help-13

[muted bear]

ln should be absolute value

[vocal patio]

but here it doesnt

[digital cliff]

if youre evaluating then it doesnt matter as long as your limits are right

[royal loom]

They are the same, but your answer is not correct if you write sin(u) where you meant sin(5x).

You have to clarify

"where u=5x"

and if you have to do that

[muted bear]

note how they changed the bounds

[royal loom]

you might aswell put it back in

[royal loom]

Yah not if you change the bounds

[vocal patio]

so ur saying that the answer would be the same so long as you change the limits and correctly sub the u values with the lims?

[muted bear]

in adefinite integral, you dont have to resub x if you change the bounds and solve the integral in terms of u

[royal loom]

if you do it in terms of u and are evaluating, then the bounds must be in terms of u and that would be equivalent

if you are going to put x back in at the end

then use the original bounds

[vocal patio]

ohh

i see

[plain ridge]

if i ask you to solve something in terms of x and you tell me the answer is "k + 1", that's not useful unless i know what k is

[royal loom]

Yes and that's what I was getting at earlier before I realized it was a definite integral

if it is indefinite integration then your answer should be in terms of the original variable certainly

[vocal patio]

understood now

thans

thanks

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[steep pelican]

What am I supposed to do?

[sturdy tide]

well you've found a,b,c and d right?

[steep pelican]

Well yeah

But how am I supposed to solve it an easier way

The last part

[sturdy tide]

for b?

[steep pelican]

B and c

[sturdy tide]

to find b since you already have a you can substiute it into g(x)=ax+b

and to find c since you already have b you can sub it into h(x)=cx+d

@steep pelican do you think you can do that

[steep pelican]

Ohh

That makes sense

Give me a couple mins

[sturdy tide]

ok np

[steep pelican]

It don't work

Bc it equals 0

@sturdy tide

[sturdy tide]

o sorry

i think you made a mistake here

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[small mural]

Could someone tell me all the step for this

[livid hound]

what exactly are you being asked to do

[small mural]

Find all integer value of m from -5 to 5 so that y min value <-1

I stopped my work at derivate the function

[covert cipher]

Min value = -1?

[small mural]

<

[covert cipher]

Oh ok

[covert cipher]

Youre on the right track

Apply the second derivative rule to check for minimas

[small mural]

ye I tend to do y'=0 then use table but I was so bad at trigonometric 🥲

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[urban mason]

I need help in complex number

[urban mason]

into the algebraic form

the hard thing i couldnt understand is conjugate thing

[low vine]

What do u not understand about that?

[urban mason]

how you conjugate

rule

[low vine]

Multiply both denominator and numerator by it's conjugate
For ex: conjugate of a+b is a-b

[urban mason]

thank you

[low vine]

Np

[urban mason]

the denominator is already conjugated right?

[low vine]

How?

The method is to rationalise it. For which u multiply both Deno and num to it's conjugate

[low vine]

Where

[urban mason]

nvm I thought so

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[peak salmon]

im so confused

so what is lim x->-infinity of sqrt(x^2)?

[plain ridge]

$\sqrt{x^2} = |x|$

[mint garnet]

when u apply - infinty squared it becomes postive

[wraith dagger]

b

[mint garnet]

sqrt of infinty is infinity

[peak salmon]

right

[peak salmon]

so why does it equal -x here?

oooooooh

cause

x is -infinity

so u get - -infinity = positive infinity

right?

[mint garnet]

yee

[peak salmon]

alright makes sense

[mint garnet]

better consider x as -h

so h tends to infinty

then you will have no problem

[peak salmon]

ok yeh that woukd be easier

thanks

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[chilly rock]

i'm not sure how to set this one up honestly

[crimson sedge]

still need help?

[chilly rock]

i got some parametric equations (set y = t) then if -1 < t < 1, you get the two points it gives you

[chilly rock]

yeh

[crimson sedge]

what even is the question?

like what is required

[chilly rock]

oh mb, the question just says evaluate the line integral

that's it

[crimson sedge]

okay, give me a few mins :>

[chilly rock]

this all i've got

[crimson sedge]

im gonna need to type the steps down cus i aint got my phone with me rn, is that okay?

[chilly rock]

np

all i need is the setup anyway

[crimson sedge]

the answer is indeed 4/15 btw

your answer is right

[chilly rock]

no the answer i got is zero, i was just writing down what the correct answer from the book is

[crimson sedge]

you can simplify it using the green's theorem

oh my this cant be typed down

i'll try to take a pic

okay im back

give me a sec to take a pic

[chilly rock]

aight

[chilly rock]

is green's theorem just the only way to make it work?

[crimson sedge]

obv not

theres many other diff ways

Another way would be just directly compute the line integral over the C

parametrization

[chilly rock]

could you tell where i went wrong with my work up there?

[crimson sedge]

i never learnt the way YOU solved tbh

thats why i had to solve it in a different way than you

theres other ways i can solve by

but idk the method you used

i have to go grocery shopping, can you wait for like half an hour?

[chilly rock]

i'm good now you've already helped me

i just didn't think to try green's theorem

[crimson sedge]

im glad :> text me private if youneed something. its always chaotic over here haha

[crimson sedge]

its complicated tbh

i am not that advanced in calculus

mostly the basics

common sense and knowledge stuff--

[chilly rock]

thanks for the help

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[soft geyser]

Hi. How do I solve this problem?

[mental quail]

log_pi (5) = log(5)/log(pi)

if that helps, using that log rule you can reduce that exponent to be a rational fraction

[neon void]

that is the change of base formula

it's equivalent to log4(5)

[soft geyser]

omg, I understand nothing, can you please elaborate?

[neon void]

$16^\log_{4}{5}$

[wraith dagger]

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

[neon void]

look up change of base formula

$(4^2)^\log_{4}{5}$

[wraith dagger]

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

[soft geyser]

can you give me a link to a website where I can read about it and see similar tasks?

[neon void]

uhh

idk

the formula should pop up

or just look up "logarithm properties"

that should do it

$4^{2\log_{4}{5}}$

[wraith dagger]

mochaccino

[neon void]

$4^{\log_{4}{25}}$

[wraith dagger]

mochaccino

[neon void]

$25$

[wraith dagger]

mochaccino

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[soft geyser]

Thanks

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[robust harbor]

how can I even go about solving this, wolframalpha and sympy both gave up

[robust harbor]

or well, wolframalpha gave me this:

but i have no idea what to make of it

[wheat moss]

E is a function

i forgot what it's called but u can evaluate it if u need to

[robust harbor]

its an elliptic integral

[wheat moss]

yeah

[robust harbor]

but like whats with the complex numbers

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[lavish pendant]

so i already got the crossproduct just not sure how to show that resulten cross product vector is perpendicular to b

[lavish pendant]

the cross product was 2i+8j+36k

[void glen]

do the dot product

if it’s equal to 0, then it must be perpendicular

[lavish pendant]

i see, give me sec

[elder lichen]

can anyone explain me why is it (56/88)(55/87)(54/86) for question c.i. please?

[cedar kiln]

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

[lavish pendant]

nah it didnt equal to zero but 8, so i guess its not perpendicular

[void glen]

well it should

can you show your work

[wicked stag]

Bro

You are doing it wrong

The question asks the resultant vector is. Perpendicular to b

2 + 32 - 36 = 0

[lavish pendant]

i am?

[void glen]

yeah that’s the dot product

[wicked stag]

Your resultant a×b • B

You need to do the dot product of b with resultant

[void glen]

oh did you do the dot product of the original vectors

[lavish pendant]

ohhh mb i didnt read the question, sry lol

i just thought it was the original two.

thanks

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[round ledge]

hi, can anybody help me sketch the graph, im pretty sure i can do the rest alone but im wondering how to graph it

[round ledge]

should i simply take the basic exp function and transform it

by moving it 3 units up

and making it 2times more narrow

?

[idle tusk]

not narrow, 2 times taller

[round ledge]

sorry

[idle tusk]

narrow would've been e^2x, you have 2e^x

[round ledge]

i was confused with a parabola

[idle tusk]

but otherwise correct

[round ledge]

so it would cross y=0 at 5 instead of 1 right ?

bc 1x2 + 3

the domain would still be R

the range would be 3;+infinity

and how can i find b without a calculator tho ?

and the horizontal asymptote is 3 right ?

[crimson sedge]

insert 1

[round ledge]

yes but i dont have a calculator

waittt

i only need the exact value

so i can keep e

as it is

its basically the same

b = 2e+3

[rugged cove]

3.0

[round ledge]

wdym ?

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[ivory token]

is this correct? is it a typo where its meant to say 1/(2Fx) or am I tripping?

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[west berry]

Hii, I'm stuck trying to solve this 1 problem from my book which is to prove $\sum^\infty e^{-n^2x}$ is of class $C^1$.\
I tried proving $\sum^\infty(e^{-n^2x})'$ is uniformly convergent but wasn't able to :c\
Is the thesis even true? Could you give me any hints?\
Thanks for having a look!

[wraith dagger]

Clippy

[cedar kiln]

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[west berry]

<@&286206848099549185> any ideas?

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[lofty epoch]

.open

[cedar kiln]

Please don't occupy multiple help channels.

[lofty epoch]

part d e and f

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[manic condor]

use log base 3 of 12 = 2.262 and log base 3 of 2 = 0.631 to evaluate the logarithm log base 3 of 32

[manic condor]

Is this possible using log base 3 of 12?

Only way I could solve was 5 (log base 3 of 2) but maybe I'm just not seeing it

[digital cliff]

with the given information only, no

[digital cliff]

youre completely right

[manic condor]

okay thank you I thought I was tripping

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[lofty epoch]

help

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[cursive crystal]

If expand this product

[cursive crystal]

Why do only those terms stay, where the exponents of the a_k are all different, while all terms with at least one same exponent cancel out?

a_k are elements of a field

[lyric widget]

by "factor out", do you mean "expand"

[cursive crystal]

yeah, sorry

its late here

im sitting on those task for hours haha

this is the last step for me to be done

[lyric widget]

so what do you know about a_k besides the fact that they belong to a field

[cursive crystal]

Nothing, but it works notheless, like look at (a-b)(a-c)(b-c)=

[lyric widget]

yes I know how to expand parenthesis

[cursive crystal]

All exponents are different, those terms with the same exponents like abc cancel

Yeah, i didnt want to assume that you dont know

sorry

[lyric widget]

no it's just 4 am and I'm trying to figure out what's going on

[cursive crystal]

its 2 am over here lol

[lyric widget]

alright I think I see what you're saying

[cursive crystal]

I dont think if that approach works:

[lyric widget]

sorry can't help good night

[cedar kiln]

@cursive crystal Has your question been resolved?

[cursive crystal]

No, my approach doesnt work.

<@&286206848099549185>

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@cursive crystal Has your question been resolved?

[tight nova]

idk if this helps but the product in question is called the Vandermode determinent

nd is the determinent of this special matrix

notice how when you calculate the determinent of this you have to kind of cross out the column and row of the element you are at- so that means no other term in that product will have the same exponent

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[forest escarp]

i need help proving this

[forest escarp]

gotten to this point, x>2

<@&286206848099549185>

[cedar kiln]

@forest escarp Has your question been resolved?

[forest escarp]

<@&286206848099549185> 😭

[winter scarab]

Try expanding left hand side

[forest escarp]

i did

what next?

ok...

[winter scarab]

Nah wait it's 1/n above

[forest escarp]

this problem is under the topic of analysis and number theory

so im guessing ill have to treat it as a continuous function and look at growth rates

cs prev problems were like that

but the derivatives are all rlu weirs

[winter scarab]

N+hn

Split n into 1s

U get (n+1)/n

N(n+1)^1/n < n^-1 *(n+1)

@forest escarp

[forest escarp]

wdym "split n into 1s"

[winter scarab]

There are n terms right

[winter scarab]

1+1+1...n times = n

[winter scarab]

U want n + h(n)

[forest escarp]

bro what

why

would you do thys

[forest escarp]

yes ik

i used induction

[winter scarab]

1 +1/n for every term

[forest escarp]

to get rid of the harmonic

[winter scarab]

That's n+1/n

U have direct comparison now

Sum of (n+1)/n

[winter scarab]

U sure abt this approach

Usually in such questions U got to twist the terms

[forest escarp]

ah i see what you mean

but how would that help

[winter scarab]

I think some sort of squeeze approx

To the summation

[forest escarp]

im not sure how much this twist of terms actually helps

ill look into it

[winter scarab]

This equality should be reversed

@forest escarp

[winter scarab]

Compare it with the upper bound

Proove (1+n)^1/n < 2

[forest escarp]

oh k thx

[winter scarab]

Should be close I think I made some simplification mistake

This should be usual approach for such qs

[forest escarp]

looking for bounds?

got it

.close

[cedar kiln]

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[winter scarab]

I got the mistake

The equality is reversed

@forest escarp

[winter scarab]

The one here

[forest escarp]

yeah i js realized that too

ru sure bounds is the method here bc the 2 are very close

.reopen

[cedar kiln]

✅

[winter scarab]

I am unable to bind them 😦 let me think more bounds

[forest escarp]

ty for the help

this problems tough

[winter scarab]

It has different behavior when n is big n small

Equality keeps changing

I get it

Right side n is discrete

We can't take continuous n on left

@forest escarp

So it n can never go close to 0

N goes from 1 to infinity

🙂

💪

I think I proved it

@forest escarp

N > 1

For n = 1 they will be equal

[forest escarp]

wait wait what

so ur saying RHS is less than n+1

[winter scarab]

I think I did l hospital wrong

I am checking the differentiation

[forest escarp]

👍

[winter scarab]

I am weak at limits...try solving the limits

This should be solution

[forest escarp]

i foudn the inf limits of both sides

oh classmate suggested binomial expansion

if you divide both sides by n and then ^n

[winter scarab]

I also thought of binomial first but this one looks simpler

[forest escarp]

im trying ur method rn

so

[winter scarab]

I got solution

[forest escarp]

oh nice

[winter scarab]

Divide both side by n+1

U can prove left is less than 1

Easily

N divided by number greater than n

[forest escarp]

wait but if you divide by n+1

you would have n*(n+1)^(1/n-1)

[winter scarab]

N+1 is not 0

[forest escarp]

which can be greater than 1

[winter scarab]

It can't be bigger than 1

@forest escarp

[forest escarp]

plug in anything greater than 1

[winter scarab]

2/3

3/4

[forest escarp]

look

blue is (n(n+1)^(1/n))/(n+1)

green is y=1

[winter scarab]

😭

Let's try expansion

Expanding left side

[forest escarp]

yep

[winter scarab]

This qs is rubbish

[forest escarp]

i have this

[forest escarp]

lmoa

[winter scarab]

Shouldn't it be monotonically increasing function

[forest escarp]

graphically it looks to be

[winter scarab]

Aaah I am dividing that's why

[forest escarp]

ohhg

[winter scarab]

(N+1)^1/n is just n^1/n

Which is 1

So it's n

And other side is n+1

[forest escarp]

what

[winter scarab]

Expand left side

Done

@forest escarp I couldn't solve simple limits for so long

I need to die asap

Here it works

No tricks involved clean sol

[forest escarp]

im havign a hard time reading that

[winter scarab]

It won't be bounded

Only way should be expansion

[forest escarp]

yeah ive been trying limits and increase/decrease stuff for a while so i dont think theres anything there

[winter scarab]

Only expansion will work

Send me the solution if u get it as well

[cedar kiln]

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[west berry]

Hii, I'm stuck trying to solve this 1 problem from my book which is to prove $\sum^\infty e^{-n^2x}$ is of class $C^1$.
I tried proving $\sum^\infty(e^{-n^2x})'$ is uniformly convergent but wasn't able to :c
Is the thesis even true? Could you give me any hints?
Thanks for having a look!

[wraith dagger]

Clippy

[dreamy void]

x is all real numbers right?

[west berry]

$x>0$, sorry forgot to mention

[wraith dagger]

Clippy

[dreamy void]

well

if we take a look at the supremum

,,\sup_{x>0} \abs{-n^2e^{-n^2x}} = \sup_{x>0} n^2e^{-n^2x}

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

e^(-n²x) is decreasing

[west berry]

but the supremum is n^2, right?

because if we take x->0 we get n^2 * e^0

[dreamy void]

yea

strange

nah i think it's correct

if x > 0 it goes suddenly

down

if your interval was x > p and p > 0 then it would be uniformly convergent because p would be bigger than 0 but constant

so the e^ term would not disappear

[west berry]

i think it might be enough tho

[dreamy void]

and e^(-nx) for any n > 0 dominates n²

[west berry]

because in that case we have\
$f(x) = \sum e^{-n^2x}$\
$\sum (-n^2e^{-n^2x})$ is uniformly convergent\
and if $f(x_0)$ is convergent for some $x_0$\
then we have that $f$ is $C^1$ in any $[p,b], 0<p<b$,\
so $f$ is $C^1$ in $(0,\infty)$

[dreamy void]

well it converges point wise for sure

[wraith dagger]

Clippy

[west berry]

true

so I think I just have to polish it but it's roughly proven? or do I not see sth?

[dreamy void]

well for x > 0 it can't be uniformly convergent because the sup is not 0

[west berry]

yeah but this is not what we're trying to prove tho

[dreamy void]

well you tried that in the beginning

that's why i was here lol

[west berry]

true, but you made me realize it suffices to prove it for [p,\infty)

at least that's what I think

[dreamy void]

if p > 0 then yea you're right

because then the e^ term isn't gone

and it will eventually -> 0

the sup

,, \sup_{x>p>0} n^2e^{-n^2x} = \frac{n^2}{e^{n^2p}} \to 0 \text{ for } p > 0

[west berry]

Because Im trying to use the theorem that states that if $\sum f_n'$ is uniformly convergent on [a,b] to g and $\sum f_n(x_0)$ is convergent, then $\sum f_n$ is convergent and f'=g

[wraith dagger]

𝔸dωn𝓲²s

Clippy

[dreamy void]

ok

well i hope that somehow helps

[west berry]

i think so, I won't close for a bit until I'll make sure I understand everything but thx!

[cedar kiln]

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[bright fiber]

So there's a line between points A and B

and you divide that line in a ration of 2:1

and the point you have at that division, for it you need to find the position vector

Very arduously

There's probably a plug in formula for this somewhere lol

Hold on, lemme cook

Are you familiar with trigonometry?

[meager flicker]

yes

[bright fiber]

So here's my approach

we have points A(x1,y1) and B(x2,y2) right?

in the generic case

[meager flicker]

yes

[bright fiber]

the distance between these points is L = sqrt((x1-x2)^2 + (y1-y2)^2))

right?

[meager flicker]

yes

[bright fiber]

Let the angle between them be p

sin(p) = (y2-y1)/L

cos(p) = (x2-x1)/L

assuming B is top right in comparison to A

the signs might change if not, not sure how the angles would behave but it should still work

This ok?

Pardon, I mean the angle that the line AB makes with the x axis

[pastel vault]

You don't need trigonometry at all
AB = (8 - 2, 8 - 5) = (6, 3)
So 2:1 means that you need a point that is 2/(2 + 1) = 2/3rds of the way
2/3 * (6, 3) = (4, 2)

Can you figure out what to do now?

[bright fiber]

oh, well I guess that works too

Except that it doesn't...

Oh sorry, 4,2

[pastel vault]

Notice that the vector (6, 3) is the same distance and direction from the origin

[pastel vault]

As the vector between A and B

You are actually one step away from the answer btw

[crimson sedge]

theres a formula if you want

[bright fiber]

Yeah you meant like this

[pastel vault]

mhm

that's it

[crimson sedge]

(x1,y2) (x2,y2) m:n -> (mx2+nx1/m+n, my2 + ny2/m+n)

[bright fiber]

I'm used to using trig lol, would've messed with angles and stuff

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[crimson sedge]

roughly made kite shape. can someone find the the remaining sides

6781 is the length of the entire diagonal btw

[bright fiber]

What remaining sides?

[crimson sedge]

the red question marks

[bright fiber]

Oh the question marks lol

[untold osprey]

pythagorean theorem

[bright fiber]

Well can't you just find the heights of the left and right triangle

Isn't there like a rule that the diagonals for these things are perpendicular

[untold osprey]

correct me if im wrong but the lines inside the kite also form right angles so you can just do pythagorean twice

[bright fiber]

Yeh

Let the top question mark be x

[crimson sedge]

but i dont know one side

[heady temple]

ur from the middle east?

[bright fiber]

the bottom one total length - x

and then go from there

[heady temple]

yes

aki

[heady temple]

u didnt help me yesterday 😦

[bright fiber]

I did you went offline, we pinged

[heady temple]

oh

how to see

how do i go thjere

[bright fiber]

follow the red circles on the side lol

[raven shard]

why are right angle marks drawn?

[crimson sedge]

idk any of these three sides

dont i need two sides atleast for pythagorean

[crimson sedge]

because theyre right angles

[bright fiber]

that's one triangle

look at the bottom one

the left and right ? are the same length

this is also a property of these things I think

[raven shard]

so the issue here is that 2232^2 + 6371^2 is not equal to 6781^2

[bright fiber]

So you really only have 2 unknowns

[raven shard]

so we can see they cannot be right angles

[crimson sedge]

2322 is an approximation

2322.136947

[raven shard]

and the other numbers?

[crimson sedge]

theyre exact

[crimson sedge]

so how do i find them

[blazing zephyr]

is that an actual kite?

[plain ridge]

2 pairs of equal sides = yes

[raven shard]

Shorter Diagonal - 4363.466889
Half of Shorter Diagonal - 2181.733445
Central Question Mark - 795.210146

[crimson sedge]

great

how

[raven shard]

u use kite area

[bright fiber]

Something like this

[raven shard]

no

[blazing zephyr]

maybe try this, Red×Blue=2232×6371 and you can get blue by pytagorean

[raven shard]

do not do that

[bright fiber]

Don't mind my art skills

[raven shard]

2232.something * 6371 * 2 = long diagonal * short diagonal

and u go from there

[crimson sedge]

simultaneous equations

thought it would be easiere

[raven shard]

.

[crimson sedge]

ok

[raven shard]

.

[crimson sedge]

how is 2232 x 6371 kite area

[raven shard]

.

[blazing zephyr]

can you post the original question?

[crimson sedge]

google says long diagonal x short diagonal /2 is kite area

[raven shard]

because you have two right triangles

[blazing zephyr]

like what was known in the original question

[blazing zephyr]

the kite is 2 of the dark red triangles, so 2×1/2×2232×6371

[crimson sedge]

but its not a rectangel

or is it

[raven shard]

the two triangles are the same

[raven shard]

so see here

[crimson sedge]

ohhh

i got area of kite

then what

[raven shard]

.

[crimson sedge]

great

thanks

i did it

thanksalot

[cedar kiln]

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[tropic timber]

Hi?

[cedar kiln]

Please don't occupy multiple help channels.

[tropic timber]

.close

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[lusty lotus]

can anybody explain the answers to me

[lusty lotus]

and what do these arrows mean

[tame knoll]

i believe the arrows mean "logically implies that"

[zenith pond]

These arrows mean "if... then..."

[lusty lotus]

ohh

[tame knoll]

a should be true due to conjugate root theorem

[lusty lotus]

i get the last one but dont get the others

[median bough]

if... then....

the one behind arrow is writtena after if... and the one after arrow is after then...

[vague rapids]

is that a theorem

[tame knoll]

complex conjugate root theorem

[lusty lotus]

oh right

[tame knoll]

if z = a + bi, then z = a - bi is a solution as long as coefficients are real

[lusty lotus]

if one conjugate is a+bi then other is a-bi

yea

[median bough]

you can't have one real and one unreal root for a quadratic equation can you?

[tame knoll]

not for this one

[lusty lotus]

yep

[tame knoll]

as coefficients are real

[lusty lotus]

complex conjugates right?

[tame knoll]

B is false

[median bough]

its not true and false it is always, sometimes, never

[tame knoll]

true

[vague rapids]

wait

[tame knoll]

you can think of counterexamples

[tame knoll]

but in general yeah its sometimes, never, always

[vague rapids]

I dont see one

[lusty lotus]

its from the answer bank

[vague rapids]

complex conjugates can never be equal

[lusty lotus]

i didnt understand

[vague rapids]

a+ib has a conjugate a-ib

now these are equal only when b=0

i.e. the root must be real (imaginary part equals zero)

[tame knoll]

oh wait coefficients are real right

[lusty lotus]

yep

i thought the same

[lusty lotus]

yea

[tame knoll]

ahh right

if the roots were imaginary

[vague rapids]

now for the 'c' one, the answer bank makes it pretty clear ig

[tame knoll]

and for the quadratic to have real coefficients

it would need to be conjugates

[lusty lotus]

yea

[tame knoll]

right

[lusty lotus]

aight

thank you guys

.close

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[somber juniper]

can someone tell me how am i supposed to find arccos -√3/2 WITHOUT checking the calc

[somber juniper]

ik its in quadrant 2 bla bla but how am i supposed to know it during an exam like buhm

without wasting time

note: this is not for an exam

[hollow totem]

Is the exam banned calculator?

[vagrant elbow]

what's the answer to $\cos \theta = -\f{\sqrt 3}{2}$ where $\theta \in [0, \pi]$

[south tundra]

In general, arccos(-x) = pi - arccos(x) and for arccos(sqrt(3)/2) you just need to know the trig ratios

[wraith dagger]

uoǝu

[somber juniper]

no but im preapring for uni and idk whther theyll allow me

[somber juniper]

do i have to memorize them?

[south tundra]

Pretty much

[somber juniper]

alr

[errant fjord]

shit

[somber juniper]

one more thing i saw a rule in the book but cant find it now

whats the thing for yk arctan(tan smthn

or tan(tan^-1x)

[tame knoll]

is x

tan and tan^-1 are inverse functions so they cancel out

[south tundra]

Not necessarily

[tame knoll]

explain

[south tundra]

Nvm I was thinking of arctan(tan(x))

[somber juniper]

what is

tan^-1(tanx)
tan(tan^-1x)

[somber juniper]

its what im askin

and the otherwise

$$tan^-1(tan3\pi)$$

[wraith dagger]

ghiolimer

[somber juniper]

$$tan(tan^-1x)$$

[wraith dagger]

ghiolimer

[somber juniper]

are these both the same

and how do i find

[south tundra]

tan(arctan(x)) = x
arctan(tan(x)) should be the equivalent of x in the interval [-pi/2, pi/2] modulo pi whenever tan(x) is defined

[tame knoll]

they are different

[wraith dagger]

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

[somber juniper]

ugh anyways

[tame knoll]

tan-1( ...)

?

[somber juniper]

its -√3

[tame knoll]

you need to know your angles

work out the tan^-1 first

[somber juniper]

i dont

[tame knoll]

then evaluate cosine of the angle

[somber juniper]

know the angles

this is one of the practice quesrtions,

[south tundra]

Knowing trig ratios for that one is not necessary

Draw a triangle where tan of one of the angles is sqrt(3), find its cosine

(Since cos(arctan(-x)) = cos(-arctan(x)) = cos(arctan(x)))

[tame knoll]

thats pretty much the trig ratio lol

[south tundra]

It doesn't require you to know the angle though

[cedar kiln]

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.close

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[smoky willow]

hello

[smoky willow]

im wondering on how to go from this

to:

and what does dy / y even mean?

[crimson sedge]

integrate on both sides

are you aware of Differential equations ?

[smoky willow]

trying to learn it

[smoky willow]

if i integrate the right side of this. i get x^2 + C

[crimson sedge]

its abuse of notation to make seperable differential equations easier to understand

[smoky willow]

but what happens if i integrate dy/y and what does "dy/y" even mean?

[crimson sedge]

yes

[neon void]

1/y * dy

[crimson sedge]

what is integral of 1/y ?

[smoky willow]

ln(|y|)

+C

[neon void]

there u go

[smoky willow]

ohhh

i didnt see it was 1/y * dy

i was confused with dy/y

thanks guys

.close

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[balmy jewel]

Topic: Combinatorics. I dont get Part 2

Idk how to do it without the deletion-contraction theorem

Deletion-Contraction results in sorta complicated polynomials that are difficult to simplify here

for example for C_5 it's k(k-1)^4 - k(k-1)^3 + k(k-1)(k-2)

A bit tedious to simplify that to (k - 1)^5 - (k - 1), unless they wanted us to do it that way which I seriously doubt

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[hot sandal]

Can somone plz look to part a, i do not sure about the answe , since yn converge to 0 its mean the two term in the sequence will be close to each other and hence it will be contractive sequence since contractive sequence is cauchy then it is converge is it true my answer? If you have another answer plz let me know

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<@&286206848099549185>

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@hot sandal Has your question been resolved?

[cerulean sail]

"since yn converge to 0 its mean the two term in the sequence will be close to each other and hence it will be contractive sequence since contractive sequence is cauchy"
for example, the sequence sqrt{n} is such that the difference in terms goes to zero, but the sequence itself doesn't converge [to a finite limit] nor is Cauchy, it's unbounded

You may (i.e. really should!) use the fact that you're also told the even terms increase and the odd terms decrease

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[left gull]

I don't understand where these equations come from
I copied it in class and I am confused rn
The way I wrote it when I was copying slightly more sense than answer key on the right image because it shows the steps but I am having trouble seeing why those steps are like that like how do I know t_A consists of t_b and t_c

[left gull]

Would be great if I could be shown visually the path

Like with this logic shouldn't t_d consist of b and c?

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[ocean orbit]

Hello

[ocean orbit]

So like I need help with these

We're not face to face and I don't quite understand 😭

Is it okay if anyone can give an example with activity A 1. And activity B 1. I'll follow up 😭

[warped cliff]

it's just finding the missing side with pytha. theorem

[ocean orbit]

So like you just apply the pythagorean theorem??

[bleak viper]

yes

[warped cliff]

yes?

[ocean orbit]

OH

[warped cliff]

a is the side opposite of angle A

and so on

you're given 2 sides already so just find the missing one with a^2 + b^2 = c^2

[ocean orbit]

So like c² = 10 + 15

[warped cliff]

no

[ocean orbit]

10²

[warped cliff]

yes

[ocean orbit]

• 15²

Sorry I forgot to add ²

OH

SO KTS JUET LOOE THAT

OKAY

Sorry it's 3 am and I am functioning slowly

So is it the same at activity 2?

[bleak viper]

nop

you complete the triangle first by finding third side using pythageorean theorem

then use trigonometric ratios to find angles

[warped cliff]

solving a triangle does mean finding all angles and sides

but i think it's just finding the missing side

since the person here isn't too familiar with pythag.

[bleak viper]

what

i said find third side then use trig ratios to find angles

isnt that true

[warped cliff]

huh yeah im agreeing with you

?

[bleak viper]

huh

[warped cliff]

"solving a triangle does mean finding all angles and sides"

[ocean orbit]

Yes

[bleak viper]

bro

[warped cliff]

but i think because the person here isn't too familiar with pythag. , so it's just finding the missing side

[ocean orbit]

It wasn't discussed much

[bleak viper]

i thought u asked the question

and was confused

my bad

[ocean orbit]

Since our schedule changed

Our classes are mixed
Modular + f2f basically it's a self study lesson but how could I fully comprehend if there is no one to teach me 😭

[warped cliff]

i mean pythag. is pretty simple

you can just use khan academy

[ocean orbit]

Yup I'm done with the first activity

What about the second one 😭

Could you give one example :')

[cedar kiln]

@ocean orbit Has your question been resolved?

[cedar kiln]

@ocean orbit Has your question been resolved?

[cursive gulch]

actuallly you have to do same thing

[cursive gulch]

this

[cedar kiln]

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[true garnet]

I found 2 different points given the whole line of intersection from 2 different plane equations but I need to find a 3rd one that isnt parallel and I dont know how to do that just given that there is a plane perpendicular to the one I am solving for

[tame knoll]

so it needs to be perpendicular to the plane x + y - 4z = 3

the normal vector of x + y - 4z = 3 is i + j -4k

as the plane needs to be perpendicular, the normal vector can be just ANY vector that is perpendicular to i + j - 4k

so let the normal vector equal ai + bj + ck and set up a dot product equation = 0

and find any values of a, b, c that work

[true garnet]

so I got <2,2,1> and then I used that combined with a point I solved for to get the equation 2x+2y+z-6=0 but it was wrong

[tame knoll]

its right

theres “infinite solutions” cause theres an infinite number of normal vectors perpendicular to i + j - 4k

[true garnet]

it didnt work when plugged in but yea there are inifinite solutions

this website supposedly gives it to u if its right even if the answer was 5 and u put 1(5)(1)(1)

so im still stuck

[tame knoll]

what did you get for the line of intersection?

[true garnet]

I just solved for 2 points instead of the line which should work but apparently not

[true garnet]

its i-3j+k

<@&286206848099549185>

2x+2y+z-6=0 is what I have and it is saying its wrong

I guess I do it another day

.close

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[smoky willow]

hello

[smoky willow]

im wondering how i would solve:
y'' + y' = e^(-x)

i know there is two steps

first solve y'' + y' = 0

and then find one solution where u "guess"

and add them together

[smoky willow]

for this i get A + Be^(-x)

[smoky willow]

but this i dont know...

[cosmic steppe]

Your guess would be Ae^(-x)

thats just like'

i mena

[smoky willow]

yeah

i tried that

they will cancel

[cosmic steppe]

okay try Axe^(-x)

just like

hunches and shit

[smoky willow]

well what made you think of that

if Axe^(-x) dont work what would u try next

[cosmic steppe]

crying and weeping

is what i would do next

[cosmic steppe]

i mean if you differentiate this

you get different terms bc of product rule

and x isnt too crazy

like im not gonna test sometjing like Asin(x)e^-x

[smoky willow]

yeah Axe^(-x) seems to work

thanks

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[crimson sedge]

what is r here again?

the vector r they use

[elder tapir]

It’s like the vector OM

The position vector

[crimson sedge]

so should it just be uh

the spherical coordinate form of it?

as in rsin(phi)cos(theta),rsin(phi)sin(theta),rcos(phi)

[elder tapir]

Whatever you want actually

Since we are talking about a sphere it might be the right solution

[crimson sedge]

alright cool ty

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[limber sparrow]

How do I go about solving this problem, they only give the tan(theta) = y/x property but idk what to do when tan(6theta) is the problem here

[smoky fiber]

what's the question

To convert r = 6 theta to rectangular?

Or otherwise?

[limber sparrow]

Convert r= 6 theta into rectangular

[smoky fiber]

Alright

do you know how to go from angle to sin(theta) and cos(theta)?

[limber sparrow]

Like by drawing a right triangle and physically working it out?

Or do you mean something else

[smoky fiber]

Yeah something like that

Wait why are u using arctan instead of tan(theta)?

[limber sparrow]

There's a video the question provided, and he applied tan to both sides

tan r = tan(sqrt(x^2 + y^2)

then the right side was tan(theta) = tan(y/x)

Then he used arctan

[smoky fiber]

Mind if you attach the video?

[limber sparrow]

It's through cengage so I'm not sure it works, but here:

https://college.cengage.com/geyser/cengage_9780538738101/html/watchit_player/?asset=larcalc9_10_04_041&prod=larcalc9

[smoky fiber]

Aight now your begining equation makes more sense

[smoky fiber]

No it works fine

try moving the 6 to the other side as r/6

you would end with (sqrt(x^2+y^2))/6 = arctan(x/y)

[limber sparrow]

Like this?

[smoky fiber]

yep

[limber sparrow]

You're able to move the coefficient out of the tan?

My PreCalc knowledge has always been really weak so I'm sorry if it sounds dumb

[smoky fiber]

No

You're moving the 6 to the other side when r=6theta

Wait is this legal?

Forget about it

[limber sparrow]

It does work!

[smoky fiber]

No way lol

[limber sparrow]

ig you can cause it's not in the parenthesis on tan yet

tysm man this was actually beating my ass for an hour

[smoky fiber]

yeah that would explain

np you're welcome

[limber sparrow]

have a great day

[smoky fiber]

Next time try to move things to the simple side of the equation

same, have a great day

[limber sparrow]

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[crimson sedge]

https://cdn.discordapp.com/attachments/496784996510466057/1232121924788686968/Screenshot_2024-04-22_at_8.12.20_PM.png?ex=66284e76&is=6626fcf6&hm=3a646ef0278beba0ad45ee692e16bd42124fadef13b3044608884141f818c488&

[crimson sedge]

how can I parameterize this surface?

I know its a triangle but I dunno how to parameterize it

[cerulean sail]

Three line segments?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

yeah

oh wait

isnt there a more efficient way though?

instead of setting up 3 different integrals for it

Like I know im supposed to use uh

Stokes theorem I think for it

[cedar kiln]

@crimson sedge Has your question been resolved?

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[viral kiln]

how would i go about finding the derivative of this? i know its simple but im just a bit cknfused

[violet flume]

nothing to do but use the rules

you got chain rule, product rule, power rule

assuming you know the derivative of cos(x)?

thats all you need

[viral kiln]

-sinx right

[violet flume]

it is

[viral kiln]

ohhh it’s x^3 * cos1/c

cos1/x *

[violet flume]

yea