help-10

[royal basin]

yes of course they are

[mighty carbon]

i forgot about the theorem

so then x would equal 16

i have one more question, is that okay?

ok

what 🥰

[obtuse pebble]

@mighty carbon Has your question been resolved?

[mighty carbon]

<@&286206848099549185>

[tacit scarab]

the height of the equilateral triangle is 3

use that to find the length

wait no

I meant 9

[mighty carbon]

what?

wait

are we doing the perimeter question

[tacit scarab]

wait what question are you trying to do

[mighty carbon]

I just wanted to check the first three and I didn't know the last one

[tacit scarab]

oh

I think they're right

[mighty carbon]

is the perimeter 9?

[tacit scarab]

nope

[mighty carbon]

i don't know

i need to go soon

[tacit scarab]

consider triangle AEC

you got all of its angles and its height

you can use trigonometry to find it's hypotenuse

[mighty carbon]

I found 9 sqrt3

hello

nvm

...

[tacit scarab]

sorry i was helping others

i think it's 18sqrt(3) though

[obtuse pebble]

@mighty carbon Has your question been resolved?

[flint hull]

Do you still need assistance on this question?

I could help while I am waiting for a helper to help with mine (:

[obtuse pebble]

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[obtuse pebble]

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[mild stag]

hi

[mild stag]

here’s my solution, can someone check b c and d?

it’s doppler effect question

[obtuse pebble]

@mild stag Has your question been resolved?

[obtuse pebble]

@mild stag Has your question been resolved?

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[urban patrol]

does this method work for every 3x3 matrix that has a determinant

[gilded needle]

sure

and every square matrix has a determinant

[urban patrol]

ah

[gilded needle]

beware that it doesn't work for 4x4 or larger matrices

[urban patrol]

yeah i got that

seems too simple to not have been taught first

if it worked for every 3x3 matrix

thank you

.close

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[languid arch]

I need help understanding the monty hall problem

[tacit scarab]

what are you understanding?

[languid arch]

What?

Did you mean to say what am I misunderstanding

[tacit scarab]

about the problem

yes

oh lol

[languid arch]

Ok

[tacit scarab]

typo

[languid arch]

So I watched the scenario with 100 doors and after the first round the host takes away 98 doors

[tacit scarab]

yes

[languid arch]

I get that on the next round the door you chose has a 99 chance of being wrong still but I dont understand how the other door wouldn't have a 99 chance too

[tacit scarab]

the other door havae 99% chance

what do you mean

the original door have 1% of being right and switching the door have 99% of being right

[languid arch]

How

[tacit scarab]

so if you choose 1 door out of 100 doors you have 1% of being correct

[languid arch]

Yes

[tacit scarab]

if you choose 1 door, open all wrong doors and switches

then you will be correct if the original door you choose is wrong

which is 99% of the time

[languid arch]

Yes but the other door only had a 1 chance of being right in the first round as well

[tacit scarab]

at the first round yes

after we open all the wrong doors, the possibility changes

[languid arch]

How

[tacit scarab]

because we eliminated all the wrong doors

if I roll a die, what is the possibility of it being 6

[languid arch]

Yes

1 of 6

[tacit scarab]

if I give you additional info, for example the result is even, it changes the probability

it is called conditional probability

[languid arch]

Yes

But

The chance would still be 1 percent for both doors out of 100

The door you choose and the one not eliminated are essentially the same right?

Like

[tacit scarab]

no

they are not

[languid arch]

It's as if he gave you 2 doors and one was right one was wrong

I don't understand

[tacit scarab]

the doors opened are the WRONG ones EXCEPT the original door and the correct door

[languid arch]

Yes

[pallid river]

hi sorry to barge in but here's a diagram to simplify:

[tacit scarab]

nice

[pallid river]

the probablility of winning if you dont switch is 2/3

[tacit scarab]

and only if you picked the correct door, a random wrong door is opened

[pallid river]

switch*

my bad

bye!

[tacit scarab]

and you have 1% of getting the right the first time

that's probably all we can teach

[languid arch]

I still dont get it

Let's say I chose the car first time

Then the middle goat is gone

It's still between the goat and the car

And vise verse

[pallid river]

initailly:
probability: 1%
probability of failure: 99%

in the end:
2 doors but the first one still has the same probability
but the second door has the remaining 99%

or you can just brute force it

[pallid river]

like in this diagram

but dont do that for 100 doors 💀

also watch a yt video!

[obtuse pebble]

@languid arch Has your question been resolved?

[obtuse pebble]

@languid arch Has your question been resolved?

[obtuse pebble]

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[flat snow]

Hello sorry i was to late to reopen the channel just had to do something suddenly Ill just post my question here again

[flat snow]

Taylor series expansion of this function about a= -3

I managed to get this answer

Here are my derivatives. sorry kinda messy

The thing that was bugging was that this was a practice problem on our textbook and its answer confuses me

[flat snow]

the textbook's answer was similar to this but its denominator had 6•3^(n+4)

i have no idea how that 6 got there

I dont know if the textbook has a wrong answer or im just dumb asf to realize what went wrong in my solution

[obtuse pebble]

@flat snow Has your question been resolved?

[tawny birch]

If u look at ur solution for example, for the first red circled its 5!

and the second circled it didn’t take into account the /3!

[flat snow]

i havent put the n! at the denominator yet

i was solving the f^(n)(x) first

This is my f^(n)(-3)

then i put this in here

[flat snow]

after simplifiying, this is what i got

i still have no idea how there is a 6 in the denominator of the supposed answer in our textbook

oh wait i think i got it

.close

[obtuse pebble]

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[hardy coral]

Could someone explain to me why Void is always a subset of any given set?

[mossy spear]

void as empty set?

[pallid river]

The empty set contains no elements. A set A is considered a subset of another set B if every element of A is also an element of B. So empty set is a subset of any set because it has no elements, and the condition for being a subset is automatically satisfied due to the absence of elements.

thats the best i can explain it

[obtuse pebble]

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[echo jackal]

im having trouble isolating a

[echo jackal]

once i substitute it into the function using f(b)-f(a)

[urban patrol]

show your work

[urban patrol]

it seems like you didnt integrate

[echo jackal]

isnt it f(x)x' - f(a)a'

[urban patrol]

,w integrate -10/x^3

[warm shale]

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[echo jackal]

what about a

[urban patrol]

5/x^2-5/a^2

because F(b)-F(a)

[echo jackal]

how come what i did was wrong

i thought it was f(x)x' - f(a)a'

[urban patrol]

how do you differentiate x

thatw ould make everything 0 cause they should be constants

[obtuse pebble]

@echo jackal Has your question been resolved?

[echo jackal]

1?

im not following

[urban patrol]

x would be a constant here

[echo jackal]

isnt it a variable tho

[urban patrol]

$\int^x_af(b)db=F(x)-F(a)$

[echo jackal]

yes

[warm shale]

Arctic

[urban patrol]

not sure where youre getting x' and a' from

unless im misinterpreting what f(x) is

[echo jackal]

are you able to tell me how u got 5/x^2

[urban patrol]

^

power rule

[obtuse pebble]

@echo jackal Has your question been resolved?

[echo jackal]

Can u go step by step for me

[urban patrol]

-10/x^3=-10x^-3

'

-10x^-2/-2

5x^-2=5/x^2

[obtuse pebble]

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[vocal verge]

You are having a conversation with your friend, who is trying to convince you that since ordinary traffic is about 60 decibels and a rock concert (with massive speakers and amplifiers) is 120 decibels, you don’t need to worry about your hearing at the concert since a rock concert is only twice as loud as normal city traffic.

Using your knowledge of the decibel scale, explain to your friend why this reasoning is wrong.

[vocal verge]

Is the difference supposed to be that big or m I tripping?

Not twice as loud, but 1 million times louder?

$$ I_{0} = 10^{-12} $$

[warm shale]

kai_funaba

[polar fossil]

"how loud is something" is kinda subjective
but yes 120dB has waves that are 1 million times stronger than 60dB

[vocal verge]

.close

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[pearl charm]

I have a few matrix questions

[pearl charm]

<@&286206848099549185>

<@&286206848099549185>

.close

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[quartz wave]

what's the proof that y^2 = ab?

[rich plume]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[quartz wave]

?

[urban patrol]

1 2 3 4 5 or 6

[rich plume]

what have you tried?

[quartz wave]

i dont have any idea, i thought about the pythagorean theory, but it doent work, cause there is multiplications, and not the additions

[rich plume]

You need to use similarity

[quartz wave]

hmm is it like:

[rich plume]

start with labelling any angle as theta and try to find rest of angles in terms of theta

[quartz wave]

$\frac{y}{a} = \frac{b}{y}`$```

[warm shale]

pl_ans_me

[rich plume]

did you find this using similarity or you're just guessing

[timid silo]

use pythagoras theorem on 3 right angled triangles, which will give you 3 eqns from there you can eliminate variables

[rich plume]

Yeah okay

Now just cross multiply

[quartz wave]

I am not really sure if i used the corresponding angles

does this makes sense?:
opposite side to the 🔵 angle opposite side to the 🔵 angle
------------------------ [left part of the triangle] == -------------------------- [Right part of the triangle]
opposite side to the 🔴angle opposite side to the 🔴 angle

does this make sure that I am using the corresponding sides?

.close

[obtuse pebble]

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[late roost]

Hi, I was wondering why we can assume that density is 1 in this example?

[glossy basalt]

to be honest, I don't think it's a good practice to do that, but since it explained that
when we are calculating M_x/M and M_y/M, the constant density would be cancelled out, therefore you can take the density to be any constant without affecting the final answer

[late roost]

Oh, it just devides out of the mass and M_x

thats kinda stupid but yeah i get it

so we just assume its one because it makes no impact on the answer as it crosses out on the final step

[glossy basalt]

we assume it's one because for "the ease of calculation" and yea, it will be crossed out in the final step

[late roost]

Ok thank you so much, that really did confuse me

.close

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[lapis hawk]

how can i increase and decrease percentages for example increase 24 by 5%

[lapis hawk]

<@&286206848099549185>

[glossy basalt]

!15m

[obtuse pebble]

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

[glossy basalt]

and also

[lapis hawk]

ok

[glossy basalt]

if you have done anything, you can post it out so that we can have a look

[lapis hawk]

24+5% of 24 5/100x24 . cancel 100 until 4

[glossy basalt]

okay, i get what you've missed

24 itself is considered as 100%

[lapis hawk]

ohh

[glossy basalt]

so, when we add 5%, we dont just add 5%, we do:

24(100%+5%)

[lapis hawk]

29?

[glossy basalt]

so it's 24 times 105%

[glossy basalt]

nope

[glossy basalt]

try and work out the math

[lapis hawk]

24x105%?

[glossy basalt]

24×105%

[lapis hawk]

ye

[glossy basalt]

24×(100%+5%)
=24×105%
=...

[lapis hawk]

2520

[glossy basalt]

did you forgot something?

[lapis hawk]

percwent

[glossy basalt]

yea

[lapis hawk]

2520%

[glossy basalt]

important note:
% = 1/100

that is

the answer would
=2520%
=2520/100
=....

[lapis hawk]

144?\

the teacher said the answer bruh

ill solve it

[lapis hawk]

.

25x100

2500

remaining 20

[glossy basalt]

yea, but do you know decimals/fraction?

[lapis hawk]

ye

[glossy basalt]

try answer in decimals 👍

[lapis hawk]

2.0

[lapis hawk]

.

[glossy basalt]

ehhhh

2520/100

[lapis hawk]

ohh

2.520\

[glossy basalt]

close

very close

2520/100

252/10

and then?

Baby Yaga~

[lapis hawk]

oh great

[glossy basalt]

yea?

[lapis hawk]

power cut sor

[glossy basalt]

no worries

[lapis hawk]

2.52

[glossy basalt]

nah

[lapis hawk]

25.2

[glossy basalt]

you moved a digit more than it should be

[lapis hawk]

tysm

.close

[obtuse pebble]

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[obtuse pebble]

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[tight wadi]

Is it correct to say that x^2-t, the minimal polynomial of \sqrt(t) on F_2(t) is not separable?
I need to show that the extension F_2(\sqrt(t))/F_2(t) is not a Galois extension.

[kind hawk]

consider the derivative

[tight wadi]

I didn't understand what you mean

[kind hawk]

do you not have a lemma about separability vs the relationship of f and f' ?

you need to argue why x^2-t is not separable

[tight wadi]

Perhaps but I didn't know it then 🤔

[kind hawk]

something about gcd(f, f') ?

I have to go now. head over to #groups-rings-fields and also ask there

[tight wadi]

ok thank you

[obtuse pebble]

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[pearl charm]

Can I get some help with matrix qudstions?

[sage geode]

Express both of b and c in terms of u and v and find their dot product

[pearl charm]

what do you mea

Oh

as in like

b = u + 2v

c = 2u + v

what do I do njow

@sage geode

[sage geode]

Take their dot product

[pearl charm]

and then what

[sage geode]

And then calculate it using the given information

[pearl charm]

ok wait

I have another question

[sage geode]

That's the same question

[pearl charm]

is it?

<@&286206848099549185>

[left basalt]

What's your question?

[candid yarrow]

where are you stuck on this question

[candid yarrow]

the most recent picture

[left basalt]

No shit

[pearl charm]

I just don't know where to start

[candid yarrow]

Do you know how to find the distance between a point and a plane

[obtuse pebble]

@pearl charm Has your question been resolved?

[pearl charm]

No not rlly

how would I do that

<@&286206848099549185>

<@&286206848099549185>

.close

[obtuse pebble]

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[lethal sand]

,rotate

[lethal sand]

something along the thought "oh the derivative of 4x^3-5 is 12x^2"

yes

[obtuse pebble]

@timid silo Has your question been resolved?

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[slender carbon]

the question asked to evaluate

[royal basin]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[slender carbon]

i dont know where to begin i guess

do u want me to show u my teacher's working

[royal basin]

if you want that explained, then sure.

if you don't then no.

[slender carbon]

uh yea sure i saw her working out and i dont get some parts of it

[cunning burrow]

Use the theorem log(a)/log(b) = log_b (a) where log(?) is base 10.Use a calculator to finish off calculations. Alternatively observe 27 = 3^3 and 4^(1/2) = 2. From there you can use the theorem (a^b)^c = a^(bc).

[royal basin]

this is your teacher's work?

[slender carbon]

i dont get why there is 4^2 on third line and also why there is 9^1/2

[royal basin]

$\log_4(8) \neq \log_4(4^2)$

[slender carbon]

yes

[warm shale]

Ann

[royal basin]

your teacher fucked up

[slender carbon]

seriously?

[royal basin]

unless this was some kind of elaborate "sPoT tHe MiStAkE!!!!" thing

8 is not 4^2

4^2 is 16

[slender carbon]

bruh lmaooo no wonder i didnt get it

[royal basin]

anyway

$\log_4(32) = \log_{2^2}(2^5)$

[warm shale]

Ann

[royal basin]

to what power do you need to raise 2^2 to get 2^5?

[slender carbon]

uh multiple 2 cubed?

[royal basin]

doesn't answer my question.

2^2 raised to what power gives 2^5?

$(2^2)^x = 2^5$, what's $x$?

[warm shale]

Ann

[slender carbon]

oh

2.5

[royal basin]

yeah, 5/2.

in the same vein, $\log_9(27) = \log_{3^2}(3^3)$, so what's that equal to

[warm shale]

Ann

[slender carbon]

what does what equal to

uh do i make it so log 3^2 (3^3) turns into 3/2 log 3 3

[slender carbon]

we cant use the calculator is there a way to do it without a calculator?

[cunning burrow]

So I have explain how to do it without a calculator but briefly. Have you read my post?

[slender carbon]

yes i read it i kind of dont get it

[high lily]

apply the same idea as before

3^2 raised to what power gives 3^3?

[slender carbon]

3/2 i think

[high lily]

yes

[slender carbon]

alr do i make it into 3/2 log 3 3

[high lily]

no

by definition
$$\log_{3^2}(3^3) = \frac32$$

[warm shale]

ℝamonov

[slender carbon]

oh okay

[high lily]

as (3^2)^(3/2) = 3^3

[slender carbon]

bringing it back to the original question i just replace log 9 (27) with 3/2

[high lily]

yes

$= \underbrace{\log_{2^2}(2^5)}{5/2} - \underbrace{\log{3^2}(3^3)}_{3/2}$

[slender carbon]

so i can basically apply this technique to the other stuff

[warm shale]

ℝamonov

[slender carbon]

ok i got it

thanks so much

.close

[obtuse pebble]

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[high furnace]

The perimeter of an isoceles right triangle is 20. Show that its area can be written as a + b sqrt (2), where a and b are integers to be determined

[high furnace]

so i assumed it was a right-angled isoceles triangle

and the hypotenuse was y

2x + y = 20

2x^2 = y^2

x^2 = y^2/2

x = y sqrt(2)/2

idk how to do from there

[glossy basalt]

sub it back into 2x+y=20?

[high furnace]

i did but

that would find the perimeter and not the area

[trim portal]

Do you know what would the area be in terms of x and y?

[high furnace]

um

wait im worng

[trim portal]

It would be x^2/2, wouldn't it?

[high furnace]

yeah

im suddenly so confused ;-;

yeah it's both x^2/2 and y^2/2 i think

[trim portal]

So maybe divide both sides by 4, to get x^2/2 = y^2/4

[trim portal]

Just x^2/2, lemme draw a diagram...

[high furnace]

it's a square

x is the side

cut diagonally

y is the hypotenuse

ye it's y^2/4

[trim portal]

I think its x^2/2 my Windows keep casting so I cant draw a diagram :/

[high furnace]

it has to be written as a + b sqrt (2) 😢

[high furnace]

yep

[trim portal]

Area of whole that is x*x

Hence the triangle is x^2/2

[high furnace]

^

[trim portal]

After you do this, try expressing y using 2x+y=20

[high furnace]

x^2/2 = y^2/4 so 2x^2 = y^2

y = sqrt(2x^2)

[trim portal]

Or just use this to find X and y and then find the area

[trim portal]

So y = sqrt(2)x

[high furnace]

yes

[trim portal]

You know: y=sqrt(2)x and 2x+y=20

Now by substitution 2x+sqrt(2)x=20

Can you solve for x?

[high furnace]

uh i'll try

[trim portal]

This is oné equation with oné variable

You dost need anyrhing else

Hint: try factoring out x

[high furnace]

im bad at algebra 😭

x(2 + sqrt(2) = 20

so x = 20/2 + sqrt(2)

so now i need to solve for x^2/2

[trim portal]

yeah, just with parenthesis, $x=\frac{20}{2+\sqrt{2}}$

[warm shale]

methisalwaysright

[trim portal]

so just square both sides and then divide by 2

[high furnace]

x^2 = 400/(6 + 4 sqrt(2) if i'm not wrong

[trim portal]

dividing by 2, $\frac{x^{2}}{2}=\frac{200}{6+4\sqrt{2}}$

[warm shale]

methisalwaysright

[trim portal]

now last step

conjugate

do you know what conjugate is?

[high furnace]

yeah

but my knowledge is very basic

[trim portal]

Great, so multiply numerator and denominator by conjugate of 6+4sqrt(2)

do you know how to do that?

[high furnace]

is the conjugate 4sqrt(2) - 6?

[trim portal]

Yes, this would work too

common convention would be writting the conjugate as 6-4sqrt(2), but it doesnt really matter

[high furnace]

800sqrt(2) - 1200/4 ?

[trim portal]

negative this

so (1200 - 800sqrt(2))/4

[high furnace]

did i multiply wrongly

[trim portal]

$\frac{200\left(4\sqrt{2}-6\right)}{\left(6+4\sqrt{2}\right)\left(4\sqrt{2}-6\right)}=\frac{800\sqrt{2}-1200}{32-36}$

[warm shale]

methisalwaysright

[trim portal]

the denominator would be -4

(a+b)(a-b)=a^2-b^2

[high furnace]

i understand where i went wrong

[trim portal]

so $\frac{1200-800\sqrt{2}}{4}$

[warm shale]

methisalwaysright

[trim portal]

just simplify this

[high furnace]

300-200sqrt(2)?

that's the answer?

[trim portal]

Which is something around 17.157

[trim portal]

That's x^2/2, which is the area

[high furnace]

tysm

[trim portal]

So you proved that the area can be written as a + bsqrt(2) where a=300 and b=-200

[high furnace]

how do i close this or smth

wait don't close yet, gimme a sec

okay, you can close this now

.close

[obtuse pebble]

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[real reef]

yeah gonna need help for this one

[real reef]

@royal basin

[royal basin]

repost the problem

[real reef]

lemme show u what I have first

one second

https://i.imgur.com/PhLeHDP.png

[royal basin]

DO NOT delete your first msg

ok

youve found the third angle is 27° yes

[real reef]

yes

i did 99+ 54= 153 something

then I did 180 minus that

then it worked out to be like 27degrees

Lemme get rough work and show u where im at

[dire plinth]

you can use sine rule

[real reef]

yes but my teacher says pairs

i have like 3 angles and 2 side numbers now

[royal basin]

yes, and the sine law relates two sides and two angles

[dire plinth]

ye

[real reef]

weird

my example looks diff than my book https://i.imgur.com/jLnvND6.png

much easier

[dire plinth]

that literally is sine rule

[real reef]

okay ty i will try to work on it

[dire plinth]

just in this case u gotta find the angle opposite to the 8cm

which u did

then use that for sine rule

[real reef]

https://i.imgur.com/l7qon9Q.png

have this right now

im guessing i use B and A?

[dire plinth]

ye

[real reef]

or all 3?

since i have numbers for all

[dire plinth]

just a and b

theres no point

[real reef]

why A and B?

what about C?

[dire plinth]

u dont even have the length of side c

but even if you did have both since its equal it wouldnt matter

[real reef]

im so confused arnt u sppose to label ur triangle?

[dire plinth]

yeah

[real reef]

https://i.imgur.com/pPiJchr.png

like if i get a question like this

whats A and whats B?

isnt the left side C?

that unknown corner

[dire plinth]

it doesnt matter

[true summit]

you can consider that yourself , use any variable to do so according to me

[dire plinth]

but the opposite should be the same letter just has to be capitalized

[real reef]

i can label anything I want A B C any point?

[dire plinth]

the side opposite to the angle should be c and C

here ill send a photo

[real reef]

yeah ty

i was gonna ask if u were gonna label it how does it look

[dire plinth]

see how the angle and its opposite side have the same letter

just lower and upper case

[real reef]

but it dosent matter where I label the circled right?

https://i.imgur.com/1vTvNnI.png

[dire plinth]

yes

[real reef]

those can be anywhere?

okay ty

the small letters yes I get that

those are opposite to angles

[dire plinth]

yep

[dire plinth]

just as long as you understand

that the label is the angle itself

anyway back to the question

[dire plinth]

so here

oh my days i didnt see the C on the left side

youd use b and c

so using the sine rule, whats ur equation?

[real reef]

im literally labeling the triangle same as u

and idk how ur using B and C

[dire plinth]

nah ur not

in ur triangle u put C on the left

[real reef]

lemme redraw what u have

to see if i get the same thing

[dire plinth]

okay

so assuming u use my labels

youll use a and c

since you have angles A and C and have side a

[real reef]

i show u what i have so far

https://i.imgur.com/yxwFM3x.png

[dire plinth]

you have angle A though

[real reef]

what are the two pairs

im not seeing it all

[dire plinth]

your equation is right

[real reef]

i must be blind

[dire plinth]

but you already have A

so idk why u have sinA

[real reef]

oh im using A and C

[dire plinth]

yeah

[real reef]

oh now i see it

okay lemme try again

[dire plinth]

okay

[real reef]

8sin54 over x sin A

[dire plinth]

no

well yes but again

you have A

[real reef]

but angle A is pointed towards 8cm

[dire plinth]

yeah ur meant to use that angle

[dire plinth]

here

you solved it back here

[real reef]

oh x over sin54

actually idk

im just gonna get marks on cosine

and forget about sin

.close

[obtuse pebble]

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[real reef]

take care

I understand cos

thats easy

just sine is wack

[obtuse pebble]

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[rough stratus]

First one

[rough stratus]

,rccw

[rough stratus]

What is wrong with this method?

,rccw

[rough stratus]

Isn't the sum -b/a?

[delicate drum]

tanAtanB=-1

[delicate drum]

yes yes mb

[rough stratus]

Ohh ok

Thanks man @delicate drum

.close

[obtuse pebble]

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[solid harness]

any one can help me

[buoyant ingot]

What

[polar fossil]

it might help to substitute the definition of f into the definition of g and simplify

[obtuse pebble]

@solid harness Has your question been resolved?

[obtuse pebble]

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[celest hazel]

Someone explain how to solve this pls

[celest hazel]

X^3-x^2<0

[royal basin]

< or ≤?

[celest hazel]

The later

Do I do the first one or the second one

[celest hazel]

I’m stuck with this pls help idk what to do

@royal basin

[royal basin]

your sign table is shoddy

make it less so

[celest hazel]

Do I need Ruffini for this

[celest hazel]

It doesn’t matter cuz I checked and by doing both I don’t get the answer

Do I need Ruffini?

[royal basin]

what's ruffini

[celest hazel]

Idk how u call it in English but it’s this

U use it when ^n>2

[royal basin]

busy now honestly

also no you dont need that

[polar fossil]

oh we usually call that synthetic division (for some reason...)

I can't really read your sign table

you have the right roots

[celest hazel]

The sign table is wrong I tried

I don’t know what to do

[polar fossil]

your equation factors as x^2 (x - 1) right?

[celest hazel]

Ye

[polar fossil]

so there's a double root at 0

it just kisses the axis it doesn't cross there

[celest hazel]

Sorry I don’t know what a root is, I study math in Italian could u pls show me

[polar fossil]

just a sec

[celest hazel]

Ty

[sacred lichen]

where the y value is 0

[celest hazel]

Ok but how do I solve the inequality

What does the double root mean in the sign table

[polar fossil]

means that y=x^2 is always either positive or 0

[celest hazel]

So in the sign table I just put +++

[polar fossil]

yeah on the top row

[celest hazel]

Instead of +—

Ok

Ty

But if the exponent is 3

For example

What do I do

Do I put +—?

[polar fossil]

yeah if the exponent is odd then it crosses

if it's even then it doesn't

[obtuse pebble]

@celest hazel Has your question been resolved?

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[obtuse pebble]

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[finite turtle]

Hello. I am a bit stumped by this question. I have no idea how I’m supposed to find roots of these partial derivatives for q9 in my textbook which is asking for the critical points and what is their nature I.e max or min or saddle and what is the value of D.

[finite turtle]

I have no idea how to pull roots from equations like this. I want to say do simultaneous eqns but I’m like that might be wrong and just complicate it further

Never solved for roots like this before with an eqn like this with a xy term together

After differentiating I mean

I don’t want to keep going encase I keep messing it up further

[spring trail]

it might be messy but

doesnt the quadratic formula work

[finite turtle]

I don’t know

Because of the combined xy term

[spring trail]

well in the first equation find x

then replace in the second

surds tho

if u dont want them then try something else

[finite turtle]

That was my first thought but we have a problem of 1/y in the eqn

If you try and pull factors n stuff to try and get rid of the xy term

For the x derivative

Well divide by y in this case

I’ve done this but I’m not sure this is correct

This is the Fx eqn btw I’m playing around with. Like I’m just unsure about this.

And that did not work

Hang on

Simultaneous eqns does work

Lemme see if this works

It does work

Thank you though.

.close

[obtuse pebble]

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[hybrid tinsel]

If (-2,3) lies on the terminal side of an angle theta whose vertex is at the origin and inital side is along the positive x-axis, then costheta = ?

[hybrid tinsel]

answer should be in fraction form as well

i tried to solve for an angle in a 2 3 sqrt 13 angle

and then subtract 180 from that angle

and then take cos of that

but the answer is not a fraction

[rotund delta]

You were on the right track initially. cos(180-theta) is known, i.e. base/hyp = 2/root(13). After that, you can find cos theta. 2/root(13) is a fraction, because fraction doesn't necessarily need to be rational, its an ambiguous term

[hybrid tinsel]

after getting -2/sqrt13 as cos of 180-theta, what should i do?

[rotund delta]

2/sqrt 13 is cos(180-theta), don't get confused by the negative x axis, you only consider length in ratio

[hybrid tinsel]

ah your right

i was thinking about it in the context of the unit circle, my fault

[rotund delta]

No worries

What do you think cos(180-theta) should be, its straight forward if you understand the unit circle

[hybrid tinsel]

oh

wouldnt it just be the same

or the negative

i mean

[rotund delta]

yes

negative

-cos theta

cos is negative in the second quadrant

[hybrid tinsel]

yeah

wait so

the answer would be -2/sqrt13

right

[rotund delta]

yes

[hybrid tinsel]

ty

[rotund delta]

wlcm

[obtuse pebble]

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[mental cosmos]

what is the derivative of f(H*G) with respect to x in H where H,G are both tensors and * denotes convolution?
i gave the chain rule an attempt but
d/dx f(H*G) = f'(H*G) d/dx H*G
what goes between f'(H*G) and d/dx H*G? it cant be scalar multiplication they're both tensors
f is a function that takes a tensor and returns a tensor

[obtuse pebble]

@mental cosmos Has your question been resolved?

[restive acorn]

Would it also be convolution?

I dont know tensor calculus but if the chain rule extends nicely to it then the * operation should be preserved

Wish i could help otherwise

[mental cosmos]

im really not sure :/

if it was convolution it would just become a dot product with one of them flipped 🤔

cuz f'(H*G) and H*G have the same dimensions

[restive acorn]

Well if * is the "equivalent of multiplication" for these tensors that would be my first guess

@mental cosmos i looked up the definitions a little and it seems im way out of my league offering advice xD

Hope you get someone that can help

[mental cosmos]

all good, thanks 😄

[obtuse pebble]

@mental cosmos Has your question been resolved?

[pseudo swift]

not so sure about that, if you think about the 1D case, f takes a vector and gets out a vector, f' would be the jacobian matrix

[mental cosmos]

i define f myself

its an activation function basically

only changes the values of the input tensor

[pseudo swift]

I mean you said it takes a tensor and outputs a tensor

I just took a special case of that to show that your logic doesn't hold

[mental cosmos]

my bad

f doesnt change dimensions of the input tensor

[pseudo swift]

yeah my comment still stands

[mental cosmos]

wdym

[mental cosmos]

this one?

[pseudo swift]

yes

[mental cosmos]

f and f' shouldnt change the dimensionality

atleast in my use cases

f could be sigmoid for example

sigmoid applied to each entry in the tensor

f' is sigmoid's derivative applied to each entry in the tensor

[pseudo swift]

you should have said that you apply your f element-wise earlier :/

[mental cosmos]

sorry, im not very familiar with tensors, i didnt know there could be cases where f doesnt?

[pseudo swift]

even then, if we're being strict about it, the derivative of a function R^n -> R^n is still a matrix

[mental cosmos]

u mean if f was some sort of linear transformation or something?

[pseudo swift]

in your case most entries will be 0 though yes (the non-diagonal ones)

[mental cosmos]

i didnt think about that

[pseudo swift]

just some function R^n -> R^n, doesn't have to be linear

[mental cosmos]

so is there a way to apply the chain rule to f(H*G)?

been looking for a few hours didnt find anything

i can provide my work so far

[mental cosmos]

i mean i was stuck at that part

but i dont think my notation is good lol

[pseudo swift]

yeah, if we stay on the 1D case for simplicity, let's say n is the simension of H*G, f'(H*G) is a nxn matrix (with a bunch on zeros true), d/dx H*G is a n-dim vector
if we do the matrix product f'(H*G) * d/dx H*G, we get the composite derivative we want (it's like using the multivariable chain rule a bunch of times)

now I'm not saying this is efficient or anything, that's just what we get when we stay on the laid-out multivar calculus path

[mental cosmos]

im not sure why f'(H*G) would be a matrix

f' applied to a 1d tensor would produce a 1d tensor

[mental cosmos]

i know, im only considering the math, not optimiziation, not yet

[pseudo swift]

the usual defn of derivative of a vector input, vector output function is that it's the matrix of partial derivatives of all the outputs wrt to all the inputs

[pseudo swift]

it is true that most entries would be 0 cause you apply the same function element-wise, so you could consider it as pretty much a "1d tensor", but you're already optimizing your computation then

[mental cosmos]

hmmmmm

what happens when the tensors are 4d?

we cant use "matrix product"

[pseudo swift]

https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant btw

this is like the generalization of the gradient when you have multiple outputs to your function

[pseudo swift]

idk prolly some kind of "higher-order" matrix-vector product

[mental cosmos]

i thought that was convolution, lol

or cross correlation

not sure

cuz convolution can be considered a matrix multiplication in the 2d case

[pseudo swift]

idk a lot about tensors tbf, just wanted to work a special case to see what we could learn from that (and maybe generalize)

[pseudo swift]

that's not even a mouthful, that's a complete bucket of stuff

why are you trying to do that by hand ? just wondering

like I've done that kind of backprop computation for more modest networks

but doing it for a convnet, that's brave

[mental cosmos]

hmmm dunno seemed like the appropriate way to do it

might try another approach tho

every blog seems to just convert convolution into nested sums

probs gonna do just that

[pseudo swift]

yeah I'm not saying it's a wrong approach, it's just gonna be hard to do by hand I guess

[obtuse pebble]

@mental cosmos Has your question been resolved?

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[obtuse pebble]

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[late roost]

Hi, I keep getting this problem wrong and I have no idea where I am messing up, can anyone take a look?

[late roost]

.close

[obtuse pebble]

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[strong python]

i dont even have any work to put here i been looking at this for like 10 minutes but i got no clue where to start

[dark stirrup]

Does you recall the definition for basis?

[strong python]

set of linearly independent vectors that span a vector space

[dark stirrup]

I will assume you know vector space and span definitions as well then

[strong python]

yeah i do

[dark stirrup]

cool

So you need x to be a vector such that, for any vector that exists in U, it can be written as a linear combination of the three vectors in your basis.

Any guesses on how to start here?

[strong python]

not really
im not entirely sure how to use the span that they gave me in the problem; ig i could write out 4 equations with the coefficients id be using for each vector from it?

[dark stirrup]

That sounds like a good start yeah.

You may need just 3. Been a while since I did this, so I could be wrong there.

[strong python]

like
-3a - b - 8c = 0
8a + 6c = 0
-3a + 2b + c = 0
2a - b - 5c = 0

maybe i solve this and determine the coefficents' relationships to each other?

[dark stirrup]

That's definitely one way I believe. You can create your own basis that way, and then maybe transform it into the basis that the answer wants, but that may take some extra work.

[strong python]

do you have another way in mind

[dark stirrup]

I'm working through it right now to see if it works.

In the meantime, here's a free <@&286206848099549185> ping while I work on it.

[strong python]

ty

[obtuse pebble]

@strong python Has your question been resolved?

[strong python]

youre still working on it right?

[dark stirrup]

Yes. But I'm at work too.

How's your progress?

[strong python]

i solved for a b and c and all i found was the trivial solution which doesnt really help me i think

[restive acorn]

Its a change of basis type of problem i think

Its been a long time since i did something like this but if you review your notes or book look for change of basis

[strong python]

i will

[restive acorn]

Im sure you'll find an example like this if you have notes

[strong python]

im looking in my book and this is actually a couple sections ahead lmao (ty prof) so i dont have any notes but i can read up on it

this actually doesnt seem like that

change of basis assumes i already have both bases

[novel grotto]

Yeah this is wrong. All the bases of a vector space have the same number of elements which is the same as the dimension of the vector space

[dark stirrup]

Yeah it's been forever

[obtuse pebble]

@strong python Has your question been resolved?

[strong python]

i got it

.close

[obtuse pebble]

Channel closed

Closed by @strong python

Use .reopen if this was a mistake.

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[pure glen]

Let x,y,z sastify the thing you can see on the first line. Don't care about part a because I just solve it. Find the minimum value of P (in part b)

[pure glen]

x,y,z is real positive numbers

[obtuse pebble]

@pure glen Has your question been resolved?

[pure glen]

<@&286206848099549185> , @left basalt

[wooden cipher]

ok im back

@pure glen you around?

[pure glen]

Yes

Can u solve part b

It is so difficult

@wooden cipher

<@&286206848099549185>

[wooden cipher]

oh sorry got distracted

first thing you wanted to do is look at 2P

like ive said before

this will get you $P=\frac{2x+2}{2x+1}+\frac{2y+2}{2y+1}+\frac{4z+6}{4z+2}$

[warm shale]

garlicbredfries

[wooden cipher]

so then you can simplify the fractions a little bit by changing the improper fractions to mixed fractions

@pure glen

[pure glen]

👍

[wooden cipher]

ok, so what do you have now?

[wooden cipher]

this should be 2P by the way, my mistake

you were close, there was just a 6 in the numerator of the last term

[pure glen]

$2P=\frac{1}{2x+1}+\frac{1}{2y+1}+\frac{4}{4z+2}+3$

[warm shale]

mendeleevpro

[wooden cipher]

now another thing you can do is the z term has a common factor 2

so you can clean it up a little bit more

[pure glen]

$2P=\frac{1}{2x+1}+\frac{1}{2y+1}+\frac{2}{2z+1}+3$

[warm shale]

mendeleevpro

[wooden cipher]

ok, now im gonna divide everything by 2

$P=\frac{\frac{1}{2x+1}+\frac{1}{2y+1}}{2}+\frac{1}{2z+1}+\frac 32$

[warm shale]

garlicbredfries

[wooden cipher]

i want you to use am-gm on the big fraction

[pure glen]

$P\geq\sqrt{\frac{1}{2x+1}+\frac{1}{2y+1}}+\frac{1}{2z+1}+\frac 3 2$

[warm shale]

mendeleevpro

[wooden cipher]

not quite

[pure glen]

hmmm ...

[wooden cipher]

remember, its (a+b)/2 >= sqrt(a*b)

[pure glen]

oh I forgot

$P\geq\sqrt{\frac{1}{2x+1}\frac{1}{2y+1}}+\frac{1}{2z+1}+\frac 3 2$

[warm shale]

mendeleevpro

[wooden cipher]

do the first two terms look familiar now?

also use parentheses please

[pure glen]

$P\geq\sqrt{\frac{1}{(2x+1)(2y+1)}+\frac{1}{2z+1}+\frac 3 2$

[warm shale]

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

[pure glen]

it sucks

[wooden cipher]

$P\geq\sqrt{\frac{1}{(2x+1)(2y+1)}}+\frac{1}{2z+1}+\frac 3 2$

[warm shale]

garlicbredfries

[wooden cipher]

there you go

im going to simplify the square root a little bit

$P\geq\frac{1}{\sqrt{(2x+1)(2y+1)}}+\frac{1}{2z+1}+\frac 3 2$

[warm shale]

garlicbredfries

[wooden cipher]

do those first terms look familiar

[pure glen]

yes

[wooden cipher]

so that means?

[pure glen]

P >= 2/3 + 3/2

[wooden cipher]

indeed

[pure glen]

so P >= 13/6

[wooden cipher]

ye

[pure glen]

we have to find the condition for P = 13/6 can happen

[pure glen]

@wooden cipher ?

[wooden cipher]

oh we have to do that too?

uhh by observation x=y=z=1 works

but i dont know how to rigorously prove that

[pure glen]

,w solve (x+1)/(2x+1)+(y+1)/(2y+1)+(2z+3)/(4z+2) = 13/6 and sqrt(1+4xy+2x+2y)+2z=5

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by mendeleevpro#0

[pure glen]

,w solve for x,y,z if (x+1)/(2x+1)+(y+1)/(2y+1)+(2z+3)/(4z+2) = 13/6 and sqrt(1+4xy+2x+2y)+2z=5

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by mendeleevpro#0

[pure glen]

I guess P = 13/6 when 1/(2x+1) = 1/(2y+1) <=> x=y and 1/(2y+1) + 1/(2z+1) = 2/3

@wooden cipher

[wooden cipher]

another thing you can do with x=y is plug it back into the original equation

then you later find out y+z=2

so i guess you can do the substitution y=2-z

then solve for y

[pure glen]

yeah

the solution for 1/(2(2-z)+1)+1/(2z+1)=2/3 is z=1

so y=2-z=2-1=1

and x=1

[wooden cipher]

cool

[pure glen]

the problem is solved

.close

[obtuse pebble]

Channel closed

Closed by @pure glen

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[brittle bobcat]

Hi! Is this channel available?

[shadow tangle]

Yep

[nocturne minnow]

No I'm stealing it... Jk

[brittle bobcat]

I have an exercise from the Schaum's of Calculus, it is about finding area under a curve, it is a case that our teachers have not explained to us, and I've been trying to solve it but i don't think i have all the knowledges that it need to solve it. I found the area of ​​the circular sector using that formula, but I don't know how to find the area by integrals. I'll send a picture now of the exercise as it is solved in the book and another picture of my progress with it, i'm from a Latin-American country so my notes are in spanish but the exercise is in english so you can help me, i hope🥲

[native inlet]

did you send the picture? :)

[brittle bobcat]

that's the exercise

this is what i did xd

[shy salmon]

Okay, so let me give you a visualization.

So, we want to find the smaller area of the circle.

Now, we want to decide between intergrating along the X or Y axis. For this, we'll choose X.

I do see that you constructed your integral, and it seems almost correct: you might want to watch out that the integral only computes the area of the area above/below the x-axis, not both.

And since the area above and below the x-axis looks pretty much the same (and in theory, it is.) We can just multiply our integral by 2.

Now, let's see the integral we have to work with here.

(I'm using styling common in North America, so it might be different from what you are used to.)

So far so good?