help-23

[loud osprey]

How do i do this

[dense prism]

A tangent being horizontal means it's slope should be 0

i.e. dy/dx at the point of contact =0

[loud osprey]

yeah

i got the derivative already

but im not sure how to find how many points there are

[dense prism]

Show where you've reached

[loud osprey]

i got (-y^3/5) / x^3/5

as the derivative

and i know that whenever i have y = 0, it would be a horizontal tangent

How do i find whenever y = 0 tho?

[tranquil ore]

Nah it’s when dy/dx=0

[tranquil ore]

So this probably

Now you have a system of equations, idk how you want to solve it

[loud osprey]

how do i solve that?

[tranquil ore]

Oh actually it’s not that bad

Cause as long as x isnt 0 you can multiply by sides by x^3/5

[loud osprey]

so i get -y^3/5

[tranquil ore]

Are you sure it’s this though

[loud osprey]

Should be

[dense prism]

Check the equation and see when can y=0, you'll see x=1 clearly satisfies

However there could be problems

With differentiability

[loud osprey]

?

wdym

[loud osprey]

+- 1?

[dense prism]

Did you set y=0 in eqn to get posible x coedinates

[dense prism]

Yes

So (±1,0) are your possible points

Where the tangent is horizontal

[loud osprey]

but it also doesnt seem differentiable at those points

[dense prism]

Yeah that's what I was bringing up here

[loud osprey]

I graphed it and got this

oh

so in that case what should i do ?

[dense prism]

We're fine, it is differentiable, the domain ends at x=1 and it's clearly continuous there and no sharp turns

You can formally check the LHD and RHD to verify

[loud osprey]

whats that

[dense prism]

In case you want to, Left hand derivative and right hand derivative,

The domain ends at 1 so we are only worried about the LHD at x=1

And RHD at x=-1

[loud osprey]

What does that mean?

Check derivative from both sides?

Like lion?

Lim?

[dense prism]

We don't need to check here since theres only side

It is differentiable at x=±1

[loud osprey]

How do u find out

[safe radish]

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[finite crypt]

For Latex, how can I add another header under the same "Company Name" field?

What's listed is:

\resumeSubheading
      {Company Name}{City}
      {Your Role}{Event dates}

I just want to add another "{Your Role}{Event dates}" field, but under the same "Company Name" field.

[junior smelt]

You may wanna try #latex-help for this one

[finite crypt]

Ah, I knew there was a latex area, but couldn't find it. Thank you.

.close

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[stuck summit]

Hello! This is a bit of a two part question. I'm hoping this is the right forum but I'm having trouble understanding how (sqrt(2)^sqrt(2))^sqrt(2)) = 2, i was hoping to understand this in terms of what is apparently an algebraic property, and likewise, where i might further look into other such algebraic properties that might be glaring holes in my knowledge of more elementary parts of mathematics

[median vigil]

,tex \exprules

[flat frigate]

cloud

[median vigil]

the third one is the relevant one

we have $\ab(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^2$

[flat frigate]

cloud

[stuck summit]

thank you for those rules as i honestly find exponents a bit difficult for some reason, do you know where i might find where this might be derived from?

[terse lichen]

exponentiation is just repeated multiplication

you can work it out easily

[stuck summit]

.close

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[remote crag]

I’m not quite sure how to solve the question when it asks the total amount of pollution entered into the lake from [0,30], and I’m not sure that I answered b correctly

[hollow otter]

B sounds good

Since you don't have a concrete function given and c only wants you to guess you can do the following

You need to "calculate" the integral by hand. So we look at every day and add that value up.

Day t ~ Pollution
0 - 7
1 - 7
2 - 7
3 - 7
4 - 7
5 - 7
Then the next one
6 - 8
7 - 8
...
11 - 8
12 - 10

[remote crag]

ohhhhhhhh

thanks!

[hollow otter]

I hope that makes sense ;D

And that will your minimum guess.
After you completed that guess you can guess the maximum as well if you want

[remote crag]

yeah it does... it's just a lot to have to do the 30 days but it's simpler than what i was thinking

[hollow otter]

Sorry for my English.. I'm native German and explaining math in a different language

Oh but you can make it easier

[remote crag]

oh you're doing great! thanks for the help

[hollow otter]

Like interval [0,5] has 7 so it's 6 entries meaning: 6×7=42

[remote crag]

ohhhh yeah

[hollow otter]

[6,11] has 8 pollution and 6 entries so
6×8

Then you're a little quicker 🙂 feel free to add me when you're stuck

[remote crag]

okay! thanks!

wait actually... i don't think i'm doing this right... @hollow otter

i got what you were saying but now- i don't ;-; like, i get the concept but actually doing it i think im doing it wrong

[hollow otter]

Hmmh

What have you done now so far

[remote crag]

well i understand that from [0,6] can be written as 6*7

OH WAIT I DID MY GRAPH WRONG

nvm i got it ahah

[hollow otter]

Wait your graph looks fine

Or am I missing something now haha

[hollow otter]

Be careful it's only the internal from [0,5]. Because the 6 has a new value again

[remote crag]

ohhhh wait yeah, and yeah i added smaller tick marks but i added too many

okay so yeah I have 6(7)+6(8)+ but am unsure of what the next one would be

[hollow otter]

6×10

[remote crag]

because like you said, [0,5] is 6 entires of 7, [6,11] is 6 entries of 8, would the next set be [12,17]?

[hollow otter]

So from 12 to 17

Yess!!

[remote crag]

ohhhh

but why would it be times 10 and not 12? cuz at 18 it is 13

[hollow otter]

Why do you want to use 12?

[remote crag]

ohhh wait, i see, cuz the graph isn't linear so the values aren't- idk, haha i see where you're coming from. like, 10 is a rough estimate of how much it changes right?

[hollow otter]

We use 10 here because that's the actual value we got at t=12. It only "updates" at t=18 again

[remote crag]

oH

im actually so silly

[hollow otter]

Yeah you're moving in the right direction

No no don't worry abou it

Your question is even more apparent in the next coming intervals but keep going!

[remote crag]

so from [18,23] would it be 6(13)

[hollow otter]

Yes

[remote crag]

so the total equation to find the estimate would be

6(7)+6(8)+6(10)+6(13)+35?

[junior smelt]

(another way you could do this is by estimating the area with rectangles (see Riemann sums), so you assume that those have width of those intervals and pick either the leftmost or rightmost endpoint as their heights)

[remote crag]

ohhh, yeah i'm kinda confused on how to use that method but ik my teacher taught us how to do it that way

[hollow otter]

Yeah we were doing exactly that but with smaller/bigger intervals depending how you want to look at it

[remote crag]

okay! thanks for the help! hope you have a great day/night

[hollow otter]

(these emojis are so cute haha)

Morning 8am here now haha

I'm glad I could help!

[remote crag]

it's almost 11pm for me ahhh

[hollow otter]

Oh you're from the west coast then

[remote crag]

and i got a crap load of hw due in the morning that i just found out about

[hollow otter]

Good luck and see you!

[remote crag]

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[honest hawk]

Can u helpe me math please ? #help-42

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[lean otter]

Any mistake I have done?

[steel stag]

yea both are good

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[vagrant hound]

In how many ways can an 11 digit number be made with only digits 1 and 2 that is divisible by 6?

[safe radish]

@vagrant hound Has your question been resolved?

[eternal carbon]

first consider divisibility rule of 6

[lean otter]

theres a pattern to it, it was in a numberphile vid i watched a couple years ago

divisible by both factors of 6 iirc

[safe radish]

@vagrant hound Has your question been resolved?

[main mural]

a number is divisible by 6 iff it is divisible by both 2 and 3

for it to be divisible by 2 it needs to be even, so your last digit is fixed

look up the divisibility rule for 3 and from there it shouldn't be that hard

[vagrant hound]

The hard part is divisibility by three. solve the whole thing please.

[main mural]

i won't solve the whole thing for you

a number is divisible by 3 iff the sum of its digits is divisible by 3

in your case you have that the last digit is 2

so you have 10 digits to choose

those 10 plus the 2 must add up to a multiple of 3

so the sum of those 10 must be 3k+1, where k is some integer

notice that if you find one such string of digits, you also find more because the sum of digits doesn't depend on their order

for example 1111111111 is a valid 10-digit string, so 11111111112 is a valid number

1111111222 is also a valid string, and so is 2221111111, so is 1121121121 etc.

[main mural]

i gave you some ideas, come back when you are stuck

[stiff patrol]

I think it's gonna be like this

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[peak imp]

i dont know where to start with this problem

[wild copper]

Just plug it in

[neat fable]

substitute x and y into the unit circle equation x^2 + y^2 = 1

it should demonstrate that the identity x^2 + y^2 = 1 holds true

now consider:

the denominator t^2 + 1 can never be zero, so there is no issue of division by 0

now examine the behaviour at limits

[wild copper]

What is the codomain of the function
$$ t \rightarrow \frac{t^2-1}{t^2+1} $$

[flat frigate]

Xwtek

[hazy elbow]

#help-13 message #help-13 message

this question has been previously asked and the doubt was successfully cleared. You may consider this as a reference.

[flat frigate]

misha

misha

[wild copper]

I don't think we need to use limits here. Property of functions suffice.

[neat fable]

just my thought process

[wild copper]

I have a feeling that this question is given before calculus

[neat fable]

hmmm apparently it was resolved due to being a trick question and there is no missing point

oh my bad i didn't see your messages

i guess we kinda reach the same conclusion

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[harsh atlas]

Could anyone help me with question C ? We never learnt about equilibria of systems of DEs, so I’m not sure where to start

[harsh atlas]

its supposed say determine stability of the point! Realized I forgot to write it all out 🥲

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[harsh atlas]

.close

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[past birch]

what did i do wrong?

this is the ans

Oh

[flat frigate]

y0shi

[inner parrot]

yeah and they just multiplied the top and bottom by 2 to get that

[past birch]

so my ans is technically correct, just not in the most simplified form?

[inner parrot]

yeah

[past birch]

got it thanks

.close

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[glad crescent]

The question is to find the length of BD using the following hint: create the perpendicular line to BC from angle E.

[glad crescent]

I've drawn some lines to help visualise it.

We're not supposed to use trig functions

[safe radish]

@glad crescent Has your question been resolved?

[neat fable]

it doesn't look possible without trigonometry

[glad crescent]

I was thinking to use the circle radius, because we can create some isosceles triangles

But I'm not sure how

[neat fable]

i have stared at this diagram for about 30 minutes now, and i cannot think of any way that drawing the line EF will help us

[hazy elbow]

are you sure you've stated the question exactly as it was in the source?

[glad crescent]

Yes

[hazy elbow]

is alpha given?

[neat fable]

can i ask about the crediblility of the hint

[neat fable]

i believe alpha is just 45/2

since ab = ac

[glad crescent]

No, but alpha can be calculated by 45/2

[hazy elbow]

oh mb yeah it can be calculated

[glad crescent]

I only could make a triangle AOE and EOC that are congruent

[neat fable]

i can't find a way to use all the information given in the question

that must be the key to solving it

[glad crescent]

I mean I've tried to use the hint to compare EOS and CEF

I concluded that they are similar, but not congruent.

Or saying that ABD and ECD are similar

[neat fable]

why did u draw that S and F are right angles

doesn't that mean OSEF is just a rectangle

[glad crescent]

I gave F a name because the hint said to create a line perpendicular to BC from E

[hazy elbow]

no

angle BEC is 90 degrees, not angle BEF

[neat fable]

but you can't have S and F at 90 if that isn't a rectangle

anyway can you resend without the markings @glad crescent

i'd like to start doodling numbers on it

[glad crescent]

Sure

[neat fable]

are you sure it's solveable without trigonometry

@glad crescent

[safe radish]

@glad crescent Has your question been resolved?

[glad crescent]

I suppose it should. But I should ask my friend about it. Because I got it from him.

I'll see if I get an answer. Should I close this channel, because I don't think we'll get any answer.

[pulsar spade]

Oo Geometry. Looks fun

Is it given that BC is a diameter here?

I guess it would have to be since A and E are both right angles.

I'm just throwing ideas out here, but ADxDC=EDxDB by the Power of a Point Theorem.

[safe radish]

@glad crescent Has your question been resolved?

[glad crescent]

Yes

[glad crescent]

But that will give 4 unknowns

[karmic hedge]

just derive half angle without trig and youre done

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@glad crescent Has your question been resolved?

[glad crescent]

Thanks for the help guys. I'll see if I can get the answer with the hint mentioned.

.close

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[twilit chasm]

hi

[twilit chasm]

can anyone help me with a very easy question?

[desert pasture]

!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/

[twilit chasm]

while doing this graphic should it be like (-infinte; 4] or [4; +infinte) does it change anything?

[eternal carbon]

yes only one is correct

(though the finite cutoff point 4 isn't)

[twilit chasm]

wdym

im sorry, english isn't my main language

[eternal carbon]

the domain is indeed either (-infinity, n] or [n, infinity)

but 4 is not the correct value of n

[twilit chasm]

?

[faint current]

Yoo I'm a helper of grade 0-7

[faint current]

That's now 7 grade, sorry can't help

[twilit chasm]

uhm

[eternal carbon]

oh that . means multiplication

[eternal carbon]

yes this is correct

[twilit chasm]

ok but, the graphic goes from (-infinte; 4] or [4; +infinite)?

that's my question

like how to do the graphic

if the 4 is considered the final point or the starting point

[eternal carbon]

you have x is less than or equal to 4

[twilit chasm]

ohhh

ok so it's -infinte

[eternal carbon]

yea

[twilit chasm]

alright

thank you

[eternal carbon]

yw mb for misinterpreting the first time

[twilit chasm]

it's ok

the pic was low quality

find a way to find numbers between 3.0001 and 3.0005
?

[candid orchid]

arent they infinite

[twilit chasm]

that's what i think, but how can i express it?

i was thinking all the real numbers between these 2

hi?

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@twilit chasm Has your question been resolved?

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[buoyant shadow]

14/3 can't be divided by 2/5

the lcm is 14

[coarse kelp]

wsnt the formula like this: LCM of numerator / HCF of denominator

[buoyant shadow]

i don't know the formula

[frigid locust]

isnt lcm 15 ?

[coarse kelp]

damnn bruh

we all confused

i havent done this since 3rd grade

[buoyant shadow]

it's not 15 because 14 is smaller

[coarse kelp]

im so ashamed to ask but i gotta

[buoyant shadow]

and 14 can be divided by all of these

[frigid locust]

but we have 7/15 , u cant convert to 14, can u?

isnt lcm a way to make all the denominators same tho?

[buoyant shadow]

no

[frigid locust]

oh i was thinking adding the fractions , my bad

[buoyant shadow]

so yeah this works

hcf of 3,5,15 is 1

[coarse kelp]

whatttt

i thought it was 3

[buoyant shadow]

yeah, so that's why

[coarse kelp]

OMFG i forgot how to do this

ok ok i remember

thanks

how about GCD

like umm GCD of 2/10 and 4/10

isnt it 1/5

[buoyant shadow]

you don't have a formula for this one?

[coarse kelp]

nope

lemme search

[coarse kelp]

nope i cant find a simple one

2/10 and 4/10
1/5 and 2/5

1/5/1/5 = 1
2/5 / 1/5 = 2

[coarse kelp]

is 1/5 wrong?

[buoyant shadow]

1/5 is right

the formula is reversed

but it says everywhere you must simplify the fractions

so i'm trying to fnid an example where it's the wrong answer if you don't simplify first

[coarse kelp]

u mean i gotta simplify as much as possible but they should have same denominator?

[buoyant shadow]

i just think it's not true

[coarse kelp]

for GCD it's easier for me to see and i dont gotta use a formula

as long as thier denominators is the same

for LCM idk I can't just see it

[safe radish]

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[coarse kelp]

i forgot to close it sorry

.close

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[lean otter]

https://math.stackexchange.com/questions/4797233/how-to-solve-functional-equation-fx2-fracfx1-x

[lean otter]

how do we do this

there is no proper solution for this equation on ms

[compact wraith]

Find all $f:(-1,1)\rightarrow\mathbb{R},$ continuous at x=0
and satisfying:
[f(x^2)=\frac{f(x)}{1+x}.]

[flat frigate]

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

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[turbid forum]

.

Hello

[safe radish]

Please don't occupy multiple help channels.

[turbid forum]

Hello

I need help with this

:

[astral lodge]

which one

and not everyone here speaks hebrew

[turbid forum]

The one top right

Yeah mb

[astral lodge]

3 aleph?

[turbid forum]

Ooh yeah

[astral lodge]

is the question asking you to find the value of x?

[turbid forum]

Yeah . How to solve it .

[astral lodge]

does it come with the answers at the back

[turbid forum]

Yeah the book has the answer

But I'm trying to solve it not to get the answer for it

[strong oxide]

just multi all in (X)(X-6) then you have it

[astral lodge]

can check the answer later

[turbid forum]

Some thing like that ?

[astral lodge]

times all by x

u will get 4 + x/x-6 = x/2

continue from here

[strong oxide]

i dont know about that but more like this:
(6/x+1/x-6=1/2)(x)(X-6)

[turbid forum]

This is what I'll get ?

[strong oxide]

yes but i miss time 2

[astral lodge]

u cant just move the 4 into the fraction

[astral lodge]

from here times everything by 2

[strong oxide]

yes

[astral lodge]

ul get 8 + 2x/x-6 = x

continue from here

[turbid forum]

Btw why did where did the 1/x-6 go ?

[astral lodge]

11?

[turbid forum]

Fixed it

[strong oxide]

when you mulit it by x-6 is gone

[astral lodge]

the 1 became 2x because we timed it with x and 2

[astral lodge]

to get this

[turbid forum]

Ok

Let me try another one

[astral lodge]

then times everything with x-6 to finally remove all fractions

[turbid forum]

Ok this one

The one on the left side high

2/x + 2/x+6 = 1/2

[strong oxide]

each one of them solve just like others

[turbid forum]

Ok let me try

Soo

The first thing that I'll do is to get x and x+6

This is the first thing that I do right ?

[stuck crest]

what r u writing at the top

[astral lodge]

cant really understand what the \ ` is

[turbid forum]

Ooh it's just like some thing if you want to time multiply some thing

[astral lodge]

do one by one

times everything by x first

[turbid forum]

I get x/2

In the right one

[astral lodge]

yes

how about the left side

[turbid forum]

Ooh I think I know

The x+6 will disappear from the 2/x+6

And it would be 1 I htink or smth

[stuck crest]

if your multiplying by just x then no

[astral lodge]

not yet

[turbid forum]

Soo what am I supposed to do from here ?

[stuck crest]

multiply each term on the left side by x

[astral lodge]

tell us what you get after multiplying the left side by x

you cant just multiply the right side only

[turbid forum]

Like this ?

[stuck crest]

no

[astral lodge]

the x is multiplied above

not below

[turbid forum]

O mb

[astral lodge]

yes

the x in 2x/x can be cancelled

thus leaving you with 2

[turbid forum]

Btw is there a reason I'm multiplying x by all of them ?

[stuck crest]

ur trying to remove the fractions

so u can solve for x

[astral lodge]

in this case, the x in 2/x

[turbid forum]

Ok

[astral lodge]

so you times the left and right side by x

[turbid forum]

Soo if it will to in the upper ones

[astral lodge]

when you multiply, yes

now we move to the x/2

how do you think can we remove the 2?

[turbid forum]

Like this ?

[astral lodge]

2x/x -> 2

[turbid forum]

Uumm

Bruh I'm soo dumb or smth

[stuck crest]

the x can cancel out

[astral lodge]

is it not explained in the book

[stuck crest]

because x/x is 1

[turbid forum]

No . I will try to find easy question

[stuck crest]

$\frac{2x}{x} = 2 \cdot \frac{x}{x} = 2$

[flat frigate]

taro

[stuck crest]

have you done like just adding fractions

with different denominators

like

$\frac{3}{4} + \frac{4}{5}$

[flat frigate]

taro

[stuck crest]

would u be able to do that

[turbid forum]

This one ?

[stuck crest]

yes

[turbid forum]

First of all I will start with the bottom

4*5

Ooh wait

I think I should do the X sign

44 and 35

What

4 * 4 AND 3 * 5

[stuck crest]

so what would ur next line look like

[turbid forum]

Uumm I don't remember ...

[stuck crest]

well with fractions

whats the condition

for you to be able to add them together

[turbid forum]

Uumm

[stuck crest]

can i just add the numerators here

[turbid forum]

I don't really understand english but I think you want from me too do 16 /15 or some thing idk

[stuck crest]

what i was trying to get to was

the denominator (number on the bottom) needs to be the same before u can add them together

[turbid forum]

Ooh I now remember

Ok

[turbid forum]

.

Soo the number at the bottom will be 20 I think cuase I did 4 * 5 and then I do some thing . I don't really remember math I was more into history biology and a lot of computer sience

[turbid forum]

Let's try some thing from here

[stuck crest]

sure

would u like a demonstration of one of them

[turbid forum]

Yes please.

[stuck crest]

okay those questions do involve quadratics

do u know how to do those

[turbid forum]

Explain

[stuck crest]

if u dont know how to do quadratics u wont be able to solve these

questions like

$x^2 + 5x + 6 = 0$

[flat frigate]

taro

[turbid forum]

Yes this is the fucking things that I know

I will show you how I do it

[stuck crest]

otherwise this will be easier

[stuck crest]

okay great then u can solve those ones

so with the first one

$10x - \frac{7}{x} = 3$

[flat frigate]

taro

[stuck crest]

i see an x in the denominator

so to remove it

ill multiply each term by x

[turbid forum]

Soo it would be

[stuck crest]

so it would now look like

$10x^2 - 7 = 3x$

[flat frigate]

taro

[stuck crest]

do u understand how i got to that step

[turbid forum]

Ooh

This is what I did

It's the same

Ooh and then I do this

Let me do this

[stuck crest]

wait no

[stuck crest]

what is this

what did u multiply everything by

[turbid forum]

This is the answer like you did

[stuck crest]

they arent the same

[turbid forum]

I answerd it in the paint app

[stuck crest]

i multipied each term by x

just x

so i could remove the denominator

[stuck crest]

then move the 3x over and solve like a quadratic

so for questions like that all you need to do is remove the denominator and then solve for x

if its a quadratic thhen well ill assume you know how to solve those

[turbid forum]

Yeah but it's the same because 7 can get in 1 . 7 times

$ A=10x^2 B = -7 C=3x $

Like this

[stuck crest]

wdym

[turbid forum]

This is what I did

[safe radish]

@turbid forum Has your question been resolved?

[safe radish]

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[forest sequoia]

Hey, is someone know how do we know when a matrix is a diagonalizable matrix?

[safe radish]

@forest sequoia Has your question been resolved?

[peak estuary]

one equivalent condition is that the minimal polynomial splits into distinct linear factors

for example when you have n different eigenvalues

or if for every eigenvalue the geom multiplicity equals the alg multiplicity

[safe radish]

@forest sequoia Has your question been resolved?

[safe radish]

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[flint agate]

Can anyone explain how they separated this equation to 2 equations?

[raven heart]

amount of v in LHS = amount of v in RHS
amount of u in LHS = amount of u in RHS
that's the two equations they got

what are u and v here ? vectors ?

@flint agate

[flint agate]

yeah

[flint agate]

that I understand, but why is it possible? and when can I use it?

[raven heart]

what are u and v specifically ?

you got more info on them ?

@flint agate

[flint agate]

yeah

sorry for the bad drawing

[raven heart]

can I see the question you're asked in full ?

[flint agate]

basically:

• D is in the middle of AB

• AE:EC = 1/2

• CB(vec) = v

• CA(vec) = u

• BF(vec) = t * BE(vec)

• CF(vec) = k * CD(vec)

• question: find t and k

this is a solution i found. But I don't understand the 3rd section(what I sent earlier)

[raven heart]

yeah ok

so essentially that simplification works because u and v aren't collinear (multiples of one another)

$$-t\vec v + \frac13 t\vec u = \left(\frac12 k-1\right)\vec v+\frac12 k\vec u$$

[flat frigate]

aPlatypus

[raven heart]

if you put the v's on one side and the u's on the other, you get that

$$\left[-t - \left(\frac12 k-1\right) \right]\vec v = \left[\frac12 k - \frac13 t\right]\vec u$$

[flat frigate]

aPlatypus

[raven heart]

you're following up to here ? @flint agate

[flint agate]

yeah

[raven heart]

ok so we end up with "some multiple of v = some multiple of u"

[upper hatch]

another way to think about this more simply is, when you have the expression (a)u + (b)v = (c)u + (d)v then what does each side mean? left hand side is a vector, where you take (a) steps in the direction of u and (b) steps in the direction of v. what about the right side? also a vector with (c) steps in the direction of u and (d) steps in the direction of v. since we have an equality, we must end up on the same point, so the steps we take on each direction have to be the same

so a=c and b=d

thats how i tihnk about it, just thought id throw it out there in case it helps

[raven heart]

and that works because u and v aren't collinear here

otherwise there are multiple combinations of u and v which get you to the same point, but yeah it's a fine explanation

[raven heart]

and u and v aren't collinear

so what do you think the multiples/number of steps can be here ?

@flint agate

[flint agate]

in this? I don't really understand the question

[raven heart]

ok I'll simplify a bit

if A*u = B*v and u and v aren't collinear, what do you think A and B can be ?

@flint agate

[flint agate]

im not sure.. I know they can't be equal

[raven heart]

they can't be equal, unless you have very specific A and B

@flint agate

[safe radish]

@flint agate Has your question been resolved?

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[wooden oyster]

how to?

[fair hound]

use properties of | | to express |an| in terms of |i-1|

(|i-1| which you have to calculate)

[wooden oyster]

root of a^2+b^2?

[fair hound]

do you know what the modulus of a complex number is ?

[wooden oyster]

yes

|a+ib| = sqrt(a^2+b^2)

[fair hound]

then, apply

[wooden oyster]

so b*sqrt(2)

raised to n

[fair hound]

yes

[wooden oyster]

|b*sqrt(2)|^n

= |a_n|

[fair hound]

now conclude when it converges or not

[wooden oyster]

it converges when b<1?

[fair hound]

no, b is not the only thing that got to the power of n

[wooden oyster]

oh yeah true

[fair hound]

also, you forgot the | |

[wooden oyster]

where

this is what i have written so far

[fair hound]

that's good

so, when does |an| goes to 0 ?

[wooden oyster]

|b*sqrt(2)|<1

[fair hound]

yes

[wooden oyster]

=1, or >1 it'll diverge to infinity i suppose

[fair hound]

= 1 is a case that needs a bit of thinking, but yeah you see that obviously |an| goes to infinity if >1

[wooden oyster]

ohg yeah nvm

1^n does neither i suppose

just a constant

[fair hound]

well I was trying to get the case <1 done before talking about the case = 1 but let's do the =1 case
|bsqrt(2)| = 1
ie |b| = 1/sqrt(2)
so an in this case is a complex number of modulus 1, but if you look at the argument, it's n(3pi/4)

in the case =1, an stays on the unit circle and its modulus is constant equal to 1

but it goes around the unit circle

[wooden oyster]

does that make it converge?

[fair hound]

no, for a sequence to converges, you must go towards something specific

if you make turns and turns around the unit circle

you don't converge

[wooden oyster]

but is 1^n not a constant

[fair hound]

1^n = 1

[wooden oyster]

yeah

[fair hound]

1 is a constant

[wooden oyster]

also yeah

[fair hound]

well that's what I was saying

(I did the drawing for b = 1/sqrt(2), for b = -1/sqrt(2) it's the same but flipped around the vertical diameter)

the modulus is a constant but an doesn't approach something specific

while in the case |b| < 1/sqrt(2), |an| goes to 0, so an does

and in the case >, it spirals outwards as the modulus goes to infinity

[wooden oyster]

i see

do i have to show that it converges uniformly

in b

ignore the n = 0 part

[fair hound]

well I don't see the word uniformly so I assume that no

[wooden oyster]

how else do i show it

[fair hound]

well you already did the work for x in (-pi/2, pi/2)

now is the question how does it behave for whatever other real

[wooden oyster]

but does it not just go in circles

[fair hound]

firstly, you found that S = 1 no matter the x in (-pi/2, pi/2)

what happenes if x = 3pi/2 ?

is S equal to 1 ?

[wooden oyster]

well, it is the same number

[fair hound]

you sure ?

[wooden oyster]

cos(x)/cos(x)

surely gives one

no matter what x is

[fair hound]

try to calculate the first few terms when x = 3pi/2

you'll be quickly convinced you're not very careful

[wooden oyster]

what exactly are you asking

[wooden oyster]

if i insert cos^2(3pi/2)/cos^2(3pi/2

[fair hound]

S is that

[wooden oyster]

this will equal to 1

how does it not equal to 1

[fair hound]

well do what I say to do and you'll see

btw look at the denominator

[wooden oyster]

so you're saying dividing the a number by iteslf does not equal 1

?

[fair hound]

I don't have time for pointless chitchat when I wrote explicitly what I wrote tbh, if you need to read it again do it but go forward in the exercise

[wooden oyster]

that was what you said, but ill assume you are on about sin/cos

[fair hound]

what's cos²(3pi/2) ?

[wooden oyster]

1/2

[fair hound]

definitely not

[wooden oyster]

oh no

0

i thought it was 4 (3/4)

[fair hound]

yeah

so S is ?

[wooden oyster]

0

[fair hound]

yeah

now you see that there are several cases

the first question was about x in (-pi/2, pi/2)

the 2nd is about R as a whole

can you split that into 2 very distinct cases ?

explicitly, so it's done in 2 cases

(my example of 3pi/2 points towards one, the first question towards the other)

[wooden oyster]

how can i be 100% sure there are only two cases

[fair hound]

(-pi/2, pi/2) has length 2pi, it's a period of cos and sine
but you're mistake was thinking that the reunion of all the (-pi/2 + k2pi, pi/2 + k2pi) was R

some points are missing, and they are only one case

[wooden oyster]

i can onyl do the first method when it is between +-pi/2

so it doesnt become 0/0

[fair hound]

3pi/2 is not +- pi/2

but it does give 0

[wooden oyster]

i know

but i meant it restricts it to that

to aovid that vase

case*

[fair hound]

the question is to show that S converges for every x in R

if S is always 0 or 1, it does always converge, that's the point

what's the case where it's 1, what's the case where it's 0 ?

and after that are you convinced that there aren't more cases ?

[safe radish]

@wooden oyster Has your question been resolved?

[safe radish]

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[lean otter]

Find: DC Length.

[upper hatch]

well, BD^2 = AD*DC

should be pretty easy from there

[lean otter]

how so?

[upper hatch]

its proven easily using similair triangles

for ABD and BDC

[lean otter]

Any good video that explains (visually) similar triangles and their proof?

[upper hatch]

well for this particular theorem, this is all you need

i dont know of any videos but ims ure you can find some

[lean otter]

Visually?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

.close

[safe radish]

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[sacred crater]

Hello i hope this also counts, i have to calculate the ip,mask,gateway,ip broadcast and the useable Ips in two different subnets. The network IP is 100.100.48.0/20
The first subnet has 80 users and the second one 20.

[sacred crater]

IF anyone is able to help suuuper appreciated

[safe radish]

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[safe radish]

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[split ether]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[indigo rover]

1

I don't know where to begin.

[split ether]

Rewriting 3^{2x} as (3^x)^2 allows you to see that the given equation is just a quadratic with 3^x as its unknown

So you can start by solving for 3^x as you would normally solve a quadratic equation

[indigo rover]

do i not do anything with the 3 at the beginning

nvm

.close

[safe radish]

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[raven hawk]

Theres atleast one error here

but i cant find what i did wrong

[inner parrot]

check the rejection region

if our rejection region starts from 1.61, that means any statistic that is greater than 1.61 gets rejected

however the probability we got for 1.61 is very far from the actual rejection region

so it wouldnt make sense if 1.61 is exactly at the cutoff

youre supposed to use the significance level to figure this out btw

[safe radish]

@raven hawk Has your question been resolved?

[raven hawk]

wait what part are you referring to

[inner parrot]

part b

[raven hawk]

how would i do it?

[inner parrot]

essentially you need to find the z score that will give us 0.005 as the right tail

so either inverse normal

or just find the z score of that using the normal distribution table

[raven hawk]

so a z that gives an area of 0.995?

[inner parrot]

yeah if thats the left hand area

[raven hawk]

i did that

but its still wrong

[inner parrot]

where are you supposed to round to?

also could be because you didn't put a space after the comma

[raven hawk]

yeah i tried all of that and i still cant get it right

:/

[inner parrot]

maybe put infty instead of infinity?

[raven hawk]

still nope...

[inner parrot]

mm are you supposed to round to the hundredths or the thousandths

cuz you did both

[raven hawk]

.close

[safe radish]

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[errant gate]

Ana places the letters A, B, C, and D on a 2x4 table. She wants that in each row and in each of the three 2x2 squares, each letter appears only once. Then, Ana can place the letters in (Answer) different ways.

[errant gate]

idk how to do it

[steel stag]

try putting ABCD into the first 2x2 box, then solve the rest

[errant gate]

now what

[steel stag]

if it worked out uniquely, then all that matters is what you fill into that first 2x2 box

which has 4! choices because you could've put it in like CBAD and done the same thing

[errant gate]

so 4! x8=24 x8=192?

[steel stag]

no need for the x8, it's just 24 as the answer

[brave wolf]

hmm

are u sure it's unique?

[steel stag]

yea I'm a sudoku god

[brave wolf]

Lmfao

i think there are 4 solutions

oh wait you are right

im not sudoku god apparently

[errant gate]

24 then

[steel stag]

👍

[errant gate]

thank you sodoku god

[steel stag]

I love it

[errant gate]

i hate it

it makes my brain hurt

yo sodoku god do you know what a polynomial is

[steel stag]

yesiree

[errant gate]

A polynomial p satisfies the relation p(x+1) = x² - x + 2p(6) for every real x. What is the sum of the coefficients of the polynomial p?

i got

-2

but its not in the answers

i have

-40
-6
12
40
-12

hold on i think is -40

[safe radish]

@errant gate Has your question been resolved?

[steel stag]

yea I think so

[errant gate]

thank you sodoku god

i will always remember you

[safe radish]

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[thick fox]

@raven heart hello

[raven heart]

yep

donc c'était quoi l'inégalité que t'avais montré déjà

[raven heart]

oui bha tu peux "gendarmer" ton e^x-x-1 là

$\frac{-x^2e^x}{6} + \frac{2x(e^x-1)}{3} \leq e^{x} - x- 1 \leq \frac{x^2}{6} + \frac{x(e^x-1)}{3}$

[flat frigate]

aPlatypus

[raven heart]

tu veux calculer la limite de (e^x-x-1)/x^2 donc autant tout diviser par x^2

et là le théorème des gendarmes va bien marcher

[thick fox]

j'ai essayé mais je ne sais pas comment résoudre les bornes

[raven heart]

ah ok

lim pour x->0 de (e^x - 1)/x c'est assez connu comme truc

ça ressemble vachement à une dérivée tiens donc

[thick fox]

you should ask in other channel in math help(available)

not in (occupied)

[wicked rapids]

oh okay sorry

ty

[raven heart]

c'est quoi le problème avec les bornes alors

[thick fox]

j'ai oublie cette lim usuelle

merci

thank you

arigato

[raven heart]

17 langues plus tard...

de rien you're welcome (jsp comment on dit en jap)

[safe radish]

@thick fox Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[vapid pebble]

i will kiss anyone who could maybe help me figure out how to prove the last formula

[vapid pebble]

the I_a in the end is the same as the one right above

I am so genuinely lsot

[safe radish]

@vapid pebble Has your question been resolved?

[burnt bison]

I don't understand any of that. None of it except that exponent of 2

Is that American notation?

Is there a difference

[vapid pebble]

<@&286206848099549185>

[vapid pebble]

Not surprised, considering you're doing stuff with discriminant, and this is something far harder

[burnt bison]

Yeah

Is this physics?

[vapid pebble]

The original is, but this is an application in data analysis which I can't seem to prove

<@&286206848099549185>

Come on helpers, help out

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

Come on guys

[safe radish]

@vapid pebble Has your question been resolved?

[vapid pebble]

<@&286206848099549185>

[rigid mesa]

i can barely understand what's on there but have you tried subtracting equation 2 and 1?
as in (equation 2 - equation 1)

[vapid pebble]

Yes, but something just doesn't work out midway and it loses me

maybe I am misusing the properties of summation

[rigid mesa]

did you factorize?

[vapid pebble]

even after you factorize, you run into problems with the pi

the problem is the pi

i can't get rid of it

[rigid mesa]

okay i get it now

lemme think

what does the sum of pi gives?

[vapid pebble]

it doesn't give 1, unfortunately

no wait

I might be fucking stupid

the sum of pi alone is 1

or is it

wait

[rigid mesa]

i hope it is

what is pi?

[vapid pebble]

imagine a matrix

okay wait this is probability

pi is basically the weight of each case

the total weight is obviously 1

[rigid mesa]

yay c:

so is it solved?

[vapid pebble]

oh my god i am so stupid

not sure actually

is summation of E (f x g) equal to summation E (f) x E(g)

[rigid mesa]

i don't think so?

[vapid pebble]

i cant

sigh

i can intuitively tell you there is something i am missing

missing8

what the hell why did my hand writethat

[rigid mesa]

you can't what?

[vapid pebble]

So basically we get something like Summation pi(xi-a)M(xi-a) - Summation pi(xi-x)M(xi-x) = Summation (x-a)M(x-a)

This would then give me summation Pi [(x - a) M (x-a)

but summation pi itself = 1

i need one more thing to make the jump to proven

[rigid mesa]

isn't it already proved?

you no longer have xi

and since summation pi = 1

you're left with (x-a)' M(x-a)

[vapid pebble]

yeah, summation pi is 1, but this is summation Pi x [(x - a) M (x-a)]

pause

HAHAHAHA

I AM SO STUPID

thank you so much kind lady

[rigid mesa]

imagine it as if you're factorizing and the factor being [(x - a) M (x - a)]

[vapid pebble]

fucking finally

yeah i got it

hours wasted on this stupid idiotic problem

[rigid mesa]

oh okok😭

[vapid pebble]

thank you so much

may huygens never find peace

[rigid mesa]

it's okayy you got it in the end

[vapid pebble]

this is such bullshit

[rigid mesa]

(he's dead)

are you majoring in physics?

[vapid pebble]

this isn't physics

if this was physics, this would have been a very easy proof

since in physics you know that there is relation between the 2

somethign something perpendicular

this is statistics

i am currently doing my masters for Comp Engineering

[rigid mesa]

parallel axes

[vapid pebble]

yeah that

[rigid mesa]

ooo that's so cool
i'm in my first year :')

[vapid pebble]

(I solved like 4 other exercises in the meantime tbh, not a complete waste of time)

[vapid pebble]

oh what are you majoring in?

[rigid mesa]

physics💀

so since i read huygens, i was intrigued

[vapid pebble]

💀

good luck

[rigid mesa]

thanks! good luck to you too
i'm glad you got done with this exercise

[vapid pebble]

this thing has been haunting for me a day or so

i have my final exam tuesday and i needed it solved

at least i am almost done with the rest of the exercises

time to close this and ask a new question

.close

[safe radish]

Channel closed

Closed by @vapid pebble

Use .reopen if this was a mistake.

[rigid mesa]

ask a new💀

[rigid mesa]

good luckk then
i also have exams on tuesday

[safe radish]

Available help channel!

Send your question here to claim the channel.

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

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

[prisma acorn]

I need to prove that lim x -> $\frac{\pi }{2}$, $\tan\left(x\right)-\sec\left(x\right)=0$

[flat frigate]

TimesZeroed

[lean otter]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[prisma acorn]

my bad

So can I just start with multiplying

by cos

to get

sin(x)-cos(x)?

[lean otter]

you can't just multiply with cos

it'll change the value

[prisma acorn]

okay

[lean otter]

I'll give you a hint

[prisma acorn]

I'll add the value to both sides

and then mulitply

sin(x)=cos(x)

does this work?

[lean otter]

write tanx=sinx/cosx and secx=1/cosx

[prisma acorn]

sin(x)/cos^2(x)

sin(x)/(1-sin^2(x))

[lean otter]

that is way off

[prisma acorn]

okay

sin(x)/cos^2(x) oh

wait

take the derivative?

like

(sin-1)/cos

[lean otter]

First tell what term you get after substituting tanx and secx

[lean otter]

correct

[prisma acorn]

yes

[lean otter]

now Lhospital

[prisma acorn]

-cos/sin

[lean otter]

that derivative rule

[prisma acorn]

oh