help-0

[pallid scarab]

yeah so you gotta account for the change in the inner function too

so here

when we look at df(x)/dg(x)

we can try to see "g(x)" as an inner function

that has its own rate of change

that's why we write

$\frac{df(x)}{dg(x)} = \frac{df(g^{-1}(g(x)))}{dg(x)}$

[ocean seal]

rafilou2003

[pallid scarab]

now chain rule (and inverse derivative rule) can be applied

gives you $f'(x)/g'(x)$

[ocean seal]

rafilou2003

[thin swallow]

got iy

it8

it*

Thank you

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[wicked kernel]

ive got the speed as 15.28 m/s

[alpine sable]

The total distance the train has to cover is 2230 m

The speed of the train in m/s is 15.2777

Then we can apply the formula speed = distance / time and get the answer

The answer would be 145.97

The question time taken by the train to cross the bridge completely

Meaning the last part of the train should also cross the bridge

Hence we have to add the length of the train to the length of the bridge

[alpine sable]

That is correct

Or we divide the time by dividing it by 15.28 then we get 145.94

Hope u understood @wicked kernel

[wicked kernel]

i got it ty

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[fluid thorn]

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

can anyone help me with this question?

@fluid thornhi

[cedar kelp]

!help

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[hot vale]

not sure how to approach b) mind is totally blank

[alpine sable]

U got a?

[hot vale]

well i think my explanation for a was poor

Basically made the horizontal like from W called l

and then i showed how both lhs and rhs = 1/l

[knotty cave]

Hi, I just joined the server because I’m having trouble with precal hw. What chat should I use for that

[cedar kelp]

!help

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[alpine sable]

So for a is correct

[hot vale]

is my a correct?

well i am cueless for b tbh

[alpine sable]

A is right, it is just basic trigo for a tan alpha = H/l ans tan 2alpha = H-X/l

So that is correct

For b, lemme think

@hot vale u know tan2alpha formula

[hot vale]

yeh

[alpine sable]

What is it

[hot vale]

2tana/1-tan^2a

[alpine sable]

We know tan a=h/x

Wait no

Tan alpha = H/l

Ohhh done

I think I got it

[hot vale]

can u explain

[alpine sable]

Yeah u want the workings

?

[hot vale]

can u explain what u did?

[alpine sable]

Yes one sec

Check this out

Substituted tan alpha as H/l

In the last second step took h as common

@hot vale

[sour mica]

!nosols @alpine sable

[lone heart]

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

[sour mica]

just guide the person along, let them solve by themselves

[alpine sable]

Ohh okay

Sorry

[hot vale]

thank u mate i understand what to do now

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[idle thorn]

Hello I am confused on how to get from the equation #1(left) to equation (right)

[idle thorn]

.rotate

[cedar kelp]

"," not "."

[idle thorn]

OPPs thanks for telling me

[idle thorn]

,rotate

[cedar kelp]

they factored out x-2 from the numerator and x+1 from the denominator

[idle thorn]

Sorry wha how

[cedar kelp]

$x^{2}(x-2)-1(x-2)=(x-2)(x^{2}-1)$

[ocean seal]

77²

[idle thorn]

Sorry sir I don’t get it

sorry I am just a bit dumb

but I want to be smarter

[cedar kelp]

have you ever factored anything?

[idle thorn]

No

[onyx swallow]

do you know the distributive property of numbers?
ac+bc=(a+b)c?

[idle thorn]

Yes

Is it like this sir

[onyx swallow]

ok now think of (x-2) as the c chunk, x^2 as the a chunk, and -1 as the b chunk

[idle thorn]

,rotate

[onyx swallow]

what they did there is exactly the distributive property

$$(x^2)(x-2)+(-1)(x-2)=((x^2)+(-1))(x-2)$$

[ocean seal]

qwertytrewq

[onyx swallow]

i put every chunk in brackets so it might be easier to follow

[idle thorn]

Thank you

[onyx swallow]

do you get it now?

[idle thorn]

I am trying to process

I processed sir

Thank you I get it now!

[onyx swallow]

the takeaway here is that the a,b,c chunk can be as complicated as you want, it still holds

[gray isle]

the -1 wasn't distributed to the (x-2) correctly

[idle thorn]

I think it’s my first time seeing this method

[gray isle]

its an application of

ac+bc=(a+b)c

[onyx swallow]

they should teach this a lot earlier then lol

[idle thorn]

Thank you sir once again! Have a lovely day or night !!👍

[gray isle]

where instead of expanding, you're going in the other direction

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[wild kindle]

Does anyone know how I should proceed with this problem?

[cedar kelp]

why are you even taking the zero

[winter light]

You should have a 3k in the sum, not 3n

[cedar kelp]

sum of first zero numbers of the form

[winter light]

Because k is the varying variable

[wild kindle]

Ohhhh thanks

[marble shale]

i swear that i can solve this when i was in elementary

...

[small lance]

Good

[small lance]

!done

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.close

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[torpid dirge]

the reason X doesn't deformation retract onto any point not on the bottom line is that we can find every neighborhood is not path connected? thats what i got from searching but i dont understand why that is a reason

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[torpid dirge]

i got it

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[alpine sable]

okay so this might be simple but still forgot how it works, it says (Rewrite as an irreducible fraction.)

[night tangle]

a is 2/11

b is 3/10

[coral flower]

u can just write the bigger line as the division sign , so a is
10/11 divided by 5

[alpine sable]

so 2 / 11?

[alpine sable]

thanks

[alpine sable]

this i dont understand tho

[night tangle]

itll be 4/5 divided by 8/3

which becomes 4/5 multiplied by 3/8

[alpine sable]

okay

so just to get it right

in the first one, you just devide the 5 with 10 adn then say 2/11

and in the other you devide 8/3 and 4/5 ??????

[night tangle]

ill send u a worked out soln

[alpine sable]

okay, thanks

[alpine sable]

thanks alot

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[torpid dirge]

now onto part b. i dont get why it can be contractible but doesn't deformation retract onto any point in Y?

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@cerulean grove Has your question been resolved?

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[lost pulsar]

the time required to finish a job is a function of the number of people working on the job. on average, p people can finish painting a house in j(p) hours, where j(p) = 75/p (assuming that they do the same amount of work). what is the domain and range of the function?

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[opaque pine]

Please help!

[exotic canopy]

okay, what's the speed of light?

(and hence the speed of the photon)

[cinder compass]

can light even bend around earth

[azure needle]

Convert the circumference to metres

[azure needle]

I think it can't

[exotic canopy]

yes, light orbits the earth due to gravity!

[cinder compass]

surely not orbit

maybe bends a bit

[opaque pine]

Please I just need the answer

[lone heart]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[exotic canopy]

oops

[azure needle]

Lol

[opaque pine]

Ok

.close

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[exotic canopy]

do you know the speed of light

oh well

[rancid scroll]

What if it were a black hole whose event horizon was the same as the circumference of earth

[tardy drift]

🤓👆

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[north grail]

hey could anyone help me with this

[cinder compass]

show ur work

[north grail]

my work?

as in?

[hoary nimbus]

Ur working till now

[north grail]

oh

i havent done it

i tried to

but i couldnt understand one thing

so for the second part 3(a^6b^3)1/3

i understood that (i think)

but i what i needed to know

is that since the 4a^2 b^3 has a negative integer

do i take it as 1/4a^2 b^3?

due to reciprocal

[versed yacht]

I need help with
"if 1A × A = 9A, A equals to?"
(Answer is 6 in answersheet but i just dont get it how)

[north grail]

.close

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[north grail]

its fine

ill solve it myself

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[frigid yarrow]

hi , i am confused on the following statement proof : the perpendicular line to the contact point of an ellipse and his tangent on that point biject the corner from the triangle FMF' where F and F' are the foci's of the ellipse and Μ a point of the tangent . In the proof teacher said we will use the slopes of the perpendicular and tangent line ?how am i procceding the proof any ideas ?

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[frigid yarrow]

<@&286206848099549185>

[alpine sable]

can sumone help me please all the ais have given up on me the missing number is 3 <@&286206848099549185>

[rocky roost]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[alpine sable]

sorrry im new here !

[rocky roost]

Np btw u can't ping helpers more than once and u can't ping helpers before 15 min

[frigid yarrow]

<@&286206848099549185>

[alpine sable]

u already told me that in another channel man we get it thanks

[frigid yarrow]

delete your messages

[frigid yarrow]

<@&286206848099549185>

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[west hollow]

How’s everyone

[junior kraken]

Fine

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[brittle lake]

67

[brittle lake]

I’m not sure where to start can I get some assistance

What does it even mean can a exist

[stark crater]

It is asking if there exists a value of a such that the limit exists, not a existing

[brittle lake]

How can I prove that it does or doesn’t?

It obviously exists

Otherwise it wouldn’t be a question

So how would I do this

[stark crater]

What happens to the denominator as x goes to 2

specifically from the left

[brittle lake]

Approaches 4

[stark crater]

Nvm I misread the limit

What happens when the denominator approaches -2

[brittle lake]

O

[stark crater]

Yeah

Now lets imagine the numerator went to some finite positive value. What would the total limit be?

[brittle lake]

Anything

[stark crater]

It would be a specific “value”

[brittle lake]

0

[stark crater]

It could possibly go to either infinity,-infinity or “both” (meaning it wouldn’t exist)

[brittle lake]

Yes

It is indeterminate

[stark crater]

The point being that we need the numerator to approach 0 if we want the limit to exist

That makes the limit have an indeterminate form

So you should find a value of “a” such that 3x^2+ax+a+3 approaches 0 when x approaches -2

[brittle lake]

Why 0 though?

Indeterminate can mean anything

[stark crater]

Yeah indeterminate can still mean the value may not exist, but it also may exist. If we have any other number (not 0) divided by 0 it means the limit does not exist for sure

So we want the numerator to go to 0 because that is the only hope for a possible value of the limit existing

And the denominator always goes to 0 no matter what our choice of a is

[brittle lake]

Would I just place values in an and x be 2?

And when it’s undefined that’s my answer?

Or when it’s 0

[stark crater]

Yeah you should let x be 2 in that part. Basically you want to solve 3(2)^2+ax+a+3=0

This will give you a possible value of a

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[earnest mantle]

the site is saying the final answer i have there is incorrect, i also tried 60-ln(27) which is also incorrect (240/4 simplified)

[proven leaf]

,w int 1 to 27 of cbrt(x)-1/x

[ocean seal]

Results provided by WolframAlpha

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

Requested by xxmrfancyu2xx

[proven leaf]

why is 60-ln27 wrong?

[tidal lichen]

Yea, (243-3)/4 is 60

[earnest mantle]

does the capital L in ln matter for the site?

its webassign

[proven leaf]

Ln and ln can mean different things

[earnest mantle]

ok that worked thanks

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[vagrant agate]

How do i do this with chain rule?

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[pastel swallow]

So i got this to be

= 1 - P(C1 ∪ C2 ∩ C3')
= 1 - P(C1) - P(C2 ∩ C3') + P(C1 ∩ C2 ∩ C3')
= 1 - 0.25 - (0.25)(0.75) + (0.25)(0.25)(0.75)
= 0.609375

Then I got a little funny and did it with another way
Let A = C1' ∩ C2'
Then P(A ∪ C3) = P(A) + P(C3) - P(A ∩ C3)
= (0.75)(0.75) + 0.25 - (0.25)(0.75)(0.75)
= 0.671875 

which of these is correct and why

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[pastel swallow]

<@&286206848099549185>

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[sour verge]

g(9) + f(9) is the same as (g+f)(9)

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[nocturne thistle]

i need help with number 2

[nocturne thistle]

this is my work and notes

answer is 926 but i got 233

[lime yew]

we are considering (2a+b)^22, not (a+b)^22. How would that change the answer?

[nocturne thistle]

are the 22*(22-1) changed to 44*(44-1)?

wait no

the a i put should be substituted for (2a)^2

thanks

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[twin crystal]

Hey can someone help me start this problem. No idea how to start the proof, how to start forming an inequality to show this result?

[sly stump]

Wait a sec

[twin crystal]

This is the arithmetic mean greater than geometric mean inequality right?

[sly stump]

This is true for $x_1, x_2, x_3, … , x_n$ positive real numbers

[ocean seal]

Randel_

[sly stump]

Yes

[twin crystal]

Ok starting from here, I suppose it suffices to prove that S greater than or equals to cubic root of S1S2S3, due to above result.

Then how to relate the areas? Should I name triangle sides as a b c and use ab.sinx formula for area or…

[grand tiger]

Ok so we know that S>S1+S2+S3

Now let S1 + S2 + S3 be x

[twin crystal]

Mm, is that correct? How can we say for sure S > S1 + S2 + S3 ?

[grand tiger]

Since S>x we can say S>x/3

And x/3 is greater or equal to 3/(1/S1 + 1/S2 + 1/S3)

[rare torrent]

That is what you have to prove to solve the problem (a way)

[grand tiger]

So S is greater than that

[grand tiger]

It's obvious

There are 4 triangles

[twin crystal]

S is the area of the middle triangle. Not full area

[grand tiger]

Oh fuck

[rare torrent]

😭

I mean you can rearrange the formula to S/S1 + S/S2 + S/S3 greater equal to 3, which means that on average, triangle s is greater or equal to the other triangles

So if you can prove that the average value of triangles S1, S2, and S3 is less or equal to that of S, you solve the problem

[twin crystal]

Average value of S1, S2 and S3 is (S1 + S2 + S3)/3 right? But that being less or equal to S, is not the same condition as S/S1 + S/S2 + S/S3 greater or equal to 3, isn’t it?

[rare torrent]

How it not?

Because lets say we average s1 s2 and s3

Then if we have S/that itll be greater or equal to one

And since its three of those, 3*1 is 3

So that would translate to the average of S1 s2 and s3 being less or equal to that of S

Get it?

[rare torrent]

Also thats not what im saying

Im saying (avg s1 s2 s3) less equal to s

[twin crystal]

When you say average of S1 S2 and S3 what exactly that is?

[rare torrent]

You add up the areas and divide by 3

[tardy drift]

am hm inequality

(S1+S2+S3)/3 ≥ 3/(1/S1 + 1/S2 + 1/S3)
S1+S2+S3=S+S1+S2+S3-S=ar ABC
S1+S2+S3 ≥ 3S
S≤S1+S2+S3 /3
from inequality
S1+S2+S3/3 ≥ 3 / 1/S1 + 1/S2 + 1/S3

[twin crystal]

Ah this makes sense now, so from am hm inequality, we get this,
S1+S2+S3/3 ≥ 3 / 1/S1 + 1/S2 + 1/S3

So, then I need to prove simply S is greater than or equal to S1 + S2 + S3 divided by 3, or the average of those three, like @rare torrent said

For some reason, it kinds of makes sense S needs to be greater than those three triangles’ average… how to show that?

[tardy drift]

consider S1 S2 S3 to be related to S through some proportion ki
where 0<ki<1 for each i, and k1+k2+k3≥3 (triangle inequality applied to areas)
S1=k1.S S2=k2.S S3=k3.S
(S1+S2+S3)/3= S(k1+k2+k3)/3
since k1+k2+k3≥1
S1+S2+S3 /3 ≤ S
this shows the area of S is atleast as large as the average of the areas S1,S2,S3

[lone heart]

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[mortal trellis]

what happens if you for example use row 2 to eliminate the a^2 in row 3

and then use the first row aswell

[lethal belfry]

Start off with column operations, perhaps

[mortal trellis]

column operations are also not a bad idea, true

[quasi vector]

$$\begin{vmatrix}
-bc&b^2+bc&c^2+bc\
a^2+ac&-ac&c^2+ac\
ab-ac&b(a+b+c)&-(c^2+ac+ab)\end{vmatrix}$$

[ocean seal]

kheerii

[quasi vector]

I tried a lot of operations and like, this doesn't look nice at all to me

[mortal trellis]

whoops made a sign mistake in my head. yeah that wont work

[quasi vector]

well the matrix is kind of symmetric so i think row and column operations should be the same

[mortal trellis]

thought so too

but seems like column ops might be different

[quasi vector]

I can cancel the bc i guess

if I add C1 to C2 and C3

$$\begin{vmatrix}
-bc&b^2&c^2\
a^2+ac&a^2&(a+c)^2\
a^2+ab&(a+b)^2&a^2\end{vmatrix}$$

[ocean seal]

kheerii

[livid sage]

are you looking for an actual proof using row operations

[quasi vector]

well yeah I need to calculate the determinant

[livid sage]

this seems very guessable

[quasi vector]

it's (ab+bc+ca)^3 i suppose

[livid sage]

i would believe it

[quasi vector]

but I do need to prove it

[livid sage]

there should be third powers in there and the final result is a 64

[livid sage]

fun fun fun

[quasi vector]

any ideas?

[livid sage]

probably nothing you haven't already tried

[quasi vector]

[livid sage]

wait no i've seen this trick before

hold on

ok yeah this is definitely legit

[quasi vector]

oh?

[livid sage]

the idea is you abuse symmetry at the start

between rows and columns

i.e. scaling a row by some factor and scaling a column by the inverse of that same factor doesn't change the determinant

[quasi vector]

right okay

so what do i do lol

[livid sage]

see multiplying the first row by a

[junior kraken]

Hii

[quasi vector]

hello there

[quasi vector]

wait so I multiply the first row by a and divide the first column by a

$$\begin{vmatrix}
-bc&ab(b+c)&ac(b+c)\
a+c&-ac&c(a+c)\
a+b&b(a+b)&-ab\end{vmatrix}$$

[ocean seal]

kheerii

[quasi vector]

i guess i can do that for all the rows and columns

okay I think I got it

$$\begin{vmatrix}
-bc&a(b+c)&a(b+c)\
b(a+c)&-ac&b(a+c)\
c(a+b)&c(a+b)&-ab\end{vmatrix}$$

[ocean seal]

kheerii

[quasi vector]

this looks more doable

[quasi vector]

that's a really interesting technique

[livid sage]

yeah that's a lot prettier lol

[junior kraken]

You can

[livid sage]

it's very novel yes

[junior kraken]

Take a,b,c common from rows and multiply along columns

You in class 11 right ?

[quasi vector]

what?

[livid sage]

just let kheerii work on it

[junior kraken]

Kk

[quasi vector]

yeah okay I got it

thanks @livid sage that was a really good method

.close

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[livid sage]

very nice also welcome 🎉

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[vale path]

im grade 4 but a + b = c and a + b = c. so ab = (ac+ aa) -ac is aa, b(a + b) = ac, ac is (b + c) and c is (a plus b) - ab is a i think??

[vale path]

i feel like im wrong

close chat kheeri

[lone heart]

@vale path Has your question been resolved?

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[misty cipher]

can someone please explain where to go from here?

[misty cipher]

I am trying to get the integral in terms of gamma

ive set u equal to the red circle

I tried getting dx but it feels like its getting overcomplicated

[autumn viper]

It's the same logic as the gaussian integral

Can't do with just u sub

It's the gaussian but just cluttered

[pallid scarab]

if you know the gaussian, just make the exponent equal to -u^2 with the appropriate u

[misty cipher]

We have not covered the gaussian integral yet and do not think we will

[pallid scarab]

then you can't answer this question without gaussian

[autumn viper]

Then this isn't an appropriate problem

[misty cipher]

I do not understand how to do the step in between, btw the method it refers to doesnt exist above

[pallid scarab]

ok then gamma function sure

after writing as e^(-u^2)

t = u^2

[pallid scarab]

or you can immediately set t = 4x^2ln(3)

[autumn viper]

Literally what is happening to that dx

Oh nvm ok

It just doesn't matter

Ur using the fact that the gamma function has 2 variables

[misty cipher]

Does this working look correct to you guys?

[autumn viper]

Why

Substitute dx as du/ ln3 8x

[pallid scarab]

to integrate with respect to u, you need to get rid of all x

[pallid scarab]

this is correct

[autumn viper]

But he's using the gamma function it doesn't matter

[pallid scarab]

well you need an integral in terms of a single variable to recognize gamma function

[misty cipher]

So this is where I got to, I'm not thinking correctly idk where to go from here

[pallid scarab]

Highly doubt you'll be able to do anything with e^(-u)/x

[autumn viper]

Gamma function has 2 variables one is the dummy variable and the other is the input

[pallid scarab]

the input is 1/2

[autumn viper]

Ig

[pallid scarab]

power rules

get the constant coefficients outside

[autumn viper]

Wait but can you do that and use 1 single variable but still call it gamma of 1/2

[misty cipher]

How do I get that u in the square root to be in the format of the gamma function

[pallid scarab]

1/u^a = u^?

[autumn viper]

Distribute the square root

Oh

Bruh

Oh nvm my bad

[misty cipher]

What about the 4ln(3), sorry I'm really rusty rightnnow

[pallid scarab]

as I said, power rules

so sqrt(ab) = sqrt(a)sqrt(b)

whatever works in your way

[misty cipher]

Ohh

[pallid scarab]

get the multiplicative constants out of the integral

[misty cipher]

Thank you I get it now

Thanks for explaining even with me being slow

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[lone heart]

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[twilit dagger]

i'm so bad at math

[modern sedge]

can you use calculator for this?

[twilit dagger]

i don't know what ^ means

[modern sedge]

it means $x^{20} = 1000$

[ocean seal]

MæthIsAlwaysRight

[raw jetty]

roblox smh

[modern sedge]

Lol

[raw jetty]

whats the game

[modern sedge]

is that roblox?

[raw jetty]

send it i wanna play!!

[twilit dagger]

yeah it's roblox lol

[raw jetty]

99% sure it is, theres been a surge of math games there i think

[modern sedge]

lmao

okay enjoy playing guys

[twilit dagger]

https://www.roblox.com/games/7392219542/Luckys-Game-Hub

[raw jetty]

thx

[modern sedge]

btw for this Q, just do 20th root of 1000

probably with calculator

,calc 1000^(1/20)

[ocean seal]

Result:

1.4125375446228

[modern sedge]

and dont forget to round it appropriately

[raw jetty]

is this a badge game 😭

[twilit dagger]

yeah but there is secret part in it like this one

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[lone heart]

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[quasi vector]

need a little help with this

[quasi vector]

let $B\in A$, then $$B=\begin{pmatrix}
a&b&c\
b&d&e\
c&e&f\end{pmatrix}$$ where ${a, d, f}={b, c, e}={-1, 0, 1}$. All permutations of the two sets work, so the number of matrices $B$ is $3!\times 3!=36$

[ocean seal]

kheerii

[quasi vector]

for B part I know I need detB = 0 but that seems a bit tedious to work through

[quasi vector]

based on this form of B we have $$\det B=adf+2bce-(ae^2+dc^2+fb^2)$$ and we have $adf=bce=0$ so that $$\det B=-(ae^2+dc^2+fb^2) := 0$$

[ocean seal]

kheerii

[sour mica]

instead of determinant, cant you go simply like two rows cant be same and so on?

coz 0, 1 and -1 dont have that many linear combinations, like restrict the values based on other values

[cedar kelp]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[quasi vector]

well I've done the second part now

[sour mica]

ah, ok nvm then

[quasi vector]

we want ae^2 + dc^2 + fb^2 = 0

and exactly one of (a, d, f) and (e, c, b) is 0

so it's only possible when corresponding variables are 0

for example a = e = 0

we will have 4 cases for each of these so there are a total of 12 matrices B with determinant 0

but then about the third part..

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[steady acorn]

what would be a useful substitution or approach for this question?

[steady acorn]

i tried u = x/(a+bx)

[lone heart]

@steady acorn Has your question been resolved?

[steady acorn]

<@&286206848099549185>

nevermind i got it

[steady acorn]

and this substitution was correct

.close

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[gloomy geyser]

i am having trouble in understanding epsilon delta defenition of limit any yt videos which teach this step by step ?

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[coral forge]

with the function f(x)=1/x , x>0 i want to find the point where the most amout of lines perpendicular to the slope of the fuction cross, somehow like this:

[coral forge]

I got the equation for the perpendicular lines:

And here im stuck because i somehow would have to compare all the possible lines for a>0 and find a point which has most solutions.

[lone heart]

@coral forge Has your question been resolved?

[boreal spire]

can u show the instruction

what do we find here

[coral forge]

there isnt any

[boreal spire]

what 😭

[coral forge]

it was the work of my demented mind at 3am

[boreal spire]

god

well what do u wanna find ther

e

[coral forge]

oh wait the work itself

1 sec

so f(x)=1/x

and f'(x) = -x^-2

so i just used these to get the line at point a

y=-x/a^2 + 2/a

[coral forge]

and the i got the perpendicular line and calculated the constant with point P(a; f(a))

so lets say we check the lines at a=1 and a=2

then x+1=4x+1/16

so x=5/16

wich means that for a=1 and a=2 the lines connect at point p(5/16; 16/5)

and i want to find a point(s) which have the most lines connecting

[lone heart]

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[young mountain]

Yo guys I need help with some propositional logic stuff

Is it possible to say that

¬(p ∧ q) ∨ r ≡ ¬(p ∨ q) ∨ r

could be reduced to

¬(p ∧ q) ≡ ¬(p ∨ q)

and then

p ∧ q ≡ p ∨ q?

[modern sedge]

¬(p ∧ q) ∨ r ≡ ¬(p ∨ q) ∨ r

This doesnt look true

[lone heart]

@young mountain Has your question been resolved?

[young mountain]

It definitely isn't it's part of a proof by contradiction

[modern sedge]

You can't imply this then ¬(p ∧ q) ≡ ¬(p ∨ q)

Can you show the proof?

[young mountain]

Lemme rephrase my question

[young mountain]

If we have p /\ q = r /\ q can we simpify to get p = r?

[modern sedge]

p <-> r?

oh wait

= is equivalent?

[young mountain]

Yea sry

On mobile

[modern sedge]

I don't think you can get p = r from that

Think about the case when q is false

then p /\ q = r /\ q
this is true immidiately

and p and r can be anything

we dont know whether they are equal or not

[young mountain]

Ah

Wb the or case

[modern sedge]

think about the case when q is true

what can you imply from it then?

as in, what can you imply from p v q = r v q

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[worldly echo]

So my teacher explained how to write the domain like -infinity is greater than x is greater than -infinity

but i dont know how to solve it when the product for everything I chose is positive

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@worldly echo Has your question been resolved?

[worldly echo]

<@&286206848099549185>

[rare knot]

is this the question?

[worldly echo]

yes

[rare knot]

or is this(whatever this is)

[worldly echo]

this is my work

[rare knot]

ohk

[worldly echo]

yeah it isnt great

[rare knot]

xD

npnp

gimme a sec

umm

The graph consists of 1 point

x=7

nothing else

[worldly echo]

💀

This gonna be a fun year

[rare knot]

Splitting this into two we get

[worldly echo]

Well its a small assignment so a bad grade really wont hurt

[rare knot]

=0, <0

[worldly echo]

😂

sure do love this

[rare knot]

typo

Factoring out the quadratic, you get (x-7)(x-7)

x-7=x-7

[worldly echo]

yeah thats what I got

[rare knot]

any number squared is never negative

therefore, x has no values for x<0

for the =0 part of the inequality

you can get x=7,7

which is the only valid answer

[worldly echo]

Well I really appreciate the help

It helps knowing someone else realizes this shouldnt work

[rare knot]

Plugged it into desmos, and yea

it doesnt have any other solutions

💀

[worldly echo]

Have a good one my friend.

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[bold latch]

thats the answer but i dont understand why

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@bold latch Has your question been resolved?

[bold latch]

halp pls

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[lone heart]

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[halcyon mango]

(x-1)(5x^4-7x^3+x) = 0

[halcyon mango]

This is quite new for me so im very confused

I'm supposed to find 2 values for X

[empty quail]

Ok

Wait nvm

I only know 1

[sour dove]

only 2 values?

[halcyon mango]

Yes

This is not very advanced

[sour dove]

okay, look at the 5x^4 - 7x^3 + x, do you see any common factors?

hint: x marks the spot

[halcyon mango]

not really 🥲

Maths isn't my strong suit

[sour dove]

look at my hint

[halcyon mango]

X?

[sour dove]

bingo

[halcyon mango]

oh lol

[sour dove]

so we can convert 5x^4 - 7x^3 + x into x(5x^3-7x^2 + 1)

So then our equation becomes (x - 1)(x)(5x^3-7x^2 + 1) = 0

Now we know two simple roots:
(x - 1) = 0
x = 0

bam, there's two answers

no need to worry about (5x^3-7x^2 + 1) = 0

[halcyon mango]

so we just skip that?

[sour dove]

yep, if the prompt is indeed only find 2

[halcyon mango]

But wait

Sorry if this is a dumb question but why

[sour dove]

So let's take a simpler example

[halcyon mango]

Okay

[sour dove]

say if we had the following: (x + 3)(x - 2) = 0

[halcyon mango]

mhm

[sour dove]

If we plug in x = 2, we get (2 + 3)(2 - 2) = 0 --> (5)(0) = 0 So since anything multiplied by 0 equals 0, then we know that x = 2 is a root

Same logic for x + 3:
Let x = -3, then (-3 + 3)(-3 - 2) = 0 --> (0)(-5) = 0. So we know that x = -3 is equal to 0

all of your factors are multiple together, so if you can find one of the factors that equals 0, then that's a root

make sense?

[halcyon mango]

ohh yeah

[sour dove]

So in this case (x-1)(x)(5x^3-7x^2 + 1) = 0

So we know that x = 0 and x = 1 will make this statement equal to 0 overall, so we know those are roots

[halcyon mango]

ic

[sour dove]

x = 0: (-1)(0)(1) = 0 --> 0 = 0
x = 1: (0)(1)(-1) = 0 --> 0 = 0

(5x^3-7x^2 + 1) is guaranteed to either have 1 real root or 3 real roots per the fundamental theorem of algebra, but we don't care since it only asked for 2

we just took a shortcut 🙂

[halcyon mango]

So if it's more than 2 then we have to solve the other part?

[sour dove]

yep, exactly. Now how many the equation truly has is a trial and error to find out at this point (if they asked for more than 2)

It's a bit more advanced, but if you were to multiply out the original equation, our highest order of power would be 5 (5x^5 in fact). Per the fundamental theory of algebra, we know that any imaginary numbers come by a factor of 2, so the fully expanded equation has to have either 1, 3, or 5 real roots.

And since we already have 2 roots, we either have 1 more or 3 more real roots.

But that's irrelevant for this questions since they only asked you for 2

This is why it's important to keep in mind what the question is asking for since you don't want to further complicate stuff for yourself

so doing things like looking for common factors is super helpful

[halcyon mango]

Okay man

Tysm this was helpful

Learnt alot

[sour dove]

yep of course!

Like I said the FTOA will be a discussion down the road so don't worry about it for now

but I wanted to show the logic of why we can more roots

[halcyon mango]

I see

Thanks :)

[sour dove]

np 👍

[lone heart]

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[agile moth]

Just asking a quick question in regards to reference number in angles in unit circles. Suppose that your terminal point is at 7rad/6, is your reference number starting at rad(180deg) positive or negative?

[charred flint]

always positive, like with the arrow pointing away from* the horizontal axis

[agile moth]

To illustrate my point, in figure B, when getting the reference number for a it's usually positive so i don't know whether figure A is negative or not

[agile moth]

So the reference number for figure A is positive?

[charred flint]

yea refernce should always be positive, somewhere between 0 and pi/2

[agile moth]

Thanks! It's a huge help

!close

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[opal jolt]

<@&268886789983436800> ?

[tardy stag]

uh, this really isn't the right place for this

[solid smelt]

oh okay, i havent seen

[bold latch]

.reopen

[lone heart]

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[clear ivy]

.close

i have 16 pages and i am 10th of math homework can some do it

[still herald]

ok

show

[jagged cobalt]

no

[clear ivy]

https://drive.google.com/file/d/1kz0zhqGKhl7YWBCxeKn2RLeTNsn9gBeh/view from page 10 to 27

[jagged cobalt]

people arent going to do all your homework for you

[clear ivy]

he said yes though

[still herald]

wait

you said, "can someone do it?"
~no lmao~

[clear ivy]

ohh

[still herald]

we can help you though

!nosol

this server is not made for providing exact answers

for help it is

meant

[clear ivy]

ohh

[lone heart]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[clear ivy]

mb

sorry

.close

[lone heart]

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[clear ivy]

.close

[still herald]

ok np

[clear ivy]

can i get help with a math problem anyone

eather im dumb as fuck or the question is fucked

.reopen

[lone heart]

✅

[clear ivy]

can i get help with a math problem anyone
eather im dumb as fuck or the question is fucked

/close

.close

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[mystic nebula]

how do i get the derivative of tan^2x

how do i apply chain rule here

[twilit dove]

To do chain rule we first identify our “inside” function

What do you think that is in this case

[mystic nebula]

is tan^2(x) the same thing as (sec^2(x))^2

[twilit dove]

No

[mystic nebula]

well the derivative of tanx is sec^2x

[twilit dove]

Yeah that is correct

But that’s not how chain rule works

[mystic nebula]

can we move the ^2 outside

[twilit dove]

What do you identify as the inside function

[mystic nebula]

so (tan(x))^2

[twilit dove]

Ok that’s a good first step

[mystic nebula]

no idea

[twilit dove]

Well, when you did this you made it more obvious what function is on the inside

So the inside function is tan(x) in this situation

And from there we can do our chain rule differentiation

[serene snow]

https://harmony-yoga1.blogspot.com/2024/08/beginners-guide-to-yoga-tips-and-basic.html

[mystic nebula]

boy stfu

[twilit dove]

For the sake of simplicity we will do a substitution

[mystic nebula]

im following

[twilit dove]

We will do u=tan(x)

Which makes our function u^2

Now I think this is much easier to take the derivative of, so what would that be

[mystic nebula]

2u

[twilit dove]

Yes exactly

However, since we are taking the derivative with respect to X

We have another step

This is where the chain rule comes in

[pallid scarab]

<@&268886789983436800>

[azure needle]

Not here ma'am

[mystic nebula]

what does "with respect to x" mean

[twilit dove]

Well u is a function of X

[mystic nebula]

i never understood the dy/dx thing with derivatives

[azure needle]

Violence 💥

[mystic nebula]

i just know its asking for the derivative when i see it

[pallid scarab]

account created same day, bot

[twilit dove]

We can pretend that u is actually just u(x)

[mystic nebula]

i see

[azure needle]

Oh

[twilit dove]

So our derivative isn’t of u itself but of x

And for that reason we need chain rule

[mystic nebula]

how do we chain rule 2u

[twilit dove]

So right now we have f’(x)=2u

[still herald]

prolly that website will hack you computer

[twilit dove]

But that’s incomplete

To finish the chain rule we multiply by the derivative of U

[still herald]

P:

[rapid thistle]

This is clearly off topic, please do not post yoga tips in a math help channel

[twilit dove]

ok @mystic nebula it’s a bit crowded here now lol but I hope you’re following

[mystic nebula]

yea im not distracted

[twilit dove]

Lmk if you are getting it so far

Ok good

[mystic nebula]

how do we do that

[twilit dove]

We simply do u’

So the derivative is 2uu’

Oops

One second

[mystic nebula]

derivative of the derivative

[twilit dove]

Um not really

[mystic nebula]

hm

[twilit dove]

Remember what u is

What did we substitute

[mystic nebula]

tanx

[twilit dove]

Yes

[mystic nebula]

ohhh i get it

i forgot about the tanx l

lmao

[twilit dove]

Ok ok good lol

So once we undo the substitution

What would we get

[mystic nebula]

2 * tanx * sec^2x

[twilit dove]

Yes exactly

Good job

[mystic nebula]

OK BUT

thats what i got initially

before i came and posted about it here

but when i subbed in pi/4

im getting -2sqrt2

[twilit dove]

Hmm

Lemme check it on my calculator

[mystic nebula]

why are they getting y =4x - pi + 1

i might have subbed it in wrong idk

[twilit dove]

I’m getting 4 as my answer

Are you sure you subbed it in correctly?

[mystic nebula]

how do i do sec^2(x) in my calculator? is it (sec(x)^2?

so 1/((cos(x)^2)

[twilit dove]

Yeah you have to do parenthesis around the whole thing because most calculators don’t understand the formatting

No

You want to square the reciprocal too

So (1/cos(x))^2

You squared it one step too early

[mystic nebula]

ohhh

okay yea i got 4

but now what

point slope formula?

[twilit dove]

Well now we can find the equation using point slope form

Yep exactly

[mystic nebula]

so this entire thing we were doing

was just finding the slope

[twilit dove]

Yeah basically

[mystic nebula]

okay one last thing

how do i graph that

[twilit dove]

The tangent line?

[mystic nebula]

do i do y = (tan(x))^2

[twilit dove]

Oh I see

Well do you know what tanx looks like

[mystic nebula]

yea

[twilit dove]

Because you can use that knowledge combined with the slope to find which graph is the answer

Ok good

Well you know that (tanx)^2 has to be always positive

So either graphs C or D

And now you can use the tangent line to get the graph

[twilit dove]

But if you wanted to graph it on Desmos or your calculator then yes you would do this

Sorry I misread the part where it told you to use a graphing utility😅

[mystic nebula]

ik this is a more technical question but im only getting the graph of the slope

[twilit dove]

What are you graphing?

[mystic nebula]

i got y1=(tan((x))^2
y2=4x-pi+1

thats what i put in

im only getting y2

[twilit dove]

Hmm what calculator are you using

[mystic nebula]

ti83

[twilit dove]

And make sure you are in radians

Oh odd

[mystic nebula]

oops thats why

[twilit dove]

It’s all good now?

[mystic nebula]

yep

thanks for all the help

[twilit dove]

Yep no problem😃

[mystic nebula]

.close

[lone heart]

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[prime hull]

can someone help me see if i understand this text correctly

we have an n+1 tuple of values/n-vector

(x, y) where x is an n vector itself

these n-x values go into the function and give us back the value y

and f(x) is the surface of the function

if an arbitrary value of y is chosen underneath the surface and the set of all x values which satisfy this forms a convex set, then function is concave

is my understanding correct?

[lone heart]

@prime hull Has your question been resolved?

[mental finch]

Well not necessarily

Apart from that you’ve basically just restated what is already said so yeah your understanding is correct if you understand it

[prime hull]

why not necessarily?

[mental finch]

There’s nothing about the value y that has to be an output of f(x), it could happen to be the case but it’s not necessary

[prime hull]

is this diagram correct?

[mental finch]

You’re assuming y is >= 0 in this diagram

[prime hull]

any point on x1 and x2 and any y such that y lies below the surface at that point, (x1,x2, y)?

yes yes just an example

to help me

[mental finch]

Yeah well in such case it’s correct

[prime hull]

so just any y?

below

[mental finch]

Yeah so long it satisfies y <= f(x)

For a particular choice of x

[prime hull]

but won't that just spit out a y on the surface

[mental finch]

Hence the tuple (x,y)

[prime hull]

of the function

[mental finch]

Hm?

[prime hull]

ah misread

i thought as long as y =f(x) you said

seems so strange

[mental finch]

No, again that could happen but it’s not necessary. And there will indeed exist y values below since for example y -1 < y

[prime hull]

so then it says the set (x,y) is convex. in this aspect they mean only the n-x values?

the set of all x such that y<f(x)

must be an interval

[mental finch]

That’s not the set, the set is all tuples (x,y) such that y <= f(x)

That’s a set in R^(n+1)

So say for example x was a 2d vector like in your diagram

Then the function is concave since the set in your diagram is convex

[prime hull]

is it defined only for a subset of Rn?

im reading in my book, they are defining a subset V of Rn to be convex if the line segment joining any two members of the subset

are contained in V

[mental finch]

What about it?

[prime hull]

if we have a concave function and it goes off to - infinity

how can we write a set for that?

just (infinity, - infinity)?

and then we can say convex?

[mental finch]

Are you asking if it’s possible to have an unbounded set?

Like in general?

I mean just consider all of R^n as a set

It’s unbounded

And also trivially convex

So I don’t see the trouble here, or atleast the question you’re trying to ask

[prime hull]

idk man

ty for your help

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[tropic dome]

If f(z) ∈ B(0) prove that for every α > ½ there exists an F_α (t) such that f(z) = (1+z)^α L[F_α], 0 < Re(z), given B(0) is the class of all functions holomorphic and bounded in Re(z) > 0 and L[f] is the Laplace transform of f

[lone heart]

@tropic dome Has your question been resolved?

[tropic dome]

<@&286206848099549185>

[lone heart]

@tropic dome Has your question been resolved?

[lone heart]

@tropic dome Has your question been resolved?

[lone heart]

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[fierce kindle]

can someone explain to me algebra

[alpine sable]

doesnt yuno gasai have good grades? ask her, she loves u

[tardy stag]

,tex .algebra lesson

[ocean seal]

hayley 🥥 🌴

[fierce kindle]

?

[placid zinc]

Wtf you really just had this ready to go?

That's awesome haha

[tardy stag]

for a long time, this is the first time i've gotten to use it haha