help-49

[radiant roost]

typo

in the denominator

[golden comet]

just need to proof ab is rational

the proof is there lol

after proving ab is rational

let a=m/n and b=p/q

a+b=m/n+p/q=(mq+pn)/nq

and that is a whole number divided by whole number

thus rational

[zealous schooner]

Thr problem is you're assuming previous knowledge about how addition and multiplication works in rationals

[twilit field]

I suppose we can assume how they work

[radiant roost]

$mn^{-1}+pq^{-1} = mqq^{-1}n^{-1}+pnn^{-1}q^{-1}=(mq+pn)q^{-1}n^{-1}$

[grand pond]

Axe

[zealous schooner]

Yeah this was what I was thinking of

You will also need a^{-1}b^{-1}=(ab)^{-1} to end it off

[radiant roost]

aha

yes

because we need to show it's the product of an integer and the inverse of an integer

[zealous schooner]

Yep

[twilit field]

Yeah

got it

Thanks

Can I close this now, then?

[radiant roost]

sure

[twilit field]

Thanks

.close

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[last slate]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[last slate]

Why js the lim x->-2 f(x) = 1, not 2?

[slender walrus]

because that's asking about what's happening around that location
and not the function value itself

[last slate]

What does the red dot above supposed to represent tho?

Also look at th graph given on teh side

Yeah i see that

[slender walrus]

the dot indicates the function value

[last slate]

Well isnt the function 2 when x approaches to negative 2?

[last slate]

Don't dots show that the function is undefined tho?

[sharp wave]

f(-2) = 2

[last slate]

My apologies if I am wrong

[golden comet]

yeah the function is 2 when it equals -2

but we are infinitely approaching -2

not AT -2

[last slate]

Ohhhhhh

[sharp wave]

it just says that f(-2) is not -1 and it has a value somewhere else

like the [x] function

[golden comet]

but

we are approaching that 1 value infinitely

againt we are not AT the -2 point

and if we are appraoching x=-2

we will get infinitely close to 1 first

before going to 2 AT -2

[last slate]

Ur better than my teacher

Ty

[golden comet]

np

[last slate]

!close

[slender walrus]

for the limit, you're considering

[golden comet]

(btw never learned calculus lol, only heard of some basics of it XD)

[last slate]

Ah

[golden comet]

pretty sure it is .close

[sharp wave]

put a . not !

[last slate]

Oh

.close

[midnight plank]

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[last slate]

👍

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[drowsy hawk]

Hello! if anybody is familiar with mathematica/wolfram (i have never touched it in my life), i just wanted to know if its feasible to graph something like this on the app? /how hard would it be/how would i do it 😔also would i need just wolfram engine or actually mathematica bc i dont wanna buy it

[sharp coral]

,w contour plot cos(x)sin(y) x from 0 to 25, y from 0 to 15

[grand pond]

Results provided by WolframAlpha

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Requested by cloudedhorizons

[midnight plank]

@drowsy hawk Has your question been resolved?

[drowsy hawk]

oh thanks!

[midnight plank]

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[tidal turret]

For how many values of the digit $A$ is it true that $63$ is divisible by $A$ and $273{,}1A2$ is divisible by $4$?

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[dreamy lichen]

just try all of the digits with A | 63

there is not too many of them

[tidal turret]

which numbers are divisible by 63?

[dreamy lichen]

63 must be divisible by A

63 is divisible by e.g. 21

but 21 isnt a single digit

so you dont even need to consider that

what else is 63 divisible by?

[tidal turret]

by 7 and by 3

[dreamy lichen]

there are 2 more possible digits

[tidal turret]

hmm

divisible by 9

[dreamy lichen]

You could just start from 1 and try every possible digit and see if it divides 63

so e.g. 1 does divide 63

2 doesnt

3 does

4 doesnt

5 doesnt

6 doesnz

7 does

...

[tidal turret]

1,3,7,9 are single digit divisors

of 63

[dreamy lichen]

yep

so now try all of these and see if
2731A2 is divisible by 4

[tidal turret]

Divisibility Rule of 4
If the last two digits of a number are divisible by 4, then that number is a multiple of 4 and is divisible by 4 completely

[slow thorn]

yes

and only 12, 32, 52, 72 and 92 are valid

odd digits

[tidal turret]

i) 273112

ii) 273132

iii)273172

iv) 273192

[slow thorn]

one more

yes

thats it i guess

[dreamy lichen]

And you can easily test that 12, 32, 72 and 92 are all divisible by 4

so that's it

now just look back at what the question asks for and answer it

[tidal turret]

thank u

.solved

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[rain wasp]

what have you tried?

[fierce canyon]

uhh i found ea=ac which =db

bruh

but*

thats basiclaly it

[rain wasp]

you basically solved the damn thing

since EBC is an isosceles triangle (angle CEA = angle CBD = 20deg), you know that EC = CB

you now are able to prove the triangle ECA is congruent to triangle BCD

the rest is obvious

[fierce canyon]

wait how so?

how do we prove congruent

[rain wasp]

wait have you learned congruent triangles?

[fierce canyon]

like sas and stuff?

[rain wasp]

yeah

[fierce canyon]

oh wait is it angle side side

[rain wasp]

side angle side, but yes

order matters

[fierce canyon]

oh oops

so is x=20?

cause theyre the same

[rain wasp]

[fierce canyon]

alr

tysm 🙂

.close

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[median arrow]

Can someone help me with this?

[median arrow]

1st question

I'm stuck here

And 2nd question

I'm here

@anyone 😃

I did 2b

But idk how to do 1) and 2)-a

.close

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[twilit field]

If $x= ( 1 2 3 )$ and $y = (1 2 )$ find $x^2 y$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I have to find (1 2 3) (1 2 3) ( 1 2)

so 2 is mapped to 1 and 1 is mapped to 2

[rough birch]

Make a map of what each number maps to. Do you understand how multiplying permutations works?

[twilit field]

Not entirely, but I'm working on it

[rough birch]

when you multiply permutations, you take each element, and see what it maps to under the rightmost permutation. take this new value. you then work your way left, and see what this now maps to on the second rightmost permutation.

continue all the way until you reach the leftmost permutation

[twilit field]

so I effectively have to find $(1 2) (1 2 3) (1 2 3)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Wait, I work right to left

do I not

[rough birch]

well when we write $(123)(123)(12)$, we first do the rightmost one, and then middle and then left

[grand pond]

Arnavutköy

[rough birch]

so it would still be written as $(123)(123)(12)$

[grand pond]

Arnavutköy

[twilit field]

yes

oh

right

[rough birch]

although a more intutive way of thinking about it is we apply $(12)$ first, and then we apply $(123)$ twice

[grand pond]

Arnavutköy

[twilit field]

so ( 1,2,2,3) is what I get

which is (1,3)

I suppose I could break it down and see what I get

[rough birch]

tell me what $1$, $2$, and $3$ go to

[grand pond]

Arnavutköy

[twilit field]

1 goes to 3

[rough birch]

no. can you work through the steps of what you did to get that so i can see what you are doing wrong?

so under $(12)$, where does $1$ go to

[grand pond]

Arnavutköy

[twilit field]

2

but 2 goes to 3, under the next cycle

and 3 goes to 1 under the next cycle

[rough birch]

yes

so $1$ goes to $1$

[grand pond]

Arnavutköy

[rough birch]

similarly, where does $2$ and $3$ go?

[grand pond]

Arnavutköy

[twilit field]

under the first cycle 2 goes to 1, under the next cycel 2 goes to 3 and under the final cycle it goes to 1

3 goes to 3 under the first cycle, to 1 under the second cycle and to 2 under the final cycle

[rough birch]

well under the first cycle $2$ goes to $1$, but then we should care about where $1$ goes for the second cycle because that was the result of the first cycle

[grand pond]

Arnavutköy

[twilit field]

what

[rough birch]

wait so like when we plugged in $1$ we had $1\to 2 \to 3\to 1$ right

[grand pond]

Arnavutköy

[rough birch]

but now, we plug in $2$

[grand pond]

Arnavutköy

[rough birch]

you did $2\to 1$

[twilit field]

yes

[grand pond]

Arnavutköy

[twilit field]

yes

[rough birch]

but then we need to use $1$ next, as that is what we are at right now.

[grand pond]

Arnavutköy

[rough birch]

so it would be $2\to 1\to 2\to 3$ in fact

[grand pond]

Arnavutköy

[twilit field]

I still don't follow

in the first cycle it's 1

in the second cycle it maps to 2

and it maps to 3 in the final sycle

oh

right

[rough birch]

so we have in the end, $1$ goes to $1$, $2$ goes to $3$, and $3$ goes to $2$

[grand pond]

Arnavutköy

[rough birch]

you will see that this is always a permutations, no $2$ terms will ever go to the same thing under the product of a bunch of cycles

[grand pond]

Arnavutköy

[twilit field]

hmm

so the permutation cycle's result would be (1 3)

[rough birch]

no, $1$ goes to $1$

[grand pond]

Arnavutköy

[rough birch]

not to $3$

[grand pond]

Arnavutköy

[twilit field]

ah right

so (2,3)

[rough birch]

yes

don't put the commas by the way

the lack of commas indicate that it is a permutation

[twilit field]

I keep forgetting

sorry

Thanks

.close

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[twilit field]

Let $S$ be a set with an associative law of composition and with an identity element. Prove that the subset consisting of the invertible elements in $S$ is a group

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

So I have to show 3 things

wait

isn't this a group by definition

The law of composition is associative, it posses an identity elements, and all elements in the set are invertible

[rough birch]

yeah so associativity is inherited

the identity is also inherited obviously

so all which is left to show is that the law of composition is well-defined (i.e. closure in the subset) and that the inverse is in the group

[twilit field]

hmm

That's going to be hard to show

[rough birch]

well first of all what is the definition of invertible we are using here

[twilit field]

$a^{-1} a =1=. aa^{-1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

right

[rough birch]

yeah so then if $a$ is in $S$, so is $a^{-1}$ right?

[grand pond]

Arnavutköy

[twilit field]

why

If it is a group, sure

but we don't know that

[rough birch]

you agree that $(a^{-1})^{-1}=a$ (set-wise right?)

[grand pond]

Arnavutköy

[twilit field]

yes

[rough birch]

therefore both $a$ and $a^{-1}$ are in the set. we now just need to show that in the group $a$ and $a^{-1}$ are actually inverses

[grand pond]

Arnavutköy

[twilit field]

hmm

Not too sure of hwo to do that

[rough birch]

okay, so i will refer to $a^{-1}$ as $c$ because we don't actually know they are inverse in the group yet and its getting confusing

[grand pond]

Arnavutköy

[rough birch]

so we have that $ac=e$ and $ca=e$

[grand pond]

Arnavutköy

[rough birch]

so now, take an arbitrary element $b$ in our new set of invertible elements.

[grand pond]

Arnavutköy

[rough birch]

actually, ignore that

[rough birch]

we have that $e$ is the identity in our new group right?

[grand pond]

Arnavutköy

[rough birch]

and $ac=e$ and $ca=e$ are literally the definition of inverses in the group

[grand pond]

Arnavutköy

[twilit field]

yes

[rough birch]

so this implies that $a$ and $a^{-1}$ are acutally inverses

[grand pond]

Arnavutköy

[rough birch]

in the group

so we now have that inverse exists

all we need to show is closure

[twilit field]

hmm, okay

thanks

[rough birch]

closure is non-trivial

we need to show that the product of two invertible elements is also invertible

[twilit field]

we also need to show that if I take two arbitrary elemets, it's in the set

do I not

[rough birch]

we need to show that the product of two arbitrary elements is also in the group yes

so suppose we have that $aa^{-1}=a^{-1}a=bb^{-1}=b^{-1}b$.

[grand pond]

Arnavutköy

[rough birch]

Show that $ab$ is also invertible

[grand pond]

Arnavutköy

[rough birch]

Hint: Show that $(ab)^{-1}=b^{-1}a^{-1}$.

[grand pond]

Arnavutköy

[twilit field]

I already have

[rough birch]

oh have you showed this? then ur done

[twilit field]

That is as?

How am I done

sorry if I'm being dense

[rough birch]

you have shown all $4$ properties

• Closure and well-definedness of set (If $a,b$ are invertible then so is $ab$
• Identity (Inherited)
• Associativity (Inherited)
• Inverse (The inverse of the element in the set is also the inverse of the element in the group)

[grand pond]

Arnavutköy

[twilit field]

Ah

Got it

Thanks!

.close

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[summer ravine]

Hi

[summer ravine]

Is this correct?
lim x-> infinity x root2 root(1-cos(180/x)) = pi

[slender walrus]

no

[rough birch]

no it is true

that is correc

[worthy wing]

Root2?

[rough birch]

yes there is a root 2

[summer ravine]

yes

[slender walrus]

180 not pi

[rough birch]

i think it was intended that cosine was calculated in terms of degrees instead of radians

[worthy wing]

The problem is not the answer, is the question

Is not correctly written

[rough birch]

assuming that cosine is calculated in terms of degrees this is still right though

[slender walrus]

in the presence of calculus and no degree symbol,
i'm going to assume the original value is in radians regardless of how special it looks

[rough birch]

it probably should just be clarified

[worthy wing]

Must*

[summer ravine]

ok thank you

so, can this be called a formula for calculating pi?

[rough birch]

?

sure, but i mean calculating cosine without using pi is pretty nontrivial

you would have to use taylor series expansions

[midnight plank]

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[twilit field]

Let $G$ be a group. Define an opposite group $G^{\circ}$ , with law of composition $a \ast b$as follows, The underlying set is the same as $G$, but the law of composition $a \ast b =ba$ Prove $G^{\circ}$ is a group
\
Proof: As the underlying set is the same the identity element belongs to the group $G^{\circ}$
\
\
as $a^{-1}a = aa^{-1}$, the group is closed under inverses
\
$a \ast ( b \ast c) = a \ast (cb) = cba$
\
$(a \ast b)\ast c = ba \ast c = cba$
\
this prove its law of composition is associative.
\
It is therefore a group

[grand pond]

ƒ(Why am. I here)=I don't Know

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

[uncut gull]

!status

[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

[twilit field]

Here is my working

4

[twilit field]

4

[runic hamlet]

thats not what closed under inverses means

and just because the identity element of G is still in G^° does not mean that it is also the identity element under the new operation

[twilit field]

hmm, okay

is the last bit fine though

[radiant roost]

the part where you show associativity is correct

[twilit field]

cool, thanks

[runic hamlet]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[twilit field]

I suppose I could start by determining what $a \ast 1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$a \ast 1 = 1a =a$. Thus It has a multiplicative identity

[grand pond]

ƒ(Why am. I here)=I don't Know

[runic hamlet]

thats not the (full) def of multiplicative identity

[twilit field]

I also need to verify that $1 \ast a=a$, which is similarly true

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$1 \ast a =a1 =. a$ Thus it has a right multiplicative identity, which is equal to its letf multiplicativ idenity

[grand pond]

ƒ(Why am. I here)=I don't Know

[radiant roost]

that shows 1 is a left identity in G°, not a right identity

[twilit field]

oops

yeah, but together these show that $1$ is an identity in $G^{\circ}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[radiant roost]

yeah

[twilit field]

I'm not too sure what closure under inverses means tbh

[radiant roost]

i think what you have to show is closure under *,
and the existence of inverses

these are 2 different things right?

[twilit field]

hmm, yeah

exsitance of inverses , I've shown

closure I haven't

[radiant roost]

you have to show for all a, b in the underlying set,
a*b is in the set

[twilit field]

We already know that in $G$ $ab$ and $ba$ are both in the set .
\
$a \ast b = ba$, which is in $G$ and thus in $G^{\circ}$ and $b \ast a = ab$, which is in $G$ and thus in $G^{\circ}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[radiant roost]

👍 it's not necessary to show both a*b and b*a

[runic hamlet]

where

[twilit field]

Oh

nvm

Thought I had

as $a^{-1}a = aa^{-1}$, the group is closed under inverses

here

[grand pond]

ƒ(Why am. I here)=I don't Know

[radiant roost]

you need to show for all a, there's some element b with a*b = b*a = 1

this should be done only after showing 1 is an identity of G°

[twilit field]

ah

I see

so $a \ast a^{-1} = a^{-1}a=1$ should come after I show $1$ is an identity in $G^{\circ}$?

[grand pond]

ƒ(Why am. I here)=I don't Know

[radiant roost]

yeah

[twilit field]

cool

That;s it then?

[radiant roost]

you should also say a^-1 * a = aa^-1 = 1

[twilit field]

yeah

[radiant roost]

i think that's it

[twilit field]

Thanks!

.close

[midnight plank]

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[craggy sequoia]

I need help with geometry, pyramids to be exact, well i got a hypothenosis and one other side of the base triangle and i got the s and i need to calculate volume, how can i get the height of the pyramid?

[craggy sequoia]

a=6cm

that is the other side of the base

and c=10cm

that is the hypothenosis

i already calculated the b and it is 8cm

i also calculated the area of the base and it is 24 square centimeters

i can get the apothem but how do i get the height

but how?

[thorn sorrel]

whats a hypothenosis

whats a b and c

[craggy sequoia]

longest side of a right angle triangle

i'm not native in english sorry

[thorn sorrel]

oh hypothenuse

[craggy sequoia]

the other sides of the triangle

[craggy sequoia]

yes!

[thorn sorrel]

so its a triangular based pyramid?

[craggy sequoia]

yes

[thorn sorrel]

whys there an apothem then

[craggy sequoia]

oh wait i think i translated it wrong

wait i will find a picture

[thorn sorrel]

that would be helpful thanks

[craggy sequoia]

The small h

the big H is height

of the pyramid

[wanton ore]

Trigo, law of cosines

[craggy sequoia]

Oh man i got a test tomorrow and we haven't learned that

[wanton ore]

Law of cosines?

[craggy sequoia]

No

[thorn sorrel]

what informstikn do you have

[thorn sorrel]

i dont think that will help here

[craggy sequoia]

I have all sides of the base

[craggy sequoia]

and the s

[thorn sorrel]

is it an equilateral triangle?

[craggy sequoia]

a what

wait imma google it

[thorn sorrel]

are the sides equal

[craggy sequoia]

no no

if they were i would know how to calculate this

[craggy sequoia]

i just used this as a reference

one side is 6cm, the other is 8cm and the hypothenuse is 10cm

[thorn sorrel]

okay,

what else do uou have

[craggy sequoia]

the s which is 13cm

I tried literally everything i could think of

[thorn sorrel]

is the apex of the pyramid directly over one of the vertices of the triangle?

[craggy sequoia]

it is

all s's are equal

[craggy sequoia]

here is the reference again

[thorn sorrel]

try sketxhing the ptramid

it might help

[craggy sequoia]

i tried but i couldn't get another side to use for pythagorian theorem

the height falls 90 degrees on the base

so if i get some line on the base that has the height on it

i could calculate it

but i don't

If i could get the R i could calculate everything

[craggy sequoia]

I sketched every side but i see no way to calculate this

<@&286206848099549185>

@midnight plank

[midnight plank]

@craggy sequoia Has your question been resolved?

[craggy sequoia]

no!?

[midnight plank]

@craggy sequoia Has your question been resolved?

[midnight plank]

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[vocal briar]

how do I find b and a?

[vocal briar]

I got c = 0 and d = 3 pretty easily

but now i did f''(1) = 0, but thats only getting me 0 = 6a + 2b

[fallow scarab]

Show your whole work

[vocal briar]

ok

i tried doing a system of equations at the bottom there

with the equation i got from the second derivative

but a != -1

nvm i think im stupid

[fallow scarab]

Where does the bottom equation here come from

[vocal briar]

yeah

uh

the origianl function

but the next step is wrong

after that

cuz i wrote 2 = a+ b

should be -4

[vocal briar]

thank you

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[obtuse totem]

i swear

im not trippin right???

u1=1/6 and u2=0 doesnt work

cause -12(1/6)+0 !=2

[vestal jungle]

yeah maybe its supposed to be -1/6

[obtuse totem]

ok

i was so confused i thought i was seeing things

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

@tribal temple hii

[tribal temple]

Hiii (kinda sleepy but wanted to say hi cause I saw you )

[obtuse totem]

lol ok

have a good sleep

[obtuse totem]

oh my goddd

smth is wrong

againn

wtf

my work, i’ll redo😓

ahh eigenvectors are wrong

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

ok imma take a short walking break

but ike damn ive got so much to review still

okok

lockin

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

why is the first vector, <1 2>

why is this wrong?

[obtuse totem]

i truly fcked myself over this time

i didnnt give myself enough time to study

[shadow hare]

Hey there do you need help?

I just had my final so this is fresh for me

[fallow scarab]

The vector (2,1) has slope 1/2

[fallow scarab]

The eigenvector has slope 2 from the graph

[obtuse totem]

yup!

I got it, thank you

also I feel alot better now

when I first looked at the practice final it was OVER

but now Im just doing each question slowly and looking at solutions, reviewing stuff I didnt review before

so its actually alright, for now

[obtuse totem]

i have a question about c)

but I can say y=Vo^4/Vo^2 - R

oh wait its not squaring entire term

its not (2gR)^2

this makes sense, cause denominator cant be 0

but this doesnt

[fallow scarab]

Do you know what g and R represent physically

[obtuse totem]

gravitational constant and radius of earth

oh

[obtuse totem]

oh wait

nono makes sense

why doesn’t this work?

@fallow scarab

wait i found roots wrong

[obtuse totem]

ok imma sleep now

byeee

.close

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[keen seal]

is the potential energy when the ball is at 40m 40x9.8x0.04?

[keen seal]

in some cases they denoted to be 30m instead

for some reason

[rough birch]

change in kinetic energy is same as change in potential energy

[keen seal]

so at the 40m

it should be 40x9.8x0.04 right

[rough birch]

yes that would be the potential energy at the top

[keen seal]

what i did is i calculated 10x9.8x0.04

but

actually yeah it gave 196

so v initial is 14 right

and 40x9.8x0.04

[rough birch]

but we are only calculating change in kinetic energy from A to B

we have no need to calculate the initial kinetic energy or anything

[keen seal]

yeah but A goes from A to the peak to B

[rough birch]

it would just be the change in potential energy from 30 meters to 0 meters

[rough birch]

yeah, so it gains potential energy and then loses it

it doesn't matter what path the ball takes assuming the energy is being converted to heat or friction or anything. only what matters is the beginning and end for potential energy

[keen seal]

okay so 0.4x9.8x30 thats like

118 or smt

x2 and /0.04

588

= v^2

it doesnt rly add up i think

[rough birch]

it should be 0.04 not 0.4

[keen seal]

wait

no like

idk why

im calculating speed

yeah its 12J

ok

[rough birch]

and that gives 11.8

[keen seal]

lol

i need uh

[rough birch]

yeah no need to calculate speed

[keen seal]

wait let me see if i can do it using the method i was using earlier

whihc is individual E_k

[keen seal]

okay yeah so basically what i did is U_k of 10m is like 3.92J therefore v initial would be 14m/s

and U_k at 40m would be 15.68

but if i put 15.68 = 1/2mv^2

the v would be like 28

wait

what am i doing

why am i calcu.ating speed again

[rough birch]

again, this is te whole point of kinetic and potential energy

they didn't want you to calculate speed, that is hard

[keen seal]

no like

im genuinely possessed

LOL

ty

.close

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[keen seal]

what evben possessed me to do all that

officer i did not kill that man i swear

[midnight plank]

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[twilit field]

My book says the klein four group, consisting of the four matices $\begin{bmatrix} \pm 1 & \&\pm 1 \end{bmatrix}$ is the simplest group that's not cyclic

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

what do we mean by simple here

like the rgular meaning

?

[rough birch]

this is not a rigorous definition

i just think they mean it is the easiest example to come up with

[twilit field]

Ah

okay

I thought it was something you could prove lol

thanks

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[keen seal]

how do i do c

[gusty falcon]

😦

[tribal temple]

Did you draw a line of best fit

If so, "the usual way", e.g. pick two points from the line of best fit and use those

[keen seal]

yeah i did

but look at the answer

How can it be 1.32 when

X moves less fast than y

[tribal temple]

Well if y moves faster than x, you’d expect the gradient to be more than 1, which it is, no…

[tribal temple]

Unless you meant to say the other way around? (In which case, )

[keen seal]

Uh

If x moves faster than y, the slope would be positive no?

I mean more than 1

[tribal temple]

It would[n’t], but it doesn’t here

E.g. at x = 1, you’re less than 1, about 0.6, but at x = 2, you’re more than 2, about 2.2

So y increased more than x does at least there

[keen seal]

Yes y increase more than x here

So why is c) 1.32 aka more than 1

That’s what i wanted to know

[tribal temple]

Oh wait I just noticed what you said, no not this

It would be less than 1

[keen seal]

Yeah….

[tribal temple]

(e.g. think about the line y = 0.5x or something)

[keen seal]

.

The answer is 1.32

See why im confused

[tribal temple]

Well how exactly do you mean “x moves faster than y” here?

For a fixed change in x, you have that y will change more than that, and that makes the gradient more than 1

If it were the other way around, for a fixed change in x that you have y changes less than that, you’d have the gradient being between 0 and 1 (assuming positive change of course)

[keen seal]

Well

Does c make sense here? Being 1.32

[keen seal]

In this

[midnight plank]

@keen seal Has your question been resolved?

[keen seal]

Turns out

I always thought for slope it was x2-x1/y2-y1

So that means

I finished calculus

And linear algebra

Without ever realizing

It was the contrary

………

How in the universe

Did that happen 😭😭😭

[midnight plank]

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[gusty falcon]

wait is it even possible for a rhombus to have diagonals 1000 and 0.001?

[gusty falcon]

😭

[thorn sorrel]

why not

[gusty falcon]

what are the restrictions on the diagonal measures?

none?

[thorn sorrel]

there are none

[gusty falcon]

okay fair enough, thanks

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[keen seal]

how do i find the first

why this extra thing

if it said I on top combines the inertia of two disks

[midnight plank]

@keen seal Has your question been resolved?

[prime hornet]

parallel axis theorem

[keen seal]

uh

[keen seal]

omg

i never had this formula what

[prime hornet]

[keen seal]

i was doing some ma=T-mg stuff

trying to derive

that

[keen seal]

but it said that I there was made of two disks of different size)

so doesnt it incorporate both? supposedtly

also

theres not even the mass of body

they didnt give

[midnight plank]

@keen seal Has your question been resolved?

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[eternal pine]

wave 9 guys, can i clear it?

[eternal pine]

it doesn't make sense to me

does this mean

$$log (12) = log (3) \cdot log(2) \cdot log(2)$$

[grand pond]

Jaeger

[eternal pine]

is this the case?

if so it would be 2x * y

for 1)

[torn compass]

$$\log (AB) = ??$$

[grand pond]

JustToPro

[eternal pine]

log(A) + log(B)

but does that help us?

[torn compass]

no , its just to spot ur mistake

[torn compass]

log 12 = log (3 * 2 * 2)
and u said log AB = log A + log B
Then how did u write that equation as log 3 * log2 * log2

[eternal pine]

dude log has so many traps wtf

[eternal pine]

do you know answer to 1?

[torn compass]

ofc

[eternal pine]

yeah then show me what i need to do

[torn compass]

basically

[eternal pine]

i know all the rules and still dunno how to act here

[torn compass]

$$\log 12 = \log (2 \cdot 2 \cdot 3)$$
$$\implies \log 12 = \log 2 + \log 2 + \log 3$$
$$\implies \log 12 = x + x + y$$

[grand pond]

JustToPro

[eternal pine]

bruuh

[torn compass]

simple stuff , no need to break log rules

[eternal pine]

lol what is up with 200 i can't think of a common multiplier where 2, 3 and 7 become 200

[torn compass]

yeah i got no clue myself xD

[eternal pine]

3. is 2z - y

[torn compass]

probably make it 2 * 10 * 10
but idk what they wants from us D:

[eternal pine]

5. y -x

4. is crazy i think it is x - x -y

@elder zephyr wave 9 is crazy

3. log (2 *7) /log 3
   so z+x - y

4 should be in theory
log (2/6)
so: log 2 / log (2 *3)
x - ( x + y)

but it is wrong i guess

yeah no this is complex stuff

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[jovial crest]

hi!

[jovial crest]

if i need to prove that for all even n i can find some k st k(2^n)+1 is composite

can i proof it like this

i assume k(2^n)+1 is prime and then i prove that if it is prime i can make a definite modification to k to make it composite?

like basically i proved that if i find k such that this sum is prime i can find a k for which the sum is composite by adding 2 to the original k for which it is prime

[west iron]

i would just look at the residues mod 3

[jovial crest]

i did residues mod 6

basically all primes ≡ 1 mod 6 or 5 mod 6 right

[west iron]

well aside from 2 and 3 yes

[jovial crest]

what i proved is that if this number is ≡ 1 mod 6 or ≡ 5 mod 6 for some k, i can make both of these ≡ 3mod6 by increasing k by 2

and any number ≡ 3mod6 can never be prime right

in essence the same thing

[west iron]

i feel like this is just false

[jovial crest]

why?

i ve proved it hold on

[west iron]

ok like n=2

k = 1 and 3

[jovial crest]

wait lmao this was a case i took where n>=4 i proved for n=2 separately

[midnight plank]

@jovial crest Has your question been resolved?

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[twilit field]

$\lim_{(x, y) \to (0, 0)} \frac{y^4}{x^4 + 3y^4}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I have to determine if this exists or not

I suspect not

we first approach along the x-axis , in which case the limit doesn't exist

we then approach it along the y-axis, in which case the limit is 1/3

[subtle blaze]

That’s it

[twilit field]

got it

thanks

[radiant roost]

hm?

approaching along the x-axis, isn't the limit 0?

[twilit field]

0/0 at 0

[radiant roost]

yes so it's undefined at (0,0)

but as you approach (0,0) it's 0

0/x^4 is 0 for x!=0

[twilit field]

right

my bad

[west iron]

If it didn't exist then that would automatically be enough to say the whole limit doesn't exist

[twilit field]

[
\lim_{(x,y) \to (1,0)} \ln \left( \frac{1 + y^2}{x^2 + xy} \right)
]

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Here, I was thinking we approach it along x=-1 and the x-axis]

[radiant roost]

you can't approach along x=-1 but you can approach along x=1

[twilit field]

isn't (-1,y) in the domain

nvm

bad question

[radiant roost]

the line must intersect (1,0)

[twilit field]

okay

right

so approaching it along x =1 , we get $\lim_{y\to 0} ln(\frac{1+y^2}{1+y})$$

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

which is 0

now along x =y+1

Okay, I think the limit exists and it;s 0

[atomic cargo]

mx plus b

[twilit field]

y=mx+b?

tried y=x-1, it gives the same limit

Sorry, I've got to sleep now

thanks for the help

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[proud scaffold]

$\theta \in \mathbb{R}, n \in \mathbb{N}$ calculate the remainder of the division of the polynomial $(\cos{\theta} + x \sin{\theta})^n \in \mathbb{R}[x]$ by $x^2 + 1$

[grand pond]

isomorphic to god

[proud scaffold]

is it $\cos{\theta} + x \sin{\theta}$?

[grand pond]

isomorphic to god

[proud scaffold]

i did the calculations in $\mathbb{C}[x]$, dividing by $x-i$ and $x+i$ and since the degree of the remainder is 1 or 0 we can find it in $\mathbb{C}[x]$

[grand pond]

isomorphic to god

[proud scaffold]

<@&286206848099549185>

[midnight plank]

@proud scaffold Has your question been resolved?

[proud scaffold]

<@&286206848099549185>

cmon helpers, easy question

helpers of mathematics when they see a question that isn't about calculus or probability: 😱

[fallow scarab]

maybe don't be a dick

[proud scaffold]

useless comment

answer my question or get off

[fallow scarab]

useful because volunteers don't want to help people who are dicks

ergo, don't be a dick

[proud scaffold]

get off man

go study

[agile cliff]

no it’s cos(nθ)+xsin(nθ)

use the complex trig identities and the remainder theorem

you’re welcome. now act your age

[midnight plank]

@proud scaffold Has your question been resolved?

[spiral osprey]

if it's an easy question then why are you here asking about it

[midnight plank]

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[noble jewel]

Anyone got a video on how to solve these

Ik the limit is x--> 0 but idk what the F(X) exists is supposed to help me with

[mellow sand]

means the limit to 0 from left and right is the same

[noble jewel]

Wym left and right

[mellow sand]

going to 0+ and 0-

[noble jewel]

Oh

U got a vid on how to solve these tho? Going through the entire process

What do I search on yt

[mellow sand]

uhhh

honestly not sure

[noble jewel]

All I know is that these are piecewise calculus limit questions?💀

Dang

Ight ty

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[fringe vector]

For this equation I am starting by making the common denominator across the equation 4x which is getting me 6x^2 - 4x + 16x = 52 - x - 6x^2 but I'm wondering if that isn't what I should be doing for the first step since I find it difficult to complete the equation after that.

[midnight plank]

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[viral dagger]

how do i do ii?

[viral dagger]

for (i) i got 2xln(2x)+x

for (ii) i couldnt think of anything using u sub

[fallow scarab]

,w diff x^2 * log(2x)

[grand pond]

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Requested by rie.mann

[digital crag]

U must integrate by parts

[fallow scarab]

solve for x log(2x) in part i) and then integrate both sides

$\int \frac{dy}{dx} dx = y + C$

[grand pond]

riemann

[viral dagger]

$\frac{\dv{x}(x^2\ln(2x))-x}{2}=x\ln(2x)$

[grand pond]

Skissue ping4response

[viral dagger]

like this lol?

[fallow scarab]

$\int 2x\log(2x) dx = \int \lp \frac{dy}{dx} - x \rp dx$

[lavish venture]

there should be a 2 though as a constant multiple for the left side

[fallow scarab]

oh i left a 2 out yes

[viral dagger]

so it becomes $y-\frac{x^2}{2}+C$

[grand pond]

Skissue ping4response

[viral dagger]

?

[grand pond]

riemann

[lavish venture]

yea then divide everyone by 2 and convert y back to what it was

[fallow scarab]

divide everyone by 2

[viral dagger]

$\frac{x^3\ln(2x)}{4}$

[grand pond]

Skissue ping4response

[viral dagger]

wait

[lavish venture]

why is there an x^3

[viral dagger]

$\frac{2x^2\ln(2x)-x^2}{4}$

fire as in im correct or im so far off its hillarious

[lavish venture]

ln(2x)

[viral dagger]

ugj

[lavish venture]

close though

[grand pond]

Skissue ping4response

[lavish venture]

🤌🏻

[fallow scarab]

,w int xlog(2x)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[viral dagger]

ok nice

tyy

[fallow scarab]

0 points. forgot to factor out x^2

[viral dagger]

.close

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[viral dagger]

bonk

[midnight plank]

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[fluid bronze]

is this solution wrong? i thought the derriv of cosx was -sinx

where did sec come from?

[proud scaffold]

from the derivative rule

division rule

and then the composite function derivative rule

[fluid bronze]

uhh im still a bit confused

divison rule as in quotient rule?

[daring dagger]

yes, basically

you can rewrite $f$ as $f(x)=\cos(x^6)\sec^5(x)$ and then use the chain rule, product rule and the fact that $\frac{\dd }{\dd x}\sec(x)=\sin(x)\sec^2(x)$

[grand pond]

derivada.schwarziana

[midnight plank]

@fluid bronze Has your question been resolved?

[midnight plank]

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[grave creek]

I need help solving A-D please step by step

[floral apex]

what have you tried?

[grave creek]

Oh wait sorry it’s multiple choice so

I haven’t try anything since I didn’t get it

[floral apex]

do you know how to use a unit circle

[grave creek]

Nope

[floral apex]

do you know what it is

[grand pond]

jan Niku

[grave creek]

Yeah I heard of it before

[floral apex]

so you know what it tells you

[grave creek]

I’m not sure how to use it

[floral apex]

it tells you sine, or cosine, of an angle

[grave creek]

Right what do I look for then?

[floral apex]

do you know tan is defined

$\tan \theta = \frac{\sin \theta}{ \cos \theta}$

[grand pond]

jan Niku

[grave creek]

Yeah

[floral apex]

so we are almost in business

you want $\tan \qty( \frac{-7\pi}{4} )$

[grand pond]

jan Niku

[grave creek]

Yeah

[floral apex]

so if we can find $\frac{-7\pi}{4}$ on this unit circle we are basically done

[grand pond]

jan Niku

[grave creek]

Lemme see

Is it the orange?

[floral apex]

thats positive

[grave creek]

Ohhh

[floral apex]

all the angles on the unit circle are nonnegative

so we need to make -7pi/4 into ap ositive angle

[grave creek]

How do we do that?

[floral apex]

https://www.youtube.com/watch?v=6BRtPfofXog

if you are totally new

basically: you add 360 degrees / 2 pi

[grave creek]

Yeah

[grave creek]

Would it be D then?

[floral apex]

i dont think so

wait

[grave creek]

Wait can you show me how you would solve it please? I’m a visual learner so 😭

[floral apex]

can you draw where -7pi/4 is?

or, do you know what you get when you add 2 pi to it?

[grave creek]

Mmmm

No sorry my bad I’m multitasking rn

I think I got it tho I’ll js watch the video

[floral apex]

sure sorry i cant really understand it for you

you have to interact with it, youll get it

[midnight plank]

@grave creek Has your question been resolved?

[midnight plank]

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[barren onyx]

how are the coordinates (2acostheta,0) and (0,2asin theta)

[barren onyx]

what

[livid python]

It’s makes a right angled triangle
With hypotenuse 2a and an angle θ
sin(θ)=opp/hyp=opp/2a
Multiply by 2a
2asin(θ)=opp
cos(θ)=adj/hyp=adj/2a
multiply by 2a
2acos(θ)=adj

[barren onyx]

my teacher said to solve via parametric

do yk how to solve it by parametric?

[livid python]

Not sure about that sorry

[barren onyx]

oh

<@&286206848099549185>

[midnight plank]

@barren onyx Has your question been resolved?

[barren onyx]

.

[barren onyx]

.close

[midnight plank]

Channel closed

Closed by @barren onyx

Use .reopen if this was a mistake.