help-13

[livid hound]

a^2 - b^2 = 5a-5b
a^2 * b^2 = (5a-3)(5b-3)

actually wait

alpha and beta are just the roots of the quadratic equation
x^2 - 5x + 3 = 0

[lyric cargo]

yeah

and

alpha * beta = c/a

[livid hound]

and alpha^2 + beta^2 can be determined using
(alpha +beta) and (alpha * beta)

[lyric cargo]

yup

but which one would u use

wait u need both

im stupid

mb

[inland ocean]

Wait aren't the both equations same so aplha/beta or beta/alpha will be 1?

Oh

Nvm

Alpha is not

Equal

To beta

[lyric cargo]

it's c i'm pretty sure

cuz alpha= 1/beta

[inland ocean]

Should be c

Imo

[cedar kiln]

@crimson sedge Has your question been resolved?

[inland ocean]

Well c is correct cause i got

$x^2 - \frac{19}{3}x + 1 = 0$

[wraith dagger]

coldtee

[lyric cargo]

yeah it's d for sure

really?

i got a negative number at the denominator

[inland ocean]

Yep there are two ways to solve it

[lyric cargo]

and u can't root that so

[inland ocean]

x^2 -5x + 3 = 0

alpha and beta are the roots

alpha + beta = 5 , alpha × beta = 3

$x^2 - (\alpha/\beta + \beta/\alpha) + 1 = 0$

[wraith dagger]

coldtee

[lyric cargo]

how'd u get this equation tho? or is it the final answer u got

[inland ocean]

.

[vague rapids]

The thing is OP is 👻

[inland ocean]

We are the OP NOW

[lyric cargo]

ahhhhhh makes snse

[inland ocean]

So if we solve that we get $x^2 - \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha × \beta}x+ 1= 0$

[wraith dagger]

coldtee

[inland ocean]

Then just subbed the values

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[stray otter]

Idk how to start

[tropic oxide]

,rccw

[limpid plume]

First differentiate wrt variable t

You have all the information given already

[tropic oxide]

dz/dt at (a, y) = (1, 3)?

[stray otter]

X is a

[limpid plume]

The a is probably x

[stray otter]

Yea

[stray otter]

By implicit diff?

[limpid plume]

Yes

[stray otter]

Like this?

[limpid plume]

Yes

[stray otter]

Ok

[limpid plume]

Now put everything you have in corresponding places

[stray otter]

Then we sub

[limpid plume]

Yes

You might have to write z in terms of x and y also

[limpid plume]

Perfect

[stray otter]

Ok

Thx

Cuz someone said it was wrong just now

.close

[limpid plume]

Wait what

[cedar kiln]

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[limpid plume]

.reopen

[cedar kiln]

✅

[limpid plume]

Wait a minute

[stray otter]

Yea

[limpid plume]

What did they say?

[stray otter]

In #help-4

It’s not that far up

[limpid plume]

Oh got it

Yeah in that one you did wrong

[stray otter]

Yea

[limpid plume]

This one is correct

[stray otter]

Just saw

Ok

Thx

[limpid plume]

No problems, glad to help

[stray otter]

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[versed pine]

do anyone know, what formula to use, soo i can know, what is this unit of this red line in the middle ?

[versed pine]

this is my case

assume that circle goes to 1

[crimson sedge]

https://en.wikipedia.org/wiki/Mean_of_a_function

[versed pine]

and this is ?

[cedar kiln]

@versed pine Has your question been resolved?

[versed pine]

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[simple bane]

$\frac{6 - x}{4x+3}$

[wraith dagger]

紅卫兵，周

[simple bane]

why cant i factorise the $3*2=6\ and 3 on the bottom?$

[tropic oxide]

you mean cancel?

[wraith dagger]

紅卫兵，周

[simple bane]

yes

[tropic oxide]

3 is not a factor of the top

and 3 is not a factor of the bottom

if you tried to divide the top and bottom by 3 you'd have $\frac{2 - \frac{x}{3}}{\frac{4x}{3}+1}$

[wraith dagger]

Ann

[tropic oxide]

which does not help you, most likely

[simple bane]

oh so that means the 4x must be 12x? and x = 3x?

[tropic oxide]

...

not at all how i'd put it

i mean, i guess?

[simple bane]

so 3 must be a common factor

oh ok thanks

nvm

thanks for the help

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[tough karma]

37b

[tough karma]

So I find the inverses of A and B and go on from there?

[crystal raptor]

Can't hurt to try

[tough karma]

The problem is the hint they gave

So Idk what order of multiplication I should do

[crystal raptor]

Well what are you going to do with the inverses

[tough karma]

F=A^-1 * C * B^-1

But Idk what order to do it

So first A^-1 or B^-1 or something else

[crystal raptor]

Are you asking whether you should do (XY)Z or X(YZ)?

[crystal raptor]

Well think about what you doing to both sides of the equation AFB = C to isolate the F

[tough karma]

Sorry I'm new to this

What does A^-1 * A give?

Or any inverse actually

[crystal raptor]

The identity matrix

[tough karma]

Oh

So U multiply both sides by the inverse?

[tough karma]

I'm not sure

[crystal raptor]

Yes but you have to decide on which side you're going to multiply by the inverse

As that matters, since the multiplication is not commutative

[tough karma]

Ok

Wdym which side tho

I have to multiply the whole equation so both sides by the inverse right?

[crystal raptor]

You multiply both sides of the equations by an inverse yes

But it matters whether you are multiplying from the left or from the right

Like the hint says XY is not the same as YX in general

[cedar kiln]

@tough karma Has your question been resolved?

[tough karma]

Ah ok thnx

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[open yarrow]

I have this function. I should calculate the fourier series and show that it converges to f (pointwise).

[open yarrow]

I calculated this fourier seires but idk how I should proof that this converges to f

?

How is that supposed to help?

[gritty galleon]

<@&268886789983436800> he's been spamming the question

[dire geode]

https://tutorial.math.lamar.edu/classes/de/convergencefourierseries.aspx

it's just a theorem/result you were supposed to be given in class

similar statement here
https://web.math.ucsb.edu/~grigoryan/124B/lecs/lec5.pdf

[open yarrow]

Well, I've seen a definition like this but I did not know how to implement it. It's written quite complicated

Thank you !

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[sleek mantle]

So to solve these sort of questions you differentiate first

Since we already know that it’s increasing >0

Then we find x is that correct?

[runic garnet]

Find where f’ > 0

That’s where it’s increasing

@sleek mantle

[sleek mantle]

In the first the derivative is your 6x-8

I have to make x alone right?

So it would be x>8/6

[runic garnet]

6x + 8

Not 6x - 8

Then set 6x + 8 = 0

Then find x, then perform sign analysis

@sleek mantle

[sleek mantle]

Sign analysis?

How can I do that

[runic garnet]

Ok

6x + 8 = 0

Solve

[sleek mantle]

X=4/3

[runic garnet]

No

[sleek mantle]

Sorry

-4/3

[runic garnet]

Yes

Now draw a number line, and have a single marking for -4/3

Test a point to the left of -4/3 and right of -4/3 by plugging it into f’ and see if it’s positive or negative

From there, you will see where it’s increasing

[sleek mantle]

Can’t I just do it like this

6x+8>0
X>-4/3

[runic garnet]

No. How do u know it’s not x< -4/3 ?

[sleek mantle]

It’s increasing

So the f’(x)>0

[runic garnet]

Right, but that doesnt tell u where it’s increasing

f’(x) > 0 on what interval

Can’t just assume it’s on x > -4/3

It can be on x < -4/3

[sleek mantle]

Alright gotcha

For f,g and h

F I completely don’t know what to do so let’s starting with that

I got the derivative don’t know what to do after

[runic garnet]

What did u get for a tho

[sleek mantle]

X>-4/3

[runic garnet]

Yea

Ok so what did u get for f derivative

[sleek mantle]

15x^2+12

[runic garnet]

Set equal to 0 and solve

[sleek mantle]

X^2=12/15

[runic garnet]

No

[sleek mantle]

U can’t really square it cause square root a negative isn’t a real number

[runic garnet]

Yea

So it’s either increasing on its entire domain or decreasing on its entire domain

Check two points to see

If a>b and f(a) > f(b), you know it’s increasing

If not, it’s decreasing

[sleek mantle]

Sorry I don’t get it

We already know it’s increasing we just need to prove it nah?

[runic garnet]

Oh we can assume that?

[sleek mantle]

It says it’s an increasing function

In the question

[runic garnet]

I mean it says find where it’s increasing, but yea ig so then

So it’s just the entire domain then

[sleek mantle]

What’s a domain 😭

[runic garnet]

The possible values of x you can plug into the original function so that the original function will still be defined

[sleek mantle]

What do you mean still be defined

[runic garnet]

For the function f(x) = sqrt(x), are there any values of x u can’t plug in and get a real number?

U shd know what a domain is if ur doing calculus lol

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[velvet mason]

I don’t understand this at all

Where did the 2pi come from

It looks like it just spawned in to me

[crimson sedge]

there's a physics server in #old-network

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[haughty cloud]

is it not 1/2?

[haughty cloud]

the derivative of ln(2)

[dire geode]

no log(2) is not 1/2

,calc log(2)

[wraith dagger]

Result:

0.69314718055995

[haughty cloud]

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[open yarrow]

I calculated the limit in x=0 and this is 0 from both sides, thus it's continous, right?
And thus, it is converging, or is it?

[open yarrow]

@dire geode you helped me earlier with the definition and I found some in my script. One for continous functions and one for none continous functions. The task says I should show it's converging pointwise. But the defintion for the continous function is saying it's uniformly converges

so, I assume it's wrong that f is continous, maybe you can tell me why. Then I'd use the (f(g+) + f(g-) )/ 2 for this?

[dire geode]

Can you not

[open yarrow]

not what

[cedar kiln]

@open yarrow Has your question been resolved?

[slate lintel]

don't ping individual helpers, even if they've helped you before

[open yarrow]

Sorry for that

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@open yarrow Has your question been resolved?

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[sweet glen]

hey guys I need a help to implement this algorithm using sage

K is an extension of a finite prime field

[sweet glen]

the thing I am missing is how to compute the invariant factor decomposition in sagemath

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@sweet glen Has your question been resolved?

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[lost token]

The values of p
and q
for which the function f(x)=[sin(p+1)x+sinx]/x , x<0
q , x=0
[√(x+x²) - √x]/x^3/2 , x>0
is continuous for all x
in R

[lost token]

i got q=p+2 after equating the first two

[cedar kiln]

@lost token Has your question been resolved?

[lost token]

nope

[vernal shell]

Hello. Maybe try finding the limit of f x→0+ because it doesn't have p or q in it

In that way you'll get a single number, which must match with q for continuity

[lost token]

ooo wait ill do that

i got 1/2

oh so ill put q=1/2 and then ohhhhhh okay got it

thank youu

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[oblique helm]

Hey i have got a quick question
If a+b = x+y then under what conditions can we say that a is equal to x and y is equal to b

[dusty hazel]

a-b = x-y

[oblique helm]

Okaay now lets start once again
If a+b = x+y AND ab=xy then can we say x equals a and y equals b?

[dusty hazel]

No.

[oblique helm]

How??

[dusty hazel]

Try using x=b and y =a

It'll still work.

[oblique helm]

Oh wait you are right
ab= xy under what conditions a equals x and y equals b ?

[dusty hazel]

You basically need a second equation right?

So that you can solve them together and then say, x=a, y =b

[oblique helm]

Yess

[dusty hazel]

so the second equation could be

x=a

[oblique helm]

💀

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[weary vessel]

@dull dirge

[dull dirge]

Yes

[weary vessel]

I'm sorry, both are false

The first is not true

Like we said earlier

[dull dirge]

Hmmm, gotta rethink the logic then

[flint plinth]

P = power set?

[dull dirge]

thank you

Yes

[flint plinth]

why is b false?

[weary vessel]

What is power set

P is probability

?!

[flint plinth]

wha

probabilities are numbers, they aren't subsets of each other

nor can you take unions of them

[dull dirge]

Yeah, it's power set

[weary vessel]

They mean + by union and . by intersection, no?

lol I been treating them as probabilities

[flint plinth]

b is in fact true

[vague rapids]

power set of a set S is the set of all subsets of S

[dull dirge]

Both are true?

[flint plinth]

nope, a is false

[weary vessel]

Uhhh my god 😭 😆

[vague rapids]

only bottom one is true

[dull dirge]

Hmm, i got things confused here

[flint plinth]

a subset of AUB isn't necessarily a subset of A or a subset of B

[vague rapids]

#(P(A))=2^x
where x is the number of elements in set A

[dull dirge]

Thank you for the help, i shall redo the exercise

[weary vessel]

Never mind what I was saying

I been treating them as probabilities

[dull dirge]

Don't worry about it 🙂

[weary vessel]

With union being OR (+) and intersection being AND (*)

😆

[dull dirge]

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[weary vessel]

.close

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[topaz pewter]

hello

can you help me?

[flint plinth]

you need to open a new channel, see #❓how-to-get-help

[vague rapids]

This channel is not available for help (for now)
Seek help from #❓how-to-get-help

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[topaz pewter]

ok thx

[vague rapids]

nvm it is avaibale now

Proceed to send your question here

[gritty galleon]

.close

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[topaz pewter]

.close

[gritty galleon]

bruh why close this?

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[inland ocean]

I tried putting theta = pi/6 but

Didint get any of them right

[cedar kiln]

@inland ocean Has your question been resolved?

[vague rapids]

Yeah pretty much seems all options are wrong

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[sharp shoal]

Hi guys. I had a small doubt. For this equation , we know that it has a HA at y=1 , so how is the middle portion of the graph allowed to cut that?

[crimson sedge]

asymptotes is the behaviour at infinity

so we can ignore say (-6,6)

also, in general a curve can cut across the asymptote (even at infinity)

[crimson sedge]

(simple example: e^-x sin(x))

[sharp shoal]

didn't understand this

i undeersstand the vertical asymtote part really well

without a gdc , how do i know what the graph is gonna look llike? I know there a point (0,-1) (1,0) and HA at y = 1

[crimson sedge]

horizontal asymptote means we only care about big x. small x doesn't matter

[sharp shoal]

what's big x?

[crimson sedge]

large x

x>>0

(and very negative x, x<<0)

[sharp shoal]

okay

but like with respect to this quetion , how do i go about drawing the graph?

I mark the asymtotes and the intercepts

[crimson sedge]

yes

then fill in the gaps

we can join up the middle 3 red dots

then make them tend towards the vertical asymptotes

[sharp shoal]

okay

[sharp shoal]

so for VA = 3 , i take like x = 2.5 and see where f(x) is tnending to?

[crimson sedge]

yeah, that works

[sharp shoal]

oh so for VA its always goign to infinity - one side negative other side ositive

[crimson sedge]

yes, they go to infinity

but its not always +inf on one side and -inf on the other

you should check to make sure

[sharp shoal]

example?

[sharp shoal]

which function graph would this be?

[crimson sedge]

,w plot 1/|x|

[wraith dagger]

Results provided by WolframAlpha

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

Requested by toby____#0

[sharp shoal]

Thanks

rational and mod functions are really confusing

you know any good resources to get better at this?

[crimson sedge]

practice is the best way to learn these things

use desmos to graph functions as well

[sharp shoal]

yeah

[crimson sedge]

khan academy probably has some videos on it too

[sharp shoal]

nah too boring

i prefer organic chem tutor when it comes to math.

anyways thank you so much

.close

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[prisma gull]

Is here anything specific about the choice of taking modulo 13

can I just took modulo 10, so 19^19 congruent to -1 mod 10?

[crimson sedge]

you can choose any mod that works

my guess is that the author tried small primes, and 13 was the first one that worked

[prisma gull]

worked as in?

reduce to negative congruent numbers?

[crimson sedge]

gives an argument that arrives at a contradiction

the important part of the proof is the second paragraph

[prisma gull]

my reasoning is that, when we take modulo m, any given number should belong to one of the congruent classes of mod m, 0 to (m-1)

[crimson sedge]

yes, thats right

we want to show that x^3+y^4 = 19^19 mod n has no solutions for a nice choice of n

n=13 happens to work

[prisma gull]

@crimson sedge

we can choose any n for this matter right?

the computation is the only thing that is hectic here i think

[cedar kiln]

@prisma gull Has your question been resolved?

[cedar kiln]

@prisma gull Has your question been resolved?

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[magic solar]

question is to verify the solution by substituting. but y' = e^2 - 1 which when i substitute on the differential equation on left, gives me Ce^x-1 .

[magic solar]

is it simply saying that the left side can be a solution if C = 1

[valid yacht]

how is y' = e² - 1

[magic solar]

i meant

x

e^x

[valid yacht]

y' = Ce^x - 1

why did u remove C

wait

[magic solar]

oh wait lmao im so dumb

soz

[valid yacht]

yeah okay, was just confirming what i said

[magic solar]

yeah you right

touching back on diff equations because i forgot them and i cant even differentiate anymore

😭

thanks

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[split pier]

How to test the convergence for the series of the sequence 1/[n(n+1)(n+2)]

[split pier]

Ratio test failed and i thought i could simply use comparison to say that 1/[n(n+1)(n+2)] < 1/n but then since 1/n diverges we can't exactly say anything about our main series

That series is convergent apparently

Ah dumb dumb me, i can just use cauchy, fuck

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[old hornet]

The values are 360,432,540,684.852,1020,1210
what unit scale should i use if im making a bar graph

[dreamy sleet]

you could use 100 to 1300 if you have space

[old hornet]

oh

wait let me think

oh yeah that works. thanks bro

[dreamy sleet]

Next time you can find this scale by considering the highest and lowest numbers

[old hornet]

what am

i supposed to with the highest and lowest values

[dreamy sleet]

they have 3 to 4 digits

[old hornet]

like how am i gonna know a good scale from them

[old hornet]

yeah

[dreamy sleet]

that indicates you should choose something like 100,1000,10000

[old hornet]

oh

makes sense

[dreamy sleet]

but 10000 is way too big

[old hornet]

yeah

[dreamy sleet]

so you shorten it to 2000 or 1500

[old hornet]

the boxes wont be enough

[old hornet]

mhm

[dreamy sleet]

the lower one, 100, is about right, it tells you how big the boxes are

[old hornet]

yeah i see that

now

thanks brother. you solved my problem for the whole bar graph thing

have a nice day/night! thanks

[dreamy sleet]

you too!

[old hornet]

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[old hornet]

:)

[cedar kiln]

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[white vault]

i've a really simple question, look at this equation

there's no negative number on the left side of the equation, the minus sign is a operation right? So why when i simplify this expression, i get this?

[white vault]

ignore the right side of the equation

it's all about the left side

my question is why the add operation is still there

[fair mortar]

the + 0?

[round wharf]

cant the equation be rewritten as,

a + (-5/9) + 5/9 = (-8/9) + 5/9

[fair mortar]

wdym rewrite, it is the same equation

[white vault]

there's no -5

there's a - 5/9

[fair mortar]

can you be more clear

[white vault]

ok

[severe thicket]

what's the question?

[round wharf]

its in the pinned messages

[white vault]

in equations, we need to isolate the icognita, removing the terms with inverse operations, so i did add the + 5/9 on both sides of the equation.

why it is 0?

[severe thicket]

I saw that but what's the point of adding zero

[white vault]

if add it, it would be 10/9

[round wharf]

when you add the (-5/9) and 5/9 on the left side of the equation you get 0

[white vault]

and not 0

but it is -5/9 or a operation of subtraction?

[round wharf]

it is -5/9 which you add with 5/9 and get a 0

making the equation a + 0 = -3/9 or a = -3/9

[white vault]

but if so, it would look like (-5/9)

does not it?

[round wharf]

they are the same thing according to me

[white vault]

i've read a book that said that it does not

that's why i'm confuse

[round wharf]

I don't really know about that

according to what I have studied

it should mean the same thing

[white vault]

the book said that -5 is a negative number, and it would be represent as (-5)

and the operation a - 5

'a' and '5' are positive numbers

[round wharf]

yes they are

[round wharf]

if you follow that

then you get

a + (-5)/9

[white vault]

the only mean of it to work is if these operations "swap"

there's some property on it?

[open nacelle]

For any a and b, a - b = a + (-b)

So, a - 5/9 = a + (-5/9)

And -5/9 + 5/9 = 0

[open nacelle]

They are the same

What did the book say?

[white vault]

the book said that i need to be cautious on it, because operations and negative numbers aren't the same thing

[open nacelle]

Hm

[white vault]

but i understood now, thanks to this

[open nacelle]

I wonder what they meant by that

Awesome

[white vault]

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[rich blaze]

I can't calculate this

[rich blaze]

if I apply barrow, I get negative numbers in the natural logarithm

[dusty hazel]

Well,

It's ln|2+x|

[rich blaze]

worst mistake in my life

I forgot about that

Thanks friend

[dusty hazel]

Atleast you're not repeating this one, I hope.

👍

[rich blaze]

👍

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[crimson sedge]

i dont really understand how theyve gone from L1 to L2

specifcally where -cosx + cos^2x has come from and where -1 has gone

[open nacelle]

$$ -1 = \frac{-\cos x (1 - \cos x)}{\cos x (1 - \cos x)}$$

[wraith dagger]

wakebloom

[open nacelle]

$$= \frac{-\cos x + \cos^2 x}{\cos x (1 - \cos x)}$$

[wraith dagger]

wakebloom

[crimson sedge]

sorry the image was poor

[open nacelle]

Yes

[crimson sedge]

let me think

[open nacelle]

Does that make sense?

Its just common denominator

[crimson sedge]

i dont think it makes sense

could you explain it again please

[open nacelle]

So

Ok hold on

$$ -1 = \frac{-\cos x (1 - \cos x)}{\cos x (1 - \cos x)}$$

Well do you get this

[wraith dagger]

wakebloom

[crimson sedge]

yes

[open nacelle]

$$= \frac{-\cos x + \cos^2 x}{\cos x (1 - \cos x)}$$

[wraith dagger]

wakebloom

[open nacelle]

And then this?

This is just distributing the cos x in the numerator

[crimson sedge]

i understand that

[open nacelle]

So then you just add that to sin^2(x)/(cos(x)(1-cos(x))

They have the same denomiator so you just add the numerators

And you get l2

[crimson sedge]

i think i understand

thank you

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[rugged cosmos]

Hi all, I'm having issues with solving this limit. I'm not sure if the second line is done ok. Thanks in advance

[buoyant latch]

(3+x)-(1+3x) = 2-2x no?

[rugged cosmos]

Ah, yes.. didn't realize that

Thanks @buoyant latch

. close

[dire geode]

.close

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[oblique prawn]

how to do $\lim_{x\to -\infty}{\frac{3-2^x}{4-5^x}}$

[wraith dagger]

jashxdlol

[hard ember]

Just put the limit value

This doesn't even require manipulation

[oblique prawn]

(3-2^-∞)/(4-5^-∞)

[hard ember]

Yeah

x^-a = 1/x^a

Use this

[oblique prawn]

oh 2^-∞=0

[hard ember]

Yes

[cedar kiln]

@oblique prawn Has your question been resolved?

[cedar kiln]

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[runic sage]

All the roots of
[x^2 + px + q = 0]are real, where $p$ and $q$ are real numbers. Prove that all the roots of
[x^2 + px + q + (x + a)(2x + p) = 0]are real, for any real number $a$.

[wraith dagger]

dabbbingpotato

[runic sage]

I think I solved it, can someone listen to my proof and double check it?

[autumn grove]

go for it

[runic sage]

$[x^2 + px + q + (x + a)(2x + p) = 0]$

[wraith dagger]

dabbbingpotato

[runic sage]

okay so we can simply this down by expanding $(x+a)(2x+p)$

[wraith dagger]

dabbbingpotato

[runic sage]

and turn that into $2x^2+2ax+xp+ap$

[wraith dagger]

dabbbingpotato

[runic sage]

then we add it to $x^2 + px + q$

[wraith dagger]

dabbbingpotato

[runic sage]

and we get $3x^2 + 2ax + 2px + ap + q$

[wraith dagger]

dabbbingpotato

[runic sage]

we can factor out the x

to get

$3x^2 + (2a+2p)x + ap + q$

[wraith dagger]

dabbbingpotato

[runic sage]

we are given from the problem that p, q, and a are all real numbers

we are already given that $[x^2 + px + q = 0]$ has real roots

[wraith dagger]

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

[runic sage]

so therefore $3x^2 + (2a+2p)x + ap + q$ must have real roots as well

[wraith dagger]

dabbbingpotato

[runic sage]

is that okay

[autumn grove]

Why is that last statement true? Everything before that made sense

[runic sage]

oh

wait

$3x^2 + (2a+2p)x + ap + q$

[wraith dagger]

dabbbingpotato

[runic sage]

factor out the 3

from this

to make it

$3(x^2 + (\frac{2a+2p}{3})x + \frac{ap + q}{3})$

[wraith dagger]

dabbbingpotato

[runic sage]

so basically $\frac{2a + 2p}{3}$ is taking place of p in the original equation

[wraith dagger]

dabbbingpotato

[runic sage]

and $\frac{ap+p}{3}$ is taking place of q

[wraith dagger]

dabbbingpotato

[runic sage]

and the question gives us that All the roots of
$x^2 + px + q = 0$ are real, where $p$ and $q$ are real numbers

[wraith dagger]

dabbbingpotato

[runic sage]

so doesnt that mean this also has real roots

or am i tripping

[autumn grove]

$\frac{ap+q}{3}$ for the second one... I think more is needed to prove it.

[wraith dagger]

epstmlgy

[autumn grove]

Here's a question: how do you tell whether a quadratic equation has only real roots?

[runic sage]

if the discriminant is not negative

[autumn grove]

Yup, so that's the missing step. Use the fact that the first equation has nonnegative discriminant to prove the second equation does as well

[runic sage]

the discriminant of the first equation is $p^2 - 4q$

right

?

[wraith dagger]

dabbbingpotato

[autumn grove]

yup!

[runic sage]

for the second equation
b = (2a+2p)
c = ap + q
a = 3

right?

[autumn grove]

that's right

[runic sage]

so the discrimnant of the second equation is

$4a^2 - 4ap +4p^2 + 12p$

[wraith dagger]

dabbbingpotato

[runic sage]

i think

[autumn grove]

I have $4a^2 - 4ap + 4p^2 - 12q$ here

[wraith dagger]

epstmlgy

[runic sage]

oh right

thank you

your right

okay

$4a^2 - 4ap + 4p^2 - 12q$ is the discrimnant

[wraith dagger]

dabbbingpotato

[runic sage]

so we have to prove this is positive

[autumn grove]

nonnegative, yup

[runic sage]

to clean up factor out 4

$4(a^2 - ap + p^2 - 3q)$

[wraith dagger]

dabbbingpotato

[runic sage]

😵‍💫

[autumn grove]

So by itself this isn't helpful, which is to be expected because we don't know enough about a, p, q

So you should use the other fact now, that

$p^2 - 4q \geq 0$

[wraith dagger]

epstmlgy

[runic sage]

right

perhaps substituting $p^2 - 3q$ with -q?

[wraith dagger]

dabbbingpotato

[runic sage]

because the lowest that value can be is 0

and we are just trying to find if its possitive or not

[autumn grove]

Couldn't q positive in some cases?

I would say try getting rid of p or q so there are fewer variables

[runic sage]

oh right

would it be just +p

$4(a^2 - ap + p)$

[wraith dagger]

dabbbingpotato

[runic sage]

so you would end up with this

[autumn grove]

hmm, I'm not sure where that comes from

looking at $4a^2 - 4ap + 4p^2 - 12q$ again, you could subtract a multiple of $p^2-4q$ and get rid of $q$ that way, since we know that can't increase the value

[wraith dagger]

epstmlgy

[runic sage]

oh lmao i'm blind

$4a^2 - 4ap + p^2$

[wraith dagger]

dabbbingpotato

[autumn grove]

Yup! Can that take on negative values?

[runic sage]

i think so

but i can't find a way to prove it

[autumn grove]

looks nonnegative everywhere to me c: I would try factoring it

[runic sage]

(2a - p)^2

so it has to be nonnegative

:D

thank you so much

[autumn grove]

Yup! Sure thing 😎

[runic sage]

you were so much help 🙏

[cedar kiln]

@runic sage Has your question been resolved?

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[old monolith]

Hey math people

[old monolith]

I'm having trouble focusing today, lots of stuff going on

And I need a sanity check XD

About the zero vector for polynomials. For "the zero vector to exist in S," do I need to show that for SOME (at least one) polynomial in S it is true that p(x) = 0?

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[finite badger]

Hey is this a valid way for the answer?

[cedar kiln]

@finite badger Has your question been resolved?

[finite badger]

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[cedar kiln]

Please don't occupy multiple help channels.

[ocean garnet]

having some trouble with this problem

not sure how to do the derivative of this

if i try to do ln(2y) and treat the 2y as a constant

[upper abyss]

You're looking for both partial derivatives

[ocean garnet]

i get 1/0

[ocean garnet]

yes I understand I need both

[upper abyss]

Tell me, what's the derivative of 2x?

[ocean garnet]

2

[upper abyss]

But the derivative of 2 is 0?

[ocean garnet]

yes

[upper abyss]

That's the mistake your making. ln(2y) is a constant multiple

[ocean garnet]

let me work it out rq and show u

this is the issue i am having in specific

when i work it out

it becomes like this

[ocean garnet]

i am not sure what you mean by this

like in this case

i know i need both partial derivatives

but right now I am just finding the one for x

then i will do y after

[upper abyss]

If I had asked:
What's the derivative of
2e^(4x/17)

Could you do it?

Now replace the constant 2, with the constant ln(2y).

[ocean garnet]

oh

so you are saying ln is a constant

yea i could do the derivative of that

[upper abyss]

ln(2y) is, yeah. Does not change with x.

[ocean garnet]

is this because ln is dependent on the inside, and since it has y and we are doing partial derivative of x we should treat ln(2y) as a whole a constant

so then i just multiply it by the 4e^(4x/17)/17 i got then right

ohw ait

i would get the right answer

wow

that wa sway simpler

than i was thinking

thank you

if i do partial in relation to y would e^(4x/17) be a constant to then

no that is a really stupid thing i said

nevermind

.close

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[chilly blaze]

Hey

[chilly blaze]

Im wondering how I got this wrong

Could it be that it’s a system error ?

[smoky idol]

does the red corner thingy mean you got it wrong?

[chilly blaze]

Yes

[smoky idol]

then that's weird

[chilly blaze]

Looks right ? Cause I’m thinking it could just be the website

[smoky idol]

$10^{25} \cdot 10^{11} = 10^{25+11} = 10^{36}$

[wraith dagger]

imtyp0

[upper abyss]

Does the website understand the inputs?

[chilly blaze]

Im not 100 percent sure my teacher just started using this website

But it does look right ?

[upper abyss]

Yes

[chilly blaze]

Okay thank you I’ll just email my teacher then have a good day you might see more of me today though

[upper abyss]

Wait, are the "as large as" and "smaller than" options all you get?

[chilly blaze]

Yea

For some reason

[upper abyss]

[chilly blaze]

The as large as one be tripping me

[upper abyss]

"as large as" is weirdly worded, but eh

[chilly blaze]

Because I would expect higher than

[chilly blaze]

Fr

[smoky idol]

and are the numbers multiple choice questions or do you have to type it out?

[chilly blaze]

Type

[smoky idol]

ah then maybe what Kaynex said is right

it might not understand the way you typed it

can you reanswer the question or is it over?

[chilly blaze]

Im retaking it again

[smoky idol]

then try answering with the ^ notation

[upper abyss]

I'm wondering if you just typed 10^11?

[chilly blaze]

But it doesn’t let me check untill I finish answering everything

[smoky idol]

welp

[chilly blaze]

I did lol I think I should maybe use the options they give me instead of typing out some stuff

[upper abyss]

Who knows lol

[chilly blaze]

That board I mean because I just be using my keyboard but maybe I gotta use that

[upper abyss]

Maybe "exponential notation" means 1×10^11

I'm reaching

[smoky idol]

I mean if the site only accepts one type of input that's kinda dumb

you can't expect everyone to think the same

[chilly blaze]

Oh

Maybe actually

So how would I answer the second one tho

Would it be like 1*10^-10

.close

[cedar kiln]

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[odd torrent]

how is this solved?

[untold torrent]

I'm assuming you want an intuitive answer

Just look at the highest degree terms

[clear berry]

Divide by the biggest power of x

[odd torrent]

so just divide by x^3?

[clear berry]

I mean divide the numerator and denominator by the biggest power of x

Yess

[odd torrent]

do I just ignore the 24 and 1

[crimson sedge]

do this if you need to show the steps

[crimson sedge]

do this if this is an mcq and you need to be quick

[crimson sedge]

and this, for mcq

[odd torrent]

well the numerator becomes 1-x^2/x^3 - x^3/x^3

so 1-1/x - 1 right

[untold torrent]

yeah, the division thing is for being more formal I guess

[odd torrent]

and denominator becomes 25?

kinda confused how this simplifies

[crimson sedge]

u shd divide 1 by x^3 as well?

[odd torrent]

so 1/x^3 - 1/x - 1

[crimson sedge]

yes

[odd torrent]

I see

so it becomes 0 - 0 - 1

over 1

thanks

[crimson sedge]

luckily he did it here

[cedar kiln]

@odd torrent Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[true nebula]

If I have a set A = {∅, 1, 2} would the P(A) = {{∅}, {1}, {2}, {∅, 1}, {∅, 2}, {1,2}, {∅, 2, 3}}?

Does that mean even though ∅ is explicitly stated in the set that it is counted weirdly when calculating the power set?

[crimson sedge]

check if {∅, 2, 3} is a typo

[true nebula]

It's not

[crimson sedge]

hmm where do u get 3

[true nebula]

ohhh sorry didn't see the 3

i thought youw ere asking about the power set symbol. yes it is a typo

[open nacelle]

You're missing the empty set itself i believe

[true nebula]

last one should be {∅, 1, 2}

[true nebula]

Isn't that covered by {∅}?

[open nacelle]

No

{∅} = {{}}

∅ = {}

Set containing the empty set is not the same as the empty set

[true nebula]

Ohhhhhh okay! I see now, that makes much more sense

[open nacelle]

Also, if you wanted to check your work you can use the fact that cardinality of the power set is 2 to the cardinality of the original set

So |A| = 3 so |P(A)| = 8

[true nebula]

So in this case, the ∅ symbol in the set declaration can be treated as any other number technically

[open nacelle]

Yes

Pretty much

[true nebula]

Got it. Thank you both for the help!

[open nacelle]

Of course

[true nebula]

.close

[cedar kiln]

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[violet flume]

The acreage of Hopetown, a town in Lewis County, is equal to 20% of the acreage of the region that is in Lewis County but not in Hopetown.

can you share the original problem?

does it define these quantities?

where is 18 coming from

[lucid skiff]

Can I just pick numbers for this

This is a Gre style question

So I can choose A which means the value there is bigger

B where that value is bigger

C where both A and B are equal

And D means no relationship

[violet flume]

im tripping over the wording on this problem

what does "of the region" mean here

[lucid skiff]

That's what I don't understand

[violet flume]

can you share the original?

[lucid skiff]

My tutor gave me this

[violet flume]

so are they saying hope county takes up 20% of lewis county?

[lucid skiff]

So yes

It's okay ill ask my tutor

Lol it's confusing

[cedar kiln]

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[lucid skiff]

.close

[cedar kiln]

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[amber axle]

Show that then Catalan numbers Cn count the number of ways one can triangulate a convex (n + 2)-gon. (For n = 0, there is no such thing as a regular 2-gon, so let us say there is a unique way of triangulating this non-entity, namely the empty triangulation.)

[long swan]

Define "triangulation"

[amber axle]

like

wait

something along these lines

not even that

like

triangulation is basically the number of ways you can cut a shape up into triangles

wuthout flipping the axis

so like for this square it will. be 2

@dapper warren

[long swan]

How many triangles?

[cedar kiln]

@amber axle Has your question been resolved?

[cedar kiln]

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[near wing]

can someone explain to me how to do this please

[near wing]

appreciate it

damn

yes

[tropic oxide]

#chill

i was answering the question posed by your gif

and redirecting you there at the same time

[near wing]

damn i blocked her but don’t remember doing it wtf

[tropic oxide]

if you don't want your questions answered then maybe do not ask them in the first place

[near wing]

damn idk when i blocked you Ann that’s crazy

or why

but this problem tho

[inland ocean]

Maybe you guys could come to my help channel, atleast it's more interesting.

I wasn't talking to you

[near wing]

degrees fs

damn what happened to him

[normal cave]

heyy can someone help me?

[near wing]

bruh

this is my channel

[south tundra]

How would you simplify sin(2pi - x) and cos(2pi - x)?

[tropic oxide]

please open your own channel. read #❓how-to-get-help

[near wing]

explain

[south tundra]

If you were given those expressions, would you be able to simplify them? Ending up rewriting them in terms of sinx and cosx

[near wing]

okay i’m lost @south tundra

[south tundra]

Generally
sin(2pi - x) = -sinx
cos(2pi - x) = cosx
If I were to rewrite 5pi/3 as 2pi - pi/3, would you be able to evaluate sin(5pi/3) and cos(5pi/3)?

[near wing]

how did you get sin(5pi/6)and cos(5pi/6)? @south tundra

did you multiply?

[south tundra]

Oh, sorry, made a typo

I meant sin(5pi/3) and cos(5pi/3)

[near wing]

so i would turn them into degrees right

wait no

it’s not asking for that

[south tundra]

Given this information

(Plug in x = pi/3 into the equations)

Well, if you are way more comfortable with degrees, then you may do that

[near wing]

so it would be Sin(2i-5pi/3) then

@south tundra

i finally understand it

i got them right

.close

[cedar kiln]

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[thick kiln]

Working on part b.

[thick kiln]

In part b, I think the sampling here is with replacement. But, there's a restriction on a consecutive restaurant.

I'm thinking it's with replacement because Fred is fine with eating at the same restaurant within that week(but just not twice in a row). Does thay sound right so far?

Would I just need to subtract off those days where they could be twice in a row? So, (Monday, Tuesday), (Tuesday, Wednesday) , (Wednesday, Thursday), (Thursday, Friday)

[cedar kiln]

@thick kiln Has your question been resolved?

[cedar kiln]

@thick kiln Has your question been resolved?

[white sage]

i don't have pen and paper to actually do this but my first thoughts are 10*9^4

because for each night for Tuesday to Friday he can eat at any restaurant except the one he ate at last night

so 10 on Monday, 9 on each day for tue, wed, thu, fri

[cedar kiln]

@thick kiln Has your question been resolved?

[marble forum]

Ok so

On Monday he has 10 options

Then next day he has 9 as he can't go to same as monday

Next day also 9 as he can't go same as Tuesday