precalculus

[stuck lark]

very good. use parens in typing though. 3x(x+4)/((x+3)(x-2))

[odd helm]

Ok

[viscid thistle]

I have a triangle

that ends up as sin55/a = sinb/7 = sin65/c

how do i create an equation thats solvable if they all have a missing element

[patent beacon]

Neither give enough information to determine anything

Well, you can say things about b maybe

[jade oar]

yea if f,g, were differentiable u could use lhopital's rule and say something about it using the derivative for (a)

[hoary valley]

1s

[jade oar]

if denominator goes to 0 your fraction goes to infty

(with numerator constant)

[hoary valley]

Hopefully this helps.

[patent beacon]

What's the difference? You put your opinion on them?

[hoary valley]

That's the question written like that in the book. no one can asnwer it,.

[patent beacon]

There's no one answer. Take (x² - 25)/(x - 5). The limit is 10. Instead, take 0/(x - 5), the limit is 0

[jade oar]

you don't have enough infomration to say a) doesn't exist, and you are wrong for b)

[hoary valley]

Told you, it's a really hard question. even my teacher doesn't know the answer.. but It's literally written like that in my book,

It's either true or false.

[jade oar]

It's not

whoever wrote your book did a bad job I guess

[patent beacon]

Okay, then we go with false

[jade oar]

Teachers and bookwriters are not infallible

[hoary valley]

James Stewart - Calculus_ Early Transcendentals-2015

@patent beacon on both (a) and (b) ?

[jade oar]

For a) it exists if the functions are both differentiable, which is something that is unknown (L'Hopital's rule). b) it goes to infinity

for example, consider 1/10^(-1) = 10. and 1/10^(-99999) = 10^99999

[patent beacon]

Can't promise infinity. Might be negative infinity from one direction

[jade oar]

true it could be +- infinity, it may not exist too for that reason. bleh

[patent beacon]

In either case, the function might not be defined to the left or right of 5, so you still get no promises

[jade oar]

i don't think it's important to be defined on both sides, the main problem is since we dont know anything about the function going to 0, it could be going to both 0 from negative and 0 from positive, like some stepwise function that changes sign and decreases in abs value

when taking x to the limit point

[viscid thistle]

how do you solve sinθ = 24/25, 0 < θ < pi/2 find cos (2θ)

im doing the double angle form

still not making sense

[rich flint]

@viscid thistle can you post what you have

[viscid thistle]

1 - 2 (24/25)^2

[rich flint]

Take a pic of your work and post it

[viscid thistle]

its all jibberish

i think he put the problem wrong

[rich flint]

if that is what you got as your answer I don’t see the issue

[viscid thistle]

the answer doesn't make sense

the answer is supposed to be -527/625

[rich flint]

Ok so they told you 24/25 is for sin theta?

[viscid thistle]

yea

and were supposed to find cos(2theta)

i used double angle

form

[rich flint]

You used the formula and you plugged it in

Post the original question then

[viscid thistle]

sin θ = 24/25, 0 < θ < pi/2, cos (2θ)

[rich flint]

Yep I got the same answer as the solutions

So since I can’t see your work I have no clue what needs to be corrected

Only thing I can think of is maybe when you simplify your fractions or when you square the sin theta

So for sin(x) your suppose to square both sides

[viscid thistle]

so 1 - 2sin(24/25)^2 == 527/625?

[rich flint]

Equals -576/625

Did you square the sine before you plugged it in

[viscid thistle]

no

what does that do

[rich flint]

Well they gave you sin(x) and the formula asks you to plug in (sin(x))^2

[viscid thistle]

isnt it 1 - 2(527/625)

[rich flint]

Yes that is correct so maybe you went wrong when simplifying the answer

You multiply the 2 by the fraction then you get the LCD which is 625

[viscid thistle]

oh

its 625 / 625

• 1054 / 625

[rich flint]

No clue what your saying

[viscid thistle]

alright

[rich flint]

2 times 576 is what?

[viscid thistle]

1 - (1054 / 625)

OOp

[rich flint]

1054 is wrong try again

[viscid thistle]

I see

1152

[rich flint]

Correct now what do you do to the 1 in front?

[viscid thistle]

625/ 625

[rich flint]

Correct

[viscid thistle]

thanks, what formula do i use for sin θ/2 *

[rich flint]

It matters what the question is asking and what information is given

I’ve never seen what you just typed there is usually an angle with sin

Sin and cos can’t just equal that

[viscid thistle]

**

sin θ = 1/4 in Q 1 or 4 find sin θ/2

[rich flint]

So what are they asking you to do?

[viscid thistle]

what formula do i use

oh

[rich flint]

Are they ask for the degree?

They have a chart for half angle formulas

[viscid thistle]

ah half angle

how do you decide which one to use

[rich flint]

You mean when dealing with signs?

[viscid thistle]

oh nvm

you know how cos 2x can be translated into multiple things

how do you decide which one to use

its just situational

ig

[rich flint]

It is what information is given you can find the solutions with different methods so yes it is situational

Each formula gives you the same answer

[viscid thistle]

alright thanks

[slender siren]

@flint stirrup

[flint stirrup]

yo

do you have a specific question ?

or problem

[slender siren]

Both

[flint stirrup]

like a homework problem

[slender siren]

So in level 2 of factoring trinomials right and yes

[flint stirrup]

it looks like this right

[slender siren]

Yes

But it’s level 2

[flint stirrup]

wdym level 2

[slender siren]

Hold on I’ll show you

Sorry if you can’t see this my camera is blurry

[flint stirrup]

its ko

[slender siren]

You know how to do this one

[flint stirrup]

yes

[slender siren]

You divide the first number by all the numbers right?

[flint stirrup]

yes

you factor out a number, which is 3 in this case

so take 3 out of each of the terms

what would you get?

[slender siren]

x to the second power + x to the second power -42?

[flint stirrup]

yes

[slender siren]

I meant minus

[flint stirrup]

but dont forget the 3

it looks like this

$3(x^2-x-42)$

[obsidian monolith]

Meir:

[flint stirrup]

since we factored out the 3

[slender siren]

Oh we got to put the the 3 out of the parenthesis?

[flint stirrup]

yes

[slender siren]

Alright hold on I’m going to come back to you in less than 3 minutes

[flint stirrup]

ok

[slender siren]

@flint stirrup like this?

[viscid thistle]

how do i put this in stardard form

(1 + i) ^ 20

[slender siren]

@flint stirrup

[brisk frigate]

Does anyone know how to solve this? I tried using arcsin(3/5) on my calculator to get the value of "t" and then substituted that into my calculator to find the rest

As far as I know arcsin turns the fraction into a degree and sin turns the degree into a fraction, so if in this case t is the degree that you have to input onto sin(t) to get 3/5 then shouldnt using arcsin(3/5) give me the value of t?

[patent beacon]

Your calculator will give you the Q1 angle

You don't need a calc here either way. Use
sin²θ + cos²θ = 1
To get cos. All of the rest follow

[brisk frigate]

oh i see, btw is there a way to turn the Q1 angle the calc gives me into the next quadrant?

wasnt it like 180-the calc

[patent beacon]

Perfect, that's it

[brisk frigate]

i took trig a while ago so i forgot alot

oh ok thnx

[patent beacon]

Imagining the unit circle is key

[brisk frigate]

yeah, it helps to figure things out

thnx for the help

[patent beacon]

Np, good luck with it

[harsh cipher]

hi

the part where it says -3 fits this condition but not x>0

what does that mean?

x>0 still fits the condition x>-4 ?

lol

[uncut mulch]

-3 is greater than -4
but it isn't greater than 0

x needs to be both:
greater than 0 and greater than 4
the intersections of those is x > 0

[harsh cipher]

......thinking......

[uncut mulch]

(and then the additional restriction of x != 1 is also applied)

[harsh cipher]

so its in between x>-3 but less than 0?

-__-

[viscid thistle]

how do i solve sqr2 ^20

[uncut mulch]

no,
the -3 was just an example of what doesn't work

at x=-3,
the argument, x + 4 = 1
however the base will be -3 which is negative which will make it undefined

[viscid thistle]

How do find the inverse of trig functions

like f(x) = 3sin x - 2 | f^-1(x) = ?

[rigid sun]

arc functions, are the closest you’ll ever get to an inverse

you’ll have to do the math from there

x=3sin(y)-2

x+2=3sin(y)

(x+2)/3=sin(y)

arcsin((x+2)/3)=y

,w plot arcsin((x+2)/3)=y

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

Ewww

😩

[viscid thistle]

you ok

[rigid sun]

Having flashbacks

To sophomore year

[brisk frigate]

I'm having a lot of trouble understanding this one, and YouTube isn't helping much, was hoping someone could explain

I was thinking about doing "2Pi-5" and using that as the answer but shouldnt it be represented with a fraction containing Pi like the rest?

[proud jetty]

https://i.gyazo.com/bfc29efedc2823e4c8abc6b53cd10745.png

can ya all explain how you get that other half of cos

i was only given the left half.

[stuck lark]

identity $\sin^2(x)+\cos^2(x)=1$

[obsidian monolith]

RokettoJanpu:

[proud jetty]

so could you explain it

[rigid sun]

m9

Solve for 1-sin^2

[stuck lark]

the identity can be rearranged to suit your problem

[proud jetty]

roketto i get it wut u mean but I still don't understand how the other half is same as the left half. ik im dumb

phd how would u slove dat

[rigid sun]

identity $\sin^2(x)+\cos^2(x)=1$

[obsidian monolith]

maleb1964:

[rigid sun]

solve for 1-sin^2

$cos^2(x)=1-sin^2(x)$

[obsidian monolith]

maleb1964:

[rigid sun]

Now substitute

[brisk frigate]

lol

[rigid sun]

you had 1-sin^2(x) on the bottom right?

replace that with cos^2 since we just proved that

[proud jetty]

yes

[brisk frigate]

I havent gotten to trig identities yet, do you think i should start memorizing them? Or is that not needed

A friend showed me his formula paper and it was filled with them

[rigid sun]

Yes

[brisk frigate]

rip, feels like there were so many

[rigid sun]

There are many

[proud jetty]

wait so why do you put a cosx / cosx^2

[rigid sun]

But just memorize the fundamental ones

[proud jetty]

like ik we just proved the bottom half but

[brisk frigate]

Sounds good man, good thing I asked so i have more time for that

[rigid sun]

? MLG god what?

[proud jetty]

lol i got quiz tomorrow and i got no clue wuts happening

[brisk frigate]

yeah we can tell xD

you substitute the bottom half

and then you realise both sides are the same

and say "wow, im done, im so good"

thats it

[proud jetty]

how do you end up with 1/ cos x

shit bro I legit need help.

[rigid sun]

make sure you say “wow, I’m done, I’m so good” out loud During tests

[brisk frigate]

HAHAHA

yeah dont forget that

[rigid sun]

Dude

[brisk frigate]

$cos^2(x)=1-sin^2(x)$

[obsidian monolith]

Mekbot:

[rigid sun]

cos/(cos*cos)=1/cos

[proud jetty]

shit nvm i see

there are two cos x at the bottom and one at top

[brisk frigate]

just substitute "1-sin^2(x)" to "cos^2(x)" and thats it

and then you realise both sides are the same
and say "wow, im done, im so good"

[proud jetty]

WOW, IM DONE, IM SO GOOD

[brisk frigate]

:D

HE GOT IT

poggers

[proud jetty]

^if only dat was true imma go do 3 days worth homework and than review everything

thx alot for help ya all

love ya no homo

[brisk frigate]

lmao, should prob avoid leaving it till last sec

<3

gl ma dude

[proud jetty]

lol gud tip but works messing me up

and thx

[brisk frigate]

How do I go about doing this? I've never seen anything but f(x) so idk what to do when they add something next to the x

Does P(t+pi) mean I have to replace cos(t) with cos(t+pi) ?

Or maybe into cos(t)+pi?

[stuck lark]

P(t+pi) replaces each t in P(t) with t+pi

[brisk frigate]

oh thnx

[obsidian monolith]

MLG GOD:

[proud jetty]

Like how does cos(x) =0 at bottom make the denominator = 0

[uncut mulch]

$\frac{\sin(x)\cos(x)}{1+\cos(x)}$?

[obsidian monolith]

ramonov:

[proud jetty]

yea like you know how denominator can't equal to 0?

if cos(x) was -1. it would make the bottom equal to zero which isn't permitted.

But if cos(x) is equal to 0 does it still affect the bottom half.

[uncut mulch]

who's saying cos(x) can't be 0?

[proud jetty]

https://i.gyazo.com/6d0c5d3dcd9e8293e7087ff31def60ce.png

ignore the right

[uncut mulch]

when 1 + cos x = 0

[proud jetty]

._. i read that wrong

didn't I?

[uncut mulch]

yes

[proud jetty]

thx >.<

https://i.gyazo.com/0cb98db7cfcf8ec00312747cf08aebf9.png

shouldn't the values be root2/2 for 1/root2 since pie/4 = 45degree

[uncut mulch]

1/sqrt(2) = sqrt(2)/2

[proud jetty]

why is it written like dat?

is there a purpose for the way its written?

[uncut mulch]

certain calculations could be simpler

[proud jetty]

oh ic thx alot

[opal swallow]

anone can help me

[willow bear]

with what

[hoary valley]

Why the limit does not exist? isn't +infty??

[willow bear]

it sounds like your book only considers a limit to exist when it is finite

this needs some context though

[viscid thistle]

@willow bear is there such a thing as an infinite limit?

can you please give me an example?

and non example?

[rigid beacon]

I think some books say limits that go to ±infinity are non-existent and some books will say that ±infinity are good answers for limits

[viscid thistle]

that does make sense

[willow bear]

what do you mean by "is there such a thing as an infinite limit"

[viscid thistle]

disregard, you stated "it sounds like your book only considers a limit to exist when it is finite" and i thought you implied that there is such a thing as an infinite limit, i have no idea what an infinite limit is supposed to mean. i was just confused please forgive

[willow bear]

$\lim_{x \to 0^+} \frac1x = +\infty$

[obsidian monolith]

Ann:

[willow bear]

that's an example of an infinite limit

[viscid thistle]

yo......

[frank nexus]

yo

[blissful wadi]

Any help with this please

Mq = show that

Convergente is convergent i guess in english

[frank nexus]

that looks like japanese to me... cant wait for the later years in school..

[willow bear]

it's not japanese, it's poorly written french

[frank nexus]

LOL

is it really?

[willow bear]

yes

[frank nexus]

okay, i didnt know why i couldn't read it

[blissful wadi]

😂😂😂 stop talking shit about my hand writing its beautiful for em

Me*

[frank nexus]

myn is trash to

too*

[blissful wadi]

Any helo tho i tried calculating U(n+1) - U(n)

To get an idea where it would be both increasing and majored

Or decreseasing and minored ( ?)

But didnt help

[willow bear]

$u_{n+1} = \sqrt{2 - u_n}$?

[obsidian monolith]

Ann:

[blissful wadi]

yes

[blissful wadi]

Any help ?

[willow bear]

let $f \colon ]-\infty, 2] \to \bbR$ be given by $\forall x \in ]-\infty, 2], f(x) = \sqrt{2-x}$

[obsidian monolith]

Ann:

[willow bear]

yuck

the spacing is all weird

screw you, french notation

[blissful wadi]

= ( french is col

[willow bear]

then $f([0,2]) = [0, \sqrt{2}]$

[obsidian monolith]

Ann:

[willow bear]

$f([0, \sqrt{2}]) = [\sqrt{2 - \sqrt{2}}, \sqrt{2}]$ ... i think?

[obsidian monolith]

Ann:

[willow bear]

maybe it'll come in handy that $f$ is differentiable on $[0, \sqrt{2}]$ and that $|f'| < 1$ on $[0, \sqrt{2}]$

[obsidian monolith]

Ann:

[willow bear]

even though i don't have the energy rn to notate that in a way that would please Tuong

[blissful wadi]

hmm wait let me try this

[willow bear]

well

$\forall x \in [0, \sqrt{2}], |f'(x)| < 1$

[obsidian monolith]

Ann:

[willow bear]

i hope it's obvious that your sequence is obtained by iterating f starting with 0

[blissful wadi]

whats next ?

i tried doing something but it didnt turns out

i tried to prove that $f(U_{n} = U_{n+1} ; f is continous , U_{0} \in [0,2] ; f([0,2]) \subset [0;2] and find that \lim U_{n} = x whichverifies the equations f(x)=x $

$f(U{n}) = U{n+1} ; f is continous , U{0} \in [0,2] ; f([0,2]) \subset [0;2] and find that \lim U{n} = x whichverifies the equations f(x)=x $

[obsidian monolith]

dvaix:

[willow bear]

hhhhf bad tex

[blissful wadi]

uhetuehuyehteuht

anyway

i wanted to show all that

and then conclude that Un must converge

but i cant get $\lim U_{n}$

[obsidian monolith]

dvaix:

[willow bear]

lim u_n is the unique fixed point of f

[blissful wadi]

huh

[willow bear]

fixed point

point fixe

[blissful wadi]

yeah and ?

what was your method ?

[willow bear]

https://en.wikipedia.org/wiki/Banach_fixed-point_theorem

i guess you could apply this

[blissful wadi]

nah i cant

( didnt study it yet so cant use it in my homework :p )

[odd helm]

Why is there a (x+2) in the numerator

-2 is a VA so (x+2) on denominator but

It’s not a zero so idk why it’s on the numerator

[hoary valley]

Is arctan continuous on R ?

[willow bear]

yes

it's even differentiable

[hoary valley]

It is not discontinuous on -pi/2,pi/2 ... ?

[willow bear]

arctan is not tan

[hoary valley]

Oh It's a RESTRICTION of tan !

[willow bear]

no, arctan isn't a restriction of tan either

it's the inverse of a restriction of tan

[hoary valley]

Oh Gotcha

[hoary valley]

What's (b) ?

Constant?

or an interval ? or what

A point?

[willow bear]

what do you mean by (b)

[hoary valley]

b , in the theorem

[willow bear]

it's a point

[long pond]

I don't remember doing limits in precalc

[blissful laurel]

[hoary valley]

Does this graph have any vertical asymptotes or horizontal asymptotes?

I see that it has Vertical asymptotes at x=-1 and x=2

Does it have any horizontal asymptotes?

[willow bear]

the way the graph looks suggests that it has two of those

one at y=2, the other at y=4

[hoary valley]

Right, thanks

This problem is not normal

I'm stuck here...

[serene heath]

Divide top n bottom by x

[willow bear]

multiply it by $\frac{x^{-1}}{x^{-1}}$

[obsidian monolith]

Ann:

[hoary valley]

Do I divide each part inside the sqrt by x like 16x^2/x and +3x/x ?

[willow bear]

no

$x^{-1} \sqrt{16x^2 + 3x} = \sqrt{x^{-2}}\sqrt{16x^2 + 3x}$

[obsidian monolith]

Ann:

[hoary valley]

And how do you solve this?

[stuck lark]

$\sqrt{ab}=\sqrt{a}\sqrt{b}$

[obsidian monolith]

RokettoJanpu:

[willow bear]

wdym "solve"

there's nothing to "solve", this isn't an equation

[stuck lark]

ig in some people's vocabularies, solve is synonymous with algebraically manipulate

[hoary valley]

I meant simplify.

@stuck lark Thanks

I'm stuck here..

[willow bear]

uh

you made two mistakes there

$x^{-1}(4x + \sqrt{16x^2 + 3x}) = 4 + \sqrt{16 + 3x^{-2}}$, not $4\sqrt{16x + \frac{1}{3x}}$

[obsidian monolith]

Ann:

[willow bear]

you forgot something

[hoary valley]

I meant this

[willow bear]

how is anyone supposed to know

how is anyone supposed to know when you meant to write a radical symbol but chose not to do so

[manic hemlock]

is here where i ask questions?

I just need help graphing a conic

[hoary valley]

This is arctan it has 2 horizontal asymptotes as you can see its graph doesn't touch the horizontal asymptotes..

How come in this function the graph of the function touched the horizontal asymptotes... yet we say it has 2 horizontal asymptotes at y=4 and y=2 ?

[stuck lark]

horiz asymptotes just tell you the function's end behavior, ie what it does for big positive x and big negative x, not what's happening in between

[hoary valley]

@stuck lark So in this example above, the function will never intersect with the horizontal asymptotes y=4 and y=2 ?

In big positive x and big negative x ?

[stuck lark]

rather, think of 4 as the y value the function will converge to as x approaches +inf

[clear glade]

how do i solve : $1 + sinx = 2cos^2x

$1 +sinx = 2cos^x$

[obsidian monolith]

Waterblade:

[stuck lark]

if that's the graph of a function f, lim x to infty f(x)=4

[hoary valley]

@stuck lark Thanks man, I think you definitely deserve the HONORABLE role.

[stuck lark]

no prob dude thanks for the kind words

[hoary valley]

I'm stuck here..

[viscid thistle]

,rotate 270

[scenic musk]

monkaS -5 points

u didnt write lim at each step

[hollow pollen]

monkaS

[rigid sun]

,rotate 1

[viscid thistle]

How come this simplified to this?

[proud sparrow]

consider multiplying numerator and denominator by $q_1^{0.6}q_2^{0.4}$

[obsidian monolith]

Element118:

[proud sparrow]

@viscid thistle

[viscid thistle]

I get dis

[proud sparrow]

what did you do?

show all your working please

[viscid thistle]

I divided the numerator and denominator

Oh nvm

I get it now

I messed up and switched up where the numerator and denominator should be

Ty

[clear glade]

I have a problem I want to make sure I did right

this is my sketch

my answer was tan(theta) = 53/38 so theta was equal to tan inverse (53/38), is that right

[undone drift]

yes it's correct

[clear glade]

ok thank you

[mighty pine]

So my opp = 24, adj = -7 and hyp = 25

is my logic correct?

[upbeat prairie]

I understood everything up till the length of the Latus Rectum

How is it 4p?

and how would I find it?

[hazy hull]

https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A3d9745da-ff64-4792-a3a7-a06de39af6eb so 6 is that the solution to f(0) gets passed to the division problem?

[plucky storm]

Can I have some insight on where to go from here?

I was thinking of bringing the cos over and factoring, but I don't think that's actually the next step?

[plucky storm]

I'm guessing in doing something wrong here too

Second one of right, there's no real solutions

[hazy hull]

@plucky storm more like 4*i

or 2i

you already featured that

[tardy plinth]

So I got this equilateral triangle with sides of 7m, in which an ant from point C goes towards point A, while another ant goes from A to B. I need to find a function distance between them in terms of y (d(y) in the picture). I tried similarities of triangles, but didn't worked, didn't get the function as showed and not sure from were to start developing this function.

[blazing parrot]

At any point in time, CD=AE (assuming both ants move at the same speed)
Then AD would be AC-CD
You know the angle at one of the corners of triangle DAE, you can use that to find the side DE, the result of the function d(y)

[tardy plinth]

I used cosine law to find d(y), still getting sqrt(49-21y+3y^2)

I'm retarded or burned out

Could the answer given be wrong?

[native timber]

,w simplify \sqrt{(7 - y)^2 + y^2 - y(7 - y)}

[obsidian monolith]

Display results online and refine query

Requested by Jiramide#7358

[tardy plinth]

So the answer in the book wrong then

[native timber]

seems like it

i mean its already suspicious that they managed to get 7y^2 considering that there's only 3 terms

[tardy plinth]

Didn't even want to consider it to be wrong, as it is taken from one of my countries exams

[native timber]

literature is written by humans, mistakes are always possible

[tardy plinth]

Agree 100% now

[willow bear]

y=0

y=e would be a horizontal tangent line

without the "never touches" requirement

a graph can cross its own asymptote as many times as you wish

even infinitely many times

eg sin(x)/x

[languid crane]

which function

[formal dome]

Sorry wrong picture

The limit of this function exists, and it is +infinity right?

Then isn't theorum 1.2.4 b wrong?

Bec in the case of the 1st picture, p(a) ≠ 0 and q(a) == 0 but limit does exist

[willow bear]

your book may consider an infinite limit as not existing

depending on the multiplicity of a as a zero of q, the limit might be ±∞ (even) or genuinely fail to exist (odd)

[formal dome]

@willow bear o idk what you mean by limit may be even or odd but i think you mean(what my book explained) that the limit may be
A) +infinity from right and -infinity from left(or vice versa) hence limit does exist
B) plus or minus from both left and right so limit exists

[willow bear]

by limit may be even or odd
no

that's not what i meant

[formal dome]

And infact it shows a few examples behind with curves similar to this and says limit exists

Oh

[willow bear]

let me rephrase

depending on the multiplicity of a as a root of q, the limit might be ±∞ (if the multiplicity is even, and depending on the signs of p(x) and q(x) for points close to a) or genuinely fail to exist (if the multiplicity is odd)

[formal dome]

I gtg somewhere. Will check your answee when i come back I.A. idk tthe meaning of multiplicacy and zero of something lol

[willow bear]

root

[odd helm]

For this one why do you divide by 12?

I thought you only divide when it's compounded monthly but the question says annually

[willow bear]

low res

can't read the blue printed text

[odd helm]

Alright I'll try to zoom in

[candid onyx]

Is this ok ?

[rigid sun]

,rotate

[scenic musk]

if a function has an oblique asymptote then it can't have a vertical asymptote correct?

[spring thunder]

nope

think about x -> x + 1/x for example @scenic musk

[scenic musk]

o ok tyvm

[viscid thistle]

Okay guys, I can feel am not doing this correctly

[dim jungle]

Not inverse's

[viscid thistle]

So why can I clear the x on the first one?

[dim jungle]

Something without x - something with x stays separate

You can't do 7/4-3x/4=4x/4

[viscid thistle]

Oooh okay okay

That's what I thought

What about this one?

[dim jungle]

Not inverse's

[viscid thistle]

Am master pro

What do I do now tho

Can I 4n/4=n?

[viscid thistle]

<@&286206848099549185>

[limber bone]

yes

[viscid thistle]

I think I'm not doing this correctly

[barren hedge]

@viscid thistle (5) should look more like this.

And for future reference,

[formal dome]

@willow bear (sorry for the late reply, im refering to the question i asked you earlier with two pics about limits)
Ok so i understood you partially that limit may be ±infinity or even fail to exist depending on situations but according to this picture, the limit does exist right? Because limit from left = limit from right = +infinity. And if the 2 one sided limits are equal, it means the actual limit exists and is equal to the one sided limits

[willow bear]

@formal dome again your book seems to only consider a limit to exist when it is finite

[viscid thistle]

@willow bear Hey

[willow bear]

uh

hello

[viscid thistle]

How do you rationalize or conjugate a fraction with a radical in the denominator

of which the index is not 2

[willow bear]

like $\frac{1}{\sqrt[3]{2} - 1}$?

[obsidian monolith]

Ann:

[viscid thistle]

yes

[willow bear]

hh

well in this particular case the "conjugate"\ is $(\sqrt[3]{2})^2 + \sqrt[3]{2} + 1$

[obsidian monolith]

Ann:

[willow bear]

i guess

[viscid thistle]

wth

uhh

so I would multiply the top and bottom by?

[willow bear]

by that

the basic idea is $(x-1)(x^2 + x + 1) = x^3 - 1$

[obsidian monolith]

Ann:

[willow bear]

or perhaps more generally $(x-a)(x^2 + ax + a^2) = x^3 - a^3$

[obsidian monolith]

Ann:

[viscid thistle]

difference of cubes?

[willow bear]

yes

[viscid thistle]

thank you

[viscid thistle]

$\frac{1}{2 - \sqrt[3]{x}}$

[obsidian monolith]

Successful:

[viscid thistle]

@willow bear I am a little confused on how you would get the "conjugate"

I know it is (2x-1)(4x^2 + 2x +1)

but I don't understand how you come up with it

[willow bear]

not quite

it'd be $4 + 2x^{1/3} + x^{2/3}$

[obsidian monolith]

Ann:

[willow bear]

so that the new denominator is $8-x$

[obsidian monolith]

Ann:

[viscid thistle]

so my teacher was wrong? Lol

[willow bear]

:?

what exactly did the teacher say

[viscid thistle]

I only have the notes from his lesson

[odd helm]

I didn't get that as my answer when I typed it into my calculator

I got 0.0001212

Also idk why the key says divided by 1575

[odd helm]

<@&286206848099549185>

[fading token]

The image you sent clearly contains a typo

,calc log(0.5)/1575

[obsidian monolith]

Result:

-4.4009344797457e-4

[fading token]

two typos, even

And I can confirm your answer is correct, @odd helm

[odd helm]

Alright thank you

[hollow bluff]

How does my teacher expand this?

why is that a valid answer and how do I get it

[tardy crescent]

That's the change of base formula

Log_b (x) = Log_a (x) / Log_a (b)

[willow bear]

^

[hollow bluff]

oh ok

[smoky vigil]

pretty sure this is where this goes

I need to find what x is

if y = 2/(x-1)

[dense oar]

Multiply by (x-1)

Y(x-1) = 2

[smoky vigil]

x-1=2y

x=2y+1?

[dense oar]

x-1 = 2/y

[smoky vigil]

ahh

oh

x=2/y-1

[dense oar]

Are you trying to find the equation in terms of x

[smoky vigil]

ye

[dense oar]

$x=\frac{2}{y}+1$

[obsidian monolith]

AutisticAri:

[smoky vigil]

thanks, more simple than I expected 😄

Ok I need more help xd

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

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

[obsidian monolith]

NotLigmaMaster69:

[smoky vigil]

need to find what x is again

[uncut mulch]

what's your first step?

[smoky vigil]

I just got to x=(x+y)/y

don't know what else to do cus otherwise I get that 1=y/y

which is true

,but not what I need

[uncut mulch]

how are you getting the right side?

[smoky vigil]

x-1=x/y

[uncut mulch]

oh. you should leave the y on the left

[smoky vigil]

hmm

[uncut mulch]

x=y(x-1)

then distribute and
combine your x terms

[smoky vigil]

x=-y+1?

[uncut mulch]

can you show your work?

[smoky vigil]

sec, gotta rewrite it then

kinda wrote it in a weird way

oh btw the 2nd line it's x=y(x-1)

[uncut mulch]

what's the line below that?

[smoky vigil]

1+(-x)=y

[uncut mulch]

dunno where you're getting that
what do you get after distributing the y?

[smoky vigil]

I just did x/x+x/-1

that's how I got that

[uncut mulch]

I'm still not following what ur doing.
what are you doing to both sides to reach that?

you are starting with
x = y*(x-1)

and the next step would be to distribute the y

[smoky vigil]

elaborate please

[uncut mulch]

a(b+c) = ab + ac

[smoky vigil]

oh

yx-y

x=yx-y

[uncut mulch]

move all your x terms to one side

[smoky vigil]

x-yx=-y

[uncut mulch]

then factorise

[smoky vigil]

x(y-1)=-y

oops

x(1-y)=y

-y

(

x=y/1-y?

[uncut mulch]

go back a step and rewrite it more clearly first

[smoky vigil]

x-xy=-y

x(1-y)=-y

x=-y/(1-y)

Thanks

[uncut mulch]

simplify the negative to make it look nicer

[smoky vigil]

x=y/-(1-y)

[uncut mulch]

which is:

[smoky vigil]

x=y/(-1+y)

[uncut mulch]

parentheses

x=y/(y-1)

[smoky vigil]

Many thanks

[uncut mulch]

np

[viscid thistle]

Hello

How do I write square root in discord?

[uncut mulch]

unless you know tex, do:
sqrt( stuff )

[viscid thistle]

Ah okay ty

So I had a question about this

When I solve this equation in the interval [0,2pi)

3secx+sqrt(3)=3sqrt(3)

Would I square it first?

[uncut mulch]

no, you shouldn't be squaring anything for this question

move constants to one side

[viscid thistle]

Ok so

[uncut mulch]

write sec in terms of cos

[viscid thistle]

Ok

[uncut mulch]

you should get 2 nice solutions

[viscid thistle]

3/cosx+sqrt(3)-3sqrt(3)

?

[uncut mulch]

what happened to your equal sign?
you want to reach something resembling
trig function = constant

[viscid thistle]

Ah okay

So I would just add an = 0

at the end

[uncut mulch]

but then you'll have the trig function and constant on the same side,
so you should've moved the constants to the other side

[viscid thistle]

Ahh

So I should keep all the trig functions

On one side?

[uncut mulch]

usually for these types of questions, yes

[viscid thistle]

And then I end up with

3/cosx = 2sqrt(3)

Couldnt I just have it in 3secx and divide by 3 on both sides?

[uncut mulch]

you can,
but its usually easier to work in sin/cos/tan instead of their recipricals

[viscid thistle]

Ok

I only got one solution

cosx=2sqrt(3)/3

[uncut mulch]

you didn't manipulate that properly

what operations were you performing to isolate cos(x)?

[viscid thistle]

I kept secx and divided 3 on both sides

And then I just did the reciprocal and got 1/cosx

[uncut mulch]

if you were working in sec, then it would be
sec(x)=2sqrt(3)/3

[viscid thistle]

Ah okay

But even then, thats the only solution I got

[uncut mulch]

that's the equation you need to solve,
there will be 2 solutions for x in [0,2pi)

unless you are very familiar with the sec
take the reciprical of both sides

[viscid thistle]

Oh I get it

The cosx of that is

sqrt(3)/2

And the solutions would be

pi/6, 11pi/6

[uncut mulch]

cosx of that is?

[viscid thistle]

When you turn the secx into the cosx

[uncut mulch]

you mean
cos(x) is sqrt(3)/2

?

[viscid thistle]

Yes

Thats what I said

Anyways

I believe pi/6 and 11pi/6 are the answers?

[uncut mulch]

yeh

[viscid thistle]

Ok ty for helping

[brisk frigate]

I need help with this, I looked up a guide and it still says the answer is wrong pls help

This is the method I used, but maybe there is a simpler way?

[uncut mulch]

did you draw a diagram?
do you know what quadrant your point is in?

what is your "y"?

[brisk frigate]

Yeah, unless im mistaken its on the fourth

y = -5

x = 4

i did pythagoras

[uncut mulch]

r = ?

[brisk frigate]

to get the radius, then used that to get angle

sqrt(41) = r

but since it was decimal i left it as sqrt

[uncut mulch]

sin(theta) = ?

[brisk frigate]

(5)(sqrt(41))/41

[uncut mulch]

the ratio of which 2 sides?

[brisk frigate]

wym by that?

[uncut mulch]

sin(theta) = opp/hyp = y/r right?

[brisk frigate]

so i did, sin= y/r, so if y=5 and r=sqrt(41) then sin(5/sqrt(41)

[uncut mulch]

y = -5

[brisk frigate]

ye

[uncut mulch]

-

[brisk frigate]

well then

xD

[uncut mulch]

also note that its Q4,
so sin(theta) should be negative

[brisk frigate]

damn

you right

such a simple mistake, yeah i got the answer now thnx bro

feel kinda dumb for that one, been at it for a while and keep going back to check it

[odd helm]

Why is it raised to the 0.7 when the rate is 0.0685

[brisk frigate]

Does anyone know why this is wrong? I feel like I'm missing something

After factoting the equation I got: cos(t)=1 and cos(t)= -1/2

[serene heath]

Ur missing a solution

[brisk frigate]

Ok, I have 3 attempts left, the first 3 solutions i got them through a video I watched bc teacher hasnt explained this yet, does anyone know how to solve it?

All I know how to do from the video is, get the values of: cos(t)=1 and cos(t)= -1/2

Then I know that cos = x

so (1, 0) and (-1/2, 0) are the answers in the unitary circle

since (1,0) is on the same axis as "2pi" in the unitary circle, thats my first answer, and since (-1/2, 0) has "2pi/3" and "4pi/3" on the same value of "x" then those two are also answers

But thats as far as youtube video knowladge gets me on this one

[serene heath]

t=0 is also a solution

[odd helm]

Could someone check the thing I posted above

[brisk frigate]

where does t=0 come from? in case i get it on my test

Oh bc you know cos(0)=1

Will I need to memorize the unit circle?

[stuck lark]

it will be useful for you later on if you do

[brisk frigate]

Alright

Im still curious as to how he got that t=0, apart from him just knowing cos(0)=1, like did he just know or can that be "found" using the unit circle like I did with the rest of my answers?

[stuck lark]

unit circle helps

for real values of t, cos(t)=1 only when t is an integer multiple of 2pi

[brisk frigate]

oh, and I suppose that goes for all of them not just cos

Then if I get that problem in a test I would have to use the unit circle for those, and then how would I find the ones not in it?

[stuck lark]

wdym not in it

[brisk frigate]

Oh I thought u meant that "0" as the answer couldnt be found by looking at the unit circle

So I thought in other cases there might be more values of "t" that I wont be able to find by looking at the unit circle

[stuck lark]

t=0 is ONE solution but there are infinitely many

[brisk frigate]

huh I see

thats why they limited it to [0,2pi]

and the unit circle only goes from 0 to 2pi for a total of 360 degrees

I think I understand now

[stuck lark]

the circle goes on and on. but after 2pi you start repeating stuff

[brisk frigate]

Yeah, now it makes sense why they limited the question from [0,2pi]

What im getting from this is that from now on I'll use the circle to get all values from (0,2pi] then substitute t=0 and check if that is also an answer

and if they ask for values after 2pi, then I keep going with the circle repeating stuff like you mentioned

[stuck lark]

@brisk frigate certainly if they want a few more solutions beyond t=2pi then do that, but i don't recommend blindly plugging in t=0, i only used that because i know cos(0+2pi*n)=1 where n is an integer

[brisk frigate]

okok Im not sure i understand the cos(0+2pi*n)=1 but prob bc im just starting, ill keep that in mind

[stuck lark]

$\cos(0)=\cos(2\pi)+\cos(4\pi)=...=\cos(2\pi n)=1$ where n is some integer

[obsidian monolith]

RokettoJanpu:

[scenic musk]

how do i know that cos(pi - 2x) is in the second quadrant

[stuck lark]

feel like you left out some details

[finite nimbus]

My teacher assigning this as extra credit

I have to find out the amount of space my turkey hand occupies

What I’m thinking right now is box the entire thing and calculate the area and then calculate the areas that have gaps

Like between the fingers

Would that be the best way to do it?

[scenic musk]

turn ur hand into a square

easy A+

or

divide ur hand into triangles

@finite nimbus

[finite nimbus]

Wouldn't the triangle method not work for the rounded edges of my finger?@scenic musk

[scenic musk]

i mean theres going to be approximations no matter which method u use

for ur fingers make rectangles

and if u really want

u can make a circle

for ur finger tips

ull need triangles in between to fill in gaps

its a lot of drawing and see what works

[finite nimbus]

okay thanks

[raw gorge]

I gotta question ab finances

You borrow $33,850 to buy a car. You take a 6 year loan at 5.75% annual interest. For the first 2 years, you make the
minimum payment of $557.01. After two years, your financial institutions offer you the option to refinance for free (roll
the existing loan into a new loan with no added cost). Your new loan requires you to make monthly payments of
$425.62 for 5 years. What is the annual interest rate on your new loan? How much did you pay in interest of the 7
years?

I tried using future value to calculate how much is left for the final loan and got $36225.42

but then i used present value and gotta negative interest rate

[rigid beacon]

@finite nimbus cut the hand out and weigh the paper

You can find the density of paper

And solve from there

[rich ocean]

engineer spotted

[green zenith]

how do you find the point where two graphs or equations intercept?

i was figuring out how on the equations of both celsius and fahrenheit

for example, i have f=(9/5)C+32 for fahrenheit and the inverse of it for celsius.

[viscid thistle]

@green zenith set those two equations equal to each other

[green zenith]

and then?

[viscid thistle]

then solve for x?

and then y

[green zenith]

so i look for the equation of both?

what do i then after I find both x and y?

[viscid thistle]

then you have the intercept at (x, y)

@green zenith do you have the two equations?

[green zenith]

yes i do

i have to say that this two are inverse functions

[viscid thistle]

paste them in here

[tardy crescent]

Hello @vague zephyr, what have you tried so far?

[vague zephyr]

I have not tried it at all due to my confusion as to where I should start

[tardy crescent]

So since it's a polynomial of degree 3, it can have at most 3 roots. In this problem they list all the roots, by saying that there's a root with multiplicity 2 at x = 2 and a root of multiplicity 1 at x = -2

So your factors are (x - 2)(x - 2)(x + 2)

[vague zephyr]

so from this i am supposed to create a polynomial equation

[tardy crescent]

Yes, you can multiply the factors together and you'll get a polynomial. Although this polynomial will have a y-intercept of 0, so we'd have to shift it by -5.6 after all the factors are multiplied together

(x - 2) * (x - 2) = x^2 - 4x + 4

And then multiplying that by (x + 2), we get: (x^3 - 2x^2 - 4x + 8)

[vague zephyr]

and then shift it by -5.6

[tardy crescent]

Yep,

So we get: x^3 - 2x^2 - 4x + 2.4

[vague zephyr]

so x^3 - 2x^2 - 4x + 2.4

[tardy crescent]

Yes, nice

[vague zephyr]

alright cool, thank you

[tardy crescent]

yeah np

[vague zephyr]

it says its wrong

[tardy crescent]

Oh crap, sorry

The shift was wrong, that's all

It's supposed to be x^3 - 2x^2 -4x - 5.6

So the constant term has to be -5.6

x = 0 has to give you y = -5.6.

[vague zephyr]

its still incorrect

[tardy crescent]

I'm kind of tired and must be making some mistakes, really sorry for messing that up. Let me see what went wrong

[vague zephyr]

its all good

[tardy crescent]

Oh okay

No constant shift, what we needed was a constant factor

so A*(x - 2)(x - 2)(x + 2) = y, we need to find A when x = 0 and y = -5.6

-5.6 = (-2)(-2)(2) * A

-5.6 = 8A

[vague zephyr]

-5.6/8

[tardy crescent]

Yup

[vague zephyr]

so -.7

[tardy crescent]

That's your function: (-5.6/8)(x - 2)(x - 2)(x + 2)

Yes

So you can also multiply them into that polynomial and then apply the * -0.7, or you might even be able to keep it all in factored form like this: -0.7 (x - 2)(x - 2)(x + 2)

[vague zephyr]

−0.7x3+1.4x2+2.8x−5.6

−0.7x^3+1.4x^2+2.8x−5.6

[tardy crescent]

Yep, that's it

[vague zephyr]

thanks for the help

[tardy crescent]

Sorry about that. So what I did earlier, adding a constant after finding the roots, actually changes your roots, so never do that haha

[vague zephyr]

Yeah its no big deal

last one left

[tardy crescent]

So whatever the polynomial f(x) is, it can be written as ((x - 6) * Q(x)) - 1

Where Q(x) is the quotient that you get by dividing f(x) by (x -6)

[vague zephyr]

so then in this instance x-6 is being divided by f(6)

with a remainder of -1

so ((x-6)/6)-1?

or 6-6/6

-1

so would the answer just be -1

[tardy crescent]

f(6) means you substitute 6 for all the x's, and since f(x) can be written as ((x - 6)*(Q(x)) - 1, you can plug in 6 for the x's and you get (0 * Q(6))) - 1, which is just -1

[proud jetty]

https://gyazo.com/e327868585fe025b959f789faf4e9b0a.png

how do we get the 2?

[tardy crescent]

(1 - sin^2 A) - sin^2 A = 1 - sin^2 A - sin^2 A = 1 - 2 sin^2 A

[proud jetty]

there 2 comes from the remaining sin?

[tardy crescent]

Yes

[proud jetty]

why aren't they subtracted tho? like (1-1=0)

[tardy crescent]

They are subtracted, cos^2 A = (1 - sin^2 A), so both sin^2 A's have a negative sign

It's kind of the same thing as -(sin^2 A + sin^2 A)

[proud jetty]

ah ic thx.

[tardy crescent]

yw

[vague zephyr]

what would we do without vital

Thank you vital

[tardy crescent]

You're welcome 🙂

[vague zephyr]

We appreciate what you do m8

[tardy crescent]

Thanks

There's a lot of great people and great helpers on this discord

[vague zephyr]

Indeed there are however, I only have you to thank at the moment, they will get their turn too lol

[tardy crescent]

haha thanks. That's it for me tonight, have a good night 🙂

[vague zephyr]

Good night pal

[tidal rain]

Hey I need some guidance with this problem

[native sequoia]

$\tan^2\alpha$ can be arbitrarily large

[obsidian monolith]

EpicGuy4227:

[proud jetty]

https://gyazo.com/2a14b539819a0a21450de33889d12ea1.png

what happened to the square root

and why did they move -1 into the bracket

[stuck lark]

identity $2\cos^2(x)-1=\cos(2x)$

[obsidian monolith]

RokettoJanpu:

[proud jetty]

thx

[stuck lark]

np

[lethal oracle]

Hey is there anyone around that can help me with a few things?

[abstract dome]

whatcha need

[proud jetty]

https://gyazo.com/9cd19c685944db3526d6b8228418af84.png

can someone help me with what happened to the 2 at the bottom on the right side

[lethal oracle]

The sin^2?

[abstract dome]

sin^2 - 2sin^2 = -1 * sin^2

If that is what you're asking for

[proud jetty]

(1-2sin^2x)+sin^2x

[abstract dome]

yeah

1 - 2x + x = 1 - x

[lethal oracle]

Also I’m having trouble with complex numbers raised to the given power

[uncut mulch]

combine like terms

[abstract dome]

which problem do you have, alpaka

[proud jetty]

oh shit. I forgot the -2 in front means the whole term is a negative right?

[abstract dome]

yes

[abstract dome]

-2 * (sin^2(x))

[proud jetty]

00f. thx boys and girls and other genders

[lethal oracle]

The answer is -8 but idk what I’m doing wrong

[abstract dome]

-8 for the first or second problem

[lethal oracle]

They’re the same one

I’m just showing my work

[abstract dome]

o

[lethal oracle]

I changed it to polar

[abstract dome]

Did you expand the cube

(1 - sqrt(3)i) * (1 - sqrt(3)i) * (1 - sqrt(3)i)

[lethal oracle]

We’re supposed to use De Moivres theorem

To solve

[abstract dome]

Oh, I'm unfamiliar with that

[lethal oracle]

Oh

[uncut mulch]

you you would also need to apply the power to the CiS

[lethal oracle]

Oh alright

Don’t I multiply the value of cos60 by 3

And the value of sin60 by 3

[uncut mulch]

multiply their argument by 3

[lethal oracle]

Like this?

[uncut mulch]

no

can you state DeM's theorem?

[lethal oracle]

r^n(cos(nx)+isin(nx)

x being theta

Oh so i multiply my angle by 3

[uncut mulch]

only the right side of the theorem but ok.

[lethal oracle]

z^n=(r(cosx+isinx))^n

Is the other side sorry

[uncut mulch]

what do you end up getting?

[lethal oracle]

I’m getting -8

[uncut mulch]

that looks good.

[lethal oracle]

I know it’s hard to see but for 23

Why is only sin negative

And not cos

[uncut mulch]

although it still works, (due to odd an evenness of cos and sin)
the sign in between should be +

[lethal oracle]

Oh

My book says it’s -

And for the problem below that it says the answer is just pi

Not 3pi

[uncut mulch]

which #?

[lethal oracle]

But 5pi-2pi is 3pi

25

[uncut mulch]

what was the original question asking for?

[lethal oracle]

Polar form of the quotient

Oh do my radians need to be in between 0 and 2pi?

So I need to change 3pi to pi

[uncut mulch]

did they specify a domain?

[lethal oracle]

No

[uncut mulch]

2CiS( pi + 2kpi) would have the same value

from just dividing (and nothing else), your expression is fine

[lethal oracle]

I might have done this completely wrong

Would this be correct though, the book gives it in a different form but my instructor said I could do it this way but idk if I’m writing it correctly

[uncut mulch]

is that meant to be the power of 3 or 1/3?

[lethal oracle]

The question is asking to find the cube roots

[lethal oracle]

I’m not sure what I’m doing wrong on 55

[willow bear]

what are you asked to do

[lethal oracle]

Find the nth roots of the complex numbers for the specified value of n

[rigid sun]

😩

جذور تمتص كثيرا. وأنا أعلم أيضا أنك وضعت هذا في جوجل ترجمة. تعتقد أن لدي دماغ صغير؟

[lethal oracle]

What’s wrong with my roots

The fact that they’re not simplified?

[rigid sun]

I can’t be bothered to do nth roots