help-13

[crimson sedge]

2x

[gusty vine]

what do you have against your ex

[crimson sedge]

And then the 3

I meant the now -3

Which is now

6

So then

Y-4 = 2x - 6

Add 4

Y = 2x-2

[gusty vine]

yay

[crimson sedge]

Imma

Kms

[gusty vine]

wait why

[crimson sedge]

Last attempt

LAST ATTEMPT

[gusty vine]

YOU TYPO’D HAHAH

OMG

😭

[crimson sedge]

Its fine if i get rest right i still get A

Good practice i guess

:sku

[gusty vine]

💪

[crimson sedge]

I got 2 more questions, ill just show them then do them

[gusty vine]

okie

[crimson sedge]

Y - 6 = -4/1(x-2)

Distritbute: X becomes -4/1 and the 2 becomes -8

Add -6 on both sides

I meant

6

-2

Y = -4/1 - 2

X

Y = -4/1x - 2

[gusty vine]

hii

sry was away for a bit

[crimson sedge]

Alg

[gusty vine]

be careful

also why is it a 2 inside

the bracket

It passes through (6,-6)

[crimson sedge]

Am i tripping

[gusty vine]

HAHAHA

[crimson sedge]

BRUHHH

Its 6

[gusty vine]

You might be tired

[crimson sedge]

That goes in there

Im exhausted

[gusty vine]

Yeahhh

[crimson sedge]

Let me redo

So

It would be

[gusty vine]

okie

[crimson sedge]

Y - 6 = -4/1 (x - 6)

X becomes -4/1

-4/1 x 6/1 = -20/1

Aka 20

Add 6 on both sides

-14

So

Y = -4/1x - 14

[gusty vine]

okay so first

[gusty vine]

be careful of the negatives

second

what’s 6•4

[crimson sedge]

We

Dont

Talk

About it

24

[gusty vine]

HAHAHAHHAA

shhhh

what’s happens in this channel stays in this channel

[crimson sedge]

So then that would be -18 when subtracted

So then

Y = -4/1x - 18

[gusty vine]

well

[gusty vine]

remember it’s -4•-6

also

[crimson sedge]

So its positive

[gusty vine]

your first step was already careless

rmb the negatives

can you spot the mistake

[gusty vine]

here

it passes through (6,-6)

(6,-6)

[crimson sedge]

But aren’t the equations y-y1 = _ (x - x1)

[gusty vine]

y-(-6)= -4(x-6)

[crimson sedge]

B
R

U

H

[gusty vine]

HAHAHAHA

[crimson sedge]

i never simplified

[gusty vine]

BEISAONDHDS

[crimson sedge]

So it is y = -4x - 18

[gusty vine]

well

remember it’s

-4•-6

[crimson sedge]

+18

+19

+18

[gusty vine]

yay

hahahahha

[crimson sedge]

Im

Gonna

Kms

Again

[gusty vine]

o-o

what happened

another typo?

[gusty vine]

OMG

HAHAHAHA

STOP

I AM LAUGHINF TOO KUCH

I CANT

HAHAHAHA

pls go rest

[crimson sedge]

no this is due tonight

Onto the next one

[gusty vine]

😭

[crimson sedge]

Y - 2 = -8 (x - 3)

Y - 2 = -8 - 24

[gusty vine]

slowly

[crimson sedge]

Wait shouldn’t the 24

[gusty vine]

check the first step

slowly

remember is (3,-2)

[crimson sedge]

So do i do +3

So (x+3)

[gusty vine]

what number is bolded in this message

[crimson sedge]

-2

[gusty vine]

look at your 2

missing anything?

[crimson sedge]

Uh

Isn’t it negative

[gusty vine]

https://tenor.com/view/you-can-do-it-gif-25785004

[gusty vine]

yesh

[crimson sedge]

Y - (-2)?

[gusty vine]

soooo…

[gusty vine]

YAS

[crimson sedge]

I knew it was negative tho

Would that 24 be positive?

Y - 2 = -8 - 24

[gusty vine]

but y-2 is not the same as y-(-2)

[crimson sedge]

Breh

[gusty vine]

y-(-2) becomes y+2

[crimson sedge]

Why

[gusty vine]

double negatives

[crimson sedge]

Ohhh

[gusty vine]

negative • negative = positive

[crimson sedge]

So shouldn’t that 24 be positive

So -8 + 24

Y + 2 = -8 + 24

[gusty vine]

yeah

forgot the x btw

[crimson sedge]

-2 -2

Y = -8x + 22

[gusty vine]

yay

[crimson sedge]

I usually just add the X at the end

Bad habits

[gusty vine]

HAHAHA

yesh bad habit

[crimson sedge]

O

I

I’d

Did

It

Again

..

[gusty vine]

BYE

[crimson sedge]

But it let me

Submit it

Again

So i got it

[gusty vine]

okay phew

[crimson sedge]

Last one

I did y = before it

Y - (-8) = 3/5 (x-(-3)

So this becomes

Y + 8 = 3/5 (x+3)

Then you have 3 * x

Which becomes

3/5

Then 3/5 x 3/1

Then i think you do common denominator

Or no

Imma go with no

So then

9/5

Y + 8 = 3/5 + 9/5

No

Its definitly

Common denominator

Yep definitly

15/5

3 x 15 = 45/5

9

Y + 8 = 3/5 - 9

• 9 *

-8 -8

Y = 3/5 + 1

[gusty vine]

okie so

you were doing perfectly up until you said common denominator

[gusty vine]

This was right

[crimson sedge]

How tf do i subtract 8 on both sides then

Am i over thinking it

[gusty vine]

Y= 3/5 x + 9/5 -8

NOW you can do common denominator

or u can just use a calculator

[crimson sedge]

Me big confusion

[gusty vine]

Are you allowed a calculator?

[crimson sedge]

Yea

But how did the 3/5

Just like

[gusty vine]

Oh then just use

[crimson sedge]

Teleport

To the other side

[gusty vine]

It was always on the right side

[crimson sedge]

You put it on the left

[gusty vine]

It’s on the right

[crimson sedge]

Y + 8 = 3/5 + 9/5

[gusty vine]

Yeah

I only moved the 8 to the left

Right*

[crimson sedge]

Oh

So then

-40/5

Is uh

-8

[gusty vine]

You can just use your calculator HAHA

[crimson sedge]

Ong wait

Omg

I never

Multiplied

The 9

X

8

40

[gusty vine]

If you don’t want to use your calculator then you can just do this

[gusty vine]

Multiplied to 9?

[crimson sedge]

9/5 x 8/1

Common demonitor

-8 i meant

So -40/5

9 x - 40

[gusty vine]

Yeah multiply 8 by the 5

[gusty vine]

just 9-40

[crimson sedge]

Oh

How hot

Tho

[gusty vine]

^

[crimson sedge]

💀

[gusty vine]

$\frac95 -8$=$\frac95 -\frac{40}{5}$ = $\frac{9-40}{5}$

[wraith dagger]

swaggofishballs

[crimson sedge]

What in Tarantino

Tarnation

[gusty vine]

just use calculator

what’s 9/5 -8

[crimson sedge]

We got it wrong

[gusty vine]

huh how

was it a calculation mistake

[gusty vine]

IT SAID PARALLEL

[crimson sedge]

WHAT

[gusty vine]

THE QN SAID PARALLEL

omg

[crimson sedge]

here we go again’

Y + 9 = 2/3 (x+2)

2/3 + 4/3

Wait

Not

Its already wrong

Parallel

Y + 9 = -3/2 (x+2)

-3/2 + -6/4

Y + 9 = -3/2 + -6/4

Move the 0

9

Common denominator

-6/4 - 36/4

[gusty vine]

why is -3/2(2) = -6/4

[crimson sedge]

?

Oh

Bc

2/1

-3/2 * 2/1

[gusty vine]

yeah

and what does that equal

[crimson sedge]

:sku

-6/2

-3

[gusty vine]

yepp

[crimson sedge]

So

Y + 9 = -3/2 + 3

Then i subtract 9 on both sides

Y = -3/2x - 6

Is that right

[gusty vine]

what did you type here

[gusty vine]

and what did you type here

[crimson sedge]

O

Its + 6

Y = -3/2x + 6

[gusty vine]

it’s y+9 = -3/2 -3

[crimson sedge]

Oh

So y = -3/2 - 12

@gusty vine

Got it

Thank you

!close

.close

[cedar kiln]

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[sturdy ledge]

can someone help? with solutions

[golden raven]

are you currently taking the quiz right now?

[sturdy ledge]

nope

[gray blade]

,rotate

[golden raven]

@gray blade can you help

[gray blade]

yea

so

basically divide each term in the first part

by 9p^2r^2

[sturdy ledge]

yup

i can do first part

im struggling on 2 and 3

[gray blade]

ah

have you learned polynomial long division?

[sturdy ledge]

yup, but i dont get it that much

[gray blade]

hm okay have you tried starting this problem?

and what part of it are you stuck

[sturdy ledge]

not yet, but my answers are always wrong lol

[gray blade]

hey we can work that out, could you send your work?

[sturdy ledge]

wait let me try to solve it

[cedar kiln]

@sturdy ledge Has your question been resolved?

[cedar kiln]

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[sudden grail]

Why is b false?

[golden raven]

um because one requirement for 'spanning' a vector space is that the vectors aren't redundant

however, S has another vector v_p

and this vector could add extra linear combinations possible in S which are outside V

[sudden grail]

oh, so the vectors should be independent to each other if they span?

[golden raven]

yeah

plus they can only span V they cannot span anything "bigger" than V

[sudden grail]

I see, thanks!

.close

[cedar kiln]

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[cedar kiln]

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[still spoke]

<@&286206848099549185>

[dire geode]

That's what they're proving. Which sentence in the proof exactly do you get lost first?

[cedar kiln]

@still spoke Has your question been resolved?

[still spoke]

So, they say using the implicit function theorem it gaurantees the unique solvability will hold for (x,y) near (a, b) (which im also assuming the neighbourhoods are M, N for a, b). And I know that the IFT when applied to F(x,y) for 'x' means there is some function of 'y' such that F(f(y), y)=0 where f: N->M but how does this mean it has an inverse

but im worried im understanding how they are using the implicity function wrong

[dire geode]

would help if you shared the statement of the implicit function theorem

[still spoke]

right

let me do that

[dire geode]

Have you tried writing the conclusions of the implicit function theorem with f(x)=y like it says in theorem 3.18?

[still spoke]

Yes

so

[dire geode]

I think x and y are supposed to be reversed here like it says in the proof

[still spoke]

at which part

the beginning?

I ended up switching them at the end

oh you know what

i think we need the more general version

but im not sure

.close

[cedar kiln]

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[dim mortar]

it is a function if there is only one output for every input

example

(1,2), (2,3)
this relation is a function because every "input" (the first number) only has 1 "output"(the second number)

but in the case of

(1,2), (1,3)
this is not a function because the number 1 has 2 different "outputs"

you talking about this?
(1,3), (1,2)

if this then yes(but preferably put it in order)

makes it easier to read

check if there are any inputs in this case x have multiple outputs in this case y

if the same x value doesnt occur more than once, its a function

dont care about y

just care about the x for now

yes

not a function

thats it

no problem

[cedar kiln]

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[crimson sedge]

If "one and a half boy" can eat one and a half hot dog in one and a half minute, how many hot dogs can 6 boys eat in 6 minutes?

[crimson sedge]

thats a riddle in a game I'm on

so?

lol

you gonna help me or not?

[solid juniper]

lmaooo

[crimson sedge]

lol, no I have to give a digit as input

Numeric digits only. E.g: '10' if you want to say 10 hot dogs.

those are the rules the game gives

I'll solve it myself the(i don't have something to write on)

[sacred grail]

,w 1/1.5

[crimson sedge]

1.5 boy eats 1.5 hotdog in 1.5 minutes

[dire geode]

How dare you stress wolfie with such hard queries

[crimson sedge]

,w 1.5/6

no

[crimson sedge]

I'm stupid

its 6 boys

bc its 1 hotdog per minute

answer is 6

[solid juniper]

i'm trying to imagine 1.5 boys getting together to eat 1.5 hotdog in 1.5 minutes

[crimson sedge]

I needed to see it in numerals thats why I struggled

[crimson sedge]

well math problem kids are weird

I've accepted that by now

Ok, 6 is wrong

[solid juniper]

in 1.5 minutes, 6 boys would eat 6 hotdogs, right?

[crimson sedge]

ohhhh

[solid juniper]

so def not 6

[crimson sedge]

ok

24?

[solid juniper]

sounds right to me

[crimson sedge]

answer was 24

thnx

[solid juniper]

np ^-^

[sacred grail]

,w 1.5/(1.5 * 1.5) = 24/(6 * 6)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by 夢雪#2250

[crimson sedge]

.close

[cedar kiln]

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[spice tundra]

need help

my thought process is that

Well first of all

the total possibilities

should be

12C3

220

the first ball can be any colour, so there are 12 options

yeah and this is where i'm lost

[solid juniper]

might have to be a little more careful there :c

[spice tundra]

The total possibilities is 12C3 right?

My first thoughts are that it would be

$\frac{3 * 4 * 5}{12C3}$

[wraith dagger]

splooze

[spice tundra]

but it's just a guess

oh wtf that's the answer

how

[solid juniper]

hmm

you want the combinations of 3 that have distinct colors

[spice tundra]

I guess it kind of makes sense

because

if we have the 3 spots, i'll just represent them as O's

O O O

The first spot, if it's white has 3 options

Second spot would have to have 4 if it's red

and the third sport would have to have 5 if it's black

3 * 4 * 5

[solid juniper]

yea pretty much

[spice tundra]

divide by total possibilities

lol

what about b

https://media.discordapp.net/attachments/903430334681608232/1047390183617740820/image.png

[solid juniper]

you can go through the 3 cases

[spice tundra]

so for white

[solid juniper]

only 1 possibility if they are all white

[spice tundra]

3 * 2 * 1

for red

4 * 3 * 2

for black

5 * 4 * 3

add them all

[solid juniper]

hold up what are those numbers?

[spice tundra]

so if white has 3 balls

then my thoughts are

if the first pick is white

then the second pick would have two options

because white only has 3 balls

then the last pick would be 1

if there is 4 red balls

first pick would be 4

then you take the ball out

you have 3 left

let me see if it's right lmao

[solid juniper]

ok you could start with that but i have a feeling you're going to get the wrong answer here

[spice tundra]

oh

it's just

$\frac{3C3 + 4C3 + 5C3}{12C3}$

[wraith dagger]

splooze

[solid juniper]

i think your other idea works if you change 12C3 to 12C3 * 3!

but yea that's more what i was thinking

[spice tundra]

every question feels so different

whole new level of lost

[solid juniper]

lol

[spice tundra]

i like to start with total possibilities

12C7

oh ok

so i would have to account for

4 teachers 3 students

5 teachers 2 students

6 teachers 1 student

7 teachers 0 students

i think

[solid juniper]

sure

[spice tundra]

something along the lines of

7C4 * 5C3 + 7C5 * 5C2 ... + 7C7 + 5C0

??

all divided by total possibilities

lets see if it's right

[spice tundra]

oops, should of been a multiplication sign at the end instead of plus

oh my god

it's right

wowz

that feels incredible

okay, think i'm good for now

ty @solid juniper

.close

[cedar kiln]

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[solid juniper]

lol np

[cedar kiln]

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[spice tundra]

@solid juniper help

denominator should be

[solid juniper]

you can think of it as 2 different problems to start

[spice tundra]

6C3 * 7C4

total possibilities

[spice tundra]

how do you mean

[solid juniper]

find the probability a particular judge gets chosen, then find the probability a particular lawyer gets chosen

aren't they independent?

[crimson sedge]

Yep

[spice tundra]

6C1 and 7C1 ?

[solid juniper]

no?

[spice tundra]

i don't know how to get the prob of a particular judge

or lawyer

[solid juniper]

there are 6 judges (let's say judge 1, judge 2, etc.) and 3 of them are being selected

[spice tundra]

ye

[solid juniper]

what's the probability judge 1 gets selected?

[spice tundra]

1/2

[spice tundra]

[solid juniper]

soooo?

[spice tundra]

3/6 * 4/7?

[solid juniper]

yea i think so

[spice tundra]

the answers are sayin

$\frac{5C2 * 6C3}{6C3 * 7C4}$

[wraith dagger]

splooze

[spice tundra]

which i guess makes sense because you're excluding one from each

jesus

pain

[solid juniper]

that's the same as 3/6 * 4/7

[spice tundra]

oh

i did that all over 6C3 * 7C4

lmao

[solid juniper]

honestly sounds like they overcomplicated it lol

[spice tundra]

why why why

okay

total options

52C5

we can have 3 aces

or 4 aces

4C3 * 49C2?

does that make sense for the first part?

then 4C4 * 48C1 ?

for the second part

[solid juniper]

why 49?

[spice tundra]

because we selected 3 cards

which leaves 49 left

that's accounting for the 3 aces

[solid juniper]

aren't you counting the number of hands with exactly 3 aces?

[spice tundra]

at least three aces it says

[solid juniper]

yea but

you're counting that in that part?

[spice tundra]

so ideally the hand can either be

3 ACES 2 CARDS

4 ACES 1 CARD

for the aces it should be

4(total aces)C3

now that picked 3 cards

we're left with 2

49C2

[solid juniper]

no

[spice tundra]

hm

[solid juniper]

what are you counting here?

[spice tundra]

Well if there's at least 3 aces

It means we can have a hand with

3 ACES and 2 CARDS

and to calculate this i did

[solid juniper]

wait

give me the cases first

[spice tundra]

4C3 * 49C2

okay

[solid juniper]

no calculations yet

[spice tundra]

there should just be two cases

3 ACES 2 Cards

4 Aces 1 card

[solid juniper]

what are cards?

[spice tundra]

Well anything but an ace

[solid juniper]

and are there 49 cards left in the deck that are anything but an ace when you take out 3 aces?

[spice tundra]

oh

no

there could be an ace there

shit

i don't know how to account for that

[solid juniper]

ez

there 48 cards left in the deck that are anything but an ace when you take out 3 aces

[spice tundra]

oh

wtf i'm an idiot

so answer should be

$\frac{4C3 * 48C2 + 4C4 * 48C1}{52C5}$

[wraith dagger]

splooze

[solid juniper]

yep 🙂

[spice tundra]

and it is

lovely

ty

that should be all for now

.close

[cedar kiln]

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[fast burrow]

(all of them equal 4)

[fast burrow]

Solve for the real solutions of xy+yz+xy

[cedar kiln]

@fast burrow Has your question been resolved?

[fast burrow]

<@&286206848099549185>

[night vale]

is this calc?

[fast burrow]

Calc as in calculus?

No I don't think so

[night vale]

yea

also wdym all of them equal 4

like x y and z equal 4

[fast burrow]

all of the statements are equal to 4

and equal to each other

[night vale]

i have no idea how to solve this, sorry

[fast burrow]

😭 Aaaa im fucked

I have an exam tommorow

[fast burrow]

<@&286206848099549185> You guys...its been 45 mins

What I tried was contract the chonky third expression

That landed on xy(x+y) + yz(y+z) + zx(x+2) = 12

so how do i solve for xy+yz+zx from there

[cedar kiln]

@fast burrow Has your question been resolved?

[pine brook]

Try to solve

With Diophantine equation

Maybe it could help you, im not sure though

[cedar kiln]

@fast burrow Has your question been resolved?

[fast burrow]

How do I do that

[crimson sedge]

so what you gotta do here is

expression the system in terms of the (x+y+z) powers

[fast burrow]

i solved itttt

x^3 + y^3 + z^3 - 3xyz = -8
(x + y + z) * (x^2 + y^2 + z^2 - xy - xz - yz) = -8
(x + y + z) * ((x + y + z)^2 - 3(xy + xz + yz)) = -8
(x + y + z)^3 - 3(x + y + z)(xy + xz + yz) = -8
(x + y + z)^3 - 3(12 + 12) = -8
x + y + z = 4

if my working is correct

and from there it should be pre self explanatory

[cedar kiln]

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[gilded blade]

yall is thus correct

[tropic oxide]

,rcw

[tropic oxide]

correct but incomplete

[gilded blade]

wdym?

[livid hound]

was your end goal to find 2m?

if yes, then you're done,
if not then you have more work to do

[gilded blade]

to find the equation

[tropic oxide]

then you definitely stopped about two steps short

[gilded blade]

in order to find the equation i need to know the 2n

2m

[tropic oxide]

you need to know m itself, presumably.

[gilded blade]

is m 1

wait

the equation said mx

[tropic oxide]

yes, m is 1.

the issue is that your work stopped short of that.

[gilded blade]

so the final equation is 5 = 4 - 2 + 3 ?

ohhh i underst

understau

understau

understand

.close

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[ebon magnet]

I'm correct or not?

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[short token]

in a ring where multiplication is commutative is it true that if a is element of that ring and a is divisor of 0 then a has no inverse element?

[tropic oxide]

yes. if a^-1 existed then 1 would be a zero-divisor.

[short token]

oh

righ ty

.close

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[rough hamlet]

Does this séquence converge diverge or dont existe ?

[silk thistle]

Well, write it down and calculate the first terms

You'll see your answer

@rough hamlet if you don't know how to do it, ask

[broken mist]

Or even check out a graph of cosine, see if that gives you some intuition as to its end behaviour

[rough hamlet]

I DON’T KNOW HOW TO DO IT

[gaunt hamlet]

What's the very first term?

[rough hamlet]

There no term its just that we have to do limite to infini to know if it converge or not

But the limite of infini of this focntion dont existe

[gaunt hamlet]

It's a sequence

A sequence is a series of terms

The first term is when n = 0

The second term is n = 1

The third is n = 2

Etc

[rough hamlet]

Then it converge to 1

[gaunt hamlet]

Yes

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[charred remnant]

Simplify.

[crimson sedge]

Use the ideas of $(a^m)^n = a^{m \cdot n}, (\frac{a}{b})^{-m} = (\frac{b}{a})^m, \frac{a^m}{a^n} = a^{m-n}$

[wraith dagger]

♡LexQa♡

[crimson sedge]

I guess $(\frac{a^m}{b^c})^n = (\frac{a^{m\cdot n}}{b^{c\cdot n}})$ as an extension of the first

[wraith dagger]

♡LexQa♡

[charred remnant]

So to start, would you multiply each power contained in the brackets by -3/5?

[crimson sedge]

I would apply the second rule here for ease of calculations first

[charred remnant]

With this rule, if a and b both already have exponents, do they become opposite when switching? ie: b's exponent becomes negative when b is flipped to the top, and vice versa?

[gusty vine]

yayaya

[charred remnant]

I’m doing something wrong but am not sure what

I think this is closer to how it should be but there’s definitely still funky things afoot

[cedar kiln]

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[plush surge]

hi

[plush surge]

😨

um

i need help with this graph??

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[storm citrus]

I need help evaluating this integral:

[storm citrus]

Did I do my substitution right? If so, I'm confused on what to do next.

[crystal raptor]

you should probably set u = tan^(-1)(9x)

[crimson sedge]

^

[storm citrus]

Let me give it a shot

[cosmic steppe]

u = arctan(9x)

This question's rigged to work out nicely

[dire geode]

u=1/cot(9x)

[cosmic steppe]

Like awfully rigged

[storm citrus]

I believe this is still wrong

[crystal raptor]

,w diff 9/2 * arctan(9x)^2

[wraith dagger]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Nov 30, 2022, 8:47 PM

[crystal raptor]

,w diff 9/2*(arctan(9x))^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iCaird#6258

[storm citrus]

Yeah so I did do It wrong, I am trying to locate my mistake

[crystal raptor]

when replacing dx with du

you don't have that factor of 9 in your integrand

so you would need to times by 9/9

[storm citrus]

I was able to complete it, thank you @crystal raptor

.close

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[crimson sedge]

use pythagoras

[nova barn]

Is there even much to simplify for this expression?

[crystal raptor]

yes

[nova barn]

Would it just be 1-sinx?

[crystal raptor]

no

[cedar kiln]

@nova barn Has your question been resolved?

[nova barn]

-sinx?

[crimson sedge]

yes

[cedar kiln]

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[brazen bolt]

taking pic of work 1s

[violet night]

u sub

[brazen bolt]

yea

i almost got it

just a small ting

1s

taking picture

[violet night]

ok

[brazen bolt]

first part

i appearntly almost got it according to symbolab

that is symbolbs answer

@violet night coulda ya help me out real quick

ya there?

[violet night]

its at the start I believe

second line

at the end

[brazen bolt]

hmm

[violet night]

wait nvm

lemme keep looking

👀

[brazen bolt]

😄

[violet night]

I dont think that 1/2 is supposed to be there, not sitting write with me ;-;

[brazen bolt]

well i just took it out

it looks to be correct as it ultimatley becomes 1/3

[violet night]

it supposed to be a 1/4

[brazen bolt]

which is what symbol says

[violet night]

because when you sub for dx

you forgot to put denom 2

as well

[brazen bolt]

where

[violet night]

dx = du/2

and when you sub dx

you just put du

and then take the 1/2 from the u sub out

[dire geode]

you missed a 2 here

[violet night]

ye

[brazen bolt]

ah ok

[violet night]

also

why do you write the C before you have final answer

nvm

[brazen bolt]

@dire geode can u find it?

[dire geode]

i told you?

[dire geode]

.

[brazen bolt]

yea but

[dire geode]

and clarkie told you?

[brazen bolt]

the answer is still diffent

[dire geode]

fix it and reupload

[brazen bolt]

ok

[dire geode]

you fixed it incorrectly

solve for dx before substituting

[brazen bolt]

dx = du/2

[dire geode]

.

[brazen bolt]

ok i did that

what will that change tho

[dire geode]

the entire answer?

did you do it yet?

it's just a constant

[brazen bolt]

yes

[dire geode]

shouldn't be too hard to fix

[dire geode]

what's your final answer

[brazen bolt]

this is what symbolab says

tho

[dire geode]

that's not the new answer

you didn't finish with your fix

[dire geode]

redo the integral with this fix

[brazen bolt]

oh

i got it

d/2 * 2x +2 = x +1

rite?

[cedar kiln]

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[cedar kiln]

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[devout ledge]

how do i approach this question?

[modern compass]

find the power series of arctan(x)

[devout ledge]

do i perhaps split the 1/x and the arctan(x) then differentiate just the arctan?

[devout ledge]

ah so youre saying integrate again?

[modern compass]

when did you integrate the first time?

[devout ledge]

theres already an integral on the expression

[modern compass]

write arctan as a power seriers
divide by x
integrate

[devout ledge]

right but to turn arctan into a power series you have to differentiate it, no?

[modern compass]

it has an infinite power series. Check your notes or your textbook, it probably has a table of common power series

[devout ledge]

my book doesnt have that. this is how they do it

so im thinking i split the 1/x and the arctan(x) and just do this to the arctan and deal with the 1/x after that

[dire geode]

divide this series by x

to get

[devout ledge]

ok sure, but how does the integral bar come into play

[modern compass]

that says $\arctan(x) = C +x-\frac{x^3}{3} +\frac{x^5}{5}+\cdots$\
which is equal to $C+\sum_{n=0}^\infty (-1)^{n}\frac{x^{2n+1}}{2n+1}$

so you have your power series for arctan(x), no need to integrate anything at this point

divide that by x, then integrate to get the power series you're looking for

[wraith dagger]

Zybikron

[devout ledge]

ok, so then i would have that 2nd expression there with the numberator being x^2n to account for the x being divided, then i integrate that?

[modern compass]

yes

[cedar kiln]

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[clever comet]

not sure how to do this

[cedar kiln]

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[clever comet]

.close

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[spice totem]

So I need to Determine magnitude, direction and sense of the moment of the force relative to point O but I don't have a clue how to begin, as I was sick for most of the lessons and none of my classmates want to explain it to me.

[spice totem]

I have the formula MoF = F x l

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[spice totem]

.close need to go to bed e

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[muted bear]

is $\int (dx+dx+dx)$ a legal way to write $\int 3dx$?

[wraith dagger]

GarlicBredFries

[long swan]

Yes

Linearity of the integral operator:

$$= \int dx + \int dx + \int dx$$

[wraith dagger]

Mr. Gamer

[muted bear]

fair

thanks

[long swan]

You can also write $3 \int dx$

[wraith dagger]

Mr. Gamer

[long swan]

This is again just the linearity of the integral operator

[muted bear]

yeah i got that

im just not used to seeing more than one dx per integral sign

[long swan]

Really you won't

[muted bear]

thanks for clarifying

[long swan]

But there's nothing particularly incorrect about writing it that way if you are so inclined

[muted bear]

alright

thanks again

[cedar kiln]

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[muted bear]

oops

.close

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[gleaming schooner]

Do we solve it by partial fractions?

[livid hound]

yes

[gleaming schooner]

how do we recognize

when to use partial fractions and when to use trig sub

[upper abyss]

Mostly the √

[gleaming schooner]

for trig?

[upper abyss]

Trig sub takes advantage of the √ to simplify the integral

[gleaming schooner]

ohl

ohk

thanks

[upper abyss]

Trig sub wouldn't help here

[gleaming schooner]

ye

.close

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[fading scroll]

how is this even possible? how can i get angle cab = cba when i can only seem to prove mea and mfb. (need to use hypotenuse leg theorem)

[modern compass]

since M is the midpoint, you AM = MB

[fading scroll]

how is that useful

i know that

i can prove triangle MEA and triangle MFA but how does that prove angle CAB and angke CBA

[modern compass]

so you have that ME = MF and AM = BM.
so the hypotenuse and leg of two right triangles are congruent

use hypotenuse-leg theorem

[fading scroll]

?

how do i prove the angles

[modern compass]

what does hypotenuse-leg theorem say

[fading scroll]

If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, the two right triangles are congruent.

[modern compass]

so what do you know about triangles AEM, and BFM?

[fading scroll]

they are congruent but i want to prove angle CAB and angle CBA

[modern compass]

so angle EAM and FBM are congruent to each other by congruent parts of congruent triangles.

[fading scroll]

can it really be corresponding parts when line CE isnt in like EAM?

[modern compass]

the small triangles are congruent. so their parts are congruent

[fading scroll]

the only parts are line AE. how will CE come into this

[modern compass]

the length of the lines don't impact how big the angle between them is

AE and AC are colinear, AM and AB are colinear
angle CAB is the same as angle EAM

[fading scroll]

but it triangle AEM

am i actually allow to just say triangle AEM and triangle MEB are congruent so angle CAB and angle CBA are to

is that all

i don’t need to say like stuff are colinear right

[modern compass]

i don't see why you would

i only said it because you kept ignoring that it's the same angle.

[fading scroll]

wait how can like CAB cover both triangles base

.close

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[upbeat pond]

how do i solve this

[upbeat pond]

I converted tan to sin over cos

idk what

next

oh wait nvm i know

.close

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[restive oriole]

how do I find out the logistic function from the initial value, limit to growth, and the inflection point the graph passes through?

[restive oriole]

<@&286206848099549185>

[dire geode]

Too vague to answer

Do you know the formula for the general form of logistic equation?

[restive oriole]

If I have the point (0,10), limit to growth of 40, and the inflection point at (1,20) how do I find out the equation for the graph

so far I got 40/(1+9^[unkown])

[dire geode]

https://socratic.org/questions/how-do-you-find-the-inflection-point-of-a-logistic-function

They derive that formula by finding where the second derivative is zero
https://en.m.wikipedia.org/wiki/Logistic_function

[restive oriole]

thanks

.close

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[crimson sedge]

Why is it that a complex number / its conjugate = 1 ?