help-13

[fallen moat]

although I've told you the answer lol

[mental crag]

I’m sorry I’m dumb lolll

[fallen moat]

like the first one, K237
is just =-E237+H237

this is because we wanna just add the original $ up
so you can also just write
=H215

the hard part is the remaining ones

as i have said
K238 should be
=K237-E238+H238
this is because, we want
the latest value (K237),
minus the money used to buy (E238),
plus the money gained from selling (H238)

[mental crag]

Could I do it rn and show you the screenshot

[fallen moat]

yea

[mental crag]

Lmk if I did it right

Ok wait hehe

I’m confused

How do you know which to subtract and what to add

[fallen moat]

think about where would the money go

instead of the word sale and purchase

[mental crag]

For purchase, it would deduct your $

For sale, you would gain $

[fallen moat]

yep

[mental crag]

Okay so question

For the first sale which is below the beginning inv., I got total $ as $3900 on the far right side

[fallen moat]

ohhh, you found the problem 👍

[mental crag]

And for the first purchase just below the first sale, I got $1425 for total $ on the far right side

[fallen moat]

please continue, what will be your question?

[mental crag]

I’ll continue hehe

Also, will the units be the same formula?

[fallen moat]

nope

the units are actually harder

for 100 units, we have $1500

and we sold 80units

we have $3900

that means the $3900 is worth 180 units

instead of 20

see if you get this first

[mental crag]

This is what I got so far

[mental crag]

For the Total $ on the far right side

[fallen moat]

hmmm

[mental crag]

Correct, right?😂

[fallen moat]

the formulas somehow messed up in the middle

are you sure you just autofill them?

[mental crag]

I didn’t autofill this time

I autofilled rn and it gave the same answer

[fallen moat]

it's so weird, the negative and positive flipped

they (total $) should be all positive

[mental crag]

This is what I have

[fallen moat]

is the format ok? or is it fixed red words?

[mental crag]

I made the text red, oops

[fallen moat]

lol

that's why it's negative lol

now it's all positive

[mental crag]

Yesss lol

[fallen moat]

next time you can use data formatting for negative inside brackets and red instead of manually adding red colour

[mental crag]

Will keep that in mind

It’s my first time using Excel hehe

[fallen moat]

ohhhh then that's pretty good!

[mental crag]

Thank you😆

[fallen moat]

i thought it's like a mid-course or something already

[mental crag]

Omggg nooo

Intro to Accounting

[fallen moat]

cool~

[mental crag]

Okay so, I don’t understand the units

[fallen moat]

then i think your prof really wanna trick you guys lol

[mental crag]

OMG YES

He literally teaches us the easy questions

But would test us on hard questions that he NEVER taught

[fallen moat]

dang

anyways for units, here's my concept:
100 units at first
(given)
100+80 units for second line
(that's because the $ is worth 180 units)
100+80-150 units for the third line
(that's because the money worth that many units)

[mental crag]

So we add running # units from bgn inv. with # units from the sale at cost, and then add # units from acquisition?

[fallen moat]

running unit from the previous inv, not beginning inv

and this is where i am confused, units running should be just the units that are remaining in your inv or ??m

[mental crag]

I have no idea either😭

[fallen moat]

i think you better ask your classmates about that too , see if they have any conclusions

[mental crag]

Will do

[fallen moat]

hope that helps!

[mental crag]

Can I autofill since I did the first formula

Or would it be different

I got 30 units for the first sale (below beginning inventory)

You know what I’m thinking

[fallen moat]

ahhh

[mental crag]

What if units also have the same formula as Total $

[fallen moat]

it feels weird now

[mental crag]

The same formula you taught me for Total $

Because my prof. did questions (easy ones nothing compared to this) and both the total $ and units had the same formula

[mental crag]

What does😂

[fallen moat]

can you show me his "easy version"

[mental crag]

Yes

Let me find it

[fallen moat]

reading

[mental crag]

He would give us questions where for each sale, the units are only given but not the cost/unit

[fallen moat]

grrrr

[mental crag]

Why grrr

[fallen moat]

i think my concept is not what your prof was thinking

[mental crag]

Oh danggg

What concept is my prof thinking

[fallen moat]

now I'm puzzled

😱

[mental crag]

But do you understand what I mean

He would show us how to do easy questions

[fallen moat]

yea, i totally get that

[mental crag]

If I do my prof.'s method, I would get a negative number which is wrong for this kind of question

I think the method you told me for Total $ is right tho

But for the units, Idk what to do LOL

If I do the same method as the Total $, do you think it would be correct

[fallen moat]

hmmm

what i think what your prof is thinking is:
only consider those units that we have in hand, and their current worth

so for this, the units part are correct

the money part will be a bit more complicated

[mental crag]

So I should use the original units that I had at the very beginning

[fallen moat]

yea

lemme think a bit

[mental crag]

Is the Total $ method that you mentioned to me is okay, right

Like we don't have an issue on Total $

[fallen moat]

since this i think it is more like just $/unit times unit

so the hardest would be $/unit

[mental crag]

For the $/unit on the far right side, my prof. said you would have to just divide Total $ by # of units

[fallen moat]

i see, so it works on the total$

lemme explain using the concept your Prof use

Total $

first date:
1500
(total worth of beginning units)

second date:
1500-80×15
(total worth of remaining units)

third date:
(1500-80×15)+2475
(total worth of the previously remaining units + total worth of the newly purchased units)

forth date:
( (1500-80×15) +2475) - 120×(third date money/unit)
(total worth of the previous units - worth of the sold units disregarding the money gain)

[mental crag]

For the second date, where did you get x15 from

[fallen moat]

(with explaination now) (check the edited explaination)

[mental crag]

Take your timeee

[fallen moat]

get it?

[mental crag]

Reading rn

Not gonna lie I'm confused lolll

Is there another method similar to my prof.'s

Where we just add, add, and subtract

[fallen moat]

starting from the second:
=H237-F238*J238+E238

[mental crag]

I did exactly what you said and this is what I got

[fallen moat]

ohhh i get why it's wrong

[mental crag]

How come

[fallen moat]

lemme Rethink

[fallen moat]

should be J237 instead

and then autofill

[mental crag]

I got -1200

I got closing inv. = $0.00

[fallen moat]

can i have a look?

[mental crag]

Yes

[fallen moat]

first
=H215
second
=H237-F238*J237+E238
third
=H238-F239*J238+E239
.
.

[mental crag]

I have a question

What if we ignore what's given in the question for each Sale

[fallen moat]

i bet you missed H237 in the first place

[mental crag]

We just use the # units given for sales and we don't include $/unit

[fallen moat]

yep, that's what we are doing

[fallen moat]

hmmm

[fallen moat]

well, i think that's for formatting?

[fallen moat]

oh dear....

i see what's happening

why add brackets lol

[mental crag]

Even if I remove the brackets, it gives the same answer of -1200

[fallen moat]

2 things

1. i thought K is H
2. it's hard to know what's wrong without a full picture the lol

first
=1500
second
=K237-F238*J237+E238
third
=K238-F239*J238+E239
.
.

[mental crag]

I'm so sorry

[fallen moat]

no brackets

[fallen moat]

now this range of sight is much better 😁

[mental crag]

I got 300 now 😛

[fallen moat]

did you fill column J?

[mental crag]

Nope

I only have column J filled for bgn inv

[fallen moat]

then you'll have to fill them with the formula given by your prof

[mental crag]

And for the units, is it the same formula?

[fallen moat]

yep, the units are correct!

[fallen moat]

how's it 😺

[mental crag]

This is what I got😭

[fallen moat]

now that's looks gooooooooooooood

[mental crag]

I have a question

[fallen moat]

ohhh, you did screenshot! nice

[fallen moat]

yea?

[mental crag]

My prof. said that you can check if you're doing it right by using these formulas: Units Avail. - Units Sold

And the other formula is GAFS - COGS

[fallen moat]

hmmm?

[mental crag]

He said that if those answers = what you got above, then you're doing it correct

When I subtract GAFS with COGS, it get a diff answer to the closing inventory

[fallen moat]

I'm reading your prof's example

and i think it's because he used another formula for
$/Unit
under Sale at Cost

[mental crag]

You know the method you told me for Total $

Would that be the only way to answer this particular question

Or would my prof. use his way

[fallen moat]

i think that's the way for this question specifically

to be honest, when your brain feels better, you can try and work out the reason behind this method

i bet it's painful for your brain to continue investigating in this question

(because mine is )

[mental crag]

LOL SAME

And for the units, did I do it correct

For the units, I used my prof's method

[fallen moat]

yea

[mental crag]

And for the method you taught me for Total $, is there a way for me to check that I did it right

[fallen moat]

under sale:
$/Unit should be
=J(same row number)

i can only tell you an estimate if it right or not

you see, from the calculated
Running: $/Unit
the numbers are very much similar

[mental crag]

I just checked and they don't equal

[fallen moat]

and it only changes when there is Acquisition

[fallen moat]

they are equal in your Prof.'s example

[fallen moat]

so, if you wanna check if the Total is correct, check if the $/Units are similar or not

[mental crag]

Omggg I think I did it wrong bc mine don't equal one another

[mental crag]

Referring to this

[fallen moat]

for Sale: $/Unit
just do
=J237 (for row 237)
=J238 (for row 238)
.
.

[mental crag]

I don't get it

Like this?

So I ignore what was given earlier in the question?

[fallen moat]

yes, this is what i meant

[mental crag]

What I did was copy and paste from J

[fallen moat]

and no, i am not sure if this correct

since

i dont know what is GAFS and COGS

also

[mental crag]

GAFS is Goods Available for Sale and COGS is Cost of Goods Sold

[fallen moat]

of this

i see

[fallen moat]

then yea, this is correct then

for better presentation, you can delete the excess info

like the lines without info under sales

[mental crag]

But when I do the sum for each, and use the GAFS - COGS, I don't get the same answer as the closing inventory

[fallen moat]

because,

you need the change the values of the total as well

[mental crag]

How do I change the values of the total

[fallen moat]

total under sale should be
=units * $/unit

[mental crag]

Oh, I got it

[fallen moat]

side note: excel is dynamic, so don't copy and paste only values

unless it's really the case lol

[mental crag]

I mean, I did = sign on Excel

My bad lolll

Thank you so much

Do I do the same for the LIFO method?

[fallen moat]

dunno

[mental crag]

Damnnn

[fallen moat]

I don't know what is LIFO lol

[mental crag]

Because LIFO method is the first-in, first out

[fallen moat]

i see

[fathom night]

??

LIFO is last in first out

The most recent inventory is sold off

[mental crag]

I mean last in last out, my bad LOL

I mean FIFO

[mental crag]

Good eye lol

[fallen moat]

I'm not sure about the difference between the 2 methods in this case

I'm thinking it'll be a lot more useful and significant in more complicated cases

[fathom night]

it's just 2 different inventory methods in accounting/finance

lifo/fifo that is

[mental crag]

This is an example of how my prof. did the FIFO method

So basically, I just have to copy what we did from the weighted-average?

[fallen moat]

haha, my brain is fried

i bet I'll stop here

[mental crag]

LOL I feel you

Thank you again

I OWE YOU

[fallen moat]

can't process anymore

[mental crag]

GOD BLESS YOU!!!

[fallen moat]

you too!

[mental crag]

Sorry to bother you but do you know how I should do the FIFO method after figuring out the weighted-average

[fathom night]

I'll take a look in a bit

[mental crag]

Thank youuu and take your time

[fathom night]

You can dm since it'll probably be a lengthy process

[mental crag]

Added and dm'd 🙂

[cedar kiln]

@mental crag Has your question been resolved?

[cedar kiln]

@mental crag Has your question been resolved?

[cedar kiln]

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[real phoenix]

Can someone please solve this for me

[cedar kiln]

Please don't occupy multiple help channels.

@real phoenix Has your question been resolved?

[real phoenix]

<@&286206848099549185>

[wise monolith]

they cant just solve the question

u need to add a solution

or u need to say where ur stuck so they can teach u how

and then u can confirm if ur correct or not

[cedar kiln]

@real phoenix Has your question been resolved?

[real phoenix]

I'm stuck in whole question question can't figure out how to manage out inequality

[wise monolith]

<@&286206848099549185>

[fallen moat]

I just got online do you still need help? @real phoenix

[real phoenix]

Ya

I still need help

[fallen moat]

any progress so far?

[real phoenix]

No I left doing it

[fallen moat]

i see, okay so, S1 and S3 are easier

let's do S1 first

[real phoenix]

S1 is a circle inner part right?

[fallen moat]

yep

circle of radius ....?

centred at ... ?

[real phoenix]

3rd is X>0 ?

[fallen moat]

just for your reference

[fallen moat]

yes

[real phoenix]

(0,0)

[fallen moat]

now S2

[real phoenix]

Ya

[fallen moat]

simplify it a bit first before plugging x+yi into it

[fallen moat]

it'll look easier if so

[real phoenix]

4 but we can't include boundary part of circle as z does not lie on it

[fallen moat]

very good

[real phoenix]

Sorry I can't type properly im from suffering palmar hyperhidrosis

[fallen moat]

oh ok, what is it about?

btw did you figure out S2?

[real phoenix]

Yeah actually I did

[wise monolith]

https://media.discordapp.net/attachments/651115852812386306/1131600180949561387/869991107344367727.gif

[fallen moat]

oh nice

so now, with all 3 parts combined

what shape do we have here?

[fallen moat]

it's ok

[real phoenix]

It is a circular region

[fallen moat]

yeah, it's a part of a circle

it's a sector, do you know how to calculate the area of a sector?

[real phoenix]

Theta/360 pie r square is it correct?

[fallen moat]

yea, but you may have to use radians

[real phoenix]

Answer is given In radian

[fallen moat]

which is using the formula
r²theta/2

[real phoenix]

My answer is coming 20pie /3 which is correct

Thank you

Should I send my solution to ?

[fallen moat]

i didn't calculate lol, i just know the way

[fallen moat]

you're welcome!

[fallen moat]

?

oh

you don't have to

glad that i can help!

[real phoenix]

I do plenty of advance question in math If i face any difficulties I will ask it here okay ?

[fallen moat]

nah, you should open a new channel for each question, but note that only on channel at a time

i see you have another problem in #help-11

is that what you mean?

[real phoenix]

I know this I was just saying generally

[fallen moat]

i see, then yea, just find an avaliable channel and ask, it'll be cool

[real phoenix]

Ya still my that question is still not solved yet

Please help if you can

[fallen moat]

I'll head there, if you're done with this q, you can close it 🙂

[real phoenix]

.close

[cedar kiln]

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[obsidian umbra]

@rain drift hello, so the channel got taken or whatever. So I was able to get cos(-t-8pi) to be equal to cos(t+8pi). and 8pi is 4 rotations. So we are just left with t, and cos(t) = b.

[obsidian umbra]

but for tan

but for tan,

-tan(-t-7pi) is same as tan(t+7pi), it made 3 rotations and a 1/2 roation, so we are legt with tan(t + pi)

[rain drift]

bingo 🙂

[rain drift]

yep :). Now consdier the period of tan(x)

[obsidian umbra]

so it would be 5pi/6 radians

the tan of that is

[rain drift]

not quite

[rain drift]

you had it correct here

[obsidian umbra]

oh wait

my bad

it would just be pi

tan (pi) = 0 / 1 = 0

well

-1

but it would stilll be 0

[rain drift]

no no sorry

remember what I'm asking with the period

how many rotations does it take to get back to the same value?

[obsidian umbra]

3 1/2?

[rain drift]

now how many pi radians?

[obsidian umbra]

that would be 6pi + pi so 7pi / 2?

[rain drift]

no

watch this video real quick

https://www.youtube.com/watch?v=JgD6fmsqD3s&ab_channel=ProfessorHeatherPierce

it'll clear up what I mean

[obsidian umbra]

oh

so

if pi is one full rotation for tangent

this would've went 7 rotations

leaving us with tan(t)

which is just

c

is that right?

[rain drift]

oh I'm sorry I was helping someone else and it took some time

yes exactly!

[obsidian umbra]

ok thanks

especially thanks for going step by step

[rain drift]

so then we get -sin(-t - 6pi) + cos(-t - 6pi) - tan(-t - 7pi) = sin(t + 6pi) + cos(t + 6pi) + tan(t + 7pi) = sin(t) + cos(t) + tan(t) = a + b + c 🙂

[rain drift]

yeah of course!

[obsidian umbra]

.close

[cedar kiln]

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[cedar kiln]

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[frail condor]

Need help solving this, I tried applying the rank-sum inequality, but of no avail. I'm unable to think of a way to get rid of the inequality

[frail condor]

Also, this is the Reference mentioned

This is my working

I’m unable to proceed after this

<@&286206848099549185>

[cedar kiln]

@frail condor Has your question been resolved?

[frail condor]

I feel that the question is incorrect in some sense, given I can make a counter example like that

thx

.close

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[thorny orchid]

Hello!

[thorny orchid]

We have to determine all possible cuts. I did all of these except the cut on the bottom left corner (cutting AE and ED).

I don't see how that cut would cut the total flow - couldn't there still be a path from A-B-C and A-B-D-C?

(This is the topic of networks, specifically critical path analysis.)

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[jolly sable]

Suppose in an ordered pair = (2,2)
This is an antisymmetric relation right?

[crimson sedge]

Antisymmetry cannot be applied here?

(a,b) have to be a distinct pair

[jolly sable]

Suppose relation is defined

[wraith dagger]

kebesque

[jolly sable]

Y=x

My focus is to learn the antisymmetric relation

And i am sure this is antisymmetric

Because 2,2 only when 2=2

This is not asymmetric because a,b is defined but b,a should not be defined but here it exists

This is symmetric too

Every asymmetric is antisymmetric

I want to understand this help

[iron saffron]

So asymmetric is antisymmetric and irreflexive, that means every asymetric is antisimmetric too

[jolly sable]

But how asymmetric is antisymmetric?

Does it mean (a to be then b to a shouldn't be there right?

But a,a can be there

So it is antisymmetric and irreflexive

[cedar kiln]

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[fair copper]

what happens here?

[spice kraken]

there are multiple possible x values for trig functions

so the question should specify the range of answer

[fair copper]

yea it was ffrom 0<x<2pi

[cedar kiln]

@fair copper Has your question been resolved?

[elfin otter]

When tan is a negative value, there r 2 ans (depending on range)
Which is in quad 2 & 4

.
.
1st quad: all trigo is positive
2nd quad: ONLY sin + || tan & cos -||
3rd quad: ONLY tan + || sin & cos -||
4th quad: ONLY cos + || sin & tan - ||

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@fair copper Has your question been resolved?

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[molten pasture]

a4−3a2+9=(a2+3+3a)(a2+3−3a). how pls tell

[lilac totem]

$a^4 -3a^2 +9$

[wraith dagger]

thickduckplayz

[lilac totem]

Is this your question?

[molten pasture]

yah

[lilac totem]

You know how to factorise a quadratic?

[molten pasture]

yeah

but it shows a comolex roota

roots*

complex roots*

[vague rapids]

try to break this into a perfect square+ other terms

[molten pasture]

did it

but it didnt

[lilac totem]

What did you do?

[vague rapids]

show your steps

[molten pasture]

wait lemme show

$3a^2-3a^2+9$

no see

[vague rapids]

wait what

[molten pasture]

a4-3a2+9

then

a4-4a2+9+a2

but it leads to a dead end

i tried first

[vague rapids]

$a^{4}-3a^{2}+9=a^{4}+6a^{2}+9-9a^{2}$

[molten pasture]

letting a2=k

[wraith dagger]

beard420

[vague rapids]

Try to proceed after this

[molten pasture]

oh my god

thank you

it was so simple i didnt see it.......

how do i close this part

like this post

[vague rapids]

.close

[molten pasture]

.close

[cedar kiln]

Channel closed

Closed by @molten pasture

Use .reopen if this was a mistake.

[molten pasture]

also

.reopen i had another doubt

[cedar kiln]

✅

[molten pasture]

i seem to get stuck after getting thak

pqk= 300

after taking lcm and

simplyfying

[cedar kiln]

@molten pasture Has your question been resolved?

[molten pasture]

<@&286206848099549185>

[cedar kiln]

@molten pasture Has your question been resolved?

[molten pasture]

<@&286206848099549185> help pls

[thin roost]

I'm not sure but I thino

[wraith dagger]

_basudev

[thin roost]

You gotta think this way I belive...

[wraith dagger]

_basudev

[lunar lynx]

Here, a and b are primes?

[thin roost]

Ye

I'm not sure if this is the way

[molten pasture]

ohh

lemme see

[thin roost]

What exactly is the ans to this question?

@molten pasture

What are the pairs?

[molten pasture]

they answer given is that there are 24 pairs to exist

sadly i dont have the pairs

[molten pasture]

am hm inequality

[thin roost]

Ye...

I don't think then it is the way to go..

[lunar lynx]

question says that there'll be only 24 such pairs ?

[lunar lynx]

Oh.

Why am i getting infinite pairs ?

[thin roost]

Actually..m

Yeah

Infinite makes sense...

[lunar lynx]

I mean if you take any pair of form (2, q) where q is a prime bigger than 100, it always satisfies.

Wait. NO.

Oh. Okay. I got it. Makes sense.

[thin roost]

Ahh.... so is it 24?

If so how...

[lunar lynx]

Less than 100.

[lunar lynx]

I didn't calculate but it has to be finite.

[thin roost]

Alr

[lunar lynx]

Basically,
Let p and q be two primes such that p < q.

Now, we are given:
$\frac {1}{p} + \frac {1}{q} > \frac{51}{100}$

Since $p < q$, we have $\frac {1}{p} > \frac {1}{q}$

      $\frac {1}{p} + \frac {1}{q} > \frac{51}{100}$

which gives, $\frac {1}{p} + \frac {1}{p} > \frac{51}{100}$
Which gives, $\frac {2}{p} > \frac{51}{100}$

So, $p < 3.92$

thus, only possible values for p are 2 and 3.

[wraith dagger]

.enemagneto

[lunar lynx]

Finally.

[thin roost]

And now perhaps take each case with p=2 or 3 then calculate q?

[molten pasture]

yeah

i was thinking more of coming up with an inequality

or a range

[lunar lynx]

Now, we are given:
$\frac {1}{p} + \frac {1}{q} < \frac{5}{6}$

Since $p < q$, we have $\frac {1}{p} > \frac {1}{q}$

$$\frac {1}{p} + \frac {1}{q} < \frac{5}{6}$$
$$\frac {1}{q} + \frac {1}{q} < \frac{5}{6}$$
$$\frac {2}{q} < \frac{5}{6}$$

So, $q > 2.4$

Now, when p is 2,
$$\frac {1}{2} + \frac {1}{q} > 51/100$$

$$\frac {1}{q} > 1/100$$

Which gives,
$ q < 100 $

Thus, $$2 < q < 100$$ and q is a prime.
Thus, 23 values of q are possible when p is 2.

Similarly do for p=3.

[molten pasture]

but this workd

[lunar lynx]

Similarly do for p= 3

[molten pasture]

oh ok

thanks so much

[thin roost]

I was grinding my mind into Am Hm ineq more....

And

Here we go

[molten pasture]

oooo

i wanna hear this

[lunar lynx]

How the hell do i go to next line in Latex? Solution will look more elegant without extraneous "which gives".

[wraith dagger]

_basudev

$$x=2$$
$$x=7$$

[thin roost]

Or

[lunar lynx]

Oh

[wraith dagger]

_basudev

$x=7 \\
Y=7$

[molten pasture]

wait you got them all???????

by the inequality or by your trial and error

daamn brooo

[thin roost]

Better

[lunar lynx]

It's not trial and error. You just constrict the possibilities using inequalities.

[wraith dagger]

.enemagneto

[molten pasture]

yeah ik but still nvm good work guys

[thin roost]

I'ma sleep now

[lunar lynx]

Okay. This should be final.

Now, we are given:
$\frac {1}{p} + \frac {1}{q} < \frac{5}{6}$

Since $p < q$, we have $\frac {1}{p} > \frac {1}{q}$

$$\frac {1}{p} + \frac {1}{q} < \frac{5}{6}$$
$$\frac {1}{q} + \frac {1}{q} < \frac{5}{6}$$
$$\frac {2}{q} < \frac{5}{6}$$

So, $q > 2.4$

Now, when p is 2,
$$\frac {1}{2} + \frac {1}{q} > 51/100$$

$$\frac {1}{q} > 1/100$$

Which gives,
$ q < 100 $

Thus, $$2 < q < 100$$ and q is a prime.
Thus, 23 values of q are possible when p is 2.

Similarly do for p=3.

[wraith dagger]

.enemagneto

[cedar kiln]

@molten pasture Has your question been resolved?

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[cedar kiln]

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[slim thicket]

Hi

[slim thicket]

I got x=12 and x=2 as my answer

but im supposed to plug it in and check and i got the steps mixed up

idrk how to check my answers

[spice kraken]

how is x = 12 a answer

[slim thicket]

let me send a pic of my work

The thing on the side is me attempting to check my work but I got -2 so idk if I did it right

[novel patrol]

When you checked your work on the bottom you forgot to plug 12 into the other x value

[slim thicket]

And then subtract it by 5?

Cuz I did that in class and my teacher said it was wrong

[tropic oxide]

no you're

fucking up the process of checking whether a number solves an equation

if you wanna know whether x=12 solves your equation you need to replace ALL instances of x with 12 at once

[novel patrol]

So essentially what you're doing there is verifying your solutions to determine if the answers are true to the equation or not

[tropic oxide]

you cannot be selective about what you do and don't replace

[slim thicket]

So like this?

[tropic oxide]

personally i would put questionmarks above all these equals signs, but yes.

[slim thicket]

Why?

What does that do

Or represent

That you’re just checking?

[tropic oxide]

exactly

[slim thicket]

Oh okay

[tropic oxide]

so it doesn't look to the reader (and importantly, to YOU) that you're asserting this equality as true

[slim thicket]

So now that I have 7=-7 what is this supposed to represent?

That they’re equal?

[tropic oxide]

$7 \overset?= -7$

[wraith dagger]

Ann

[slim thicket]

And that makes my solution right?

[tropic oxide]

idin, answer me this,

and i know this question is gonna sound stupid or obvious,

but here it is anyway:

are 7 and -7 the same number?

[slim thicket]

Yes

[tropic oxide]

no

no, they are not

they are a whole fourteen units apart on the number line

[slim thicket]

Oh

[tropic oxide]

-7 is not a seven that's cosplaying something, -7 is -7

[slim thicket]

Okay make sense

[tropic oxide]

so stop denying negative numbers their sense of identity

[slim thicket]

K

[cedar cypress]

I love how you phrase your sentences

[tropic oxide]

thanks that one was on purpose

also @slim thicket while i'm out here giving my opinion/advice/whatever on the way you write things:

cross your sevens

this digit 7 feels empty inside and is very easy to confuse for a one

and the quicker you write the more likely it will be that you'll eventually confuse it in exactly this manner

[slim thicket]

Aight

But when checking did I do my steps right?

That’s all I wanna know essentially

[tropic oxide]

you did now, aside from the un-question-marked equals signs and the incorrect conclusion

upon seeing $7 \overset?= -7$ you should have concluded that no, the two sides aren't equal, so $x=12$ is not a solution

[slim thicket]

K

[wraith dagger]

Ann

[slim thicket]

OHHH

I see it now

On the other side I got 3=3 and that means they are the same number because they have the same distance to 0 on the number line

But 7 and -7 isn’t

[tropic oxide]

they are the same number because they have the same distance to 0 on the number line

no

[slim thicket]

Making only X=2 the correct solution

Wait yea I worded that wrong

[tropic oxide]

"they are the same number because they have the same distance to 0 on the number line"

[slim thicket]

Cuz then both would be equal

Yea I get you

Aight I’m all good

Thanks

[tropic oxide]

i was gonna say that you keep denying negative numbers their sense of identity

[slim thicket]

Idk what you mean by that

[tropic oxide]

well you just expressed your belief that if two numbers have the same modulus then they are in fact the same number

[slim thicket]

Yes but that’s wrong

[tropic oxide]

and my whole point is that that's (a) wrong and (b) a mindset you should train yourself out of

[slim thicket]

can you just explain again how i determine which answer is right?

not which

but like how you know

in like simple terms

[tropic oxide]

you verify them all one by one

@sly drift if this is a serious question please open your own channel, if you're trolling would you please be so kind as to fuck off

[slim thicket]

thats disgusting

im outta here the more i say shit the more garbage this kids gonna say

.close

[cedar kiln]

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[twilit badger]

what is the integral of (3/2)*x²

[twilit badger]

what i thought it was is 1/2t^4

[vagrant elbow]

Wait my namesake

[tropic oxide]

why t and why ^4

[vagrant elbow]

Ann no I help this man

[tropic oxide]

ok as you wish

[vagrant elbow]

[twilit badger]

lol the name

hello quite a rare coincidence

anyway

[vagrant elbow]

Indeed

yeah Ann's questions are still valid

Why ^4, and why t

[twilit badger]

you can relplace the t with x

[vagrant elbow]

well that's only in the case of indefinite integration

Unless you made the sub x = t for whatever reason

[twilit badger]

i used t because it's a physics issue

but an algebra method

[vagrant elbow]

I see

how did you arrive at your answer

[twilit badger]

and using the power rule my answer was 1/2*t^4

so to go from (3/2)*t²

[vagrant elbow]

why power 4

you add 1 to the power and divide by the new power

Not 2

[twilit badger]

x n+1/n+1

[vagrant elbow]

indeed

[twilit badger]

yes

[vagrant elbow]

You seemed to have done x^(n + 2)/(n + 1)

[twilit badger]

t^3/3

[vagrant elbow]

yes...

[twilit badger]

3/2t

[vagrant elbow]

no don't multiply them lol

[twilit badger]

gimme a sec its tough with the keyboard

i will show a photo

and refer to it

[vagrant elbow]

sure

[twilit badger]

few minutes if you mind

so using that rule

i depart from 3/2 and go to 3t/2

and t² to t^3/3

and (3t/2)*t^3/3

you get (3t^4)/6

1/2*t^4

but my problem is is that the course im following says that the answer is t^3/2 which confuses me

[vagrant elbow]

Why are you multiplying

3/2 and t^2 doesn't get integrated seperately

[twilit badger]

how so

[vagrant elbow]

$\int x^n \dd x = \frac{x^{n + 1}}{n + 1}$

[wraith dagger]

neonperseus

[vagrant elbow]

This is the rule

You can't distribute an integral over a product

$\int uv \neq \int u \times \int v$

[wraith dagger]

neonperseus

[twilit badger]

so i just adapt the rule onto 3t²/2

a single factor

[vagrant elbow]

yes

Much like with derivatives, the 3/2 can simply be pulled out of the integration

$\int \frac 32 t^2 \dd t = \frac 32 \int t^2 \dd t$

[wraith dagger]

neonperseus

[twilit badger]

a i though you where supposed to do number * x

here for example 3/2*t

[vagrant elbow]

nah

I hope you get it now

[twilit badger]

but then one of my previous answers would make sense

3*t

wait nevermind

3t^2/2

[vagrant elbow]

yes

[twilit badger]

so if it is 3/2*t²

then what do you do to integrate

$\int \frac 32 t^2$

[wraith dagger]

neonel.

[vagrant elbow]

Only apply the power rule to t^2

[twilit badger]

and nothing happens to 3/2

[vagrant elbow]

Nothing

[twilit badger]

so then i get 1/2*t³

3t³/6

but that as far as a know isnt the expected

t^3/2

[vagrant elbow]

What's the original question

[vagrant elbow]

that is the answer

It even says on the right

[twilit badger]

oh im a doofus

i thought it meant t to the power of 1.5

sorry

[vagrant elbow]

it's alright not really your fault

Parenthesis exist

[twilit badger]

ye still

thanks for being helpful

have a great rest of your day

https://tenor.com/view/pacha-okay-great-meme-just-perfect-gif-16605517

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[crimson sedge]

Hello, I'm having a problem with this question. We're supposed to devise a system of equations and solve.

[crimson sedge]

For the most part, I understand that the first equation would be:

x + y + z = 28

For the rest, I think it's supposed to be

x - y = 0

and

x - 5z = 0

To represent that the same number of touchdowns and extra-point kicks. and then 5 times the amount of touchdowns as field goals.

But, I'm curious how I would add in how many points each are worth.

I tried 6x + 1y + 3z = 28, but that fractionalized the results.

[drifting marlin]

Equate it to 38, not 28

[crimson sedge]

ok

ok

That solved me.

Thank you!

.close

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[oblique prawn]

how to solve a system of equation with one implicit equation and one linear equation

[oblique prawn]

basically im given that $2y^3+6x^2 y-12x^2+6y=1$ and $\dv{y}{x}=\frac{4x-2xy}{x^2+y^2+1}$

[wraith dagger]

jashxdlol

[oblique prawn]

and

and y=-x is tangent to the curve at point P

and im asked to find the x and y of point P

so i set dy/dx=-1

and i need some other equation to solve the system

so i was thinking i could find the intersection between the curve and the line y=-x

wait cant i just do substitution

[cedar kiln]

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[alpine quiver]

$\frac{2(n!)}{(2n)!} = \frac{1}{2^{n-1}(2n-1)!!}$

[wraith dagger]

nusuntrares

[cedar kiln]

Please don't occupy multiple help channels.

[alpine quiver]

help me proof that

[cedar kiln]

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

[wanton sail]

As the bot said, please don't occupy multiple help channels with the same question.

Please go back to your original channel

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[distant pecan]

i used the A=P*e^rt formula since its continuous

[distant pecan]

and i made P=500,000

was that wrong

should A=500,000

[spice kraken]

yes

[distant pecan]

i still got it worng

heres my equation

500,000=P*e^40 times 0.04

[fathom night]

Use PV formula

[distant pecan]

huh

whats that

its continous tho

so it should be $/P*e^rt$

[wraith dagger]

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

[fathom night]

$PV=\frac{FV}{(1+r)^n}$

[distant pecan]

bruh

[wraith dagger]

akumaenjeru

[distant pecan]

i never learned that

so im not supposed to use ot

it

[spice kraken]

the compund interest is continuous though

[distant pecan]

yeah

so it should be

P*e^rt

but i used that and it didnt work

[spice kraken]

hmm

that's weird

[distant pecan]

you wanna try?

[fathom night]

What answer do you get

[distant pecan]

like

ohhh

it says round to the nearest penny

so is that a decimal?

[fathom night]

Use $A=P(1+\frac{r}{n})^{nt}$

[wraith dagger]

akumaenjeru

[distant pecan]

uhuh

[fathom night]

You should get the right answer using the formula

[fair geyser]

i get 10095126

[fathom night]

I don't think that's the answer

[distant pecan]

i got that

[fair geyser]

wait maybe they want it in dollars actually

[distant pecan]

they want in pennues

[fair geyser]

100951.26

[distant pecan]

it says

yeah

thats it

[fair geyser]

no it says it weridly

[distant pecan]

i didnt use a decimal

[fathom night]

You should be getting $104,144.52

[fair geyser]

it could be interpreted both ways

[distant pecan]

bruh

[fathom night]

Use this or the PV formula

Both will get you the answer

[distant pecan]

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[oblique prawn]

can u use lhopitals

[oblique prawn]

on lim as x→-∞ of 2xe^x

since it’s a -∞*0

[spice kraken]

when you have 0*inf divide them

[oblique prawn]

can u only use lhopitals when it’s a fraction?

[spice kraken]

yes

[oblique prawn]

isn’t 0*∞ also an indeterminate form tho

[spice kraken]

we change it into fraction

$2xe^x=\frac{2e^x}{x^{-1}}$

[wraith dagger]

_wherewolf_

[spice kraken]

and they are both infinite

[oblique prawn]

can you also do (2x)/(e^-x)

[spice kraken]

yep

it's the same

[oblique prawn]

so then u take derivative

$(2e^x)/(-1/x²)$

[spice kraken]

oof mine isn't really gonna work xD

[wraith dagger]

jashxdlol

[oblique prawn]

how do u know if it’s gonna work or not

[spice kraken]

it's going to be x^-n no matter how many times we defferentiate it

[oblique prawn]

ye

so you do (2x)/(e^-x)

and then differentiate

2/(-e^-x)

and then plug-in so you get 2/-∞=0

do you always want one of the sides to be constant after you take derivative?

[spice kraken]

umm we just want it not be indeterminate after differentiating

[oblique prawn]

ok ty

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[frail needle]

How do I find the rate of growth of $a_n = \lceil e^{-\gamma+\sum_{k=1}^{n-1} \left( a_k + \frac{1}{a_k}\right) } \rceil$

[wraith dagger]

a8d_

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[grand kernel]

hello, i had a question concerning a self-adjoint endomorphism from an euclidian space.

[grand kernel]

I saw in course the following theorem : f is self-adjoint iff it's matrix M in an orthonormal basis is symetric, which is quite easy to see

but the matrix is actually symetric if each basis isn't it? because for another basis B, we have [M]B = PMP^T for some matrix P invertible and PMP^T is symetric

or did i misunderstand something?

[cedar kiln]

@grand kernel Has your question been resolved?

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[tawdry mesa]

develop and count: -3(4-x)(-1/3 -x)

[spice kraken]

what do you mean?

[tawdry mesa]

on a book the answer was: -3x^2 +11x+4 but chatGPT said it was -12 + 6x - 9x^2 and i got 11x+12-13^2

[obsidian coral]

Don't use chatgpt to do math

It's not good at math

[tawdry mesa]

i was confused

[obsidian coral]

!show

[cedar kiln]

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

[tawdry mesa]

11x+12-13^2

cant 😦 am on my laptop

PC

[obsidian coral]

Doesn't really help us if we don't know what you did

[rain drift]

Are you just trying to expand this out?

[tawdry mesa]

Yes! thats the word

-3(4-x)(-1/3 -x)
first i did: -3(4-x)= (-12+3x)

and then: (-12+3x)((-1/3 -x)

[rain drift]

oh, if that's the case then I would first use the FOIL method on $(4 - x)\left(-\frac{1}{3} - x\right)$ and the multiply your polynomial by $-3$

[wraith dagger]

mellowdramallama

[obsidian coral]

This is fine too

[rain drift]

that's fine as well

[obsidian coral]

Continue

[rain drift]

lol nice

[tawdry mesa]

i got 11x+12-13^2

but the book says otherwise

-3x^2 +11x+4

i dont understand where did the 4 come from

maybe 3*4?

[tawdry mesa]

What's the foil method?

sorry but i haven't studied this yet

guys?

[obsidian coral]

FOIL is just an acronym to distribute two binomials, First, Inner, Outer, Last

[tawdry mesa]

do i have to not turn 36/3 into 12 to expand it or? in multiplying -12*-1/3

ooh thank you so much! but i already used this technique, i used the board tactic

its basically like this

[obsidian coral]

No, you're thinking about adding/subtracting fractions

That's just multiplication that you're doing

[obsidian coral]

Try that with your problem

[tawdry mesa]

i alr did thx

[obsidian coral]

Did you get the correct answer now

[tawdry mesa]

im just wondering why im wrong

[obsidian coral]

.

You don't need a common denominator

[tawdry mesa]

oopssssss

wow thankss sm

lol totally forgott

/finish

[obsidian coral]

.close

[tawdry mesa]

👍🏼

.close

[cedar kiln]

Channel closed

Closed by @tawdry mesa

Use .reopen if this was a mistake.

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[frail iris]

For A do I just add all the areas?

[spice kraken]

yes

[frail iris]

ok

what about B?

[spice kraken]

don't forget the negative area though

[frail iris]

is it negative if its under the curve or in the negative y?

[spice kraken]

substitude try to substitude f(x) = F'(x) into (a)

[spice kraken]

under x axis

[frail iris]

ok so it would be 3+4+(-9)

[frail iris]

how?

[spice kraken]

yes

[frail iris]

I am confused on b

[spice kraken]

$\int_{0}^3f(x)dx = \int_{0}^3F'(x)dx$

[wraith dagger]

_wherewolf_

[spice kraken]

then use the fundamental theorem of calculus

[frail iris]

huh?

so

$\int{0}^3(3+4+(-9))dx

[spice kraken]

no no no

[frail iris]

thats the first one right?

you cant take F' of constants though?

it would just be 0

[spice kraken]

do you know how to evaluate $\int f'(x)dx$?

[wraith dagger]

_wherewolf_

[frail iris]

not really

we are saying from one value to another in a specific function

right?

[spice kraken]

can you explain what you mean

[frail iris]

like we are evaluating the function between 2 values

or in this case the deriv function

[spice kraken]

yes we evalate definite integral of two values

and how exactly do we do that??

[frail iris]

plug each value into the function and solve?

but if we dont know the function how can we do that?

[spice kraken]

which function though

when we integrate f'(x)

[frail iris]

uhm

idk tbh

[frail iris]

the parent function?

[spice kraken]

what is the antiderivatice of f'(x)

[frail iris]

idk bc I dont have the function

I am so confused

wait

is it using this

[spice kraken]

the fundamental theorem of calculus states that integrals and derivatives are inverses

it means they cancel each other

[frail iris]

so does F(0)=7 turn into f(0)=7x?

[spice kraken]

nope

[frail iris]

huh

[spice kraken]

$\int \frac{d}{dx}f(x)dx = f(x)$

[wraith dagger]

_wherewolf_

[spice kraken]

i forgot the +C but you get the idea

[frail iris]

or in this case =F(x)?

[spice kraken]

yes

exactly

[frail iris]

ok

so

[spice kraken]

that's how we solve the question

[frail iris]

now what do I do with this though?

[spice kraken]

$\int_{0}^3F'(x)dx = F(x) |_{0}^3$

[wraith dagger]

_wherewolf_