid just silly mistakes!

[uneven crest]

wait but why is this 0 and not -inf

wait no it is inf?

ans key is wrong

[hidden frigate]

it should be zero

[austere ember]

it's 0 because lim as x tends to -inf of f(x) is the same as lim as u tends to -inf of f(u)

x and u are just dummy variables

the limiting values and the functions are the same

[uneven crest]

wait no its 0

[austere ember]

if you make another substitution, u = x, you'll get back to the original limit they gave you

[hidden frigate]

change of basis!

[uneven crest]

i am so lost...

[austere ember]

@uneven crest you know that f of something tends to 0 as that something tends to -infinity

they gave you that

[uneven crest]

ok

i see

[austere ember]

you can rewrite the second limit here as the limit of f(u) as u tends to -infinity

you're just calling the something x vs u

but it's the same problem

[astral bay]

the way i'd think of it is just

if you have some really small negative number $\varepsilon$

[wraith dagger]

bee [it/its]

[astral bay]

like -0.000001 or something

then $f\left(\frac1\varepsilon\right)$ is $f$ of a really big negative number

[wraith dagger]

bee [it/its]

[astral bay]

in the example of -0.000001 you get -1000000

[uneven crest]

why is this

we dont know what the function is

or at least im not seeing something

[astral bay]

and we know that f of a really big negative number is 0 (ish)

[uneven crest]

oh nvm

i see what ur saying

[astral bay]

well because $\frac1\varepsilon$ is a really big negative number

[wraith dagger]

bee [it/its]

[uneven crest]

ok so we get 1/sinx as x->0 = inf

and we see in the previous equation that

[astral bay]

...?

is this just an unrelated question now?

since i never mentioned sin(x) or g

[uneven crest]

oh sry lemme give ful context

here is question and work

[hidden frigate]

it might help to think about csc(x) as a number evaluated at "+0", which ends up being some really large number we call "infinity"

and if the limit at infinity for the function g(x) is infinity, then evaluating the function g(x) at this "csc(0+)"="infinity" will also give you infinity, which is why the limit is infinity

[uneven crest]

yes

that is waht i was thinking

tysm for confirming that i was unsure

.close

[cedar kiln]

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[rocky surge]

i need help

<@&286206848099549185>

i know how to graph the paraboloid but i dont get when it says "from -4 <x < 4 around x axis "

confuses me

[remote zinc]

It’s something like this

Just with the function you were given

As it says to “rotate around the x-axis”

[rocky surge]

My answer key dosent show that, and we have to graph in 2d not 3d.

this is the answer to the problem but i dont understand it

[remote zinc]

That’s what they did

The arrows are showing how it’s rotated around the x axis

[rocky surge]

Ill show what i got

[remote zinc]

Which axis is x and which axis is y

[rocky surge]

horizontal is x

up is y

its not drawn to scale

[remote zinc]

So first of all I think you rotated around y axis instead of x

Notice the circles go around the y axis

[rocky surge]

what does rotating around [blank] axis mean anyway?

[rocky surge]

in my drawing?

[remote zinc]

Yeah

[rocky surge]

Ok

Yea i see that

[remote zinc]

Rotating around an axis means imagine adding a third dimension

[rocky surge]

uhh

[remote zinc]

Your drawing a circle around the axis line where the axis line goes through the center of the circle

[rocky surge]

uhm

i think im just getting confused

[rocky surge]

how to get to this

[remote zinc]

So first of all just graph the function they have given

Which I can see you did initially

See how you drew the circles around the y axis?

[rocky surge]

Yes

this stuff right?

[remote zinc]

Yep

Your meant to do them around the x axis instead

[rocky surge]

Ohh

So if it says around y axis, I should do it over y axis?

[remote zinc]

Yep

[rocky surge]

Wow that makes sense

Ok

I got that part down

Now

oh wait

nvm

I understand it from there

you draw circles where the hyperbola extends to

whether it be on the y axis or x axis depending on the problem ( in this case x axis )

[remote zinc]

Yep

What you did is the correct solution if it was y axis

[rocky surge]

Oh yeah right

Great, but I also have another question if you have the time

[remote zinc]

Yeah no worries

[rocky surge]

This is a question on graphign 3d equations

im not sure if you can see, sorry for bad quality it says : x^2 = -z

[remote zinc]

Yeah

[rocky surge]

Yeah for this question

[remote zinc]

Have you tried anything?

[rocky surge]

I had like no thought process i was so confused but I did do one thing, I did identify the traces for the graph.

Traces:

Parallel to XY = Parabola
Paralel to XZ = Parabola
Parallel to YZ = Line

I am not sure if this is right

[remote zinc]

Yeah so id immediately rewrite it in terms of z

So pretty much z=-x^2+0y if you’d like

So you can see the z value is the same regardless of the y value

So your kinda left with a parabola which is just extended out in the y direction

[rocky surge]

What do you mean?

[remote zinc]

If you have two points say

(-4,100) and (-4,-100), or any other y value, the z value is the same

Assuming x is the same

[rocky surge]

Right

Wait

hmm

How would the z value still be the same?

[remote zinc]

Well plug the points into the equation

z=(-4)^2+0*100

z=(-4)^2+0*-100

[rocky surge]

ohhhh

[remote zinc]

The y axis isn’t taken into account

[rocky surge]

Well algebrically its right yes, its just that I can't visualize it in my head. Thats fine

[remote zinc]

What your left with is like a parabola in the XZ plane, that is just extended infinitely in the y direction

[rocky surge]

Sorry

How am i left with that parabola

I understand y axis isn't taken into account

[rocky surge]

in here

Don't we have to use these traces in account?

Sorry for these questions, kind of slow on this topic lol

Well

looks like my helper left

<@&286206848099549185>

can someone else help me

@remote zinc u still there?

<@&286206848099549185>

please anyone

[rocky surge]

<@&286206848099549185>

please?

[versed zenith]

Whats going on

[rocky surge]

Hey

Finally!

[versed zenith]

I doubt ik it unfortunately

I’ll try tho

[rocky surge]

Oh...

[versed zenith]

Hollon

[rocky surge]

Okay

So

How would I graph x^2 = -z

[versed zenith]

Wait

Is that it?

[rocky surge]

Thats just one of the standard question

[versed zenith]

O

Js make the arrows point to infinity

[rocky surge]

No rush but i want to go to sleep quick and its like 12 am for me rn and i have test tmr 💀

[rocky surge]

huh

[versed zenith]

Not point but goes

[rocky surge]

like

[versed zenith]

Its 3 am for me

Im studying physics rn not good time

[rocky surge]

oh dam

[versed zenith]

Tryna understand chaos theory

[rocky surge]

i have to graph it like this

[versed zenith]

The shortest explanation

Is on

Google

I think

I saw one like years ago

That was

1-3 mins

[rocky surge]

Oh

[versed zenith]

Wait

Did u not try google

https://youtu.be/fJ9Rh0ow0B0?si=erhUo8CB-bmJHmXw

[rocky surge]

lmao not like that

[versed zenith]

Ik

But

It’s practically the same besides a few additions

[rocky surge]

im kinda new graphing 3d

[versed zenith]

Same

I dont use graphs much

Besides parabola

Cant help mb g

[rocky surge]

k

[cedar kiln]

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[crimson sedge]

Help

[blazing zephyr]

this becomes trivial if we can get p^2q^2

oh wait

$$(p^2+q^2)^2=\sqrt{64}$$
$$p^4+q^4+2p^2q^2=8$$

[wraith dagger]

Skill_Issue

[blazing zephyr]

$$(p^2-q^2)^2=\sqrt{81}$$
$$p^4+q^4-2p^2q^2=9$$

[wraith dagger]

Skill_Issue

[blazing zephyr]

$$4p^2q^2=-1$$

[wraith dagger]

Skill_Issue

[blazing zephyr]

now we can write $p^4+q^4$ as $(p^2+q^2)^2-2p^2q^2$
so
$$8p^2q^2((p^2+q^2)^2-2p^2q^2)$$

[wraith dagger]

Skill_Issue

[blazing zephyr]

lets subtitute $(p^2+q^2)^2=8$ and $2p^2q^2=\frac{1}{2}$

[wraith dagger]

Skill_Issue

[blazing zephyr]

$$2(8-\frac{1}{2})$$
$$16-1=15$$

[wraith dagger]

Skill_Issue

[crimson sedge]

That's -17

@blazing zephyr am I making some mistakes

[blazing zephyr]

oh yea sorry

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[cerulean bloom]

hello, I've been trying this one but im stuck 😓

[crimson sedge]

do you know the relation between cscx and cotx?

using the pythagores theorem, we get an identity

can you recall that?

[cerulean bloom]

yeah

1+cot^2x=csc^2x

right?

[crimson sedge]

yep!

now, with that, the only issue left is of the sign right?

[cerulean bloom]

yea

[crimson sedge]

notice,

in the 4th quadrant

we have that sin -> negative

[cerulean bloom]

ok

[crimson sedge]

and what about cos?

[cerulean bloom]

positive

?

[crimson sedge]

yes

so cotx = cosx/sinx right?

so cot x will be negative right?

[cerulean bloom]

that looks right

[crimson sedge]

so now you just need to do the computation

you should have no problem

[cerulean bloom]

ok thanks

[crimson sedge]

np

[cerulean bloom]

im sorry could i have further help with the computation? 😓

[crimson sedge]

wait, you arent' given a calculator?

they expect you to calculate (3.544)^2 by hand?!?!

[cerulean bloom]

no thank god lol

im just stuck on hoiw to finish it up

[crimson sedge]

can you show the computation somehow?

[cerulean bloom]

so i have 1+cot^2x=-3.544^2 keeping that cotx is negative in mind

yea sure

[crimson sedge]

yea, so, (-3.544)^2 = 12.559936

[cerulean bloom]

mhm

[crimson sedge]

now, we get:

cot^2 (x) = 11.559936

[cerulean bloom]

why?

[crimson sedge]

we subtract 1 on both sides

[cerulean bloom]

oh got it lol

ok

[crimson sedge]

and then take sqrt

[cerulean bloom]

alr

[crimson sedge]

np : )

[cerulean bloom]

thanks so much

.close

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[winged wharf]

how to do question b?

This is my work

what should I do next

[winged wharf]

<@&286206848099549185> please help.

[cedar kiln]

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[chilly inlet]

x.9.9

[chilly inlet]

.close

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[slate lintel]

good talk

[cedar kiln]

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[mint raft]

Hey I managed to do the trig question I’d just like some feedback too see if I’m correct or not.

[mint raft]

It’s correct I’m certain.

.close

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[whole vortex]

What's the most correct way to address this expression?

[modern compass]

the y-th root

[cedar kiln]

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[half glacier]

f

[half glacier]

In the Belgian Lotto, 6 numbered balls are drawn from 42 (without
discount), then we classify these numbers in ascending order.
What is the number of prints possible?

need help for this .

how i can know if its in order or no ?

when i drawn balls

[cedar kiln]

@half glacier Has your question been resolved?

[half glacier]

<@&286206848099549185>

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[lyric narwhal]

The task is to find the product $$\prod_{n=2}^{\infty}\left(1-4\sin^2\left(\frac{\pi}{3\times 2^n}\right)\right)$$

[wraith dagger]

kheerii

[lyric narwhal]

asking this for the third time lmao, still no progress

by half angle identity this becomes $$\prod_{n=2}^{\infty}\left(2\cos\left(\frac{\pi}{3\times 2^{n-1}}\right)-1\right)$$

[wraith dagger]

kheerii

[lyric narwhal]

but what after this?

[mighty shuttle]

maybe polar coordinates ?

[lyric narwhal]

you mean complex numbers?

[mighty shuttle]

yes

[sand cradle]

Maybe something similar to https://math.stackexchange.com/questions/2145511/how-does-one-show-that-prod-n-4-infty-left1-2-sin-left-pi-over-2n?noredirect=1?

[mighty shuttle]

I mean this

[lyric narwhal]

I tried that, but the argument of the angle has a power of n within it

so it's not making a GP or anythng

[sand cradle]

Very similar to what OP on MSE had

[lyric narwhal]

hmm, they converted it completely into cos here

without the +

[lyric narwhal]

that made it way easier

the + is the bad part

or the - rather

[mighty shuttle]

are you sure it has a closed form ?

[sand cradle]

Yes, seems to be 2/3

[lyric narwhal]

yes

indeed

[mighty shuttle]

https://math.stackexchange.com/questions/3884804/evaluate-prod-k-1-infty-left2-cos-left-frac-pi3k1-right-1-righ

[lyric narwhal]

$$\prod_{n=2}^{\infty}\left(1+2\sin\left(\frac{\pi}{3\times 2^n}\right)\left(1-2\sin\left(\frac{\pi}{3\times 2^n}\right)\right)$$

[wraith dagger]

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

[mighty shuttle]

this looks similar

[lyric narwhal]

does this help?

[lyric narwhal]

ah yes this is quite similar

let me take a look

[mighty shuttle]

this stuff is scary NGL

all your questions are

sorry

[lyric narwhal]

this can become $$\prod_{n=2}^{\infty}\left(\frac{\cos\frac{\pi}{2^n}}{\cos\frac{\pi}{3\times 2^n}}\right)$$

[wraith dagger]

kheerii

[lyric narwhal]

does this help at all?

[lyric narwhal]

[mighty shuttle]

this is in the form (cos(3x)/cos(x))

[lyric narwhal]

yes, that's how I got this

$\cos 3x=\cos x(2\cos 2x -1)$

[mighty shuttle]

hmm

[wraith dagger]

kheerii

[mighty shuttle]

let me try somrthing

why are you even solving such problems

[lyric narwhal]

it's for jee lol

[mighty shuttle]

yeah, my idea doesn't work

💀 , first time I'm seeing these

in that case there will be some elementary trick

[lyric narwhal]

yes that's what I thought

[mighty shuttle]

hmm

multiply and divide by 2sin(pi/2^n) each time

I remember there being a formula for the product of the num

[lyric narwhal]

hmm

oh yeah you're right lmao

that's a pretty easy product

[mighty shuttle]

my idea actually worked?

wow I'm suprised

[lyric narwhal]

i think so yeah

lemme work it out

[lyric narwhal]

yes it actually did work

this is an awesome method

wow

[mighty shuttle]

could you share your working

don;t have a nb with me rn

sorry

[lyric narwhal]

it's just that identity

wait

$$\prod_{k=0}^n \cos(2^n\theta)=\frac{\sin(2^{n+1}\theta)}{2^n\sin\theta}$$

[wraith dagger]

kheerii

[mighty shuttle]

ah yeah

makes sense

thanks

[lyric narwhal]

yeah

[mighty shuttle]

I should have actually prepared for jee

[lyric narwhal]

lmaooo

it's ok

thank you so much

[mighty shuttle]

no problem

[lyric narwhal]

i've been stuck with this question all day

.close

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[slate garden]

How do i do this

[drifting marlin]

what do you know about the large data set?

[cedar kiln]

@slate garden Has your question been resolved?

[slate garden]

nothing

but it says u dont need to memorise the values

[drifting marlin]

if you know literally nothing then you can't answer the question

[slate garden]

oh shit

ok

I thought this implies

u don't need to

[drifting marlin]

well if you have never seen the data set, then you don't know what's sensible

just because you don't have to memorize values doesn't mean you can't know nothing about it

[slate garden]

ok thx

[cedar kiln]

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[proud sable]

Is there someway I can calculate (t) with the calculator (or not) because I dont understand How I can put 1,4 as a base

[hollow trail]

you can use the change of base formula [ \log_b(a) = \frac{\log a}{\log b} ]

[wraith dagger]

cloud

[proud sable]

Oh yes thank you !

[knotty stone]

hi everyone

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@proud sable Has your question been resolved?

[knotty stone]

whats up

[cedar kiln]

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[old sedge]

Sum and products of roots question. Where should I start

[cedar kiln]

@old sedge Has your question been resolved?

[quiet plover]

can you figure out the sum and product of zeroes here?

[eager wraith]

no

[cedar kiln]

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[quick flume]

I don't get the -root of x

how does it become 1/2 root of 2

[runic garnet]

@quick flume see this?

[quick flume]

yes

oh okay

I get it

I'm just tired lol

[runic garnet]

yes, rule 2 on ur formula sheet

[runic garnet]

try taking a break

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[hoary parrot]

Can someone give me the starting steps for this

I have no idea how to solve it

I don't want an answer, just a nudge in the right direction

<@&286206848099549185>

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[fast jewel]

so all i have trouble with is finding the subfunction for b

I know that you need to find the slope, multiplied

its just will i have the (x-x) + 1 in parentheses?

then is comma x>=7

.close

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[dreamy void]

do you still need help?

[fast jewel]

no thank you

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[peak bough]

what do they mean by this

[upper abyss]

Is this supposed to be a bad math source?

[cedar kiln]

@peak bough Has your question been resolved?

[peak bough]

I saw it on google

[cedar kiln]

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[marble smelt]

ok so

[marble smelt]

I'm not sure how to graph this one

because like theres no way of determining the period/frequency with only one time point and no period/frequency given

yknow?

i'm assuming that the period is 24 so the difference between the low and high is 12 hours but thats just not how temperature works like at all

[sly abyss]

just make a horiz tranls of 5 units

for example like sin(t-5)

diff between low and high is 12 ig bc its sinusoidal but

but like the max temp would be at t=17 hours

[earnest wadi]

What's 1+1

[marble smelt]

thats not the period/frequency though @sly abyss