help-23

[worthy estuary]

y r u using h and k for this part

[leaden harness]

to get the 2nd h and k equation

Is that not how ur supposed to do it?

[worthy estuary]

(h,k) is the centre of the circle

[leaden harness]

yeah

But I'm getting the equation so I can use the other equation to eliminate so I get the center?

Idk

[worthy estuary]

wait no ur right

how do uk that something is wrong

[leaden harness]

I'm not really sure but I suspect it because the h is a fraction, I do know its possible for fraction terms

but

[worthy estuary]

fractional terms should be fine

[leaden harness]

So its correct?

[worthy estuary]

i think so

[leaden harness]

okok

aight thanks

.close

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[worthy estuary]

mb man i found the error

u shouldve added the 2 equations to cancel the k terms

[leaden harness]

oh

h = -26?

[worthy estuary]

yes

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[leaden harness]

wait sec

[leaden harness]

k = 21/2?

[worthy estuary]

yes

[leaden harness]

oh

nice alr

alr man thanks

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[silent palm]

Hi everyone.

I'm stuck with this: $lim_{x\to0} \frac{\log_a(x+2) - \log_a 2}{x}$

[flat frigate]

Shadow91518

[quick crater]

maybe you can combine the things on the top?

[ocean crypt]

Try to bring yourself back to the notable limit of the logarithm

[severe otter]

$lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

[flat frigate]

Krezio

[quick crater]

$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

[flat frigate]

Skill_Issue

[quick crater]

:p

[severe otter]

thanks

[severe otter]

You're just finding the derivative of the function, try to use change of bases and chain rule

[quick crater]

im split on weather he knows or doesent, cause like what if hes trying to use the basic to find the derivative of limits, and this is for a specific case which will get expanded into a general case later

[safe radish]

@silent palm Has your question been resolved?

[silent palm]

Thank you both for the suggestions, but I still haven't got to the derivatives of functions.

Where can I find insights into these topics or some guiding exercises similar to this?

[neat yew]

You need to know all the logarithm formulas (change of base, particularly)

And the list of important limits as x->0, such as

sinx/x

(e^x -1) /x

ln(1+x)/x

[silent palm]

change of base is this $\log_a(x) = \frac{\log_b(x)}{log_b (a)}$ ?

[flat frigate]

Shadow91518

[neat yew]

Yes

[silent palm]

Thank you very much, I'll try to solve this exercise

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[pure vector]

Question 12

Is there anything wrong with my calculation?

[quick crater]

,rccw

[pure vector]

I'm dumb now

there must be something wrong with the calculation I think

[quick crater]

fuck im dumb sorry

[quick crater]

your calculations were for question 13?

[pure vector]

oh shit

[quick crater]

lmaooo i did the same mistake aswell

😭

[pure vector]

nvm

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[pale steeple]

In discrete math, how do i know when an argument is valid or invalid? for example this one, i tried with inference rules and couldn't get to the conclusion. Do I assume then it is invalid?

[lean otter]

what do you mean when you say an argument is invalid? as in https://en.wikipedia.org/wiki/Validity_(logic) ?

[pale steeple]

yes, i'm being asked to give a derivation (idk if that's the word in english) if the argument is valid, but in the case that it is not valid, then show why

[brave wolf]

Are you given a derivation?

Or do you have to make it?

[pale steeple]

i tried with modus ponens and modus tollendo ponens in premises 2 and 3, but i got stuck there

[pale steeple]

i have to make it

[brave wolf]

Okay, so what do you get when you apply modus ponens in premise 2

[pale steeple]

I get S --> T

[brave wolf]

and what do you get by applying modus tollendo ponens?

[pale steeple]

P V T

[brave wolf]

Hmm this is mildly weird actually

I don't think you can reach the conclusion

yeah, you definitely can't. I just found a truth-valuation under which all the premises are true and the conclusion is false

[pale steeple]

ooh so do i just say the argument is not valid? do i show it with truth tables? or is there a faster way?

[brave wolf]

Truth tables is a decent way, I think it would suffice to only show the 1 row where it's invalid

And if you look at it, the only way it would be invalid is that premises are all true and conclusion is false

that means that R has to be false (otherwise premise -R would be false), Q and S have to be false too (otherwise the conclusion would be true)

[pale steeple]

ok i get it now thank you so much!!

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[worn forge]

hello

[worn forge]

i need help with class 10 cbse

india

can anyone help?

[thin bridge]

just send your question
people will chime in if they feel like they can assist

[worn forge]

oh alright

[cunning pasture]

Send your question.

!da2a

[safe radish]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[worn forge]

so here is the bpt/thales theorem the triangles chapter and my questions is that why is it ar(ADE)/ar(DBE)

why is it specifically that traingles

[cunning pasture]

you have to get the ratio for AD and DB so you take as that

area of ADE / area of DBE = AD/DB

because the heights are the same

[worn forge]

ohhh

[cunning pasture]

so the ratio will be equal

[worn forge]

thank you so much

[cunning pasture]

welcoe

chime?

[worn forge]

what is chime

[cunning pasture]

,w chime

[flat frigate]

Results provided by WolframAlpha

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

Requested by anshulyolo121

[cunning pasture]

musical instrument

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[faint seal]

,rccw

[faint seal]

@dull sequoia @obsidian oracle sorry about that, i had to leave unexpectedly last time

@gusty inlet you too

Sorry I pribably shouldn't be pinging you all like this

So so far I have that det(M1) = det(M2) = 0, which gives us part A

To prove/disprove part D I would need to say something about det(M1+M2) which seems hard..

[peak estuary]

question 4?

[faint seal]

Yes

It's a multi correct question btw

[peak estuary]

I mean if M is already not invertible...

[faint seal]

Yeah so D is trivial

[faint seal]

Because of this

The problem is with B and C

[peak estuary]

well you said you had a problem with D

[faint seal]

I also need that for part B

[lean otter]

fitjee module ?

[faint seal]

Yea

[lean otter]

what did you try

[faint seal]

^

[faint seal]

Is there some sort of addition property for determinants?

Something that can help me relate det(A+B) with detA and detB

[lean otter]

how do you deal with B

[peak estuary]

no

[faint seal]

Hmm...

[peak estuary]

the det is multilinear in the columns/rows but that doesnt help

addition inside det is always yuck

[faint seal]

yeah it is

Where can I even get M1+M2 from like

It doesn't even make sense

[peak estuary]

I am still thinking about how many things you can get from M1^2=-M2^2. that seems so strong

[faint seal]

It really does

[lean otter]

the most we can do is add M1M2 + M2Ṁ1 both sides

[faint seal]

And yet I can't figure out anything

[faint seal]

What does that do?

[lean otter]

we wanna comment on M1 + M2

[faint seal]

Yes

[lean otter]

we should try to get an expression in M1 + M2

first

right

[faint seal]

Ah so you're saying (M1+M2)^2 = M1M2+M2M1

[lean otter]

yes

[faint seal]

Hmm

Ohhh hold on now

We have (M1+M2)^2 + (M1-M2)^2 = 0

Which means both of these determinants also have to be zero

Ahhhhh

Damn that was nice

[lean otter]

yes

lol

[faint seal]

So in fact every linear combination of M1 and M2 will have determinant 0

That's pretty

[lean otter]

c

[faint seal]

Yes indeed

Thank you for the help!

[lean otter]

yw

[faint seal]

Once you said M1M2+M2M1 it clicked for me

.close

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[honest bone]

i dont understand this

we’re using the limit comparison test

but i have two

actually three questions

why did we choose b(k) as so

why is the limit 1

and how do i choose b(k) correctly

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[honest bone]

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[cunning robin]

(x-y+z)(x+y-z)=8
(y-z+x)(y+z-x)=12
(z-x+y)(z+x-y)=24

Is there anything in this system that suggests it only has 2 solutions? If i combine the equations i get to (z-x+y)^2=36

[solar hazel]

i would look at divisors of 8, 12, and 24

[cunning robin]

To be clear, I have the solutions, but I am asked to show that I've found them all

[soft matrix]

ahh so u have to show whether ur cases are exhaustive

[solar hazel]

you can prove you’ve found all solutions this way

[cunning robin]

How would I go about doing that?🤔

[solar hazel]

(x-y+z) must divide 8

not many possibilities there

[eternal carbon]

are variables restricted to integers

[solar hazel]

and (x+y-z) = 8/(x-y+z)

[solar hazel]

oh…

[cunning robin]

Not explicitly

[solar hazel]

i don’t know why i thought that

[cunning robin]

But I think that is a fair assumption, it's a pre uni course

[solar hazel]

ignore everything i just said

[cunning robin]

This is how i solved the system, then I got a hint from my TA to factor it like I did above to get to the conclusion that the system only has 2 solutions, ill attach the original equations as well

[eternal carbon]

notice that you have repeated terms in these products

i.e. you can say a = - x + y + z; b = x - y + z; c = x + y - z (you don't actually have to make this substitution, but i'm writing it out to show you it's there)

try to see how that can help you (if you tried and still can't see, ||multiply the 3 equations together||)

[cunning robin]

I tried that, but without the substitution and got to to (z-x+y)^2=36, i divided 3 by 1 and then multiplied 2

[eternal carbon]

so you were able to solve for (- x + y + z)^2; that's nice. you can also solve for (x - y + z)^2 and (x + y - z)^2

[cunning robin]

I get (-x+y+z)^2=36, (x-y+z)^2=16, (x+y-z)^2=4

I do see that if you take the root you get 2 solutions, is that it?

[eternal carbon]

no

it is true each equation can be split into 2 parts after taking the root, but this does not imply that the overall system only has 2 solutions

[eternal carbon]

for example, if you take the positive root for each equation, you have this system:

• x + y + z = 6; x - y + z = 4; x + y - z = 2
  however, you can also take the negative root for each equation, giving you this different system:
• x + y + z = - 6; x - y + z = - 4; x + y - z = - 2
  and every other combination of positive/negative may also be possible

[cunning robin]

Okay, so to show there are only 2 roots i need to disprove every other case of these permutations?

[eternal carbon]

🤔

[eternal carbon]

uh

[cunning robin]

Oh

I have tunnelvision deluxe

[eternal carbon]

sorry let me just make sure i didn't say something completely silly

[eternal carbon]

yes this is correct

[cunning robin]

Either way, i need to check that the combinations of + and - do not yield additional solutions, I was dialed in on a completly separate solution and didn't realise that is what the TA meant

Thanks for the help and your patience mate , greatly appreciated

[eternal carbon]

yes but thankfully this does not take long because the original system already gave that the 3 pairwise products of a, b, and c were all positive

[eternal carbon]

you're welcome 🙂 glad that helped

[shadow verge]

What have you tried ?

@cunning robin ?

Your approach is quite tiresome bro

NVM , you could have tried working with ratios

x:y:z can be easily found

[cunning robin]

I have the answers already, just need to be sure I've found them all(didn't mean to reply to that chat in this one)

[cunning robin]

I do see from the original equation that we get the same solution if we negate the solution, but how does this help us discount certain combinations? All we know is that all the left hand expressions need to be the same sign, right?

[eternal carbon]

say we have
bc = 8
ac = 12
ab = 24

is it possible for any of these variables to have different signs?

[cunning robin]

No

[eternal carbon]

right, and that's the reason we can discount 6 of the systems without actually solving them

[cunning robin]

Oh, because if you have different signs in the (z-x+y)=+-6... equations, then they will no longer satisfy those equations

Great, thank you again!

[eternal carbon]

welcome! 🎉

[cunning robin]

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[lost wharf]

is "for every" the same as "for all" the same as $\forall$?
in other words is $\text{for every } e \in E$ the same as $\forall_e \in E$?

[flat frigate]

PolloTundra | Aidan

[rich elm]

in my exams they are used interchangeably

i'm not sure if that is universal tho

[red delta]

Yes they are used interchangeably

Every means : All of a countable group (considered individually), without exception

[rich elm]

ig it might be weird for ppl who dont have english as first language?

[lost wharf]

^ yes this

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[wet gust]

Okay so just need to fact check this real quick

[wet gust]

So I have the problem Vc=Vo(1-e^t/T) and I have to get the t by itself to do that I believe I take ln of both sideds

And then just moving everything over and factor out the ln to get t on the left

Does that sound about right

[tardy mango]

I’d divide both sides by $V_0$ first but you don’t have to

[flat frigate]

Civil Service Pigeon

[tardy mango]

Also “factor out the ln” is uh ….

ln is a function

You don’t factor out functions

[wet gust]

So it can’t just be applied to things

Like ln3 + ln 6x isn’t the same as ln (3+6x)

[tardy mango]

No

[wet gust]

Damn

That’s a shame

[tardy mango]

Look up the “universal law of linearity” (this is not an actual law, more so a figure of speech)

[wet gust]

Okie lol but other than that part was it almost right haha

It’s the first week of class and the homework we have no example problems for so I’m not even sure what the right answer should look like

[tardy mango]

I have no idea

You’re better off just doing it and sending it so someone can check

Your wording isn’t very precise, so it’s hard to tell exactly what you mean

[wet gust]

Yea I do be doing the typing and driving lmao

[tardy mango]

._.

Go back to driving

😭

[wet gust]

Thx for some help haha I’ll check back n a bit and write it properly out

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[idle schooner]

idk where to start with this, we were never taught this

[idle schooner]

im sorry but im still really confused

idk where the hell to start

[lean otter]

recall the definition of continuity

[idle schooner]

basically no abrupt changes

but like

i still have nooooo clue

[lean otter]

$\lim_{x\to a} f(x) = f(a)$ if $f$ is continuous at $a$

[flat frigate]

worthless loser

[idle schooner]

so x would be 2?

[lean otter]

a=2

[idle schooner]

ok so i would be able to plug 2 into the function for x?

[lean otter]

yes, assuming continuity

[idle schooner]

so where would i go from here?

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[glossy dock]

I have no idea where to go from here

[glossy dock]

Do I need to take the inverse cosine twice or something

[vapid iron]

for these problems with a lot of exponents its rarely going to help to write csc^2(x) as 1/sin^2(x) or stuff like that

look there are a bunch of powers of 2

do you know any trig identities that help to simplify those

[glossy dock]

I mean I could use cos^2 x = 1 - sin^2 x but that also has a square

[vapid iron]

look at the csc^2(x) and cot^2(x)

is there anything you can do with those

[glossy dock]

Cancel them out and leave a 1?

That sounds right

[vapid iron]

whats the identity for csc^2(x) and cot^2(x)

[glossy dock]

1 + one another

Is that not right?

[vapid iron]

yeah so if you move the cot^2(x) to the right

youll get csc^2(x)-cot^2(x) which simplies to 1

[glossy dock]

It would cancel out and leave a 1

[vapid iron]

so what are we left with

[glossy dock]

Everything left equal to 0

[vapid iron]

whats the stuff thats left

[vapid iron]

its cos^4(x) not cos^2(x)

[glossy dock]

Oh oops

[vapid iron]

are there any terms you think we can group together

[glossy dock]

Cos^2 x + 1

[vapid iron]

no

look at the cos^4(x) + sin^4(x) + 2sin^2(x)cos^2(x) part

and think about how you know cos^2(x)+sin^2(x) = 1

[glossy dock]

So would cos^4(x)+sin^4(x) also equal 1 since you can just square everything?

[vapid iron]

ah but what is (cos^2(x)+sin^2(x))^2

because its not cos^4(x) + sin^4(x)

remember FOIL

[glossy dock]

wait so all of it equals 1?

because if you foil it out its just cos^4(x) + sin^4(x) + 2sin^2(x)cos^2(x)

so it just simplifies down to tanx = 1

which is just all the pi/4's

or the ones in quadrant 1 and 3 i mean

so 1 and 5 since its between 0 and 2pi

alright got it

thanks

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[agile pewter]

Can someone help me with math

[agile pewter]

please

[glossy dock]

https://dontasktoask.com

[thin narwhal]

!da2a

[safe radish]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[agile pewter]

Hi

[thin narwhal]

Damn too late

[glossy dock]

ez

[agile pewter]

Can I send a picture of the question

This

5 a and b

[thin narwhal]

.rccw

You get my point

[agile pewter]

what

[thin narwhal]

!rccw

What have you tried

[agile pewter]

I don’t know it at all

I’m so confused

[modest hazel]

the number inside the () moves the parabola in the x axis, meanwhile the number outside the parenthesis moves the parabola in the y axis

[agile pewter]

what.

[modest hazel]

have you tried visualizing this in a graphing calculator?

[agile pewter]

nope

[modest hazel]

try doing:

[agile pewter]

I’ve never used a graphic calculator

I don’t have one either

r there online ones

[thin narwhal]

yes

try desmos

[agile pewter]

I have calculate84

[modest hazel]

what can you tell me about this graphs:

[agile pewter]

I don’t know parabola

[modest hazel]

parabola is the name of the shape

[agile pewter]

ok

[agile pewter]

I mean the parabola thing moves when u add 1

And moves when u subtract 1

[modest hazel]

notice the diference between adding 1 inside and outside the parenthesis

[agile pewter]

Yes

[modest hazel]

what happens in x2 + 1? blue one

[agile pewter]

it goes up

And the green one goes right

[modest hazel]

lets do a

[agile pewter]

okay

[modest hazel]

lets do it transformation by transformation. What happens when you add 1 inside the parenthesis?

the (x+1)square

[agile pewter]

Uhh

it will go left

it will move on the x axis

[modest hazel]

correct

a horizontal translation in the x axis to the left (1 unit)

now

what does the "-2" part means?

[agile pewter]

I don’t know

does it go right or smth

[modest hazel]

Hint: y axis

[agile pewter]

it goes down

On the y

[modest hazel]

here you can visualize the transformations

[agile pewter]

Yes

[modest hazel]

yeah, so now can you answer the question a?

[agile pewter]

but how would u figure out where the blue one would go

Oh wait

Nvm

[agile pewter]

But how would I answer it

Would I say how it translates

[modest hazel]

For example: Traslation in x axis 1 unit to the left. Vertical traslation in y axis downwards

you can write it in many ways, but that's how I would do it

[agile pewter]

ohh okay

I’m writing it rn

[modest hazel]

k

[agile pewter]

Is this good

[modest hazel]

yeah, exellent

[agile pewter]

Thank youu

what about b

[modest hazel]

the same, but the 1/2 means that the parabola is wider

[agile pewter]

It’s -1/2

Does that make a difference

[modest hazel]

oh yes

absolutely

[agile pewter]

how

[modest hazel]

if you see a minus sign there, that means that the parabola is upsidedown

see the poing where both functions collide?

[agile pewter]

Yes

[modest hazel]

there is like a mirror there

[agile pewter]

it’s mirrored

yes

[modest hazel]

yeah

[agile pewter]

So the translations of the normal one without the negative sign would be the opposite with the negative sign?

Since it’s mirrored

[modest hazel]

no

[agile pewter]

Oh

Then what

[modest hazel]

look that in both graphs, you moved the parabola to the right and downwards

[agile pewter]

Yes..

[modest hazel]

then the horizontal and vertial translations are the same

[modest hazel]

the equations are the same. just the simbol is diferent

[agile pewter]

Ohh okay

So In that case

it wld translate 2 to the right on the x axis and 3 down on the y axis?

[modest hazel]

yeah

remember the -1/2 transformation

[agile pewter]

yes

[agile pewter]

so wait what about it

should I include it in my answer

[modest hazel]

yes

[agile pewter]

what do I say

[modest hazel]

thats a transformation too

um

[agile pewter]

oh okay

[modest hazel]

the -1/2 can be decomposed in 2 transformations

first, the parabola is wider, you stretch it in the x axis

then, it was made a reflection

[agile pewter]

Wait so is this transforming from the previous problem?

[modest hazel]

there are 4

[agile pewter]

huh

4 of what

[modest hazel]

oh no, Im drunk, wait. I misread the question

[agile pewter]

Okay

[modest hazel]

no, there was no reflection or stretch in the previous

[agile pewter]

Wait are u actually drunk?

[modest hazel]

maybe

[agile pewter]

uh

ok

Anyway

So what would I write for this problem

[modest hazel]

name the 4 transformations u made

[agile pewter]

OHHH

ok so

It became wider

Then got mirrored

Then would go 2 units to the right

Then down 3 units

[modest hazel]

yes

[agile pewter]

is there a fancier way or smth to write “became wider” and “got mirrored”

[modest hazel]

yes, u can say "a stretch in the x axis by a ratio of 2 (2 times wider)" and "a reflection in the horizontal axis"

something like that

[agile pewter]

oh so like if it were 1/3 instead of 1/2 it wld get 3 times wider?

[modest hazel]

correct

[agile pewter]

Okay gotcha

I’ll write it now

idk if you can read it

but is this good

okay

[modest hazel]

yeah, its good

[agile pewter]

okay thank you!

sorry if I gave u a hard time

but if u don’t mind I have some other questions if you are willing to still help me

if not that’s okay too

[modest hazel]

no problem, I was just like u some time ago with quadratics

[agile pewter]

I’m new to quadratics so I don’t know much of it 😭

yet my whole packet is based off of it.

[modest hazel]

It's ok, tell me if you need any help later👍

[agile pewter]

can I add you on discord?

[modest hazel]

ok

[agile pewter]

thank you so muchh

.close

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[rich wadi]

Hey can anyone explain to me how to find the range of a function? But without the graph

[stark spire]

uh

what do you mean

[devout shale]

Do you have any functions in mind?

[safe radish]

@rich wadi Has your question been resolved?

[rich wadi]

Like domain and range

[rich wadi]

Like (x-2)/x^2-4

And for different types of functions

Like for different types of equations as well

Is there steps to finding the range or do I just have to visualize it

[stark spire]

oh

[old chasm]

you can sort of algebraically solve it. Here you essentially have $y = \frac{1}{x+2}$ knowing that $x \neq -2, 2$. Substitute that in should give you range

[flat frigate]

annoying

[old chasm]

so there are two mains way:

• find the inverse of the function, the domain of the inverse is range of original function
• find the removable discontinuity point and simplify your original equation, put that point in and that should be the point outside your range

by should be I meant not always

[safe radish]

@rich wadi Has your question been resolved?

[rich wadi]

this gives the domain right and then from there visualize it and then find range?

or do you avoid those numbers on range too

[old chasm]

you should know immediately that the domain is $x \neq -2,2$ because you can't divide by zero

[flat frigate]

annoying

[rich wadi]

yes

[old chasm]

you use that knowledge then to simplify that function knowing that there are two points that will affect the range

because think about it, something like y = x/x^2 is just y = 1/x but with the problem where x = 0

you can try plotting it to convince yourself

[rich wadi]

yea this is confusing..

i might js have to draw it lol

[old chasm]

basically, what you have is $y = \frac{x-2}{x^2-4} =\frac{1}{x+2}$ where $x \neq -2,2$

[flat frigate]

annoying

[rich wadi]

yeah i get that

and the domain part but using that to find range?

[old chasm]

so you put what x can't be into the simplified function

[rich wadi]

so plug in -2 or 2

[old chasm]

yeah. you can't really put x = -2 in

[rich wadi]

get 0

oh 1/4oh

oohh

OHh

1/4

[old chasm]

yeah and that is one of the value that your y can't be

[rich wadi]

like (-inf, 1/4)u(1/4, inf)?

for range

[old chasm]

not really because you have another point to look at

you need to consider what happens if x = -2

[rich wadi]

it cant

so it doesnt touch -2

?

[old chasm]

now you use inverse method where you inverse $f(x) = \frac{1}{x+2}$ to be $f^{-1}(x) = \frac{1}{x} - 2$

[flat frigate]

annoying

[rich wadi]

huh

oh my i did not learn this in class

[old chasm]

and notice that the domain of the inverse function can't be zero, then on top of what we have ($x \neq 2 \implies y \neq \frac{1}{4}$), another condition is $y$ can't be 0. We have 2 points and that should be all the problematic range that we have

[flat frigate]

annoying

[old chasm]

yeah, that is how people do it without visualization.

it is just series of arguments to the conclusions. There is not strict step really

[rich wadi]

(0,1/4) u(1/4,2)u(2,inf)...?

[rich wadi]

oh okay but u do plug in what x doesnt equal into the eq to find what doesnt equal y

[old chasm]

it should be $(-\infty, 0) \cup (0,\frac{1}{4}) \cup (\frac{1}{4}, \infty) = \mathbb{R} \setminus {0,\frac{1}{4}}$

[flat frigate]

annoying

[rich wadi]

oh yea i forgot tht - inf

what about the x doesnt equal 2?

we ignore that for range?

[old chasm]

essentially yes. you find the way to plug it in or argument your way that y can't be this and that

[old chasm]

$x \neq 2 \implies y \neq \frac{1}{4}$

[flat frigate]

annoying

[rich wadi]

mm arguments.. okay

make a bunch of rules then like solve

[rich wadi]

oh so yea ignore it for range

[old chasm]

yeah, like you can't divide by zero, square root can't be negative bla bla bla

[rich wadi]

yea

thanks i got it now

ty for ur time :)

.close

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[worldly venture]

can somebody explain why the modulus of z is equal to 1 for this question? thats what the answers said and i dont undertsand

[lone hornet]

because |z| = |a + bi| = sqrt(a² + b²)
and
sqrt(cos²(theta) + sin²(theta)) = sqrt(1) = 1

[worldly venture]

OHHH

i forgot abt the pythag rule thing

thank you so much

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[brave wolf]

How do I know x^2 will be -1 (mod p) here?

[safe radish]

@brave wolf Has your question been resolved?

[soft matrix]

x² can be either of the form 4k Or 4k+1
So they are taking a square of the form 4k hence it will be -1 mod p

[soft matrix]

Here x is a variable dependent on p

[brave wolf]

How do we even know this?

x = g^(1/4 (p-1)) where g is a primitive root

how does one get x mod 4 from this

[mossy lotus]

so if you calculate x^2, youd get g^((1/2)(p-1))

[brave wolf]

right

[mossy lotus]

and you can apply fermats little theorem to it

[brave wolf]

hmm, how exactly

[mossy lotus]

since g is primitive root, you know g^p is g mod p

or g^p-1 is 1 mod p

[brave wolf]

and g^(p-1) = 1

right

[mossy lotus]

and x^2 is sqrt(1modp) or +/-1 mod p

[brave wolf]

sqrt exists in modulo

that's new to me

[mossy lotus]

its just multiplying two numbers

[brave wolf]

can't there be some other number that = 1 when squared?

[mossy lotus]

1modp * 1modp is always 1modp

[brave wolf]

right

[mossy lotus]

and we know two numbers are equal in this case coz sqrt

[brave wolf]

But the converse is weird, if a = 1 and b * b = a, how do I know that b = 1 or b = -1?

[mossy lotus]

the number is of form (ap+k) ^ 2 = k^2modp

so, if its a primitive, you cant always get that k^2modp = 1

[brave wolf]

oh

[mossy lotus]

yeah, so we basically know x^2 is mod +/- 1 mod p

[brave wolf]

And it cant be 1

[mossy lotus]

and since p is 1mod4, you cant have 1

exactly

[brave wolf]

lest the order would be smaller

g wouldnt be primitive root i think

[mossy lotus]

yes, thats why we need g as primitive root

[brave wolf]

I see

thanks

I get it now

.close

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[mossy lotus]

Also, from this you can see the converse is true too

[brave wolf]

Yeah, I see it now

[soft matrix]

Well no can be either of the form 2k Or 2k+1
Square both those forms
4k² and 4k² + 4k +1 - >4k(k+1) +1

sorry for late reply was busy ig bacter already did the job

[shadow verge]

🥲

Can you help me display ?

[soft matrix]

Il try my best

[shadow verge]

help-27 😭

I've said it there

[safe radish]

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[vapid knoll]

Why is the result of 0.4^2 * 0.6 multiplied by 3? (Referencing X = 1)

[vapid knoll]

It makes sense to me that getting 1 head is the same as getting no heads twice then 1 head once

which is 0.4 * 0.4 * 0.6

or 0.4^2 * 0.6

= 0.096

but why is the result *3 (0.288) not just 0.096

oh is it because there are 3 outcomes (HTT, THT, TTH) of it

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[magic junco]

Set alpha and beta are the roots of the above equation

Find alpha+beta
Given solution

1. -5
2. -3
3. 1
4. 3
5. 5

[desert pasture]

sure it isn't $2^{2(x+5)}$?

[flat frigate]

Veni, vidi, perii

[brave wolf]

I think both work equally

[magic junco]

no

[brave wolf]

can you rewrite it as a quadratic equation?

[magic junco]

I feel like approaching it geometrically, but I’m a bit stuck

[magic junco]

Like substitute it with t?

[brave wolf]

that's gonna be a bit painful

[brave wolf]

yeah, sure

[magic junco]

ah nvm, I got it

Tysm

[brave wolf]

be careful though

sum of roots of the quadratic equation isnt gonna be the same

[magic junco]

For sure

yeah, I’m rusty tho, haven’t done Math for almost a month

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[dim wraith]

what does that mean?

[wise saddle]

Context ?

[dim wraith]

wdym?

[wise saddle]

In what context did you cet that notation ?

[dim wraith]

Calculus

Indefinite integral

[wise saddle]

Yes but whtas the question surrounding it etc

[dim wraith]

im preparing for an exam from questions about, how to say, integration techniques etc

sauce not on english

[wise saddle]

Send it anyway

Or like the question

[dim wraith]

i'm reading a digital lecture outline, and it's a Q+A on how to integrate certain things

[wise saddle]

Ok im sorry ive done 2 years of uni never seen that notation

[dim wraith]

i just dont understand wha the R

and why R(sinx, cosx)

nonsense

crazy

[wise saddle]

If no one here knows

Try chatgpt its good at broad knowledge stuff

[dim wraith]

i asked already

they don't know

maybe i asked wrong

but very strange notation

would something happen if i ping Helpers?

[wise saddle]

Try it

[dim wraith]

<@&286206848099549185>

[edgy isle]

I've seen it before as just meaning R(x, y) = p/q

coming right after a section showing you how every rational function can be integrated

[dim wraith]

not sure p/q

[twin atlas]

My best bet would be something related to the Rational numbers

Thing

[unreal shoal]

Can't help

Jeez

[dim wraith]

hopefully it's supposed to be any rational function composition of two functions

[unreal shoal]

Ye

X=34

[dim wraith]

only in my case it's a rational function of functions

[edgy isle]

oh wait I meant R(x, y) = p(x)/q(y)

[dim wraith]

yeah

[edgy isle]

for any polynomials

[dim wraith]

now i understand

q(x) can be q

you were absolutely right

[safe radish]

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[gentle summit]

Hello, I'm not really good in these types of questions, honestly don't know where to start
(fyi, english isn't my first language so I may have a hard time trying to understand some words)

[compact rampart]

do you know what tangent means

[red delta]

Join O to R

[red delta]

@gentle summit

[gentle summit]

Wait

[red delta]

Ok

[gentle summit]

OH

Yea tangent's a single line that touches the circle

Gotta refresh my memory

[red delta]

And is perpendicular to the normal (which is one of the radii)

[gentle summit]

Wait is the tangent going to be in cm?

Okay wait I'm still lost here 😭

Joined O to R

[safe radish]

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[tame trout]

how is this solved and graphed on a number line? i need help thanks

[severe pond]

graphically you would look at the graphs of both functions and see for which intervals the left hand sides graph is higher than the right hand sides graph

you’ll need to use algebra to isolate x to solve it explicitly

[tame trout]

wait i looked at mathway and this is it

what do i do after zero is on the other side already?

[lime dust]

Rewrite so you have a single fraction > 0

[tame trout]

Ahh i see.

[lime dust]

And then check for the critical points and the points close ti them

[tame trout]

Can I get help as to how this is rewritten?

[lime dust]

In both sides

[tame trout]

I am confused haha

[lime dust]

Right and left of esch point

Do you know how to add fractions?

[tame trout]

yes, but i am confused here because of (x-5) and (x-6)

[lime dust]

If so, start by adding 11/(x-6) to both sides

They have different denominator

So what do you do when they are different?

If i say 1/2+1/3 what do u do?

Here it is the same procedure

[tame trout]

wait

[tame trout]

x 3? in both numerator and denominator

[lime dust]

You do that to 1/2, but to the 1/3 you do with 2 in num and denom

That way u have same denom

[tame trout]

ahhh

wait

so

i multiply x-6 to first term

and x-5 to second

?

[lime dust]

U multiply and divide remember

The idea is that u dont change what u have

[tame trout]

wait i forgor the dviding part haha wait

[lime dust]

If you have

5 for example

And u multiply by 3, u change the number and now u have 15

But if u multiply and divide by 3

U will have 15/3

This way u didnt change the number

Just wrote in a different way

[tame trout]

wait

the multiplying part, is it numerator and denominator?

[lime dust]

Yes, you want to multiply by 1

And 1 = a/a for any a different from 0

Now u replace that a with whatever u need

In this case (x-6)/(x-6)

[tame trout]

like this?

[lime dust]

Exactly

[tame trout]

okay wait

so after this, what do i do

to carry on with the same denominator

do i solve the numerators, then use this fixed denominator

[lime dust]

Wait u have a typo

[tame trout]

Oh

[lime dust]

You need parenthesis in both denominators

[tame trout]

okay okay

[lime dust]

For (x-5) and (x-6)

U solve numerators now

And denominators no need to expand

[tame trout]

like this right

[lime dust]

Yes

Now expand numerator

And simplify

[tame trout]

wait give me a moment real quick haha

[lime dust]

Wait

You have made a mistake

[tame trout]

Okay

[lime dust]

You have a sign error

The minus from the beginning affects only to the first numerator

But u put in a way that now affects both

[tame trout]

Ahh wait

so its jut beside 10

[lime dust]

To avoid this problem, solve the sign problem before adding the fractions

Yes

[tame trout]

Wait

got it! haha!!!

[lime dust]

Ok what u got

[lime dust]

Perfect

[tame trout]

wait so after this

i have

-5, 5, 6

?

[lime dust]

You have to check the following

[tame trout]

wait

[lime dust]

Ok

I recommended you to do a table of signs

[tame trout]

wait, so if i plot this on the number line

-5, 5, 6
with intervals a,b,c,d

[lime dust]

No you dont need to plot this

[tame trout]

Ohh okay

[lime dust]

You have to do a sign table

And check for signs on the important points

[tame trout]

Okay thank you

[lime dust]

5 from the left and from the right, 6 from the left and right

[tame trout]

I will try something right now

give me 5 minutes haha i hope this works

[lime dust]

Ok

[tame trout]

i got

(-5,5) and (6,∞)

[lime dust]

Looks good

[tame trout]

Thank you

I have one last problem I have trouble determining

would you help me with it Haha

[lime dust]

It is better you close this channel and open a new one with the new question cause i am leaving now

[tame trout]

Alright.

thank you

.close

[safe radish]

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[sacred sentinel]

i have been stuck on this problem for a while now, but i can't find what i'm doing wrong. i know it's incorrect because when i differentiate it with wolframalpha at x=1 it doesn't give the same value as when i ask it to compute the integral

eta is dirichet eta btw

it can look like a 'n' sometimes

[safe radish]

@sacred sentinel Has your question been resolved?

[safe radish]

@sacred sentinel Has your question been resolved?

[safe radish]

@sacred sentinel Has your question been resolved?

[safe radish]

@sacred sentinel Has your question been resolved?

[sacred sentinel]

<@&286206848099549185>

[sacred sentinel]

<@&286206848099549185>

see pinned message btw because some stupid kid needed attention

[hard crest]

please do not troll in help channels, thanks!

[safe radish]

@sacred sentinel Has your question been resolved?

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[sacred sentinel]

i have been stuck on this problem for a while now, but i can't find what i'm doing wrong. i know it's incorrect because when i differentiate it with wolframalpha at x=1 it doesn't give the same value as when i ask it to compute the integral

[sacred sentinel]

eta is dirichet eta

[safe radish]

@sacred sentinel Has your question been resolved?

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@sacred sentinel Has your question been resolved?

[sacred sentinel]

<@&286206848099549185>

[fathom jewel]

I see I and I(a), can you explain what you are evaluating exactly? Is this the Feynman technique or what exactly are you trying to do?

[sacred sentinel]

I is the integral i'm evaluating

and i made a new integral I(a) with a parameter

such that I'(1)=I

a is a alpha btw

[fathom jewel]

Ok I see

Ok I think I get it

Ok then you turned 1/(1-e^(-2x)) into a geometric series

[sacred sentinel]

yeah

[fathom jewel]

The pulling out the sum is a bit sus

[sacred sentinel]

i didn't fully write out actually pulling x^alpha and e^-x into the sum

but why do you think it's sus?

[fathom jewel]

I am thinking it had to do that all f(x)_n = [e^(-2x)]^n have to be integrable but that might be the case here

and that the converges uniformly

But I think that should be the case here so let me continue

[sacred sentinel]

yeah it's almost the gamma function but with 1+2n and alpha>0 so it will converge

[fathom jewel]

Ok up before you use that dirichet eta function i couldnt find a mistake so far

[sacred sentinel]

this is my proof for replacing that sum with eta and zeta function btw

[fathom jewel]

,w Sum[1/(1+2n)^(α+1),{n,0,inf}]

[flat frigate]

Results provided by WolframAlpha

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

Requested by adoniszgks

[fathom jewel]

Did you try to pull out 1/2^s ?

[sacred sentinel]

yeah

[fathom jewel]

Ah I see the minus

now

2^(-s)

or I suppose it's still supposed to be the α, but anyway ok

[sacred sentinel]

in this case i use s-es, because they usually do that for zeta and eta functions, later (in my main image) i sub in alpha+1 for s

[fathom jewel]

But

Ah no

Starts at n = 0

[sacred sentinel]

yes

[fathom jewel]

haha ok you are pretty good

up here i cant find to seem anything weird let me see further

You replace now ξ with η

[sacred sentinel]

found this on wikipedia

[fathom jewel]

Ι guess thats similar to wolfram here

[sacred sentinel]

so $\eta(\alpha+1)=(1-2^{-\alpha})\zeta(\alpha+1)$

[flat frigate]

pixel

[sacred sentinel]

right now i'm trying to see whether i what i have is the same as wolfram

but i'm having trouble

[fathom jewel]

Cant you equate the results and let wolfram check?

[sacred sentinel]

im checking rn

with wfa

[fathom jewel]

May you explain this pls

[sacred sentinel]

those 2 lines aren't related

maybe i should have put more space

between them

[fathom jewel]

Ah ok

Ok I think I got the next step too

Ok you corrected it

Was about to say here

η(α+1)

Damn last two lines

You factored η(α+1)

Multiplied Γ(α+1) in

[sacred sentinel]

oooohohuawhduiwahduiwadhuaw

im so dumb

i see now

[fathom jewel]

Didnt you differentiate it or what haha

Since it was I(a) the whole time

[sacred sentinel]

noo

its the gamma

i didn't multiply 2^-alpha/that by gamma

the second term

[fathom jewel]

Ah ok

[sacred sentinel]

in the parentheses

[fathom jewel]

I was about to ask how you got there

FINALLY

[sacred sentinel]

i'll see whether my answer is right now

TYSM

[fathom jewel]

but you are pretty smart

[sacred sentinel]

thanks

[fathom jewel]

really cool solution

[sacred sentinel]

i'll close this channel now

have a very nice day

.close

[fathom jewel]

u2

[safe radish]

Channel closed

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[fathom jewel]

goated

[safe radish]

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[formal barn]

I ve tried several times but always get wrong result at the end while comparing.

[safe radish]

@formal barn Has your question been resolved?

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@formal barn Has your question been resolved?

[plucky elk]

show your attempt

[formal barn]

Wait I am not at home rn will send after 1 hour

[plucky elk]

just re-open when you get home then and don't hog a channel if you're not ready to work

[formal barn]

I was ready at that time but I ve to go for some work

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[safe radish]

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[safe radish]

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[rustic goblet]

hmm

what did Kepe ask

[manic radish]

[lone arch]

I figured it out lol

[rustic goblet]

ah, nice

[lone arch]

It was about counting subspaces of the dual of a finite field

[safe radish]

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[rustic goblet]

.oooh, nice!

[twin skiff]

<@&286206848099549185> i just have a conceptual question about which subsets of R (the real numbers) are "totally bounded," ie for all eps > 0, E can be covered by finitely many balls of radius eps?

so for example, how do you verify that [0,1] is totally bounded? because it is certainly "normally" bounded (normal in the colloquial sense)