precalculus

[fleet yew]

You got standard form

[sharp marsh]

oh

wait what

[fleet yew]

How does your teacher define standard form

Because im getting mixed signals here

[sharp marsh]

wait I did

Web assign is dumb

she's probably right

[fleet yew]

You good now?

[sharp marsh]

excuse me wtf?

[fleet yew]

Lol your teacher dumb

She calls "standard form" what everyone else calls vertex form

So you didnt even need to do all that expanding

[sharp marsh]

.

I think web assign sets up the problems

I'm not sure

[uncut mulch]

depends on where you are

[sharp marsh]

my brain hurty

[fleet yew]

y = 2(x-2)^2-1

This is probably what she wanted

The answer you got earlier

[sharp marsh]

why is math like this

juSt aGreE on A foRM

i'm fuming

[fleet yew]

What country you in

[sharp marsh]

USA

oh

that explains

alot

[fleet yew]

No

[sharp marsh]

oh

[fleet yew]

Im american

I do not call that standard form

I thought you were from some uncivilized country like britain that called things different from us

[sharp marsh]

oh

I mean US uses feet and lb cause we special like that

So I just assumed we pulled something weird again

[fleet yew]

What state

Not tryna dox you i swear

[sharp marsh]

Penslyvania

lol

[fleet yew]

Yeah thats weird

Your teacher is just a crazy

Now uh whats your moms maiden name

[sharp marsh]

LOL

[fleet yew]

Just curious lol

[sharp marsh]

ho

damn

just ask me for my social security already

and my credit card number why don't you

also last 3 digits

[fleet yew]

Yeah sure thatd be great

Also the date

[sharp marsh]

LMAO

just don't go to the notes on my computer

cursed stuff

[fleet yew]

I just want your money i dont care about your notes

No offense

[sharp marsh]

Not even wanting to know things about me :(((

I see how it is

Just take my money like those giRls lAst nIght

[fleet yew]

What sort of girls are you hanging out with that just take your money

You're getting played son

[sharp marsh]

Noted

I have a story not about these imaginary girls but something else

I'll tell in general

[gilded mirage]

Does everything look ok

[willow bear]

what's that last line

[gilded mirage]

Top or bottom

[willow bear]

bottom

log base ab of what

[gilded mirage]

Oh no it’s just log base a of b

[willow bear]

no on the rhs

[gilded mirage]

Yea

Just lowered the b too much

[willow bear]

...so you wrote "1 + log_a(b) = 1 + log_a(b)"

[gilded mirage]

Yea

[willow bear]

anyway are you sure you copied the thing to be proved correctly bc as written $\frac{\log_a(x)}{\log_{ab}(x)} = 1 + \log_a(b)$ does not appear to be true

[obsidian monolith]

Ann:

[gilded mirage]

Yea that was the problem

[willow bear]

let me just real quick make sure i'm not screwing up

[fleet yew]

Untrue statement

[gilded mirage]

Really

[fleet yew]

Yeah

[gilded mirage]

If the a and b on the rhs was switched would it be true

[willow bear]

gimme a sec.

[stuck lark]

seems true for values of a,b,x that make sense to plug in

[willow bear]

the caffeine has yet to reach my thinkpan

imma dip for a bit

[fleet yew]

Nvm its true

Just proved it

[gilded mirage]

I just had to do the reverse of change of base right

[stuck lark]

seems that it's sleepy math hours for both amd and ann

yeah welfare you're good

[gilded mirage]

omg so I did get a extension problem right

hope is not lost

[stuck lark]

my ONLY nitpick is that your work lacks

[gilded mirage]

oh yea I just fixed that

[fleet yew]

You're missing a few steps on your page that tripped me up

[gilded mirage]

what did I miss

I just thought of change of base

[stuck lark]

you kinda glossed over applying change of base but it's whatever

[gilded mirage]

do you guys have any tips on like doing problems like these

most of the time I don't know what to manipulate first

[willow bear]

a general rule of thumb is start with the side that looks more complicated

not a hard and fast rule by any means but it's easier to simplify than to complexify

[fleet yew]

Try random bullshit till something works

[gilded mirage]

i mean my teacher gave us strategies but thats all she really told us

like log properties, manipulating fractions with 1

[fleet yew]

Just be smart and see patterns

[stuck lark]

easier said than done

[gilded mirage]

wish I could be smart

[fleet yew]

Its not something that can really be taught

Thats why IQ tests are all patterns and shit

[gilded mirage]

so I should just practice a lot

and then it'll become second nature

[fleet yew]

Sure

[willow bear]

yes

practice makes perfect

[fleet yew]

Except with integration

[stuck lark]

integration needs practice too

[fleet yew]

i mean sure it does but it doesn't help much

[stuck lark]

you seem like you're speaking from your own experience which doesn't say anything about others

[fleet yew]

xkcd can back me up here

[stuck lark]

gotcha you're just meming

[fleet yew]

only a little bit

[pale bison]

Just be smart and see patterns
thanks man i'm going to solve RH

[stuck lark]

glhf

[valid violet]

I'll never be able to do integrals like the arccot(sqrt(golden ratio)) one so why try at all

https://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1/565626#565626

[formal dome]

I get the english definition of slope of tangent that if we want to find slope at point P and PQ is a secant line then slope of tangent is slope of PQ as Q tends to P where P remains fixed and Q is variable

But i don't get the mathematical definition. How is the slope of tangent equal to that equation on the right in above picture? If lim ∆x -> 0 happens then according to me, PQ should become a vertical line, because its not travelling in x at all

@serene heath? @viscid thistle? @mystic coyote ?

[pale bison]

nice pings

[viscid thistle]

No because even if it is a curve the limit is saying that when the height and width of that curve approach an infintesimally small value close to 0 the curve will become a straight line and u can use the definition of the gradient

[formal dome]

@pale bison they're my friends.. it's okay, they don't mind my pings

[pale bison]

i see

[formal dome]

@viscid thistle does ∆x approaching 0 imply that ∆f is approaching 0 too?

Ok that seems to point to that English definition. If ∆f and ∆x become zero that Q becomes P. But then it'll just become a point. How does a point have gradient?

[serene heath]

I have no idea who this person is

[formal dome]

@serene heath 😔😭

This'll just become 0/0 😩

[warped ether]

good god

[formal dome]

Ayyy explain please @warped ether

Or let me back into the server. I really need it for this online course😩

And will need it for when uni restarts too

Orrr someone else may help me too, I won't mind :harold:

[pale bison]

this does become 0/0

but you need to look at it geometrically

now, consider a smaller delta X

now, if you use the power of imagination, imagine as delta x approach 0
remember that, at no point are you setting delta x = 0!

then, you'll obtain the slope of this purple tangent line

@formal dome i hope this helps

[formal dome]

Hmm yeah it sort of does @pale bison

Thanks a lot 😊

[formal dome]

Why is it that when the circled term is expanded, only first two terms are written and then O(∆x)^2 is written instead of more terms?

And O(∆x)^2 means that its a sequence with powers of ∆x from two onwards and some constant, like a0∆x^2, a1∆x^3, a2∆x^4,..... right?

(ping me if you answer)

[pale bison]

you don't have to worry about that big O notation, what is important is the expansion of (x + delta x)^n
look up binomial theorem
@formal dome

[fluid shore]

Lmao he even says in the video that it's precisely because those are the terms that go to 0 when you actually evaluate the limit.

[formal dome]

you don't have to worry about that big O notation, what is important is the expansion of (x + delta x)^n
look up binomial theorem
@formal dome
Oh ik about binomial expansion. I was just wondering that why he didn't write more terms. And Abhijeet's answer was good enough, that they become 0 when limit is evaluated

And O(∆x)^2 means that its a sequence with powers of ∆x from two onwards and some constant, like a0∆x^2, a1∆x^3, a2∆x^4,..... right?

Ps, is my understanding of O((∆x)^2) correct? It's the first time ive heard this term

[remote veldt]

@formal dome Saying $f(x) \in O(g(x))$ or $f(x)$ is $O(g(x))$ means that (informally)

$$\lim f(x) \leq \lim Mg(x)$$ for some fixed constant M. Usually, the limit is to infinity, but sometimes we consider the limit to some other number (like 0). $f(x)$ and $g(x)$ don't need to be polynomials - they can be arbitrary function

[obsidian monolith]

Nicholas:

[formal dome]

[remote veldt]

@formal dome basically, in this case, x^2 + x^4 + x^6... as x approaches zero is affected mostly by x^2. Like, x^4, x^6 etc. go to 0 way more quickly

[formal dome]

Ohh yeah kind of makes sense

Ok i get that limit of O((∆x)^2) would be zero but in example of binomial theorem, O((∆x)^2) represents term with (∆x)^2 onwards, right?

[remote veldt]

the formal definition of towards infinity would be "you can select an x_0 such that f(x) <= Mg(x) for every x > x_0" - like, once you go beyond a point, f(x) is smaller than Mg(x). In this case, you're exactly right about what it represents

since we're going towards zero

you can basically say "oh ok so x^2 + x^4 + x^6 is dominated by x^2 so we'll gorup all these terms"

[formal dome]

Ok i get that limit of O((∆x)^2) would be zero but in example of binomial theorem, O((∆x)^2) represents term with (∆x)^2 onwards, right?
Then the powers increase, like (∆x)^3, (∆x)^4 and so on

you can basically say "oh ok so x^2 + x^4 + x^6 is dominated by x^2 so we'll gorup all these terms"
Ahhh this was good

Thanks a lot :'))

[warped jungle]

hey i need some help with the cursed trig identities

):

[stuck lark]

$\trig$

[obsidian monolith]

RokettoJanpu:

[warped jungle]

i have those but i feel like i did something wrong in my steps

[stuck lark]

Post

[serene heath]

malone

[warped jungle]

i combined tan with sin to get sin^2x over cosx

[stuck lark]

Show steps

[lucid plume]

cosine is an even function

[stuck lark]

^ that's where you messed up. this means:

$\cos(-x)\equiv\cos(x)$

[obsidian monolith]

RokettoJanpu:

[warped jungle]

so the bottom half of tan is cos x

?

[stuck lark]

ye

https://i.gyazo.com/aed804cd8fa1a853488bcb5c0f2f4669.png this isn't true either but w/e

[warped jungle]

ok

would it make sense if i made the number, 1-sin^2x, to cos^2x

?

[stuck lark]

if you look at my sheet we have $\sin^2(x)+\cos^2(x)\equiv1$ so whaddya think?

[obsidian monolith]

RokettoJanpu:

[sharp marsh]

Did I do this wrong?

[uncut mulch]

it probably didn't like the unnecessary characters

wait

why do you have a y= in there

[sharp marsh]

wait

idk

i'm dumb

fasdfssdf

Speaking of which

If y = k and x = 11

Then like what is y and x when it's in the other form

[sharp marsh]

Is this right?

[forest kiln]

what would the period be for this?

[fleet yew]

@forest kiln period of a trig function is the distance between minimums

Or maximums

[forest kiln]

omg it's ok i figured it out

could u double check my function for the graph though?

i got y=3sin1/2pi(x+pi)

idk it doesnt seem right tho

[fleet yew]

I would use cosine

[forest kiln]

can i ask why

[fleet yew]

It simplifies the equation quitea bit

The cosine function starts at its maximum

And so does this

So you dont have to worry about horizontal shifts

[sharp marsh]

What's the easiest way to find the X intercept

[fleet yew]

@sharp marsh

What is an x intercept

[sharp marsh]

um

When it's touching X axis

and Y = 0

@fleet yew

So like for 0= (x+6)^2- 81

What do I do?

To get 2 x intercepts

[clever inlet]

It's just a regular algebraic equation

What would you normally do?

[sharp marsh]

errr

0= (x^2+12x-45)?

[valid violet]

If you want to make it harder that's a good strategy

[sharp marsh]

What's the easier way

[clever inlet]

You have a squared term

Why not add 81 to both sides

And take square roots

[sharp marsh]

oh

wouldn't it be harder to square root the right side

[clever inlet]

Why?

[sharp marsh]

wait tnvm

wait

Would it be 9 = x+6 then?

[clever inlet]

That gives one solution

Sure

[sharp marsh]

I'm lost

[valid violet]

Consider a simpler example first

$z^2-16=0$

[obsidian monolith]

gfauxpas:

[valid violet]

How would you solve for z

[sharp marsh]

square both sides

but the -16

[valid violet]

$z^4 -32z^2+16^2=0$

[obsidian monolith]

gfauxpas:

[valid violet]

Now what

[sharp marsh]

Oh I meant

square root

[valid violet]

Show me what you mean

[sharp marsh]

Wait can I square root -16 or no?

[valid violet]

[sharp marsh]

wait my brain

[valid violet]

$z^2-16=0$ you need to know how to solve this before doing a more complicated case

[obsidian monolith]

gfauxpas:

[sharp marsh]

could you square root z and the 0?

then move 16 to the other side or no?

[valid violet]

$\sqrt{z^2-16}=0$

You mean like this?

[obsidian monolith]

gfauxpas:

[sharp marsh]

uh

i think

but like

Just the z^2

wait

o

Do you just move 16 over

and square root it?

If you square root it

Why do you have to square root the z^2 and the -16

Cause when you move the number over you just add that specific number

the opposite of it

or like

uh

ik but I'm saying

why when you square root something

you do it to the whole problem

And not the one thing

[valid violet]

@broken light no factoring, we're preparing him to do the case (x+3/2)^2-2=0

[sharp marsh]

I'm lost now

• or - 4

by squaring

[valid violet]

I dunno, I'm just here for free snacks

[sharp marsh]

lmao

wait

my brain just expanded

cause square rooty 81

u can do -9

Wait so for f(x) = -1 (x+2)^2 + 5

Did I get the X intercept wrong?

[clever inlet]

It looks reasonable

[sharp marsh]

Apparently one of the X intercepts

Is

(square root of 5) - 2

Isn't that the same as -2 + ( square root of 5 though) .-.

[clever inlet]

Yes

[sharp marsh]

Ok I guess the website is just dumb then

Thanks

[livid umbra]

Square root of 5 minus 2 is not the same thing as -2 times the square root of 5

[sharp marsh]

oh?

I meant + or - though

[livid umbra]

yeah thats right

[sharp marsh]

Thanks

[vernal spindle]

If sin(t)=3/8 and t is in the 2nd quadrant how can i find cos(t)

[merry sphinx]

You can draw a triangle to figure it out

Also, cause t is in the 2nd quadrant, what do you know about cos

[vernal spindle]

It's negative?

[merry sphinx]

Yeah

Ping me again if stuck

[vernal spindle]

I would use sinx+cosx=1 right

[merry sphinx]

No

You are overthinking this

Also, incorrect identity

sin^2(x) +cos^2(x) = 1

(which is not needed here btw)

Sin(x) is the ratio of what 2 sides

[vernal spindle]

opp/hyp

[merry sphinx]

Cos(x) is what ratio

[vernal spindle]

ajd/hyp

[merry sphinx]

I encourage you to draw an arbitrary right triangle and label the sides you know

[vernal spindle]

dumb question, how do i know which is the adj, hyp, and opp, i keep mixing them all up

[merry sphinx]

Ignore the bad drawing but this is basic diagram

So the hypotenuse is always the longest side of a right triangle

[fluid shore]

Just eat triangular pizzas and you’ll be able to memorize it

[merry sphinx]

The opposite side is the side opposite of the angle (in this case x)

Adj is always the side "next" to the angle that is not the hypotenuse

[vernal spindle]

which one is "the angle", it's which ever one im currently finding?

[merry sphinx]

Doesn't matter as long as you label sides correctly

You should learn about the unit circle if you have not already

[fluid shore]

Dude you might as well eat a pizza if you’re gonna learn about the unit circle

[fleet yew]

It has to have a radius of 1 meter tho

[vernal spindle]

I'm learning the circle right now, just struggling to understand it for some reason

[fleet yew]

Dude its just a fucking circle

Whats so difficult

[vernal spindle]

I don't mean the circle itself just all the angles and sides and formulas

[merry sphinx]

Yo STFU amd

[fleet yew]

Dude its just trig

[merry sphinx]

People struggle with this and this degrading does not help

[fleet yew]

No im actually genuinely asking what he finds so difficult

Usually when people say that it means that theres something important they're missing about the concept

[fluid shore]

@vernal spindle The first thing you do

is to forget about memorizing any formulas

Don’t even think about memory work here

[fleet yew]

^

[fluid shore]

Get a good book that will bring you through the derivations of each of the most prominent results

Then, do some hard problems to get yourself accustomed to using those results.

[vernal spindle]

I got this so far

[fluid shore]

Draw a picture of the situation

[willow bear]

,rccw

[fluid shore]

Eyy haii ann

[vernal spindle]

I drew it wrong?

[fluid shore]

No. You didn’t draw a diagram of the situation.

[vernal spindle]

so just a mirror of the triangle to the left?

[fluid shore]

Nope

What’s the context under which the term ‘2nd quadrant’ is used?

[vernal spindle]

What do you mean the context

[fluid shore]

Well, what is the term ‘2nd quadrant’ referring to?

[vernal spindle]

the -cos,+sin section of the circle?

[fleet yew]

It doesn't make sense for it to be on unit circle because the radius isn't 1. It probably means cartesian coordinates

[vernal spindle]

Not second quadrant like this?

[fleet yew]

Yeah its the same thing dw about it

You set it up right

Just find the adjacent side

[vernal spindle]

That's what I'm trying to figure out how to calculate right now

[willow bear]

sin^2(t) + cos^2(t) = 1

aka the Pythagorean theorem

[vernal spindle]

3/8+cos^2(t)=1 ?

[fleet yew]

sin^2(t) means sin(t)*sin(t)

Its probably easier to do from the triangle lol

Because integers

Just do normal pythag

[willow bear]

sin**^2**(t) + cos^2(t) = 1

(3/8)^2 is not the same as 3/8

@fleet yew the triangle does not show the quadrant that the angle is in

[fleet yew]

Doesnt really matter because t itself doesnt matter. Just make sure the signs are right

[vernal spindle]

I need to square 3/8 ?

[willow bear]

if sin(t) = 3/8 then what do you think sin^2(t) could possibly be if not (3/8)^2?

[harsh cipher]

Hello

[willow bear]

...

can we uh

not have two convos in one channel

maybe

[harsh cipher]

Hi, Ann

[willow bear]

either you or kingbluesapphire will have to move

[harsh cipher]

I will move

where can I move to

[willow bear]

#❓how-to-get-help

[harsh cipher]

exact location

[fleet yew]

#help-1 is free

[harsh cipher]

okay

[vernal spindle]

Ok so sin^2(t)+cos^2(t)=1

[willow bear]

yes

you know the value of sin(t), so you can isolate cos^2(t) and then find cos(t) by taking into account the quadrant that t is in

[vernal spindle]

I don't square 3/8 when replacing it for sin^2(t) right

I did 3/8-cos^2(t)=1
-3/8 from both sides
cos^2(t)=.625
sqrt(cos^2(t))=sqrt(.625)
= .791

but i dont think thats right

[uncut mulch]

did you ignore the comments being made?

sin(t) = 3/8
sin^2(t) = (3/8)^2

[vernal spindle]

So .927?

[fleet yew]

(a/b)^2=a^2/b^2

leave it in rational form

[vernal spindle]

How do I choose which is a, b, c, from opp, adj, and hyd?

[fleet yew]

no those were just arbitrary numbers

what i just said there had nothing specifically to do with triangles, it's just an algebraic rule

basically (some number/some other number)^2=(some number^2)/(some other number^2)

in math we often use letters of the alphabet to represent "some number"

[vernal spindle]

Oh so just leave it as a faction without actually dividing it?

[fleet yew]

no lol

read what i just said

how do you simplify (3/8)^2 if you know that (a/b)^2=(a^2/b^2)

where a and b are just some numbers

any numbers

[vernal spindle]

You mean like (3/8)^2 to 9/64

[fleet yew]

yes

[vernal spindle]

then 1-9/64 which is 55/64?

[fleet yew]

yes

that is cos^2(t)

yeah i see the problem now

[uncut mulch]

you don't need arccos

[fleet yew]

yeah mb said the wrong thing

[uncut mulch]

cos^2(t) = 55/64

[vernal spindle]

then cos(t)=sqrt(55)/8 ?

[uncut mulch]

well when applying the sqrt it would be cos(t) = ±sqrt(55)/8
then take the one with the correct sign based on the quadrant

[vernal spindle]

oh yeah forgot the +-

It would be - since Q2

[uncut mulch]

yep

[copper vigil]

hi

is this channel open

[vernal spindle]

I'm done

[copper vigil]

k thx

how do i use the limit definiton of the derivative to find the derivative of x

i mean ik it's 1 but i'm kinda stuck

[uncut mulch]

do you know the limit definition?

[copper vigil]

uh

lim as h approches 0 of (f(x+h)-f(x)/h)

right?

what is h though

that part kind of confuses me

[uncut mulch]

parentheses

just a variable

[copper vigil]

but i thought x was the variable

[willow bear]

as h approaches 0

h isn't anything

[plush heron]

they're both variables. When you're finding the limit as h approaches 0 you're finding an infinitely small difference between f(x+h) and f(x), the derivative

[willow bear]

besides maybe an arbitrary number that is close to zero but is not zero

[copper vigil]

i remember my teacher saying it had something to do with slope formula or something

but what does this have to do with deltay/deltax

[willow bear]

this is Δy/Δx that you're talking the limit of.

[copper vigil]

?

[willow bear]

h can be seen as Δx.
f(x+h) - f(x) is then the corresponding Δy

yknow, if y = f(x)

[copper vigil]

ok i think i sort of get it. so you have to find the slope of the tangent over a really small distance

h

so it's asking what happens as h get's small

but a straight line like y=x doesn't really have a tangent?

[plush heron]

It's a straight line but it goes linear

With a slope of 1.

upwards

[copper vigil]

wait so tangent line of x is itself?

what

[plush heron]

What other possibility could it be?

[copper vigil]

but tangent only touches at one point right?

like the tangent of a circle

a line touches itself like a lot

[plush heron]

The derivative is about the slope of the tangent line, not the actual line

[copper vigil]

ok but what is the tangent line

[plush heron]

Just an arbitrary line tangent to the curve

[copper vigil]

but x can't be tangent to x

[plush heron]

you're overthinking this tangent thing

[copper vigil]

ok then

what am i "supposed" to think

[plush heron]

The derivative is not the same thing as a tangent line. Instead, the derivative is a tool for measuring the slope of the tangent line at any particular point

[copper vigil]

ok

ooh

just realized something

the tangent line can actually pass through the curve multiple times right?

but its behavior only matters close to that specific point

like a cubic

that does the curvy thing

not all cubics do that but

@plush heron

right?

<@&286206848099549185>

im still kinda stuck

guess the americans are sleeping

[fleet yew]

Yes we are

Now stop pinging unnecessarily

[jolly silo]

Yes

[runic tapir]

Yes

[sour plinth]

well i guess im not american so i can't answer this question so i shall leave it to the americans

[left lance]

what's this definition of a derivative? can someone explain?

i've learnt this

ive understood the bottom one

is it same as the above one?

[fleet yew]

Yeah

[left lance]

what is that first onee 😩

[fleet yew]

The same thing

[left lance]

ik but...

[fleet yew]

But what

[left lance]

is it derived the same way? deltaf/deltax?

[fleet yew]

Yea

[left lance]

o i think i see it now

[fleet yew]

Uh huh

[rigid beacon]

They are not the same thing what

@left lance the black picture you sent will give you the derivative at x = a only

The pink picture will give you a generalized function that you can use to find the derivative at every point on the function

[odd helm]

How would I find HA for this?

[willow bear]

have you tried entering "y=1"

this looks like a formatting issue

[odd helm]

Oh that might have been it I guess

How would I figure out oblique when I’m only given the graph?

[fleet yew]

@odd helm oblique asymptote is the slanted line

That the function never touches as it approaches +-infinity

[prime yoke]

how do u do this thing

how do u find K

and how is finding the equation of cosine diff from sine

[vernal spindle]

Which formula do I use to calculate a side if i have 1 side and all 3 angles in a triangle

[empty fulcrum]

Law of Sines

But think about why that is for a sec

[vernal spindle]

Is it since cos^2+sin^2=1

[empty fulcrum]

Nope, the idea is you only want to have to solve for one variable

Notice how the law of cosines requires two "legs" and an angle, whereas law of sines needs only two angles and a leg

[vernal spindle]

this?

[empty fulcrum]

Mhm!

Have you used the law of cosines yet?

[vernal spindle]

I don't think so

[empty fulcrum]

Oh nvm then

It's a similar formula that can be used to solve for different sides/angles depending on whatcha want

For now just stick to the law of sines

[vernal spindle]

So for this

I'd want to find 9/sinc = b/sinB

[empty fulcrum]

Yeye

You can get angle B pretty easily, so just do some algebra and solve for b

Then plug and chug

[vernal spindle]

9/90=b/79 ?

[empty fulcrum]

Not quite

sine of those angles

The sine of 90 degrees isn't 90, it's 1

You'll have to use your calculator for 79

[vernal spindle]

Why when I put sin(90 in a calc i got .893

[empty fulcrum]

Are you in radians mode

[vernal spindle]

Oh yeah I forgot to change that

Wait it is in radians

I need degrees?

[empty fulcrum]

Yes

Take a step back and think about it when stuff like that happens

For instance, you should have sin(90) memorized, so if you get something like that out it should tip you off that something's wrong

[vernal spindle]

Should I leave it in degrees or will I need radians for anything with triangles

[empty fulcrum]

Depends on the question

In this case angles are in terms of degrees

Mostly you'll need rads though

[vernal spindle]

I got it right!

The law of sine is pretty fun

[empty fulcrum]

Good job!

Yeah it's a fun one
You'll learn some more about it soonish, I assume you juuuust started with it

[vernal spindle]

My problem is remembering all these formulas for tests

[odd helm]

Wouldn’t you take the sqrt of 1/4 on the last step but instead he said that -1/4 < x < 1/4 what happens to the x^2 why does it just become x?

[pale bison]

looks like a mistake

it should be -1/2 < x < 1/2 after taking the sqrt

[viscid thistle]

If someone could help me with this that would be amazing

[rigid dune]

What the madness is that

What the madness is that

What the madness is that

Discord be glitching me out

[small kayak]

does anyone know how to add vectors that arent on a plane

[harsh cipher]

Hello

how do I solve question a?

i can't get past 2cosx - 3 x 1/cosx - 1 = 0

[dim jungle]

multiply by cos(x)

quadratic in cosx

[harsh cipher]

so I'm supposed to multiply 1/cosx by cos x to make denominator the same

?

[willow bear]

DO

NOT

USE

THE

LETTER

X

FOR

[harsh cipher]

I feel so dumb because I finished the lesson and didn't have problems but when I get to the assignment

[willow bear]

MULTIPLICATION

[harsh cipher]

Hi Ann!

[empty fulcrum]

Multiply the whoooole mess by cosine

[fiery wave]

1/cosx x cosx = 1

[willow bear]

[harsh cipher]

okay I got it now

phew.....

[jovial horizon]

@willow bear what do we use instead of x for multiplication

[jovial horizon]

epic

[runic blade]

Limits is precalculus right?

Can someone check pls. I'm getting k to be infinity

Are you getting it as infinity as well?

[fading token]

Did you just delete the messages just to type them again

[runic blade]

I thought I was in a different server

Lol

[fading token]

[runic blade]

[fading token]

Anyhow you can literally plug in x = pi/2 in that expression

[runic blade]

Cos90 is zero

K becomes infinity.

[fluid shore]

No

[runic blade]

What

[fluid shore]

It just means that, apparently, $k \cdot 0 = 3$, which is false for all $k \in \bR$.

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

There's no further information given about k

[runic blade]

The question is wrong then?

@fluid shore

[fading token]

I don't think that's the entire question

[runic blade]

Yes it is

Why don't you think so

[fluid shore]

Well, then there seems to be a problem there

[runic blade]

Maybe the denominator was supposed to be 90-x?

[fluid shore]

Like i said, k.0 = 3 is just false for all k belonging to R.

Why would it be 90 radians - x?

[fading token]

I think they meant 90 degrees

[runic blade]

90 degrees

[fluid shore]

I know lol, i'm just messing around

Stick to the units being used for angle measure in the question

[runic blade]

π/2 - x

[fading token]

I think the FULL question is determine the value of k such that f is continuous everywhere?

I wouldn't be able to make sense out of it otherwise

[runic blade]

@fading token f is continuous everywhere if it is continuous at that point anyways

[fluid shore]

Riji was just clarifying that that's what the question was

[fading token]

Continuity is never mentioned in your screenshot

So that's why I was wondering

[runic blade]

Ohh

I got you

[fading token]

Wait

I can't read

It is mentioned

[willow bear]

f is guaranteed to be discontinuous at pi/2 no matter what

[fluid shore]

Well, okay, they're sorta kinda implying it

[fading token]

[runic blade]

They say it inside the brace thing

Whatever it is called

I mean after the definition of f(x) not inside the brace

By the way. Why do we only put one brace? Why not two braces?

[willow bear]

the fact that it's said inside the brace thing like like bad typesetting

but the brace thing and the fact that it's a single brace is just... that's how the notation is defined?

[full garden]

hey guys I have a small question

h(x)=mx2+2 , evaluate h(−3a+1)

am i doing this properly

because I never saw this a^2m before

[modern kettle]

$-6am+m\neq -5am$

[obsidian monolith]

ChrisIntegral:

[modern kettle]

in general

[full garden]

oof this is embarassing

thank you Chris thank you man

[modern kettle]

👌

[viscid thistle]

I'm so confused but, I'll put this here in hopes someone can explain.

Given an inequality like:

(2x+1)/(x-3)<4

Why is it necessary to consider both cases for when x-3>0 and x-3<0?
Why is it when I consider the case when x<3, (when the denominator is negative), that I get a result like x<13/2. I try to use 4, which is less than 13/2, but it doesn't work. Why?
It seems to only work when x>13/2 or x<3. But Shouldn't 4 work if it is less than 13/2? But then it clearly doesn't, which is the source of my confusion.

[vernal spindle]

I'm confused on how to start this problem

[clever inlet]

Well

What does the notation mean

[vernal spindle]

wait do I just plugin 5pi/6 for x?

[vernal moon]

yes

@vernal spindle

[vernal spindle]

I got sin(5pi/3)

[stuck lark]

evaluate that

[vernal spindle]

sin(300)

[vernal spindle]

I am trying to calculate a missing angle in a triangle with 12.53/1 = 6/sinb but I'm stuck what to do next to isolate b

[clever inlet]

Well

Why not try flip both sides for a start

[vernal spindle]

1/12.53 = sinb/6 ?

[clever inlet]

Yes

Now solve for sin(b)

[vernal spindle]

Didn't know you could do that

[clever inlet]

You're just taking the reciprocal of both sides

[vernal spindle]

I got sinb=.47885

[clever inlet]

Probably shouldn't round now

But whatever

[vernal spindle]

I used a program to get the answers and none show .47885 so i did something wrong

[clever inlet]

Take arcsin

You have sin(b)

You want b

[vernal spindle]

Ohh sin^-1

[willow bear]

do not confuse an angle with its sine

[vernal spindle]

in 5=b/(sqrt(3)/2) do i just multiply both sides by sqrt(3)/2 or do i flip it to 2/sqrt(3)

[willow bear]

you can do whatever you want so long as you do not mess up in doing it

you can multiply both sides by sqrt(3)/2, or you can multiply both sides by 2/sqrt(3), or by 4647/3222245, or by 7π+1

[proud jetty]

need sum help with composite function

https://gyazo.com/43cc267c3654017a11e5c1282a8f26f6.png

what happens after f(2)? why is final answer 3?

[long tartan]

you're given f(2)=3

that's just applying what they gave u

[proud jetty]

what happens to g(3)=2. Its inside the bracket.

[long tartan]

you're taking f(g(3))

imagine g(3)=x

so it's like f(x)

but since g(3)=2 x=2

so it's f(2)

makes sense

[proud jetty]

yea but why is second one zero?

[long tartan]

you're given f(5)=0

that's at the top right

also ur given g(2)=5

[proud jetty]

oh

nvm i see

thx

[long tartan]

np

[full garden]

guys can someone please tell me what like finding the reference is going to make us accomplish or like what other information would it help us obtain ?

[remote veldt]

@full garden do you mean finding the reference angle, like in a trig context?

[full garden]

sorry bro yes

like finding it was kinda easy but I don't really understand why we find it

like for coterminal makes sense

but refrence angle I kinda don't know why we use it

[remote veldt]

ok. The reference angle gives you the magnitude of whatever trig function. So, the reference angle tells you the angle with the x-axis. Remember, sin is opposite/hypotenuse, cos is adjacent/hypotenuse, tan is opposite/adjacent. The angle your triangle makes with the x-axis determines the ratios of the side lengths

like here, the red triangle and the blue triangle are in different quadrants, but would have the same reference angle

(excuse the shitty paint)

[full garden]

bruh what do you mean love you for this amazing explanation bro

bro but does that mean the like the cos and the sin will be the same

no? right

[remote veldt]

no, but they'll be same magnitude

[full garden]

I think only tan is + in third

[remote veldt]

yes! only tan is + in the third

that's exactly what you use the CAST rule for

the reference angle tells you how big cos/sin/tan are

the quadrant tells you the sign

[full garden]

thank you

[proud jetty]

can someone explain this

https://gyazo.com/20306b6e5804dd39ac3dd7687c1880aa.png (ans:C )

[patent beacon]

If I take x² and stretch it vertically by 1/2, the new function would be 1/2 x²

If I take x² and stretch it horizontally by 1/4, the new function would be (4x)²

Note that 1/2 x² ≠ (4x)² so A isn't the answer

@proud jetty

[proud jetty]

Yes I see that but

[patent beacon]

You want to know why C is the right answer?

[proud jetty]

Since the answer is C. 2sqX and sq4 isn't same right?

yea

[patent beacon]

Simply because
√[4x] = √[4]√[x] = 2√[x]

[proud jetty]

ah The basics

thx thx

[patent beacon]

Np. Knowing that √ and ² can split over multiplication is cool knowledge

[proud jetty]

I coulda had also checked using graphing as well?

indeed. Who knows when it could come handy

[patent beacon]

Yeah, just need to show that √[4x] and 2√[x] are the same function

[proud jetty]

mk thx mate

I got unit test tom wish me luck

[patent beacon]

Or take the √x function and stretch it vertically, it should seem reasonable that this also looks like a horizontal compression

Good luck!

[proud jetty]

thx

[left lance]

@left lance the black picture you sent will give you the derivative at x = a only
@rigid beacon ohhk ty(sorry for the late reply lol)

[sturdy haven]

How to simplify complex fractions???!??!!????!?????

[neon garden]

Can anyone explain how holes work in question 3? I know that it’s (6, -22/3) but I don’t get how that works. Wouldn’t f(6) just be 0?

[raw night]

Nope, because it is 0/0 which is undefined

[neon garden]

Yeah undefined, but how is it a point then?

[raw night]

It's not a point, it's a hole

It is not a part of the graph of the function

[neon garden]

Oh

Ok

[viscid thistle]

How would you attempt this question? I've tried a couple of different ways but I haven't seemed to have got it

[clever inlet]

Hmm

What have you tried

[viscid thistle]

Firstly, I have tried plugging in ath(2x/1+x^2) into ath(x)

[clever inlet]

Ok

[viscid thistle]

Didn't work

[clever inlet]

Ok

The algebra for this

Is

Honestly not that bad

[viscid thistle]

Next, I tried plugging ath(x) into 2ath(x), don't think I did it right

[clever inlet]

What did you get when calculating ath(2x/(1 + x^2) exactly?

In fact

Just ignore the log for now

Just look at the fraction

What did you get for that?

[viscid thistle]

Gotta do the working out for that again hol' up

At this point I feel as if I’m doing something horribly wrong

[clever inlet]

nope

it's perfect

(a^2)/(b^2) = (a/b)^2 right?

[viscid thistle]

Yeah

[clever inlet]

additionally (x - 1)^2 = (1 - x)^2

right?

[viscid thistle]

Aren't they 2 different things tho

Oh wait

nvm

yes they are

[clever inlet]

(x - 1)^2 = (-(1 - x))^2 = (-1)^2(1 - x)^2 = (1 - x)^2

so we have

(1 + x)^2 / (1 - x)^2 = ((1 + x) / (1 - x))^2 right?

[viscid thistle]

yeah

[clever inlet]

so that's the argument to our function

what log laws can you apply

[viscid thistle]

Ah I see

you can bring the 2 to the front

but then hold up

Like this?

[clever inlet]

exactly

yes

[viscid thistle]

Oh wait that's the final step yooo thanks so much

[clever inlet]

np

[fading token]

Chitoge from Nisekoi

[lucid hearth]

worst girl

#TeamOnodera

[viscid thistle]

Misaka Mikoto

[novel dirge]

I have to find the parameter m for which P wont be empty

I got 2 but the solution says 7

[willow bear]

you are correct

[novel dirge]

So the solution is wrong?

[willow bear]

i don't know, are you sure you're looking in the right place in the answer key

[novel dirge]

Yes

[willow bear]

are you sure that m and not x is what you are asked for

[novel dirge]

It even says that P is empty for every x not 2

[willow bear]

can you show where it says that

[novel dirge]

M*

[willow bear]

i know it's in another language but i can tell it's a Slavic language

[novel dirge]

Yes

[willow bear]

so i can probably figure it out

[novel dirge]

I just realized that it is indeed looking for x

But yeah, my mistake. I know what to do now

[lucid hearth]

,rotate

[fluid shore]

Oooh is that your worksheet from school or uni?

[pale bison]

it's from ur mom

jk

prank

relax

memes

[fluid shore]

Rekt

I got pranked

[novel dirge]

School

But I am preparing for uni

[fluid shore]

What the fuckkkkkkkk

They teach logical notation in your school then?

Jeez

[novel dirge]

Thats first year of hs

[fluid shore]

That's actually pretty amazing

[novel dirge]

But the sad thing is that our teacher did a shitty job and almost nobody learned it

[fluid shore]

Have fun with it, it's pretty cool stuff

Oh

[novel dirge]

And we only did simple prpblems

Like this one

[fluid shore]

I remember helping you out once with Flynn on integration problems

[novel dirge]

Or this

Yes

[fluid shore]

Oh

That's pretty sad. There are resources you can use to learn this stuff properly, though

[novel dirge]

I mostly do understand this though

Just need practoce

[fluid shore]

Ahh

[fluid shore]

Wow i'm actually pretty impressed though. Logic & Set Theory aren't covered well, if at all, in the IB. I learnt them almost entirely on my own.

[novel dirge]

Any ide how I can approach this?

Well, I also have to learn it on my own😢

[fluid shore]

,rotate

[novel dirge]

I have to get the total number

[fluid shore]

I would suggest drawing out a venn diagram, actually

Then, work through the problem by finding each bounded region in the venn diagram.

[novel dirge]

Hmm

Ill try

Having a bit of a hard time vizualising this in ny head

What exactly I want to do

[neon garden]

Can anyone help me graph 15? I understand 7-13 when it’s linear, but I’m not sure what to do when the degree > 1

[fluid shore]

Don't visualize it in your head. The whole point is to draw the diagram so you don't have to visualize it in your head

[neon garden]

I tried, I have a graph with the vertical asymptotes and intercepts

[fluid shore]

Pandastrafes, $\frac{x-2}{x^2-x-6} = \frac{x-2}{(x-3)(x+2)}$

[obsidian monolith]

Abhijeet Vats:

[neon garden]

I understand that

[fluid shore]

Uh i was talking to mr pancake with the previous remark i made

[novel dirge]

Cant he also solve it as a quadratic equation and get x1 and x2?

[neon garden]

Oh lol

[novel dirge]

But that would probably be useless

[fluid shore]

But anyways, you can see that there are two vertical asymptotes at x = 3 and x = -2

[lucid hearth]

1. Find x-intercepts (y=0)
2. Find y-intercepts (x=0)
3. Find horizontal asymptote (by limit as x->∞ and -∞)
4. Find y-asymptote (by limit as y->∞ and -∞)
5. Find maximum/minimum point by f’(x)
6. Using the coords u found in the previous steps, find out if the graph curves up or down on every portion in between them by using f’’(x) [positive -> U, negative -> n]
7. Find any graph holes that makes graph not continuous

[neon garden]

This is what I have so far

[fluid shore]

Alright, now consider the roots of f(x)

Clearly, f(2) = 0

So you have one point down

Now, you need to find the behaviour of f(x) near x = -2 and x = 3

[neon garden]

So I do f(2.9) and stuff like that?

[fluid shore]

That would be a good start, yes.

Like, okay, just trying some points out to get a feel for what it's like is fine

But in general, let x -> 3 and x<3. Then, clearly, x-2>0 and x +2 > 0. However, x-3<0. So, you're dividing a positive number by a negative number. That means that f(x) goes to negative infinity as x ->3 from the left.

[willow bear]

x -> 3 and x<3
that's called x -> 3-

[fluid shore]

I wasn't sure that they were familiar with that notation

[neon garden]

Ok so f(2.9) = -1.8 roughly

[fluid shore]

But anyways, don't interrupt

Yeap

[lucid hearth]

lim(x→3⁻)

[fluid shore]

Alright so it's negative right?

[neon garden]

Yeah

[fluid shore]

Did you understand what i was saying above?

[neon garden]

Kind of

[fluid shore]

The idea is that it becomes more and more negative. It goes to negative infinity.

You can try it with another number that's close to 3

[novel dirge]

When you finish with him can you help me a bit?

[fluid shore]

Sure

[novel dirge]

Im nkt making any progress

[fluid shore]

But pandastrafes, can you see why it goes to negative infinity?

[neon garden]

Because the y values keep getting smaller but never hit the asymptotes?

[fluid shore]

Now, suppose that x -> 3 and x>3. Then, x-2>0 and x+2>0. In this case, x-3>0 as well. So, you're dividing a positive number by the product of positive numbers. The result is going to be a positive number and it's going to get extremely large the closer you go to x=3 from the right.

Well, they keep blowing up as negative numbers but they do not hit the asymptotes because f(x) is undefined at the asymptote

[neon garden]

do I just keep plugging in values into f(x) until a get a general shape of the graph?

For the middle part I know how to do the ends

[fluid shore]

No. I've given you a way to reason out how to get the general shape of the graph

Similarly, Kelfran has given you a way to work it out as well.

Attempt to use the reasoning i've used above to determine the general shape of the curve near the asymptotes

[lucid hearth]

(also I hope you put a hole at x=2 on question 13)

[neon garden]

I did

[lucid hearth]

ok nice

[fluid shore]

Work through it again and look through the approaches that Kelfran and I have presented. Really, really put in some effort into it. Don't just give up after a few minutes. Then, come here if you have additional questions.

[neon garden]

So as x approaches -2, f(x) -> positive infinity. And as x approaches 3, f(x) -> -infinity

is that kinda what you are looking for?

[fluid shore]

Not quite. See, the direction of your approach does matter.

But you're getting somewhat close to the general idea

[lucid hearth]

Well, you can just find the positions of the asymptotes, then later see if it’s increasing or decreasing using f’(x)

To see which goes down to -∞ and which goes up to ∞

[fluid shore]

I don't think they know how to use derivatives 😦

Or do they?

[neon garden]

No

[lucid hearth]

Okay..

[fluid shore]

Anyways, pandastrafes, look through both paragraphs that i wrote for you

And try to replicate the reasoning used for x = -2

[neon garden]

ok

[fluid shore]

Then, come back here when you're really stuck

[novel dirge]

@fluid shore my turn now?

[fluid shore]

Indeed

Post the exact question you have

[lucid hearth]

,rotate

[novel dirge]

Ina school 330 students learn french, 470 english, 420 russian

140 f and e

180 f and r

250 e and r

And 120 all 3

[fluid shore]

Aight so what's the question?

[novel dirge]

How many students in total

[fluid shore]

Well, there's a rather interesting formula in question 46 that might just apply here

But rather than relying on formulas, the first thing you should do is to draw a venn diagram to visualize what's happening

[novel dirge]

Did it and didnt help

[fluid shore]

So there will be 3 circles you need to draw. Each represents the set of students that learn french, english and russian

Show me the diagram you drew

[fluid shore]

Ah that's where you've goofed

See there are 330 students who learn french in TOTAL

That includes those who learn french only, those who learn french & (whatever) as well as those who learn all three languages

Also, labelling your diagrams is a very good thing to do

Your diagram should also be about half a page before it will do you any good. It's for your own sake, it helps to make things more clear

[novel dirge]

Did it the first time

[fluid shore]

Anyways, did you understand the point i made above?

[novel dirge]

Yes

[fluid shore]

Similarly, 180 students learn french and russian. This also includes those who learn french, russian & english

[novel dirge]

But now idk how to draw that

[fluid shore]

So, as long as they learn both french and russian, they're included in the set of those who learn both languages.

It's the same diagram

However, begin from the inside and move outwards

I'll work you through it, okay?

[novel dirge]

Ok

Do I put that in a circle or an intersection?

[fluid shore]

We are told that 120 students learn all 3 languages. Furthermore, we're told that 250 learn english & russian. So, how many just learn english & russian (without french)

Put what in a circle or an intersection?

120 is exactly where you intially placed it

[novel dirge]

All 3?

Ok

[fluid shore]

Anyways, answer my question

[novel dirge]

130?

[fluid shore]

Good

Then, we're told that 140 learn french & english

How many just learn french & english only