help-49

[glossy pagoda]

oh alr

[steep hinge]

negation of (4 does not divide a and b)

[glossy pagoda]

hmhm

[steep hinge]

is (4 divides a or 4 divides b)

[glossy pagoda]

yh

[steep hinge]

"and" becomes "or"

[glossy pagoda]

yh

[steep hinge]

wdym with "yh"?

[glossy pagoda]

um how to say xD

ackknowledge

your response

[steep hinge]

i don t know english very well, but i think its a compliment

[glossy pagoda]

basically i let you know that i read your message

imma send a pic of the proof i wrote, tell me what you think

[steep hinge]

this lokks fine

[glossy pagoda]

alr cool

actually i think i made a small mistake

[steep hinge]

but idk if the first statement is the converse of the second one

[glossy pagoda]

so

the first statement is original

the second is converse

the third statement is the assumption for the converse

[steep hinge]

the first statement is clearly false

[glossy pagoda]

yh

[steep hinge]

4 does not divide 2 and 4 does not divide 2 but 4 divides 2*2=4

[glossy pagoda]

hmhm

[steep hinge]

2=a=b

[glossy pagoda]

oh

i didnt give a counterexample where i had 2 = a =b

i did a =6 and b =2

but yh all is good

[steep hinge]

the first statement is true only for prime numbers

[glossy pagoda]

really?

[steep hinge]

yes

but now i have to go

[glossy pagoda]

since when did 4 divide prime is true

ah alr

thanks for the help

.close

[midnight plank]

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[steep hinge]

the statement with a prime number in place of 4

.close

[glossy pagoda]

hmhm

alr

[midnight plank]

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[leaden matrix]

Im guessing the answer is E since the graph of k/x has an asymptote, not sure if my thinking is correct here tho

[tropic nexus]

well what if the x-1 were to somehow cancel from numerator and denominator

is that possible with some c?

[leaden matrix]

if we set C equal to 3 then we'd be left with 1/x tho

and that has an asymptote at the y axis so the limit doesn't exist since the limit isn't the same from the right and left

[tropic nexus]

but you're taking the limit at x=1 not 0

so it's fine

[leaden matrix]

I see, so it's just 3 then?

[tropic nexus]

yes

[leaden matrix]

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[graceful olive]

why are these wrong

[grizzled gazelle]

Because its just slowing down until v=0 and after that its accelerating in the inverse direction

slowing down means that the absolute value of v is decreasing

[graceful olive]

so for the first image, youre saying that the particle is actually speeding up on (0,1) and (2,3)

and its slowing down on (1, 2)

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[zenith raft]

I'm not completely sure what I did wrong here

[last slate]

can you show your work so we can tell what's wrong?

[midnight plank]

@zenith raft Has your question been resolved?

[zenith raft]

<@&286206848099549185>

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[iron cloud]

hello

[hearty rune]

Ello

[iron cloud]

so\

[hearty rune]

Do tell

[hearty rune]

,rccw

[hearty rune]

What you wrote so far seems alright from what i can see

[iron cloud]

yes but idk where to go from here

it gives me an impossible sqrt

5 in the index, √96^2

and there's nothing you can do for multiply the index by it

liek idk what to do

i get that 96^2/5 and u have to turn it into the sqrt

so i did that but

[tacit rose]

The helper made a mistake, you did it wrong

You don't raise to the 2/5 before dividng by 3

[iron cloud]

please do tell

[tacit rose]

Because it's 3x^(5/2) = 96, you want to isolate for x

[iron cloud]

OH I SEE

so first divide by 3

[tacit rose]

Yes

[iron cloud]

then mult by 2/5 right

[tacit rose]

Not multiply but to the 2/5 power

[iron cloud]

YES

I KNEW IT

thank u thak u tnak

thank

you

ill go try it out

[tacit rose]

When you solve for x, think about reverse PEMDAS

Or BODMAS

You work backwards to isolate for x, you get rid of addition/subtraction first, then multiplication/division, then exponents, then whatever you would have in parentheses

[hearty rune]

Oof i apologise for that blunder on my part

[iron cloud]

pawn to c3

help please

i have aother one

[hearty rune]

what did you have before?

[iron cloud]

well, i knew that 25 went into 100

so i figured, why not 8 and 25

but there's no 3 pairs in either of those

bc 2√2 is 8 simplified, right

[hearty rune]

this is sensible

[iron cloud]

and the 25 simplified is simply 5

[hearty rune]

its cube root, not square root

[iron cloud]

so idk what to do

[hearty rune]

you have $(8\cdot25)^{\frac{1}{3}}=8^{\frac{1}{3}}\cdot25^{\frac{1}{3}}$

[grand pond]

AℤØ

[hearty rune]

try simplify that

[iron cloud]

wait but why is it 1/3

[hearty rune]

cube root

[hearty rune]

i assume we are looking at the left one first?

[iron cloud]

yes!

but where does 1/3 come from jiust curoius

[hearty rune]

oh right

$x^{\frac{1}{a}}=\sqrt[a]{x}$

[grand pond]

AℤØ

[hearty rune]

just as a rule

last one

[iron cloud]

ahhh

\i see what you did there

you reversed it

okay okay i see your logic

continue pls

[hearty rune]

when it asks you to simplify what is is generally asking for exactly? Exponents of prime numbers?

or just simplified surd?

[iron cloud]

just simplified essentially

[hearty rune]

okay, lets just write this as
$\sqrt[3]{8}\cdot \sqrt[3]{25}$

[grand pond]

AℤØ

[hearty rune]

then i suppose

how would you simplify that?

[iron cloud]

hmmm

well 2√2 is 8

[hearty rune]

is it?

[iron cloud]

yes

thats what it simplifies to

[hearty rune]

youre square rooting

this is cube root

[iron cloud]

hm?

what do i do

i mean it can only be simlifie by 2

[hearty rune]

yeah, cube root of 8 is 2

[iron cloud]

OHHHHHHHHH

IM STUPID

thats right

2 times 2 times 2

2 times itself 3 times

is 8

[hearty rune]

you have $2\sqrt[3]{25}$ or maybe written as $2\sqrt[3]{5^2}$

[grand pond]

AℤØ

[hearty rune]

thats about as simple as it can be

[iron cloud]

wait u ant multiply anything by itself 3 times

to get 25

[hearty rune]

well, you can, just not an integer

[iron cloud]

i suppose not bc is multiple of 5 right

so it wouldn't really

wait but also

does 25 just simplify to 5

[hearty rune]

when doing what?

[iron cloud]

oh wait nah ur right

bc the 25 cant go anywhere

then there's nothing inside the uhh thingy

[hearty rune]

i suppose?

[iron cloud]

ok ok

so same thing for the second one right

ill see what happens and ill brb

[hearty rune]

okie dokie

[midnight plank]

@iron cloud Has your question been resolved?

[midnight plank]

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[iron cloud]

hello

[hearty rune]

hello

[iron cloud]

so

the secons one

[iron cloud]

idk how to start

[hearty rune]

do essentially what you did before

25 and 8 seem fine

[iron cloud]

but for 4

for 25

[hearty rune]

yeah $\sqrt[4]{25}\cdot\sqrt[4]{8}$ just simplify those

[grand pond]

AℤØ

[iron cloud]

but 25

oh wait

wait a minute

[iron cloud]

i dont get it

8 simplifies to

[hearty rune]

just write 25 and 8 as exponents of primes and simplify

[iron cloud]

4^2√2

but

[hearty rune]

dont put surds in surds

[iron cloud]

25 can't do nothing

wdym by that

idk what surds mean

[hearty rune]

roots

25 is 5^2
the fourth root of 5^2 is the square root of 5

thats a simplification

[iron cloud]

but the root is 4 tho

[hearty rune]

i said that

$\sqrt[4]{5^2}=\sqrt{5}$

[grand pond]

AℤØ

[iron cloud]

but

wouldnt oit be

^4√5

[hearty rune]

no

why would it be that

[iron cloud]

wait hold on

how did u get that tho

cuz simplifying 25 givew you 5

5 goes outside of the radical

but in this case its still inside the radical

[hearty rune]

what? no

wdym simplifying 25 to 5

i just square rooted, the 5 is still in the sqrt

because its a fourth root

$\sqrt[4]{5^2}=5^{\frac{2}{4}}=5^{\frac{1}{2}}=\sqrt{5}$

[grand pond]

AℤØ

[hearty rune]

is an exponent argument

but its just intuitive

[iron cloud]

shit

i forgot that rule

the flippy reverse radical

that's what makes that possible

i need to write that down

[hearty rune]

i guess, you dont need that to see it though
its intuitive that the fourth root of 5^2 is the sqrt of 5

[iron cloud]

yea

but i always saw it as

there has to be smth multiplied by itself 4 times or such

otherwise its impossible

so 8 just simplifies to 8 1/4, right

[hearty rune]

id just leave it as $\sqrt[4]{2^3}$

[grand pond]

AℤØ

[iron cloud]

ah

okay

[iron cloud]

how do i solve one with a variable in it

[hearty rune]

such as?

[iron cloud]

^4 √162x^11

[hearty rune]

just focus on the 162

[iron cloud]

ok i got it

wait i dont mess with the variable?

cuz i got

[hearty rune]

theres nothing you can do to it

[iron cloud]

because its uneven right

[hearty rune]

yeah

[iron cloud]

ahhh right its coming back to me

if you suspect there's nothing you can do with it and you cant divide it into 4

then you leave it as is

so that means its

3^4√2x^11

i got a lot of studying to do

[hearty rune]

yeah

[iron cloud]

my test is tomorrow

but im not failing that

also btw

um there's this symbol

a + and -

and i think the teacher said it had to do with an even root or something

can you help explain that

pls

[hearty rune]

do you have an example?

[iron cloud]

like 2^√25

he said "the answer can be either positive or negative", so we use this symbol

[hearty rune]

you mean $\pm$

[iron cloud]

because -5 times -5 equals 25

[grand pond]

AℤØ

[iron cloud]

but 5 times 5 also equals 25

so there's two answers

but idk what to do with that

[hearty rune]

the sqrt function will only return the positive value
but indeed if you were asked to solve x^2=25, then x can be either 5 or -5

[iron cloud]

these m&ms taste expired

anyway

[iron cloud]

so both are included in the answer

[hearty rune]

theyre both possible x's

$x=\pm5$

[grand pond]

AℤØ

[iron cloud]

oh

ok ok

[hearty rune]

i wrote solving x^2=25 gives x=5 or x=-5

if i saw just x=sqrt(25), x would only be 5, at least in my eyes anyway

be mindful of that

[iron cloud]

√32 equals

2 times 2 times 2 times 2 times √2 right

[hearty rune]

no

that would be 16√2=√512

√32=√2^5 (or √16 * √2=4√2)

=4√2

[midnight plank]

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[south rampart]

yo

[south rampart]

How do u do this question?

[echo forge]

can you get the equation of the 2nd graph?

[midnight plank]

@south rampart Has your question been resolved?

[south rampart]

Idk how to do that

[slender walrus]

the graph on the right is a straight line

[midnight plank]

@south rampart Has your question been resolved?

[midnight plank]

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[lofty walrus]

Question 1

[lofty walrus]

Exercise 5.10

Answer says y = -5x + 8

I got y = -5/x + 8

Is the textbook wrong?

[tulip mason]

no

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[sharp coral]

that equation is not a line

[lofty walrus]

Here

Sorry got better quality

[tulip mason]

oh

m1m2=-1

the slope is -5

so -5x+8

[lofty walrus]

Explain this

My brain is trying

[marble pumice]

yeah if y=ax+b // y = a'x + b' , a.a' = -1

[lofty walrus]

Pls

[tulip mason]

you don't need to take reciprocal of x

[marble pumice]

u can prove this pretty easily

[tulip mason]

the slopes multiply to -1

[lofty walrus]

What is m1m2

[marble pumice]

An equation of a line can be expressed as y = ax + b

[tulip mason]

I used m to represent slope

[lofty walrus]

Yeah

[lofty walrus]

But what is 1 and 2

[tulip mason]

$m_1m_2=-1$

[grand pond]

WhereWolf(ping if needed)

[marble pumice]

so lets call the undefine line y1 = a'x + b'

[lofty walrus]

What does ‘ represent

Oh

Jsut nothing ?

[marble pumice]

Yeah

[lofty walrus]

To represent the undefined line

Ok

[marble pumice]

Just a variable

[lofty walrus]

Yeah

[marble pumice]

so the line

Passes 1,3

[lofty walrus]

Yeah

[marble pumice]

lets keep it there

[lofty walrus]

Wait

[marble pumice]

oh wait

[lofty walrus]

Where the x/5 go

[marble pumice]

sr im dumb

[lofty walrus]

And a

Ok

[marble pumice]

We pluh it in and ge

3 = a' + b'

[lofty walrus]

Yeah

[marble pumice]

next up

the y = a'x + b' // with y = x/5 + 3

[lofty walrus]

Yeah

[marble pumice]

So we have this thing

if a line pendi a line

a.a' = -1

[lofty walrus]

Wait

What does // represent

Perpendicular ?

[marble pumice]

oh my bad yes

[lofty walrus]

Yeah

The gradients multiplied by the gradient is -1

[marble pumice]

yeah yeah

So we have a' . 1/5 = -1

since a of that equation is 1/5

[lofty walrus]

Wait

Yeah

You right

What now @marble pumice

[marble pumice]

right so

[lofty walrus]

Wait

wiar it’s clicking

Just the gradient becomes reciprocal

Not x aswell

X/5/-1

[marble pumice]

what

lets plug that in

[lofty walrus]

is how it should be right

like it’s not the answer

but yk it’s different

[marble pumice]

We can calculate a' = -5, plug that here and get the answer

[midnight plank]

@lofty walrus Has your question been resolved?

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[primal heron]

Why limit to inf an is not diverge? Or do i forgot to add something to my answer?

is it not diverge? And is it infinitely osculate between 1 and -1 when n is even?

[jagged saffron]

The limit does not exist yes

If this is your work then you likely got marked wrong for writing $-\infty / \infty$

[grand pond]

ΣΑCu

[midnight plank]

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[grand rock]

What's sin theta then

Can you express r from here

Can you move things around

So you get

r=...

(yes)

[fickle ravine]

why would you do that

[grand rock]

sin theta = r / r+a

[fickle ravine]

cant it become r+a(sintheta)=r

[plush prism]

no

almost

[grand rock]

(r+a)sin theta = r
rsin theta + a sin theta - r = 0
r(sin theta - 1) = -a sin theta
r= -asin theta / (sin theta - 1)

[fickle ravine]

then its in terms of r

[plush prism]

sin = r / (r+a)
(r+a) * sin = r

[grand rock]

Pretty simple

No

[plush prism]

distributive rule

[grand rock]

Thays why thats not the end

[fickle ravine]

im still in rational function in pre calc i thought id try 😭

[grand rock]

Distribute the brackets and move r to the left side

No we do not

What???

[plush prism]

no we dont

use distributive rule

[fickle ravine]

they want the answer in terms of r

and a

[plush prism]

also no

in terms of a and theta

[fickle ravine]

oh right my bad

[plush prism]

get everything with r on one side

and the rest on the other

[plush prism]

try this

[grand rock]

Ok now move r to the left side

[plush prism]

uhm

a+b=c
then we cant follow with
a=b/c

that is what you did there

we have r * sin + a * sin = r

move everything with an r to one side

and the rest to the other

using the inverse functions
to move a summand to the other side, we subtract

dont dividee

good

[grand rock]

So now factor out the r

[plush prism]

now use the distributive rule again to factor out the r

ab+ac=a(b+c)

works in both ways

good

[grand rock]

Now what do you think

You should do

To have only r

Yes

[plush prism]

coorect

[grand rock]

Awesome

You win

[plush prism]

in case you wanna make it even smaller, you can change the fraction on the left

but definitely not needed

but i am 99.9% sure you are not supposed to do that here haha

[midnight plank]

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[fossil fossil]

i got 10 000 +80pi/9

but the answer key is 10000 + 80000pi/9

[midnight plank]

@fossil fossil Has your question been resolved?

[midnight plank]

@fossil fossil Has your question been resolved?

[fossil fossil]

.close

[midnight plank]

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[hasty bridge]

Can someone explain me what's done here:

[round parcel]

I assume they use the double angle identities.

[robust isle]

Well they're showing the double angle identites using complex exponentials

[hasty bridge]

but why can they cancel the isin(20), thi sis the part i dont understand

[round parcel]

Well, let's see.

cos(2 theta) = cos^2(theta) - sin^2(theta).

[robust isle]

They're just comparing the real and imaginary parts on both sides

[round parcel]

sin(2 theta) = 2 sin(theta) cos(theta).

[hasty bridge]

Ah I think I understand this

[round parcel]

So, cos(2 theta) + i sin(2 theta) = cos^2(theta) - sin^2(theta) + 2i sin(theta) cos(theta)

So, that's how the finish the first equation.

[hasty bridge]

I already see it, Thanks (:

[round parcel]

Oh, OK.

[visual tiger]

they're doing it the other way around

[round parcel]

Oh, I see.

[hasty bridge]

they just ignore the imaginary part

smart move

.close

[midnight plank]

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[unborn loom]

Shouldn't the phi go to pi?

[midnight plank]

@unborn loom Has your question been resolved?

[unborn loom]

<@&286206848099549185>

[fallow scarab]

Which angle is phi

Azimuthal?

Actually both of them should be 0 to pi

So yes you're right

[unborn loom]

Why is $\theta$ from $0$ to $\pi$? We have $0\leq x\leq 4$ and $0\leq y \leq \sqrt{16-x^2}$.

[grand pond]

Zander

[unborn loom]

That's the first quadrant

[midnight plank]

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[weak idol]

can anyone help with this question??it's trig and I cannot figure it out. ty

[tribal temple]

"hence" what's the part that comes before that?

[craggy flicker]

Can you use half angle identity?

[tribal temple]

Comparing to double angle might help a bit in this case(!)

[weak idol]

I used this for d

[weak idol]

wait how

cuz pi/16 is not on the unit circle and idk how to solve it

[tribal temple]

Look at what you're working with here, $\frac{5 - 5\tan^2(\theta)}{\tan(\theta)}$

[grand pond]

@tribal temple

[tribal temple]

Do you remember the tan double angle formula?

[weak idol]

2tanx/1-tan²x

[tribal temple]

Ye, that's tan(2x) - now compare that to what you're trying to find...

[weak idol]

is that inversed??

[tribal temple]

Yep [reciprocal], but also two other changes too...

[weak idol]

oh yeah reciprocal

pi/16 × 2 = pi/8

would it be 5/2tan2x

[tribal temple]

Yep

[weak idol]

oh then sub 2x with (sqrt2 -1) ???

i mean tan2x sorry

[tribal temple]

Actually wait one moment that should be 10/tan(2x)

[weak idol]

oh

yeah

[tribal temple]

But yep

[weak idol]

cuz 1/2

omg ty I got it now

.close

[midnight plank]

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[oblique torrent]

G is a commutative group. ord(<x>)=3 and ord(<y>)=5. z=xy. How do i show that 3 divides ord(<z>)?

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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7. None of the above

[fresh sparrow]

big hint: z^2 = xyxy = x^2y^2

[oblique torrent]

(z^3)^5 = (y^3)^5 = y^15= e so it means that ord(z)=15?

[fresh sparrow]

yes

[oblique torrent]

ok but now we immediately found ord(z)

[oblique torrent]

shouldn't i say that ord(<z>) divides 15 instead because y^15=e doesn't mean that ord(<y>)=15?

[fresh sparrow]

is that wrong

you can also just go by the argument that

like

z=xy, and x has order 3

idk

ord(z) = lcm(ord(x),ord(y))

[fresh sparrow]

well it divides both 15 and 3

15/15 = 1 ✅
15/3 = 3 ✅

[worn halo]

Hmm, not sure how to solve in general, I think you are right in saying ord(<z>) divides 15. If you would have something like z^4 = e, then x^4y^4 = e -> x=y, which is a contradiction. You can do similar things for other powers below 15, but that's probably not the quickest way

[oblique torrent]

does this formula only work in commutative groups?

[fresh sparrow]

i think so

if a group isn't abelian then xyxy ≠ x^2 y^2, generally

so this argument isn't applicable

[oblique torrent]

so is ord(<z>)=15 or we just know that its order divides 15?

[oblique torrent]

yes it makes sense

[oblique torrent]

if ord(<z>) divides 15, we can only say it's either 1, 3, 5 or 15

[worn halo]

Yeah

you might have to prove that as well though

[oblique torrent]

and 3 does not divide 5?

[worn halo]

z^5 = e would imply x^5 * y^5 = e, or x^2 = e

so it would mean ord(<x>) = 2

I feel like the proof might involve bezout's lemma

[oblique torrent]

there's something i don't get @fresh sparrow

[oblique torrent]

yes to what?

[worn halo]

second one ord(<z>) divides 15

[oblique torrent]

oh ok

so we can't say ord(<z>)=15?

[worn halo]

No, I think we might even still have to prove that ord(z) | 15

because it hasn't really been proven

[oblique torrent]

the seond quastion in the exercise was show that 15 divides ord(<z>)

[worn halo]

z^15 = e, suppose z^k = e with gcd(k, 15) = 1, then by bezout z^1 = e, so that's some progress

Oh, wait

suppose z^k = e, with gcd(k, 3) = 1. then z^1 = e

so 3 | k

nvm, that just works for x

[oblique torrent]

does gcd(k, 3)=1 mean that 3 divides k?

[fresh sparrow]

i don't get why the order of z isn't 15

[worn halo]

it is, but we have to prove it

[fresh sparrow]

z^15 = x^15y^15 = e*e

[worn halo]

yeah, but what if z^3 = e?

then z^15 = (z^3)^5 = e^5 = e

[fresh sparrow]

z^3 = y^3

which isn't e because ord(y)=5

[worn halo]

yeah, but how does that proof extend to other integers than 3?

[oblique torrent]

yeah so that would'nt work for 5 either so it's 15

[fresh sparrow]

does it have to

z^k = e only when x^k = e and y^k = e, which happens for the first time for k=lcm(ord(x), ord(y))

[oblique torrent]

why does it ask me to show that 3 divides ord(z) and then that 15 divides ord(z) if we can immediately know that it's 15

[fresh sparrow]

i feel like i am missing something very simple but i genuinely don't see what's wrong with my approach

[worn halo]

why does z^k = e only happen when x^k = e and y^k = e?

x^k = e and y^k = e imply z^k = e, but does it go the other way around too

[fresh sparrow]

🧐

okay..

[fresh sparrow]

so a possible counterexample is when y=x^-1 right

y^l = x^-1 for example

but it must be 15 in our case because it's already a product of 2 primes

[worn halo]

y = x^-1 won't work because y^3 = (x^3)^-1 = e

You can probably go down the list of 1 ... 14 and show each one individually with a bit of algebra, but that's not the intended solution I think

[fresh sparrow]

so essentially your argument boils down to the fact that for some m and n, both respectively less than ord(x) and ord(y), x^m y^n may be e?

[worn halo]

I think so, m = n in this case

[fresh sparrow]

i see

[oblique torrent]

[worn halo]

Hmm, weird argument but: suppose z^k = e, with gcd(k, 15) = 1, then 3 and 5 don't divide k. And z^k = x^k * y^k = e, then raise both sides to the 5th power, so x^5k = e

that gives a contradiction though, since 5k can be reduced to something less then 3 by subtracting 3 multiple times

Oh suppose z^k = e and 3 doesn't divide k, then x^k * y^k = e, x^5k * y^5k = x^5k = e

same contradiction

[oblique torrent]

wait im lost

we're still on question 1 right?

[worn halo]

yeah

I'll go step by step

suppose z^k = e and 3 doesn't divide k, then x^k * y^k = e

right?

[oblique torrent]

yes

[worn halo]

And x^(5k) * y^(5k) = e as well

[oblique torrent]

yes

[worn halo]

so x^(5k) = e

and x^(5k - 3) = e

and x^(5k - 6) = e

etc..

x^(5k - 3a) = e

[oblique torrent]

i don't understand this

[worn halo]

x^(5k) / x^3 = e / e = e

[oblique torrent]

i prefer not to use division when we alrady use the multiplicative notation

so it's x^(5k)*x^(-3)?

[worn halo]

x^(5k) * (x^3)^-1 = e * e^-1 = e * e = e

[worn halo]

yeah

[oblique torrent]

ok

[worn halo]

so x^(5k - 3a) = e, right? (a is an integer)

[oblique torrent]

yes

[worn halo]

but 3 doesn't divide 5k - 3a, suppose it would: then 3 | 5k -> 3 | k, but we assumed 3 didn't divide k

[oblique torrent]

3 doesnt divide 5k - 3a because 3 doesn't divide k?

[worn halo]

yeah

and 3 doesn't divide 5

[oblique torrent]

yes i see

[worn halo]

so for some a, 5k - 3a = 1 or 5k - 3a = 2

and x^1 = e, or x^2 = e, which are contradictions

[oblique torrent]

so i do the exact smae thing with 15 in the second question?

[worn halo]

Not with 15, but with 5

so we get that 3 | ord(z) and 5 | ord(z)

so 15 | ord(z)

[oblique torrent]

wait, i start with z^k=e and 15 doesnt divide k right?

uhm yes

i see

[worn halo]

for (1) you start with z^k = e and "3" doesn't divide k

for (2) you start with z^k = e and 5 doesn't divide k

[oblique torrent]

ok

[worn halo]

so you can conclude that 3 | k and 5 | k which would mean 15 | k

[oblique torrent]

ok thanks and in the last question it says: show that G has a subgroup of order 15. So we now 15 divides ord(z) but is lagrange theorem enough to say that a subgroup of order 15 exists?

[oblique torrent]

by the way, do the 2 questions imply that ord(z)=15? if yes, why?

[worn halo]

(1) implies 3 | ord(z), and by replacing all the 3's in the proof with 5's you can get 5 | ord(z)

so yes

[worn halo]

Maybe you can use the subgroup generated by z

since we know that z^15 = e

the subgroup will have order 15

[oblique torrent]

ok thanks

.close

[midnight plank]

Channel closed

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[oblique torrent]

.reopen

[midnight plank]

✅

[oblique torrent]

i also asked to someone yesterday and they disnt give me the answer but they told me to use $<x> \cap <y>$. Do you know how it could've worked out if we used it?

[grand pond]

lilisworld

[oblique torrent]

i mean the first question, not the last one

@worn halo

@fresh sparrow

<@&286206848099549185>

[worn halo]

No, not sure

[oblique torrent]

ok np

.close

[midnight plank]

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[midnight plank]

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[nocturne oasis]

Any idea of how to solve this ?

[visual tiger]

Find 4 paths first :

• One that goes from 0 to 1,
• One that goes from 1 to 1+i,
• One that goes from 1+i to i,
• One that goes from i to 0.

[nocturne oasis]

so is it always z = x+iy correct?

[visual tiger]

yes, and you're asked to find a path here

so z(t) = x(t) + iy(t)

[nocturne oasis]

so should I integrate from 0 to 1 or 1 to 0?

[visual tiger]

Have you found the paths yet?

[visual tiger]

those 4 paths

you can name them f1,...,f4 or anything else you'd like

[nocturne oasis]

if Im doing it right

[visual tiger]

yes sure, but are you not asked in question a) to explicit the path as a single one?

[nocturne oasis]

I thought it meant to be like find the four sides of the square aka the 4 traces

because what I have done in the previous exercise was something similar but English wise better explanted

[midnight plank]

@nocturne oasis Has your question been resolved?

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[bitter tartan]

How do I start this

[tribal temple]

The title there is probably a good suggestion

Do you have any identities in mind?

[bitter tartan]

sin+cos=1?

[nocturne oasis]

just convert it to sin x

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

[bitter tartan]

-(1-sin^2 x)?

why -

[nocturne oasis]

you need to review the trig identities if u wanna solve it

[bitter tartan]

why u put negative sign on outside

oh

ok/

[nocturne oasis]

sorry typo

than u multiply by 2 and move it on the other side

u will have a quadratic equation, just solve it

[bitter tartan]

?

[nocturne oasis]

correct

[bitter tartan]

and move that to the left?

like subtract 2 from 1

[nocturne oasis]

yeah so u get sin^2 x +sinx -1 =0

[bitter tartan]

can i do this for the quadratic formula

[nocturne oasis]

you can use U-substitution

where u = sin x

[bitter tartan]

do i do 2u^2

[nocturne oasis]

why 2u

[bitter tartan]

2 sin?

[nocturne oasis]

yeah 2u^2

[bitter tartan]

does sinx = 1/2 and -1?

is this right

actuallu

i did that wrong

u cant sqrt a negative number

wait

nvm

-1-3 -4/4 = -1

im looking at my professors notes and she got postive 1

so idk what i did

[meager path]

YOU WROTE THE FORMULA WRONG

ITS B^2 - 4AC

[bitter tartan]

oops

wait

i wrote it here

isnt b 1

1^2

is 1

oh

the +

ik thats wrong

when i rewrote it

i subtracted

[meager path]

+1 FOR WHAT

[bitter tartan]

for the other sin value

[meager path]

IF WE CHANGE COS^2(X) INTO SIN^2(X)

1 + SINX = 2( 1 - (SINX)^2 )

AND IF WE SUBSTITUTE SINX = 1 INTO THAT WE GET 2=0

WHICH IS WRONG

MAYBE UR TEACHER MADE MISTAKE

[bitter tartan]

Wym 2 = 0

I was asking if my work was correct

I got what i outlined in green

[meager path]

IT IS CORRECT

I WAS JUST CHECKING YOUR TEACHER'S ANSWER

YOUR TEACHER SAID SINX = 1
I SUBSTITUTED THAT INTO THE ORIGINAL EQUATION

AND GOT 2=0 , WHICH MAKES NO SENSE, MEANING SINX = 1 IS WRONG

[midnight plank]

@bitter tartan Has your question been resolved?

[bitter tartan]

Ohh i see yeah i was wondering how she got 1

lemme send u her notes

wait

Was it cuz she factored and i did the quadratic formula ?

does that change anything

[bitter tartan]

<@&286206848099549185>

[meager path]

SHE FACTORED WRONG ON THE LEFT SIDE, SHE FACTORED CORRECT ON THE RIGHT SIDE

CORRECTION:

[bitter tartan]

Damn she fr uploaded the notes like that😭 it was making me confused ty tho

[meager path]

ALSO, FACTORING AND USING QUADRATIC FORMULA ALWAYS RETURNS THE SAM ANSWERS

[bitter tartan]

Alright ty

.close

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[zenith ledge]

...

I remember this one!?

...

we are!?

[midnight plank]

@last slate Has your question been resolved?

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[red marsh]

i don't really know where to start

i mean i know an upper bound can be established by doing x^3/6, right?

[midnight plank]

@red marsh Has your question been resolved?

[red marsh]

<@&286206848099549185>

[midnight plank]

@red marsh Has your question been resolved?

[red marsh]

<@&286206848099549185>

[red marsh]

<@&286206848099549185>

i found the polynomial

but i don't know how to do the rest

[red marsh]

<@&286206848099549185>

[red marsh]

<@&286206848099549185>

[red marsh]

<@&286206848099549185>

[red marsh]

<@&286206848099549185>

[red marsh]

<@&286206848099549185>

[red marsh]

<@&286206848099549185>

[last slate]

wow

thats alot.. idk how to help ): sorry

[quiet flame]

write out the polynomial

p(x) - 1/1-x is the error

[red marsh]

i found p4

[quiet flame]

just fjnd the largest (absolute) value for |x|<0.1

[red marsh]

like the 4th derivative evaluated at x = 0 is 24. Thus the 4th term of the polynomial is x^4, right?

[quiet flame]

but you dont need that?

ur right

but thats not important

[red marsh]

wait why isn't it important? I thought the error must be less than the next term?

[quiet flame]

idk idc

[quiet flame]

.

thats easy enough

[red marsh]

so the error is the same as the original function?

[quiet flame]

no…

the error is defined as the difference between the polynomial and the original function

[red marsh]

oh i see. So im just seeingwhat value of 0 < x < .1 in which the value of p(x) - 1/1-x is the greatest?

[quiet flame]

-.1<x<.1

but yes

[red marsh]

but isn't the error going to be greatest at x= 0.1 since that's the furthest from the center?

[quiet flame]

yeah thats good intuition

prove it

[red marsh]

i don't really understand this. in this problem, e^x - (1 + x + x^2/2) for x = 1/3 is 0.0067 and x=-1/3 is -0.0057. But why is the answer in the problem 0.0123?

<@&286206848099549185>

[midnight plank]

@red marsh Has your question been resolved?

[midnight plank]

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[last slate]

hello

[last slate]

<@&286206848099549185>

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[fresh sparrow]

can you send a better quality pic

[last slate]

nvm

not that

Ethan has a points card for a movie theater. He receives 55 rewards points just for signing up. He earns 12.5 points for each visit to the movie theater. He needs at least 150 points for a free movie ticket. Write and solve an inequality which can be used to determine x,the number of visits Ethan can make to eran his first free movie ticket

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[last slate]

1

[fresh sparrow]

okay

[fresh sparrow]

let x be number of visits

[last slate]

yea

[fresh sparrow]

we have 55 points initially

[last slate]

uhuh

[fresh sparrow]

we get 12.5 per visit

[last slate]

correct

[fresh sparrow]

so it's 12.5*x per x many visits

[last slate]

yea

got it

[fresh sparrow]

so we have 55+12.5x

and the threshold is 150

[last slate]

for the inequality

[fresh sparrow]

$55+12.5x \leq 150$

[grand pond]

artemetra

[fresh sparrow]

there's your inequality

[last slate]

ok

[fresh sparrow]

now, how would you go about solving it

[last slate]

so the number of visits is 8 to get a free movie ticket

is that correct?

[fresh sparrow]

yes

[last slate]

so what about x

is it<,> or what

or the one with a dash under it

[fresh sparrow]

the question says first free ticket

so it's just x=8

[last slate]

sorry if its bad quality

[fresh sparrow]

yes

wait

[fresh sparrow]

x = 8

=

[last slate]

right next to 8 theres something there

what do i put

[fresh sparrow]

not ≤

[fresh sparrow]

^

[last slate]

ok is it the other one?

[fresh sparrow]

x "=" 8

literally

equals

not <, not >

[last slate]

so =

[fresh sparrow]

not ≤, not ≥

[fresh sparrow]

yes

=

i've been saying that lol

[last slate]

it wont
let me put that

[fresh sparrow]

then ≥

[last slate]

ok

[fresh sparrow]

also it's ≥150

[fresh sparrow]

i made a typo, sorry

[last slate]

incorrect

it was 7.6

on the visits

[fresh sparrow]

bruh

but

there can't be 7.6 visits......

[last slate]

now i know at least

it said that idk

kinda dumb

thx for the help tho

[fresh sparrow]

np

[last slate]

have a good one

.close

[fresh sparrow]

you too

[midnight plank]

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[naive lily]

How to calculate P(X>2) when my cumulative distribution function is

[nimble copper]

Well what information does a cumulative distribution function give us?

[naive lily]

the probability accumulated to that point ?

[nimble copper]

Sure

So what would be the meaning of, say, F(2)?

[midnight plank]

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[midnight plank]

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[kind finch]

i understand the idea of the taylor series but i just keep getting it wrong, and i dont understand what the problem is asking

sorry I should probably use this one as there is better work for it

[midnight plank]

@kind finch Has your question been resolved?

[hollow tundra]

You entered 5 terms

Idk if that’s the only error

[midnight plank]

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[pine brook]

Please could I have some help with this differential equation

I tried dividing through by 2ylnx to try and use the integrating factor approach but I have a function of y and x on the RHS, so that doesn't work. I don't think separation of variables will work either as I've tried that

[halcyon trellis]

Show ur work

Hm

Yeah that doesn’t work does it

No wait it does

[pine brook]

I just did scribbles it isn't very readable

[halcyon trellis]

Nvm

Hm

Yeah okay

[pine brook]

I'm confused because all I've been taught is integrating factor and separation of variables so it would have to be one of those methods I would assume

[halcyon trellis]

Integrating factor should work

[atomic magnet]

notice that d/dy(y(x)^2) is 2ydy/dx and d/dx(ln(x)) is 1/x, you can simplify the LHS with this

[pine brook]

The question then is how I can get rid of the 'y's so they don't end up on the RHS at all

But since there are two of increasing power I'm not sure how, I've tried factorising it but it leads nowhere seemingly

[atomic magnet]

Notice that the LHS is a disguised application of the product rule where y is a function of x

[pine brook]

@atomic magnet that's a good point

[atomic magnet]

if you notice this you wont have to deal with the algebraic pain of integrating factor

[pine brook]

Let me try that

[atomic magnet]

good luck!

[pine brook]

I'm struggling to do it, do you know a method to "reverse the product rule" I suppose in this instance?

[atomic magnet]

in a formal sense, that is what integrating factor does - consider what happens if you take the derivative of y^2ln(x) with respect to x, where y is a function of x

[pine brook]

You would get 2yln(x) dy/dx + y^2/x

Ohhhhhh

Hahaaa

I forgot I could differentiate a function of y with respect to x

[atomic magnet]

yes!

[pine brook]

Man I'm rusty 😂😂

How did you go about getting that result

What did you do

[atomic magnet]

haha thats okay uve got it in the end which is important

well its somewhat just understanding the equation as a whole, i noticed how the y terms (2y and y^2) were related and the x terms (ln(x) and 1/x) were related too

and then one realises that its just the product rule

[pine brook]

Ohhh I managed to do it

[atomic magnet]

what were your steps

[atomic magnet]

yep cool

[pine brook]

Is that something I can do everytime

[pine brook]

Wait a minute I have a two in there I'm not supposed to have

Maybe my method doesn't work

[atomic magnet]

oh yeah so sorry didnt notice that, it should be d/dx(y^2ln(x)) = e^(-2x)

[pine brook]

No no worries at all it's my mistake

[atomic magnet]

the u = y^2, du = 2y, v = ln(x), dv = 1/x

[pine brook]

So would you just take two very obvious differentials and say they are the the u and du/dx, so for instance the ln(x) and 1/x in this case

And then you know what the remaining terms have to be

[atomic magnet]

yeah, it isnt a coincidence that they both occur in the equation, its so that the LHS can be simplified

yes

[pine brook]

Perfect

And I take it when the equation meets the form of integrating factor, I just do that

Man thank you so much for your help

I do really appreciate it it all makes perfect sense